geometry
trapezoid
circumcircle
algebra
polynomial
function
geometric transformation

Chinese Team Selecting Test 2006:
Problem set 1, March 19th. Time given: 4 hours. Each problem worths 7 points.

Problem 1: http://www.mathlinks.ro/Forum/viewtopic.php?p=462115#p462115
$ABCD$ is a trapezoid with $AB \parallel CD$. There are two circles $\omega_{1}$ and $\omega_{2}$ is the trapezoid such that $\omega_{1}$ is tangent to $DA$, $AB$, $BC$ and $\omega_{2}$ is tangent to $BC$, $CD$, $DA$. Let $l_{1}$ be a line passing trought $A$ and tangent to $\omega_{2}$(other than $AD$), Let $l_{2}$ be a line passing trought $C$ and tangent to $\omega_{1}$(other than $CB$).

Prove that $l_{1}\parallel l_{2}$.

Problem 2: http://www.mathlinks.ro/Forum/viewtopic.php?p=462117#p462117
Find all positive integer pairs $(a,n)$ such that $\frac{(a+1)^{n}-a^{n}}{n}$ is an integer.

Problem 3: http://www.mathlinks.ro/Forum/viewtopic.php?p=462118#p462118
Given $n$ real numbers $a_{1}$, $a_{2}$ $\ldots$ $a_{n}$. ($n\geq 1$).  Prove that there exists real numbers $b_{1}$, $b_{2}$ $\ldots$ $b_{n}$ satisfying:
(a) For any $1 \leq i \leq n$, $a_{i}-b_{i}$ is a positive integer.

(b)$\sum_{1 \leq i < j \leq n}(b_{i}-b_{j})^{2}\leq \frac{n^{2}-1}{12}$

Yes,jin.Please continue it.Thanks  :)  :)

Well, there was a post by Mr. Feizhenpeng here:
http://www.aoshoo.com/bbs1/dispbbs.asp?boardid=66&replyid=14519&id=3021&page=1&skin=0&Star=2
I hope that you can read it and tell me why Mr. Fei said so......

Thanks

[quote="mecrazywong"]Maybe (just maybe) some of them problems are from shortlist...[/quote]

Not sure

I don't know why Mr.Fei said so but I also don't think that it is because of the shortlist problems. I guess that Mr.Fei has other meanings and that's why I stopped posting the rest of the problems.

[quote="jin"]Oh,I don't go to aoshoo very often.[/quote]
That doesn't mateter.

I think that Mr.Fei is worrying about the coryright problem......

[quote="jin"]I think the problems are from Li Shenghong and other Chinese couches,Only the first three is from Feng Zuming[/quote]
I think you mis-understand it......

There were some guys who claim to have the copyright of the English versions of CMO problems. That's why the past CMO we translated are still in the Chinese Community. So I guess Mr.Fei may by worrying about that......

By the way, jin, what was your score at the TST?

I think more than 200,I don't know the exactly score.But I soluted the last six problem

[quote="jin"]I think the problems are from Li Shenghong and other Chinese couches,Only the first three is from Feng Zuming[/quote]

just want to clarify: there were 5 problems on the tests was given to Mr. Li via me. But I am not the original anthors for some of these 5 problems.

[quote="jin"]I think the problems are from Li Shenghong and other Chinese couches,Only the first three is from Feng Zuming[/quote]

Coaches in China and I constantly exchanges files/problems. We agree to release all the files after the IMO for a simple reason. When we give students as practices 3-problem mock (muo(3) ni (3) ke (1) shi (4) ) IMO tests -- not real selection tests -- we hope that students have not seen them before. Otherwise, it makes the practice less meaningful. That is why I ask all of you guys to hold these for a while, say, till June -- because we do many practices in June for both teams' training.  If you really really really wants to post problems, please DO NOT post solutions, so at least students needs to solve them on their own. Otherwise, it defeats all the purpose of IMO -- it becoming a memeorzation game and will not be good to anyone of us in the MO community.

Chinese Team Selecting Test 2006:
Problem set 2, March 20th. Time given: 4 hours. Each problem worths 7 points.

Problem 1: http://www.mathlinks.ro/Forum/viewtopic.php?t=97483
Two sequences $\{ a_n \}$ and $\{ b_n \}$ satisfy:
(a): $a_0=1 \geq a_1$, $a_n(b_{n+1}+b_{n-1})=a_{n-1}b_{n-1}+a_{n+1}b_{n+1}$, $n \geq 1$.

(b): $\sum_{i=1}^{n} b_i \leq n^{\frac{3}{2}}$, $n \geq 1$.

Find the general term of $\{ a_n \}$.

Problem 2: http://www.mathlinks.ro/Forum/viewtopic.php?t=97484
Let $\omega$ be the circumcircle of $\triangle{ABC}$. $P$ is an interior point of $\triangle{ABC}$. $A_1, B_1, C_1$ are the intersections of $AP, BP, CP$ respectively and $A_2, B_2, C_2$ are the semetrical points of $A_1, B_1, C_1$ with respect to the midpoints of side $BC, CA, AB$.
Show that the circumcircle of $\triangle{A_2B_2C_2}$ passes through the othorcentre of $\triangle{ABC}$.

Problem 3: http://www.mathlinks.ro/Forum/viewtopic.php?t=97486
Let $a_i$ and $b_i$ ($i=1,2, \cdots, n$) be rational numbers such that for any real number $x$ there is:
\[ x^2+x+4=\sum_{i=1}^{n} (a_ix+b)^2 \]
Find the least possible value of $n$.

Chinese Team Selecting Test 2006:
Problem set 3, March 22nd. Time given: 4 hours. Each problem worths 7 points.

Problem 1: http://www.mathlinks.ro/Forum/viewtopic.php?t=97487
The centre of the circumcircle of quadrilateral $ABCD$ is $O$ and $O$ is not on any of the sides of $ABCD$. $P=AC \cap BD$. The circumecentres of $\triangle{OAB}$, $\triangle{OBC}$, $\triangle{OCD}$ and $\triangle{ODA}$ are $O_1$, $O_2$, $O_3$ and $O_4$ respectively.

Prove that $O_1O_3$, $O_2O_4$ and $OP$ are concurrent.

Problem 2: http://www.mathlinks.ro/Forum/viewtopic.php?t=97488
$x_1, x_2, \cdots, x_n$ are positive numbers such that $\sum_{i=1}^{n} x_i = 1$.

Prove: $\left( \sum_{i=1}^{n}\sqrt{x_i} \right) \left( \sum_{i=1}^{n}\frac{1}{\sqrt{1+x_i}} \right) \leq \frac{n^2}{\sqrt{n+1}}$

Problem 3: http://www.mathlinks.ro/Forum/viewtopic.php?t=97489
$d$ and $n$ are positive integers such that $d \mid n$. The n-number sets $(x_1, x_2, \cdots x_n)$ satisfy the following condition:
(1) $0 \leq x_1 \leq x_2 \leq \cdots \leq x_n \leq n$

(2) $d \mid (x_1+x_2+ \cdots x_n)$

Prove that in all the n-number sets that meet the conditions, there are exactly half satisfy $x_n=n$.

Chinese Team Selecting Test 2006:
Problem set 4, March 24th. Time given: 4 hours. Each problem worths 7 points.

Problem 1: http://www.mathlinks.ro/Forum/viewtopic.php?t=97493
Let $K$ and $M$ be on side $AB$ of $\triangle{ABC}$, $L$ and $N$ be on side $AC$. $K$ is between $M$ and $B$, $L$ is between $N$ and $C$. If $\frac{BK}{KM}=\frac{CL}{LN}$, then prove that the orthocentres of $\triangle{ABC}$, $\triangle{AKL}$ and $\triangle{AMN}$ lie on the same line.

Problem 2: http://www.mathlinks.ro/Forum/viewtopic.php?t=97495
$x$, $y$ and $z$ are positive real numbers such that $x+y+z=1$, prove:
\[ \frac{xy}{\sqrt{xy+yz}}+\frac{yz}{\sqrt{yz+zx}}+\frac{zx}{\sqrt{zx+xy}}\leq \frac{\sqrt{2}}{2} \]

Problem 3: http://www.mathlinks.ro/Forum/viewtopic.php?t=97496
Find all second degree polynomial $d(x)=x^2+ax+b$ with integer coefficients, so that there exists an integer coefficient polynomial $p(x)$ and a non-zero integer coefficient polynomial $q(x)$ that satisfy:
\[ \left( p(x) \right)^2 - d(x) \left( q(x) \right)^2=1, \qquad \forall x \in R \]

Chinese Team Selecting Test 2006:
Problem set 5, March 26th. Time given: 4 hours. Each problem worths 7 points.

Problem 1: http://www.mathlinks.ro/Forum/viewtopic.php?t=97497
Let $A$ be a non-empty subset of the set of all positive integers $N^*$. If any sufficient big positive integer can be expressed as the sum of $2$ elements in $A$(The two integers do not have to be different), then we call that $A$ is a divalent radical. For $x \geq 1$, let $A(x)$ be the set of all elements in $A$ that do not exceed $x$, prove that there exist a divalent radical $A$ and a constant number $C$ so that for every $x \geq 1$, there is always $\left| A(x) \right| \leq C \sqrt{x}$.

Problem 2: http://www.mathlinks.ro/Forum/viewtopic.php?t=97498
A function $f(n)$ satisfy $f(0)=0$, $f(n)=n-f \left( f(n-1) \right)$, $n=1,2,3 \cdots$. Please find all real coefficient polynomials $g(x)$ so that:
\[ f(n)= [ g(n) ], \qquad n=0,1,2 \cdots \]
Where $[ g(n) ]$ denote the greatest integer that does not exceed $g(n)$.

Problem 3: http://www.mathlinks.ro/Forum/viewtopic.php?t=97500
Given positive integers $m$ and $n$ so there is a chessboard with $mn$ $1 \times 1$ grids. Colour the grids into red and blue (Grids that have a common side are not the same colour and the grid in the left corner at the bottom is red). Now the diagnol that goes from the left corner at the bottom to the top right corner is coloured into red and blue segments (Every segment has the same colour with the grid that contains it). Find the sum of the length of all the red segments.

Chinese Team Selecting Test 2006:
Problem set 6, March 28th. Time given: 4 hours. Each problem worths 7 points.

Problem 1: http://www.mathlinks.ro/Forum/viewtopic.php?t=97501
Let the intersections of $\odot O_1$ and $\odot O_2$ be $A$ and $B$. Point $R$ is on arc $AB$ of $\odot O_1$ and $T$ is on arc $AB$ on $\odot O_2$. $AR$ and $BR$ meet $\odot O_2$ at $C$ and $D$; $AT$ and $BT$ meet $\odot O_1$ at $Q$ and $P$. If $PR$ and $TD$ meet at $E$ and $QR$ and $TC$ meet at $F$, then prove: $AE \cdot BT \cdot BR = BF \cdot AT \cdot AR$.

Problem 2: http://www.mathlinks.ro/Forum/viewtopic.php?t=97502
Prove that for any given positive integer $m$ and $n$, there is always a positive integer $k$ so that $2^k-m$ has at least $n$ different prime divisors.

Problem 3: http://www.mathlinks.ro/Forum/viewtopic.php?t=97503
$k$ and $n$ are positive integers that are greater than $1$. $N$ is the set of positive integers. $A_1, A_2, \cdots A_k$ are pairwise not-intersecting subsets of $N$ and $A_1 \cup A_2 \cup \cdots \cup A_k = N$.

Prove that for some $i \in \{ 1,2,\cdots,k \}$, there exsits infinity many non-factorable n-th degree polynomials so that coefficients of one polynomial are pairwise distinct and all the coeficients are in $A_i$.

Chinese Team Selecting Test 2006:
Problem set 7, March 31st. Time given: [b]4.5[/b] hours. Each problem worths [b]21[/b] points.

Problem 1: http://www.mathlinks.ro/Forum/viewtopic.php?t=97504
$H$ is the orthocentre of $\triangle{ABC}$. $D$, $E$, $F$ are on the circumcircle of $\triangle{ABC}$ such that $AD \parallel BE \parallel CF$. $S$, $T$, $U$ are the semetrical points of $D$, $E$, $F$ with respect to $BC$, $CA$, $AB$. Show that $S, T, U, H$ lie on the same circle.

Problem 2: http://www.mathlinks.ro/Forum/viewtopic.php?t=97506
Given positive integer $n$, find the biggest real number $C$ which satisfy the condition that if the sum of the reciprocals of a set of integers (They can be the same.) that are greater than $1$ is less than $C$, then we can divide the set of numbers into no more than $n$ groups so that the sum of reciprocals of every gruop is less than $1$.

Problem 3: http://www.mathlinks.ro/Forum/viewtopic.php?t=97507
For a positive integer $M$, if there exist integers $a$, $b$, $c$ and $d$ so that:
\[ M \leq a \leq b \leq c \leq d \leq M+49, \qquad ad=bc \]
then we call $M$ a GOOD number, if not then $M$ is BAD. Please find the greatest GOOD number and the smallest BAD number.

Chinese Team Selecting Test 2006:
Problem set 8, April 1st. Time given: [b]4.5[/b] hours. Each problem worths [b]21[/b] points.

Problem 1: http://www.mathlinks.ro/Forum/viewtopic.php?t=97508
$k$ is an odd number that is greater or equal than $3$. Prove that there exist a $k$-th degree integer valued polynomial with non-integer coefficients that has the following properties:
(1) $f(0)=0$, $f(1)=1$.
(2) There exist infinity many positive integer $n$ so that if the equation:
\[ n= f(x_1)+\cdots+f(x_s) \]
has integer solutions $x_1 \cdots x_s$, then $s \geq 2^k-1$.

Problem 2: http://www.mathlinks.ro/Forum/viewtopic.php?t=97510
Given positive integers $m$, $a$, $b$, $(a,b)=1$. $A$ is a non-empty subset of the set of all positive integers, so that for every positive integer $n$ there is $an \in A$ and $bn \in A$. For all $A$ that satisfy the above condition, find the minimum of the value of $\left| A \cap \{ 1,2, \cdots,m \} \right|$

Problem 3: http://www.mathlinks.ro/Forum/viewtopic.php?t=97511
$\triangle{ABC}$ can cover a convex polygon $M$.Prove that there exsit a triangle which is congruent to $\triangle{ABC}$ such that it can also cover $M$ and has one side line paralel to or superpose one side line of $M$.

That's all the problems of China TST 2006.

[quote="ptung"]Could you please check the following problems again, shobber?

http://www.mathlinks.ro/Forum/viewtopic.php?t=97486

$b_i$ or $b$?

http://www.mathlinks.ro/Forum/viewtopic.php?t=97507

the sign $\leq$  with $a,b,c,d$

http://www.mathlinks.ro/Forum/viewtopic.php?t=97510[/quote]

For topic 97486, it's $b_i$.

For topic 97507, the statement is right.

For topic 97510, it's
"... so that for every positive integer n, there is $an \in A$ [color=red][b]or[/b][/color] $bn \in A$. ...

[quote="ptung"]In the case $a= b, c= d$, it is true for every $M$.[/quote]

Oh, I look through the chinese version again, it is
$M \leq a < b \leq c < d \leq M+49$

sorry!

Thank you very much for your help, zhaoli and shobber !
If you have China TST from another years, please post them but not their solutions. It is very good source for students to pratice before IMOs, because of its difficulty level as well as non-posted solutions.

[quote="Gabriel(fr)"]I have a question:

the problem '3' of the the 'problem set 5' need more detail or it was writen correctly?[/quote]

The pattern of coloring is just like normal chessboard.

The question is to find the sum of red segements in the diagonal .

This problem is something like Balkan 2003-4.

I am sorry for all the mistakes I made in the statements of problems. My English isn't good enough to do the translation work I suppose......But anyway, any comments and corrections are welcome!

[quote="orl"]@all Chinese speakers with access to past Chinese TST problems. As you may have noticed [url=http://www.mathlinks.ro/Forum/resources.php?c=37&cid=47]here[/url] I only have the 6 main problems of those respective years. Can anybody provide me/us with those remaining tests (problem sets) for those years you are capable of retrieving?  :)[/quote]
I think zhaoli has already posted the problems of TST 2003, 2004,2005 on the forum. But they are only in the PDF file. I shall help put the problems in the Olympiad Sections.


Prove
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\geq\frac{a\plus{}b\plus{}c}{\sqrt [3]{abc}}$
For: a,b,c>=0


logarithms
inequalities
rearrangement inequality
Divisibility Theory

Find four positive integers, each not exceeding $70000$ and each having more than $100$ divisors.

[quote="nicetry007"][quote="Peter"][quote="nicetry007"]The numbers are

2^5. 3^2. 5^2. 7^1 ,
2^6. 3^3. 5^1. 7^1 ,
2^2. 3^2. 5^2. 7^1. 11^1 ,
2^4. 3^2. 5^1. 7^1. 11^1.

I'm not posting my solution here for it is not a solution but a mere enumeration of all possible cases. If there is a nice way of doing the problem, please post your ideas. Otherwise, I feel this isn't a problem that deserves to be on the IMO shortlist.[/quote]I don't see the problem? You gave a correct example, that is all that the question asks. How can you expect a shorter solution than just 4 numbers? :)[/quote]I only provided the answers and not the solution. In other words, I didn't outline a method for identifying the numbers. The reason for not doing so is that it is very tedious.[/quote]

I will outline here an approach for the general problem:
Find the positive integers $ n\in\overline{2,m}$ for which $ \tau(n)\ge t$, where $ \tau(n)$ represents the number of positive divisors of $ n$.
Assume that $ n$ is a number satisfying the conditions. Write $ n \equal{} p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}...p_{r}^{\alpha_{r}}$, where $ p_{1}< p_{2}< ... < p_{r}$ are primes and $ \alpha_{1},\alpha_{2}, ...,\alpha_{r}$ are positive integers. So $ \tau(n) \equal{} (\alpha_{1}\plus{}1)(\alpha_{2}\plus{}1)\cdots (\alpha_{r}\plus{}1)\ge t$.
Now, notice that $ m\ge n\ge p_{1}p_{2}...p_{r}\ge\mathcal{P}(r)$, where $ \mathcal{P}(r)$ is the product of the first $ r$ prime numbers. This gives a first upper bound for $ r$: denote it by $ r_{1}\in\mathbb{Z}$. Next, by AM-GM, $ \alpha_{1}\plus{}\alpha_{2}\plus{}...\plus{}\alpha_{r}\ge r(\sqrt [r]{t}\minus{}1)$, so $ m\ge n\ge 2^{\alpha_{1}\plus{}...\plus{}\alpha_{r}}\ge 2^{r(\sqrt [r]{t}\minus{}1)}$, which shows that $ r(\sqrt [r]{t}\minus{}1)\le\log_{2}(m)$. We infer some lower bound for $ r$, call it $ r_{0}\ge 1$.
Now, we take each $ r\in [r_{0}, r_{1}]$ and try to find the corresponding set of solutions. One idea is to use the fact that $ m\ge n\ge p_{1}p_{2}...p_{r}\cdot 2^{\alpha_{1}\plus{}...\plus{}\alpha_{r}\minus{}r}$. Now we get that $ \alpha_{1}\plus{}\alpha_{2}\plus{}...\plus{}\alpha_{r}\le k$ for some positive integer $ k$ (for example, $ m \equal{} 70000$ and $ r \equal{} 4$ implies $ \alpha_{1}\plus{}...\plus{}\alpha_{r}\le 12$). Thus, if $ \beta_{1}, ...,\beta_{r}$ are the $ \alpha$s in increasing order, by the condition $ \tau(n)\ge t$ we see that $ (\beta_{1}, ...,\beta_{r})$ is in some set $ S$, and by rearrangement inequality $ m\ge n\ge q_{1}^{\beta_{r}}...q_{r}^{\beta_{1}}$, where $ q_{1}, ..., q_{r}$ are the first $ r$ primes, $ q_{1}< q_{2}< ... < q_{r}$. Eventually we will get a few $ r$-uples to deal with.
In our case, $ m \equal{} 70000$, $ t \equal{} 100$, we get $ n\in\{ 45360, 50400, 55440, 60480, 65520, 69300\}$.
The problem would be more suitable with lesser and more adequate bounds, still for the students not to try "computational" methods.


Enough of my stupid solve-in-one-post riddles. It's time for some (hopefully) harder ones.

Denise died at sea while Harry died on land. People were glad that Harry had died and were even more glad that Denise had died. why?

[quote="hsiljak"]Are they codenames for storms or winds, hurricanes? :)[/quote]

I thoroughly agree  :yup:

If that someone isn't the correct answer, perhaps they were rulers that waged war in the sea and on land?

[quote="1=2"]They were diseases? That's the only other explanation I can think of.[/quote]

I haven't heard of diseases being named common human names, only "The Black Plague" and the likes.

Why would they appreciate Denise dying at sea more when people are in land, anyway?

[quote="hsiljak"]Are they codenames for storms or winds, hurricanes? :)[/quote]
That makes the most sense.  If Harry died on land, it already caused some damage.  If Denise died at sea, it didn't get a chance to cause damage.

I could be a stickler for accuracy and say that Harry cannot be found [url=http://www.nhc.noaa.gov/aboutnames.shtml]here[/url] or [url=http://www.nhc.noaa.gov/retirednames.shtml]here[/url].

Yes, they were hurricanes.

(i must get a REALLY tough one now.


decimal representation
number theory
prime numbers
Digits
IMO Shortlist

Find all natural numbers $ n$ for which every natural number whose decimal representation has $ n \minus{} 1$ digits $ 1$ and one digit $ 7$ is prime.

[hide]$ n \equal{} 1$ works, since $ 7$ is prime.
$ n \equal{} 2$ works, since $ 17$ and $ 71$ are prime.
$ n \equal{} 3$ does not work, since $ 117$ is divisible by $ 3$.
$ n \equal{} 4$ does not work, since $ 1711 \equal{} 29\cdot 59$.
$ n \equal{} 5$ does not work, since $ 11711 \equal{} 7\cdot 1587$.
$ n \equal{} 6$ does not work, since $ 111117$ is divisible by $ 3$.

For $ n\ge 7$, each number with $ n \minus{} 1$ digits $ 1$ and one digit $ 7$ is of the form

$ x \plus{} 6\cdot 10^k$

for integers $ 0\le k\le n \minus{} 1$ where $ x$ is the n-digit number made only of the digit $ 1$. As $ k$ goes from $ 0$ to $ 6$, the term $ 6\cdot 10^k$ takes on all values $ \mod 7$, so at least one of the numbers $ x \plus{} 6\cdot 10^k$ must be divisible by $ 7$. Hence no $ n\ge 7$ can work, so the only values of $ n$ are $ n \equal{} 1$ and $ n \equal{} 2$.[/hide]

[quote="tjhance"]As $ k$ goes from $ 0$ to $ 6$, the term $ 6\cdot 10^k$ takes on all values $ \mod 7$, so at least one of the numbers $ x \plus{} 6\cdot 10^k$ must be divisible by $ 7$. Hence no $ n\ge 7$ can work, so the only values of $ n$ are $ n \equal{} 1$ and $ n \equal{} 2$.[/quote]

$ 6\cdot 10^k$ never takes on the value $ 0$, so none of the numbers $ x \plus{} 6\cdot 10^k$ is necessarily divisible by 7. Take, for instance, $ x \equal{} 111111111111$ or as in the example you gave for $ n \equal{} 6$ , which are 0 mod 7. Each of the permutations of the digits $ 111111111117$ is not divisible by 7. The same is true for any other number of $ n$ digits 1, as long as $ n \equal{} 6q \plus{} 6$. But in those cases, we have a number of $ 6q \plus{} 5$ digits 1 and one $ 7$, which are always divisible by $ 3$.

[hide="Yay big numbers"]
So, clearly $n=1,2$ work.

Also, no $n$ divisible by $3$ can work, as all such $n$-digit numbers are divisible by $3$.

For odd $n$, consider
  $71...1=6\cdot10^{n-1}+\frac{10^n-1}{9}=\frac{54\cdot10^{n-1}+10^n-1}{9}=\frac{64\cdot10^{n-1}-1}{9}=\left(\frac{8\cdot10^{\frac{n-1}{2}}+1}{9}\right)\left(8\cdot10^{\frac{n-1}{2}}-1\right)$, which is thus not prime.

So now we've shown that $1,2$ work, and that $3,5,6,7,9,11,12...$ don't. The only numbers left to disprove are those of the form $6k+4$ and $6k+8$ for natural $k$.

In the former case, consider the number $1....17111=1...1\cdot10^4+7111$. This can be expressed as

$111111\cdot(\text{a big number } b)+7111=13(8547b)+13(547)$ which is divisible by $13$ and not prime.

As for the latter case, consider $1...1\cdot10^8 + 11111711=111111(b)+11111711=13(8547b)+13(854747)$ which is again divisible by $13$.

Thus, only $\boxed{n=1,2}$ work.[/hide]


Solve  the inequality:$ \arcsin x\leq \arctan x\leq\arccos x\leq\arctan x$
____________________________
Azerbaijan Land of the Fire   :play_ball:  :first:

[quote="Pirkuliyev Rovsen"]Solve  the inequality:$ \arcsin x\leq \boxed{\arctan x}\leq\arccos x\leq\boxed{\arctan x}$
____________________________
Azerbaijan Land of the Fire   :play_ball:  :first:[/quote]

this must be wrong


[quote="jli"]Theres this contest called Odessey of the Mind

anybody doing it?[/quote]

Where can it be found?  :?

It's offered in my school, and our team has gotten third at world finals before.


Find all function $ f: R \rightarrow R$ that satisfy for all $ x,y \neq 0$:
$ y\cdot f(x) \minus{}x\cdot f(y)\equal{}f(\frac{x}{y})$

Let $ x\equal{}0$, we get $ f(0)\equal{}0$.
Let $ x\equal{}y$, we get $ f(1)\equal{}0$.
Let $ u\equal{}\frac xy\ne0$.
$ f(x)\equal{}f(uy)\equal{}uf(y)\plus{}\frac1yf(u)$.
Switch $ u$ and $ y$, we have
$ uf(y)\plus{}\frac1yf(u)\equal{}yf(u)\plus{}\frac1uf(y)$.
Let $ y\equal{}\minus{}1$ here, we see $ f(\minus{}1)\equal{}0$.
For any $ u,y\ne\pm1,0$,
$ \frac{f(u)}{\frac1u\minus{}u}\equal{}\frac{f(y)}{\frac1y\minus{}y}$.

Let $ c$ be a constant,
$ f(y)\equal{}c(\frac1y\minus{}y),\forall y\ne0;f(0)\equal{}0$,


Let $ E_k \equal{} a_1 \plus{} a_2 2^k \plus{} \ldots \plus{} a_m m^k$. Then if $ P(x) \equal{} c_n x^n \plus{} c_{n \minus{} 1} x^{n \minus{} 1} \plus{} \ldots \plus{} c_0$ for arbitrary reals $ c_0, c_1, \ldots c_n$, we can find that $ c_0 E_0 \plus{} c_1 E_1 \plus{} \ldots \plus{} c_n E_n \equal{} a_1 P(1) \plus{} a_2 P(2) \plus{} \ldots \plus{} a_m P(m)$. Since $ E_0 \equal{} E_1 \equal{} \ldots \equal{} E_n \equal{} 0$, we have
\[ a_1 P(1) \plus{} a_2 P(2) \plus{} \ldots \plus{} a_m P(m) \equal{} 0,
\]
for any polynomial $ P$ of degree at most $ n$ with real coefficients.

Suppose that $ \{a_i\}$ has at most $ n$ pairs of consecutive terms with opposite signs, say $ i_1, i_2, \ldots i_p$ with $ p \leq n$ so that $ a_{i_j}a_{i_j \plus{} 1} < 0$ for $ j \equal{} 1, 2, \ldots p$. Pick $ P(x) \equal{} (i_1 \plus{} 1/2 \minus{} x)(i_2 \plus{} 1/2 \minus{} x)\cdots (i_p \plus{} 1/2 \minus{} x)$. This is allowed since $ p \leq n$ so $ P$ is at most degree $ n$. Then all terms in $ a_1 P(1) \plus{} a_2 P(2) \plus{} \ldots \plus{} a_m P(m)$ are positive. Contradiction.

We only need the $ a_i$ to be nonzero reals, by the way.


The cards in a stack of $2n$ cards are numbered consecutively from $1$ through $2n$ from top to bottom. The top $n$ cards are removed, kept in order, and form pile $A$. The remaining cards form pile $B$. The cards are then restacked by taking cards alternately from the tops of pile $B$ and $A$, respectively. In this process, card number $(n+1)$ becomes the bottom card of the new stack, card number $1$ is on top of this card, and so on, until piles $A$ and $B$ are exhausted. If, after the restacking process, at least one card from each pile occupies the same position that it occupied in the original stack, the stack is named [i]magical[/i]. Find the number of cards in the magical stack in which card number $131$ retains its original position.

[hide="Answer"]Card $131$ must come from $A$, and since they are placed alternately, card $131$ is $n-65$, so $n=196$. Since $A$ is the top pile, this fits that $n$ is in the first half, so we have $2n=\boxed{392}$.[/hide]

I just did simple case-checking for if 131 is in A or B.  If 131 is in A, then when the reshuffling gets to 131, the new stack has 261 cards.  So 131 becomes the 262nd (from the bottom).  There must be 130 above it to make it the 131st card, so A and B each have 65 cards left, and the sum is 262 + 130 = 392.  I just got lucky the first case I tried worked.

Oops, I kept forgetting that it was numbered from the top. :oops:

(Apologies for reviving an old thread.)

[hide]Create an array as follows. The first column represents the original stack, and the second column represents the result of restacking.

$\begin{array}[b]{c|c} \text{Original} & \text{New} \\ 1 & n \\ 2 & 2n \\ 3 & n - 1 \\ 4 & 2n - 1 \\ 5 & n - 2 \\ 6 & 2n - 2 \\ 7 & n - 3 \\ 8 & 2n - 3 \\ 9 & n - 4 \\ 10 & 2n - 4 \\ \vdots & \vdots \\ 2n-1 & 1 \\ 2n & n+1 \\ \end{array}$

Now we have two cases.
Case 1: Card $131$ comes from the $1$ through $n$ stack, which is represented by $n, n-1, n-2, \ldots$ on the right column. We notice that for every number of the form $n - a$, the original card is $2a + 1$. Hence: $n - a = 2a + 1$
But we have the original card number = 131, namely $2a + 1 = 131$. Thus we have
$n - \frac{131-1}{2} = 131 \Rightarrow 2n = 392$.

Case 2: Card $131$ comes from the $(n+1)$ through $(2n)$ stack, which is represented by $2n, 2n-1, 2n-2, \ldots$ on the right column. We notice that for every form $2n -a$, the original card is of the form $2a + 2$. Hence $2n - a = 2a + 2$.
As in Case1, we have the original card number = 131, namely $2a + 2 = 131$. Thus we have
$n - \frac{131-2}{2} = 131 \Rightarrow 2n = 262 + 129 = 391$,
which is, errg, quite impossible.

Thus only Case 1 works, and our answer is $\boxed{392}$.[/hide]

I thought this topic was revive worthy because my solution is a bit different.
Edit: Uh oh, I just read through more carefully and deej21's solution is almost exactly the same, sorry guys.

Solution:
Starting from the bottom of the stack, the cards go n+1,1,n+2,2... 2n,n.  Since the total number of cards in the stack is even, n+1 holds an even place. Therefore, all cards numbered n+1 through 2n hold an even place in the new stack, while all cards numbered 1 through n hold an odd place. 131 is odd, so obviously it is from the 1 through n stack. Counting from the bottom, the 1 through n cards occupy even places in the stack, so 131 is in place #262 from the bottom. For card #131 to be the 131st card in th new stack, there must be 130 cards above it. It is the 262nd card from the bottom, so there are 130 cards above it and 261 cards below it. This means the total stack has 130 + 1 + 261 = 392 cards, which is the solution.

[quote=OlympusHero]Question about the problem statement (misread earlier oops): If n+1 is the bottom card in the stack that means it's in pile B but pile A is cards from n+1 to 2n and pile B is everything else. So how does that work?[/quote]

Uhm, this was made over a decade and seven years ago, I don't think he will respond.

I'm not talking to OP, I'm asking a question that anyone can answer.

Umm... It was very was easy firstly I thought that it might take time but it didn't
So for a card to be magical the position that it retains should be equal to the original position
Speaking mathematically that for 1 to be a magical card its position is 1 and after position is 2n-1 which means 2n-1=1
n=2
Similarly for 2 its after position is 2n-3
And for 3 it's  2n-5
And so on
For 131 it would be 2n-261=131 hence 2n=392


Hey everyone ... I wrote this problem last night and wanted to know what you think.  Calculators are allowed once you figure out how to do it:

A circle has chords of legnth 2sqrt(3), 12, and 12sqrt(3).  The chords can be positioned such that the distance between the shortest chord and the center is the sum of the distances between the longest and the center and the middle and the center.  Find the radius of the circle.

Again, Calculators are allowed once you figure out how to proceed.

JB

[quote="shobber"]It seems to become:
$\sqrt{R^2-3}=\sqrt{R^2-36}+\sqrt{R^2-108}$[/quote]

Oof ... that seems to be a much easier method than what I was using ....

But yes, that works ... What a beautiful solution to what I thought was a messy problem.

Thank you shobber
JB


calculus
integration
calculus computations

[img]http://www.geocities.com/paolayeye/ej1.jpg[/img]

please, don't give me only the result, i need to understand all the procedure... Thanks :)


trigonometry
geometry
geometric transformation
reflection
projective geometry
geometry solved

let (C) be a circle with center O.and let AB be the diameter of it.X' and X" are two points on AB,inwich :OX'=OX"

and X is an arbitrary point on the circle .

XO intersects(C) in point C.

XX' intersect (C) in point M.

XX" intersects (C) in point N.

prove that:     MN and AB and, tangent to circle(C) in point C,are concurrent.

I think this problem is wrong. Fix $M,N,C$ on the circle and let $MN$ and tangent in $C$ meet in $S$. Let $X$ change on the diameter $CO$. There is a diameter through $O$ that the corresponding $X',X''$ exist such that $X'O=X''O$. You say $X'X''$ passes through $S$. Now change position of $X$ and let $SO$ meet  $XM,XN$ at $K,L$. It must be $OK=OL$. So the lines $XK,XL,XO$ and the line through $X$ parallel to $OS$($l$) must be harmonic. Project these line on the line $MN$ and the points $M,N, CO%Error. "capMN" is a bad command.
, l%Error. "capMN" is a bad command.
$ must be harmonic. But three of these points are fixed and one changes. So the problem is wrong.

[u]A equivalent enunciation:[/u]

$\triangle ABC-acute,\ E\in AC,\ F\in AB,\ O\in EF,\ OE=OF,$

$\{A,M\}\subset AO\cap C(O,R)\ \Longrightarrow BC\cap EF\cap MM\ne\emptyset.$

($XX$ is the tangent in the point $X\in w$ to the the circle $w$) The proposed problem (v.v.very nice) is [u]correctly[/u].

[u]The my solution[/u]. I note $L\in EF\cap BC,\ L'\in BC\cap MM,\ D\in BC\cap AO,\ x=m(\widehat {BAD}),\ y=m(\widehat {CAD})$.

$\frac{L'B}{L'C}=\frac{MB}{MC}\cdot \frac{\sin x}{\sin y}=\left( \frac{\sin x}{\sin y}\right)^2=\left( \frac bc \cdot \frac{DB}{DC}\right)^2\Longrightarrow \frac{L'B}{L'C}=$ $\left(\frac{\sin x}{\sin y}\right)^2\ \ (1)\ \ \ \ \ \frac{EA}{FA}=\frac{\sin x}{\sin y}$

$\frac{FB}{FA}=\frac{MB}{MA}\cdot \frac{\sin (C-y)}{\sin y}=\frac{MB}{2R}\cdot \frac{\cos A}{\sin y},\ \frac{EC}{EA}=\frac{MC}{MA}\cdot \frac{\sin (B-x)}{\sin x}=$ $\frac{MC}{2R}\cdot \frac{\cos A}{\sin x}\Longrightarrow$

$\frac{FB}{EC}=\frac{FA}{EA}\cdot \frac{MB}{MC}\cdot \frac{\sin x}{\sin y}=\frac{FA}{EA}\cdot \left( \frac{\sin x}{\sin y}\right)^2$

$\frac{LB}{LC}\cdot \frac{EC}{EA}\cdot \frac{FA}{FB}=1\Longrightarrow \frac{LB}{LC}=\frac{EA}{FA}\cdot \frac{FB}{EC}\Longrightarrow \frac{LB}{LC}=\left(\frac{\sin x}{\sin y}\right)^2\ \ (2)$

From the relations (1) and (2) results $\frac{LB}{LC}=\frac{L'B}{L'C}\Longrightarrow B\equiv C\Longrightarrow BC\cap EF\cap MM\ne \emptyset.$

[u]1. Remark.[/u] I used often this following Lemma:

\[ \underline {\overline {\left| \blacksquare\ \triangle ABC,\ M\in BC,\ x=m(\widehat {MAB}),\ y=m(\widehat {MAC})\Longrightarrow \frac{MB}{MC}=\frac{AB}{AC}\cdot \frac{\sin x}{\sin y}\ \ \blacksquare\right| }}  \]

[u]2.Remark.[/u] $S\in EF,\ MS\perp EF\Longrightarrow \widehat {MSB}\equiv \widehat {MSC}.$

A very nice problem! I didn't hope against hope that this problem be true ! It is true.

If your solution is correct, I must have mistaken because points were not defined correctly in the problem.

the problem is true ali.u can see(ketabe amadegi baraye olampiade riazi.yahya tabesh.) ;)

I shall be delighted ! :clap: [i]Ph-An (past Levi)[/i]

Here is how I understood Ashegh's problem (we don't need A and B):

[color=blue][b]Problem.[/b] Let k be a circle with center O, and let X' and X" be two points in the plane such that the point O is the midpoint of the segment X'X".

Let X be an arbitrary point on the circle k, and let the lines XO, XX' and XX" intersect the circle k at the points C, M and N, respectively (apart from X).

Prove that the lines MN and X'X" and the tangent to the circle k at the point C are concurrent.[/color]

My [i]solution[/i]: Since the points X and C lie on the circle k and the line XC passes through its center O, the segment XC must be a diameter of the circle k. Thus, the center O of the circle k is the midpoint of this segment XC. On the other hand, let M' be the point diametrically opposite to the point M on the circle k. Then, the segment MM' is a diameter of the circle k, and thus it passes through the center O of the circle k, and this center O is the midpoint of this segment.

Now, since the point O is the midpoint of the segments XC, X'X" and MM', the reflection with respect to the point O maps the points X, X' and M to the points C, X" and M', respectively. Since the points X, X' and M are collinear, their images C, X" and M' must also be collinear; in other words, the point X" lies on the line CM'. But we know that the point X" lies on the line NX. Thus, $X^{\prime\prime}=CM^{\prime}\cap NX$. Also, as we know, both lines M'M and XC pass through the point O, so that $O=M^{\prime}M\cap XC$.

Now, the Pascal theorem, applied to the (degenerate) cyclic hexagon CCM'MNX (inscribed into the circle k) yields that the points $CC\cap MN$, $CM^{\prime}\cap NX$ and $M^{\prime}M\cap XC$ are collinear, where the line CC has to be understood as the tangent to the circle k at the point C. In other words, the points $CC\cap MN$, X" and O are collinear. Equivalently, the point $CC\cap MN$ lies on the line X"O. But since O is the midpoint of the segment X'X", the line X"O is the line X'X". Hence, the point $CC\cap MN$ lies on the line X'X". In other words, the lines MN, X'X" and CC are concurrent. This solves the problem.

By the way, Ashegh, please translate the sentence in Farsi you wrote. It looks like it contains information on the source of the problem.

@ Ali.Kh: The point X must lie on the circle k.

  darij

thanks alot darij.the source that i said in persian, was  a persian book.and i taught it doesnt matter wether i said that

or not ;)

and thanks to u levi too :D

I thought X is [b]in[/b] the circle. Here is my solution:
Let $MN$ meet the tangent at $C$ in $K$ and $OK$ meet $XM,XN$ at $A,B$. We prove $O$ is the midpoint of $AB$. And then we are done. We show the lines $XA,XB,XO$ and the line $l$ through $X$ parallel to $AB$ are a harmonic system. Project these lines on the circle so we prove $M,N,C$ and $T=l\cap C$ are harmonic. See these points from $C$ on the circle then it is equivalent to show that $CM,CN,CO,CK$ are harmonic. Project these lines on $MN$ and $B,C,K,CT \cap MN$ are harmonic. This is true if $CT$ is the polar line of $K$. This is true because $OK || AT$ is perpendicular to $CT$


before I give the solution I think I should let you have another try at it by giving you a tip:

try multiplying everything out then moving all the terms with 'x' in it to one side, and all the junk to the other side, and see what you can do from there

[quote="DaReaper"]how do u solve for x in :
\[ a(3tx \minus{} 2b) \equal{} c(dx \minus{} 2)
\]
Also state any restrictions on the variables..[/quote]

Do you need to solve it in terms of another variable or something?

$ a(3tx \minus{} 2b) \equal{} c(dx \minus{} 2)$
$ \Longleftrightarrow (3at\minus{}cd)x\equal{}2(ab\minus{}c)$

ohhh, whoops  :oops: ... thanks, i got it.

here is the solution:
$ a(3tx\minus{}2b) \equal{} c(dx\minus{}2)$
$ 3atx\minus{}2ab \equal{} cdx \minus{} 2c$
$ 3atx\minus{}cdx \equal{} \minus{}2c\plus{}2ab$
$ x(3at\minus{}cd) \equal{} \minus{}2c\plus{}2ab$
$ x \equal{} \frac{\minus{}2c\plus{}2ab}{3at\minus{}cd}$

but that means that $ 3at\minus{}cd \not\equal{} 0$

I hate problems like this.. They look complicated but in reality are very simple..

no, when $ 3at\minus{}cd\equal{}0$ then x can be arbitrary, but we also have $ \minus{}2c\plus{}2ab\equal{}0$

Just remember that you need to isolate the given variable. After that, it's just a little factoring and division.


m,n adaadi tabii and be tori ke   m^2  bar  n^3+1  va  n  bar  m^2+1  bakhsh pazirand   m,n ra biabid?????

m^2|n^3+1

n|m^2+1

 :idea:


What is the smallest positive integer divisible by all of the first ten positive integers?

Prime factorize, and then take the highest prime powers used.

First 10 positive integers:

$ 1$
$ 2$
$ 3$
$ 4\equal{}2^2$
$ 5$
$ 6\equal{}2\times 3$
$ 7$
$ 8\equal{}2^3$
$ 9\equal{}3^2$
$ 10\equal{}2\times 5$

$ 2^2\times 3^2\times 5\times 7\equal{}4\times 9\times 5\times 7\equal{}\boxed{1260}$

$ 2^3$ is the highest power of $ 2$, so it should be $ 8\times{9}\times{5}\times{7}\equal{}\boxed{2520}$.


conics
parabola
geometry
geometric transformation
reflection
function
analytic geometry

A parabola with equation $y=ax^{2}+bx+c$ is reflected about the $x$ axis, then both it and it's reflection are translated horizontally in opposite directions $5$ units to become the graphs $y=f(x)$ and $y=g(x)$. What describes the graph $y=(f+g)(x)$?

What do you mean by what describes the graph? It's a line...

Yeah, but why is it a line? I don't know what f+g means...

edit: thought you said y-axis :blush: i really need to read these more carefully...

tjhance, when a graph $f(x)$ is reflected across the x-axis, the new graph is $g(x) =-f(x)$

[hide]
$f(x) = a(x-5)^{2}+b(x-5)+c$
$g(x) =-a(x+5)^{2}-b(x+5)-c$

Adding together gives $y =-20ax-10b$
[/hide]

When a graph $f(x)$ such as this one is reflected across the x-axis, for a certain x input, the output, or value of the function at x is negated, therefore the reflection across the x-axis is $-f(x)$. When you do a horizontal translaction of $h$ you plug in $(x-h)$ for x.

When you wanted to find $(f+g)(x)=f(x)+g(x)$ you just plug it in and add the functions.

I have a question for this problem. Heres how I did it.
Original parabola $y=ax^2+bx+c$. Let this be $y=a(x-h)^2+k$. Reflection becomes  $y=-a(x-h)^2+k$. Since it doesn't matter which way as long as they're opposite, the horizontal translation gives. $y=a(x-h+5)^2+k$. and $y=-a(x-h-5)^2+k$.

Now expansion. $y=a(x-h+5)(x-h+5)+k=a(x^2-xh+5x-xh+h^2-5h+5x-5h+25)+k=a(x^2+h^2-2xh+10x-10h+25)+k$.
Reflection. $y=-a(x-h-5)(x-h-5)+k=-a(x^2-xh-5x-xh+h^2+5h-5x+5h+25)+k=-a(x^2+h^2-2xh-10x+10h+25)+k$

Now we can add get rid like terms. $2y=ax^2+ah^2-2axh+10ax-10ah+25a+k-ax^2-ah^2+2axh+10ax-10ah-25a+k=20ax-20ah+2k$
Now we get $y=10ax-10ah+2k$. So 10a is slope, (-10ah+2k) is constant but we dont care about that.

[b]Here where I'm Confused[/b]. $10a$ is not zero, or else original equation would be line not parabola which is explicitly stated in problem. Horizontal line is in form $y=k$ ($k$ is constant) and slope is $0$ so theres no $x$ term. But $10a$ is not zero so this must be a non-horizontal line answer $D$, which is what I got but apparently answer is $C$, horizontal line. Can someone explain thanks.

P.S. Other wrong answer explanation: $A$-its not a parabola, $B$-its not a parabola, $C$-its not horizontal, $E$-its not cubic.

Can someone please explain to me my question (see previous post). In the AOPS wiki for this problem, it says answer is C even though final equation they got has a non-zero slope as well which means should be D, non-horizontal line instead of C (horizontal line). I really want to know what I did wrong. Thanks alot!

[quote="zhuzhu88"]Reflection becomes  $y=-a(x-h)^2+k$.[/quote]
It should be $y=-a(x-h)^2-k$

Thanks, LeFromage. But if redid entire problem (since I put reflection as $y=-a(x-h)^2-k$ instead of $y=-a(x-h)^2+k$) with that equation, the final equation would still be $y=10ax-10ah$ where $-10ah$ is constant, $10a$ is non zero slope since $a$ is non zero. So the answer should still be D, why is it C?


00:30 am is my limit. My parents do not allow me to stay up beyond that.  :(
[img]http://www.slocartoon.net/slike/recenzije/brother_bear/brother_bear5.jpg[/img]

Usually like 10-11. past few days its been 12 though. :(

[quote="Koda"]00:30 am is my limit. My parents do not allow me to stay up beyond that.  :(
[/quote]
huh you're at a college aren't you?

me:around 11:30-12:30

What do you guys usually do staying up so late :ninja:  :huh:  :roll:

11:30-12:30 sounds like homework.

3:30 sounds like fooling around  :wink:

i usually go to bed.... like 9-10.. it depends.. but sometimes i stay on till 11..

on fridays i don't go to sleep :rotfl:  i stay up all night..

u stay up all night on fridays?
why?
-jorian

9:00 my mom freaks out after that (yes, she is really paranoid 'bout my health)
unless im doing hmw...den she sometimes lets me stay up to about 10 before she....yeah
of course...once in a while i stay up till 3 playing video games (wen i play w/ my friends)

10 to 11...  If I stay up any longer it's impossible to get out of bed in time to get to school. :P

Same here, 10 to 11. I'm trying to write a book, but it's not going so well, so I've been spending about an hour a night on that.

Hmmm... I fall asleep naturally around 10:30. :(


[quote="myc"]${n\choose0}+{n\choose1}+{n\choose2}+{n\choose3}...{n\choose n}$ prove that this equals $2^{n}$.[/quote]
Old and classic.
[hide]From Newton's binomial :
$2^{n}= (1+1)^{n}= \sum_{k=0}^{n}\binom{n}{k}1^{k}\cdot1^{n-k}={n\choose0}+{n\choose1}+\ldots+{n\choose n}$.
It can be also done by elementar combinatorics: the left-hand side is the number of subsets of a $n$-element set (number of $0$-element subsets + number of $1$-element subsets + $\ldots$ + number of $n$-element subsets). But each subset can be written as a $n$-term binary sequence in which the $i$-th term tells us if the $i$-the elemts of our $n$-element set is involved in the subset. There are $\underbrace{2\cdot\ldots\cdot 2}_{n}= 2^{n}$ such sequences, so ${n\choose0}+{n\choose1}+\ldots+{n\choose n}= \text{the number of subsets}= 2^{n}$. [/hide]


Let n be a positive integer.Prove ([b]using decimal expansion [/b]of $\dfrac{1}{n}$)that there exists a mutiple of n of the form 99...900...0
(An alternative proof is obtained by pigeon hole principle.)

[quote="The Captain"][b]using decimal expansion [/b]of $\dfrac t {1}{n}$[/quote]

What do you mean? :huh:

I've just copied this statement from my book.I myself did'nt get the question. :)


Ah, I see.  Oops.  :oops:
Thanks for pointing it out!

Good job
Especially the number one was my favorite one because most people don't really know how to work out complex fraction with no defined valuable and don't know whether they are ended or not..
Number two, I can't remember answer since it's quite time ago but that looks right..
Except.. How do you work out that LaTEX thingie?

There is another forum on Latex.  It's at the bottom of the page with all the forums.


geometry
angle bisector
geometry proposed

Let the tangents at A and B meet at T.
Note that $\angle AEM = \angle AEC = \angle DEB$, therefore the lines $EM,ED$ are isogonal conjugates in triangle $EAB$. But EM is a median, therefore ED is a symmedian, and so T,E,D are collinear.

Now $\angle TAO = \angle TBO = \angle TNO = 90^{\circ}$, therefore T,A,B,N is cyclic. Now it is obvious: $\angle ANE = \angle ANT = \angle BNT = \angle BNE$.

The proof in fact could be shorter:
$\angle EMB = \angle DMB$ (easy), then the quadrilateral EADB is harmonic, then $\angle ENA = \angle ENB$  :wink:

[quote="Litlle 1000t"]$AB$ and $CD$ are parallel chords of circle $S$.$M$ is midpoint of $AB$. $CM$ intersects with $S$ at $E$. $N$ is midpoint of $DE$. Prove that $\angle ANE=\angle BNE.$[/quote]

It is easy to show that $\angle{CEA}=\angle{DEB}$ (tha follows directly by the parallel lines).So $ET$ is the symmedian of the triangle $EAB$, so $\frac{TA}{TB}=\frac{EA^{2}}{EB^{2}}$ [b](1)[/b].

Denote by $T$ the intersection of lines $AB$ and $ED$.
We have: $\frac{(EAD}{(EBD)}=\frac{h_{1}}{h_{2}}=\frac{TA}{TB}$
But $\frac{(EAD)}{(EBD)}=\frac{AE \cdot AD}{BE \cdot BD}$ and finally we get:

 $\frac{TA}{TB}=\frac{AE \cdot AD}{BE \cdot BD}$ [b](2)[/b]

From relations [b](1)[/b] and [b](2)[/b] we get $\frac{EA}{EB}=\frac{AD}{BD}\Longrightarrow \frac{TA}{TB}=\frac{AD^{2}}{BD^{2}}$ and thus $DT$ is the symmedian of the triangle $ABD$.

Similarly we prove that $BT$ is the symmedian of the triangle $EBD$ and $AT$ is the symmedian of the triangle $EAD$. So $\angle{NAD}=\angle{BAE}=\angle{EDB}$ and $\angle{NBD}=\angle{EBA}=\angle{EDA}$

But $\angle{ANE}=\angle{NAD}+\angle{NDA}$

and $\angle{BNE}=\angle{NBD}+\angle{NDB}$

Hence $EN$ is the angle bisector of the angle $\angle{ANB}$


The roots of two quadratic polynomials are negative integers and the polynomials have one root in common.

Can the values of the polynomials at some positive integer be 19 and 98 ?

\begin{eqnarray*}f(x) = \frac {98}{2\cdot3}(x + 1)(x + 2) & \qquad & g(x) = \frac {19}{2\cdot4}(x + 1)(x + 3) \\
f( - 1) = g( - 1) = 0 & \qquad & f( - 2) = g( - 3) = 0 \\
f(1) = 98 & \qquad & g(1) = 19\end{eqnarray*}
If all coefficients are integers, then it's not possible. Let \[ f(x)=a(x+r_1)(x+r_2)\qquad g(x)=b(x+r_1)(x+r_3),\] where $ r_1,r_2,r_3$ are positive integers (therefore, $ -r_1,-r_2,-r_3$ are negative integer roots) and $ a$ and $ b$ are nonzero integers. For $ f(n)=19$ and $ g(n)=98$ for some positive integer $ n$, then $ (n+r_1)$ divides both $ 19$ and $ 98$. Since $ (n+r_1)>1$, this is not possible.


Hello. Can anybody post Problems from Bulgarian National Olympiad in Mathematics 2002 and 2003 and Team Selection Tests for IMO. And just for Curiosity enclose the hardest problem to have ever appeared in Bulgarian OLympiad.

                                                  THANKS

hey, I'll try to request some tough problems from the Bulgarian Olympiads.


The pressure exerted on a copper block, with mass $ m = 0.010\,kg$, is increased from $ 0$ to $ 10^{8}$ Pa, while keeping the temperature constant. The density of copper is $ \rho = 8900\, kg/m^{3}$, and its compressibility is $ k_{T}= 31.5 \times 10^{-11}\, Pa^{-1}$ (both these quantities are constant). Compute the work done on this process.

If everything is fine, then the work done on the block is
$ W=\int_{0}^{P}{pdV}$. Also if $ k_{T}$ is the coefficient of linear compressibility then $ \triangle V = V (k_{T}p)^{3}$, thus $ dV=3Vk_{T}^{3}p^{2}dp$, so $ W=\int_{0}^{P}{3Vk_{T}^{3}p^{3}dp}=\frac{3}{4}Vk_{T}^{3}p^{4}$.
Is this correct?

How is exactly compressibility defined???

The coefficient of isothermal compressibility is defined as $ k_{T}=-\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_{T}$.

Then I guess the thing I did wrong there was $ \triangle V= V(k_{T}\cdot p)^{3}$, it should have been $ \triangle V =-Vk_{T}p$ I guess, or $ dV=-Vk_{T}dp$, thus the work done will be $ W=\int_{0}^{P}p^{2}Vk_{T}dp= Vk_{T}P^{3}/3$.

Oh, sorry man :blush:, it's just stupid , I don't know why I put $ p^{2}$ in the integral  :huh:  it was supposed to be the $ \int_{0}^{P}pdV$;
$ W= \int_{0}^{P}{pVk_{T}dp}= \frac{Vk_{T}P^{2}}{2}$.
Don't tell me that this is incorrect.

You can't integrate in that way because $ V$ in $ dV=-Vk_{T}dp$ is not constant...

Did you look at the value of $ k_{T}P$ its of power $ -3$ then the maximal change in volume is negligibly small...

To make myself more concrete I'll perform the calculation...
As we saw $ dV =-Vk_{T}dp\Rightarrow \frac{dV}{V}= k_{T}dp \Rightarrow d(\ln \frac{V}{V_{0}})= k_{T}p \Rightarrow V=V_{0}e^{k_{T}p}$.
$ k_{T}P$ is a very small number of power $ -3$, and $ e^{-1000}$ can be considered $ 1$, I think :lol:

I guess the expression I got is with the right sign. You see, the potential energy of the block is increased, thus the external forces did positive work...

Here's what I meant:

$ dV =-Vk_{T}dp \Rightarrow \frac{dV}{V}=-k_{T}dp \Rightarrow \int^{V}_{V_{0}}\frac{dV}{V}=-\int^{P}_{P_{0}}k_{T}dp$

$ \Rightarrow \ln\left(\frac{V}{V_{0}}\right) =-k_{T}P \Rightarrow V = V_{0}e^{-k_{T}P}$

as $ P_{0}= 0$. Now let's not make approximations and do the exact calculation.


says some things like latin thread.
actually i have a question i know there are different japanese alphabets(is that what you call them?): hiragana,katakana and another one(i think i forget) which one is better to study or is it normal to study them all? JW

no-i can't speak it if that's what your trying to do...i just started to study it on a HP site(you'd be amazed how much they can teach you!) but decided I needed a proper curriculum.

konichiwa :maybe:

random japanese guy: Hwa chuching ya! Chi nakeguma kurinami hwa hung?

me: uuummmm... konichiwa?

 :D  :rotfl:

Hiragana and katakana are syllabic scripts that you use for foreign words and names (katakana) and for Japanese words that aren't from Chinese words (hiragana). You use kanji (which are pretty much just Chinese characters) for most other stuff. If you don't know the kanji for a word you can write in hiragana and people will understand it. That's useful for when you're still learning kanji (it's not easy).

You need to learn all of them if you want to be fluent in Japanese. Kanji are probably least important at the start since you can get by writing and speaking without them, but if you ever want to read stuff you'll need to learn a lot of them eventually.

[quote="junggi"][quote="Iversonfan2007"]tampura, nissan, subaru, suzuki, toyota, saki, sushi, sashimi![/quote]
is that from scary movie 4?[/quote]

 :wink:  :wink:  :wink:

[quote="Iversonfan2007"][quote="junggi"][quote="Iversonfan2007"]tampura, nissan, subaru, suzuki, toyota, saki, sushi, sashimi![/quote]
is that from scary movie 4?[/quote]

  :wink:[/quote]
 :rotfl: that was my fav scen from the movie :rotfl:

shizugani.

nihongo ga hanasenai hito ga [b]spammers[/b] da ne.  nihongo wo benkyoushite, renshyuu ga hoshii da kara, kono [i]post[/i] ga aru.  ashita wa gakkou no hajime hi da shi, nihongo no AP ga toru tsumori da shi, natsu yasumi ni zen zen benkyoushinakattan do shi, boku wa [b]screwed[/b] da   :(

And that dash is the word for tomororw, which they censored...


The subject sounds philosophical, but why does

(e^x - e^-x)/(e^x + e^-x) = (e^(2x) - 1)/(e^(2x) + 1) ?

hello, you will get it by multiplying the numerator and denominator with $ e^x$:
$ \frac{(e^x\minus{}e^{\minus{}x})\cdot e^x}{(e^x\plus{}e^{\minus{}x})\cdot e^x}\equal{}\frac{e^{2x}\minus{}1}{e^{2x}\plus{}1}$
Sonnhard.


bobby shen won a close match over david yang. david had him but answered half a second too late. david was being rude the whole time though. good job bobby! did anyone on aops make cd?

bobby was 3rd(somehow) and david was 1st. gah. he was just so rude!


function
inequalities
geometry unsolved
geometry

It is the third time,I write this folowing sentences: It's problem 4 ,Mathematics and Youth,month 8,2007(issuse).Sorry

The deadline is over; the problem can be discussed again.

  darij

[quote="nmtvn"]Let $ P$ be an arbitrary point in a quadrilateral $ ABCD$ such that  $ \widehat{PAB}$, $ \widehat{PBA}$, $ \widehat{PBC}$, $ \widehat{PCB}$, $ \widehat{PCD}$, $ \widehat{PDC}$, $ \widehat{PAD}$, $ \widehat{PDA}$ are all acute angles. Let $ M$, $ N$, $ K$, $ L$ be the feet of the perpendiculars from the point $ P$ to the lines $ AB$, $ BC$, $ CD$, $ DA$. Find the smallest possible value of $ \frac {AB}{PM} \plus{} \frac {BC}{PN} \plus{} \frac {CD}{PK} \plus{} \frac {DA}{PL}$.[/quote]

By using convex function inequalities, I think!
Due to: $ AB.PM \plus{} BC.PN \plus{} CD.PK \plus{} DA.PL \equal{} constant (2S_{ABCD})$

I 've tried to find the position where the sum gets its minimum value. And I conjecture that P can be constructed as follow: Let E, F be the intersection of AB,CD and BC,AD. Then P is the intersection of the angle bisectors of $ \angle{AEC}$ and $ \angle{DFC}$.

Hope that it helps solving the problem.

[quote="nmtvn"]Let $ P$ be an arbitrary point in a interior of a given quadrilateral $ ABCD$. Let $ M$, $ N$,
$ K$, $ L$ be the projections of the point $ P$ to the lines $ AB$, $ BC$, $ CD$, $ DA$ respectively. Find the smallest possible value of $ \frac {AB}{PM} \plus{} \frac {BC}{PN} \plus{} \frac {CD}{PK} \plus{} \frac {DA}{PL}$.[/quote]
I"ll apply the C.S.B.'s inequality over $ \mathcal R^*_ \plus{}$ : $ \sum_{k \equal{} 1}^n\frac {x_k^2}{y_k}\cdot\sum_{k \equal{} 1}^ny_k\ge\left ( \sum _{k \equal{} 1}^n x_k \right )^2\ .$

Therefore, $ \sum\frac {AB}{PM} \equal{} \sum\frac {AB^2}{AB\cdot PM} \equal{} \sum\frac {AB^2}{2[APB]}\ge \frac {\left(\sum AB\right)^2}{2[ABCD]}$.

We have equality iff $ \frac {AB}{PM} \equal{} \frac {BC}{PN} \equal{} \frac {CD}{PK} \equal{} \frac {DA}{PL}\ \ (*)$.

The main problem is when realizes $ (*)\ ?!$

I think that you missed something in your proof.

The equality in C.B.S occurs iff  $ \frac{x_i}{y_i} \equal{} \frac {x_j}{y_j} \forall i,j$

So the equality (in your proof) must occur when $ PM\equal{}PN\equal{}PK\equal{}PL$ which is not possible if ABCD is not an inscribed circle.


Hi guys can you help me out? I am stuck on these two questions:
1. A fatty acid has the formula C16 H32 O2 is A.saturated B. Unsaturated C. Branched

2. Three molecules of the fatty acid C16 H32 O2 are joined to a molecule of glycerol C3H8O3. The resulting formula A. C48H46O6 B. C48H98O9   C.   C51H98O61 D. C51H98O61 or E. C51H104O9.

Thanks guys!

1. We know that we first have a nice long carbon chain, 16 carbons long. At one end, we have a double bonded oxygen and a hydroxyl group (OH). So, that takes care of one hydrogen, both oxygens, and one carbon. We have 15 carbons left; 14 of them can make two more bonds, 1 can make 3, for a total of 31 bonds. We also have 31 hydrogens left. So, all the hydrogens bond to all the spots on the carbons, and there is no way for the carbons do undergo additional bonding. So, it's saturated.

2. Three molecules of fatty acid consist of a total of $C_{48} H_{96} O_6$. Add the glycerol part, and we get $C_{51} H_{104} O_9$. However, this is not the answer. Remember that whenever the fatty acid bonds to the glycerol, it undergoes dehydration synthesis, and loses a water molecule, or 2 hydrogens and one oxygen. It does this three times, so it loses $H_6 O_3$. Thus, the resulting molecule has formula $C_{51} H_{98} O_6$. (Is there a typo? Choice C and D look the same...)

Thank you very much for the explanation and answer. I appreciate it.
Have a great day! :lol:


What do you mean? The most you can do (without plugging away in a calculator, of course) is factor out the 3 - the radicals themselves cannot be simplified.

$ 3\sqrt6\minus{}3\sqrt{2}\equal{}3\left(\sqrt{6}\minus{}\frac{\sqrt{6}}{\sqrt{3}}\right)\equal{}3\sqrt{6}\left(1\minus{}\frac{1}{\sqrt{3}}\right)$

That's about all I can do with it.


abstract algebra
group theory
superior algebra
superior algebra unsolved

let $ G$ be a abelian group , and $ \phi : G\longrightarrow G$ be a homorphism , Prove $ G\equal{}\text{Ker}\phi \times \text{Im} \phi$

For every $ i \in \mbox{Im }\phi$, we pick a fixed element $ g(i) \in G$ such that $ g(i) ^{\phi} \equal{} i$.
Now, define $ \theta$ like this:

\[ \theta : \mbox{Ker }\phi \times \mbox{Im }\phi : (k,i) \mapsto k\cdot g(i).\]

Clearly, $ \theta$ is a morphism since $ G$ is abelian. Now, $ \theta$ is also injective. Assume $ k\cdot g(i) \equal{} k'\cdot g(i')$, then it follows that
\[ i' \equal{} g(i')^{\theta} \equal{} (k'\cdot g(i'))^{\theta} \equal{} (k\cdot g(i))^{\theta} \equal{} i.\]
Now clearly, also $ k \equal{} k'$.

$ \theta$ is also surjective, since $ i^{\theta^{\minus{}1}} \equal{} g(i)\cdot \mbox{Ker }\phi$.

Therefore, we have proven that $ \mbox{Ker }\phi \times \mbox{Im }\phi \cong G$.

[b]Remarks:[/b]
[list]
[*]I denote $ g^{\phi}$ for the action of $ \phi$ on $ g$.
[*]Note that it follows from the first isomorphism theorem that $ \left|G\right| \equal{} \left|\mbox{Ker }\phi\right|\cdot \left|\mbox{Im }\phi\right|$, hence it actually suffices to prove surjectivity, but in case you shouldn't know this theorem, the above proof doesn't use it.[/list]

[quote="Jan"]Clearly, $ \theta$ is a morphism since $ G$ is abelian.[/quote]$ \theta$ is a homomorphism iff $ g$ is. How do you arrange that $ g$ is a homomorphism? BTW: You cannot in general. ;)
[quote]Note that it follows from the first isomorphism theorem that $ \left|G\right| \equal{} \left|\mbox{Ker }\phi\right|\cdot \left|\mbox{Im }\phi\right|$, hence it actually suffices to prove surjectivity, but in case you shouldn't know this theorem, the above proof doesn't use it.[/quote]This can only be applied if $ G$ is assumed to be finite.

Of course you are right, I didn't think things through carefully. Thanks for pointing out!


Find all funtions such that the following equality holds for all $ x;y \in R$
$ f(x^{3}-y)+2y(3f^{2}(x)+y^{2})=f(y+f(x))$

You've already posted this problem here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159278[/url]

Do not double post.  Find the right forum, and then post only once in that forum.


modular arithmetic
algebra
difference of squares
special factorizations
number theory
prime factorization

Question 1 of today's (Friday) paper:

Find four prime numbers less than $100$ that are factors of $3^{32}-2^{32}$.

[hide]
It factors into this:
\[(3+2)(3^{2}+2^{2})(3^{4}+2^{4})(3^{8}+2^{8})(3^{16}+2^{16})\]
The first three factors are the primes 5, 13, and 97. Now for the tricky part: suppose $p$ is a prime that divides $3^{8}+2^{8}$. Then $\frac{3p+3}{2}$ is an 8th root of -1 modulo $p$, so it is a root of order 16 modulo $p$. Therefore, we find that $p-1$ is divisible by 16; similarly, if $q$ is a prime dividing $3^{16}+2^{16}$, then $q-1$ is divisible by 32. This makes 17 the only other possibility, and in fact, $3^{8}\equiv-1, 2^{8}\equiv 1\pmod{17}$ so 17 works, making the primes $\boxed{5,13,17,97}$.
[/hide]

[quote="nutz_for2.718281828"][hide="solution"]We can do some expanding: (3-2)(3+2)(3^2+2^2)(3^4+2^4)(3^8+2^8)(3^16+2^16)

We can take the prime factorization of that:

5*13*5*7*97*17*401*(something really big)

The only prime factors below 100 are 5, 7, 13, 17, 97[/hide]

I decided to find a fifth.[/quote]

$7$ does not divide $3^{32}-2^{22}$.  There is a remainder of $5$.

[quote="nutz_for2.718281828"][hide="solution"]We can do some expanding: (3-2)(3+2)(3^2+2^2)(3^4+2^4)(3^8+2^8)(3^16+2^16)

We can take the prime factorization of that:

5*13*5*7*97*17*401*(something really big)

The only prime factors below 100 are 5, 7, 13, 17, 97[/hide]

I decided to find a fifth.[/quote]

Nope, 7 doesn't work; $3^{32}\equiv 2$ and $2^{32}\equiv 4$ giving a remainder of 5. And, as I posted above, there is no fifth prime less than 100 that works.

Decided to find the sixth (and seventh)  :)
[hide]As  scorpius[b]19 [/b]said [quote]if q is a prime dividing $3^{16}+2^{16}$, then $q-1$ is divisible by $32$[/quote]
If we wanted, we can take $19$ from the name, multiply by $5$ (digits on my hand) and multiply by $32$ to get q-1 and find 3041 is indeed a factor) :) So other factors (other than 401) are 3041 and 14177 [/hide]

[quote="rodgab"][quote="pianoforte"][hide]Using difference of squares, it factors to
$(3-2)(3+2)(3^{2}+2^{2})(3^{4}+2^{4})(3^{8}+2^{8})(3^{1}6+2^{1}6)$

$3+2=\boxed{5}$
$3^{2}+2^{2}=\boxed{13}$
$3^{4}+2^{4}=\boxed{97}$
$3^{8}+2^{8}=6817=401\times \boxed{17}$.[/hide][/quote]

How did you prime factorise 3^8 +2^8 without use of a calculator?[/quote]
Guess and check works. Note that 17 is the 7th prime so it wouldn't even make a difference.


My school's math club has 6 boys and 8 girls.  I need to select a team to send to the state math competition.  We want 6 people on the team.  In how many ways can I select the team to have 3 boys and 3 girls?

$ \binom{6}{3} \times \binom{8}{3} \equal{} 20 \times 56 \equal{} \boxed{1120}$.
boys are $ \binom{6}{3}$
and girls are $ \binom{8}{3}$


[quote="mdk"]Show that, if n is odd, then 1^n + 2^n + · · · + n^n is divisible by n^2.[/quote]

Since $ k^{n}+(n-k)^{n}\equiv 0 (n^{2})$, $ 1^{n}+2^{n}+\cdots+n^{n}\equiv n^{n}(n^{2})$. Clearly, $ n^{2}|n^{n}$, so we are done.


Soon I'm getting a Mac and I am going to want to get [tex]\LaTeX[/tex] for it.  What distributions are good?

http://itexmac.sourceforge.net

or

http://www.uoregon.edu/~koch/texshop/texshop.html

I prefer the former. You'll need a teTeX distribution, however.


[color=darkblue]Let $w$ be a circle with the diameter $[AB]$and let $M,N$ be two points which belong to the circle $w$. For two points $X\in BM$, $Y\in BN$ so that $AX=AY$ define two points $U\in AM$, $V\in AN$ so that $XU\perp XA$ and $YV\perp YA$. Prove that $UV\perp AB\ .$[/color]

Just show that the cuadrylateral VNMU is cyclic. Then it's easy to show the statement. Another special case of Perpendicularity I :D .

Your imagination is awesome.


superior algebra
superior algebra unsolved

Let $ (A,\plus{},.)$ Be a ring with the unit element $ 1_k$

and $ \forall x \in A$   $ x^{12}\equal{}x$

Prove that:
$ \forall x \in A$   $ x\equal{}\minus{}x$

$ 1 \equal{} ( \minus{} 1)^{12} \equal{} \minus{} 1$ gives $ x \equal{} \minus{} x$ for all $ x\in A$.
In fact it's easy to see that $ A$ must be a commutative $ \mathbb{F}_2$-algebra
containing only idempotents. (i.e. $ x^2 \equal{} x$ for all $ x\in A$).

Ok, now can someone prove

1
$ x^8 \plus{} x^4 \equal{} 0$
using the fact that $ (x \plus{} 1)^{12} \equal{} x \plus{} 1$
2
$ x^2 \equal{} x$

no one??

[i]The message is too small. Please make the message longer before submitting.[/i]

Try starting a new thread.  And be a little more clear; I can't tell what quantifiers you mean to use in your last question.


A function $f:N \rightarrow N$ satisfies the following:

     (i) $f(xy)= f(x)+f(y)-1$
     (ii) there exist a finite number of $x$, such that $f(x)=1$
     (iii) $f(30)=4$

Determine $f(14400)$.

Note: $N=\{1,2,3,...\}$

This was one of the warmup problems at the canadian winter camp..

Btw, do I know you?


[Moderator edit: These problems all have extra topics now: http://www.artofproblemsolving.com/Forum/resources.php?c=174&cid=95&year=1994 ]

1. For n  N, let a_n denote the closest integer to  :sqrt: n.
    Evaluate the sum :Sigma: 1/(a_n)^3 , (from 1 to n)

2. n blue lines, no two parallel and no three intersecting in the same point , are drawn in a plane. An intersection of two blue lines is called a blue point. Through two blue points, not already joined by a blue line, a red line is drawn. An intersection point of two red lines is called a red point and an intersection point of a red line and a blue line is called a purple point. What is the maximal number of purple points ?

Dear $matematikator$ have you got Turkey 2005 National Olympiad  problems.If yes can you post them in National Olympiads section. :)

3. Let f : R^+ ---> R^+ be un increasing function. For each u in R^+, denote the greatest lower bound of the set { f(t) + ( u / t ) : t>0 } by g(u). Show that
a) If x<= g(xy), then x <= 2f(2y).
b) If x <= f(y), then x <= g(xy).

4. The incircle of a triangle ABC touches the sides BC and AC at the points D and E respectively.
K is a point on CB with CK = 3BD, and L is a point on CA with AE= CL. Let AK intersect BL at P. If Q is the midpoint of BC, I is the incenter and G is the center of gravity of ABC. Show that
a) IQ || AK.
b) Area (AIG) = Area (QPG).


inequalities
limit
calculus
calculus computations

we have the following sequence $ {x_n}$ :
\[ x_0\equal{}\sqrt{2}\]
\[ x_{n \plus{} 1}  \equal{} \frac{{x_n (x_n^2  \plus{} 3a)}}{{3x_n^2  \plus{} a}},a > 2
\]
Find limit of $ {x_n}$

You can obtain an inequality such that $ |x_{n\plus{}1}\minus{}\sqrt{a}|<r|x_n\minus{}\sqrt{a}|\ (|r|<1).$

$ x_{n\plus{}1}\minus{}\sqrt{a}\equal{}\frac{(x_n\minus{}\sqrt{a})^3}{3x_n^2\plus{}a}$ then because $ x_0 <\sqrt{a}\longrightarrow x_n< \sqrt{a},\forall n\in\mathbb{N}$ .

$ x_{n\plus{}1}\minus{}x_n\equal{}\frac{2x_n(a\minus{}x_n^2)}{3x_n^2\plus{}a}> 0$

So the sequence has limit $ b$. Solving the equation : $ b\equal{}\frac{b(b^2\plus{}a}{3b^2\plus{}a},b\ge \sqrt{2}$ we have $ b\equal{}\sqrt{a}$

Hence :

$ \lim_{n\to\infty}x_n\equal{}\sqrt{a}$

[quote="TTsphn"][b]Other[/b] Find all $ a$ for this sequence converge.[/quote]

For all $ a\ge 0$

Please post solution and find it limit in some case :wink:
It is not very hard.

oh yes, TTsphn , but in other case, we also use the same method, i used to before :wink:

Oh yes but I want complete solution.
For example when $ a<0$ you can not consider $ x_n\minus{}\sqrt{a}$
I know that you solved it but please post a complete solution.

This is an easy problem. You may find such problems in a book by Nguyen Luu Cuong Vol. 1. I do not know exactly its name and I read it last 7 years.

[quote="zaizai-hoang"]yeah, it is also my solution. I think it is not hard to prove. I used to derivate method and a little knowledge about limit sequence  :)
Quang, can you post the solution? I don't have much time to post this one into forum  :P[/quote]

Me too, Vu Hoang


inequalities
geometry
circumcircle
inradius
trigonometry
geometry proposed

Let $ABC$ be a triangle and $R$, $r$ be respectively the circumradius and inradius of $ABC$. Prove that for every point $P$ in the plane of triangle $ABC$, we have $PA+PB+PC \geq 2\sqrt{r\left(4R+r\right)}$.

May be used $\sum (s-a)(s-b)=r(4R+r)$...

Here is an extremely ugly solution. First of all, the inequality is obviously true if $\max\{A,B,C\}\ge120^\circ$.
Assume $\max\{A,B,C\}<2\pi/3$. Thus we may take P to the "normal" Fermat point.
Recall the well-known fact that FA+FB+FC=AD, where D is the point outside $\triangle ABC$ s.t. DBC is equilateral.
Using cosine law, we find that $AD^2=2\sqrt{3}S+\frac{a^2+b^2+c^2}{2}$, where S is the area of $\triangle ABC$.
As usual, make use of the incircle subsitution(a=y+z,etc.). The original inequality is equivalent to \[2\sqrt{3xyz(x+y+z)}+x^2+y^2+z^2\ge3(xy+yz+zx)\]Note that we have $2\sum xy-2\sqrt{3xyz(x+y+z)}=2\frac{\sum yz(x-y)(x-z)}{\sum xy+\sqrt{3xyz(x+y+z)}}$ and $\sum x^2-\sum xy=\sum(x-y)(x-z)$. Thus it suffices to prove \[\sum(1-2\frac{yz}{\sum xy+\sqrt{3xyz(x+y+z)}})(x-y)(x-z)\ge0\]This obviously follows from generalized Schur/Vornicu Schur using $\max\{A,B,C\}<2\pi/3$.

Very cool, mecrazywong. Thanks for this nice solution.
 You can solve this problem by using the following inequality:
   $\sum \frac{PA.PB}{ab} \geq 1$


Let $ f(n)$ ($ n$ positive integer) be the minimum $ i$ such

$ 1-\frac{1}{n}=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{i}}$ with $ a_{j}$ positive integers $ 1\leq j\leq i$

i) Prove that $ f(n) < n$

ii) Find all $ n$ such $ f(n)=1$

iii) Find all $ n$ such $ f(n)=2$

iv) Find $ f(2007)$

1) Because $ 1-\frac{1}{n}=\sum_{i=1}^{n-1}\frac{1}{i(i+1)}$ therefore $ f(n)\leq n-1<n$.
2) If $ f(n)=1$ then $ \frac{n-1}{n}=\frac{1}{a}(a\in\mathbb N)$ therefore $ \frac{n}{n-1}\in\mathbb N$, so $ n=2$. If $ n=2$ then $ 1-\frac{1}{2}$ write as $ \frac{1}{2}.$

iv) I got an example for $ f(2007)$ such as $ f(2007) \le 7$. is it equal?

[quote=Rust]...If n is prime, obviosly $ f(p)=p-1$. It is not hard ...[/quote]

Unfortunately it is not true, for example:
$ 1-\frac{1}{7}= \frac{1}{2}+\frac{1}{3}+\frac{1}{42}$
$ f(7)=3$
However, $ f(p)\le p-1$

Anyone?
[quote="mszew"]Let $ f(n)$ ($ n$ positive integer) be the minimum $ i$ such

$ 1 \minus{} \frac {1}{n} \equal{} \frac {1}{a_{1}} \plus{} \frac {1}{a_{2}} \plus{} \cdots \plus{} \frac {1}{a_{i}}$ with $ a_{j}$ positive integers $ 1\leq j\leq i$

iv) Find $ f(2007)$[/quote]


Why does [tex]\frac{dy}{dx}=\frac{dy}{dz}\cdot\frac{dz}{dx}[/tex] mean the chain rule (I get the chain rule in terms of the f' notation)? They define it so that [tex]z=g(x)[/tex] and [tex]y=f(z)[/tex], thus [tex]y=f(g(x))[/tex].

You have a little typo there.  The basic idea is, "The change of the compound function with respect to the independent variable is equal to the products of the changes of the two functions involved, each with respect to their domain."  That probably doesn't help too much, but try making substitutions.  The key point is that
dy/dz is the change in f(g(x)) with respect to g(x), which is equal to f'(g(x)).

right, JBL I think that's a pretty good explanation... this is a hard one for most to understand... To give an image of what JBL said:

When taking the derivative of [tex]h(x)=f(g(x))[/tex], the [tex]f'(g(x))[/tex] is the same as [tex]\displaystyle \frac{dh(x)}{dg(x)}[/tex] and the [tex]g'(x)[/tex] is the [tex]\displaystyle \frac{dg(x)}{dx}[/tex] which comes out as [tex]\displaystyle \frac{dh(x)}{dg(x)}\cdot \frac{dg(x)}{dx} = \frac{dh(x)}{dx}[/tex] which is [tex]h'(x)[/tex]...

Not sure if that helped, but that's the idea.

I understand it now, thanks for your replies.

By the way, notice that in your post about parametric equations and velocities and such, you had an expression which was just the chain rule written by dividing one term onto the other side.


ratio college email induction algebra polynomial

Consider the product

$P = (1+x)(1+x^{2})(1+x^{4})(1+x^{8})...(1+x^{2^{n}})$.

Here $n$ is an integer.  $x$ is a real number.

Part 1: Find a simple closed form for this product

Part 2: Write this product as a geometric sequence.

[quote="junggi"]Do you mean the product
$(1+x)(1+x^{2})\dots(1+x^{2^{n}})$?
otherwise for some values of n(like n=3), the product would be messed up...
assuming that's the case:
[hide]Part 1
Multiply (1-x) on both sides.
then we can multiply the product on the RHS out to get:
$(1-x)P=1-x^{2\cdot2^{n}}$
$P=\frac{1-x^{2\cdot2^{n}}}{1-x}$
Part 2
The product is like a sum of a geom. series.
The formula for sum of geom. series of m+1 terms(starting with one and ratio x):
$\frac{1-x^{m+1}}{1-x}$
in this case, $m+1=2\cdot2^{n}$
so there are $2\cdot2^{n}-1$ terms in the series.
this would be written as:
$1+x+x^{2}+\dots+x^{2\cdot2^{n}-1}$[/hide][/quote]

yes, sorry I typed that in wrong. (I'll edit it)
btw your solutions are correct (edit: I said something else earlier, nvm)

However there's another neat way to do part 2 without doing part 1 first, I'd like to see if anyone can find it

Hey Berkely! here I'll do [hide="part 2:"]Just multiply out: there is one way to get a 1. one way to get an x, 1 way to get x^2, 1 way to get x^3, etc. ($\forall n$ one way to get $x^{n}$... you can get any $n$ the same way you convert numbers to binary). So $P=\sum_{i=0}^{2^{n+1}-1}{x^{n}}=1+x+x^{2}+x^{3}+...x^{2^{n+1}-1}$[/hide]
Is that it? See you Berkely (I'm not sure I'll be able to email Jenn because I will be at my grandparents house...)

[quote="calculuslover800"]Part 2: Write this product as a geometric sequence.[/quote]

[hide="Solution"] It's trivial to show by induction that

$P = 1+x+x^{2}+...+x^{2^{n+1}-1}$

Alternately:  Every number is expressible as a sum of distinct powers of two in exactly one way (the binary expansion); ergo, the powers in the polynomial expansion are simply the set of numbers that have $n+1$ (not necessarily nonzero) binary digits.  [/hide]


A normal clock (analog) is set at the correct time at 11:48 PM on December 31st, 2003 in order to celebrate the new year on the correct time. Little did they know the clock moved slower than the real time, 2.5 minutes per day to be exact. When will the clock show the correct time again?

The "real time" reference made it sound like an episode of "The Twilight Zone" :lol:, go figure :lol:

Depends. If the clock doesn't show AM or PM, at 11:48 AM on September 14, 2004. If it does, at 11:48 PM on July 29, 2005. I think. :?

Mnnn, it seems a bit too small...try writing out each step on a piece of paper with the corresponding unit (for example minutes, hours, years, etc.), it helped a lot with me.

[hide=answer]
Each day the clock goes back $\frac{5}{2}$ minutes, and a day is 1440 minutes. So the answer is $1440\div\frac{5}{2}=576$, or $\boxed{\text{July 29, 2005}}$.


Does anybody here have that book?  I bought it like a month ago and thought it was kinda advanced.  Is it a good book to prepare for the USAMO or would it be more useful for IMO-level problems?

-interesting_move

I have it. I must say that I basically only used it to solve countless functional equations while on vacation last summer. (Yes, I pass vacations doing math.) I think it's quite a good book from what I can tell.

[quote]Is it a good book to prepare for the USAMO or would it be more useful for IMO-level problems?[/quote]

There isn't a big enough difference in difficulty for that to be a really meaningful distinction.  Anything useful for studying for one is useful for studying for the otehr.

Best book ever. Definitely my favourite. Only thing is the geometry isn't really the stuff you find in imo/other similar tests.. strange.

hey im in middle school now so do you think that book is too advanced for mathcounts? if not then where could i get it?
thanks

I, too, really like this book as a source of challenging problems.

But it is way, way, way too advanced for MATHCOUNTS.

You can get it at Amazon when the time comes.  There's a link right to it from our Resources section.  I'd suggest starting with the Math Circles book in middle school for this type of problem solving (assuming you've already knocked off AoPS1).

I wouldn't recommend this book until you think you'll qualify for the USAMO. If you qualify for the USAMO in 8th grade then get it when you are in eighth grade, but not very many people do that. But as soon as you think you have a shot at USAMO, see if you can get the book.


topology
vector
integration
induction
rotation
real analysis
advanced fields

Show that the complement of a finite set of points in $\mathbb{R}^{n}$ is simply-connected if $n \geq 3$

Let $d$ be the minimal distance between the points of a finite set $A=\{a_{1},\dots,a_{m}\}$. Any continuous map $f\colon S^{1}\to\mathbb R^{n}\setminus\{a_{1},\dots,a_{m}\}$ is homotopic to a $f_{1}\colon S^{1}\to\mathbb R^{n}\setminus\bigcup_{k}B(a_{i},d/3)$ (use radial projection from inside $B(a_{i},d/3)$ onto its boundary). Using an appropriate mollifier, obtain a smooth map $f_{2}\colon S^{1}\to\mathbb R^{n}\setminus\bigcup_{k}B(a_{i},d/4)$, still homotopic to $f$. Let $F\colon S^{1}\times [0,1]\to\mathbb R^{n}$ be a straight-line homotopy between $f_{2}$ and a constant map. Let $E$ be the image of $F$: it is a compact set of $n$-dimensional measure zero*. Given a vector $x\in\mathbb R^{n}$, let $E+x$ be the shift of $E$ by $x$. Note that $f_{2}$ is homotopic to $f_{2}+x$ whenever $|x|\le d/5$. Since
\[\int_{x\in B(0,d/5)}\#\{i\colon a_{i}\in E+x\}\,dx=\sum_{i=1}^{n}|E\cap (B(0,d/5)+a_{i})|=0, \]
there is $x\in B(0,d/5)$ such that $E+x$ does not meet $A$.

(*) this is where $n\ge 3$ is used.

Question: is $\mathbb R^{n}\setminus A$ simply-connected for any closed countable set $A$? [url=http://mathworld.wolfram.com/AntoinesNecklace.html]Antoine's necklace[/url] is uncountable.

why not using van kampen?
i think it's pretty easy showing that automorphism of $\mathbb{R}^{n}$ (as a topological manifolds, as well as a differentiable manifold) are transitive on subsets of same (finite) cardinality: which is, any $m$ points can be sent to the points $(k,0)$ (induction? something even easier? i'm sure it's true, just would like to find a nice and quick proof).

now apply van kampen to sets $\{x_{0}<2/3\}$ and $\{x_{0}>1/3\}$.

of course, this doesn't work when the set is countable...

[quote="ma_go"]i'm sure it's true, just would like to find a nice and quick proof[/quote]
I posted a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=128183]proof[/url] a while ago. But you don't actually need such a homeomorphism: any finite set can be easily separated by a hyperplane.

[quote="mlok"]
Question: is $\mathbb R^{n}\setminus A$ simply-connected for any closed countable set $A$?[/quote]

I believe it is.

Let $\alpha: I=[0,1]\to\mathbb R^{n}\setminus A$ be a loop. We can make $\alpha$ piecewise linear through homotopy inside a neighborhood $V$ of $\alpha(I)$ disjoint from $A$. Now $\alpha$ will simply be a concatenation of segments $[A_{i},A_{i+1}]\subset V,\ i=\overline{0,k}$ (with $A_{k+1}=A_{0}$). Moreover, since $A$ is countable, we can move each $A_{i}$ slightly inside $V$ so as to make sure that none of the lines $A_{i}A_{j}$ intersect $A$. We can now argue by induction on $k$, the base case $k=2$ being trivial: keep $A_{0},A_{2}$ fixed, and rotate the plane $\pi=A_{0}A_{1}A_{2}$ around $A_{0}A_{2}$ until it reaches a position where it doesn't cut $A$. This is possible because the uncountably many positions of $\pi$ are mutually disjoint except for the line $A_{0}A_{2}$, and we know that $A_{0}A_{2}\cap A=\emptyset$. When you reach such a position, move $A_{1}$ towards $A_{0}A_{2}$ inside $\pi$ until it hits $A_{0}A_{2}$. The number of segments in $\alpha(I)$ has now decreased by one, and we can apply the induction hypothesis.

another possible approach: take a path $\gamma$ and for any $\gamma(t)$ an hyperplane $H_{t}$ containing it, and disjoint from $A$. now, let's restrict ourselves to the closed ball of radius $2{\rm diam }\gamma$: we can separate every hyperplane from $A$, since we're in a metric space, so we have an open covering $U_{t}$, which we can choose to be homotopic equivalent to a ball in $\R^{n}$, of ${\rm supp}\gamma$: we can extract a finite subset, and work on induction, applying van kampen on the number of hyperplanes.

i think it's pretty similar to grobber's one, though :)


hey y'all shouldnt the quiz be out by now??

When they say it'll be posted "in January" they mean "by the end of January".  :P

oh.. :oops:  well all d best to all anyway..

Also note that the "Info for 2010 Applicants" link indicates that the QQ will be posted in February, rather than January. So... yeah.

Nope, we [url=http://www.mathcamp.org/quiz.php]made it in January[/url]. :)

Proper announcement of its posting coming shortly.[/url]


A triangle is formed with sticks of length 8, 15 and 17 inches. Pieces of the same integer length are cut from each stick so that the three remaining pieces can no longer form a triangle. What is the least number of inches that could have been removed from each stick?

Even when part of the sticks are cut off, the 3 lengths will be x, x+7, x+9.
Using the triangle inequality theorem, 2x+7>x+9.
(2 sides' sum must be > than the other 1).
Solving get x>2.
x must be 2 for the lengths to not form a triangle (least amount cut off), so the answer is 8-2=6.

[quote="isabella2296"]I know the answer is $ 6$, but I don't know how to get it.[/quote]

Er... izzy, Jongao said $ 6$.

He edited it as I was entering the post.

Or you could just do $8-x+15-x=17-x$ to make it not work under triangle inequality theorem.

From there you get $23-2x=17-x$ which give you that $\boxed{x=6}$

Given the lengths $a$, $b$, and $c$; if following inequality is true, the lengths can be the sides of a triangle.
$|a-b|<c<a+b$

The sides of the triangle are 8, 15, and 17.
If the inches they were all reduced by is $x$, the new sides are $8-x$, $15-x$, and $17-x$.

We want to find the minimum value of x for which the following inequality is false:
$|(17-x)-(15-x)|<8-x<(17-x)+(8-x)$

Simplifying:
$|(17-x)-(15-x)|<8-x<(17-x)+(8-x)$
$|(17-x-15+x|<8-x<17-x+8-x$
$|17-15|<8-x<25-2x$
$2<8-x<25-2x$

Let's find the minimum value for which $2<8-x$ is false.
$2<8-x$
$2+x<8$
$x<8-2=6$
$x<6$

$x$ must be less than $6$ for the inequality to be true so the least possible value of $x$ for which the inequality is false is $x=6$

Let's now find the minimum value for which $8-x<25-2x$ is false as see if it's less.
$8-x<25-2x$
$8-x+2x<25$
$8+x<25$
$x<25-8=17$
$x<17$

$x$ must be less than $7$ for the inequality to be true so the least possible value of $x$ for which the inequality is false is $x=7$

We can now see that $6<7$ making $x=\boxed{6}$ the correct answer.


Between the year 2000 and 2005, US Postal rates for first-class letters have increased
from:  “34 cents for the first ounce and 21 cents for each additional ounce (or part of an ounce)”
   to:  “37 cents for the first ounce and 23 cents for each additional ounce (or part of an ounce)”.

What is the increase in postage for mailing a first-class letter that weighs 5.2 ounces?

A.  10 cents	B.  11 cents	C.  12 cents	D.  13 cents	E.  15 cents

[hide] There's a difference of $ 3$ cents fro the first ounce and $ 2(5)\equal{}10$ cents for the next five ounces or part ounces, and $ 3\plus{}10\equal{}13$. Hence, the answer is D.[/hide]

Hey sri are you getting these problems from somewhere or r u making them up???


calculus
integration
trigonometry
function
LaTeX
algebra
rational function

Evaluate [tex] \displaystyle {\int_{0}^{\pi} \frac {x \sin x}{1 + \cos^2 x} dx} [/tex].

Well, I came up with one solution that I'm not especially fond of, but here it is:

[hide]Let [tex]\displaystyle I=\int_0^\pi \frac{x\sin x}{1+\cos^2 x}\; dx.[/tex] Then substituting [tex]y=\pi-x[/tex] gives

[tex]\displaystyle I=\int_0^\pi \frac{(\pi-y)\sin y}{1+\cos^2 y}\; dy.[/tex]

Changing the ys back to xs and adding the two expressions for I together gives

[tex]\displaystyle \frac{2I}{\pi}=\int_0^\pi \frac{\sin x}{1+\cos^2 x}\; dx.[/tex]

But this function has an antiderivative! We use what Spivak calls ``surely the world's sneakiest substitution'' (if I recall correctly) [tex]x=2\tan^{-1}(t)[/tex] to change it into a rational function, which can certainly be integrated. The antiderivative comes out to be [tex]-\tan^{-1}(\cos(x))+C[/tex], so now we plug in 0 and [tex]\pi[/tex] to get

[tex]\displaystyle \frac{2I}{\pi}=\frac{\pi}{2},[/tex]

so

[tex]\displaystyle \boxed{I=\frac{\pi^2}{4}.}[/tex][/hide]

I know it's hard with LaTex, but it would be nice if you could use spoiler text so that other people can try it too.

[hide]

You found the key idea to the problem, which is to let y = :pi: - x; but there's an easier way to integrate [tex] \displaystyle \int_0^\pi \frac{\sin x}{1+\cos^2 x}\; dx [/tex]. Just make the substitution u = cos x, and you'll get a very familiar integral.[/hide]

The "sneaky" substitution that Spivak refers to is to let x = arctan (u/2). It converts any rational expression in sines and cosines into a rational expression in u. In this case, though, that substitution is not necessary.

Ah yes, your substitution is much simpler.


conics
ellipse
ratio
vector
analytic geometry
symmetry
graphing lines

An ellipse has center O and the ratio of the lengths of the axes 2 + 3^0.5. If P is a point on the ellipse, prove that the acute angle between the tangent to the ellipse at P and the radius vector PO is at least 30.

I don't understand why nobody has tried this beautiful and not so hard problem yet. I will prove a generalization:

[i]Let O be the center of an ellipse, let a be its major halfaxis, and let b be its minor halfaxis. If P is a point on the ellipse, then the acute angle between the tangent to the ellipse at P and the line PO is greater or equal to $\arctan \frac{2ab}{a^2-b^2}$.[/i]

For the proof, consider a Cartesian coordinate system, whose center coincides with the center O of the ellipse, whose x-axis is the major axis of the ellipse, and whose y-axis is the minor axis of the ellipse. Then, the equation of this ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Because of the symmetry of the ellipse with respect to both of its axes, we can WLOG assume that the point P lies in the first quadrant of our coordinate system; hence, if P ( X | Y ) are the Cartesian coordinates of the point P, then both X and Y are nonnegative.

Since a is the major half-axis of the ellipse, and b is the minor half-axis, we have $a\geq b$, thus $a^2-b^2\geq 0$.

Let t be the tangent to the ellipse at P. We have to prove that the acute angle between the lines t and PO is greater or equal to $\arctan \frac{2ab}{a^2-b^2}$.

In order to calculate the angle between the lines t and PO, we need the slopes of these lines. The slope of the tangent t to the ellipse at P is $m_t=-\frac{b^2X}{a^2Y}$ (well-known formula), and the slope of the line PO is $m_{PO}=\frac{Y}{X}$ (this is clear from the coordinates of P). Hence, the acute angle u between the lines t and PO satisfies

$\tan u=\left\vert \frac{m_{PO}-m_{t}}{1+m_{t}m_{PO}}\right\vert =\left\vert \frac{\frac{Y}{X}-\left( -\frac{b^2X}{a^2Y}\right) }{1+\frac{Y}{X}\cdot \left( -\frac{b^2X}{a^2Y}\right) }\right\vert =\left\vert \frac{a^2Y^2+b^2X^2}{\left( a^2-b^2\right) XY}\right\vert =\frac{a^2Y^2+b^2X^2}{\left( a^2-b^2\right) XY}$

(the last step is because $a^2-b^2\geq 0$). Since u is an acute angle, and since the tangens function is monotonically increasing on [0; 90], in order to prove that $u \geq \arctan \frac{2ab}{a^2-b^2}$, it is enough to show that $\tan u \geq \frac{2ab}{a^2-b^2}$. In other words, we have to prove

$\frac{a^2Y^2+b^2X^2}{\left( a^2-b^2\right) XY} \geq \frac{2ab}{a^2-b^2}$.

Since $a^2-b^2\geq 0$, this is equivalent to proving

$\frac{a^2Y^2+b^2X^2}{XY} \geq 2ab$,

i. e. to proving $a^2Y^2+b^2X^2 \geq 2abXY$. But this is clear, since $a^2Y^2+b^2X^2-2abXY = \left(aY-bX\right)^2 \geq 0$. Proof complete. Note that equality holds if and only if aY = bX.

Now, the problem posted by Orl is a particular case of this one, since we can set $a=2+\sqrt3$ and b = 1 and conclude that the angle between the tangent to the ellipse at P and the line PO is greater or equal than

$\arctan \frac{2ab}{a^2-b^2}=\arctan \frac{2\cdot \left( 2+\sqrt3\right) \cdot 1}{\left( 2+\sqrt3\right) ^2-1^2}=\arctan \frac{2\cdot\left( 2+\sqrt3\right) }{6+4\sqrt3}$
$=\arctan \frac{2+\sqrt3}{3+2\sqrt3}=\arctan \frac{1}{\sqrt3}=30^{\circ }$.

  Darij

I did try to solve it, but I don't know what you found so nice about it :). From what I can see here, that's not the case.. I thought I could solve it in a non-analytical manner, but failed :(.

I found the problem for itself quite nice, not the solution (which is not too messy, by the way)...

  Darij


calculus
integration
function
LaTeX
limit
real analysis
calculus computations

I do not understand why

$lim_{n->\infty} \frac{1}{n} \Sigma_{i=1}^n ([2n/i] - 2[n/i]) = \int_0^1 ([2/x] - 2[1/x]) dx$,

where $[x]$ is the floor function.

If someone could help explain why, that would be great. Thanks.

The function $f(x)=\left[\frac2x\right]-2\left[\frac1x\right]$ is a bounded and piecewise continuous function on $[0,1]$; it alternates between 0 and 1 for values. Therefore, it is Riemann integrable.

Now, the sum that we want to take the limit of is a Riemann sum for $\int_0^1f(x)\,dx$, evaluating at points $\frac in$. Since the lengths of the intervals go to 0, the sum converges to the integral.

For another proof using this technique of interpreting a sum as a Riemann sum, and a demonstration of just how powerful the technique is, see my solution to this year's B5 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20512]here[/url].

(A note on $\LaTeX$: \sum looks better than \Sigma, and you want \lim_{n\to\infty})

Thanks for the clarification. I understand now.


[quote="wonki_noh"]1.         x+2y+3z=0
                x-y+z=0
   x^2+y^2+z^2=152

2.        x+y+z=12
      xy+yz+zx=41
               xyz=42[/quote]

Solution 1
[hide]rearrange the equations and we get: $x =-5z/3$, $y =-2z/3$

put in last equation and we get, $z = \pm6$
thus, $x = \mp10$ and $y = \mp4$[/hide]

Solution 2
[hide]Too tired....
One solution is $(x, y, z) = (2, 3, 7)$[/hide]

[quote="wonki_noh"]Solve for real numbers $x, y, z$: \[\left\{\begin{array}{rcl}x+y+z & = & 12 \\ xy+yz+zx & = & 41 \\ xyz & = & 42\end{array}\right.\] [/quote]

[hide]Observe that $x, y$ and $z$ are roots of the equation $t^{3}-12t^{2}+41t-42 = 0$. But after some trials, we get that $t^{3}-12t^{2}+41t-42 = (t-2)(t-3)(t-7)$. Therefore, $(2, 3, 7)$ and all it's permutations are the unique solutions. [/hide]


Prove that for $ \ x,y,z\ge 0$
$ \ \frac{1}{(a\minus{}b)^2}\plus{}\frac{1}{(b\minus{}c)^2}\plus{}\frac{1}{(c\minus{}a)^2}\ge \frac{11\plus{}5\sqrt{5}}{a^2\plus{}b^2\plus{}c^2}$

[quote="duongptnkvn"]Prove that for $ \ x,y,z\ge 0$
$ \ \frac {1}{(a \minus{} b)^2} \plus{} \frac {1}{(b \minus{} c)^2} \plus{} \frac {1}{(c \minus{} a)^2}\ge \frac {11 \plus{} 5\sqrt {5}}{a^2 \plus{} b^2 \plus{} c^2}$[/quote]

this problem is false (c=0;a=-b)
we have ∑1/(a-b)^2≥9/(2∑a^2)

I'm sorry,The condition is $ \ a,b,c> 0$

[quote="quykhtn-qa1"]
this problem is false (c=0;a=-b)
we have ∑1/(a-b)^2≥9/(2∑a^2)[/quote]
Notice that $ a, b, c \geq\ 0.$ Your example isn't effect.
But indeed, this ineq was wrong. Try with $ a\equal{}2$; $ b\equal{}1$; $ c\equal{}0$
 :wink:


I got the following question from Elias Saab's MathCounts page:

[quote]
 4.  (Team) How many four-digit integers greater than  3000 for which the thousands digit equals the sum of the other three digits?
[/quote]

[hide="Here is their solution"]
Here is one way of doing this problem:
Let us try to count such four digit numbers. Let us count them this way:
1) Count all those whose thousands digit is equal to 3, you get
(3+1)*(3+2)/2.
2)Count all those whose thousands digit is equal to 4, you get
(4+1)*(4+2)/2.
3) Count all those whose thousands digit is equal to 5, you get
(5+1)*(5+2)/2.
And so on until you
count all those whose thousands digit is equal to 9, you get
(9+1)*(10+1)/2.
So the answer is:
(3+1)*(3+2)/2 +(4+1)*(4+2)/2 + (5+1)*(5+2)/2 + ... + (9+1)*(10+1)/2 =210.
[/hide]

I did get the right answer, but I had a much more laborious solution, using organized lists of the ways one might make such 4-digit numbers, as demonstrated
[hide="here"]
Thousands digit = 3:
Other digit combinations possible are
003 => 3 possible arrangements
012 => 6 possible arrangements
111 => 1 possible arrangement
total: 10 such numbers with thousands digit = 3.

With some persistence, I did manage to work through all the possible solutions this way, and came up with the correct solution of 210.[/hide]

My question is: their solution is so much simpler and more elegant than mine... but, I don't see why it works!  There seems to be a formula being used for the number of three-digit numbers with a particular sum with which I am not familiar.  Can someone show me the derivation of that formula?

[quote="RC-7th"]if the sum is $a$ and there are $b$ numbers, then imagine that we have a+b-1 symbols in a line. Then we change b-1 of them into lines which seperate the symbols into b groups.[/quote]

Ah, that's cool, thanks.


when is roundd 2 prolems available and when can we start discussing round 1?

Round 2 problems: probably by the end of this week.  Join the USAMTS mailing list by clicking Stay Informed on the sidebar to receive USAMTS notifications.

Discuss Round 1: please wait until Monday.  We will host a Math Jam on Wednesday to discuss the problems.  The official answers will be posted then.  Full official solutions will be posted after grading.


$ a_{1}\equal{}a$
$ a_{n\plus{}1}\equal{}a_{n}\plus{}a_{n}^{b}$
$ a,b\geq 0$
Find c such that the sequence $ \frac{a_{n}}{n^{c}}$ converge to a point different from 0

You are wrong. If $ b>0$ sequense $ a_{n}$ increase very fast (about $ e^{a^{n}}$).
If $ b<0$ $ a_{n}\equal{}((1\minus{}\frac{1}{b})n)^{b/(b\minus{}1)}(1\plus{}o(1))$.

I'm sorry when I post it was 2 Am in my country.

if we suppose $ a>0$ and $ b<0$ we can find such $ c$


geometry
rectangle
perimeter
Pythagorean Theorem

The area of rectangle $ PQRS$ is 48 sq cm. $ PQ \equal{} \frac {1} {3} QR$. $ U$ is a point on $ QR$ such that $ QU \equal{} UR$. $ T$ is a point on $ PS$ such that $ PT \equal{} 2TS$. Find the perimeter of the quadrilateral $ RSTU$.

davidkim2106, if you think the answer is 14+2sqrt5, you should explain your method so people can understand the question and how you obtained your answer.

[hide="Not sure"]The sides must be $12$ and $4$. So $UR =6$ and $TS = 4$. $SR = 4$ Then we use Pythagorean theorem to figure out $TU$
We do $\sqrt{6^2 + 4^2}$ = $\sqrt{36+16}$=$\sqrt{52}$ = $2\sqrt{13}$ So the answer is $6+4+4+2\sqrt{13}$ = $\boxed{14 + 2 \sqrt{13}}$ ?[/hide]

[hide][hide="Diagram"][asy]
pair P=(0, 0), Q=(4,0), R=(4,-12), S=(0,-12);
draw(P--Q--R--S--cycle);
pair U=midpoint(Q--R);
dot(U);
label("$U$",U,E);
label("$P$",P,NW);
label("$Q$",Q,NE);
label("$R$",R,SE);
label("$S$",S,SW);
pair T=(0,-8);
dot(T);
label("$T$",T,W);
[/asy][/hide]
Since $PQ=\frac{1}{3}QR$ and $PQ \cdot QR=48,$ it can easily seen that $PQ=4$ and $QR=12.$ Then, $RS=4, TS=\frac{1}{3}PS=4, UR=\frac{1}{2}QR=6, \text{ and } TU=\sqrt{(UR-TS)^2+4^2}.$ Then, the perimeter of $RSTU$ is $4+4+6+\sqrt{20} = \boxed{14+2\sqrt{5}}.$[/hide]


geometry
geometric transformation
ratio
algebra unsolved
algebra

If $A^{m}=1$ then $A=cis2k*pi/{m}$ with $k=1,2.... or (m-1)$ and if $(A+1)^{m}=1$ then $A+1=cis2l*pi/{m}$ with $l=1,2.... or (m-1)$. Then we now that $A+1$ is a translation of $A$ , such that both are in the unite circunference then they form a angle of $60$ degree, then we now that $2l*pi/{m}-2k*pi/{m}=pi/{3}$ $\Rightarrow$ $m$ is a multiple of $6$

A and A+1 are in the circunference of center $(0;0)$ and ratio $1$ then they form a angle of 60 degree and the segment that join both have length 1 and is pependicular to the y axis then it's easy to see that A form a angle of $120$ degree, then $A^{3}$ form a angle of $360$ degree, then $A^{3}=1$.

[quote="Till"]Let $A\in\mathbb{C}\ ,\ m\in\mathbb{Z}^{+}$ such that $A^{m}=1\ ,\ (A+1)^{m}=1$

Prove that 6 divides $m$ and $A^{3}=1$[/quote]

[hide="Posted before; well known I think. Here is an alternate method"]
$a+bi=A;$ Then $a^{2}+b^{2}=1=(a+1)^{2}+b^{2}\implies 2a+1=0\implies a=-\frac12\implies b=\pm\frac{\sqrt{3}}{2}$
The rest is clear now.
[/hide]
Overall the nicest solution is with equilateral triangles; it is also the most natural. I think stuart had that solution but I'm not sure.


Hoang, it should take this form
\[\frac14\left(\frac{a^{2}}b+\frac{b^{2}}c+\frac{c^{2}}a+\frac{d^{2}}a\right)\geq\sqrt[4]{\frac{a^{4}+b^{4}+c^{4}+d^{4}}4}. \]
Have you got a new technique for it? Hoang

Does it work for
\[\frac14\left(\frac{a^{2}}b+\frac{b^{2}}c+\frac{c^{2}}a+\frac{d^{2}}a\right)\geq\sqrt[5]{\frac{a^{5}+b^{5}+c^{5}+d^{5}}4}. \]


geometry
geometric transformation
reflection

In an acute triangle $ABC$, $E, F$ are feet of altitudes from $B,C$. $G,H$ are projections of $B,C$ on $EF$. Show: $HE = FG$

[hide]Let $K$ be the feet of the altitude from $A$.
For a well-known fact (it's only angle chasing), $\Delta AFE \sim \Delta ACB$. Then also $\Delta HCE \sim \Delta FCB$ and $\Delta GFB \sim KCA$. So $\frac{BF}{FG}= \frac{AC}{CK}$ and $\frac{BC}{BF}= \frac{CE}{EH}$.
Combinig these we get $\frac{AC \cdot FG}{CK}= \frac{BC \cdot EH}{CE}$. On the other hand $\Delta ACK \sim \Delta BCE$, from which the thesis.[/hide]

Here's an alternative solution. (Not that there's anything wrong with Andreas' solution, it's the first thing I thought of too)

Call $I$ the midpoint of $EF$. It's clear that $CEFB$ is cyclic, so the midpoint of that circle $\omega$ is $O$, the midpoint of $BC$.

Now, it's clear that the reflection of $C$ in $OI$ lies on $BG$. Therefor, $G$ is the reflection of $H$ in $OI$. The reflection of $\omega$ is the identic circle. The power of the points $G$ and $H$ is equal because they are each others reflections. Therefor $HE=GF$  :lol:


So, the 2009 USAMO is one month left - seems just around the corner :D

So I've qualified to the USAMO two years in a row.....and I don't want to repeat the score

I got on last year's USAMO as a 9th grader. (*Sigh)

I've been practicing by solving national olypmiad problems of Korea and Poland and Canada and such,

and some random contest problems that I find challenging enough.

I want to get some advice on specifically WHICH CONTESTS I should look for in order to get my hands

on good problems that shall prepare me for the USAMO....

and more importantly, are there any "Theory Books - not compilations of problems" that I can study

to practice advanced mathematical proofs?? Since the USAMO questions are all "proof" questions, I want

to study it in depth.....the various kinds and proof questions and how to approach each type.

Finally, for all those who have advanced, or will advance to the 2009 USAMO,

let's get our gears up  :D   Good Luck to you all!

im in the same predicament. trying to make red mop sigh.. im just doing canada mo problems and usamo 1&4's

USAMO problems are good practice for USAMO. Do the 1s and 4s.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70654]Book for Geometry[/url]

You'll notice that on most USAMOs, there is always at least one geometry problem, and it is usually not number 1 or 4. Since you're in 10th grade, it's necessary to get 3+ problems, and 3 is assuming you have perfect solutions.
First, you should definitely work through all the previous #1s and #4s to help prepare for those. And if you're aiming for Blue MOP, you're going to need to be able to the #2s and 5s as well.

You should be able to get #1/#4 consistently. Then, you could luck out and randomly get #2 and make blue. It works :P

seriously though, you should be able to get #2/#5s at least half the time if you expect to make blue


Dear Math Enthusiast,

The 2005 Ciphering Time Trials took place on Thursday, December 8th. This competition consists of ten rounds in which individuals are challenged to solve three problems in three minutes. The first-place winners in each division are:

School: Case High School, Racine, WI
Middle School Student: Alan Chou, Grissom High School, Huntsville, AL
9th-grade Student: Alex Ryan, Klein High School, Klein, TX
10th-grade Student: Jeffrey Chen, A&M Consolidated High School, College Station, TX
11th-grade Student: Young Jeong, Thomas Jefferson High School, Auburn, WA
12th-grade Student: Anderthan Hsieh, Saratoga High School, Saratoga, CA

The recognized scores in each division were:

Schools: 160-362
Middle School Students: 19-68
9th-grade Students: 36-87
10th-grade Students: 57-88
11th-grade Students: 62-94
12th-grade Students: 64-97

Complete results for the competition (the top 25 scores in each division) are available on our website; take a look to see who in your area you can congratulate!

Thank you,

Tom Clymer
Director of Academic Competitions
National Assessment & Testing
http://www.natassessment.com

P.S. Our next contest, the 2006 Four-by-Four Competition, is very similar.  It consists of ten rounds in which teams of four students are challenged to solve four problems in three minutes.  Problems in all rounds cover a broad range of topics and difficulties, offering confidence and challenges to students of all abilities. The 2006 Four-by-Four Competition will occur on Thursday, February 9th, 2006, and has a registration fee of fifty dollars.

Hmm... I knew we would be too late. Is there any way you could tell us what our scores were? (Vestavia)

[quote="tom_clymer"]
11th-grade Student: Young Jeong, Thomas Jefferson High School, Auburn, WA.[/quote]

Yay!

blah ;)
10th in 12th grade...

good job Young.  :)