calculus
function
geometry
3D geometry
modular arithmetic
number theory

As one of the question writers for this competition, I would like to know: Which questions did you like most/least?

For number theory, make the questions harder. :)

For algebra, make some of the questions have less contrived solutions (#7) and please don't just straight up steal them out of acops (#5).

I liked 9 the most in advanced topics, and 6 the second most if you consider it in the manner you wrote it as opposed to the manner the solution seems to indicate that you intended it.

I disliked 2 due to the fact that is killed by Calculus, and I suppose somewhat disliked 7 because it has been posted like 4 times on the WOOT board so it gave WOOTers an unfair advantage (although you might have been unaware of that).

More, well, [i]Advanced[/i] topics (Generating Functions, more advaned Inequalities/Series/whatever), might be cool, though that might be unfair to some people so I'll leave that up to you.

Number 1 seemed kind of random for a number 1. And a bit tough to do rigorously. I personally just found a way to get 30 and assumed there was nothing better.

The random painting problem was also pretty cool.

For geo test:
I liked #9 and #10.
#7 was also alright.
I slightly disliked #8 & #5 because they were trig.
I disliked #3, since it could be easily solved if you remembered that cos(36)=(sq(5)+1)/4, which I did, but then I got a nested radical in my answer and thought that wasn't simplified for some reason and couldn't figure out how to simplify it, so I started doubting my memory and ended up putting something wrong.

Also for relay, 1-5 and 2-4 were way too guessable; we ended up getting them both without having all the previous answers correct =/.

The team questions were nice.

The power questions were kind of weird but I don't have any real complaints.

For geometry, can we get more cyclic quadrilateral questions?

[quote="JSteinhardt"]Yeah Haitao, you and I had some pretty awesome relay predicting going there.

(Unfortunately we only got it at the 12 minute mark both times, but I passed him the correct answer within 60 seconds in both cases, without having the previous answer.)

The first time I then passed him the incorrect answer >_<, and the second time he misread the question and passed the incorrect answer in at the 6-minute mark.[/quote]

lol, I was person 3 for my team and since both problems were really easy, I was having fun watching the other teams work on the problems. Every time I looked at TJ #3, he always was furiously scratching at his paper even though the problems were trivial. :rotfl:

I thought Relay 2 was funny. Just because the most common mistake went something like this:

Person 1: Oh. It's just 1,2 and -1,4. *passes 2*

Person 2: An integer sided right triangle with side length 2? He must mean three....*passes correct answer*

Person 1: OH SHOOT! 3,0 works too *passes 3 EXTREMELY quickly*

Person 2: Hey...look at that, he did mean 3.

Also, relay one was like
Person 2: uhhhh....what?
*5 minutes later*
Person 2: Oh hey! look at that! The y's cancel out...I didn't even need problem 1's answer!

As for my individual test, I have to say the # theory test was not that easy, but rather insanely guessable. Let's take a look at the one which said to find the maximum value of x+y+z.
Me: Hmmmm...I have 4 minutes left. I know zero is possible, so I'll just write that.

And the one that wanted you to find all n such that 7n^3-3n^2-3n-1 is a cube
(not) me: -1 works, 0 works, 1 works....I don't think there'll be any more solutions just cuz they'll be UBER big. *writes -1, 0, 1*
(btw, actually me thought that 1 didnt work due to some freak brain fart so I only put -1,0)

And the one where you had to find the number that couldn't be made from 12s 21s and 28s.
Me: Hmm, it'll be 83+12k for some integer k....83 can't be made....I think 95 can't be made....I think that 107 can't be made.....I think 119 can't be made....Oh wait, actually 119 can be made....*erases answer for the 5th time and rewrites 107*

Also, a lot of them were just huge computation and not really testing any number theory skill.
For example: find the last two digits of $2003^{2005^{2007^{2009}}}$. It's so easy. As everyone knows, $\varphi(100)=40$ and $\varphi(\varphi(100))=16$ and $\varphi(\varphi(\varphi(100)))=8$ so we take 2003 mod 100. 2005 mod 40, 2007 mod 16 and 2009 mod 8 so we get $3^{5^{7^{9}}}$ A bit of computation reveals that the answer is then $43$. I even skipped the first two to go to this one just because it was mindless computation.

[quote="Hamster1800"]
Also, a lot of them were just huge computation and not really testing any number theory skill.
For example: find the last two digits of $2003^{2005^{2007^{2009}}}$. It's so easy. As everyone knows, $\varphi(100)=40$ and $\varphi(\varphi(100))=16$ and $\varphi(\varphi(\varphi(100)))=8$ so we take 2003 mod 100. 2005 mod 40, 2007 mod 16 and 2009 mod 8 so we get $3^{5^{7^{9}}}$ A bit of computation reveals that the answer is then $43$. I even skipped the first two to go to this one just because it was mindless computation.[/quote]

well..there really wasn't too much computation necessary on that one. first taking it mod 100 gives $3^{{2005}^{{2007}^{2009}}}$. And then since $\varphi(100)=40$, $3^{40}\equiv 1 \pmod{100}$, so we need to find ${2005}^{{2007}^{2009}}$ $\pmod{40}$, which is trivialized when you note that $5^{1}$ is $5 \pmod{40}$, $5^{2}$ is $25 \pmod{40}$, $5^{3}$ is $5 \pmod{40}$...and so from that you see that ${2005}^{{2007}^{2009}}$ reduces to $5 \pmod{40}$ because it is odd, and thus the last two digits are the last two digits of $3^{5}$, which is $43$.

but yes i do agree that some of the NT problems were kind of guessable (especially the one that involved fermat's last theorem...i was pleased that i figured it out during the test but disappointed in that i knew that most people would probably be able to guess the only solutions.)

[quote="Hamster1800"]
Also, a lot of them were just huge computation and not really testing any number theory skill.
For example: find the last two digits of $2003^{2005^{2007^{2009}}}$. It's so easy. As everyone knows, $\varphi(100)=40$ and $\varphi(\varphi(100))=16$ and $\varphi(\varphi(\varphi(100)))=8$ so we take 2003 mod 100. 2005 mod 40, 2007 mod 16 and 2009 mod 8 so we get $3^{5^{7^{9}}}$ A bit of computation reveals that the answer is then $43$. I even skipped the first two to go to this one just because it was mindless computation.[/quote]

just take it mod 4 and 25...


I'm probably just over-thinking it, but this one is giving me trouble.

In Boston, the mean yearly temperature is 58 degrees Fahrenheit and the total mean temperature variation is 56 degrees. Assume the temperature on April 1st is the mean yearly temperature so C = 0 and t is the number of days past April 1. Find a complete graph of this problem situation and determine about what month and day is it the hottest and the coldest in Boston?

$58+56\sin{\frac{2\pi}{365}x}$

Assuming that this sine wave models the temperature, the hottest day will be 3 months later, and the coldest day will be 3 months before.


function
Putnam
geometry
geometric transformation
homothety
ratio
real analysis

hi all

Perhaps you all know this pb but I find it very nice:
let f a continuous fonction such that for all real a>0 we have:
limit (f(n*a), n->+infinity)=0
Prove that limit(f(x),x->+infinity)=0

bye.

Here's a question strongly related to it (don't get me wrong: I haven't solved any of them :)):

Can we find a sequence of disjoint (for simplicity) intervals $I_n=[a_n,b_n]\subset (0,\infty)$ s.t. $a_n\to\infty$ and for any $a>0$ only finitely many elements of $\{na|n\in\mathbb N^*\}$ lie in $\bigcup_{n\ge 1} I_n$?

My guess (suggested by the problem posted here) is that the answer is "no", but I don't know if the problems are perfectly equivalent, so I can't be sure.

Sorry for a posting this potentially useless non-solution message :).

I think I solved it, but I am not sure. The idea is to consider the sets A_n formed by all x nonnegatives such that for all k greater than n we have f(kx) in (-epsilon, epsilon), where epsilon is fixed. All these sets are closed and the interiour of their union is non-empty, thus there is one that has on-empty interiour. After that everything is easy. Where is this problem from? I find it super-cool. I knew this problem if the function is uniformly continuous, but the proof is much simpler. And also, is there a more normal solution?

It's B3 from [url=http://www.kalva.demon.co.uk/putnam/putn64.html]Putnam 1964[/url].

I see you used Baire's theorem. I think I have a solution using Cantor's theorem: a descending sequence of closed intervals has non-void intersection, but I'm not sure either. I thought it would be easy to explain, but it turns out not to be that easy :).

Assume there is a sequence $x_n$ s.t. $f(x_n)>\alpha>0$. Because of the continuity of $f$ we also have $f(x)>\alpha,\ \forall x\in I_n$, where $I_n$ is a non-degenerate closed interval around $x_n$. This means that for any $a>0$ the intersection $\{na|n\in\mathbb N^*\}\cap\bigcup_{n\ge 1} I_n$ is finite. I think this is a contradiction (this is the problem I mentioned in my previous message).

Take a random closed interval $J_1=[a,b]$. If no $n_1J_1$ (the interval obtained from $J_1$ by a homothety of center $0$ and ratio $n_1\in\mathbb N^*$) encloses an $I_n$, then we either have infinitely many $na$ or infinitely many $nb$ in $\bigcup_{n\ge 1} I_n$, and this is a contradiction. This means that we can find $n_1,k_1\in\mathbb N^*$ s.t. $n_1J_1\supseteq I_{k_1}$. Take $J_2=\frac 1{n_1}I_{k_1}\subseteq J_1$. We do for $J_2$ exactly what we did with $J_1$ (this time we find an $n_2>n_1$), and we get $J_1\supseteq J_2\supseteq J_3\supseteq\ldots$, i.e. a descending sequence of closed intervals. Take $x$ from their non-void intersection. Then $n_tx\in I_{k_t},\ \forall t\in\mathbb N^*$, and this is a contradiction, because there can't be infinitely many values of the form $nx$ where $f$ takes values $>\alpha$.

I hope it's correct..

in fact I found this exercice in several books. It is a classical problem to show one application of Baire's theorem

Yes, that seems to be the nicest solution. The one on Kalva is essentially the same. I've seen other solutions, though. If mine is correct (:)) it's such an example, and a classmate showed me another one.


algebra
polynomial
number theory unsolved
number theory

Let f(n)=the number of odd terms in the nth row of Pascal's Triangle.

Find a formula/algorithm for f(n)

Hint
[hide]
It has something to do with the binary representation of n.
[/hide]

Answer w/o solution
[hide]
Let b(n)=the number of 1's in the binary representation of n.
$f(n)=2^{b(n)}$

See also:
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=pascal+odd&t=54624

Unfortunately, I didn't understand the above ^ link, that's why I posted this problem.  Could someone please explain the solution?
[/hide]

Lucas Theorem can be applied as follows:

For $n,i\in\mathbb{N}$ and $0\le i\le n$, let $n=a_{j}a_{j-1}\dots a_{2}a_{1}$ and $i=b_{k}b_{k-1}\dots b_{2}b_{1}$ be the binary representations of these two numbers.  Then:

${2^{n}\choose i}\equiv{a_{1}\choose b_{1}}{a_{2}\choose b_{2}}\dots{a_{k}\choose b_{k}}{a_{k+1}\choose 0}\dots{a_{j}\choose 0}$ mod $2$.

Now, the RHS is $0$ if and only if one of the digits in the binary representation of $n$ is $0$ and the corresponding digit in $i$ is $1$.  So all the digits $a_{m}$ that equal $0$ must have $b_{m}=0$ also.  Only the other digits can be $0$ or $1$, and hence the number of odd combinations what you surmised in your post.

Oh yeah, and you can also do this problem with a counting argument that uses the fact ${n-1\choose i-1}+{n-1\choose i}={n\choose i}$.

Thanks, I understand the method via Lucas' Theorem now.  But what would the counting argument look like?  I think the different positions of odds make it hard.

It is Problem 9 from the LVII Polish Mathematical Olympiad (there it was said in form: given $n=2^{k_{1}}+...+2^{k_{m}}$, where $k_{1}<...<k_{m}$, then find the number of odd coefficients of the polynomial $(x+1)^{n}$). You used method similar to my method; the official solution started with $2\mid{{2^{k}}\choose{t}}$ for $1\leq t\leq 2^{k}$, then there was used factorisation of polynomial $(x+1)^{n}$ to polynomials $(x+1)^{2^{k_{i}}}$.


Hi!

I'm just posting some easy AHSMEs (AMC same yeah) problems. Enjoy! I don't have time to collect them all but just few ones I like:

1982 AHSME #3

Evaluate \[(x^x)^{(x^x)}\]
at
\[x=2\]

[hide]Simplifying within the parantheses, you'll get 4^4 = 256[/hide]

AHSME #1 1981

If \[\sqrt {x+2} = 2\], then \[(x+2)^2\]equals:

[hide]The relationship between sqrt x+2 and x+2 thing that's whole squared is that it's fourth powered. So, we have to take fourth power of 2 to get the answer, which is 16.[/hide]

Sorry for bad explanation. ;)

1) Um... I guess that is:

[hide]2^8 = 256.[/hide]

2) That's:

[hide]2^4 = 16?[/hide]


The number of terms of the series 54 + 51 + 48 +… Such that the sum is 513 is
a.	18
b.	19
c.	both a & b
d.	none of these

I get the solution as 18 however the given solution is a & b... weird??? how is it possible??? kindly explain....

How many terms are there in the GP 5, 20, 80,…. 20, 480?
a.	6
b.	5
c.	7
d.	8

and how about this one... i'm not getting anywhere...

Try to find out what's the 19th term and you'll figure it out in an instant :)

[quote="fizsle"]

How many terms are there in the GP 5, 20, 80,…. 20, 480?
a.	6
b.	5
c.	7
d.	8

and how about this one... i'm not getting anywhere...[/quote]So geometric progressions are like $a$, $ar$, $ar^{2}$, etc.

So to find the $n$th term of a geometric sequence $x$ you use the formula $x_{n}=ar^{n-1}$...,plug $r=4$ and $a=5$ into this formula and set it equal to 480, then solve for $n$.  That should be your answer.

[quote="eryaman"](...)set it equal to 480 (...)[/quote]That wouldn't be an integer. But I think he meant $20480 = 5\cdot 4^{6}$. So the answer is 7.

oops... thats it... it is 20480... and i was wondering how 20, 480 would fit into the series...and to add to my worries...20, was printed on the first line and the 480 was on the next line in the book! lol... as for the first problem thanks... 0 as a term could change number of terms! Thanks guys...


linear algebra
matrix
Putnam
symmetry
linear algebra unsolved

Let A be real 3x3 matrix which is initially empty. Players I and II make moves alternately - in each turn player I writes 1 in an empty cell and player II writes 0 elsewhere. When the A is filled, player I wins if detA<>0 and loses otherwise. Player I makes the first move. Who has the winning strategy?

[quote="neworder"]. When the A is filled, player I wins if detA<>0 and loses otherwiseegy?[/quote]
I don't get the part about the value of detA.

This is Putnam 2002 A4. That "<>" is $\neq$.

The solution requires brute force casework; I think the second player wins.

Indeed, the solution isn't very cute - "brute force" case analysis. I hoped that maybe someone would come up with a trick or something, but the problem seems to be rather straightforward...

Here's my writeup for the benefit of my class that year. No, it isn't pretty.

[b]A4:[/b] In the game of Determinant Tic-Tac-Toe, player 1, who plays first, is trying to produce a non-zero determinant; player 0, who plays second, is trying to produce a zero determinant. Given optimal play, player 0 must always win.

First: since any permutation of rows and any permutation of columns does not change whether a determinant is zero, all starting positions are equivalent. WLOG player 1 starts in the center. We will have player 0 respond in the upper left corner. From this position, player 1 has (up to symmetry) 4 possible replies. We will deal with each of these possible replies in turn. We note that any row or column of all 0’s will result in a victory for player 0, so in analyzing the game we will assume that player 1 will act to block a row or column that already has two 0’s.

Case I: $\begin{bmatrix}0&1&\\&1&\\&&\end{bmatrix}$ leads to  $\begin{bmatrix}0&1&0\\&1&\\&&\end{bmatrix},$ which has (up to symmetry) three replies:

IA: $\begin{bmatrix}0&1&0\\&1&\\&1&\end{bmatrix}$ followed by $\begin{bmatrix}0&1&0\\0&1&\\&1&\end{bmatrix}$ leads to  $\begin{bmatrix}0&1&0\\0&1&0\\1&1&1\end{bmatrix}.$

IB: $\begin{bmatrix}0&1&0\\&1&\\1&&\end{bmatrix}$ followed by $\begin{bmatrix}0&1&0\\&1&0\\1&&\end{bmatrix}$ leads to  $\begin{bmatrix}0&1&0\\0&1&0\\1&1&1\end{bmatrix}.$

IC: $\begin{bmatrix}0&1&0\\1&1&\\&&\end{bmatrix}$ followed by $\begin{bmatrix}0&1&0\\1&1&\\&&0\end{bmatrix}$ leads to  $\begin{bmatrix}0&1&0\\1&1&1\\0&1&0\end{bmatrix}.$

Case II: $\begin{bmatrix}0&&1\\&1&\\&&\end{bmatrix}$ leads to $\begin{bmatrix}0&&1\\&1&\\0&&\end{bmatrix}$ leads to $\begin{bmatrix}0&&1\\1&1&\\0&0&\end{bmatrix}$ leads to $\begin{bmatrix}0&0&1\\1&1&1\\0&0&1\end{bmatrix}.$

Case III: $\begin{bmatrix}0&&\\&1&1\\&&\end{bmatrix}$ leads to $\begin{bmatrix}0&&\\&1&1\\0&&\end{bmatrix}$ leads to $\begin{bmatrix}0&&\\1&1&1\\0&&\end{bmatrix},$ which continues as in case 1A, rotated.

Case IV: $\begin{bmatrix}0&&\\&1&\\&&1\end{bmatrix}$ leads to $\begin{bmatrix}0&&0\\&1&\\&&1\end{bmatrix},$ which continues as in case II.

This exhausts the possibilities; player 0 can always win.

It's actually rather pretty compared to my casework!
How hard is this compared to other Putnam problems?


How do you do a mid-line tilde in LaTeX to represent a similarity symbol, i.e. ABC ~ DEF?  If I just put the symbol in directly, it doesn't show up in the compiled document.

Also, how do you put an arc symbol over an arc of a circle?

[quote="solafidefarms"]\widehat{ABC}=$\widehat{ABC}$[/quote]
Hmm, that's not quite what I'm thinking about.  I want an actual arc of a circle (like the $\frown$ command) on top of an arc that I want to designate as such.

[quote="joml88"]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=56552

You can put \usepackage{tipa} in the preamble of your document and then use the command \textroundcap.[/quote]
Unfortunately, that produces a very small arc that doesn't stretch to cover several characters  :(  I ended up just using widehat, but unfortunately there's no nice-looking way to produce the arc, which is very odd.

Hmm, that's interesting.  I've never had to use the arc symbol so I just assumed what they suggested there would work.  Oh well.

Just use \widehat and make sure in beginning, say like $\widehat{ABC}$ means arc $ABC$. I remember Valentin and couple other people wondering why the people who made Latex didn't make the one for arc sign. I'm not sure why neither.

I don't know if you guys still need this but I found a way to put the arc sign above more than 3 letters. The command is \wideparen and you need to have \usepackage{yhmath} at the beginning of the document.

Hope this helps.


from aops book 2:

1.  If log 36 = $a$ and log 125 = $b$, express log(1/12) in terms of $a$ and $b$.

2.  Given that $\log_{4n}40\sqrt{3} = \log_{3n}45$, find $n^3$.

3.  Prove the following equalities.
i.  sin10*sin20*sin30 = sin10*sin10*sin100
ii.  sin20*sin20*sin*30 = sin10*sin20*sin80
iii.  sin20*sin30*sin30 = sin10*sin40*sin50

from 1999 HMMT algebra test:

4.  Find $x - y$, given that $x^4 = y^4 + 24, x^2 + y^2 = 6$, and $x + y = 3$.

5.  A polynomial $P$ has four roots, $\frac{1}{4}, \frac{1}{2}, 2, 4$.  The product of the roots is 1, and $P(1) = 1$.  Find $P(0)$.

[hide="Number 4"]i don't think u need the first equation.
$x^2+y^2=6$
$x+y=3$
$x^2+2xy+y^2=9$
$2xy=3$
$4xy=6$
$x^2-2xy+y^2=3$
$x-y=\sqrt{3}$[/hide]


[quote="#H34N1"]I'm still in 8th grade, about four years away from college, but I'd like to start planning out a way to make it into an ivy league. My SAT score is atm 2040 (Mainly due to reading inability, which I'm working to improve on).[/quote]
way too early?
[quote]

What would being a MOPper have an effect on college selections? What about a perfect score on the AMCs? What colleges will those count.
[/quote]
if we answered "no colleges dont care about MOP or AMC perfect scores" then would stop doing math?
[quote]
Also, is an SAT score of 2300, with combination of some advancement in the math olympiad process help me?
[/quote]
a better question to ask: would those things harm you in the admission process?

so yeah, its probably early for you to start worrying about college, and its better not to worry at all. just do what you want (like being good at math) instead of asking around what colleges like and trying to fit yourself into their admission criteria

[quote]so yeah, its probably early for you to start worrying about college, and its better not to worry at all. just do what you want (like being good at math) instead of asking around what colleges like and trying to fit yourself into their admission criteria[/quote]Couldn't agree with you more.

i also can't stress this more
doing what you like is most important in not just high school, but also life.
if you really want to, you can spend time on SATs, but try to take them only once and get them over with. also, junggi put it a great way when he asked "would this hurt you"
there really isn't a way to guarantee admission or something, but there are definitely things they call "red flags"

Yes, higher scores do have an influence but only to a certain, relatively small extent.  Colleges looks for more than test scores and even grades; they look for what a student can brign to their campus and their potential.  They consider how the student has challanged themselves and how they utilized the resources around them.  I agree with the poster above.

The only thing you should really be worrying about at this stage is exploring things that interest you. I assume that math is currently one such interest--you can set some goals and spend some time each day working towards them, or if you'd rather just do some free exploration, that's fine too. If you just work on it regularly, the experience you gain will help give you perspective by the time you're a senior.

Something to note: don't ever be motivated by achievements for the reason that they will help you get into college. Enjoyment and self-fulfillment come first. Other motivations generally lead to ineffective training methods (e.g., rote memorizing vocabulary for the SAT).

Also, make an honest effort to learn in all your subjects, not just making it through with a good grade (applicable only to classes that have sufficiently engaging material... this could be a problem at some schools).

Well obvioulsy you need a 2250+. My score of a 2080 won't get me into a college up to par with MIT (typo on teh first post there), regardless of experience etc

Actually, that's not true.  SAT scores do not have "cut off"  lines like that.  Plenty of people have gotten into MIT with around that.  Also, you have 4 more years to take the SAT, so relax.   Get through high school first!

I will somewhat repeat what others have said, but you should start picking things you like to do, and achieve what you can through those interests (math, clubs, music, w/e else).   Then, when it comes time (junior year) to start looking at colleges, find which colleges match your interests and accomplishments.  In this way, you determine what you do and what college you go to based on your interests and what makes you happy, instead of letting your dream of a college determine your life (because in the end, no matter what you do, that college may still reject you.   MIT rejects throngs of kids whose dream it was to attend there and had the stats to back it too).

Have fun.

[quote="dauntless"]SAT scores do not have "cut off"  lines like that.[/quote]

Specifically, see [url]http://www.mitadmissions.org/topics/apply/standardized_test_requirements/index.shtml[/url].

I know you're Indian and your parents have laid out the three life options: Ivy, MIT, failure.

Chill. Take the time to find out what your interests are, to shape your habits, to spend time with friends, and enrich yourself. Have a taste of everything. Figure out what you like, and who and what you are. And then run with it. If you get into an Ivy+, then great.

But I don't think that getting into an Ivy+ school should not be anyone's principle goal -- I think that's a superficial take on life.

Do whatever you want to do, make the most of your life in the present. When your parents go insane after you don't get into the desired college of their choice, you can say that you "tried your best." They'll sympathize, either that or you're screwed, in which case you should probably go hide in Mexico or something until they cool down.

Along the lines of doing a few things that you like and doing them really well, you should note that you don't necessarily have to have one thing right now that you're really passionate about.  Instead, I read a scientific article not too long ago about how not only does being passionate about something generally correlate to more perseverance in doing that activity but persevering through an activity can, likewise, instill a passion for it in you.  My advice is to try to find things you're passionate about.  If you can't find any, try to make yourself passionate about things.

[quote="joml88"] Instead, I read a scientific article not too long ago about how not only does being passionate about something generally correlate to more perseverance in doing that activity but persevering through an activity can, likewise, instill a passion for it in you.  [/quote]

Personal experience aside, this idea was also given in <i>The Wedding Planner</i> (or some romantic comedy on cable) with respect to people and arranged marriages: tolerance builds into respect, which escalates into friendship and romantic love. Less rigorous, but relevant.

I also remember reading an article about extracurricular activities in China, how schoolkids picked an activity in first grade that they would do every afternoon until at least sixth grade, making a commitment of length unimaginable to Americans. The writer noted that after seeing the progress that they made from, kids were encouraged--they began to like the activity, no matter what initial misgivings.

[quote="#H34N1"]I'm still in 8th grade, about four years away from college, but I'd like to start planning out a way to make it into an ivy league. My SAT score is atm 2040 (Mainly due to reading inability, which I'm working to improve on).

What would being a MOPper have an effect on college selections? What about a perfect score on the AMCs? What colleges will those count.

Also, is an SAT score of 2300, with combination of some advancement in the math olympiad process help me?

Thanks.[/quote]

Hey that sounds a bit like me. I'm also in 8th grade, but with a bit less in SAT and also anticipating college (and sweating about it). But I do know that colleges will consider your achievements if they make you considerably special, namely MOP. Also, places like MIT do want your AMC scores, so it's a good idea to have those well done.
About the SAT, I think most people like you do end up achieving high scores (my aim is 2350+, but I'll take 2300+ if all else fails), despite how bad they might start off, although your current score is on a good track. It's good that you begin thinking early, but just don't kill yourself doing so like me. I have to thank my English teacher for pulling back to Earth everytime I get too concerned about the future  :P .

The original post mentioned "Ivy League," so I'll link to the Wikipedia article about that sports conference. Not all good colleges are part of the Ivy League, and arguably at least a few of the eight Ivy League colleges are less desirable for many AoPSers than several other colleges.

http://en.wikipedia.org/wiki/Ivy_League

It's very sad to me that 8th graders are stressing about getting into college.  If it makes you feel any better, the demographics of the US indicate that admissions should be easier rather than harder in 4 years.


Find the smallest natural number so that its factorial finishes in 2006^2006 zero

[quote="surge"]2. the answer is always 2.[/quote]

Wrong. 42 is the answer.

The answer's some huge number I don't want to calculate. It's 6712876532*6712876531*...*2*1.

that number would be so big, it'd probably not fit on my computer screen

[quote="stancioiu sorin"][quote="vicseb"]what is the answer to 6712876532! ?[/quote]
     Yeah, it's very big. :? But I've composed a problem with factorials...find the smallest natural number so that its factorial finishes in 2006^2006 zero.How about this?It's easier than yours, but it's incredible to calculate that factorial... :ninja:[/quote]

yeah I'll edit the problem and change it to that

[quote="surge"][quote="vicseb"]Its factorial... how could it be 2, or 42?[/quote]

I'll mail you the exact answer if you mail me back a large cookie.[/quote]

(forgot to reply to this) *gives surge a cookie* now you give me the exact problem to 6712876532!

don't feel like giving my address

[quote="stancioiu sorin"]find the smallest natural number so that its factorial finishes in 2006^2006 zero[/quote]
[hide]$5*2006^2006$???[/hide]

[hide]well the number of zeros in the factorial is dependent on the number of 5s that factor into it so it would be the first number with 2006^2006 5s, so at first I would say its 5 x 2006^2006 because it gains another 5 every five terms, however every 25 terms it gets two, and every 125 it gets 3, ect.

so hmmm how do we get to 2006^2006 5s? ill come back to this later[/hide]

I don't think working hard to try and solve this problem is a good idea: the answer is the smallest integer $N$ such that $\sum_{k=1}^\infty \left\lfloor \frac{N}{5^k} \right\rfloor =2006^{2006}$, which is somewhere between $4\cdot 2006^{2006}$ and $5\cdot 2006^{2006}$

However, those interested could try finding the smallest integer $N$ such that $N!$ ends in $5^{2006}$ 0's

[hide="which is"]$4\cdot 5^{2006}+5$[/hide]


The following problem is giving me a little trouble

What  is  the  probability that aninteger in the set ${1,2,3,...,100}$ is divisible by 2 and not divisible by 3?

i  used a counting approach but came up one short. I believe its somewhere near the end but not quite sure...anyway help would be appreciated.[/hide]

Well, $\frac{67}{100}$ are not divisible by 3, and $\frac{50}{100}$ are divisible by 2. Multiply the two conditions together, $\frac{67}{200}$ is your probability. (Not entirely sure on the correctness of this.)

[hide]I got 17/50 cause 1/2 of the numbers from 1-100 are divisible by 100. That is obviously 50 intergers. However, we need to get rid of the ones divisible by 3. Therefore, our even integers cannot be divisible by 6. There are 16 of those from 1-100 so we subtract that from 50. We get 34 numbers out of 100 total integers, which makes the probability 17/50.[/hide]

Sounds better to me. (i.e. ignore my solution)

we have 50 numbers divisible by 2, but we counted numbers that are also divisible by 6, so 100/6=16.66...  16 numbers overcounted, so we have 34/100 or 17/50

In {1, 2, 3, 4, 5, 6}, there are 2 that work. Modulo 6, all the other consecutive sets of 6 (like {7, 8, 9, 10, 11, 12}, etc.) just reduce to this, except the last one, {97, 98, 99, 100}. There are 2 that work in here, and there are 96/6=16 of the other sets, so (16*2+2)/100=34/100=17/50.


Let n be three digit number and divisible by 11 ,denote  n=ax10^2+bx10 + c ,and n=11(a^2+b^2+c^).Find the all possble solutions of n.

You should learn LaTeX to write math on this website. Did you mean this?

Let $ n$ be three digit number and divisible by $ 11$ , denote $ n\equal{}a\times10^2\plus{}b\times10 \plus{} c$ , and $ n\equal{}11(a^2\plus{}b^2\plus{}c^2)$ . Find the all possible solutions of n.

If this is what you mean,

[hide]We get $ 100a\plus{}10b\plus{}c\equal{}11a^2\plus{}11b^2\plus{}11c^2$ by substituting $ n$. Also, we get $ a\minus{}b\plus{}c\equal{}0$, since the 3-digit number $ abc$ is divisible by 11.

By plugging in $ b\equal{}a\plus{}c$ into the first equation, we get $ 2a^2\plus{}2c^2\minus{}10a\minus{}c\plus{}ac\equal{}0$.

In addition to the fact $ a,c<10$ and $ a\neq0$, we also get that $ c$ has to be even in order for $ 2c^2\minus{}c$ to be even and quadratic of $ a$ to have integer roots. Also, $ c<5$ because you cannot get positive roots from a quadratic with all positive terms. The only possible $ c$ values are 0, 2, 4. However, only $ c\equal{}0$ can give you a valid solution:

$ a\equal{}5, c\equal{}0$ => $ b\equal{}5$ by $ a\minus{}b\plus{}c\equal{}0$

So, there is only one solution: [b]550[/b][/hide]

This problem is from an old IMO (2nd ?).  :wink:

The above proof are elegant.
But when $ a\plus{}c\minus{}b\equal{}11$ , How about the possible solution to $ n$?

:o Aw... That's a good point. I guess I will have to consider that case as well. (boxedexe - I don't think this problem is half as hard enough to be on IMO, I mean I haven't done good enough on USAMO to be on MOP yet)

This is going to be slightly messier...

[hide]
[b]Case 2[/b]

Plugging $ b\equal{}a\plus{}c\minus{}11$ into $ 11a^2\plus{}11b^2\plus{}11c^2\equal{}100a\plus{}10b\plus{}c$, we get $ 2a^2\minus{}32a\plus{}2c^2\minus{}23c\plus{}131\plus{}2ac\equal{}0$.
Writing this into a quadratic in $ a$, we get $ 2a^2\plus{}(2c\minus{}32)a\plus{}(2c^2\minus{}23c\plus{}131)\equal{}0$. Since $ (2c^2\minus{}23c\plus{}131)$ has no real root and is above the x-axis, we can be sure that this will always be positive. Also, $ 2c\minus{}32$ can only become positive when $ c>16$ which is impossible with decimal numbers. Therefore, This equation will never have all positive terms.

Now, let's look at the parity of the constant term. Since we already know that $ 2c^2$ has to be even and $ 131$ is odd, $ 23c$ must be odd for the constant term to be even. This narrows possible values of $ c$ to 1,3,5,7,9. By trying these possibilities, we get

$ a\equal{}5,8$ for $ c\equal{}3$

However, $ a\equal{}5, c\equal{}3$ would not work, since $ b$ cannot be negative in $ a\minus{}b\plus{}c\equal{}11$. Thus, we only have one solution for this case: [b]803[/b]
[/hide]


Five positive integers $ a,b,c,d,e>1$ , satisfy the following conditions:

$ a(b \plus{} c \plus{} d \plus{} e) \equal{} 128$

$ b(a \plus{} c \plus{} d \plus{} e) \equal{} 155$

$ c(a \plus{} b \plus{} d \plus{} e) \equal{} 203$

$ d(a \plus{} b \plus{} c \plus{} e) \equal{} 243$

$ e(a \plus{} b \plus{} c \plus{} d) \equal{} 275$

What's the sum of $ a\plus{}b\plus{}c\plus{}d\plus{}e$ ?

A) 9
B) 16
C) 25
D) 36
E) 49

hello, i have found
$ a\equal{}4,b\equal{}5,c\equal{}7,d\equal{}9,e\equal{}11$ and their sum is $ 36$.
Sonnhard.

That's correct ;)
But how do you do it ?

[hide]$ c(a\plus{}b\plus{}d\plus{}e) \equal{} 203 \equal{} 7 \times 29$

Since $ a,b,c,d,e$ are integers greater than one the smallest possible value of $ a\plus{}b\plus{}d\plus{}e$ is $ 8$, which means we must have $ c\equal{}7$ and $ a\plus{}b\plus{}d\plus{}e\equal{}29$ which gives $ a\plus{}b\plus{}c\plus{}d\plus{}e \equal{} 36$ so the answer is $ \boxed{D}$.[/hide]

$ 128 \equal{} 2^7 \not\in prime \times prime$
$ 155 \equal{} 5 \times 31 \in prime \times prime$
$ 203 \equal{} 7 \times 29 \in prime \times prime$
$ 243 \equal{} 3^5 \not\in prime \times prime$
$ 275 \equal{} 5^2 \times 11 \not\in prime \times prime$

So, $ a\plus{}b\plus{}c\plus{}d\plus{}e\equal{}5\plus{}31\equal{}7\plus{}29\equal{}\boxed{36}$.
And, $ b\equal{}5, c\equal{}7$.

$ b\plus{}c\equal{}12 \iff a\plus{}b\plus{}c\plus{}d \ge 14$.
So, $ a\plus{}b\plus{}c\plus{}d\equal{}5^2\equal{}25, e\equal{}11$.

$ b\plus{}c\plus{}e\equal{}23 \iff 24 \le b\plus{}c\plus{}d\plus{}e \equal{} 2^x \le 35$.
So, $ b\plus{}c\plus{}d\plus{}e\equal{}32 \iff d\equal{}9, a\equal{}36\minus{}32\equal{}4$.

Thus, $ (a, b, c, d, e)\equal{}(4, 5, 7, 9, 11)$.


$ ABCDE$ is a convex pentagon inscribed in a circle of radius $ 1$ unit with $ AE$ as a diameter. If $ AB\equal{}a, BC\equal{}b, CD\equal{}c and DE\equal{}d$, prove that
$ a^2\plus{}b^2\plus{}c^2\plus{}d^2 \plus{} abc\plus{} bcd <4$.

well i cant help much except for posting the diagram

Well I have got a solution to this already using cosine rule.
But i want to see a pure geometry soln for this. :oops:

let AC=x, EC=y, $ 4\equal{}AC^2\plus{}EC^2\equal{}a^2\plus{}b^2\plus{}aby\plus{}c^2\plus{}d^2\plus{}cdx>a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}abc\plus{}bcd$

[quote="thaithuan_GC"]AC^2 + EC^2 = a^2 + b^2 + aby + c^2 + d^2 + cdx[/quote]

Can you explain please?


For the 2007 AMC 10 B administered on Feb 21:  The 1% AIME invitation score is 115.5.

For the 2007 AMC 12 B administered on Feb 21:  The AIME invitation score is 100.0.  About 8% to 9% of 2007 AMC 12 B contest takers scores achieved this score or higher.

Steve Dunbar
Director, American Mathematics Competitions


vector
analytic geometry
graphing lines
slope
real analysis
real analysis unsolved

[quote="lajekahr"]Maybe its just cause I was up until 4am last night but for the life of me I can't recall how to get a line equation... :(

If $f(x,y)=x^{2}+4y^{2}$, find the gradient vector $\nabla f(2,1)$ and use it to find the tangent line to the level curve f(x,y)=8 at the point (2,1)

So far I got $\nabla f=\left< 2x, 8y \right> \Rightarrow \nabla f(2,1)=\left<4,8\right>$
Any vector in the family of $\left< -2k, k \right>$ is perpendicular to the gradient. So should I just use the point (2,1) and the point gotten by adding that vector to it? Argh, most not stay up so late anymore....[/quote]

Ok so you know that the tangent vector is of the form $<-2k, k>$ therefore the slope of such a vector is $\frac{k}{-2k} = \frac{-1}{2}$.  This means your tangent line is of the form $y-1 = \frac{-1}{2}(x-2) \rightarrow 2y+x = 4$.

Thanks, I feel kinda silly having asked that, now that I look at it with a fresh set of eyes. It seems almost too simple to condense that easily, heh.


Alli rolls a standard 6-sided dice twice. What is the probability of rolling integers that differ by 2 on her first two rolls? Express your answer as a common fraction.


Reduce the expression :
 $S_n = {n\choose 1}.{{n-1}\choose {p-1}}.{{n-p}\choose {q-1}}+{n\choose 2}.{{n-2}\choose {p-2}}.{{n-p}\choose {p-2}}+...+{n\choose k}.{{n-k}\choose {p-k}}.{{n-p}\choose {q-k}}+...+{n\choose q}.{{n-q}\choose {p-q}}.{{n-p}\choose {0}}$  , where $n,p,q$ are positive integers and $n\geq p\geq q$

I will show that

[color=green][b](1)[/b][/color] $\sum_{k=0}^q\binom{n}{k}\cdot\binom{n-k}{p-k}\cdot\binom{n-p}{q-k}=\binom{n}{p}\cdot\binom{n}{q}$.

Then, your sum will easily follow:

$\sum_{k=1}^q\binom{n}{k}\cdot\binom{n-k}{p-k}\cdot\binom{n-p}{q-k}$
$=\sum_{k=0}^q\binom{n}{k}\cdot\binom{n-k}{p-k}\cdot\binom{n-p}{q-k}-\binom{n}{0}\cdot\binom{n-0}{p-0}\cdot\binom{n-p}{q-0}$
$=\binom{n}{p}\cdot\binom{n}{q}-1\cdot\binom{n}{p}\cdot\binom{n-p}{q}=\binom{n}{p}\cdot\binom{n}{q}-\binom{n}{p}\cdot\binom{n-p}{q}$.

We will prove [color=green][b](1)[/b][/color] by double counting:

Given the set S = {1; 2; ...; n}, how many ways are there to choose two subsets A and B of S such that A has p elements and B has q elements?

On the one hand, there are $\binom{n}{p}\cdot\binom{n}{q}$ ways to do this, since there are $\binom{n}{p}$ ways to choose the subset A with its p elements and $\binom{n}{q}$ ways to choose the subset B with its q elements.

On the other hand, we can first choose how many elements the sets A and B will have in common. Let's say they have k elements in common (where k can be an arbitrary integer such that $0\leq k\leq q$). Now, we choose the k elements that the sets A and B will have in common; this is possible in $\binom{n}{k}$ ways. Then, we choose the p - k elements which the set A has, but the set B doesn't have; this gives us $\binom{n-k}{p-k}$ possibilities. Then, we choose the q - k elements which the set B has, but the set A doesn't have; there are $\binom{n-p}{q-k}$ ways to do this choice. Thus, there are $\binom{n}{k}\cdot\binom{n-k}{p-k}\cdot\binom{n-p}{q-k}$ possibilities for a given k, so that, summing over all possible k, we get $\sum_{k=0}^q\binom{n}{k}\cdot\binom{n-k}{p-k}\cdot\binom{n-p}{q-k}$ ways to choose two subsets A and B of S such that A has p elements and B has q elements.

Comparing our two results, we obtain $\sum_{k=0}^q\binom{n}{k}\cdot\binom{n-k}{p-k}\cdot\binom{n-p}{q-k}=\binom{n}{p}\cdot\binom{n}{q}$, and [color=green][b](1)[/b][/color] is proven.

  Darij


What does your signature mean? I'll let ya all have a guess at what mine means :)

AoPS style users may see it on 2 lines, just do as like it were on one line (your style makes your screen very small)

mine is some quotes from a friend of mine

Mine is my name, plus some nicknames I've accumulated.

Favorite famous last words from one of the most chilling short stories of all time.

Mine is self-explanitory.  And also kind of true...a lot of math gets done with guesses being made and then proved.

hehe yeah let's make a site http://www.artofluckyguessing.com :D

besides, anyone an attemt to decrypt mine? :cool:

[quote="nr1337"]_________________
I need a new signature.[/quote]
You definitely do! :shock: lol

nr1337 wrote:I figured out Peter VDD's sig!  (in spoiler)
[hide]
**N**0**1**7**P**Y**R**C**N**3**Y**7**5**4* *N**5**1**H**7**3**3**5**U**0**Y**N**4**C**
[/hide]
 The real question is what does it mean ?

Indeed :)

Besides, Valentin, what does yours mean? It looks like a quote... but about what? :P

Peter, the thing about the spoiler was immediately clear to me, but I think I have no chance to decipher the text inside the sig since I really have never deciphered any nontrivial code ("trivial" means Caesar). As for Valentin, I am wondering what his sig means, too; probably it's about the MathLinks project (well, maybe it's also about the op02 project, cuz nobody seems to have a clue how it will end...).

And finally, I change my signature quite often. My current signature is from a Russian poem by Dmitrii Suharev. The former one was from a song by Julii Kim (a Russian bard). Before, I had a quote from Iosif Stalin ("He is not just a negative value, he is a negative value squared" - said by Stalin about one of his political opponents) showing his great mathematical talent. Sometimes, my signatures even have to do with mathematics :D - for instance, one of these signatures was "The sum of the minima is less or equal to the minimum of the sum", a [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15125]trivial but powerful principle[/url] in proving inequalities. I had learnt this principle at a lecture Arthur Engel gave at the IMO training seminar in Bad Homburg. I seem to have had some other signatures, actually I don't remember what they said. Remind me and I'll explain them to you :-)

  Darij

[quote="darij grinberg"]I think I have no chance to decipher the text inside the sig since I really have never deciphered any nontrivial code ("trivial" means Caesar).[/quote]

Heck no! :P

This one is way easier than Caesar, it isn't really a strong encryption, actually it isn't even really an encryption :)

But just like any good olympiad problem... you need to see it :cool:

[quote="grobber"]If I'm not mistaken, Valentin's signature is a quote from the movie "Matrix". In fact, it looks like one oh Morpheus's lines :).[/quote] You're correct ;)

[quote="sunnyray"]i thought u said it was ur school's motto...

did u kno switchfoot is considered a christian band?

(just a lil info... :lol: )[/quote]

My sig used to be my school's fight song.  :D

Yup, switchfoot is Christian Rock. I'm not Christian though, I just like some songs.

it doesnt need one....as long as were well respected and devote time and patience to our learning experience........which um.......i gess we are....except for the well respected part on this forum...but thats okay. miller is a UNIQUE school...heh heh.

no it is not

and mine says that i am in seventh grade cuz two threads started asking
one kid at my school thought i was in eighth
and that all the mathcounts forum moderators are past the age of mathcounts so they might not be continuing to moderate there

i made my sig [b][i][u]RATE THIS POST A 6!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!![/u][/i][/b] so i could increase  my post rating

1. ^That won't work

2. This is like a 7 year revival

My signatures meaning is easy to figure out.
________________
:o penny13 revived a topic! :o

mine is some questions that stump really good AOPSERS.

srsly? This has to be revived again after it has been revived for a 7 year period? I dont get why people feel the need to do this.

You should justt put revival in the title.

I am sorry my brother must have gotten on my account and pulled this back, I am very sorry He did. Mods please lock!!!
I never pull back old topics, my brother will pay.


I think Phuong write uncorect problem :?

It is easy to prove, that equation had not solution $x\ge 1$. I think solution is x=0.

I guess we are looking for real solutions?

I suggest the similar problem in mind my is corect problem:
$x-2\sqrt{x-1}-(x-1)\sqrt{x}+\sqrt{x^2-x}=0$
$\Leftrightarrow\left(\sqrt{x-1}-1\right)^2-\sqrt{x(x-1)}.\left(\sqrt{x-1}-1\right)=0$
$\Leftrightarrow\left(\sqrt{x-1}-1\right).\left(\sqrt{x-1}-1-\sqrt{x(x-1)}\right)=0$
$\Leftrightarrow x = 1$


A Senate committee consists of 5 Republicans, 6 Democrats, and 2 Independents.  A subcommittee of 3 members is randomly chosen.  What is the probability that the subcommittee consists of 1 Republican, 1 Democrat, and 1 Independent?

One of the most important things in combinatorics is called a [i]combination[/i]. It represents the number of ways to choose $ k$ distinguishable objects from a set of $ n$ without regard to order. It is denoted $ \binom{n}{k}$ and can be calculated by $ \binom{n}{k}\equal{}\frac{n!}{k! \cdot (n\minus{}k)!}$.

Another very important principle of combinatorics is the multiplication principle, which states that if there are $ a$ ways to do something, and $ b$ ways to do another thing, the number of ways to do both is $ a\cdot b$.

To find a probability, we must divide the number of desired outcomes by the number of possible outcomes. The number of desired outcomes is the number of ways to choose 1 Republican, 1 Democrat, and 1 Independent. This is $ 5\cdot 6 \cdot 2$ by the Multiplication Principle. The number of possible outcomes is $ \binom{5\plus{}6\plus{}2}{3}$. So the final answer is their quotient, or $ \frac{30}{143}$.


What would be the final temperature when a 0.500 kg chunk of ice at 0 degree Celsius is put into a liter of 0.0 degree Celsius water? (Does all the ice melt?)

I am pretty sure the answer is 0 Celcius.  There will be no net change in the ice-water system--should some water change into ice, by conservation of energy an equivalent amount of ice must change into water, and vice-versa.


a,b,c>0
prove that:
$ \prod (\sqrt{3*(c\plus{}a)(c\plus{}b)}\minus{}\sqrt{c(a\plus{}b\plus{}c)})\ge \sqrt{abc(a\plus{}b\plus{}c)^3}$


AoPSwiki algebra function domain LaTeX \/closed

The code windows for the AoPS Wiki and the forum are different. Shouldn't the forum switch to the better AoPSWiki model?

Plus, between every two characters in the source of the wiki code window's source, there is a strange ASCII character resembling a square. Is this supposed to be the case?

EDIT: For the first, the updated code window appears only for new posts in the forum, old ones (before about June or so 2007) have the old code window.

EDIT2: Mathlinks skin on forum, AoPS on wiki, AoPS domain, Firefox 3, Windows XP, Dell Dimension, Pentium 3 Processor.

You should know better by now: browser, skin(s) that you are using, etc. I can't reproduce any ASCII bug.

I'm very sure there is an ASCII bug. In fact, screenshot:

EDIT:
Directions: Go on an XP Media machine, click on any clickable LaTeX image. Screenshot is scrn2.png.

Then, right-click -> view source, to get scrn.png

I really do not understand how did you get that source window. Post more screenshots and describe exactly what you are doing.


complex numbers
superior algebra
superior algebra unsolved

Does there exist an algebraically closed field which "strictly" contains $\mathbb{C}$?

Well, yes.

I am no expert in the complex numbers and certainly not in their wacky automorphisms, but as I recall, when you accept the axiom of choice, every field has an algebraic closure.

So what do you do here, first extend $\mathbb{C}$ with $\mathbb{C}[x]$, then take the algebraic closure of that field

Also using the axiom of choice, one can also prove that every field of characteristic $0$ is a union of subfields of $\mathbb{C}$...
So define 'strictly contain' ;)

A field $E$ strictly contains a field $F$ if $F$ is a subfield of $E$ and $F \neq E$.

Anyone? This question makes me really confused... I'm still not sure about the answer.

The problem is: this question can't be answered, we can get $\mathbb{C} \subset \mathbb{C}(x) \subset \mathbb{C}$ with proper embedings, and the same should hold for every other field that contains $\mathbb{C}$.
But the most courious example should be that $\mathbb{C}$ can be seen as a proper subfield of itself ;)

small edit: some of these properties require the AC, so it could also depend...

[quote="ZetaX"]The problem is: this question can't be answered, we can get $\mathbb{C}\subset \mathbb{C}(x) \subset \mathbb{C}$ with proper embedings, and the same should hold for every other field that contains $\mathbb{C}$.
But the most courious example should be that $\mathbb{C}$ can be seen as a proper subfield of itself ;)

small edit: some of these properties require the AC, so it could also depend...[/quote]

*bump*

I had a class today about this subject and I asked myself the very same question Arne asked himself (same professor). I'm totally confused and find this last thing contradictionary. Does someone want to be so kind to elaborate on this please (assuming that I'm following my first algebra course)

i'm still not understanding a thing of it

what's the algebraic closure of $\mathbb{C}(X)$ ?

erm... it's the algebraic closure. Would else should be?

By the way, I got this question during my oral examination, one year ago  :D

Er... how is that possible, you posted before the exam period started?  :maybe:

Yes, I know. The professor giving this course mentioned this question in class, and afterwards he asked it as a small question ("bijvraag") during the oral examination.

[quote="Gockel"]@ZetaX:
The question is about subsets not about embeddings and hence $\mathbb{C}\subsetneq\mathbb{C}[x]$.

But even if we consider embeddings... just take the algebraic closure of $\mathbb{C}(X)$ where X is set of indeterminants with $|X|>|\mathbb{C}|$. Of course many subfields can be embedded into $\mathbb{C}$ but certainly not the whole field.[/quote]

Yes, I forgot that the cardinality has to be the same as that of $\mathbb C$. But then it should be true.

But the answer still depends on the term "contains...


If $ n!5! \equal{} 6!$, then what is value of $ n$ ?

We know $ \frac{6!}{5!}\equal{}6$, so we want to find a number $ n$ such that $ n!\equal{}6$. (hopefully you memorized the first $ 10$ factorials. If you haven't, it's good idea). We find $ n\equal{}\boxed{3}$


ratio
vector
trigonometry
analytic geometry
exterior angle
geometry
similar triangles

Let $ ABC$ be a right triangle with $ \angle A \equal{} 90^\circ$. Let $ D$ be the midpoint of $ AB$ and let $ E$ be a point on segment $ AC$ such that $ AD \equal{} AE$. Let $ BE$ meet $ CD$ at $ F$. If $ \angle BFC \equal{} 135^\circ$, determine $ BC / AB$.

I liked this problem because of the numerous distinct approaches :)

[hide="Vector Solution"]Assign $ A(0,0)$, $ B(0,2)$, $ C(k,0)$, $ D(0,1)$ and $ E(1,0)$.

We know that $ \vec{BE} \equal{} ( \minus{} 2,1)$ forms a 135 degree angle with $ \vec{DC} \equal{} (k, \minus{} 1)$.

From the dot-product formula, $ \frac {(1, \minus{} 2) \cdot (k, \minus{} 1)}{\| ( \minus{} 2,1) \| \| (k, \minus{} 1) \|} \equal{} \cos 135^{\circ}$.

$ \frac {k \plus{} 2}{\sqrt {5} \sqrt {k^2 \plus{} 1}} \equal{} \frac {1}{\sqrt {2}} \implies \frac {(k \plus{} 2)^2}{5(k^2 \plus{} 1)} \equal{} \frac {1}{2}$ $ \implies 3k^2 \minus{} 8k \minus{} 3 \equal{} 0$

Solving in the positives, we get $ k \equal{} 3$. So, the coordinates of $ C$ are $ C(3,0)$, from where it is easy to see that $ BC/BA \equal{} \frac {\sqrt {3^2 \plus{} 2^2}}{2} \equal{} \boxed{\frac {\sqrt {13}}{2}}$[/hide]
[hide="Hint for Geometric Solution"]Exterior Angle Theorem[/hide]

[hide="trigonometric solution"]
The angles in quadrilateral AEFD must add up to 360, so since we know angleEFD=135, angles ADF and AEF must add up to 360-90-135=135. WLOG, we can let AD=DB=AE=1. Let EC=n. Then, we have ADC+AEB=135, so arctan(n+1)+arctan(2)=135. Taking the tan of both sides and using the formula tan(a+b)=(tan a+tan b)/(1-(tan A tan b)) and solving gives n=2, from which it follows by Pythag that the answer is $ \sqrt {13}/2$. [/hide]

[hide="similar triangles"]
probably too close to the previous solutions, whatever.

DEF=BED obviously because they are the same angle.
because BDE+ADE=180, BDE=135, so BDE is similar to DFE
we use this similarity, knowing BD:DE, to find DF:FE, $ \frac{1}{\sqrt{2}}$
we use similar triangles again to find that DFE is similar to DEC.
we already know DF:FE, so we can find DE:EC
so if AD=BD=AE=x, EC=2x, and BC=$ \sqrt{13}x$
so the answer is $ \frac{\sqrt{13}}{2}$

[/hide]


Prove that the only solution (>1) to the following equation is:  (a,b,c)=(2,2,3)

$3^{a}-b^{c}=1$

[hide="starting point"]

Lemma:
Only (m,n) (>1) satisfying $3^{m}-2^{n}=1$ is (2,3)

Proof:
if m>2 then n>3
$\Rightarrow 3^{m}\equiv 1 \pmod{8}\Rightarrow m=2k$
then
$2^{n}=(3^{k}-1)(3^{k}+1)$
by unique factorization:
$3^{k}+1=2^{r}$
but this is impossible
since, then $3^{k}-2^{r}=-1\Rightarrow 3^{k}\equiv-1\pmod{8}$
contradiction, since $3^{k}\equiv 1,3 \pmod{8}$ depending if $k$ is even or odd.
$\blacksquare$[/hide]

$b^{c}=3^{a}-1\cong-1 mod(3)$, so $b\cong-1 mod(3)$ and $c$ is odd.
Then, $b+1|b^{c}+1=3^{a}$ and there exists a positive integer $n$ such that $b=3^{n}-1$.
Because the post of amirhtlusa, we can suppose $n\geq 2$.
$3^{a}=(3^{n}-1)^{c}+1>(2.3^{n-1})^{c}>3^{(n-1)c+1}$
Let $d$, $k$ nonnegative integers such that $c=d.3^{k}$ and $(d,3)=1$.
We can prove: (by induction over $k$)
$(3^{n}-1)^{c}\cong d.3^{n+k}-1 mod(3^{2n+k})$
Then,
$n+k\geq a>(n-1)c+1\geq (n-1)(k+1)+1=n+(n-1)k\geq n+k$
wich is absurd.


vector
linear algebra
linear algebra unsolved

Say $ T$ is a set of $ s$ vectors of an Euclidean space $ V$ such that the set $ A$ of its nonnegative combinations doesn't contain $ w$ and $ \minus{}w$ for any nonzero vector $ w$. Prove that there exist a set $ R$ of $ r$ independent vectors such that $ A$ is contained in the set of positive combinations of $ R$. Where $ r$ is the maximum number of linearly independent vectors in $ T$.


1)Simplify the expression:
$ A\equal{} cotx\minus{}tanx\minus{}2tan2x\minus{}4tan4x$
2)Let ABC be a triangle, prove that:
$ a)\sin 3A.\cos ^3 (B \minus{} C) \plus{} \sin 3B.\cos ^3 (C \minus{} A) \plus{} \sin 3C.\cos ^3 (A \minus{} B) \equal{} 3\sin A.\sin B.\sin c$
b)$ \sin 3A \plus{} \sin 3B \plus{} \sin 3C \equal{}  \minus{} 4\cos \frac{{3A}}{2}.\cos \frac{{3B}}{2}.\cos \frac{{3C}}{2}$

[hide="Problem 1"]\begin{eqnarray*}A\tan x &=& 1-\tan^2x-2\tan x\tan 2x-4\tan x\tan 4x\\ &=& {2\tan x\over\tan 2x}-2\tan x\tan 2x-4\tan x\tan 4x\\ &=& {2\tan x(1-\tan^22x)\over\tan 2x}-4\tan x\tan 4x\\ &=& \frac{2\tan x\cdot{2\tan 2x\over\tan 4x}}{\tan 2x}-4\tan x\tan 4x\\ &=& {4\tan x\over\tan 4x}-4\tan x\tan 4x\\ &=& {4\tan x(1-\tan^24x)\over\tan 4x}\\ &=& \frac{4\tan x\cdot{2\tan 4x\over\tan 8x}}{\tan 4x}\\ &=& {8\tan x\over\tan 8x}\end{eqnarray*}

Hence $ A=8\cot 8x$[/hide]

I think a couple of signs are wrong (and cancel each other out).

I make it:

$ \sin 3A \plus{} \sin 3B \plus{} \sin 3C$

$ \equal{} 2 \sin \frac{3(A\plus{}B)}{2}\cos \frac{3(A\minus{}B)}{2} \plus{} 2 \sin \frac{3C}{2} \cos\frac{3C}{2}$

$ \equal{} \minus{}2\cos \frac{3C}{2} \cos\frac{3(A\minus{}B)}{2} \plus{} 2 \sin \frac{3C}{2} \cos \frac{3C}{2}$

$ \equal{} 2\cos \frac{3C}{2} [\sin\frac{3C}{2} \minus{} \cos \frac{3(A\minus{}B)}{2}]$

$ \equal{}2\cos\frac{3C}{2}[\minus{}\cos\frac{3(A\plus{}B)}{2} \minus{} \cos \frac {3(A\minus{}B)}{2}]$

$ \equal{}\minus{}2\cos\frac{3C}{2}[2\cos\frac{3A}{2}\cos\frac{3B}{2}]$

$ \equal{} \minus{}4 \cos\frac{3A}{2}\cos\frac{3B}{2}\cos\frac{3C}{2}$

[b][u]First problem[/u].[/b] Observe that $ \tan 2x \equal{} \frac {2\tan x}{1 \minus{} \tan^2x}$ $ \Longleftrightarrow$ $ \frac {2}{\tan 2x} \equal{} \frac {1 \minus{} \tan^2x}{\tan x}$ $ \Longleftrightarrow$ $ \boxed {\ 2\cot 2x \equal{} \cot x \minus{} \tan x\ }\ (*)$ .

Apply the relation $ (*)\ : \ \left\|\begin{array}{c} \cot x \minus{} \tan x \equal{} 2\cot 2x \\
 \\
2\left(\cot 2x \minus{} \tan 2x\right) \equal{} 4\cot 4x \\
 \\
4\left(\cot 4x \minus{} \tan 4x\right) \equal{} 8\cot 8x\end{array}\right\|\ \bigoplus\ \Longrightarrow$ $ \cot x \minus{} \tan x \minus{} 2\tan 2x \minus{} 4\tan 4x \equal{} 8\cot 8x$ .

[b]An easy exrtension[/b]. For any $ n\in \mathbb N^*$ we have $ \boxed {\ 2^n\cot\left(2^nx\right) \plus{} \sum_{k \equal{} 0}^{n \minus{} 1}2^k\tan\left(2^kx\right) \equal{} \cot x\ }$ (independent of $ n$).

[b][u]Second problem[/u].[/b] $ \boxed {\ \sin 3A\cdot\cos ^3 (B \minus{} C) \plus{} \sin 3B\cdot\cos ^3 (C \minus{} A) \plus{} \sin 3C\cdot\cos ^3 (A \minus{} B) \equal{} \sin 3A\cdot\sin 3B\cdot\sin 3C\ }$ .

[b][u]Proof[/u].[/b] You can prove easily with  third problem (following). I"ll come back here after one week. Success !

[b][u]Third problem[/u].[/b] This problem is a particular case of the following general conditioned trigonometrical identities, where $ n\in \mathbb N$ :

$ \boxed {\ \begin{array}{cccccc} 1\blacktriangleright & x \plus{} y \plus{} z \equal{} (2n \plus{} 1)\pi & \implies & \sin x \plus{} \sin y \plus{} \sin z & \equal{} & ( \minus{} 1)^n4\cos\frac x2\cos\frac y2\cos\frac z2 \\
 \\
2\blacktriangleright & x \plus{} y \plus{} z \equal{} 2n\pi & \implies & \sin x \plus{} \sin y \plus{} \sin z & \equal{} & ( \minus{} 1)^{n \minus{} 1}4\sin\frac x2\sin\frac y2\sin\frac z2\end{array}\ }$ .

[b][u]Proof of[/u] $ (1)$ .[/b] $ \sin x \plus{} \sin y \plus{} \sin z \equal{}$ $ \sin\left[(2n \plus{} 1)\pi \minus{} (y \plus{} z)\right] \plus{} \sin y \plus{} \sin z \equal{}$ $ \sin (y \plus{} z) \plus{} \sin y \plus{} \sin z \equal{}$

$ 2\sin\frac {y \plus{} z}{2}\cos\frac {y \plus{} z}{2} \plus{} 2\sin\frac {y \plus{} z}{2}\cos\frac {y \minus{} z}{2} \equal{}$ $ 2\sin\frac {y \plus{} z}{2}\left(\cos\frac {y \plus{} z}{2} \plus{} \cos\frac {y \minus{} z}{2}\right) \equal{}$ $ 4( \minus{} 1)^n\cos\frac x2\cos \frac y2\cos\frac z2$

because $ \frac {y \plus{} z}{2} \equal{} n\pi \plus{} \frac {\pi}{2} \minus{} \frac x2$ $ \implies$ $ \sin \frac {y \plus{} z}{2} \equal{}$ $ ( \minus{} 1)^n\sin\left(\frac {\pi}{2} \minus{} \frac x2\right) \equal{} ( \minus{} 1)^n\cos\frac x2$ . You can show easily and second identity.


geometry
3D geometry
sphere
tetrahedron
ratio
incenter

Source: College of Charleston Math Meet 2004

Find the length of the radius of a circle in which a chord of length $6$ is twice as far from the center than a chord of length $12$.

Using Pythagoras theorem twice we find r=3*5^(1/2)
Is this ok? :)

that's what i got.  Just let the distance from the center to the chord of length 12 be x, then use the Pythagorean  Theorem.

Yep that's how I did it, you know that you have two triangles with the same hypotenuse length (because it's the radius), so you can let the distance from chord $6$ to the center be $2x$ and the distance for chord $12$ be $x$, thus the equation would be $(2x)^2+3^2=x^2+6^2$ so $x=3$. Then plug in $x$ for either one of the triangles and use the pythagoreum theorem again to get the hypotenuse (radius) of $45^{1/2}$ or $3(5^{1/2})$

Okay i'll put two, one 2d, one 3d.

1. There are two concentric circles, $C_1$ and $C_2$, with $C_1$ in the interior of $C_2$.  A chord of $C_2$ has length 17 and is tangent to $C_1$.  The area inside $C_2$ but outside $C_1$ is shaded.  Find the area of that region. (easy)

2 (not so easy).  You have a sphere, $S_1$.  Inside it is inscribed a regular tetrahedron, $T_1$.  Inside that is inscribed another sphere, $S_2$.  Then another tetrahedron, $T_2$.  So on to infinity.  The radius of $S_1$ is 1.  Find
a) the volume inside each tetrahedron which is, for each $T_n$, outside $S_{n+1}$
b) the radius of $S_n$ in terms of n

Number 1, you mean the annulus right? Okay so the chord is $17$ and 1/2 of it is $17/2$, so let's call $x$ the radius of the small circle, and $y$ the radius of the big circle. So since it's a triangle with $17/2$ and $x$ as legs and $y$ as the hypotenuse, $x^2+289/4=y^2$, so then $y^2-x^2=289/4$ so $289/4pi$ is the answer.

Well I'm assuming that the pi isn't in the denominator, and that you're right...but write it as \pi instead of pi
$pi$
$\pi$
See?
As for the fraction...if you want to make it two numbers vertically seperated, you'd put it in the form \frac{x}{y} for the fraction x/y
$x/y$
$\frac{x}{y}$

Can you do number 2?

[hide="small hint"]Draw the perpendicular from the apex of the tetrahedron down to the base.  Now draw triangles off of that.  Also, use the fact that the ratio of the radii between one sphere and the next smaller one is the same as the ratio between the distance from the incenter of the tetrahedron to a vertex to the distance from the incenter to a face.[/hide]

[hide="bigger hint"] Inscribe a sphere in a tetrahedron and, using the length of the perpendicular from the apex, solve for its radius using the pythagorean theorem.  REMEMBER TO USE THE FACT THAT EVERY SIDE OF THE TETRAHEDRON IS AN EQUILATERAL TRIANGLE![/hide]

I'm working on $2$, it's a lot more difficult than the first.


I've managed to con my AP Chemistry teacher from last year into doing an independent study in organic chemistry second semester with at least one other student.  Any advice about a rudimentary syllabus, and perhaps a textbook?  We want this to be challenging, but not too challenging.  After all, it is second semester senior year.

*does dark magic to resurrect this post*

exactly how do you con your ap chem teacher to let you do independent study?
im trying to get my teacher to let me do independent study


Set $d_1=3$ and all others $=1$.
And don't post in the Solved section in the future.

[quote="pavel kozlov"][quote="ZetaX"][quote="Number1"]Thank you and you shoud learn some manner ZetaX.[/quote]
 :|  :?  :o  :huh: [/quote]
It is field for the mathematics :!:[/quote]
Can you explain what that triplepost wants to say¿
And Number1: explain your post, too!


algebra
polynomial
vector
linear algebra
matrix
real analysis
real analysis unsolved

Let P(x) be a polynomial with integers coefficients.
Let m be a positive integer.
Let q(x) be a polynomial whose roots are (root of p(x))^m.
Is it true that q(x) has integers coeficients?
What is the proof of this??

Yes, it's true. It is an easy consequence of the theorem which states that elementary symmetric polynomials of $ n$ variables generate all the symmetric polynomial of $ n$ variables. (Although this theorem itself may not be easy to prove...)

It looks like you mean "monic" polynomial.  Here we have to show that the product of two [url=http://en.wikipedia.org/wiki/Algebraic_integer]algebraic integers[/url] is again an algebraic integer (so it follows that if $ P(r) \equal{} 0$ then $ r^m$ is an algebraic integer) and that the field conjugates of $ r^m$ over $ \mathbb{Q}$ are precisely the field conjugates of $ r$, all raised to the $ m^{th}$ power.

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=202382]this thread[/url] from last month.
Interpret multiplication by $ r$ as a linear operator on the vector space over $ \mathbb{Q}$ spanned by the powers of $ r$. With a sensible choice of basis ($ 1,r,r^2,\dots$), this operator has an integer matrix $ A$. If $ p$ was irreducible, we have in addition that the characteristic polynomial of $ A$ is exactly $ p$.

What is the matrix of multiplication by $ r^m$? It's $ A^m$. This matrix has integer entries, and its eigenvalues are the roots of $ p$ raised to the $ n$th power (assuming irreducibility of $ p$). The characteristic polynomial is a monic integer polynomial, which must be $ q$.
If $ p$ isn't irreducible, break it up into a product of irreducibles first and apply the above to each part.

This is either linear algebra or superior algebra; it's definitely not calculus.


inequalities
algebra
polynomial
inequalities proposed
109

This ineq is very useful to solve the ineq that has different degree

 x 2 +y 2 +z 2 +2xyz+1  >= 2(xy+yz+zx)

One very first question: Which interesting inequality with different degrees are you referring to :?
Well I hope that one shall prove your quite trivial inequality...
Otherwise, I apology as I misunderstood your post.

I suppose that $x,y,z$ are positive reals. If they weren't it would be easy to find an counterexample. Am I right

But I think this is quite trivial, as $2xyz+1 \geq 0$ and as this stronger inequality holds:
$x^2+y^2+z^2 \geq 2(xy+yz+xz)$ which is equivalent to:
$(x-y)^2+(y-z)^2+(x-z)^2 \geq 0$

First, DanielKirchner you are wrong, your "stronger inequality" is false.
Second, this inequality was posted before.
Vasc provided an ingenious proof by multiplying both sides by x+y+z then use schur and completing the square.
Another possible solution is to prove a stronger inequality,
x^2+y^2+z^2+3(xyz)^(2/3) >= 2(xy+yz+zx) by muirhead.
Of course, it is also possible to prove the original inequality by muirhead.

[b]To DanielKirchner:[/b]

Unluckily, your proof is wrong:

[quote="DanielKirchner"]But I think this is quite trivial, as $2xyz+1 \geq 0$ and as this stronger inequality holds:
$x^2+y^2+z^2 \geq 2(xy+yz+xz)$ which is equivalent to:
$(x-y)^2+(y-z)^2+(x-z)^2 \geq 0$[/quote]

In fact, the inequality $x^2+y^2+z^2 \geq 2(xy+yz+xz)$ is incorrect in general (it is [i]not[/i] equivalent to $(x-y)^2+(y-z)^2+(x-z)^2 \geq 0$; you made a calculation mistake). Never mind, that's quite a usual error when one tries to prove inequalities.

[b]To Suihochung:[/b]

[quote="siuhochung"]Second, this inequality was posted before.
Vasc provided an ingenious proof by multiplying both sides by x+y+z then use schur and completing the square.[/quote]

Hmm, well, do you mean this thread: http://www.mathlinks.ro/Forum/viewtopic.php?t=19666 ? Well, just to note that I am not Vasc in this case  ;) ...

[quote="siuhochung"]Another possible solution is to prove a stronger inequality,
x^2+y^2+z^2+3(xyz)^(2/3) >= 2(xy+yz+zx) by muirhead.[/quote]

Nice proof indeed, I see it's a variation of ThAzN1's proof given in http://www.mathlinks.ro/Forum/viewtopic.php?t=19666 .

  Darij

oh i thought it's vasc who posted the proof.... sorry...  :blush:

Just for vocabulary interests, a polynomial inequality of a non-constant degree is usually called [i]inhomogenous[/i].

A common mistake amongs beginners of inequalities is to assume something like $a+b+c=1$ when the inequalitiy is inhomogenous.

Thanks to all who replied!

I'm very sorry for my mistake that x,y,z must be non-negative real number.

And I saw wondeful proof about this ineq.

Thanks for letting me know about that.

And This ineq is very useful to solve 2004 APMO #5 ineq. ^^

Pretty much any way you do it, APMO 2003 comes down to Schur.  See

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21835

for a solution using only Schur and AM-GM.

I think It has a very simple solution which didn't use Schur Ineq or any diffirence inequality
 :)

[quote="Chang Woo-JIn"]$ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz \plus{} 1\geq 2(yz \plus{} zx \plus{} xy),$

where $ x,y,z$ are nonnegative real numbers.[/quote]$ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz \plus{} 1 \minus{} 2(yz \plus{} zx \plus{} xy)$

$ \equal{} \frac { \sum{xyz(2x \plus{} 1)(y \plus{} z \minus{} 2)^2}\plus{}\sum{(y \minus{} z)^2\left[x \plus{} yz(y \plus{} z)\right]} }{4xyz\plus{}(y \plus{} z)(z \plus{} x)(x \plus{} y)}\geq0.$

By the way, $ a^4 \plus{} b^4 \plus{} c^4 \plus{} 2a^2b^2c^2 \plus{} 1 \minus{} 2(b^2c^2 \plus{} c^2a^2 \plus{} a^2b^2)$ cannot be represented as a sum of squares of other polynomials.

[quote="Ji Chen"]

By the way, $ a^4 \plus{} b^4 \plus{} c^4 \plus{} 2a^2b^2c^2 \plus{} 1 \minus{} 2(b^2c^2 \plus{} c^2a^2 \plus{} a^2b^2)$ cannot be represented as a sum of squares of other polynomials.[/quote]

Can you give a proof of why not? Thanks :maybe:

this toic is very old, this is my solution t the problems, it's easy you can see it here in e post 277,
http://www.mathlinks.ro/viewtopic.php?t=299899&start=260


ive been trying different stuff to set this up and im not sure how.

The postion of a rocket is given by x(t)=t^2+2t+3

Compute average velocity between time t and 1 for t=.5, t=.9, t=.99, t=.999

Posted in the wrong place and with the wrong attitude. We are not here to solve your homework problems...  :?

If $x(t)=t^2+2t+3$, then the average velocity over $[t,1]$ is given by

$\frac {x(1)-x(t)}{1-t}$

You can crunch the numbers from here.

Wrong place?

I believe a calculus problem should be listed in the calculus category.

Wrong attitude?

OK, don't understand that one, the last I checked math is a serious, down to the point subject. I'm sorry if I didn't include a heart warming story in my problem.

Thanks for the help blahblah.
At least someone would rather help me than criticize.

blueray, he meant that this problem should've gone in the Calculus Computations and Tutorials section.  I will be moving  it there.


a b c  triangle lengths ribs hs   x,y,z
x+y+z=1
proved that
x^2 +y^2+z^2 +4xyz <1\2


Max must take 10 steps to go the same distance as three of his dad's steps. His dad takes 30 steps to walk down the hall. How many steps must Max take to walk down the same hall?

Since the ratio of Max's steps to his dad's steps is 3/1, and if the dad takes 30 steps, the ratio = 3*30/30, so isn't the answer 90?

No.

Let $ s \equal{} \text{the number of steps Max takes}.$
We have the ratio
\[ \frac {10\text{ Max steps}}{3\text{ Dad steps}} \equal{} \frac {m}{30\text{ Dad steps}}.
\]
Solving for $ m$ gives
\[ m \equal{} (30\text{ Dad steps})\left(\frac {10\text{ Max steps}}{3\text{ Dad steps}}\right)\Rightarrow m \equal{} \boxed{100\text{ Max steps}}.
\]
(of course the answer would just be 100.)

RoFlLoLcOpT is right, but for a countdown i would cut out the stuff he wrote and just divide 30 by 3 to get 10, then multiply it by 10 to get 100 steps.

I'm just explaining it more thoroughly for someone who might not know what to do at the beginning. If this were countdown, it'd be pretty easy to reason the way you did, but this isn't :P this is a forum


What integer is closest to $ \frac{12.55 \cdot 8 \cdot 61}{24.88 \cdot 82}$?


Define A $ \triangle$ B=$ a^2-b$

What is the value of $ (2^{4\triangle13})\triangle(3^{3\triangle5)}$?

PS: the $ \text{\LaTeX}$ is hard for this one...

This is straight plugging in.

$ 4 \triangle 13 \equal{} 4^2 \minus{} 13 \equal{} 16 \minus{} 13 \equal{} 3$.
$ 3 \triangle 5 \equal{} 3^2 \minus{} 5 \equal{} 9 \minus{} 5 \equal{} 4$.

So you're calculating $ 2^3 \triangle 3^4$.

This is $ 2^6 \minus{} 3^4 \equal{} 64 \minus{} 81 \equal{} \minus{}17$.

Try not to get intimidated with these problems.  They may look scary at first, but if you just plug in the numbers, they are not so bad.

What was your train of thought of solving this problem?

When you have an operation such as $ a^2 \triangle b$, you want to know which numbers to plug in and where to start.

Obviously, you would need to square the number to the left of the triangle and subtract the number to the right of the triangle.  And in this case, you would probably want to start with the exponents.


1)  J. Van Hengel's inequality

    For  n,r natural numbers with n \geq 3 or r \geq 3 prove that

                               (n+r)^n<n^(n+r)

[i]unknown source[/i]

2)Goormaghtigh's inequality

   For n,r,p natural numbers such that n \geq 3 and p<r prove that

                       (n+r)^n<(n+p)^n*(n)^(r-p)

[i]unknown source[/i]

The first one is just wellknown fact:
If m > n >=3 then  m^n < n^m. We only have to prove that
(n+1)^n < n^(n+1). But is equivalent to (1+1/n)^n < n which is trivial because (1+1/n)^n < e < n.

The condition r>=3 was not used, just r>=1.

Namdung


In the $ xy$ plane, consider the circle $ C: x^2 \plus{} y^2 \equal{} 1$ and the line $ l: 2x \minus{} y \equal{} 0$.

(1) Find the image of $ C$ by the linear transformation expressed by $ \left( \begin{array}{cc} 1 & \minus{} 1 \\
1 & 1 \end{array} \right)$.

(2) Find the intersection points of $ C$ and $ l$.

(3) Find all matrices $ \left( \begin{array}{cc} a & b \\
c & d \end{array} \right)$ by which $ C$ mapps to $ C$ and $ l$ mapps to $ l$.

For (3):
$ A\equal{}\begin{pmatrix}a&b\\c&d\end{pmatrix}$ maps $ \mathcal{C}$ to $ \mathcal{C}$, that is if $ x^Tx\equal{}1$ then $ (Ax)^TAx\equal{}1$. Thus for all $ x$: $ x^Tx\equal{}x^T(A^TA)x$; then $ A^TA\equal{}I$ and $ A$ is an orthogonal matrix.
$ A$ maps $ l$ to $ l$, that is if $ [2,\minus{}1]x\equal{}0$ then $ [2,\minus{}1]Ax\equal{}0$. Thus $ [1,2]$ is an eigenvector of $ A$.

I forget the conclusion
For (3):
$ A\equal{}\begin{pmatrix}a&b\\c&d\end{pmatrix}$ maps $ \mathcal{C}$ to $ \mathcal{C}$, that is if $ x^Tx\equal{}1$ then $ (Ax)^TAx\equal{}1$. Thus for all $ x$: $ x^Tx\equal{}x^T(A^TA)x$; then $ A^TA\equal{}I$ and $ A$ is an orthogonal matrix.
$ A$ maps $ l$ to $ l$, that is if $ [2,\minus{}1]x\equal{}0$ then $ [2,\minus{}1]Ax\equal{}0$. Thus $ [1,2]^T$ is an eigenvector of $ A$. The condition is $ 2a\plus{}4b\minus{}c\minus{}2d\equal{}0$.
Conclusion:
if $ det(A)\equal{}1$ then $ A\equal{}\pm{I}$.
If $ det(A)\equal{}\minus{}1$ then $ A\equal{}\pm\begin{pmatrix}3/5&\minus{}4/5\\\minus{}4/5&\minus{}3/5\end{pmatrix}$.


trigonometry
real analysis
real analysis unsolved

\[ \tan \left( {x \minus{} 1} \right)\tan \left( {x \plus{} 3} \right) \equal{} 2\tan \left( x \right)
\]

what is x.

thx

hello, applying the addition formulas we get
$ \frac{(\tan(x)-\tan(1))(\tan(x)+\tan(3))}{(1+\tan(x)\tan(1))(1-\tan(x)\tan(3))}-2\tan(x)=0$
simplifying this we have to solve
$ {-2\tan(1)\tan(3)\tan(x)^3-\tan(x)^2(1+2\tan(3)-2\tan(1))+\tan(x)(2-\tan(3)+\tan(1))+\tan(1)\tan(3)=0}$
Sonnhard.

[b]but how do we further solve that equation ??[/b]

hello, setting $ \tan(x)\equal{}t$ you will get a cubic equation in $ t$ to solve.
Sonnhard.

[quote="Dr Sonnhard Graubner"]
hello, setting $ \tan(x) \equal{} t$ you will get a cubic equation in $ t$ to solve.
[/quote]

well, i tried solving it that way Sonnhard but wasn't able to get the value of x .... can you please post the solution ?  :|

Thank You !


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $G$ be a group of order 144. Prove that $G$ is not simple.

(hint: unlike one may expect, there's no normal sylow subgroup. this one is different. try counting the subgroups.)

$G$ acts on the set $S$ of its $3$-Sylows by inner automorphisms, where $|S|=4$. This induces an homomorphism $G \rightarrow \sigma(S)$ where $|\sigma(S)|=4!$. The kernel of this homomorphism is nontrivial (since $9$ does not divide $4!$), proper and normal, which solves the problem.

[quote="julien_santini"]$G$ acts on the set $S$ of its $3$-Sylows by inner automorphisms, where $|S|=4$.[/quote]Why do you think $|S|=4$? I think $|S|\not=4$. I think $|S|=16$. But since $9|16!$, I'm afraid your proof is wrong...

Assume simplicity

there are 9 sylow2groups
and 16 sylow3groups

now of those last kind it is impossible that they all have trivial intersection, as that would make the elements of order a power of two a subgroup of 16 elements, which would be characteristich and thus normal

so suppose $A$ and $B$ intersect in $C$
now the centralisor has an order at least 15, being a multiple of 9, and dividing 144

thus it must be 18,36,72,144
however all cases disappear as we would have a permutation on at most 4 elements otherwise which is transitive

but then i am stuck on the eighteen case :(

18 is kind of similar to 36, no? Anyway how do you 'know' that |Syl3|=16 and not 4, that |Syl2|=9 and not 3, etc?

same standard trick
when $G$ is simple , the number n of sylow p groups is always at least such that the order of G divides $n!$ otherwise you have a transitive permutation on a nontrivial set, with a kernel that is nontrivial, that would make the kernel a normal subgroup not the total or trivial group

Okay I should be ashamed not having found the solution being so close

If that centraliser has eighteen elements, it contains two groups of order nine, which is impossible by sylow on that eighteen order group

However, as I saw it on the blackboard , Peter, you probably already have it? :lol:


limit trigonometry LaTeX calculus algebra polynomial function

Compute [tex]\lim _{n\rightarrow \infty } \left( \sin \left( \pi \,\sqrt {{n}^{200}
+{n}^{100}+1} \right)  \right) ^{2}
[/tex]
Without a Caculator
Wow! I can latex now.

I'm pretty sure this method is flawed.

[hide]lim n->inf n^200 + n^100 + 1 can be justified as equal to lim n->inf n^200 by factoring out n^200 and seeing that the remaining expression becomes 1 as n increases w/o bound.

From there, :sqrt: n^200 = n^100. sin :pi: k is going to be 0 for all integers k. Thus, the limit is 0.[/hide]

Yep it is flawed. the answer isn't 0. Hint: [hide]Think of the period of sin^2[/hide]

Anybody have any calculus tips for me. Because I am going to take the Calculus Challenge tomorrow which is a undergraduate 3 hour competition at the university. (the questions are just like the one I posted)

Wrong Again. I will post a solution on Sunday if no one solves it.

My answer:

[hide]

Since polynomials behave like their highest degree term as it approaches infinity, (n^200 + n^100 + 1) will behave as if it were just n^200.

The problem can then be rewritten as:

lim(n-> :inf: ) (sin  (:pi:  :sqrt: n^200)) :^2:

Which is then simplified to

lim(n-> :inf: ) (sin ( :pi: n^100)) :^2:

Since lim(n-> :inf: )  :pi: n^100 tends toward infinity, and the nature of the sin :^2: x function dicates that it will continuously oscillate between 0 and 1 as x approaches infinity, it will never tend toward a specific number, therefore the limit does not exist.

[/hide]

Are you taking n to be a natural number, or any real, churchilljrhigh? I assume (from the question/answer) you mean natural.. but you should really say..

[quote="scm007"]Look, because of continuity of limit of sine, we get

Sin^2 lim  :pi:  :sqrt: (n^200 + n^100 +1)

 lim :sqrt:  :pi: (n^200 +n^100 +1) as n approaches infinity

obviously equals infinity.

Therefore there is no limit.[/quote]

I agree, besides I checked it with Maple to be sure.

Never mind, I just checked it by hand, and [tex]\sqrt{n^{200}+n^{100}+1}[/tex] aproaches  [tex]n^100+\frac{1}{2} [/tex], justified by the fact that  [tex]n^{200}+n^{100}+1=\left(n^{100}+\frac{1}{2}\right)^2+\frac{3}{4}[/tex] so, the value approaches [tex]\sin^2(\pi(m+.5))[/tex], where m is [tex]n^{100}[/tex], since sin :^2:  has a period of  :pi: , m will not play any significance, and [tex]\sin^2\left(\frac{\pi}{2}\right)=1^2=1[/tex]

[quote="scm007"]I'm getting anwers that it is 1, but I think my methods are flawed, this doesn't seem right...[/quote]

your methods aren't flawed. and you are right. triplem's comment is right. congratulations to triplem.

here's the solution.
btw i took seventh at the univerity's calc challenge with 16.5/30. dang it. there is always next year.

What university do you attend? And do you think you could post some questions that were on the contest? I'm interested to see what level the test consisted of.


number theory
least common multiple
number theory proposed

The positive integers $a_1, a_2, ..., a_n$ have 1 as their highest common factor. Any two of them, moreover, have the same least common multiple. Prove that there exists an integer p such that, for any integer u, exactly one of the numbers u and p-u can be expressed in the form$ a_1.x_1+a_2.x_2+...+a_n.x_n$ with a suitable choice of nonnegative integers $x_1, x_2, ..., x_n$.

I can't help but notice that this is probably a tiny bit more than the generalization of IMO 1983/A3: For pairwisely coprime $a,b,c$, show that $2abc-(ab+bc+ca)$ is the largest integer which can't be written in the form $xab+ybc+zca,\ x,y,z\in \mathbb N$. It follows from the fact that $ab-a-b$ is the largest integer which can't be written as $xa+yb,\ x,y\in \mathbb N$.

I'm pretty sure that for $a_i$ satisfying the given conditions we can find pairwisely coprime $x_i,\ i\in \overline{1,n}$ s.t. $\displaystyle a_i=\frac{\prod_{k=1}^nx_k}{x_i}$ and I think that the number we're looking for, $p$, is $\displaystyle (n-1)\prod_{i=1}^nx_i-\sum_{i=1}^na_i$.

This is all just a hunch, of course :). I hope you people will be understanding in case this is just a big mistake :D.

The problem (the initial one) is easy if we find two coprime $a_i,a_j$, so let's just assume that each two have a common factor $>1$.

Here's a proof of my first assertion:

If $a_1,\ldots,a_n$ have no common factor $>1$ and each two have the same lcm, then there exist pairwisely coprime $x_i,\ i\in \overline{1,n}$ s.t. $\displaystyle a_i=\frac {\prod_{k=1}^nx_k}{x_i}$.

It's obvious for $n=2$, so let's assume $n\ge 3$. We first show that $(\frac {a_i}{(a_i,a_j)},\frac {a_j}{(a_i,a_j)})=1$. Assume $a_i=dd'a,a_j=db$, with $(a,b)=(a,d)=(b,d)=1,d'|d$. The common lcm for each pair of numbers is $dd'ab$. In this case when we select another $a_k$, it's not hard to show that it should be $dd'ab$, which is true for any other $a_k$, so all the terms would have the common factor $d$, which is false.

Now let $x_i=\frac {a_j}{(a_i,a_j)}$. It's clear that these are all equal for all $j\ne i$, because $\frac {a_ia_j}{(a_i,a_j)}=[a_i,a_j]$, which is always the same. We can show that $(x_i,x_j)=1,i\ne j$: $(x_i,x_j)=(\frac {a_i}{(a_i,a_j)},\frac {a_j}{(a_i,a_j)})=1$, as we have shown above.

Now we have to show that $\displaystyle a_i=\frac {\prod_{k=1}^nx_k}{x_i}$. We have $a_ix_i=L=[a_k,a_l],\forall i,k\ne l\in \overline{1,n}\ (1)$. This means that $\prod x_i|L\Rightarrow L=d\prod x_i$. From $(1)$ and from the condition that $a_i$ are coprime (not pairwisely) we get the conclusion.

This is just the beginning :).


$ a,b,c$ nonnegative numbers such that $ a\plus{}b\plus{}c\plus{}abc\equal{}4$ then
\[ \frac{a}{\sqrt{b\plus{}c}}\plus{}\frac{b}{\sqrt{a\plus{}c}}\plus{}\frac{c}{\sqrt{a\plus{}b}}\geq \frac{\sqrt{2}}{2}(a\plus{}b\plus{}c)\]

[quote="arqady"][quote="Sunjee"]$ a,b,c$ nonnegative numbers such that $ a \plus{} b \plus{} c \plus{} abc \equal{} 4$ then
\[ \frac {a}{\sqrt {b \plus{} c}} \plus{} \frac {b}{\sqrt {a \plus{} c}} \plus{} \frac {c}{\sqrt {a \plus{} b}}\geq \frac {\sqrt {2}}{2}(a \plus{} b \plus{} c)
\]
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127956[/quote]
thank you!!! mr.arqady


$D$ is a point on side $BC$ of triangle $ABC$. Can it be that $\frac{AB}{BC}=\frac{AD}{DC}$?

Only if point $D$ is on point $B$. If that's the case, then the equation reduces to $\frac{AB}{BC}=\frac{AB}{BC}$, which is obviously true.  If you move point $D$ down side $BC$, $AD$ increases and $DC$ decreases, meaning the $\frac{AD}{DC}$ term only gets larger, while the other side of the equation stays the same.

Therefore, they can not be equal unless $B=D$.

Such point always exists. If $\angle C=90^\circ$, then $D\equiv B$. If the triangle is acute, then $D$ is between $B$ and $C$. If $\angle C$ is obtuse, then $D$ is outside the segment $BC$.


algebra
polynomial
complex numbers
algebra proposed

Find all polynomials satisfying : $P(x).P(x+1)=P(x^2+1)$

This one appeared over and over... ;)

Pierre.

Could you please give me a link , Pierre ?

I think we can use complex numbers to solve this problem but.. I'm not sure
  ;)

I'm not good at complex numbers , so I believe this problem can be solved without complex numbers .
  I have an idea (please help me to complete it):
   If $a$ is a root of $P(x)$ then $(a^2+1)$ is also a root of $P(x)$
  We have : $a^2 +1 \geq a \forall a$
 So $P(x)$ has infinite roots : contradiction !
 This means deg $P = 2n$  ....


[quote="buzzer11"]I currently own the 5th edition of the AoPS books (v1 and v2), and I was wondering, what changes in between editions? Does the content of the book change at all?[/quote]
I'm pretty sure only typos are fixed.

[quote="solafidefarms"]One of the editions changed the cover size, typeface, cover, ... to match the Intro series books. I like the old editions more than the new.[/quote]

The old edition seemed more like a dictionary, because of its size. I could flip through it really quickly. The new one's pages are slightly harder to flip through, and the pages feel different. I don't know if this is a good thing or not, but the new cover definitely looks more interesting. I didn't really notice any changes, other than the cover, between the 8.5 by 11 and the older ones.

[quote="pianoforte"]i like the old one, more compact and easier to wield[/quote]

So do I! I have the old V1 and V2 books, and the newer introduction series... I like the size of the V1 and V2 books much better.

Does anyone like the new one's format more than the old?

[quote="pianoforte"]too blue[/quote]

As in the color or the feel? :)


quadratics
modular arithmetic
number theory

Prove $2^\frac{p-1}{2}\equiv{\pm{1}}{ mod p}$

Source:  "Number Theory" by George E. Andrews

EDIT:  DUH!!!  $p =$ odd prime

By Euler's Criterion of quadratic residues, $\pm 1=\left(\frac{2}{p}\right)\equiv 2^{\frac{p-1}{2}}\pmod p$.

Or easier:

$(2^{\frac{p-1}{2}})^{2}\equiv 2^{p-1}\equiv 1 \pmod{p}$ by Fermat's Little Theorem.

Then, we already know two roots of $1$, $\pm 1$, and a prime mod can only have two square roots of any residue (consider the equation $x^{2}\equiv c \pmod{p}$. It can only have 0 or 2 roots), so $2^{\frac{p-1}{2}}\equiv \sqrt{1}\equiv \pm 1 \pmod{p}$.

Also,  useful, well known  and not too hard to prove: (for odd prime $p$)

$2^{\frac{p-1}{2}}\equiv (-1)^{\frac{p^{2}-1}{8}}\bmod p$ or in other words:

if $p =\pm 1 \bmod 8$ then $2^{\frac{p-1}{2}}\equiv+1$ otherwise it is $-1$
(That is for primes like $3, 11,13$ etc it is $-1$ while for primes like $7,17$ etc it is $+1$)


modular arithmetic
number theory
relatively prime
number theory proposed

Let $ a > 1$ be an integer.Show that the set $ {a^2\plus{}a\minus{}1,a^3\plus{}a^2\minus{}1,...}$
contains an infinite subset,any two members of which are relatively prime.

Let $ a_n \equal{} a^n \plus{} a^{n \minus{} 1} \minus{} 1$. Suppose that we have $ m$ elements of sequence $ (a_n)$: $ b_1, b_2, ..., b_m$ such that any two of them are relatively prime and suppose that $ p_1, p_2, ..., p_k$ are all primes which divide at least one of the $ b_i's$. Then we take $ b_{m \plus{} 1} \equal{} a_{(p_1 \minus{} 1)(p_2 \minus{} 1)...(p_k \minus{} 1) \plus{} 1}$. From the Fermat's Little Theorem:
\[ b_{m \plus{} 1} \equiv a^{(p_1 \minus{} 1)(p_2 \minus{} 1)...(p_k \minus{} 1) \plus{} 1} \plus{} a^{(p_1 \minus{} 1)(p_2 \minus{} 1)...(p_k \minus{} 1)} \minus{} 1 \equiv a \plus{} 1 \minus{} 1 \equiv a \pmod{p_i}
\]
and $ a$ is obviously relatively prime to every $ p_i$.


Inscribed in a circle of radius $ R$ is a square, a circle is inscribed in the square , a new square in the circle and so on for $ n$ times....
The find:
The sum of the areas of all circles.
The limit of the sum of areas of all squares as $ n$ tends to infinity.
the limit of the circles as $ n$ tends to infinity.

The radius of the $ n$th circle is $ R2^{\minus{}\frac{n}{2}}$.
The side of the $ n$th square is $ R2^{1\minus{}\frac{n}{2}}$


Since $E^{2}=E$, $(I+E)(2I-E)=2I+E-E^{2}=2I$. If $2\neq 0$ (our field doesn't have characteristic 2), $I-\frac12E$ is an inverse for $I+E$.

I didnt understand, where did $(E+I)(2I-E)$ came from?

I actually didnt understand the proof  :)  :o

It came from seeing what the answer should be. Constructing an explicit inverse is the simplest possible proof of invertibility.

All of this works for any idempotent element in any ring, as long as we can divide by 2.

[quote="RealTime"]Let $E$ be a projection.

Prove $E+I$ is invertible and calculate $(E+I)^{-1}$ in terms of $E$.

Thanks.[/quote]

If $E$ is projection, $E=E^{2}$.
[b]
Lemme1:[/b] $Ker(Id+E)\cap Ker(E)=\{\Theta\}$

[u]Proof:[/u] Let $v\in V: v\in Ker(Id+E))\cap Ker(E)$. then, $v+E(v)=\Theta$ and $E(v)=\Theta$, $\therefore v=\Theta\blacksquare$

[b]Lemme 2[/b] $Ker(Id+E)\subseteq Ker(E)$

[u]Proof:[/u]  Let $v\in ker(Id+E)$, The hipotesis says $v=-E(v)\Rightarrow E(v)=-E^{2}(v)$, but $E=E^{2}$. $\therefore E(v)=-E(v)\Rightarrow E(v)=\Theta \blacksquare$

****
$(Id+E)^{-1}$ exists

proof: For the lemme 1, $Ker(Id+E)\cap Ker(E)=\{\Theta\}$, and for the llemme 2 $\Rightarrow Ker(Id+E)=\Theta$, hence $(Id+E)$ is injective$\blacksquare$

The inverse is $I-\frac{1}{2}E$


induction
Subsets
number theory
Additive Number Theory
IMO Shortlist
Divisibility

Show that for any finite set $S$ of distinct positive integers, we can find a set $T$ ⊇ $S$ such that every member of $T$ divides the sum of all the members of $T$.

[b]Original Statement:[/b]

A finite set of (distinct) positive integers is called a [b]DS-set[/b] if each of the integers divides the sum of them all. Prove that every finite set of positive integers is a subset of some [b]DS-set[/b].

[b]Lemma 1[/b] Given $1,t>1$, we can find a set with the mentioned properties which contains $1$ and $t$ amd such that all its terms except for $1,t$ are divisible by $t+1$.

We prove this by induction on $t$. For $t=2$ we take the set $\{1,2,3\}$. Now assume it's proven for $t-1$ and we want to prove it for $t$. If the construction for $t-1$ was $1,t-1,a_1,\ldots,a_k$ with $t|a_i,\ \forall i$, then the construction for $t$ will be $1,t,(t-1)(t+1),a_1(t+1),\ldots,a_k(t+1)$. It's easy to see that it satisfies all the requirements.

[b]Lemma 2[/b] Given $a,b\in\mathbb N^*$, we can construct a set with the properties required by the problem which contains $a,b$ and such that $a,b$ are its smallest elements.

Obviously, we can assume WLOG that $(a,b)=1$. We find some $t>a,b$ s.t. $a|b+t,b|a+t$, and then, if the construction resulting from Lemma 1 for $t-1$ is $1,t-1,a_1,\ldots,a_k$, then our set will be $a,b,t,(t-1)(a+b+t),a_1(a+b+t),\ldots,a_k(a+b+t)$.

Now for the problem itself, we use induction on the number of elements. Suppose we have $n$ elements $x_1,\ldots,x_n$, and we've shown that the statement of the problem holds for $\le n-1$ numbers. We consider a set $\mathcal F$ of positive integers with the required properties which contains $x_1,\ldots,x_{n-1}$. If $S$ is the sum of all the elements in $\mathcal F$, then the set we get by adding $S,2S,4S,8S,\ldots$ to $\mathcal F$ still has the desired property, so we may very well assume that $S>x_n$ (what's important here is that $S\ne x_n$). Now we apply Lemma 2 to the pair of numbers $S,x_n$ to get a set $\{S,x_n,a_1,\ldots,a_k\}$. The set we want for the numbers $x_1,\ldots,x_n$ will be $\mathcal F\cup\{x_n,a_1,\ldots,a_k\}$.

 [quote=GoldenFrog1618]Let $n=\max S$.  I will prove that there exists a DS-set that contains all the numbers from $1$ through $n$.

If $n\ge 6$, then
\[1,2,\cdots, n,\frac{n(n-3)}{2},(n)(n-1)(n-3),(n)(n-1)(n-2)(n-4),\cdots \]
\[(n)(n-1)\cdots (3)(1)\] has a sum of $n!$ (the first $n+1$ elements sum to $n(n-1)$), all of its elements are distinct, and every element divides $n!$, as desired.  If $n<6$, then just use the sequence for $n=6$[/quote]

Guess you should prove why this sum equals $n!$.


How should I prepare if I am finishing up alg. II and am going onto geom/trig/precalc(I know weird school, funcia) but how would you suggest preparing for the qualification quiz? :?:

[quote="E^(pi*i)=-1"]You don't really need to "prepare" for the qualifying quiz.  Remember that you have a few months to work on it, and you can use books/online resources.  Mainly what you need is persistence and clear thinking.[/quote]

Seconded -- Especially the persistence part.  The problems will be tough if you haven't had much experience with the type of problems on there (more thinking, less equation juggling).  At the very least read all of the problems early on (as the above poster *waves at Mr. A. K.* said, you get several months to do the quiz), and keep them cycling around the back of your mind.  I know I solved at least two problems on last year's quiz amidst various particularly uninteresting lectures at school.  Keep at the problems, you never know when your subconcious will come up with a solution for you.

Also-my family had an over 100,000 income but I have 4 siblings...do you take this into account?

Funcia

We take all reasonable financial considerations into account.  The automatic tuition discount is there so that we can say right away to some families that they'll get some aid, but most of our help comes in the form of special scholarships, which are decided by real human beings who take into account whatever information you send us.  In recent years we've been able to provide sufficient scholarship funds that no one has been unable to come for financial reasons.


limit
integration
real analysis
real analysis unsolved

Prove that \[ \lim_{n \to \infty} n \left( \frac{\pi}{4} - n \int_0^1 \frac{x^n}{1+x^{2n}} \, dx \right) = \int_0^1 f(x) \, dx ,  \] where $f(x) = \frac{\arctan x}{x}$ if $x \in \left( 0,1 \right]$ and $f(0)=1$.

[i]Dorin Andrica, Mihai Piticari[/i]

[quote="perfect_radio"]Prove that \[ \lim_{n \to \infty} n \left( \frac{\pi}{4} - n \int_0^1 \frac{x^n}{1+x^{2n}} \, dx \right) = \int_0^1 f(x) \, dx ,  \] where $f(x) = \frac{\arctan x}{x}$ if $x \in \left( 0,1 \right]$ and $f(0)=1$.[/quote]

Write ${{n \int_0^1 \frac{x^n}{1+x^{2n}} = \int_0^1 \frac{x d(x^n)}{1+(x^n)^2 }}dx = \int_0^1 u^{\frac{1}{n}}d(\arctan u) = \frac{\pi}{4}- \frac{1}{n} \int_0^1 \frac{\arctan u}{u^{1-\frac{1}{n} } }}du$, so  the left limit is ${\lim_{n \to \infty} \int_0^1 \frac{\arctan u}{u^{1-\frac{1}{n} } }} du$ which tends to $\int_0^1 f(x)dx$ by Lebesgue dominated theorem.

Are Lebesgue's theorems (dominated & monotone) considered as known at the Romanian National Olympiad?

No there aren't. The ideea is the same with eugene's, but to finish it requires some easy work with $\epsilon$ and $\delta$..


Fie $A$ o mulţime oarecare. Există o lege de compoziţie $\cdot : A \times A \to A$ astfel încât $\left( A, \cdot \right)$ să fie grup?

vazusem eu ceva in genu insa nu mai stiu f bine...

[hide="solutie"]pt inceput

lema :
fie $(G,\cdot)$ un grup si $H$ o multime in bijectie cu $G$. atunci exista $\times$ lege de compozitie pe $H$ ce formeaza o structura de grup.

demonstratie :
fie $f: G\rightarrow H$ bijectie. fie $x,y\in H$. voi defini $x\times y$ ca fiind egal cu $f\big(f^{-1}(x)\cdot f^{-1}(y)\big)$. atunci se verifica imediat ca $(H,\times)$ este grup izomorf cu $(G,\cdot)$.

daca $A$ este finita, atunci poate fi construit un grup izomorf cu $\mathbb({Z}_{n},+)$, unde $n=|A|$.

daca $A$ e numarabila, atunci poate fi construit un grup izomorf cu $(\mathbb{Z},+)$.

acuma.... daca se admite GCH, atunci e usor prin inductie dupa $k$ unde $|A|=\aleph_{k}$, folosind faptul ca $\big(\mathcal{P}(A),\Delta\big)$ e grup.

fara GCH...cred ca merge daca consider $\mathcal{F}$ familia de submultimi finite ale lui $A$, unde $|A|=\aleph_{k}$. evident $\mathcal{F}$ este inchisa la operatiiile cu multimi. voi face o partitie a lui $\mathcal{F}$ dupa numarul de elemente al unei submultimi. pentru un $n$ dat, cardinalul unei clase este chiar $\aleph_{k}$, si cum exista un numar numarabil de clase, rezulta si ca chiar $|\mathcal{F}|=\aleph_{k}$.
se arata imediat ca $(\mathcal{F},\Delta)$ este grup, pt ca $A\Delta A=\emptyset$ si $A\Delta \emptyset=A$.
$\mathcal{F}$ e in bijectie cu $A$, si deci aplic lema si gata.[/hide]


symmetry
binomial coefficients
combinatorics unsolved
combinatorics

For n and k which are natural numbers, I need to give a combinatorial proof of this.

$ \sum_{k \equal{} 1}^{n}$k$ \binom{n}{k}^2$ = n$ \binom{2n \minus{} 1}{n \minus{} 1}$

I know the right side can be understood as the number of sets of size n ouch of 2n people in which (1/2)n are a 'leader'.

I don't know how to interpret the left side though, any help would be greatly appreciated.

Consider a group of $ n$ boys and $ n$ girls and count the number of team of $ n$ people with a leader which has to be a boy.
You may select the leader among the $ n$ boys and then select the $ n\minus{}1$ other persons of the group among the $ 2n\minus{}1$ others. This gives the RHS.
In another hand, you may select $ k$ boys and $ n\minus{}k$ girls to form the group. Then among the selected $ k$ boys, you have to choose the leader. Using the symmetry of the binomial coefficients and summing over $ k$ gives the total number of desired teams.

Pierre.

Thanks, I was making it much harder than it needed to be.


Show that in each set of ten consecutive integers there is one that is coprime with each of the other integers. (For example, in the set $ \{ 114,115,...,123 \}$ there are two such numbers: $ 119$ and $ 121.)$

let the consecutive integers be $ k,k\plus{}1\dots ,k\plus{}9$ let $ k\equiv0\mod2$ (if the k is odd we do analogously)

let's pick $ k\plus{}5$ then $ \gcd(k\plus{}5,k_i)\equal{}1,\dots ,4 ,$  for $ i\equal{}0,\dots,9$  where $ k_i\equal{}k\plus{}i$ if $ k\plus{}5\equiv0\mod3$

then we take $ k\plus{}3$ if  $ k\plus{}3\neq0\mod5$ then we are done cause

$ k\plus{}3\equiv1\mod3$ and $ \gcd(k\plus{}3,k_i)\equal{}1,\dots,6$

if $ k\plus{}3\equiv0\mod5$ then we pick
$ k\plus{}1$ now $ \gcd(k\plus{}1,k_i)\equal{}1,\dots ,8$ if

$ k\plus{}1\neq0\mod7$ then we are done

samely else we pick $ k\plus{}7$ we have from the previous assumptions that

$ k\plus{}7\equiv2mod3$
$ k\plus{}7\equiv6mod7$
$ k\plus{}7\equiv4mod5$

and since since $ \gcd(k\plus{}7,k_i)\equal{}1,\dots ,7$
$ \gcd(k\plus{}7,k_i)\equal{}1$

A relatively standard problem.

In any set of ten consecutive integers, we must have 5 even integers, at most 2 odd integers divisible by 3, at most 1 odd integer divisible by 5 and at most 1 odd integer divisible by 7. Thus there must be at least one integer that is not divisible by any of those primes, and is thus coprime with the other integers in the set.


vector
Functional Analysis
advanced fields
advanced fields unsolved

$H$ a complex Hilbert space and $U: H \to  H$ a normal operator. Prove that if $\lambda$ is an isolated point of the spectrum of $U$, then $\lambda$ is an eigenvalue.

It's a consequence of the spectral theorem. If $E$ is a resolution of the identity corresponding to $U$, and $\lambda$ is the isolated point of the spectrum, then $\{\lambda\}$ is an open subset of the spectrum, so $E(\{\lambda\})$ is a non-zero projection. Any vector in the range of this non-zero projection will be a $\lambda$-eigenvector of $U$.