There are 100 lamps , 60 lamps belong to (serie 1) , 40 lamps belong to (serie 2). The probability that 1 lamp of (serie 1) lights in a moment $ T$ is $ \frac{1}{2}$ while the probability that a lamp of (serie 2) lights in a moment $ T$ is $ \frac{1}{8}$.
Find the probability that one lamp (took casualy from the 100) lights in a moment $ T$.

isn't it just the weighted average?

$ \frac{\frac{1}{2}\times 60 \plus{} \frac{1}{8} \times 40}{100} \equal{} \frac{7}{20}$

no, the weighted average would be the expected proportion of lamps lit.  The probability that one lamp in series 1 is lit is $ 1 \minus{} (\frac {1}{2})^{60}$ (probability that at least one lamp is lit: not that each lamp is not lit 60 times) and the probability of a series 2 lamp is lit is $ 1 \minus{} (\frac {7}{8})^{40}$ so the total probability is $ 1 \minus{} (\frac {1}{2})^{60} \plus{} (\frac {1}{2})^{60}(1 \minus{} (\frac {7}{8})^{40})$ (we only care if a series 2 lamp is lit when a series 1 lamp isn't).  This probability is approximately 0.99999999999999999999584546539215.

Oh I suppose I didn't quite answer the question - I gave the probability that at least one lamp lit.  The probability that exactly one lamp is lit is:

for series 1: $ 60(1/2)^{60}$ (60 ways to choose the one to light) and a $ (1/2)^{60}$ chance none are lit.

for series 2: $ 40(7/8)^{39}(1/8)$ and a $ (7/8)^{40}$ chance none will be lit.

Here we must have none lit in series 1 with 1 in series 2 or vice versa (to keep the 1 lit restriction) for a total probability of $ 60(1/2)^{60} \cdot (7/8)^{40} \plus{} (1/2)^{60} \cdot 40(7/8)^{39}(1/8)$ which is about $ 2.73012274 \cdot 10^{\minus{}19}$ - really small, because its more likely a larger number will be lit.

So depending on the exact question intended we have 2 completely different answers.

:maybe: I think the question was
"Pick an arbitrary lamp. What is the probability that it is lit in moment T?"
not
"What is the probability of at least one lamp being lit"
or
"What is the probability of exactly one lamp being lit"

The answer is obviously between 1/8 and 1/2 so none of the answers you gave are plausible.

Yeah I guess you were right then.  I interpreted the problem differently.

[quote="Elio (n)"](took casualy from the 100)[/quote]

This tipped me off  :lol:


If $ a$ and $ b$ are positive numbers such that $ a^b \equal{} b^a$ and $ b \equal{} 9a$, then the value of $ a$ is:

$ \textbf{(A)}\ 9\qquad
\textbf{(B)}\ \frac {1}{9}\qquad
\textbf{(C)}\ \sqrt [9] {9}\qquad
\textbf{(D)}\ \sqrt [3] {9}\qquad
\textbf{(E)}\ \sqrt [4] {3}$

[quote="thebrokenlight"][hide="Solution"]
$a^{9a}=9^{a}a^{a}\\ a^{9}=9^{a}\\ a=9$

a=9 works, but I'm not exactly sure how to get from my second line to my third line.
[/hide][/quote]
Are you sure? I got a different answer...
[hide="My answer"]
$a^{9a}=(9a)^{a}$

$\log a^{9a}=\log (9a)^{a}$

$9a(\log a)=a(\log 9a)$

$a(9\log a-\log (9a))=0$

$9\log a=\log {9a}$

$9\log a=\log a+\log 9$

$8\log a=\log 9$

$\log a^8=\log 9$

$a^8=9$

$\therefore a=\sqrt[8]{9}=\sqrt[4]{3}$

$\boxed{\text{(E)}}$[/hide]

[hide="Answer"]$a^{9a}=(9a)^a\Rightarrow (a^9)^a=(9a)^a \Rightarrow a^9=9a$
$\log 9a=9\log a\Rightarrow \log 9+\log a=9\log a\Rightarrow \log 9=8\log a=\log a^8$
$a=\sqrt[8]{9}=\sqrt[4]{3}\Rightarrow \boxed{E}$[/hide]

[quote="shadysaysurspammed"]One small request silverfalcon,could you post all the questions at once rather than making different topics for each because then it is more convenient rather than opening ne,then going back and opening the other one......[/quote]

The answer is no because I'm adding these problems to Olympiad Resources page.

[hide]$\frac{a^{b}}{b^{a}}=1 \\\Leftrightarrow \frac{a^{9a}}{{(9a)}^{a}}=1 \\\Leftrightarrow \frac{a^{9}}{9a}=1 \\\Leftrightarrow \frac{a^{8}}{9}=1 \\\Leftrightarrow a=\sqrt[8]{9}=\sqrt[4]{3}$[/hide]

[quote="thebrokenlight"][hide="Solution"]
$a^{9a}=9^{a}a^{a}\\ a^{9}=9^{a}\\ a=9$

a=9 works, but I'm not exactly sure how to get from my second line to my third line.
[/hide][/quote]

you should be getting $a^9=9a$.


quadratics
function
algebra proposed
algebra

[quote="Tiger100"]Let $ a,b,c$ integers such that the quadratic function $ ax^2 \plus{} bx \plus{} c$ has two distinct zeros in the interval $ (0,1)$. Find the least valuo of $ a,b,c$[/quote]

You should be more precise in your question since there is no well known total order in $ \mathbb Z^3$

For example, between $ 6x^2\minus{}5x\plus{}1$ and $ 8x^2\minus{}6x\plus{}1$, what triple $ (a,b,c)$ has the least value since we have $ 6<8$ but $ \minus{}6<\minus{}5$ ?


[quote="fizix"]In the computer game NetHack, your alignment can be one of three things: lawful, neutral, or chaotic.  Wearing a helm of opposite alignment will
(a) if you are lawful, switch your alignment to chaotic
(b) if you are chaotic, switch your alignment to lawful
(c) if you are neutral, 50% chance of switching to chaotic and 50% chance of switching to lawful.

You come across some random guy wearing a helm of opposite alignment.  He is currently lawful.  What's the probability that he was neutral before donning the helm?  (We assume that each starting alignment has an equal chance of being generated.)[/quote]
[hide="this probability question is like too hard for me so don't sue me when i get it wrong"]you have 1/3 of being either but knoqing you are lawful there is a 1.5 in total  for the things that could be lawful after alignment and .5 of this is neutral
so .5/1.5=1/3
probably wrong, PM me if you feel like it :blush: [/hide]


function
number theory unsolved
number theory

Let $ n$ be a non-negative integer, which ends written in decimal notation on exactly $ k$ zeros, but which is bigger than $ 10^k$.
 For a $ n$ is only $ k\equal{}k(n)\geq2$ known. In how many different ways (as a function of $ k\equal{}k(n)\geq2$) can $ n$ be written as difference of two squares of non-negative integers at least?

Let $ n\equal{}a^2\minus{}b^2\equal{}(a\minus{}b)(a\plus{}b)$ , where $ (a\minus{}b)$ and $ (a\plus{}b)$ both odd or both even.
As 4|n , we have $ a\minus{}b\equal{}2u$ and $ a\plus{}b\equal{}2v$, where $ m,n$ are positive integers and $ v \ge u$.
So now let $ n\equal{}x*5^k*2^{k\minus{}2}*4\equal{}4uv$, the number of solution $ \equal{}\frac{the \;\; number \;\; of \;\; factor \;\; of \;\; x*5^k*2^{k\minus{}2}}{2}$
$ \ge \frac{[(k\plus{}1)\plus{}1][(k\minus{}2)\plus{}1]}{2}\equal{}\frac{k^2\plus{}k\minus{}2}{2}$, and the least number of ways attains iff $ x\equal{}5$


hi im new to your site but would like help with a problem and some hints and tips with how to deal with such problems.

as the title suggests the problem genre is inequalities. the problem itslef is find the minimum value of \[(2x-5)^{5}(7-x)^{2}\] on the interval \[-5/2<x<7\]

thank you very much

to clarify, what i mean when i say this type of problem is where you have to find a maximum or minimum value of an equation given set conditions

This polynomial has all real roots; it has an order-2 root at 7 and an order-5 root at 5/2.

For $5/2>x>7$, the polynomial is positive
For $x=5/2$, the polynomial is zero
for $-5/2>x>5/2$, the polynomial is negative.

The minimum's gonna be in the negative range.

Now, you have that the polynomial is equal to $-32 (5/2-x)^{5}(7-x)^{2}$
For $x<5/2$, $(5/2-x)^{5}$ and $(7-x)^{2}$ both get more positive as $x$ decreases
So, $-32 (5/2-x)^{5}(7-x)^{2}$ gets more negative as $x$ decreases.

The minimum value has to be at $x=-5/2$, so the minimum value is $(-10)^{5}(19/2)^{2}=-9025000$

a second part of the quuestion which i could also not do but made some proggress with was to find the maximum but i forgot to post this. can you please help me with this. thanks for your help

[hide="for the maximum..."]
First note that $x>\frac{5}{2}$ for the expression to have a chance of being positive. Now we have the AM-GM inequality: given some nonnegative real numbers, their arithmetic mean
\[\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\]
is greater than or equal to their geometric mean
\[\sqrt[n]{a_{1}a_{2}\cdots a_{n}}\]
Here we apply the AM-GM in a clever way to make the $x$'s cancel out. If we apply AM-GM on 5 copies of $2x-5$ and 2 copies of $k(7-x)$ for some positive $k$, their geometric mean is
\[\sqrt[7]{k^{2}(2x-5)^{5}(7-x)^{2}}\]
(once we figure out the good value of $k$, we will get the thing we're trying to maximize.) Their arithmetic mean is
\[\frac{5(2x-5)+2k(7-x)}{7}\]
We want the $x$'s to drop, so $k=5$. We end up with
\[\sqrt[7]{25(2x-5)^{5}(7-x)^{2}}\leq \frac{45}{7}\]
so that the maximum value of $(2x-5)^{5}(7-x)^{2}$ is $\frac{(45/7)^{7}}{25}$, or $\boxed{\frac{3^{14}\cdot 5^{5}}{7^{7}}}$. The maximum occurs when $2x-5=5(7-x)$, or $x=\frac{40}{7}$.
[/hide]


I hear about all these problem sets and packets. Unless I'm mistaken, the only packet I got was on the box principle.

I did test C and I got an induction packet. Perhaps you should email one of the program cooridnators about the other packet. Nobody here really seems to understand the motivation for giving students different packets.

[quote="Altheman"]Does anybody want to get some AIM sessions going where we work on the problems? It is fun to work with other people on problems. Furthermore, it is benefical to have other minds working at the problem at the same time. I feel like some of the problems are hard to get started on...

--Alex

my sn is in my profile[/quote]

we could ask one of the admins to open a classroom because i dont have aim

See my reply in the "let's clear this up" topic

-Zac

[quote="Altheman"]Okay. We are not required to do the problems. However, I still think that they are interesting; that motivates me to work on them. Furthermore, it is enjoyable to work on problems with others in real time. So...

aim: altheman1[/quote]
cept you're like constantly offline
edit wait nvm you just signed on ok

[quote="Altheman"][quote="junggi"][quote="Altheman"]Okay. We are not required to do the problems. However, I still think that they are interesting; that motivates me to work on them. Furthermore, it is enjoyable to work on problems with others in real time. So...

aim: altheman1[/quote]
cept you're like constantly offline[/quote]

Okay you just like IMed me...[/quote]
no i didnt


We consider the following triangular array

\[ \begin{array}{cccccccc}
  0 & 1 & 1 & 2 & 3 & 5 & 8 & \ldots \\
  \ & 0 & 1 & 1 & 2 & 3 & 5 & \ldots \\
  \ & \ & 2 & 3 & 5 & 8 & 13 & \ldots \\
  \ & \ & \ & 4 & 7 & 11 & 18 & \ldots \\
  \ & \ & \ & \ & 12 & 19 & 31 & \ldots  \\
 \end{array} \]

which is defined by the conditions

 i) on the first two lines, each element, starting with the third one, is the sum of the preceding two elements;

ii) on the other lines each element is the sum of the two numbers found on the same column above it.

a) Prove that all the lines satisfy the first condition i);

b) Let $a,b,c,d$ be the first elements of 4 consecutive lines in the array. Find $d$ as a function of $a,b,c$.

Part a) is a typical problem for induction. If you have 2 lines satisfiing the Fibonacci condition with numbers b_0,b_1,... and c_0,c_1,... then you have in the next line you have numbers of the form b_i+c_(i-1). So you have to prove: b_i+c_(i-1)+b_(i+1)+c_i=b_(i+2)+c_(i+1) for all i. That follows because of the Fibonacci structure of the previous lines, q.e.d.

Now, I try to solve b) but it seems to be more difficult!

Actually, even part [b]b)[/b] is not really difficult...

I call "horizontal construction law" the law that every number equals the sum of its two precedents in its row (this was established in part [b]a)[/b] of the problem), and I call "vertical construction law" the law that every number equals the sum of its two precedents in its column (this was given in the problem).

If you consider just the four consecutive lines starting with a, b, c, d, then you get the following partial array:

[code]
          a          ?          ?          ?          ...
                     b          ?          ?          ...
                                c          ?          ...
                                           d          ...
[/code]

Now, if we call u the second number in the first line, then we can fill up the first line by the horizontal construction law:

[code]
          a          u         a+u        a+2u        ...
                     b          ?          ?          ...
                                c          ?          ...
                                           d          ...
[/code]

Now, by the vertical construction law, it follows that

  (a+u) + (the second number in the second row) = c,

hence the second number in the second row is c - (a+u). Thus the partial table takes the form

[code]
          a          u         a+u        a+2u        ...
                     b        c-(a+u)      ?          ...
                                c          ?          ...
                                           d          ...
[/code]

Now, by the horizontal construction law in the second line, this becomes

[code]
          a          u         a+u        a+2u        ...
                     b        c-(a+u)   b+(c-(a+u))   ...
                                c          ?          ...
                                           d          ...
[/code]

Now, if we call T the remaining second element of the third line:

[code]
          a          u         a+u        a+2u        ...
                     b        c-(a+u)   b+(c-(a+u))   ...
                                c          T          ...
                                           d          ...
[/code]

then the vertical construction law yields both

  (a+2u) + (b+(c-(a+u))) = T

and

  (b+(c-(a+u))) + T = d.

Thus,

  d - (b+(c-(a+u))) = T = (a+2u) + (b+(c-(a+u))).

After some simplification, this becomes d + a = 2b + 2c, and thus d = 2b + 2c - a. This is what we wanted, if I have not made some hundreds of mistakes on the way. And even if it's right, I guess this solution will be rated the worst-written solution of all MathLinks; yes, I'm sorry, I'm very tired now and besides, I'm not used to write up such solutions (and before Valentin complains about my rudimentary use of [ code ] tags instead of LaTeX, I must say I haven't succeeded to write the arrays with LaTeX without ending up with [unparseable and potentially dangerous formula]s all the time). Puh, it's time to go to bed.

  Darij

Darij, your solution is really true, I've checked it now.

Good night, I go to bed too!

Misha, thank you for the verification. One final thing I would like to note: If one wouldn't restrict oneself on the "upper right" part of the table only, and one would consider the whole table, placing a 1 on the leftmost place of the second line, then one would get the following "completed" table:

[code]
      0      1      1      2      3      5      8
      1      0      1      1      2      3      5
      1      1      2      3      5      8     13
      2      1      3      4      7     11     18
      3      2      5      7     12     19     31
      5      3      8     11     19     30     49
      8      5     13     18     31     49     80
[/code]

I hope I've made no calculation mistakes here, but anyway the moral is clear: The table is symmetric with respect to its main diagonal, i. e. the n-th number of the m-th row equals the m-th number of the n-th row. But this is really a kind of trivial.

  Darij


Let $a,b,c>0$ such that:$a^2+b^2+c^2=1$.Prove that:
\[a+b+c+\frac{1}{abc} \geq 4\sqrt 3\]

Homogenize the inequality, we have to prove,
\[
\begin{array}{l}
 \left( {a + b + c + \frac{{\left( {a^2  + b^2  + c^2 } \right)^2 }}{{abc}}} \right)^2  \ge 48\left( {a^2  + b^2  + c^2 } \right) \\
  \Leftrightarrow \left( {abc(a + b + c) + \left( {a^2  + b^2  + c^2 } \right)^2 } \right)^2  \ge 48a^2 b^2 c^2 (a^2  + b^2  + c^2 ) \\
 \end{array}
\]
then i think muirhead and schur should work.

a simpler method:
by amgm
\[
1 \ge 3\sqrt[3]{{a^2 b^2 c^2 }},\frac{1}{{3\sqrt 3 }} \ge abc
\]

\[
a + b + c + \frac{1}{{9abc}} + ... + \frac{1}{{9abc}} \ge 12\sqrt[{12}]{{\frac{1}{{9^9 a^8 b^8 c^8 }}}} \ge 4\sqrt 3
\]

As  far as  i know  this  one was  a  problem  to  Kazakhstan MO , 2000  and  posted  before  in this  forum

[quote=sqing]Let $a,b,c>0$ and  $a^2+b^2+c^2=1$.Prove that\[a+b+c+\frac{1}{15abc} \geq \frac{6\sqrt 3}{5}.\][/quote]

We can use this inequality to prove yours:

$ abc\leq \frac{(ab+bc+ac)^2}{3(a+b+c)}$

Let $a$, $b$, $c$ be positive real numbers such that $ab+bc+ca=1.$  Prove that:
$$(a+b+c)^2 \geq 4\left(\frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1}\right).$$


counting
distinguishability
vid999 used alcumus sol
9 plus 10 equals 21

Alex has 5 pieces of paper, each with a different math problem. In how many ways can he give these problems to his 10 friends (he can give each friend more than one problem)?

I don't see anything wrong with not giving at least 5 friends a problem...
But I think that your answer may be incorrect.  Since each friend could get more than 1 problem, we have $ 10$ choices to give the first problem, $ 10$ choices still to give the second problem, and so on.  So our final answer is \[ 10\cdot10\cdot10\cdot10\cdot10\equal{}\boxed{100000}\]

Well, i think people are distinguishable PowerofPi.  :P

If PowerofPi is correct then it's just a simple sticks and stones problem with $\binom{5+10-1}{5}=\boxed{2002}$ (assuming the friends are distinguishable :P).

If you can give problems to more than one person, then the answer will be $(2^{10}-1)^5 = \boxed{1023^5}.$

For the first paper, there are $10$ friends that the problem can be given to. For the second, again there are $10$ choices, as friends can receive more than one problem. This is true for all $5$ of the papers, so the answer is $10^5=\boxed{10,000}$

For each of Alex's five problems, there are 10 friends he can give it to. Therefore, there are $10^5=\boxed{100,\!000}$ ways for Alex to distribute the problems.

You know what? I am really bad. This is how I solved the problem:
If we give all 5 problems to 1 friend, there are 10 friends to give it to. There are $10$ ways here.

If we give 4 problems to 1 friend and 1 to another, there are 10 choose 2 = 45 ways here. There are 2! ways to order the friends, so it becomes 90. We can choose 1 of the 5 problems to give to the friend with 1 problem, so there are $90*5=450$ ways in total here.

If we give 3 problems to a friend and 2 to another, there are 10 choose 2 = 45 friends to pick the friends. There are 5 choose 2 = 10 ways to choose 2 of the 5 problems to give to the friend with 2 problems. There are $90*10=900$ ways in total here.

...etc, in the end I worked out seven cases.

Adding these up, we have $10+450+900+7200+10800+50400+30240=100000$ cases. (In my head, this is the point where I go "WOW! WHAT A COINCIDENCE!" Then I see the two sentence solution and I'm like o.o

That's how I solved it too, only I didn't get it right. :wallbash: That's quite the problem!

[quote=PowerOfPi]It depends if the friends are distinguishable. If they are, then jxl28 is right. If not, [hide]
5
4-1
3-2
3-1-1
2-2-1
2-1-1-1
1-1-1-1-1
There are 6 possibilities.
[/hide].[/quote]

um people are distinguishable

[quote=DuoDuoling0][quote=myyellowducky82]This doesn't make sense. Wouldn't (at the least) 5 friends not have a problem?
I did $ 10\cdot9\cdot8\cdot7\cdot6$[/quote]

That's actually right

Answer is 30240[/quote]

This is wrong because it says one person can get [size=150][i][u][b]more than one piece of paper.[/b][/u][/i][/size]

Like this comment if those of you, like me, over-thought this problem wayyyy to much!

[quote=myyellowducky82]This doesn't make sense. Wouldn't (at the least) 5 friends not have a problem?
I did $ 10\cdot9\cdot8\cdot7\cdot6$[/quote]

friends can have multiple problems so u just divide the distinguishable problems to the distinguishable friends each problem has 10 choices and there are 5 problems so 10x10x10x10x10=100000. (kind of like the tourists and travelers problem)

I thought that the question stated that he can give one friend more than one problem, like giving 3 out of the five problems to one person. I didn't think that meant that you can split the problem among multiple friends... :(

[quote=PowerOfPi]It depends if the friends are distinguishable. If they are, then jxl28 is right. If not, [hide]
5
4-1
3-2
3-1-1
2-2-1
2-1-1-1
1-1-1-1-1
There are 6 possibilities.
[/hide].[/quote]

Think gamebot said that it has a different math problem on it.

[quote=LeonidasTheConquerer]Why wouldn't stars and bars work for this? Does it have to do with distinguishability?[/quote]

yes, but stars and bars does not work for this problem
the people are distinguishable, but the math problems are distinguishable, not indistinguishable here. So, we can’t use stars and bars here.

[quote=ShrewdBunny][hide=sol]$10^5=100,000$[/hide]

am I right please correct me if I am not[/quote]

[hide= A more explained sol]Since there are ten people and each person can get more than one piece of paper. There are ten options for each slip to who the slip will go to. Therefor, there are: $$10^5=10\cdot10\cdot10\cdot10\cdot10=\boxed{100,000}$$ ways for Alex to give the slips of paper to his friends.[/hide]

[quote=ultraman_aiden]your wright the answer is 100,000[/quote]

Thanks.

[quote]ShrewdBunny wrote thanks[/quote]                  YOUR WELCOME


geometry
circumcircle
perimeter
geometry proposed

P is a point inside acute-angled the triangle ABC. The perpendiculars from P meet the sides BC, CA, AB at D, E, F, respectively. Show that P is the circumcenter iff each of the triangles AEF, BDF, CDE has perimeter not exceeding that of DEF.

This is ISL 2001, G7. See here: [url]http://www.mathlinks.ro/viewtopic.php?t=17468[/url]


Let be $ a,b,c\in \mathbb{N}^*$ such that $ b > a^2 \plus{} c^2$ . If $ x_1,x_2$ are the roots of the equation $ ax^2 \plus{} bx \plus{} c \equal{} 0$ , prove that $ x_1,x_2\in \mathbb{R}\backslash \mathbb{Q}$

I think its not true. Take $ a \equal{} c \equal{} \frac {1}{3}$ and $ b \equal{} \frac {2}{3}$.
edit: :blush: Oh, I'm sorry. I didn't  pay attention to: $ a,b,c \in\mathbb N^*$.

[quote="Ognjen09"]What is $ \mathbb N^*$?[/quote]

$ \mathbb{N}$ is the notation for the set of the natural numbers and $ \mathbb{N}^*$ means the set of the natural numbers not equals with zhero .

You can see [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1483522#1483522]here[/url] the notations for the other sets at post 107#

$ b>a^2\plus{}b^2$ is impossible in naturals

[quote="Bugi"]$ b > a^2 \plus{} b^2$ is impossible in naturals[/quote]

Sorry , my bad again  :wallbash: , i changed  :)

Solutions are real becouse $ b^2\minus{}4ac>4a^2c^2\minus{}4ac>0$.

In order to prove that they are irrational we will prove that $ (b\minus{}1)^2<b^2\minus{}4ac<b^2$ for $ a\neq c$. Right-hand inequality is obvious, lets prove the left one.

$ (b\minus{}1)^2<b^2\minus{}4ac\Leftrightarrow 2b\minus{}4ac\minus{}1>0$

$ 2b\minus{}4ac\minus{}1>2(a^2\plus{}c^2)\minus{}4ac\minus{}1\equal{}2(a\minus{}c)^2\minus{}1>0$ for $ a\neq c$.

If $ a\equal{}c$ its enough to prove that $ \frac{b}{a}>2$, which is true becouse $ \frac{b}{a}>\frac{2a^2}{a}\geq2$.


Once we've submitted our application, can we discuss the qualifying quiz with other people who have already applied?  And if the answer turns out to be yes, is anyone interested?

Right; I mean only with people who have already submitted their apps.

I did find a quote from an old Mathcamp Math Jam and I wonder if it still applies:

[quote]Yasha (19:44:54)
Can we discuss the quiz outside the forum?

ztbb (19:47:06)
Sure. [/quote]

It's certainly okay to discuss the quiz with others who have already applied in non-public forums.  By which I mean, please don't discuss them on your blog, since that's Googleable and we've actually had problems with this in the past.  But if you're just IMing a friend or talking in real life, by all means you're welcome to discuss it.  After all, part of the fun of doing a good math problem is getting to discuss different solutions with friends!

i'm down for some quiz discussion.  here, let's agree on an easy-to-access venue (i hope you all are familiar with irc): irc.efnet.net, #mathcamp - now let's figure out a time.  evenings are always good for me.

edit: now that i think about it, that's a really stupid idea since anyone who hasn't done the quiz yet can just pretend like they have...

better ideas?

[quote]anyone who hasn't done the quiz yet can just pretend like they have[/quote]

Hopefully people who sincerely want to go to MathCamp would have the integrity and intellectual curiosity to solve the problems on their own.

i agree, and i'm certain that if we started an irc room, nothing bad would happen.  the application process, however, exists to filter out such n00blets, and thus there's some level of risk.  i guess i'll just loaf around #mathcamp until someone comes on and hope i'm not subverting the program.

How about a private members-only forum, where you only get to join if you agree to give us the username/password for the application page, so we can confirm you already submitted. It's not like there's anything important there post submission anyways.

Of course, there are 2 problems:

1) People may be unwilling to give out their passwords, especially if they already used them somewhere else
2) Not everyone applied online

Hmm... how about a screenshot of you applying. Then, the only way to cheat would be either to make a copy of the application page (which would be more work than anyone would reasonably want to do) or to create a dummy account and submit nothing, which I doubt people would want to bother with.

(edit: that solves problem 1, of course, but not 2). Maybe Mathcamp wants to create a forum like the one proposed, where you get access only after applying.


How many ordered triples $(x,y,z)$ of integers satisfy the system of equations below?

\[ \begin{array}{l} x^2-3xy+2yz-z^2=31 \\ -x^2+6yz+2z^2=44 \\ x^2+xy+8z^2=100\\ \end{array}  \]
$\text{(A)} \ 0 \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ \text{a finite number greater than 2} \qquad \text{(E)} \ \text{infinately many}$

[hide="Solution"]If we multiply the first equation by $3$, the second by $-1$ and the third by $9$ and add them all up, we get:

$13x^2+67z^2=949 \Leftrightarrow x^2+{67 \over 13}z^2=73$

This means that $z$ is divisible by $13$, so let's put $z=13t$. We get:

$x^2+871t^2=73$.

Since neither $x^2$ nor $t^2$ can be negative, the only possible value for $t^2$ is zero - otherwise the LHS would exceed $73$. But for $t=0$ we have $x^2=73$, which has no integer solution.

Answer: $\boxed{A}$[/hide]


trigonometry
geometry
similar triangles
geometry solved

Show that

$ \angle B =  2\angle A \Rightarrow ac = (a+b)(b-a)$

$ \angle B =  3\angle A \Rightarrow ac^2 = (a+b)(b-a)^2$

$ \angle B =  4\angle A \Rightarrow ab^2c^3 = (b^2 - a^2 + ac)(a^2 - b^2 + ac)^2$

The first one is easy through similar triangles, but it gets a lot messier with the last two. Does anyone have any neat methods? And can this be generalized?

It is my first time to post.
Please bear my poor English. :blush:
I'll prove when  :ang: B=3 :ang: A, ac :^2: =(b+a)(b-a) :^2:
assume that D,E are on AC and  :ang: ABD= :ang: DBE= :ang: EBC= :ang: A
(i wonder how to make a picture.)
extend AC to F ,making BC=CF=a. And since  :ang:  BDC= :ang: DBC=2 :ang: A, BC=DC=a. so  :ang: DBF is a right angle.
in triangle ABD, we have c/(b-a)=sin2 :ang: A/sin :ang: A
so (c/(b-a)) :^2: =(sin2 :ang: A/sin :ang: A) :^2: =4(cos :ang: A) :^2: =2(1+cos2 :ang: A)
because cos2 :ang: A=BD/DF=(b-a)/2a. therefore 2(1+cos2 :ang: A)=2(1+(b-a)/2a)=(b+a)/a
so ac :^2: =(b+a)(b-a) :^2:

Third one:
Extend AB to B' such that BB'=BC.
We have  [tex]\angle AB'C=2\angle A[/tex], make use of the first formula and  [tex]\cos 2x=2\cos^2 x-1[/tex]. Then the result follows.


Did you intend the third $f$ at the end?  Normally, I think this word is spelled with two $f$s at the end.

yes, actually i did intend for the third $f$.

[quote="DaReaper"]How many distinct permutations of the word hufflepufff exist?[/quote]
[hide]$\frac{11!}{2!5!}=\boxed{166320}$[/hide]

For you guys that don't understand what happened, they just took the factorial of all 11 numbers (letters in this case) and divided it by the letters that appear more than one time (because they overlap.)  So because u happens twice you factorial 2, and since there are 5 f's, you factorial 5.


hallo,
solve in positive integers:
$x^{3}-y^{3}=xy+61$.  :)

[hide="hint"]Suppose that $x>y>0$
then $x^{2}+61\geq xy+61\geq x^{3}-y^{3}\geq 3x^{2}-3x+1$so there is finit $x$
the other case is similar [/hide]

It give $x\le 6$, from $x^{3}>61$ we have $x\ge 4$.
Solution is only $x=6,y=5$.


analytic geometry
algebra proposed
algebra

Let $X=\mathbb{Z}^3$ denote the set of all triples $(a,b,c)$ of integers. Define $f: X \to X$ by \[ f(a,b,c) = (a+b+c, ab+bc+ca, abc) .  \]
Find all triples $(a,b,c)$  such that \[ f(f(a,b,c)) = (a,b,c) . \]

By the definition of $f$ and after some easy algebra, we find that $f(f((a,b,c)))=(a,b,c)$ is equivalent to the system

$\begin{eqnarray*}a+b+c+ab+ac+bc+abc &=& a\\(a+b+c)(ab+ac+bc)+abc(a+b+c+ab+ac+bc) &=& b\\ abc(a+b+c)(ab+ac+bc) &=& c$

We'll discuss the first equation, which is equivalent to $(a+1)(b+1)(c+1)=a+1$, and that splits up to two cases: $(b+1)(c+1)=1$ or $a+1=0$

[b]First case.[/b] $(b+1)(c+1)=1$ results in $b=c=0$ or $b=c=-2.$ It's obvious that the triplet $(a,0,0)$ is a solution for any $a\in\mathbb{Z}$, and as for $(a,-2,-2)$, we easily find that $f(f((a,-2,-2)))=(a, -4a^2+8a-16, 16a(a-4)(1-a))$, but the equation $-4a^2+8a-16=-2$ has no real solution.

[b]Second case.[/b] $a+1=0$ gives a triplet $(-1,b,c)$, and we can easily calculate that $f(f((-1,b,c)))=(-1,(b+c-1)(bc-b-c)-bc(bc-1),bc(b+c-bc)(b+c-1))$. Here we'll discuss the third coordinate: $bc(b+c-bc)(b+c-1)=c$ splits up to the following two cases:

$1^\circ\quad$ $c=0$ gives the triplet $(-1,b,0)$, and $f(f((-1,b,0)))=(-1, b(1-b),0)$. The equation $b(1-b)=b$ has the unique solution $b=0$, but the triplet $(-1,0,0)$ is already covered for in the First case. Hence, no new solution here.

$2^\circ\quad$ $b(b+c-bc)(b+c-1)=1$ means that $b=1$ or $b=-1$. For $b=1$ the equation becomes $c=1$ identically, but the triplet $(-1,1,1)$ isn't a solution, since $f(f((-1,1,1)))=(-1,-1,1)$. For $b=-1$ the equation becomes $(2c-1)(c-2)=-1$, which has the unique integer solution $c=1$, and the triplet $(-1,-1,1)$ is a solution as $f(f((-1,-1,1)))=(-1,-1,1).$

[b]Conclusion.[/b] The solution set for the given equation is $\{(a,0,0)|a\in\mathbb{Z}\}\cup\{(-1,-1,1)\}$

set $\alpha=\sum a, \beta=\sum ab, \gamma=abc$\\

so $f(f(a,b,c))=f(\alpha,\beta,\gamma)=\left(\sum \alpha, \sum \alpha \beta, \alpha \beta \gamma \right)$\\

which gives $a+b+c+ab+bc+ca+abc=a \implies (a+1)(b+1)(c+1)=a+1$\\

[color=#f00][b]Case 1:-[/b][/color] $a\neq -1$\\

then $(b+1)(c+1)=1 \implies b=0, c=0$ or $b=-2 , c=-2$\\

[color=#00f][b]Subcase 1.1:-[/b][/color] $b=0, c=0$ \\

then we observe that for any $a\in \mathbb{Z}$ works\\

[color=#00f][b]Subcase 1.2:- [/b][/color]$b=-2 , c=-2$\\

$(a-4)(4-4a)(4a)=-2$ so for $a\in \mathbb{Z}$ we don't get any such a satisfying this equation, hence this case doesn't work \\

[color=#f00][b]Case 2:-[/b][/color] $a=-1$\\

clearly gives $(bc-(b+c))(b+c-1)(-bc)=c$\\

[color=#00f][b]Subcase 2.1:-[/b][/color] $c \neq 0$\\

In this case we get $(bc-(b+c))(b+c-1)(-b)=1$\\

[color=#f0f][b]Subcase 2.1.1:-[/b][/color] $b=1$\\

gives $(-1)(c)(-1)=1 \implies c=1$\\

[color=#f0f][b]Subcase 2.1.2:-[/b][/color] $b=-1$\\

we get $(1-2c)(c-2)=1$ which for $c \in \mathbb{Z}$ gives $c=1$ to be the only possible solution\\

[color=#00f][b]Subcase 2.2:-[/b][/color] $c=0$\\

then this give $(b-1)b=-b$ which on solving gives solution $(-1,0,0)$ as the possible solution.\\

[color=#ff9a00][b]Conclusion:-[/b][/color]The solution set for the given equation is $\{(a,0,0)|a\in\mathbb{Z}\}\cup\{(-1,-1,1)\}$ $\blacksquare$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Given a subset $ T\subset S_{n}$. Let $ G_{T}$ be a graph where {1,2,...,n} is the vertex set an {$ {{\sigma, \tau}|(\sigma, \tau) \in T}$} is the edge set.

Show that if $ T$ is a subgroup of $ S_{n}$, then the connected components of $ G_{T}$ are complete subgraphs.

My idea (I wasn't sure if it is a correct argument):
Choose $ v_{1}, v_{2} \in$ {1,2,3,...,n}, which are connected by a path P in $ G_{T}$.
We must show that these two vertices are connected by an edge.
Suppose there is no edge connecting $ v_{1}$ and $ v_{2}$.
Then isolate them by deleting every edge connecting each of them to the remaining $ n \minus{} 2$ vertices. Now the permutation $ \sigma \in T$ sends  $ v_{1} \to v_{1}$ and $ v_{2} \to v_{2}$. But $ \sigma(v_{1})\sigma(v_{2}) \in T \implies (v_{1}v_{2}) \in T \implies (v_{1}, v_{2})$ is an edge, a contradiction.

That argument makes no sense to me- what do you mean by "isolate them"? You don't get to change the graph.

Just construct an edge. If $ \sigma$ is the transposition $ (v_1,v_2)$ and $ \tau$ is the transposition $ (v_2,v_3)$, $ \sigma\tau\sigma$ is the transposition $ (v_1,v_3)$. That collapses two edges into one; repeat for paths of arbitrary length.

Thanks jmerry. I had a feeling that my argument was nonsense. But thank you for pointing it out.


I the sum of 2 numbers is 2, and the product of these same two numbers is 3, find the sum of the reciprocals of the numbers. The problem should be fairly easy if you do it a certain way.

[quote="math92"][hide]
x+y=2
xy=3
solve the system:
x=3/y
3/y+y=2
y^2/y+3/y=2
y^2+3=2y
y^2-2y+3=0
i gotta go, i'll edit my post when i come back.[/hide][/quote]

there is a MUCH simpler way to solve it!!!

I'll give a hint if no one else  post the solution I'm looking for.

[quote="pianoforte"]hmm...
[hide]
$x+y=2$
$xy=3$
$\frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}=\frac{2}{3}$
 :D [/hide][/quote]

YEAH!!! nice job.. :)  [hide]I like that problem becuz some people start trying to find x and y when they don't really have to.[/hide]

good job pianoforte. :D
i would have never done that. :lol: [/hide]


There are five research labs on Mars. Is it always possible to divide Mars to five connected congruent regions such that each region contains exactly on research lab.
[img]http://i37.tinypic.com/f2iq8g.png[/img]

i remember visiting this question a while back and there was a long discussion going on... why was it deleted?
anyways i tried to come up with a solution to this and would like to see some other approaches..

create an axis of rotational symmetry and assign a top and bottom to the axis where it intersects the sphere. draw a line from top to bottom s.t the line never reverses direction, except perhaps horizontally (*for short distance) create four more copys of this line and rotate them about the axis of symmetry and space them evenly s.t 5 congruent regions are formed, each line belongs to one of the regions (*sometimes only partially). now for a more flexible creation we can consider the surface of the sphere to be composed of infinitesimally thin rings (slice the sphere horizontally from each point of the axis of symmetry) divide each slice/ring into 5 segments of equal length to form 5 pieces of our congruent regions, one of each. instead of using lines to form 5 congruent regions we can take each ring and color the 5 segments a different color (always in same order) and rotate each about the axis s.t each ring touches the ring below and above with at least one point of the same color. this makes sure the regions are connected.

now suppose we have any 5 points on the sphere, we place an axis through it s.t no 2 of the 5 points are on the same horizontal line (same ring) this is easy to do. call the points $ A,B,C,D,E$ and the regions $ 1,2,3,4,5$. attach a different number to each point to tell us the region its in eg: $ A1,B2...$ to each point there corresponds a point in each different region (in the same ring) so we have the points $ A1,...,A5,B1,...,B5,...$ as long as the points with the same letter follow the same order of the numbers and direction (clockwise/anti) around their respective rings then we can create 5 congruent regions that satisfy our goal. all we have to do is insert the intermediate colored rings, and we can easily connect two points which span a distance of half the sphere beacuse the rings are infinitesimally thin.

to show that we can always find an orientation s.t no 2 points are on the same horizontal line (on a surface) we can take $ n$ points and draw all lines connecting any 2 of them. take a point on the surface now and draw all lines through it that are perpendicular to a line we drew between two points. there are only a finite amount of these lines, so we can chose a line through the chosen point s.t no 2 points from our set are on the same horizontal.

one thing that bothers me is what about the points at the top and bottom of the axis of symmetry of the sphere, they do not have 4 corresponding points each...?
and i was also wondering if you can split the surface of the sphere into congruent regions that do not have an axis of symmetry? i think not, but how can this be proved.

   I think not always.Assign axis for Mars like the earth's axis.Also define the north and south pole on Mars.Draw a sufficiently small circle centered at the north pole,put the research lab one by one an half of the circle.And any $2$ of the labs are sufficiently close to each other.Suppose there are $5$ connected congruent region whose union covers the Mars and each one of them contain exactly $1$ lab.Look at the front elevation,from left to right,denote the $5$ labs be $A,B,C,D,E$,draw $2$ vertical lines $l_1,l_2$ each from $A,E$.Call the area between the $2$ lines a strip,call the region covers $A,B,C,D,E$ be $S_1,……,S_5$.We know that $S_1,……,S_5$ pairwise disjoint,so if $S_2$ bent out of the strip at some point on $l_1$ called $P$,either it brokes into parts or not congruent with $S_1$ or have common point with $S_1$,since $A,B$ are not concided on the circle around the pole.So $S_2,S_3,S_4$ all in strip,but $S_1,S_5$ must have portion out of strip,they cannot be congruent.
   Maybe somewhere wrong,oooop


function
complex numbers
real analysis
real analysis unsolved

Let $(c_{n})_{n\in \mathbb{Z}}$ be complex numbers satisfying $|c_{n}|\leq |n|^{-3}$ for all $n\neq 0$.
Show that
\[f(x) : =\sum_{-\infty}^{\infty}c_{n}e^{inx}\]
defines a continuously differentiable function on R.

Hint: use Weierstrass M-test to show uniform convergence, including uniform convergence for the derivatives.


does anyone know when the octm site is going to have the results for 2007, with the rankings of top scoring students and schools? Does anyone remember about how long after the test the results came out last year?

The results don't appear to be on the web-site, but I have results in front of me.  I was the only person to score a 33 on the test (bracketed by 8 32's and 10 34's :huh: ), so I don't need to worry about how they break the tie (not that that matters much at 27th :roll: ), and was top in my school and at the site (by one point for both :| ).  My school (Revere) scored highest at our site and 8th overall :D .

The highest score was a 39 (it would be mathgenius if he were in Ohio :wink: ) and the cut-off for OHMIO was 27.

I have no clue, except that they did not take the OCTM at Akron U (or Copley HS for that matter).

[quote="tjhance"]
who got the 39?[/quote]

Tong Zhan, the same guy that got a 39 last year. I just found out from my teacher that I'm the highest scoring freshman, tied with 2 seniors for 3rd place.

Nice!  O.K., it's official, the best math students are usually AoPSers (not that we didn't know that :wink: ).

By the way, is Tong Zhan an AoPSer?

Yes he is an AOPSer.  He goes to my school and won't tell me his screename.  So he could be sitting right in front of us and we wouldn't know.


inequalities
function
floor function
inequalities unsolved

Given $n$ real number $x_1,x_2,...,x_n\in [0,1]$ . Prove the following inequality
$\sum_{i=1}^n x_i-\sum_{i=1}^n x_ix_{i+1}\le [\frac{n}{2}]$

(I will assume that the indices permute cyclically mod $n$.) The function is linear in each variable and thus attains extrema at the endpoints, so if $\{x_i\}_{i=1}^n$ maximizes $f = \sum_{i=1}^n x_i - x_ix_{i+1}$, we have $x_i \in \{0, 1\} \forall i$. If $\exists k$ such that $x_{k-1} = x_k = 1$, consider the effect of replacing the sequence $\{x_i\}$ by $\{x'_i\}$, defined by $x'_i = x_i, \forall i \ne k, x'_k = 0$. If $x_{k+1} = 1$, the value of $f$ increases by $1$, and if $x_{k+1} = 0$, the value of $f$ does not change. Thus, performing this operation does not decrease the value of $f$, so continue performing it (it must stop eventually) until one reaches the sequence $\{a_i\}_{i=1}^n$. There are no consecutive $1$'s in this sequence, so $\sum_{i=1}^n a_ia_{i+1} = 0$. Thus, we seek the maximum of $\sum_{i=1}^n a_i$, and as there are no consecutive $1$'s, this is $\lfloor \frac{n}{2} \rfloor$. Indeed, this is attained for the sequence $a_{2i} = 1, a_{2i-1} = 0$. If we seek the cases of equality, it is fairly simple to reverse the argument of the process of increasing $f$ to see that for even $n$, the sequence must alternate $1$'s and $0$'s, while for odd $n$, the sequence must alternate $1$'s and $0$'s except for one time, at which point there can be either two neighboring $1$'s or two neighboring $0$'s.


inequalities
search
inequalities proposed

This problem appeared on the 1998 Korean Olympiad: For positive real numbers $ a, b, c$ such that $ a \plus{} b \plus{} c \equal{} abc$ prove

$ \frac {1}{\sqrt {1 \plus{} a^2}} \plus{} \frac {1}{\sqrt {1 \plus{} b^2}}\plus{} \frac {1}{\sqrt {1 \plus{} c^2}} \le \frac {3}{2}.$

So in general if $ x_i$, $ 1\le i\le n$ are positive real numbers and $ \sum x_i \equal{} \prod x_i$ is it true and if so prove:

$ \sum \frac {1}{\sqrt {1 \plus{} x_i^2}} \le \frac {n}{\sqrt {1 \plus{} n^{2/(n \minus{} 1)}}}$.

with the substituation a=tanA,b=tanB,c=tanC ,the first inequality is easy now,

I know, but there is no similar identity (at least that I know) that works for more than three variables.


So we CAN mail our solutions?  I'm a little wary about using all this latex stuff since I don't really know how to.  Maybe I'll figure it out this weekend, but can we mail our solutions if we want?

Yes, you have three choices:
- regular mail: USAMTS, PO Box 2090, Alpine, CA 91903-2090
- fax: (619) 445-2379
- email: solutions@usamts.org

All the details are here:
[url]http://www.artofproblemsolving.com/USAMTS/AoPS_U_AbPart.php[/url]

If you submit by mail or fax, please make sure that your name, your USAMTS ID#, and the round and problem number are on [u]every[/u] sheet of paper that you submit.

After logging in to the forum, go to the USAMTS pages.  Click on My USAMTS.  Your User ID is on both the Scores page and the Profile page.

Yeah, you can switch between styles, and you can send the entry form separately (by fax or snail mail).  Either way, always remember to put your Username and User ID number on [b][u]each page [/u][/b]of your solutions.

[quote="Simfish"]Since my handwriting is rather messy, I have no choice but to Tex my solutions.

Question: As for the entry form, can it be sumitted via fax seperately from the solutions?[/quote]

You don't have to use TeX.  You can always create your solution in MS Word (DON'T e-mail your solutions in that format), and send it in by fax or snail-mail with your printed result (my handwriting is probably messier than yours is; that is what I did in years past).

Isn't using MS Word without doing it in the computer suggesting that this is not really a computer user that just does it?

I mean, I have always sent mine handwritten; it's usually easier that way. Still, It could be a little messy. Sotty if this is offensive.


\[ \mathbb{DAS} \ \mathbb{WEDDER}\]

[u][b]Kreisteilungspolynome[/b][/u]

[b]Definition:[/b]
Eine komplexe Zahl $\zeta$ heißt $n$-te Einheitswurzel, wenn $\zeta ^n =1$ ist. Sie heißt primitive $n$-te Einheitswurzel, wenn alle $n$-ten Einheitswurzeln als Potenzen von $\zeta$ darstellbar sind. Insbesondere ist die Zahl $\zeta_n = e^{\frac{2 \pi i}{n}}$ eine $n$-te Einheitswurzel.

[b]Lemma 1:[/b]
Die $n$-ten Einheitswurzeln sind genau die Zahlen $\zeta_n^k = e^{\frac{2 k \pi i}{n}}$ mit $k \in \{ 0, 1, 2, ... , n-1 \}$. Darunter sind genau diejenigen primitiv, bei denen $k$ zu $n$ teilerfremd ist.

Beweis: Die erste Aussage ergibt sich trivial aus der Polarkoordinatenschreibweise komplexer Zahlen.
Offensichtlich ist aber $\zeta _n$ primitiv und es gilt:
$\zeta = \zeta _n ^k$ primitiv $\Leftrightarrow $ $\exists a \in\mathbb{Z}$ mit $\zeta _n ^{k\cdot a} = \zeta ^a = \zeta _n$ $\Leftrightarrow $ $\exists a \in\mathbb{Z}$ mit $a\cdot k \equiv 1 \mod n$ $\Leftrightarrow $ $ggT (k,n)=1$

[b]Korollar:[/b]
Es gibt genau $n$ verschiedene $n$-te Einheitswurzeln und darunter $\phi (n)$ primitive

[b]Definition:[/b]
Das Polynom $\Phi_n (x) := \prod\limits_{\substack{\zeta ^n =1  \zeta primitiv}} (x- \zeta )$ heiße $n$-tes Kreisteilungspolynom.

[b]Lemma 2:[/b]
$x^n - 1 = \prod\limits_{d \mid n} \Phi_d (x)$

Beweis: jede Einheitswurzel $\zeta$ ist für genau ein $d \mid n$ (das kleinste $d$ mit $\zeta ^d = 1$, dieses $d$ wird auch die 'Ordnung' von $\zeta$ genannt) eine primitive Einheitswurzel (Beweis der Teilbarkeitsaussage: ähnlich Lemma 11). Also haben beide Polynome die selben Nullstellen und sind aufgrund des selben Leitkoeffizienten also identisch.

[b]Lemma 3:[/b]
Die Koeffizienten von $\Phi_n$ sind ganzzahlig

Beweis (Induktion nach $n$):
$n=1$: trivial
$n>1$: nach Induktionsvoraussetzung kann das Polynom $p(x) := \prod\limits_{\substack{d \mid n  d \neq n}} \Phi _d (x) $ als Produkt ganzzahliger Polynome selbst als ganzzahlig angenommen werden. Da aber $x^n - 1 = \Phi _n (x) \cdot p(x)$ ist (Lemma 2), besitzt $\Phi_n$ (wie man anhand der Polynomdivision sieht) ganze Koeffizienten.

[u][b]Klassenformel[/b][/u]

[b]Definition:[/b]
Ist $G$ ein Gruppe (mit neutralem Element $e$) und $M$ eine Menge, so nennt man eine Verknüpfung $G \times M \rightarrow M, (g,m) \mapsto g \cdot m$ die zusätzlich noch $e\cdot m =m$ und $x\cdot (y\cdot m ) = (xy)\cdot m $ für all $x,y \in G, m \in M$ erfüllt, eine Operation (von $G$ auf $M$).
Im folgenden operiere immer die Gruppe $G$ auf der Menge $M$.

[b]Definition:[/b]
Man nennt $G\cdot m:= \{ x\cdot m | x \in G\}$ die Bahn / den Orbit und $G_m := \{ x \in G | x\cdot m =m \}$ die Fixgruppe / den Stabilisator von $m \in M$.

[b]Lemma 4:[/b]
Seien $s,t \in M$. Die Bahnen $G \cdot s$ und $G \cdot t$ sind entweder identisch oder disjunkt.

Beweis:
Seien die beiden Bahnen nicht disjunkt, es gibt also $x,y \in G$ mit $x\cdot s = y\cdot t \Rightarrow s = e \cdot s = (x^{-1} x) \cdot s = x^{-1} \cdot (x \cdot s) = x^{-1} \cdot (y \cdot t) = (x^{-1}y) \cdot t \Rightarrow s \in G \cdot t$. Nun ist $G\cdot s \subseteq G\cdot t$ und genauso zeigt sich auch $G\cdot t \subseteq G\cdot s$ woraus unmittelbar $G\cdot s = G \cdot t$ folgt, also Gleichheit der Bahnen.

[b]Korollar:[/b]
$M$ wird durch die Operation von $G$ in Bahnen zerlegt oder mit anderen Worten: es gibt $s_i$, so dass $M$ disjunkte Vereinigung der Bahnen $G\cdot s_i$ ist.

[b]Lemma 5:[/b]
Für beliebiges $s \in M$ gilt: $|G \cdot s| \cdot |G_s| = |G|$ (diese Gleichung bleibt auch für unendliches $G$ gültig).

Beweis:
Für jedes $t \in G \cdot s$ wähle man ein $x_t \in G$ mit $t = x_t \cdot s $. Dann ist die Abbildung $\kappa : (G \cdot s) \times G_s \rightarrow G , (t , w) \mapsto x_t w $ bijektiv:
injektiv: $x_t a = x_u b \Rightarrow x_t a \cdot s = x_u b \cdot s \Rightarrow x_t  \cdot s = x_u \cdot s \Rightarrow t=u \Rightarrow x_t = x_u \Rightarrow a=b $
surjektiv: sei $y \in G$: $ \Rightarrow x_{y\cdot s}^{-1} \cdot (y \cdot s) = x_{y\cdot s} ^ {-1} \cdot (x_{y\cdot s} \cdot s) = s \Rightarrow x_{y\cdot s}^{-1}y \in G_s \Rightarrow \kappa (x_{y\cdot s} , x_{y\cdot s}^{-1} y ) = x_{y\cdot s} x_{y\cdot s}^{-1} y =y $

[b]Korollar:[/b]
Ist $M$ endlich, seien $B_1 , B_2 , ... , B_k$ die verschiedenen Bahnen und $m_i \in B_i$ 'Repräsentanten' dieser Bahnen, so gilt die Bahnenformel: $|M| = \sum\limits_{i=1}^k |B_i| = \sum\limits_{i=1}^k |G\cdot m_i| = \sum\limits_{i=1}^k \frac{|G|}{|G_{m_i}|}$

[b]Definition:[/b]
Setzt man nun $G=M$ und nimmt als Operation die Konjugation $x\cdot y := x y x^{-1}$ (die Operationseigenschaften können leicht nachgeprüft werden), so nennt man die Fixgruppe von $y \in G$ hier den Zentralisator $C_y = G_y = \{ x \in G | xy=yx \}$. Desweiteren nennt man die Menge $Z:= \{ x \in G | xy=yx \forall y \in G \}$ das Zentrum von $G$ (welches dann genau der Durchschnitt aller Zentralisatoren ist).

[b]Satz:[/b]
(Klassengleichung) Seien $x_1 , x_2 , ... , x_k \in G$ Repräsentanten der verschiedenen, durch die Konjugation gegebenen, Bahnen mit mehr als einem Element (!), so gilt: $|G| = |Z| + \sum\limits_{i=1}^k \frac{|G|}{|C_{m_i}|}$.

Beweis:
Da im Zentrum genau die Elemente aus $G$ liegen, deren Bahn unter der Konjugation die Länge $1$ besitzt (nach Definition), bzw. deren Zentralistoren die gesamte Gruppe ausmachen, ergibt sich die Klassengleichung direkt aus der Bahnengleichung.

[u][b]Der Satz von Wedderburn[/b][/u]

[b]Definition:[/b]
Bei einem Schiefkörper handelt es sich um eine Menge mit zwei Verknüpfungen ($+$, $\cdot$), von der nur gefordert wird, die Körperaxiome bis auf die Kommutativität der Multiplikation zu erfüllen. Diese darf natürlich trotzdem gelten.

[b]Definition:[/b]
Wie im Falle von Gruppen definiert man in einem Schiefkörper $S$ zu einem $y \in S$ den Zentralisator $C_y := \{ x \in S | xy=yx \}$ sowie das Zentrum $Z:= \{ x\in S | xy=yx \forall y \in S\}$.

[b]Lemma 6:[/b]
Die Zentralisatoren eines Schiefkörpers $S$ bilden mit den von $S$ 'geerbten' Verknüpfungen wieder einen Schiefkörper. Entsprechend bildet das Zentrum sogar einen Körper(ist also insbesondere kommutativ).

Beweis: nachrechnen (Übungsaufgabe)

[b]Lemma 7:[/b]
Ist $K$ ein Körper, enthalten in einem endlichen Schiefkörper $L$, so ist $|L|$ eine Potenz von $|K|$.

Beweis: Zuerst ist $K$ als Teilmenge einer endlichen Menge selbst endlich. Man kann aber $L$ als Vektorraum über $K$ auffassen. Dieser ist natürlich endlichdimensional mit Dimension $n$ und es folgt (man wähle eine Basis von $L$ aus und betrachte dann die Zahl der $n$-Tupel von Elementen aus $K$) $|L|=|K|^n$.

[b]Lemma 8:[/b]
Ist $S$ ein endlicher Schiefkörper, $x\in S$ und $Z$ dessen Zentrum, so gibt es $n, n_x \in \mathbb{N}$ mit $|S|=|Z|^n$ und $|C_x|=|Z|^{n_x}$.

Beweis: Das Zentrum ist als Unterkörper in diesen (natürlich endlichen) Schiefkörpern enthalten, so dass die Existenz ganzzahliger Exponenten direkt aus Lemma 7 folgt.

[b]Lemma 9:[/b]
für $a \in \mathbb{N}\backslash \{ 1\}$ und $m,n \in \mathbb{N}$ gilt: $ggT( a^m-1 , a^n-1 ) = a^{ggT(m,n)}-1$. Insbesondere gilt: $a^n-1|a^m-1 \Leftrightarrow n|m$.

Beweis: ObdA sei $n\leq m$. Da für $k \in \mathbb{N}_0$ ja $a^k$ und $a^m-1$ sicher teilerfremd sind, gilt $ggT( a^m-1 , a^n-1 ) = ggT( a^m-1 , a^{m-n}\cdot(a^n-1 ) ) = ggT( a^m-1 , a^m-a^{m-n} ) = $
$=ggT( a^m-1 , a^m-a^{m-n}-a^m+1 ) = ggT( a^m-1 , -a^{m-n}+1 ) = ggT( a^m-1 , a^{m-n}-1 )$, so dass man einen euklidischen Algorithmus 'in den Exponenten' durchführen kann.

[b]Satz:[/b]
(Wedderburn, 1905): Jeder endliche Schiefkörper ist kommutativ

Beweis: Sei $z:=|Z|$ ($\geq 2$, da mindestens $0$ und $1$ enthalten sind). Die Klassenformel bezüglich der Konjugationsoperation von $S'$ ($:=S\backslash \{0\}$) auf sich liefert bei Wahl eines Repräsentantensystems $x_i$ der nicht einelementigen Bahnen: $|S'| = |Z'| + \sum\limits_{i=1}^k \frac{|S'|}{|C_{x_i}'|}$ bzw. (Lemma 8, mit $n_i := n_{x_i}$) $z^n - 1 = z-1 + \sum\limits_{i=1}^k \frac{z^n - 1}{z^{n_i} - 1}$. Nun ist jeder der Summanden aber ganzzahlig (man beachte die Herleitung der Klassenformel), also auch $\frac{z^n - 1}{z^{n_i} - 1}$, so dass also $z^{n_i} - 1$ nun $z^n - 1$ und damit nach Lemma 9 $n_i$ auch $n$ teilt. Aber nun teilt $\Phi_n(z)$ sowohl $z^n - 1 = \Phi_n(z) \cdot \prod\limits_{\substack{d \mid n , d \neq n }} \Phi_d (x)$ als auch (man beachte $n_i<n$, da die von $x_i$ erzeugte Bahn nicht einelementig ist) $\frac{z^n - 1}{z^{n_i} - 1} = \Phi_n(z) \cdot \prod\limits_{\substack{d \mid n , d \nmid n_i , d \neq n }} \Phi_d (x)$ und infolgedessen auch $z-1$. Dies ist für $n>1$ aber absurd, da dann (wie eine einfache Rechnung zeigt) gelten würde: $|\Phi_n(z)| = |\prod\limits_{\substack{\zeta ^n =1,\;  \zeta\text{ primitiv}}} (z- \zeta )| = \underbrace{|(z-\zeta_n)|}_{> z-1} \prod\limits_{\substack{\zeta ^n =1,\;  \zeta \neq \zeta_n \text{ primitiv}}} \underbrace{|(z- \zeta )|}_{\geq 1} > z-1$.
Somit ist $n=1$ und damit $S=Z$, was nichts anderes bedeutet, als dass $S$ ein echter Körper ist.

[u][b]Der Dirichletsche Primzahlsatz[/b][/u]

[b]Lemma 10:[/b]
Das Polynom $x^n - 1$ hat modulo der Primzahl $p$ betrachtet nur dann doppelte Nullstellen, wenn $p|n$.

Beweis: Man kann ein Polynom $a(x) = a_n x^n + ... + a_1 x + a_0 $ mit Koeffizienten in einem beliebigen Körper 'formal' ableiten (dann wäre also $a'(x) = n a_n x^{n-1} + ... + a_1$), wobei dann auch die Summen- und Produktregeln gelten (Beweis durch Nachrechnen). Ist also $z$ Nullstelle, so ist $a$ faktorisierbar als $a(x)=(x-z)b(x)$ und es folgt $a'(x) = b(x) + (x-z)b'(x)$, so dass $a'(x)$ genau dann $z$ als Nullstelle hat, wenn dies auf $b(x)$ zutrifft. Nun ist aber die so gebildete Ableitung von $a(x)=x^n-1$ ja genau $n x^{n-1}$, was modulo $p$ nur für $z=0$ oder $p|n$ gleich Null wird, wobai aber $0$ sicher nicht Nullstelle von $a(x)$ ist.

[b]Definition:[/b]
Ist $p$ Primzahl sowie $a$ ganzzahlig und nicht durch $p$ teilbar, so nennt man die kleinste natürliche Zahl $o=ord_p(a)$ mit $a^o \equiv 1 \mod p$ die Ordnung von $a$ modulo $p$. (eine solche existiert nach dem kleinen Satz von Fermat immer)

[b]Lemma 11:[/b]
Ist die ganze Zahl $a$ nicht durch die Primzahl $p$ teilbar und erfüllt die natürliche Zahl $k$ nun $a^k \equiv 1 \mod p$, so gilt $ord_p(a) | k$.

Beweis:
Division mit Rest von $k$ durch $o=ord_p(a)$ ergibt $k=qo+r$ mit $0\leq r <o$. Aber wäre $r > 0$, so würde $1 \equiv a^k = a^{qo+r} = (a^o)^q \cdot a^r \equiv a^r \mod p$ gelten, was der Minimalität von $o$ wiederspräche. Also ist $r=0$ und $o$ ein Teiler von $k$.

[b]Lemma 12:[/b]
Für beliebige natürliche $k$ und $n$ gilt: jeder Primteiler $p$ von $\Phi_n(k)$ teilt entweder $n$ oder ist $\equiv 1 \mod n$.

Beweis:
Sei $p$ ein Primteiler von $\Phi_n(k)$, der nicht $n$ teilt. Dann gilt $p|\Phi_n(k) | k^n - 1 $ und somit $k^n \equiv 1 \mod p$ (insbesondere ist $k$ zu $p$ teilerfremd). Wäre nun $o=ord_p(k)|n$ (nach Lemma 11) ungleich $n$ selbst, so würde das Polynom $x^n - 1 = \Phi_n(x) \cdot (x^o-1) \cdot \prod\limits_{d \mid n , d\nmid o , d \neq n} \Phi_d(x)$ modulo $p$ die doppelte Nullstelle $k$ haben, was Lemma 10 widerspricht. Also ist $o=n$ und es gilt somit (da ja $k^{p-1} \equiv 1 \mod p$): $n|p-1 \Leftrightarrow p \equiv 1 \mod n$.

[b]Satz:[/b]
(Dirichlet, 1837): Für jede natürliche Zahl $n$ gibt es unendlich viele Primzahlen kongruent $1 \mod n$.

Beweis:
Angenommen, es gäbe nur endlich viele. Sei dann $r$ deren Produkt und $y$ eine ganze Zahl. Da $\Phi_{nr}(x)$ als Polynom in $x$ nicht konstant ist, geht mit $y$ auch $|\Phi_{nr}(nry)|$ gegen unendlich (und nimmt nur ganzzahlige Werte an), besitzt also für groß genuges $y$ einen Primteiler $p$. Da dieser ja $p|\Phi_{nr}(nry)|(nry)^{nr}-1$ erfüllt, ist $p$ kein Teiler von $rn$ und erfüllt also nach Lemma 12 nun $p \equiv 1 \mod (nr) \Rightarrow p \equiv 1 \mod n$, so dass eine 'neue' Primzahl aus dieser Restklasse gefunden wäre. Es gibt also unendlich viele Primzahlen der geforderten Art.

[b]Verallgemeinerung:[/b]
Allgemein liegen in jeder zu $n$ primen Restklasse unendlich viele Primzahlen (sogar 'gleichmäßig' verteilt).

[u][b]Aufgaben[/b][/u]

[b]Aufgabe:[/b]
Zeige (mit Lemma 2): $n= \sum\limits_{d|n} \varphi(d)$

[b]Aufgabe:[/b]
Zeige für prime $p,p_1,p_2,...,p_k$ und natürliches $n$:

 $n$ ungerade $\Rightarrow \Phi_{2n} (x) = \Phi_n (-x)$
 $p \mid n \Rightarrow \Phi_{pn} (x) = \Phi_n(x^p)$
 $p \nmid n \Rightarrow \Phi_{pn} (x) \cdot \Phi_n (x)= \Phi_n(x^p)$

[b]Aufgabe:[/b]
Es ist $\Phi_n(x) >0$ für alle reellen $x$ und $n\geq 3$

[b]Aufgabe:[/b]
Bestimme die Werte von $\Phi_n(0)$, $\Phi_n'(0)$, $\Phi_n(1)$, $\Phi_n(-1)$

[b]Aufgabe:[/b]
Zeige für $n\geq 2$: $\Phi_n'(1) = \frac{\varphi (n)}{2} \Phi_n(1)$, $\Phi_n'(-1) =- \frac{\varphi (n)}{2} \Phi_n(-1)$

[b]Aufgabe:[/b]
$\Phi_n(x) \neq 0 $ für alle $x\in \mathbb{C},|x|=1$

[b]Aufgabe:[/b]
Sei das Polynom $\Phi_{-1}(x) := x$ hier auch als 'Kreisteilungspolynom' aufgefasst. Finde alle $n$, so dass $\Phi_n (x) - 1$ als Produkt von 'Kreisteilungspolynomen' geschrieben werden kann.

[b]Aufgabe:[/b]
Sind $a,b$ rational, so ist $z=a+bi$ nur Einheitswurzel wenn $z \in \{ 1 ; -1 ; i ; -i \}$

[b]Aufgabe:[/b]
Zu jedem natürlichen $m$ gibt es ein Kreisteilungspolynom $\Phi_n(x)$, so dass $m$ als Koeffizient darin vorkommt.

[b]Aufgabe:[/b]
Die Kreisteilungspolynome sind irreduzibel, das heißt: schreibt man $\Phi_n(x) = a(x)b(x)$ als Produkt rationaler Polynome, so ist einer der Faktoren ein konstantes Polynom.

teuer ZetaX

wie konnen ich finden diese in englisch?

danke

correction for grammar:  Wo kann ich in englisch dass finden?


limit
number theory
greatest common divisor
prime factorization

Let $a$ and $b$ be positive integers such that $a|b^{2}, b^{2}|a^{3}, a^{3}|b^{4}, b^{4}| a^{5}, ...$

Prove that $a=b$.

Clearly every prime in the prime factorization of $a$ is also in the prime factorization of $b$, and vice-versa. We must show that each prime appears the same number of times in each factorization.

Suppose a prime power $p^{k}$ divides $a$ (though $p^{k+1}$ does not divide $a$), but $p^{m}$ divides $b$ (though $p^{m+1}$ does not divide $b$).  Let $\frac{m}{k}=t$

By the given condition, $k \leq 2m \leq 3k \leq 4m \leq 5k...$.

Considering each relation in turn, and dividing through by $m$ or $k$:
$t \geq \frac{1}{2}$, $t \leq \frac{3}{2}$, $t \geq \frac{3}{4}$, $t \leq \frac{4}{3}$ $...$

It is clear that since these conditions go on, no value of $t$ is possible except $1$, and so $m=k$.

Hence each prime divides $a$ and $b$ the same number of times; they are equal.

Has been posted before, [hide="here is a bit simpler solution..."] In fact, we can replace $|$ by $\le$, a weaker condition.

Consider $r = \frac{a}{b}$.  We are given

$a^{k}\le b^{k+1}\implies r^{k}\le b$
$b^{k}\le a^{k+1}\implies \frac{1}{a}\le r^{k}$

Put together,

$\frac{1}{a}\le r^{k}\le b$

Obviously we can only have $r = 1$.  Otherwise, consider $\lim_{k \to \infty}r^{k}$.  For $r < 1$ it tends to $0 < \frac{1}{a}$ and for $r > 1$ it tends to $\infty > b$. [/hide]

let $\frac{a}{b}=t=\frac{p}{q}$ with $(p,q)=1$, looking at the seqence, we have that $\frac{p^{k}b}{q^{k}}$, let $q_{0}$ be a prime factor of $q$, due to the gcd condition, $q_{0}^{k}|b$ forall $k$, but that means $b$ is not finite, contradiction, so $q=1$, likewise, looking at the other half of the sequence, $p=1$, so $\frac{a}{b}=1\implies a=b$

Sorry to be annoying, but what is the | notation that you are using?

It means "is a divisor of"

Like 2|4 and 13|39


geometry
trapezoid
symmetry
LaTeX
cyclic quadrilateral

Problem 3

Given a circle with ceneter $O$ and an inscribed trapezium $AB\Gamma\Delta$ $(AB\parallel\Gamma\Delta)$ with $AB<\Gamma\Delta$. If $P$ is a point of arc $\Gamma\Delta$ on which $A$ and $B$ do not belong to, and $P_{1},P_{2},P_{3}$ and $P_{4}$ are the projections of $R$ on the lines $A\Delta$, $B\Gamma$, $B\Delta$, and $A\Gamma$ respectively, show that:

(i) The circumscribed circles of the triangles $PP_{1}\Delta$ and $PP_{2}\Gamma$ intersect on the side $\Gamma\Delta$.
(ii) The points $P_{1},P_{2},P_{3}$ and $P_{4}$ are concyclic.
(iii) If $AB=\alpha$, $\Gamma\Delta=\beta$ and the distance between the parallel chords is $h$ find all the points of the axis of symmetry of the trapezium $AB\Gamma\Delta$ that can "see"* at right angle the non parallel sides and calculate their distance form $AB$ and $\Gamma\Delta$ in terms of $\alpha$, $\beta$ and $h$.
Examinate if  such pints always exist.
(Draw a separate diagram for part (iii)).

[size=75]
*An example what I mean: In any cyclic quadrilateral $ABCD$ the point $A$ and $B$ "see" the side $CD$ at the same angle.[/size]

[quote="neelnal"]I would put a solution, but how do I use Latex? Do I have to download it? Please help.[/quote]

See [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]Latex Help[/url]
You can write latex in the forum by puting $\$$ at the beginning and the ending, eg. if you write $\$$\Gamma \Delta$\$$ it will show $\Gamma\Delta$.

Given a circle with ceneter $O$ and an inscribed trapezium $AB\Gamma\Delta$ $(AB\parallel\Gamma\Delta)$ with $AB<\Gamma\Delta$. If $P$ is a point of arc $\Gamma\Delta$ on which $A$ and $B$ do not belong to, and $P_{1},P_{2},P_{3}$ and $P_{4}$ are the projections of $R$ on the lines $A\Delta$, $B\Gamma$, $B\Delta$, and $A\Gamma$ respectively, show that:
(i) The circumscribed circles of the triangles $PP_{1}\Delta$ and $PP_{2}\Gamma$ intersect on the side $\Gamma\Delta$.
(ii) The points $P_{1},P_{2},P_{3}$ and $P_{4}$ are concyclic.
(iii) If $AB=\alpha$, $\Gamma\Delta=\beta$ and the distance between the parallel chords is $h$ find all the points of the axis of symmetry of the trapezium $AB\Gamma\Delta$ that can "see"  at right angle the non parallel sides and calculate their distance from $AB$ and $\Gamma\Delta$ in terms of $\alpha$, $\beta$ and $h$.
Examinate if  such points always exist. (Draw a separate diagram for part (iii)).


USAMTS search trigonometry AMC AIME geometry USA(J)MO

As everyone should know by now, USAMTS is starting now. I was just interested in knowing how many of you guys are doing it this year. Good Luck to all of you.

It stands for USA Math Talent Search. If you aren't in the US (which you aren't), you can do the International Math Talent Search, which has the same questions. It's basically an olympiad that you do at home, and you have about a month to work on the problems and write up solutions and then submit them.

im doing usamts this year. for you guys who have done it for the last few years, are the questions in this first round easier than usual? they seemed pretty easy to me.

[quote]The first round does tend to be a bit easier than the other three.
[/quote]

Uh oh...

It says on my letter which says "open to all United States high school students"
a.) Technically, am I still a middle schooler even if I take math over at a high school?
b.) If so, am I still eligible?

I'm asking this here since the USAMTS website isn't working.

Yes Tare. I did it for my last 2 yrs of middle school. Although u'll probly find that there's some thiings on there u havent learned yet (trig for example) but hey...its a good learning experience.

When something says it's open to all high school students, that generally means that older kids can't take it, not that younger ones can't.

I can't go this time I'm do young :cry: .

how old are u A+?...and u should be able to...

I believe there's a page on the site (accessible by clicking the word "high schoolers") that says that younger students are encouraged to participate as well, although the questions are geared towards high school students.

In general, you can participate in things above but not below your level.

They mailed the letter to me [b][i][u]15[/u][/i][/b] days late!  I only have like 10 days to do it now..  Wait, if you didn't finish the 1st round, can you still do the others? :lol:

[color=cyan]Yes, you can.  And you don't need to send in all 5 solutions -- you can just send as many as you get.[/color]

So are the prizes generally good?  :lol:

the prizes are mostly books and stuff...

and i believe u can qualify for AIME or something thru USAMTS...i forget but its on the site too

and yes, u can do the other rounds without doing the first...i probly wont be doing the first 2 rounds this yr b/c debate is so demanding right now...

how hard is it to get a commendation on a problem?

[quote="MithsApprentice"]and i believe u can qualify for AIME or something thru USAMTS...i forget but its on the site too[/quote]
Correct.
You need at least two solutions.
Is there a prize?

i only did it last year.  There are book prizes if you are a winner (bronze ... grand prize winner).

I was only commended for one solution, and i wrote it up at ~noon the morning that I had to send the solutions in to the post office.  I wasnt sure how to do the problem, but I decided that i'd just try it and see what happened.  I figured out the problem, but I bet it wasnt very well-written or worthy of commendation.  This was the only solution that I was commended for, but I felt that I had others that were much stronger yet were not commended.

[quote] I figured out the problem, but I bet it wasnt very well-written or worthy of commendation. This was the only solution that I was commended for, but I felt that I had others that were much stronger yet were not commended.[/quote]

Perhaps your other solutions were to easy problems for which many other students also had excellent solutions, and the solution you got commended on was for a difficult problem for which few students also had good solutions.

maybe, but i still believe it to be pretty random...

Or maybe they just pick the one that attracted their attention the most...maybe I should type mine up in bold and italics and see what happens :)

[quote="cats..."]Andover was....one of the best at math in MA, along with Lexington.  quote]

Lexington ALWAYS outperforms Andover by a big margin. How come Andover is along with Lexington?

[quote="Tare"][quote="MithsApprentice"]and i believe u can qualify for AIME or something thru USAMTS...i forget but its on the site too[/quote]
Correct.
[/quote]

While it's true that you can qualify for the AIME through USAMTS, I wouldn't recommend that you plan on qualifying for the AIME solely through it.  That is to say, if you want to qualify for the AIME, your best bet is the AMC competitions.

Why?  Because if you qualify for the AIME through the USAMTS, it's as if you got a 100 on the AMC12.  That means that you'd need between a 13 and a 15 (around there) to qualify for the USAMO.  By any standards, that is a difficult feat.  By contrast, if you got a 140 on the AMC10/12, you would need to get 4 fewer questions right on the AIME, which is much more manageable.

There are numerous discussions on the USAMTS --> AIME qualifying scheme, some on the http://www.mathcounts.org forum.

[quote="Tare"]Oh, olympiad is this weird test that we take that's much harder than AMC. It might have something to do with the IMO, but I'm not sure.
I can't even ask the math dept. since the head retired and the system/competitions might change. I'll ask my former teachers though.[/quote]

Math Olympiad is (confusingly) the official name of a contest for Elementary and Middle Schoolers.  See [url]http://www.moems.org/[/url] for information.  I'm not sure if this is what Tare is referring to, but it's worth a try...

As for any other "Olympiad" 's relation to the IMO, I doubt it.  The only olympiad in America that is related to the IMO is the USAMO.  (Correct me if I'm wrong here).

Why the word "olympiad" is confusing?  Olympiad (with respect to math, at least) generally refers to problems involving proofs.  From the looks of the problems on the site (for instance, "It is now the month of November.  What month will it be 100 months from now?"  - and you get 4 minutes to answer that!) the problems are [i]not[/i] proofs.

And to cats' post from a while back, they sent out invitations to students who did well on the AMC, I believe.  (I got one and the only contests I participated in were mathcounts (up through nationals) and AMC10A/12B).

Nope, I'm talking about the MAML olympiad for high-schoolers.

...why is there order form for calculus books on this website? :)

Why couldn't they have just called them "Olympics" instead? We changed the name to "International Math Olympics" when we translated IMO into Japanese :)

I don't know why they didn't call them "Olympics".  You should write them a nasty e-mail explaining that their questions are [i]definitely[/i] not true olympiad problems, because that involves proofs, and that you are (just pretend this) a very gifted fourth-grader who was expecting proof problems and now you want your money back.  </jk>


limit
trigonometry
geometry
calculus
congruent triangles
calculus computations

Discuss the convergence/divergence of the following sequence and find the limit of the same in case it's convergent.

 x_n = nSin(1/n) where n belongs to [b]N[/b]

here is a good thing to look at...this is similar to your idea...
$\lim_{n-> \inf}    x * \sin ( \frac{180}{n} )$
your thing converges to $\frac{\pi}{180}$

i suppose the better question is: Why does this happen?
this is actually a problem that i have given to my friends and it stumpted them

hint
[hide]think of a cyclic polygon with n sides[/hide]

[quote="Altheman"]solution

[hide]if you think of a polygon inscribed in a unit circle, a polygon with n sides can be found by
$\frac{n * \sin \frac{360}{n} * 1^{2}}{2}$

this is because drawing the radii, divides the polygon into n congruent triangles, with sides of $1$ and since $360$ total degrees, included angle is $\frac{360}{n}$

so if you have infinite sides to the polygon, it becomes a circle, and we know that a unit circle has an area of $\pi$

then let x = 360n, you have
$180* x* \sin \frac{1}{x} \approx \pi$ when $\lim_{x-> \infty }$
or
$x* \sin \frac{1}{x} \approx \frac{\pi}{180}$ when $\lim_{x-> \infty }$
[/hide][/quote]

Guys..pl. note that here we r dealing with RADIANS only and not degrees

oh well, it converges to 1 then...i was wondering if i was doing something wrong when i did the limit

$\lim_{x -> \infty}  \frac {\sin ( \frac{1}{x} )}{\frac{1}{x}}$

using l'hopital's rule...not supposed to really but oh well

$\frac{(\frac{-1}{x^2}) * \cos (\frac{1}{x})}{(\frac{-1}{x^2})}$ or

$\cos {(\frac{1}{x})}$ which obviously is $1$

but if it is in degrees, the limit would be $\frac{\pi}{180}$, even though the conversion would make it go the opposite...i dont really see why

[quote="metru"]$\mathop {\lim }\limits_{n \to  + \infty } \left( {n\sin \frac{1}{n}} \right) = \mathop {\lim }\limits_{n \to  + \infty } \frac{{\sin \frac{1}{n}}}{{\frac{1}{n}}} = 0$[/quote]

The ans. is 1 for lim sin(x)/x as x->0 = 1...Here we are trying to find why the limit is 1

 We can prove , by some analytical arguments ( may be by using calculus) that the fn.  sin(x)/(x) is monotonically decreasing

 Therefore, as 1/(n+1)  < 1/n for any natural n,

               sin(1/(n+1))/(1/(n+1)) > sin(1/(n))/(1/(n))

             => (n+1)Sin(1/(n+1))  >  nSin(1/n)

              => xn+1 > xn

So, the sequence is INCREASING.

 Now, let's try to prove that it's bounded.

Indeed  nSin(1/n) > 0 as     0< 1/n <= 1 < pi/2 where sin(x) is always +ve

 Also nSin(1/n) < n(1/n) = 1  because in (0,1), sin(x) < x

So  , we conclude that         0 <  xn < 1 and INCREASING.

So   , we have  0 < x1 < x2 < x3 <....<xn < 1

Now ,using a lil intution, it's not hard to see that xn converges to 1 as n-> infinity.

Actually the above fact is confirmed by Bolzano-Weirstraas theorem which says that a [b]monotonic [/b](increasing or decreasing) and [b]bounded [/b]sequence is convergent

Does this infinite product converge?

$\prod_{n=1}^{\infty}n\sin\left(\frac1n\right)$

We use the following Test:
If $a_n>0(<0)$ and $\sum a_n$ is convergent, then $\prod (1+a_n)$ is convergent.

Let $a_n = n\sin\left(\frac1n\right)-1$. It is clear that $a_n < 0$. And $\sum_{n=1}^{\infty} a_n$ is convergent, because $n\sin\left(\frac1n\right)-1 \sim n\left(\frac{1}{n}-\frac{1}{6n^3} + o\left(\frac{1}{n}\right)\right)-1 = -\frac{1}{6n^2}+o(1)$.

Now we conclude that $\prod_{n=1}^{\infty} n\sin\left(\frac1n\right)$ is convergent from the Test.

[quote="liyi"] $n\sin\left(\frac1n\right)-1 \sim n\left(\frac{1}{n}-\frac{1}{6n^3} + o\left(\frac{1}{n^3}\right)\right)-1 = -\frac{1}{6n^2}+o(\frac{1}{n^2})$ because with $o(1)$ it's not convergent  :huh:

[quote="metru"]$\mathop {\lim }\limits_{n \to  + \infty } \left( {n\sin \frac{1}{n}} \right) = \mathop {\lim }\limits_{n \to  + \infty } \frac{{\sin \frac{1}{n}}}{{\frac{1}{n}}} = 0$[/quote]

 Let us call  $\frac{1}{n}$ as $x$.  As $n \to \infty$, so $x \to 0$

Therefore,

$\mathop {Lt}\limits_{n \to \infty } nSin\frac{1}{n} = \mathop {Lt}\limits_{x \to 0} \frac{{Sinx}}{x} = 1$


topology
geometry
geometric transformation
reflection
advanced fields
advanced fields unsolved

I got the following lemma when I tried to prove that $ RP^n$ for n even is not orientable. I would like to know whether this statement is correct.

Lemma:
Let $ M$ be a connected smooth manifold. An open cover of $ M$ is called irreducible if for each open set in the cover, there exists at least one point which is covered by this open set only. Suppose that $ M$ is orientable and suppose that $ C$ is an atlas that is irreducible. Then $ C$ can be modified - compose with reflection so that the orientation locally is inverted if necessary- so that $ det(d(g^ \minus{} 1 f)) > 0$ for any two parametrizations in the collection.

My idea is that if $ M$ is orientable, then there are two orientations, say positive and negative. I try to divide the parametrizations in $ C$ into two groups, one is compatible with positive orientation and one for negative orientation. Then I compose those that has negative orientation with a reflection map to rectify the orientation. Then I obtain an atlas that is oriented.

Am I right?

Looks fine to me- you're saying that we can tweak the atlas slightly so that the orientation is the standard $ \mathbb{R}^n$ orientation in coordinates. That's easy, and we automatically get positively oriented transition maps by the global orientation.


There are two points A and B on the surface of a sphere. The shortest path between them is the [url=http://wikipedia.org/wiki/Great_circle]great circle[/url] that passes through them. If you have a third point C, how can you find the point D, on the great circle that passes through A and B, that is closest to C?

If A and B define a great circle, they essentially define a plane, which is also defined by the normal vector.  If the normal vector is set through point C, we can use point-distance formula to find D.


quadratics
LaTeX
number theory unsolved
number theory

Let c be odd and not =1 mod 8. Show with elementary methods that there are infinitely many primes q such that q=1 mod 2^k and (c/q)=-1 for any k.
The case of c=1 mod 8 seems to be more difficult. All I managed to show was there are infinitely many q with q=1 mod 8, (c/q)=-1.
If anyone could show a better result for c=1 mod 8, I would be grateful.

[quote="Abel"]Let $c$ be odd and $\not\equiv 1 \bmod 8$. Show with elementary methods that there are infinitely many primes $q$ such that $q\equiv 1 \bmod 2^{k}$ and $\left(\frac{c}{q}\right)=-1$ for any $k$.[/quote]
I'm proving the following (much more general) statement: "[i]for any $k \in \mathbb{N}^{+}$ and any $c \in \mathbb{Z}$ such that $|c| \ge 2$, there exist infinitely many positive (rational) primes $q \equiv 1 \bmod 2^{k}$ such that $\left(\frac{c}{q}\right) =-1$[/i]."

Proof: let $a_{k}= (c^{2^{k+1}}-1)/(c^{2^{k}}-1) = c^{2^{k}}+1$. Then by the $\mbox{FTA}$ there exist $n$ pairwise distinct [i]odd[/i] primes $q_{1}, q_{2}, \ldots, q_{n}\in \mathbb{N}$ and positive integers $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}\in \mathbb{N}$ such that $a_{k}= 2^{r}\cdot\prod_{i=1}^{n}q_{i}^{\alpha_{i}}$, with $r = 0$ if $c$ is even and $r = 1$ otherwise. Note that $n \in\mathbb{N}^{+}$, since $a_{k}\ge 3$ and $v_{2}(a_{k}) \le 1$, where $v_{2}(\cdot)$ is a $2$-adic valuation (here, simply the greatest integer $v \in \mathbb{N}$ such that $2^{v}\mid a_{k}$). Now $q_{i}\equiv 1 \bmod 2^{k+1}$ and $q_{i}\nmid (c^{2^{k}}-1)$, for any $i = 1, 2, \ldots, n$, as $\gcd(c^{2^{k}}-1, c^{2^{k}}+1) \le 2$. Thus $c$ isn't a quadratic residue $\mod q_{i}$, for $i = 1, 2, \ldots, n$. So finally the conclusion comes up observing $\gcd(c^{2^{m}}+1, c^{2^{n}}+1) \le 2$, whenever $m, n \in \mathbb{N}$ are distinct.

Yeah, but what if the q(subscript i) is 1 mod 2^(k+2). Isn't that a gap in the argument? Actually my approach was similar but this is what I couldn't prove.
(I'm sorry I don't know any Latex)

You're absolutely right. I will edit, waiting for any idea to patch my previous arguments.

I think I've solved it. I would appreciate if someone approves or refuses what I did.

Be C = p1*.......*pt with pi not necesarily different. By quadratic reciprocity we know that: (pi/q)={(-1)^[(pi-1)/2*(q-1)/2]}*(q/pi) we also know that (C/q)=(p1/q)*....*(pt/q) so, substituting the formula from QR it yields: (C/q)=[(-1)^{[(q-1)/2]*[(p1-1)/2+.....+(pt -1)/2]}]*(q/p1)*....*(q/pt).

It's clear that if k=1 we have an infinite number of solutions if c isn't a square (if C is a square it's obvious that we'll never find q such that (c/q)=-1), because if we can find solutions for k>1 they'll also be for k=1.

So, lets supose k>=2. Then (q-1)/2 is even, the expresion on the top becomes: (C/q)=(q/p1)*....*(q/pt). At least one of the pi appears odd times (if not, C wolud be a square) lets choose one of them and name it ps. We choose rs any non quadratic residue modulo ps, and define as ri for i≠s any quadratic residue modulo pi. If q fulfills the following congruences:
q=ri modulo pi, for all pi.
then by the formula (C/q)=-1
So, we have that q follows the next congruences:
q=1 modulo 2^k, q=ri modulo pi, for all pi. (we exclude all repeated pi)
By Chineese remainder theorem, these congruences have a unique solution modulo Y=2^k*y1*.....*yx (where yi are all non repeated primes from C). Be this solution D  modulo Y, as all ri≠0 modulo pi, and 1≠0 modulo 2^k; (D, Y)=1. By Dirichlet there are an infinite number of primes of the form Yx+D if (Y,D)=1. So we are done.

paqui, your proof is correct. But in the last line you assumed a result that has no elementary proof. I wanted a solution with elementary methods.

s.tringali has proved the case for c=3 or 5 mod8.
Let c=-1 mod8.
Let f(x)=(x^2^k)+1.
Let x be such that (f(x)/c)=-1. Such a x exists since c=-1 mod 4.
Then there is a prime q such that (q/c)=-1.
But q=1 mod 2^(k+1).
Also, (c/q)=-1 since (q/c)=-1.
So q satisfies the condition.
As is easily seen there are infinitely many such q.


geometry
geometric transformation
rotation

The radius of the Earth is approximately 4000 miles at the equator. How fast, in miles per hour, is a person standing on the equator moving due to the rotation of the Eart?

A. $ \frac{1000\pi}{3}$

B. $ 1200\pi$

C. $ \frac{4000\pi}{3}$

D. $ 600\pi$

E. $ 2400\pi$

[hide="Solution"]Using $ 4000\text{ mi}$ as an approximation for the radius, the circumference of the Earth is $ 8000\pi\text{ mi}$.

The Earth rotates through those $ 8000\pi\text{ mi}$ every $ 24$ hours, so $ r \equal{} \frac {8000\pi\text{ mi}}{24\text{ hr}}\equal{} \frac {1000\pi\text{ mi}}{3\text{ hr}}\quad\Longrightarrow\quad\text{A}$.[/hide]

[hide]Thus, the answer is $ \boxed{A}$[/hide]


So, I had been wondering how to factor:

(2x^4 + 5x^3 - 29x^2 - 17x + 15)

I seriously can't factor it in any smart way.

Thanks for any suggestions,
Drake

Thanks t0rajir0u for the link. However, I have books with all the root theorems, but with this one they just dont get me anywhere (I know they should). It's a bug:)

Thanks for any further help,
Drake

Then I advise you to look at the rational root theorems again. All I can say is that It factors in the rationals, so you should be able to find solutions.


algebra
polynomial
geometry
system of equations
algebra unsolved

Let $P(x,y)$ and $Q(x,y)$ be two coprime polynomials with degrees $\deg{P}=m$ and $\deg{Q}=n$. Prove that the system of equations
\[
\left\{
\begin{array}{c}
P(x,y)=0,\\
Q(x,y)=0,
\end{array}
\right.
\]
have at most $m \cdot n$ roots.

This is a result in algebraic geometry known as Bezout's theorem:
[url=http://mathworld.wolfram.com/BezoutsTheorem.html]mathworld.wolfram.com/BezoutsTheorem.html[/url]

I believe the proof is quite advanced.


geometry
3D geometry
AMC
AIME
geometric transformation
reflection

Given positive integers a, b, c, d, and e, which are consecutive, and a < b < c < d < e.

a+b+c+e is a perfect cube and b+c+d is a perfect square.

Find the least value of c.

Let the numbers be $ n,n\plus{}1,n\plus{}2,n\plus{}3,n\plus{}4$. What happens when you express these numbers has the equation wants.

I actually tried this but could not figure out a "n" for which -

3(n+2) is a square --- (As obvious, i tried out squares of numbers which are multiples of 3)

and that same very "n" which satisfies

4n + 7 --- perfect cube

Might have missed out an obvious thing in a hurry :(..help!

There is a typo (it should say $ a\plus{}b\plus{}c\plus{}d\plus{}e$ is a perfect cube); see [[1989 AIME Problems/Problem 4]].

Call the numbers as follows:

$ a \equal{} a$

$ b \equal{} a \plus{} 1$

and so on until

$ e \equal{} a \plus{} 4$

We can write two equations.

(1) $ 3(a \plus{} 2) \equal{} x^2$
(2) $ 5(a \plus{} 2) \equal{} y^3$

So 3 times $ c$ yields a perfect square, and 5 times $ c$ yields a perfect cube. So let's make a list of the first few cubes divisible by 5.

125, 1000, 3375

$ 125 \div 5 \equal{} 25$ and $ 25 \times 3 \equal{} 75$, which isn't a perfect square. Next number.

$ 1000 \div 5 \equal{} 200$ and $ 200 \times 3 \equal{} 600$, which still isn't a perfect square.

$ 3375 \div 5 \equal{} 675$ and $ 675 \times 3 \equal{} 2025$, which is a perfect square! Thus our answer is $ \boxed{675}$.

Ah!!!

I was wondering:)...Yep, now it made all the difference!

[quote="TZF"]Haha, that change made [b]all[/b] the difference!

[hide]$ 3c \equal{} n^2$ and $ 5c \equal{} m^3$

To find the minimal $ c$ value, we will involve only** the prime factors $ 3$ and $ 5$. So let $ c \equal{} 3^a 5^b$

We want $ 3c \equal{} 3^{a \plus{} 1} 5^b$ to be a perfect square, so $ a \plus{} 1$ and $ b$ are 2.
We want $ 5c \equal{} 3^{a} 5^{b \plus{} 1}$ to be a perfect cube, so $ a$ and $ b \plus{} 1$ are divisible by 3.

The least $ a$ such that $ a$ is divisible by $ 3$ and $ a \plus{} 1$ is even is simply $ 3$
The least $ b$ such that $ b$ is divisible by $ 2$ and $ b \plus{} 1$ is divisible by $ 3$ is simply $ 2$

So now we have $ c \equal{} 3^3 5^2 \equal{} 675$

[b]** Reflection:[/b] How do we [i]know[/i] that we can do this? Prove it.[/hide][/quote]

Okay, thank you! But why did we have $ 3^{a \plus{} 1} 5^b$ and $ 3^{a} 5^{b \plus{} 1}$ ? What is with the $ ^{\plus{}1}$'s?

Well, $ 3c\equal{}3\cdot 3^a5^b\equal{}3^{a\plus{}1}5^b$, and $ 5c\equal{}5\cdot3^a5^b\equal{}3^a5^{b\plus{}1}$.

 :wink:


can anyone post explanations and problems on pell equations?thank you.
Sorry if i shouldn´t be posting it here...


Prove that the equation $x^3+11^3=y^3$ has no solution in positive integers $x$ and $y$.

[hide]
fermat's last theorem :P
but i guess that doesnt count[/hide]

[hide="better"]
we have $11^3=x^3-y^3=(x-y)(x^2+xy+y^2)$

the factors or $11^3$ are 1,11,121,1331.
so $x-y$ is one of $\{1,11,121,1331\}$
while $x^2+xy+y^2$ is the other factor so that the product is $11^3$

lets use
$x-y=1$
and $x^2+xy+y^2=1331$
so $1331=(x-y)^2+3xy$ so $xy=1330$ we find all of its prome factors $\{2,5,7,3,3\}$ and see that there is no combination so that $x-y=1$ and $xy=1330$

we basically repeat this for all other cases except for those where $xy<0$.
[/hide]

$y-x=1, y^2+xy+x^2=1331$

$\implies y=x+1$

$\implies (x+1)^2+x(x+1)+x^2=3x^2+3x+1=(x+1)^3-x^3$

$\implies (x+1)^3-x^3=11^3$

Now let us work mod 9,

cubic residues are, $1, 8, 0, 1, 8, 0, 1, 8, 0, 1, 8, 0...$

$\implies (x+1)^3-x^3 \equiv 8 \pmod 9$

contradiction, as only possible values mod 9 for the difference between two consectuvie cubes are $7, 1$

$y-x=11, x^2+xy+y^2=121$

$\implies y=11+x$

$\implies x^2+x(11+x)+(x+11)^2=x^2+11x+x^2+x^2+22x+121=121$

$\implies 3x^2+33x=0$

$\implies x=0, -11$

no solns in positive integers.

$y-x=1331, x^2+xy+y^2=1$

$x^2+xy+y^2 \geq 3xy \implies 1 \geq 3xy \implies xy \leq \frac{1}{3}$

no solutions in positive integers, hence no solutions.

[hide]We can rewrite the given eqn as $y^3 - x^3 = (y - x) (y^2 + xy + x^2) = 11^3$.
Since x, y are positive integers, $y - x \leq y^2 + xy + x^2$. Since the only possible factors of $11^3$ are 1, 11, 121, and 1331, we only have to consider the cases $y - x = 1$ and $y - x = 11$.

i) If $y - x = 1$, $y=x + 1$ and $y^3 = x^3 + 3x^2 + 3x + 1$, which implies that $3x^2 + 3x + 1 = 1331$. Since $1331 \neq 1 \pmod 3$, no integer solutions.

ii) If $y - x = 11$, $y = x+11$ and $y^3 = x^3 + 33x^2 + 363x + 1331$, which implies that $33x^2 + 363x = 0$. No positive integer solutions.

yay :)[/hide]

short comment on sky9073
[hide][quote="sky9073"]
Since x, y are positive integers, $y - x \leq y^2 + xy + x^2$. Since the only possible factors of $11^3$ are 1, 11, 121, and 1331, we only have to consider the cases $y - x = 1$ and $y - x = 11$.
 [/quote]
Actually, since $(y - x)^2 < y^2 + xy + x^2$ you need to consider  the case $y-x=1$ only. [/hide]

[quote="t0rajir0u"]Euler has a very nice proof for the special case $n=3$.  It uses infinite descent.[/quote]

could someone show a proof of this. i have been trying but have not been able to.

http://mathpages.com/home/kmath009.htm
I don't know if it's euler's proof, ask him  ;)

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.CMO1972Problem5

Vo Duc Dien

[quote=Altheman][quote="maokid7"][hide]
fermat's last theorem :P
but i guess that doesnt count[/hide]

...[/quote]

well, i think that is the best solution, however, Andrew Wiles had not proven fermat's last theorem by 1972, so during the contest, people could not reference it[/quote]
They could. Case $n=3$ was already proven. That doesn't change a fact that surely they got lesser number of points for such solution.

I remember that the proof (I had seen the proof long time ago) for $n=3$ involved Eisenstein integers and some properties of $\mathbb{Z}[\omega]$ like it is a UFD etc.

$n=4$ case was proved by Fermat himself (using infinite descent). And it was perhaps Kummer, who proved fermat's last theorem till $n=100$ (except some $n$ in this range), by using Ideal numbers and realizing that Unique Factorization cannot be taken for granted. :)


Sucks.  Especially when you take it for 7 hours a day.

Huh? 7hrs a day? Dang.

It's better than like english or something.  What do you guys do for 7 hours?

Actually, it sounds like it could be either really fun or really boring.

Yeah, if you are using time efficiently that would be pretty cool. Sure it would hurt some, but you would get into good shape pretty soon. I have one question though. If you havn't been working out (out of shape) then your body can't handle much more than a few hours of physical activity without being really sore the next day. You could strain or pull muscles easily.

How did you get into that class?

Well, I'd rather have 7 hours of english than this, since you won't hurt at all, and it's not as hot.  It's not really that bad, but I guess I'm really out of shape, and most people usually take 3.5 hours, so that's why it sucks more for me.

But we just sign up for the class.

and exactly why did you sign up for gym for 7 hours?

Because I can't do it at any other time besides now.

I actually think that's kinda cool.  What kind of stuff are you going to be doing?

My school offers some fun stuff, not all physical during gym class. Like we go to the pool hall and play pool for a few hours, or bowling.  We've gone canoing and hiking too, which was absolutely awesome.

However, if you're just running around for 7 hours, hahaha yeah, that would suck.

7 hours! I just excercise 1 hour (thanks god at home)


Can anyone explain how to get the parametric solution for
$25e+23d = 4$ ?

The answer is[hide]$d= 25n-2$
$e=-23n+2$[/hide], but I don't understand how they got that.
also how  do you know when to stop subsituting variables in for the fractions, when can you tell what the parametric solutions will be?
Thank you

Let's let $e=an+b$ and $d=pn+q$.

Sub these into the equation to get
$25an+23pn+25b+23q=4$
Note that the RHS is independent of $n$, so we know that the two $n$s must cancel. So, let $p=25$ and $a=-23$ (we could pick anything that cancels). Sub this back in, and we get $25b+23q=4$. Again, we can pick any values for $b$ and $q$, but $b=2$, $q=-2$ is one possibility.

So, we get $e=-23n+2$ and $d=25n-2$.

Rewrite the equation as
\[23d = 4-25 e \]
Looking at that equation $\bmod{25}$,
\[23 d \equiv 4 \bmod{25}\]
And then adding multiples of $25$ to the RHS until you get
\[\begin{eqnarray*}23d &\equiv& 4+21(25) \bmod{25}\\ d &\equiv& 23 \bmod{25}\]
So $d=25k+23$, for some integer $k$, which is the same as $d=25j-2$.


Let $ O$ be the intersection point of the diagonals of the convex quadrilateral $ ABCD$, $ AO \equal{} CO$. Points $ P$ and $ Q$ are marked on the segments $ AO$ and $ CO$, respectively, such that $ PO \equal{} OQ$. Let $ N$ and $ K$ be the intersection points of the sides $ AB$, $ CD$, and the lines $ DP$ and $ BQ$ respectively. Prove that the points $ N$, $ O$, and $ K$ are colinear.

[size=117][color=darkred][b][u]Very nice problem, thank you April for it ![/b][/u][/color][/size]

[quote="April"]$ \{\begin{array}{c} \mathrm {convex\ quadrilateral\ }ABCD \\
\ = = = = = = = = = = = = = = = = = = = \\
\ O\in AC\cap BD\ ,\ OA = OC \\
\ P\in (OA)\ ,\ Q\in (OC)\ ,\ OP = OQ \\
\ N\in AB\cap DP\ ,\ K\in CD\cap BQ\end{array}\|$ $ \implies O\in NK\ .$[/quote]
[color=darkblue][b]Proof.[/b] Apply the [b]Menelaus' theorem[/b] to the transversals :

${ \{\begin{array}{cc} \overline {BQK}\mathrm {\ in\ }\triangle {COD}\ : & \frac {BO}{BD}\cdot\frac {KD}{KC}\cdot\frac {QC}{QO} = 1 \\
 \\{
\overline {DPN}\mathrm {\ in\ }\triangle AOB}\ : & \frac {DO}{DB}\cdot\frac {NB}{NA}\cdot\frac {PA}{PO} = 1\end{array}\|\ \boxed {\div}$ $ \implies$ $ \frac {OB}{OD}\cdot\frac {NA}{NB} = \frac {KC}{KD}\ \ (*)\ .$

Denote $ \{\begin{array}{c} R\in ON\cap AD \\
\ S\in OK\in AD\end{array}$ and apply the [b]Menelaus' theorem[/b] to the transversals :

$ \{\begin{array}{cc} \overline {RNO}\mathrm {\ in\ }\triangle ABD\ : & \frac {RA}{RD}\cdot\frac {OD}{OB}\cdot\frac {NB}{NA} = 1 \\
 \\
\overline {SKO}\mathrm {\ in\ }\triangle ACD\ : & \frac {SA}{SD}\cdot\frac {KD}{KC}\cdot\frac {OC}{OA} = 1\end{array}\|$ $ \implies$ $ \{\begin{array}{c} \frac {RA}{RD} = \frac {OB}{OD}\cdot\frac {NA}{NB}\ . \\
 \\
\frac {SA}{SD} = \frac {KC}{KD}\ .\end{array}$

From the relation $ (*)$ obtain $ \frac {RA}{RD} = \frac {SA}{SD}\implies R\equiv S\implies O\in NK\ .$
[/color]

We denote the points $ E\equiv AB\cap CD,$ $ F\equiv AD\cap BC$ and it is easy to show that the quadrilateral $ ACEF$ is a trapezium, with $ AC\parallel EF,$ because of the line connecting the point $ D,$ as the intersection point of its opposite sidelines $ AF,$ $ CE,$ with the point $ B,$ as the intersection point of its diagonals, passes through the midpoint $ O,$ of its side segment $ AC.$

Let $ L,$ $ M,$ the intersection points of $ EF,$ from the line segments $ DP,$ $ DQ,$ respectively and it is easy to prove that the line segments $ BQ,$ $ BP,$ pass through the points $ L,$ $ M,$ respectively.

$ ($ from $ \frac {LF}{LE} \equal{} \frac {PA}{PC} \equal{} \frac {QC}{QA}$ we conclude the collinearity of $ Q,$ $ B,$ $ L$ and so, the line segment $ BQ\equiv BK,$ passes through the point $ L$ and similarly the line segment $ BP,$ passes through the point $ M.$

$ \bullet$ Because of now, the collinearity of $ L,$ $ M,$ $ E,$ by [b][size=100]Desarques theorem[/size][/b], we conclude that the triangles $ \bigtriangleup NBP,$ $ \bigtriangleup KDQ,$ are perspective $ ($ $ L\equiv NP\cap KQ,$ $ M\equiv BP\cap DQ,$ $ E\equiv NB\cap KD$ $ ).$

So, the line segments $ NK,$ $ BD,$ $ PC,$ are concurrent at the point  $ O\equiv BD\cap PC.$

Hence the points $ N,$ $ O,$ $ K.$ are collinear and the proof is completed.

Kostas Vittas.

Thank you dear [b]Virgil Nicula[/b], and dear [b]Kostas Vittas[/b] for your solutions. (I liked [b]Kostas' solution[/b] best  :lol: )

$ \bullet$ This problem can be done by use double cross ratio. Here it goes

Denote $ E$ the intersection of $ AB$ and $ CD$.

We have $ (EBNA)\equal{}D(EBNA)\equal{}(COPA)$ (by intersecting with the segment $ AC$).
On the other hand, $ (COPA)\equal{}\frac{OC}{OP}: \frac{AC}{AP}\equal{}\frac{OA}{OQ}: \frac{CA}{CQ}\equal{}(AOQC)\equal{}B(AOQC)\equal{}(EDKC)$ (by intersecting with line $ CD$)

Hence we have $ (EBNA)\equal{}(EDKC)$, therefore $ N$, $ O$ and $ K$ are collinear.


A container contains 24.5 g of N2 and 28.0 g of O2, both in gas form.

If it develops a pinhole-sized leak and some of the gaseous mixture escapes, would the ratio N2 : O2 in the container increase, decrease, or remain the same?

consider their effusion rates which are proportional to the inverse square root of the molar mass. If I recall properly this is known as grahams law.


probability
trigonometry
real analysis
real analysis solved

A fair die is thrown n times. Find those values of n for which the probability that the total score is divisible by 4 is exactly 1/4.

This is NOT an analysis problem.
A die is another word for dice. When "the die is cast" we can thus get a number from the set {1,2,3,4,5,6}.

Define $N_k=\left(\begin{array}{cccc} n_{k,0} \\ n_{k,1} \\ n_{k,2} \\ n_{k,3}\end{array}\right)$, for k>0.
Let $N_0=\left( \begin{array}{cccc} 1\\ 0\\ 0\\ 0\end{array}\right)$.

$n_{k,i}$ is the number of combinations of k throwings of the dice that give a total score =i(mod 4).

A recursive argument shows that $N_{k+1}=\left(\begin{array}{cccc} 1&1&2&2 \\ 2&1&1&2 \\ 2&2&1&1 \\ 1&2&2&1 \end{array}\right) \cdot N_k$.
Notice that this is also true for k=0.

We are required to find the numbers n such that $(3,\ -1,\ -1,\ -1)\cdot N_n=0$

Let $A=\left(\begin{array}{cccc} 1&1&2&2 \\ 2&1&1&2 \\ 2&2&1&1 \\ 1&2&2&1 \end{array}\right)$.
Let $E=\left(\begin{array}{cccc} 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \\ 1&0&0&0 \end{array}\right)$.
We have $E^2=\left(\begin{array}{cccc} 0&0&1&0 \\ 0&0&0&1 \\ 1&0&0&0 \\ 0&1&0&0 \end{array}\right)$.
$E^3=\left(\begin{array}{cccc} 0&0&0&1 \\ 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \end{array}\right)$.
$E^4=I_4$.
We have $A=(I_4+E)(I_4+2E^2)$. Some clever computations will give
$A^n=(a_nI_4+b_nE+c_nE^2+d_nE^3)\cdot (A_nI_4+B_nE^2)$, where
$a_n=\frac{2^{n-1}+\frac{u^n+v^n}{2}}{2}$
$b_n=\frac{2^{n-1}+\frac{u^n-v^n}{2i}}{2}$
$c_n=\frac{2^{n-1}-\frac{u^n+v^n}{2}}{2}$
$d_n=\frac{2^{n-1}-\frac{u^n-v^n}{2i}}{2}$
$A_n=\frac{3^n+(-1)^n}{2}$
$B_n=\frac{3^n-(-1)^n}{2}$

Up there, $u=(1+i)^n$ and $v=(1-i)^n$.

The equation we have to solve in n is $(3,\ -1,\ -1,\ -1)\cdot A^n\cdot N_0=0$

After some rewritings, this comes to:
$(3a_n-b_n-c_n-d_n)\cdot A_n+(-a_n-b_n+3c_n-d_n)\cdot B_n=0$.
After some more computations this comes to $(u^n+v^n)(A_n-B_n)=0$. Because always $A_n \neq B_n$, this is equivalent to
$u^n=-(v)^n\Leftrightarrow (1+i)^n=-(1-i)^n\Leftrightarrow \cos(\frac{n\pi}4)=0 \Leftrightarrow n\equiv 2(mod\ 4)$

because I am high school boy so I can't distinguish right forum in college playground  :D  :D  :blush:  any way thanx for your solution,amfulger.


trigonometry
function
real analysis
real analysis unsolved

can you find these 2 sums ? :D

$ \textbf{(i)}\;\;\sum_{k\equal{}1}^{\infty}\;\;\frac{\sin\;k\theta}{k^{3}}\;\;\;\quad 0\leq\theta\leq 2\pi$

and $ \textbf{(ii)}\;\;\sum_{k\equal{}1}^{\infty}\;\;\frac{\cos\;k\theta}{k^{3}}\;\;\;\quad 0\leq\theta\leq 2\pi$

$ \mathbf{(i)}\quad\sum_{k\equal{}1}^{\infty}\frac{\sin k\theta}{k}$ $ \equal{}\frac{\pi\minus{}\theta}{2}\quad\Rightarrow\quad\sum_{k\equal{}1}^{\infty}\frac{\cos k\theta}{k^{2}}$ $ \equal{}\left(\frac{\pi\minus{}\theta}{2}\right)^{2}\minus{}\frac{\pi^{2}}{12}\quad\Rightarrow\quad\sum_{k\equal{}1}^{\infty}\frac{\sin k\theta}{k^{3}}\equal{}\frac{\theta (\theta\minus{}\pi) (\theta\minus{}2\pi)}{12}\quad\Rightarrow\quad ...$

$ \mathbf{(ii)}\quad$ I guess $ \sum_{k\equal{}1}^{\infty}\frac{\cos k\theta}{k^{3}}$ can't be expressed with elementary functions alone. Therefor you need artificial functions like $ \text{polylog}$.


inequalities
induction
function
limit
real analysis
real analysis theorems

$1\leq p\leq \infty$. If $f_k\in L^p(E)$ and $\sum\limits_{i=1}^\infty f_i(x)$ converges almost everywhere
Prove that
$\left\|\sum_{k=1}^\infty f_k\right\|_p \leq \sum_{k=1}^\infty \|f_k\|_p.$

This is all about size. Since $|\sum f_k|\le \sum|f_k|$, we lose nothing by assuming that $f_k\ge 0.$

Assume as known $||f+g||_p\le||f||_p+||g||_p.$ Then it follows by induction that for all $n$, $\left|\left|\sum_{k=1}^nf_k\right|\right|_p\le \sum_{k=1}^n||f_k||_p.$ (*)

Since we're assuming the functions are postive, it follow by the Monotone Convergence Theorem that

$\lim_{n\to\infty}\left|\left|\sum_{k=1}^nf_k\right|\right|_p= \left|\left|\sum_{k=1}^{\infty}f_k\right|\right|_p.$

Now, if we just take limits on both sides of (*), we have the conclusion.