algorithm algebra function domain polynomial

Prove that for any integers $ a,b$, $ a\neq0$, there exists a unique pair $ (q,r)$ of integers such that $ b\equal{}aq\plus{}r$ and $ 0\le r<|a|$.

How is this an "extension" of the division algorithm?  As far as I know this result is just called the division algorithm.  (Unless you're referring to the distinction between "integer" and "positive integer," which is a cosmetic change and doesn't represent a significant extension of the algorithm by any standard.)

I guess the book referred to this as the extended version because it called the statement with "positive" integers as the division algorithm... can the proof of this statement with integers be easily modified from the proof with the case of positive integers?

Yes; they're pretty much equivalent.  This is clearer if you are familiar with the generalization of the division algorithm to other Euclidean domains.

A good example is a ring of polynomials in one variable over a field, such as $ \mathbb{R}[x]$.  This is a situation in which it's clear that the distinction between "positive" and "negative" is irrelevant, since all one cares about is the degree of the polynomials in question and the division algorithm carries through regardless of the leading coefficients involved.


function
inequalities
algebra solved
algebra

2. Let a, b, c, d be positive real numbers, which satisfys ab+cd=1, points Pi (xi, yi) (i=1,2,3,4) are on the unit circle.

Prove (ay1+by2+cy3+dy4)2+(ax4+bx3+cx2+dx1)2 \leq 2[(a2+b2)/ab+(c2+d2)/cd)]

One may come up with an idea as one might think of considering a convex function in four variables to define the left hand side of the inequality. Unfortunately that function is not convex, and that can be proven by computing the second derivative. I believe I have found a solution to your problem, which is "quite" elementary:

Firstly looking at the left hand side, we make by Cauchy the following adjustment :) :
(ay1+by2+cy3+dy4)2+(ax4+by3+x+cx2+dx1)2 \leq (a2+b2+c2+d2)(y12+y22+y32+y42+x12+x22+x32+x42)=4 \sum a2 (1)

Still by Cauchy  we are taking the right hand side into the zoom:
2((a+b)2/ab+(c+d)2/cd)(ab+cd) \geq  2(a+b+c+d)2 (2)

Back to our problem, we should prove that given ab+cd=1 then (\sum a)2 \geq 2 \sum a2
This I have done using Sturm's method for solving inequalities. I will have an article soon up on Mathlinks about this.
For now consider the function f: (0,+ \infty )4 -> [b]R[/b] given by f(a,b,c,d)=( \sum )2-2 \sum a2 and suppose not all the numbers a,b,c,d are equal. WLOG suppose that a  1 such that a/b \geq e2 (i.e. a/e \geq  be). Computing the difference f(a,b,c,d)-f(a/e,be,c,d) we get:
 (1-e)(be-a)((1+e)(a- be)+2cde)/e2 \geq 0
because 1-e < 0 and be-a \leq 0. Now it is clear that the function will attain it's minimum when a=b and c=d (not necessarily a=b=c=d). But along with ab+cd=1 which now comes into a2+c2=1 we get that:
f(a,b,c,d) \geq  f(a,a,c,c)=4(a+c)2-4a2-4c2 = 4ac \geq  0 and by putting this together with (1) and (2) we are done.

Any questions? :D

[quote]siutz: For the inequality are there any mistake in the proof, i can find a counterexample take a=5, b=0.1, c=0.2, d=2.5[/quote]

yes siutz you are right   ... this is a good counterexample ... i'll have to re think the solution ...  :?

maybe you can share with us how did you came up with such an example?

i still belive that we must prove that
\sum a2 \leq 2((a+b)2/ab+(c+d)2/cd)

any other ideas are welcomed, because that's what this forum is about :)

since  (\sum a) ^2 - 2 \sum a ^2 =2-(a-c) ^2 -(b-d) ^2 +2ad+2bc
just try to make a-c and b-d larger,

in fact, many of my teamates discover that the inequality (1) does not work during the competition

another thing may be useful: we may assume the four P_i are in the first quadrant

well .. the counter example works for
 \suma2 \leq  2((a+b)2/ab+(c+d)2/cd) too :(

guess will have to find another way around this ...


calculus
induction
geometry
function
trigonometry
AoPS Books
linear algebra

I am planning on becoming a Math major if my plans to attend a different school does not go through.  I really do love math, I find it very fun to learn and the outcome/solution to some problems amaze me.  My main problem is that I have huge gaps and they'ved continued to build since I am now in Calculus III.  I started studying math just last summer, and it was the first time I ever took Calculus, or more importantly, took a math class seriously.  I have been tutoring math for the past year, courses from Algebra to Calculus II, though only a few Calculus students come now and then.  I have a decision to make on whether to take summer courses, or to use the time during the summer to continue working as a tutor to practice what I've learned and catch up on what I did not learn.

I am already self-studying material that I did not learn or learn well enough to feel satisfied, such as Geometry.  A few things that really hurt me is not having been even exposed to mathematical induction, sequences & series, and simple proof writing.

Books I am currently using for self-studying:

Geometry, Peter Selby
Algebra & Functions and Graphs & Trigonometry, I. M. Gelfand
Precalculus, Cohen
Calculus, James Stewart

I also have Spivak's Calculus book, but obviously that book is too hard for me.  I can get by doing some of his problems, but not the ones that ask me to prove stuff.  I am very application driven, I love being able to apply the tools I've learned, but I also realize that as being a math major, I will have to face courses that are abstract and demand me to prove and disprove things.  I'm a good student and highly disciplined.  I'm willing to spend days/weeks on a problem before I give in and ask for help or look at my solution manual.

When I am self-studying, I re-write the proofs and even do the derivations of the formulas myself.  I really want to learn math, and it has really helped me to do well in other courses.  I have learned to analyze my answers more thoroughly, and overall, think logically.  Math has boosted my confidence, I make less errors each day whether tutoring or doing my homework, and I know this skill is truly priceless and it will definitely carry my far into my professional career.

I have been thinking of purchasing the AoPS books to learn more, but I am not sure if I will be able to finish them all in an adequate time.

Well the AoPS books won't help you with higher-level math (Calculus, Linear Algebra,...) but they will give you a great foundation in the basics of algebra, geometry, and everything else.

For most math majors, courses like linear algebra serve as the introduction to proofs and greater abstraction. A discrete math course would also be good; those cover a wide range of material, but induction and related techniques are central.
You're not falling behind. The next courses you take will naturally address the perceived holes in your knowledge.

wow!  This thread is really old, hehe.  Thank you two for your replies!  It's been long over-dued ;)  I'm a Private tutor now and I earn pretty good money.  I'll be transferring to the Univ. of Houston, and I hope I can get a job there so that my popularity will go up even more :p

I think I've become addicted to tutoring, lol!


so hxtung ... could you tell us where are you from? :)

well.. when we join this forum there is a gap that we have to fill or not.. with our location.. I think everyone should write where they are from because no boggy man gonna them :D :)  don't you think it would be better for everyone to know where you are from? ... just me and my ideas! :D

bye bye!

[quote]don't you think it would be better for everyone to know where you are from?[/quote]

i think "would be useful" is a more appropriate wording in this particular situtation :)

helloo
valentin
  is there any way we cud change our profile or more appropriately add somthing to it"
regards

there is a button near the "You have no new messages" button (in its left) called "Profile" ... you can edit your account from there ...

if you are meaning if you can add some more forms (i.e. say a line for the school you are attending, etc.) I will have to check further in the code of phpbb2 (I'll probably end a programmer at phpbb comunity after all this :D )

oh and I forgot to mention - if you would like new items in your profile such as gender, studies, etc. please let me know - I will very soon move the board to a new location and I will upgrade it with that occasion.

best regards,
Valentin

PS where did you find those functional equations you are postin' in the unsolved problems for collge (you say something about Russian problems, could you be more specific?) ?

yeah thanks ... I figured you guys will like the flags ... :D

hey all,

im from glasgow, scotland (so im disappointed to see there is no scottish flag available! ;) but i dont mind really... hehee)

i didnt get into the UK IMO team, but i did go to the IBO (latvia 2002, gold) and the IPhO (bali 2002, bronze), and i've been part of the UK IMO training/selection programme for two years now (well, not any more, since im going to university in october).

and i have an olympiad website :)

erica

if think you should double check that flag thing (i've added the scotland flag - hope it's correct :D)  - btw bronze at the IPhO is super

yes ,bronze in ipho is really gr8,erica.and i think i cud ask for  some advice from u :D (i'll appear for the nationals physics olympiad come nov 23rd) and it wud be gr8 if i make to the ipho.
cheers

hey thanks for the scotland flag!  very cool :)

and galois if you want any help on physics i can try... but i think my bronze was really just lucky!!  good luck for your national olympiad, i hope you get into the IPhO :)

Well... Could I have a Hong Kong flag?  :D

isn't that the hong-kong flag? the one you have selected ?

Yes, it was just that I had forgotten to choose it in my profile...

[quote=samrocksnature]Did anyone else get reported by AoPSSheriff for bumping? LMAO[/quote]

no

what is the bigest bump ever is it this?

[quote=ZhengLiu]wow this is old[/quote]

not anymore.

[quote=ZhengLiu]anyway is any going to say there locatons i am at cordinents [hide][hide][hide]did... you really think... i was going.. to tell you?[/hide][/hide][/hide][/quote]

No. $           $

[quote=ObjectZ]isn't search engine older?[/quote]

he said SECOND.

[quote=samrocksnature]Heh. Had to evacuate because of Cali fires :dry:[/quote]

NOOO NOT SAM! dont get burned r u ok???


probability
integration
calculus
limit
probability and stats

Hi all,

  suppose that $ B_t$ is a brownian motion. Suppose that I know that $ B_1(\omega) = 1$. What is the probability $ P(x,a)$ that $ \int_0^1 %Error. "mathbold" is a bad command.
{1}_{[a;\infty[}(B_s(\omega)) ds$ is greater than $ x$. (in short, I am looking for the distribution of the time length above the level $ a$ knowing the value of the brownian motion $ B_t$ at time $ t = 1$)

Any ideas?

Hi, alekk!

The only thing I can say right away is that if we could answer your question in reasonable terms, we would be able to get some nice expression for the integral $ \int_0^1e^{\minus{}\frac\alpha x\minus{}\frac\beta{1\minus{}x}}\frac{dx}{\sqrt{x(1\minus{}x)}}$, which is the integral that arises if one tries to compute the expectation of that measure. I'm afraid that this integral is already not something that is easy to find, so I do not expect any decent solution to your problem either (I may mistake of course and I'll be happy to be proved wrong :)). I'll ask a couple of probabilists if they know anything like that. Meanwhile, can you tell us what exactly you need from your distribution? Perhaps, we'll be able to figure something out without knowing it explicitly.

Hi fedja!

  I may have been too optimistic because I know for example that if there is no condition at time $ t\equal{}1$ (one does not know the value of $ B_1$), then the random variable
$ X(\omega) \equal{} \int_0^1 1_{B_s(\omega) > 0} ds$
has a nice density, namely:
$ d(x) \equal{} \frac{1}{\pi} \frac{1}{\sqrt{x(1\minus{}x)}} 1_{]0;1[}(x)$
(wich is quite interesting: $ X$ is quite far from being uniform on $ [0;1]$)

Nevertheless, the extra condition at the end seems to make everything much more complicate ?

consider the set $ S^n_k$ of every sequences $ U = (a_1, a_2, \ldots, a_n) \in \{-1, +1\}^n$ such that $ \sum_1^n a_i = k$.

To each sequence $ U \in S^n_k$, one can associate the number:
$ f(U) = 1{}_{a{}_1 \geq 0} + 1{}_{a{}_1 + a_2 \geq 0} + \ldots + 1{}_{a{}_1 + a_2 _ \ldots + a_n \geq 0}$
Could you find (a good approximation of) the cardinal $ \phi(n,k,m)$ of $ \{ U \in S_n^k : f(U) \leq m \}$ ?

Ps: I am indeed looking for the limit of $ F_z(x) = \lim_{n \to \infty} \frac{ \phi(n, z \sqrt{n}, x) }{n}$ where $ 0<x<1$ and $ z \in \mathbb{R}$ wich should be equal to the distribution of the amount of time that a brownian motion stays above zero when one knows that $ B_1=z$.
Some simulations seem to show that $ F_0(x) = x$, but I cannot guess the distribution for $ z \neq 0$.

I cannot do it either but here is a piece of advice. If you ever want a formula for anything related to the Brownian motion, go to the library and ask for the book by Andrei Borodin and Paavo Salminen titled "Handbook of Brownian motion - Facts and Formulae". I don't know what state of mind one should be in to write a book like that (once you see it, you'll understand what I mean) but as a reference source it is priceless. :)

[url=http://books.google.ca/books?id=8cZrllyXfOUC&dq=handbook+of+brownian+motion+facts+and+formulae&pg=PP1&ots=VztvylAWl5&sig=HF5vz7WYZ9B73rxc-FkORsPo3Wk&hl=en&prev=http://www.google.ca/search%3Fhl%3Den%26q%3DHandbook%2Bof%2BBrownian%2Bmotion%2B-%2BFacts%2Band%2BFormulae%26btnG%3DGoogle%2BSearch&sa=X&oi=print&ct=book-thumbnail&cad=one-book-with-thumbnail#PPA9,M1]Handbook of Brownian Motion: Facts and Formulae[/url]

Who doesn't enjoy online books? :)

[color=blue][Please use the url tags when posting such a long link. -mathmanman][/color]


Hello all,

I need to teach myself Bayesian statistics.  Are there any book(s) you would recommend?  I haven't done much statistics recently, so something which doesn't require a thorough knowledge of statistics would be slightly less intimidating!   :D

Thanks,
Osud

Im just curious, since I never taken advanced probability. What is Bayesian statistics, I heard this term before but no idea what it means. Wikipedia article is confusing. In basic terms what is Bayesian?

Well...when I find a book and read it, I'll tell you!

Well...I wonder if Osud mentioned about Bayesian decision theory in statistics?


Dear Math Enthusiast,

     Our first competition of the year, the 2006 Fall Startup Event, took place on Thursday, September 28th.  Because students are asked to solve 100 problems in just 30 minutes, the problems were designed to be fast and easy for those familiar with them, requiring students to quickly recall common equations and relations, and to skip problems that they did not immediately know how to do.  The first-place winners in each division are:

School:  Academy for the Advancement of Science & Technology, Hackensack, NJ
Middle School Student:  Colby Royce, Lake Placid Christian School, Lake Placid, FL
9th-grade Student:  Damien Jiang, Cary Academy, Cary, NC
10th-grade Student:  Alex Ryan, Klein High School, Klein, TX
11th-grade Student:  Jeffrey Chen, A&M Consolidated High School, College Station, TX
12th-grade Student:  Adam Hesterberg, Garfield High School, Seattle, WA

The recognized scores in each division were:

School:  139-240
Middle School Student:  17-41
9th-grade Student:  34-60
10th-grade Student:  40-70
11th-grade Student:  49-80
12th-grade Student:  51-80

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

Thank you,

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

OH YEAH!!!

3 points lower than I thought though...

Still have LOTS of room for improvement

errr... I still have lots of room for improvement. I got a 28 =( 15th in middle school division

oh well 4 more years to go, time for preparing for geometry tournaments. First, vestavia and AMC 8

Im not gonna prepare for amc8, just vestavia tho

[quote="Elemennop"]Bleh

carelessness

coulda done better[/quote]

Eh, wtf?  You got beat by Hesterberg.  I'm sure he could have done better also.  [i]Everyone[/i] could have done better.

yea, but I bombed the first column; I skipped stuff like rt{1960} cause I thought it would take too long

I think I got 23 on the third column and 21 on the first column...it's not really that great to do better on the third column than first

but whatever

lol $196=14^{2}$.   :rotfl:  :rotfl:  :rotfl:

But seriously, that's the whole point  of the contest.  The person to make the least stupid little mistakes wins.  It's not like the winner wins on a question that no one else knows how to do.  For the top 25 or so the test is trivial.  It comes down to speed (obviously), accuracy, and the whims of that certain day.

But congrats man.  There's no shame in getting second by so few points to a former nat'l mc champion.

[quote="joml88"]lol $196=14^{2}$.   :rotfl:  :rotfl:  :rotfl: [/quote]

ya ya ya I know dont rub it in

[quote]But seriously, that's the whole point  of the contest.  The person to make the least stupid little mistakes wins.  It's not like the winner wins on a question that no one else knows how to do.  For the top 25 or so the test is trivial.  It comes down to speed (obviously), accuracy, and the whims of that certain day.

But congrats man.  There's no shame in getting second by so few points to a former nat'l mc champion.[/quote]

Yea, I know i'm whining for a stupid reason, but whatever

Does anyone know where to get a copy of the Questions for this Fall Startup.

they said that they can't be distributed or something because they are going to make a book of past tests someday. dunno but thats what i heard


If the factors of twenty-four and the multiples of four are removed from the set of integers from 1 through 50, how many integers remain in the set?

There are $ 12$ multiples of $ 4$ (those from $ 4$ to $ 48$) that need to be removed. There are $ 8$ factors of $ 24$, but four of these ($ 4$, $ 8$, $ 12$, and $ 24$) are multiples of $ 4$ and they have already been removed. So, we have $ 50 \minus{} 12 \minus{} 4 \equal{} \boxed{34}$.

[quote="gaussintraining"]There are $ 12$ multiples of $ 4$ (those from $ 4$ to $ 48$) that need to be removed. There are $ 8$ factors of $ 24$, but two of these ($ 4$ and $ 8$) are multiples of $ 4$ and they have already been removed. So, we have $ 50 \minus{} 12 \minus{} 6 \equal{} \boxed{32}$.[/quote]
what about 12 and 24? they're multiples of 4 too :)
30

Thanks, I simply overlooked those.  :oops:. But shouldn't it be $ 34$ instead of $ 30$?


[quote="math92"]A box with height (x+5) cm has a square base with side x cm.  A second box with height (x+2) cm has a square base with side (x+1) cm.  If the two boxes have the same volume , find x.[/quote]
[hide]
$(x+5)x^{2}=(x+2)(x+1)^{2}$
$x^{3}+5x^{2}=x^{3}+4x^{2}+5x+2$
$x^{2}-5x-2=0$
uh oh
$x=2.5+\sqrt{8.25}$[/hide]


At how many points do the graphs of the equations $ y \equal{} x^3$ and
$ y \equal{} x^2\plus{} 4x$ intersect?

This is asking how many real solutions does the equation $ x^3=x^2+4x$ have, which, if you had to guess, is $ \boxed{3}$.

If we want to be sure there are no double roots, we can solve:
\begin{align*}
x^3=x^2+4x &\Rightarrow x^3-x^2-4x =0 \\
&\Rightarrow x\left(x-\frac{\sqrt{17}+1}{2}\right) \left(x-\frac{1-\sqrt{17}}{2}\right)=0
\end{align*}
This has $ \boxed{3}$ solutions, which we got before.


factorial
inequalities
number theory unsolved
number theory

Find all primes $ a,b,c,d$ such that $ a!\equal{}b!\plus{}c!\plus{}d!$

suppose
$ b\geq c\geq d$
easy to check $ a > max\{b,c,d\}$
Then
$ \frac {a!}{d!} \equal{} \frac {b!}{d!} \plus{} \frac {c!}{d!} \plus{} 1$
 If $ c\geq\d \plus{} 2$ then tradition.
So $ c \equal{} d \plus{} 1$ or $ c \equal{} d$
Case1:$ c \equal{} d$
$ \frac {a!}{d!} \equal{} \frac {b!}{d!} \plus{} 2$
Then $ b\leq d \plus{} 2$
If $ b \equal{} d$ then
$ {a!}{d!} \equal{} 3$ then
$ a \equal{} 3,b \equal{} c \equal{} d \equal{} 2$
If $ b \equal{} d \plus{} 1$
$ \frac {a!}{d!} \equal{} d \plus{} 3$
then use inequalities it has no solution.
If $ b \equal{} d \plus{} 2$ then
$ \frac {a!}{d!} \equal{} (d \plus{} 1)(d \plus{} 2) \plus{} 2$
$ (d \plus{} 1)..a \equal{} (d \plus{} 1)(d \plus{} 2) \plus{} 2$
It has no solution
Case 2:$ c \equal{} d \plus{} 1$
$ \frac {a!}{d!} \equal{} \frac {b!}{d!} \plus{} d \plus{} 2$
so $ b|d \plus{} 2$ so $ b \equal{} d \plus{} 2$
Solve saminal obove.
Note :$ (a,b,c,d)$ is prime not use .

[quote="bambang"]Find all primes $ a,b,c,d$ such that $ a! \equal{} b! \plus{} c! \plus{} d!$[/quote]

Wlog say $ a>b\geq c\geq d$.
If $ c>d$, $ c!$ divides $ a!$, $ b!$ and $ c!$ but not $ d!$ so impossibility. So $ c\equal{}d$ and we have :

$ a!\equal{}b!\plus{}2c!$ with $ a>b\geq c$
Then, if $ b\equal{}c$, we have $ a!\equal{}3b!$ and the only solution $ 3!\equal{}2!\plus{}2!\plus{}2!$
Else, if $ b>c$, then $ b!$ divides $ a!$ and $ b!$ and so $ b!$ divides $ 2c!$ and $ c!<b!\leq 2c!$ and so $ b\equal{}2$ and $ c\equal{}1$ but then $ b!\plus{}2c!\equal{}4$ and no $ a$ exists.

So we have a unique solution $ 3!\equal{}2!\plus{}2!\plus{}2!$

I have found a result
Consider equation
$ (x_1)!\equal{}(x_2)!\plus{}..\plus{}(x_n)!$
where $ x_i\in N$
a)Prove that for each n the equation has a solution
b)$ n>2$ prove that the equation has finite solution .

[quote="TTsphn"]I have found a result
Consider equation
$ (x_1)! \equal{} (x_2)! \plus{} .. \plus{} (x_n)!$
where $ x_i\in N$
a)Prove that for each n the equation has a solution
b)$ n > 2$ prove that the equation has finite solution .[/quote]
[hide="a"]
$ x_{1}\equal{}n\minus{}1$, and $ x_{k}\equal{}n\minus{}2$ for $ 2 \leq k \leq n$ always works.

[/hide]
[hide="b"]
For $ x_{1}>n$, clearly all the other $ x_{k}<x_{1}$.
The maximum value of the RHS is $ (n\minus{}1)(x_{n}\minus{}1)!<x_{n}!$.
So, there is a limit of $ n^{n}$ solutions (a very weak upper bound, but we only need to show that there is a bound)[/hide]

More
Let $ (a_2,...,a_{n})\in N$
$ (x_1)!\equal{}a_2 (x_2!)\plus{}...\plus{}a_n (x_n!)$
a)Prove that the equation  has solution
b)Prove that the equation has finite solution.

we can obivously assume $ x_{1}! > x_{2}\geq x_{3}...\geq x_{n}$

a)
for a set of $ (a_{2},..,a_{n})$ such that $ a_{2} \plus{} a_{3} \plus{} ... \plus{} a_{n} \equal{} k$ there is always a solution
$ x_{1} \equal{} k$ and $ x_{2} \equal{} x_{3} \equal{} ... \equal{} x_{n} \equal{} k \minus{} 1$

b)
 if $ a_{2} \plus{} a_{3} \plus{} ... \plus{} a_{n} < x_{2}\plus{}1$ than:
$ x_{1}!\geq (x_{2} \plus{} 1)! \equal{} (x_{2} \plus{} 1)x_{2}! > (a_{2} \plus{} a_{3} \plus{} ... \plus{} a_{n})x_{2}!\geq a_{2}x_{2}! \plus{} ... \plus{} a_{n}x_{n}!$
in short terms than $ x_{1}! > a_{2}x_{2}! \plus{} ... \plus{} a_{n}x_{n}!$ which means that max of the value $ x_{2}$ is limited by the set $ (a_{2},..,a_{n})$. and since $ x_{2}\geq x_{3}...\geq x_{n}$ there is only a finite nubmer of possible solutions which means finite solutions.


geometry
ratio
trigonometry
circumcircle
angle bisector

ABC a triangle, D a point on AB, E and F projections of D on AB and AC; P intersection between BF and CE; if AP is perpendicular to BC, then AD bisects A. How to demonstrate?
Thank you all!

Projections in what sense?  I don't quite understand the problem.  Also, AD bisects A?  Do you mean that  D is the midpoint of AB?

I mean that DE is perpendicular to AB and DF is perpenpendicular to AC. And with bisects I mean that angle BAD=angle CAD.
Thank you for your answer anyway.

This attached picture is approximate, but it gives an idea...

Oh...D is on BC, not AB...that clears it up...
And I don't know how to prove that one...

$p \equiv$ "$AP$ is the altitude from the vertex $A$."
$q \equiv$ "$AD$ is the bisector of the angle $\angle A$."

One way to prove the implication $p \Rightarrow q$ is to show $\text{not}\ q \Rightarrow \text{not}\ p$. This in itself may not be easy either, but it may become easy after showing $q \Rightarrow p$, if possible.

[color=blue][b]q => p:[/b][/color] Assume that $AD$ bisects the angle $\angle A$. The angle bisector $AD$ divides the opposite side $a = BC$ in the ratio of the adjacent sides:

$\frac{BD}{CD} = \frac{AB}{AC} = \frac c b$

$BD = \frac{ac}{b + c},\ \ \ CD = \frac{ab}{b + c}$

Since $E, F$ are perpendicular projections of the point $D$ to the sides $c = AB,\ b = AC$,

$BE = BD \cos{\widehat B} = \frac{ac}{b + c}\ \cos{\widehat B},\ \ \ CF = CD \cos{\widehat C} = \frac{ab}{b + c}\ \cos{\widehat C}$

$\frac{BE}{CF} = \frac{ac \cos{\widehat B}}{b + c} \cdot \frac{b + c}{ab \cos{\widehat C}} =  \frac{c \cos{\widehat B}}{b \cos{\widehat C}}$

The right angle triangles $\triangle ADE \cong \triangle ADF$ are congruent by ASA, because the side $AD$ is common, the angles $\angle DAE = \angle DAF = \frac{\angle A}{2}$ are equal and the angles $\angle ADE = \angle ADF = 90^o - \frac{\angle A}{2}$ are also equal. Hence, $AE = AF$. Let $AP$ meet the side $BC$ at a point $Q$. Using Ceva's theorem for the cevians $AQ, BF, CE$ concurrent at the point $P$ yields

$\frac{CQ}{BQ} \cdot \frac{BE}{AE} \cdot \frac{AF}{CF} = 1$

Substituting $\frac{BE}{CF} = \frac{c \cos{\widehat B}}{b \cos{\widehat C}}$ and $\frac{AF}{AE} = 1$, we get

$\frac{CQ}{BQ} \cdot \frac{c \cos{\widehat B}}{b \cos{\widehat C}} = 1,\ \ \frac{CQ}{BQ} = \frac{b \cos{\widehat C}}{c \cos{\widehat B}}$

which means that $AQ \equiv AP$ is the altitude from the vertex $A$, $AP \perp BC$.

[color=blue][b]not q => not p:[/b][/color] Assume that $AD$ does not bisect the angle $\angle A$. Let $D'$ be the intersection of the bisector of the angle $\angle A$ with the side $BC$. Assume, for example, that the point $D$ is on the segment $BD'$. Then $BD < BD',\ CD > CD'$ and

$\frac{BE}{CF} = \frac{BD \cos{\widehat B}}{CD \cos{\widehat C}} < \frac{BD' \cos{\widehat B}}{CD' \cos{\widehat C}} = \frac{c \cos{\widehat B}}{b \cos{\widehat C}}$

Also, $\angle DAE < \angle DAF$. Then in the right angle triangles $\triangle ADE, \triangle ADF$ with the common side $AD$, $\angle ADE = 90^o - \angle DAE > 90^o - \angle DAF = \angle ADF$ and consequently, $AE > AF$. Let $AP$ meet the side $BC$ at a point $Q$. Again, using Ceva's theorem for the cevians $AQ, BF, CE$ concurrent at the point $P$ yields

$\frac{CQ}{BQ} \cdot \frac{BE}{AE} \cdot \frac{AF}{CF} = 1$

Substituting $\frac{BE}{CF} < \frac{c \cos{\widehat B}}{b \cos{\widehat C}}$ and $\frac{AF}{AE} < 1$, we get

$\frac{CQ}{BQ} \cdot \frac{c \cos{\widehat B}}{b \cos{\widehat C}} > 1,\ \ \frac{CQ}{BQ} > \frac{b \cos{\widehat C}}{c \cos{\widehat B}}$

which means that $AQ \equiv AP$ is not the the altitude from the vertex $A$, i.e., $AP$ is not perpendicular to $BC$. The possibilty that $AP$ is perpendicular to $BC$, when the point $D$ is on the segment $CD'$ is disproved similarly, by reversing the direction of all inequalities.

[quote="davi24"]How can we say $\frac{BD}{CD} = \frac{AB}{AC}$? Is it a trig theoreme?[/quote]
An internal angle bisector cuts the opposite side of a triangle in the ratio of the adjacent sides. This is a well known fact, frequently used. For example, using the sine theorem for the triangles $\triangle ABD, \triangle ACD$,

$\frac{AB}{BD} = \frac{\sin{\widehat{BDA}}}{\sin{\frac{\widehat A}{2}}},\ \ \frac{AC}{CD} = \frac{\sin{\widehat{CDA}}}{\sin{\frac{\widehat A}{2}}}$

because $AD$ bisects the angle $\angle A$. The angles $\angle BDA + \angle CDA = 180^o$ add up to a straight angle, hence, $\sin{\widehat{BDA}} = \sin{\widehat{CDA}}$. As a result,

$\frac{AB}{BD} = \frac{AC}{CD}\ \Rightarrow\ \frac{AB}{AC} = \frac{BD}{CD}$

[quote="davi24"]$BE = BD \cos{\widehat B}$
And also this, where did it come from??[/quote]
The segment $BE$ is a perpendicular projection of the segment $BD$ to the line $AB$. The angle $\angle BED = 90^o$ is right. Look at the figure.

Another prooof for the bisector theorem:
$\frac{BD}{CD}=\frac{\triangle{BDA}}{\triangle{CDA}}=\frac{AB \cdot AD}{AC \cdot AD}=\frac{AB}{AC}$

Really sorry for the stupid questions. In my picture I scrambled E and F...
Thank you all for your help!

yetti, why not prove directly that p => q ? I mean:

($T$=foot of $AP$ on $BC$)

Let $\alpha = \measuredangle BAC$, $\beta = \measuredangle ABC$, $\gamma = \measuredangle BCA$ and $a=BC$, $b=AC$, $c=AB$. We have $\cot \beta = \dfrac{EB}{ED}$, so $EB=ED \cot \beta$, and, similarly, $FC = FD \cot \gamma$. Since $\cos \beta=\dfrac{BT}{AB}$, it follows that $BT=AB \cos \beta$, and, similarly, $CT=AC \cos \gamma$.

By Ceva we have \[ \dfrac{EA}{EB} \cdot \dfrac{TB}{TC} \cdot \dfrac{FC}{FA} = 1 .\]

$\Leftrightarrow \dfrac{EA}{ED \cot \beta} \cdot \dfrac{AB \cos \beta}{AC \cos \gamma} \cdot \dfrac{FD \cot \gamma}{FA} = 1$

$\Leftrightarrow \dfrac{EA \sin \beta}{ED \cos \beta} \cdot \dfrac{AB \cos \beta}{AC \cos \gamma} \cdot \dfrac{FD \cos
\gamma}{FA \sin \gamma} = 1$

$\Leftrightarrow \dfrac{EA \sin \beta}{ED} \cdot \dfrac{AB}{AC} \cdot \dfrac{FD}{FA \sin \gamma} = 1$

$\Leftrightarrow \dfrac{EA}{ED} \cdot \dfrac{AB}{AC} \cdot \dfrac{\sin \beta}{\sin \gamma} \cdot \dfrac{FD}{FA} = 1$

By the Sine Theorem, we have $\dfrac{\sin \beta}{\sin \gamma}=\dfrac{AC}{AB}$.

$\Leftrightarrow \dfrac{EA}{ED} \cdot \dfrac{AB}{AC} \cdot \dfrac{AC}{AB} \cdot \dfrac{FD}{FA} = 1$

$\Leftrightarrow \dfrac{EA}{ED} \cdot \dfrac{FD}{FA} = 1$

$\Leftrightarrow \dfrac{FD}{ED} = \dfrac{FA}{EA} = \delta$

Therefore $FD=ED \cdot \delta$ and $FA=EA \cdot \delta$. Plugging this into $EA^2+ED^2=AD^2=FA^2+FD^2$, we get $EA^2+ED^2=EA^2 \cdot \delta ^2+ED^2 \cdot \delta ^2$, which yields $\delta = 1$, so $FD=ED$. This is equivalent to the fact that $D$ lies on the bisector of $\widehat{A}$.

is it ok ?   :?

It is a nice direct proof. I said that to show [b]not q => not p[/b] is one way to prove [b]p => q[/b], not the only way.

Regards, Yetti

[quote="shobber"]Another prooof for the bisector theorem: ...[/quote]
There are a number of different ways. For example, let $P$ be the intersection of the bisector of the angle $\angle A$ with the triangle circumcircle $(O)$. Since the angles $\angle PAB = \angle PAC = \frac{\angle A}{2}$ are equal, the chords $BP = CP$ are also equal. The quadrilateral $ABPC$ is cyclic, hence, the triangles $\triangle ABD \sim \triangle CPD$ and the triangles $\triangle ACD \sim \triangle BPD$ are similar:

$\frac{AB}{BD} = \frac{CP}{PD} = \frac{BP}{PD} = \frac{AC}{CD},\ \ \Rightarrow\ \frac{AB}{AC} = \frac{BD}{CD}$

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=5460 .

  darij

Another proof of the bisector theorem:

Let $P \in AB$ s.t. $AP=AC$. ($B$ and $P$ lie on different sides wrt $A$)

We have $\angle PAC = \pi - \angle BAC$, so $\angle APC = \frac{A}{2}$. Thus, $AD \| PC$. By Thales we get $\displaystyle \frac{AD}{DC}=\frac{AB}{AC}$.


Find all functions $ f: \mathbb {R}\to\mathbb {R}$ for which we have
\[ (x\plus{}y)(f(x)\minus{}f(y))\equal{}(x\minus{}y)f(x\plus{}y), \quad \forall x,y \in \mathbb R.\]

[quote="Gaoranger"]Find all functions $ f: \mathbb {R}\mapsto \mathbb {R}$ for which we have:

$ (x \plus{} y)(f(x) \minus{} f(y)) \equal{} (x \minus{} y)f(x \plus{} y)$[/quote]

Let $ P(x,y)$ be the assertion $ (x\plus{}y)(f(x)\minus{}f(y))\equal{}(x\minus{}y)f(x\plus{}y)$

Let $ x\in\mathbb R^*$ :

$ P(x\plus{}1,x\minus{}1)$ $ \implies$ $ f(x\plus{}1)\minus{}f(x\minus{}1)\equal{}\frac{f(2x)}x$

$ P(x\minus{}1,3\minus{}x)$ $ \implies$ $ f(x\minus{}1)\minus{}f(3\minus{}x)\equal{}(x\minus{}2)f(2)$

$ P(3\minus{}x,x\plus{}1)$ $ \implies$ $ f(3\minus{}x)\minus{}f(x\plus{}1)\equal{}(1\minus{}x)\frac{f(4)}2$

Summing these three lines implies : $ \frac{f(2x)}x\plus{}(x\minus{}2)f(2)\plus{}(1\minus{}x)\frac{f(4)}2\equal{}0$ and so $ f(2x)\equal{}\left(\frac{f(4)}2\minus{}f(2)\right)x^2\plus{}\left(2f(2)\minus{}\frac{f(4)}2\right)x$

And so $ f(x)\equal{}ax^2\plus{}bx$ $ \forall x\ne 0$.

Plugging back in the original equation, we get that this indeed is a solution. And since $ P(1,\minus{}1)$ $ \implies$ $ f(0)\equal{}0$, we get :

$ \boxed{f(x)\equal{}ax^2\plus{}bx}$ $ \forall x$


geometry
incenter
circumcircle
analytic geometry
geometry proposed

In a triangle $ ABC,$let $ BC\equal{}a,CA\equal{}b,AB\equal{}c,I$ be the incenter of the triangle.Prove that
$ a(IA)^2\plus{}b(IB)^2\plus{}c(IC)^2\equal{}abc$

$ a(IA)^2\equal{} a \frac{(p\minus{}a)^2}{cos^2{\frac{A}{2}}} \equal{} \frac{abc(p\minus{}a)}{p}$ then

$ a(IA)^2 \plus{} b(IB)^2 \plus{} c(IC)^2 \equal{} abc \frac{p\minus{}a\plus{}p\minus{}b\plus{}p\minus{}c}{p} \equal{} abc$

From the general [b]Lagrange[/b]'s identity $ (*)$ obtain $ a\cdot XA^2 + b\cdot XB^2 + c\cdot XC^2 = (a + b + c)\cdot \left[\ XI^2 - p_w(I)\ ]$ ,

where $ X$ is a point of the plane $ (ABC)$ and $ p_w(I) = - 2Rr$ is the power of the point $ I$ w.r.t. the circumcircle $ w$ of $ \triangle ABC$ .

For $ X: = I$ results $ a\cdot IA^2 + b\cdot IB^2 + c\cdot IC^2 = abc$ because $ 4Rpr = abc$ .

[color=darkred]$ (*)$ [b][u]The general Lagrange's identity.[/u][/b] For any point $ X$ of the plane $ (ABC)$ exists the relation

 $ \alpha\cdot XA^2 + \beta\cdot XB^2 + \gamma\cdot XC^2 = XP^2 - p_w(P)$ , where the fixed point $ P$ has the barycentrical coordinates

$ (\alpha , \beta , \gamma)$ with $ \alpha + \beta + \gamma = 1$ and $ p_w(P)$ is the power of the point $ P$ w.r.t. the circumcircle $ w$ of $ \triangle ABC$ .[/color]

In the particular case $ P: = I$ and $ X: = I$ obtain the required identity, where $ I(\frac {a}{2p},\frac {b}{2p},\frac {c}{2p})$ and $ p_w(I) = - 2Rr$ .


Find the last two digits of the following sum: \[ 5! \plus{} 10! \plus{} 15! \plus{} \ldots \plus{} 100!\]

$ 10!$ and every factorial above it has 2 multiples of 2 and a 2 multiples of 5 in it, therefore, they will all end in 2 zeros (since $ 2*2*5*5\equal{}100$).

So all you need to worry about is $ 5!$, which equals $ 5*4*3*2*1 \equal{} 120$, so the answer is $ 20$.

[quote="iin77"]$ 10!$ and every factorial above it has 2 multiples of 2 and a 2 multiples of 5 in it, therefore, they will all end in 2 zeros (since $ 2*2*5*5 \equal{} 100$).

So all you need to worry about is $ 5!$, which equals $ 5*4*3*2*1 \equal{} 120$, so the answer is $ 20$.[/quote]

you forgot about 6!, 7!, 8!. and 9!

6! ends in 20

7! ends in 40

8! ends in 20

and 9! ends in 80

20+20+40+20+80=80

[quote="AIME15"]mz, it's onlyh 5!+10!+..., not 5!+6!+...[/quote]

righttttt...i knew that... :oops:


Se considera un poligon convex $A_{1}A_{2}A_{3}...A_{n}$, si notam $A_{n+1}=A_{1}$. Fie $P$ un punct in interiorul poligonului. Demonstrati ca exista doua segmente $s_{1},s_{2}$ cu capetele in $\{A_{1},A_{2},...,A_{n}\}$ cu proprietatea ca \[\sum_{i=1}^{n} d(P,A_{i}A_{i+1}) < 2\pi \frac{s_{1}^{2}}{s_{2}}.\]

Fie s2 cea mai mica latura a poligonului si S aria poligonului. Atunci suma d(P, AiAj)*s2*1\2 <= S. (1)Acum fie s1 cea mai mare diagonala a poligonului, AiAj. Considerand cercul cu centrul in Ai si raza AiAj, acesta va contine in interior sau pe circumferinta toate celelalte varfuri ale poligonului fiindca AiAk<=AiAj. => S<= aria cercului=pi*s1*s1.(2)
Din (1) si (2)=> concluzia.

Sper ca ati inteles, si sper sa fie corect...Promit sa invat Latex cat mai curand :blush:  :blush:  :blush:


does anyone know when the US team will be selected? This is the end of the first week at MOP, seems like it should be about now. Has anyone heard anything?

I think the second half of the TST is today.  I'm sure AoPS moppers will update us when the team is picked.

[quote="darktreb"]Results tomorrow. Top couple of slots are probably assured. Will be interesting though.[/quote]I hope you guys also post in the [url=http://www.mathlinks.ro/Forum/index.php?f=87]IMO Forums[/url] the details, right? ;)

Good luck Zach Abel, Zhou Fan and everybody!

I thought he was taking the test as one of the highest HMs?

[quote="Valentin Vornicu"][quote="Geehoon"]Good luck Zach Abel[/quote]You're wishing him good luck for next year perhaps? ;)[/quote]

Well he could be selected this year.  Remember he got to take the TST because some winners were not US team candidates (citizenship or whatever else).

Valentin is right. I did come to MOP in black group as one of the highest HMs, and I did take the TST (which ended a few hours ago), but I am going to RSI and not IMO. But like Albert said, results tomorrow..... We'll keep y'all updated.

[quote="rrusczyk"][quote="Valentin Vornicu"]As far as I know, he already has chosen to go to RSI, so he cannot be at the IMO at the same time ;)[/quote]

I don't know - I've seen Zach pull off some pretty amazing contortions.  If anyone could pull this one off, it's Zach.[/quote]

I'll be willing to serve as your double at the IMO if you disguise me well. :)

[quote="probability1.01"]I'll be willing to serve as your double at the IMO if you disguise me well. :)[/quote] Sorry, too late. The team has been chosen.

[quote]Sorry, too late. The team has been chosen.[/quote]

Haha.

Anyway, Zach is right and the results are out. However we're not necessarily at liberty to divulge this information, as of yet. Patience :).

[quote="darktreb"]Patience :).[/quote]From your tone above, I think I know at least one person in the team :D :D

[quote="darktreb"]In my opinion, the selected team was the strongest possible team (with all possible due respect to every candidate) and certainly has a lot of potential to surprise as all this year (I certainly hope :)). If you had asked me to pick 6 without using a TST, this would have been my choice exactly (just ask the other MOPpers who I had predicted before the results were revealed)[/quote]

Do you also have a name as you have so much knowledge? Then tell us please about your expectations for the USA IMO Team!  :)

Wait.... Albert, I was hoping you to get picked.... after how you rebounded from being Non-MOPper last year to a Winner this year.... your story still inspires me to work harder. By the way, are you going to RSI too??? (that might have been the justification for you not being picked, just like Zach, Zhou, and Yi)

Does it come to as a surprise to anybody that Rishi Gupta did not make it on to the team?

[quote="xxreddevilzxx"]By the way, are you going to RSI too???[/quote]

I believe that he said he's graduating early and is thus a senior this year.

is he going to MIT? That would be the only place I would think that an early colllege entrance person will go to.

[quote="jb05"][quote="qweretyq"]wow two sophomores
makes me feel like a lazy !@#$%!#[/quote]

There are actually a whole lot of people who go multiple times. Therefore, there are often non-seniors on the team. As far as I know, the age distribution is just about normal this year.[/quote]Yeah, think about Reid, Caroll, Tiankai - all 3 went in 9th grade first time :) All 3 got gold in 9th grade also! :D

Well I think I should explain the whole "early entrance" to college thing, since I certainly don't recommend going to college directly out of junior year (except in certain extremal circumstances I suppose, but that's not for me to judge).

Basically, after my 8th grade year, I went to IMSA, a 3-year boarding high school in Illinois. IMSA is a sophomore-junior-senior school, with no freshmen. However, generally about 10% of each incoming class consists of students coming directly from 8th grade as opposed to their freshman year. Originally, I had "planned" (loosely used word here) to attend Exeter after my junior year at IMSA (equivalently my second year of high school). I even applied and visited. However, in the end I decided not to go.

Since I've effectively been "on record" with AMC since 8th grade as a USAMO qualifier that year, I put 9th grade as my grade level during my first year in high school, and 10th grade my second year. This year, since I knew I was going to college next year, I did elect to put 12th grader instead of 11th (not a big deal since USAMO qualification is equivalent and being a top 12er it didn't affect my MOP status either). However, I guess AMC didn't really look at that, and just have me labeled as an 11th grader. Not a big deal either way.

Oh yeah, I am going to MIT (as are Thomas, Eric, and Hyunsoo), though it's not for the same reason as theorized earlier (since I'm a "true" senior).

I'm under the impression that it is a magnet school.

[quote="qvc122"]darktreb, is IMSA a public boading school or a private one?[/quote]

Its a public boarding magnet school (public - in that its free and funded by the State, boarding - because you have to live there, and magnet - because it focuses on science and mathematics and you have to apply to get in). There are 16 schools like it in the country.

The first of its kind was the [url=http://www.ncssm.edu/]North Carolina School for Science in Math[/url] which opened in 1980. Most of them are two year schools, for the Junior and Senior year, but IMSA is unique in that it is a three year school.

Thanks. I sort of guessed. I assume that Thomas Jefferson High School of Science and Technology is one of the 16.

No, its not actually. Thomas Jefferson is not residential, students have to commute.

What sets these 16 schools apart is that they are residential. Most of them are run like little colleges. Students live there and attend classes on a college-like schedule - with maybe 3 or 4 classes a day, each for 3 or 4 hours a week. The teachers usually hold at least a masters, and most a doctorate.

I went to one of the 16 schools, The South Carolina Governor's School for Science and Mathematics (not to be confused with the summer Governor's School programs found in various states). I've only been to 3 of the other schools - IMSA, NCSSM, and the Alabama School for Science and Mathematics, all of which had very nice campuses and facilities. I know there are also schools in Mississippi and Arkansas. I don't know where the other 10 are. Schools like these are mostly in rural states, where the population density isn't high enough to have math and science magnet schools that people can commute to.

All of these magnet schools, both the residential and non-residential, are part of the National Consortium of Specialized Secondary Schools for Science, Mathematics, and Technology ([url=http://www.ncsssmst.org/membership/institutions.htm]NCSSSSMT[/url]).


A five-digit number is created using the digits 1-5 each once. What is the probability that the number is odd? Express your answer as a common fraction.

Last digit must be 1,3,5

Possible last digits: 1,2,3,4,5

Probability is $ \boxed{\frac{3}{5}}$.

*Notice that we did not have to count all the numbers formed with each last digit (why?).


trigonometry
calculus
derivative
inequalities
real analysis
real analysis unsolved

[quote="zhaobin"]Find the minimum of
$\cos^2x \sin 2x-(1+\sqrt 3)\sin^2x \cos 2x$,when $x \in [\frac{\pi}{12},\frac{\pi}{6}]$
It seems easy,but I can't do it except use the computer.
Can anybody do it without computer
Thanks[/quote]
The derivative turns out to be
$f'(x) = 2 \cos^2 x \left(3 - 4 \cos^2 x\right)\left((1 + \sqrt{3})\sin x - \cos x \right)$.
The first factor does not vanish on $[\frac{\pi}{12},\frac{\pi}{6}]$, the second one vanishes at $x = \frac{\pi}{6}$, the third one at $x = \arctan (1+\sqrt{3})^{-1}$, which is in the interval and turns out to be a maximum. Thus the minimum must be at one of the endpoints. It turns out that $f(\frac{\pi}{12}) = f(\frac{\pi}{6}) = \frac{2\sqrt{3}-1}{8}$.

What is the point of doing this without a computer?

[quote="hpe"][quote="zhaobin"]Find the minimum of
$\cos^2x \sin 2x-(1+\sqrt 3)\sin^2x \cos 2x$,when $x \in [\frac{\pi}{12},\frac{\pi}{6}]$
It seems easy,but I can't do it except use the computer.
Can anybody do it without computer
Thanks[/quote]
The derivative turns out to be
$f'(x) = 2 \cos^2 x \left(3 - 4 \cos^2 x\right)\left((1 + \sqrt{3})\sin x - \cos x \right)$.[/quote]

This seems wrong to me. I obtain

$f^{\prime}\left(x\right)=-\left(1+\sqrt3\right)\left(2\sin x-\left(\sqrt3-1\right)\cos x\right)\cos 3x$.

The conclusion about the minimum is, however, the same.

Anyway, you can verify that the minimum is indeed obtained for $x=\frac{\pi}{12}$ and $x=\frac{\pi}{6}$ without using derivatives by noting that $f\left(\frac{\pi}{12}\right)=f\left(\frac{\pi}{6}\right)=\frac{2\sqrt3-1}{8}$ and rewriting f(x) as

$f\left(x\right)=\cos^2 x\sin\left(2x\right)-\left(1+\sqrt3\right)\sin^2 x\cos\left(2x\right)$
$=\frac12\left(\left(1+\cos\left(2x\right)\right)\sin\left(2x\right)-\left(1+\sqrt3\right)\left(1-\cos\left(2x\right)\right)\cos\left(2x\right)\right)$
$=\frac12\left(\left(1+u\right)\sqrt{1-u^2}-\left(1+\sqrt3\right)\left(1-u\right)u\right)$

with $u=\cos\left(2x\right)$, and then verifying the inequality $f\left(x\right)\geq\frac{2\sqrt3-1}{8}$ by squaring and factoring (it is ugly, but it works).

  Darij

[u]Define[/u] $x\ .s.s.\ y\Longleftrightarrow x=y=0\ \vee\ xy>0\ ($x,y have [b][u]s[/u][/b]ame [b][u]s[/b][/u]ign $)\ .$

\begin{eqnarray*} 4f(x) & = & 4\cos^2 x\sin 2x-4(1+\sqrt 3)\sin ^2x\cos 2x \\ &=& 2\sin 2x(1+\cos 2x)-2(1+\sqrt 3)\cos 2x(1-\cos 2x)\\ &=& \sin 4x+2\sin 2x+(1+\sqrt 3)(1+\cos 4x)-2(1+\sqrt 3)\cos 2x \ \ \Longrightarrow\end{eqnarray*}

\begin{eqnarray*} f'(x) &=& \cos 4x+\cos 2x-(1+\sqrt 3)\sin 4x +(1+\sqrt 3)\sin 2x \\ &=& (2\cos 2x-1)(\cos 2x+1)-(1+\sqrt 3)\sin 2x(2\cos 2x-1)\\ &=& 2\cos ^2 x(2\cos 2x-1)-2(1+\sqrt 3)\sin x\cos x(2\cos 2x-1) \\ &=& 2\cos x(2\cos 2x-1)[\cos x-(1+\sqrt 3)\sin x]\ \Longrightarrow \end{eqnarray*}

$\phi \in \left(\frac{\pi}{12},\frac{\pi}{8}\right),\ \tan \phi=\frac{\sqrt 3 -1}{2}\Longrightarrow \left(\forall\right)x\in \left[\ \frac{\pi}{12}\ ,\ \frac{\pi}{6}\ \right]\ ,\ f'(x)\ .s.s.\ (\phi -x)\ .$

Therefore, $\left(\forall\right)x\in \left[\frac{\pi}{12},\frac{\pi}{6}\right],\ f(\phi )\ge f(x)\ge f\left(\frac{\pi}{12}\right)=f\left(\frac{\pi}{6}\right)=\frac{2\sqrt 3-1}{8}\ .$

[b]Generalization.[/b] $m\in \left[\sqrt 3,2+\sqrt 3\right],\ x\in D=\left[\frac{\pi}{12},\frac{\pi}{6}\right],$
$f(x)=\cos^2x\sin 2x-m\sin^2x\cos 2x\Longrightarrow\min_{x\in D} f(x)=\frac{3\sqrt 3-m}{8}\ .$ Exactly,
$\frac{\pi}{12}\le x\le \frac{\pi}{6}\Longrightarrow f\left(\frac{\pi}{12}\right)=f\left(\frac{\pi}{6}\right)=\frac{3\sqrt 3-m}{8}\le f(x)\le f(\phi)$, where $\tan \phi =\frac 1m\ .$

Thanks all of you.I think darij and Virgil Nicula's must be right.(In fact,they are same)
and hpe may forget something,you only have an extra$-\cos x$.
Thanks again


blogs
LaTeX
geometry
3D geometry
MATHCOUNTS
AMC
AIME

I borrowed this from another forum. The purpose of this game is to make a statement about the person who will respond below you. The person must say whether it's true or false, then make a statement about the person who will reply below them. For instance:

Me: The person below me has blue eyes.
Next: False. The person below me loves sushi.
Next: True. The person below me...

I'll start:

The person below me likes pizza.

True. The person below me loves basketball.

Very true. The person below me is a hardcore Milwaukee Bucks fan! :D

False. The person below me is a Notre Dame Fighting Irish football fan.

Very false! The person below me is a rib fanatic

very false!

The person below me is some sort of addict.

True!!

The person below is nutz.......

Edit: Mustafa beet me to it.........

[quote="mustafa"]very true! I'm a n-addict

The person below me has a dog[/quote]

What is an n-addict.

True. The person below me is over 180 lbs.

False. I weigh 110. The person below me loves sports.

true!!!!!!!!!!!!!!

the prson below me is a bot.

False. I wish. The person below me goes to middle school.

True.(grade 7) The person below me likes the mavs

true, under 150

the person under me will use 4 smilies in their post

jorian

True. :!:  :!:  :!:  :!:

The person below me will say the letter O.[color=white][hide="lalaala"],skjhgkdhggdgfdfgfdgdfg[/hide][/color]

True. The person below has tasted [u]O[/u]xi Clean

not really

the person below me reads MegaTokyo

False.

The person below me is not bald.

true

the person below me is in 8th grade

false.

the person below me likes english class

False.

The person below me is not CheeseIsGood.

true

the person below me wishes they were not logged on to aim in their dreams

jorian

falshe.

the person above me stole my library card.

false

The person below me had his library card stolen

False

The person below me has never had an  operation


[quote="Michael Niland"]Without using calculus prove that

$\frac{1}{2-x}$ $<$ $x^{x}$ $<$ $x^{2}-x+1$ when x satisfies  $0<x<1$[/quote]
I don't think that it's possible without using calculus. :maybe:


$ (x_n)$ such that :$ x_1 \equal{} 5,x_{n \plus{} 1} \equal{} 6x_n \minus{} x_{n \minus{} 1}$
find $ limit x_n*{\sqrt{2}*x_n}$

hello, what is with $ x_2\equal{}$?.
Sonnhard.

[quote="Math pro"] $ limit x_n*{\sqrt {2}*x_n}$[/quote]
and what is it,please check your $ LaTex$


Consider the operation of $ ax^2\plus{}bx\plus{}c$ in $ P_2$ to $ |a|$ in $ P_0$.
Show that it does not correspond to a linear transformation by showing that there is no matrix that maps $ (a,b,c)$ in $ \mathbb{R}^3$ to $ |a|$ in $ \mathbb{R}$.

Suppose there is a matrix $ A \in M_{1 \times 3} (\mathbb{R})$ such that, for $ u \equal{} [a \,\,\,b\,\,\,c]^T \in \mathbb{R}^3$ we have $ Au \equal{} |a|$. Then $ A \equal{} [x \,\,\, y \,\,\, z]$, and defining $ v \equal{} [ \minus{} a \,\,\,b\,\,\,c]^T$, we would have

$ Au \equal{} Av \equal{} |a| \equal{} > ax \plus{} by \plus{} cz \equal{} \minus{} ax \plus{} by \plus{} cz \equal{} > ax \equal{} \minus{} ax$  (1)

For the matrix A to operate on any vector in R^3 we must have $ x \neq 0$, but then (1) implies a = 0, which means such a matrix can operate only on a vector of the form $ u \equal{} [0 \,\,\,b\,\,\,c]^T$. Therefore, there is no matrix A such that $ Au \equal{} |a|$ for any $ u \in \mathbb{R}^3$, and so the given map is not linear.

We can also see that the given map $ f: P_2 \to P_0$ is not linear in another way. If $ p, q \in P_2$ then $ p \equal{} ax^2\plus{}bx\plus{}c$, $ q \equal{} dx^2 \plus{} ex \plus{} f$, and $ f(p \plus{} q) \equal{} f[(a\plus{}d)x^2 \plus{} (b\plus{}e)x \plus{} (c\plus{}f)] \equal{} |a\plus{}d|$, and $ f(p) \plus{} f(q) \equal{} |a| \plus{} |d|$, and we don't have necessarily $ |a\plus{}d| \equal{} |a| \plus{} |d|$.


limit
function
real analysis
real analysis unsolved

Let $ \alpha$ be a fixed positive real number and define
\[ x_{n \plus{} 1} \equal{} \frac {x_{n}^{3} \plus{} 3\alpha x_{n}}{3x_{n}^{2} \plus{} \alpha},\quad n \equal{} 0,1,\ldots
\]
Assume $ x_{0}$ is so that the sequence $ \{x_{n}\}$ is convergent.
1) Find $ \lim_{n\to\infty}x_{n}$.
2) Determine the order of convergence of the sequence $ \{x_{n}\}$.

What is "the order of convergence" of a sequence?

Suppose you have a sequence $ \{x^k\}$ such that $ x^k\rightarrow x^*$ as $ k\rightarrow\infty$.  Define the function
\[ C(p) \equal{} \limsup_{k\rightarrow\infty} \frac {\left\| x^{k \plus{} 1} \minus{} x^* \right\| }{\left\| x^k \minus{} x^* \right\|^p}.
\]
Then there exists a value $ 0\leq p^* \leq \infty$ such that for $ p > p*$, $ C(p) \equal{} \infty$ and for $ p\leq p^*$, $ C(p) \equal{} 0$. The value $ p^*$ is called the order of convergence.

[hide="1"]If the sequence converges to $ L$, then $ L\equal{}\frac{L^3\plus{}3\alpha L}{3L^2\plus{}\alpha}\Rightarrow L(L^2\minus{}\alpha )\equal{}0$.

If $ x_0>0$, then $ x_n>\frac{1}{3}$ for all positive $ n$, since $ 3\alpha x_n>\frac{1}{3}\alpha x_n$. So the limit must be $ \sqrt{\alpha}$.

If $ x_0\equal{}0$, then the sequence is everywhere 0, and if $ x_0<0$, then if $ y_n$ satisfies the same recurrence with $ x_0\equal{}\minus{}y_0$, then $ x_n\equal{}\minus{}y_n$ so the limit is $ \minus{}\sqrt{\alpha}$.[/hide]

[hide="2"]I'll just consider $ x_0$ positive. The negative case is equivalent and the zero case is just silly (a constant sequence).
\[ x_{n\plus{}1}\minus{}\sqrt{\alpha} \equal{}\frac{\left( x_n\minus{}\sqrt{\alpha } \right)^3}{3x_n^2\plus{}\alpha}\]
Then we have
\[ \lim_{n\rightarrow \infty} \frac{4\alpha \left( x_{n\plus{}1}\minus{}\sqrt{\alpha}\right)}{\left( x_n\minus{}\sqrt{\alpha }\right)^3}\equal{}1\]
so the order of convergence is 3.[/hide]


I'm looking for a theme for my diploma's thesis.
Can some  one help me.

The Proof of the Riemman Hypothesys should suffice. :D


In the [i]Unlucky Lottery[/i], all the prizes were powers of $\$$13, namely $\$$1, $\$$13, $\$$169, etc., and the total prize money given away was $\$$1 million.  What was the least possible number of prizes in the [i]Unlucky Lottery[/i]?

No Calculators

[hide="Hint"] Find $1000000$ in base $13$. :D [/hide]

[hide]I got $1000000=290221_{13}$ but that's alot of multiplying and subtracting with out a calculator so this in probably wrong.

Add those together to get 2+9+2+2+1= 16 prizes

16 prizes[/hide]

I got the same answer but I accidentaly used a calculator  :blush:

Both of your answers are correct.

Next time try to read whether or not you can use a calculator before solving the question.


The right-hand side can be improved to $ 3k^2\plus{}3k\plus{}3$:

$ \frac{a^{2}\plus{}kb^{2}\plus{}k^{2}c^{2}}{bc}\plus{}\frac{b^{2}\plus{}kc^{2}\plus{}k^{2}a^{2}}{ca}\plus{}\frac{c^{2}\plus{}ka^{2}\plus{}k^{2}b^{2}}{ab}\equal{}\sum_{cyc} \frac{a^2}{bc}\plus{}k\sum_{cyc} \frac{b}{c}\plus{}k^2\sum_{cyc} \frac{c}{b}$
$ \geq 3\plus{}3k\plus{}3k^2$ by AM-GM


All of
$ x,y,z$
are members of $ R$(real number set).
here are 3 conditions.
$ x \plus{} y \minus{} z \equal{} 1$
$ x \minus{} 5y \plus{} 2z \equal{} 16$
$ 3y \plus{} z \equal{} 5a$
When
$ T \equal{} x^2 \plus{} y^2 \plus{} z^2$
is minimal, Find
$ T$
and a rational number
$ a$

key? express $ x,y,z$ using $ a$.

hello, multiplying the first equation by $ \minus{}1$ and adding the second equation we get
$ \minus{}6y\plus{}3z\equal{}15$
or
$ \minus{}2y\plus{}z\equal{}5a$
Multiplying this equation by $ \minus{}1$ and adding the third equation we get $ y\equal{}a\minus{}1$ and then$ z\equal{}2a\plus{}3$,$ x\equal{}5\plus{}a$, inserting these terms in $ T(x,y,z)$ we get
$ T\equal{}(5\plus{}a)^2\plus{}(a\minus{}1)^2\plus{}(2a\plus{}3)^2\equal{}6a^2\plus{}20a\plus{}35$
So we have the searched global minimum as $ \frac{55}{3}$ for $ a\equal{}\minus{}\frac{5}{3}$.
Sonnhard.


function
inequalities
number theory proposed
number theory

One defines a function on the nonnegative integers as follows:

$f(0) = 0$
$f(2n + 1) = f(2n) + 1$
$f(4n + 2) = f(4n) + 2$
$f(4n) = 2f(2n) - f(n)$

(a) Show that $f(3n) \leq 2f(n) + \sqrt f(n)$ for all positive integers $n$.
(b) For which $n$ does the above inequality become an equality?

I tried to solve this one a long time ago. Looking at the binary representation of numbers helps a lot!


If a,b,c are non-negative reals such that a+b+c=3, prove
a/(1+b^2) + b/(1+c^2) + c/(1+a^2) >= 3/2

I think we had this at http://www.mathlinks.ro/Forum/viewtopic.php?t=13943 .

  Darij


What is the arithmetic mean of the first ten numbers in the
arithmetic sequence $ 5, 11, 17, 23, \ldots$?

The arithmetic mean of the first 10 numbers in the sequence is the same as the mean of the middle 2 numbers, the 5th number and the 6th. Because the common ratio is 6, the answer is $ 6 \times 4 \plus{} 5 \plus{} \frac {6}{2} \equal{} 32$

The numbers are $5$, $5+6*1$, $5+6*2$, $5+6*3$, $5+6*4$, and so on.

$tr (n)$ is triangular number n.

The sum of the first $n$ numbers is $5n + 6 * tr (n - 1)$

In this case $n=10$ so $5*10 + 6 * tr (9) = 50 + 6 * \frac{9*10}{2} = 50 + \frac{6*9*10}{2} = 50 + 270 = 320$

There are 10 numbers so the arithmetic mean is $\frac{320}{10} = \boxed{32}$


This information can be googled somewhere else and this topic is way off from the [i]purpose[/i] of this forum. This forum is about [b]problem solving[/b], not about answering conceptual questions relating to your review for state exam.

[quote="Interval"]In what type of geometry would this be true?

Given a line L and a point P, there is an infinite number of lines
passing through P parallel to L.

In this geometry, what is the sum of the angle measures of triangle?
Will it add up to 180 degrees?[/quote]

i think its spherical.  the angles dont add up to any specific value though

No, I dont believe spherical works, in fact in spherical, elliptical, and Riemannian (which may the the same as elliptical) geometries there are no parrellel lines.  This describes Hyperbolic geometry I beleive, and if I recall correctly the angles of a triangle in hyperbolic geometry are always less than 180, but approach 180 as the triangle gets larger.


I'm doing workouts and warmups. How about you guys?

Warm-Ups and Workouts are too easy.

My coach has a good collection of old nats that we've been reviewing.

Still, to get to states in my chapter you need a 44-46.

Not sure thats happening.

Well, the qualifications for state are much more relaxed for our state (Top 5 teams, individuals, in chapter

I'm not doing anything yet. Maybe I'll do a few old chapter and state tests. Maybe a few of the challenge problems in the Intro books.

Actually, we're doing practice chapter tests for the final one or two weeks...


http://www.mathcounts.org/webarticles/articlefiles/536-State_Winners_06.pdf
Did anyone notice South Dakota's team is 4 8th graders from the same school with the coach from that school as well? Crazy  :wacko:
A 6th grader won Vermont and Guam :o

I think the 6th grader placed 2nd in Puerto Rico.
There's a grand total of 6 6th graders going this year.

Hey, State Department people aren't on there yet :o

Wow...South Dakota is crazy.
Wow...I haven't posted in such a long time.
Wow...Things have changed greatly since santa's little spammer left. :P
Wow...Go to my site on my sig.

Here is an updated analysis of the gradewise composition of the each state if anyone wanted to see.

[hide]
Total:
165-8th
57-7th
6-6th

4-8th:
AL, CA, CO, GA, IL, IN, KY, ME, MT, NM, OK, SC, SD, N. Mariana Islands, Virgin Islands

3-8th, 1-7th:
AK, DE, FL, IA, LA, MD, MI, MN, NV, NH, NY, OH, PA, RI, TN, TX, VA, WA, WV, WI, WY, State Department

2-8th, 2-7th:
AZ, AR, CT, DC, HI, ID, KS, MS, MO, NJ, NC

1-8th, 3-7th:
MA, NE

4-7th:
UT

3-8th, 1-6th:
ND, Defence Department, Peuto Rico

1-8th, 2-7th, 1-6th:
Guam

1-8th, 1-7th, 2-6th:
VT

[/hide]

It seems that Mr. Hite isn't coaching GA, although his students are half the team.  :(

Is is just me or are the S's and C's that indicate whether a person is a student or a coach slightly misaligned for FL and GA?

Also, Everyone on WV's team is from the same school.

[quote="Xantos C. Guin"]Here is an updated analysis of the gradewise composition of the each state if anyone wanted to see.
The State Department needs to post their list going to nationals.

[hide]
Total:
162-8th
56-7th
6-6th
4-not known yet.

4-8th:
AL, CA, CO, GA, IL, IN, KY, ME, MT, NM, OK, SC, SD, N. Mariana Islands, Virgin Islands

3-8th, 1-7th:
AK, DE, FL, IA, LA, MD, MI, MN, NV, NH, NY, OH, PA, RI, TN, TX, VA, WA, WV, WI, WY

2-8th, 2-7th:
AZ, AR, CT, DC, HI, ID, KS, MS, MO, NJ, NC

1-8th, 3-7th:
MA, NE

4-7th:
UT

3-8th, 1-6th:
ND, Defence Department, Peuto Rico

1-8th, 2-7th, 1-6th:
Guam

1-8th, 1-7th, 2-6th:
VT

[/hide]

It seems that Mr. Hite isn't coaching GA, although his students are half the team.  :(

Is is just me or are the S's and C's that indicate whether a person is a student or a coach slightly misaligned for FL and GA?

Also, Everyone on WV's team is from the same school.[/quote]

Hehe, my state is the same. Beddoes isn't coaching, even though her students make half the team (even though i never had her as a teacher).

Indiana: 3 Honey Creekers all under Mr. Fischer, and the 4th person works with him too. So does the 5th one.

[quote="Xantos C. Guin"]Here is an updated analysis of the gradewise composition of the each state if anyone wanted to see.
The State Department needs to post their list going to nationals.

[hide]
Total:
162-8th
56-7th
6-6th
4-not known yet.

4-8th:
AL, CA, CO, GA, IL, IN, KY, ME, MT, NM, OK, SC, SD, N. Mariana Islands, Virgin Islands

3-8th, 1-7th:
AK, DE, FL, IA, LA, MD, MI, MN, NV, NH, NY, OH, PA, RI, TN, TX, VA, WA, WV, WI, WY

2-8th, 2-7th:
AZ, AR, CT, DC, HI, ID, KS, MS, MO, NJ, NC

1-8th, 3-7th:
MA, NE

4-7th:
UT

3-8th, 1-6th:
ND, Defence Department, Peuto Rico

1-8th, 2-7th, 1-6th:
Guam

1-8th, 1-7th, 2-6th:
VT

[/hide]

It seems that Mr. Hite isn't coaching GA, although his students are half the team.  :(

Is is just me or are the S's and C's that indicate whether a person is a student or a coach slightly misaligned for FL and GA?

Also, Everyone on WV's team is from the same school.[/quote] Fl and GA are misaligned because they were added in late.

to fix problems like fanatics, why doesnt mathcounts let power house states to send two teams, or if u would have been eligible in 5 other states let you in or something like that

Then most countdowners would make it. Anyone pretty much who is good enough to make top 10 in a good state can make it in Guam, DE, Marino Islands, Defense Department, and Virgin Islands

I guess then it is in the best interest of the state, not the best interest of the people in the state.  If it was like the U.S. House of Representatives, I would feel bad if I had to make "top 1" to get to nationals...

[quote="biffanddoc"]eeek all of this arguing. It should be top 4 from each state. The end.
If your not satisfied (fanatic/natmc) you can move to a different state.[/quote]  :rotfl:  :rotfl:  :rotfl: :rotfl: :rotfl:  :rotfl: I am [b]so [/b]glad I'm in FL where people can't count and making nationals is easy. Just look at the 2000 presidential election.

[quote="easyas3.14159..."]Speaking of that, does anyone know the Alaska qualifiers' screennames (either from this year or last)?[/quote]

Are you speaking of Brent Woodhouse, of 05 Nats? His screenname is the same as his name.

[quote="noneoftheabove"][quote="pgpatel"][quote="Klebian"]
Jeffrey chan is noneoftheabove, he got 8th written, 9th countdown.[/quote]
Well that explains why Jeffery Chan didn't make it. He left AoPS. By the way, does anyone know why Nathan Benjamin (I_Would_Say) stopped coming here? Do you know fanatic?[/quote]

I didn't leave AoPS. I just left the Mathcounts section for various other sections like the AMC. :|[/quote]
Oh, okay. I hadn't seen you post in a while so I figured that you left.

[quote="pgpatel"][quote="easyas3.14159..."]To add to your list, a sixth grader won North Dakota and Puerto Rico, too.[/quote]
The winner of ND is an aopser. PhilB[/quote]

How do you know who everyone is and every mathcounts AoPSer and other random stuff about Mathcounts?

i have a feeling a 6th grader will win.... :P

[quote="jclarkemathL314159"]i have a feeling i won't win...[/quote]
My feelings, exactly.  :(

[quote="frost13"][quote="biffanddoc"]eeek all of this arguing. It should be top 4 from each state. The end.
If your not satisfied (fanatic/natmc) you can move to a different state.[/quote]

Ok, while I don't think people should go moving, I have to chime in here :)

While I am sure it is frustrating to score a 39 or even higher at state and not make nationals and know that 4 kids from another state all made it with lower scores I'd like to give some actual data as to why it is important each state send 4 kids.

Last year the #1 kid in NH got a 30.  I am sure in states like Texas and Indiana, the top 2 states at Nationals last year, there were probably 20 or more kids with a score like that.

At Nationals last year that same kid from NH took 31st missing the Countdown by just 3 points.  Two Texas kids came in 29th and 30th(probably tied with him), and an Indiana kid came in 54th.  My kid didn't even win his chapter, got a 30 at state, and took 31st at Nationals!  Each state needs representation because the top kids in every state can do well if they work hard.  Don't knock the small states I believe we are 41st in population and have placed 22nd and 15th in the last 3 years.  We may never win, but we try :)

It would be great if 10 kids from every state could go to Nationals, but 4 is still more than the National Spelling Bee I believe.  Congrats to everyone that made the top 4 in their state and to everyone that didn't.  Going to Nationals isn't going to change anyone's life, but what you learn from MATHCOUNTS and math in general probably will.

Oh yeah, and a kid from tiny little Rhode Island took 20th!![/quote]

Well, just a thing about the National Spelling Bee...they send as many people as the newspapers can sponser.  Last year, CA had about 70 people, I believe.  They just conduct more preliminary rounds.  So if you live in CA, the Spelling Bee may be a better bet than MATHCOUNTS :p :ddr:
Yay for DDR!

Well I'm in 7th grade this year, and we have 3 7th graders (including me)

So if we get like, top 5 this year, we have a real shot at winning next year.

Oh and by the way, is it just a rumor that someone in some state came in 5th 3 years in a row?

[quote="pgpatel"]Well, piano1321 claims to have done so in AR. Maybe morgoth can clear this up... Anyways, he said he came as a guest all three years (including this one). So we're going to have to find out from him at nats.[/quote]

I thought that piano was female...? ...he...he....him.... Think you might have malfunctioning s and r keys :D.

[quote="piano1321"]I DIDN'T MAKE IT :o

i think i just made history by having 5th place at state THREE years in a row. lol! I really thought it would be easy to make it in Arkansas.  I didnt really care this year because the state coach is such an a**.  Also, i would NOT wanna go to nationals with no girls. it just wouldn't be fun.  I went to nationals (as a guest) both last two years, so i really dont mind.[/quote]
I thought that it was an accidental double negative but when I looked at pianos post there is a female symbol next to her name.

[quote="biffanddoc"]Very, very weird this year. Did anyone notice that two top-30 people from 7th grade are not returning? (Ian Osborne from NJ and Jeffrey Chan from TX)
This will be interesting with 3 6th grade state winners...
3 states have all top 4 from the top 4!!!!!!![/quote]

Ian Osborn placed 6th at states. Must've been overconfident :p .

On a side note, there's a coach in NJ who coached three of the four team members, however, she's not coach. Furthermore, NJ's coach has NO people at all in the top four, or even top six if I remember straight...


Αποδείξτε ότι η διοφαντική εξίσωση $ x^3\plus{}y^3\plus{}z^3\plus{}x^2y\plus{}y^2z\plus{}z^2x\plus{}xyz\equal{}0$ δεν έχει μη μηδενικές ακέραιες λύσεις.

Έχουμε ότι $ x^3\plus{}y^3\plus{}z^3\plus{}x^2y\plus{}y^2z\plus{}z^2x\plus{}xyz\equiv x\plus{}y\plus{}z\plus{}xy\plus{}yz\plus{}zx\plus{}xyz\equiv0\bmod2$.
Αν $ 2\nmid x,y,z$ άτοπο. Τότε χωρίς περιορισμό της γενικότητας $ 2\mid x$. Άρα $ y\plus{}z\plus{}yz\equiv0\bmod2$
Αν $ 2\nmid y,z$ άτοπο.Τότε χωρίς περιορισμό της γενικότητας $ 2\mid y$. Άρα $ 2\mid z$.
Άρα έχουμε ότι αν $ (x_1,y_1,z_1)$ είναι λύση της εξίσωσης στους ακεραίους τότε και η $ (\frac{x_1}{2},\frac{y_1}{2},\frac{z_1}{2})$ είναι λύση στους ακεραίους ... τότε και η $ (\frac{x_1}{2^n},\frac{y_1}{2^n},\frac{z_1}{2^n})$ είναι λύση στους ακεραίους(άπειρη κάθοδος).
Άρα θα πρέπει $ x\equal{}0$ και $ y\equal{}0$ και $ z\equal{}0$ και η λύση $ (0,0,0)$ επαληθεύει.


function
induction
algebra proposed
algebra

Find function $f: \mathbb N^{*}\to [0;\infty)$ with
$f(2)=4$ ,$(n+1)f(n)\leq nf(n+1)$ and $f(m\cdot n )=f(n)\cdot f(m)$ ,$\forall m,n \in \mathbb N^{*}$

By induction $f(n)=n^{2}$. It is true for p=2. Let $f(p)=a$ p - prime.
Then let $m/n<log_{2}p<k/l$ are 2 good rational approah. Consider $f(2^{m})=4^{m}<f(p^{n})=f(p)^{n}$ and $f(p^{l})=f(p)^{l}<f(2^{k})=4^{k}$.


$\sec 50^\circ + \cot 80^\circ = \frac{1}{\sin 40^\circ}+\frac{\cos 80^\circ}{\sin 80^\circ} = \frac{2\cos 40^\circ+\cos 80^\circ}{\sin 80^\circ}$
Write $40^\circ = 60^\circ - 20^\circ$ and $80^\circ = 60^\circ + 20^\circ$
we have
$2\cos 40^\circ+\cos 80^\circ = \frac{3}{2}\cos 20^\circ + \frac{\sqrt 3}{2}\sin 20^\circ = \sqrt 3\left(\frac{\sqrt 3}{2}\cos 20^\circ + \frac{1}{2}\sin 20^\circ \right) = \sqrt 3 \sin 80^\circ$
hence
$\sec 50^\circ + \tan 80^\circ = \sqrt 3$

$\tan{9^\circ}+\cot{117^\circ}-\tan{243^\circ}-\cot{351^\circ} = \tan 9^\circ + \cot 9^\circ - (\tan 63^\circ + \cot 63^\circ) = \frac{2}{\sin 18^\circ}-\frac{2}{\cos 36^\circ}$
because $\tan x+\cot x = \frac{2}{\sin 2x}$

$\frac{1}{\sin 18^\circ}-\frac{1}{\cos 36^\circ} = \frac{1}{\cos 72^\circ}-\frac{1}{\cos 36^\circ} = \frac{\cos 36^\circ - \cos 72^\circ}{\cos 36^\circ\cos 72^\circ} = \frac{2\sin 54^\circ\sin 18^\circ}{\cos 36^\circ\cos 72^\circ} = 2$

hence $\tan{9^\circ}+\cot{117^\circ}-\tan{243^\circ}-\cot{351^\circ}=4$


topology
real analysis
real analysis unsolved

Show that every open subset of $ \mathbb{R}^n$ is a countable union of sets of the form $ (a_1, b_1) \times \cdots \times (a_n, b_n)$.

Open set $ X$.
For every $ x \in X$
Then there is an $ y \in X$ with rational coëfficients and a hyperblock $ (a_1,b_1)\times\cdots\times(a_n,b_n)$ in $ X$ with $ y$ and $ x$ within the hyperblock with all $ a_i$ and $ b_i$ rational.
The countability results from the countability of $ Q^{3n}$


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let G be a nonabelian group of order p^3 where p is prime. Prove that Z(G)=D(G), Z(G) is center and D(G) is commutator of group G.

This may help: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=162041

the center has order $ p$ or $ p^{2}$, so $ G/Z(G)$ is a group of order $ p^{2}$ or $ p$. in particular it's abelian, i.e. $ D(G)\subseteq Z(G)$. but since $ G$ is not abelian, $ G/Z(G)$ can't by cyclic, so that $ Z(G)$ has order $ p$. since $ D(G)$ is a nontrivial subgroup of $ Z(G)$, we have $ D(G)\equal{}Z(G)$.

[b]-oo-[/b], i like your solution, very nice. But i had to prove that the group of order $ p^{2}$ is abelian, it's a good exercise.


$A,B,C$are three angles of the triangle$ABC$show that if $\sin^2{A}+\sin^2{B}=\sin{c}$. then $\hat{A}=\frac{\Pi}{2}$
i have solved this probleme but i search another solution nice than the mine :lol:

[quote="Davron"]I think something is wrong there  ;)

Davron[/quote]

I think he meant $C=\frac{\pi}{2}$ but even then $\sin^2A+\sin^2B=\sin C=\sin(A+B)=\sin A\cos B+\sin B\cos A$ so if we would get to our desired conclusion we must have $\tan^2A+\cot^2A=\tan A+\cot A$ but $\displaystyle x^2+\frac{1}{x^2}=x+\frac{1}{x}\Leftrightarrow \left(x+\frac{1}{x}-\frac{1}{2}\right)^2=2+\frac{1}{4}\Rightarrow x+\frac{1}{x}=2\Rightarrow x=1$, hence this is valid only if $A=B=\frac{\pi}{4},C=\frac{\pi}{2}$.

So there is definitely something wrong about it ;).

:)  sure you are right and i had the same solution with yours ...

Davron

[quote="akhrif"]it s true i m sur!! :mad:
i have solved it!!![/quote]

But how that can be true ? Ok you may show your solution  :)

Davron

Try it with a 30-60-90 traingle.  It' doesn't work!

EDIT: Why is this in the unsolved problems? :?
Masoud Zargar

If the problem is
[quote="akhrif"]$A,B,C$ are three angles of the triangle $ABC$ show that if $\sin^2{A}+\sin^2{B}=\sin{C}$ then ${C}=\frac{\pi}{2}$
[/quote]

then the problem works for any right triangle.

one has to prove that if $C \neq 90^o$ then the formula $\sin^2 A + \sin^2 B = \sin C$ doesn't hold

[b]akhrif[/b], which is the problem, $C$ or $A$ is the right angle ?

Salam Masoud.
Do you see akhrif your problem is INCORRECT  :lol:

Davron

Salam Davron,

If we have $A=\frac{\pi}{2}$, then we also have
$\\\sin^2 A+\sin^2 B=1+\sin^2(90-C)\\ =1+(\sin 90\cos C-\cos 90\sin C)^2=1+\cos^2C\neq \sin C$
Therefore, your problem is incorrect for sure.

Now, if $C=\frac{\pi}{2}$, then we have
$\\\sin^2 A+\sin^2 B\\ \sin^2A+\sin^2(90-A)=\sin^2A+\cos^2A\\ =\sin\frac{\pi}{2}=1$

If $B=\frac{\pi}{2}$, then we have
$\sin^2A+1\neq\sin C=\sin(90-A)=\cos A$
$\therefore$ The only possibility is if $C=\frac{\pi}{2}$.

Masoud Zargar

I'm sorry, akhrif, but this problem is not correct
If $A$ is less than about $18^o$ then we can't be sure that $C$ is right angle

$\sin^2A+ \sin^2B -sinC = 0$

$\pi-C = A+B \Rightarrow \sin C = \sin(A+B)$

set $A=a$, constant parameter
and $x=B$

Consider the function $f(x) = \sin^2a+ \sin^2x -sin(a+x)$

One root of $f$ is $\frac{\pi}{2}-a$

I found that, there is a number $a_0 \approx 0,223297$ (about 18 degrees) such that if $a\leq a_0$ then $f$ has root(s) different than $\frac{\pi}{2}-a$

[color=blue]I didn't prove it; I just used the zeros of the derivative.
The dots on the x-axis are the zeros of $f'$
Accurately, if $a_0=3\theta$ then $\theta$ satisfies the equation[/color] $6 \sin^2{\theta}+4\sin^2{3\theta}+3\sqrt 3 \sin {2\theta}=1$
We don't need a proof for this. Just take any $a<a_0$ and you'll see that there are two roots in $(\pi,2\pi-a)$

The graph below is for $a = a_0$, and there in only one root in $(\pi,2\pi-a)$.

If $B$ is one root of $f$, then $C=\pi-a-B$

Since $f$ has two or three roots, $C$ is not unique

[quote="akhrif"]$ A,B,C$are three angles of the  $ \triangle{ABC},$ show that if $ \sin^2{A} \plus{} \sin^2{B} \equal{} \sin{C}$ and $ \max\{\angle{A},\angle{B}\}\leq\frac {\pi}{2},$ then $ \angle{C} \equal{} \frac {\pi}{2}.$ [/quote]Let $ \sin{A} \equal{} \frac {s^2 \plus{} 2s}{s^2 \plus{} 2s \plus{} 2},\sin{B} \equal{} \frac {2t \plus{} 2}{t^2 \plus{} 2t \plus{} 2},$ where $ s,t\geq0.$ Then

$ 0 \equal{} \sin^2{A} \plus{} \sin^2{B} \minus{} \sin{C} \equal{} \frac {2(s \minus{} t)[s^2t^2(3s \plus{} t) \plus{} 2st(3s^2 \plus{} 6st \plus{} t^2) \plus{} 4s(s \plus{} t)(s \plus{} 4t) \plus{} 4(3s^2 \plus{} 6st \plus{} t^2) \plus{} 4(3s \plus{} t)]}{(s^2 \plus{} 2s \plus{} 2)^2(t^2 \plus{} 2t \plus{} 2)^2}$

$ \Longleftrightarrow s \equal{} t\Longleftrightarrow \sin^2{A} \plus{} \sin^2{B} \equal{} 1\Longleftrightarrow \angle{C} \equal{} \frac {\pi}{2}.$

[b]Remark[/b]  $ \max\{\angle{A},\angle{B}\}\leq\frac {\pi}{2}$ can be insteaded of $ \max\{A \minus{} 3B,B \minus{} 3A\}\leq\frac {\pi}{2}.$


Euler
number theory unsolved
number theory

prove that
\[ \prod_{n\equal{}1}^{\infty}(1\minus{}k^{\minus{}i})\]
is irrational for every $ k\in\mathbb{Z}\backslash\{\minus{}1,0,1\}$.

hint: Euler's identity may be useful
\[ \prod_{n\equal{}1}^{\infty}\left(1\minus{}\frac{1}{a^n}\right)\equal{}\sum_{\minus{}\infty}^{\plus{}\infty}(\minus{}1)^na^{\minus{}\frac{1}{2}(3n^2\plus{}n)}\]
for complex $ a$ such that $ |a|<1$.

So use $ a\equal{}k$ and $ n\equal{}i$ in Eulers identity, so a is a an integer other than -1,0 or 1

Consider the representation of the number in base $ |k|$. If it were rational the representaion would have to terminate or repeat but from Euler's identity it can't do because $ \frac{1}{2}(3n^2 \plus{} n)$ is an increasing integer sequence whose consecutive differences do not repeat.

a small doubt...

what about negative terms?

$ \frac {1}{2}(3n^2 \plus{} n)$ isn't an increasing sequence (here $ n$ is integer), it decreases on $ \minus{}\mathbb{N}$ and increases on $ \mathbb{N}$

Just take $ k\equal{}10$ as an example and work our how the number looks in decimal after the first few terms.
The number is going to have sequences of mostly zeros and nines of growing length.
Then prove what you find and generalise to all $ k$ to get the result.

There may be ways to simplify the proof.


[hide="da answer"]325...see i answerd a question, related to math, correctly[/hide]

26*25/2=325

how about the first 25 triangle number(pyramid number)

[hide]
The sums of the first [i]n[/i] numbers is n(n+1)/2
..so we can plug 25 in to get
25(25+1)/2
=25(26)/2
=325
[/hide]

[quote="MCrawford"]Find the sum of the counting numbers from 1 to 25 inclusive.  In other words, if $S = 1 + 2 + 3 + \cdots + 24 + 25$, find the value of $S$.[/quote]

$S=\frac{n(n+1)}{2}=\frac{25\cdot 26}{2}=325$

[hide="jli"]you made a careless error, its 325 not 425[/hide]

Is that just a memorized formula, or do you know why it works?

Well, 1+25=26, 2+24=26, and so on. There are 25 numbers but you divide by 2. so 26*25/2

[quote="mathmom"]So, what if you don't start at 1?  What is the sum of the counting numbers from 100 to 999 inclusive?  (This was a MATHCOUNTS handbook problem from this past year.)[/quote]
[hide]
$999*1000/2 = 499500$
$99*100/2 = 4950$
$499500 - 4950 = 494500$
[/hide]

ccy, that's a good approach.  :)  I use a slightly different approach (that I'll post here in a while if no one else does).


calculus
integration
trigonometry
calculus computations

[quote="DON FAVOR"][quote="Moubinool"]Calculate with 10% relative error

$\int_{R}\cos(100(x^4-x))dx$[/quote]

what number on the top of integral?[/quote]
It's over all real numbers, so the integral goes from $-\infty$ to $\infty$.  (I think).

Do you not write the Taylor series and then approximate with values until the approximate value $< .1$ error from the actual value?

No, you dont. This integral is not absolutely convergent, so you need to manipulate it before applying the standard numerical methods. Also, ordinary power series cannot be used to estimate an integral on an unbounded interval.

I would recommend integration by parts to generate an absolutely convergent integral. There might be a useful asymptotic series here, but I haven't really done any of the work.

[quote="jmerry"]I would recommend integration by parts to generate an absolutely convergent integral.[/quote]
Partly right, you ignored the fact that $x^4-x$ has extreme points, so we can't do I. by P. , try steepest descent method, or somebody's(sorry I can't remember this famous name) method,and you will get a DIVERGENT series.
The secret of an asymptotic series is, it diverges, but "converges" much FASTER than convergent series does! :P

You can integrate by parts- you don't have to do it on the entire line. Repeated integration by parts applied to a tail end integral produces an asymptotic series.

We have
$\int_c^\infty\cos(p(x))\, dx=-\frac{\sin(p(c))}{p'(c)} +\int_c^\infty\frac{\sin(p(x))\cdot p''(x)}{(p'(x))^2}\, dx$
$\int_c^\infty\cos(p(x))\, dx=-\frac{\sin(p(c))}{p'(c)} +\frac{\cos p(c)\cdot p''(c)}{(p'(c))^3}+ \int_c^\infty\frac{\cos(p(x))\cdot(p'''(x)p'(x)-3(p''(x))^2)}{(p'(x))^4}\, dx$

Similarly,
$\int_{-\infty}^b\cos(p(x))\, dx=\frac{\sin(p(b))}{p'(b)}-\frac{\cos p(b)\cdot p''(b)}{(p'(b))^3}+ \int_{-\infty}^b\frac{\cos(p(x))\cdot(p'''(x)p'(x)-3(p''(x))^2)}{(p'(x))^4}\, dx$

If we use $b=-1$ and $c=1$ here, we can estimate the remainder integrals quite accurately:
$|\int_1^\infty\frac{\cos(100(x^4-x))\cdot(24x\cdot(4x^3-1)-3(12x^2)^2)} {100^2\cdot(4x^3-1)^4}\, dx|$
$\le|\int_1^\infty\frac{-24x\cdot(14x^3+1)}{10^4\cdot(4x^3-1)^4}\, dx|$
$=| \frac{12x^2}{(4x^3-1)^3\cdot10^4}]_1^\infty|= \frac{12}{27\cdot10^4}$
Similarly, the remainder integral from $-\infty$ to $-1$ is less than $\frac{12}{125\cdot10^4}$.
Therefore, with error no more than $5.4\cdot10^{-5}$, the desired integral over $\mathbb{R}$ is
$\int_{-1}^1\cos(100(x^4-x))\, dx+\frac{12}{27\cdot10^4}+\frac{\sin 200}{-500}-\frac{12\cos 200}{-125\cdot 10^4}$

I told my calculator to evaluate that integral. Several minutes later, it came up with the answer $-.092158$. Adding in the other pieces, the whole integral is approximately $-.090362$, with the error confined to the last two places.

I wonder how large error one will get just applying the stationary phase method without much thinking. On the other hand, even in this case one will have to compute $\cos$ of some unpleasant number, so I do not see how a normal person is supposed to solve it with just pen and paper, which is why I like this problem much less than the $\int_0^{10} x^x$ one (actually, I don't like it at all  :( )

Don't you see the problem? The integral mainly comes from region near $x_0=2^{-\frac{2}{3}}$, you just leave this integral in your result, that's not allowed.
The first order approximation is $\frac{1}{x_0}\sqrt{\frac{2\pi}{100*12}}\cos \left(100(x_0^4-x_0)-\frac{\pi}{4}\right)\approx -0.0706$

We could also try to make the integral simpler, decreasing the burden on the calculator. If we reduce the interval to $[.5,.75]$, the integral runs much faster. Here the estimates are a little more complicated; the integral estimate from $-\infty$ to $\frac12$ has to be done in three parts, but it comes out to less than $\frac{26}{10^4}$. The other tail integral is less than $\frac{10}{10^4}$. Therefore,
$\int_{\mathbb{R}}\cos(100(x^4-x))\, dx\approx \int_{-1}^1\cos(100(x^4-x))\, dx-\frac{\sin(100(\frac{81}{256}-\frac34))}{100(\frac{27}{16}-1)}$
$+ \frac{12\cdot\frac{9}{16}\cdot\cos(100(\frac{81}{256}-\frac34))} {10^4(\frac{27}{16}-1)^3}+ \frac{\sin(100(\frac1{16}-\frac12))}{100(\frac{1}{2}-1)}- \frac{12\cdot\frac14\cdot\cos(100(\frac1{16}-\frac12))} {10^4(\frac12-1)^3}$
with error less than $3.6\cdot10^{-3}$.
This works out to $-.081144-.008486+.001687-.004604+.002336=-.089078$.

This is less accurate, but still within the range asked for, and the calculator did the integral in a few seconds.

I don't think we're playing by the same rules, Buffalo. The integral near $\sqrt[3]{\frac14}$ can be handled by ordinary numerical integration methods. Rather than doing this by hand, I left it to the calculator.

When an ugly numerical question comes up, I will use the tools at hand. What good is a computer if you never use it to compute?


What fraction in the interval $1/2 < x < 7/11$ has the smallest denominator?

[hide]$\frac12=0.5$ and $\frac7{11}=0.6363...$.  I guess since the interval is so large, just guess and check.  2, 3,  and 4 don't work, but 5 works because $\frac35=0.6$, so the answer is $5$.[/hide]

[hide="Answer"]
The answer to this question is $\boxed{5}$.
[/hide]

[hide="Solution"]
The answer to this question is $\boxed{5}$. First, you calculate the equivalent decimals for the two fractions given. $\displaystyle \frac{1}{2}=0.5$ and $\displaystyle \frac{7}{11}=0.\bar{63}$. Now, you list the possible denominators above 2. Start with three and get the following results:
$\displaystyle \frac{1}{3}=0.3\bar{3}$
$\displaystyle \frac{2}{3}=0.6\bar{6}$
$\displaystyle \frac{1}{4}=0.25$
$\displaystyle \frac{2}{4}=\frac{1}{2}=0.5$
$\displaystyle \frac{3}{4}=0.75 \text{ which is over the limit}$
$\displaystyle \frac{1}{5}=0.2$
$\displaystyle \frac{2}{5}=0.4$
$\displaystyle \frac{3}{5}=0.6 \text{, which is between the two numbers.}$
We have listed all the possible denominators for the problems and have thus arrived at the answer. The answer is now five, because the denomitor of $\displaystyle \frac{3}{5}$ is $\boxed{5}$.
[/hide]


number theory
relatively prime
number theory unsolved

#1 relocated [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=21999]here[/url]

#2 relocated [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=22000]here[/url]

#3 relocated [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=22001]here[/url]

#4 relocated [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=22002]here[/url]

#5 relocated [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=22003]here[/url]

For each integer n such that $n=p_1p_2p_3p_4 $where $p_1,p_2,p_3,p_4$ are distinct primes, let:
$1=d_1<d_2<....<d_{16}=n$
be the sixteen positive integers which devide n. Prove that if n<1995 then $ d_9 -d_8\neq 22.$

[edited to comply with forum rules].

I haven't solved some problems :( . Can anybody help me? ;)

[quote="harazi"]Pretty nice, I remeber that hxtund posted these problems not long time ago. So, try to find the solutions in the solved section .[/quote]
Yes,you solved problem 3 but the hardest problem is problem 2(in my opinion),problem 6 is not easy(at least for me).Can anybody solve problem 2 and problem 6?

6) Suppose contrary. It is clear $d_8=\frac{n}{d_9}$. So $d_9^2-22d_9-n=0$. Therefore, $11^2+n$ is a perfect square.
Since $d_9-\frac{n}{d_9}$ is even we conclude $n$ is odd, by the similar reason, $n$ is not divisible by 11.
Moreover, if $n$ doesn't contain 3, 5 or 7 then $n$ at least $3\cdot5\cdot 11\cdot 13>1995$. So $n=3\cdot 5\cdot 7\cdot 13$ or $n=3\cdot 5\cdot 7\cdot 17$. But both these numbers don't satisfy $121+n$ is perfect square.

Ok, here are some steps that can give the solution of this problem, which I dont particularly like. First of all, both $ d_6, d_7$ have at most 2 prime divisors ( since if for example $d_7$ has as prime factors $ p_1, p_2, p_3$ then all numbers $ 1, p_1, p_2, p_3, p_1\cdot p_2, p_2\cdot p_3, p_3\cdot p_1 $ are divisors of n smaller than $ d_7$) and the two divisors are relatively prime. Suppose that none of $ d_6, d_7$ is a prime power. Then the exponent of any prime number in $ d_6, d_7$ is at most 2 (if for example $ p^3$ divides $ d_7$ then if q is any other prime factor, the numbers $ 1, p, p^2, p^3, q, pq, p^2q$ are 7 divisors of n smaller then $d_7$). Suppose for example that there is p prime such that $ p^2|d_7$. Let q any other prime factor. Then $ 1, p, p^2, q, pq, p^2q $ are divisors of n and thus $ d_6$ is not relatively prime with $ d_7$. Thus in this case it follows that both $ d_6, d_7$ are products of 2 prime numbers. Let $ d_6= p_1\cdot p_2$ and $ d_7=p_3\cdot p_4$. Then the numbers $ 1, p_1, p_2, p_3, p_4, p_1\cdot p_2, p_3\cdot p_4$ are the first 7 divisors of n and it follows that $ p_1\cdot p_3> p_3\cdot p_4$ and $ p_2\cdot p_4> p_3\cdot p_4$, from where we find $ d_6>d_7 $, false.
   Thus at least one of $ d_6, d_7$ is a prime power. After this result it is easy but long to prove that $ d_6$ is not a prime power. Also, it is not difficult at all to discuss the case when $ d_7$ is a prime power. It is just long and boring. In the end you find that $ d_7$ must be a power of 2 and after that that $ d_7=32$ and $ d_6=31$. Thus $n=1984$. Bad problem


Ok this is a story I wrote with my friend:

-----------------------------------------------------------------

me: let's write another story
8:45 PM Nivedita: story??
 me: yes story
  about a prince and a princess :D
8:46 PM Nivedita: :D ??..ok there was aprince
 me: and a princess
 Nivedita: the prince died
 me: no he didn't
8:47 PM Nivedita: yes he did!:-)
 me: no he didn't
 Nivedita: ok then the princess died
 me: no nobody died
8:48 PM Nivedita: !!!!!
 me: she was abducted by the bad guy
 Nivedita: and the prince saved her
8:49 PM me: and while he was saving her
 Nivedita: another bad guy came
8:50 PM me: and while they were fighting any idea how to reverse a linked list without using three pointers?
 Nivedita: hmm..what's a linked list?:-)
8:51 PM me: horrendous hitchhikers!
8:52 PM you don't know _linked lists_?
8:53 PM Nivedita: no...is it some dynamic list or something..
8:54 PM me: it is something fascinating
  i'll teach you linked lists some time
 Nivedita: ok that will be grt
8:55 PM me: but i think the prince has disposed of both the bad guys by now and is now untying the princess
 Nivedita: yeah but the princess is bleeding to death:-)
8:56 PM me: so the prince tears his shirt and ties the wounds up, and then they kiss :)
 Nivedita: yuck!
  ok then they marry have kids and die
 me: not so fas
  fast
8:57 PM they go back and marry and have a large marriage party
 Nivedita: ok
  then the cake is poisened
8:58 PM me: and a faithful servant discovers this
 Nivedita: and eats the cake himself
8:59 PM me: and the prince vows to avenge his death
9:00 PM Nivedita: and discovers that the princess's servants husband's sisters brother is the culprit
9:01 PM me: so the culprit gets a huge kick in the seat of his pants
 Nivedita: but princess gets mad because the guiy is somewhat related to her
9:02 PM me: so they have a quarrel
9:04 PM Nivedita: and the prince in a age shoots princess
  age->rage
 me: no he doesn't

-----------------------------------------------------------------

Comment please. The Prince and The Princess 2 will come soon ...

[quote="Koda"] any idea how to reverse a linked list without using three pointers?[/quote]

Actually, I was just doing that the other day...I know how...:ninja:

[quote="Koda"][quote="jli"]how did that take 20 mins[/quote]

Not because of slow typing I assure you.
TPATP 2 has taken more than an hour  :roll:

[quote="Ignite168"]No incest = lose.[/quote]
Huh? :huh:[/quote]
Don't worry about Ignite, his [color=#999999]fantasies[/color] [b]opinions[/b] are not usually that of the general population.


linear algebra
matrix
combinatorics unsolved
combinatorics

If $0 < d \leq i$, prove that

$\displaystyle\sum_{j=0}^{i-d}\begin{pmatrix} i \\ j \\ \end{pmatrix}(-1)^j(q^{i-j-d+1}-1) = (q-1)\displaystyle\sum_{j=0}^{i-d}\begin{pmatrix} i-1 \\ j \\ \end{pmatrix}(-1)^jq^{i-j-d}$

Alexandros

Hint: Use the identity $\begin{pmatrix} i \\ j \\ \end{pmatrix} = \begin{pmatrix} i-1 \\ j-1 \\ \end{pmatrix} + \begin{pmatrix} i-1 \\ j \\ \end{pmatrix}$

[quote="ehsan2004"]It's a very ugly identity  ;)[/quote]

I agree, but uses in smart way the sum to get the result.
Find out the way...

Alexandros


In triangle $ ABC$, medians $ AD$, $ BE$, and $ CF$ are drawn, intersecting at the centroid, G.

Show that if the incircles of triangles $ AEG$, $ BFG$, and $ CDG$ are congruent, then triangle $ ABC$ is equilateral.

It's easy to see that medians split the triangle into 6 area equal triangles.

As $ P\equal{}rs$ for all triangles, we have $ EG\plus{}GA\plus{}AE\equal{}BF\plus{}FG\plus{}BG\equal{}CD\plus{}DG\plus{}CG$.

Now do a lot (warning, a lot) of computation to prove the triangle is equilateral.

[quote="Bugi"]It's easy to see that medians split the triangle into 6 area equal triangles.[/quote]

Ooh, whoops, I forgot about that identity.

I'll do the computation by myself; I once spent a whole hour trying to understand one step of a solution, so this is no different.