vector
linear algebra
matrix
induction
algebra
polynomial
linear algebra unsolved

Let $ V$ be a finite dimensional (real or complex) vector space.
If $ A, B$ are triangularizable linear operators such that $ AB \equal{} BA$,
prove that there is a basis of $ V$ on which $ A$ and $ B$ are triangular.

I would appreciate a proof as elementary as possible.

Thanks!

Saying the a matrix is triangularizable is saying that all of its eigenvalues belong to the field of scalars. That's automatic for complex matrices, but it is saying something in the case of real matrices. In what follows, it won't matter what the field is - including that we can consider fields other than $ \mathbb{R}$ and $ \mathbb{C}$ as well - as long as we have that hypothesis that the matrices can be triangularized.

The fact that $ A$ and $ B$ commute tells us that they share eigenvectors. To prove this, assume that $ v\ne 0$ and $ Av=\lambda v.$ Then $ \lambda Bv=B(\lambda v)=BAv=A(Bv).$ That means that $ Bv$ belongs to the same eigenspace of $ A$ as $ v.$ What we've really shown is this: if $ E_{\lambda}=\{v: Av=\lambda v\}$ then $ B$ maps $ E_\lambda$ into $ E_{\lambda}.$ If $ E_{\lambda}$ is one-dimensional, we're done. If not, then because $ B$ is triangularizable, its restriction to $ E_{\lambda}$ is also triangularizable and $ B$ has an eigenvector in $ E_{\lambda}.$

Now for the main theorem. We proceed by induction on $ n,$ for $ n\times n$ matrices. The theorem is trivially true for $ n=1.$ Suppose it is true for $ n\time n$ matrices and let $ A$ and $ B$ be $ (n+1)\times(n+1)$ matrices satisfying the hypotheses.

Find a vector $ v$ that is an eigenvector of both $ A$ and $ B.$ Extend this to a basis, which is to say find an invertible matrix $ P$ such that $ v$ is the first column. Then $ AP$ is a matrix whose first column is $ \lambda v.$ Since $ P^{-1}P=I,$ it follows that $ P^{-1}v$ is a column with $ 1$ at the top and zeros everywhere else. Putting this together, we have the following block-structured form:
\[ P^{-1}AP=\begin{pmatrix}\lambda&*\\0&A'\end{pmatrix}.\]
By the same argument, $ P^{-1}BP=\begin{pmatrix}\mu&*\\0&B'\end{pmatrix}.$

Now multiply both of those together in both orders and we discover that $ A'B'=B'A'.$ By the inductive hypothesis, $ A'$ and $ B'$ can be simultaneously triangularized, which finishes the proof.

Lele-Br: to complete the proof of Kent Merryfield, try to prove that if $ A$ is triangularizable, then its restriction to any stable subspace is still triangularizable. That is a good exercice if you do not know much about matrix theory.

Taking that up myself:

We know that $ A$ is triangularizable if and only if its characteristic polynomial splits (factors completely into linear factors). The proof of this proceeds by induction on $ n$ and resembles in some ways the argument that I gave above.

Now, suppose $ A$ has a stable subspace $ V.$ Find a basis for $ V$ and then extend that to a basis for the whole space $ \mathbb{F}^n.$ That is, find an invertible matrix $ P$ such that the first $ k$ columns are a basis for $ V.$ Then the fact that $ V$ is stable under $ A$ (that is, $ A(V)\subseteq V$) gives us that in block form,
\[ P^{\minus{}1}AP\equal{}\begin{pmatrix}B&*\\0&C\end{pmatrix}\]
Use that the characteristic polynomial of $ P^{\minus{}1}AP$ is identical to the characteristic polynomial of $ A,$ and that the determinant (hence, characteristic polynomial) of a block-diagonal matrix is the product of the determinants (hence, characteristic polynomials) of the blocks on the diagonal. It follows that $ P_A(x)\equal{}P_B(x)P_C(x).$ And now by the unique factorization of polynomials, the fact that $ P_A$ splits implies that both $ P_B$ and $ P_C$ split. Hence, $ B$ is triangularizable, which is the result that alekk is asking for.

or more simple: a matrix $ A$ is triangularizable iff there exists a polynomial (not necessary its characteristic polynomial) $ P$, that factors completely into linear factors, and that satisfies $ P(A)\equal{}0$


[u][b]Problem1:[/b][/u]Let a,b,c,k be positive reals.Where k is constant.
Prove that:
               $\frac{ab+(k-3)bc+ca}{(b-c)^2+kbc}+\frac{bc+(k-3)ca+ab}{(c-a)^2+kca}+\frac{ca+(k-3)ab+bc}{(a-b)^2+kab}\geq\frac{3(k-1)}{k}$

[u][b]Problem2:[/b][/u].Let $n\geq3$ and $x_1,x_2,...,x_n\geq 0$ show that:$\sum\limits_{i=1}^n x_i=1$.Find the max value of expression:
      $E=x_1^3x_2^2+x_2^3x_3^2+x_3^3x_4^2+....+x_n^3x_1^2+(n-1)^{2(n-1)}x_1^3x_2^3...x_n^3$

[quote="Frey"][u][b]Problem1:[/b][/u]Let a,b,c,k be positive reals.Where k is constant.
Prove that:
               $\frac{ab+(k-3)bc+ca}{(b-c)^{2}+kbc}+\frac{bc+(k-3)ca+ab}{(c-a)^{2}+kca}+\frac{ca+(k-3)ab+bc}{(a-b)^{2}+kab}\geq\frac{3(k-1)}{k}$
[/quote]
See here the soarer's post:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26102&start=20


If $ x$ was a real positive number then what is the value of $ x$ in the equation:
$ x^{x^4 }  \equal{} 4$
------------------------
How can it be known?  :(

$ \rightarrow x$

FOUND IT! :P

[hide="only solution I know of"]Note that $ 1^{1^4}\equal{}1$ and $ 2^{2^4}$ is way too big. So we are looking for a value between 1 and 2.

Now note that if $ x^4\equal{}4$, then the equation is satisfied, because $ x^{x^4}\equal{}x^4\equal{}4$. Therefore $ x\equal{}\sqrt[4]{4}\equal{}\sqrt[2]{2}$ is a solution.[/hide]

I hope this helps.

Do you check the answer?

I think its a bit confusing.  :rotfl:

I think x can be equal -$ \sqrt[2]{2}$
I hope my English isn't too bad :D

[quote="SuperDragon"]I think x can be equal -$ \sqrt [2]{2}$
I hope my English isn't too bad :D[/quote]
No its good. At least better than mine  :blush:
Can you put the method please? How you get that answer?

[quote="SuperDragon"]I think x can be equal -$ \sqrt [2]{2}$
I hope my English isn't too bad :D[/quote]
x is positive , so can not be $ \minus{}\sqrt {2}$

[quote="enndb0x"]hasan4444 are you sure that the problem is in real numbers and not in [b]integers[/b][/quote]
As far as integers is a part of real numbers so it might be integers.
Integers is a part from real numbers.

[quote="enndb0x"]
x is positive , so can not be $ \minus{} \sqrt {2}$[/quote]
Oh, I'm sorry  :D

[quote="enndb0x"]Yes,but some problems can be solved just for some integers (a part of real numbers).By the way ,there's no integer solution for that equation.[/quote]
Maybe, he say real number so it can be $ \sqrt{x}$ even.

Let $ f(x)\equal{}\ln x^{x^4}\equal{}x^4 \ln x$. We want to find $ x$ such that $ f(x)\equal{}\ln 4$. We have $ f'(x)\equal{}4x^3\ln x\plus{}x^4\cdot\frac{1}{x}\equal{}x^3(4\ln x\plus{}1)$. $ f(1)\equal{}0<\ln 4$, and we see that $ f$ is increasing for $ x\geq 1$, so there can only be one solution to $ f(x)\equal{}\ln 4$. Using [b]1=2[/b]'s solution, we have $ \sqrt{2}$ as the unique solution for $ x$.

Sorry for the calculus, but I cannot think of any other way to prove it rigorously. (Maybe graphing it? >.>)

[quote="hasan4444"]There something I can't figure if I substitute $ \sqrt 2 ^{\sqrt 2 ^4 }$ in a calculator the answer will not come to $ 4$  :([/quote]
Are you doing $ \sqrt{2}^{(\sqrt{2}^4)}$ (sqrt(2)^(sqrt(2)^4), or $ (\sqrt{2}^{\sqrt{2}})^4$ (sqrt(2)^(sqrt^2)^4)?

Sorry. Yes I think the calculator was entering parenthesis automatically.  :blush:
You are right  :D its $ \sqrt{2}$

Something interesting I found out of this question...

$ x^{x^{x^{x\cdots^n}}}\equal{}n$
Then $ \sqrt[n]{n}$ is a possible solution for $ x$, because when it's plugged in, each term cancels out the $ n$th root leaving $ n$ at the end, I'm not sure yet if it's the only solution.

[quote="tonypr"]Something interesting I found out of this question...

$ x^{x^{x^{x\cdots^n}}} \equal{} n$
Then $ \sqrt [n]{n}$ is a possible solution for $ x$, because when it's plugged in, each term cancels out the $ n$th root leaving $ n$ at the end, I'm not sure yet if it's the only solution.[/quote]
You are the expert :)


[quote="quykhtn-qa1"]Let 0 ≤ x≤ y≤ z≤ 1  and    3x+2y+z ≤4.Prove that 3(3x^2+2y^2+z^2) ≤10[/quote]
Change the notation a bit: $ x = \frac{\alpha}{3}$, $ y = \frac{\beta}{2}$, and $ z = \gamma$.
Then $ 0 \le \frac{\alpha}{3} \le \frac{\beta}{2} \le \gamma \le 1$, $ \alpha + \beta + \gamma \le 4$ and we have to prove that:
$ f(\alpha, \beta, \gamma) = \frac{1}{3} \alpha^2 + \frac{1}{2} \beta^2 + \gamma^2 \le \frac{10}{3}$.

First we see that the max of $ f(\alpha, \beta, \gamma)$ is obtained iff $ \alpha + \beta + \gamma = 4$, since $ f(\alpha, \beta, \gamma)$ is increasing in all variables.

First $ \frac{\Delta f(\alpha,\beta,\gamma)}{\Delta \gamma} - \frac{\Delta f(\alpha,\beta,\gamma)}{\Delta \alpha} = 2(\gamma - \frac{\alpha}{3}) \ge 0$, by the definition. So $ f(\alpha,\beta,\gamma) \le f(\alpha-\epsilon,\beta,\gamma+\epsilon)$ when $ \alpha-\epsilon \ge 0$. Thus the max is either obtained when $ \gamma = 1$ or $ \alpha = 0$. But $ \alpha =0$ makes no sense, because then $ \alpha + \beta + \gamma = 4$ cannot be fulfilled. Thus $ \gamma = 1$.

${ \frac{\Delta f(\alpha,\beta,\gamma)}{\Delta \beta} - \frac{\Delta f(\alpha,\beta,\gamma)}{\Delta \alpha} = 2(\frac{1}{2}\beta - \frac{1}{3} \alpha} \ge 0$ by definition. Thus $ f(\alpha, \beta, \gamma) \le f(\alpha-\epsilon,\beta+\epsilon,\gamma)$ when $ \alpha-\epsilon \ge 0$. By the same reasoning as before we get $ \beta = 2$, and thus max is obtained when $ (\alpha,\beta,\gamma) = (1,2,1)$.
Then $ f(\alpha,\beta,\gamma) \le f(1,2,1) = \frac{1}{3} + 2 + 1 = \frac{10}{3}$. And we are done! QED

[quote="quykhtn-qa1"]Let 0 ≤ x≤ y≤ z≤ 1  and    3x+2y+z ≤4.Prove that 3(3x^2+2y^2+z^2) ≤10[/quote]
Wow. Notice that $ (1,1,1,\frac {1}{3},\frac {1}{3},\frac {1}{3}) \succ (z,y,y,x,x,x)$. (At least when $ 3x\plus{}2y\plus{}z\equal{}4$ which is the only case we need to consider) And note that $ x \to x^2$ is convex.
Then $ \frac {10}{3} \ge 3x^2 \plus{} 2y^2 \plus{} z^2$ and we are done!


What is the arithmetic mean of $ 1^2, 2^2, 3^2, 4^2$ and $ 5^2$?

The sum of the first $ n$ positive perfect squares is equal to $ \frac{n(n\plus{}1)(2n\plus{}1)}{6}$.  Therefore, the sum of the first $ 5$ positive perfect squares is $ \frac{5(6)(11)}{6}\equal{}55$.  Since there are five numbers, the arithmetic mean of them is $ \frac{55}{5}\equal{}\boxed{11}$.


limit
real analysis
real analysis unsolved

Let the series $x_{n}=\frac{1}{n}\left(\sqrt{\frac{n+2}{n+1}}+\sqrt{\frac{n+3}{n+2}}+...+\sqrt{\frac{2n+1}{2n}}\right)$. Compute: $\lim_{n \rightarrow\infty}{x_{n}}$ and $\lim_{n\rightarrow\infty}{(x_{n})^{n}}$.


[b]Problem[/b]

[i]$a_1,...,a_n$ are non-negative reals.
$A_{i,j}=\frac{\sum _{k=1}^n a_k-(a_i+a_j)}{n-2}$
$G_{i,j}=\left(\frac{\prod _{k=1}^n a_k}{a_ia_j}\right)^{1/(n-2)}$

Prove the following inequality: \[{\left(\prod _{i\not = j} A_{i,j}\right)^{\frac{2}{n(n-1)}} \geq \frac{2}{n(n-1)}} \sum _{i\not = j}G_{i,j}\][/i]

P.S. I hope you enjoy it  :D

[i]Edited by Myth[/i]

This statement is not true.  The case n=3 simplifies to $^3\sqrt {a_1a_2a_3}\ge \frac{1}{3}(a_1+a_2+a_3)$ which clearly goes against AM-GM.  Try rewording your problem. :D

By the way, I did enjoy it.  It was about twenty minutes before I found this flaw.

Hm...I think the inequality is correct though I should write $n \geq 4$  :) - or do you manage to find a counterexample for $n=4$ as well ?  :P

It's wrong for n = 3, and for n = 4, it's even wronger. :D  In fact, let $a_1=a_2=0$ and $a_3=a_4=1$. Now it's an easy exercise to see why this is a counterexample.

  darij

That would be wrong either; take $a_1=0$ and $a_2=a_3=a_4=1$.

  darij

Ok - you wanted that  :D - I'm posting inequality which special case here I presented - maybe I've done something wrong in the statement [though I failed to find my mistake  :blush: ] or you've just destroyed someone's paper ;) :

[b]Theorem[/b]
Let $X= (x_1,...,x_n | x_i \geq 0 , i=1,2,...,n)$. For $A\subset X$, let $a_A$ and $g_G$ denote the arithmetic and geometric mean of all elements of $A$ respectively. If $k$ is an integer in $(\frac{n}{2},n]$, then:

$(\prod _{|A|=k; A \subset X} a_A)^\frac{1}{C_n ^k} \geq \frac{1}{C_n ^k}(\sum _{|A|=k; A \subset X} g_A)$

Hmm, well, look at this:

[quote="Megus"]If $k$ is an integer in $(\frac{n}{2},n]$,[/quote]

In particular, $k>\frac{n}{2}$. Thus, your "special case" is a special case only for $n\geq 5$.

By the way, what paper is the inequality from?

  darij

Thanks Darij for pointing my mistake

This paper is "Mixed-mean inequality for subsets" and you can download it on AMS


graph theory
combinatorics proposed
combinatorics

There are $ n\geq 1$ cities in planet Math with exactly one one-way road between any two cities.
a)	Prove that, to each city one can assign a unique (=different from other cities) label between $ 1$ and $ n$ inclusive such that the direction of the road between the city with label $ i$ and the city with label $ i\plus{}1$ ($ i\equal{}1,2,\cdots, n\minus{}1$) is from the former city to the latter one.
b)	A trip of length $ k\geq 1$ from the city $ A$ to the city $ B$ is a trip which one starts from the city $ A$ and visits $ k\minus{}1$ cities (not necessary distinct) before ending up in the city $ B$ (Example: going from $ A$ to a city $ S$ and then to $ B$ is a trip of length $ 2$ from $ A$ to $ B$). Find the largest value $ k$ for which, there are two cities $ A$ and $ B$ (not necessary distinct) in planet Math such that there exists a trip of length $ i$ between $ A$ and $ B$ for all $ 1\leq i\leq k$

a) is Redei's theorem that any tournament contains a Hamiltonian path, see
[url]http://wapedia.mobi/en/Tournament_(graph_theory)[/url]


function
algebra
linear equation
algebra proposed

Consider the function $ f: \mathbb{Z} \rightarrow \mathbb{R}$ with the property that $ f(n) \ge \frac{f(n\plus{}1)\plus{}f(n\minus{}1)}{2}, \forall n \in \mathbb{Z}.$ Prove that if there is $ m \in \mathbb{R}$ so that $ f(x) \ge m, \forall x \in \mathbb{Z}$ then $ f$ is constant.

It is clear that $ f(0),f(1)$ uniquely determine the function. If $ f(0)\equal{}a, f(1)\equal{}b$, then $ f(n)\equal{}(b\minus{}a)n\plus{}a$ satisfies the equation. The linear equation can only be bounded below if it is a constant function, implying our result.


Wow guys the site is super slow today. I keep getting network timeouts also. It isn't happening on any other sites else for me...

It's like a trippy ride to the '98...

Anyway I'm running XP with the beta firefox...

Works just fine for me and it has been working fine all day. Must be your internet connection.

ehh you were right. Just my crappy internet connection... -.-

I created a topic on this on accident without seeing this... It happens to me to about 2-3 times every day. And other sites work just fine while this is happening. Is it perhaps because I love this site too much (and therefore click a lot :) )?

Yes, MathLinks is very often very slow (and that's definitely not caused by my side of the connection only: other people have the same problems at the same times, too).

We will upgrade our servers soon to fix these problems. It has just outgrown our current setup :)


Earth rotates on an axis through its poles. The distance from the axis to a location on Earth 40 degrees north latitude is about 3033.5 miles. Thus, a location on Earth 40 degrees north is spinning on a circle of radius 3033.5 miles. Compute the linear speed of the surface of the Earth at 40 degrees north latitude.

This involves angular velocity $ \omega$, which relies on linear velocity $ v$ and radius $ r$.  You know the angular velocity (1 rev/24 hours) which can be converted into proper units, and you can find the radius, so use the formula $ v\equal{}\omega r$ and you'll get your answer.

That is linear speed @ that point on earth.  The total linear speed of earth in respect to its comological position to the sun, @ a 93,000,000 mile distance is also a problem of the form v=wr.  We know that the earth orbits the sun every 365.25 days, now looking at w, we can break this down knowing that 1 circular orbit is equal to 360 degrees, or an exact number of 6.283185307 radians.  W = Angle(6.283185307)/Time(365.25days) = .0172024238/day.  Now going back to our V= (Linear Speed).....   V=WR we can use V=(.0172024238/day)(93,000,000mi) = 1,599,825.417mi/day and   1,599,825.417 X 1day/24hours = 66,659.39239mi/hr = V.  This is the linear speed of our planet, not synonymous with surface speed.


Suppose that $ a$ is a real number such that all numbers $ 1^a;2^a;3^a;.....$ are integers, Prove that $ a$ is also integer.

S e a r c h:
http://www.mathlinks.ro/viewtopic.php?t=290825
http://www.mathlinks.ro/viewtopic.php?t=252647
http://www.mathlinks.ro/viewtopic.php?t=24042
http://www.mathlinks.ro/viewtopic.php?t=29085
http://www.mathlinks.ro/viewtopic.php?t=5607


geometry
area of a triangle
combinatorics unsolved
combinatorics

Given n points in the plane, let A be the minimum area of a triangle with vertices among these n points. Let G be the graph such that the edges of G are the sides of all triangles of area A. Prove that G can always be colored in 2 colors such that there is no monochromatic triangle in the graph.

You mean no monochromatic triangle of area $A$? (Because otherwise it is obviously false: take any 5 points no 3 of which lie on one line and for each pair of points choose a point far away such that the resulting triangle has a fixed small area, which will be your $A$).

Seems I know a counterexample to this too. But why don't we continue in the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=28973]minimum area[/url] thread where it all started? I'm not sure that creating spin-offs like this is a good practice (No offence meant, manuel... :)  It just confused me where to post now).


Using infinite geometric series:
$ 2.357357357357...\equal{}2\plus{}\frac{357}{1000}\plus{}\frac{357}{1000000}\plus{}...\equal{}2\plus{}\frac{\frac{357}{1000}}{1\minus{}\frac{1}{1000}}\equal{}2\plus{}\frac{357}{999}\equal{}\frac{2355}{999}$ which is basically what Kouichi Nakagawa had but in a longer version xD

[quote="Kouichi Nakagawa"]$ 2.357357\cdots \equal{} 2 \plus{} \frac {357}{999} \equal{} \boxed{\frac {2355}{999}}$.
So answer is [b]C[/b].[/quote]

To see that the repeating part is $ \frac {357}{999}$, let $ n\equal{}.357357357...$. Then subtract $ n$ from $ 1000n$, to get $ 999n\equal{}357$.  Solving for $ n$, get $ n\equal{}\frac {357}{999}$.

In general, when there is a repeating decimal, say $ 0.abcdabcdabcd...$ it can be expressed as $ \dfrac{abcd}{9999}$.  Therefore, the repeating part is $ \dfrac{357}{999}$, so $ 357/999 \plus{} 1998/999 \Rightarrow \dfrac{2355}{9999}$


Since this forum is basically like dead, why not post something.
So a poll will do. Lmao. Vote for whether there will be a question in Game Theory on the COMC

LMFAO i totally forgot about the NO option  :P

NOOOOOOOOOOOOOOOOOOOOOOOOO
THEY DIDN'T HAVE GAME THEORY :( :(
oh whatever :P hahaha

[quote="Bacteria"]No Llamas either  :([/quote]

I wanted another game :(

i spent a few hours working on game theory and was ready to pwn #3/4 for game theory =( instead, number theory owned me


I'm beginning calculus, and I trust the textbooks, but somehow I thought that AoPSers will be better at teaching calculus.  Just put all the basics here.  thnx

1. Many textbooks have been written. This forum is not an efficient way to deliver textbook material.

2. The regular posters here don't respond to generalities, but do tend to respond to specific questions, if interesting enough.

Maybe you should let your teacher teach you...and come here either to have an attempt at some further material, or to post problems you think the community would enjoy trying  :lol: .


function
calculus
derivative
superior algebra
superior algebra unsolved

1. Prove that :
The order of A group $ G$ is odd if and only if, for each $ b$ in $ G$, there exist $ a$ so that $ b\equal{}a^2$, in other words, an element is a square of the other elements. Thank you for your answer before.
2. If $ R$ is a finite ring, prove that the following statement is equivalent :
a. $ 1\plus{}1\equal{}0$
b. If $ f(x) \in R[x]$, then $ f(x^2)$ is reducible (over $ R[x]$).

For 1),(I just consider the finite group,Because you can easily find a infinite group has the property.)
Suppose the order of a group $ G$ is odd,then the order of each element $ b$ is order,then we have $ b^m \equal{} e$,where $ e$ is the identity of $ G$,and $ m$ is odd,then $ b \equal{} b^{m \plus{} 1}$,set $ a \equal{} b^{\frac {m \plus{} 1}2}$

Suppose the order of a group $ G$ is even,we must to find an element of $ G$ isn't a square of other element.First define a function of $ G$,$ f: G \to \mathbb {N}$ by $ f(a) \equal{} \max \{k \in \mathbb{N}: 2^k|order(a)\}$,Let $ f(b) \equal{} \max\{f(x): x \in G\}$,try to prove there is no element $ a$ of $ G$ such that $ b \equal{} a^2$(Left an exercise for you) :)

Slightly simpler argument for the second part of 1):

Suppose that every element is a square, then the homomorphism $ g \to g^2$ is surjective, hence injective, so there are no elements $ g$ such that $ g\neq 1$, $ g^2 \equal{} 1$.  But every group of even order has an element of order $ 2$, contradiction.

2) If $ 1\plus{}1\equal{}0$, then $ f(x^2)\equal{}(f(x))^2$.

Conversely, if $ x^2\minus{}a$ is reducible for all $ a$, then every element of $ R$ is a square.  As above, the map $ a \to a^2$ is injective.  Since $ (\minus{}1)^2\equal{}1^2$, we must have $ \minus{}1 \equal{} 1$.

Make sure you are clear on why this argument fails when $ R$ is infinite.

How can we conclude that : If $ 1\plus{}1\equal{}0$ then $ f(x^2)\equal{}(f(x))^2$. I am really really sorry but I don't understand. Please explain to me. Thank you for your explanation before.

I'm not sure LydianRain's solution is quite right.

If $ R$ has characteristic $ 2$ then $ x^2 \minus{} a$ has repeated roots (its derivative is zero) for every $ a$, so $ x^2 \minus{} a \equal{} (x \minus{} b)^2 \equal{} x^2 \minus{} b^2 \Leftrightarrow a \equal{} b^2$ and every element is a square.  Let $ f(x) \equal{} \sum a_k x^k$ and let $ b_k$ be such that $ a_k \equal{} b_k^2$, and let $ g(x) \equal{} \sum b_k x^k$.  The lemma $ (a \plus{} b)^2 \equal{} a^2 \plus{} b^2$ lets us prove that $ f(x) \equal{} g(x)^2$.

[quote="zhentil"]LydianRain's solution is not correct for #1, since $ \phi : g \rightarrow g^2$ is not a homomorphism if the group is nonabelian.[/quote]

I did write that a bit hastily, but it's still fine if you replace the word "homomorphism" by "map".


Find all functions  $ f: \mathbb R \to \mathbb R$ such that
i) $ f(2)\equal{}2$, and
ii) $ f(xy)\equal{}yf(x)\plus{}f(xy\minus{}xf(y)), \forall x;y \in  \mathbb R$.

I'm not sure for my answer....

Setting $ z\equal{}x\minus{}f(x)$ the equation becomes:
$ f(xy)\equal{}yf(x)\plus{}f(xz)$

Setting $ z\equal{}y$ thw equation becomes:
$ zf(x)\equal{}0$.
But $ f(x) \not \equal{}0$ so $ z\equal{}0 \Longleftrightarrow f(x)\equal{}x$,which true by checking.

[quote="Dimitris X"]I'm not sure for my answer....

Setting $ z \equal{} x \minus{} f(x)$ the equation becomes:
$ f(xy) \equal{} yf(x) \plus{} f(xz)$

Setting $ z \equal{} y$ thw equation becomes:
$ zf(x) \equal{} 0$.
But $ f(x) \not \equal{} 0$ so $ z \equal{} 0 \Longleftrightarrow f(x) \equal{} x$,which true by checking.[/quote]

You cant write once $ z\equal{}y\minus{}f(y)$ and two lines later $ z\equal{}y$ !!

So your solution is wrong.

And, btw, $ f(x)\equal{}x$ is not the unique solution of this equation.

[quote="Allnames"][quote="pco"]

And, btw, $ f(x) \equal{} x$ is not the unique solution of this equation.[/quote]
Yes pco  :lol: .The solution is $ f(x) \equal{} \frac {x}{1 \plus{} cx}$ with $ c$ is a real number.
And dear pco; I guess that  you had a nice proof for it  :)[/quote]

This is surely not the solution : any simple test prove this is wrong.

For example $ f(2)\neq 2$ if $ c\neq 0$

For example too, it's immediate to show that $ f(1) \equal{} 1$, so $ f(x) \equal{} \frac x{1 \plus{} cx}$ doest not fit the equation if $ c\neq 0$

:blush:  :blush:
My god! Today; I am very tired so I made many many mistakes.Indeed, That function should be solution of the equation in the topic  http://www.mathlinks.ro/viewtopic.php?t=305815 .And that is my confusion! :oops:
So dear pco, Can you show us your idea!
.

[quote="Allnames"]Find all functions  $ f: \mathbb R \to \mathbb R$ such that:
i) $ f(2) \equal{} 2$
ii) $ f(xy) \equal{} yf(x) \plus{} f(xy \minus{} xf(y)) \forall x;y \in \mathbb R$[/quote]

Let $ P(x,y)$ be the assertion $ f(xy)\equal{}yf(x)\plus{}f(xy\minus{}xf(y))$

$ P(0,1)$ $ \implies$ $ f(0)\equal{}0$. Then, if $ f(u)\equal{}0$, then $ P(2,u)$ $ \implies$ $ f(2u)\equal{}2u\plus{}f(2u)$ and so $ u\equal{}0$ and so $ \boxed{f(x)\equal{}0\iff x\equal{}0}$

$ P(1,1)$ $ \implies$ $ f(1)\equal{}f(1)\plus{}f(1\minus{}f(1))$ and so $ f(1\minus{}f(1))\equal{}0$ and so $ \boxed{f(1)\equal{}1}$

$ P(x,2)$ $ \implies$ $ f(2x)\equal{}2f(x)$

Suppose $ \exists u$ such that $ f(u)\equal{}\minus{}u$. Then $ P(x,u)$ $ \implies$ $ f(ux)\equal{}uf(x)\plus{}f(2ux)$ and so $ f(ux)\equal{}\minus{}uf(x)$
Then $ P(1,y)$ $ \implies$ $ f(y)\equal{}y\plus{}f(y\minus{}f(y))$, so $ f(y\minus{}f(y))\equal{}\minus{}(y\minus{}f(y))$ and so, using the line above, $ f((y\minus{}f(y))x)\equal{}\minus{}(y\minus{}f(y))f(x)$

But $ f(xy\minus{}xf(y))\equal{}f(xy)\minus{}yf(x)$ and so $ f(xy)\minus{}yf(x)\equal{}\minus{}yf(x)\plus{}f(x)f(y)$ and so $ \boxed{f(xy)\equal{}f(x)f(y)}$
Setting $ x\equal{}y\equal{}\minus{}1$ in this equation, we get $ f(\minus{}1)^2\equal{}1$ and so two cases :

1) $ f(\minus{}1)\equal{}\minus{}1$
Then $ P(x,\minus{}1)$ $ \implies$ $ f(\minus{}x)\equal{}\minus{}f(x)$
And since $ f(xy)\equal{}f(x)f(y)$ implies that $ f(x)>0$ $ \forall x>0$, we get $ f(x)$ has the same sign as $ x$ $ \forall x\neq 0$
But $ P(1,x)$ $ \implies$ $ f(x\minus{}f(x))\equal{}\minus{}(x\minus{}f(x))$ and so $ f(x)\equal{}x$ $ \forall x$ else we would have $ f(u)\equal{}\minus{}u$ and a contradiction with the previous line.

So $ f(x)\equal{}x$ $ \forall x$ and it is easy to check that this is indeed a solution

2) $ f(\minus{}1)\equal{}1$
Then $ P(x,\minus{}1)$ $ \implies$ $ f(\minus{}x)\equal{}f(x)$
And since $ f(xy)\equal{}f(x)f(y)$ implies that $ f(x)>0$ $ \forall x>0$, we get $ f(x)>0$ $ \forall x\neq 0$
But $ P(1,x)$ $ \implies$ $ f(x\minus{}f(x))\equal{}\minus{}(x\minus{}f(x))$ and so $ f(x)\geq x$ $ \forall x$

Let then $ x>0$. If $ f(x)>x$, and since $ f(\frac 1x)\geq \frac 1x$, we get $ f(x)f(\frac 1x)>1$, which is impossible since $ f(x)f(\frac 1x)\equal{}f(x\frac 1x)\equal{}f(1)\equal{}1$

So $ f(x)\equal{}x$ $ \forall x>0$
And so, since $ f(\minus{}x)\equal{}f(x)$ :  $ f(x)\equal{}|x|$ $ \forall x$ and it is easy to check that this is indeed a solution

3) Synthesis of solutions :
We got two solutions :
$ f(x)\equal{}x$ $ \forall x$
$ f(x)\equal{}|x|$ $ \forall x$

[quote="pco"]

Suppose $ \exists u$ such that $ f(u) \equal{} \minus{} u$.[/quote]
Here is the unique sentence that I don't understand in your nice proof dear pco.Can you explain more !
I also proved $ f(xy) \equal{} f(x)f(y)$ but by the other way!.Maybe it is a bit shorter !
We have $ P(x;y \minus{} f(y))$ implies
$ f(x(y \minus{} f(y)) \equal{} [y \minus{} f(y)]f(x) \plus{} f([y \minus{} f(y)]x \minus{} x[f(y) \minus{} y])$
$ \equal{} yf(x) \minus{} f(x)f(y) \plus{} f(2x(y \minus{} f(y))$
But note $ f(2x) \equal{} 2(x)$ thus $ f(xy \minus{} xf(y)) \equal{} f(x)f(y) \minus{} yf(x) \equal{} f(xy) \minus{} yf(x)$.Then $ f(x)f(y) \equal{} f(xy)$.
You finished this problem too well so I don't post my last parts!
Thank you!

[quote="Allnames"][quote="pco"]

Suppose $ \exists u$ such that $ f(u) \equal{} \minus{} u$.[/quote]
Here is the unique sentence that I don't understand in your nice proof dear pco.Can you explain more !
[/quote]

I dont understand what you dont understand :)

I demonstrated that if  $ \exists u$ such that $ f(u) \equal{} \minus{} u$, then $ f(ux)\equal{}\minus{}uf(x)$ for all $ x$

Then i demonstrated that $ f(y\minus{}f(y))\equal{}\minus{}(y\minus{}f(y))$ and so, (considering $ u\equal{}y\minus{}f(y)$), we get $ f((y\minus{}f(y))x)\equal{}\minus{}(y\minus{}f(y)f(x)$

What is the part not clear for you ?

[quote="pco"]
I demonstrated that if  $ \exists u$ such that $ f(u) \equal{} \minus{} u$, then $ f(ux) \equal{} \minus{} uf(x)$ for all $ x$

Then i demonstrated that $ f(y \minus{} f(y)) \equal{} \minus{} (y \minus{} f(y))$ and so, (considering $ u \equal{} y \minus{} f(y)$), we get $ f((y \minus{} f(y))x) \equal{} \minus{} (y \minus{} f(y)f(x)$

[/quote]
 :blush:  :blush:  :blush: .
Yes pco, Sorry for my silly question above. Now, I really realized that your proof was very nice and skillful.
Thank you!
How about my problem I linked at #6?


[quote="linker"]Solve the equation :

$ \sin^7{x} \plus{} \cos^7{x} \equal{} 1$ .[/quote]
[hide="Solution"]
Notice that $ \sin x\le 1$ and $ \cos x\le 1$.  Hence, $ \sin^2 x\ge \sin^7 x$ and $ \cos^2 x\ge \cos^7 x$ with equality if and only if $ \cos x \equal{} 0,1$ or $ \sin x\equal{}0,1$  Hence, $ 1 \equal{} \sin^7 x \plus{} \cos^7 x\le \sin^2 x \plus{} \cos^2 x \equal{} 1\implies 1\le 1$.  Hence, we have the equality case, meaning that $ \sin^2 x\equal{}\sin^7 x\implies \sin x\equal{}0,1$ and $ \cos^2 x\equal{}\cos^7 x\implies \cos x\equal{}0,1$.  Thus, $ x \equal{} 90 \plus{} 360k$ or $ x \equal{} 360k$ for an integer $ k$.  [/hide]


trigonometry
calculus
calculus computations

a)  Prove if $ \sum_{ \minus{} \infty}^{\infty}c_ne^{inx}$ converges absolutely at a point $ x$, then $ \sum_{ \minus{} \infty}^{\infty}|c_n|$ converges.

b)  Show that the series $ \sum_{n \equal{} 1}^{\infty}\sin{nx}$ which is of the form $ \sum_{k \equal{} 1}^{\infty}a_k\sin{kx}$, converges absolutely at some points $ x$ and yet $ \sum_{k \equal{} 1}^{\infty}|a_k|$ does not converge.

(a) Just write down the definition.
(b) $ x\equal{}0$.


$24)$ Two distinct points are selected at random from the nine point set $S$. $S={(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)}$. What is the probability that their midpoint belongs to $S$?

oh wait, i did this one.
you just have to count it out, it's not that big of a list

OH OK I GOT IT THANKS!!! :lol:

$\frac{3+1+2+1+0+0+1+0}{8+7+6+5+4+3+2+1}=\frac{8}{36}=\frac{2}{9}$

I just looked what kind of numbers they had to be to match... example Odd Odd for 3 and 3 to stay even to be able to divide into an integer. It worked quickly starting from the first coordinate.  :lol:

The total number of possibilities is $\binom{9}{2}$

The total number of desired must have the same parity for the coordinates.

This can be done in $\binom{4}{2}+\binom{2}{2}+\binom{2}{2}+\binom{1}{2}=8$.

So the answer is $\frac{2}{9}$


calculus
derivative
function
algebra
polynomial
real analysis
real analysis theorems

[quote="Ijo07"]Could someone explain to me how to use the power/chain rule short cut to solve derivative problems?

I get the xn=nxn-1 for simple equations like y=x5, but how do you use it for more complex equations, such as 2000+100x-0.1x2?[/quote]

2000 + 100x - 0.1x = 2000 + 99.9x

this can be written (to show you how the rule works) as

= 2000x^0 + 99.9x^1

It is a general rule that the derivative of the sum of the two functions is equal to the sum of the derivatives of those functions, formally written:

d/dx (v + u) = dv/dx + du/dx

so by letting v = 2000x^0 and u=99.9x^1,

dv/dx = 0*2000x^(-1) = 0
du/dx = 1*99.9x^(1-1) = 99.9

so the derivative of the expression is 0 + 99.9

...

in general, the derivative with respect to x of any constant term is zero, and the derivative with respect to x of "ax", where a is a constant, is a.

so your rule of:

x^n ---> nx^(n-1)

can be generalised to:

ax^n ---> anx^(n-1)

if you want pracitce questions, just write down a polynomial of degree 5 or higher and find the first derivatives, eg. if y = f(x), find f'(x), f''(x), f'''(x) and so on. n can be any real constant, so if

y = 7x^(-pi)
dy/dx = -pi*7x^(-pi-1)

still works..

The chain rule is different to the power rule, and states that for a function y(x), if the function can be written:

y(x) = v(u(x))

y'(x) = [v'(x)][u'(x)]

so for example,

y(x) = 3(x^2 - 3)^17

if you let:

v = 3u^17, then v' = 51u^16,
u = x^2 - 3, then u' = 2x

and y'(x) = 102x(u^16) = 102x(x^2 - 3)^16

which is much simpler than multiplying out and differentiating each term seperately.

[quote]2000 + 100x - 0.1x = 2000 + 99.9x [/quote]

^True, but my problem was 2000 + 100x-0.1x[u]2[/u].

But I think I still get it... you'd take the derivative of 2000 (which is 0) and add it to the derivative of 100x (which is 100) and add [i]that[/i] to the derivative of 0.1x2 (which is 0.2x)... so I'd get 100 +0.2x, right?


Is it OK to send just 5a for the USAMTS round 3, or can i not send it at all without having both?

I don't see why not.  The worst that will happen is that you don't get any credit for the question.  But I don't think that will be the case.

you never know. They [i]might[/i] give you negative points!!!  :P

Please don't say stuff that isn't true like that because it's not clear that it's sarcasm.  They will never give negative points in USAMTS.  For final clarification, if you have at least two solutions to send, by all means send a partial solution to #5.  You can probably get at least 1 point for 5a, and I'm guessing that they'll set the rubric up as 2 points for 5a and 3 points for 5b.

[quote="ZzZzZzZzZzZz"]Please don't say stuff that isn't true like that because it's not clear that it's sarcasm.[/quote]
The emoticon shows that it's sarcasm.

Um... so... stop arguing about whether it's sarcasm or not. It was not a very useful statement, and was kind of condescending.


geometry
3D geometry
tetrahedron
sphere
geometry solved

hey guys help
The tetrahedron ABCD is incribed in a sphere with center O,with radius R ,such that O lines inside the tetrahedron.The lines OA,OB,OC,OD cut repectively the faces BCD,CDA,DAB,ABC at A' ,B' , C',D'.Prove that AA'+BB'+CC'+DD'>=16/3R.

I think this problem is similary this problem VMO 1999 .We can use
combinatorics and prove it is  long :?  :D


algebra
polynomial
linear algebra
linear algebra unsolved

Proof that similar matrixes A and B has the same charactstics polynomial and minimal polynomial. :!:
Cheers,

 :D

[quote="alekk"]$A$ and $B$ represent the same endomorphism. So it's OK.[/quote]

Endomorphism? Can you explain? (i'm not native english speaker...)

Thanks


The task begins now. The rules again:

This task lasts 72 hours. Everyone in the game must post here at least once. No double posting is allowed. In the first 48 hours, consecutive posts must be at most 3 hours apart. In the last 24, they can be up to 5 hours apart. Discuss what you want here.

If you are not part of this game, please refrain from posting. Thanks.

I will renew ourselves with 8 more minutes to post the next post.

So:

We need someone to post:
8:30 PDT/9:30 MDT/10:30 CDT/11:30 EDT
11:00 PDT/12:00 MDT/1:00 CDT/2:00 EDT [I can probably do this]
1:30 PDT/2:30 MDT/3:30 CDT/4:30 EDT [We need somebody in the east coast]
4:00 PDT/5:00 MDT/6:00 CDT/7:00 EDT [again, from the east coast]
...

Detailed Schedule to come

someone just post right as they go to bed (after midnight), if you live in cali

let's just talk about some subject so this seems less like something you have to do and more like something you want to do

Let's play 20 Questions!

OK it's a thing.

Is it a solid?

@James: 1:30 PDT/2:30 MDT/3:30 CDT/4:30 EDT - will be the hardest one, but it'll certainly be easier for those on the west coast. If it's not possible, then just do it as late as possible, and I'll get up as early as possible and make another post.

Yes it is solid, 18 left

I will be going to bed now, so all you west coasters are going to have to help out here...

I'll wake up briefly for a post as early as I can, if I can force myself.

New deadline:
1:20 PDT/2:20 MDT/3:20 CDT/4:20 EDT

Just saw this... I'll be sleeping soon, so I might as well post this now. I'll also be gone all day tomorrow until about 6:00 PST... sorry :(

I suppose I am allowed to post again? ... not too helpful though

EDIT: I'm going to bed soon too, but I'll make a post when I leave.
EDIT again: This thread should probably go in fun factory.

and another post

bruteforce i am going to go to bed

I've been wondering too.

Nope he isn't active Maf34.

Oh, I probably won't be sleeping that late today...

How bout 11:30 PM

then someone in the east coast can pick up at 7:00 AM

Night.
/moan /complain (had to stay up... even though it's before midnight... want Zzz's...)

just woke up - have a good day everyone!

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

Because the task ends in less than 4 hours, you have successfully made it through the 72 hours. However, mathelete9999 did not post.

I've decided to give you all a break. mathelete9999 told me that he was hit by a power outage due to Hurricane Ike, and said, with his limited access, that he would be ready to play again in a few days.

So $ \$20,000$ has been added to the pot, congratulations.


There are $ n$ people in a class. The first person forgets his seat, and just picks one randomly. The next $ n\minus{}1$ people operate as follows:
If their seat is vacant, then they will take it.
If their seat is taken, then they will pick a random one.
What is the likelihood that the last arriving person will sit in his or her assigned seat?

1/2.  The nth person sits in their own seat iff one of the first (n-1) people sit in their own seat before anyone has sat in the nth person's seat.  By symmetry, this happens with probability 1/2.

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=929961680&t=169528

And an extension was recently posted

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=929961680&t=277835

(I found these by searching for "randomly seat.")

[quote="ayush_2008"]what is the probabilty that rth person where 0<r<=n will sit on his seat[/quote]

Posts (at least) 5 & 6 here http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=929961680&t=169528 are both helpful, but the counting approach below is a bit more like post 6 here http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=929961680&t=277835. Works for me, and I hope it helps.

Firstly, for r =n, count the possible seatings...

The first person sits in his own seat... 1

He sits in the nth person's seat... 1

So far that's two possibilities, one of them in the last (i.e. nth) person's favour.

The first person sits in the (n-1)th person's seat... another 2 possibilities, one of which (again) unites the nth person with his seat.

The first person sits in the (n-2)th person's seat... another 4 possibilities, 2 of which unite the nth person with his seat.

The first person sits in the (n-3)th person's seat... another 8 possibilities, 4 of which unite the nth person with his seat.

Etc.

E.g., the above five cases for n = 7, where numbers are people in number order of choosing a seat, and the position of a number indicates whose seat the person has chosen. We have...

1 2 3 4 5 6 7 ... everyone has chosen their own

7 2 3 4 5 6 1 ... first has chosen the seat belonging to the nth

6 2 3 4 5 1 7
7 2 3 4 5 1 6 ... first has chosen the seat belonging to the (n-1)th

5 2 3 4 1 6 7
6 2 3 4 1 5 7
7 2 3 4 1 6 5
7 2 3 4 1 5 6 ... first has chosen the seat belonging to the (n-2)th

4 2 3 1 5 6 7
5 2 3 1 4 6 7
6 2 3 1 5 4 7
6 2 3 1 4 5 7
7 2 3 1 5 6 4
7 2 3 1 4 6 5
7 2 3 1 5 4 6
7 2 3 1 4 5 6 ... first has chosen the seat belonging to the (n-3)th

Treating each block of people from person 2 to person 1 as one person you can reduce the numbers as shown below, and proceed (after case one) on the assumption that the first person/block chooses seat 2. And this shows that the sets of possibilities are bound to double.

1 2 (everyone has chosen their own)

2 1 (first has chosen the seat belonging to the nth)

2 1 3
3 1 2 first has chosen the seat belonging to the (n-1)th

2 1 3 4
3 1 2 4
4 1 3 2
4 1 2 3 first has chosen the seat belonging to the (n-2)th

2 1 3 4 5
3 1 2 4 5
4 1 3 2 5
4 1 2 3 5
5 1 3 4 2
5 1 2 4 3
5 1 3 2 4
5 1 2 3 4 first has chosen the seat belonging to the (n-3)th

Notice each new set of seatings takes the previous and adds a new successor on the right, then duplicates the set but switches left and right ends round. So you have one version of the previous set with the nth person satisfied, and another equal set where he loses. So wherever the first person sits (apart from seats one and n), the nth person has an equal chance of sitting in his own or the first person's place (and zero chance of sitting anywhere else).

For r < n, the rth person may have the opportunity to take the seat of a later chooser - if, that is, the first person chose the seat of an earlier chooser than r.  But such opportunities are already counted in the above tables, from which you can see that the rth person still has an even chance of getting his own seat, in these cases (i.e. when the first person steals the seat of an earlier chooser than r). The other possibilities are also divided equally between satisfaction and dis-satisfaction for r, although all of the second kind are caused by the first person taking his seat.

So, everyone has the same chance (evens) of finding their own seat.

Instead of finding the probability, can anybody find the number of ways in which all can be seated provided the last one occupies his assigned seat?

Consider the case for $4$ people.
Thus we must consider the cases where $1,2$ and $3$ must sit in their places.
The arrangements are
$(1,2,3,4)$
$(2,1,3,4)$
$(3,1,2,4)$
$(3,2,1,4)$
Now let the subsets of the set $S=({2,3})$ represent the arrangements where the elements of the subsets denote that the persons with the numbers in the elements do not sit in their own seat. We are not considering where $1$ sits. The number of subsets of $S$ denotes the number of arrangements.
For example, the powerset of $S$ comprises of $((\phi),(2),(3),(2,3))$. $(\phi)$ denotes that all the people from $1$ to $3$ seat in their seats, i.e. arrangement $1$. $(2)$ denotes that only $2$ does not sit in his seat and $3$ does, i.e. arrangement $2$. While $(2,3)$ denotes that both $2$ and $3$ do not sit in their seats, i.e. arrangement $4$.
So generalizing this principle for $n$ persons we get the number to be $2^{(n-2)}$ arrangements.


you can post again and again.

I'll start off:

I'm spectacular because I am a person who posts on AoPS.

I am spectacular because I just finished one year on AoPS.

im spectacular cuz 1+1=2

duh...its a no brainer.

but then meybe 1+1 isnt 2.

I'm spectacular cuz I said so. Since I'm spectacular, whatever I said must be right, so I'm spectacular.

i'm spectacular because my psychiatrist says so

I'm spectacular because I can carry on a conversation with you longer than you can carry on a conversation with me. (Try it! AIM: Randomdragoon)

I'm spectacular because my mommy said so

I am spectacular because I am the CheeseLord.

[quote="footballrocks41237"]I am spectacular because I know Wisconsin will win.[/quote]
i'm spectacular because i know osu will pwn wi

I'm spectacular because i read the post above me, so i know that too. knowledge made both of us spectacular :D

I'm not that spectacular but my friend Umbreon sure is!

He's on vacation right now ... he might come back later.

My friend is not spectacular, but I sure am!


First one:
[img]8021[/img]
[hide="Remark."]
Use this construction to provide a general construction for the blue circle:
[img]8022[/img]
[/hide]

Is the green circle touching the blue line(one)

The first one:
In the part of the plane which is denoted by $m,l$ and not contains $B$ the two lines $x,y$ that parallel with $m,l$, respectively, and both of the distance between $x$ and $m$ ;$y$ and $l$ are equal to the radius of $B$. We are going to construct the circle which tangents $x,y$ and passes through the center $O_{1}$ of $B$. It's easy, let $f$ is the inversion having $O_{1}$ for pole so $f(C)$ is the common tangents of $f(x)$ and $f(y)$. So the center of this circle is the center of $C$.
The second:
Let $f$ is the inversion having an intersection of two arbitrary orange circle for pole. So we converted this problem to the first one  :wink:

Very nice solution!
[img]8061[/img]
For the second I think you have the same idea as me except I'm not sure what you mean by pole of an inversion, the way I did it involved constructing the circle of inversion and took several more steps than yours...

I solved the following easy problem with your "method" from the first one:

[i]Construct the green circle[/i]
[img]8062[/img]

Ah, I mean that from 3 given orange circle, arbitrary select 2 circle and let one of two intersection of them is the pole of the inversion  :D
About your new problem, my idea is let two given points  are $A,B$ and the given line is $d$, so if $f$ is the inversion having $A$ for pole and $AB^{2}$ for constant, in order to construct the green circle $C$ we will construct $f(C)$. It's the line pass $B$ and touches the circle $f(d)$
P/s: Can you guide me how to insert a picture looks like you please? I see that the pictures I atteched to forum isn't nice  :( . Thank you  :blush:

After looking at some of your other posts I understand what you mean by pole etc., you have a different way of looking at inversion than me (you specify pole and constance, I specify reference circle.

I'm not going to explain it but here is a picture of my solution... it doesn't use any inversion.

[img]8085[/img]

The way to insert nice images is to start a personal gallery, in mathlinks mode, and upload images to it. Be sure to make them small so that they look best.


For what value of $ n$ is $ i\plus{}2i^2\plus{}3i^3\plus{}\cdots\plus{}ni^n\equal{}48\plus{}49i$?

Note: here $ i\equal{}\sqrt{\minus{}1}$.

$ \textbf{(A)}\ 24 \qquad
\textbf{(B)}\ 48 \qquad
\textbf{(C)}\ 49 \qquad
\textbf{(D)}\ 97 \qquad
\textbf{(E)}\ 98$

The sum of every four terms starting from the first is $ 2\minus{}2i$. $ (2\minus{}2i)24 \plus{} 97i \equal{} 48 \plus{} 49i$, so it's the 97th term, D.

[hide="Barely Different From First"]
$ m$ block of $ 4$ terms gives you $ 2\plus{}2i$ if you skip the first term. Obviously you need $ 24$ blocks to achiever $ 48\plus{}48i$ and the extra $ i$ at the beginning makes it $ 48\plus{}49i$. The final $ n^{th}$ term in block $ m$ is $ 4m\plus{}1\equal{}4(24)\plus{}1\equal{}96\plus{}1\equal{}97 \implies \boxed{D}$
[/hide]

or you can do like:
since it's 1i-2-3i+4+5i....
sort the i terms together:
set it as x+1 terms in it, so it will look like (1-3+5-7+9-11...)i=49i
since 1-3=5-7, so set up your equation: -2(x+1-1)/2+2(x+1)-1=49, and you get x+1=49
since the term without the i is one less than the i terms, so the whole is 49+49-1=97


Is there any FREE[b] LITERATURES in the website that gives information about Artificial[u] intelligence?

                                                             Respectfully
                                                             Arthur eddington [/u][/b]

Tons of it. Just google [i]free books on artificial intelligence[/i] or something like that to get the links. It is not easy to tell where wild dreaming ends and real science begins in this subject nowdays so the quality of the related literature varies quite a lot but its availability is not a problem.


The natural numbers 1 through 25 are arranged in a 5 by 5 magic square.  The sum of the numbers in each row is the same.  What is that sum?
\[ \text{(A)}\ 325\qquad \text{(B)}\ 75\qquad \text{(C)}\ 65\qquad \text{(D)}\ 55\qquad \text{(E)}\ 26 \]

[hide]we know that all natural numbers from 1-25 are in the magic square so the sum of all the numbers is (26*25)/2=325. Since there are 5 rows in a 5x5 square and we know each row is equal, the sum of an individual row is 325/5=65.[/hide]

Also, if you have consecutive numbers like that, you can just fill them in across the magic square:
 1     2 3    4  5
 6    7  8   9  10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25

When you add up either of the diagonals, you get 65. (C)

Since you have the numbers 1+2......+25, to find out the sum of this you do (26X25)/2, which would give you 325, then you have to find out the sum of only 1 row, so to do this you have to divide 325 by 5, resulting in 65.


calculus
integration
function
limit
real analysis
real analysis theorems

Is the following statement true?:

Let $\left\{f_{n}\right\}_{n\in\mathbb{N}}$ be a sequence of Riemann integrable functions on $\left[a,b\right]\subset\mathbb{R}$ converging pointwise to a function $f$ on $\left[a,b\right]\subset\mathbb{R}$.  Then it follows that

\[\int_{a}^{b}f\,dx=\lim_{n\rightarrow\infty}\int_{a}^{b}f_{n}\,dx.\]

It is certainly false. I'll even make the $f_{n}$ in my example continuous.

Let $f_{n}$ for $n\ge 2$ be the piecewise linear continuous function on $[0,1].$ whose graph has vertices at $(0,0),\left(\frac1n,n\right),\left(\frac2n,0\right),$ and $(1,0).$

$\lim_{n\to\infty}f_{n}(x)=0$ for every $x\in[0,1],$ but

$\lim\int_{0}^{1}f_{n}(x)\,dx=1.$

Make the "spike" taller, and we could have the limit of the integrals go to infinity if we wanted. Pointwise convergence, even pointwise convergence everywhere, is such a weak form of convergence that it gives you very little control.


Can someone plz explain balls and urns..thanks!!!

Try looking here:
http://www.artofproblemsolving.com/Wiki/index.php/Balls_and_urns

Or here, i had the same exact question few wks ago..lol:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=264138[/url]

ohhh..lol...i knew what this thing was....but i didnt know it had a name:"balls and urns"...lol.


Obviously $f(2) = 0$ and $f(-2) = 12$. That gives you two equations for $a$ and $b$... :)

[hide=" ansewr"]  $a=-7$  and $b=2$ [/hide]

$(x-2)n=x^3+x^2+ax+b$
$(x+2)n+12=x^3+x^2+ax+b$

Plug in $2$ for the first equation and $-2$ for the second one and we get:

$0=2^3+2^2+2a+b \Longrightarrow 2a+b=-12$
$12=(-2)^3+(-2)^2-2a+b \Longrightarrow 16=-2a+b$

Thus using sys of eq, we get $b=2, a=-7$, so we have $x^3+x^2-7x+2$. From there simply use synthetic division to find the roots  :).

We get $(x-2)(x^2+3x-1)$ so the roots are: $\boxed{x={2,\frac{-3+-\sqrt{13}}{2}}}$

[quote="ifai"]If $x^3+x^2+ax+b$ is divisible by x-2, and the remainder is 12 when
it is divided by x+2, determine a and b, and also, find the roots of the equation $x^3+x^2+ax+b=0$[/quote]
Use factor theorem...that should get you somewhere. :)


[quote="Howdiddlydoo"]A roll of paper contains 440 sheets each 9cm x 9cm. The outside diameter of the roll is 14cm and the inside diameter is 4cm. Find the thickness of each sheet to the nearest hundredth of a cm.

(Use 22/7 as an approximation for Pi)

I have the answer, but I can't figure out how to get it.

ANSWER: [hide] .04 cm [/hide][/quote]

This problem has been previously posted by #H34N1 at
[url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1199108518&t=137608[/url]

[hide="solution"]Let the thickness be $x$.
When seen from the side, there are two concentric circles with diameters 14cm and 4cm. The area of the tissues being shown is the area in between the two circles, which is equal to $7^{2}\pi-2^{2}\pi = 45 \pi$. Each of the 440 sheets have the surface area of $9x$ when seen on this side. From this,

$45 \pi = 440 \cdot 9x$

$45 \cdot \frac{22}{7}=440 \cdot 9x$

$x= \frac{45 \cdot \frac{22}{7}}{440 \cdot 9}\approx \boxed{0.04}$ (cm).[/hide]


inequalities
limit
algebra proposed
algebra
Fixed point

some pro can solve by inequality  [tex]\lim{r^n}=0[/tex] for real number r such that |r|<1
FOR example, we have some pro:
E-1: (VMO-02-TABLE B): for real number [tex]t\neq{0}[/tex], define {[tex]x_n[/tex]}:
[tex] x_1=t[/tex]
[tex]x_{n+1}(x_{n}+t)=t+1[/tex]
find [tex]\lim{x_n}[/tex]


Something seems to be wrong. Surely $ x\plus{}y < x^n\plus{}y^n$ and the coprimality is never needed.

Sorry, I made a typo. I have just edited.
 :oops:
kunny

You mean $ x^n\plus{}y^n$ is not divisible by $ x\plus{}y$, I think.

[quote="kunny"]Let $ x,\ y$ be coprime positive integers such that $ xy\neq 1$. If $ n$ is positive even, then prove that $ x^n \plus{} y^n$ is not divisible by $ x \plus{} y$.[/quote]
$ d\equal{}\gcd(x^n\plus{}y^n,x\plus{}y)$
Then $ d|x\plus{}y$ so $ x\equiv \minus{}y (\mod d )$
But from $ d|x^n\plus{}y^n$ so $ d|2x^n$
Because $ \gcd(x,y)\equal{}1$ so $ d|2$
From $ xy>1$  so $ x\plus{}y>2$ it means that $ x\plus{}y\not |x^n\plus{}y^n$


Find the maximum area of a quadrilateral with sides $3, 4, 5, 6$.

Let's say that $ABCD$ is the quadrilateral with side length you gave. Let the intersection of $AC$ and $BD$ be $O$. Hence: $S_{ABCD}= \frac12 \cdot AC \cdot BD \sin{AOB} \leq \frac12 AC \cdot BD$.

Now use Ptolemy Theorem: $ AC \cdot BD \leq AB \cdot CD + AD \cdot BC $ and we get the answer.

So this quadrilateral must be a cyclic with AC $\perp$ BD.

Correct. Now use Brahmagupta's formula to find the area. :)

[quote="chess64"]Brahmagupta's formula[/quote]
What's this? I think that we have already get the answer: $21$. Right?

[quote="nr1337"]http://mathworld.wolfram.com/BrahmaguptasFormula.html

After we find that the quadrilateral is cyclic, we find that the area is sqrt(6*5*4*3) = 6 sqrt(10)[/quote]

Looks right to me. :)

This should be easy when applying the Brahmagupta formula, which should give you the maximum area:

$S=9$

$\sqrt{(9-3)(9-4)(9-5)(9-6)}=\sqrt{360}=\boxed{6\sqrt{10}}$.

Hmmm... :(  What's wrong with my proof? Is that the two equality can't hold at the same time?

For $ABCD$ to be both cyclic and $AC\perp BD$ we would have $a^2+b^2=c^2+d^2$ for $\{a,b,c,d\}$, a permutation of $\{3,4,5,6\}.$  But there are no solutions to that equation.  Thus we cannot achieve that area with the given side lengths.

AC $\perp$ BD does not imply $a^2+b^2=c^2+d^2$.

O, sry...I'm still waking up  :blush:

But, I think that the flaw should be along the same lines.  Because there are only a few possibilities for the quadrilateral if you know it's cyclic, you probably can't have the diagonals perpendicular also.


linear algebra
matrix
vector
superior algebra
superior algebra solved

[quote="alekk"]$-I_n$ is not the product of $4$ real symetric positive matrices.[/quote]"Positive"?

You mean positive definite or something? :)

Or when is a matrix positive?

Assume that $ ABCD = -I $, where $ A,B,C,D $ are real positive symmetric.
Then, because of $ A $ is real positive symmetric, we can write $ A = M^{2} $,
where $ M $ is real positive symmetric. Then we have
$ MMBCD = -I $ ,so $ MBCDM = -I $, so $ (MBM)(M^{-1}CM^{-1})(MDM) = -I $, so if
we define $ X = MBM $, $ Y = M^{-1}CM^{-1} $ and $ Z = MDM $, then they are symmetric
and positive , since, for example, for each nonzero vector $ v $, $ (MBMv,v) = (BMv,Mv) > 0 $.
So we get that $ XYZ = -I $. Taking transprose, we get $ ZYX = -I $. Therefore
$ YZ = -X^{-1} = ZY $. So $ Z,Y $ commute, therefore there exists common eigenvector
$ v $. If we assume that $ Yv = \lambda v $ and $ Zv = \mu v $, then $ \lambda,\mu > 0 $,
so we get $ XYZv = \lambda \mu Xv = -v $, so $ Xv = (-1/ \lambda \mu) v $. So
$ -1/ \lambda \mu $ is an eigenvalue of $ X $, but $ X $ is positive,
and we come to contradiction.

I hope it is correct.

I still don't get the problem?

If they are positive definite, then their eigenvalues are positive.

if explicit_eigenvalues(A)={a1..an} and explicit_eigenvalues(B)={b1..bn}, then we have that explicit_eigenvalues(AB)={a1b1,..,anbn} right?

So far from gettin -In, we can't even get any matrix with a negative eigenvalue :?:

(with explicit_eigenvalues() I just mean: if m appears 3 times, then write m,m,m instead of m. and write all eigenvalues in the order they would appear in the jordanform)

Peter, you just can't multiply the eigenvalues as you want!!!

Why not, harazi? :?

Symmetric real positive definite matrices are diagonizable. That means their jordanform is a diagonal matrix.

That means the eigenvalues of the product can easily be computated by finding the product of the 4 diagonal matrices.

And we all know how to multiply diagonal matrices :)

They are diagonalizable, but nobody tells you that you can diagonalize them by the same matrix.

Indeed harazi&grobber, you are right :) it is not the exact product. So I'll have to find another proof :)

Still, can you give me one example of 4 such matrices giving a matrix with at least one negative eigenvalue? it may not be a correct proof still I can't find any counterexamples for my initial claim...


calculus
articles
MATHCOUNTS
geometry
AMC 8
AoPS Books

anybody interested in hosting a "competition" where we pm the solutions and the guy marks it and put up hte ranks?

you mean like what toppers doing?  i think hes doing a pretty good job.  i dont really care for the competition part, but whatever makes people post more questions is good.  What level were you thinking?

i think people tried to do this over the summer, making a compilation of questions, but i dont know how it turned out.

'twould be excellent.

As a followup, what do people think of this idea:
The creation of a site somewhat like a combination of this site and mathworld.
Basically, what I was thinking was that there are a lot of tutorials/guides on the internet that will teach you to program, build desks, and weave baskets (random examples); why not have a site with an extensive series of tutorials on math concepts? Obviously the primary focus would be for competitions, but it could also include things like Calculus or really whatever, as long as it was accurate. I'd like to see a basic guideline of what people should learn and in what order they should learn it, along with at least an introduction to the topics - preferably more. Each tutorial could have a couple problems at the end with solutions - not many, because problems are pretty easy to find online. Anyway, if there's some interest in this, I'll cross post it to etc. and some other sites online, and if there's enough, I (along with some others) could get a site started.

confuted:

Well, AoPS does a pretty good job of that, don't you think? But still, if you do plan on making such a site, would it just be a compilation of various articles by various people (ie. one article on ceva's theorem, one article on diophantines of the form ax+by=n, etc...)?

Yeah, AoPS does a pretty good job of it, but there's a bit of a "gap" between V1 and V2 in terms of difficulty... or so it seems. But yeah, I was thinking of a collection of articles by various authors. I sure don't know enough to make the entire site.

[quote="JBL"]I'm up for it . . . I once did this on a math forum created by people fed up with the mathcounts forum a few years ago.  (I think that forum is long since defunct.)  Anyhow, I think it's a good idea.[/quote]

I promise, this is my last post before I turn in to get ready to bring my son to AMC 8 tomorrow  :oops: I like the idea of having some of the content bugzpodder and the others who replied mentioned online. Why not just post it here on this site? My reasoning is the basic Occam's Razor principle: entities should not be multiplied without necessity. What's posted here is, of course, up to the owner of the site, but I think posts that are CLOSELY related to mathematics problem-solving competitions could only be good building up participation on this site, which I hope for the site owners' sake will sell more books and online courses. (I write this as the owner of a Web site that doesn't make any money--it's important to help providers of good content gain something from their toil.)

So, yes, it would be interesting to see an online "competition," and to see short capsule descriptions of interesting areas of math relevant to competitions, and to see some guidance about self-study curricula in math. Wouldn't most of that be on-topic here? (I guess this implicitly is a question for the owner(s) of this site.) I LOVE to read good content online. I just hate to bop around to too many different places to read it. That cuts too severely into time for reading dead-tree books, which include valuable titles like, um, The Art of Problem Solving volumes 1 and 2.

I guess I'm up for it too . . . .

If anyone is interested in writing articles for this site, we'd be happy to include suitable ones among our resources.  'Suitable' means well-written and of interest to our Community.  Such articles would include those like those contemplated in this thread.  We would strongly prefer the articles be basically HTML-ready.  We likely won't have time to convert articles from, say, Word to HTML.  This doesn't mean you have to build a webpage or anything, it just means that we may not be able to do your math formatting or image building for you.  We can do limited conversion, so if your article doesn't have too much fancy stuff, we can take care of it.

If you are going to submit something that has images/math notation, let me know before you go through the trouble of making the images so we can let you know about colors/etc to use so your article doesn't come out funky when we put it up on the site.

With regards to the competition, I'd be happy to try and put something together.  I would need people to submit problem suggestions, of course.  Then I could make up smallish contests of a harder/more advanced and easier/less advanced variety.   Of course, this is limited by number of participants and by the amount of free time I have, but I expect neither of those would be a major limitation.  How does that sound?

Well I think I can do that on the AMC forum, have a competition with several problems.  Nice idea.  :-)

I think part of the principle that people are suggesting is that it not be on a forum, so that people's solutions will not be posted immediately and thus they can all be graded seperately.

This competition idea is great. Everyone can finally get a chance to try out questions and answer, no matter when they go online. But a few questions. Where would this be run? And furthermore, who would make/grade the solutions? What timeframe will it be on?

I've already suggested myself as an administrator, at least for the first one.  I would write some questions, and I would also request that every competitor send in a question or two (either of their own design or stolen from somewhere else).  I would then assemble 10 or more questions across a variety of levels, post it, and have everyone e-mail me responses over the next 24 hours, say.  I would be happy to grade that and what-not (if everyone doesn't mind waiting a couple of days for their scores).  Also, immediately after it was over you would all be able to talk about it, so it would be good.

that sounds really cool. when can you start it?

Sorry, I never saw the last two messages.  Okay, everyone who's interested in participating:  my e-mail address is
fuzzyjoel  at  yahoo  dot  com.
With the obvious transformations needed to convert it to an actual e-mail address.  Please e-mail me a question or two -- I will do my best to put a competition together and post it before Thanksgiving, but no promises -- it may have to wait until next week.

[quote="JBL"]Please e-mail me a question or two.[/quote]

hrm you still haven't said how hard these questions should be. and also, all my questions suck (i had to write PoWs last year for mathc clats and none were any more inventive than the problems in the book).

Any level is acceptable -- I'm going to try to make the contest cover as wide a span as possible.  I was thinking that I could also weight it, by question difficulty and by age/experience of various participants.  What do you think?

If having a variety of levels and weigthed for age isn't too much work for you, I think that would be great.

i dont think they have to be original questions, just take them out of some place that others probablly will not have seen.  Like problems from the AOPS books might not be a great idea, since most people have the books.

I'll be fine posting it here, it will just take a little work.  I believe I posted more information in Getting Started, so you might want to read what I wrote there.  Please try to send me some questions on or before Wednesday -- they don't have to be hard, they should ideally be mediumish for whatever level you are at.  But no matter what, send me some questions.  (Topper already has.  I'm waiting for the rest of you.)

For those interested, information is now up on the Intermediate forum.  The problems will be posted there, as well.


I want to collect some AMC past tests.. Especially AMC10..
I know tests were known as AHSME so, if you have any past ones that is not up on the Harold Reiter's page, please PM me with those.. I'm collecting them really eagerly..

The old AHSME just split into AMC-10 and AMC-12 four (maybe five) years ago, so there aren't very many to collect. You may already have all of them.

I suggest you buy from http://www.unl.edu/amc, since they really don't cost very much, and they're worded a lot better than Kalva's and other sites.  I think you can pretty much buy everything for about $50 (less if you find good deals).  Of course, I doubt you will need every single test and solution packet.

[quote="zscool"]you'd be surprised, I went to a non-magnet public high school and the person who organized the AMC tests had every AHSME and AIME published.[/quote]

Of course I was thinking of the many schools where there isn't anyone organizing the AMC tests.

[quote="Dnas"]Well, the school I went to to take the AMC tests doesn't seem to care a whole lot about math, but the math person still had tons of materials for me.[/quote]

That's how our local high school is...no one cares bout it...but the math department is stocked with these practices. (mostly because the state gives the school district more than they could spend and the school district gives that extra money to the schools and then the schools give it to the clubs/organizations with the most participants...Mu Alpha Theta had well over 200...so that's how the math department ended up with so many AMCs...)

Our school has tons of AMC, AIME, and USAMO practice materials.  I got about 15  past AIME examines from my school.  I also know a website that contains a lot of past AIME and USAMOs.

i think you are talking about kalva. my school has very few materials at all, altough we are/were arguably the best school for math in the state.

I don't know about you but my school's Mu Alpha Theta is very different from the math team.  Mu alpha theta has ppl who have high grades in math but aren't necessarily good problem solvers(and some of the ppl with high grades have slipped this year!).  So I think it is a little easier to get more ppl on mu alpha theta because it is less math intensive than math team(at least at my school).  Still 200 ppl is excessive.  Mine has about 20-30.  I think I'll join it next year to get some service hours or something.

[quote="tokenadult"][quote="Dnas"]First try talking to your school's math person. They probably have a lot of materials that they'll give you.[/quote]

Actually, I would expect that to be a very exceptional situation. Where have you observed schools that are well stocked with AMC practice materials?[/quote]

AAST :)

[quote="mathnovice"]How does your school get 200 people in Mu Alpha Theta? I mean, it would be nearly impossible for me to get 20 people in my school's math team, nevermind 200! What attracts people to Mu Alpha Theta? Is your school a math/ science school? I know my school's NHS has a lot of members because they send a letter to a lot of students about how they have been invited into NHS and then their parents force them to join. Is that the case with Mu Alpha Theta? 200 people = slightly less than my entire junior class.[/quote]

First off, there's about 650 students per class at the high school I was referring to. (It's not my school b/c I'm homeschooled...but the local high school) So we get more than 2500 kids...

And as joml88 said, Mu Alpha Theta  :!=: math team. The Mu Alpha Theta membership is mainly all the high-scoring students in regular math classes who registered for MAO to be able to put it on their transcripts and attend the minimal number of required competitions just to say they're a part of it. (There's also about 150 kids that take AMC each year from this school for the same reason...the high-scoring students from each math class are pulled out and asked to take it.) So yeah, it's actually not too hard to get well over 200 (almost 250 i think) kids this way. And also, they're given extra credit points and free homework passes and stuff from the math department when they go to competitions.

MAO also does it's own fundraisers and community service projects, so members can also get service hours.

The actual math team that actually places in some competitions and consists of around 40 kids...and then only about 2/3 of the actual math team qualify for AIME each year...so yeah, it's not an extremely strong math school...but they're pretty well-funded.

mu alpha theta is more of a social club. it puts a bigger emphasis on knowledge than problem solving, but this year they're also hosting a mathematical olympiad and proof test. there's a lot of stuff for not as math-oriented to do, like some artistic competitions. that's the main reason why it attracts more people, but most 'good' people do it just for the hell of it, because, well, why not. and not all tests are easy, but it really sucks when they are. it's a good way to learn 1 or 2 years worth of 'standard' math curriculum in less than a month.

Yeah, but in my state (FL) MAO is almost like a cult...I like going to the competitions for the social value (and to beat people who spend hours studying for it :twisted: ), but the problems are very standard usually...even on the "hard" tests the difficulty is because of massive calculator/dumb algebra work.  If they spent time studying for the AMC or something, something might come out of it, but now they're just wasting all their time.  Oh well... it's fun sometimes to rest the brain and do some quick 60-minute, 30-question tests.

At least MAO sponsors some things from the money it takes for registration (180+ dollars a person for the state convention!).

buy the tests. they arent that expensive, and its always good to have for future reference

as for MAO

i also live in FL, and i'll have to agree with what everyone else says.. Mu Alpha Theta very rarely tests promblem solving abilities.. much of it is suff you'd find as excercises in a text book and applications of tons of formulas. Questions are also often very calculator intensive. (with an exception to nat'ls)

but there arent any other competitions in FL, so why not? it's fun, as long as you know that Mu Alpha Theta competition isn't the ultimatum of mathematics competitions.

[quote="Silverfalcon"]Can anyone like P.M. with past AHSME questions?
Solutions aren't really necessary.. I'll try to solve them and P.M. back with answers...At least, I need to see some problems and solve it... My school doesn't have those at all, i guess..[/quote]

Hardly possible with the notations used (although I'm not sure if TeX is enabled on PMs or not) and the geometry.

As mentioned, the problems you are looking for are extremely inexpensive. I don't believe AMC charges much more than how much it costs to print them, a small fee, and shipping.

[quote="Silverfalcon"]Can anyone like P.M. with past AHSME questions?[/quote]

I don't send out copies of copyrighted materials by private message or by email, or by any other means not authorized by law.

[quote="Silverfalcon"]My school doesn't have those at all, i guess..[/quote]

I figured that was your problem, because you seem to be very unfamiliar with the AMC tests in general. The top-level initial test is now called the AMC 12, and it has had that name for years. The books Contest Problem Book IV, Contest Problem Book V, and Contest Problem Book VI have old math competition questions and solutions, and you can request those by interlibrary loan. Contest Problem Book V is especially good and useful, and lots of libraries have it, so it shouldn't be hard to request. (Your local library ought to buy it, but if they don't they can get it from another library.) I finally bought Contest Problem Book V and Contest Problem Book VI when my son qualified for AIME because I figured I ought to have those books at hand any time we want to practice.

Hope this helps!

[quote="Silverfalcon"]I've found several past tests from 1997 - 1999. But my problem was, that it was like in pieces, not as whole. Why's that?[/quote]

Uhh... where did you find them?

Hmm... i found AHSMEs from 1980-2001 w/ solutions, but it looks kinda blurry...

the link: [url=http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm]here[/url]

Thanks!

Also, it's kinda pieces but tests from 1997-1999. Numbers are weird, though...

http://search.yahoo.com/search?p=american+high+school+mathematics+examination+&ei=UTF-8&fr=fp-tab-web-t&vst=0&vs=www.as.ysu.edu

By copyright law, most likely not. On the other hand, a lot of professors have put up these questions - some who work/worked for the AMC. Theres not much you can tell them.

I'm pretty sure AMC let's people post slightly modified versions of their problems (kalva's a pretty good place).  However, as was said before, you might as well buy the tests for yourself.

[quote="mathfanatic"][quote="probability1.01"]I'm pretty sure AMC let's people post slightly modified versions of their problems (kalva's a pretty good place).[/quote]

Really?  I was under the impression that they don't want anyone to post anything, modified or not, but they don't actively prevent anyone from doing so.  Everything that Kalva says about preventing problems with copyright law by paraphrasing the questions is essentially untrue; changing the wording of a copyrighted work doesn't make posting it on the internet not stealing.  (Suppose you took Harry Potter, changed the words around a little, and posted it on the internet.  If J.K. Rowling sues you, she has a case.)[/quote]

Math problems themselves are not considered copyrightable works.  On their phrasing can be considered such.  Also, fair use seems to trump all but the most egregious offenses.

The powers did not see fit to include the mathematician as the recipient of significant monopoly powers. (And while I approve of appreciate that fact, I'm not sure I see the distinction.)

One question:

Does Kalva provide AHSME too?


Is there a sequence $\{ s_{n}\}$ with $s_{n}> 0$ for all $n$, $\limsup s_{n}= \infty$ and $\lim \sigma_{n}= 0$ where $\sigma_{n}$ is the arithmetic mean $\frac{s_{0}+\ldots+s_{n}}{n+1}$?

Something like $s_{2^{k}}=k$ and $s_{n}=1/n$ otherwise.


trigonometry
analytic geometry
limit
integration
real analysis
real analysis solved

Consider the solid [b]K[/b] of revolution generated by rotating the curve $y=\sin x\ (0\leqq x \leqq \pi)$ around the $x$-axis.
[b]K[/b] is divided into the number of $2n$ parts by the number of $2n-1$ planes ,which is perpendicular to $x$-axis,
they have all equal volume. Among the $x$-coordinate of intersection points for these plane to intersect with $x$-axis,

let $a_n$ be the $x$-coordinate of intersection point  which is less than $\frac{\pi}{2}$ and the closest  $\frac{\pi}{2}$.

Evaluate

\[\lim_{n\to\infty} n\left(\frac{\pi}{2}-a_n\right) \]

Could anyone translate this sentence in plain English. :(
Regards!

kunny

The volume of [b]K[/b] is

$\int_0^{\pi}\pi \sin^2x\,dx=\frac{\pi^2}2.$

We choose $a_n<\frac{\pi}2$ such that

$\int_{a_n}^{\frac{\pi}2}\pi\sin^2x\,dx=\frac{\pi^2}{4n}.$

Since the value of $\pi\sin^2x=\pi$ at $\frac{\pi}2$, as $a_n\to\frac{\pi}2$ we have

$\int_{a_n}^{\frac{\pi}2}\pi\sin^2x\,dx=\pi\left(\frac{\pi}2-a_n\right) + o\left(\frac{\pi}2-a_n\right).$

Hence $\left(\frac{\pi}2-a_n\right)=\frac{\pi}{4n}+o\left(\frac{\pi}2-a_n\right)$ or

$n\left(\frac{\pi}2-a_n\right)=\frac{\pi}4+o(1).$

The limit is $\frac{\pi}4.$

Thank you for your reply,Kent Merryfield. :)


LaTeX
number theory unsolved
number theory

Let $\, k_1 < k_2 < k_3 < \cdots \,$ be positive integers, no two consecutive, and let $\, s_m = k_1 + k_2 + \cdots + k_m \,$ for $\, m = 1,2,3, \ldots \; \;$.  Prove that, for each positive integer $\, n, \,$ the interval $\, [s_n, s_{n+1}) \,$ contains at least one perfect square.

We notice that $s_n = k_1 + k_2 + \ldots + k_n \geq 1 + 3 + \ldots + (2n-1) = n^2$. If $s_n = m^2$ for some $m \in \mathbb{Z}^+$, then we are done. If not, then letting $r$ be the largest non-negative integer such that ${(n+r)}^2 < s_n$, we note that the next perfect square ${(n+r+1)}^2$, which is $2n+2r+1$ greater than ${(n+r)}^2$. We now claim that this difference $2n+2r+1$ is $\leq k_{n+1}$. To see this, we suppose not, then we have $2n+2r+1 > k_{n+1} \Rightarrow 2n+2r \geq k_{n+1} \Rightarrow 2n+2r-2 \geq k_n$
$\Rightarrow k_n+k_{n-1} + \ldots + k_1 \leq (2n+2r-2) + (2n+2r-4) + \ldots + 2r = n^2 + 2nr -n$, which is less than $n^2 + 2nr + r^2 = {(n+r)}^2$. (Contradiction, since we have assumed that $s_n = k_1 + k_2 + \ldots + k_n > {(n+r)}^2$.)

Thus ${(n+r+1)}^2 \in [s_n, s_{n+1})$ and we are done.

$n(x_{n+1}-1)>=s_n+n^2>=2n\sqrt{s_n}$ so $x_{n+1}>=2\sqrt{s_n}+1$ and $s_{n+1}>=(\sqrt{s_n}+1)^2$ so we get$ \sqrt{s_{n+1}}-\sqrt{s_n}>=1$ so the interval contains a square.

[quote="Makaveli"][quote="dblues"]Erm... I believe you mean the interval $[sn, s(n+1) )$, or rather $[s_n, s_{n+1})$.[/quote]
The second one[/quote]
Well... if i've already solved the problem, it just means i definitely know it's the second one.  ;) You originally wrote $[sn, sn+1)$ without latex, which is not what you mean. I'm just trying to say that even without latex, u should write $[sn, s(n+1) )$ instead, so that we can interpret it as $[s_n, s_{n+1})$. Best thing is to go latex of course.  :P

Can't we prove this using the argument that perfect squares have differences of consecutive odds, and the interval with width $ s_m$ are larger than these odd numbers (because the base case of the sequence is 1,3,5,7,..., and every other sequence either has the same interval width or a larger one), i.e. proving the base case and then proving that it would work for any case greater than the base case?

[quote="gauss202"]that is essentially what dblues did.[/quote]

thanks, i read through his proof once more, and i see it now. :)

Dear Friends

I solved it by Apriority( if I used the right word!)

1.it's also clear that between $K$ and $K'$ which $K\leq K+2$ there is a square.by using what dear dgreenb801 said.

2.we assumed that for integer $m$, between $s_{m}$ and $s_{m+1}$ there is atleast one square.

3.we'll prove that between $s_{m+1}$ and $s_{m+2}$ there is atleast one square.

4.and it's Done.

also the third step proved by what dear dgreenb801 said.

With Regards.

[quote=dgreenb801]Note that $ s_{n \plus{} 1} \minus{} s_n \equal{} k_{n \plus{} 1} \ge k_{n} \plus{} 2$. Also, $ s_n \le k_n \plus{} (k_n \minus{} 2) \plus{} (k_n \minus{} 4) \plus{} ......$($ 2$ or $ 1$)$ \equal{} (\frac {k_n \plus{} 1}{2})^2$ if $ k_n$ is odd and $ \frac {k_n^2 \plus{} 2k_n}{4}$ if $ k_n$ is even. In either case,
$ s_n \le (\frac {k_n \plus{} 1}{2})^2$. But then the greatest possible difference between $ s_n$ and the next square number after $ s_n$ must be $ (\frac {k_n \plus{} 3}{2})^2 \minus{} (\frac {k_n \plus{} 1}{2})^2 \equal{} k_n \plus{} 2$, and since the difference between $ s_{n \plus{} 1}$ and $ s_n$ is at least this, there must be a square on the interval $ [s_n,s_{n \plus{} 1})$.[/quote]
If $s_{n+1}\le (\frac {k_n + 1}{2})^2$ then your conclusion is incorrect.

Consider a particular $n$. Say that $k^2< s_n\le (k+1)^2$ for some integer $k$. Suppose for contradiction that $[s_n,s_{n+1})$ does not contain at least one perfect square, then $s_{n+1}\le (k+1)^2$, so $s_{n+1}-s_n=k_{n+1}\le 2k$. This implies that $s_n$ is at most
\[(2k-2)+(2k-4)+\cdots+2 = 2\cdot \frac{k(k-1)}{2} = k(k-1),\]
contradiction.


geometry
circumcircle
trapezoid
geometry unsolved

In triangle ABC, the bisector of angle B intersect the circumcircle at D and AC at F so that 2AD^2 = AB^2 + BC^2.The midpoint of AC is F. Prove that DF = EF.

Hello
Where from this $ E$ came?I believe u have used $ F$ two times.
Thank u.

Correction, E is the point of intersection of AC and the bisector of angle B and F is the midpoint of AC

"In triangle ABC, the bisector of angle B intersect the circumcircle at D and AC at E so that 2AD^2 = AB^2 + BC^2.The midpoint of AC is F. Prove that DF = EF." is still wrong.

Correct version: In triangle $ \triangle ABC,$ the bisector of angle $ \angle B$ intersect the circumcircle at $ D$ and $ AC$ at $ E$ so that $ 2BD^2 \equal{} AB^2 \plus{} BC^2.$ The midpoint of $ AC$ is $ F.$ Prove that $ DF \equal{} EF.$

Parallel to $ AC$ through $ B$ cuts the circumcircle at $ B',$ $ ABB'C$ is isosceles trapezoid, $ B'C \equal{} AB$ and $ BD \equal{} B'D.$  Let $ M$ be midpoint of $ BB'$ and $ DM, CM$ C-, D-medians of $ \triangle BCB'$ and isosceles $ \triangle BDB'.$ $ B'C^2 \plus{} BC^2 \equal{} AB^2 \plus{} BC^2 \equal{} 2BD^2 \equal{} B'D^2 \plus{} BD^2$ $ \Longrightarrow$ $ CM \equal{} DM,$ which in turn implies that $ M$ is common circumcenter of $ \triangle BCB', \triangle BDB', \triangle ABC.$ $ BB'$ is a circumcircle diameter, and the angles $ \angle B'CB \equal{} \angle B'DB$ are right. $ DF \perp (AC \parallel BB')$ bisects the right angle $ \angle B'DB$ $ \Longrightarrow$ $ \angle FDE \equal{} \angle FDB \equal{} 45^\circ.$ Then the right $ \triangle DEF$ is isosceles with $ DF \equal{} EF.$

We use complex numbers. Let $a=x^2,b=y^2,c=z^2,d=-xy$. We claim that $y^4+x^2z^2=0$.
We are given that \begin{align*}
\vert y^2-x^2\vert ^2+\vert y^2-z^2\vert ^2=2\vert y^2+xy\vert ^2\Longleftrightarrow\\
\frac{x^2}{y^2}+\frac{z^2}{y^2}+\frac{y^2}{x^2}+\frac{y^2}{z^2}=\frac{-2y^2}{xz}+\frac{-2xz}{y^2}\Longleftrightarrow\\
(y^4+x^2z^2)(x+z)^2=0.
\end{align*} Note that if $x+z=0$, then $\angle BAC=90^\circ\implies \sqrt{2}BD=2R$, which implies that $B$ is the midpoint of $ND$, where $N$ is the midpoint of arc $BAC$. This implies that $B\equiv A$ or $B\equiv C$, impossible.
Hence, we have $y^4+x^2z^2=0$.

Let $B'$ be the reflection of $B$ over $O$, where $O$ is the circumcenter of $\triangle ABC$.
From the complex numbers, $BB'\parallel AC$. This means that $$\measuredangle CAB=\measuredangle B'CA=90^\circ+\measuredangle BCA\implies \measuredangle NCB=90^\circ-\frac{\measuredangle ABC}{2}-\measuredangle BCA=45^\circ,$$
and as $\measuredangle EFD=90^\circ$, we get that $FE=FD$. We are done.


calculus
integration
logarithms
calculus computations

Evaluate $ \int_0^{\ln 2} \frac {2e^x \plus{} 1}{e^{3x} \plus{} 2e^{2x} \plus{} e^{x} \minus{} e^{ \minus{} x}}\ dx.$

Set $ u\equal{}e^{2x}\plus{}e^{x}$

$ \int_{0}^{\ln 2}\frac{2e^{x}\plus{}1}{e^{3x}\plus{}2e^{2x}\plus{}e^{x}\minus{}e^{\minus{}x}}dx\equal{}\frac{1}{2}\allowbreak \int_{2}^{6}\,\left( \frac{1}{u\minus{}1}\minus{}\frac{1}{u\plus{}1}\right) du\allowbreak \equal{}\allowbreak \allowbreak \ln \sqrt{\frac{15}{7}}$

[quote="Stephen"]

Kirill, if we let $ t \equal{} e^x$, $ e^{ \minus{} x} \equal{} \frac {1}{t}$.[/quote]

Yes, and $ e^{3x} \equal{} t^3$. He multiplied top and bottom by $ e^x$ before the substitution or by $ t$ after to get the second integrand (note that $ e^x$ was taken out of the numerator in order to change to $ dt$)

Those are correct. :)

Compute $ \frac {e^{x} \plus{} e^{ \minus{} x}}{e^x \minus{} e^{ \minus{} x} \plus{} 1} \minus{} \frac {e^{x} \minus{} e^{ \minus{} x}}{e^{x} \plus{} e^{ \minus{} x} \plus{} 1}$

Nice, kunny!

I get

$ \frac{4\plus{}2e^{\minus{}x}}{e^{2x}\plus{}2e^{x}\plus{}1\minus{}e^{\minus{}2x}}$

$ \equal{}\frac{2(2e^{x}\plus{}1)}{e^{3x}\plus{}2e^{2x}\plus{}e^{x}\minus{}e^{\minus{}x}}$

$ \Rightarrow\, I \equal{} \frac{1}{2}\ln\frac{e^{x}\minus{}e^{\minus{}x}\plus{}1}{e^{x}\plus{}e^{\minus{}x}\plus{}1}$

Putting in the limits, I get $ \frac{1}{2}\ln\frac{5}{7}$


LaTeX
calculus
search
abstract algebra
limit
integration
geometry
\/closed

please put here your deepest wishes concerning this forum. maybe they will come true (I hope so, but I can't make any promisses :D)

i think that it would be nice to have a mathllinks's messenger only for mathlinks members.
---------------------------
kuba

that would be indeed a very nice wish - there is a way ... :) I will be thinking about it. wishes people, c'mon :)

How about spoilers? I mean buttons that you can push to make your writing invisible so you don't spoil the problem for ne1 else? (everyone can see what you wrote by highlighting the writing). I saw this on another forum.

[color=white] Grobber you can make spoliers here too! [/color]

when posting select the post you don't want to spoil, and then set the font colour to white.

Oh... I've said it b4: it's amazing how you can sometimes miss the most basic ideas! :D  :D  :D

The Downloads section. Please update it, especially the Romanian problems section. Tnx

Hi Valentin,

Why have you not maintained some sections like 'misc problems'?
There is no more olympiad/level section on functional equations or analysis (like sequences). Are these kind of problems supposed to be in the college sections?

Pierre.

in my point of view functional equations belong to algebra.

if they make use of continuity, they will go to college playground at the calculus section.

as for the sequences, if they as you to compute limits, etc. that is analisys - calculus- so still there.

[quote="grobber"]How about spoilers? [/quote]

I don't like spoiler !

If you don't want to look at the solution, just don't !
And spoiler often makes reading difficult (if not impossible if there are a lot of maths symbols which are not spoiled).

that is why we DON'T have spoilers (sorry grobber :) ). however I am trying to make a search for the first post only, but I'm a little stuck right now, so bear with me :D

It was just a suggestion, so it's not like I'm upset or anything! :D:D

Valentin,

please look here

http://www.mathlinks.ro/viewtopic.php?t=3551

This isn't really a wish I have, but maybe it would be useful to have a forum for new members to introduce themselves and to also ask for advice. Maybe there's already a forum just like this that I've just been missing. Just a thought to make new members more comfortable.

[quote="yif man12"]This isn't really a wish I have, but maybe it would be useful to have a forum for new members to introduce themselves and to also ask for advice.[/quote]

Asking for advice is pretty much always on-topic in Other Problem Solving Topics. Introducing ourselves can happen on Games & Fun Factory. I'll open [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15747]a thread there[/url] right now.

[quote="JSRosen3"]It'd be nice if there were a way to get rid of offensive/inappropriate posts without waiting for the moderator of the forum to find it first.[/quote]

I will make a concrete suggestion to the site owners: maybe EVERY moderator, of any forum, ought to have delete privileges for Games & Fun Factory. That would be a fast way to make that forum more civilized. Most of the limit-testing on the site occurs on that forum.

What is wrong exactly then? :? I just checked most of the topics and I hardly saw anything wrong there...

Yeah if we do that half of this site would turn into an adult site.  You can't just have part of a site like that.  It will either soon spread or it will die off.  If you want stuff like that there are thousands of sites out there for you... this site is one-of-a-kind in my opinion.  There isn't any other site like it, so we shouldn't corrupt it.

I think it would be interesting if this site had its own RSS feeds, where there was a problem that updated every few hours. For those who don't know, an RSS feed is a stream of information to a computer which can be viewed using a program (which can be easily set up on your desktop). You can use RSS feeds to get the latest headlines, sport statistics, etc, on your desktop. It also updates at regular intervals. I was thinking that this site could have a separate RSS feed for each of its difficulty levels, and then feed a problem of appropriate difficulty every so often.

I just thought it would be a nice addition to this site.

Hmm... does that exist yet?... or are you gonna make it for us :D

[quote="yif man12"]I think I speak for many people when I suggest an Ultimate Frisbee Forum.  :D[/quote]

No even better, an online game :P

like pingu used to be for the mathlinks, we can have Ultimate Frisbee Online for the merged one :P

And it can be shortened UFO if that helps ;)

[quote="JSRosen3"]It'd be nice if there were a way to get rid of offensive/inappropriate posts without waiting for the moderator of the forum to find it first. Case in point: a thread in the Fun and Games forum about links.[/quote]You can PM any of us (the admins) and/or any of the moderators of the respective forum. This should work.

With only 1 exception, in which the user didn't knew that he was accessing an xxx subject, and that is not welcomed, I have not  seen such offensive posts that you are talking about on the old MathLinks board, and I have yet to hear about such a post on the new board since the merge.

[quote="Valentin Vornicu"][quote="RC-7th"]Not that it matters much, but could you encrypt the connection?[/quote]What use would that do to anyone, besides screwing up some already fragile settings?

I am sorry, but if you are worried that someone is going to intercept someone else's solution to a problem, I think that you have nothing to be afraid about.[/quote]

Not that, but it does get annoying if your're not allowed to turn off numerous firewalls which keep bugging you. :?

You should set your firewall to accept stuff coming from this server only. This should solve your problem.


Hello, I neef an explanation for combinatorical formula of Circle Division by Chords.

http://mathworld.wolfram.com/CircleDivisionbyChords.html

Thank you all!

There are a total of n [b]circular[/b] regions between the circle and the n-gon

The inscribed n-gon is convex, of course
[url=http://mathworld.wolfram.com/PolygonDiagonal.html]Here[/url] it says that [i]The number of regions which the diagonals of a convex polygon divide its center if no three are concurrent in its interior is $\binom{n}{4}+\binom{n-1}{2}$[/i]

So, we have to calculate $g(n) = \binom{n}{4}+\binom{n-1}{2}+n$

Setting $m=n-1$ and $k=2$ in the idendity $\binom{m}{k}+\binom{m}{k-1}=\binom{m+1}{k}$ we get

$\binom{n-1}{2}+\binom{n-1}{1}=\binom{n}{2}\Rightarrow $

$\binom{n-1}{2}+n-1=\binom{n}{2}\Rightarrow $

$\binom{n-1}{2}+n=\binom{n}{2}+1 \Rightarrow $

$g(n) = \binom{n}{4}+\binom{n}{2}+1$


Let $a$ be a positive integer. Define a sequence $x_n,n\ge 0$ by:

$x_0=1;x_1=a; x_{n+1}=x_n+x_{n-1}$.

Find all $a$ such that there exists a positive integer $m$ satisfies the following condition:

forall $n=0;1;2;...;$ then  $gcd(m;x_n)=1$.

Proceed as usual to obtain explicit formula for $x_n$. After some computations it's not hard to get:

$x_n=\frac{2a+\sqrt{5}-1}{2\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n+(1-\frac{2a+\sqrt{5}-1}{2\sqrt{5}})\left(\frac{1-\sqrt{5}}{2}\right)^n$

Take some $a$ and choose prime $p$ such that $p \not =2,3$ and $a\equiv 1 (p)$ (which is possible if $a\not = 6^m-1$ for some $m$). Suppose there exists $n$ such that $(x_n,p)=p$ so also we must have $x_n=0$ in a field $\mathbb{F} _p (\sqrt{5})$. It rewrites as (everything below is in $\mathbb{F} _p (\sqrt{5})$) :

$\frac{2a+\sqrt{5}-1}{2\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n+(1-\frac{2a+\sqrt{5}-1}{2\sqrt{5}})\left(\frac{1-\sqrt{5}}{2}\right)^n=0$

multiplying by $2\sqrt{5}$ and  grouping terms we get:

$(1-2a)(\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n)=\sqrt{5}(\left(\frac{1+\sqrt{5}}{2}\right)^n+\left(\frac{1-\sqrt{5}}{2}\right)^n)$

and using binomials expansion:

$(1-2a)(\sum_{i=0} \binom{n}{2i+1} 2\sqrt{5} 5^i)=\sum_{i=0} \binom{n}{2i} 2\sqrt{5} 5^i$

dividing by $2\sqrt{5}$ (so operating on 2\sqrt{5} is unnecessary after all  :P ) and adding $\sum_{i=0} \binom{n}{2i+1} 5^i$ to both sides we get:

$2(1-a)(\sum_{i=0} \binom{n}{2i+1} 5^i)=6^n$

which is a contradiction because left side is equal to zero (due to $a\equiv 1 (p)$) and right side is not.

So we are left only with a case $a=6^m-1$... (I'll do it later....or not  ;) )


Given triangle $ ABC$. Let $ AB\equal{}c,BC\equal{}a,CA\equal{}b$ and $ m_a,m_b,m_c$ be the medians of $ \Delta ABC$.
Suppose that $ a\ge b\ge c$. Prove that:
 $ m_a\plus{}m_b\plus{}m_c\equal{}\frac{\sqrt3}{2}(a\plus{}b\plus{}c) \Leftrightarrow m_b\equal{}\frac{\sqrt3}{2}b$

Yeah, in general, $ m_a,m_b,m_c$ are the medians from vertices $ A,B,C$ to $ \overline{BC},\overline{AC},\overline{AB}$, respectively.

Nobody solves this problem ?:|
I think it's a nice problem


search
counting
derangement
MATHCOUNTS
complementary counting

A special number is made from the set 1~9 so no number takes up its 'usual' position. (ex. 1 is in any place other the the first position, 2 is in any place other than the second position....all the way to 9.)

For instance, 514327698 is an example of a disarranged number.

How many disarranged numbers are possible from the set 1~9?

[hide="Notsure..."]

Would it be $ 8!$. Because each number be in 8, 7, 6... places?
Giving an answer of 40320...

Answer: 40320
[/hide]

Search "derangements." And no, the answer is not 40320.

What if you use 2 for number 1, so for the "real" 2, there are still 1,3,4,5,6,7,8,9, so 8 numbers?
That's 8 for the first two, and what if you use 3 for number 2 and so on? I'm no sure if it will make a difference, but I don't know..
What if you use comlimentary counting?

No complemetary counting doesnt work.  The only way is using the derangement formula.  The reason complementary counting doesnt work is because order matters in choosing the spots for the number. In other words, this problem is not Mathcounts and it does not require a unique solution.

[quote="mihail911"][hide="would it be?"] $ 8^{9}$ because there's 8 possible numbers for each of the 9 places?!?!?! [/hide][/quote]

Uh No, like I said dont attempt this problem unless you know the derangement formula.  Theres no other way of doing it. (unless you want to use casework of 100+diffeernt cases)

[quote="mihail911"]ok... :|
what's the formula then?[/quote]

Do you know what $ e$ is?

One formula uses recursion,the other $ e$.

Either way you'll need a caculator.

No point in posint any more messages..

[quote="mihail911"]of course i know what e is...the mathematical one. it's euler's universal constant. approximately like 2.18..or something like that. is you need a calculator for derangements than how are you supposed to solve those types of problems???[/quote]

uh tahts was why I said the problem is kinda bad for mathcounts.....theres another way with recursions to find the 9th derangment butit also gets really big numbers

Isn't the formula the nearest number to $ \frac{n!}{e}$?

Oops :oops:  I learned about derangements a long time ago, and I guess I forgot.
The derangment formula is probably what vamathletes said, or the greatest integer less than or equal to $ \frac{n!+1}{e}$

[quote="13375P34K43V312"]Those are two different expressions. vamathlete's expression is correct, it is the value of $ {n!}{e}$ rounded to the nearest integer, either up or down. Rounding and flooring are two different things; make sure you see why.

A less calculator intensive formula, which is equivalent, is $ n!\sum_{k=0}^{n}\frac{(-1)^{k}}{k!}$, and the proof that the two expressions are equivalent involves the fact that $ \sum_{k=0}^{\infty}\frac{1}{k!}=e$.[/quote]

More like the general formula, $ \sum_{k=0}^{\infty}\frac{x^{n}}{k!}=e^{x}$.

[quote="xpmath"][quote="abacadaea"][hide]wouldnt it be 8! because if we start with the number 1 it has 8 positions, 2 has 7 postions and so on.[/hide][/quote]
No offense, but didn't you read all the posts above?[/quote]

he obviously didnt i think he just wants to get attention.

[quote="newclassic"]well, you have the numbers

123456789

and you want combinations where none of the numbers are in their original place. These are permutations without repitiion, right? Or maybe that's not what it's called. But the way I figure it is:

bahaha. This is the basic counting principle..

_ _ _ _ _ _ _ _ _
For each disarranged number, you have 9 spots where you could put a number.
In the first spot, you can put ANY number EXCEPT 1. That's 8 possibilities for that spot. For the second spot, it's the same thing. 8 possibilities.
8 * 8 * 8 * 8 * 8 * 8 * 8 * 8 * 8
Because there are 8 possible numbers that can go in each spot.
8^9=134 217 728

So 134 217 728 possible numbers.

Whoo-freaking-hoo. That was so difficult.[/quote]

Dude no.  thats not right.  ONce we use a digit, do we still have 8 possible for the next one?  Nope. And 8! isnt right because that doesnt take into consideration where and when you place the numbers.

I agree that this derangement problem with 9 numbers is way too difficult for MATHCOUNTS.  Many of the posts above are wrong in confusing ways that are hard for you to check because the numbers are just too large to verify.

However, MATHCOUNTS students can get some useful insights out of exploring the case of derangements of smaller numbers.  E.g., how many derangements of 1, 2, 3, or 4 objects?

You can use simple reasoning to work directly through these small number cases and get some sort of intuitive feel for derangements.

You can then check your results from these simple cases against the results you'd get if you used the same rule you previously applied to the 9-object case.  (E.g., by considering the number of derangements for small values of $ n$, you should be able to convince yourself that the rules suggested above of $ (n-1)!$ or $ 2^{n}$ must be incorrect.

This technique of checking whether your method works on simpler problems is a valuable tool for doing mathematics in general.

After you've worked on this problem for small numbers of objects, you might want to look at a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=158809]recent discussion in the high school basics forum on the derangements of four objects.[/url]


Who got in?  My best friend applied and got his acceptance on friday !

Haha congrats Jeff. You're probably the only one from our school

Congrats to all of you guys.  I'm applying through Regular Action.  How many of you guys are going to accept Stanford's offer and be a Cardinal?

Congratulations. I was going to apply, but decided not to (i'm content looking between the other schools i got into)

FYI, no one at Stanford says "I'm a Cardinal".  It'd more likely be "I'm Cardinal", but no one says that, either.

Hope to see you all on the Farm!

Anyway, just making a ploy to get yall up there. Also i know slysi wont like this but apply to frosoco-it has two room double, huge one room doubles and arguably some of the best food on campus. Plus you have me...joking. Anyway, you guys will have fun up here (take the 50H series...reminder take at own risk).
See yall up here at admit weekend
J.D.