function
induction
inequalities
5th edition

Could any one give me a hint(or solution) about problem 2.

Can anyone give me a solution for problem 3 ??  :?  :?  :?

Well, soln of #3 was quite ez. Here is my soln:

Don't you have it in a PDF file ? or you can just write it here ??
Because I don't have a doc. reader  :D

Ok! i am not at house. i have pdf writer at house. So when i will be at home i will try to post a pdf soln!

My solution is the same as you mahbub.
erdos, the main idea is that,
1. we can assume $abc=8$
2. perform the substitution $a=\frac{2}{x}$ and so on.
3. Cauchy rocks. :lol:

No need of such much steps. Just remark that you can write it as $\sum{\frac{\frac{1}{a^2}}{1-\frac{1}{a}+\frac{1}{a^2}}\geq 1}$. Next, apply Cauchy. It suffices to prove that $2\sum{\frac{1}{ab}}+\sum{\frac{1}{a}}$ attains its minimal value when $a=b=c=2$. But this is obvious from AM-GM and the hypothesis.

anyone got a soln for 2 (the one with the disks)?
i just could not get anywhere with it  :(

Here is the PDF file.

Well, yes that process don't need substitution but the begginers like me will prefer these kinds of solution(simple solution)! But when one can solve the problem by substituting then there is always a way by not substituting. And i knew it! But i did not go in that step :D

Here are my solutions. Can somebody please teach me P2? Thanks.

[quote="mahbub"]I don't have Telex and for that reason i can't open that .dvi file. So can u plz give that in pdf?[/quote]

Sorry, I don't know how to convert dvi into pdf. But I copied and pasted the whole thing here.

[hide="click to see"]
{\bf Problem 1:} For what positive integers $k$ there exists a function $f: N \longrightarrow N$ such that for all $n \in N$, $f^{(k)}(n)=f(n)+2$, where $f^{(1)}(n)=f(n)$, $f^{(k+1)}(n)=f(f^{(k)}(n))$

{\bf Solution:}
We will show that the only possible values for $k$ are 2 and 3. For $k=2$ or 3, simply let $f(n)=f(n)+2$ or $f(n)=f(n)+1$ respectivly.

Note that $k=1$ dosn't work. Suppose that there is $k\geq 4$ that satisfy the given conditions. There can not be any fixed points in this function. Otherwise, let the fixed point be $w$, $f^{(k)}(w)=w \neq w+2 = f(w)+2$. Now, let $G$ be the set $\{f(x) | x \in N \}$. By the given condition, for every $z \in G$, $z+2 \in G$. Moreover, $\forall n$, $f(f(n)+2)=f(f^{(k)}(n))=f^{(k)}(f(n))=f(f(n))+2$. This means that $\forall z \in G$, $f(z+2)=f(z)+2$.

Let $z_0$ be the least member of $G$. Define $z_{i+1} = z_{i}+2$. Let $f(z_0)-z_0 = t$. Since there are no fixed points, we have $t \neq 0$. Since $f(z_{i+1})-z_{i+1}=f(z_i+2)-z_i-2 = f(z_i)+2-z_i-2=f(z_i)-z_i$, an easy induction yields $\forall i \in N$, $f(z_i)-z_i=t$.

If $t$ is even, then we claim that there is a sufficiently large $j$ such that  $z_j, f(z_j), f^{(2)}(z_j), \ldots ,f^{(k)}(z_j)$ are all larger than $z_0$ and with the same parity as $z_0$. For all numbers $z$ that are larger than $z_0$ and with the same parity as $z_0$, $f(z)=z+t$. Thus, a $j$ such that $z_{j}>z_0+|t|k$ would suffice. if $t$ is positive, this is obvious. If $t$ is odd, then this can be proven by a trivial induction with the induction hypothesis $f^{(s)}(z_j)>z_0+|t|(k-s)$ $\forall 0\leq s \leq k$. The claim implies that $f^{(s+1)}(z_j)-f^{(s)}(z_j)= t$ $\forall 0\leq s \leq k-1$. Using the, we have $f(z_j)+2=f^{(k)}(z_j)=f^{(k-1)}(z_j)+t= f^{(k-2)}(z_j)+2t=\cdots = f(z_j) + (k-1)t$. This implies that $(k-1)t=2$, which is not possible since $t \neq 0$ and $k-1 \geq 3$.

If $t$ is odd, then from $f(z_i)-z_i=t$, there must exist a $y$ such that $y \in G$ and $y$ has a different parity with respect to the $z_i$'s. Let $y_0$ be the smallest $y$ satisfying the above property. Let $f(y_0)-y_0=q$. We define $y_{i+1} = y_{i}+2$. Using the exact same argument we used with $z_i$'s, we get that $\forall i$, $f(y_i)-y_i=q$. If $q$ is even, then using the argument in the previous paragraph but replacing $z_i$'s with $y_i$'s and $t$ with $q$, we get a contradition. Thus, $q$ is odd. Let $\{ x_i \}$ be the sorted union of the sequences $\{ y_i \}$ and $\{ z_i \}$ (sorted from small to large) . Note that for sufficiently large $m$, $\{ x_i \}$ contains all natural numbers larger than $m$. We claim that there is a sufficiently large $j$ such that  $x_j, f(x_j), f^{(2)}(x_j), \ldots ,f^{(k)}(x_j)$ are all larger than $m$. Using an argument similar to the one in the previous paragraph, we see that a $j$ such that $z_{j}>m+|t|k+|q|k$ would suffice. This can be provn by using a trivial induction with induction hypothesis $f^{(s)}(x_j)>m+|t|(k-s)+|q|(k-s)$. The claim implies that $f^{(s+1)}(x_j)-f^{(s)}(x_j)$ alternates from $t$ to $q$ $\forall 0\leq s \leq k-1$. Using this, it is easy to see that for even $k$ with $k=2h$, either $f^{(k)}(z_j)=f(k)+hq+(h-1)t$, or $f^{(k)}(z_j)=f(k)+(h-1)q+ht$. This which implies that either $hq+(h-1)t=2$ or $(h-1)q+ht=2$. Since $q$ and $t$ are odd, working(mod 2), we get a contradiction. For odd $k$ with $k=2h+1$, we have $f^{(k)}(z_j)=f(k)+hq+ht$. This implies that $h(q+t)=2$. Since $q+t$ is even, the only way this would work is if $h$ is 1 or -1, but since $k\geq 4$, we get a contradiction again.

Therefore, the only $k$ that work are 2 and 3.
{\bf Q.E.D}

{\bf Problem 3:} Let $a$, $b$, $c$ be positive numbers such that $abc\leq 8$. Prove that
\[{1\over a^2-a+1}+{1\over b^2-b+1}+{1\over b^2-b+1} \geq 1\].

{\bf Solution:}
	Since $a+1 \neq 0$, the given inequality is equivalent to, \[{1\over 2}\sum_{sym}{a+1 \over a^3 +1} \geq 1\]
	Multiplying out the denominators and symplifying, this is equivalent to
\[{1\over 2}\sum_{sym}{a^3b^3c}+\sum_{sym}{a^3b}+{1\over 2}\sum_{sym}{a^3}+{1\over 2}\sum_{sym}{a}+2 \geq a^3b^3c^3\]
	Applying AM-GM to each of the terms on the left, and using $abc\leq 8$,
\[{1\over 2}\sum_{sym}{a^3b^3c}+\sum_{sym}{a^3b}+{1\over 2}\sum_{sym}{a^3}+{1\over 2}\sum_{sym}{a}+2 \geq 3a^{7\over 3}b^{7\over 3}c^{7\over 3}+6a^{4\over 3}b^{4\over 3}c^{4\over 3}+3abc+3a^{1\over 3}b^{1\over 3}c^{1\over 3}+2\]
\[\geq 3a^{7\over 3}b^{7\over 3}c^{7\over 3}{a^{2\over 3}b^{2\over 3}c^{2\over 3}\over 8^{2\over 3}}+6a^{4\over 3}b^{4\over 3}c^{4\over 3}{a^{5/3}b^{5/3}c^{5/3}\over 8^{5/3}}+3abc{a^{2}b^{2}c^{2}\over 8^2}+3a^{1\over 3}b^{1\over 3}c^{1\over 3}{a^{8/3}b^{8/3}c^{8/3}\over 8^{8/3}}+2{a^3b^{3}c^{3}\over 8^3}= a^3b^3c^3\]

{\bf Q.E.D}[/hide][/hide]

the 5th edition of the ml contest will end very soon, and still i haven't a soln for problem 2. i mean, there must be someone who has solve it   :ninja:

harazi please can you explain your solution ... Thank you
 :(
Davron


Which contests exist for chemistry except IChO, mendleev,BChO ?

I've actually only heard of the first one, at least in the US.

How do you participate in the other two if you are a US citizen?


(a) What is $\sqrt{x^3}$, when x=4?
(b) What is $\sqrt{x^3}$, when x=25?
(c)  *Classroom Challenge*  (Hard for the average classroom)  What is $\sqrt{x^3}$, in simplest radical form?


m and n are fixed positive integers and k is a fixed positive real.show that the minimum value of $ {x_1^m}\plus{}{x_2^m}\plus{}...{x_n^m}$ for $ {x_i}$ satisfying  $ {x_1}\plus{}{x_2}\plus{}...\plus{}{x_n}\equal{}k$ occurs at $ {x_1}\equal{}{x_2}\equal{}...\equal{}{x_n}$

use Holder, to find that $ min\equal{}\frac{k^mn}{n^{m}}$

It is old and easy. I mean, you can proof it if you think about it ......
you know,it is just cool...

yes,I think it should be :
$ (1\plus{}1\plus{}...\plus{}1)^{\frac{m\minus{}1}{m}}({x_1}^m\plus{}{x_2}^m\plus{}...\plus{}{x_n}^m)^{\frac{1}{m}}\geq k$
thank you

[quote="xiaohui"]yes,I think it should be :
$ (1 \plus{} 1 \plus{} ... \plus{} 1)^{\frac {m \minus{} 1}{m}}({x_1}^m \plus{} {x_2}^m \plus{} ... \plus{} {x_n}^m)^{\frac {1}{m}}\geq k$
thank you[/quote]
Yes, you see, Holder is a really nice inequality. it is a generalisation of Cauchy $ \implies$ any problem that follows from Cauchy can be generalised. :rotfl:

We all know the Titu's Lemma. Can we find the maximum of
$ P\equal{}\sum \frac{x_i^k}{a_i}$ where $ x_i, k_i\in\mathbb{R}^\plus{}$. :wink:
PS: n Case of Titu, we have $ x_i\in\mathbb{R}$ as $ k\equal{}2\implies x_i^k\geq 0$
 :P


complex analysis
function
complex analysis unsolved

let $ f$ be holomorphic on $ C$ and periodic. prove that $ f(z_0)\equal{}z_0$ for  some $ z_0$

If $ f(z)\ne z$ for all $ z$, then also $ f(z)\ne z\plus{}kT$ for any $ k\in\mathbb Z$, where $ T$ is the period of $ f$. But the entire function $ f(z)\minus{}z$ can omit at most one value (Picard).


sabet konid ke har adade tabie ra mitavan be soorate majmoa jomlehaye mokhtalefe donbaleye fiboonachi nevesht  :?:

natanha har adad ra mitavan besoorat majmooe jomlehaye motemayeze donbaleye fibonachi neshan dad balke in noe namayesh yektast.

masale ra ba esteghra hal mikonim:
farz konid bekhahim n ra be in soorat namayesh bedahim. adade k ra tori dar nazar begirid ke f(k)<=n , f(k+1)>n.
be vozooh tebghe farze esteghra n-f(k) ra mitavan be soorat matloob namayesh dad. tanha chizi ke bayad sabet konim in ast ke f(k) dar namayeshe n-f(k) nayamade ast. baraye neshandadan in matlab sabet mikonim ke n-f(k)<f(k). farz konid chonin nabashad. dar in soorat n-f(k)>=f(k) ama midanim ke f(k+1)<=2f(k) (zira f(k+1)=f(k)+f(k-1) va f(k)>=f(k-1)) pas n>=2f(k)>=f(k+1) pas n>=f(k+1) ke ba nahveye entekhabe k dar tanaghoz ast. :wink:


limit
integration
logarithms
calculus
geometry
rectangle
real analysis

How do I prove that:
$ \lim_{n\to\infty} \frac{1}{n\plus{}1} \plus{} \cdots \plus{} \frac{1}{kn} \equal{} ln(k)$

(I think it is true :P)
I've tried something with $ f(x) \equal{} \frac{1}{x}$ And $ F(x) \equal{} ln(x)$ So $ \lim_{n\to\infty} \frac{1}{n\plus{}1} \plus{} \cdots \plus{} \frac{1}{kn}$ is approximatly $ ln(kn)\minus{}ln(n\plus{}1) \equal{} ln(\frac{n}{n\plus{}1}k)$ which goes to $ ln(k)$ for $ n\to\infty$

$ \lim_{n\to\infty}\frac{1}{n\plus{}1}\plus{}\frac{1}{n\plus{}2}\plus{}\cdots\plus{}\frac{1}{kn}$

$ \equal{}\lim_{n\to\infty}\frac{1}{n}\cdot\frac{n}{n\plus{}1}\plus{}\frac{1}{n}\cdot\frac{n}{n\plus{}2}\plus{}\cdots\plus{}\frac{1}{n}\cdot\frac{1}{k}$

$ \equal{}\int_1^k\frac{1}{x}dx$

$ \equal{}\ln k$

[quote="cosinator"]$ \lim_{n\to\infty}\frac {1}{n \plus{} 1} \plus{} \frac {1}{n \plus{} 2} \plus{} \cdots \plus{} \frac {1}{kn}$

$ \equal{} \lim_{n\to\infty}\frac {1}{n}\cdot\frac {n}{n \plus{} 1} \plus{} \frac {1}{n}\cdot\frac {n}{n \plus{} 2} \plus{} \cdots \plus{} \frac {1}{n}\cdot\frac {1}{k}$

$ \equal{} \int_1^k\frac {1}{x}dx$

$ \equal{} \ln k$[/quote]

I do not understand your solution at all - unfortunately. How do you get from the second line to the third ? (I understand the other steps thougt :P )

[quote="Mathias_DK"][quote="Albanian Eagle"]Do you know how to define the Riemann integral?[/quote]
No :blush:[/quote]
Come on, you know something about it if you understand 3rd to 4th line. I too didn't know the term (I knew it as definite integral). Looked it up on Wikipedia and it's pretty basic stuff. [url=http://en.wikipedia.org/wiki/Riemann_integral]Here[/url] it is.
I too don't understand 2nd to 3rd line

Consider the graph of $ f(x)\equal{}\frac{1}{x}$ between $ 1$ and $ k$.  The second line represents the sum of rectangles with width $ \frac{1}{n}$ and height $ f(i)$ for $ i\equal{}\frac{n\plus{}1}{n},\frac{n\plus{}2}{n},...,k$.  These rectangles approximate the area underneath the curve between $ 1$ and $ k$.  As $ n$ goes to $ \infty$, the approximation becomes more and more accurate to the actual value of $ \int_1^k\frac{1}{x}dx\equal{}\ln k$.

[quote="cosinator"]Consider the graph of $ f(x) \equal{} \frac {1}{x}$ between $ 1$ and $ k$.  The second line represents the sum of rectangles with width $ \frac {1}{n}$ and height $ f(i)$ for $ i \equal{} \frac {n \plus{} 1}{n},\frac {n \plus{} 2}{n},...,k$.  These rectangles approximate the area underneath the curve between $ 1$ and $ k$.  As $ n$ goes to $ \infty$, the approximation becomes more and more accurate to the actual value of $ \int_1^k\frac {1}{x}dx \equal{} \ln k$.[/quote]
Thank you for the explanation.

Actually I know what the Riemann Integral, but I forgot the name. (Actually I thought you were talking about the Lebesgue integral, which I have only heard about, but never studied it at all)


function
number theory unsolved
number theory

The sequence $ (x_n)_{n \in \mathbb{N}}$ is defined by $ x_1\equal{}2, x_2\equal{}3,$ and

$ x_{2m\plus{}1}\equal{}x_{2m}\plus{}x_{2m\minus{}1}$ for $ m \ge 1;$
$ x_{2m}\equal{}x_{2m\minus{}1}\plus{}2x_{2m\minus{}2}$ for $ m \ge 2.$

Determine $ x_n$ as a function of $ n$.

[hide="Hint (Recursive formula for n)"]$ x_{n}\equal{}4x_{n\minus{}2}\minus{}2x_{n\minus{}4}$[/hide]
[hide="Prove for that formula for even n"]$ x_{2m}\equal{}x_{2m\minus{}1}\plus{}2x_{2m\minus{}2}\equal{}3x_{2m\minus{}2}\plus{}x_{2m\minus{}3}\equal{}4x_{2m\minus{}2}\minus{}x_{2m\minus{}2}\plus{}x_{2m\minus{}3}$

$ \boxed{x_{2m}\equal{}4x_{2m\minus{}2}\minus{}2x_{2m\minus{}4}}$[/hide]
[hide="Prove for that formula for odd n"]$ x_{2m\plus{}1}\equal{}x_{2m}\plus{}x_{2m\minus{}1}\equal{}2x_{2m\minus{}1}\plus{}2x_{2m\minus{}2}\equal{}4x_{2m\minus{}1}\minus{}2x_{2m\minus{}1}\plus{}2x_{2m\minus{}2}$

$ \boxed{x_{2m\plus{}1}\equal{}4x_{2m\minus{}1}\minus{}2x_{2m\minus{}3}}$[/hide]


calculus integration function inequalities algebra domain derivative

I have big trouble with this one, any help is appreciated.

Let $f(x,y)=x^{3}+xy+y^{3}$ and $Z=\{(x,y)|f(x,y)=0\}$
Show that $Z$ is a closed subset of $R^{2}$ but $Z$ is [b]not[/b] compact.

Thanks in advance,

$f$ is continuous, so $f^{-1}(\{0\})=Z$ is a closed set.

As a closed subset of $\mathbb{R}^{2}$, $Z$ is compact iff it is bounded.  It isn't; for each fixed $t$ the equation $f(x,t) = x^{3}+xt+t^{3}=0$ has a solution $x_{0}\in\mathbb{R}$ such that $f(x_{0},t)=0$; letting $t\to\infty$ shows that $Z$ isn't bounded and that's that.

Thank you (blah)^3!

Is there a method to solve for y in $x^{3}+xy+y^{3}=0$ for any fixed non-negative x?

And there's another problem: Use the method of repeated integration to evaluate:
$\int_{0}^{1}\left[\int_{y}^{1}e^{x^{2}}dx\right]dy$
Any hint on this one?

Thank you,

Fixing $x$ and solving for $y$ in $x^{3}+xy+y^{3}=0$ requires (in general) solving a cubic, so we need the cubic formula, which is pretty horrible.  However, we know a solution exists, since for fixed $x$, as $y\to\infty$, $f(x,y) = x^{3}+xy+y^{3}\to\infty$ as well.  Similarly, as $y\to-\infty$, $f(x,y)\to-\infty$, so there is $y_{0}$ such that $f(x,y_{0})=0$ by the intermediate value theorem.

For the other problem, you need to interchange the order of integration.  Draw a diagram first and see if you get anywhere :)

Hi blahblahblah,

In the first problem, I was asked to find a continuous function g(x) such that for any $x\ge0$, $Z\cap\{x>0\}=\{(x,g(x))\}$, so I think solve for g(x) in $x^{3}+xg(x)+g(x)^{3}=0$ will do.

In the second one, how do you get around with intergrating $e^{x^{2}}dx$?

Thank you,

[quote="analysis_beginner"]In the second one, how do you get around with intergrating $e^{x^{2}}dx$?[/quote]

You don't. Listen to blahblahblah's suggestion here, and try it as an integral $dy\,dx.$

[quote]$=\int_{y}^{1}\int_{0}^{1}e^{x^{2}}\,dy\,dx$ [/quote]
This is not just wrong, it's meaningless. The variable $y$ only has meaning inside the inner integral, and hence cannot be used on that outer limit. Later on in the problem, when you encountered that meaningless $y$ out there, you turned it into $1$ for no good reason. Of course, there was also no good reason to interpret it any other way.

The set you are integrating on can be described by the chain of inequalities $0\le y\le x<1.$ Does that help? Drawing a sketch of this domain should help. (That sketch was blahblahblah's original suggestion.)

[hide="Nonsense - ignore this."]Oh, and this integral is badly divergent. To make it a more interesting problem, try this:

$\int_{0}^{\infty}\int_{y}^{\infty}e^{-x^{2}}\,dx\,dy.$[/hide]

Everything's inside the square $[0,1]^{2}$- there aren't any problems with divergence.

I'm sorry for my previous attempt. I think it should be done this way:
$\int_{0}^{1}\int_{y}^{1}e^{x^{2}}dxdy$
$=\int_{0}^{1}\int_{0}^{x}e^{x^{2}}dydx$

Can anyone give me a suggestion on solving for g(x) in the second one ($x^{3}+xg(x)+g(x)^{3}= 0$)?

My suggestion is [b]not[/b] to solve for $y$. It's a Real Analysis problem, not an Algebra problem, right? Real Analysis (specifically, the intermediate value theorem) tells us that for any value of $x$ the equation $x^{3}+xy+y^{3}=0$ has at least one solution for $y$. This is all you need to prove that $Z$ is unbounded.

Oh 'Z is unbounded' was taken care of.

I want to show that there is a continuous function g(x) for each x >= 0, so that f(x, g(x)) = 0...

For each $x>0$, $f_{x}(y)=f(x,y)=x^{3}+xy+y^{3}$ is an increasing function of $x$; its derivative is $x+3y^{2}$. This means that there is a unique real $g(x)$ satisfying $f_{x}(g(x))=0$.
By the implicit function theorem, $g(x)$ is differentiable, and its derivative at $x$ is $-\frac{3x^{2}+g(x)}{x+3(g(x))^{2}}$. A differentiable function is continuous.

You don't need to solve for $g$.


There was two box,  one box have a blue and a red marble, the other one have 2 red marbles
you randomly reach into a box and took out a red marble. what is the probability that the other marble in that box is blue?

state your answer and defend it!

[hide]SPOLIER
is it  1/2: because it doesn't matter if the first one is red or not, pick out of the two boxs
or 1/3: there was 3 red marbles, only one of them was paired with blue
or 1/4:only one out of the four total marbles was paired with blue
?
[/hide]

[hide]I would say 1/3 b/c there are 3 other marbles, each of them equally likely to be the other one in the box.[/hide]

This is known as conditional probability.  The "condition" here is that you chose a red marble.


linear algebra
matrix
logarithms
algebra
system of equations
algebra unsolved

Solve the system of equations:
$ \left\{\begin{matrix} 2\log _{6}\left ( \sqrt {x} \plus{} \sqrt [4]{x} \right ) \equal{} \log _{4}y & \\
5^{1 \plus{} x \minus{} y} \minus{} 3^{1 \plus{} x \minus{} y} \minus{} 2^{1 \plus{} x \minus{} y} \equal{} 3^{1 \minus{} x \plus{} y} \plus{} 2^{1 \minus{} x \plus{} y} \minus{} 5^{1 \minus{} x \plus{} y} & \end{matrix}\right.$


quadratics
inequalities
number theory proposed
number theory

For postive integers $ m,\ n$,  Let $ f(m,\ n)$ such that $ f(m,\ n) \equal{} \frac {1}{2}\{(m \plus{} n \minus{} 1)^2 \plus{} (m \minus{} n \plus{} 1)\}$.

(1) Find one pair of $ (m,\ n)$ such that $ f(m,\ n) \equal{} 100.$

(2) Let $ a,\ b,\ c,\ d$ be integers such that $ a^2 \plus{} b \equal{} c^2 \plus{} d.$ If $ \minus{} a < b\leqq a,\ \minus{} c < d\leqq c$ holds, then show that we have $ a \equal{} c,\ b \equal{} d$ .

(3) Show that there is one pair of $ (m,\ n)$ such that $ f(m,\ n) \equal{} k$ for any positive integers $ k$.

[hide="1)."]
Rewrite as a quadratic,
\[ m^2 \plus{} m(2n \minus{} 1) \plus{} n^2 \minus{} 3n \minus{} 198 \equal{} 0
\]
. We have $ 8n \plus{} 4(198) \plus{} 1 \equal{} k^2$ for some odd $ k$. The smallest positive $ n$ for this is $ n \equal{} 6$, although it [i]appears to have[/i] infinite solutions. Taking this value of $ n$ and substituting, our solution is $ (n,m) \equal{} (9,6)$.
[/hide]
[hide="2)."]
WLOG take $ a,c>\equal{}0$.
Note $ |b| <\equal{} a$,$ |d| <\equal{} c$. Then
\[ a^2 \plus{} b \equal{} c^2 \plus{} d\implies|a \minus{} c\parallel{}a \plus{} c| \equal{} |b \minus{} d|
\]
Suppose $ a\not \equal{} c$: then
\[ |a \minus{} c\parallel{}a \plus{} c| >\equal{} |a \plus{} c| \equal{} a \plus{} c >\equal{} |b| \plus{} |d| >\equal{} |b \minus{} d|
\]
Equality holds only when $ a \equal{} |b|,c \equal{} |d|$. However, from the bounds this implies $ a \equal{} b,c \equal{} d$. Also, $ |b| \plus{} |d| \equal{} |b \minus{} d|$ only when $ bd < 0$. But $ bd\equal{}ac$, so $ ac<0<\equal{}ac$, contadiction.
[/hide]
[hide="3)."]
\[ f(\mathbf{m}) \minus{} f(\mathbf{n}) \equal{} 0\iff a^2 \plus{} b \equal{} c^2 \plus{} d
\]
for $ a \equal{} m_x \plus{} m_y \minus{} 1$, $ b \equal{} m_x \minus{} m_y \plus{} 1$ and analogues.
Now $ a \plus{} b \equal{} 2m_x > 0$, and $ a \minus{} b \equal{} 2(m_y \minus{} 1) >\equal{} 0$, so by (2) we have uniqueness$ \square$
[/hide]

[quote="me@home"]
$ |a \minus{} c\parallel{}a \plus{} c| > \equal{} |a \plus{} c| > \equal{} |a| \plus{} |c| > \equal{} |b| \plus{} |d| > \equal{} |b \minus{} d|$[/quote]
why ? the corrrect form is $ |a \plus{} c| \le |a| \plus{} |c|$

I corrected it... in retrospect, I don't know why I even payed attention to the signs of $ a,c$ because they don't matter

Triangle inequality is powerful for this problem.

Sort of...? I'm sure many many ways could be used to yield the same solution... its not a difficult problem (I guess that's why it was on a Science and Technology exam)

Another way to do 2: $ \left( a\minus{}\frac{1}{2}\right)^2<a^2\plus{}b<\left( a\plus{}\frac{1}{2}\right)^2$.


[quote="FrankieBlueEyes"]x^2+y^2=25
x+2y=10
ahh help[/quote]

uh you can rearrange x+2y=10 into x=10-2y. then you can substitute it into the other equation and solve for y.


Let $ ABC$ be a right triangle $ \hat{A} \equal{} 90^o$ and let $ M$ lie on $ AB$ such that $ \frac{AM}{MB}\equal{}3\sqrt{3}\minus{}4$. It is known that the symmetric point of $ M$ with respect to the line $ GI$ lies on $ AC$.
( $ G$ is centroid of triangle and $ I$ is incenter of triangle )
Find $ \hat{B}$.


[quote="the future"]For how many positive integers n less than or equal to 24 is n! evenly divisible by 1+2+3+...+n?[/quote]
[hide="Solution"]
Notice that $ \frac{n!}{1\plus{}2\plus{}3\cdots n}\equal{}\frac{n!}{\frac{n(n\plus{}1)}{2}}\equal{}\frac{2(n\minus{}1)!}{n\plus{}1}$.  The only way $ (n\minus{}1)!$ is not divisible by $ n\plus{}1$ is if $ n\plus{}1$ is prime.  The primes under $ 24\plus{}1\equal{}25$ are $ 2, 3, 5, 7, 11, 13, 17, 19, 23$, but $ n\plus{}1\equal{}2$ works since $ \frac{2\cdot 1!}{2}$ is an integer.  Thus, there are $ 8$ values of $ n$ that don't work and so $ 24\minus{}8\equal{}\boxed{16}$ values of $ n$ work. [/hide]

Problems like these illustrate why they got rid of calculators :roll:


$ \tan{15^{\circ}}\tan{25^{\circ}}\tan{35^{\circ}}\equal{}\tan{\theta}$, for $ 0^{\circ}\le\theta<180^{\circ}$. Find $ \theta$.

Does this problem involve multiplying by tan 45.


hmm a function passes the vertical line test.  In short for every X value there is either 1 or no Y values.

Also a function of X "f(x)" is the same as saying Y like "f(x)=2x" and "Y=2x" mean the same thing

Abbr. f Mathematics.

   1. A variable so related to another that for each value assumed by one there is a value determined for the other.
   2. A rule of correspondence between two sets such that there is a unique element in the second set assigned to each element in the first set

I like what somedude said better... Short and sweet... But this is the long way of saying it  (http://dictionary.com).

Basically an equation that passes the vertical line test (see above). It is written as f(x) = an expression in terms of x, where f(x) replaces y.

A more general definition and more detailed explanation than what others have given: a function is something that takes an input and gives an output.  Thus, for example, we can think of "doubling" as a function -- given any number, put it into the doubling machine, and out comes twice that number.  Similarly for "adding 1."  There is a lot of notation associated with functions of different types.  The most common is the notation $f(x)$ to represent "the output we get when we apply the function f to the input x."  So, if f is the function "doubling" and x is 3, f(x) is 6.  In general, if f is the function doubling, we can say $f(x) = 2x$.  There are lots of other sorts of functions.  For example, one could consider the function which takes inputs in the Roman alphabet and outputs into the Roman alphabet by replacing a with b, b with c, c with d, and so on.  This function is known as the "Caesar cipher," I believe.  The only rule limiting what a function can do is that there can only be a single unique output.  Thus, for example, " :sqrt: " can be considered a function whenever it has inputs of real numbers greater than or equal to 0.  However, "the number which when squared equals the input" is [i]not[/i] a function for inputs of positive real numbers, since it often has more than 1 output.  (For example, if we inputted 4 into that function, we would get two outputs: 2 and -2.)  If you have a function whose inputs and outputs are real numbers, you can graph it in the Cartesian plane -- in other words, use a coordinate system.  In those circumstances, a function f can be represented by a graph, y = f(x).  However, there are more graphs than functions.  If you want to check whether a graph is the graph of a function, you must make sure that it passes the "vertical line test" that other people have mentioned.  That is, for any x value (input), if you draw a vertical line, it must intersect the graph in only one point (a unique output).

ok......i think i'll worry about functions lata but i think i have a little idea of what's goin on... :D  :blush:

As a bit of advice, I'd recommend that people responding to requests like this one consider the audience.  This is middle school classroom math.  What if the questioner hasn't done graphing yet?  How can you explain functions in simple terms.  Here might be a way to start:

A function is a machine - you put a number in, apply some rules to it, and get a number out.

For example, consider the function

f(x) = 2x.

Here, f is the name we have for our function.  The 'f(x) = 2x' tells us how the function works.  It means, for any number we stick in the machine, we double it to get the number that comes out:

f(1) = 2(1) = 2.

f(3) = 2(3) = 6

f(2.5) = 2(2.5) = 5

And so on.

We can define all sorts of functions.  We can make one that squares the input, then adds 7:

$f(x)=x^2 + 7$.

Or one that subtracts the reciprocal of the number we put in from 1:

$f(x) = 1 - \frac{1}{x}$.

And so on . . .

A function is something that passes the vertical line test. Hope this helps :)


Euler
number theory unsolved
number theory

Given a natural number $ n>1$.
Let $ F \equal{} \{ \frac{a}{b} | 1 \leq a <b \leq n \}$ prove that the number of element of set $ F$ greater than $ \frac{n^2 \minus{} 4}{8}$

I think, you calculate \[ \sum_{b\equal{}2}^n \phi (b).\]

Rust, I don't see the condition $ gcd(a,b)\equal{}1$ so the problem is just to calculate the sum $ 1 \plus{} 2 \plus{} ... \plus{} (n \minus{} 1) \equal{} \frac {(n \minus{} 1)n}{2}$ and now it's easy to see that $ \frac {(n \minus{} 1)n}{2}\ge \frac {n^2 \minus{} 4}{8}\Leftrightarrow 4(n^2 \minus{} n)\ge n^2 \minus{} 4\Leftrightarrow 3n^2 \minus{} 4n \plus{} 4\ge 0$, which I think it's obvious ;]

[quote="broniran"]Moreover, Rust, I think that if $ gcd(a,b) \equal{} 1$ is given then the problem is false since $ \frac {n^2 \minus{} 4}{8}\ge n$ for $ \forall n\ge 9$ and $ \varphi (n) < n$ for $ \forall n\in\mathbb{N}$, where $ \varphi(n)$ is Euler Totient function.[/quote]
If $ (a,b)\not \equal{} 1$, then $ \frac ab \equal{}\frac{a_1}{b_1}$ for some $ a_1 < a,b_1 < b$ and we calculate same element F before.
$ \sum_{k \equal{} 2}^n \phi(k) \equal{} \frac {3n^2}{2\pi^2}(1 \plus{} o(1)) > \frac {n^2}{8}$ for big n.


[quote="Meatman"]Calculate $ lg(tg{1}^{\circ})*lg(tg{2}^{\circ})*lg(tg{3}^{\circ})*...*lg(tg{89}^{\circ})$[/quote]

[hide="Hide"]
Note that $ \tan \theta ^\circ \equal{} \frac 1{\tan (90^\circ \minus{} \theta ^ \circ)}$.

[/hide]

[quote="Meatman"]Calculate $ lg(tg{1}^{\circ})*lg(tg{2}^{\circ})*lg(tg{3}^{\circ})*...*lg(tg{89}^{\circ})$[/quote]

Is $ lg$ is the function of $ log_{10}$ or $ log_e$?

If so,

$ \because lg( tg{45}^{\circ})\equal{}0$

$ lg(tg{1}^{\circ})*lg(tg{2}^{\circ})*lg(tg{3}^{\circ})*...*lg(tg{89}^{\circ})\equal{}0$


calculus
integration
trigonometry
probability
calculus computations

Prove that $\int_{-\pi}^{\pi}\cos(2x)\cos(3x)\cos(4x)...\cos(2005x)dx >0$

(M.Pupashenko)

Write $\cos nx=\frac12(e^{inx}+e^{-inx})$, and multiply out the sum. The integral becomes $\int_{-\pi}^{\pi}\sum_{k=-N}^{N}a_{k}e^{ikx}\,dx =\sum_{k=-N}^{N}a_{k}\int_{-\pi}^{\pi}e^{ikx}\,dx =a_{0}\int_{-\pi}^{\pi}1\,dx=2\pi a_{0}$,
where $N=2+3+\cdots+2005$ and each $a_{k}$ is nonnegative. We need to show that $a_{0}$ is actually positive.
For this, note that the term $\frac1{2^{2004}}e^{2ix}e^{-3ix}e^{4ix}e^{-5ix}\cdots e^{2002ix}e^{-2003ix}e^{-2004ix}e^{2005ix}=\frac1{2^{2004}}$ contributes positively to $a_{0}$. Many other terms do as well, and the integral is positive.

An easy parity argument shows that $a_{k}$ is zero when $k$ is odd. The $a_{k}$ give a probability distribution with mean zero and variance $v=2^{2}+3^{2}+\cdots+2005^{2}=2688727054$, so the normal approximation gives that $a_{0}\approx \int_{-1}^{1}\frac1{\sqrt{2\pi v}}e^{-x^{2}/(2v)}\,dx\approx 1.5387\cdot 10^{-5}$, so the integral should be approximately $9.668\cdot 10^{-5}$.
[Edit- fixed another minor mistake]
Can anyone verify this numerically?

With a tolerance of mach-eps MATLAB gives $9.6661\cdot10^{-5}$ using a Lobotto quadrature.

Nice. I'll take an error of one part in 5000, when all I used was a simple theoretical estimate. The $e^{-x^{2}/(2v)}$ is incredibly close to 1, so my estimate for $a_{0}$ is just the coefficient $\frac2{\sqrt{2\pi v}}$.


ratio integration logarithms function trigonometry calculus derivative

Anyone can proof this MacLaurin series :?:  :?: thanks

hm. I'm not sure how much proof you want, so I'll be somewhat pedantic about it.

lemma 1: the series for e^x takes on real values for real x.
proof: simple application of the ratio test.

lemma 2: the series (S(x)) for e^x satisfies the differential equation S'(x) = S(x)
proof: $\frac{d}{dx} \sum ^{\infty} _{n = 0} \frac{x^n}{n!}$
$= \sum ^{\infty} _{n = 0} \frac{d}{dx} \frac{x^n}{n!}$
$= \frac{d}{dx} \frac{x^0}{0!} + \sum ^{\infty} _{n = 1} \frac {x^{n-1}}{(n-1)!}$
$= 0 + \sum ^{\infty} _{n=0} \frac {x^n}{n!}$

lemma 3: all solutions of the differential equation S'(x) = S(x) are of the form A*e^x:
proof:  $\frac{dS}{dx} = S$
$\frac{dS}{S} =dx$
$\int \frac{dS}{S} = \int dx$
$ln(S) = x + C$
$S = e^{\ln(S)} = e^{x+C} = e^C * e^x = A*e^x$

so, we know the series for e^x converges to agree with a function of the form A * e^x. Point plugging at x = 0 yields A = 1, and so we're done for that one.

for sin(x) and cos(x), note that since $\frac{d}{dx} \cos(\frac{x}{i}) + i \sin(\frac{x}{i})  = \cos(\frac{x}{i}) + i \sin(\frac{x}{i})$, by lemma 3, and point plugging, we get the famous equation $e^{ix} = \cos(x) + i \sin(x)$.

now, consider the series expansion of e^ix:
$\sum _{n = 0} ^{\infty} \frac{(ix)^{n}}{n!}$
$=\sum _{n = 0} ^{\infty} \frac{(ix)^{2n}}{(2n)!} + \sum _{n = 0} ^{\infty} \frac{(ix)^{2n+1}}{(2n+1)!}$
$=\sum _{n = 0} ^{\infty} \frac{((i^2)^n){x^{2n}}}{(2n)!} + \sum _{n = 0} ^{\infty} \frac{((i^2)^n * i ){x^{2n+1}}}{(2n+1)!)}$
$=\sum _{n = 0} ^{\infty} \frac{(-1)^n}{(2n)!} x^{2n} + i \sum _{n = 0} ^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1}$
since both of these sums can be shown to converge to real values for real x, the real part must be the real part of e^ix = cos(x) + i sin(x), which is cos(x), and complex, sine.

(insert little end of proof box thingie)  :D

This actually works easily for many algebraic expressions whose derivatives cyle, or are easily known.

Other McClaurin series include:

$cos(x) = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2 n}}{(2 n)!}$

$ln(1+x)=\sum_{n=0}^{\infty} (-1)^{n-1} \frac{x^n}{n}$

Try to come up with series for  $tan^{-1}(x), \frac{1}{1+x}$ ...

Remember...some series only converge in certain intervals.


if 84 kadoodlers=7 kaboblers, and 36 kadoodlers=3 kaboblers, and 1 kabobler=5 whatchamacalits, then how many whachamacalits equal 120 kadoodlers?
(i'm really bad at making up problems, so don't be suprised if the problem is flawed)

ok i think this is right
[hide]
if 36 kadoodlers=3 kaboblers
then 12 kadoodlers=1 kabobler
so 120 kadoodlers=10 kaboblers
if 1 kabobler=5 whatchamacalits
then 10 kaboblers=$\boxed{50 whatchamacalits}$[/hide]

I hate these types of problems, mainly cuz the names are so similar >.> .

I remember I got one of these wrong at Chapter either last year or this year due to a stupid confusion between crapthing and craptheeng, or something like that. Prevented me from a 46.

[color=indigo]:rotfl: Since 36 kadoodlers=3 kaboblers, 12 kadoodlers = 1 kabobler, and 120  kadoodlers = 10 kaboblers. Now, because 1 kabobler = 5 whatchamacallits, you multiply 5 and 10 to get 50 whatchamacallits as your answer. :rotfl: [/color]

36 kadoodlers=3 kaboblers, 12 kadoodlers = 1 kabobler, and 120 kadoodlers = 10 kaboblers.
Because 1 kabobler = 5 whatchamacallits, 50 whatchamacallits as your answer.


Find the product of all values of x which satisfy: $ sin(4arctanx)\equal{}\frac{24}{25}$.
A) $ \minus{}1$
B) $ \minus{}\frac{25}{6}$
C) $ 1$
D) $ \minus{}\frac{1}{3}$
E) None of the Above

hello, we have $ x\equal{}\tan(\frac{1}{4}\arcsin(\frac{24}{25}))$ with $ x_1\equal{}\minus{}2,x_2\equal{}\frac{1}{2},x_3\equal{}\frac{1}{3},x_4\equal{}\minus{}3$, so we have $ x_1\cdot x_2\cdot x_3\cdot x_4\equal{}1$.
Sonnhard.

Dr s.Graubner
  can you be a bit more clear with x_1\cdot x_2\cdot x_3\cdot x_4=1


LaTeX
AMC
AIME
USA(J)MO
USAMO
email
AMC 10

The 2nd AMC 10 & 12 this year happens during our winter break. The school made decision not to offer this to us. There is another private school willing to let us take the test as a guest at their location. However, they want us to register with AMC directly and having the test material mailed to this private school. I looked at the AMC web site and noticed that in order to register, I need to pay $32 registration fee, plus $15 for ten sets of test (I only need one for myself), plus $12 shipping charge for a total of $59! This does not make sense to me. Should I convince my own high school to register for us or is there another way to do it at AMC? This private high school hosted such test before for guests and said we only need to pay $13 each at AMC. I am confused.

May I ask what private school this is? (I'm in the same situation as you, and I'm also from Michigan.)

"they want us to register with AMC directly and having the test material mailed to this private school. I looked at the AMC web site and noticed that in order to register, I need to pay \$32 registration fee, plus \$15 for ten sets of test (I only need one for myself), plus \$12 shipping charge for a total of \$59! This does not make sense to me.

Suggestion Number 1:  Find 9 friends at your school (or form a math club!) to take the B-date contest with you at the other school and divide the cost evenly among you.  It now is less than the cost of going to a movie.

"Should I convince my own high school to register for us "

Suggestion Number 2:   Yes, if you can get 10 total to take the contest at another school, it probably makes more sense then to take it at your own school.  I would be willing to bet you can find a teacher to help administer the B-date contest under those terms.  It doesn't have to be a math teacher, it could be a science teacher, a vice-principal, a librarian, any certified teacher.

As for it making sense, I have studied the issue just about every way I can and we charge about what it costs us to adminster the contests.  We do the same amount of work for a school or unit whether one person takes the contest, or 10 take the contest.   Only after 20 or more do the overhead costs not dominate.

"or is there another way to do it at AMC? This private high school hosted such test before for guests and said we only need to pay \$13 each at AMC. I am confused.""
Yes, we have a provision for a student unable to take a contest at their own school to take the contest at another school as a "visitor".   The hosting school must agree to host the visitor.   The visitor must pay a separate visitor fee, this year the visitor fee is \$32.  Last year it was less, but we found that handling each visitor took an enormous amount of time and we were getting killed on the costs.    I have done the cost-accounting to justify the costs, this year we should survive.

Steve Dunbar
AMC Director

Can anyone figure out how to make the Latex not take effect every time AMC director writes "$13" ?

When posting, click on the checkbox "Disable BBCode in this post".

tetrahedr0n: sorry to get back to you late. It was ICAE.

Thanks Mr. Dunbar. It looks like the AMC is more geared toward school sponsored competition vs individual participated activity. At least I am clear now with my options. Thanks again.

Question: Is it possible to take both the AMC 12-A AND the AMC 12-B? That is to say, if we fail the A, can we take the B? That seems to be a possibility from what I've been reading, but I remember my teacher last year saying you could only take one of them. Could someone clear this up for me?
Also, if it is possible to take both of them, which score will count? The highest one?

Thanks.

[quote="sran"]Question: Is it possible to take both the AMC 12-A AND the AMC 12-B?
Also, if it is possible to take both of them, which score will count? The highest one?
[/quote]

You can take both exams.  For the purposes of AIME qualification, if you qualify with either test, you qualify.  For the purposes of USAMO qualification (if you are in grades 11 or 12 -- otherwise, your AMC score is irrelevant), they take the higher score.

when are the AMC B tests? Aren't they on the 15th or something like that.
the reason our school doesn't take B tests is because they are a day or so before exams
so the most economical solution is to find like 9 more people to take AMCB's and find a convenient school which allows us and then take tests?

The "B" date in 2005 is Wednesday 16 February 2005.

http://www.unl.edu/amc/whatswhen.html

"Mr. Dunbar, I took the AMC 8 at another school because my own school
didn't let me take it. However have to pay any fees. That is allowed,
right?"

"Mr. Dunbar: I am taking AMC 10A at my high school and registered AMC
12B at another high school. Is this allowed if the other school
willing to take me? Is there any complication deciding qualification
for AIME if I took two AMC tests at two different schools?"

These situations are allowed, however these are the consequences:

1.  To us, our machine scoring, and our reporting of scores, it
    appears that you attend School A where you take the  contest, whereas in reality you attend School Z, across town.  That means:

    a.  We report your score to Teacher A at School A.  If Teacher
    A is good enough to call you up and report your scores to you,
    that is fine.  Likewise, it is good if Teacher A shares the
    solutions with you. But often students call us up or email us and ask
    us for the score.  That requires us to find your
    score under School A.  That is special handling which takes time
    from our small staff.  We do it, of course, but it is unfair under
    those circumstances to request special handling.

    b.  Suppose you qualify for an award, say the school winner pin,
    or maybe qualify for the AIME.  That award gets sent to Teacher A
    at School A.  Then Teacher A at School A has to notify you at
    School Z.  Because we believe you attend school A, we send the
    AIME automatically to school A.  If you qualify for the AIME, will
    Teacher A notify you in time?  We won't notify you, because we
    don't know you aren't at School A.

    c.  In some ways, it is unethical to take the contest at School A
    under those circumstances.  One might reasonably expect that
    students reading AoPS might do better on the AMC 12 and be the
    "school winner", perhaps preventing the "real" school winner from
    getting the award pin.   For another example, we use the school team
    score from the AMC 12 to help select the Sliffe Award for
    teachers.  If you contibute to the team score, that teacher may
    get an award which is deserved better by another teacher, perhaps
    even the teacher at your own school.

If you register as a "visitor" then we explicitly pull your answer
sheet from the hosting school and process, score, and report it
separately.  The "visitor" fee pays for the special handling and
separate reporting.

"Is there any complication deciding qualification
for AIME if I took two AMC tests at two different schools?"

Yes.  What if you qualify twice, which is entirely possible for an
AoPS reader?  Do we know if there is one John Smith or two in your
town?  If one, how can it be that one John Smith attends two high schools?
Which one is the correct high school, which one do we send the AIME
to?  What if School A wants to take the AIME on the first date, but
School Z wants to take the AIME on the Alternate date?

Again, if you register as a visitor, we pull your answer sheet and
handle it separately, and have a contact point with address, phone
number and eamil for answering all these questions.

Steve Dunbar
AMC Director

Mr. Dunbar,
How would one register for the AMC as a visitor?  I've currently made arrangements to take the AMC B at another school because my school does not offer it, but I can't seem to find anywhere on the form to indicate this.  Also, does my registration have to be sent seperately from  the other school's, and which school should I fill in on my answer sheet?
Thanks.


number theory
greatest common divisor
geometry
parallelogram
invariant
inequalities
modular arithmetic

sorry if this is not the appropriate place to poste this problem:

$ a_1 < a_2 < a_3$ are positive integers, proof that : $ \exists x_1,x_2,x_3 \in \mathbb{Z}$ st:

i)  $ |x_1| \plus{} |x_2| \plus{} |x_3| > 0$

ii)  $ a_1x_1 \plus{} a_2x_2 \plus{} a_3x_3 \equal{} 0$

iii) $ max_ {1 \le i \le 3}|x_i| \le \frac {2}{\sqrt3} \sqrt {a_3} \plus{} 1$

proof that  $ \frac {2}{\sqrt3}$is the best constant

Consider $ f(y)\equal{}a_1y_1\plus{}a_2y_2\plus{}a_3y_3,0\le y_1,y_2,y_3\le a$.
We had $ (a\plus{}1)^3$ means for $ f(y)$, when $ y_i\equal{}0,1,...,a$. All means $ f(y)$ satisfyed $ 0\le a(a_1\plus{}a_2\plus{}a_3), gcd (a_1,a_2,a_3)|f(y)$.
Therefore, if $ (a\plus{}1)^3>\frac{a(a_1\plus{}a_2\plus{}a_3)}{gcd(a_1,a_2,a_3)}$ then exist 2members $ y^{(i)}\equal{}(y_{1}^{(i)},y_2^{(i)},y_3^{(i)})$, suth that $ f(y^{(1)})\equal{}f(y^{(2)})$. Let $ z\equal{}(z_1,z_2,z_3), z_i\equal{}y_i^{(1)}\minus{}y_i^{(2)}$.
We had $ f(z)\equal{}0$ and $ max_i|z_i|\le a$.
Let $ N\equal{}\frac{a_1\plus{}a_2\plus{}a_3}{gcd(a_1,a_2,a_3)}$ and $ a$ is minimal natural, suth that $ a^2\plus{}3a\plus{}3\ge N$.
Then exist $ (x_1,x_2,x_3)$, suth that $ f(x)\equal{}0, max_i |x_i|\le a\equal{}[\sqrt{N\minus{}3/4}\minus{}\frac 12].$

[b]Rust[/b]'s argument doesn't suffice for some $a,b,c$.

It seems that geometry is the way to go here, although I haven't managed to prove that $2/\sqrt{3}$ is the best constant yet... for convenience, let $(a,b,c,x,y,z)=(a_1,a_2,a_3,x_1,x_2,x_3)$.

Of course, it suffices to prove the bound for $\gcd(a,b,c)=1$. Let $T=\{(x,y,z)\in\mathbb{R}^3\mid ax+by+cz=0\}$, $L=T\cap\mathbb{Z}^3$, and $S_t=[-t,t]^3\cap T$.

[b]Lemma 1:[/b] If $\gcd(a,b,c)=1$ for positive integers $0<a<b<c$, then $L=\{pv_1+qv_2\mid p,q\in\mathbb{Z}\}$ is a lattice for some basis $\{v_1,v_2\}$ (with $\dim(v_1)=\dim(v_2)=3$) with fundamental parallelotope $P=\{t_1v_1+t_2v_2\mid t_1,t_2\in[0,1)\}$ (which in this case is a parallelogram) of volume $\det(L)=V(P)=\sqrt{a^2+b^2+c^2}$.

[i]Proof:[/i] There's most likely a cleaner, more general way (which I asked about [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=352&t=438331]here[/url]), but I'll just give an ad hoc solution here. Let $a=ga'$ and $b=gb'$ for $g=\gcd(a,b)$, and take integers $u,v$ such that $a'u+b'v=1$. Then $L=\{(-cru+sb',-crv-sa',gr)\mid (r,s)\in\mathbb{Z}^2\}$. The volume of $P$ is invariant under the transformations $(v_1,v_2)\mapsto(v_1\pm v_2,v_2)$ and $(v_1,v_2)\mapsto(v_1,v_2\pm v_1)$ (and we can get all possible bases this way), so WLOG $v_1=\langle{-cu,-cv,g}\rangle$ and $v_2=\langle{b',-a',0}\rangle$, whence
\[V(P) = \| v_1\times v_2 \| = \| \langle{ga',gb',cua'+cvb'}\rangle \| = \| \langle{a,b,c}\rangle \| = \sqrt{a^2+b^2+c^2},\]as desired.$\blacksquare$

[b]Lemma 2:[/b] For all $0<a<b<c$ we have $|S_t|>3t^2\sqrt{a^2+b^2+c^2}/c$.

[i]Proof:[/i] WLOG $t=1$ by dilation. Let $S$ denote the points of intersection that lie on edges of the cube. It's easy to check that because $|a+c|\ge |b|$ and $|b+c|\ge |a|$, $|S|\le 8$ and the possible points satisfy
\begin{align*}
&\pm\left(1,1,\frac{-a-b}{c}\right)\in S\Longleftrightarrow a+b\le c, \\
&\pm\left(1,-1,\frac{-a+b}{c}\right)\in S, \\
&\pm\left(1,\frac{-a+c}{b},-1\right)\in S\Longleftrightarrow c-a\le b\Longleftrightarrow a+b\ge c, \\
&\pm\left(\frac{-b+c}{a},1,-1\right)\in S\Longleftrightarrow c-b\le a \Longleftrightarrow a+b\ge c.
\end{align*}We thus have two cases to consider:

[list][b]Case 1:[/b] If $a+b\le c$, then $S$ is a centrally symmetric quadrilateral (i.e. a parallelogram) centered at $(0,0,0)$ with vertices $(1,1,-(a+b)/c)$, $(1,-1,(b-a)/c)$, $(-1,-1,(a+b)/c)$, $(-1,1,-(b-a)/c)$ in cyclic order, so
\[|S| = \left\| \left\langle{0,2,-\frac{2b}{c}\right\rangle\times \left\langle{-2,0,\frac{2a}{c}\right\rangle \right\| = \left\| \left\langle -\frac{4a}{c},-\frac{4b}{c},-4\right\rangle \right\| = \frac{4\sqrt{a^2+b^2+c^2}}{c},}}\]which is clearly larger than $3\sqrt{a^2+b^2+c^2}/c$.$\blacksquare$

[b]Case 2:[/b] If $a+b\ge c$, then $S$ is a centrally symmetric hexagon centered at $O=(0,0,0)$ with vertices $V_1=(1,-1,(b-a)/c)$, $V_2=(1,(c-a)/b,-1)$, $V_3=((c-b)/a,1,-1)$, $-V_1=(-1,1,-(b-a)/c)$, $-V_2=(-1,-(c-a)/b,1)$, $-V_3=(-(c-b)/a,-1,1)$ in cyclic order, so because $V_1\to V_2\to V_3$ is a cyclic shift,
\begin{align*}
|S|
&= 2[V_1OV_2]+2[V_2OV_3]+2[V_3O(-V_1)] \\
&= \|V_1\times V_2\|+\|V_2\times V_3\|+\|V_3\times V_1\| \\
&= \sum_{\text{cyc}}\|V_1\times V_2\| \\
&= \sum_{\text{cyc}}\left\| \left\langle 1-\frac{b-a}{c}\frac{c-a}{b},\frac{b-a}{c}+1,\frac{c-a}{b}+1\right\rangle \right\| \\
&= \sum_{\text{cyc}}\frac{b+c-a}{bc}\sqrt{a^2+b^2+c^2} \\
&= \frac{2(ab+bc+ca)-(a^2+b^2+c^2)}{abc}\sqrt{a^2+b^2+c^2} \\
&> \frac{3\sqrt{a^2+b^2+c^2}}{c},
\end{align*}where the inequality becomes $uv+vw+wu>3w^2$ under the substitution $(a,b,c)=(v+w,w+u,u+v)$ for some $u>v>w\ge 0$ (because $0<a<b<c$ and $a+b\ge c$).$\blacksquare$[/list]

By Minkowski's theorem, $S_t=[-t,t]^3\cap T$ (which is obviously convex and centrally symmetric about the origin) contains a nonzero point in the lattice $L\subset T$ as long as
\[|S_t|>2^{\dim(T)}\det(L)=4\sqrt{a^2+b^2+c^2}.\]Thus by lemma 2, $|S_t\cap L\setminus\{(0,0,0)\}|\ge 1$ for $t=\sqrt{4c/3}$, as desired.

OK, here's a proof that $2/\sqrt{3}$ is the best constant, motivated by the fact that the Minkowski bound above is only close to being tight when $a\approx b\approx c$ but at the same time the solution $(b-c,c-a,a-b)$ forces $c-a=\sqrt{4c/3}+O(1)$.

Take $(a,b,c)=(3k^2-2k,3k^2-k,3k^2+1)$ for large $k$ such that $2\nmid k+1$ and $7\nmid k-3$; then
\begin{align*}
\gcd(a,c)=\gcd(3k-2,3k^2+1)
&=\gcd(3k-2,2k+1) \\
&=\gcd(k-3,2k+1)=\gcd(k-3,7)=1
\end{align*}and
\[\gcd(b,c)=\gcd(3k-1,3k^2+1)=\gcd(3k-1,k+1)=\gcd(-4,k+1)=1.\]It suffices show that if $ax+by+cz=0$, then $\max(|x|,|y|,|z|)=\sqrt{4c/3}+O(1)=2k+O(1)$, because we can take $k$ arbitrarily large.

Suppose to the contrary that $|x|,|y|,|z|\le 2k-3$. Since $k\mid a,b$, and $\gcd(k,c)=1$, $z=kz'$ for some $|z'|<2$; WLOG $z'\le 0\implies z'\in\{-1,0\}$ (negate everything if necessary). If $z'=0$, then $(3k-2)x+(3k-1)y=0\implies 2k-3\ge |x|\ge 3k-1$, which is absurd. Otherwise, $z'=-1$, so $(3k-2)x+(3k-1)y=3k^2+1$. Then because $|y|\le 2k-3$, we must have
\begin{align*}
y\equiv 3k^2+1\equiv 2k+1\pmod{3k-2}
&\implies y=-k+3 \\
&\implies x=\frac{3k^2+1+(3k-1)(k-3)}{3k-2}=2k-2,
\end{align*}contradicting our assumption that $|x|\le 2k-3$.


calculus
integration
function
trigonometry
calculus computations

\[ \int {(\sin x\sqrt {\sin 2x} )} dx
\]

hello, set $ t\equal{}\sqrt{\sin(2x)}$.
Sonnhard.


when you say that the AMC-12 PS  will have " Some material from this class comes from the AMC/AIME Problem Series offered in winter 2003-4," how much of an overlap are you implying?

I can't give an exact number, but I would guess than about half the material is new since we split the AMC/AIME pretty much in half and filled in the rest of the material.  There are more Message Board problems this time around however.


geometry
geometric transformation
reflection
circumcircle
Euler
geometry unsolved

Consider the isosceles triangle $ ABC$ with $ (AB)\equal{}(AC)$, and a circle of center $ O$ which passes through $ A$ and $ C$ so that the point $ B$ is in the interior of the circle. The height $ AD$ of the triangle $ ABC$ $ (D \in BC)$ meets the circle again in $ F$. Prove that:
$ a)$ the distances from the middle of $ (OB)$ to the middles of the sides of the triangle $ AFE$ are equal.
$ b)$ $ DG \perp EO,$ where $ G\equal{}EF \cap AB.$

According to my figure I guess that $ E \equal{} BC \cap (O)$ , C is the reflection of of B about AF , C lies on its circumcircle ---> B is orthocenter of $ AEF$ . the middle of OB is the center of the Euler circle of $ AEF$ therefore it equidist from the middles of the sides of AEF , However for the part b) with this notation is not true , could you check again the enuntiation. thanks


Five sailors are marooned on an island.  So they can survive until the day that they are (hopefully) rescued, the sailors and a native monkey spend the day gathering coconuts, which are all put in a big pile at the end of the day, after which they all fall asleep.

During the night, one of the sailors gets up, and, not trusting the other sailors to distribute the coconuts evenly, decides to take his $\textstyle\frac{1}{5}$ of the pile of coconuts for his own then and there.  When he divides the coconuts into five equal piles, however, he has one left over, so he tosses it to the monkey.  He then hides his share, puts the rest of the coconuts in one pile again, and falls asleep.

Later in the night, one of the other sailors gets up, and, also not trusting the other sailors to distribute the coconuts evenly, decides to take his $\textstyle\frac{1}{5}$ of the pile to avoid getting shortchanged in the morning.  Like the first sailor, he finds that when he divides the pile into five equal piles, there is one coconut left over, so he tosses that to the monkey.  He hides his share, puts the rest of the coconuts in one pile again, and falls asleep.

As you have probably guessed, the other three sailors do the same thing.  Each time they split up the coconuts into five equal piles; each time they find one leftover coconut, which is tossed to the monkey; and each time they take their $\textstyle\frac{1}{5}$ of the remaining coconuts before putting the rest back into a big pile, hiding the coconuts that they took, and going back to sleep.

In the morning, when the sailors get up, they all pretend that they have taken none of the coconuts during the night and split up the remaining pile as they had planned; each sailor gets $\textstyle\frac{1}{5}$ of the pile, except that when they form five equal piles, they again find that there is one coconut left over, which they throw to the monkey.

What is the least number of coconuts they could have gathered on their first day on the island?

Can you generalize?  If there are n sailors, and at each division the monkey gets k coconuts tossed to it, what is the least number of coconuts they could have gathered (in terms of n and k)?

Click [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=114986#114986]here[/url] to reach a thread with a harder version of this problem.

A slightly smart-alecky answer for the original number of coconuts: $-4$. If you can understand how that could work, you're on your way to understanding the "real" solution to this problem.

Yep, -4 is indeed a potential theoretical solution.  Can anyone else (I assume Kent Merryfield already knows the answer) use this to find the actual lowest possible answer?

one can always add  5^n (n=6 or 5 .. too lazy to read the whole problem) .. to an existing solution to get  another solution...


can help me delete this post? i will post once new

[quote="fedja"]There is no way. It is just false as stated.[/quote]

counterexample?

[quote="alekk"]if a number can be written as $ x \equal{} a \plus{} b \sqrt {2}$ with $ a,b \in \mathbb Q$, define $ f(x) \equal{} a \plus{} b$. Otherwise, define $ f(x) \equal{} 0$.[/quote]

sorry i write wrong .

not + but .

luckman168: I'm not going to delete this.

I'm also going ask you to please refrain from what you've done in this thread. Please do not make changes (other than grammar and notation corrections) to your posts after they've already drawn substantive answers. I came across this thread after you had already removed the body of your original post, and it's fairly hard to tell what is going on. Next time, just leave it up there, even if you're wrong, and if you want to retract something you've said, make another post to do that.

thank you, Kent Merryfield.

i see.

please del .


number theory
greatest common divisor
number theory proposed

consider a set of $ n$ positive integers : $ (a_1,\cdots,a_n)$.  Let $ d_k$ be the gcd of all products of $ k$ integers from the given set (where $ 2\le k\le n \minus{} 2$.  Show that $ d_k^2|d_{k \minus{} 1}d_{k \plus{} 1}$.


Ring Theory
calculus
integration
algebra
function
domain
abstract algebra

let R be a UFD, if each not zero prime ideal be maximal,prove R is PID.  :oops:

It is one of the characterizations of UFD's that they are integral domains in which every nonzero prime ideal contains a prime element. Every prime element generates a prime ideal. These two combined prove that every nonzero prime ideal of $R$ is principal.
Cohen's Theorem states that a ring in which every prime ideal is finitely generated is noetherian.

Assume $R$ is not a PID. Since it is noetherian, there exists an ideal $I$ maximal among the ideals that are not principal.
Let $I=(x_{1},\ldots,x_{n})$. This is possible because $R$ is noetherian.
Let $d=gcd(x_{1},\ldots,x_{n})$. This exists because $R$ is an UFD.

Assume $d\neq 1$.
Let $J=I/d$. This is an ideal of $R$ that strictly contains $I$ ($d$ is the gcd of all the elements in $I$ and $d^{2}$ is the gcd of all the elements of $dI$).
By the maximality assumption, $J=(e)$ for some $e\in R$. Then $I=(de)$ which contradicts $I$ not being principal.

Therefore $d=1$.
Since $I$ is maximal among the non-principal ideals of $R$, it cannot be the unit ideal. Consider a maximal ideal $m=(p)$ of $R$ such that $I\subset m$. This implies $p|x_{i}$ for all $i$, so $p|d=1$ which is a contradiction.

We have proved that $R$ is a PID.


[hide="old message"][quote="bubka"]This will be a hypothetical conversation between two or more persons. The scenario can be changed every 100 posts or so. The speech of the same person can not continue beyond one post, nor can two persons speak in the same post. [/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=84794&start=220

[quote="later, bubka"]Ok this has got a long enough break. I will soon start a new topic: "Conversation II". I got this bumped up so that people could read through the early works and so on. In fact the first piece we created was quite good.

I don't think this should be locked or anything. You can discuss general things regarding 'conversation' here. However people start spamming here badly, you may lock this topic.[/quote]

ok guys, new conversation

Person A: mental person with anger management
Person B: pychologist

Note: if this topic gets too spammy, we'll get it locked

EDIT: guys, if u want to change the characters and plotline, that is okay[/hide]
[b]
NEW MESSAGE:[/b]

This is a conversation between several people. It works in this format:[list]The person who is speaking is represented by a couple of letters (i.e. Darth Vader is DV, God is G, etc.)
This is how people talk: DV: blahblahblah (DV is the talker, blahblahblah is wat he says).
It is cool to talk in CAPS LOCKS.
Whenever the character does something, it goes like this:
[code]*J eats a rotten banana*[/code] The ** mark the action, it is easier to classify actions that way.
Whenever you want to introduce a character, you "invite" him / her into the conversation.
Whenver you want to take away a character, you make him / her die. The preffered way of making it die is like this:
[code]MT shoots rockets at A[/code] That is where A is the person you want to get rid of.
There is no plot line. You make up everything along the way.
[/list]

Okay I give this topic official endorsement for now, but if it fails, I must start one myself.

C: Since the other game did this, I'll start it off immediately by doing this.

(confused person standing nearby): What are you talking about? Also, who are you?

A: I'm A, this is B and this is C. We are duscussing on what to do.

(G(od)): They are who they are. Their names are A, B and C.

CP(confused person): Am I hearing things?

G=God
A=Amelia, the salesperson at Walmart
B=Bob, the person looking around
C=Carl, who was also looking around who is a friend of Bob
CP= Carmen Potato, Carl's pet potato
____________________________________________________

C: shut up CP

B: CP... sounds familiar...

*walks towards CP and C*

CP: I'm a confused person. I'm going home, since C think that I'm a potato, which I am. *marches home*
CP(couch potato): Do'h! I said I WAS a potato! *slams head everywhere*

C: Go home
CP:ok
C: SHUT UP
*CP goes home and dies from grief*

C: OH GOD did I just kill my best friend oh man life sucks now it is best to just end it <hangs himself>

G: what oh you're dead haha sucks

n(utz): MT, please say things we can understand and qualify to the faith in G.
MT(MysticTerminator): WHAT DID YOU SAY I'M SORRY I CAN'T HEAR YOU
n: And STOP typing in capitals all of the time.

night breaks in mafia town. the blankets are covered up almost everywhere......

MT: OH MAN AM I IN THIS STORY NOW HEY GOD WHAT UP
G: DUDE MT YOU ROCK LIKE REALLY YOU ARE OFFICIALLY THE DEPUTY GOD FOR WHEN I'M AWAY
MT: OK THANKS

n: what what that's not fair

MT: DUDE NO YOU GET STOMPED <STOMP>
G: GOOD GOING MT

T: let's talk about ga-
MC: no. ban.
T: i'll beha-
MT: I KNOW A COUPLE GAY GUYS AT SCHOOL THEY'RE WEIRD

[quote="nutz_for2.718281828"]everyone: Hi, stfu![/quote]

Again, please refrain from spamming on this website.

T: What happened to the Nazguls though?
B: Probably gone to sleep.
b: Oh the bush is still here?
N1: Word.

stfu: I know who's mafia! T is mafia, kheg_k's mafia, etc. is mafia! :rotfl:

hey nutz guess who wins the "you're not funny" award and the "let's use  :rotfl: all the time" award?

N1: Word.
b: The others are at least trying to be funny you see.

G:OMG THIS CANNOT DIE
DV:tOTalLY nOt!

P (pizza dude): OMG I HAVE THE PIZAA

N1: Guys, this conversation is getting really uninteresting
N2: Word
N3: Ok we need to stop thinking about food and mafia
Iver07: word
N4: Let's go pla golf
N5: sure
N9: I have the best swing
N15: so what
Gandalf: can we cheat and make the ball immidiately end up in the hole?
N1: it's Gandalf!!!!!
Nzguls: After him!

N4 and N7: Naa ... we are too tired.
Gandalf: Me too! Me too! Me too!
Iver07: There's a golf course beyond that bush.
B: No No not me.
b: No one's talking of you.
R(em): We are talking of that mulberry bush.
T: So shut up.
MT: HEY DUDE WHY PLAY GOLF LETS HAVE A DINNER AT GOD'S
N1: Word.

Gandalf: Hey wait! When did these additional 8 Nazgul appear? I thought there were but 7 of the Ringwraiths!

G: O SRRY GUYS YEA I FELT LIKE VISITING THE GRAVEYARD AND ACCIDENTLY *COUGH* REVIVED ALL OF THE NAZGULS

*OOPS*


Problem:
Let $ p$ be a prime and let $ x_1,x_2,...,x_p$ be nonnegative integers.Prove that if $ x_1^k\plus{}x_2^k\plus{}....\plus{}x_p^k$ is divisible by $ p$ for any $ k \in \{1,2,...,p\minus{}1\}$,then there are 2 distinct numbers $ k,l \in \{1,2,...,p\}$ such that $ x_k \equiv x_l (mod p)$.

[quote="SUPERMAN2"]Problem:
Let $ p$ be a prime and let $ x_1,x_2,...,x_p$ be nonnegative integers.Prove that if $ x_1^k \plus{} x_2^k \plus{} .... \plus{} x_p^k$ is divisible by $ p$ for any $ k \in \{1,2,...,p \minus{} 1\}$,then there are 2 distinct numbers $ k,l \in \{1,2,...,p\}$ such that $ x_k \equiv x_l (mod p)$.[/quote]
Not that hard I think. Suppose all $ x_i$ are different $ \mod p$, so we can assume wlog $ x_i\equiv i$ for all $ i$.
But it is well known that $ i^{p\minus{}1}\equiv 1(mod p)$ for all $ i\equal{}1,...,p\minus{}1$, so for $ k\equal{}p\minus{}1$, we have
$ x_1^k \plus{} x_2^k \plus{} .... \plus{} x_p^k\equiv p\minus{}1\not\equiv 0(mod p)$. contradiction :)

[quote="spanferkel"][quote="SUPERMAN2"]Problem:
Let $ p$ be a prime and let $ x_1,x_2,...,x_p$ be nonnegative integers.Prove that if $ x_1^k \plus{} x_2^k \plus{} .... \plus{} x_p^k$ is divisible by $ p$ for any $ k \in \{1,2,...,p \minus{} 1\}$,then there are 2 distinct numbers $ k,l \in \{1,2,...,p\}$ such that $ x_k \equiv x_l (mod p)$.[/quote]
Not that hard I think. Suppose all $ x_i$ are different $ \mod p$, so we can assume wlog $ x_i\equiv i$ for all $ i$.
But it is well known that $ i^{p \minus{} 1}\equiv 1(mod p)$ for all $ i \equal{} 1,...,p \minus{} 1$, so for $ k \equal{} p \minus{} 1$, we have
$ x_1^k \plus{} x_2^k \plus{} .... \plus{} x_p^k\equiv p \minus{} 1\not\equiv 0(mod p)$. contradiction :)[/quote]
nice solution :D
If we remove the condition for any  $ k\in {1,2,..,p\minus{}1}$ replace by $ k\in {1,2..,p\minus{}2}$ problem is true???

[quote="math10"][quote="spanferkel"][quote="SUPERMAN2"]Problem:
Let $ p$ be a prime and let $ x_1,x_2,...,x_p$ be nonnegative integers.Prove that if $ x_1^k \plus{} x_2^k \plus{} .... \plus{} x_p^k$ is divisible by $ p$ for any $ k \in \{1,2,...,p \minus{} 1\}$,then there are 2 distinct numbers $ k,l \in \{1,2,...,p\}$ such that $ x_k \equiv x_l (mod p)$.[/quote]
Not that hard I think. Suppose all $ x_i$ are different $ \mod p$, so we can assume wlog $ x_i\equiv i$ for all $ i$.
But it is well known that $ i^{p \minus{} 1}\equiv 1(mod p)$ for all $ i \equal{} 1,...,p \minus{} 1$, so for $ k \equal{} p \minus{} 1$, we have
$ x_1^k \plus{} x_2^k \plus{} .... \plus{} x_p^k\equiv p \minus{} 1\not\equiv 0(mod p)$. contradiction :)[/quote]
nice solution :D
If we remove the condition for any  $ k\in {1,2,..,p \minus{} 1}$ replace by $ k\in {1,2..,p \minus{} 2}$ problem is true???[/quote]
of course not. Put $ x_i\equal{}i$ and see the very recent one
http://www.mathlinks.ro/viewtopic.php?t=285383


Prove that the product of 8 consecutive integers cannot be the square of a perfect square (a perfect fourth power).

[quote="krustyteklown"][quote="chess64"][quote="cincodemayo5590"][hide="counterexample"]
$0\cdot 1 \cdot 2 \ldots \cdot 7 = 0^4$  :P [/hide][/quote]

A counter-example doesn't show anything. If a set doesn't have a certain property $p$, that doesn't show that all sets don't have property $p$.[/quote]
How does it not show anything, he proved that the product 8 consecutive integers [i]can[/i] be a perfect 4th power.

I'm pretty sure MCrawford wasn't looking for this trivial case tho...[/quote]

oh, uh i misread, i thought he was providing a counter-example that showed that the product of 8 consecutive integers wasn't a perfect 4th power

sorry :blush:

Let $p(x)=x(x+1)(x+2)(x++3)(x+4)(x+5)(x+6)(x+7)$
Then $p(x)=(x^2+7x)(x^2+7x+6)(x^2+6x+10)(x^2+7x+12)$
As stated by JBL, $p(x)<(x^2+7x+7)^4$.
If we write $u=(x^2+7x)$
Then $p(x)=u^4+28u^3+252u^2+720u$
While $(u+6)^4=u^4+24u^3+216u^2+864u$
$p(x)>(u+6)^4$ except for a finite number of values of u.....

In fact, $p(x)-(u+6)^4=4u^3+36u^2-144u=4u(u+12)(u-3)$
Therefore, $p(x)>(u+6)^4$ except for $u \leq 3$.
That is, $x^2+7x\leq 3$, which is impossible for positive integer x.

In conlusion, $(x^2+7x+6)^4<p(x)<(x^2+7x+7)^4$ for positive integer x, and hence the problem is solved.

[b]Question[/b]
[i]Is there any generalization to this problem?[/i]


AMC
AMC 12
AMC 12 A
AMC 10
AIME
USA(J)MO
USAMO

Although the AMC directors didn't put that directly on the home page under "what's new", http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2006-12a/06AMC1012dir-pict.html  the state stats is available now. Just click your state icon.
How well you did on AMC in your state? How well your school did in your state?
Do you get the first place in your state? Then you can get a plaque from the AMC!
I only get 3rd in my state in AMC12A, and I didn't participate in B. ;) need a little bit more effort!

[quote="Go Beyond"]Although the AMC directors didn't put that directly on the home page under "what's new", http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2006-12a/06AMC1012dir-pict.html  the state stats is available now. Just click your state icon.
Do you get the first place in your state? Then you can get a plaque from the AMC!
I only get 3rd in my state in AMC12. ;) need a little bit more effort![/quote]

What??!!!! You get a plaque??!!! NOOOOOO!!!!!  OK i going to prepare for math starting now! :bomb:  :wallbash_red:  :cursing:
[color=white]
oh well i still have 3 more years........[/color]

Wow, you're going crazy over a plaque? :huh: *sighs* materialistic... :D

I didn't receive a plaque. But there do have a plaque offering to each top 1 student in AMC12 A,B，　and AMC10 A,B, four divisions, in each state. (I think it true!)

Yay! plaque.
I already have enough on my teeth.  :D
In GA, it is easy to win.

How do we get the plaque.

SECOND IN WASHINGTON IN 12B GODDAMN YOU ADAM HESTERBURG  :cursing:

its hesterberg :P

i tied him on 12 A, lol  :)

[quote="t0rajir0u"]SECOND IN WASHINGTON IN 12B - YOU ADAM HESTERBURG  :cursing:[/quote]
What did you get on it? I got a 139.5 on the 12B, 1st in IN. Do I get a plaque?  I got second on the AMC 12A in IN with a 143.0. YOU DAVID BENJAMIN (150)!!! jk, lol. he's cool

lol, he got 139.5 too  :)  (adam got 150 )

Yeah, Adam hasn't missed or failed to answer an AMC problem since '04, his freshman year. He's amazing.

Hmm... Dropped from first in my grade last year to fourth this year on 10A. :? Oh well.

2nd in PA on AMC 10A, 4th in PA on the AMC 12B.

I am surprised to see that no 9th or 10th grader in PA scored higher than me on either AMC 12.   :o

[quote="nat mc"][quote="lordhighgoose"]Tied for 2nd in Minnesota with 136.  Shui Hu got 143.[/quote]
On what conteest? 10? 12? A/ B?[/quote]
Oh, sorry. I was talking about 12 in general.  On the A I did badly, on the B I got first.

You guys need to chill. Quit complaining about getting "only" second or third, think about what that does to people who don't land anywhere on the map... lame.

Oh, nooooo.... the horrid moral guilt of complaining about second or third.

Seriously, I don't think the people are taking themselves very seriously to be complaining - so it's all just banter... and amounts to nothingness. And criticizing it, well... adds more to the meaninglessness of it all. So er... I think my point is that every time people say something self-deprecating, there's always the police patrol on the forums that's like the angel whispering in the ear, "Be content, and don't be so presumptuous as to complain about being the runner up," but seriously... there's no need to make yourself the police... if it gets out of hand, a moderator or administrator will handle it.

Sorry I offended you whitehorse. It's just a maturity thing, don't worry about it. And I know it's just banter and that people aren't taking themselves seriously, but that adds to the problem because the people who would [i]kill[/i] to be second or third or twentieth in their state observe that silly behavior and feel lousy. One has to wonder: if you guys think second is so lousy, what would you think of someone who tries really hard only to get 110 and then sees people bantering about 140s?

So no need to feel like I'm policing you, just telling it like it is. Besides, I don't need to "police," because an administrator-- Mr. Rusczyk, as a matter of fact-- ran a great thread with "not complaining about your high scores" as a topic last year. So if you won't take it from me, take it from him-- you seemed to like the idea of administrators running this business anyway :)

Remember, it's all about learning.

What if you get scores on AMC below 120 *really ticking me off* (114.5 last year and 118 this year) for two straight years yet high AIME scores (11 and 8)?  Then you'd feel like an even bigger idiot...

To be more specific, you'd feel just like me.

hey now, I know some people in our state who definitely should have made AIME and could have had a run for USAMO, but they somehow messed up on AMC. And Yannik, you made USAMO last year and even if u didn't this year, those are still terrific scores :)

State winner plaques arrived here today by DHL.

[quote="mathgeniuse^ln(x)"]We still haven't received ours yet.  We are from Michigan.[/quote]
Same here, Indiana. We've gotten the team plaque (state winnners) but not individual (state winners)

School just finished, so I'll probably get my plaque in August.

Hold it. Do you get stuff for best in your grade in your state?

The state winner plaque is for the highest score in the following categories:

AMC 10A
AMC 10B
AMC 12A
AMC 12B

Aparently, If there is a tie, everyone in the tie gets a plaque.

Highest in your grade level does not get a plaque.

I thought it was only for AMC12. Hm, this means that I should get one- I live like 100 miles from my school though, so maybe they'll just give it to me next year.

I got my first place plaque for 12B about 2 weeks ago

yesterday morning I got my plaque, 10A, easiest test woohoo!!


When proving trig identities (meaning that one side is equal to the other), are we only supposed to work with one side of the equation or are we allowed to work with both?

For example, take the following problem

\[ \sin ^2 \theta \plus{} \cos ^2 \theta \equal{} \csc ^2 \theta \minus{} \cot ^2 \theta
\]

Fairly simple - because both sides is based off of the the Pythagorean Identities. My teacher, however, says that simply writing $ 1\equal{}1$ (as they both simplify to that) is [b]incorrect[/b] because you are working with both sides of the equation (which is apparently not allowed at all under no circumstances) and the correct method would be to start on the right hand side, and rewrite everything in terms of $ \sin$ and $ \cos$. [b]Is this because we cannot simply assume given identities to be correct?[/b] In that case, that would be understandable, but what about the following problem? [b]Am I not allowed to take everything in base $ e$? Am I not allowed to rewrite the left hand side for simplicity?[/b]

\[ \minus{}\ln (1\plus{} \cos \theta) \equal{} \ln (1\minus{} \cos \theta) \minus{} 2 \ln |\sin \theta|
\]

When I'm trying to prove an identity of any kind, say $ \textbf{X}\plus{}\textbf{Y}\equal{}\textbf{Z}$, (this isn't an identity, the letters are just meant to stand for some arbitrary quantity), I typically work from the left side and try to turn it into the right side (unless I'm instructed to do otherwise).  So as far as I'm concerned, the right side (or left side) is a complete mystery to me.

For the second one, no need to take anything to the base $ e$.  Note that
$ \minus{}\ln(1\plus{}\cos\theta)\equal{}\ln\left|\frac{1}{1\plus{}cos\theta}\right|\equal{}\ln\left|\frac{1\minus{}\cos\theta}{(1\minus{}\cos\theta)(1\plus{}\cos\theta)}\right|
\equal{}\ln\left|\frac{1\minus{}\cos\theta}{1\minus{}\cos^{2}\theta}\right|$
$ \equal{}\ln\left|\frac{1\minus{}\cos\theta}{\sin^{2}\theta}\right|
\equal{}\ln\left|1\minus{}\cos\theta\right|\minus{}\ln\left|\sin^{2}\theta\right|\equal{}\ln\left|1\minus{}\cos\theta\right|\minus{}2\ln\left|\sin\theta\right|$,
as desired.

[quote="JRav"]For the second one, no need to take anything to the base $ e$.[/quote]

I know; I'm just saying it as a generalization. Is it "illegal" to do that?

Two rules of thumb.

1)  Anything that is [i]reversible[/i] is logically fine.  For example, multiplying by a nonzero real number or applying another known identity.  It really doesn't matter which side of the equation you manipulate; I have no idea why high school teachers insist on this rule, except insofar as students might be less careful this way.  To denote a reversible implication, one simply writes

$ x \equal{} y \Leftrightarrow$
$ z \equal{} w$.

2)  Nevertheless, you should write things in the order that you deduce them.  That is, start from true statements and turn them into the statement you're trying to prove, not the other way around.

Haha I just got docked 2 points on a 14-point quiz for working with both sides. And all I did was multiply both sides by $ \minus{}1$ and by the denominator of one side...

Apparently your teacher doesn't want you to work with both sides of the equation-it seems you're supposed to "discover" the other side.  That's why you lost points.  What was the problem?

[quote="7h3.D3m0n.117"]by the denominator of one side...[/quote]
This isn't reversible if the denominator can be zero.  Trig identities do have the property that both sides "divide by zero" at the same values of $ x$, but it's still better to avoid this.

When you work on both sides, you should be able to change them as much as you want as long as their values each stay unchanged.

So overall, I think your first thing (changing it to $ 1 \equal{} 1$) is allowed. Your teacher probably wanted you to show it only to one side though, like this:
$ \sin ^2 \theta \plus{} \cos ^2 \theta \equal{} \csc ^2 \theta \minus{} \cot ^2 \theta$
$ 1 \equal{} \csc ^2 \theta \minus{} \cot ^2 \theta$
$ \csc ^2 \theta \minus{} \cot ^2 \theta \equal{} \csc ^2 \theta \minus{} \cot ^2 \theta$

If you want to change their values equally (such as taking base $ e$), you must make sure that the reverse of that operation can only get the values you started with.

For example, multiplying both sides by $ 0$ is not a valid change of value since there is more than one value that can be multiplied by $ 0$ to become the same value.

t0rajir0u's two rules are good :)

You should probably prove that $ \csc^{2}\theta\minus{}\cot^{2}\theta\equal{}1$.


Find the least possible value of $a+b$ if $a$ and $b$ are positive integers such that $13$ divides $a+11b$ and $11$ divides $a+13b$.

$143$ divides $a+24b$.  To minimize $a+b$, we maximize $b$; ergo $b = 5, a = 23, a+b = \boxed{28}$.


Inductive proof is quite straightforward.

Base case: $ {1\over x_1}\geqslant 4(1\minus{}x_1)\iff (2x_1\minus{}1)^2\geqslant 0$

Step: Denote $ P\equal{}x_1x_2\cdots x_{n\plus{}1}$. It's sufficient to prove

$ 4(1\minus{}P)\plus{}{P\over x_{n\plus{}2}}\geqslant 4(1\minus{}Px_{n\plus{}2})$, but that's equivalent to $ (2x_{n\plus{}2}\minus{}1)^2\geqslant 0$


I am happy to announce the winning students from yesterday's Math Prize for Girls contest, sponsored and organized by the Advantage Testing Foundation.  First, here are our 12 money winners.  (The score is the number of correct answers out of 20.)

[code] 1. Elizabeth Synge (11th grade, Boston U. Academy, MA), $20,000 (Score 14)
 2. Shijie (Joy) Zheng (11th grade, Exeter, NH), $6000 (Score 13)
 3. Lynnelle Ye (12th grade, Palo Alto HS, CA), $6000 (Score 13)
 4. Jane Wang (12th grade, Montgomery HS, NJ), $6000 (Score 13)
 5. Adisa Kruayatidee (11th grade, Exeter, NH), $1000 (Score 12)
 6. Sherry Wu (12th grade, AAST, NJ), $1000 (Score 12)
 7. Hanh Nguyen (12th grade, Loomis Chafee, CT), $1000 (Score 12)
 8. Jenny Lu (12th grade, Pomperaug HS, CT), $600 (Score 11)
 9. Kate Rudolph (12th grade, Naperville Central HS, IL), $600 (Score 11)
10. Janice Lan (12th grade, Blair HS, MD), $600 (Score 11)
11. Ramya Rangan (10th grade, Harker, CA), $600 (Score 11)
12. Victoria Xia (8th grade, Longfellow MS, VA), $600 (Score 11)
[/code]
We also presented honorable-mention awards to 19 other students.  Below are the 12 honorable-mention students with a score of 10:
[hide="Tied for 13th Place"]Christina Chen (9th grade, Newton North HS, MA)
    In Young Cho (11th grade, Exeter, NH)
    Jessie Duan (11th grade, Green Hope HS, NC)
    Alina Kononov (12th grade, IMSA, IL)
    Shiyu (Jing-Jing) Li (12th grade, Cupertino HS, CA)
    Mihika Prabhu (11th grade, West Lafayette HS, IN)
    Lauren Stephens (12th grade, North Central HS, IN)
    Gabby Studt (8th grade, Homeschool, MD)
    Mariya Toneva (12th grade, Buchholz HS, FL)
    Danielle Wang (7th grade, Moreland MS, CA)
    Iris Xu (12th grade, Madison Memorial HS, WI)
    Jenny Yung (12th grade, Thomas Jefferson HSST, VA)[/hide]

Next are the 7 honorable-mention students with a score of 9:
[hide="Tied for 25th Place"]Allison Bukys (10th grade, Lexington HS, MA)
    Rebecca Chen (10th grade, Park Tudor, IN)
    So yeun (Ashley) Cho (11th grade, Emma Willard, NY)
    Heesun Ham (11th grade, Paul VI Catholic HS, NJ)
    Christine Hong (11th grade, JH Rose HS, NC)
    Grace Huckins (10th grade, Weston HS, MA)
    Elizabeth Shen (10th grade, Myers Park HS, NC)[/hide]

Just below honorable mention, here are the 20 students with a score of 8.
[hide="Tied for 32nd Place"]Denise Chen (12th grade, Wayzata HS, MN)
    Jiechen Chen (12th grade, Fayetteville-Manlius HS, NY)
    Michelle Deng (11th grade, Harker, CA)
    Caroline Ellison (10th grade, Newton North HS, MA)
    Elaine Fang (8th grade, Roosevelt MS, OR)
    Valeria Fedyk (12th grade, Andover, MA)
    Sarah Herrmann (9th grade, La Jolla HS, CA)
    Kristina Hu (9th grade, Thomas Jefferson HSST, VA)
    Joan Kim (11th grade, Northern Highlands Regional HS, NJ)
    Kristine Kim (12th grade, Lexington HS, MA)
    Allison Koenecke (12th grade, Thomas Jefferson HSST, VA)
    Lilly Shen (9th grade, Clements HS, TX)
    Sahiba Singh (10th grade, Wayzata HS, MN)
    Aejin Song (12th grade, Langley HS, VA)
    Amanda Wang (11th grade, Andover, MA)
    Tiffany (Haotian) Wu (9th grade, Clements HS, TX)
    Elizabeth Yang (10th grade, West Windsor- Plainsboro HS South, NJ)
    Ha Kyung (Heather) Yoon (12th grade, ASMSA, AR)
    Emily Zhang (9th grade, Newport HS, WA)
    Shikun Zhang (12th grade, Williston Northampton School, MA)[/hide]

Finally, here are the 16 students with a score of 7.
[hide="Tied for 52nd Place"]
   Jane (Jenny) Chen (11th grade, Andover, MA)
   Angela Choi (12th grade, Blair HS, MD)
   Angela Chu (12th grade, Stuyvesant HS, NY)
   Bianca Homberg (11th grade, Westborough HS, MA)
   Cheska (Da-In) Hong (12th grade, Great Neck South HS, NY)
   Elaine Hou (8th grade, Young Middle Magnet School, FL)
   Sunjeong Im (12th grade, Lexington HS, MA)
   Harini Kannan (10th grade, Exeter, NH)
   Veronica Lee (8th grade, Longfellow MS, VA)
   Qingzhao Shi (11th grade, Ann Arbor Huron HS, MI)
   Rose Sloan (10th grade, IMSA, IL)
   Elena Slobodyan (11th grade, Bronx Science, NY)
   Angela Yang (10th grade, Germantown HS, WI)
   Jessica Yao (11th grade, Miramonte HS, CA)
   Amy Zhang (10th grade, Lexington HS, MA)
   Alyssa Zisk (12th grade, Sharon HS, MA)[/hide]

Congratulations to all 67 of these students!  Thanks to all the 220 students who participated.  Special thanks to the dozens of volunteers who took time out of their busy lives to make our inaugural event run smoothly.

I have posted the 20 problems in separate threads in this subforum, and have posted a PDF file of the whole contest on the [url=http://mathprize.atfoundation.org]Math Prize[/url] site.  (Click on "Past Events".)

What an impressive accomplishment--great problems, great field of contestants, great setting, great speakers!

The suffragettes who once gathered in that historic Great Hall where the awards ceremony was held would be beaming with pride if they could have seen all the young women with so much mathematical promise who gathered there yesterday.

I have a post on my blog
http://albanyareamathcircle.blogspot.com/2009/11/math-prize-for-girls-highlights.html

I'll have more later.

Hi Mary,

I can't comment on your blog, but that was a wonderful summary -- thanks!  Also, congratulations to everyone involved!

congrats to everyone! :]]]esp. amsp 08 girls :) <3

yay to iris for repping wisconsin :DDD

Will there be a posting of everyone's scores from the math prize for girls?

Eventually, yes.  First, I will make public the names and scores of a few more top girls, probably on the Math Prize site.  Second, I will email all the contestants their scores.  (The scores will be keyed by the student number given on the student's name badge.)  Third, I plan to produce a few statistics on the scores.  If everything goes well, expect these things by Friday.

Lynnelle is spelled with two ns. Could you please correct that? Thanks! :)

Sorry about that.  I just fixed it.  I hope we got it right on the oversized check.

Thanks! Don't worry about it. It's correct on the big check, though wrong in the press release :P

Sorry, I'll try to make sure it's corrected in any press going forward, though we can't undo the past.  By the way, congrats on your fine performance!

Hello, great job writing the contest! I liked the problems a lot. In particular, I thought that the algebra problems were nice.

Thanks!  Given that you have written some excellent mock contests in the past, I'll take that as high praise.

I neglected to mention that several reviewers helped me figure out which of my 60 proposed questions were any good.  I'll single out JBL for special mention -- thanks, Joel!

I updated my original post with a list of other contestants who did well, just shy of Honorable Mention.

When you have time, Ravi, it would also be nice to see the tiebreak questions posted.  I really liked them, too!

[quote="Ravi B"]Sure, I'll post the two tiebreak questions in separate threads.  They were meant to be easier than the contest questions.  The quickest contestant in tiebreaks (Joy Zheng) solved them in 18 seconds and 24 seconds, respectively.[/quote]

That's quite impressive.  By the way, it looks like Joy followed up her second place finish at MPG with an extremely impressive second place overall individual finish at PUMaC this past Saturday!

http://www.facebook.com/photo.php?pid=3023914&id=654972943&ref=mf#/photo.php?pid=3023916&id=654972943&ref=mf&fbid=181359927943

Oh wow, my photos are really getting around :P .


advanced fields
advanced fields unsolved

Let " ≤ " be transitive, reflexive and antisymmetric relation over a set $ X$.

1. Prove that there exists the transitive, reflexive and antisymmetric relation $ R$ over $ X$, such that  ≤  $ \subseteq R$ and for all transitive, reflexive and antisymmetric relation $ S$ that includes  ≤ , is included in $ R$.

2. Prove that $ R$ is a complete order (all two elements $ x$ and $ y$ we can compare them: $ x R y$ or $ y R x$).

And, please, say me, how transitive, reflexive and antisymmetric relation is called in english? (it is some order, but I don't know, how it is called correctly)

Thank you very much!

A relation that is transitive, reflexive, and antisymmetric is called a partial order. Sets with a partial ordering are sometimes called posets ("partially ordered set" abbreviated).


linear algebra
matrix
superior algebra
superior algebra solved

Proove that for any a,b,c \in R
(a+b+c)detM \geq  0 where
M=
(a b c)
(c a b)
(b c a)
---------------------------------------------
M is a permutation matrix
detM= a^3+b^3+c^3 -3abc=(a+b+c)(a+jb+jc)(a+jb+jc) so
(a+b+c)detM=(a+b+c)|a+jb+jc| with j=e^(2i \pi /3), |z| is quare
of the modulus of z \in C
Remark: the matrix M is diagonalisable in M_3(C) and the eingenvalues are (a+b+c), (a+jb+jc), (a+jb+jc).


linear algebra
matrix
linear algebra unsolved

Now we know that
\[ {T_1} \equal{} \left( {\begin{array}{*{20}{c}} a \\
b \\
1 \\
\end{array}} \right),{T_2} \equal{} \left( {\begin{array}{*{20}{c}} c \\
b \\
1 \\
\end{array}} \right),{T_3} \equal{} \left( {\begin{array}{*{20}{c}} c \\
d \\
1 \\
\end{array}} \right),\]

\[ U \equal{} \left( {\begin{array}{*{20}{c}} 1 & t & 0 \\
t & 1 & 0 \\
0 & 0 & {1 \minus{} {t^2}} \\
\end{array}} \right),V \equal{} \left( {\begin{array}{*{20}{c}} 1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & m \\
\end{array}} \right),\]

\[ A \equal{} \left( {\begin{array}{*{20}{c}} {(U{T_1}) \times (V{T_1})} & {(U{T_2}) \times (V{T_2})} & {(U{T_3}) \times (V{T_3})} \\
\end{array}} \right),\]
$ \times$ means crossprod.

For example,
\[ \left( {\begin{array}{*{20}{c}} {{x_1}} \\
{{y_1}} \\
{{z_1}} \\
\end{array}} \right) \times \left( {\begin{array}{*{20}{c}} {{x_2}} \\
{{y_2}} \\
{{z_2}} \\
\end{array}} \right) \equal{} \left( {\begin{array}{*{20}{c}} {{y_1}{z_2} \minus{} {y_2}{z_1}} \\
{{z_1}{x_2} \minus{} {z_2}{x_1}} \\
{{x_1}{y_2} \minus{} {x_2}{y_1}} \\
\end{array}} \right)\]
How to calculate
\[ \det A \equal{} \left| A \right| \equal{} ?\]
Perfactly?

Postscript:

By using maple, we know that the result is $ t*(t \minus{} 1)*(t \plus{} 1)*(t \minus{} 1 \plus{} m)*(t \plus{} 1 \minus{} m)*(b \minus{} d)*( \minus{} c \plus{} a)*(c*a \minus{} d*b)$.

BUT I WANT TO KNOW A PERFECT WAY TO SOLVE IT. I REALLY DON'T ONLY WANT TO KNOW THE RESULT!!!

A perfect way WITHOUT softwares such as Maple and Mathematica.

With a lot of multiplication.  The process is not difficult, it's just extremely long and tedious.  Fortunately, most of the terms are 0's and 1's.  As a quick example, $ UT_{1}\equal{}\begin{pmatrix}
a\plus{}bt\\
at\plus{}b\\
1\minus{}t^{2}
\end{pmatrix}$.

And the word is "perfect," not "perfact."

[quote="JRav"]With a lot of multiplication.  The process is not difficult, it's just extremely long and tedious.  Fortunately, most of the terms are 0's and 1's.  As a quick example, $ UT_{1} \equal{} \begin{pmatrix} a \plus{} bt \\
at \plus{} b \\
1 \minus{} t^{2} \end{pmatrix}$.

And the word is "perfect," not "perfact."[/quote]

I think that the way of using linear algebra may be easier and more perfect, but I don't know it well.

Only using the basic algebra may be okay, but I want to know the essence(I think that the problem's essence is about linear algebra).

I'm sorry about my poor English. I made a mixture of "perfect" and "fact".


linear algebra
matrix
geometry
geometric transformation
rotation
vector
rotation matrix

Hi,

We all know a matrix can be used to rotate a point in R^3.
For example, I found a formula [url=http://en.wikipedia.org/wiki/Rotation_matrix#Three_dimensions]here[/url].
But I did not find a proof of this or any other formula used for three-dimensional matrix rotations. Does anyone of you know the general outlines of this proof or a URL I did not find?

Thanks in advance,

RicOh

I believe you will have to check the whole thing "by hand" (i.e, brute force computations). This is funny, though, how you say that "we all know ...". I'll ask my girlfriend later  :rotfl:

For the rotation around the $x-axis,$ consider the image of the points $(1,\ 0,\ 0),\ (0,\ 1,\ 0),\ (0,\ 0,\ 1).$

I'm sorry, I'm not quite getting you there.
I was just looking for some proof that matrices can really perform a rotation, and the way the exact matrix is found - someone must have done this right?

well the rotation matrix is a standard matrix for a linear transformation. in general, you can find the standard matrix of a linear transformation by considering how the transformation acts on the standard unit vectors.
so when you find how the rotation transformation acts on $e_{1}$, $e_{2}$, and $e_{3}$, place the transformed vectors into a matrix (as columns) in that order. that will be the rotation matrix you're looking for.

I tried to do what you advised me:
If I rotate using the matrixes for single-axis rotation:
(x,0,0) around the x-axis I get (x,0,0)
(0,y,0) around the y-axis I get (0,y,0)
(0,0,z) around the z-axis I get (0,0,z)

Is this enough as a proof that those matrices rotate around the axes?

And in what order would I have to multiply those 3 matrices?

RicOh

Hi to all

In matrix rotation, there are three rotation angles namely alpha, sigma and kappa. My question is: Is there any way that we could extract these three angles? (I always read that these three angles are always given)

OK, I now found out why a matrix can rotate a point around an (1) axis.
Now I am having trouble to find how you can combine these rotations (x,y,z-axis) to get a rotation around a point.
Can anyone help here please?

RicOh


An octagon is inscribed in a circle, four of its sides are of length 3 and the other four
sides are of length 4. The side lengths alternate around the octagon. Find the area of
the octagon.

A) $\displaystyle\frac{73\sqrt{2}+48}{4\sqrt{2}}$
B) $25+24\sqrt{2}$
C) $13 + 6\sqrt{2}$
D) $9 + 24\sqrt{2}$
E) None of these

[hide="Solution"]
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=700[/img]

\begin{eqnarray*} \text{Yellow Area} &=& 3^2\\ &=& 9\\ \text{Blue Area} &=& 4\cdot \frac 12 (2\sqrt{2})^2\\ &=& 16\\ \text{Orange Area} &=& 4(3)(2\sqrt{2})\\ &=& 24\sqrt{2}\\ \Rightarrow \text{Total Area} &=& 25+24\sqrt{2} \end{eqnarray*}[/hide]

cool!  but i cant draw in real life........

you don't need to be able to draw well, just make it so you can see what's going on. when questions ask for the areas of unusual shapes i find it easier to break them up into more familiar pieces like squares, triangles, etc.


[quote="spider_boy"][quote="silouan"]Posted many times .
Just note that $\frac{a}{b^{2}+1}=a-\frac{ab^{2}}{b^{2}+1}\geq a-\frac{ab}{2}$[/quote]

but silouan wait.. he asks $\frac{a^{2}}{b^{2}+1}+...$ so your approach des not work.[/quote]
silouan meant that
$\sum_{cyc}\frac{a^{2}}{b^{2}+1}\geq\sum_{cyc}\left(a^{2}-\frac{a^{2}b}{2}\right).$
And $\sum_{cyc}\left(a^{2}-\frac{a^{2}b}{2}\right)\geq\frac{3}{2}\Leftrightarrow\frac{(a^{2}+b^{2}+c^{2})(a+b+c)}{3}-\frac{a^{2}b+b^{2}c+c^{2}a}{2}\geq\frac{(a+b+c)^{3}}{18}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(5a^{3}-6a^{2}b+3a^{2}c-2abc)\geq0\Leftrightarrow$
$\Leftrightarrow2\cdot\sum_{cyc}(a^{3}-abc)+3\cdot\sum_{cyc}(a^{3}-2a^{2}b+b^{2}a)\geq0\Leftrightarrow\sum_{cyc}(4a+b+c)(a-b)^{2}\geq0.$ :wink:


Gracie starts at her house and walks one block east. After each block she walks, she flips a fair coin. She walks one block east if the coin lands heads; otherwise one block west. She stops her walk the first time she lands back at home. What is the probability she stops her walk after walking exactly eight blocks? Express your answer as a common fraction.

Ah yes, the old #30 Nats Sprint on FTW...

Anyways, we must make sequences that match the description, with 3 E's and 4 W's.  Obviously, the first letter must be E.  Also, the last two letters must be WW, because if there was an E, it would mean that Gracie stopped at her house or west of her house.  Then, there are $ \binom{4}{2}$ ways to make sequences.  However, one of these does not work: EEWWEEWW.  So there are indeed only $ 5$ sequences that work, out of a possible $ 2^7$, so our answer is $ \boxed{\frac{5}{128}}$.

Well each time she flips she has a 1/2 chance to go east and 1/2 chance to go west

This can be expressed as a binomial distribution on a number line when the mean is where the house is.


trigonometry
inequalities unsolved
inequalities

Find the least value of the expression:
$P=2\tan^2\frac{A}{2}+3\tan^2\frac{B}{2}+4\tan^2\frac{C}{2}$ when A,B,C are the measure of three angles of a triangle

If $x=(a+b-c)(a+c-b), y=(b+a-c)(b+c-a),z=(c+a-b)(c+b-a)$ then
$P\geq k\Leftrightarrow2x^2+3y^2+4z^2\geq k(xy+xz+yz).$ ;)

Thank you,arqadi.But can someone give me a nice solution using this lemma:
$A+B+C=\pi$then $\forall x,y,z$ we have $2yz\cos A+2zx\cos B+2 xy\cos C\le x^2+y^2+z^2$

[quote="Guest"]:roll:  The least value is  2.84667202780066..., which is the unique positive root of

k^3+9*k^2-96=0.[/quote]
How can you find it,Guest.Thank you!


Let $ a,b,c\ge \minus{}1$ such that $ a^2\plus{}b^2\plus{}c^2\equal{}6$. find minA=a+b+c.

Clearly this problem satisfies all the assumptions for the use of lagrange multipliers.

First, grad f=lambda grad g, gives a=b=c, 3 root 2 here

if one of the variables is -1, then the other two by the same consideration above are root(5/2), this gives -1+root 10

if two variables are -1, the other is 2, so 0 here

if all three variables are -1, contradiction

hence the minimum is 0 and the maximum is 3 root 2

It is nice and easy! thanks, very much, Quy!

[quote="Hong Quy"][quote="thegod277"]Let $ a,b,c\ge \minus{} 1$ such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 6$. find minA=a+b+c.[/quote]
   let $ x \equal{} a \plus{} 1 , y \equal{} b \plus{} 1,z \equal{} c \plus{} 1$   then $ x,y,z$  positive real numbers
$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 6 \iff x^2 \plus{} y^2 \plus{} z^2 \minus{} 2(x \plus{} y \plus{} z) \minus{} 3 \equal{} 0$
note that : $ x^ \plus{} y^2 \plus{} z^2 \leq (x \plus{} y \plus{} z)^2$  and  let $ x \plus{} y \plus{} z \equal{} t \geq 0$
  then  we have inequality :$ t^2 \minus{} 2t \minus{} 3 \geq 0 \iff (t \plus{} 1)(t \minus{} 3) \geq 0 \iff t \geq 3$ ( because $ t \geq 0$)
 hence $ Min t \equal{} 3$  or $ Min(a \plus{} b \plus{} c) \equal{} 0$
  Occur equa if and only if : $ a \equal{} \minus{} 1 ,b \equal{} \minus{} 1$and $ c \equal{} 2$[/quote]

Thank for nice solution.


[color=green](reload)[/color]
Let $n \geq 3$ be an integer and let $x,y,z \in {\mathbb Z}$ with $x \neq y \neq z \neq0$. Show that the equation $x^{3n}+y^{3n}-z^{3n}+3x^{3n}y^{3n}z^{3n}=0$ has no solutions in ${\mathbb Z}$.

 :ninja:

$x=z\not= 0=y$ is solution satisfied your conditions.

[quote="ZetaX"]But again it's wrong: $x=y=1 , z=-1 , n=3$.[/quote]
don't forget that $x \neq y \neq z \neq 0$


trigonometry
linear algebra
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calculus
calculus computations

Define $ \bold{g}: \mathbb{R}^2\to\mathbb{R}^2$ by $ \bold{g}(x,y)\equal{}(y\cos x,(x\plus{}y)\sin y)$, and $ \bold{f}: \mathbb{R}^2\to \mathbb{R}^3$ by $ \bold{f}(x,y)\equal{}(x^2\minus{}y,3x\minus{}2y,2xy\plus{}y^2)$

(i) Shwo that $ \bold{g}$ maps a neighborhood of $ (0,\pi/2)$ bijectively to a neighborhood of $ (\pi/2,\pi/2)$

(ii) If $ \bold{h}\equal{}\bold{f}\circ\bold{g}^{\minus{}1}$, find the matrix $ \bold{h'}(\pi/2,\pi/2)$

Have you ever heard of the inverse function theorem?


Prove that there exist an infinity of triplets (x,y,z) of natural numbers so that:
x 2 +y 2 +z 2 =xyz

It's enough to show that you can generate an infinity of solutions,
starting from a given solution.

Notice that (x, y, z) = (3, 3, 3) is a solution.

Now consider the equation x - (yz) x + (y + z) = 0.
Suppose that (x, y, z,) = (a, b, c) satisfies this equation.
Then (x, y, z) = (bc - a, b, c) also satisfies it.

So starting from (3, 3, 3) we obtain the following solutions :

(6, 3, 3)
(6, 15, 3)
(6, 15, 87)
...

It's obvious that you can produce an infinity of solutions now.

Notice that in the second round of the Mathlinks Contest 2002,
Valentin Vornicu proposed a similar (but more interesting) problem.


$ AC$x$ BD \equal{} E$
M-midpoint of AE.
N-midpoint of DC.
<ABD=<DBC.
Proof that ABCD is inscribed <=> when MBCN is inscribed.

[i]The First implication:[/i]

Let's suppose that $ ABCD$ is inscribed and we 'll prove that $ MNCB$ is inscribed

Since $ \angle BAE \equal{} \angle BDC$ and $ \angle ABE \equal{} \angle EBC$ Then $ \angle AEB \equal{} \angle DCB$ And hence $ \Delta AEB$ and $ \Delta DCB$ are similar which implies that $ \frac {AB}{EB} \equal{} \frac {BD}{CB}$

Let $ \angle EBM \equal{} \alpha_1$ and $ \angle MBA \equal{} \alpha_2$

On the other hand in The triangle $ AEB$ we have $ M$ is the mid-point of $ AE$ which give us  $ \frac {\text{sin}(\alpha_1)}{\text{sin}(\alpha_2)} \equal{} \frac {AB}{EB}$$ \implies \frac {\text{sin}(\alpha_1 \plus{} \alpha_2)}{\text{tg}(\alpha_2)} \minus{} \text{cos}(\alpha_1 \plus{} \alpha_2) \equal{} \frac {AB}{EB}$

And similarly in The triangle $ DBC$ we get $ \frac {BD}{BC} \equal{} \frac {\text{sin}(\gamma_1 \plus{} \gamma_2)}{\text{tg}(\gamma_2)} \minus{} \text{cos}(\gamma_1 \plus{} \gamma_2)$ Where $ \angle DBN \equal{} \gamma_2$ and $ \angle NBC \equal{} \gamma_1$

and Since we have $ \gamma_1 \plus{} \gamma_2 \equal{} \alpha_1 \plus{} \alpha_2$ because $ \angle ABE \equal{} \angle EBC$ we obtain $ \gamma_2 \equal{} \alpha_2$ And  and Therefore  $ \alpha_1 \equal{} \gamma_1$  and Thus $ \angle MBN \equal{} \angle EBA \equal{} \angle DCA \equal{} \angle NCM \implies \angle MBN \equal{} \angle NCM$ Whis means that $ MNCB$ is inscribed

+If $ ABCD$ is cyclic.
We have $ \Delta AEB\sim \Delta DCB$ then $ \Delta MEB\sim \Delta NCB \Rightarrow \Delta NMB\sim \Delta CEB \Rightarrow \angle MNB\equal{}\angle ECB \Rightarrow MNCB$ is cyclic.
+If $ MNCB$ is cyclic.
We have $ \angle MNB\equal{}\angle ECB$. Let $ E',A'$ be a point on $ AC$ such that $ \angle E'BC\equal{}\angle NBM, E'M\equal{}A'M$
$ \Rightarrow \Delta NMB\sim \Delta CE'B \Rightarrow \Delta ME'B\sim \Delta NCB \Rightarrow \Delta A'E'B\sim \Delta DCB$
$ \Rightarrow \angle A'BE'\equal{}\angle DBC\equal{}\angle ABE$
$ \Rightarrow E'\equiv E$ and we get $ ABCD$ is cyclic.

Sorry I'm just a beginner in Geometry  :blush:  Can you tell me why :

$ \Delta AEB\sim\Delta DCB \Rightarrow \Delta MEB\sim\Delta NCB$

Thank you  :)

[quote="Evariste-Galois"]Sorry I'm just a beginner in Geometry  :blush:  Can you tell me why :

$ \Delta AEB\sim\Delta DCB \Rightarrow \Delta MEB\sim\Delta NCB$[/quote]

$ \Delta AEB\sim \Delta DCB \Rightarrow \angle AEB\equal{}\angle DCB$ and $ \frac{BE}{BC}\equal{}\frac{AE}{DC}\equal{}\frac{ME}{NC} \Rightarrow \Delta MEB\sim \Delta NCB$

Oh yes !! Thank you Very much my friend  :blush:  My mind is  malfunctioning Nowdays Since I've missed something simpler like that  :(

I'm sorry but i need a solution without trigonometry,similar triangles etc.