number theory
least common multiple
number theory unsolved

Let $ m\geq 2$ be a positive integer. Let $ A_m\equal{}\{m\plus{}1,3m\plus{}2,5m\plus{}3,7m\plus{}4,9m\plus{}5,...\}$

(i) For all $ m\geq 2$, prove that there exists a natural number $ a$ such that $ 1\leq a<m$ and either $ 2^a$ or $ 2^a\plus{}1$ belongs to $ A_m$.

(ii) For some $ m\geq 2$, suppose there exist $ a,b$ such that $ 2^a$ and $ 2^b\plus{}1$ are elements in $ A_m$. Let $ a_0$ be the least of such $ a$, and $ b_0$ the least of such $ b$. Find the relationship between $ a_0$ and $ b_0$.

Obviosly $ n\in A_m$ if and only if $ 2m \plus{} 1|2n \minus{} 1$. Therefore $ 2^a\in A_m\to 2^{a \plus{} 1} \equal{} 1\mod 2m \plus{} 1$, $ 2^a \plus{} 1\in A_m\to 2^{a \plus{} 1} \equal{} \minus{} 1\mod 2m \plus{} 1$. All statements followed from it.
ii)$ a_0\equal{}2b_0$.

[quote="Rust"]All statements followed from it.[/quote]

I don't think it's that easy. Please show...

1)$ Per(2)\mod 2m\plus{}1 \le \phi(2m\plus{}1)\le 2m$, therefore $ a\plus{}1\le 2m/2$ or $ a<m$.

[quote="outback"]Let $ m\geq 2$ be a positive integer. Let $ A_m = \{m + 1,3m + 2,5m + 3,7m + 4,9m + 5,...\}$

(i) For all $ m\geq 2$, prove that there exists a natural number $ a$ such that $ 1\leq a < m$ and either $ 2^a$ or $ 2^a + 1$ belongs to $ A_m$.

(ii) For some $ m\geq 2$, suppose there exist $ a,b$ such that $ 2^a$ and $ 2^b + 1$ are elements in $ A_m$. Let $ a_0$ be the least of such $ a$, and $ b_0$ the least of such $ b$. Find the relationship between $ a_0$ and $ b_0$.[/quote]
[hide="i"]
We notice that a number of the $ A$ sequence may be represented in the form of $ (2k + 1)m + k + 1$ for a nonnegative integer $ k$.  Then, we want this value to equal $ 2^a$ or $ 2^a + 1$ for $ 1\le a < m$, so we want $ (2k + 1)2m + 2k + 1 + 1$ to be equal to $ 2^b$ or $ 2^b + 2$ when $ 2\le b\le m$ and $ b = a + 1$.  This yields that $ (2k + 1)(2m + 1)$ is equal to either $ 2^b - 1$ or $ 2^b + 1$.  This is true if and only if $ 2m + 1$ is a factor of $ 2^b + 1$ or $ 2^b - 1$.  Now, if $ 2m + 1$ is a prime, then we get that $ 2^{2m} - 1\equiv 0\bmod (2m + 1)$, so $ (2^m + 1)(2^m - 1)\equiv 0\bmod (2m + 1)$.  Since $ 2^m + 1$ and $ 2^m - 1$ are coprime, we have that $ 2m + 1$ will be a factor of exactly one of them.  Either way, we set $ b = m$ and get the desired result.  Now, if $ 2m + 1$ is composite, then we let $ 2m + 1 = \prod_{i = 1}^{x} p_i^{e_i}$ when $ p_i$ are primes and $ 2m + 1$ has $ x$ prime factors.  Then, we can let $ b$ be the LCM of $ \phi (p_i^{e_i})$ for all $ i$, giving  that $ 2^b\equiv 1\bmod p_i^e_i$ for all $ i$, meaning that since $ p_i$ are primes, we get that the Chinese Remainder Theorem yields exactly one solution for $ 2^b$ in $ \bmod \prod_{i = 1}^{x} p_i^{e_i}$, so a solution in $ \bmod (2m + 1)$.  Notice that $ 2^b\equiv 1\bmod (2m + 1)$ works, so that is the solution.  We now must verify that $ b\le m$.  Since $ 2m + 1$ is odd, we get that all of the prime factors of $ 2m + 1$ are odd, meaning that $ \phi (p_i^{e_i})$ all are even.  This implies that the LCM of all of these totients is less than or equal to\[ \frac {\prod_{i = 1}^{x} {\phi (p_i^{e_i})}}{2} = \frac {\phi (2m + 1)}{2}\le \frac {2m}{2} = m\]as desired. [/hide]

[hide="ii"]
From the above, we have that $ a_0$ is the order of $ 2\bmod (2m + 1)$ and $ b_0$ is the lowest integer so that $ 2^{b_0}\equiv - 1\bmod (2m + 1)$, so $ 2^{2b_0}\equiv 1\bmod (2m + 1)$.   Then, it is well-known that the order $ 2\bmod (2m+1)$ must be a factor of $ k$ if $ 2^k\equiv 1\bmod (2m+1)$, so $ a_0|2b_0$.  I am not quite sure if this is the relationship that is wanted... [/hide]

[quote="The QuattoMaster 6000"]
From the above, we have that $ a_0$ is the order of $ 2\bmod (2m \plus{} 1)$ and $ b_0$ is the lowest integer so that $ 2^{b_0}\equiv \minus{} 1\bmod (2m \plus{} 1)$, so $ 2^{2b_0}\equiv 1\bmod (2m \plus{} 1)$.   Then, it is well-known that the order $ 2\bmod (2m \plus{} 1)$ must be a factor of $ k$ if $ 2^k\equiv 1\bmod (2m \plus{} 1)$, so $ a_0|2b_0$.  I am not quite sure if this is the relationship that is wanted... [/quote]

I would think there is a more specific relationship  :)

[quote="outback"]I would think there is a more specific relationship  :)[/quote]
Sorry about that.  It is in fact that $ 2b_0 \equal{} a_0$, which follows immediately from the following Lemma:

[i]Lemma:[/i] We let $ k$ be the order of $ x\bmod a$ and let $ l$ be the lowest positive integer so that $ x^l\equiv \minus{} 1\bmod a$ (assume that it exists and that $ a\ne 2$.)  Then, $ 2l \equal{} k$.

[b]Proof:[/b] We clearly have that $ x^{2l}\equiv 1\bmod a$, so $ k|2l$.  Now, assume that $ 2l > k$, so $ 2l\ge 2k$, so $ l\ge k$ (yet, $ l\ne k$ since $ x^l\equiv \minus{} 1\bmod a$ and $ x^k\equiv 1\bmod a$.)  Thus, $ l > k$.  Then, we have that $ x^{l \minus{} x}\equiv \minus{} 1\bmod a$.  It follows that $ 0 < l \minus{} x < l$, contradicting the minimality of $ l$.  We then have that $ 2l\equal{}k$.


geometry
circumcircle
geometric transformation
reflection
ratio
homothety
geometry unsolved

In  triangle ABC , D is a point on segment BC.Let  O1,O2 be the circumcenter of  ABD,ACD  respectively. O' be the center of circle passing point  A,  O1,O2. Let N be the nine-point circle of ABC .Point E on line BC such that O'E square corner with BC.
Prove that: NE//AD

This result is well-known: $ \Delta AO_1O_2\sim \Delta ABC$  :)
Let O'' be the reflection of N through $ JK. NO''\cap JK\equal{}\{I\}, S$ be the reflection of E through I. $ L,K,J$ are the midpoints of $ BC, CA, AB$, respectively. It's easy to see that I is the midpoint of AL. So $ AS//BC$
Let $ SO''\cap O'E\equal{}\{Q\}$. Because $ O''I//OE$ and $ SI\equal{}EI$ then O'' is the midpoint of $ SQ. (1)$
First, we will prove that if $ AD\cap JK\equal{}\{V\}$ then $ V, O_1, O_2$ are collinear: $ (*)$
We have $ \angle AVJ\equal{}\angle ADB\equal{}\frac{1}{2} \angle AO_2C\equal{}\angle AO_2K$ then $ AVKO_2$ is cyclic.
$ \Rightarrow \angle AVO_2\equal{}\angle AKO_2\equal{}90^o$
But $ AD\perp O_1O_2$ so $ V, O_1, O_2$ are collinear.
Second, since $ O''$ is the circumcenter of triangle AJK, let $ (O'')\cap (O_2)\equal{}\{T\}$. We will show that O'' is the midpoint of AT:
Let $ (O'')\cap O_2K\equal{}\{T'\}$. We have $ \angle AT'O_2\equal{}\angle AJK\equal{}\angle AO_1O_2 \Rightarrow AO_1TO_2$ is cyclic
$ \Rightarrow T'\equiv T.$
We get $ \angle ATK\equal{}90^o \Rightarrow O''$ is the midpoint of AT. $ (2)$
From $ (1)$ and $ (2)$ we obtain $ AS//TQ$
$ \Rightarrow \angle NED\equal{}\angle ASQ\equal{}\angle O''QT\equal{}180^o\minus{}\angle O''O'T$ (because $ O'O''QT$ is cyclic)
From (*) we have $ \angle VDE\equal{}90^o\minus{}\angle VUD\equal{}90^o\minus{}\angle O_2VK$
To have $ NE//AD$ we need to show that $ \angle O_2VK\equal{}\angle O'AT$
Since $ \angle AVO_2\equal{}\angle AKO_2\equal{}90^o \Rightarrow AVKO_2$ is cyclic $ \Rightarrow \angle O_2VK\equal{}\angle O_2AK\equal{}90^o\minus{}\angle AO_2T\equal{}90^o\minus{}\angle AO'O''\equal{}\angle O'AT$
$ \rightarrow$ QED

$B',C'$ are the midpoint of $AB,AC$ . $K$ is circumcenter of $AB'C'$ .
WLOG :$O_1,C$ are on the same half-plane wrt $AB$
$\angle O_1AB'=\alpha$ ( we can prove that :$\alpha =90-\angle ADB$ )
It's easy to see that :$\triangle AO_1O_2 \sim \triangle \triangle AB'C'$ with the ratio $\frac{1}{cos\alpha}$
and $\angle O_1AB'=\angle O_2AC'=\alpha \Rightarrow O'A=\frac{OA}{cos\alpha} ;\angle O'AK=\alpha$ (*)
$\Rightarrow \angle O'KA=90$ (1)
$F$ is the reflection of $E$ wrt $B'C'$.
Since $OE\perp BC $ and $B',C'$ are midpoint of $AB,AC \Rightarrow AF ||BC$ and $\angle AFO'=90$(2)
From (1),(2) we get:$A,F,O',K$ are concyclic
$\Rightarrow \alpha=\angle KAO'=\angle KFO'=\angle KFE$
Note that :$K$ is the reflection of $N$ wrt $B'C$'
hence : $\angle NEF=\angle KFE=\alpha= 90-\angle ADB$
$\Rightarrow \angle NEB=\angle ADB \Rightarrow NE||AD$

Solution :
Let A' is midpoint of BC
Easy to see that A'N || OA , A'N = OA/2 , were O is circumcenter of ABC
CBAO~O_2O_1AO' , so OO'A ~ BB_1A
Let point N' is given , such that A'N'N ~ OO'A
Reflect O wrt BC and get point H
Easy to see that H is homotety center of A'N'N , OO'A , N' is midpoint of HO'
Reflect A' wrt N' and get point X
XO' || HA' , let XO' intersect BC at point T
Angle A'TX = TA'H = 90 , TN' = A'N' = N'X = N'N , so angle A'TN = A'N'N/2 = OO'A/2 = CDA . done


Let $ \alpha > 1$, $ x_{1}>\sqrt{\alpha}$, and define $ x_{n\plus{}1}\equal{}\frac{\alpha\plus{}x_{n}}{1\plus{}x_{n}} \equal{} x_{n}\plus{}\frac{\alpha\minus{}x_{n}^{2}}{1\plus{}x_{n}}$. Prove that $ \lim x_{n} \equal{} \sqrt{\alpha}$.

I've shown that $ x_{1}>x_{3}>x_{5}>\cdots$ and that all the odd terms are bounded below by $ \sqrt{\alpha}$; similarly I've shown that $ x_{2}<x_{4}<x_{6}\cdots$ and that all the even terms are bounded above by $ \sqrt{\alpha}$. How do I show that $ \lim x_{n} \equal{} \sqrt{\alpha}$?

Express $ x_{n\plus{}1}$ in terms of $ x_{n\minus{}1}$.

[hide]
$ x_{n\plus{}1}\equal{}\frac{\alpha\plus{}x_{n}}{1\plus{}x_{n}}\equal{}\frac{\alpha\plus{}\frac{x_{n\minus{}1}\plus{}\alpha}{x_{n\minus{}1}\plus{}1}}{1\plus{}\frac{x_{n\minus{}1}\plus{}\alpha}{x_{n\minus{}1}\plus{}1}}\equal{}\frac{\alpha(x_{n\minus{}1}\plus{}1)\plus{}x_{n\minus{}1}\plus{}\alpha}{(x_{n\minus{}1}\plus{}1)\plus{}x_{n\minus{}1}\plus{}\alpha}$

Plug $ x_{n\plus{}1}\equal{}x_{n\minus{}1}\equal{}L$ and get $ L\equal{}\pm \sqrt{\alpha}$.
[/hide]


Beginning from last night (a little after the math jam ended), I started having some trouble viewing the forum. I can't seem to see the backgrounds on the inner tables of each page (ie the table that lists each forum on the main forum page). Initially I thought that it was a problem with the forum or server, but when I checked at school earlier today, the forum looked fine.

Any suggestions?

I'm guessing you already tried this, but try hitting Refresh/Reload.  You may have something cached that's causing the problem.

Yeah, I tried refreshing but that didn't work. As for your other suggestion, I deleted the temporary cahced internet files on my computer and then returned to the forum, but I still have the same problem. :(

It should be fixed now; the forum went haywire for me too but it starting working fine after like 10 minutes.

What exactly is happening?  I don't know what you mean by not being able to see table backgrounds.

Yesterday evening, a similar problem happened to me.  The left, darker part of the AoPS pages had expanded to take over the majority of the browser window, leaving just a little room at the right edge for the content at the right.  The problem went away after about 20 minutes.  It's hard for me to describe the problem better.  If it's still happening to Fiery, maybe he can save the page as an html file, and send it to you?

yeah i had this too yesterday, and a couple of times before.  I'll take a screenshot next time and send it in.

Mr. Rusczyk,

It is difficult to describe what I am seeing. Perhaps you could PM me an e-mail address to which I could send a .html or screenshot of the page to.

Um I'll try: The dark background on the menu somehow "duplicated" to form layers which consists of a wide strip of the dark color in the aforementioned background with thin strips of whiteish creamy color background the forum has in between. The main screen (The part that is other than the menu and the title) is way over to the right so you have to scroll to the right in order to see it and it is about half the size it normally is thus screwing up some layout. (at one point "Forum Index" tried to cover the same space as Profile, Private Message, and FAQ (The buttons were grouped differently; the first 3 formed the first row, the next 3 formed the second, and Forum Index was in the third row all on its own) that it stretched making the button unusually smushed vertically) The problem ("Start with 4 congruent triangles...") is dislocated and is way off (I forgot exactly where it went) and so did the quotes (I think it went to the far left bottom corner of the screen). That's about it. (also it didn't go back to normal even after I went to another website and closed the browser and came back again)

Hmm...Tare...that's not quite what mine looks like, I think. Eh, best described with a screenshot.

I haven't seen it, but if anyone that notices the problem notices that it happens when going from one specific page to another specific page on the site, it would probably be very beneficial for finding the problem.

It wasn't a specific page; I visited Fun and Games and The Round Table I think but I know that it wasn't specific to a specific transition because it happened in all the pages I saw during that period of time.

[quote="JBL"]I experienced what Tare described, but it went away after a few minutes.  Actually, there were also some other weird things (the white background being replaced with dark blue when I entered a topic).  But I only experienced them that one night.  I unfortunately don't remember what I was doing when it happened.[/quote]

I've seen what JBL described - white blue became dark blue, but it went away when I went to a different page so I suppose it's just a little bug.

I've also seen at least part of what Tare described, where the buttons that are normally on the sidebar (Books, Order, Online Classes, Enroll, Community, etc...) appear at the top of the screen in three rows, but it's usually only as a page loads; that is, the buttons assume their proper place after the page finishes loading.  Usually it's when the connection becomes extremely slow, probably on the AoPS end.


group theory
abstract algebra
superior algebra
superior algebra unsolved

how to find normal subgroup of order 4 of A4.........

Look for it.

Have you even bothered to try the exercise before asking?


Does anyone know the dates of next year's USAMO (or have an idea of what the dates will be)? The reason why I need to know this is because I'm planning a trip in mid-April next year, but do not want to miss the USAMO because of it if I qualify. I realize that the dates may not have been set yet, but any hunches or predictions would be helpful too.

The USAMO is supposed to be the last week of April or the first week of May.

I have proposed to the USAMO Committee that the
2005 USAMO be adminstered on Tuesday, April 19, 2005
and Wednesday,  April 20, 2005 in the same time frames
as the last two years.  I have not yet received approval and
confirmation from the Committee, but I anticipate approval.

These dates avoid all national, state and religious holidays
known to me.  They are also consistent with the spacing between the
AMC 10/12 and the AIME, as in previous years.

Please note that the AMC 10A/12A date is February 1, 2005
AMC 10B/12B is February 16, 2005,
AIME I is March 8, 2005,
AIME II is March 22, 2005.

Steve Dunbar
AMC Director

Any one from AIME cares to comment why are we not allowed to appear for both AIME 1 and AIME 2.  It should be the same logic as you could do both AMC A and B. If it is a question of money (or time) to hand grade the AIME papers, I am positive that the number of additional students would not be that high.

[quote="justdudxit"]Any one from AIME cares to comment why are we not allowed to appear for both AIME 1 and AIME 2.[/quote]

I'm not from AMC, but one bit of logic I can see is that the examination is the American INVITATIONAL mathematics examination. If your friend invites you to dinner, and then invites another friend to dinner on a separate occasion, it's not standard practice for you to attend BOTH dinners. The preliminary tests (at this level, the AMC10/12) are tests offered to all eligible comers who pay the fee, and they can use those tests to qualify for an invitation (singular) to take the AIME, but the AIME is meant to be a one-shot chance each year at the USAMO, and I think it should remain so.

Meanwhile, you have all year to prepare, and there are plenty of old AIME exams and official solutions available from AMC to help you get ready.

Good luck.

[quote="tokenadult"]... but the AIME is meant to be a one-shot chance each year at the USAMO, and I think it should remain so...[/quote]

In a previous thread, Mr. Dunbar had requested opinions from the AoPS members regarding possible improvements to the USAMO selection process. So, not wanting to repeat the many passionate opinions, here's the link to the old thread. I am certain that I am speaking for many others that Mr. Dunbar's comments towards all those suggestions will be greatly appreciated.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12034&postdays=0&postorder=asc&&start=0


An integer is called snakelike if its decimal representation $a_1a_2a_3\cdots a_k$ satisfies $a_i<a_{i+1}$ if $i$ is odd and $a_i>a_{i+1}$ if $i$ is even. How many snakelike integers between 1000 and 9999 have four distinct digits?

[quote="JesusFreak197"]I think you mistyped something for the number.[/quote]

I left out a space in the $\LaTeX$ code.  It's fixed now.

[hide="solution"]
let the four digits be $a,b,c,d$. assume: $a<b<c<d$.  from these digits we can see that the possible snakelike numbers are: $acdb,adbc, bcad, bdac, cdab$ (this is if $a\not =0$, if $a=0$, the only possible ones are: $bcad, bdac, cdab$).  if $a\not =0$ there are $5\binom{9}{4}$ numbers. and if $a=0$. there are $3\binom{9}{3}$ snakelike numbers.
so the total would be: $5\cdot 126 +3\cdot 84=\boxed{882}$[/hide]

[hide]How many numbers would there be if the question was rephrased to how many four digit snakelike integers are there?

Note: I don't think the answer is an AIME answer. [/hide]

That's odd... I was just practicing this problem today and it comes up on the forums. I haven't been able to get an answer yet though.

Follow up:
Almost this exact question was on yesterday's AIME II, and in the same position (both number 6). What a coincidence...

Can you guys figure out what's wrong with my solution?  Sorry for reviving an old thread.

[hide="Wrong Solution"]
Let the 4 digit number be $abcd$, so thus due to the restrictions $a<b>c<d$.
Also, $a$ can't be $0$.

So... I split this into subcases:
Case 1: $a>c$
      Case 1a: $b>d$  Basically this is $c<a<d<b$.  Let's say we choose 4 different numbers from the set (0,1,2,3,4.....,8,9).  There is exactly one way to line up the numbers from least to greatest, so those would correspond to each digit.  For example, we choose 1, 3, 7, and 8 (already in least to greatest order).  Thus, $c=1$, $a=3$, etc.  Therefore, there are $\dbinom{10}{4}$ ways to pick 4 numbers from that list.  There is no way that $a=0$, since if $0$ is one the numbers picked, it will be $c$ since $c<a$.
     Case 1b: $d>b$  This is the same as Case 1a, except we switched $d$ and $b$.  There is no overlap with Case 1a, since $d$ and $b$ are now switched.  Thus there are $\dbinom{10}{4}$ numbers.

Case 2: $c<a$
     Case 2a: $b>d$.  This is $a<c<d<b$.  This is the same as the cases above, except we can only choose from the digits (1,2,3....8,9) since if 0 is included, $a=0$ since it is the smallest.  Thus there are $\dbinom{9}{4}$ numbers.
     Case 2b: $d>b$.  Same as above.

I don't think there is any overlap, so our answer is
$\dbinom{10}{4}+\dbinom{10}{4}+\dbinom{9}{4}+\dbinom{9}{4}=672$
[/hide]

[hide="@above"]
No, there is no overlap in your solution.  In fact, what you have so far is correct.  I think you have forgotten about something: if $a > d$.  It says no such thing in the problem about this restriction.  For example, you can have $8912$.

$c < a < d < b$ is $210$ ways.
$c < a < b < d$ is $210$ ways.
$c < d < a < b$ is $210$ ways.
$a < c < d < b$ is $126$ ways.
$a < c < b < d$ is $126$ ways.

So $210 \cdot 3 + 126 \cdot 2 = 882$.  Bingo!
[/hide]

[color=#FF0000][size=89]Inaccurate information deleted. [/size][/color]

hi can someone please explain what's wrong with this:

First off there are $6$ possible ways to arrange the digits of any four digit number with different digits to make it snakelike. The amount of ways to pick $4$ distinct digits where the first cannot be $0$, not counting the $4!=24$ arrangements, is $\frac{9 \cdot 9 \cdot 8 \cdot 7}{24} = 189$, and multiplying by $6$ gives $1134$.

[quote=pog]1. there aren't 6 possible ways to arrange the digits of any four digit number with different digits to make it snakelike
2. one of the digits can be 0 anyways[/quote]

Ok there are 5 ways, sorry --> 945?

Regarding second point, I know - first digit has 9 ways (not zero) second digit has 9 ways (can be zero, but cannot be first digit) and so on

If one of the digits is $0$, there are only $4$ ways in which you can rearrange the number to be snakelike. (You have to consider two cases.)

Oh, that's right! I won't do any math today so can't fix it until tomorrow, but thank you.

We take cases.

Case 1: No zeros in the number. There are then $5$ ways to rearrange it to be snakelike, so that gives $\binom{9}{4} \cdot 5 = 630$ ways.

Case 2: One zero in the number. Then there are $3$ places to put the $0$ and only $1$ way to arrange the number, so that gives $3 \cdot \binom{9}{4} = 252$ ways.

The total is $\boxed{882}$.


Round 1 results, the top 20 teams:

\begin{tabular}{aaa}1. & Colorado Math Circle & 37 \\
2. & Thomas Jefferson HS & 36 \\
3. & Georgia ARML & 35 \\
4. & Utah ARML & 32 \\
5. & Long Beach Math Circle & 30 \\
6T. & Lynbrook HS & 28 \\
6T. & New Mexico Math League & 28 \\
8. & Western Washington & 27 \\
9. & Western MASS A & 26.5 \\
10T. & Illinois Math and Science Academy & 26 \\
10T. & Stuyvesant HS & 26 \\
12. & Waterloo Collegiate Institute & 25 \\
13T. & Naperville North HS & 23 \\ 13T.&Belmont HS&23\\
15. & Henry M. Gunn HS & 22 \\
16. & Sofia Math Circle & 21 \\
17. & Evanston HS & 19 \\
18. & Wayzata HS & 18 \\
19T. & Iowa City West HS & 17 \\
19T. & West Tennessee & 17  \end{tabular}

Hmm... 6 of the top 8 teams will surely be present at ARML West in Las Vegas come May. And yes, these scores ran low.

In hindsight, Western Washington's group was a little too small, and we lost a few points too many on the first few questions.  (It didn't help that I misinterpreted #8...)  We'll invite more people for Round 2 :)

Final standings.

$ \begin{tabular}{ccccc}Place & Team & Round 1 & Round 2 & Total \\
1. & Georgia ARML & 35 & 34 & 69 \\
2T. & Colorado Math Circle & 37 & 31 & 68 \\
2T. & Thomas Jefferson HS & 36 & 32 & 68 \\
4T. & Long Beach Math Circle & 30 & 37 & 67 \\
4T. & Lynbrook HS & 28 & 39 & 67 \\
6. & AAST \#1 & 36 & 30.5 & 66.5 \\
7T. & Utah ARML & 32 & 33 & 65 \\
7T. & Western Washington & 27 & 38 & 65 \\
9. & Belmont HS & 23 & 36 & 59 \\
10T. & Stuyvesant HS & 26 & 30.5 & 56.5 \\
10T. & Western MASS A & 26.5 & 30 & 56.5 \\
12. & New Mexico Math League & 28 & 28 & 56 \\
13. & Illinois Math and Science Acad. & 26 & 27.5 & 53.5 \\
14. & Sofia Math Circle (Bulgaria) & 21 & 31 & 52 \\
15T. & Albany Area HS & 11 & 34 & 45 \\
15T. & Hon. Vincent Massey SS & 13 & 32 & 45 \\
17T. & Montgomery County & 18 & 26 & 44 \\
17T. & Naperville North HS & 23 & 21 & 44 \\
19T. & Eden Prairie HS & 11 & 32 & 43 \\
19T. & Walter Payton College Prep & 16 & 27 & 43\end{tabular}$

Well, hooray for (edit: pretty good) improvement between rounds tied with Lynbrook  :D

It looks like Albany has you beat there?

I'm going to repeat Kent's post with the 21st and 22nd placers also included for the unsubtle reason that our team is on there.  :-D  The team had an average of about 6 students over the two rounds.

Catherine Asaro, Coach
Howard Area Homeschoolers
(Also Chesapeake ARML, but the ARML team didn't compete)

[quote="Kent Merryfield"]Final standings.

$ \begin{tabular}{ccccc}Place & Team & Round 1 & Round 2 & Total \\
1. & Georgia ARML & 35 & 34 & 69 \\
2T. & Colorado Math Circle & 37 & 31 & 68 \\
2T. & Thomas Jefferson HS & 36 & 32 & 68 \\
4T. & Long Beach Math Circle & 30 & 37 & 67 \\
4T. & Lynbrook HS & 28 & 39 & 67 \\
6. & AAST \#1 & 36 & 30.5 & 66.5 \\
7T. & Utah ARML & 32 & 33 & 65 \\
7T. & Western Washington & 27 & 38 & 65 \\
9. & Belmont HS & 23 & 36 & 59 \\
10T. & Stuyvesant HS & 26 & 30.5 & 56.5 \\
10T. & Western MASS A & 26.5 & 30 & 56.5 \\
12. & New Mexico Math League & 28 & 28 & 56 \\
13. & Illinois Math and Science Acad. & 26 & 27.5 & 53.5 \\
14. & Sofia Math Circle (Bulgaria) & 21 & 31 & 52 \\
15T. & Albany Area HS & 11 & 34 & 45 \\
15T. & Hon. Vincent Massey SS & 13 & 32 & 45 \\
17T. & Montgomery County & 18 & 26 & 44 \\
17T. & Naperville North HS & 23 & 21 & 44 \\
19T. & Eden Prairie HS & 11 & 32 & 43 \\
19T. & Walter Payton College Prep & 16 & 27 & 43\\
21. & Olimpiadas Columbianas De Mathematicas & 11 & 31.5 & 42.5 \\
22. & Howard Area Homeschoolers & 22 & 19 & 41
\end{tabular}$[/quote]

Our school really messed up the second one.  I think the only points we got were for drawing the paths of the ball... :|

We (Western WA) also have a small team, about 8 for each round.  Only 5 students are among the core that consistently participated in weekend practice sessions.  We did not organize / allocate work well enough for round 1 but had a nice come back in round 2.  Big thanks to t0rajir0u for his lead as team captain.


algebra
polynomial
combinatorics unsolved
combinatorics

A circle is divided into $ m$ equal sectors, where $ m$ is a natural number.
How many different ways can these $ m$ sectors be colored with $ n$ given colors if two colorings are considered different only when neither can ve obtained from the other by rotating the circle?
(If $ m$ is a prime number then answer to this problem is $ \frac{n^m\minus{}n}{m}\plus{}n$.)

This is a standard application of [url=http://en.wikipedia.org/wiki/Burnside%27s_lemma]the lemma formerly known as Burnside's[/url].  The group $ G$ is cyclic of order $ m$ and has a conjugacy class for each of its divisors; count how many colorings are fixed by each conjugacy class.  Equivalently, determine the [url=http://en.wikipedia.org/wiki/Cycle_index]cycle index polynomial[/url] of a cyclic group.


geometry
perimeter
rhombus
Pythagorean Theorem

Sorry for those very easy question but I'm a very beginner in Geometry.

1) Find the perimeter  [geogebra]abc92fe435a3e5796aa4bdc82d697b96630e0271[/geogebra]

2) Find $ x^\circ$  [geogebra]c5f0be6a5947a9aa159d465fd2f5ac8b0d696e22[/geogebra]

[hide="Hint For Number One"]Use the Pythagorean Theorem.[/hide]

[hide="Solution For Number One"]Okay, so we have a $ 6\minus{}8\minus{}x$ right triangle, where $ x$ is the hypotenuse.  So we have $ 6^2 \plus{} 8^2 \equal{} x^2 \implies 36 \plus{} 64 \equal{} x^2 \implies 100 \equal{} x^2 \implies 10 \equal{} x$.  Since this shape is a rhombus, the perimeter is simply $ 10 \times 4$ or $ \boxed{40}$.

* Note, I am assuming that the angle in the middle is a right angle, which means that the diagonals are perpendicular bisectors, which means that the shape is a rhombus.[/hide]

[hide="Hint For Number Two"]Can you find a relationship between those two triangles?[/hide]

Thanks a lot I really didn't see the Pythagorean for that one  :blush:
For number 2 do you mean congruence or such thing  :| ? If yes how?

[hide="Solutions/Hints"]For number one, is it a rhombus? Only in that case are the diagonals perpendicular and Pythagorean Theorem applicable.

For number two, I will assume that the point not on the circle is the center. In this case, it uses the property of an "inscribed angle". If you have AoPS Volume 1, it is on page 88.

We see that the $ 50^\circ$ angle and the $ x^\circ$ subtend the same arc. Assuming that the $ 50^\circ$ angle is at the center of the circle, and since we know that the measure of the inscribed angle is half the measure of the arc, $ x\equal{}\frac{50}{2}\equal{}\boxed{25^\circ}$.

Someone might be able to explain this better than I did.
[/hide]

Thanks. In number one yes it is a rhombus.
And for number 2 I get what you mean when I entered http://en.wikipedia.org/wiki/Inscribed_angle_theorem

Thanks a lot


function
algebra
polynomial
complex numbers
number theory
relatively prime

Where should this be put?  Also, has it been posted before?

Find a formula for this in terms of n and b.  $\binom{n}{0}+\binom{n}{b}+...+\binom{n}{n}$
where b is a divisor of n, n>b, and both are positive integers.

How about if b wasn't a divisor and we just summed it up to $\binom{n}{kb}$ where n>kb but n<(k+1)b.  Of course, k is a natural number.

[quote="dengmi"]
Find a formula for this in terms of n and b.  $\binom{n}{0}+\binom{n}{b}+...+\binom{n}{n}$
where b is a divisor of n, n>b, and both are positive integers.
[/quote]

Maybe I didn't understand the problem, but we have that $\sum_{i=0}^n \binom ni=2^n$, so don't we have that
$\binom n0 + \sum_{i=b}^n \binom ni=2^n-2^{b-1}+1$

The notion is "going up by $b$ each time," not "skipping directly to $b$ and then going up by 1 each time."  Also, even if it were the latter, you sould reconsider your evaluation of that sum.

[quote="Elemennop"]I'm pretty sure the key is the expansion of the bth roots of unity.[/quote]
Indeed.  Here are two hints that should get you going:
[hide="Hint 1"]Consider the sum $\omega_1^m + \omega_2^m + \ldots + \omega_b^m$ where $\omega_i$ are the b-th roots of unity and $m$ is some integer.[/hide]
[hide="Hint 2"]Consider the expansion of $(1 + \omega)^n$, where $\omega$ is a b-th root of unity.[/hide]

so wait, say $n=21$, then our sum would be

$\binom{21}{0} + \binom{21}{1} + \binom{21}{3} + \binom{21}{7} + \binom{21}{21}$

right? hmm im still not quite sure how the roots of unity come into play.... since the difference in the roots of the unity stay the same and that isnt true of $b$ in the problem

Also, $0 \not | 21$.  Although, the question you've suggested could be interesting (but maybe very hard).

your original question is pretty difficult, although it isolates a concept that i like very much, namely,

let $f(x)$ be a polynomial, let the distinct solutions to $x^r=1$ be $x_1,x_2,...x_r$, the sum of the coefficients of the terms in the form $x^{7k}$ for $k\in\mathbb{N}+\{0\}$ is given by:

$\frac{1}{r}\sum_{i=1}^rf(i)$

JBL's method is essentially the same because taking one nonreal root to certain powers gives all the roots of unity anyway

--

your problem is essentially the sum of the coefficients that are multiples of $b$ for $(1+x)^n$...and this is a direct application of the above theorem...but i am not sure that it is easy to evaluate

[quote="Altheman"]but i am not sure that it is easy to evaluate[/quote]

It depends what you mean -- it's just a sum of fairly simple-to-compute powers of complex numbers, so in that sense any given case of the problem is easy.  On the other hand, there isn't a nice, closed-form way to express that sum in general, as far as I can tell.

I think the question gotztah raised is still interesting.  Also, its counterpart: the sum over all terms when the bottom is relatively prime to $n$.  I suspect neither of these has nice expressions, but they seem kind of cool anyhow.


Most of it depends where you what you want to study.

Math would probably be MIT or Caltech. But Caltech right now is in more of an up-and-coming place which has a large need of well-trained mathmatechians (spelled it wrong), so it gets the slight edge. Medicine is probably John Hopkins. Law: most people think Harvard, but Stanford is probably better, and, as someone said before, don't put too much stock in the U.S. News and World Report rankings. Also, much like Caltech, Stanford is located in Palos Altos, near Silicon Valley, home of a lot of new businesses, so if you want to be a corporate lawyer, Stanford is the obvious choice. Business is probably Penn State (U.S. News got this one right).

were you trying to respond to the graduate school thread?


ratio
geometry
vector
complex numbers
projective geometry
geometry proposed

Let $A$, $B$ and $C$ be collinear points. Show how to construct a point $D$ which satisfies $(A, B, C, D) = 3$.

${A=0,B=b,C=c,D=\alpha+i \beta }$

${\frac{(A-C) (B-D)}{(A-D) (B-C)}=3\Rightarrow (A-C) (B-D)-3 (A-D) (B-C)=0}$
${(-b c-2 \alpha c+3 b \alpha )+i (3 b \beta-2 c \beta )=0}$

${(3 b \beta-2 c \beta )=0}$
${(-b c-2 \alpha c+3 b \alpha )=0}$

${\beta =0,\alpha =\frac{b c}{3 b-2 c}}$

I didnt understant well your cross ratio. but...

I use complex numbers just assumind D is not colinear, maybe. but the proof says it is colinear too.

so the construction is very simple

MN is the length you want

i made this image simplicist like that to become more easy

but there is a fast way to do that. but for that you need to define Cross-ratio well.. humpf

take a look http://en.wikipedia.org/wiki/Cross-ratio

Let $A,$ $B,$ $C,$ the three collinear points, belong to the line $(e).$

Through an arbitrary point $P,$ not lies on $(e),$ we draw the lines $(a),$ $(b),$ $(c),$ pass through the points $A,$ $B,$ $C,$ respectively.

Through the point $C,$ we draw an arbitrary line $(e_{1}),$ which intersects the segment line $PB$, at one point so be it $B_{1}.$

We define the points $D_{1},$ $A_{1},$ on the segment line $CB_{1}\equiv (e_{1}),$ such that $CB_{1}= B_{1}D_{1}= D_{1}A_{1}$ $($ $B_{1},$ between $C,$ $D_{1}$ and $D_{1},$ between $B_{1},$ $A_{1}$ $).$

Through the point $A_{1},$ we draw a line parallel to $BB_{1}\equiv PB,$ which intersects the segment line $PA\equiv (a),$ at one point so be it $A'.$

We denote as $B',$ the intersection point of $PB,$ $CA'$ and as $D',$ the intersection point of the segment line $(e')\equiv CA',$ from the line trough $D_{1}$ and parallel to $A_{1}A'\parallel PB.$

The segment line $PD',$ intersects the line $(e)$ at one point so be it $D',$ has the wanted property.

Really we have $(A,B,C,D) = (A',B',C,D') = (A_{1},B_{1},C,D_{1}) = 3$ and the solution is completed.

Kostas Vittas.

PS. We say the cross ratio as $(A,B,C,D) = \frac{CA}{CB}\div \frac{DA}{DB}.$

Well, this is projective geometry, so in fact you should give a construction which doesn't use lengths/parallel lines.

OK, dear [b][size=100]Arne.[/size][/b]

If I understood what you mean, we can define the point $P,$ as the intersection point of the segment lines $B_{1}B$ and $A_{1}A$ ( as the notations in my drawing ).

So, we can find the wanted point $D,$ as the intersection point of the line $(e),$ from the segment line $PD_{1}.$

Thank you for your remark, Kostas Vittas.

No, I mean that your construction uses lengths and parallel lines explicitly (for example "we draw a line parallel to...") But this is projective geometry, so parallel lines do not exist ;) You should give a construction which is valid for all projective situations. Your solution is Euclidean.

So you say that this is projective geometry, hence vittasko can't use parallel lines in order to solve a particular case of your problem? I find that totally absurd. By the way, I consider that you're being a little [b]vague[/b] when talking about constructing (when I see "construct", I automatically think about "compass and ruler") some point given $4$ points, some of which may be points at infinity. How do you seize those? I hope that after your reply I'll understand what the problem wants from us.

It is indeed not clear what is meant by "construction" here. I hope this fits your definition of construction.

Note that eventhough this is $PG(2,\mathbb R)$, it works for all fields $K$. Pick your reference base in the underlying affine geometry $AG(3,K)$ such that $A=\infty,B=0,C=1$. Then with the classical calculations we have that $D=3$, so $D=C+2(C-B)$, so adding twice the vector $\overrightarrow{BC}$ to $C$ gives a construction for $D$.

[quote="Arne"]Let $A$, $B$ and $C$ be collinear points. Show how to construct a point $D$ which satisfies $(A, B, C, D)\equiv\frac{BA}{BC}\ : \ \frac{DA}{DB}= 3$.[/quote]
[color=darkblue][b][u]Construction.[/u][/b] Denote the line $d\equiv\overline{ABC}$ and choose a point $P\not\in d$. Construct :

[b]1. The middlepoint $X$ of the segment $[PB]$, i.e. $XB=XP$.

2. The middlepoint $M$ of the segment $[PX]$, i.e. $MB=3\cdot MP$.

3. The point $S\in AX\cap PC$ and the point $N\in PA\cap BS$, i.e.

(euclidian) $NC\parallel PB$ $\Longleftrightarrow$ $\frac{NA}{NP}=\frac{CA}{CB}$.

4. The point $R\in AM\cap BN$.

5. The points $D_{1}\in BC\cap PR$ and $D_{2}\cap BC\in MN$.[/b]

The required points are $D_{1}$ and $D_{2}$.  Indeed :

From the Ceva's theorem applied to the point $R$ and the triangle $PAB$ obtain :

$\frac{D_{1}A}{D_{1}B}\cdot\frac{MB}{MP}\cdot\frac{NP}{NA}=1$ $\implies$ $\frac{D_{1}A}{D_{1}B}=\frac{MP}{MB}\cdot\frac{NA}{NP}=\frac{1}{3}\cdot\frac{CA}{CB}$ $\implies$ $\frac{DA}{DB}=\frac{1}{3}\cdot\frac{CA}{CB}$.

From the Menelaus" theorem applied to the transversal $\overline{MND_{2}}$ of the triangle $PAB$ obtain similarly $\frac{D_{2}A}{D_{2}B}=\frac{1}{3}\cdot\frac{CA}{CB}$.

[b]Remark.[/b] Given are two segments with the lengths $m,n$. Construct on the line $d=ABC$ the points $D$ for which $\frac{DA}{DB}=\frac{m}{n}\cdot\frac{CA}{CB}$.[/color]


Let $ a,b,c \geq\ 0$.Prove that:
$ \frac{a^2\plus{}b^2}{a^2\plus{}ab\plus{}b^2}\plus{}\frac{b^2\plus{}c^2}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{c^2\plus{}a^2}{c^2\plus{}ca\plus{}a^2}\plus{}\frac{4(ab\plus{}bc\plus{}ca)}{3(a^2\plus{}b^2\plus{}c^2)}\leq\frac{10}{3}$

Easy with SOS.

$ \sum_{cyc}\frac{(a\minus{}b)^2 (a\plus{}b\minus{}c)}{a^2\plus{}b^2\plus{}ab}\geq 0$

now order $ a,b,c$  :wink:


Is the cutoff score for qualifying AIME 100?

I believe it is the top 50%.  Although I am pretty sure you can take the Mock AIME even if you don't 'qualify'.


geometry
3D geometry
circumcircle
inequalities
tetrahedron
sphere
geometry proposed

Prove that if the distance from a point inside a convex polyhedra with $n$ faces to the vertices of the polyhedra is at most 1, then the sum of the distances from this point to the faces of the polyhedra is smaller than $n-2$.

[i]Calin Popescu[/i]

I dont understand the problem....So a point inside a convex poliedra satisfies, its distance to the faces of the poliedra is at most $1$? and we have to show the sum is at most $n-2$??? what happens then with the center of a cube with side 2? its distance to the faces is 1, but the sum is $6$....whats wrong?

Yes, he missed some words...
How many time we will wait for the correction?...

A friend of mine told me that it said the distances from the point to the vertices of the polyhedron are all $\le 1$ (actually, he told me they were all $1$, but it was over the phone, so maybe I didn't hear well :)).

I'll solve the problem with equality (i.e. the point in question is the circumcenter of the polyhedron, which has unit circumradius). I think the one with inequality instead of equality can be reduced to this one.

For each plane which contains a face of the tetrahedron, consider the cap $\mathcal C_i$ determined by that plane which lies on the opposite side of the plane from the center of the sphere. If the height of $\mathcal C_i$ (i.e. the largest distance from a point on it to the plane containign the face $i$) is $h_i$, then we have to show that $\sum h_i>2$. The area $S_i$ of $\mathcal C_i$ is $2\pi h_i$, so $\sum h_i=\frac{\sum S_i}{2\pi}\ (*)$. However, the caps do not cover the sphere perfectly (i.e. they cover it, but they overlap), so $\sum S_i>4\pi$. this, together with $(*)$, gives the desired result: $\sum h_i>2$.

A beautiful solution, grobber! It works word by word for the general case too (you have never used the assumption that the vertices are on the unit sphere)  :)

I've corrected the statement  :blush: sorry for the mistake.

Of course he used it fedja, but it can be easily seen that the problem can be reduced to that case.

[quote="ikap"]Of course he used it fedja.[/quote]
Where? It seems to me that all he needed was that the polyhedron was completely contained inside the sphere of radius 1 centered at the given point inside the polyhedron. Or am I missing something?  :?

fedja, "caps" imply boundaries on the sphere and these are given by the points ON the sphere and their circumscribed circles....

Actually, ikap, I have to agree with fedja here. All you need are a few planes which cut the sphere and determine a cap each. You never use any points on the sphere, just the fact that the planes of the faces cut the sphere so that the caps they determine cover surface of the sphere completely.

oh, yes, of course, i wasn't looking at it from that point of view  :)  and I also didn't read word by word your solution grobber, that's why I was always refering in mind to the steps I followed to the solution, allthough this was after the contest.. My fault, sorry.


algebra
polynomial
vector
algebra unsolved

Let $ p(x)$ be a monic polynomial with integer coefficients such that all its zeros lie on the unit circle. Prove all the zeros of $ p(x)$ are roots of unity.

thanks!
But just wondering though:

there should be an elementary proof, because this is just an olympiad problem. Does one exist?

[quote="t0rajir0u"]Points on the unit circle that are not roots of unity are transcendental (the vector space they generate cannot have finite dimension).[/quote]
That's wrong, for example $ \frac {3 \plus{} 4i}5$ is on the unit circle, algebraic, but not a root of unity.
Nonetheless the original statement is true. It's equivalent to:
An algebraic integer such that all it's conjugates are on the unit circle is a root of unity.
(see my last post in http://www.mathlinks.ro/Forum/viewtopic.php?t=179773 )

Proof 1:
Let $ x$ be as required (algebraic integer and all conjugates on the unit circle) and root of a monic minimal integer polynomial of degree $ n$. Then for any $ m$ we find a monic minimal integer polynomial of degree at most $ n$ having $ x^m$ as a root. Looking further, we get bounds on all the coefficients. In other words, there are only a finite number of such polynomials, thus only a finite number of powers of $ x$. This gives that $ x$ is a root of unity.

Proof 2:
Repair t0rajir0u's argument.

Hardcore proof 3:
Embed the set $ X$ of powers of $ x$ into minkowski space $ S$. As all infinite valuations are $ 1$, we find that this set is closed and bounded. As the set of algebraic integers is discrete in $ S$, we get that $ X$ is a compact discrete set, thus finite. So $ x$ is a root of unity.

:oops:

I noticed a problem when I tried to formalize the argument...


topology
linear algebra
matrix
vector
algebra
polynomial
calculus

Let E be the set of (n,n)-dimensional real symmetric matrices A s.t.  Spectrum(A) is included in [a,b]. Show that E is homeomorphic to the compact ball in  R^(n(n+1)/2).

it is homeomorphic to $[a;b]^n \times O_n$ thanks to the spectral theorem. From here, it should be easy.

Sorry Alekk,
[a,b]^n x On is not isomorphic to B the n(n+1)/2 ball;
1°) On is not connected; then we take F=[a,b]^n x SO(n,R) who is connected.
2°) the F fundamental group (1-dimentional homotopy group) is Z if n=2, Z/2Z if n>2;
the B fundamental group is {1}.
Thus B and F are not homeomorphic.
   But my statement is correct (I think and hope so).
     Friendly.

The isomorphism $\varphi$ from the space $S^n$ of $n\times n$ real symmetric matrices to $\mathbb R^{\frac{n(n+1)}2}$ which sends a matrix to the vector formed by its entries above and on the main diagonal (in some predefined order we agree upon) sends the subset $S^n([a,b])$ of $S^n$ with spectrum contained in $[a,b]$ onto a closed, bounded (i.e. compact), convex subset of $\mathbb R^{\frac{n(n+1)}2}$ with non-empty interior (call this assertion $(*)$), and it's obvious that a convex body in $\mathbb R^k$ is homeomorphic to the closed unit ball in $\mathbb R^k$.

The only slightly non-obvious assertion of $(*)$ would be the convexity of $\varphi(S^n([a,b]))$. To prove this convexity, just use the well-known characterization for symmetric real matrices $A$ of the lower and upper bounds $\min\sigma(A),\ \max\sigma(A)$ of the spectrum of $A$: $\min\sigma(A)=\min_{\|x\|=1}(Ax,x),\ \max\sigma(A)=\max_{\|x\|=1}(Ax,x)$. Now, if $A,B\in S^n([a,b])$, and $\lambda\in[0,1]$, then $\max_{\|x\|=1}((\lambda A+(1-\lambda)B)x,x)$ equals $\max_{\|x\|=1}(\lambda(Ax,x)+(1-\lambda)(Bx,x))\le b$ (and we do something similar for $\min$). This proves that $\lambda A+(1-\lambda)B\in S^n([a,b])$, and we're done.

slightly off topic possibly, what happens if you map each matrix onto its spectrum
let us agree we order then from the lowest to the biggest, and we always get an ntuple (so we respect multiplicities)

Is this map continuous?

Hi fredbel,
let p be a complex polynomial and u0 a root of p; let q be another polynomial; for a real t (near 0) p+tq has a root u(t) s.t. u(0)=u0 and u is a continuous function. Moreover  if u0 is a simple root of p then u is analytic.
   Yet the calculus of the roots may be ill-conditioned: if p=(x-1)....(x-20) then Maple, with 12 digits, gives 19+0.8.I for one of the roots!
    If A is a normal matrix (AA*=A*A), then the research of the eigenvalues is well-conditioned: a small disturbance of the A-coefficients  involves a disturbance of the same order of the eigenvalues.
   There is no such result for the eigenvectors.
      Friendly


calculus
integration
analytic geometry
algebra
polynomial
Ring Theory
function

How do you go about figuring out an integral closure? Is there a general technique to fall back on, or is it very case specific?

For example, find the integral closure (in the field of fractions) of $ R \equal{} k[x,y]/(x^2 \minus{} y^3)$. So it isn't hard to figure out that it isn't integrally closed, since $ x/y$ is in the field of fractions and $ (x/y)^2 \minus{} y \equal{} x^2/y^2 \minus{} y \equal{} y^3/y^2 \minus{} y \equal{} y \minus{} y \equal{} 0$, so it satisfies a monic polynomial.

So we want to somehow adjoin this element, which can be done in various ways, for instance $ k[x,y,z]/(x^2 \minus{} y^3, z \minus{} x/y)$.

My guess would be that this is the integral closure of $ R$, but I know of no techniques to prove that that was all we needed to tack on.

There is probably some nice geometric way to think about this that I'm missing, since it is a coordinate ring...

Edit. I guess the more general questions still stand, but I just figured out how to prove it. You can show that $ R$ gives a ring isomorphic to $ k[t^2, t^3]$, so adjoining $ x/y$ under the iso gives $ k[t]$ and hence is a polynomial ring and so UFD and hence integrally closed.

For domains that are Z-algebras and finitely generated as abelian groups, there are algorithms that find the integral closure (Cohen's CCANT has a description of a few).

In your case, it helps to randomly recognize your "intermediate" step as the final answer:

[quote="HilbertThm90"]So we want to somehow adjoin this element, which can be done in various ways, for instance $ k[x,y,z]/(x^2 \minus{} y^3, z \minus{} x/y)$[/quote]

If z=x/y, then zz=xx/yy = yyy/yy = y and zzz=yz=yx/y=x, so k[x,y,z]/(x^2-y^3,z-x/y)=k[z], and presumably it is easy to see that k[z] is integrally closed (there is some fiddling in identifying the quotient field k(z) and the original quotient field in terms of x and y).

I don't know anything about the geometry, but you might look for "resolution of singularities" or so.

I haven't looked at it yet, but maybe Eisenbud's book "Commutative Algebra with a View Toward Algebraic Geometry" has some answers. Chapter 11 is on the subject and there is a brief passage about what happens if the rings in questions are affine rings.

Normalisation gives a canonical resolution of singularities for $ 1$-dimensional varieties. It could be calculated by blowing up the singular locus until the curve is regular. A proof of the latter can be found in Liu's book in the chapter on blow-ups. For a stronger theorem (reduced excellent $ 1$-dimensional schemes confer the appendix of S. Saito and K. Sato,  A finite theorem for zero-cycles over $ p$-adic fields, to appear in Annals of Mathematics (2009).

Picture of the normalisation morphism:

[code]
----------------------------- k[t]
                |
                |
                V
--------------\   /------------------ k[t^2,t^3]
                \/[/code]


algebra polynomial function limit integration ratio calculus

These question arent competition questions, but I am studying for the AP test and these were sample free-response questions from the past years on the Calc BC AP test (I got these questions from collegeboard.com in the sample-free response AP section)

The answers, however, did not come with the questions, so I was wondeirng if anyone knew the answers (you can just post answers if you just want since I will just try to see if I am doing it right), or if you really want, you can post a solution with the answer.

Thanks

2001 #6
A function f is defined by $ f(x) = \frac {1}{3} + \frac {2}{3^2}x + ... + \frac {n + 1}{3^{n + 1}}x^n$ for all x in the interval of convergence of the given power seires.

a.  Find the interval of convergence for this power series.
b.  Find $ \lim_{x = > 0}\frac {f(x) - \frac {1}{3}}{x}$ (sorry its a limit as x approaches 0, cant do latex...)
c. Write the first three nonzero terms and the general term for an infinite series that represents $ \int_0^1 f(x) dx$
d. Find the sum of the series determined in part c.

2002 #6
The Maclaurin series for ${ ln(\frac {1}{1 - x}})$ is $ \sum_{n = 1}^\infty \frac {x^n}{n}$ with interval of convergence -1 <= x < 1

a. Find the Maclaurin series for ${ ln(\frac {1}{1 + 3x}})$ and determine the interval of convergence.
b. Find the value of $ \sum_{n = 1}^\infty \frac {( - 1)^n}{n}$
c. Give a value of p such that $ \sum_{n = 1}^\infty \frac {( - 1)^n}{n^p}$ converges, but $ \sum_{n = 1}^\infty \frac {( - 1)^n}{n^{2p}}$ diverges.
d. Give a value of p such that $ \sum_{n = 1}^\infty \frac {( - 1)^n}{n^p}$diverges, but $ \sum_{n = 1}^\infty \frac {( - 1)^n}{n^{2p}}$ converges.

a) We can "eyeball" this. In looking for the radius of convergence, the polynomial $ n \plus{} 1$ is insignificant; powerless to change the radius of convergence. We're only looking for things that grow or shrink geometrically fast or faster. All we see is $ \frac {x^n}{3^n}.$ That's going to be in balance when $ |x| \equal{} 3,$ so the radius of convergence for this power series (centered at zero) is $ 3.$ [If you don't trust this reasoning, go ahead and confirm it with the root test or ratio test. But knowing the answer in advance should make your computation more accurate.] As for convergence at the endpoints: that's when the $ n \plus{} 1$ matters. If $ |x| \equal{} 3,$ then the terms don't go to zero and the series diverges. So the interval of convergence is $ ( \minus{} 3,3).$

b) Just work with the series. Subtracting $ \frac13$ precisely kills the original constant term. Then dividing by $ x$ drops the degree of each remaining term by $ 1.$ Hence:

$ \frac {f(x) \minus{} \frac13}{x} \equal{} \frac {2}{3^2} \plus{} \frac {3x}{3^3} \plus{} \cdots \plus{} \frac {(n \plus{} 1)x^{n \minus{} 1}}{3^{n \plus{} 1}} \plus{} \cdots$

You should be able to see what happens if you let $ x\to0$ in that series.

As for c) and d), I'll just say this: Integrate term by term, as if the series were a polynomial. As long as you stay within the interior of the the interval of convergence (and $ [0,1]$ is well inside $ ( \minus{} 3,3)$,) then this is justified. The resulting series has a very nice pattern, and should be an instance of a series you should know about. (Probably the most important of all specific series.)

If it were my problem, I'd add parts e) and f):

e) Write out the series for $ g(x) \equal{} 1 \plus{} \int_0^xf(t)\,dt.$ (Valid for $ |x| < 3.$) On the basis of this series, identify $ g(x)$ by giving a specific closed form formula for it. If you can do d), you ought to be able to do e).

f) Give a formula for $ f(x).$

----

@the future: Go ahead and post what you think the answers are. Some one of the flock of calculus teachers who post here will tell you whether you're right.

Ill try and do that in the next few hours when I get bored on homework (tons before spring break)

but for 1a, I got about the same solution
Ill try and post all my solutions as clear as possible but sorry if its confusing.
1a.

So I got (n+1)x^n/3^n+1 as a summation from n=0 to infinity.

So I used the ratio test and eventually got that it will be |3/x|<1 because as the n goes to infinity, it will be basically 1 and that the ratio test converges as
-3<x<3 where R=3 and its centered at x=0.  Then I need to test for endpoints

So I plug back in 3 for x and I got (n+1)3^n/3^n+1=(n+1)/3, which obviously diverges as n goes to infinity.
So then I plug in back -3 for x and I got (n+1)(-3)^n/3^n+1, which is the same as rewritting it as (n+1)(-1)^n/3.  So since this is an alternating series, I have to apply the anternating series test.  But when n is infinity, that thing clearly diverges because we will have a huge number* either negative or positive.

So I got $ ( \minus{} 3,3)$.

1b. alright so that one seemed quite easy which means i probalby overestimated it.  If you just subtract it, it seems to be the only constant term is 2/3^2=2/9.  Everything else seems to go to 0 as x>0

1c. save...

1c.  This is the part where I dont really get...
Clearly by integrating, its a geometric series.  The 0 part will have a sum of 0, and the 1 part will just be 1/3+ 1/9+ 1/27... for x=1.  The general term seems to be (1/3x)^n+1 where n starts at 0 (do i word it like this).

1d.  The sum is  just 1/3/(1-1/3)=)(1/3)/(2/3)=1/2

so does anyone wana confirm this?

thanks.

Those are the correct answers for c) and d), although your notation is sloppy and hard to read. Be careful with parentheses, and make sure you're saying what you mean to say.

Can you do the parts e) and f) that I added?

Dang forgot about these two below

e.  Hmm just looking at it it seems like its $ 1 \plus{} ({\frac {1}{3}})^x \plus{} ({\frac {1}{3}})^{2x}... \plus{} ({\frac {1}{3}})^{nx} \plus{} ...$

$ g(x) \equal{} \sum_{n \equal{} 0}^{\infty}({\frac {1}{3}})^{nx}$

f. I dont think this is good but I just got it from the orignal given equation so $ f(x) \equal{} \sum_{n \equal{} 0}^{\infty}\frac {n \plus{} 1}{3^{n \plus{} 1}}x^n$

sorry here are my new answers.

e. $ \frac {1}{1 \minus{} \frac {1}{3}x}$

f. $ S \equal{} \frac {\frac {\frac {1}{3}}{1 \minus{} \frac {1}{3}x}}{1 \minus{} \frac {1}{3}x} \equal{} \frac {\frac {1}{3}}{{(1 \minus{} \frac {1}{3}x)}^{2}}$

I really hope when you mean closed form, you mean just finidng the inifite sum in terms of x  :).

I have the solutions below your post.

[quote="Kent Merryfield"]That's bad notation on (d). There should not be $ x$ in any exponents.

The intent for (e) was a "closed form" formula - that is something that doesn't require any infinite processes (no infinite sums.)[/quote]
k thanks ...
wait first i meant e and f not d and e.
but even that i think i did it wrong...let me try again.

but how can I changeg e to having something without x, because the problem you gave told me to find the integral from 0 to x, so woundlnt it have to depend on x?

Do you mean im suppose to find the sum of that formula.

I get like [b]e.[/b] $ \frac {1}{1 \minus{} \frac {1}{3}x}$ depending on what x is.
whoa wait i get what you mean by no x, i meant $ 1 \plus{} \frac {1}{3}x \plus{} (\frac {1}{3}x)^n$ and so forth oops...forgot what i was actually doing.

and sorry i dont quite get what you are asking for f to find f(x), if you are asking for what the sum would be like [b]f.[/b]

$ f(x) \equal{} \sum_{n \equal{} 0}^{\infty}\frac {n \plus{} 1}{3^{n \plus{} 1}}x^n$
then you would have to manipulate the sum a bit.

$ S \equal{} \frac {1}{3} \plus{} \frac {2}{9}x \plus{} \frac {3}{27}x^2 \plus{} \frac {4}{81}x^3...$
$ \frac {S}{3}x \equal{} \frac {1}{9}x \plus{} \frac {2}{27}x^2 \plus{} \frac {3}{81}x^3 \plus{} ...$
$ S \minus{} \frac {S}{3}x \equal{} \frac {\frac {1}{3}}{1 \minus{} \frac {1}{3}x}$ so yea
$ S(1 \minus{} \frac {1}{3}x) \equal{} \frac {\frac {1}{3}}{1 \minus{} \frac {1}{3}x}$,
$ S \equal{} \frac {\frac {\frac {1}{3}}{1 \minus{} \frac {1}{3}x}}{1 \minus{} \frac {1}{3}x} \equal{} \frac {\frac {1}{3}}{{(1 \minus{} \frac {1}{3}x)}^{2}}$

I hope this is what you are asking for (but i am not positive because f. really required no calculus, just more of an "AMC/AIME" summation trick) I have bad interpretation problems sorry.

you are correct that the answer in part e must be in terms of x. I think what Kent was objecting to was that you had x in the exponents which is incorrect.

f looks correct to me


Alia's digital clock read 7:15 a.m. when she left for school. When she returned home 7 hours and 15 minutes later, the clock read 5:55 a.m. because the power had gone off during the day. If her clock automatically reset to 12:00 a.m. when power was restored, at what time that morning did the power return?

It should have been 7,5 hours later, but the clock has moved 5 and 55 mins from 12, or it was out of power for 1h, 35 min. Power returned at 7:15+1:35=8:50

[hide]
Since the clock resets to $12:00$, and her clock read $5:55$, that must mean that the clock has ran for $5$ hours and $55$ minutes after the power came back. The time that she returned home was $2:30$. That minus $5:55$ would be $\boxed{8:35}$.
[/hide]

Chenmichael is right. A faster solution is to just use military time and say she gets back at 14:30 (because that's way easier to calculate). Subtracting 6 hours and adding 5 minutes you get 8:35.

Question: Hmm.. do we assume that the power just went off and back on in a matter of a second? Or does it go off, then turn back on a while later?

[quote=annieliu12]No, that's not necessary because we don't need to know when the power went out. After the power came back, the clock started counting time from $12:00$ am. All we need to do is find the time when the power came back by finding the time Alia came home, which is $7:15+7:15=14:30=2:30$ pm,and subtract $5:55$ from that time.[/quote]

Ah, makes sense. Thanks!


Lem I:

agar $abc = 1$ va  $a, b, c > 0$

angah :

${a \over{2a+1}}+{b \over{2b+1}}+{c \over{2c+1}}\leq  1$

esbat : tarafeyn vasateyn konid bad sade konid, mishavad
$3 \leq  a+b+c$  va chon $abc = 1$, dorosti in ham badihiye!

hala rah hale khode masale :

${a \over a^{2}+2abc}+{b \over b^{2}+2abc}+{c \over c^{2}+2abc}={1 \over a+2bc }+{1 \over b+2ac}+{1 \over c+2ab}$

--> ${1 \over 2\sqrt{abc}+bc}+{1 \over 2\sqrt{abc}+ac}+{1 \over 2\sqrt{abc}+ab}={a \over 2a+1}+{b \over 2b+1}+{c \over 2c+1}$
ke hamoon Lem I, va hokm masale sabet shod!

PS. in soal ke badihi bood! :P harcki shakhe oon tabeyiye post
ghablo hal kone. :D fek konam ke chine, 2003 ya 2004.


number theory
greatest common divisor
number theory unsolved

I'm trying to prove the following:
[u]Theorem:[/u]
Let $ a,b,c\in\mathbb{Z}$. If $ \gcd(a,c) \equal{} \gcd(b,c) \equal{} 1$, then $ \gcd(ab,c) \equal{} 1$.
[i]Proof:[/i]
First I'm going to state a lemma of which the converse is obvious, but which I hadn't realized being true itself until today:

[u]Lemma:[/u]
If $ 1 \equal{} ax \plus{} by$ for some integers $ x,y$, then $ \gcd(a,b) \equal{} 1$.
[i]Proof:[/i]
Let $ d \equal{} \gcd(a,b)$. Then by definition, $ d|a$ and $ d|b$, hence $ d|(ax \plus{} by)$ for [u]all[/u] integers $ x,y$ (not both zero). Therefore, $ d|1$, and because the gcd is by definition positive we must have $ d \equal{} 1\ \ \ \ \square$

Back to the theorem: write $ 1 \equal{} ax_1 \plus{} cy_1$ and $ 1 \equal{} bx_2 \plus{} cy_2$ and multiply:
$ 1 \equal{} (ax_1 \plus{} cy_1)(bx_2 \plus{} cy_2) \equal{} (x_1x_2)ab \plus{} (x_1y_1a \plus{} x_2y_1b \plus{} y_1y_2c)c$, so by the previous lemma we have $ \gcd(ab,c) \equal{} 1$.

Is this correct? Thank you for your time.

That is correct, except that you made a minor error, $ (ax_1\plus{}cy_1)(bx_2\plus{}cy_2)\equal{}(x_1x_2)ab\plus{}(x_1y_2a \plus{} x_2y_1b \plus{} y_1y_2c)c$, not $ (x_1x_2)ab \plus{} (x_1y_1a \plus{} x_2y_1b \plus{} y_1y_2c)c$.

Ah, I see, a typo. Thanks for your reply. I see that the converse of the theorem is also true: if $ \gcd(ab,c) \equal{} 1$ then $ 1 \equal{} xab \plus{} yc$ for some integers $ x,y$. But it is now directly clear that $ \gcd(a,c) \equal{} 1$ and $ \gcd(b,c) \equal{} 1$, since $ 1 \equal{} (xa)b \plus{} (y)c \equal{} (xb)a \plus{} (y)c$.

The Lemma in my first post turns out to be pretty useful, making the next proof pretty straightforward:

[u]Theorem[/u]: If $ \gcd(a,b) \equal{} 1$ and $ c|(a \plus{} b)$, then $ \gcd(a,c) \equal{} \gcd(b,c) \equal{} 1$.
[i]Proof:[/i]
$ c|(a \plus{} b)$ gives $ a \plus{} b \equal{} kc$ for some integer $ k$. Now $ \gcd(a,b) \equal{} 1$ gives $ 1 \equal{} ax \plus{} by$ for some integers $ x,y$. We make two similar manipulations:

1) $ 1 \equal{} ax \plus{} by \equal{} ax \plus{} (kc \minus{} a)y \equal{} (ky)c \plus{} (x \minus{} y)a$, therefore $ \gcd(a,c) \equal{} 1$
2) $ 1 \equal{} ax \plus{} by \equal{} (kc \minus{} b)x \plus{} by \equal{} (kx)c \plus{} (y \minus{} x)b$, therefore $ \gcd(b,c) \equal{} 1\ \ \ \ \ \square$

It's always satisfactory to come up with a proof on your own :) Thanks again.

Well, usage of this lemma seems too [i]brutal[/i]. A slightly easier way exists:

1st:
if $ (ab,c)>1$, then there is $ p\in\mathbb{P}$ such that $ p|c$ and $ p|ab$, it follows $ p$ divides either $ a$ or $ b$.

But then either $ (a,c)$ or $ (b,c)$ is more than $ 1$, contradiction.

Precisely the same method works for the second problem.

That is a nice, short proof. I tried this method too, but I got stuck on the following: why, assuming $ (ab,c) \equal{} d > 1$, does there exist a [u]prime [/u]that divides both $ ab$ and $ c$?
I see it now (it's quite obvious): if $ d$ is prime, then $ d \equal{} p$. If $ d$ is not prime, it has a prime factor $ p$. Then we have $ p|d$, $ d|ab$ and $ d|c$, hence $ p|ab$ and $ p|c$.

Btw: you said "a contradiction", but actually your proof was by contrapositive, right?

Let me try your method on the last theorem:
Assume $ (a,c) \equal{} d > 1$, then there is a prime $ p$ such that $ p|a$ and $ p|c$. Because $ c|(a \plus{} b)$, we have $ p|(a \plus{} b)$. Because $ p|a$, we have $ p|((a \plus{} b) \minus{} a)$ or $ p|b$. Therefore, $ p|a$ and $ p|b$, hence $ (a,b)\geq p > 1$, a contradiction. (Not contrapositive because I used that $ c$ divides $ a \plus{} b$.)

[quote="Erken"]Well, usage of this lemma seems too [i]brutal[/i]. A slightly easier way exists:

1st:
if $ (ab,c) > 1$, then there is $ p\in\mathbb{P}$ such that $ p|c$ and $ p|ab$, it follows $ p$ divides either $ a$ or $ b$.

But then either $ (a,c)$ or $ (b,c)$ is more than $ 1$, contradiction.

Precisely the same method works for the second problem.[/quote]

This theorem is usually used for the purpose of proving the unique factorization theorem, but your proof assumes the unique factorization theorem, so even though your proof seems easier, the first proof is more correct.

[quote="Erken"]the lemma states:

if there exist $ x,y$ such that...blah blah blah... then $ (a,b) \equal{} 1$,

But [b]BlueVelvet [/b] said: "Since $ (a,b) \equal{} 1$ we're allowed to assume that there exist $ x,y$...blah blah blah..." Does that make sense?
[/quote]It does make sense, but maybe I wasn't too clear about it.

Like I said before the Lemma: [quote]First I'm going to state a lemma [b]of which the converse is obvious[/b], but which I hadn't realized being true itself until today: [/quote]
I assumed that the fact that if $ d \equal{} \gcd(a,b)$ then there are $ x,y$ such that $ d \equal{} xa \plus{} yc$ (call this fact 1), is known (it was to me). But the [i]converse[/i], for the special case $ d \equal{} 1$, was not known to me, that's why I stated and proved the lemma.
So by saying "Since (a,b)=1 there are x and y such that 1+xa+by", I am not using the lemma but fact 1. After this, I multiply and get 1=(ky)c+(x-y)a. It is [i]here [/i]where I'm using the lemma to conclude that (a,c)=1. I don't see any mistake in this reasoning (if you do, please point it out).

[i]Twin Prime Conjecture[/i]'s objection to your solution is "but your proof assumes the unique factorization theorem". I don't know whether this is a problem. I guess your solution is equally correct, but that it's only a matter of taste which one you prefer.

[quote="Erken"]dude, if you were more attentive, you'd probably notice that the first proof is actually incorrect. Due to the lemma, which states precisely the opposite fact. I mean the lemma states:

if there exist $ x,y$ such that...blah blah blah... then $ (a,b) \equal{} 1$,

But [b]BlueVelvet [/b] said: "Since $ (a,b) \equal{} 1$ we're allowed to assume that there exist $ x,y$...blah blah blah..." Does that make sense?

now tell me what's wrong with my solution?

if $ (ab,c) \equal{} d > 1$ it follows that $ d$ has a prime factor $ p$, besides that, $ d|c$ and $ d|ab$, therefore $ p|c$ and $ p|ab$?

Or do you consider the fact that $ p|ab$ implies $ p|a$ or $ p|b$, as a not trivial one?

But if $ p$ doesn't divide neither of $ a$ and $ b$, then it doesn't divide $ ab$ - I am convinced we are allowed to use this fact without a proof.[/quote]
If you are assuming the unique factorization theorem, there is nothing wrong with your proof. What I am trying to say is that if you want to prove the unique factorization theorem, then we use this theorem, but if we use your way of proving it, the whole proof becomes circular.


geometry
incenter
trigonometry
invariant
circumcircle
inequalities
analytic geometry

The sides of triangle ABC have lengths AB=7, BC=8, and AC=9.

What is the distance between the incenter and orthocenter of triangle ABC?

Use cosine rule and find cosB.
Use it to find tanB.
Now consider a co-ordinate system centred at B and BC along the X axis.
Use tanB to find co-ordinates of A.
Find the co-ordinates of incentre and orthocentre and find distance between them.

(Note: You colud have also found cosC or cosA and proceeded but the calculations would have probably become tedious.)

[hide]
$ OI\equal{}\sqrt{R(R\minus{}2r)}$
$ R\equal{}\frac{7 \cdot 8 \cdot 9}{4 \cdot \sqrt{12 \cdot 3 \cdot 4 \cdot 5}}$
$ R\equal{}\frac{7 \cdot 8 \cdot 9}{16 \cdot 3 \sqrt{5}}$
$ R\equal{}\frac{21\sqrt{5}}{10}$
$ r\equal{}\frac{12\sqrt{5}}{12}\equal{}\sqrt{5}$
$ OI\equal{}\sqrt{\frac{441\minus{}420}{20}}\equal{}\frac{\sqrt{105}}{10}$
[/hide]

proof of diophantient's invariant, which you might be able to use without citing

[hide]
We need to show that $ OI^2 \equal{} R^2 \minus{} 2Rr$.  This is not that hard to do.  Let $ ABC$ be a triangle with circumcenter $ O$, circumcircle $ \Omega$, incenter $ I$, and incircle $ \omega$.  Extend $ AI$ to meet $ \Omega$ at $ T$.  Then obviously $ OI^2 \minus{} R^2 \equal{} IA\cdot IT$ by power of a point.  But since $ I$ is in the interior of $ ABC$, which is inside of $ \Omega$, $ A,I,T$ are collinear in that order.  Hence we write $ R^2 \minus{} OI^2 \equal{} IA\cdot IT$ with all quantities now happily positive.

$ IA \equal{} r\csc{\frac{\angle A}{2}}$, because the points of tangency of $ \omega$ with $ AB, AC$ form right triangles with $ A$ and the incenter $ I$.

Now we need only compute $ IT$.  We want to show that $ IT \equal{} BT \equal{} CT$.  Thankfully, this is not hard to do.  Note that $ \angle CIT \equal{} 180^{\circ} \minus{} \angle AIC \equal{} 90 \minus{} \frac{\angle B}{2}$.  But $ \angle ICT \equal{} \angle ICB \plus{} \angle BCT \equal{} \frac{\angle A}{2} \plus{} \frac{\angle C}{2}$.  Obviously $ \angle ICT \equal{} \angle CIT$, hence $ IT \equal{} CT$.  In addition, triangle $ BCT$ is obviously isosceles, hence $ CT \equal{} BT$.  Thus $ IT \equal{} BT \equal{} CT$ as desired.  Note that $ \angle BOT \equal{} \angle A$, and $ BO \equal{} TO \equal{} R$.  Hence $ BT^2 \equal{} 2R^2 \minus{}2R^2\cos{\angle A}$.  It follows that $ BT \equal{} 2R\sqrt{\frac{1\minus{}\cos{\angle A}}{2}} \equal{} 2R\sin{\frac{\angle A}{2}}$.

Thus $ IA\cdot IT \equal{} 2Rr$.  Hence $ OI^2 \equal{} R^2\minus{}2Rr$.  [/hide]

You are mistaking cirumcenter with orthocenter.  That $ \sqrt {R^2 \minus{} 2Rr}$ is for the incenter with circumcenter

The right answer is not $ \frac {\sqrt {105}}{10}$.

I actually manage to do this one at school just today and it requires parallel lines for the orthocenter to one of the sides with the incenter to one of the sides.

Then you cna actually find one of the heights, the radius of the incircle, length of tangent segments from two of the sides, and using trig find the part where it is the length of the radius minus the length of the segment from the orthocenter to the side.  Then use  pythagorean theorm. The answer is actually a pretty simplified answer.

[quote="Teki-Teki"]proof of diophantient's invariant, which you might be able to use without citing

[hide]
We need to show that $ OI^2 \equal{} R^2 \minus{} 2Rr$.  This is not that hard to do.  Let $ ABC$ be a triangle with circumcenter $ O$, circumcircle $ \Omega$, incenter $ I$, and incircle $ \omega$.  Extend $ AI$ to meet $ \Omega$ at $ T$.  Then obviously $ OI^2 \minus{} R^2 \equal{} IA\cdot IT$ by power of a point.  But since $ I$ is in the interior of $ ABC$, which is inside of $ \Omega$, $ A,I,T$ are collinear in that order.  Hence we write $ R^2 \minus{} OI^2 \equal{} IA\cdot IT$ with all quantities now happily positive.

$ IA \equal{} r\csc{\frac {\angle A}{2}}$, because the points of tangency of $ \omega$ with $ AB, AC$ form right triangles with $ A$ and the incenter $ I$.

Now we need only compute $ IT$.  We want to show that $ IT \equal{} BT \equal{} CT$.  Thankfully, this is not hard to do.  Note that $ \angle CIT \equal{} 180^{\circ} \minus{} \angle AIC \equal{} 90 \minus{} \frac {\angle B}{2}$.  But $ \angle ICT \equal{} \angle ICB \plus{} \angle BCT \equal{} \frac {\angle A}{2} \plus{} \frac {\angle C}{2}$.  Obviously $ \angle ICT \equal{} \angle CIT$, hence $ IT \equal{} CT$.  In addition, triangle $ BCT$ is obviously isosceles, hence $ CT \equal{} BT$.  Thus $ IT \equal{} BT \equal{} CT$ as desired.  Note that $ \angle BOT \equal{} \angle A$, and $ BO \equal{} TO \equal{} R$.  Hence $ BT^2 \equal{} 2R^2 \minus{} 2R^2\cos{\angle A}$.  It follows that $ BT \equal{} 2R\sqrt {\frac {1 \minus{} \cos{\angle A}}{2}} \equal{} 2R\sin{\frac {\angle A}{2}}$.

Thus $ IA\cdot IT \equal{} 2Rr$.  Hence $ OI^2 \equal{} R^2 \minus{} 2Rr$.  [/hide][/quote]

In addition, Stewart's bash works.

The equality easily gives rise to Euler's inequality, $ R\ge2r.$.

[color=darkblue][b][u]Generally[/u][/b], for a fixed point $ M$ with the barycentrical coordinates $ (u,v,w)\ ,$

$ (u + v + w = 1)$ w.r.t. $ \triangle ABC$ and any point $ X$  exists the [b]Lagrange's identity[/b] :

$ \boxed {\ u\cdot XA^2 + v\cdot XB^2 + w\cdot XC^2 = XM^2 - p_w(M)\ }$ .

[b]Particular case.[/b] For the points $ X: = H$ and $ M: = I(\ \frac {a}{2p}\ ,\ \frac {b}{2p}\ ,\ \frac {c}{2p}\ )$ ,

$ p_w(I) = - 2Rr$ obtain $ \sum aHA^2 = 2pHI^2 + 4Rpr$ a.s.o. In conclusion,

$ \boxed {HI^2 = 4R(R + r) + 3r^2 - p^2}$ and $ HI^2\ge 0$ $ \implies$ $ \boxed {p^2\le 4R(R + r) + 3r^2}$ .

[b]Remark.[/b] You can use $ \{\begin{array}{ccc} 2p & = & a + b + c \\
\ 2(p - a) & = & b + c - a \\
\ 2(p - b) & = & c + a - b \\
\ 2(p - c) & = & a + b - c\end{array}\|$ $ \implies$

$ S = \sqrt {p(p - a)(p - b)(p - c)}$ $ \implies$ $ \{\begin{array}{c} R = \frac {abc}{4S}\ . \\
 \\
r = \frac Sp\ .\end{array}$  [/color]

[quote="the future"]You are mistaking cirumcenter with orthocenter.  That $ \sqrt {R^2 \minus{} 2Rr}$ is for the incenter with circumcenter.[/quote]

Well, hey, it's a nice proof anyways.  And I think there is a nice way to find a good invariant for $ IH$ as well.  Coming soon...


find all functions with reel numbers $x$ such that
$f(2x)=\frac{2f(x)}{1+f(x)^{2}}$. :)

$f(x)=tg(ax)$. Your equation equavalent only $a(x)=a(x*2^{m}) \ \forall m\in Z$. Therefore $a$ can be $a\not = const$.

[quote="Rust"]$f(x)=tg(ax)$.[/quote]
I believe you mean $f(x)=\tanh (ax)={e^{ax}-e^{-ax}\over e^{ax}+e^{-ax}}$.

(*) $f(2x)={2f(x)\over 1+f(x)^{2}}$ for all $x\in\mathbb{R}$

Then $|f(2x)|={2|f(x)|\over 1+f(x)^{2}}=1-{(1-|f(x)|)^{2}\over 1+f(x)^{2}}\leq 1$ for all $x\in\mathbb{R}$ and $f: \mathbb{R}\to [-1,1]$.
$x=0$ in (*) gives $f(0)={2f(0)\over 1+f(0)^{2}}$ and $f(0)\in\{0,\pm 1\}$.
$x\mapsto{2x\over 1+x^{2}}$ is a bijection $[-1,1]\to [-1,1]$.
So every map $f: (-2,-1]\cup[1,2)\to [-1,1]$ has a unique extension $f: \mathbb{R}^{*}\to [-1,1]$ satisfying (*),
and therefore exactly $3$ extensions $f: \mathbb{R}\to\mathbb{R}$ satisfying (*).

[quote="olorin"]
$ x\mapsto{2x\over 1 \plus{} x^{2}}$ is a bijection $ [ \minus{} 1,1]\to [ \minus{} 1,1]$.
So every map $ f: ( \minus{} 2, \minus{} 1]\cup[1,2)\to [ \minus{} 1,1]$ has a unique extension $ f: \mathbb{R}^{*}\to [ \minus{} 1,1]$ satisfying (*),
and therefore exactly $ 3$ extensions $ f: \mathbb{R}\to\mathbb{R}$ satisfying (*).[/quote]

I didnt understand this step
could you clarify more

Thanks  :)

Put $ x \equal{} 0$ and let $ c \equal{} f(0)$ and $ c$ is not $ 0$.
$ {1} \plus{} c^2 \equal{} 2$ so $ c \equal{} 1$ or $ {\minus{}1}$. Now assume we have a function $ h(x) \equal{} \frac{f(x)}{x}$

...
...

i don't have much time to talk.

[quote="FOURRIER"][quote="olorin"]
$ x\mapsto{2x\over 1 + x^{2}}$ is a bijection $ [ - 1,1]\to [ - 1,1]$.
So every map $ f: ( - 2, - 1]\cup[1,2)\to [ - 1,1]$ has a unique extension $ f: \mathbb{R}^{*}\to [ - 1,1]$ satisfying (*),
and therefore exactly $ 3$ extensions $ f: \mathbb{R}\to\mathbb{R}$ satisfying (*).[/quote]

I didnt understand this step
could you clarify more

Thanks  :)[/quote]
Sorry for the delay.

$ x = 0$ in $ (*)$ gives $ f(0) = {2f(0)\over 1 + f(0)^2}\Rightarrow f(0)\in\{0,\pm 1\}$.
Also every $ x\in\mathbb{R}^*$ has a unique expression $ x = 2^ny$ with $ n\in\mathbb{Z}$ and $ y\in ( - 2, - 1]\cup [1,2)$.
If $ \sigma: [ - 1,1]\to [ - 1,1],x\mapsto{2x\over 1 + x^{2}}$ and
$ g: ( - 2, - 1]\cup [1,2)\to [ - 1,1]$ arbitrary, then

$ f(x) = \{ \begin{array}{ll} \sigma ^n(g(y)) & \text{for } x = 2^ny\text{ with }n\in\mathbb{Z}\text{ and }y\in ( - 2, - 1]\cup [1,2) \\
0\text{ or }1\text{ or } - 1 & \text{ for }x = 0\end{array}$

are $ 3$ well-defined solutions of $ (*)$ with $ f(x) = g(x)$ for all $ x\in ( - 2, - 1]\cup [1,2)$,
and these are all the solutions of $ (*)$.

Here $ \sigma^0(x): = x$
and $ \sigma^n(x): = \sigma(\cdots \sigma(x)\cdots ), n\text{ times }\sigma$
and $ \sigma^{ - n}(x): = \sigma^{ - 1}(\cdots \sigma^{ - 1}(x)\cdots ), n\text{ times }\sigma^{ - 1}$
for all $ n\in\mathbb{N}$ and $ x\in[ - 1,1]$.


geometry
3D geometry
tetrahedron
trigonometry
calculus
derivative

Let $ ABCD$ be a regular tetrahedron, and let $ O$ be the centroid of triangle $ BCD$. Consider the point $ P$ on $ AO$ such that $ P$ minimizes $ PA \plus{} 2(PB \plus{} PC \plus{} PD)$. Find $ \sin \angle PBO$.

[hide="General Hint"]$ AO \equal{} AP \plus{} PO$ is fixed, so minimizing $ PA \plus{} 2(PB \plus{} PC \plus{} PD)$ is the same as minimizing $ 2(PB \plus{} PC \plus{} PD) \minus{} PO$.

Also, note that $ PB \equal{} PC \equal{} PD$, so we really have to minimize $ 6PB \minus{} PO$.

Now, we no longer have to deal with a tetrahedron, but with right triangle $ \triangle POB$, where $ \angle O$ is right, $ OB$ is fixed, and $ OP$ is variable.[/hide]
[hide="Calculus Solution"]If we set $ OP \equal{} 1$, we now want to minimize $ \frac {6 \minus{} \sin \angle PBO}{\cos \angle PBO} \equal{} K(\angle PBO)$

[i]For cleanliness, let us denote $ s \equal{} \sin \angle PBO$, $ c \equal{} \cos \angle PBO$, and $ k \equal{} K(\angle PBO)$[/i]

Multiplying by $ s$ gives $ 6 \minus{} s \equal{} kc$. Take the derivative to get $ \minus{} c \equal{} \minus{} k s \plus{} k' c$.

Since we want an extreme value, we use $ k' \equal{} 0 \implies k \equal{} \frac {c}{s}$.

Equating, $ \frac {6 \minus{} s}{c} \equal{} \frac {c}{s} \implies 6 s \minus{} s ^2 \equal{} c ^2 \equal{} 1 \minus{} s^2 \implies 6s \equal{} 1 \implies \boxed{s \equal{} \frac {1}{6}}$[/hide]

[b]Comment:[/b] [i]When written in the manner above, this problem becomes analogous to the well-known problem:[/i]
"If you are located at a distance $ d$ from a linear road, and can travel at a speed of $ r$ on-road and a speed of $ s$ off road, then what is the quickest path to some far-away point on the road?"


function
search
induction
inequalities
calculus
calculus computations

Let $ f$ be a real valued function defined on $ [a,b]$. Prove that if $ \frac{1}{2}f(x)\plus{}\frac{1}{2}f(y)\ge f\left(\frac{x\plus{}y}{2}\right)$ for all $ x,y\in [a,b]$, prove that $ f$ is convex on this interval.

I think you need an extra hypothesis like continuous or measurable. Otherwise you'd have midpoint convex implies convex which isn't true in general.

HilbertThm90, can you give the counterexample?

The Axiom of Choice is required to "construct" the counterexample. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=191855]Here[/url] is one thread where it is discussed - I found it by using the search function to find "convex & Hamel".

So how would you go about solving the problem if you were to assume that $ f$ was also continuous?

Suppose we have an interval $ [x(0),x(1)]$ and define $ x(r) \equal{} (1 \minus{} r)x(0) \plus{} rx(1)$ for $ r\in[0,1].$

Claim: for $ r \equal{} 2^{ \minus{} n}m,$ where $ n$ and $ m$ are nonnegative integers and $ 0\le m\le2^n,$ then $ f(x(r))\le (1 \minus{} r)f(x(0)) \plus{} rf(x(1)).$

We prove this by induction on $ n.$ Assume the statement is true for a particular $ n$ and all appropriate $ m.$ Now consider $ r \equal{} 2^{ \minus{} (n \plus{} 1)}m.$ If $ m$ is even, once we reduce the fraction, the proposition is already true under the inductive hypothesis. Hence, assume $ m$ is odd. We can write $ r \equal{} \frac12\left(2^{ \minus{} n}(m \minus{} 1)/2 \plus{} 2^{ \minus{} n}(m \plus{} 1)/2\right).$ By the midpoint-convex hypothesis:
\[ f(x(r))\le\frac12f(x(2^{ \minus{} n}(m \minus{} 1)/2)) \plus{} \frac12f(x(2^{ \minus{} n}(m \plus{} 1)/2))
\]
Now, apply the inductive hypothesis to each of the quantities on the right hand side and do (off-stage) the appropriate arithmetic, and we verify the claim for this $ r.$

Once we have that, we have that $ f(x(r))\le(1 \minus{} r)f(x(0)) \plus{} rf(x(1))$ for a set of points $ x(r)$ that are dense in the interval. Use the continuity of $ f$ to take limits to extend this to the entire interval.

The following is meant as an example to understand the induction. Let the interval be $ [0,1]:$

$ f\left(\frac58\right)\le\frac12f\left(\frac12\right)\plus{}\frac12f\left(\frac34\right)$

$ \le \frac12f\left(\frac12\right)\plus{}\frac14f\left(\frac12\right)\plus{}\frac14f(1)$

$ \le \frac34f\left(\frac12\right)\plus{}\frac14f(1)$

$ \le\frac38f(0)\plus{}\frac38f(1)\plus{}\frac14f(1)$

$ \le\frac38f(0)\plus{}\frac58f(1).$

[quote]Now, apply the inductive hypothesis to each of the quantities on the right hand side and do (off-stage) the appropriate arithmetic, and we verify the claim for this $ r$[/quote]
I don't see how to use the inductive hypothesis to get beyond this point.

You want me to bring that arithmetic back on stage? OK. The inductive hypothesis was that for $ r$ rational with denominator $ 2^n,$ $ f(x(r))\le(1\minus{}r)f(x(0))\plus{}rf(x(1)).$ Recall also that I am assuming that $ m$ is odd, so $ (m\minus{}1)/2$ and $ (m\plus{}1)/2$ are integers.

$ f(x(2^{\minus{}n}(m\minus{}1)/2))\le[1\minus{}2^{\minus{}n}(m\minus{}1)/2]f(x(0))\plus{}2^n(m\minus{}1)/2f(x(1)).$

$ f(x(2^{\minus{}n}(m\plus{}1)/2))\le[1\minus{}2^{\minus{}n}(m\minus{}1)/2]f(x(0))\plus{}2^n(m\plus{}1)/2f(x(1)).$

Now we add these two inequalities together, and divide by $ 2.$ The key step:

$ \frac12(2^{\minus{}n}(m\minus{}1)/2)\plus{}\frac12(2^{\minus{}n}(m\plus{}1)/2)\equal{}2^{\minus{}n}m/2\equal{}2^{\minus{}(n\plus{}1)}m.$

Hence,

$ \frac12\left(f(x(2^{\minus{}n}(m\minus{}1)/2))\plus{}f(x(2^{\minus{}n}(m\minus{}1)/2))\right)$
\[ \le [1\minus{}2^{\minus{}(n\plus{}1)}m]f(x(0))\plus{}2^{\minus{}(n\plus{}1)}f(x(1)).\]

This is precisely what I need for the induction.


Follow these steps to show that the sum $ C\plus{}C\equal{} \{x\plus{}y \mid x,y \in C \}$ is equal to the closed interval $ [0,2]$.
The observation that $ \{x\plus{}y \mid x,y \in C \} \subset [0,2]$ passes for obvious, so we only need to prove the reverse inclusion $ [0,2] \subset \{x\plus{}y \mid x,y \in C \}$. Thus, given $ s \in [0,2]$, we must find two elements $ x,y \in C$ satisfying $ x\plus{}y\equal{}s$.
(a) Show that for an arbitary $ n \in \mathbb{N}$, we can always find $ x_n, y_n \in C_n$ for which $ x_n\plus{}y_n\equal{}s$.
(b) Keeping in mind that the sequence $ (x_n)$ and $ (y_n)$ do not necessarily converge, show how they can nevertheless be used to produce desired $ x$ and $ y$ in $ C$ satisfying $ x\plus{}y\equal{}s$.

And what have you tried/which part is giving you trouble?


In trapezoid $ ABCD$, $ AB \equal{} CD \equal{} 2$ cm, $ BC \equal{} 6$ cm, segment $ AD$ is parallel to segment $ BC$, and $ m \angle BAD \equal{} 60^{\circ}$. What is the area of the trapezoid? Express your answer in simplest radical form.

[asy]pair A,B,C,D;

A = (0,0);
B = 2*dir(60);
C = B+(6,0);
D = 2*dir(300)+C;

draw(A--B--C--D--cycle);
label("A",A,W);
label("B",B,N);
label("C",C,N);
label("D",D,E);[/asy]

let's draw a foot of perpendicular from B and C to segment AD, and make them B' and C'.

because AB=2, AB'=1, BB'=$ \sqrt{3}$.

so the height of the trapezoid is $ \sqrt{3}$.

and AD=1+1+6=8.

so it becomes $ (8\plus{}6)*\frac{\sqrt{3}}{2}$, which is $ 7\sqrt{3}$.

answer : $ 7\sqrt{3}$


geometry
3D geometry
probability
circumcircle
angle bisector
AMC

#23: Each face of a cube is painted either red or blue, each with probabilty 1/2. The color orf each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color.

#24: In triangle ABC, we have AB = 7, AC = 8, and BC = 9. Point D is on the circumscribed circle of the triangle so that AD bisects angle BAC. What is the value of AD/CD?

Thanks.

[hide="24"]First step is to draw the diagram, next draw the lines to create $\overline{CD}$ and $\overline{BD}$.  Since quadrilateral $ACDB$ is cyclic we can use Ptolemy's and find that $7*\overline{CD}+8*\overline{BD}=9*\overline{AD}$ thus $AD=\frac{7*\overline{CD}+8*\overline{BD}}{9}$ and $\frac{\overline{AD}}{\overline{CD}}=\frac{7*\overline{CD}+{8}*\overline{BD}}{9*\overline{CD}}=\frac{7}{9}+\frac{8*\overline{BD}}{9*\overline{CD}}$

However, since $\overline{AD}$ is the angle bisector of $\angle BAC$ then $\overline{BD}=\overline{CD}$ (equal arcs leads to equal arc segments) thus $\frac{\overline{AD}}{\overline{CD}}=\frac{7}{9}+\frac{8}{9}=\boxed{\frac{5}{3}}$
[/hide]

[hide="23"]There are 3 different ways to choose 4 vertical sides, and the probability that any 4 sides are all the same color is $1*1/2*1/2*1/2=1/8$, so I think that it would be 3/8. Is that one of the choices?[/hide]

[hide="23"]Require the cube to have a fixed orientation. The cube in question has either 4, 5, or 6 sides of the same color. If the cube has four sides of the same color, the two faces which are the other color must be opposite each other. $2\cdot 3=6$ cubes have 4 sides of the same color and meet the above requirement. $2\cdot 6=12$ cubes have five sides of the same color. 2 cubes have six sides of the same color. There are a total of $2^{6}=64$ cubes. Thus the requested probability is $\frac{6+12+2}{64}=\boxed{\frac{5}{16}}$[/hide]

[quote="mustafa"][quote="Fanatic"]How are there 3 ways to choose the 4 vertical sides?[/quote]

Anything is possible when you're sleep-deprived and braindead :P[/quote]

I just hope that doesn't happen to me on the AMC A and/or B.

Is there another way to do number 23?


What is the length of the longest pipe that can be carried level around a right-angled corner at the intersection of two corridors of width 4 feet and 6 feet?

An [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=24520]old thread[/url] with several solutions to the general version.


Prove that
$ \left[ {\frac {{4\cos ^2 C \minus{} 1}}{{4\cos C}} \minus{} \frac {{\sin B \minus{} \sin AcosC}}{{\cos ^2 A}}(\frac {{\sin B}}{{2\cos C}} \minus{} \frac {{\sin A}}{2})} \right]\frac {1}{{2\cos ^2C \minus{} 1}} \ge \frac {3}{4}$(C<90).

[quote="TRAN THAI HUNG"]Prove that
$ \left[ {\frac {{4\cos ^2 C \minus{} 1}}{{4\cos C}} \minus{} \frac {{\sin B \minus{} \sin AcosC}}{{\cos ^2 A}}(\frac {{\sin B}}{{2\cos C}} \minus{} \frac {{\sin A}}{2})} \right]\frac {1}{{2\cos ^2 \minus{} 1}} \ge \frac {3}{4}$(C<90).[/quote]
It should have the condition
$ A \ne 90$ and $ C \ne 45$


topology
vector
Functional Analysis
advanced fields
advanced fields unsolved

Let H be a Hilbert space and $(T_n)\subset BL(H)$. Suppose that
rank($T_n$) is finite for each $n \in \mathbf{N}$ and that
$\parallel T_n - T\parallel \rightarrow 0$,
where $T \in BL(H)$.Show that the closure of $T(S)$ is compact where S:=\{x$\in$H:$\parallel x\parallel \leq 1$\}. show that 0$\in\sigma(T)$ and that $I-T$ is one-to-one if and only if it is onto (kind of rank nullity theorem).

Please prove the last part.

I suppose $H$ must be infinite-dimensional? Otherwise it's not necessarily true that $0\in\sigma(T)$, since all operators $T$ on a finite-dimensional Hilbert space are limits of operators of finite rank (since they all have finite rank), but not all of them are non-invertible.

In the first part we have to show that a limit in the norm topology of operators with finite rank is compact. Every operator with finite rank is compact, so if we show that in the norm topology the set of compact operators is closed, we're done.

We have to prove that the closure of $T(S)$ is compact, which ammounts to saying that it's complete and totally bounded. Being a closed subset of a complete space, $H$, it certainly is complete, so we only need to prove that it's totally bounded, which is equivalent to $T(S)$ being totally bounded (in every metric space the total boundedness of a set is equivalent to the total boundedness of the closure of the respective set).

Fix some $\varepsilon>0$. There is an $n$ s.t. $\|T_nx-Tx\|\le\frac\varepsilon 3,\ \forall x\in S$. Now, we can find $x_1,\ldots,x_k\in S$ s.t. $T_n(S)\subset\bigcup_{i=1}^kS\left(T_nx_i,\frac\varepsilon 3\right)$ because $T_n$ is compact ($S(x,r)$ denotes the ball centered at $x$ of radius $r$). Now take any $x\in S$. There is some $i\in\overline{1,k}$ with $\|T_nx-T_nx_i\|\le\frac\varepsilon 3$. We thus have $\|Tx-Tx_i\|\le\|Tx-T_nx\|+\|T_nx-T_nx_i\|+\|T_nx_i-Tx_i\|\le 3\cdot\frac\varepsilon 3=\varepsilon$, which means that $T(S)$ is covered by the balls $S(Tx_i,\varepsilon),i=\overline{1,k}$. $\varepsilon>0$ was chosen arbitrarily, so this is exactly what we want: $T(S)$ is totaly bounded.

The second part, stating that $0$ belongs to the spectrum of $T$, follows from the fact that $T$ is compact, for this implies that $T$ isn't even onto, let alone invertible with continuous inverse. This is because if we assume $T$ to be onto, we get $H=\bigcup_{n\ge 1}T(nS)=\bigcup_{n\ge 1}nT(S)$. The closure of $T(S)$ is compact, so it must have empty interior (no closed ball in an infinite-dimensional Hilbert space can be compact), so the same must be true for $nT(S),\ \forall n\ge 1$, which makes $H$, a complete metric space, a countable union of nowhere dense sets, i.e. a set of first category, contradicting Baire's Theorem.

I can prove at least one implication of the last assertion, namely that if $I-T$ is one-to-one then it's onto and even invertible (i.e. its inverse is also continuous, or, in other words, it's also bounded from below), but while trying to type it it became too long for my taste, so I don't feel like posting it right now :). I'll modify this post if I decide to write it.

Sorry for the double post, but this is way too long to just add it at the end of the previous message.

We prove that if $I-T$ is one-to-one, then it is invertible (i.e. not only is it onto, but it’s also bounded from below).

[i]Step 1[/i]: If $I-T$ is one-to-one, then it is bounded from below.

Suppose we can find a sequence of vectors with norm $1,\ (x_n)_n$, s.t. $(I-T)x_n\to 0$. We have a convergent subsequence $(Tx_{k_n})_n$ because $T$ is compact, and then $x_{k_n}$ will also be convergent to some $x$ with norm $1$. In turn, this makes $Tx_{k_n}$ converge to $Tx$, which gives, in the end, $x-Tx=0,\ x\ne 0$, contradiction. We have thus shown that $I-T$ is indeed bounded from below, i.e. there is some $m>0$ with $\|(I-T)x\|\ge m\|x\|,\ \forall x$.

[i]Step 2[/i]: We prove that in the given setting, if $I-T$ is one-to-one, then $I-T_n$ will be one-to-one for large enough $n$.

If we assume the contrary, this would mean that we can find a subsequence $(T_{k_n})_n$ s.t. $I-T_{k_n}$ are not one-to-one. To keep things simple, we assume $I-T_n$ is not one-to-one for any $n$. Then we can find $(x_n)_n,\ \|x_n\|=1,\ \forall n$ s.t. $(I-T_n)x_n=0$ but then, since $\|(I-T_n)-(I-T)\|\to 0$, we get $(I-T)x_n\to 0$, contradicting step 1.

[i]Step 3[/i]: The desired result holds for operators of finite rank, i.e. if an operator $S$ has finite rank and $I-S$ is one-to-one, then it is onto.

Let $(v_i)_{i=1}^n$ be an orthonormal basis of the finite-dimensional range $\mathcal R(S)$ of $S$. $x\mapsto (I-S)x$ will then be written as $x\mapsto x-\sum_{i=1}^n(x,u_i)v_i$ for some vectors $u_i$ in our space.

When restricted to $\mathcal R(S),\ I-S$ is an one-to-one and thus onto linear transformation, so $\mathcal R(S)$ is contained in the range of $I-S$. If we prove that the orthogonal complement $\mathcal R(S)^\perp$ of $\mathcal R(S)$ is also contained in the range of $I-S$, we’re done.

Let $y\in\mathcal R(S)^\perp$. We have to prove that there is an $x$ s.t. $x=y+\sum_{i=1}^n(x,u_i)v_i$. Since $x=y+\sum_{i=1}^n(x,v_i)v_i$, this amounts to finding $v\in\mathcal R(S)$ s.t. $(y+v,u_i)=(y+v,v_i),\ \forall i$, which is equivalent to $(v,v_i)-(v,u_i)=(y,u_i),\ \forall i$ (since $y\in\mathcal R(S)^\perp\Rightarrow(y,v_i)=0,\ \forall i$). This is indeed always possible, since the restriction of $I-S$ to $\mathcal R(S)$ is onto, so we can find $v\in\mathcal R(S)$ s.t. $\sum_{i=1}^n(y,u_i)v_i=(I-S)v=\sum_{i=1}^n[(v,v_i)-(v,u_i)]v_i$

[i]Step 4[/i]:

By step 2, we can assume $I-T_n$ to be one-to-one for all $n$. By steps 1,3, all $I-T_n$ are invertible. All the $I-T_n$ are bounded from below, so for each $n$, let $0<m_n=\inf_{\|x\|=1}\|(I-T_n)x\|$. Assume now the sequence $(m_n)_n$ is not bounded away from $0$. Then, for simplicity, we assume $m_n\to 0$. This means that we can find a sequence $(x_n)_n$ of vectors of norm $1$ s.t. $(I-T_n)x_n\to 0$, which means that $(I-T)x_n\to 0$, contradicting step 1.

[i]Step 5[/i]: Finally, we prove the result.

By step 2, we can assume all $I-T_n$ are one-to-one. By steps 1,3 we find that $I-T_n$ are invertible, and by step 4 the $m_n$’s defined there form a set which is bounded from below by some $m>0$. Since $I-T_n$ are onto, given some $y\ne 0$ in our space, we can find $x_n$ s.t. $(I-T_n)x_n=y,\ \forall n$. Now, since $\|y\|=\|(I-T_n)x_n\|\ge m\|x_n\|$, we find $\|x_n\|\le\frac 1m\cdot\|y\|$, i.e. the $x_n$ lie in some closed ball around the origin. This means that $\|y-(I-T)x_n\|=\|(I-T_n)x_n -(I-T)x_n \|\to 0$. Since $y$ was arbitrary, we have found the image of $I-T$ to be dense in our space.

We conclude that $I-T$ is invertible by using the well-known characterization of invertible operators: a bounded linear operator on a Hilbert space is invertible iff it’s bounded from below and it has a dense image.


Again Holder!We have $(\sum_{cyc}\sqrt[3]{x^{2}y^{2}(1-z)^{2}})^{3}\leq(\sum_{cyc}xy)(\sum_{cyc}xy-3xyz)(\sum_{cyc}1-x)$.Now we come back to $(\sum_{cyc}xy)(\sum_{cyc}xy-3xyz)\leq\frac{2}{27}$ wich i proved. :lol:

You can also generalise from here.We have $\sum_{cyc}\sqrt[n]{x^{n-1}y^{n-1}(1-z)^{2}}\leq\frac{\sqrt[n]{4}}{3}$ :lol:


let S(n) denote the sum of the decimal digits of  the enteger n
for example s(23)=5
find the smollest integers such
$S(n)+S(S(n)+S(S(S(n)))=2007$

$S(n)$ is a number from this problem:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=750071#p750071

[hide="From there..."] We want the smallest integer whose digits sum to $1977, 1980, 1983, 2001$.

Clearly we should use a lot of $9$s.  In fact $1980$ is divisible by $9$, so we can use $n = 999...$ (220 times).  However, with a digit sum of $1977$ we can change the first digit to a $6$, so that is our smallest answer. [/hide]

torajirou, he/she tricked you.

It is sum of digits first, not n.

Did you read my solution?  I'm aware of that.

If $k$ is an integer satisfying $k+S(k)+S(S(k)) = 2007$ then $k$ is one of the integers from the problem I linked to, and we want to find the smallest $n$ such that $S(n) = k$, which is exactly what I did.

Oh sh**

Sorry, I did, read the answers at top.

YOu are correct.

I bow down to you.


Find integers $a$ and $b$ that satisfy the equation
\[2a+ab+\frac{a}{b}=243 \]

[hide]
$a=24$
$b=8$
From $2a+ab+\frac{a}{b}=243$, I multiplied by $b$ to get
$2ab+ab^{2}+a=243b$ Then divide by $a$ to get $2b+b^{2}+1=\frac{243b}{a}$ This can be factored into $(b+1)(b+1)=\frac{243b}{a}$ Multiply by $a$ and divide by $b$ to get $\frac{a}{b}(b+1)(b+1)=243$ Now you find
$243=3^{5}$
This means $b+1=3$ or $b+1=9$
Trying out values you get $b=8$
Solving for $a$, you get $a=24$
[/hide]

you guys got it right good job!! :first:


I don't know whether this belongs on the Intermediate or not but oh well. If it's too easy, remove this post from here, moderator.

In a certain tetrahedron, each vertex is incident to three edges of length 5, $\sqrt {41}$ and $\sqrt {34}$, respectively. Determine the length of each segment between the midpoints of some two opposite edges of the tetrahedron. :)

It's 3,4 and 5. Just consider a rectangular parallelepiped ABCDEFGH with AB=3, BC=4, CD=5. Then DBGE is your tetrahedron.


function
number theory proposed
number theory

we are given k.
Can the set of natural numbers be partitioned into k number of arithemtic progressions of the form a(i)+ n*t(i) ( n=0,1,2,...), where a(i) and t(i) are natural numbers (i=1,...,k) where all the t(i) are pairwise different?

as I remember we can solve it with generating function and complex number (I don't know any other solutions)

[quote="lhvietbao"]we are given $k$.
Can the set of natural numbers be partitioned into $k$ number of arithemtic progressions of the form $a_i + nt_i$ ($n=0, 1, 2, ...$), where $a_i$ and $t_i$ are natural numbers ($i=1, ..., k$) where all the $t_i$ are pairwise different?[/quote]

 :D

UPINT means Unsolved Problems In Number Theory. ;)


invariant
real analysis
real analysis unsolved

If we consider the map $ T: [0,1) \to [0,1)$, $ x \mapsto 2x \mod 1$ then the following set $ C\equal{}\{ \frac{n}{2^m}: n,m \in \mathbb N\}$ is invariant in the sense that $ T(C)\equal{}C$. My question is: how to find the invariant set if we chang the map $ T$ to, for example, $ T(x)\equal{}\frac{1\minus{}x}{2}$?

For this particular example, $ T(x)\equal{}\frac{1\minus{}x}{2}$, we have $ T([0,1)) \equal{} (0, \frac {1} {2} ]$, then $ T(T([0,1))) \equal{} [\frac {1} {4}, \frac {1} {2} )$, so $ \bigcap_{n\geq 1} T^n([0,1)) \equal{} \emptyset$.

Therefore there exists no "invariant" set but $ \emptyset$.

[quote="mavropnevma"]Therefore there exists no "invariant" set but $ \emptyset$.[/quote]

Oh thanks, that's not good example. OK, let's consider the following new one

$ T : x \mapsto \frac{x}{p}$ if $ 0 \leq x < p$ and $ x \mapsto \frac{x\minus{}p}{1\minus{}p}$ otherwise where $ p \in (0,1)$.

A simple calculation shows that $ T^{\minus{}1}([0,p)) \equal{} [0,p)$ and $ T^{\minus{}1}([p,1)) \equal{} [p,1)$. So I think the invariant set for $ T$ consists of two parts: the first part consists of numbers $ \frac{n}{p^m}$ but I don't know how to find the invariant set for $ T$ since I am confusing how can we deal with the map $ x \mapsto \frac{x\minus{}p}{1\minus{}p}$  :( .


modular arithmetic
algebra unsolved
algebra

find $(m,n)$   is positive integers   satisfy  condition :
$\frac{(3+m)^n+1}{3m}$ is a integer

\[ \frac{(3+m)^n+1}{3m} = \frac{3^n+ {n \choose 1} 3^{n-1}m + ... + m^n + 1}{3m} \]
So the question simplyfies to: find $(m,n) \in \mathbb{N}^2$ so that
\[ \frac{3^n+m^n+1}{3m} \] is an integer.

So $m^n + 1 \equiv 0 \pmod{3}$
and $3^n + 1 \equiv 0 \pmod{m}$

Since I'm not such a hero when it comes to congruences, I'm stuck here.


I forgot to mention that $ \angle ADC \equal{} \angle BAE$! The statement should be correct now.

[quote="Mathias_DK"]In $ \triangle ABC$, let $ D$ be the midpoint of $ AB$ and $ E$ a point on the side $ BC$ such that $ \frac {BE}{EC} \equal{} k$ and $ \angle ADC \equal{} \angle BAE$.
Prove that:
$ \cos \left ( \angle BAC \right ) \cdot \frac {AC}{AB} \equal{} \frac {k \minus{} 2}{4k}$

As a corollary: http://www.mathlinks.ro/viewtopic.php?t=42760[/quote]
Let $ F$ be the foot of the perpendicular from $ C$ on $ AB$ and let $ B'$ be the point on the line $ AB$ such that $ FB'\equal{} FB$, $ B\ne B'$. Triangles $ \triangle ABE$ and $ \triangle DB'C$ are similar. Hence
\[ \frac {AB}{BE} \equal{} \frac {DB'}{B'C} \equal{} \frac {\frac {3}{2}AB \minus{} 2AF}{BC} \Leftrightarrow \frac {BC}{2BE} \equal{} \frac {3}{4} \minus{} \frac {AF}{AB}.\]
Therefore, the right-hand of the equation equals
\[ \frac {1}{4} \minus{} \frac {EC}{2BE} \equal{} \frac {3}{4} \minus{} \frac {BC}{2BE} \equal{} \frac {AF}{AB},\]
unsing the relation above.

However, we have $ \cos(\angle BAC) \equal{} AF/AC$. Hence the left-hand side of the equation equals $ AF/AB$, which is what we had to prove.


Is there any strategy to finding the general form for a recursive function of the following form, in terms of $ r$, $ s$?
$ A(0) \equal{} r, A(1) \equal{} s, A(n) \equal{} A(n \minus{} 1) \plus{} A(n \minus{} 2)$

I know that when $ r$ and $ s$ are 1, $ A(n) \equal{} \frac {1}{\sqrt {5}}\cdot{({(\frac {1 \plus{} \sqrt {5}}{2})^{n\plus{}1} \minus{} (\frac {1 \minus{} \sqrt {5}}{2})^{n\plus{}1}})}$

$ A(n) \equal{} a \left( \frac{1 \plus{} \sqrt{5}}{2} \right)^n \plus{} b \left( \frac{1 \minus{} \sqrt{5}}{2} \right)^n$ for appropriate $ a, b$.  In fact, this strategy can be generalized to find the general form for the recursion

$ A(n) \equal{} p A(n\minus{}1) \plus{} q A(n\minus{}2)$.

But even without that very general strategy, note that

$ A(2) \equal{} r \plus{} s$
$ A(3) \equal{} r \plus{} 2s$
$ A(4) \equal{} 2r \plus{} 3s$
$ A(5) \equal{} 3r \plus{} 5s$
$ A(6) \equal{} 5r \plus{} 8s$
...
$ A(n) \equal{} F_{n\minus{}1} r \plus{} F_n s$

where $ F_0 \equal{} 0, F_1 \equal{} 1$ are the familiar Fibonacci numbers.  :)

Well, ignore the first part of what I said.  Do you understand the second argument?  Let's try some extra terms.

$ A(7) \equal{} 8r \plus{} 13s$
$ A(8) \equal{} 13r \plus{} 21s$
$ A(9) \equal{} 21r \plus{} 34s$
...