MATHCOUNTS
probability
geometry
number theory

Hi, my son is entering 6th grade and he's considering joining his schools math team (they are enrolled in mathcounts). I just started doing some research on this and found this website. I'm considering to buy a mathcounts handbook for him.

A few questions about mathcounts handbooks:
[1] It says that the handbooks will be distributed to middle schools - so do students get the books from schools when they join the math team? I don't want to be in a situation where I bought the book and later found out that the school gave him the same book.
[2] There are 05-06 and 06-07 school handbooks, each contains about 300 problems and solutions - are 06-07 problems largely the same problems with minor changes or they are completely different questions?

I just registered and this is my first post, if this should be posted to another forum, please kindly let me know.

Thanks,

- glacier

I don't think you get the handbooks from your school - I think they mean each coach gets one handbook for himself. Every school is different, but I doubt the teacher would buy that many copies of the handbook when he doesn't even know how many students are going to join. I'd recommend buying the handbook.

The handbook is posted on the internet, for free download. The problems are different every year.

Yes, each coach gets one, but it is better for a student to work at his or her own pace by him or herself at the problems.

The main difference between the online handbook and the paper handbook is that the paper one has solutions (not just answers) to all the problems.

Welcome to the forum, glacier!

Although the students don't get the handbook, the coach should be providing lots of the problems and other information to the students.  I would suggest you talk to the coach, but I would think that it would better to buy other math material to complement these problems.

I suggest you purchase the various Art of Problem Solving books.  Some of the topics may be hard initially, but your son can skip sections until he's ready for the various topics.  These book not only help in Mathcounts, but for all the other competitions.

I also suggest you look into finding problems from other competitions.  The best help is to work on many and different problems, so I think it helps if you give him different problems from the coach.

The handbooks are very different. You can download them and post questions that are problematic here. We have some very capable problems here.

Wow, I really like this new community that I just joined, the comments and suggestions are very helpful.

I wasn't aware that the handbook is downloadable. Now that I knew it, I'd put my money to better use by getting some AoPS books.

I noticed that there are a few new "Intro" series books from AoPS, judging from the table of contents, it appears to me that they are at an easier level than AoPS Basics although all these books are said to target 7-10th graders. Did someone happen to know these new intro books? How do they compare to AoPS v1? In particular, I'm looking at the intros for Number Theory and Counting & Probability, kind of trying to decide if I'll get one of these or the AoPS v1: Basics, which has a proven record. Please advice. Remember, this is for a 6th grader who has limited exposure to algebra (He can do simple 2-variable linear equations, but that's about it).

Thanks,

- glacier

The Intro books start at an easier level than Aops Volume 1, but by the time they finish they are about the same level.  I am going into 7th grade, and they are just the right level for me.  In the Counting and Probability book, the only chapter that might be hard is the Binomial Theorem.

I have a daughter entering 7th grade who is working her way through the new Intro series.

I would definitely recommend the new "Intro to Number Theory" and "Intro to Counting" books for your son.  They are more accessible to 6th graders than the AoPS Basics book is.  The new "Intro to Geometry" book is also good, though it is a little more sophisticated than the other two Intro books.  (It involves more proofs than the other two.)

I highly recommend AoPS vol. 1.  I am also going into 7th grade and I understand most of what is covered in AoPS volume 1.

[quote="jclarkemathL314159"]The great thing about AoPS volume 1 and the Intro series is that they assume very little knowledge, other than really basic algebra concepts. The Intro series goes into further depth, however.[/quote]
 Much, much further depth...

http://www.goodlymaths.0catch.com

Try the mathcounts archive

[quote="kheg_k"][quote="jclarkemathL314159"][b]The handbook is posted on the internet, for free download. The problems are different every year.[/b]

Yes, each coach gets one, but it is better for a student to work at his or her own pace by him or herself at the problems.[/quote]

where can I download it?[/quote]

http://www.mathcounts.org
:lol:

Thank everyone for the help. I have ordered AoPS Intro of Number Theory and Counting & Probability based upon your comments and the fact that my son is just entering 6th grade (better get materials that are easy to get into) - the excerpts pdfs on the intro books also gave me good idea about the books, unfortunately no excerpts are available for AoPS v1 - I wanted to compare them but after reading the excerpts of the intro books, I really like them.

Thanks,

- glacier


The difference between a $ 6.5 \%$ sales tax and a $ 6 \%$ sales tax on an item priced at $ \$20$ before tax is

\[ \textbf{(A)}\ \$.01 \qquad
\textbf{(B)}\ \$.10 \qquad
\textbf{(C)}\ \$.50 \qquad
\textbf{(D)}\ \$1 \qquad
\textbf{(E)}\ \$10
\]

$ .06\times 20\equal{}1.2$ and $ .065\times 20\equal{}1.3$ so there is a difference of $ .10$.


inequalities
induction
inequalities unsolved

Let $a,b,c>0$ such that $a^{b}b^{c}c^{a}\ge 1$. Prove that for every positive integer $n$, $a^{n}+b^{n}+c^{n}\ge a+b+c$

what, cant anyone solve this easy problem? i'm waiting to see your solution.

[quote="nayel"][quote="ali666"]my solution is long
lemma.if $a+b+c\ge{3}$ and $n\in{\mathbb{N}}$then $a^{n}+b^{n}+c^{n}\ge{a+b+c}$
proof:by chebyshev inequality:
$a^{n}+b^{n}+c^{n}\ge{\frac{(a+b+c)(a^{n-1}+b^{n-1}+c^{n-1})}{3}\ge a^{n-1}+b^{n-1}+c^{n-1}}$
and rest is easy by induction.

$a^{b}b^{c}c^{a}\ge 1\Rightarrow{a^{\frac{b}{a+b+c}}b^{\frac{c}{a+b+c}}c^{\frac{a}{a+b+c}}\ge 1}\\ a(\frac{b}{a+b+c})+b(\frac{c}{a+b+c})+c(\frac{a}{a+b+c})\ge{a^{\frac{b}{a+b+c}}b^{\frac{c}{a+b+c}}c^{\frac{a}{a+b+c}}\ge 1\\ ab+bc+ca\ge 1\Rightarrow(a+b+c)^{2}\ge 3(ab+ac+bc)\ge3(a+b+c)\\ Rightarrow(a+b+c)\ge 3}$[/quote]
how did you get that $a+b+c\ge{3}$?[/quote]
see my post again  :D

One more solution:
From the inequality
\[\left(\frac{a+b+c}{3}\right)^{a+b+c}\geqslant a^{b}b^{c}c^{a}\]
(see [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=102691[/url]) we get $a+b+c\geqslant 3$. By the power mean inequality and $t^{n}\geqslant t$ for $t\geqslant 1$:
\[a^{n}+b^{n}+c^{n}= 3\cdot\left( \sqrt[n]{\frac{a^{n}+b^{n}+c^{n}}{3}}\right)^{n}\geqslant 3\cdot\left( \frac{a+b+c}{3}\right)^{n}\geqslant 3\cdot \frac{a+b+c}{3}= a+b+c. \]
Edit:
[quote="nayel"]assume that $a\ge b\ge c$.[/quote]
You can not assume this, because the inequality is not symmetric.

[quote="Rzeszut"]
Edit:
[quote="nayel"]assume that $a\ge b\ge c$.[/quote]
You can not assume this, because the inequality is not symmetric.[/quote]
Right, but Nayel's solution can be completed by noting that in the other case, we can assume $a\le b\le c$, and
$a^{a-b}b^{b-c}\ge c^{a-b}c^{b-c}$ still holds.

[quote="ali666"][quote="weiquan"][quote="ali666"]my solution is long
lemma.if $a+b+c\ge{3}$ and $n\in{\mathbb{N}}$then $a^{n}+b^{n}+c^{n}\ge{a+b+c}$
proof:by chebyshev inequality:
$a^{n}+b^{n}+c^{n}\ge{\frac{(a+b+c)(a^{n-1}+b^{n-1}+c^{n-1})}{3}\ge a^{n-1}+b^{n-1}+c^{n-1}}$
and rest is easy by induction.

$a^{b}b^{c}c^{a}\ge{1}\Rightarrow{a^{\frac{b}{a+b+c}}b^{\frac{c}{a+b+c}}c^{\frac{a}{a+b+c}}}\ge{1}\\ a(\frac{b}{a+b+c})+b(\frac{c}{a+b+c})+c(\frac{a}{a+b+c})\ge{a^{\frac{b}{a+b+c}}b^{\frac{c}{a+b+c}}c^{\frac{a}{a+b+c}}}\ge{1}\\ ab+bc+ca\ge{a+b+c}\Rightarrow{(a+b+c)^{2}\ge{3(ab+ac+bc)}\ge{3(a+b+c)}}\\ \Rightarrow{(a+b+c)\ge 3}$[/quote]

Hmm... why do you need induction? If you just used Power Mean, you could prove the inequality for all real $n \geq 1$! ... and it would be only 2 lines in total.[/quote]because I dont know power mean inequality  :blush:[/quote]

Power mean inequality is...

$f(r) = \left( \frac{1}{n}\sum_{i=1}^{n}{{a_{i}}^{r}}\right)^{\frac{1}{r}}$ is strictly increasing for positive $a_{i}$s :)

[quote="weiquan"][/quote][quote="ali666"][quote="weiquan"][quote="ali666"]my solution is long
lemma.if $a+b+c\ge{3}$ and $n\in{\mathbb{N}}$then $a^{n}+b^{n}+c^{n}\ge{a+b+c}$
proof:by chebyshev inequality:
$a^{n}+b^{n}+c^{n}\ge{\frac{(a+b+c)(a^{n-1}+b^{n-1}+c^{n-1})}{3}\ge a^{n-1}+b^{n-1}+c^{n-1}}$
and rest is easy by induction.

$a^{b}b^{c}c^{a}\ge{1}\Rightarrow{a^{\frac{b}{a+b+c}}b^{\frac{c}{a+b+c}}c^{\frac{a}{a+b+c}}}\ge{1}\\ a(\frac{b}{a+b+c})+b(\frac{c}{a+b+c})+c(\frac{a}{a+b+c})\ge{a^{\frac{b}{a+b+c}}b^{\frac{c}{a+b+c}}c^{\frac{a}{a+b+c}}}\ge{1}\\ ab+bc+ca\ge{a+b+c}\Rightarrow{(a+b+c)^{2}\ge{3(ab+ac+bc)}\ge{3(a+b+c)}}\\ \Rightarrow{(a+b+c)\ge 3}$[/quote]

Hmm... why do you need induction? If you just used Power Mean, you could prove the inequality for all real $n \geq 1$! ... and it would be only 2 lines in total.[/quote]because I dont know power mean inequality  :blush:[/quote][quote="weiquan"]

Power mean inequality is...

$f(r) = \left( \frac{1}{n}\sum_{i=1}^{n}{{a_{i}}^{r}}\right)^{\frac{1}{r}}$ is strictly increasing for positive $a_{i}$s :)[/quote]
Thank you very much.  :)


in Scott Adams's book Dilbert and the Way of the Weasel, Mr. Adams states:

1 weasel = 10 dollars
weasel being defined as a person who isn't trustworthy but not stays within the law. weasels use many sneaky, but legal, tactics to avoid work and put the blame on others.

he explains that wherever anything worth more than 10 dollars is, there is a weasel. It increases proportionately, e.g. if there were 20 dollars somewhere, there would be 2 weasels, 30 dollars, 3 weasels, etc.

The Pointy-Haired Boss, a weasel, is giving raises on this particular day.
Wally, a weasel, has a [u]yearly[/u] pre-raise salary of 40,500 dollars, and gets a 2.25% raise.
Dilbert, a weasel, has a [u]yearly[/u] pre-raise salary of 60,400 dollars, and gets a 1.5% raise.

At lunch break, Dilbert and Wally walk down the street to a somewhat good deli to get some sandwiches.  They each have their new [u]monthly[/u] paychecks in their wallets.

Assuming The Weasel Rule is true, how many weasels should Wally and Dilbert expect to meet on the way to the deli? don't forget that dilbert and wally also count as weasels.  :lol:

edit: sorry, it wouldn't tke the dollar sign.  it took me multiple edits to find that.  i took dilbert's bonus out because bonuses come at the end of the year so it would be irrelevant.  :(
also, the number of total weasels there includes wally and dilbert, so don't include them in your answer!
another edit:  my dad asked "what about taxes?" they live in a tax-free utopia.

also, round [u]down[/u] to the nearest weasel since you can't have a fraction of a person.

in Scott Adams's book [i]Dilbert and the Way of the Weasel[/i] Mr. Adams states:

1 weasel = $\$$10 weasel being defined as a person who isn't trustworthy but not stays within the law. weasels use many sneaky, but legal, tactics to avoid work and put the blame on others. he explains that wherever anything worth more than $\$$10 is, there is a weasel. It increases proportionately, e.g. if there were $\$$20 somewhere, there would be 2 weasels,$\$$30, 3 weasels, etc.

The Pointy-Haired Boss, a weasel, is giving raises on this particular day.  Wally, a weasel, has a pre-raise salary of $\$$40,500, and gets a 2.25% raise. Dilbert, a weasel, has a pre-raise salary of 60,400, and gets a 1.5% raise. In addition, he gets a $\$$500 bonus for finishing a project.

At lunch break, Dilbert and Wally walk down the street to a somewhat good deli to get some sandwiches.  They each have their new [u]monthly[/u] paychecks in their wallets.

Assuming The Weasel Rule is true, how many weasels should Wally and Dilbert expect to meet on the way to the deli? don't forget that dilbert and wally also count as weasels.  :lol:

Watch your dollar signs. I fixed the latex for you. :)

[quote="Captain Sugar"]thanks. i have all the edits (seven or eight) in the first post.  my parents were reading over my shoulder and said "what about this? And this? and this and this and this?" but i finally finished.[/quote]
lol :rotfl: .

for future reference, this is the code for dollar signs:
[code]$\$$[/code]

and finally, a solution:
[hide]Wally's new paycheck is worth $40500*1.0225=41411.25$, and Dilbert's is worth $60400*1.015=61306$. Add the extra $\$$500 to get $\$$103217.25 total. Thus, we would expect them to see10321-2=10319 weasels. That's a lot of weasels![/hide]

[quote="easyas3.14159..."][code]$\$$[/code][/quote]
Nice easyaspi. You can either click quote on his thing to see the real code or remove the spaces from S \ S S. Substitute the S with the dollar sign  :rotfl:  :ninja:

Sorry :oops: well this is the code anyways:$\$$

sorry, easyaspi, you're wrong. they only have their [u]monthly[/u] paychecks in their wallets.

[hide]
Wally:
40500*1.0225=41411.25
divide by 12: 3450.9375
Dilbert:
60400*1.015=61306
divide by 12: 5108.83333333333333333333333333333333333333333
add them together and get
8559.7708333333333333333333333333333333333333333333333333
then since there is a weasel for every 10 dollars divide by 10 to get
855.97708333333333333333333333333333333333333333333333333
round down since the problem says to and we can't have parts of weasels.
855
then subtract two for dilbert and wally, since they are weasels, but they don't meet themselves walking down the street!
the final answer is 853
[/hide]

[hide]$40500\cdot\frac{1.0225}{12}=3450.93$ dollars.  We add the other weasal's $60400\cdot\frac{1.015}{12}=5108.83$  Add them both together...$3450.90+5108.83=8559.73$  We divide by 10, and get 855.97 weasels-wally and dilbert=853 weasels[/hide]


Let $ V$ be the vector space off all real $ nxn$ matrices and let :
  $ A\equal{}\left[\begin{array}{cccc} \lambda_1 & 0 & ... & 0 \\
0 & \lambda_2 & ... & 0 \\
... & ... & ... & ... \\
0 & 0 & ... & \lambda_n \end {array} \right]$
Here $ \lambda_k > 0 ,\forall k \in$ {1,2,...,n}
Known that : the linear transformation of $ T$ : $ V \to V$ defined $ T(X)\equal{}\frac{1}{2}(AX\plus{}XA)$
1) Slove equation $ T(X)\equal{}A^{2009}$
2) Calculate the determinant of the linear transformation $ T$ on $ V$

1) Clearly, $ Ker T=\{0\}$ holds, since $ \forall i,j: T(X)_{ij}=\frac{\lambda_i+\lambda_j}{2}X_{ij}$. But then $ X=A^{2008}$ is the only solution.

2) From 1) and using the standard basis ${ \{E(r,s)}\}_{r,s=1,\ldots,n}$ of $ M_n(\mathbb{R})$ we calculate $ T(E(r,s))=\frac{\lambda_r+\lambda_s}{2}E(r,s)$, i.e. $ E(r,s)$ is an eigenvector for the eigenvector $ \frac{\lambda_r+\lambda_s}{2}$ and $ T$ is diagonal for the standard basis with determinant $ \prod_{r,s=1,\ldots,n}\frac{\lambda_r+\lambda_s}{2}$

With $ A \equal{} \left[\begin{array}{cccc} \lambda_1 & 0 & ... & 0 \\
0 & \lambda_2 & ... & 0 \\
... & ... & ... & ... \\
0 & 0 & ... & \lambda_n \end {array} \right]$
Here $ \lambda_k \neq 0 ,\forall k \in$ {1,2,...,n}
Known that : the linear transformation of $ T$ : $ V \to V$ defined $ T(X) \equal{} (AX \minus{} XA)$
Do we have  $ Rank(T)\equal{}n^2\minus{}n$ ?

Again, use the standard basis $ E(r,s)$ and check $ T(E(r,s))\equal{}(\lambda_r\minus{}\lambda_s)E(r,s)$, which implies $ 0\equal{}T\left(\sum_{r,s}c_{rs}E(r,s)\right)$ iff $ \forall r,s: (\lambda_r\minus{}\lambda_s)c_{rs}\equal{}0$. If we define $ {\cal L}\equal{}\{\lambda_1,\ldots,\lambda_n\}$ and for $ \lambda\in{\cal L}$ the set $ S(\lambda)\equal{}\{i\equal{}1,\ldots,n|\lambda_i\equal{}\lambda\}$, then we get $ Ker(T)\equal{}\left\{\sum_{\lambda\in{\cal L}}\sum_{r,s\in S(\lambda)}c_{rs}E(r,s)\Big | c_{rs}\in\mathbb{R}\right\}$. In particular, $ rank(T)\equal{}n^2\minus{}\dim (Ker(T))\equal{}n^2\minus{}\sum_{\lambda\in{\cal L}}|S(\lambda)|^2\leq n^2\minus{}n$, the last inequality since $ n\equal{}\sum_{\lambda\in{\cal L}}|S(\lambda)|$ holds.

P.S. from this description it is obvious that your criterion $ \lambda_i\not\equal{}0$ has no influence, since only the multiplicities of the $ \lambda_i$ are relevant. The extreme cases are (1) $ \lambda_i$ pairwise different, i.e. $ |S(\lambda)|\equal{}1$ and $ rank(T)\equal{}n^2\minus{}n$ and (2) $ \lambda_i$ all equal, i.e $ |S(\lambda)|\equal{}n$ and $ rank(T)\equal{}0$


integration
trigonometry
calculus
calculus computations

Hi,

I usually do not ask homework questions but this is one question that I don't know what to do. I solved other ones with ease but this one seems be out of my reach. Please give help except I only know basic things about integral.

\[ \displaystyle \int^{\frac{\pi}{2}}_{\frac{\pi}{3}} (\sin z - 2 \cos (2z)) dz \]

$\int \sin z dz = -\cos z+C$

$\int 2\cos 2z dz = \sin 2z + C$

Now use the fundamental theorem of calculus.


The digits 1, 2, 3, 4, 5 can be arranged to form more than one hundred different five digit numbers.  If these five-digit numbers are arranged from least to greatest, in what position is 42513?

5*4*3*2*1=120 arrangements.
group them up.
the ones starting with 4 are 73-96
those with ones are 73-78
then in those the 2's are 79-84
then with those
42135#79
42153#80
42315#81
42351#82
42513#83

83rd

hmm i got something differernt from jimli

[hide]

5! = 120 different numbers,120/5 = 24 for every ten thousand.  so we're already no better than rank 24 cuz we're in the 40000's.

in the 40000's, there are 24 numbers; 45000's have 6, 44000's have 6, and 43000's have 6. thus we're no better than 24 + 18 = 42.  in the 42000's, there are 2 42500's, and we are the lower of the 2 (42513 as opposed to 42531).  thus our rank is

42 + 2 = 44.

I didn't get an answer that's someone else's as well.  :?
[hide]
5!=120 different numbers.
120/5=24 numbers for every ten thousand
45000 has 6, 44000 has 6, 43000 has 6, 41000 has 6, 42000 has 6, 42500 has 2.
24+24+24+6+1=$\framebox{73}$ [/hide]

Nice title. Maybe you should make them a little more informative....?  ;)

its the 83rd, i made a list and counted the :blush: m all up


I have a problem. For some reason, my password no longer works on the usamts site, and I've never changed it, and I'm pretty sure I'm typing it in right. When I typed in the thing for sending a new password, it sent to the account that I no longer use, and has been inactivated.

Is the old email hotmail or yahoo or gmail?  If so, just reactivate it.

I don't have the password to that either.  :(

My User ID is 7803, username is jclarkeLHS, current email address is minesweeperman337@yahoo.com.


linear algebra
matrix
vector
Gauss
abstract algebra
induction
algebra

It is well known that a square matrix $A$ is invertible if and only if the equation system $Ax = y$ has a unique solution for every column vector $y$. However, it is intuitively clear that if $Ax = u$ has a unique solution for some (fixed) column vector $u$ then $Ax = y$ must have a unique solution for every y. This is intuitively clear because that $Ax = u$ has a unique solution means that you can find it by guass-elimination and thereby diagonalize the equation system. This can be done even if we change $u$ to something else.

However, I don't feel that this is a formal argument. How can it be made stringent? I have somehow managed to forget a lot of linear algebra, so there might be something obvious I am missing here. I know about the basics of eigenvectors such as the spectral theorem but I do not know about more advanced things like Gauss-Jordan and Cayley-Hamilton (so if there is a simple argument not using advanced things I'd like to see it).

Consider the kernel $\ker A$ of $A$: the set of vectors $u$ s.t. $Au= 0$.
If $x$ is a solution of the system $A x= y$ and $u\in \ker A$ then $x+u$ is also a solution of the system (Why?). Therefore, if the system has exactly one solution, then $\ker A = \{0\}$; that is no (non-trivial) linear combination of columns of $A$ equals to $0$. We conclude that $A$ is not singular.

[quote="Kalle"]This is intuitively clear because that $Ax = u$ has a unique solution means that you can find it by guass-elimination and thereby diagonalize the equation system.[/quote]
What does this really mean? Let's look at what the process of Gaussian elimination does.

At each step, we apply an elementary row operation to both $A$ and $u$, systematically reducing $A$ to echelon form and then clearing the elements above each pivot. There will be at most 1 solution if each column has a pivot, and at least 1 solution if each row has a pivot. If any column does not have a pivot, there is a free variable and we cannot have exactly 1 solution. For a square matrix, this means we have exactly one solution if and only if the reduced row-echelon form of $A$ is the identity, independent of the choice of $u$.
Now, each row operation we applied can be represented by multiplication $A_{i+1}=E_iA_i$ (with $A_0=A,A_m=I$) and $u_{i+1}=E_iu_i$ (with $u_0=u,u_m=x$). If we set $B=E_{m-1}E_{m-2}\cdots E_0$, then $BA=I$ and $Bu=x$

Since each $E_i$ is invertible (a routine calculation), we can write down a two-sided inverse to $B$: $B^{-1}=E_0^{-1}E_1^{-1}\cdots E_{m-1}^{-1}$. Generic algebra now gives us that $A$ and $B$ are two-sided inverses, and $A$ is invertible.

Gaussian elimination is how we show that $A$ is invertible when $Ax=u$ has exactly one solution. The proof works even if we know this only for one particular $u$. Given that, it's more efficient to just quote invertibility of $A$ in trying to show that $Ax=v$ has exactly one solution for every $v$; this solution is $A^{-1}v$.

Thanks! I think Yury's response was more what I was looking for. jmerry, you kind of skip my problem in the first place:

[quote="jmerry"] For a square matrix, this means we have exactly one solution if and only if the reduced row-echelon form of $A$ is the identity, independent of the choice of $u$.[/quote]
I think this statement is what I don't quite believe in, and was really my original question (i.e. unique solution for one u is equivalent to unique solution for all u?). I've never seen a formal presentation of the whole deal wth echelon form. Should it be proved with induction over rows?

Since you were so nice to show me a proof, I thought I could show you the proof I know to prove that the $n\times n$ matrix $A$ is invertible iff $Ax = y$ has a unique solution for every $y$. You might already know it, but I think I prefer this proof:

Consider the matrix equation $AX = C$ where $C$ is a given $n\times n$-matrix. This equation is equivalent to the system of equations $AX_k = C_k$, $k=1,2,\cdots,n$ where $C_k$ means the k:th column in C. That $Ax = y$ has a unique solution for every $y$ then means that this system has a unique solution. That is, the equation $AX = C$ has a unique solution. Specifically with $C = I$ there exists a $B$ such that $AB = I$.

Now to prove that $BA = I$ we can use that a linear map is injective iff it is surjective which in matrix language translates $AB = I$ to $BA = I$. However, we might also utilize a trick that I really like:

We know that $A(BA) = A$,  i.e. $BA$ is a solution to the equation $AX = A$. However, we also know that $X = I$ solves this system. But since we know the system has a unique solution we then must have $BA = I$ and we are done.


ratio
trigonometry
geometry proposed
geometry

New problem:
a) Constuct triangle ABC by given angles: CAB, CBA and median from C to AB (let say - CM - M is from AB such that AM=MB).
b) Find the Surface/Area of the triangle by these elements.

For a), construct a triangle $ \Delta A'B'C'$ which is similar to triangle $ ABC$ from 2 angles.
Then scale $ \Delta A'B'C'$ with an approriate ratio to get triangle $ ABC$.

[color=darkblue]The relations $ \left\|\begin{array}{c} \frac {a}{\sin A} \equal{} \frac {b}{\sin B} \equal{} \frac {c}{\sin C} \equal{} 2R \\
 \\
S \equal{} 2R^2\sin A\sin B\sin C \\
 \\
4m_c^2 \equal{} 2(a^2 \plus{} b^2) \minus{} c^2\end{array}\right\|$ are well-known.

Thus, $ m_c^2 \equal{} R^2\cdot\left[2\left(\sin^2A \plus{} \sin^2B\right) \minus{} \sin^2(A \plus{} B)\right]$ $ \implies$[/color]

$ \boxed {S \equal{} \frac {2m_c^2\sin A\sin B\sin (A \plus{} B)}{2\left(\sin^2A \plus{} \sin^2B\right) \minus{} \sin^2(A \plus{} B)}}\ .$

:)  :)  :)  :)  I have in mind a different idea by using your approach is more intuitive. My idea were to prove that:
$ cotg \phi \equal{} 0.5(cotgA\minus{}cotgB)$ where $ \phi$ denotes the angle between median and AB. Then using Sine law for AMC and BMC. I'm disapointed - this problem is too easy ... but I expected it.

The your problem is very easily and isn't the case to calculate the angle $ \phi\ !$ If you wish, please to present here how you ascertain it.

It is easy. Draw a height HD to AB. Look at the triangles ADC, BDC, CDM. M is the midpoint of AB. Exclude all segments (HD, MD, AD, BD) from the equalities for cotg's. It is a problem from 80's for Bulgarian Math Olympiad II round.

I don't say it is hardly. More, in the my opinion is and noninterestingly.


Let $ a_1 \equal{} 1$; $ a_2 \equal{} 2$; $ a_3 \equal{} 3$. For each integer $ n\geq3$, define $ a_{n\plus{}1}$ by the formula\[ a_{n\plus{}1}\equal{}a_n\minus{}a_{n\minus{}1}\plus{}\frac{a_n^2}{a_{n\minus{}2}}\]Prove that $ a_n$ is a positive integer and is not divisible by $ 4$.
[hide]
We need a manageable recursion relation: divide through by $ a_{n\minus{}1}$ and factor: \[ 1\plus{}\frac{a_{n\plus{}1}}{a_{n\minus{}1}}\equal{}\frac{a_n}{a_{n\minus{}1}}\left(1\plus{}\frac{a_n}{a_{n\minus{}2}}\right)\]Now, define $ b_n\equal{}\frac{a_{n\plus{}2}}{a_n}$ for $ n\geq1$. Notice that \[ \frac{1\plus{}b_{n\minus{}1}}{1\plus{}b_{n\minus{}2}}\equal{}\frac{a_n}{a_{n\minus{}1}}\implies\frac{1\plus{}b_{n}}{1\plus{}b_{n\minus{}1}}\equal{}\frac{a_{n\plus{}1}}{a_{n}}\]
Multiply the two together: \[ \frac{1\plus{}b_n}{1\plus{}b_{n\minus{}2}}\equal{}b_{n\minus{}1}\iff b_n\equal{}b_{n\minus{}1}(1\plus{}b_{n\minus{}2})\minus{}1\]Since $ b_1\equal{}3,b_2\equal{}5$ we find that $ b_n$ is an odd positive integer by induction. Thus:\[ a_{2n}\equal{}\left(\prod_{i\equal{}1}^{n}b_{2n}\right)a_2\equiv(\pm1)(2)\equiv2\mod4\\a_{2n\minus{}1}\equal{}\left(\prod_{i\equal{}1}^{n}b_{2n\minus{}1}\right)a_1\equiv1\mod2\]
Thus $ a_n$ is a positive integer not divisible by $ 4$.
[/hide]

I have a solution wich uses 3 inductions  :lol:

I'll post it when I have more time.

 :)


I think this is an excellent problem. It's not too difficult, but it is tricky. (Let beginners have a shot at this first).

Find the simplified expression for [tex]\sqrt{111\ldots1 - 222\ldots2}[/tex], where there are [tex]n[/tex] 2's and [tex]2n[/tex] 1's in the expression.

Good Luck!!!  :)

JS157- your solution doesn't come out as nicely as mine...i think you made a few errors. first, explain how you got from your first post to your second post. its not clear, but i think you end up saying that x^2n=(x^n)(x^2), which is definitely not true. plus...i don't really see how your solution is much simpler than the original problem.

Jeffery - okay, I fixed it.  He asked for a expression w/n in it...a formula w/o the ellipses.  That's what I gave.  I actually don't understand your solution too well so...

From the last expression in your first post, notice that 10^2n-10^n+1
is simply (10^n-1)^2. After that the final expression is truly simpler than the original.

i just introduced my own notation for the problem. my solution is equivalent to yours...i just didn't rely on that infinite decimal trickery

This probably isn't a valid solution but if you notice the pattern...
[tex]
\sqrt{11-2}=3\\
\sqrt{1111-22}=33\\
\sqrt{111111-222}=333[/tex]

You can easily see the answer.  But then...that's not much of a proof is it :lol:

Mikewd - [hide]You did it the same way I did, but you made a mistake:

1/9 = 0.11111.....

But if you multiply it by 10^3 you don't get 111 you get 111.111111...... so you have to get rid of that .111... at the end.[/hide]

[hide]Ah I see. Also I realised (when explaining my answer to someone, the usual time when I see a mistake) that 1111....1 (2n digits) =/= (1/9)^2n because if the ending digits were removed, still, the correct form would be (1/9)*10^2n. hmm..

So 111111.....1 (2n digits) = (1/9)*10^2n - (1/9)

and 222222.....2 (n digits) = (2/9)*10^n - (2/9)

111111.....1 (2n digits) - 222222.....2 (n digits) = (1/9)*10^2n - (1/9) - ((2/9)*10^n - (2/9))

= (1/9)*10^2n - (1/9) + (2/9) - (2/9)*10^n

= (1/9) + (10^2n)/9 - (20^n)/9     <~~~~~(im a bit skeptical about this step... perhaps I'm not allowed to do this)

= (10^2n - (10*2)^n + 1)/9

= (10^2n - (10^n)(2^n) + 1)/9

Hmm... this looks like a quadratic in disguise, but I'm a bit stuck here.[/hide]


How do you convert a fraction such as $\frac{1}{3}$ into any other base besides 10?

Let's say you want to convert $\frac{1}{3}$ into base 6.

Multiply by $\frac{6}{6}$ and get $\frac{6(\frac{1}{3})}{6}=\frac{2}{6}$. Since the first place after the decimal point has a value of $\frac{1}{6}$ in base ten, $\frac{1}{3}_{10}=0.2_6$.

How about base 7?

Multiply by $\frac{7}{7}$ to have $\frac{7(\frac{1}{3})}{7}=\frac{2\frac{1}{3}}{7}=\frac{2}{7}+\frac{\frac{1}{3}}{7}$. Since the first place after the decimal point has a value of $\frac{1}{7}$, we have 0.2 ....  (because of the $\frac{2}{7}$).

Now we look at the $\frac{\frac{1}{3}}{7}$ that's left (since we already considered $\frac{2}{7}$). Multiply again by $\frac{7}{7}$ and we have $\frac{7(\frac{1}{3})}{49}=\frac{2\frac{1}{3}}{49}=\frac{2}{49}+\frac{\frac{1}{3}}{49}$. The second place after the decimal point has a value of $\frac{1}{49}$, so we have 0.22 .... (because of the $\frac{2}{49}$).

Now there's $\frac{\frac{1}{3}}{49}$ left. It seems like that if we multiply by $\frac{7}{7}$ again, another 2 will appear. We can safely assume that the other digits will be a series of 2's, so the result is

$\frac{1}{3}_{10}=0.\overline{2}_7=\frac{2}{9}_7$

[quote="236factorial"]

$\frac{1}{3}_{10}=0.\overline{2}_7=\frac{2}{9}_7$[/quote]

This last part is incorrect - It should be $\frac{2}{6}$.  Here's a proof.

All computations below can be assumed to be in base 7.

Let's say we have some number with a repeating decimal in base $7$, like $0.\overline{1}$.  Let $x=$ that number.  If we multiply the number by $7$, it essentially moves the decimal place to the right one digit, so that $7x=1.\overline{1}$.  Therefore,
$7x-x=1.\overline{1}-0.\overline{1}\Rightarrow 6x=1\Rightarrow x=\frac{1}{6}$
In general, $0.\overline{a}_b=\frac{a}{b-1}$.  Also, if there was a string of $n$ repeating digits, the denominator of the fraction would be $b^n-1$.


abstract algebra
group theory
superior algebra
superior algebra theorems

Suppose that $ G$ is a finite group whose order is divisible by $ p$.  Then we let $ X \equal{} G\times G\times\cdots\times G$ ($ p$-times) where $ g_{1}g_{2}\cdots g_{n} \equal{} 1$ for $ \left(g_{1},g_{2},\dots ,g_{n}\right)\in X$.  Then let $ \left < \sigma\right >$ be a cyclic group and define the group action $ f: \left < \sigma\right > \times X\rightarrow X$ by $ \sigma*\left(g_{1},g_{2},\dots ,g_{n}\right) \equal{} \left(g_{n},g_{1},\dots ,g_{n \minus{} 1}\right)$.  This is easily verified as a group action.  Now, my question is, how do we show that a fixed point in $ X$ contains "entries" of order $ p$?  I know that if $ \left(g_{1},g_{2},\dots g_{n}\right)$ is a fixed point, then $ g_{1} \equal{} g_{2} \equal{} \cdots \equal{} g_{n}$ and $ g_{1}g_{2}\cdots g_{n} \equal{} 1$, but couldn't $ g_{1} \equal{} e$?

Also, a notation question:  I have seen the set $ \left\{\left(a,b\right)\ |\ a\in A,\ b\in B\right\}$ written as $ A\times B$ (as above) and $ A\oplus B$, although I though that the former was used to indicate an [i]internal[/i] direct product rather than an external direct product.  When is it appropriate to use each?

Thank you  :)

$ g_1$ could be $ e$. But after all, the number of elements which is of order $ p$ is divisible by $ p$, so there is at least 1, other than $ e$.

[quote="amcavoy"]
Also, a notation question:  I have seen the set $ \left\{\left(a,b\right)\ |\ a\in A,\ b\in B\right\}$ written as $ A\times B$ (as above) and $ A\oplus B$, although I though that the former was used to indicate an [i]internal[/i] direct product rather than an external direct product.  When is it appropriate to use each?
Thank you  :)[/quote]

$ \times$ is direct product which always coincides with direct product as sets.

$ \oplus$ is direct sum. This can vary. If you make a direct sum of finitely many abelian groups, you get the same as with direct product.
For example $ \bigoplus_{\mathbb Z}\mathbb  Z$ si different from $ \mathbb Z^{\mathbb Z}$.
And if you work in the category of not necessarily abelian groups, the direct sum of G and H is the free product $ G*H$ which is gigantic compared to $ G$ and $ H$. For example $ \mathbb Z/2\mathbb Z*\mathbb Z/2\mathbb Z$ is infinite.
And so on.


What would be the cutoff in terms of ques/marks for INMO 2007 considering the levelof paper which seemingly is tougher :?: ?

I solved four of them in around three hours and was wondering what to do...Anyone have an elegant proof for the 6 th one?And how are the cut off's this year and how much are the problems worth each?

Cheers
Bhargav

Where can I find all INMO 2007 Problems??
Please, help me.

[quote="Sathej"]Hello,
    Best of luck to everyone writing the INMO 2007. Solutions to INMO 2007 and other related issues can be discussed in this thread.It would be advisable to put up the questions in the respective forums also so as to spur solutions from others.
Sathej[/quote]
You not post all problem of INMO 2007 in National Olympiad box, Why?  :maybe:

They generally are out anytime around the last week of February and the first week of March.
  AR

did any one solve the inequality(prob 6) n how?2,3,4 looked the easiest... :wallbash_red:

Hello,
    Just check out the Inequalities Unsolved Problems Forum.The 6th Problem has been solved using two different approaches.
Sathej