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Hi the AoPS team,

Can you create a mathematics research class for us that want to do mathematical research? It will be a great help to the community! Thanks!

This should be in Questions, Suggestions and Announcements near the bottom of the forum list...

Huh. I don't know how feasible a class would be, but I'd be willing to host a regular open classroom hour about mathematical research. Depending on the topics people are interested in, I might even be able to do more than guide people to good references... ;)

It is by no means obvious what such a class can be about. If all you are looking for are some research type problems, you do not need a special class for that. Just search the net, post your request in the subforum on the subject you want to do your research on, talk to your teachers, etc. If you wonder whether the mathematical research is any different from olympiad problem solving, the main difference is that usually you have have neither an exact statement of the problem (what you deal with is rather some general theme), nor the guarantee that it is solvable. Otherwise, it is pretty much the same.

The way the undergraduate research is usually organized is that the teacher offers a few themes, the students choose one and try to do as much as they can. But this rather calls for a special subforum than for a class. We can certainly try to run something like online REU on AoPS and I guess almost every teacher/professor/advanced student will be willing to organize the discussions of the themes he is interested in (I'll certainly be willing to take a shot at it myself). The only thing that would require is creating one more subforum somewhere and giving a few "leaders" the powers to create subsubfora in it  when they want to start a new project, which doesn't seem to be a lot of work for admins. But are there, indeed, sufficiently many people interested in something like that?

Or you can go to a place like PROMYS or the Ross program, since they are meant for that kind of stuff. If you live in the US, I'd look into it :).

I'm fairly certain that there used to be an independent study course like this actually, but it got canceled.

I'm inclined to agree with fedja that a culture or community in this respect would be more useful than a formal class.  But we'd need a signficant number of mentors to make it work.

@rrusczyk: So, will you be able to do as what fedja suggested? Lots of students here will be happy if you do!  :lol:

[quote="fedja"]The only thing that would require is creating one more subforum somewhere and giving a few "leaders" the powers to create subsubfora in it  when they want to start a new project, which doesn't seem to be a lot of work for admins.[/quote]Actually, it is rather complicated right now to be able to do that (we would have to operate extensive modification onto the code, which I doubt we will have time for any time soon).

[quote="The Visionary"]Well, it looks like a lot of students are gonna be sad!  :gathering:
@ Valentin: When will you have the time to operate the modification onto the code then?  :D[/quote]I think you guys should design a system using only one forum (perhaps one topic per topic of study or something), rather than multiple subfora.

[quote="The Visionary"]For the research thingi, can you tell me how can we design a system using only one forum? Please explain. Should we used any existing AoPS subforum (Is this even a word?  :rotfl: ) or do you have anything else in mind, like creating a forum outside of AoPS website? Moreover, can you tell me how did some students (like the Intel STS and Siemens Westinghouse Winners for example) get professors (some from MIT, Harvard, etc I might add!) to mentor them?[/quote]

I certainly cannot speak for Intel and Siemens contestants, but in my experience if you have an interesting problem and you respectfully approach the professor, then they usually will assist you :).

Also it helps to have connections. I'm sure some of the students that you see from the bigger schools have really nice teachers with professor friends.

[quote="Anaxerzia"]connections[/quote]
Yes.

Regarding the original topic: I would be willing to help organize this. I am probably not expert enough in any field of maths to be a mentor, but I am happy to help with moderating, giving people references, etc.

[quote="Valentin Vornicu"]I think you guys should design a system using only one forum (perhaps one topic per topic of study or something), rather than multiple subfora.[/quote]
I wonder if it is possible/easy to write a script which allows to display only the topics with a given 4 symbol combination in the beginning of their name (or, at least, to sort the topics so that those with this particular combination appear on the top of the list) that can be entered upon the entrance to the forum. This way we can assign a code to each project and keep everything in just one forum. It will require certain discipline in naming the threads, of course, (though even this can be automatized: just request this 4 symbol combination upon the entrance and automatically add it to the name) but in all other respects, it will be just as good as multiple subfora. Or I completely misunderstand what the real difficulty is? (I think every database has some simple tool that does pretty much what I described but I'm not an expert :))

I still do not understand why 1 (or 2-3 topics) can't be used for this purpose. You can arrange them by author if you want for example (have the student post the first post). I'm really reluctant of writing additional code for something that can be easily made through conventions (you can arrange by topic name also btw).

I would rather expect 20-30 topics for each project if it works fine. The point is that long topics (with more than 15 posts) are very hard to read and discussion of several things simultaneously in one topic creates a lot of confusion (that's why we have our "one problem per post" policy) and manual search through several pages for the topic you want is a bit tiresome.

Essentially the existing tools (say, the regular search) can work for the beginning if we use the code system I proposed (just search the topics containing the code in the title in the particular forum and you'll get the list of relevant posts arranged by their time), so there is no urgent need to write some new code tomorrow. It seems like we are looking for a system that

a) can somehow (though maybe a bit inefficiently) work using just the existing tools
(so there will be no need for Valentin to write any extra code until he has more time and until it is clear that it is really needed)
b) if needed, can be easily upgraded to something that is for all practical purposes equivalent to the multiple subfora system.

The best I can offer in this venue is the 4 symbol code system but I do not insist it is the best solution. If somebody else can invent something better, it will be great.

@Fedja: Where would you post if you wanted to try your hand at some original problem? I don't know too much mathematics to do too much advanced research but I am rather keen on trying some. In any case, could we have the forum up and running. We could worry about subfora as the time comes couldn't we?

Alright fedja, where do you want it to be? In the College Playground area somewhere?

Honestly, that's a question about which I didn't think at all. :lol: College Playground is OK, if you think that such placing won't scare younger AoPSers off immediately. Perhaps sticking it between the Olympiad Section and College Playground is a bit more logical but if you disagree with me in this respect, just do as you think appropriate. I rely upon your common sense here :). Give all the "mentors" the moderating powers in it, and we'll try to start.

[quote]
In any case, could we have the forum up and running. We could worry about subfora as the time comes couldn't we?
[/quote]
All right, but let us still use some precautions not to make a total mess from the beginning. I suggest that we use a six letter code for each project, which may be either mnemonic or just arbitrary and start all topics related to a given project with this code (6 letters rather than 4 come from the fact that our search doesn't work with short quieries; also, 6 letters give more room for inventing meaningful codes). So, if you want to post into the existing project, add the corresponding code to the beginning of your title and put a space after it. If you want to start a new one, choose the code not in use yet. It may sound like an unnecessary bureaucracy but, if the time ever comes to sort things into subfora or write some scripts for displaying only topics related to something particular, this code system will come very handy.

OK, enough blubbering about how to do everything. Let's try to start. :)

Maybe Other Resources/Programs.  If it's in the college area, the students you're aiming for will be less likely to see it.  Stick it in there with the forums dedicated to the various summer programs, and it will be visible to your intended audience.

Is it possible to put a link to it in the College Playground that would make it look like it was a subforum there but would actually just lead to the other place? Alternatively we could just put a permanent announcement in there with a link and short explanation.

[quote="fedja"][quote="rrusczyk"]Maybe Other Resources/Programs. [/quote]
Erm..., Where is that? :? I couldn't locate the forum with exactly this name :blush: By the way, it would be nice to make it visible to college students too. :)[/quote]

I meant in either of the "Other" sections under either High School or Contests & Programs.  And I think it's easy to have shadow forums in other places that point to this one.


Find the entire solutions  $ {x}^{3}\plus{}{y}^{3}\minus{}3xy\equal{}2$

You mean the integer solutions, yes?

[hide="Solution"] $ x^{3}\plus{}y^{3}\plus{}1\minus{}3xy \equal{} (x\plus{}y\plus{}1)(x^{2}\plus{}y^{2}\plus{}1\minus{}xy\minus{}x\minus{}y) \equal{} 3$

Case:  $ x\plus{}y\plus{}1 \equal{} 1$.  Then $ y \equal{}\minus{}x\implies 3x^{2}\equal{} 2$, no integer solutions.

Case:  $ x\plus{}y\plus{}1 \equal{} 3$.  Then $ y \equal{} 2\minus{}x$, so we have the quadratic

$ x^{2}\plus{}4\minus{}4x\plus{}x^{2}\plus{}1\minus{}2x\plus{}x^{2}\minus{}x\minus{}2\plus{}x \equal{} 1\Leftrightarrow$
$ 3x^{2}\minus{}6x\plus{}3 \equal{} 1\Leftrightarrow$
$ (x\minus{}1)^{2}\equal{}\frac{1}{3}$

Which also has no integer solutions. [/hide]

Your equation is

[quote="t0raji0u"]$ x^{3}\plus{}y^{3}\plus{}1\minus{}3xy$[/quote]
How do you get this? The original equation is

[quote="Elio (n)"]$ x^{3}\plus{}y^{3}\minus{}3xy\equal{}2$[/quote]
the two are not equivalent.

[quote="AstroPhys"]Look closer
$ x^{3}\plus{}y^{3}\minus{}3xy \equal{} 2$
$ x^{3}\plus{}y^{3}\plus{}1\minus{}3xy \equal{} 3$.[/quote]
My bad  :blush:

My apologies; of course I actually missed two cases.

[hide="Finishing up..."] Case:  $ x\plus{}y\plus{}1 \equal{}\minus{}1$.  Then $ y \equal{}\minus{}x\minus{}2$ and we have the quadratic

$ x^{2}\plus{}4\plus{}4x\plus{}x^{2}\plus{}1\plus{}x^{2}\plus{}2x\minus{}x\plus{}x\plus{}2 \equal{} 3x^{2}\plus{}6x\plus{}7 \equal{}\minus{}3\Leftrightarrow$
$ (x\plus{}1)^{2}\equal{}\minus{}\frac{7}{3}$

Which has no integer solutions.

Case:  $ x\plus{}y\plus{}1 \equal{}\minus{}3$.  Then $ y \equal{}\minus{}x\minus{}4$ and we have the quadratic

$ x^{2}\plus{}16\plus{}8x\plus{}x^{2}\plus{}1\plus{}x^{2}\plus{}4x\minus{}x\plus{}x\plus{}4 \equal{} 3x^{2}\plus{}12x\plus{}21 \equal{}\minus{}1\leftrightarrow$
$ (x\plus{}2)^{2}\equal{}\minus{}\frac{10}{3}$

Again, no integer solutions. [/hide]
So much for the integer solutions.  The rational solutions will either require more tricks or a transformation into Weierstrass form.


Find the summation of the series
$ \frac{1}{1!}\plus{}\frac{1}{1!\plus{}2!}\plus{}\frac{1}{1!\plus{}2!\plus{}3!}\plus{}\cdots\plus{}\frac{1}{1!\plus{}2!\plus{}3!\plus{}\cdots\plus{}n!}?$

I very much doubt there is a closed formula for either the $ n$-terms sum, or the infinite series...

Hi joshisunny16.

I asked a friend of mine who is a mathematics professor. He says it is a well-known problem, that no closed form is known and that it is believed that no closed form exists.  Hope this helps.

NO, the series is convergent (just a simple look at it shows it is majorated by the series for $ e$).

Posting it in LaTeX is not going to suddenly make it have a nice answer:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=286645&search_id=1511183158

[quote]NO, the series is convergent (just a simple look at it shows it is majorated by the series for ).[/quote]
ok fine.but then why are we not getting the sum in terms of e?

Why should you get the sums in terms of $ e$?  The standard series for $ e$ bounds this one from above, as mavropnevma noted, but there's no nice transformation that lets you translate between them and no reason at all to suppose that this should converge to a nice value.  (Of course, whatever value it does converge to, it does so very quickly -- to get 8 or 10 digits of accuracy you only need about 13 or 15 terms of the series.)


Let there be $ z_k \in \mathbb{C}$ with $ |z_k|= r  \geq 0$ , $ \forall k=\overline{1,3}$ and $ z_1 + z_2 + z_3 \neq 0$ .
Consider the points :

$ A \left( \frac{z_2 + z_3}{z_1} \right) , B  \left( \frac{z_3 + z_1}{z_2} \right)$ and $ C  \left( \frac{z_1 + z_2}{z_3} \right)$ .
Find the affix of the circumcenter of the triangle $ \triangle ABC$ .

[hide="hint"]Consider the three numbers $ (z_1 \plus{} z_2 \plus{} z_3)/z_1$, $ (z_1 \plus{} z_2 \plus{} z_3)/z_2$, and $ (z_1 \plus{} z_2 \plus{} z_3)/z_3$.[/hide]

Yes , the main ideea is to translate the triangle $ \triangle{ABC}$ with $ 1$ unity .

Application :

Find the necessary condition for $ \triangle{ABC}$ to be equilateral .

oh...silly me  :blush: I didn't see the hint...sorry


Eliminate theta

(1)  $ \frac {x\cos \theta}{a}\plus{}\frac {y\sin \theta}{b}\equal{}1$
   $ x\sin \theta\minus{}y\cos \theta\equal{}\sqrt {a^2\sin ^2 \theta\plus{} b^2\cos ^2 \theta}$

(2)  $ \sin \theta\minus{}\cos \theta\equal{}p$
   $ \csc \theta\minus{}\sin \theta\equal{}q$

Eliminate theta and phi

(3)  $ \frac {x\cos \theta}{a}\plus{}\frac {y\sin \theta}{b}\equal{}1$
   $ \frac {x\cos \phi}{a}\plus{}\frac {y\sin \phi}{b}\equal{}1$
   $ a^2 \sin \frac {\theta}{2} \sin \frac {\phi}{2} \plus{} b^2 \cos \frac {\theta}{2} \sin \frac {\phi}{2} \equal{} c^2$

(4)   $ a\cos ^2 \theta\plus{}b\sin ^2 \theta \equal{} c$
   $ b\cos ^2 \phi\plus{}a\sin ^2 \phi \equal{} d$
   $ a\tan \theta \equal{} b\tan \phi$

lemme brief the method?
1, get $ \sin \theta$ in terms of $ \cos \theta$ n all and substitute in 2nd eqn n solve fer $ \cos \theta$ n then put $ (\sin \theta)^2 \plus{} (\cos \theta)^2 \equal{} 1$  :)

For 2 do similarly instead of sine do fer cos .

varunarasu: yes

madness: ur solution is not very clear. can u explain better? if u writ sin in terms of cos or other way round and substitute, u dont always get the answer.


A set of tiles numbered 1 through 100 is modified repeatedly by the following operation: remove all tiles numbered with a perfect square, and renumber the remaining tiles consecutively starting with 1. How many times must the operation be performed to reduce the number of tiles in the set to 1?

[hide="Hint"]From 1 to 100, there are $\lfloor \sqrt{100}\rfloor = 10$ squares.  Subtract them and we have 90 numbers left.  $\lfloor \sqrt{90}\rfloor = 9$ so we have 81 left, after 2 applications.  Keep going...[/hide]

[hide]
100
90
81
72
64
asdf
49
asdf
36
...

I think you get the pattern now.[/hide]

using mysmartmouth's hint.......
i get 18......
which happen's to be the right answer...
well, thanks
the solution in book made no sense

What 13375P34K43V312 was saying:

The pattern was:

$100 = 10^{2}$
$90$
$81 = 9^{2}$
$72$
$64 = 8^{2}$
$56$
$49 = 7^{2}$
$42$
$36 = 6^{2}$
.....

So we notice that after 2 applications, we get $9^{2}$.  After 4 applications, we get $8^{2}$.  After 6 applications, we get $7^{2}$.  Hmm, this seems like a linear function almost...  After basic algebra, we can see that there is an explicit function, $y=-2x+20$, where x is the square root of the target number, and y is the number of applications.  To get to 1, we plug in 1 for x and get $\boxed{18}$ for y.


floor function
algebra
polynomial
function
combinatorics proposed
combinatorics

Let $ n$ be a positive integer.

Show that
\[ \sum_{k \equal{} 0}^{n}2^{k}\binom{n}{k}\binom{n \minus{} k}{\lfloor\frac {n \minus{} k}{2}\rfloor} \equal{} \binom{2n \plus{} 1}{n}
\]

You can easy get the result after computing in two ways the sum of the coefficients of $x^{n-1}$ and $x^{n}$ in the polynomial $(x+1)^{2n}$
     However, i 'met' this problem earlier and i saw a more insightful solution given by Jian Gu. The basic idea was that we chose n girls, n boys and their teacher and we pair each boy with a girl. Afterwards, we count in two ways how we can assign n tickets to them.

also

http://www.mathlinks.ro/Forum/viewtopic.php?t=48585

  darij

Your solution is much simpler Negative_3...I love it :lol:


Interval, I've noticed that you're posting a lot of algebra and geometry problems. Now I'm not criticizing that, but could you please give us an idea about how much math experience you have?

Also, you could consider moving some of your posts to the Classroom Math forum. Just a suggestion.

[hide="For the problem"]You're going to spend $40 for the DJ. You make$2 every time you sell a ticket. So...[/hide]

The equation is the whole thing. and seriously, about 1/2 ur posts should be in classroom math.

[quote="Interval"]The senior class is sponsoring a dance. The cost of a student disk jockey is 40 dollars, and tickets sell for 2 dollars each. Write a linear equation and, graph the equation to represent the relationship
between the number of tickets sold and the profit from the dance. Then
find how many tickets must be sold to break even.[/quote]

[quote="Interval"]Set up the equation and I'll take it from there.[/quote]

Anyone besides me find this juxtaposition amusing?

Seriously, Interval, if you're having trouble setting up equations for word problems, it wouldn't help you to post all of them here and watch them be solved. You would need more practice setting up equations.


:maybe:  Whenever I press the next topic button, I get a blank page with "a:0:{}}}'); ?>" on top.

Egosearch too. See the topic in Support.  :)


Four couples play chess together. For the match, they're paired as follows:  ("man Clara" indicates the husband of Clara, etc.)
\[Bea \Longleftrightarrow Eddy\]
\[An \Longleftrightarrow man\ Clara\]
\[Freddy \Longleftrightarrow woman\ Guy\]
\[Debby \Longleftrightarrow man\ An\]
\[Guy \Longleftrightarrow woman\ Eddy\]
Who is $Hubert$ married to?

I'm not giving 10 names.

Look, we got:[list][*][b]A[/b]n[*][b]B[/b]ea[*][b]C[/b]lara[*][b]D[/b]ebby[*][b]E[/b]ddy[*][b]F[/b]reddy[*][b]G[/b]uy[*][b]H[/b]ubert[/list]

The first 4 female, the last 4 male.

devious_ wrote:You said 4 couples and gave 10 names (5 couples).

Two of the pairings are the same. That's why there are 5 pairings, but really only 4 couples.

[hide]Hubert can only be Clara's husband, or An's husband, because he would not fit into the other three pairings. Also, he can't be Clara's husband because if he were, then Clara doesn't show up on any of the matches. If person A was married to person B, then the only way you could write person A in a pairing is either A or man/woman B. Thus, Hubert can only be married to An.[/hide]

Yifman:

[hide]

If Hubert is with An, who is with Eddy?

[/hide]

The solution I have involves

[hide]

Hubert with Bea... with the first match being Eddy/Bea, the second and third both being An/Freddy and the fourth and fifth being Debby/Guy, with the Clara/Hubert match being unlisted.

[/hide]

Hehe I won't tell yet!

Just try to give a little "mathematical" solution, you see it's even easier than try-and-error here! :)

Comon, I can't believe no one has made an attempt to kill this problem :D

I'm going to "promote" the challenge problem threads that don't get solved here in Getting Started. Look for them in Intermediate Topics.

Ok. Strange, when I saw this first (in 9th grade) I needed exactly 2 minutes to solve it :?

actually it isn't even mathematics, just some trivia that could appear in any book...

I will post the solution and like this the problem can stay where he belongs, the getting started forum.

let c(X) be the couple of X.
the games are:
B - E
A - c(C)
c(G) - F
D - c(A)
c(E) - G  in the left we have the women and in the right the men.

[tex]c(E), c(G) \neq B[/tex] and [tex]c(C), c(A) \neq E[/tex]
so c(E) = D. from now on is easy.
c(A) = G, c(C) = F and c(B) = H.

Yep, perfect.

Or in getting-started wordings:

Eddy can't be married to Bea since Guy plays with Eddy's wife.
Eddy can't be married to An since Debby plays with An's husband.
Eddy can't be married to Clara since An plays with Clara's husband.
[b]So Eddy is married to Debby.[/b]

Guy plays with Debby and Debby plays with An's husband.
[b]So Guy is married to An.[/b]

Freddy plays with An and An plays with Clara's husband.
[b]So Freddy is married to Clara.[/b]

If 3 out of 4 couples are known in a group of 8 persons, we know the 4th couple as well.
[b]So Hubert is married to Bea.[/b]

Peter VDD wrote:This should rather have appeared in some trivia book than on an olympiad, but anyway:

Four couples play chess together. For the match, they're paired as follows:  ("man Clara" indicates the husband of Clara, etc.)

Who is  married to?

[hide]I'll use more compact notation here.  Represent An through Hubert with A through H, and let S(X) be the spouse of X.  Thus, the given equations become

B - E

A - S(C)

S(G) - F

D - S(A)

S(E) - G

Consider B-E and A-S(C).  Since A and B are obviously distinct, the same is true of E and S(C), i.e. E and C are not a pair.  Similarly, comparing B-E and D-S(A), we conclude that E and A are not a pair.  Thus, E is either married to B or D.

Now consider B-E and S(E)-G.  As before, E and G are distinct, so B and S(E) are distinct; B and E are not a pair.  But earlier, we found that E can only be married to B or D; since he can't be married to B, he must be married to D.  In our notation, then, E = S(D) and D = S(E).

Substituting S(E) for D in last original equation yields D-G, so D played against G.  But the equation above that has D-S(A), so G = S(A).  Thus, A = S(G); substituting A for S(G) in the third original equation yields A-F.  But the equation above that had A-S(C), so F = S(C).  Now everyone is paired up (D with E, A with G, C with F) except B and H, so they must be a couple.  Thus, B (Bea) is the wife of Hubert.[/hide]

Whoa, just realized that this problem is a month old.  I should read Getting Started more.

[quote="devious_"]You said 4 couples and gave 10 names (5 couples).[/quote]

'woman' and 'man' ain't names.
4 couples with 8 names, nothing wrong.


Let $n$ be an odd integer greater than 1. Prove that the sequence: $\binom{n}{1}, \binom{n}{2}, \ldots, \binom{n}{\frac{n-1}{2}}$ contains an odd number of odd numbers.

[hide="Solution"]

Consider the sum of the sequence modulo $2$. If the sum is congruent to $1 \bmod{2}$, then we must have an odd number of odd numbers. Since $\binom{n}{k}= \binom{n}{n-k}$ and $\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n}= 2^{n}$, we have the sum of the above sequence to be:

$\frac{2^{n}-\binom{n}{0}-\binom{n}{n}}{2}= 2^{n-1}-1$

Clearly, this sum is always going to be odd, as:

$2^{n-1}-1 \equiv-1 \equiv 1 \bmod{2}$

Since the sum of the sequence is odd, there are an odd number of odd numbers in the sequence.

$QED$

[/hide]

using Lucas's theorem, supose number of 1's in binary representation of $n$ is $r$. We have number of odd number in sequence $C_{n}^{k},k=0,1,...,n$ is $2^{r}$ and we have done!  :wink:


[hide="Solution"]
We know that $n^3$ is an integer.  $m$ must be an integer larger than $n^3$, so we want the smallest value of $n$ such that $(n+r)^3 - n^3 \ge 1$.

Expanding, $3n^2r + 3nr^2 + r^3 \ge 1$.  Set $r = \frac{1}{1000}$.  Then, after multiplying, we have $3(1000^2)n^2 + 3(1000)n + 1 \ge (1000)^3$.  Ignore the last two terms and consider $3(1000^2)n^2 \approx 1000^3 \implies 3n^2 \approx 1000 \implies n^2 \approx 333$.  This will give a value of $n = \boxed{19}$.
[/hide]

[quote="Scrambled"][quote="chess64"]Let $m$ be the smallest integer whose cube root is of the form $n+r$, where $n$ is a positive integer and $r$ is a positive real number less than $1/1000$. Find $n$.[/quote]
[hide](n+1/1000)^3=n^3+3n^2/1000+3n/1,000,000 +1/1000,000,000)>=m>=n^3, Thus we want m=1+n^3 for the least m, and so 3n^2/1000+3n/1,000,000+1/1000,000,000>=1 and since the 3n/1000,000 is utterly insignificant, we want the largest value of 3n^2 that will exceed 1000. Thus our answer is 19[[/hide][/quote]

$\LaTeX$ed for ease of reading :roll:

[hide]${\displaystyle \left( n + \frac{1}{1000}\right) ^3 = n^3+\frac{3n^2}{1000}+\frac{3n}{1,000,000} +\frac{1}{1000,000,000} \ge m \ge n^3}$. Thus we want $m=1+n^3$ for the least $m$, and so \[ \frac{3n^2}{1000}+\frac{3n}{1,000,000}+\frac{1}{1000,000,000} \ge 1, \] and since the $3n/1000000$ is utterly insignificant, we want the largest [smallest??] value of $3n^2$ that will exceed 1000. Thus our answer is 19.[/hide]

Nice solution, BTW. :)

Why can you guys use (1/1000) as r even though the problem suggested that r is less than (1/1000)?

We can get arbitrarily close to that, though.

This is the optimal value for us to pick because the value of $ r$ needed to make $ (x \plus{} r)^3 \minus{} x^3 \equal{} 1$ increases as $ x$ decreases.

Now, if it had turned out that $ 19\plus{}1/1000$ cubed actually equals $ 19^3\plus{}1$, then we would have to say, ok, we can't quite reach $ 1/1000$, so our minimum $ x$ is $ 20$. But that's not the case here.

[quote="TZF"]Now, if it had turned out that $ 19 \plus{} 1/1000$ cubed actually equals $ 19^3 \plus{} 1$, then we would have to say, ok, we can't quite reach $ 1/1000$, so our minimum $ x$ is $ 20$. But that's not the case here.[/quote]
And clearly, any integer + .001 can ever be the cube root of an integer, so we would never have to worry about this.

[quote="Brut3Forc3"][quote="TZF"]Now, if it had turned out that $ 19 \plus{} 1/1000$ cubed actually equals $ 19^3 \plus{} 1$, then we would have to say, ok, we can't quite reach $ 1/1000$, so our minimum $ x$ is $ 20$. But that's not the case here.[/quote]
And clearly, any integer + .001 can never be the cube root of an integer, so we would never have to worry about this.[/quote]

[hide="Solution"]After solving the problem in much the same way as previous posters did, I was tempted to use the generalized binomial theorem:

$ \frac{1}{1000} \plus{}n > \sqrt[3]{(n^3\plus{}1)} \equal{} n \plus{} \frac{\binom{1/3}{0}}{n^2} \plus{} \frac{\binom{1/3}{1}}{n^5} \plus{} \cdots \equal{} n \plus{} \frac{1}{3n^2} \minus{} \frac{1}{9n^5} \plus{} \cdots$

So $ \frac{1}{1000} > \frac{1}{3n^2} \minus{} \frac{1}{9n^5} \plus{} \cdots$. We see that the terms after $ \frac{1}{3n^2}$ have no appreciable effect on the inequality, especially after we approximate $ n$:

$ \frac{1}{1000} > \frac{1}{3n^2} \Rightarrow n^2 > \frac{1000}{3} \Rightarrow n^2 > 324 \equal{} 18^2 \Rightarrow n\equal{}19$.

The advantage is that $ 3n^2 > 1000$ is immediate, without worrying about canceling powers of $ 1000$.[/hide]

p4: (i refuse to add details of the nasty expanding) To try to minimize the value of $m$, we would take the cube root of $n^3+1$ and subtract off $n$, so we want $\sqrt[3]{n^3+1}-n < \frac{1}{1000}$. Adding $n$ to both sides, squaring, cancelling $n^3$ and using the quadratic formula we get $n > 18$ after rounding down, so the minimum is $\boxed{19}$ which indeed works.

Okay, I'm gonna skip the nasty stuff I did, we do some nasty expanding to get $\sqrt[3]{n^3+1}-n < \frac{1}{1000}$, we get $\boxed{19}$, by further computations.


a) Prove that there are infinitely many prime numbers.
b) In view of part a, prove that there are infinitely many prime numbers of the form 3k+2, k being a nonnegative integer.

Ooh, I can do this one :-)

a) Assume that there is a largest prime number P
Then, multiply all the integers up to P, and add 1. Call this number N.

Clearly, N is not divisible by any of the primes up to and including P (since, for some number A < P, N mod A = 1). So, either it is divisible by some prime number larger than P, or it is a prime itself.

However, either of this cases is condradictory with our assumption, so clearly P is not the largest prime, so there must be infinitely many primes.

b) Any integer is 0, 1, or 2 mod 3. However, if some integer A is 0 mod then it is divisible by three, and thus is not prime (unless it is prime). If some numer is 1 mod 3, then it is even, and so is not prime (unless it is 2). Numbers of the form 3k+2 are all of the numbers which are 2 mod 3. We know that there are infinitely many primes, and know that there are only a finite number which are 0 or 1 mod 3, so there must be infinitely many which are 2 mod 3, that is, there must be infinitely many primes of the form 3k+2

[quote="swiftstream"]If some numer is 1 mod 3, then it is even, and so is not prime (unless it is 2).[/quote]

Is it? how about 7? 13? 19? comon man, check what u wrote :)

I think this proof is not very consistent thus...

[quote]We know that there are infinitely many primes, and know that there are only a finite number which are 0 or 1 mod 3, so there must be infinitely many which are 2 mod 3[/quote]
The logic behind this statement is not correct. What IF there are only a finite number of them? Be careful when you use the law of excluded middle (ruling out other choices). Also, as pointed out by Peter, what about 7,13,and 19? The proof to part b can be done similarly like part a with a little twist into it.

A topological proof of infinitely many primes, have a look here
http://www.cut-the-knot.org/proofs/Furstenberg.shtml

Whoops...  :blush:

But... 3X.lich... I don't understand what your other point is. If set P has infinitely many objects, and set A has finitely many, then the set P - A must have infinitely many... that is what I am saying...

[quote="Peter VDD"]no i don't get his second part either

probably he forgot that you already proved a) or something :s[/quote]
Nah, I didn't forget he proved part a, I was only suggesting a way to prove part b because the technique is similar. He knows the loophole of his proof in part b, that I know too because he jumped into conclusion too quickly. Again, I was only suggesting a way, you don't have to prove it this way. Is this clear?

No, it was the 'law of the excluded middle' thing that had me confused...

Anyway, I'll try to prove the second part again...


integration
logarithms
calculus
real analysis
real analysis unsolved

Let
$ u_k = \int_{0}^{\pi}e^{iv}\{\frac{1}{|v + 2k\pi|} - \frac{1}{|v + \pi + 2k\pi|}\} dv$
I want to show
$ \sum\limits_{k\geq 1} |u_k|$ convergent.
Can anyone help me? thanks a lot!

Why do you have absolue values on terms that are clearly positive? I would just note $ |u_k| \leq \int_{0}^{\pi}\{\frac {1}{v + 2k\pi} - \frac {1}{v + \pi + 2k\pi}\} dv$, do the algebra, and you'll see the integrand $ \leq \frac{1}{k^2}$. No integration necessary

[quote="mathmean"]thank you !
but what if here $ |k| \geq 1$ instead of $ k \geq 1$?[/quote]
the same. i don't see the problem. :maybe:

[quote="WWW"]Why do you have absolue values on terms that are clearly positive? I would just note $ |u_k| \leq \int_{0}^{\pi}\{\frac {1}{v + 2k\pi} - \frac {1}{v + \pi + 2k\pi}\} dv$, do the algebra, and you'll see the integrand $ \leq \frac {1}{k^2}$. No integration necessary[/quote]
how you prove that $ \int ... < 1/k^2$ ??

[quote="anik_andrew"]

[quote="WWW"]Why do you have absolue values on terms that are clearly positive? I would just note $ |u_k| \leq \int_{0}^{\pi}\{\frac {1}{v + 2k\pi} - \frac {1}{v + \pi + 2k\pi}\} dv$, do the algebra, and you'll see the integrand $ \leq \frac {1}{k^2}$. No integration necessary[/quote]
how you prove that $ \int ... < 1/k^2$ ??[/quote]

The expression in braces is $ \frac {\pi}{(v + 2k\pi)(v + \pi + 2k\pi)} \leq \frac{\pi}{(2k\pi)^2}$, so the integral $ \leq \frac{1}{4k^2}$.

[quote="mathmean"]

Problem is for $ k \leq \minus{} 1$,

$ \int_{0}^{\pi}\left|\left\{\frac {1}{|v \plus{} 2k\pi|} \minus{} \frac {1}{|v \plus{} \pi \plus{} 2k\pi|}\right\}\right| \ dv \equal{} \int_{0}^{\pi}\frac {1}{|v \plus{} 2k\pi|} \minus{} \frac {1}{|v \plus{} \pi \plus{} 2k\pi|} \ dv$
doesn't hold[/quote]

Well $ u_k$ is not even defined for $ k \equal{} \minus{} 1$ because of the second term. For $ k < \minus{} 1$, let's note $ |\frac {1}{|x|} \minus{} \frac {1}{|y|}| \equal{} |\frac {|x| \minus{} |y|}{|x\parallel{}y|}| \leq \frac {|y \minus{} x|}{|x\parallel{}y|}$. Apply this to the integrand and then ask yourself how small $ |x|$ and $ |y|$ can be here. Answer: when $ v \equal{} \pi$. This leads to $ |u_k| \approx \frac {1}{k^2}$. Again, no integration necessary.

[quote="mathmean"]

Problem is for $ k \leq \minus{} 1$,

$ \int_{0}^{\pi}\left|\left\{\frac {1}{|v \plus{} 2k\pi|} \minus{} \frac {1}{|v \plus{} \pi \plus{} 2k\pi|}\right\}\right| \ dv \equal{} \int_{0}^{\pi}\frac {1}{|v \plus{} 2k\pi|} \minus{} \frac {1}{|v \plus{} \pi \plus{} 2k\pi|} \ dv$
doesn't hold[/quote]

siply write the magnitude ($ |\cdot|$) on all expressions in my proof.

$ \left|\int_{0}^{\pi}\left|\left\{\frac {1}{|v \plus{} 2k\pi|} \minus{} \frac {1}{|v \plus{} \pi \plus{} 2k\pi|}\right\}\right| \ dv\right| \equal{} \left|\int_{0}^{\pi}\frac {1}{|v \plus{} 2k\pi|} \minus{} \frac {1}{|v \plus{} \pi \plus{} 2k\pi|} \ dv\right|\equal{}...$


calculus
integration
trigonometry
LaTeX
logarithms
function
real analysis

Could someone help me solve the integrals at the bottom of the post.
it shouldn't be so difficult, but I just couldn't figure out how to approach this.
there was a hint sugesting the use of trigonometric/hyperbolic identities.
and these are the only ones that I know of:

For $ \int \sqrt {a^2 \minus{} x^2}\,dx$ make $ x \equal{} a \sin t$ (alternatively, make $ t \equal{} \sqrt{a^2\minus{}x^2}$).

For $ \int (x^2 \minus{} a^2)^{ \minus{} 3/2}\,dx$ make $ x \equal{} a \sec t$ (alternatively, you can also make the substitution $ x \equal{} \frac{a}{\sqrt{1\minus{}t^2}}$).

am I doing it right? somehow I don't think so...

You forgot to also substitute dx. Considering $ x \equal{} a \sin t$ then $ \frac{dx}{dt} \equal{} a \cos t \equal{}> dx \equal{} a \cos t\, dt$, and so

$ P \equal{} \int \sqrt{a^2\minus{}x^2}\,dx \equal{} \int \sqrt{a^2\minus{}a^2 \sin^2t} \cdot a \cos t\, dt$.

thank you.

p.s. how do you add mathematical symbols in here?

Use Latex (look [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:About]here[/url]).

hello again,

would you mind showing me the full solution to second one?
I got some ugly results, and maybe I'm doing something wrong...
here's what I did:

$ P \equal{} \int \frac {1}{(x^2 \minus{} a^2)^{3/2}}\,dx \equal{} \int \frac {a \tan t \sec t}{(a^2 \sec^2t \minus{} a^2)^{3/2}}\,dt \equal{} \frac {1}{a^2} \int \frac {\tan t \sec t}{\tan^3 t}\,dt$

$ \equal{} \frac {1}{a^2} \int \frac {\sec t}{\tan^2t}\,dt \equal{} \frac {1}{a^2} \int \frac {\cos^2t}{\cos t \sin^2t}\,dt \equal{} \frac {1}{a^2} \int \cos t (\sin t)^{ \minus{} 2}\,dt$
$ \equal{}> P \equal{} \minus{} \frac {1}{a^2 \sin t} \plus{} C$.

[quote="Gilad.R"]Thanks for the replies.

Carcul, isn't that just what I did?
scorpius119, that's a nice solution, but I'm supposed to use trigonometric identities...

A question for both of you:
How did you know which substitutions to do?[/quote]

The substitution $ \frac{a}{2}\left( t\plus{}\frac{1}{t}\right)$ is very similar to the use of the hyperbolic identity $ \cosh^2 x\minus{}\sinh^2 x\equal{}1$, in fact, $ \frac{a}{2}\left( t\plus{}\frac{1}{t}\right)$ is actually equal to $ a\cosh (\ln t)$, and $ \frac{a}{2}\left( t\minus{}\frac{1}{t}\right) \equal{}a\sinh (\ln t)$. The major difference is that the exponential function appears in the hyperbolic substitution, which can be nice for simpler expressions but messy for complicated ones.

The key to using trig/hyperbolic identities to simplify an integral with a radical is to be familiar with the identities that involve squares. The following all equal 1:
\[ \cos^2 x\plus{}\sin^2 x\]
\[ \sec^2 x\minus{}\tan^2 x\]
\[ \cosh^2 x\minus{}\sinh^2 x\]
\[ \text{sech}^2 x\plus{}\tanh^2 x\]
(The second and fourth follow from multiplying the first by $ \sec^2 x$ and multiplying the third by $ \text{sech}^2 x$.)

So common trig substitutions are $ a\sin \theta$ when $ \sqrt{a^2\minus{}x^2}$ appears, $ a\tan \theta$ when $ \sqrt{a^2\plus{}x^2}$ appears, and $ a\sec \theta$ when $ \sqrt{x^2\minus{}a^2}$ appears. From the hyperbolic identities above, there are similar hyperbolic/algebraic substitutions.


Question :

Ten marbles are inserted in a box. A coin was flipped to decide the color of the marbles. For every heads, there was a white marble placed in. For every tails, a black marble. Suppose you draw out a marble, and observe its color. You then put it back in. You shake the box to mix up the marbles. You then draw again, etc. Suppose you draw 10 times, each time putting the marble back. What is the probability, to the nearest whole percent, that
all 10 marbles in the box are white?

do you mean the probability that the marbles that you draw are all the same color, because that is really different then all of the marbles in the box being white

[hide]if what symmetry said is what he meant, then it's $(\frac{1}{2})^{10}$ bcuz each $\frac{1}{2}$ is the probability of getting a heads, or white marble. $(\frac{1}{2})^{10}$ is the same as $\frac{1}{1024}$.

if what altheman said is what symmetry meant, then the probability that all he draws is white is...well again $(\frac{1}{2})^{10}$ bcuz there is an equal probability that all marbles in the box are white than all are black, so thye cancel out. same with 9white 1 black= 9 black 1 white, etc, until you get 5 white  5 black and there is $\frac{1}{2}$ chance that white will be picked out once so if he wants to pick it out ten times it is again $(\frac{1}{2})^{10}$ or $\frac{1}{1024}$. [/hide]

i am not sure that they cancel out, i will go through the case work when i have time(not that hard really), i.e. what is the prob of getting n whites, then choosing all white is (n/10)^10


I can't seem to figure this out:

Given a number (say 20), what is the maximal product of positive real numbers that add to that number? It is obvious that the maximum for positive integers is 2*3^6 and I can even find the maximum for integers with any integer (not just 20), but I don't know for real numbers. This is my work:

[hide]3 is the [i]best[/i] integer for this because each 1 that is added contributes the most to the product. Each is worth 1 when added, but 3^(1/3) when multiplied. Therefore, the question for reals becomes: what is the maximum value of x^(1/x)? The answer, of course, is [i]e[/i]. So we need to add as many [i]e[/i]s as we can. But if you add too many, the result may bring the product down! I want to know how many [i]e[/i]s you should involve and what other numbers to use when you're done.[/hide]

I think the answer is

$n= \lfloor \frac{m}{e}+\frac{1}{2} \rfloor$, and then set all the $n$ numbers equal.


trigonometry
geometry
geometric transformation
reflection
trig identities
Law of Cosines
geometry unsolved

hello:
Consider a triangle $ ABC$ such that: $ BC \equal{} a$ $ AB \equal{} c$  $ AC \equal{} b$  and $ G$ is the barycenter of this triangle.
1)Prove that:
$ \vec{GB}.\vec{GC} \equal{} \frac {1}{18}(b^2 \plus{} c^2 \minus{} 5a^2)$

$ \vec{BC}\equal{}\vec{GC}\minus{}\vec{GB}$
$ \equal{}>BC^2\equal{}GB^2\plus{}GC^2\minus{}2\vec{GB}\vec{GC}$
$ \equal{}>\vec{GB}\vec{GC}\equal{}\frac{1}{2}\left(GB^2\plus{}GC^2\minus{}BC^2\right)\equal{}\frac{1}{2}\left(\frac{4}{9}m_b^2\plus{}\frac{4}{9}m_c^2\minus{}a^2\right)$.
Note that $ m_b^2\equal{}\frac{c^2\plus{}a^2}{2}\minus{}\frac{b^2}{4}$ and $ m_c^2\equal{}\frac{a^2\plus{}b^2}{2}\minus{}\frac{c^2}{4}$, we will have it.

[quote="sami.d"][quote="ppt"]$ \vec{BC} \equal{} \vec{GC} \minus{} \vec{GB}$
$ \equal{} > BC^2 \equal{} GB^2 \plus{} GC^2 \minus{} 2\vec{GB}\vec{GC}$
$ \equal{} > \vec{GB}\vec{GC} \equal{} \frac {1}{2}\left(GB^2 \plus{} GC^2 \minus{} BC^2\right) \equal{} \frac {1}{2}\left(\frac {4}{9}m_b^2 \plus{} \frac {4}{9}m_c^2 \minus{} a^2\right)$.
Note that $ m_b^2 \equal{} \frac {c^2 \plus{} a^2}{2} \minus{} \frac {b^2}{4}$ and $ m_c^2 \equal{} \frac {a^2 \plus{} b^2}{2} \minus{} \frac {c^2}{4}$, we will have it.[/quote]
Helo
I do not understand why $ m_b^2 \equal{} \frac {c^2 \plus{} a^2}{2} \minus{} \frac {b^2}{4}$[/quote]
Here is the proof for why $ m_b^2 \equal{} \frac {c^2 \plus{} a^2}{2} \minus{} \frac {b^2}{4}$
We call the triangle $ ABC$ and we have median $ BE$.  Allow $ m_b \equal{} BE$, $ c \equal{} AB$, $ b \equal{} AC$, and $ a \equal{} BC$.  We apply the Law of Cosines on $ \triangle ABE$ to get that
\[ \frac {AB^2 \plus{} AE^2 \minus{} BE^2}{2(AB)(AE)} \equal{} \cos A\implies \frac {c^2 \plus{} \frac {b^2}{4} \minus{} m_b^2}{2(\frac {b}{2})(a)} \equal{} \cos A\]
Applying the Law of Cosines on $ \triangle ABC$, we get that\[ \frac {AB^2 \plus{} AC^2 \minus{} BC^2}{2AB(AC)} \equal{} \cos A\implies \frac {c^2 \plus{} b^2 \minus{} a^2}{2ab} \equal{} \cos A\]
Thus,\[ \frac {c^2 \plus{} b^2 \minus{} a^2}{2ab} \equal{} \cos C \equal{} \frac {c^2 \plus{} \frac {b^2}{4} \minus{} m_b^2}{ab}
\]\[ 2c^2 \plus{} \frac {b^2}{2} \minus{} 2m_b^2 \equal{} c^2 \plus{} b^2 \minus{} a^2
\]\[ 2m_b^2 \equal{} c^2 \minus{} \frac {b^2}{2} \plus{} a^2
\]\[ \boxed{m_b^2 \equal{} \frac {a^2 \plus{} c^2}{2} \minus{} \frac {b^2}{4}}
\]
I hope this helps  :)

This is another proof for $ m_b^2 \equal{} \frac {c^2 \plus{} a^2}{2} \minus{} \frac {b^2}{4}$:
Let M be midpoint of AC, we have:
$ \left(\vec{BA} \plus{} \vec{BC}\right)^2 \equal{} BA^2 \plus{} BC^2 \plus{} 2\vec{BA}.\vec{BC}$
<=>$ \left(2\vec{BM}\right)^2 \equal{} BA^2 \plus{} BC^2 \plus{} 2\left(\vec{BM} \plus{} \vec{MA}\right).\left(\vec{BM} \plus{} \vec{MC}\right)$
<=>$ \left(2\vec{BM}\right)^2 \equal{} BA^2 \plus{} BC^2 \plus{} 2\left(\vec{BM} \plus{} \vec{MA}\right).\left(\vec{BM} \minus{} \vec{MA}\right)$
<=>$ 4BM^2 \equal{} BA^2 \plus{} BC^2 \plus{} 2(BM^2 \minus{} MA^2)$<=>$ 4BM^2 \equal{} BA^2 \plus{} BC^2 \plus{} 2BM^2 \minus{} 2MA^2$
<=>$ 2BM^2 \equal{} BA^2 \plus{} BC^2 \minus{} \frac {AC^2}{2}$<=>$ BM^2 \equal{} \frac {BA^2 \plus{} BC^2}{2} \minus{} \frac {AC^2}{4}$
<=> $ m_b^2 \equal{} \frac {c^2 \plus{} a^2}{2} \minus{} \frac {b^2}{4}$

[color=darkblue][b][u]An another proof of the median theorem :[/u][/b] $ 4m_a^2 \equal{} 2(b^2 \plus{} c^2) \minus{} a^2$ .

Denote the reflection $ N$ of the vertex $ A$ w.r.t. the middlepoint $ M$ of the side $ [BC]$.

Observe that the quadrilateral $ ABNC$ is a parallelogram. Hence

$ AB \equal{} c$ , $ BN \equal{} b$ , $ AN \equal{} 2m_a$ and $ m(\widehat {ABN}) \equal{} 180^{\circ} \minus{} A$ .

Apply the [b]generalized Phytagoras' theorem[/b] to the side $ [AN]$ in the triangle $ ABN$ :

$ AN^2 \equal{} BN^2 \plus{} BA^2 \minus{} 2\cdot BN\cdot BA\cdot\cos\widehat {ABN}$ , i.e. $ \boxed {4m_a^2 \equal{} b^2 \plus{} c^2 \plus{} 2bc\cdot\cos A}$ .

But $ a^2 \equal{} b^2 \plus{} c^2 \minus{} 2bc\cdot\cos A$. In conclusion, $ 4m_a^2 \plus{} a^2 \equal{} 2(b^2 \plus{} c^2)$ .[/color]


$n=2k+1$. $n\geq13$.
There is a group of succesive numbers from $1$ to $n$.
We choose any sub-group  of $k$ numbers - group $A$.
Prove that in group $A$ exist $k-2$ numbers such that the sum of them is divisible by $n$.

Is this group's binary operator defined by modular addition?

We can chose pairs $\{a_{1},a_{n-1}\},\{a_{2},a_{n-2}\},...$ and $a_{0}$, were $a_{i}=i(mod \ n)$. We have $C_{k}^{[k/2]}$ numbers of A with summ divisible n (if k is odd we add to [k/2] pairs one element $a_{0}$). If $n\ge 13$ $C_{k}^{[k/2]}>n>k-2.$


function
modular arithmetic
algebra proposed
algebra

Find the number of  functions satisfying :
1)
\[ f: \mathbb N \to \mathbb N
\]

\[ f(f(n)) \equal{} n \plus{} 2k
\]
where $ k\in N$
2)

\[ f: \mathbb N \to \mathbb N
\]

\[ f(f(n)) \equal{} 4n
\]
P/s::o . A right result appears  only 30 minutes since I posted. I see that it is too easy for you. How about b)

[quote="Allnames"]Find the number of  functions satisfying :
1)
\[ f: \mathbb N \to \mathbb N
\]

\[ f(f(n)) \equal{} n \plus{} 2k
\]
where $ k\in N$
[/quote]

Hello Allnames. Thanks for dedicating this problem :)

As in other post about $ f(f(n))\equal{}pn$, we quickly see :
- that $ f(n)$ is injective
- that $ f(n\plus{}2k)\equal{}f(n)\plus{}2k$ and so that knowledge over $ [1,2k]$ is enough.
- that $ [1,2k]$ may be split in two equipotent subsets (each of cardinal $ k$) $ A\equal{}\{x\in[1,2k]$ such that $ f(x)\in[1,2k]\}$ and $ B\equal{}\{x\in[1,2k]$ such that $ f(x)>2k\}$

And the number of solution  is the number of $ (A,B,h(x))$ where $ h(x)$ is a bijection from $ A\to B$

We have $ \frac{(2k)!}{(k!)^2}$ different couples $ (A,B)$ and, for each couple, $ k!$ bijection from $ A\to B$

And so $ \frac{(2k)!}{k!}$ functions from $ \mathbb N\to\mathbb N$ such that $ f(f(n))\equal{}n\plus{}2k$

IMHO

[quote="Allnames"]Find the number of  functions satisfying :
\[ f: \mathbb N \to \mathbb N
\]

\[ f(f(n)) \equal{} 4n
\]
[/quote]

And here, we have infinitely many solutions.

Example :
Let $ u$ an odd integer. Let define $ f(x)$ such that :

$ f(1) \equal{} 2u$
$ f(u) \equal{} 2$
$ f(n) \equal{} 2n$ for any odd $ n\neq u$ and $ n\neq 1$
$ f(2) \equal{} 4u$
$ f(2u) \equal{} 4$
$ f(2n) \equal{} 4n$ for any odd $ n\neq u$ and $ n\neq 1$

$ f(4^pn) \equal{} 4^pf(n)$  for any $ n\neq 0\pmod 4$

Each $ u$ defines a different solution, so there is infinitely many such solutions.


calculus
derivative
function
trigonometry
logarithms
calculus computations

1) Find the derivative of the function $ f(x)\equal{}(\cos x)^{\sin x}$.

2) Find the derivative of the function $ f(x)\equal{}x^{\sin x}$.

3) Use implicit differentiation to find $ \frac{dy}{dx}$ if $ x^{y}\equal{}y^{x}$.

hello, writing your first expression in the form
$ f(x)\equal{}e^{\sin(x)\ln(\cos(x))}$ we get by the chain and the product rule
$ f'(x)\equal{}e^{\sin(x)\ln(\cos(x))}\cdot\left(\cos(x)\ln(\cos(x))\plus{}\frac{\sin(x)}{\cos(x)}\cdot(\minus{}\sin(x))\right)$
Sonnhard.

hello, rewriting your second equation in the form like above
we have
$ f(x)\equal{}e^{\sin(x)\ln(x)}$
getting the first derivative as
$ f'(x)\equal{}e^{\sin(x)\ln(x)}\cdot\left(\cos(x)\ln(x)\plus{}\frac{\sin(x)}{x}\right)$
Sonnhard.

hello, for 3) rewriting your equation in the form
$ y\ln(x)\equal{}x\ln(y)$
differentiating both sides with respect to $ x$ we get
$ y'\ln(x)\plus{}\frac{y}{x}\equal{}\ln(y)\plus{}\frac{x}{y}\cdot y'$
$ \frac{y}{x}\minus{}\ln(y)\equal{}y'\left(\frac{x}{y}\minus{}\ln(x)\right)$
finally we have
$ y'\equal{}\frac{y(y\minus{}x\ln(y))}{x(x\minus{}y\ln(x))}$
additionally it must be $ x>0,y>0$ and $ x\ne y\ln(x)$
Sonnhard.


inequalities
calculus
geometry
3D geometry
function

Given positive reals $x,y,z$, prove that

$((x-1)^2+11)((y-1)^2+11)((z-1)^2+11) > 8(xyz+27)^{\frac{3}{2}}$

EDIT: Actually, I think it's strict...

Use holders, we need now $\sqrt[3]{\prod (x+3)}  + \sqrt[3]{\prod (x^2 - 3x + 9)} \geq 2\sqrt{xyz + 27}$

Amgm then it is $\sqrt[3]{\prod(x+3)(x^2 - 3x + 9)} \geq xyz + 27$

which is $\sqrt[3]{\prod (x^3 + 27)} \geq xyz + 27$

obviously true

strict because we can check equalities

Nice, that's pretty much it.

$\prod{((x-1)^2+11)} = \prod{((x+3)+(x^2-3x+9))} \ge 8\prod{(\sqrt{(x+3)(x^2-3x+9)})} = 8\prod{(\sqrt{x^3+27})} \ge 8(xyz+37)^\frac{3}{2}$

by AM-GM and Holder. And since $x+3 \not = x^2-3x+9$ we have a strict inequality on the AM-GM.

does holder require calculus... because I thoughti could use CS here, but it ended up being of third power... :(
basically what i did was do AM-GM for each term on the LHS, then multiply together... then I boiled everything down to:

$(x^3+27)(y^3+27)(z^3+27) \ge (xyz+27)^3$
$(x^3+3^3)(y^3+3^3)(z^3+3^3) \ge (x\cdot y\cdot z+ 3\cdot3\cdot3)^3$
which reminded me of CS, but there is not such CS for power of 3.

BTW, I knew the hint before paladin actually posted... but for the above reason, I could not post my partial solution until someone did theirs.

[quote="paladin8"]No, Holder doesn't need calculus:

The two-term one is $\left(\sum{a_i^{\frac{1}{p}}}\right)^p\left(\sum{b_i^{\frac{1}{q}}}\right)^q \ge \sum{a_ib_i}$ where $p+q=1$, but it can be extended to as many terms as you want. Note that when $p=q=\frac{1}{2}$ it is Cauchy.[/quote]

so then could you break down Singular's solution? I dont see the holders in his solution.

So he uses it in his first step (after cube-rooting both sides):

If you let $a_1 = x+3$, $a_2 = x^2-3x+9$ and define $b_1,b_2,c_1,c_2$ similarly for $y$ and $z$, we have

$\sqrt[3]{(a_1+a_2)(b_1+b_2)(c_1+c_2)} \ge \sqrt[3]{a_1b_1c_1}+\sqrt[3]{a_2b_2c_2} = \sqrt[3]{\prod (x+3)} + \sqrt[3]{\prod (x^2 - 3x + 9)}$

[quote="blahblahblah"]Inspired by APMO 2005?[/quote]

Umm, I think the AM-GM to get the sum of 2 cubes might have come from that? Other than that, it was mine (though that is a big part of it I guess). I didn't refer to it before writing the question, but I probably unconsciously remembered it from that.

Well, the two cubes part is the only hard part of the problem; you don't even need to know holder (AM-GM and multiplying out suffices)

I think Holder can be made from several Cauchy's... don't quote me on that.

I won't, because it's wrong.

Holder is more generally a statement about convex functions, whereas Cauchy follows from some basic algebra.  Plus you can get non-rational exponents in Holder - it would be hard to get those from repeated applications of Cauchy.  I'll post a proof of Holder when I get to my office.

A proof of holder

Set $1/q+1/p=1$ and $\sum_i a^p=\sum_i b^q=1$

By the inequality $ab\leq a^p/p+b^q/q$, we get

$\sum_i ab\leq \frac {1}{p}\sum_i a^p+\frac {1}{q}\sum_i b^q=1/p+1/q=1=\sum_i a^p\cdot \sum_i b^q$

Exponentiating gives the desired result from here.

The problem is the middle inequality - it's easy to show for $p=q=2$ (it is just AM-GM) but not so much in other cases (especially with irrationals).  Convexity takes care of it easily though.


Hello, I'm new to these boards.  I live in Texas and am going to the Texas State Competition. (March 27th).  Could any of you tell me how to prepare and what score I need to get to nationals?  Also, will the Countdown round be official this year in Texas?

you need to be toP 4 from what i know and i dont think countdown counts at all, it's a separate trophy (in calif) prepare by doing old problems


limit
ratio
calculus
function
inequalities
geometric sequence
geometric series

Suppose $p_n$ for $n=0.. \infty$ is a sequence that converges to $p$ with $p_n \neq p$ for all $n$. If positive constant $\lambda$ and $\alpha$ exists with
\[ \lim_{n \rightarrow \infty} \frac{| p_{n+1}-p |}{|p_n - p|^{\alpha}} = \lambda  \]
then the sequaence $p_n$ [b]converges to $p$ with order $\alpha$. with asymptotic error constant $\lambda$ [/b]

OK I know this is just a defenition.. :roll:  But I've to explain this with "human language" to someone who don't have a strong mathematical background.
I'm just thinking to explain this with asymptotic analysis, but it will be more complicated for her. :ninja:
She just understand the "defenition" of limit.

Reply this as soon as posibble please
Thanks in advance.

We have an iterative process that is producing succesive estimates $p_n$ for a desired result $p.$ That makes $|p_n-p|$ the error. If we're asking what a single step got us, $|p_{n+1}-p|$ is the "new error", while $|p_n-p|$ is the "old error."

A process is first order (or "linear") is the new error is (at worst) proportional to the old error. If that's so, the constant of proportionality had better be less than one, or we're not making any progress. A good example of a first order process would be the successive bisection method for finding a root of an equation. This method has the advantage of being "foolproof" (in the sense that an abstract infinite version of it is a perfectly good proof of the Intermediate Value Theorem), but how fast does it work? We have to adjust our meaning of error slightly, to make it the width of the interval in which we've confined our answer. It's clear then that the new error is half of the old error. If we start by knowing the answer to the nearest integer, but want an answer good to twelve decimal digits, we know it's going to take us 40 steps. ($2^{40}\approx 10^{12}.$)

This rate of convergence, first order or "linear" convergence, is the rate of convergence of a geometric sequence or series. The ratio does matter (the $\lambda$ of your definition above); the closer to zero, the faster, the closer to one, the slower. In a Calculus II class, converging like a geometric series is considered fast, but in a numerical analysis class, it's considered slow.

Newton's method of root finding is a second order process. That means that the new error is proportional to the square of the old error. If the old error is too large, then the square makes things worse, and we could be wandering around aimlessly, but if the old error is small, the new error can be much smaller. Once we get close, we zip in to the answer quickly.

Here's an idea to illustrate this. I'm going to approximate $p=1$ by two processes. The first will be a first order process: functional iteration of the function $f(x)=\sqrt{x}.$ The second will be a second order process: Newton's method applied to $g(x)=x^2-1.$ What I want you to do is to draw a line marking the edge of the correct digits (easy to spot becuase the correct digits are zeros). Look at it as if it were the graph of a function; look at its shape.

The first order method:[code]1.400000000000
1.183215956620
1.087757305937
1.042956042188
1.021252193236
1.010570231718
1.005271222963
1.002632147382
1.001315208804
1.000657388322
1.000328640159
1.000164306581
1.000082149916
1.000041074115
1.000020536846
1.000010268370
1.000005134172
1.000002567083
1.000001283541
1.000000641770
1.000000320885
1.000000160442
1.000000080221
1.000000040111
1.000000020055
1.000000010028
1.000000005014
1.000000002507
1.000000001253
1.000000000627
1.000000000313
1.000000000157
1.000000000078
1.000000000039
1.000000000020
1.000000000010
1.000000000005
1.000000000002
1.000000000001
[/code]
Now, the second order method:[code]3.000000000000000
1.666666666666670
1.133333333333330
1.007843137254900
1.000030518043790
1.000000000465660
1.000000000000000[/code]

[quote="Beginner"]Suppose $(p_n)$  converges to $p$ ...  If positive constant $\lambda$ and $\alpha$ exists with
\[ (*)\; \; \; \; \; \lim_{n \rightarrow \infty} \frac{| p_{n+1}-p |}{|p_n - p|^{\alpha}} = \lambda  \]
then the sequaence $p_n$ [b]converges to $p$ with order $\alpha$. with asymptotic error constant $\lambda$ [/b]
[/quote]
Some problems regarfing the order of convergence , perhaps of interest from didactical point of view, are :
=================
[b](P.1) [/b] For  $A>0 \; ,\; \alpha>1 \; ,\; I=\left[\sqrt{A},\frac{A+1}{2}\right]$  let  $F_{\alpha}$ $: I\to {\mathbb R}$ be defined by \[ \begin{array}{|c|} \hline \\ \displaystyle F_{\alpha}(x)=\displaystyle \sqrt{A}\frac{(x+\sqrt{A})^{\alpha}+(x-\sqrt{A})^{\alpha}}{(x+\sqrt{A})^{\alpha}-(x-\sqrt{A})^{\alpha}}\\ \\ \hline \end{array} \; \; ,\; \; x\in I\; . \] Prove that:
1) $F_{\alpha}(\sqrt{A})=\sqrt{A}$
2) $F_{\alpha}(I)\subset I$
3) $F_{\alpha}\left(F_{\gamma}(x)\right))= F_{\gamma}\left(F_{\alpha}(x)\right))=F_{\alpha\gamma}(x)\; ,\; \forall x\in I\; .$
======================
[b](P.2.)[/b]Consider  the sequences $\left(p_n\right)_{n=0}^{\infty}\; ,\; \left(q_n\right)_{n=0}^{\infty}$ defined by $p_0=q_0=\displaystyle \frac{A+1}{2}$ and
\[ \begin{array}{|lcl|} \hline &&\\ p_{n+1}&=& F_2(p_n)=\displaystyle \frac{1}{2}\left(p_n+\frac{A}{p_n}\right)\\ &&\\ q_{n+1}&=& F_5(q_n)=\displaystyle q_n\cdot\frac{q_n^4+10Aq_n^2+5A^2}{5q_n^4+10Aq_n^2+A^2}\\ &&\\ \hline \end{array}\; \; \; ,\; \; \; \forall n\in {\mathbb N}.  \]
Prove that :
a) $\lim\limits_{n\to \infty} p_n=\sqrt{A}$ with order $\alpha=$[b]2 [/b]  of convergence.
b) $\left(q_n\right)$  converges to $\sqrt{A}$, with order [b]5[/b] of convergence.   More exactly, prove/ or disprove that
\[ \lim\limits_{n\to \infty} \frac{|p_{n+1}-\sqrt{A}|}{\left|p_n-\sqrt{A}\right|^{2}}=\frac{1}{2\sqrt{A}}\; \; ,\; \; \; \lim\limits_{n\to \infty} \frac{|q_{n+1}-\sqrt{A}|}{\left|q_n-\sqrt{A}\right|^{5}}=\frac{1}{16A^2}\; .  \]
================
[b](P.3.)[/b] In the following let $\begin{array}{|c|}\hline A=2\\ \hline \end{array}\;$ and  $P_0=q_0=\frac{3}{2}\; .$
i) Using a PC , find $p_k$ and $q_k$ for $k\in \{1,2,3,4,5\}.$
ii) Show that

\[ \begin{array}{lcl} \displaystyle p_n&=&\displaystyle F_{{}_{2^n}}(\beta)=\displaystyle \sqrt{2}\frac{1+\beta^{2^n}}{1-\beta^{2^n}}\\ &&\\ \displaystyle q_n&=&\displaystyle F_{{}_{5^n}}(\beta)=\displaystyle \sqrt{2}\frac{1+\beta^{5^n}}{1-\beta^{5^n}} \end{array}\; \; , \; \; \; \beta: =\frac{3-2\sqrt{2}}{3+2\sqrt{2}}=\left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)^2< \frac{1}{5^2}\; .  \]
iii) Prove the inequalities
\[ \begin{array}{lcl} 0<p_5-\sqrt{2}&<& \displaystyle \frac{1}{{10}^{42}} \\ &&\\ 0<q_5-\sqrt{2}&<& \displaystyle \frac{1}{{10}^{10416}} \\ \end{array}\; .  \]
This means that $p_5 \approx \sqrt{2}$ , both numbers having in  common at least [b]41[/b] decimals. On the other hand,   $q_5$ approximate $\sqrt{2}$ with at least [b]10415[/b] exact decimals !
Note that $\sqrt{2}=1.4142\; 13562\; 37309\; 50488...$

[b]REMARK:[/b]
 A better [b]definition of order of convergence:[/b] Let $(p_n)_{n=0}^{\infty}$ such that $\lim\limits_{n\to \infty} p_n = p\; .$  If   there exists $m,M,\varepsilon ,\; \alpha ,$ with   $0<m<M \; ,\; \varepsilon \in (0,1) , \; \alpha> 1$ and such   that
\[ (**)\; \; \; \; \; \; \; \begin{array}{|c|} \hline \\ m\cdot \varepsilon^{\displaystyle {}^{\alpha^n}} \le \left|p_n-p\right|\le M\cdot \varepsilon^{\displaystyle {}^{\alpha^n}} \\ \\ \hline \end{array}\; , \; \; \; \; \forall n\in {\mathbb N}^*\; ,  \]
[b]then we  say that  $\left(p_n\right)_{n=0}^{\infty}$ has order of convergence $\alpha \; .$ [/b]

[b](P.4.)[/b] Find a connection between (*) and (**)  : which is stronger ?
[b](P.5.)[/b] Give a sequence which approximate $\sqrt{A}$ having order of convergence $=$ [b]7[/b] .
[b](P.6.)[/b] The sequence generated by $p_{n+1}=F_2(p_n)$ is a particular case of Newton method (iterations for $\sqrt{A}$ ). However the sequence is known under the name of [b]Heron[/b]'s  method  . Other assert that the method was known in ...[b]Babyloon[/b] , that is   $(p_n)$ is sometimes the sequence $(p_n)$  defines  [b]the  Babylonian method[/b] .
Question: Find a justification  how was possible in Babyloon to discover this  method !


Indeed not difficult!
$\sum_{cyc}\frac{x^{2}-y^{2}}{x^{2}-xy+y^{2}}=\frac{(x-y)(y-z)(x-z)(\sum_{sym}x^{2}y-xyz)}{\prod_{cyc}(x^{2}-xy+y^{2})}$
and everything follows...

My solution is the same. Thank you Albanian Eagle !


Mine we probably be a Chopin song, I like them all, but the list would be toooo long to put.
I also am a big fan of the 3rd movement of the Moonlight Sonata by Beethoven.

Well I don't know any [i]songs[/i] for piano as songs are generally sung. Some of my favorite [i]pieces[/i] are the Chopin etudes.

Well, yeah, I guess more specifically known as favorite pieces.

I love all the pieces from the Romantic period. They were so soothing...

I like Chopin etudes Opus 10 No.4 and No.5. I also like the Chopin Trio. I played it, and considering that it was the only trio Chopin wrote, I like it very much. Bach English Suite No.3 in G Minor is also one of my favorites.

I really liked playing "Night Thoughts" by Copland, very fun piece.  Sort of obscure, very dissonant.
Some of the Chopin Nocturnes and Preludes are really beautiful.
Brahms has written some good short pieces.
I also liked 6 Hungarian Dances from Mikrokosmos-Bela Bartok.  Haven't played them, heard my piano teacher perform them in concert though.
Also, some Rachmaninoff preludes are cool.

Sorry, this list is sort of all over the place...

I just listened to the Blue Danube;
it's so wonderful...
my new favorite :)

I really enjoy pieces that go from the major key into the relative minor key in the middle. Sad, but beautiful pieces. I can't really put my finger on one at the moment....

@ cottontail: Listen to Moonlight Rhapsody by Melody Bober. It is in a minor key but goes to a major key, then back to minor. Kind of the opposite of what you meant but I think it might be what you are looking for.

I love interesting peices. STACATTO! FORZANDO! very softly very softly BANG!!!!!
That's me for you.
:)

haha...im not a classical girl...i like FEARLESS by taylor swift...my piano teacher cant stand it though...i teach myself pop music


geometry
geometric transformation
rotation
vector
linear algebra
matrix
trigonometry

If we choose $ \{1,i\}$ as the basis of $ \mathbb{R}^2$

then the Transform: $ \cal{A}$ : $ z \rightarrow e^{i\alpha}z$

can be written as $ \begin{bmatrix}cos\alpha&\minus{}sin\alpha\\sin\alpha&cos\alpha\end{bmatrix}$

Ok, we usually call this transformation as "the rotation of $ \mathbb{R}^2$ of angle $ \alpha$"

Now I want to know "the rotation of $ \mathbb{R}^n$ of angle $ \alpha$ is ?"

and its corresponding matrix is? :roll:

It takes far more than one parameter. One angle isn't enough to specify a rotation in more than two dimensions.

What do rotation matrices in general look like? They're the special orthogonal group of matrices $ Q$ satisfying $ Q^TQ\equal{}I$ and $ \det Q\equal{}1$.


Care este numarul minim de rasturnari (momentul in care incepe sa curga nisipul in clepsidra) necesare pentru a masura trecerea a $ 9$ minute folosind doua clepsidre cu nisip in care scurgerea dintr-un compartiment in altul masoara $ 7$ min respectiv $ 4$ min?

$ \boxed A \; 3 \qquad \boxed B \;  4\qquad  \boxed C \; 5 \qquad \boxed D \; 6 \qquad \boxed E \; 7$

Intoarcem amandoua clepsidrele.Se va termina primul cu 4 minute(il notam cu (2)) deci mai raman 3 minute pentru (1).Il lasam pe (1)sa se scurga in totalitate si simultan pe (2) il inversam consumanduse inca 3 minute (deci sau scurs 7 si au ramas 2).A ramas 1 minut pentru (2) si in timp ce
2 se scurge il inversa iar pe 1 (trece inca un minut),si il inversa  din nou pe 1 si se mai scurge 1 minut
deci avem 4 operatii.9=4+3+1+1.


inequalities
linear algebra
matrix
vector

Hi, I need some help of proving the following inequalities, I guess I might need to use the Cauchy-Schwarz Inequalty or might not, please help, Thanks in advance.

$1.$ $\Vert A \Vert_1 \leq \sqrt{n} \Vert A \Vert _2 \leq n \Vert A \Vert _1$

$2.$ $\Vert A \Vert_\infty \leq \sqrt{n} \Vert A \Vert _2 \leq n \Vert A \Vert _\infty$

As I said, I dont know if I need to use the Cauchy-Schwarz, but I got this Lemma if it helps...
Lemma: $\Vert x \Vert_\infty \leq \Vert x \Vert_2 \leq \Vert x \Vert _1 \leq \sqrt{n} \Vert x \Vert _2 \leq n \Vert x \Vert _\infty$

Are these matrix norms?  If so, you use Cauchy-Schwarz to prove that $\|A\|_1\leq \sqrt{n}\|A\|_2$.  The other estimates can be done far more crudely.

if $A \in R^{n \times n}$, how will you apply the Cauchy-Schwarz? and what about the infinite one, does it work in a similar way?

It should look something like
\[ \left(\sum_{i,j}|a_{ij}|\right)^2\leq \left(\sum_{i,j}|1|\right)\left(\sum_{i,j}|a_{ij}|^2\right)  \]
Note that
\[ \sum_{i,j}|1|=n^2  \]
because $i,j$ run over $1,2,\ldots,n$.  So for matrix norms you actually want the inequality
\[ \|A\|_1\leq n\|A\|_2  \]
if I'm not mistaken.  Your inequalities apply to vectors in $\mathbb{R}^n$.  The infinity norm is even easier.  Note that in a sum of nonnegative numbers $a_1,a_2,\ldots,a_n$, we have the trivial inequalities
\[ \max_i a_i\leq \sum_i a_i\leq n\max_i a_i  \]

Sorry for keep asking since I want to know how these inequalities come from and then be able to read the following sections, please help.

I guess I am still confused some points...
I see the first inequality is used by Cauchy-Schwarz, but how did u get the $\Vert A \Vert_1 \leq n^2 \Vert A \Vert _2$ then to provide the inequality of $\Vert A \Vert_1 \leq \sqrt{n} \Vert A \Vert _2$ and do u do the similar way to show that $\sqrt{n} \Vert A \Vert _2 \leq n \Vert A \Vert_1$

then to the $\infty$ one, could u say more about it...?

Sorry for asking that much... I'm just confused :blush:

$\Vert A \Vert_1= max \frac{\Vert Ax \Vert_1}{\Vert x \Vert_1}$
$\Vert A \Vert_2= max \frac{\Vert Ax \Vert_2}{\Vert x \Vert_2}$and along with the Lemma $\Vert x \Vert_\infty \leq \Vert x \Vert_2 \leq \Vert x \Vert _1 \leq \sqrt{n} \Vert x \Vert _2 \leq n \Vert x \Vert _\infty$
how will you show that $\Vert A \Vert_1 \leq \sqrt{n} \Vert A \Vert _2 \leq n \Vert A \Vert _1$

And what is defintion of $\Vert A \Vert_\infty$

Prove the following Inequalities:
$1. \Vert A \Vert_1 \leq \sqrt{n} \Vert A \Vert _2 \leq n \Vert A \Vert _1$
$2. \Vert A \Vert_\infty \leq \sqrt{n} \Vert A \Vert _2 \leq n \Vert A \Vert _\infty$

Lemma $\Vert x \Vert_\infty \leq \Vert x \Vert_2 \leq \Vert x \Vert _1 \leq \sqrt{n} \Vert x \Vert _2 \leq n \Vert x \Vert _\infty$

I think I got the ideas, but dont know if it's right, could anyone check?

$\Vert A \Vert_1= max \frac{\Vert Ax \Vert_1}{\Vert x \Vert_1}$

$\Vert A \Vert_2= max \frac{\Vert Ax \Vert_2}{\Vert x \Vert_2}$

and from the Lemma $\Vert x \Vert_\infty \leq \Vert x \Vert_2 \leq \Vert x \Vert _1 \leq \sqrt{n} \Vert x \Vert _2 \leq n \Vert x \Vert _\infty$
know that
$\Vert x \Vert_\infty \leq \Vert x \Vert_1\leq \sqrt{n} \Vert x \Vert_2$ then
$\Vert Ax \Vert_\infty \leq \Vert Ax \Vert_1 \leq \sqrt{n} \Vert Ax \Vert_2$ then
from Lemma again that $\frac{1}{\Vert x \Vert_1} \leq \frac{1}{\Vert x \Vert_2} \leq \frac{1}{\Vert x \Vert _\infty}$
then $\frac{\Vert Ax \Vert_\infty}{\Vert x \Vert_\infty} \leq \frac{\Vert Ax \Vert_1}{\Vert x \Vert_1}\leq \frac{\sqrt{n}\Vert Ax \Vert _2}{\Vert Ax \Vert _2}$ and to get the inequality of $\Vert A \Vert_\infty \leq \sqrt{n} \Vert A \Vert _2$

Is this right? I hope so!!
I am still confused of the inequality of $\sqrt{n} \Vert A \Vert _2 \leq n \Vert A \Vert _1$, could anyone tell me what to start with?


Let $ a,b,c,d$ positive real numbers.
a) Show that $ a^6\plus{}b^6\plus{}c^6\plus{}d^6\minus{}6abcd \ge \minus{}2$
b) For which positive integers $ k$ does an inequality of the form $ a^k\plus{}b^k\plus{}c^k\plus{}d^k\minus{}kabcd  \ge M_k$ exist? Determine the largest integer $ M_k$ for every $ k$ and the cases of equality.

[quote="FelixD"]Let $ a,b,c,d$ positive real numbers.
a) Show that $ a^6 \plus{} b^6 \plus{} c^6 \plus{} d^6 \minus{} 6abcd \ge \minus{} 2$
b) For which positive integers $ k$ does an inequality of the form $ a^k \plus{} b^k \plus{} c^k \plus{} d^k \minus{} kabcd \ge M_k$ exist? Determine the largest integer $ M_k$ for every $ k$ and the cases of equality.[/quote]
a) $ a^6 \plus{} b^6 \plus{} c^6 \plus{} d^6 \plus{} 1 \plus{} 1 \ge 6 \cdot \sqrt [6]{a^6 \cdot b^6 \cdot c^6 \cdot d^6 \cdot 1 \cdot 1} \equal{} 6abcd$ and the result follows.
b) Obviously $ k \ge 4$ (else let $ a,b,c,d \to \infty$). $ a^k \plus{} b^k \plus{} c^k \plus{} d^k \plus{} (k \minus{} 4) \cdot 1 \ge k \cdot \sqrt [k]{a^kb^kc^kd^k1^{k \minus{} 4}} \equal{} kabcd$. So $ M_k \equal{} 4 \minus{} k$. Equality iff $ a \equal{} b \equal{} c \equal{} d \equal{} 1$. (unless $ k\equal{}4$ where $ a\equal{}b\equal{}c\equal{}d$ suffices)


geometry
circumcircle
IMO Shortlist
Polars
Brocard
projective geometry
power of a point

Given a cyclic quadrilateral $ABCD$, let $M$ be the midpoint of the side $CD$, and let $N$ be a point on the circumcircle of triangle $ABM$. Assume that the point $N$ is different from the point $M$ and satisfies $\frac{AN}{BN}=\frac{AM}{BM}$. Prove that the points $E$, $F$, $N$ are collinear, where $E=AC\cap BD$ and $F=BC\cap DA$.

[i]Proposed by Dusan Dukic, Serbia and Montenegro[/i]

I like the problem very much :).

Let $P$ be the second point of intersection between $CD$ and the circle $(ABM)$, and let $G=AB\cap CD$. A very simple computation, based on the fact that $GD\cdot GC=GA\cdot GB=GM\cdot GP$ and $M$ is the midpoint of $CD$ will show that $P$ is, in fact, the harmonic conjugate of $C,D$ art $G$, so it belongs to $EF$.

$ANBM$ is a harmonic quadrilateral, so $(PA,PB;PN,PM)=-1$, and since $PM=PG$, so $(PA,PB;PE,PM)=-1$ as well, we get $PE=PN$, as desired.

A very cute problem !
  Here is another solution (hope it isn't the same as yours, Darij  ;) )
  Let $G$ be the intersection of $AB$ and $CD$. Let $J$ be the midpoint of $AB$. Let $U$ be the second intersection of $LN$ and the circumcircle $ABM$.
  Notice that $\measuredangle{BMN} = \measuredangle{JMA}$ since $ANBM$ is a harmonic quadrilateral. Therefore, $\measuredangle{MJG} = \measuredangle{MJB} = \measuredangle{MAN}$. Since $\measuredangle{OMG} = \measuredangle{OJG} = 90^0$, the quadrilateral $OJGM$ is cyclic. Hence, $\measuredangle{MJG} = \measuredangle{MOG}$. This implies $\measuredangle{MOG} = \measuredangle{MAN}= \measuredangle{MUG}$. Thus, $OUGM$ is cyclic, or $OU$ is perpendicular to $NG$.
   Now, the inversion with center $G$, constant $GA.GB$ maps the two circles into themselves, and maps the circle $(S)$ with diameter $GN$ into the line $OU$. This imples that $(S)$ is orthogonal to $(O)$, or $N$ lies on the polar of $G$ wrt $(O)$, which is $EF$.
  PS Why is Victor Thebault related to this problem ?

can someone explain me what is harmonic quadrilateral? (appeared in Grobber's solution)  :?

[quote="Megus"][quote="treegoner"]Let me explain like this :
   I give you 4 points $A, B, C, D$ on a circle. An arbitrary point S lies on that circle. Then $(SA, SB,SC, SD)$ is a constant.
  If $(SA, SB,SC, SD) = -1$, then $ABCD$ is a harmonic quadrilateral.[/quote]

Thanks Treegoner but I still have no idea what $(SA, SB,SC, SD)$ could mean?[/quote]
that means harmonic ratio.

Are you sure it is harmonic ratio? Harmonic ratio says about four collinear points, and here we are regarding four concurrent lines  :? Maybe I'm wrong? So I'll be thankful if someone actually explain this term to me - not just name but also properties - a link would be enough. Thanks in advance

Thanks Grobber very much - that's what I needed. Sorry for askin silly questions  :oops:

Your solutions are both very nice. There is really no need in posting mine, since it is just unreasonably long.

A few technical comments to Treegoner's solution:

[quote="treegoner"]  Let $G$ be the intersection of $AB$ and $CD$. Let $J$ be the midpoint of $AB$. Let $U$ be the second intersection of $LN$ and the circumcircle $ABM$.[/quote]

I guess this should mean: "Let U be the second intersection of GN and the circumcircle of ABM."

[quote="treegoner"]  Notice that $\measuredangle{BMN}= \measuredangle{JMA}$ since $ANBM$ is a harmonic quadrilateral. Therefore, $\measuredangle{MJG}= \measuredangle{MJB}= \measuredangle{MAN}$. Since $\measuredangle{OMG}= \measuredangle{OJG}= 90^{0}$, the quadrilateral $OJGM$ is cyclic. Hence, $\measuredangle{MJG}= \measuredangle{MOG}$. This implies $\measuredangle{MOG}= \measuredangle{MAN}= \measuredangle{MUG}$. Thus, $OUGM$ is cyclic, or $OU$ is perpendicular to $NG$.[/quote]

You have forgotten to say that the point O is defined as the circumcenter of the cyclic quadrilateral ABCD.

[quote="treegoner"]   Now, the inversion with center $G$, constant $GA.GB$ maps the two circles into themselves, and maps the circle $(S)$ with diameter $GN$ into the line $OU$. This imples that $(S)$ is orthogonal to $(O)$, or $N$ lies on the polar of $G$ wrt $(O)$, which is $EF$.
  PS Why is Victor Thebault related to this problem ?[/quote]

Thebault is not directly related to this problem; I just wanted to say that the problem is very much in his style, and, more generally, in the style of the old school synthetic geometers. I am glad that problem like this have become popular again on shortlists...

[b]PS.[/b]

[hide="comment"] I think everybody knows now that the ISL 2004 is avaliable in Vietnamese ([url=http://www.ddtoanhoc.net/forum/index.php?act=Print&client=printer&f=92&t=2793]number theory[/url], [url=http://www.ddtoanhoc.net/forum/index.php?act=Print&client=printer&f=92&t=2794]geometry[/url], [url=http://www.ddtoanhoc.net/forum/index.php?act=Print&client=printer&f=92&t=2795]combinatorics[/url], [url=http://www.ddtoanhoc.net/forum/index.php?act=Print&client=printer&f=92&t=2796]algebra[/url]). [/hide]

  greetings from oberwolfach,
  darij

Well, how come this problem was arranged as G8? Everything becomes trivial after noticing the second intersection of the circumcircle of ABM and CD lies on EF, as what Grobber did.
I don't mean it is too easy to be in shortlist, but at least IMO04\5 is surely harder than this one. In my mind it should be around G4-G5.

A very simple complex numbers solution (found by dysfunctionalequations and me):

Let $U = AA \cap BB$ and $V = DD \cap CC$. By Pascal's theorem on $AACBBD$ and $ADDBCC$, $U$, $V$, $E$, and $F$ are collinear, so it is sufficient to show that $U$, $V$, and $N$ are collinear.

Without loss of generality, let $ABCD$ be the unit circle, and let $a$, $b$, $c$, $d$, $m$, $n$, $u$, and $v$ be the affixes of $A$, $B$, $C$, $D$, $M$, $N$, $U$, and $V$, respectively. Because $AMBN$ is cyclic, $Q = \frac{(a-n)/(n-b)}{(m-a)/(m-b)}$ must be real. Since $\frac{NA}{NB} = \frac{MB}{MA}$, $Q$ must have magnitude 1. Hence, $Q = \pm 1$. If $Q = -1$, then $(a-n)(m-b) = (a-m)(n-b)$, which clearly has as its only solution $m=n$. But $m \neq n$, so $Q = 1$. Hence, $(a-n)(m-b) = (m-a)(n-b)$, so $am - ab - mn + nb = mn - bm - an + ab$, so $n(m-a + m-b) = am + bm - 2ab$, so $n = \frac{m(a+b) - 2ab}{2m-(a+b)}.$

We wish to show that $\frac{u-n}{v-n}$ is real. It is well-known that $u = \frac{2ab}{a+b}$ and $v = \frac{2cd}{c+d}$. Hence,
\begin{align*}
\frac{u-n}{u-v}
&= \frac{\frac{2ab}{a+b} - n}{\frac{2ab}{a+b} - \frac{2cd}{c+d}} \\
&= \frac{2ab(c+d) - n(a+b)(c+d)}{2ab(c+d) - 2cd(a+b)} \\
&= \frac{2ab(c+d) - \frac{m(a+b) - 2ab}{2m-(a+b)} (a+b)(c+d)}{2ab(c+d) - 2cd(a+b)} \\
&= \frac{2ab(c+d)(2m-(a+b)) - (a+b)(c+d)(m(a+b) - 2ab)}{(2m-(a+b))(2abc + 2abd - 2acd - 2bcd))} \\
&= \frac{m(4ab(c+d) - (a+b)^2 (c+d)) - 2ab(a+b)(c+d) + 2ab(a+b)(c+d)}{(c+d-(a+b))(2ab(c+d) - 2cd(a+b))} \\
&= \frac{- \left(\frac{c+d}{2}\right) (c+d) (a-b)^2 }{(c+d-(a+b))(2abc + 2abd - 2acd - 2bcd)} \\
&= -\frac{1}{4} \frac{(c+d)^2 (a-b)^2 }{(c+d-(a+b))(abc+abd-acd-bcd)}.
\end{align*}

Recalling that $\overline{z} = \frac{1}{z}$ for complex numbers $z$ on the unit circle, we have
\begin{align*}
\left( \frac{u-n}{u-v} \right) \left( \frac{\overline{u-v}}{\overline{u-n}} \right)
&= \left( \frac{(c+d)^2 (a-b)^2 }{(c+d-(a+b))(abc+abd-acd-bcd)} \right) \left( \frac{(\frac{1}{c}+\frac{1}{d}-(\frac{1}{a}+\frac{1}{b}))(\frac{1}{abc}+\frac{1}{abd}-\frac{1}{acd}-\frac{1}{bcd})}{(\frac{1}{c}+\frac{1}{d})^2 (\frac{1}{a}-\frac{1}{b})^2 } \right) \\
&= \left( \frac{(c+d)^2 (a-b)^2 }{(c+d-(a+b))(abc+abd-acd-bcd)} \right) \left( \frac{(\frac{abc + abd - acd - bcd}{abcd})(\frac{c+d-(a+b)}{abcd})}{\frac{(c+d)^2 (a-b)^2}{a^2 b^2 c^2 d^2}} \right) \\
&= 1, \end{align*}

so $\frac{u-n}{u-v} = \overline{\left(\frac{u-v}{u-n}\right)}$. It follows that $\frac{u-n}{u-v}$ is real, which was what we wanted.

[b]Lemma 1:-[/b]
[i]In a triangle $ ABC $, the A-appolonius circle intersects the circumcircle of $ ABC $ at $ X $(other than $ A $). Then $ X $ is the intersection of the A-symmedian of $ ABC $ and $ \odot ABC $. [/i]

[b]Proof:-[/b]
If we invert the figure wrt $ A $, then $ B,C $ goes to some points $ B',C' $ and $ \odot ABC $ goes to $ B'C' $. If $ AL,AM $ ($ L,M $ lying on $ \odot AB'C' $) are the angle-bisectors of $ \angle B'AC' $, then image of $ X $ will be the intersection point of $ LM $ and $ B'C' $. Clearly $ LM $ bisects $ B'C' $, so $ AX $ is a median of $ AB'C' $. Since $ BC $ is anti-parallel to $ B'C' $, so $ AX $ is a symmedian of $ ABC $.

[b]Lemma 2:-[/b]
[i]If $ AM,AP $ are isogonal conjugates wrt $ \angle BAC $ of $ \Delta ABC $, with $ M $ lying on $ BC $ and $ N $ lying on $ \odot ABC $, then consider any circle passing through $ A,M $ which intersects $ BC $ and $ \odot ABC $ at $ X,Y $. Then $ XY $ will pass through $ N $.[/i]

[b]Proof:-[/b]
Suppose, $ NX $ intersects $ \odot ABC $ at $ Y' $. $ \angle AY'X=\angle ACN=\angle ACB+\angle BAN=\angle ACB+\angle CAM=180-\angle AMC $. So $ AMCY' $ is cyclic. So $ Y\equiv Y' $.

[b]Back to Main Proof:-[/b]
Suppose, $ FE\cap DC=U, FE\cap AB=V, DC\cap AB=K $. $ (KU,DC)=-1 $. So $ KU.KM=KD.KC=KA.KB $. So $ AUMB $ is cyclic.
Similarly $ UMM'V $ is cyclic(where $ M' $ is the mid-point of $ AB $). By lemma 1, $ N $ is the intersection of A-symmedian of $ ABC $ at $ \odot ABC $. So by lemma 2, $ U,V,N $ are collinear. So done.

Solution :
Construct points Y and X on line CD such that BCY ~ BAD and ABC ~ ADX
YC = BC*AD/BA = DX
Let circle (ABE) intersects FE at points G and E
J is Miquel point of lines AD , BC , BD , AC
Easy to see that GBJ ~ MBY and AGJ ~ AMX
Construc point Z on GJ such that GZ/ZJ = MC/CY = MD/DX
ZBJ ~ MBY , AZJ ~ AMX , so angle BZA = BMC + AMD = 180 - BMA  , so Z is on (BAM)
MZ/BM = ZJ/MY = ZJ/MX = ZA/MA , so Z = N . done

Let $G=AB\cap CD$, $XX=EF\cap CD$ and $N'=EF\cap(ABM)$.

[b][color=#f00]Claim[/color][/b]: $D\in (ABM)$
[b]Pf[/b]: $(D, C; G, X)=-1$, so $GX\cdot GM=GC\cdot GD=GA\cdot GB$ $\blacksquare$

To prove that $N'=N$, we will show that $(ABMN)$ is harmonic:
$$(A, B; N, M)\overset{X}{=} (A, B; EF\cap AB, G)\overset{F}{=}(D, C; X, G)=-1$$
And we are done. $\square$

Let $T=\overline{EF} \cap \overline{CD}$ and $G=\overline{CD} \cap \overline{AB}$. The key claim is that that $T \in (AMB)$. Here are two proofs.

[b]Proof 1:[/b] Let $S=\overline{AB} \cap \overline{EF}$; it is well-known that $(G,S;A,B)=-1$, hence by projecting through $F$ we obtian $(G,T;D,C)=-1$. I now claim that $GT\cdot GM=GD\cdot GC$. Unlike everyone else, I will actually show the algebra. Let $GD=x,GC=y$, so $DC=y-x$, and we thus have $DT=\tfrac{x(y-x)}{x+y}$ and $TC=\tfrac{y(y-x)}{x+y}$. Thus,
$$GT\cdot GM=\left(x+\frac{x(y-x)}{x+y}\right)\left(\frac{x+y}{2}\right)=\frac{x(x+y)+x(y-x)}{2}=xy=GD\cdot GC.$$
Then by power of a point on $ABCD$, we have $GT\cdot GM=GD\cdot GC=GA\cdot GB$, hence $ABMT$ is cyclic as desired. $\blacksquare$ (Note: apparently this is also a well-known lemma)

[b]Proof 2:[/b] Let $O$ be the center of $ABCD$. Note that $\overline{OM} \perp \overline{CD}$. Further, let $P$ be the foot of the perpendicular from $O$ to $\overline{FG}$ and $Q$ the foot of the perpendicular from $O$ to $\overline{EF}$, so $TMOQ$ and $FPOQ$ are cyclic. Thus, $GT\cdot GM=GQ\cdot GO=GP\cdot GF$. On the other hand, it is well-known that $P$ is the Miquel point of $ABCD$, so $P$ lies on $(FDC)$, hence $GP \cdot GF=GD\cdot GC=GA\cdot GB$, and we conclude that $ABMT$ is cyclic. $\blacksquare$

To finish, define $S$ as we did in proof 1. We have
$$-1=(G,S;A,B)\stackrel{T}{=}(M,\overline{EF} \cap (AMB);A,B),$$
hence $\overline{EF} \cap (AMB)=N$, as $MNAB$ is harmonic as well, which finishes the problem. $\blacksquare$

Let $\omega =\odot (ABM)$ meet $CD$ again at $K, G=AB\cap CD,$ $L=KN\cap AB.$ By PoP $$|GC|\cdot |GD|=|GA|\cdot |GB|=|GK|\cdot |GM|\implies (BCGK)=-1=(ABMN)_{\omega}\stackrel{K}{=}(ABGL),$$ what implies that $\overline{KLN}$ is the polar of $G$ wrt $\omega$ and hence passes through $E,F.$

It's equivalent to prove that if $N'$ is the intersection of $EF$ with $(ABM)$, then $AN'BM$ is a harmonic quadrilateral. So, we will prove this.

Let $G$ be the intersection of $EF$ with $CD$. We claim $GMAB$ is cyclic.

Let $X$ be the intersection of $AB$ and $CD$. WLOG $CX<DX$. Note that $D, G, C, X$ is a harmonic bundle. So, $CG=CD\cdot\frac{CX}{CX+DX}=\frac{CX\cdot DX-CX^2}{CX+DX}$. So, $GX=CX+CG=CX+\frac{CX\cdot DX-CX^2}{CX+DX}=\frac{2CX\cdot DX}{CX+DX}$. Also, we have $MX=\frac{CX+DX}2$. So, $$GX\cdot MX= CX\cdot DX=AX\cdot BX.$$
Therefore, $GMAB$ is cyclic.

Let $Q$ be the intersection of $EF$ with AB$. It is well known that $A, Q, B, X$ is a harmonic bundle. So, $GA, GQ, GB, GX$ is a harmonic bundle. Taking perspectivity onto $(ABM)$, we find that $AN'BM$ is a harmonic quadrilateral.

OTIS problem :)

We will use phantom points. That is, let $EF$ intersect $(ABM)$ at two points, say $N'$ and $H.$ We wish to prove that $ABMN'$ is harmonic, which implies the result.

To do so, we will need the following claim:
[b]Claim:[/b] $C,H,D$ are collinear.
[i]Proof:[/i] We will also use phantom points to prove this. Let $EF$ intersect $CD$ at a point $H'.$ We wish to prove that $ABH'M$ is cyclic. If $G = AB \cap CD,$ then by Power of a Point, it suffices to show that $GA \cdot GB = GC \cdot GD = GH' \cdot GM.$ It is well-known that $(GH';CD) = -1,$ so the midpoint harmonic bundle lemma finishes. $\blacksquare$

Let $EF$ intersect $AB$ at a point $X.$ It is again well-known that $(GX;BA) = -1.$ Thus projecting through $H$ from $AB$ to $(ABN'H)$ yields the result.

Let $(ABM) \cap CD=Q$ and $AB \cap CD=T$
From $PoP$ we know from $TD.TC=TQ.TM$
It's well known that $(TQ;DC)=-1$ so $F,E,Q$ are collinear.
We should prove that $(ABNM)$ is harmonic quadrilateral.
Let $QE \cap (ABM)=N'$ from $Q(TN;AB)=-1$ we have $(ABNM)$ is harmonic quadrilateral.

Let $G = AB\cap CD$ and $P$ the other intersection of $(MAB)$ with $CD$. From $GP\cdot GM = GB\cdot GA = GC\cdot GD$ we know $(CD;PG)$ is harmonic -- in particular $P\in EF$. Now we have

$$-1 = (AB;MN) \overset{P}{=} (AB;G,PN\cap AB)$$

But it is well known that $K = EF\cap AB$ also satisfies $(AB;GK)=-1$ so we're done.

Let $EF \cap DC = P, G = AB \cap CD, J = PA \cap BC$.

Then by Brokard's theorem on $ABCD$ and $AFCP$, $(D, C; P, G) = (J, B; F, C) = -1$ are harmonic bundles. Observe also that $ANBM$ is harmonic (and completely unrelated to the problem, that $MN$ is the symmedian of $MAB$ and $NAB$).

[b]Claim: [/b] $P$ lies on $(ANBM)$.
[b]Proof: [/b] We have $GA \cdot GB = GD \cdot GC = (GM - MD)(GM + MD) = GM^2 - MD^2$. Therefore it suffices to show $GP \cdot GM = GM^2 - MD^2$: As $(D, C; P, G) = -1$,
\begin{align*}
\frac{GC}{CP} &= \frac{GD}{DP} \\
\frac{GM + MD}{MD + PM} &= \frac{GM - MD}{MD - PM} \\
\implies \frac{GM}{MD} &= \frac{MD}{MP} \\
\implies GM \cdot MP &= MD^2 \\
\implies GM \cdot GP &= GM^2 - MD^2, \text{as desired.} \square
\end{align*}

Now take perspective of $ANBM$ at $P$ onto $BC$ to obtain \[-1 = P(A, B; N, M) \stackrel{BC}{\longrightarrow} (J, B; PN \cap BC, C) = (J, B; F', C) = -1 = (J, B; F, C).\] Therefore $F = F'$, and so $P, F, E, N$ are collinear.

	The condition gives $(A,B;M,N)=-1$. As $EF$ is the polar of $G$ wrt to $(ABCD)$, we have $(G,P;C,D)=-1$, where $P\coloneq EF\cap CD$. By the midpoint lemma, $GP\cdot GM=GC\cdot GD=GB\cdot GA$, so $P\in (ABM)$. Letting $Q\coloneq PF\cap AB$ and $N'=PF\cap (ABM)$, we have by first projecting through $E$ and then through $P$ that
	\[-1=(G,P;C,D)=(G,Q;B,A)=(M,N';B,A),\]
	so $N'=N$. Thus, $EFP$ and $PFN$ are the same lines, finishing the problem.

[color=#edbe00][b]Motivation:[/b][/color] The ratio condition gives a harmonic quad, so it is natural to search for harmonic bundles. Knowing Brocards theorem motivates using the polar $EF$. Letting it intersect $CD$ at $P$ gives a nicer framework as we observe the harmonic bundle $(G,P;C,D)$. Now we need to project this onto $ABMN$, which is only possible through a point on $(ABM)$ as otherwise the harmonic bundle isn't preserved. Then we notice that $P$ lies on $(ABM)$, from which we can finish the problem.

We utilize phantom points. Assign $N_1$ as the intersection of $EF$ with $(ABM)$ at $N_1$. We also define $P = AB \cap EF$, $Q = AB \cap CD$, and $R = CD \cap EF$.

[color = red][b]Lemma:[/color][/b] Quadrilateral $ABMR$ is cyclic.

From the Ceva-Menelaus Lemma we know $(QR;DC) = -1$. Since $M$ is the midpoint of $CD$, it is well known that
\[QR \cdot QM = QC \cdot QD = QA \cdot AB.~{\color{red} \Box}\]

[color = blue][b]Claim:[/color][/b] $(AB;MN_1)$ is harmonic.

From Ceva-Menelaus and perspectivity, we know
\[-1 = (AB;QP) \overset{R}{=} (AB;MN_1).~{\color{blue} \Box}\]

Therefore $\frac{AN_1}{BN_1} = \frac{AM}{BM}$, so $N = N_1$, and $E$, $F$, and $N$ are collinear. $\blacksquare$

[asy]
size(260); defaultpen(fontsize(8)+linewidth(.35));

pair A, B, C, D, E, F, Q, P, R, N;
A = dir(190);
B = dir(350);
C = dir(80);
D = dir(145);
E = extension(A, C, B, D);
F = extension(B, C, D, A);
Q = extension(A, B, C, D);
P = extension(E, F, A, B);
R = extension(E, F, C, D);
N = IP(E--3P-2E, circumcircle(A, B, R));

draw(Q--B--F--A--Q--C);
draw(A--C^^B--D^^F--N);
draw(circumcircle(A, B, R), blue);

dot("$A$", A, SW);
dot("$B$", B, SE);
dot("$C$", C, NE);
dot("$D$", D, NW);
dot("$E$", E, dir(300));
dot("$F$", F, dir(90));
dot("$P$", P, SE);
dot("$Q$", Q, SW);
dot("$R$", R, NW);
dot("$M$", .5C + .5D, dir(90));
dot("$N_1$", N, S);
[/asy]

Posting this for storage.

Let $G=(AB)\cap(DC)$,$Y = (AB) \cap (EF)$and $Z=(EF) \cap (DC)$.

Since $(CD;ZG) = -1$, then $GZ \cdot GM = GD \cdot GC = GA \cdot GB$, thus  $ABMZ$ is cyclic. Let $N'$ be the second intersection of the $(\odot ABM)$ with $(ZE)$

Now we have:
$$(AB;N'M) \stackrel{Z}{=}(AB;YG) = -1$$
So $N'=N$ and this ends the problem.

 $$\mathbb{Q.E.D.}$$

Let $FE$ intersect $CD$ at $T$,then we know that $(PT,CD)=-1 $

therefore by midpoint length $PT.PM=PC.PD=PA.PB$ => $ABTM is cyclic$
Now $(P,T;C,D)=A=(B,W;E,D)=-1$, where $W$ is intersection of $TA$ with $BD$

Let $FE$ intersects $(ABTM)$ at $N$ , $(B,W;E,D)=X=(B,N;A,M)=-1$

therefore $ANBM$ is [color=#00f]harmonic[/color] so $AN.BM=BN.AM ==> AN/BN=AM/BM$
so $N$ is [color=#f00]desired point[/color].

We first note that the given ratio condition is equivalent to $(AB;MN)=1$ or $(AB;MN)=-1$. Thus, there can exists at most two points which satisfy the given ratio condition. Clearly, $M$ satisfies the first cross ratio. So only one other point $N$ satisfies the required ratio. This is the point $N$ such that $AMBN$ is a harmonic quadrilateral.\\

Let $G=\overline{AB} \cap \overline{CD}$. Now, let $X=\overline{EF} \cap \overline{CD}$, $Y = \overline{EF} \cap \overline{AB}$ and $Z= \overline{GE} \cap \overline{AD}$. Let $N'=\overline{EF} \cap (ABM)$. Now, we first make the following observation.

[b][color=#f00]Claim : Point $X$ lies on $(AMB)$.[/color][/b]

    Note that $(DC;GX)=-1$. Thus,
    \[GA \cdot GB = GC \cdot GD = GX \cdot GM\]
    and thus $ABXM$ must be cyclic as needed.

Let $P= \overline{BX} \cap \overline{AD}$. Now, note that,
\[-1=(DA;ZF)\overset{G}{=}(XY;EF)\overset{B}{=}(PA;DF)\overset{X}{=}(BA;MN')\]
Thus, $AN'BM$ is indeed harmonic. This means $N'=N$ and indeed  points $E,F,N$ are collinear as was required.

Let $AB$ intersect $CD$ at $P$, $(ABM)$ intersect $CD$ at $K\neq M$, $BK$ intersect $AC$ at $L$, and $KE$ intersect $(ABM)$ at $N'$. Note that $PK\cdot PM=PA\cdot PB=PC\cdot PD$ which implies $(P,K;D,C)=-1$. By configuration, we know that $K$, $E$, $F$ are collinear. Additionally, we have
\[-1=(P,K;D,C)\stackrel{B}{=}(A,L;E,C)\stackrel{K}{=}(A,B;N',M)\]
So $AN'BM$ is harmonic. We also know that $ANBM$ is harmonic, so $N=N'$, and so $N$ lies on $KE$ and thus $EF$.

Call $G = AB \cap CD$, $X = EF \cap CD$, $Y = EF \cap AB$, $N^* = EF \cap (ABM)$.

We first prove that $X$ lies on $(ABM)$, which is true since
$$GA \cdot GB = GC \cdot GD = GX \cdot GM$$
where the last equality follows from the well known lemma about midpoints in harmonic bundles.

Now, we have
$$-1 = (D, C; X, G) \stackrel F= (B, A; Y, G) \stackrel X= (B, A; N^*, M)$$
However, $(B, A; N, M) = -1$, thus $N = N^*$ and we are done.


A set $M \in \mathbb{N}$ is called [b]gappy [/b], if $\forall m \in M$ its successor $(m+1) \not \in M.$ The empty set is defined as gappy. Find an algorithm in order to determine all gappy subsets of $M = \{1,2,3 \ldots, n\}$ and determine how many subsets there are for each $n.$

eum...
the only subsets are the empty one and {n} ? :D

Not true.. there are a lot more than that.. {1,3,5} would be an example. Don't get mixed up between the M that was used in general in the first sentence and the specific M in the last.

[quote="orl"]A set $M \in \mathbb{N}$ is called [b]gappy [/b], if $\forall m \in M$ its successor $(m+1) \not \in M.$ The empty set is defined as gappy. Find an algorithm in order to determine all gappy [b]subsets of [/b]$M = \{1,2,3 \ldots, n\}$ and determine how many subsets there are for each $n.$[/quote]

Reading that wording, I conclude that they mustn't be in the original M.

Your version of the problem is better of course, but my solution is for what Orl posted, right?

Anyway your way is pretty easy as well: Any set in which there are no 2 consequtive numbers.

[quote="Peter VDD"]

Reading that wording, I conclude that they mustn't be in the original M.

Your version of the problem is better of course, but my solution is for what Orl posted, right?
[/quote]

Nope, that wording doesn't mean they can't be in the original M.

"A set M is called gappy if for all m in M m+1 is not in M."

This doesn't include any mention of the set of which M is a subset of. It always talks about M by itself.
[quote="Peter VDD"]

Anyway your way is pretty easy as well: Any set in which there are no 2 consequtive numbers.
[/quote]

So solve it then :)

Fibonacci again  :D  :D  :D

Let's denote the answer by $a_n$.

Suppose we have gappy sets $\subset \{1,2,\dots,n\}$. We divide them into two kinds: those which contain $n$ and those which don't.

For the kind that contain $n$, after removing $n$ from each gappy set, we see that they are gappy sets of $\{1,2,\dots,n-2\}$. So the number of sets of this kind is $a_{n-2}$.

For the kind that don't contain $n$, they are gappy sets of $\{1,2,\dots,n-1\}$. So the number of sets of this kind is $a_{n-1}$.

Thus $a_n = a_{n-1} + a_{n-2}$..
-----------------
Orlando likes Fibonacci very much?  :maybe:

oh they also ask how many

yes, that's the fibonnaci sequence with a_0=1, a_1=2

Hey Orl, this is just the same problem as the Hamburger Schülerzirkel one you posted at [url]http://artofproblemsolving.com/Forum/viewtopic.php?t=14770[/url] . Indeed, if you consider the set of all floors colored blue, then it is gappy, and all gappy subsets of $\left\{1, 2, ..., n\right\}$ correspond to such colorings...

  Darij


Here is a cute little problem which has a very quick solution, probably known:

The wrong way of adding fractions.

Let $ a$, $ b$, $ c$ and $ d$ be positive integers. Prove that the equation

$ \frac{a}{b}\plus{}\frac{c}{d}\equal{}\frac{a\plus{}c}{b\plus{}d}$,

is impossible.

Assume to the contrary that there exist $ a,b,c,d\in\mathbb{N}$ such that $ \frac{a}{b}\plus{}\frac{c}{d}\equal{}\frac{a\plus{}c}{b\plus{}d}$.  But the equation is equativalent to $ b^2c\plus{}ad^2\equal{}0$, which is impossible since $ a,b,c,d\geq 1$. $ \Box$

This problem should be moved to the Intermediate forum.  :wink:

[hide]
$ \frac {a}{b} > \frac {a}{b \plus{} d}$ , $ \frac {c}{d}>\frac{c}{b \plus{} d} \Rightarrow \frac {a}{b} \plus{} \frac {c}{d} > \frac {a \plus{} c}{b \plus{} d}$. Contradiction.[/hide]

It also turns out that if $ \dfrac{a}{b}<\dfrac{c}{d}$, $ \dfrac{a}{b}<\dfrac{a\plus{}c}{b\plus{}d}<\dfrac{c}{d}$.


Exercise 2-3: Find cos 215

I have no way of finding this, and looking at the answer gives -sqrt2/2 which I believe is cos 225.

Anyone have any help?

Oops, that's a typo in the book.  You're right, it should be cos 225.

It's just cos(180+45) which is in the third quadrant. which makes cos(180+45)= -cos 45. -cos(45)= -sqrt2/2 :yoda:


I don't know if this is a problem with the site or my computer, though other sites seem to be working fine.  I've had this problem once or twice before (see image).

I had that problem when I was using Internet Explorer. When Internet Explorer started causing me too many problems, I began using Mozilla, which has been working perfectly for me. So I think it might be some weird Internet Explorer caching problem.

Me too, when internet explorer was fine in the beginning but when i installe dthe java virtual machine new version the virtual classroom thing stopped working. so i have to use FIREFOX.

Does hitting the reload page make it go away?  You might have hit the site right in the middle of our changing something.

No, i hit refresh and it didn't help, so I then exited my browser and went to the site again and it still came up like that... all the buttons on the left were also x's, though I missed those in the picture.

Have you tried clearing your browser's cache?  In Internet Explorer, go to Tools | Internet Options | Temporary Internet files | Delete files.  Then restart your browser.

I always clear my cache when clearing history, though i'm not exactly sure what it is.  But I don't have this problem now - it was only last night and some other times before, so  I guess it must be my computer if no one else had this problem last night.


Let $ABCD$, $A'B'C'D'$ be quadrilaterals incsribed in the same circle $O$. The four diagonals all pass through a point $P$. The angle between the diagonals of the first quadrilateral is equal to that of the second quadrilateral. Moreover, $AB=A'B'$ and $CD=C'D'$. Show that the two quadriaterals must be congruent.

By a bit of angle chasing (in particular angle in same segment) it can be shown that A'PD' is similar to APD and B'PC' to BPC. In addition B'P/D'P=BP/DP. This is not very rigorous, but this shows that ABCD is similar to A'B'C'D', and, since they have the same circumradius, therefore they are congruent. I think this is valid...

I can't quite see how you show the similarity. Please give a more detailed proof.  :)

We know that APD is sim to A'PD'. In addition we know BP/DP = B'P/D'P. This effectively means that BD is DP x some constant, and B'D' = D'P x same constant. Also <ADP=<APB=<A'PB'. Therefore, by SAS similarity criterion ABD is similar to A'B'D'. You can do the same thing for BCD and B'C'D', and hence the quadrilateral is similar. Sorry for my vagueness...


Mitch invested m dollars at 9% and n dollars at 7.5%. Write an algebraic expression for each situation:
- the total amount of money Mitch invested
- the interest Mitch earned at 7.5% in one year
- the total interest Mitch earned in one year

...anyone?? :?

Not exactly.

The interest Mitch earned in one year at 7.5% is $0.075n$. Since it's asking for interest, no need to add the original $n$ in there. For the third one, it's the same reasoning, no need to add the additional $n$ and $m$.

[quote="lamefif"]Total amount Mitch invested
(m+n=?)

The interest Mitch earned in one year at 7.5%.
(n*7.5%+n=?)

The total interest Mitch earned in one year.
((n*7.5%+n)+(m*9%+m)=?)

There ur homework done, at least make the effort to understand.[/quote]In addition to what Zyloch said, it asks for algebraic [i]expressions[/i], so there is no need to make them into equations with "=?" after each one.


In the past, there are students who qualified for AIME through USAMTS.  It is not clear to me what is the path for these students to USAMO or USAJMO.  Anyone knows?

Your USAMO/USAJMO index is calculated with a 100 as your AMC 12 score.