geometry
3D geometry
sphere
modular arithmetic
parameterization
number theory
number theory theorems

I found the formula :
a^2 + ^b2+ c^2 = d^2
a=m^2 + n^2 - p^2 - q^2
b=2(mp + nq)
c=2(mp - nq)
d=m^2 + n^2 + p^2 + q^2

Would anyone care to explain how the solutions are obtained?, or preferably give a link to the proof? Thanks.

I believe a derivation can be found using [url=http://en.wikipedia.org/wiki/Quaternion]quaternions[/url] (specifically, the [url=http://en.wikipedia.org/wiki/Hurwitz_quaternion]Hurwitz integers[/url]).  There is a corresponding derivation of of the formula for primitive Pythagorean [i]triples[/i] using the [url=http://en.wikipedia.org/wiki/Gaussian_integer]Gaussian integers[/url].

There may be an "elementary" solution.  Specifically, we are trying to find rational points on the unit sphere $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$, which we can do by rationally parameterizing it.  There are well-known techniques for this, although they go into the subject of [url=http://en.wikipedia.org/wiki/Algebraic_geometry]algebraic geometry[/url].

Sorry I was not specific enough.
This is a problem from my elementary number theory course.
And I never used any geometry in this course till now, so I guess there's a solution without using geometry.
Can someone give me that proof? Or the geometry one? I would be grateful.

[hide="Algebra"] Consider parity for a moment.  Since squares are congruent to $ 0, 1 \bmod 4$ we can easily deduce that for a primitive triple exactly one of $ a, b, c$ is odd (and so is $ d$).  Let $ a$ be odd WLOG.  Then

$ b^2 \plus{} c^2 \equal{} d^2 \minus{} a^2$

After dividing by $ 4$, we have

$ \left( \frac {b}{2} \right)^2 \plus{} \left( \frac {c}{2} \right)^2 \equal{} \left( \frac {d \plus{} a}{2} \right) \left( \frac {d \minus{} a}{2} \right)$

Suppose a prime $ p \equiv 3 \bmod 4$ divides $ d \plus{} a$.  Then $ p | b^2 \plus{} c^2 \implies p | b, c$.  This implies $ p^2 | (d \plus{} a)(d \minus{} a)$.  Then either $ p | d \minus{} a$ (in which case $ p | a, b, c, d$ - contradiction) or $ p^2 | d \plus{} a$.  We conclude that any prime $ \equiv 3 \bmod 4$ dividing $ d \plus{} a$ does so with even multiplicity.

We conclude that $ \frac {d \plus{} a}{2}, \frac {d \minus{} a}{2}$ are both themselves a sum of squares.  (This is a well-known lemma.)

Let $ d \plus{} a \equal{} 2(m^2 \plus{} n^2), d \minus{} a \equal{} 2(p^2 \plus{} q^2)$.  Then

$ d \equal{} m^2 \plus{} n^2 \plus{} p^2 \plus{} q^2$
$ a \equal{} m^2 \plus{} n^2 \minus{} p^2 \minus{} q^2$

Moreover, we have the identity

$ (m^2 \plus{} n^2)(p^2 \plus{} q^2) \equal{} (mp \plus{} nq)^2 \plus{} (mq \minus{} np)^2$

so we conclude

$ b \equal{} 2(mp \plus{} nq)$
$ c \equal{} 2(mq \minus{} np)$

Because $ m, n, p, q$ were chosen generically (they can be any representation of $ \frac {d \plus{} a}{2}, \frac {d \minus{} a}{2}$ as a sum of squares), this approach describes all primitive Pythagorean quadruples. [/hide]

Thanks! That's exactly what I was looking for..

[quote="t0rajir0u"]

Moreover, we have the identity

$ (m^2 \plus{} n^2)(p^2 \plus{} q^2) \equal{} (mp \plus{} nq)^2 \plus{} (mq \minus{} np)^2$

so we conclude

$ b \equal{} 2(mp \plus{} nq)$
$ c \equal{} 2(mq \minus{} np)$

[/hide][/quote]

that part is unclear to me. of course we also have the representation $ (mq\plus{}np)^2\plus{}(mp\minus{}nq)^2$, and i understand this other one is already included as a solution by symmetry. but how do we know there aren't other representations??

[quote="Muhigai"]I have one more question :
$ p | b^2 \plus{} c^2 \implies p | b, c$

what happens if p divides neither b nor c ?[/quote]

this part is okay. $ p|b^2\plus{}c^2$ means that $ b^2\equiv \minus{}c^2 \pmod{p}$. so if $ p$ divides neither $ b$ nor $ c$, we get that $ \left(\frac{\minus{}1}{p}\right) \equal{} 1$, which is false for $ p\equiv 3\pmod{3}$.

The parametric solution is actually fairly simple.
[hide]Every rational point on $ x^2\plus{}y^2\plus{}z^2\equal{}1$ other than $ (\minus{}1,0,0)$ can be written in the form $ (\minus{}1\plus{}t,ut,vt)$, where we have
\[ t\equal{}x\plus{}1\neq 0,u\equal{}\frac{y}{t},v\equal{}\frac{z}{t}\]
so that $ x,y,z$ rational implies $ t,u,v$ rational. Conversely, if $ t,u,v$ rational, then $ x,y,z$ rational. Now for every rational $ u,v$, we can find the values of $ t$  such that $ (\minus{}1\plus{}t)^2\plus{}(ut)^2\plus{}(vt)^2\equal{}1$, namely $ t\equal{}0$ and $ t\equal{}\frac{2}{1\plus{}u^2\plus{}v^2}$. $ t\equal{}0$ gives $ (\minus{}1,0,0)$, the other value of $ t$ gives the rational solutions
\[ x\equal{}\frac{1\minus{}u^2\minus{}v^2}{1\plus{}u^2\plus{}v^2},y\equal{}\frac{2u}{1\plus{}u^2\plus{}v^2},z\equal{}\frac{2v}{1\plus{}u^2\plus{}v^2}\]
From there, Pythagorean quadruples can be found. [/hide]

[quote="pleurestique"][quote="t0rajir0u"]

Moreover, we have the identity

$ (m^2 \plus{} n^2)(p^2 \plus{} q^2) \equal{} (mp \plus{} nq)^2 \plus{} (mq \minus{} np)^2$

so we conclude

$ b \equal{} 2(mp \plus{} nq)$
$ c \equal{} 2(mq \minus{} np)$

[/hide][/quote]

that part is unclear to me. of course we also have the representation $ (mq \plus{} np)^2 \plus{} (mp \minus{} nq)^2$, and i understand this other one is already included as a solution by symmetry. but how do we know there aren't other representations??[/quote]

Every other representation is given by a different choice of $ m, n, p, q$.  The proof of this is actually not elementary, and I omitted it because I believe it boils down to unique factorization in the Gaussian integers.

scorpius:  the solution may be elementary once the parameterization itself has been found, but general techniques exist for finding rational parameterizations that are not elementary (not that they are difficult).


I am a freshman physics major at a top university in India. I was considering (as a really long shot) applying to REU programs in the US or Europe. What places should I consider? Obviously, places like Caltech SURF come to mind, but do I have a better chance at a less glamorous program?

Some background: I have previously worked on a vacation program at a radio telescope facility and I have done a reading project at a major research institute in India. Along with my regular physics courses, I am also taking some math, (Measure Theory last semester and Topology this semester). I am expecting my grades to be decent.

(I get this fellowship from our government which makes it relatively easy for me to get summer internships in Indian institutes.)

Thanks

The NSF keeps a list of REUs in the US here: http://www.nsf.gov/crssprgm/reu/reu_search.cfm

As you can see, there are many to choose from.

NSF Sponsored REU programs might be difficult for an international student though.  Generally the NSF has restrictions on who can be supported by their funds, so many of the programs are U.S. Citizen/Permanent resident only.


Prove that every integer $ k>1$ has a multiple less than $ k^4$ whose decimal expansion has at most four distinct digits.

I don't undestand. If $ k\equal{}2$, multiple $ k|n<2^4\equal{}16$ had only 1 or 2 digits, were  4 distinct digits?

The question states 'at most 4 distinct digits', so 1 or 2 digits is acceptable.

Let $ k \equal{} a_0 \plus{} a_1*2 \plus{} .. \plus{} a_m*2^m, a_i \equal{} 0,1$ and $ N \equal{} a_0 \plus{} a_1*10 \plus{} ... \plus{} a_m*10^m$.
In the set of numbers M, whose digits only 0 or 1 and $ 0\le M\le N$, we had 2, suth that $ M_1\equiv M_2\mod k$. Therefore
$ k\parallel{}M_1 \minus{} M_2|\le N\le k^{log_2{10}}$. Number $ M_1\minus{}M_2|$ had only digits $ 0,1,8,9$.

[quote="Rust"]Let $ k \equal{} a_0 \plus{} a_1*2 \plus{} .. \plus{} a_m*2^m, a_i \equal{} 0,1$ and $ N \equal{} a_0 \plus{} a_1*10 \plus{} ... \plus{} a_m*10^m$.
In the set of numbers M, whose digits only 0 or 1 and $ 0\le M\le N$, we had 2, suth that $ M_1\equiv M_2\mod k$. Therefore
$ k\parallel{}M_1 \minus{} M_2|\le N\le k^{log_2{10}}$. Number $ M_1 \minus{} M_2|$ had only digits $ 0,1,8,9$.[/quote]

Wow, quite nice !!!!


Determine all ordered pairs $(m, n)$ of positive integers such that \[\frac{n^{3}+1}{mn-1}\] is an integer.

$ mn\minus{}1|n^{3}\plus{}1$  $ \implies$  $ mn\minus{}1|n^{2}\plus{}m$

$ \implies$  $ k(mn\minus{}1)\equal{}n^{2}\plus{}m$  $ \implies$  $ n^{2}\minus{}kmn\plus{}m\plus{}k\equal{}0$

$ \Delta\equal{}k^{2}m^{2}\minus{}4(k\plus{}m)$  $ \implies$  $ (km\minus{}1)^{2}<k^{2}m^{2}\minus{}4(k\plus{}m)<(km)^{2}$

for  $ 2km>4(k\plus{}m)\plus{}1$

if $ 2km\leq 4(k\plus{}m)\plus{}1$

[quote="cnyd"]$ mn \minus{} 1|n^{3} \plus{} 1$  $ \implies$  $ mn \minus{} 1|n^{2} \plus{} m$

$ \implies$  $ k(mn \minus{} 1) \equal{} n^{2} \plus{} m$  $ \implies$  $ n^{2} \minus{} kmn \plus{} m \plus{} k \equal{} 0$

$ \Delta \equal{} k^{2}m^{2} \minus{} 4(k \plus{} m)$  $ \implies$  $ (km \minus{} 1)^{2} < k^{2}m^{2} \minus{} 4(k \plus{} m) < (km)^{2}$

for  $ 2km > 4(k \plus{} m) \plus{} 1$

if $ 2km\leq 4(k \plus{} m) \plus{} 1$[/quote]
Actually since $ (km \minus{} 1)^2 \equal{} k^2m^2 \minus{} 4(k \plus{} m) \Rightarrow \minus{} 2km \plus{} 1 \equal{} \minus{} 4(k \plus{} m) \Rightarrow 1 \equal{} 0 \bmod 2$ we just have to check the cases where $ (km \minus{} 2)^2 \ge k^2m^2 \minus{} 4(k \plus{} m) \iff (m \minus{} 1)(k \minus{} 1) \le 2$, giving $ k \equal{} 1$ or $ m \in \{1,2,3\}$.
$ k\equal{}1$: $ (m\minus{}2)^2\minus{}8$ is a square and hence $ m\equal{}5$ giving $ (m,n)\equal{}(5,2);(5,3)$.
$ mn\minus{}1 \mid n^3 \plus{} 1 \iff mn\minus{}1 \mid m^3\plus{}1$ so:
$ m\equal{}1$: $ n\minus{}1 \mid 2 \iff n\equal{}1,2$
$ m\equal{}2$: $ 2n\minus{}1 \mid 9 \iff n\equal{}1,2,5$
$ m\equal{}3$: $ 3n\minus{}1 \equal{} 28 \iff n\equal{}1,5$


Let be $ x,y,z > 0$ . Prove that :

$ \frac{2}{9}\left(\frac {(x \plus{} 2y)(y \plus{} 2z)}{x \plus{} 2z \plus{} 3y} \plus{} \frac {(y \plus{} 2z)(z \plus{} 2x)}{y \plus{} 2x \plus{} 3z} \plus{} \frac {(z \plus{} 2x)(x \plus{} 2y)}{z \plus{} 2y \plus{} 3x}\right)\le \frac {x^3}{x^2 \plus{} xy \plus{} y^2} \plus{} \frac {y^3}{y^2 \plus{} yz \plus{} z^2} \plus{} \frac {z^3}{z^2 \plus{} zx \plus{} x^2}$

How many inequalities do you create?
That's amazing  :P

edit: But this ineuqality is not true ;) did you forget a condition?

[quote="FelixD"]How many inequalities do you create?
That's amazing  :P[/quote]

  :) Well , how many i can . Anyway , i think i'll never be able to surpass Can_hang2007 or quykhtn-qa1 .  :P

My bad . You were right . I changed .

[quote="alex2008"]Let be $ x,y,z > 0$ . Prove that :

$ \frac {2}{9}\left(\frac {(x \plus{} 2y)(y \plus{} 2z)}{x \plus{} 2z \plus{} 3y} \plus{} \frac {(y \plus{} 2z)(z \plus{} 2x)}{y \plus{} 2x \plus{} 3z} \plus{} \frac {(z \plus{} 2x)(x \plus{} 2y)}{z \plus{} 2y \plus{} 3x}\right)\le \frac {x^3}{x^2 \plus{} xy \plus{} y^2} \plus{} \frac {y^3}{y^2 \plus{} yz \plus{} z^2} \plus{} \frac {z^3}{z^2 \plus{} zx \plus{} x^2}$[/quote]
It must be
We have $ \frac{(x\plus{}2y)(y\plus{}2z)}{x\plus{}2z\plus{}3y} \le \frac{x\plus{}2z\plus{}3y}{4}$
Then $ LHS \le \frac{x\plus{}y\plus{}z}{3}$
But $ RHS \ge \frac{(x^2\plus{}y^2\plus{}z^2)^2}{x(x^2\plus{}xy\plus{}y^2)\plus{}y(y^2\plus{}yz\plus{}z^2)\plus{}z(z^2\plus{}zx\plus{}x^2)} \equal{}\frac{x^2\plus{}y^2\plus{}z^2}{x\plus{}y\plus{}z} \ge \frac{x\plus{}y\plus{}z}{3}$
So,we have Q.E.D :)


Calculate $\sqrt{\sqrt 5-\sqrt{3-\sqrt{29-12\sqrt5}}}$.
(Of cause the calculator is NOT acceptable)

$29-12\sqrt5 = (\sqrt{5}-3)^2$
$6-2\sqrt{5} = (\sqrt{5} - 1)^2$

So it's just a complicated way to write $1$  :P


geometry
inequalities
perimeter
triangle inequality

given a quadrilateral ABCD circumscribe a circle center O and the radius r
prove that $ AB\plus{}CD \geq 4r$

I think you got the equality sign reversed.

In which case, apply the triangle inequality to $ OAB$ and $ OCD$ to get $ 2r >\equal{} CD$ and $ 2r >\equal{} AB$, and add them to get $ 4r >\equal{} AB \plus{} CD$.

I think the problem is saying that quadrilateral $ ABCD$ has an incircle with radius $ r$.

yes , the problem talking about the quadrilateral's incircle

[hide]
The area of our quadrilateral $ rs_1$ where $ s_1$ is the semiperimeter of the quadrilateral.    Consider circumscribing a square around the circle.  Then the area of the square is equal to $ rs_2$ where $ s_2$ is the semiperimeter of the square. Since the square is the quadrilateral of minimal perimeter, we can say that $ rs_1 \ge rs_2$.
But, since our quadrilateral has an incircle, $ AB\plus{}CD\equal{}s$.  So the area of the quadrilateral is $ r(AB\plus{}CD)$.
Also, the area of the square is $ (2r)^2\equal{}4r^2$.
Thus,
$ r(AB\plus{}CD) \ge 4r^2$ so
$ AB \plus{} CD \ge 4r$.[/hide]


[quote="zaizai-hoang"][b]Problem (zaizai-hoang):[/b]
Let $a,b,c$ are positive number. Prove that:
$2(a^{3}+b^{3}+c^{3})+3(ab\sqrt{ab}+bc\sqrt{bc}+ca\sqrt{ca}) \\ \ge (a-b)(b-c)(c-a)+3(ab^{2}+bc^{2}+ca^{2})+2(a^{2}\sqrt{ac}+b^{2}\sqrt{ab}+c^{2}\sqrt{bc})$

how about your solution? is it shorter than mine? :roll:[/quote]
Is your solution with AM-GM?
2500 already?!?!?!?!

No, I don't use AM-GM inequality  :P My short solution used to a different method  :D This is $x^{2}\ge 0$


in graph $G$, every vertex has degree at least $k\geq 2$.  show that G has a path of length at least $k$ and a cycle of length at least $k+1$.

Assume that $v_{0}v_{1}...v_{l}$ is a path of $G$ , where its length isn't less than length of all another paths and $l<k$, by $\deg (v_{l})\geq k>l=|\{v_{0},v_{1},...,v_{l-1}\}|$ we have there exist a vertex $u\in G-\{v_{0},v_{1},...,v_{l-1}\}$ such that $u$ connect to $v_{l}$, so we have $v_{0}v_{1}...v_{l}u$ is path has length greater than $l$, contradiction! 2nd part of problem is similary.  :maybe:

Im sorry, but i dont see how to do the second part  :(

You need only choose a cycle of length greater than or equal to length of all another cycles.

I dont think it works :(   or maybe i just cant get it.

the graph must be simple (otherwise the claim about the cycle is wrong: for example take 2 vertices and connect them together with 3 edges~there is no cycle of length four).

if its a simple graph then look at the maximum path and its end point...it must be connected to k-1 other points other than the one directly before it on the path; since the path is maximal; all other k-1 points must also be on the path...from here its easy to see there must be a cycle of at least k+1.


function
real analysis
real analysis unsolved

Suppose that $f: \mathbf{R}\rightarrow \mathbf{R}^{2}$ is continuously differentiable. Prove that the planar outer measure of its image $f(\mathbf{R})$ is 0.

Thanks!

It suffices to prove that $f(I)$ has measure zero for every closed interval $I$. $f'$ bounded by some constant $M$ on $I$. Divide $I$ into $N$ equal subintervals $I_{k}$. Since the diameter of $f(I_{k})$ does not exceed $\frac{M|I|}{N}$, the outer measure of $f(I_{k})$ does not exceed $\frac{\pi M^{2}|I|^{2}}{N^{2}}$. Summing over $k$, you get $\frac{\pi M^{2}|I|^{2}}{N}$, and this can be made arbitrarily small.

I think that the conclusion is still true if $f$ is just differentiable ..


thalai i told ya  :D
And congrats to apek too  :D

congrats evrybody and to rak and jothi o course...

shre, u couldnt gimme back the one u owed sigh.....  :(  :(

Congratulations guys! :) And looks like Vihang got through to Physics. Congrats to him too

congratss guys!
and is abishek kar the only guy to have qualified for both?

well i just slipped of and forgot to give the paper

Congrats to all those who got thru nd more specifically for those from Chennai  :lol:


Prove that in any triangle ABC, we have:
$ cosA.cosB.cosC \leq \frac{1}{8}cos(B\minus{}C).cos(C\minus{}A).cos(A\minus{}B)$

[quote="nmtvn"]Prove that in any triangle ABC, we have:
$ cosA.cosB.cosC \leq \frac {1}{8}cos(B \minus{} C).cos(C \minus{} A).cos(A \minus{} B)$[/quote]
It's true for all reals $ A,$ $ B$ and $ C$ such that $ A\plus{}B\plus{}C\equal{}\pi.$ :wink:

[quote="nmtvn"]Prove that in any triangle ABC, we have:
$ cosA.cosB.cosC \leq \frac {1}{8}cos(B \minus{} C).cos(C \minus{} A).cos(A \minus{} B)$[/quote]
YOU SHOULDN'T POST THIS PROBLEM IN THIS FORUM. :mad:  :mad:  :mad:  :mad:  :mad:
IT IN MATHEMATICS AND YOUTH MAGAZINE IN VIETNAM.

This is insane of you to post ONGOING CONTEST PROBLEMS, so, if you do not mind, I have informed the mod :mad:


geometry
3D geometry
MATHCOUNTS
Purple Comet

Has anyone taken/going to take the purple comet math tests?

purplecomet.org

Here is a sample problem:

A cubic inch of the newly discovered material madelbromium weighs 5 ounces.  How many
pounds will a cubic yard of madelbromium weigh? (easy)

Another one:

Eight identical cubes with of size 1 × 1 × 1 each have the numbers 1 through 6 written on their
faces with the number 1 written on the face opposite number 2, number 3 written on the face
opposite number 5, and number 4 written on the face opposite number 6.  The eight cubes are
stacked into a single 2 × 2 × 2 cube.  Add all of the numbers appearing on the outer surface of the
new cube.  Let  M  be the maximum possible value for this sum, and  N  be the minimum
possible value for this sum.  Find M – N.
(harder)

I might take it if it does not conflict with any of my other competitions. So the format is similar to the Itest right?

No, it is 60 minutes, 15 problems (i think, theyv changed it before)

6 members on a team.  there's a middle school test and a hs test.  have to be from the same school.

[quote="Poincare"]i hate competitions like that "only from the same school". At my school people are obsessed with dating and sex jokes :mad: :mad: :wallbash_red:[/quote]

Same at my school! Although all of the people at my school (on the math team) who make those jokes are actually (semi-)smart. [hide="cough"]math154 and tinytim[/hide]

Yes, my school will. Last year our name was "No 8th graders allowed"

Lol...our coach missed the deadline by a day, because he didn't know it was in GMT. That's what people get for procrastinating. :wink:

How hard do you guys think the test is?  Hard MC - AMC 10 - AMC 12?

Looked at the site, goes from easy amc to easy aime/hard amc.

[quote="AIME15"]Huh?  Not even AMC?[/quote]
No.

And a lot of middle schools do AMC?? Jeez, that's soooo above my school's standards.

MY SCHOOL IS FULL OF ****ing SEX FREAKS!

Therefore, my school would fail the Purple Comet test too. (Except for me) :P

[quote="gauss1181"]My school doesn't even know about any math competitions other than MATHCOUNTS. :mad:[/quote]

Same here. :bomb:

i took the purple comet test last year w/ a few other kids from my mathteam team
some of the questions were harder than the others
we did okay...

OK being how many out of 15?

During math on friday, i had 40 minutes to take the 2007 version by myself, and i got 8/15.

That's 20 minutes less than I'm supposed to get with 5 buddies.

I have to say this is a very good test; to add to it, the writers, I think, include Titu Andrescu  :omighty:

Coach Frost and I both gave it to our kids last year.  We like it so much that we are going to start giving it to the kids at the end of the year.

It's a very good test!

yea...our school did purple comet xcept there was a website bug that day, so all our answers didn't get submitted, and basically, we did all the probs for nothing.    :mad:  :mad:  :mad:

no, had fun when doing the problems and learned something new; that's not for nothing.

[quote="shentang"][quote="AIME15"]I assume your eighth graders are...stupid?

What was your score?[/quote]

no there were two, no eighth graders allowed and eighth graders allowed. It was supposed to be for fun since our coach found the website at the last minute. One team tied for first and another got second or third, I forgot.[/quote]

Actually your team tied for third(with 2 wrong) and the 8th grade team tied for 1st(with one wrong).

For those of you who hate "all from the same school" competitions, purple comet does allow mixed school teams, you just have to get together to take it at the same time in the same place.  You just end up competing in a different division.  There were two mixed middle school team from the Phillipines that had perfect scores and won the mixed school division.

It is held in the Spring after almost everything else is over.  About a week or so before MATHCOUNTS Nationals.

[quote="mathemagician1729"]So will any prizes be awarded for top scoring teams?[/quote]

I don't think so.  We had two of top 3 scores and never got anything.

Coach Boyd, did you get anything(we tied for 1st)?

When I was in sixth grade in my old school, we won in the large school division I believe. I don't think we got anything.

[quote="JBoyd"]Didn't you get that 5000 dollar grant?   We didn't either!  We did not even get a certificate but I still like the test.  We will take it just for the academic challenge![/quote]

5000 dollar grant for winning purple comet!??

If we could convince evan odorney to come to our school...5000 bucks a year!

EDIT: math154, i can use sarcasm too.

OMG!!! Please know that sometimes people use sarcasm. :wink:

My school won't sign up, even though i told them it's free...

[quote="purplecomet.org"]
 The high school competition will consist of 25 mathematics problems each with a non-negative integer answer. The problems will have a wide range of difficulty. The middle school competition will be similar, but with 15 problems. There will be different questions for middle school teams than for high school teams (although there may be some overlap). The contest will be administered via the Internet. Teams of up to six members each select the time they wish to start the contest. Teams may log in to the contest at any time between 12:00 AM UTC Monday, April 27, 2009 and 12:00 AM UTC Monday, May 04, 2009, keeping in mind that final answers must be submitted by 12:00 AM UTC Monday. Once the team logs in to the contest, the problems will become available on the contest web site. Teams fill in answers to the questions over the next 60 (MS) or 90 (HS) minutes following their start time, but no later than 12:00 AM UTC Monday, May 04, 2009. See official rules for details. A five problem practice contest is provided.[/quote]

I don't see why not...

Why did they say they don't wanna do it?  I know you guys have an MC coach at ur school...


Find all values of $ y$ such that $ \left | \frac{2\minus{}3y}{3\plus{}y} \right | \leq \frac{1}{y}$.

hello, we will distinguish several cases, it must be $ y\ne\minus{}3$ and $ y\ne 0$, then your inequality is equivalent with $ |3y\minus{}2|\le \frac{1}{y}|y\plus{}3|$
Let $ y\geq\frac{2}{3}$ then our inequality is equivalent with $ y^2\minus{}y\minus{}1\le0$
Solving this we get the solution $ \frac{2}{3} \le y \le \frac{1}{2}\plus{}\frac{\sqrt{5}}{2}$ (1)
Now let $ \minus{}3<y<\frac{2}{3}$ After a short computation we get the solution
$ 0<y<\frac{2}{3}$ (2). The last case $ y<\minus{}3$ gives no solution, so get all $ y$
which fulfills our inequality as $ 0<y\le \frac{1}{2}\plus{}\frac{\sqrt{5}}{2}$
Sonnhard.


$ f(x)\equal{}|1\plus{}x|\minus{}|1\minus{}x|$, we define that $ f^{(2)}(x)\equal{}f(f(x))$, etc. Please find $ f^{(n)}(x)$

if $ x\ge1$, $ f^n(x)\equal{}(1\plus{}x)\minus{}(x\minus{}1)\equal{}2$
if $ x\le \minus{}1$, $ f^n(x)\equal{}(x\minus{}1)\minus{}(x\minus{}1)\equal{}\minus{}2$

if $ \minus{}1<x<1$, $ f(x)\equal{}(1\plus{}x)\minus{}(1\minus{}x)\equal{}2x$
So:
For $ n<\frac{1}{2^nx}$, $ f^n(x)\equal{}2^nx$
For $ n\ge\frac{1}{2^nx}$, $ f^n(x)\equal{}\frac{2x}{|x|}$


Okay.. This is like "easy" stuffs for me but I just get confusion on its wording and well, I just didn't get this well..

It said:

Determine the quadrant(s) in which (x,y) is located so that the condition(s) is (are) satisfied.

1. (x,-y) is in the second quadrant.
2. (-x,y) is in the fourth quadrant.

This should be easy but I just got little bit confusion on wording.

Thanks for helping me!

[hide]From the conditions, both are negative, because if any of them were positive, the conditions would not be satisfied.[/hide]

Ok...

Um... I'm confused still...

Simply, more explanation please..

Obviously, (x,y) cannot be in the second and fourth quadrant. (If you can't convince yourself, It's not too hard to prove)

We actually only need one condition. Let's use the first.

If (a,b) is in the second quadrant, a is negative and b is positive. Therefore x is negative and -y is positive. So x and y are both negative.

Then (x,y) is in the third quadrant.


Hello,

Is it allowd to use any theorem in IMO's competition ?

what kind of  "theorems" do you mean?

Naphthalin

Theorems like

In [b]inequalities[/b]:
Cauchy-shwartz
Chebychev
Weighted AM-GM
...

In [b]Gemetry[/b]
Ceva
Melanuis
.....

[size=150][b]Can we use all of them or there's a restriction?[/b][/size]

The ones mentioned you can use.  What I was told by someone who was at IMO last year:

"You can use anything which is sufficiently well-known to have a name, but if you're not sure, then give a proof in your paper.  However, it's generally a bad idea to use something which trivializes the problem completely."

You probably can't use anything with a name...
There are some properties widely known or accepted, those you can use.

But I'm pretty sure using Chebotarev's theorem is not allowed or at least very much disliked ;)

The theorems accepted are most likely those that sound like "elementary" properties at olympiad level.

[quote="Valentin Vornicu"][quote="K81o7"]However, it's generally a bad idea to use something which trivializes the problem completely."[/quote]
It is a bad idea to have such a problem in the IMO to begin with ;)[/quote]

This is true.  The reason this came up was that a problem came up in a test at MOP (US training camp) which became almost trivial if one used well-known facts about continued fractions (for example the form of their convergents).  And many people invoked that fact, saying it's well known (which it is, after all): but that made the problem very straightforward, so those people lost points.

[quote="FOURRIER"][quote="K81o7"]The ones mentioned you can use.  What I was told by someone who was at IMO last year:

"...  However, it's generally a bad idea to use something which trivializes the problem completely."[/quote]

Write the proof  :wink:[/quote]

Or even don't write that your going to use the theorem :P, just write the proof of it inside the solution.


prove that if x>y then x + 1/y(x-y) /ge 3

Maybe some restrictions on the range of $ x$ and $ y$? I guess it's $ \mathbb{R^\plus{}}$. Substitute $ x\equal{}y\plus{}z$ for $ z>0$ and the inequality follows from AmGm.

[quote="FelixD"]Maybe some restrictions on the range of $ x$ and $ y$? I guess it's $ \mathbb{R^ \plus{} }$. Substitute $ x \equal{} y \plus{} z$ for $ z > 0$ and the inequality follows from AmGm.[/quote]
wow good one

By AM-GM
$ LHS\equal{}(x\minus{}y)\plus{}y\plus{}\frac{1}{y(x\minus{}y)}\ge 3$,


trigonometry
algebra
polynomial
difference of squares
special factorizations
algebra proposed

Evaluate $ \prod_{x \equal{} 1}^{\infty}\frac{ (4x\minus{}2)^{2}}{ (4x\minus{}3) (4x\minus{}1) }$.

I think you can re-write the expression as:
$ \prod_{x \equal{} 1}^{\infty}{\frac {(\frac {(4x \minus{} 1) \plus{} (4x \minus{} 3)}{2})^2}{(4x \minus{} 1)(4x \minus{} 3)}}$

$ \equal{} \prod_{x \equal{} 1}^{\infty} \frac {1}{4} \cdot \left({2 \cdot \left(\frac {1}{4x \minus{} 3} \minus{} \frac {1}{4x \minus{} 1}\right) \plus{} 4}\right)$

$ \equal{} \prod_{x \equal{} 1}^{\infty} \frac {1}{4} \cdot \left({\frac {4}{(4x \minus{} 3)(4x \minus{} 1)} \plus{} 4}\right)$

$ \equal{} \prod_{x \equal{} 1}^{\infty}{\frac {(1 \plus{} (4x \minus{} 3)(4x \minus{} 1))}{(4x \minus{} 3)(4x \minus{} 1)}}$

[hide]
After this I didn't know how to proceed ( :blush: ) so I used MATLAB for $ 10^5$ iterations and got the answer as 1.4142 upto 4 decimal places. Does anyone know a purely mathematical method?
[/hide]

--EDITED--

In the prev sol. the step in which 1/4 is taken common is not correct bcoz in product 1/4 will be coming as many times as the number of terms:
Let us try in anather way :
the product simplifies to give :
((2^2)/1.3)*((6^2)/5.7)*((10^2)/9.11)* ... (inf)
=> ((2^2)/1.3)*(((2^2)*(3^2))/5.7)*(((2^2)*(5^2))/9.11)*... (inf)

ot can be seen clearly that 1 of the denominators cancel 1 power in numerator (3,5,7,9 ..etc)

so now we have:
((2^2)(1)*(2^2)(3)*(2^2)(5)* ... inf) upon (7)(11)(15)...inf

(a)--->numerator has AP type terms: 1,3,5,7,9...
(b)--->denominator has AP type terms: 7,11,15,19...

we know (b) is contained in (a)
so that denominator vanishes completely and we are left with a infinite product.
hence the product is infinite...

[b]but now an anomaly:

this product can be converted in to the form:
prod. ((n^2)/(n-1)(n+1)) from n=2 to inf.

=> prod. ((n^2)/(n^2 -1))
=> prod. (1+(1/2)/(n^2 -1))
=> prod.(1)+prod.(1/2(n^2 -1)
=>1+ (productof reciprocal of numbers which are increasing)

as the number of terms increase the 2nd part will increasingly tend to 0 so that we can say that the product will become 1 .

I donot know which is correct....[/b]

[quote="madness"]I think you can re-write the expression as:
$ \prod_{x \equal{} 1}^{\infty}{\frac {(\frac {(4x \minus{} 1) \plus{} (4x \minus{} 3)}{2})^2}{(4x \minus{} 1)(4x \minus{} 3)}}$[/quote]

Nice!

It looks like $ \sqrt{2}$.  (See the [url=http://en.wikipedia.org/wiki/Square_root_of_2#Series_and_product_representations]Wikipedia[/url] entry.)

[quote="pratique_21"]In the prev sol. the step in which 1/4 is taken common is not correct bcoz in product 1/4 will be coming as many times as the number of terms:
[/quote]

Sorry, that $ \frac{1}{4}$ should have been inside the product symbol and not outside.

[quote="nsato"]It looks like $ \sqrt {2}$.  (See the [url=http://en.wikipedia.org/wiki/Square_root_of_2#Series_and_product_representations]Wikipedia[/url] entry.)[/quote]

Has anyone got a purely mathematical proof for it?

[hide]The roots of $ \cos(x)$ are in the form $ \pm \dfrac{(2k \minus{} 1)\pi}{2}$ for some $ k \in \mathbb{N}$.

Expressing $ \cos(x)$ as a constant $ C$ times the product of $ (1 \minus{} \frac {x}{r_i})$ over all roots $ r_i$:
$ \cos(x) \equal{} C\left(\prod_{k \equal{} 1}^{\infty}\left(1 \minus{} \dfrac{2x}{(2k \minus{} 1)\pi}\right)\right)\left(\prod_{k \equal{} 1}^{\infty}\left(1 \plus{} \dfrac{2x}{(2k \minus{} 1)\pi}\right)\right)$ $ \equal{} C\prod_{k \equal{} 1}^{\infty}\left(1 \minus{} \dfrac{4x^2}{(2k \minus{} 1)^2\pi^2}\right)$

Where $ C \in \mathbb{R}$. Plugging in $ x \equal{} 0$ yields $ \cos(0) \equal{} C \prod_{k \equal{} 1}^{\infty}1 \Rightarrow C \equal{} 1$.

Thus:
$ \cos(x) \equal{} \prod_{k \equal{} 1}^{\infty}\left(1 \minus{} \dfrac{4x^2}{(2k \minus{} 1)^2\pi^2}\right)$

Plug in $ x \equal{} \dfrac{\pi}{4}$ to get:
$ \cos\left(\dfrac{\pi}{4}\right) \equal{} \prod_{k \equal{} 1}^{\infty}\left(1 \minus{} \dfrac{4\left(\dfrac{\pi}{4}\right)^2}{(2k \minus{} 1)^2\pi^2}\right)$

$ \dfrac{1}{\sqrt {2}} \equal{} \prod_{k \equal{} 1}^{\infty}\left(1 \minus{} \dfrac{1}{4(2k \minus{} 1)^2}\right)$

$ \dfrac{1}{\sqrt {2}} \equal{} \prod_{k \equal{} 1}^{\infty}\left(\dfrac{(4k \minus{} 2)^2 \minus{} 1}{(4k \minus{} 2)^2}\right)$

$ \dfrac{1}{\sqrt {2}} \equal{} \prod_{k \equal{} 1}^{\infty}\left(\dfrac{(4k \minus{} 1)(4k \minus{} 3)}{(4k \minus{} 2)^2}\right)$

Take the reciprocal of both sides to get:
 $ \prod_{k \equal{} 1}^{\infty}\left(\dfrac{(4k \minus{} 2)^2}{(4k \minus{} 1)(4k \minus{} 3)}\right) \equal{} \sqrt {2}$[/hide]

hello, we have $ \prod_{x\equal{}1}^n\frac{(4x\minus{}2)^2}{(4x\minus{}3)(4x\minus{}1)}\equal{}\frac{4^n\Gamma(n\plus{}\frac{1}{2})^2}{\sqrt{\pi}\Gamma(2n\plus{}\frac{1}{2})}$.
Sonnhard.

[quote]Expressing  as a constant  times the product of  over all roots :

can you please explain this step...[/quote]

[hide="Traditional method to write a polynomial given the roots"]Suppose a polynomial, $ P(x)$ has $ 1,2,3,4$ as its only roots, (and each root is of multiplicity $ 1$). (I probably should have proved this part for $ \cos(x)$).

Then we could write $ P(x) \equal{} K(x \minus{} 1)(x \minus{} 2)(x \minus{} 3)(x \minus{} 4)$ where $ K$ is a real, non-zero constant.

Notice that plugging in $ x \equal{} 1$, $ x \equal{} 2$, $ x \equal{} 3$, $ x \equal{} 4$ makes $ (x \minus{} 1)$, $ (x \minus{} 2)$, $ (x \minus{} 3)$, $ (x \minus{} 4)$ (respectively) equal to $ 0$, and thus, $ P(x) \equal{} 0$ for those values of $ x$.

In general, if the roots of a polynomial are $ r_1,r_2, \ldots, r_{n \minus{} 1}, r_n$, then we can write the polynomial as:
$ P(x) \equal{} K(x \minus{} r_1)(x \minus{} r_2) \cdots (x \minus{} r_{n \minus{} 1})(x \minus{} r_n)$.

And for $ i \equal{} 1,2,\ldots,n \minus{} 1,n$ plugging in $ x \equal{} r_i$ makes $ (x \minus{} r_i) \equal{} 0$, and thus, $ P(r_i) \equal{} 0$.

So, we can write $ P(x)$ as a constant $ K$ times the product of $ (x \minus{} r_i)$ over all roots $ r_i$. [/hide]
----------------------------------------------------------------------------------------------------
[hide="Alternative method to write a polynomial given the roots"]Alternatively, if $ P(x)$ has $ 1,2,3,4$ as its only roots, (and each root is of multiplicity $ 1$), we could write $ P(x) \equal{} C(1 \minus{} \frac {x}{1})(1 \minus{} \frac {x}{2})(1 \minus{} \frac {x}{3})(1 \minus{} \frac {x}{4})$ where $ C$ is a real, non-zero constant.

Notice that plugging in $ x \equal{} 1$, $ x \equal{} 2$, $ x \equal{} 3$, $ x \equal{} 4$ makes $ (1 \minus{} \frac {x}{1})$, $ (1 \minus{} \frac {x}{2})$, $ (1 \minus{} \frac {x}{3})$, $ (1 \minus{} \frac {x}{4})$ (respectively) equal to $ 0$, and thus, $ P(x) \equal{} 0$ for those values of $ x$.

In general, if the roots of a polynomial are $ r_1,r_2, \ldots, r_{n\minus{}1}, r_n$, then we can write the polynomial as $ P(x) \equal{} C(1 \minus{} \frac {x}{r_1})(1 \minus{} \frac {x}{r_2})\cdots(1 \minus{} \frac {x}{r_{n \minus{} 1}})(1 \minus{} \frac {x}{r_n})$.

And for $ i \equal{} 1,2,\ldots,n \minus{} 1,n$ plugging in $ x \equal{} r_i$ makes $ (1 \minus{} \frac {x}{r_i}) \equal{} 0$, and thus, $ P(r_i) \equal{} 0$.

So, we can write $ P(x)$ as a constant $ C$ times the product of $ (1 \minus{} \frac {x}{r_i})$ over all roots $ r_i$. [/hide]
----------------------------------------------------------------------------------------------------
[hide="Applying this alternative method to cos(x)"]The roots of $ \cos(x)$ are $ \pm \dfrac{1}{2}\pi$, $ \pm \dfrac{3}{2}\pi$, $ \pm \dfrac{5}{2}\pi$, ....

For all $ x$ that are roots of $ \cos(x)$: $ \dfrac{d}{dx}\cos(x) \equal{} \minus{} sin(x) \equal{} \pm1 \neq 0$. Thus, all roots of $ \cos(x)$ are of multiplicity $ 1$.

Since $ \cos(x)$ has infinitely many roots, it is not a polynomial. However, we can try to write it as a polynomial with infinitely many zeros:

$ \cos(x) \equal{} C\left(1 \minus{} \frac {x}{\frac {\pi}{2}}\right)\left(1 \minus{} \frac {x}{ \minus{} \frac {\pi}{2}}\right)\left(1 \minus{} \frac {x}{\frac {3\pi}{2}}\right)\left(1 \minus{} \frac {x}{ \minus{} \frac {3\pi}{2}}\right)\left(1 \minus{} \frac {x}{\frac {5\pi}{2}}\right)\left(1 \minus{} \frac {x}{ \minus{} \frac {5\pi}{2}}\right)\cdots$

Some easy simplification yields:
$ \cos(x) \equal{} C\left(1 \minus{} \frac {2x}{\pi}\right)\left(1 \plus{} \frac {2x}{\pi}\right)\left(1 \minus{} \frac {2x}{3\pi}\right)\left(1 \plus{} \frac {2x}{3\pi}\right)\left(1 \minus{} \frac {2x}{5\pi}\right)\left(1 \plus{} \frac {2x}{5\pi}\right)\cdots$

Using the product notation, this becomes:
$ \cos(x) \equal{} C\left(\prod_{k \equal{} 1}^{\infty}\left(1 \minus{} \frac {2x}{(2k \minus{} 1)\pi}\right)\right)\left(\prod_{k \equal{} 1}^{\infty}\left(1 \plus{} \frac {2x}{(2k \minus{} 1)\pi}\right)\right)$

We can combine the two products into one product by using the difference of squares identity for corresponding values of $ k$:
$ \cos(x) \equal{} C\prod_{k \equal{} 1}^{\infty}\left(1 \minus{} \frac {4x^{2}}{(2k \minus{} 1)^{2}\pi^{2}}\right)$[/hide]

The ideas are nice (the same as the "Eulerian" proof of $ \zeta(2)\equal{}\pi^2/6$) but it is no where near rigorous. I think that actually proving the infinite product is equal to $ \cos x$ is hard.


It can't be \[\underbrace{888\cdots8}_{2005}\] either ;)

[quote="joml88"]This is a challenge problem...leave it for someone else if it is too easy for you.

Can the digital sum of a perfect square ever be 888?

The digital sum of a number is just the sum of its digits e.g. the digital sum of 4554 is 4+5+5+4=18.[/quote]
is this like the sum of the coefficients in a square?
(ie, 1+2+1)?

i think this could help:

[hide="solution"]

any number 121, 14641,...., e.g. the numbers formed by the even rows of the Pascal's triangle, are squares in base 10 even if they are in any other base. That is because when we try to turn those numbers from a base A to base 10,

it becomes $A^2+2A+1$ and $A^4+4A^3+6A^2+4A+1$, which are squares.

adding those digits, we have to have some forms of 2^n, which 888 is not...
[/hide]

nvm.. .that didnt cover EVERY SINGLE square.

[quote="paladin8"][hide="hint"]
look at divisibility
[hide="solution"]
if the sum of the digits is 888, then the number is divisible by 3. however, it also means that the number is not divisible by 9. thus it cannot be a square.

[/hide][/hide][/quote]
yes, but what about gyans question?


Consider two arbitrary  $2\times 2 $  matrices $  A, B . $ Find an algorithm to multiply these  matrices using only [b]seven multiplications[/b].
 Find an instance  $   A_0,B_0 $ of such matrices for which the product $A_0B_0$ can'nt be found by means of only  [b] six multiplications.[/b]

[quote="fleeting_guest"]There is a much stronger sense in which Strassen's method is (recently proved to be)
optimal:
http://www.math.tamu.edu/~jlands/Lmatrix1104.pdf[/quote]
Thank you very much for this reference ! Regards,Alex


Hi,

I never had time to take a CS course, so I have no background in CS whatsoever. (I have NO idea how a computer works, even though i'm currently touching one).

Is there a really good textbook I can use to learn how to program computers?  And to learn how a computer works on the hardware level?

Are there good websites I can use to learn how to program?  I'd like to master programming this summer.
Also, what kind of languages should I learn first?

Thanks.

You can't actually master programming by sitting through the summer, :lol: However, you can gradually bolster your programming skills.

[quote] Also, what kind of languages should I learn first?
[/quote]

In regards to this,IMO, I would advise- you would be better of by first learning 'C' programming language.

[quote="Aristocrat"]Actually, learn JAVA first, then move on to C after you get used to the process of programming.[/quote]

Nay, too many errors in Java. C is safe too, if you either know what you're doing or don't do much with it. For starters, it won't be much of a problem.

Perhaps learn Visual Basic. (I'm waiting for the beating...)

Well, I know I need to learn a language, but how?

Where do I start? What should I start doing?

Things I know I should do:
1st: buy or download a compiler

Other than that, I don't really know what to start with.  I mean with physics or math, I never had a hard time knowing where to start, because... I don't know it's just obvious how to study it, you know?  With CS, I don't even know what studying CS really means.

My opinion on this is probably a little un-orthodox (and to be honest, I've never been in a programming contest and have no advice on how to excel in that discipline) but if you're feeling overwhelmed and really have NO idea where to start, go here: download, read, program:
[url]http://www.htdp.org/[/url]

Once you've wet your feet a bit, pick up a good book on programming in C (I don't know of any exceptional ones, I learned out of "C programming: A Modern Approach" by K.N. King which I thought was pretty good).

If HtDP feels a little too mickey-mouse for you (but to be honest, is really an excellent start IMO) then you can go here instead:

[url]http://mitpress.mit.edu/sicp/[/url]

dude knuth's art of computer programming is really good but really hardcore
also just try out coding small programs for experience and get a reference book to look stuff up.

and about assembler language?Could you give me any advice for the beginer.I am ready quite fluent with C language,but It seems to me very diffcult to understand something about system of computers,and assembler.All my book written in Vietnamese are very incomprehensible,
sorry for breaking into your discussion :oops:

I second the SICP recommendation. It is still the gold standard for an intro CS text.

All you need;
http://www.amazon.com/Programming-Challenges-Steven-S-Skiena/dp/0387001638/ref=sr_1_1?ie=UTF8&s=books&qid=1275302539&sr=8-1

Good luck,
Robert Otome

For learning how a computer works, try out [url=http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-004-computation-structures-spring-2009/index.htm]MIT's 6.004 on MIT OCW[/url].  Awesome class.  It had a textbook like 25 yrs ago but that hasn't been used in recent years: http://books.google.com/books?id=m1973nU88yoC&printsec=frontcover&dq=computation+structures+ward+halstead&hl=en&ei=XykHTJqvMYK0lQeK0YGKDg&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCgQ6AEwAA#v=onepage&q&f=false


Find all solutions in integers of $x^{3}+2y^{3}=4z^{3}$.

A typical application of the infinite descent.

We will show that $(0, 0, 0)$ is the only solution.
Let us assume that $(x, y, z)$ is a solution not of that form.
Then we notice that $x$ must be even, so we may let $x = 2k,$ for some $k \in \mathbb{Z}.$ Hence, our equation rewrites
\[(2k)^{3}+2y^{3}= 4z^{3},\]
or
\[(-y)^{3}+2z^{3}= 4k^{3},\]
so that $(-y, z, \frac{x}{2})$ is also a solution. Continuing like this, we get that $(-z, \frac{x}{2},-\frac{y}{2}),$ and so on and so forth, and we'll eventually get a solution containing an odd number, which is the desired contradiction.

for none of $ x,y,z\equal{}0$
we may assume that $ gcd(x,y,z) \equal{} 1$
and from $ a^3 \equal{} 0, \minus{} 1,1 (mod 9)$ will lead to the solution such that there is no $ (x,y,z)$ possible if none of $ x,y,z$ is equal to 0
if one of them is 0 the rest is trivial


i am wondering what are some books that i can learn inequalities from.
i have AoPS and and can do a few of the inequality problems in there. i also have Art and Craft of Problem Solving, but really do not like it since i find that it does not provide enough in terms of examples of when u would want to use a certain inequality (really i find the book as a whole  to be ok not as good as AoPS). well in ACPS i am able to do about 1 or 2 problems in its inequality section.

so could you please provide the names of books that start at about AoPS level and work up.


abstract algebra
group theory
topology
function
superior algebra
superior algebra solved

Is it possible to show that a group $G$ of order $p^{2}$ is abelian without using the class-equation? Our professor challenged us to find an elementary proof for this. :D

-oo: How do you prove |Z(G)|=p "without using the class-equation"?

Simply re-prove the equation in this special case?

ups, i forgot the case Z(G)=1 :-). in deed, it's interesting whether there's an "elementary" proof ;-).

If we can put a non-trivial element in the center, aren't we done then? Or on which other manner can we attack this problem?

An hint:
If $G$ is monogenous, then it must be cyclic, so it is abelian.
Otherwise, every element of $G$ has order $p.$

[b]LemmaI.[/b] Let $G$ be group of order $p^{2}$, then every normal subgroup of order $p$ is included in the center of $G$. (I do not post the proof here, it is quite straightforward.)

[b]LemmaII.[/b] Let $G$ be a finite group and $H$ be a normal subgroup of $G$ such that $|G|\nmid [G: H]!$, then $H$ contains a normal and nontrivial subgroup of $G$.(hence $G$ is not simple.)

Every group, $G$ of order $p^{2}$ is abelian. [b]Proof[/b]:Suppose the opposite, then every nonzero element of $G$ is of order $p$. Let $H=(a)\ne (e)$ then $H$ is a normal subgroup of $G$(You can prove it using [b]LemmaII[/b]). and $b\in G-H$, then $K=(b)$ is another normal subgroup of $G$. It is clear that $H\cap K=(e)$.[b]Lemma I [/b]yields $H\subset Z(G),\; K\subset Z(G)$, so $HK\subset Z(G)$
hence $|Z(G)|>p$, or $Z(G)=G$.

[quote="The Captain"][b]LemmaII.[/b] Let $G$ be a finite group and $H$ be a normal subgroup of $G$ such that $|G|\nmid [G: H]!$, then $H$ contains a normal and nontrivial subgroup of $G$.(hence $G$ is not simple.)
[/quote]

Sorry, But I have no clue how you came to this... I'm in a position where I've just learned the class equation. Can I look this theorem up somewhere on wikipedia? Since I suppose that your $!$ stands for 'factorial', it must have  to do something to do with the symmetric group. And that might not be out of reach.

This is a famous theorem usually stated after Cayley's theorem, see for example [i]Topics in algebra [/i]by[i] Herstein[/i]. It follows directly from a theorem that is somehow a generalization of Cayley's theorem. I state it here and you may be able to prove it yourself folowing a procedure we've learned from the usual proof of Cayley's theorem.

[b]Theorem.[/b] Let $G$ be a group, $H$ one of its subgroups and $S$ be the set of all right cosets of $H$ in $G$, then there exists a homeomorphism from $G$ into $A(S)$ with its kernel being the largest normal subgroup of $G$ contained in $H$.
$A(S)$ is the alternating group formed by $S$'s elements ie the set of all functions from $S$ onto $S$. Note that in the case $H=(e)$ Cayley's theorem follows. I hope this would help. :)


Does anyone have any idea where all these people are going to stay in Madrid? Will it be like in Athens where everyone stays at a different site?
As far as I know us the coordinators will stay at the Convencion Hotel ...

It seems nobody knows anything and, sadly, IMO website doesn't contain any information... It doesn't say what's the venue for the competition itself, it doesn't say what the excursions are, it doesn't say anything about accomodation...

Yeah that makes me kind of sad. But it's not long now and we are going to find out all of these details.

I only know that contestants will stay in a school... don't have its full name...

a Highschool to be exact... at least is want I understood... as in spanish it said "colegio"

It would be a bit weird, staying in a high-school, wouldn't it? Could you even imagine that? I'm sure it's a dormitory of some kind...

In Spain the university residences are called "Colegios Mayores"

28121941, do you know where the contestants will be staying? Per http://www.consejocolegiosmayores.es/ccmm_uni.htm , there are more than a dozen of colegios mayores in Madrid.


$ a^3 \minus{} b^3 \equal{} 3b^2 \plus{} 3b \plus{} 1$

$ a^3 \equal{} b^3 \plus{} 3b^2 \plus{} 3b \plus{} 1$

$ a^3 \equal{} (b\plus{}1)^3$

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

This means that the identity only holds if $ a \equal{} b\plus{}1$.


limit
logarithms
algebra proposed
algebra

Let $ u_n$ defined by:
$ u_0>0,u_{n\plus{}1}\equal{}u_n\minus{}e^{\minus{}\frac{1}{u_n^2}}$, $ \forall n \in N$
Prove that \[ \mathop {\lim }\limits_{x \to \infty } ({u_n}^2\ln n) \equal{} 1\]

[quote="Victory.US"]Let $ u_n$ defined by:
$ u_0 > 0,u_{n \plus{} 1} \equal{} u_n \minus{} e^{ \minus{} \frac {1}{u_n^2}}$, $ \forall n \in N$
Prove that
\[ \mathop {\lim }\limits_{x \to \infty } ({u_n}^2\ln n) \equal{} 1\]
[/quote]

who can slove it  :P

[quote="Moubinool"]It seems that you can change 2 by a >1 an integer

$ u_0 > 0,u_{n \plus{} 1} \equal{} u_n \minus{} e^{ \minus{} \frac {1}{u_n^a}}$, $ \forall n \in N$

we have

$ \mathop {\lim }\limits_{x \to \infty } ({u_n}^a\ln n) \equal{} 1$[/quote]

yes, can you post your solution?


Let $ \mathcal{H}_n$ be the set of all numbers of the form $ 2 \pm\sqrt{2 \pm\sqrt{2 \pm\ldots\pm\sqrt 2}}$ where "root signs" appear $ n$ times.
(a) Prove that all the elements of $ \mathcal{H}_n$ are real.
(b) Computer the product of the elements of $ \mathcal{H}_n$.
(c) The elements of $ \mathcal{H}_{11}$ are arranged in a row, and are sorted by size in an ascending order. Find the position in that row, of the elements of $ \mathcal{H}_{11}$ that corresponds to the following combination of $ \pm$ signs: \[ \plus{}\plus{}\plus{}\plus{}\plus{}\minus{}\plus{}\plus{}\minus{}\plus{}\minus{}\]

If $ y=2+2\sin{x}$ term of $ H_{n-1}$, then $ 2\pm \sqrt{y}=2\pm 2\sin{(\frac{\pi}{4}-\frac x2 )}$  terms of $ H_n$. If $ y=2-2\sin{x}$ term of $ H_{n-1}$, then $ 2\pm \sqrt y =2\pm \sin{(\frac{\pi}{4}-\frac x2)}$ terms of $ H_n$.
Therefore all terms of $ H_n$ is $ 2\pm 2\sin{(\pm (\frac{\pi}{4} \pm (\frac{\pi}{8} (\pm ...\pm \frac{\pi}{2^{n+1}})...)}$.
b)For all terms $ 2+y$ of $ H_{n-1}$, there are 2 terms $ H_n$ - $ 2+\sqrt{2+y}, 2-\sqrt{2+y}$ with product $ 2-y$ is term $ Y_{n-1}$. Therefore all products of $ H_n$ is term $ H_0$ or $ 2$.
c)For $ N=x_{n-1}*2^{n-1}+x_{n-2}*2{n-2}+...x_0, x_i=0 \ or \ 1$ we construct term $ y_N$ from $ H_n$ by $ i$-th sign =$ +$ if $ x_{n-i}=1$ else $ -$. It give monotonic sequence of terms $ y_i>y_j$ if and only if $ i>j$.


how does one change the spacing in the itemize command and the spacing in general from line to line? thanks.


A dealer wishes to produce 300 gallons of oil worth 0.40/quart. He has oil worth 0.36/quart and oil worth 0.52/quart. How many quarts each should be he mix?

____qts of 0.36
____qts of 0.52

I understand that this is set up using the idea of first mixture + second mixture equals total mixture.

This is my set up:

x0.40 + x0.36 = 300(2x)

Is this the correct set up?

[quote="sharkman"] $ x0.40 \plus{} x0.36 \equal{} 300(2x)$

Is this the correct set up?[/quote]

I don't think this set up is appropriate, for I don't see how this directly relates to what the question is asking.

However:
[hide="Hint"]Think ratios and averages.[/hide]

[hide="what you probably want to do is..."]
Form a system of equations (one hint to do this is to see that you are asked to find 2 things). Let's call the 0.36 oil x and the 0.52 oil y.

We know they total add up to 300 gallons = 1200 quarts, so we know our first equation is

x + y = 1200.

Next, we can form another equation based on the fact that each quart is 0.40.

0.36x + 0.52y = 480

Now we just solve. We multiply the second equation first by 100, getting 36x + 52y = 48000. Then, we divide by 36, getting x + 2y = 1333 + 1/3. Now subtracting the equations, y = 133 1/3. Now we plug it back in to equation one and solve for x. Therefore

$ \boxed{1066\plus{}2/3\text{ qts of 0.36 and }133\plus{}1/3\text{ qts of 0.52}}$[/hide]

[quote="Nerd_of_the_Ages"][hide="what you probably want to do is..."]
Form a system of equations (one hint to do this is to see that you are asked to find 2 things). Let's call the 0.36 oil x and the 0.52 oil y.

We know they total add up to 300 gallons = 1200 quarts, so we know our first equation is

x + y = 1200.

Next, we can form another equation based on the fact that each quart is 0.40.

0.36x + 0.52y = 480

Now we just solve. We multiply the second equation first by 100, getting 36x + 52y = 48000. Then, we divide by 36, getting x + 2y =1333 + 1/3. Now subtracting the equations, y = 133 1/3. Now we plug it back in to equation one and solve for x. Therefore

$ \boxed{1066 \plus{} 2/3\text{ qts of 0.36 and }133 \plus{} 1/3\text{ qts of 0.52}}$[/hide][/quote]

[hide="One Small Mistake"]
[quote]getting 36x+52y=48000. Then, we divide by 36, getting x + 2y =1333 + 1/3[/quote]
$ \frac {52}{36} \not \equal{} 2$[/hide]

[hide="Another Way"]
Let $ x_1$ be the number of quarts for $ .36$ and $ x_2$ be the number of quarts for $ .52$.

If the price is determined by being the average value of the entire set, then it follows that
[list]$ \frac {.36x_1 \plus{} .52x_2}{x_1 \plus{} x_2} \equal{} .40 \Longleftrightarrow .12x_2 \equal{} .04x_1 \Leftrightarrow \frac {x_1}{x_2} \equal{} \frac {3}{1}$.

Knowing that the total amount of quarts is $ x_1 \plus{} x_2 \equal{} 1200$ and the ratio of $ x_1: x_2$ is $ 3: 1$, it's possible to find the values of $ x_1$ and $ x_2$ very quickly.[/list]
I'll allow you to finish the calculations.[/hide]


limit
integration
floor function
calculus
real analysis
real analysis unsolved

Calculate: $ \lim_{N\to\infty}{(N - 2\sum_{k=1}^{N}{k\arctan(\frac{1}{2k})}}).$

I've transformed this into

$ \int_0^{\frac12}\frac{2t^2}{1\plus{}t^2}\cdot\left\lfloor\frac1{2t}\right\rfloor\,dt$

where $ \lfloor\cdot\rfloor$ is the greatest integer function.

It's clear that this converges, but I don't yet see the next step.

I tried to use series for $ \arctan$ and get this [url]http://www.mathlinks.ro/viewtopic.php?p=1213994#1213994[/url] for the value of the limit, but don't know what to do with it. All this beguns from misan's integral [url]http://www.mathlinks.ro/viewtopic.php?t=204318[/url]  :D.

[quote="Yuriy Solovyov"]Calculate: $ \lim_{N\to\infty}{(N - 2\sum_{k = 1}^{N}{k\arctan(\frac {1}{2k})}}).$[/quote]
The answer is $ \frac{\pi^2}{72}$, not?

[quote="Kirill"][quote="Yuriy Solovyov"]Calculate: $ \lim_{N\to\infty}{(N - 2\sum_{k = 1}^{N}{k\arctan(\frac {1}{2k})}}).$[/quote]
The answer is $ \frac {\pi^2}{72}$, not?[/quote]
O, no, I have mistaken...

Numerically, the limit is $ \approx0.1254572736,$ whereas $ \frac{\pi^2}{72}\approx0.137077839.$ If $ L$ is the limit here, then $ \frac{\pi^2}{L}\approx 78.669.$

Cross-reference: also posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=218894]here.[/url]

[quote="Kent Merryfield"]I've transformed this into

$ \int_0^{\frac12}\frac {2t^2}{1 \plus{} t^2}\cdot\left\lfloor\frac1{2t}\right\rfloor\,dt$

where $ \lfloor\cdot\rfloor$ is the greatest integer function.[/quote]
Can you show me how did you get that?

How'd you evaluate the remaining integral?

To answer the second question: I didn't evaluate that integral. This is just a reformulation, and perhaps not a particularly useful one. Furthermore, I've now realized that it's wrong, so I'll have to fix it now. (I left a $ k$ out somewhere where it belongs.)

$ \arctan x\equal{}\int_0^x\frac{1}{1\plus{}t^2}\,dt\equal{} \int_0^x\left(1\minus{}\frac{t^2}{1\plus{}t^2}\right)\,dt$

$ \equal{}x\minus{}\int_0^x\frac{t^2}{1\plus{}t^2}\,dt$

Hence, $ 2k\arctan\left(\frac{1}{2k}\right)\equal{} 1\minus{}\int_0^{\frac{1}{2k}}\frac{2kt^2}{1\plus{}t^2}\,dt.$

$ N\minus{}\sum_{k\minus{}1}^N2k\arctan\left(\frac{1}{2k}\right)\equal{} N\minus{}N\plus{}\sum_{k\equal{}1}^N\int_0^{\frac1{2k}}\frac{2kt^2}{1\plus{}t^2}\,dt$

We now interchange the sum and the integral, noting that $ t\le\frac1{2k}$ whenever $ k\le\frac1{2t}.$ For each fixed $ t,$ the sum inside the integral has $ \left\lfloor\frac1{2t}\right\rfloor$ terms (or $ N$ terms, whichever is smaller), all of which are the same.

$ \sum_{k\equal{}1}^{m}2k\equal{}m(m\plus{}1),$ so if $ m\equal{}\min\left(N,\left\lfloor\frac1{2t}\right\rfloor\right)$ and take the limit as $ N\to\infty$ we get $ \left\lfloor\frac1{2t}\right\rfloor\left(\left\lfloor\frac1{2t} \right\rfloor\plus{}1\right)$

Thus,

$ \lim_{N\to\infty}\left(N\minus{}\sum_{k\minus{}1}^N2k \arctan\left(\frac{1}{2k}\right)\right)\equal{} \int_0^{\frac12}\left\lfloor\frac1{2t}\right\rfloor\left(\left\lfloor\frac1{2t} \right\rfloor\plus{}1\right) \frac{t^2}{1\plus{}t^2}\,dt$

Taking the limit as $ N\to\infty$ then produces the form I gave. This is clearly a convergent integral, as it it bounded above by $ \int_0^{\frac12}\frac{1\plus{}2t}{4(1\plus{}t^2)}\,dt,$ but it isn't easy to evaluate.

In this [url=http://pi.physik.uni-bonn.de/~dieckman/InfProd/InfProd.html#SeriesxofxInversexTangentsxxxArcustangentxx]table[/url] you find the formula

$ \sum_{k\equal{}1}^\infty \left(x\minus{}k\arctan\frac{x}{k}\right)\equal{}\frac{\pi}{24}\minus{}\frac{x}{2}\plus{}\frac{\pi x^2}{4}\plus{}\frac{x}{2}\log(1\minus{}e^{\minus{}2\pi x})\minus{}\frac{1}{4\pi}\text{PolyLog}[2,e^{\minus{}2\pi x}]$ .

For $ x\equal{}\frac12$ we don't get rid of the PolyLog function.


As an application of the Acetylene Zipper Reaction, propose a synthesis of 7-octyn-1-ol from 1-heptyne:

$ CH_3CH_2CH_2CH_2CH_2C \equiv CH \longrightarrow HC \equiv CCH_2CH_2CH_2CH_2CH_2CH_2OH$

[Note: this problem was based on a Natural Product Chemistry article which features a synthesis of the sex pheromone of the european grape vine moth ([i]Lobesia botrana[/i]), E,Z-7,9-dodecadien-1-yl acetate, from 1-heptyne.]

1-heptyne + nanh2 + formaldehyde->ch3(ch2)4c=c-ch2oh
use THP ether to protect oh group foll by nanh2 +heat(zipper reaction)

will this work???

Actually, its not necessary to protect the -OH group. And the base used should be $ H_2NCH_2CH_2CH_2NHK$.

wont meyer schuster rearrangement( rearrangement into alpha beta unsaturated compound)  take place if i dont protect the oh group. or is it only acid catalyzed rearrangement. btw, can you post the answers for some of the problems that you posted recently.

The Maeyer-Schuster rearrangement occurs preferantly with tertiary propargyl alcohols and under acid catalysis.


Perhaps do you want to mean "What is the product of this reaction?". By the way, what is PO?


calculus
integration
symmetry
logarithms
calculus computations

Well some of you might not find this hard.
I would really like to solve this problem on my own but am unable to.  If you could would you please point me in the right direction or give a hint to what I should do next if I am headed in the wrong direction.

[hide="Here is the question"]
$\displaystyle\int_2^4\frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}-\sqrt{\ln(3+x)}}dx$
[/hide]

[hide="here is what I have so far"]
well since i didnt like the sqrare roots in the denominator i first multiplied by $\sqrt{\ln(9-x)}-\sqrt{\ln(3+x)}$ and got

$\displaystyle\int_2^4\frac{\ln(9-x)-\sqrt{\ln(9-x)}\sqrt{\ln(3+x)}}{\ln(9-x)-\ln(3+x)}dx$.

from here i thought about trying $v=\ln(3+x)$ or $v=\ln(9-x)$ but after doing so i came up with an integral that was just as difficult as the first to evaluate.
[/hide]

any ideas or hints would be greatly appreciatted.  :)

[quote="maokid7"]
any ideas or hints would be greatly appreciatted.  :)[/quote]

Symmetry.

I think i have an idea about what you mean...ill go and try it.

well here is what i got.
[hide="haha I like this answer"]

so after the hint to look at symmetry i went back and did both substitutions of $v=\ln(3+x)$ and $v=\ln(9-x)$

*note: $a=\ln 5$ and $b=\ln 7$

1) $v=\ln(9-x)$,  $dx=-e^{v}dv$ and $\ln(x+3)=\ln(12-e^v)$ subbing this into the original problem we have

$\displaystyle\int_a^b\frac{v-\sqrt{v\ln(12-e^v}}{v-\ln(12-e^v)}\cdot e^vdv=I$.

2) $v=\ln(x+3)$  $dx=e^vdv$ and $\ln(9-x)=\ln(12-e^v)$ now we have

$\displaystyle\int_a^b\frac{\ln(12-e^v)-\sqrt{v\ln(12-e^v)}}{\ln(12-e^v)-v}\cdot e^vdv=$
$\displaystyle\int_a^b\frac{\sqrt{v\ln(12-e^v)}-\ln(12-e^v)}{v-\ln(12-e^v)}\cdot e^vdv=I$

3) so we have $2I=\displaystyle\int_a^b\frac{v-\sqrt{v\ln(12-e^v)}+\sqrt{v\ln(12-e^v)}-\ln(12-e^v)}{v-\ln(12-e^v)}\cdot e^vdv=\int_a^be^vdv=2$

so $I=1$
[/hide]
I am pretty sure this is right.  Thanks blahblahblah   :lol:


calculus derivative trigonometry quadratics

Try to solve it.
Find the maximum and the minimum values:
[b]a)[/b]$\frac{4xy-5x^{2}}{x^{2}+y^{2}}$
[b]b)[/b]$\frac{4xy-9x^{2}}{7x^{2}+3y^{2}}$

In both cases it is clear that $x=0$ is not relevant for the extrem points so we can assume $x\neq0$ and $y=kx$ for some real $k$. Then you have to mximize/minimize terms like:

$\frac{4k-5}{k^2+1}$ and $\frac{4k-9}{7+3k^2}$

which is a standard school problem which I leave to fans of derivates :lol: .

Misha

Yes,[b]Misha[/b] the problem is easy if we use a strong weapon of derivative,but I
posted the problem in the Pre-Olympiad section,and I think that it would be better the problem to be solved not by routine derivative counting,but by nice [i]olympic[/i] way(without using derivatives).So I suggest to solve the problem by putting
$x=r\cos{\phi},y=r\sin{\phi}$,or $\sin{\phi}=\frac{x}{\sqrt{x^{2}+y^{2}}},\cos{\phi}=\frac{y}{\sqrt{x^{2}+y^{2}}}$
Using this we will get more beautiful solution.Do you agree with me [b]Misha[/b]? :)

Why?  There's a simple solution that doesn't involve some non-intuitive substitution (at least, it's non-intuitive to me).

First, I'll assume that x and y are both real; otherwise, there is no maximum.

It is clear that, for an ordered pair (a,b) that yields the maximum in either of those expressions, any ordered pair (ka,kb) where k is a real number will also yield the maximum.  Thus, we can scale it so x=1, and we are left with, for the first problem, the following expression to maximize:

\[\frac{4y-5}{y^2+1}\]

Maximizing this is equivalent to finding the largest value of n for which the following has a real solution for y:

\[\frac{4y-5}{y^2+1}=n\]
\[4y-5=ny^2+n\]
\[ny^2-4y+n+5=0\]

A value of n for the above is attainable if and only if the discriminant of the above quadratic is nonnegative.  The discriminant is

\[(-4)^2-4n(n+5)\]
\[-4n^2-20n+16\]

Which is nonnegative if and only if $\frac{-5-\sqrt{41}}{2}\leq n\leq\frac{-5+\sqrt{41}}{2}$.  Thus, the maximum possible value for the first expression is $\frac{-5+\sqrt{41}}{2}$.

We can apply exactly the same method to the other problem.

Of course I agree with you Levon but I was so tired yesterday night that I didn't see anything except the straightforward solution (with derivates).

Misha


Hi ,

     I started to use Latex recently, I really enjoy using  it and i use to compare it with Word.

     I need to set the document to Arial Font. and i need to set the top , bottom and other measures.

    How i can set the font to arial and do all these settings. I am using WinEdt.

for the arial font, try this in the preambule

\usepackage{pslatex}
\renewcommand{\familydefault}{\sfdefault}

it sould be in arial.

to set the margins, look in "the not so short introduction to $\LaTeXe$" page 111 (the number on the page)

if you don't have this document, look for it on google, you'll find it and it's very usefull

hope it will help

ledurt

[quote="Ledurt"]for the arial font, try this in the preambule

\usepackage{pslatex}
\renewcommand{\familydefault}{\sfdefault}

it sould be in arial.[/quote]

Very good to know! For clarification, what does sf in \sfdefault stand for, and what other fonts can we get? Thanks

well, i found the answer on this site

http://www.geocities.com/mobrien_12/latex.htm

in the section "Common changes to latex default behaviour"

you sould look because i just copied and pasted...;)

anyway, "sf" stands for sans-shérif and with the package, the sans-shérif default font is arial.  this is what I understood but i'm far from being a pro

did you try to edit your margins?

ledurt


I would like you to notice that it is equivalent to $(a+b-c)(b+c-a)(c+a-b)\leq a^{2}b^{2}c^{2}$ which is equivalent to $(3-2a)(3-2b)(3-2c)\leq a^{2}b^{2}c^{2}$.Note that at most one of a,b,c is bigger then $\frac{3}{2}$.So we can have the product negative which solves the problem or we have a,b,c sidelenghts of a triangle .We have in a triangle $9R^{2}\geq \sum a^{2}$,which is equivalent to $9 a^{2}b^{2}c^{2}\geq 3(3-2a)(3-2b)(3-2c)(\sum a^{2})$(QED) :lol:

[quote="Fitim"]Why if $F(a,b,c)-F\left(a,\frac{b+c}{2},\frac{b+c}{2}\right)$>=0; then $F(a,b,c)$>=0 ????!  :|[/quote]
If $F(a,b,c)-F\left(a,\frac{b+c}{2},\frac{b+c}{2}\right)\geq0$
and $F(a,b,b)\geq0$ for all $a\geq0,$  $b\geq0$ then $F(a,b,c)\geq0.$
Since, $F(a,b,c)\geq F\left(a,\frac{b+c}{2},\frac{b+c}{2}\right)\geq0.$ :wink:


calculus
integration
function
real analysis
probability and stats

X~Gamma(a,p)
Y~Gamma(b,p)
X&Y are independent

HOW to prove X+Y~Gamma(a+b,p)?
[url]http://img695.imageshack.us/img695/1406/200911181024.jpg[/url]

Is there any wrong step in the above computation?
I dont know how to continue....

You are correct.

The expression you obtained contains a beta integral

$ B(a,b) \equal{} \int_{x\equal{}0}^1 x^{a\minus{}1} (1\minus{}x)^{b\minus{}1} \, dx \equal{} \frac{\Gamma(a)\Gamma(b)}{\Gamma(a\plus{}b)}$.

This is seen by performing the substitution

$ u \equal{} x/z, \quad du \equal{} dx/z$,

which gives

$ \int_{x\equal{}0}^z x^{a\minus{}1} (z\minus{}x)^{b\minus{}1} \, dx \equal{} \int_{u\equal{}0}^1 (uz)^{a\minus{}1} (z(1\minus{}u))^{b\minus{}1} z \, du \equal{} z^{a\plus{}b\minus{}1} B(a,b)$.

Continuing with your calculation then yields

$ f_{X\plus{}Y}(z) \equal{} \frac{p^{a\plus{}b} z^{a\plus{}b\minus{}1} e^{\minus{}\lambda z}}{\Gamma(a\plus{}b)} \sim {\rm Gamma}(a\plus{}b,p)$,

as claimed.  Therefore, all that remains is to prove the beta integral identity.  See http://mathworld.wolfram.com/BetaFunction.html for more information.

Must i learn Beta funtion?
Is there a easier proof if moment generating funtion is not used?

[quote="Tarasenya"]First of all note that your statement is true when  $ X$ and $ Y$ are independent!  :!:

The easiest method is using characteristic functions. This is the easiest in case of you know what it is.

Moment generating functions are not used neither in your solution ,nor in mine.

[quote]Must i learn Beta funtion?[/quote] Yes  :D[/quote]

searched in Wiki, is characteristic function  E(exp{tX})?
I get a wrong name  :|
My "Moment generating functions" means E(exp{tX})...

Actually, I want to prove it by using [b]convolution[/b] but I havent learn Beta funtion........

Normally, characteristic function means $ E(e^{itX})$ while moment generating function means $ E(e^{tX}).$ The former is a Fourier transform, while the latter is a Laplace transform. They are, of course, quite closely related. There are theoretical differences in when they can be applied and what for, but those differences are irrelevant here. Either of them would work in approximately the same way (and very simply) for the narrow purposes of showing that the sum of two independent gamma distributions is a gamma distribution.

Fourier/Laplace transforms convert convolution into pointwise multiplication. That's why they work so well for this particular problem.


logarithms
floor function
number theory theorems
number theory

$ f_n \equal{} \frac {p^n \minus{} q^n}{\sqrt {5}}$ - n-th Fibonacci number, $ p \equal{} \frac {1 \plus{} \sqrt {5}}{2}$, $ pq \equal{} \minus{} 1$.

prove $ \ln f_1f_2\dots f_n\sim \frac {n^2}{2}\ln p$

$ F_n \approx \frac{p^n}{\sqrt{5}}$.
So,
\begin{eqnarray*}
\ln \prod_{i=1}^n F_i
& \approx & \ln \prod_{i=1}^n \frac{p^i}{\sqrt{5}} \\
& = & \left(\sum_{i=1}^n i\right)\ln p - \frac{n}{2} ln 5 \\
& = & \left(\frac{n^2}{2} + \frac{n}{2}\right)\ln p - \frac{n}{2} \ln 5 \\
& = & \frac{n^2}{2} \ln p + \frac{n}{2}\ln \frac{p}{5}
\end{eqnarray*}

You're welcome!
I apply this evidence and intend to announce it at next study meeting of Japan Fibonacci Association.
Thank you :lol:

May I ask one question about it ?

As I see, when you multiplied:
[quote="Kouichi Nakagawa"]$ F_n \approx \frac {p^n}{\sqrt {5}}$.
So,
\[ \ln \prod_{i = 1}^n F_i \& \approx \& \ln \prod_{i = 1}^n \frac {p^i}{\sqrt {5}}
\]
[/quote]
you used the fact $ x_n\sim y_n\Rightarrow x_1\dots x_n\sim y_1\dots y_n$.
But it isn't true (not always is true) - for example $ n\sim n+1$ but not $ n^n\sim (n+1)^n$.

How do you know that you can multiply here ?

[quote="knoppix"]May I ask one question about it ?

As I see, when you multiplied:
[quote="Kouichi Nakagawa"]$ F_n \approx \frac {p^n}{\sqrt {5}}$.
So,
\[ \ln \prod_{i = 1}^n F_i \& \approx \& \ln \prod_{i = 1}^n \frac {p^i}{\sqrt {5}}
\]
[/quote]
you used the fact $ x_n\sim y_n\Rightarrow x_1\dots x_n\sim y_1\dots y_n$.
But it isn't true (not always is true) - for example $ n\sim n + 1$ but not $ n^n\sim (n + 1)^n$.

How do you know that you can multiply here ?[/quote]

Really?
$ (n+1)^n=n^n+\mathit{O}(n^{n-1})$.
So, $ (n+1)^n \approx n^n$.

This is a problem of the order.

no, I think you had to use $ \approx$ in the meaning $ a_n\approx b_n\iff \frac{a_n}{b_n}\to 1$

then $ (n\plus{}1)^n\approx n^n$ is not true

That's because we can say much more than $ F_n \sim \frac {\phi^n}{\sqrt {5} }$.  It is the case that not only their quotients go to $ 1$, but their differences go to $ 0$:  in fact, we have precisely

$ F_n \equal{} \left\lfloor \frac {\phi^n}{\sqrt {5} } \plus{} \frac {1}{2} \right\rfloor$

which means the estimate for $ \ln F_n$ is very, very accurate.

Edit:  One has $ F_n \equal{} \frac{\phi^n}{\sqrt{5}} \left( 1 \plus{} \frac{1}{\phi^{2n}} \right)$ hence $ \ln F_n \equal{} n \ln \phi \minus{} n \sqrt{5} \plus{} O \left( \frac{1}{\phi^{2n}} \right)$, so the error term is $ O(1)$ for the sum.


Does this inequality always hold for $ a\ge b\ge c>0$???

$ ac^2\ge b^3$.

 :(

Suppose you make $ c$ very tiny.  Seems like you'll have trouble then.

For example, if $ a \equal{} b \equal{} 1$, the statement becomes: $ c^2 \geq 1$.  This is false if $ c$ is less than $ 1$ (and positive).

[hide] $ a < b < c \implies c^2 > b^2$ multiply by a to get $ ac^2 > b^2a$ however  $ b^3 > b^2a$ so its not neccessarily true.[/hide]

its for the other side
a>b>c not a<b<c  :maybe:

The left hand side depends on $ c$, but the right hand side doesn't.  So, by making $ c$ very small, you can make the left hand side as small as you like, and the right hand side will not change.  Therefore, this inequality will fail for small enough values of $ c$.

If you try different values, you can find some counterexamples pretty easily.  For example, what if $ a \equal{} 3, b \equal{} 2, c \equal{} 1$.  The inequality fails in that case.

If you're not sure if something is true, you can do some experiments like this---just try some examples and see if it works.


Find : $ x$ is prime number and y is interger number, satisfy : $ y^3\equal{}x^2\plus{}x\plus{}1$

$ y \perp x$
$ y^3 \equiv 1 \bmod{x}$
x is prime, so $ d | 3$. d -> order.
Then $ d \equal{} 1$ or $ d \equal{} 3$.
1. If $ d \equal{} 3$ then from Fermat $ 3 | x \minus{} 1 \Rightarrow 3 | x^2 \plus{} x \plus{} 1$. But you can easy check that $ 9 \nmid x^2 \plus{} x \plus{} 1$, thus contradiction (as $ y^3 \equal{} x^2 \plus{} x \plus{} 1$).
2. Hence $ d \equal{} 1 \Rightarrow y \equal{} kx \plus{} 1$ Case with $ k \equal{} 0$ is obvious. $ k > 0$, therefore $ y^3 > x^2 \plus{} x \plus{} 1$
There's no solution.


[quote="Silverfalcon"]The official email has not gone out yet because I have a lab final on Monday.

But, this thread is still official and feel free to spread the word.

The Hidden Discoveries Math Olympiad FIRST ROUND will be held on December 15th to December 21st. All problems will be sent by email, and answers must be received to me by 11:59:59 PM on December 21st EST.

Thanks.[/quote]

But how do we register?

i want to register. Please send the stuff at fordhaivat@gmail.PLEASENOSPAMcom

[color=green]No spambots please. -mod[/color]

Poincare, you might want to edit your post so that your email doesn't get spammed by bots.
You can change the "@" to a "&" or something like that.

I apologize for inability to respond. You can still register but please look at the other sticky thread in this forum because it tells you exactly how to register. Thank you.

For those who already registered, the official email will go out next Thursday. I was planning to do it earlier but my studies have been overwhelming and with finals next week, I need to survive first.  :)

The official announcement was just sent out. If you do not receive the email, then please send me a PRIVATE MESSAGE. There were some contestants that email did not work so I'll fix that through private message then.

Thanks.  :)

The official email with problems is OUT! There were three contestants that their emails failed to work so if you registered, got my other documents, but not this one, CONTACT ME VIA PM, and mention all the information and other alternate email that can work!

Also, some questions will look blur to you. They are because I took the images from texer and used them. But still, they are readable so do not panic! :)

Sincerely,

I got the email today and it doesn't say anywhere in the first two pages how long we have to take the test.

It stated in the second page that all rules applying to AMC by MAA will exist in this contest as well. So, with little thoughts, it can be inferred that you have 1 hour and 15 minutes (no returning later once started) to finish the exam.

[quote="mathemonster"][quote="Silverfalcon"]I don't know. Like I said, I don't request names of Art of Problem Solving users. I go by the real name. Send me private message with your name, grade, school, etc (it's in the other thread), and I'll let you know if you're in my database.[/quote]

why do u need that stuff instead of just the email?[/quote]

It is for security issue. It is such an easy task to find someone else's email on internet and pretend to be yours. But other private information are much harder to find, and I'll be using the area and other stuffs for statistics reason. And for any further comment like this, please use that for PM. It is highly recommended that you use the function of private message to ask any direct question relating to the format of test or questions relating to the problems in the test. I do not like this announcement thread to be spammed.

There is a typo in the question #24.

It should read:

"Find the maximum possible value of
$ (x^{4}_1 \minus{} x^{3}_n) \plus{} (x^4_2 \minus{} x^{3}_{n \minus{} 1}) \plus{} \cdots \plus{} (x^{4}_{n \minus{} 1} \minus{} x^{3}_2) \plus{} (x^{4}_n \minus{} x^{3}_{1})$"

(The sent document had $ (x^4_2 \plus{} x^{3}_{n \minus{} 1})$)

Answer choices stay same.

Just to let everyone know, the deadline for submitting your answers is TODAY! I have received only about ~31 answers so far so, please keep sending them! Don't forget!  :)


induction
combinatorics solved
combinatorics
Recurrence

A recurrence relation is defined like that:

(1) a_1 = 1
(2) a_n+1 = 1/16 * (1 + 4a_n +  \sqrt (1 + 24a_n))

and n \in N

Determine the explicit formula and prove that it is correct !

We have 1+24*a_n=(3+1/2^(n-2))^2. I just calculated the first 6 terms and tried to see what happens. I didn't do anything smart or of that kind. From here on it's just plain induction. You can see that this isn't an explicit formula for a_n, but the explicit formula for a_n is obvious from here. I just thought it looked nicer like this.

Yeah - induction is trivial but finding the formula requires an intriguing investigation of the sequence properties !  :D

The explicit formula for:

(1) a_1 = 1
(2) a_n+1 = 1/16 * (1 + 4a_n + (1 + 24a_n)^0.5)

is:

a_{n} = [(2^{n}+2)(2^{n}+1)]/(3*2^{2n})

Try to figure out the explicit formula for yourself. It requires several systematic attempts. I omit the induction as it is straight forward.


Prove: if a trapezoid $ ABCD$ HAS an inscribed circle.
Then we have $ AB \plus{} CD \equal{} AD \plus{} BC$.

[quote="unimpossible"][hide="hint"]common tangents are congruent[/hide][/quote]

really? why?

[url]http://www.artofproblemsolving.com/Wiki/index.php/Power_of_a_Point_Theorem[/url]
Look in the additional  notes section.


calculus
integration
function
calculus computations

Do you know how the product rule for functions works?
\[ d(fg) \equal{} gdf \plus{} fdg\] Or, \[ \int f dg \equal{} fg \minus{} \int g df\] Letting \[ u(x) \equal{} f(x), v(x) \equal{} g(x)\] we get, \[ \int u(x) dv(x) \equal{} u(x)v(x) \minus{} \int v(x) du(x)\] Which is another notation.


IN space are given $n$ points in general position( no three collinear and no four coplanar) Through any three of them is drawn a plae. Show that for any $n-3$ points in space there exists one of the drawn planes not passing through any of these points.

I think there is a mistake in your problem (or perhaps I misunderstood?). Otherwise, for any $n-3$ points we simply take the plane through the remaining 3 points.

The statement concerns any $n-3$ points in space, not any $n-3$ among the $n$ given.

I think I can remember solving this some time ago (not on the forum). If I remember my proof, I'll repost this message.

Edit:

Done :). I've remembered it. Myth told me about this problem a long time ago, via Messenger. I knew it looked familiar.. :)

Take any $n-3$ points in space, and take any two of our $n$ points. They determine $n-2$ planes. If all of them are to pass through at least one of the $n-3$ points, then one of the $n-3$ points must lie on the line determined by the two chosen points. This holds for any two of the $n$ points we have, and the conclusion is easy to draw from here.

Nice solution! Sorry about my mistake above...


inequalities
geometry
cyclic quadrilateral
inequalities unsolved

Let $O$ be the intersection of diagonals of quadrilateral $ABCD$. Then

\[ \frac{AB}{CD}+\frac{CD}{AB}+\frac{BC}{AD}+\frac{AD}{BC}\leq \frac{OA}{OC}+\frac{OC}{OA}+\frac{OB}{OD}+\frac{OD}{OB}. \]

[quote="Mircea Lascu"]Let $O$ be the intersection of diagonals of quadrilateral $ABCD$. Then
\[\frac{AB}{CD}+\frac{CD}{AB}+\frac{BC}{AD}+\frac{AD}{BC}\leq \frac{OA}{OC}+\frac{OC}{OA}+\frac{OB}{OD}+\frac{OD}{OB}. \]
[/quote]

Generally, this inequality is not true. (Try, for instance, a quadrilateral ABCD with $AC\perp BD$, OA = 5, OB = 1, OC = 12, OD = 1.)

However, the inequality is true if the quadrilateral ABCD is cyclic. In this case, the proof is easy:

[color=blue][b]Problem.[/b] The diagonals $AC$ and $BD$ of a convex cyclic quadrilateral $ABCD$ intersect each other at a point $O$. Prove that

$\frac{AB}{CD}+\frac{CD}{AB}+\frac{BC}{AD}+\frac{AD}{BC}\leq \frac{OA}{OC}+\frac{OC}{OA}+\frac{OB}{OD}+\frac{OD}{OB}$.[/color]

[i]Solution.[/i] Since the quadrilateral ABCD is cyclic, < ABD = < ACD. In other words, < ABO = < DCO. Similarly, < BAO = < CDO. Thus, the triangles AOB and DOC are similar; hence, $\frac{OA}{OD}=\frac{AB}{CD}$ and $\frac{OB}{OC}=\frac{AB}{CD}$. Thus, by the AM-GM inequality,

$\frac{OA}{OC}+\frac{OB}{OD}\geq 2\cdot\sqrt{\frac{OA}{OC}\cdot\frac{OB}{OD}}=2\cdot\sqrt{\frac{OA}{OD}\cdot\frac{OB}{OC}}=2\cdot\sqrt{\frac{AB}{CD}\cdot\frac{AB}{CD}}=2\cdot\frac{AB}{CD}$.

Similarly,

$\frac{OB}{OD}+\frac{OC}{OA}\geq 2\cdot\frac{BC}{DA}$;
$\frac{OC}{OA}+\frac{OD}{OB}\geq 2\cdot\frac{CD}{AB}$;
$\frac{OD}{OB}+\frac{OA}{OC}\geq 2\cdot\frac{DA}{BC}$.

Summing up these four inequalities, we get

$2\cdot\left(\frac{OA}{OC}+\frac{OB}{OD}+\frac{OC}{OA}+\frac{OD}{OB}\right)\geq 2\cdot\left(\frac{AB}{CD}+\frac{BC}{DA}+\frac{CD}{AB}+\frac{DA}{BC}\right)$.

Thus,

$\frac{OA}{OC}+\frac{OB}{OD}+\frac{OC}{OA}+\frac{OD}{OB}\geq \frac{AB}{CD}+\frac{BC}{DA}+\frac{CD}{AB}+\frac{DA}{BC}$,

so the proof is complete.

  Darij


[quote="Yimin Ge"]Let $0<x<y$ be positive reals and let
\[ H=\frac{2xy}{x+y}, G=\sqrt{xy}, A=\frac{x+y}2, R=\sqrt{\frac{x^2+y^2}2}  \]
It is well known that \[ H<G<A<R  \]
Sort the intervals $[H,G], [G,A], [A,R]$ increasingly by their length![/quote]

Let $GH,AG,RA$ be the lengths of the intervals $[G,H],\ [A,G],\ [R,A]$, respectively.
We have:
$GH=G-H=\sqrt{xy}-\frac{2xy}{x+y}=\frac{\sqrt{xy}}{x+y}\cdot (\sqrt{x}-\sqrt{y})^2$
$AG=A-G=\frac{x+y}{2}-\sqrt{xy}=\frac{(\sqrt{x}-\sqrt{y})^2}{2}$
and $RA=R-A=\sqrt{\frac{x^2+y^2}{2}}-\frac{x+y}{2}=\frac12 \cdot \frac{(x-y)^2}{\sqrt{2(x^2+y^2)}+x+y}$.
Now, consider diferences:
$AG-GH=(\sqrt{x}-\sqrt{y})^2\cdot \left(\frac12-\frac{\sqrt{xy}}{x+y}\right)=\frac{(\sqrt{x}-\sqrt{y})^4}{2(x+y)}>0$
$\implies AG>GH$

Similarly,
$RA-AG=(\sqrt{x}-\sqrt{y})^2 \left(\frac{(\sqrt{x}-\sqrt{y})^2}{2\sqrt{2(x^2+y^2}+2(x+y)}-\frac12\right)=-2\sqrt{2}(\sqrt{x}-\sqrt{y})^2 \cdot \frac{(x-y)^2}{\sqrt{2xy}+\sqrt{x^2+y^2}}<0$
$\implies RA<AG$

And:
$RA-GH=\frac12 \cdot \frac{(x-y)^2}{\sqrt{2(x^2+y^2)}+x+y}-\frac{\sqrt{xy}}{x+y}\cdot (\sqrt{x}-\sqrt{y})^2=\frac{(\sqrt{x}-\sqrt{y})^2}{2}\cdot \left(\frac{(\sqrt{x}+\sqrt{y})^2}{2\sqrt{x^2+y^2}+x+y}-\frac{2\sqrt{xy}}{x+y}\right)=\frac{(\sqrt{x}-\sqrt{y})^2}{2}\cdot \frac{x^2+2xy+y^2-2\sqrt{2xy(x^2+y^2)}}{\cdots}$
By AM-GM:
$x^2+y^2+2xy \ge 2\sqrt{2xy(x^2+y^2)}$
which follows that:
$RA-GH \ge 0$
$\implies RA>GH$
The equality doesn't occur.

As a result of this:
\[ \boxed{[G,H]<[R,A]<[A,G]}  \]


algebra
polynomial
search
sum of roots
product of roots

Should post in Prepair Olympiad , not in number theory. :wink:

[quote="Moh3en_DDD"][hide="2"]
analize this :

a^10+a^5+1

[/hide][/quote]

Monic, sum of roots is 0, product of roots is 1, no rational roots, not in bold, italics, or underlined, black color, etc.

What do you mean "analize"?

[hide="1"]Put in terms of a, b, and c, simplify:

$ \frac{(b\plus{}c\minus{}a)^3}{8}\plus{}\frac{(b\plus{}a\minus{}c)^3}{8}\plus{}\frac{(a\plus{}c\minus{}b)^3}{8}\plus{}\frac{24abc}{8}$

After combining into one fraction and bashing,

$ \frac{a^2\plus{}b^2\plus{}c^3\plus{}3(a^2b\plus{}a^2c\plus{}b^2a\plus{}b^2c\plus{}c^2a\plus{}c^2b)\plus{}6abc}{8}$

Since $ p\equal{}\frac{a\plus{}b\plus{}c}{2}$ and $ (a\plus{}b\plus{}c)^3\equal{}a^2\plus{}b^2\plus{}c^3\plus{}3(a^2b\plus{}a^2c\plus{}b^2a\plus{}b^2c\plus{}c^2a\plus{}c^2b)\plus{}6abc$, then $ (p\minus{}a)^3\plus{}(p\minus{}b)^3\plus{}(p\minus{}c)^3\plus{}3abc\equal{}p^3$.
[/hide]

[hide="1"]
Note that
$ 3abc\equal{}\sum{abc}\equal{}\sum{ab(p\minus{}a)\plus{}ab(p\minus{}b)}\equal{}\sum{a(p\minus{}a)(b\plus{}c)}\Rightarrow$
$ \sum{(p\minus{}a)^3\plus{}abc}\equal{}\sum{(p\minus{}a)^3\plus{}a(p\minus{}a)(b\plus{}c)}\equal{}\sum{(p\minus{}a)(p^2\minus{}2pa\plus{}a^2\plus{}b\plus{}c)}\equal{}\sum{(p\minus{}a)(p^2\plus{}a(a\plus{}b\plus{}c\minus{}2p))}\equal{}\sum{(p\minus{}a)p^2}\equal{}p^3$

 :)
[/hide]
[hide="2"]
What exactly do you want, the factorization? Besides...
This polynomial has appeared many times in the forum you shold search for it.

 :)
[/hide]

 :)


algebra
polynomial
trigonometry
complex analysis
complex analysis unsolved

let n be a positive integer and $ \omega\equal{}\cos \frac{2\pi}{n}\plus{}i\sin \frac{2\pi}{n}$.the $ n^{th}\minus{}cyclotomic\ polynomial$ defined by $ \phi_n(x)\equal{}\prod_{1\le k\le n\minus{}1,gcd(k,n)\equal{}1}(x\minus{}\omega^k)$ then prove the following:
$ 1)$ if $ q>1$ is an odd integer then:$ \phi_{2q}(x)\equal{}\phi_q(\minus{}x)$
$ 2)$ if $ n>1$ then:
$ \phi(1)\equal{}p$ when $ n$ is a power of prime $ p$,otherwise $ \phi(1)\equal{}1$

1) is very easy.
2) see http://www.mathlinks.ro/Forum/viewtopic.php?t=25794 .

You may also want to try http://www.mathlinks.ro/Forum/viewtopic.php?t=43028 and http://www.mathlinks.ro/Forum/viewtopic.php?t=42609 .


probability
function
combinatorics proposed
combinatorics

A machine generates $ 0$ or $ 1$ randomly, with equal probability.
It is generating a sequence of binary digits. It only stops when the number of zeroes printed are twice the number of ones.
Find the probability of an output of $ n$ digits.

Thank you very much, Mr.[b]JBL[/b]. Sorry for my idiot mistake :blush:

The number of sequences of $ 2k$ 0s and $ k$ 1s is $ \binom{3k}{k}$, not $ \binom{3k}{k}^2$ as you've written.  Thus, the probability that a random string of length $ 3k$ has twice as many 0s as 1s is $ \frac{\binom{3k}{k}}{2^{3k}}$, not whatever you wrote.  Finally, this isn't the answer to the question being asked, since we also need that the process does not end earlier, i.e. that there is no $ j < k$ such that there are twice as many 0s as 1s in the first $ 3j$ terms of the sequence.

Initially this looked to me like a generalized Catalan problem, but it's actually not.  Instead, we should have that if $ a_0 \equal{} 0$, $ a_k$ is the number of these things of length $ 3k$ for $ k \geq 1$, and $ F(x) \equal{} \sum_{k \geq 0} a_k x^k$ then
\[ \sum_{k \geq 0} \binom{3k}{k} x^k \equal{} \frac{1}{1 \minus{} F(x)}.\]
(The point here is that $ F(x)$ counts the number of sequences of 1s and 0s with twice as many 0s as 1s such that no prefix has this property; $ F(x)^2 \equal{} a_1^2 x^2 \plus{} (a_1 a_2 \plus{} a_2 a_1)x^3 \plus{} (a_1 a_3 \plus{} a_2^2 \plus{} a_3a_1)x^4 \plus{} \ldots$ counts the number of sequences of 0s and 1s with twice as many 0s as 1s with exactly one prefix that also has exactly twice as many 0s as 1s; similarly $ F(x)^3$ counts those that can be broken up into three pieces, and so $ \frac{1}{1 \minus{} F(x)} \equal{} 1 \plus{} F(x) \plus{} F(x)^2 \plus{} \ldots$ counts all of them.)

Now we can try to evaluate the generating function on the left-hand side and solve for $ F(x)$ and so for its coefficients.


[quote="Karth"]does anyone know a Connie Wu?[/quote]
I think i do. or at least a connie.

Karth Er, if you're her, or if you know her, could you ask her (or yourself) to reveal the Beta Convention Math results? Particularly, does she (or you) remember who placed in the top 3?

Posted: Mon Jun 25, 2007 4:04 pm

but I cant do that.

[quote="Karth"]does anyone know a Connie Wu?[/quote]
 no! connie who


geometry
circumcircle
geometric transformation
homothety
ratio
modular arithmetic
radical axis

Let $ ABC$ be a triangle, and let $ D$, $ E$, and $ F$ be the tangency points of its incircle with $ BC$, $ CA$, and $ AB$, respectively. Let $ I$ be the center of this incircle. Let $ X_1$ and $ X_2$ be the intersections of line $ EF$ with the circumcircle of triangle $ ABC$. Similarly, define $ Y_1$ and $ Y_2$ as well as $ Z_1$ and $ Z_2$. Prove that the radical center of the circles $ DX_1X_2$, $ EY_1Y_2$, and $ FZ_1Z_2$ lies on the line $ OI$, where $ O$ is the circumcenter of triangle $ ABC$.

[img]http://img15.imageshack.us/img15/6281/81687109.jpg[/img]

Very nice problem: http://www.mathlinks.ro/viewtopic.php?t=253661

[b]Here is my full solution to this problem[/b]:
Consider the homothety with center $ H$, ratio $ \frac {R}{r} \equal{} k$ (here $ R,r$ is understoof as its well-known name). It is an obvious fact that $ \mathcal H(H,k): (I)\mapsto (O)$. Denote by $ C_1,B_1$,respectively, be the midpoints of smaller arc $ AB,AC$. $ \mathcal H(H,k): E\mapsto C_1$, $ F\mapsto B_1\Longrightarrow EF//B_1C_1$. Let $ O_d$ be the midpoint of $ B_1C_1$, hence $ O_d$ will lie on the bisector of $ X_1X_2$. Now, we will prove that $ O_d$ is also the circumcentre of $ (DX_1X_2)$. Indeed, let $ O'_d$ be the circumcenter of $ \triangle DX_1X_2$, we have:
$ (O'_dC_1,O'_dB_1)\equiv (O'_dC_1,O'_dO_d) \plus{} (O'_dO_d,O'_dB_1)\equiv (O'_dC_1,O'_dO_d) \plus{} (O'_dO,O'_dC_1) \plus{} (O'_dO_d,O'_dB_1) \minus{} (O'_dO,O'_dC_1) \pmod \pi \equiv (O'_dO,O'_dO_d) \plus{} (O'_dO_d,O'_dB_1) \plus{} (O'_dB_1,O'_dO)\equiv 0\pmod \pi \Longrightarrow \overline {O'_d,B_1,C_1}\Longrightarrow O'_d\equiv O_d$. Arguing the same for $ O_e,O_f$.
Now, we have $ \mathcal P(D/(O)) \equal{} \overline {DY_1}.\overline{DY_2} \equal{} \overline {DZ_1}.\overline {DZ_2} \Longrightarrow \mathcal P(D/(O)) \equal{} \mathcal P(D/(O_f))$. Hence $ D$ lies on the radical axis of $ (O),(O_f)$. Denote $ H_i$ is the radical center of $ 3$ circles $ (O_e), (O_d), (O_f)\Longrightarrow DH_i\bot O_eO_f$. But $ O_eO_f//EF\Longrightarrow DH_i\bot EF$. Argue the same for $ EH_i,FH_i$. Hence, we conclude, $ H_i$ is the orthocenter of $ \triangle DEF$. Therefore, we need to prove that $ H_i,O,I$ are collinear, you can see my proof here:[u][b]http://www.mathlinks.ro/viewtopic.php?t=152601[/b][/u], which implies to the result of the problem.
Our proof is completed. :lol:
[b]p/s: $ O_d, O_e, O_f$ is the circumcenter of triangles $ DX_1X_2$, $ EY_1Y_2$, $ FZ_1Z_2$[/b]
[b]@Hophinhan[/b]: May I ask you where do you get this problem?? Of, course, it is not from my previous post, isn't it? :maybe:


The reals a, b satisfy a 3  - 3a 2  + 5a - 17 = 0, b 3  - 3b 2  + 5b + 11 = 0. Find a+b.

Here is a simmilar problem! :D

http://www.mathlinks.ro/phpBB/viewtopic.php?t=1396

cheers! :D


Probelm 1

What is the largest positive integer $n$ for which $n^3+100$ is divisible by $n+10$?

Problem 2

The pages of a book are numbered 1 through n. When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in the incorrect sum of 1986. What was the number of the page that was added twice?

Tetrahedr0n: $k\geq n$....

[hide]Switch 'em[/hide]

I don't get either of your solutions. :?

Here is how I learned from Mysticterminator  :)

[hide="Sol"]

Since we know that $a+b|a^3+b^3$ and that question says $n+10|n^3+100$, we can imply the first one and say that $n+10|n^3+1000$. So, just looking at this part:

$n^3+100$ and $n^3+1000$, we see that if we subtract them, 900 is the result. And 900 has to be divisible by $n+10$ or the whole thing wouldn't be divisible by $n+10$. So:

$900 = n+1 \Rightarrow \fbox {n = 890}$[/hide]

Yeah, your answer's right, tetrahedr0n.

[hide="Me"]

This is same as saying:

$1+2+3...n + k = \frac {n(n+1)}{2} + k$

Then $\frac {n(n+1)}{2}$ has to be largest triangular number. Plug in 62 and it works, yielding $\fbox {k = 33}$.[/hide]

Just same answer as he did.

[quote="Silverfalcon"]I don't get either of your solutions. :?

Here is how I learned from Mysticterminator  :)

[hide="Sol"]

Since we know that $a+b|a^3+b^3$ and that question says $n+10|n^3+100$, we can imply the first one and say that $n+10|n^3+1000$. So, just looking at this part:

$n^3+100$ and $n^3+1000$, we see that if we subtract them, 900 is the result. And 900 has to be divisible by $n+1$ or the whole thing wouldn't be divisible by $n+1$. So:

$900 = n+1 \Rightarrow \fbox {n = 890}$[/hide][/quote]

origamimasterjared just used long division.

Cool way to to do Problem 1:

[hide]Let n+10 = x .
So we have ((x-10)  3 + 100 )/x
Expanding, (x 3 -30x 2 +300x - 1000 + 100)/x.
Simplifying: (x 3 -30x 2 +300x - 900)/x.

Insight: x 3 -30x 2 +300x  will be divisible by x, so this problem is nothing more thann finding the largest positive x that 900 is divisible by.
Obviously x= 900, so [b]n =890.[/b][/hide]

Anyone thought 1986 AIME was too easy? My score: 5.

Yep, I thought it was easier than current AIMEs. I did far better than I do on more recent tests.

Thanks, Rep.

It was typo.

I think joml88 answered my question though. ;)

yes, 1986 was extremely easy.  i just took it and got my first double-digit score of 10...would have been 11 if i was more careful counting the cases :)

I also got a 10. I didn't think it was way easy though. I can get 10 on most AIMEs from the 80s (my average is just under 10 from 83-90 so far). However, I've also done most of them once or twice, so they aren't entirely representative of my score.

AIME 1986 did feature some quite easy problems as long as you knew a few tricks.

i envy all of you lower termers.  dam, i wish i had found out about aops earlier...much earlier.


did I study my latex commands bad, or is there something wrong:

]0;1[ gives $]0;1[$
but [0;1] gives $[0;1]$
and [0;1[ even results in $[0;1[$ !!!!

And i checked hte help, there are no different commands for [ and ], so how should I denote my intervals???

It must be something funny with the LaTeX renderer.  One workaround is to write {[} instead of [.

One possibility: if you have an open bracket immediately after a command which can take an optional argument, then LaTeX will parse the left bracket as the start of that optional argument.  For example, something like:

\\[0;1]

would produce the kind of behavior that you describe.

An easy fix is to use \left[ and \right]

test: $[0;1[$

hmm not nice with the spacing but it works

thx :)


A point charge is set in motion in the space containing magnetic($B$),electric($E$) and gravitational fields($g$) with initial velocity$v_{0}$ in the direction x.The directions are shown in the attachment.Find the components $v_{x},v_{y},v_{z}$of velocity of the point charge as a function of time.

I got
\[v_{x}= \left(v_{0}+\frac{mg}{qB}\right) \cos\omega t-\frac{mg}{qB}, \]

\[v_{y}= 0, \]

\[v_{z}= \frac{E}{B}(1-\cos\omega t), \]
where $\omega = qB/m$. Is it ok?

That's correct.But I want to ask you something.Did you use differential equations as me or you solved it in different way?


When $ x^{13} \plus{} 1$ is divided by $ x \minus{} 1$, the remainder is:

$\textbf{(A)}\ 1\qquad
\textbf{(B)}\ -1 \qquad
\textbf{(C)}\ 0 \qquad
\textbf{(D)}\ 2 \qquad
\textbf{(E)}\ \text{None of these answers}$

$ f(x)\equal{}x^{13}\plus{}1$
By Remainder Theorem:
$ f(1)\equal{}(1)^{13}\plus{}1$
The Remainder is D 2

[hide="Alternate Way"]

Remainder Theorem is the most direct way but we can also solve it by factoring.

$ x^{13} \plus{} 1 \equal{} (x^{13} \minus{} 1) \plus{} 2$

Clearly, $ x \minus{} 1|x^n \minus{} 1$ so the remainder is just 2. D. [/hide]

How exactly do you get the fact

$x-1$ is divisible by $x^n - 1$.

In addition, how valid is this method?

We can say that $x-1 = 0$ so $x = 1$

$1^{13} + 1 = 2$

Dividing by 0, our remainder is $2$.

This method only works for $x \ge 2$.
I like to think of it as a number in base x.
Let the digit d be $x-1$.
$x^n-1=dddd\ldots dddd_x$, where there are n d's.
$\frac{x^n-1}{x-1} = \frac{x^n-1}{d} = 1111\ldots1111_x$, where there are n 1's.
This is clearly an integer in base x, so it is proven.
(Note this is also the proof of the sum $1+x^1+x^2+\ldots+x^{n-1}$, just look at the end result in base x)

$x-1|x^n-1$ means that $x-1$ is a factor of $x^n-1$, not that $x^n-1$ is a factor of $x-1$ ;)
Anyway, you can tell that this is a factor by substituting $x=1$ into $x^n-1$.
As for your second method, that is exactly what you do in the remainder theorem. Say $x^{13}+1 = (x-1)q(x) + r(x)$, where $q, r$ are polynomials. We know that $\deg(r) < \deg(x-1) = 1$, so $r$ is a constant. Then if we set $x=1$ into this equation, we get $2 = (1-1)q(x) + r(x)$ so the remainder is $2$. I don't see why you'd divide by 0 at any point in this process.

Even if x=2, 2^13=8192 which added to one would be 8193, a multiple of x-1

Here's a concrete proof that $x-1$ is divisible by $x^n -1$.
We can multiply $x-1$ by $x^{n-1}+x^{n-2}...+x^2+x+1$. When we expand this using the distributive property, we get $x(x^{n-1}+x^{n-2}...+x^2+x+1)-(x^{n-1}+x^{n-2}...+x^2+x+1)=(x^n+x^{n-1}+x^{n-2}...+x^2+x)-(x^{n-1}+x^{n-2}...+x^2+x+1)$. All the terms cancel except for the $x^n$ and the $-1$, which together make $x^n -1$. Therefore, we can divide $x^n -1$ by $x^{n-1}+x^{n-2}+...+x^2+x+1$ to get $x-1$, implying that $x-1$ is divisible by $x^n -1$. $\square$


superior algebra
superior algebra unsolved

let $ B$ be tne ring $ R[x]/(1\plus{}x\plus{}...x^{p\minus{}1})$ p- prime number.To simplifity notations, we shall write x for the image of x in $ B$. hten $ x^p\equal{}1$in B. Why $ p\in (x^j\minus{}1)B$ where $ j<p$ is any positive integer.

HINT $ (x^j\minus{}1,x^p\minus{}1) \equal{} x^{(j,p)}\minus{}1 \equal{} x\minus{}1$ see http://www.mathlinks.ro/Forum/viewtopic.php?p=1426109


$ a,b,c$ positive numbers $ ab\plus{}ac\plus{}bc\equal{}1$ . Prove :

$ 3\sqrt[3]{\frac{1}{abc}\plus{}6(a\plus{}b\plus{}c)}\leq\frac{\sqrt[3]{3}}{abc}$

(a+b+c=p) and (ab+bc+ca=q) and (abc=r)

 now we have (q=1) and we know that (3pr<=1) after using this inequality we must prove that  (27r^2<=1) which is clear by AM-GM :wink: