geometry Support calculus AMC USA(J)MO USAMO MIT

Anybody here doing the IB Programme? I am deciding whether to do it or not? Golly Gee what a tough descision!

(Edit note from moderator:) I edited this post only to correct the spelling of the subject heading.

if your school doesn't offer ap's maybe, but I think that IB is overratted because you have to take certain classes, unlike AP classes, in which you get to choose which APs you want to take.

[quote="churchilljrhigh"]Anybody here doing the IB Programme? I am deciding whether to do it or not?[/quote]

What reasons do you hear about why NOT to do it?

By the way, how long has the IB program in your area been established? I hear my local high school is planning to set up an IB program, but I don't know how soon that will get done. Probably by then my son will be taking lots of university classes.

Your schedule for 11th is practically set down. And you have to a year of TOK which is like philsophy. And you have to take two years of art (music or drwaing) (they don't offer the computer science option)

I am personally not a fan of the IB system.  It seems far too rigid to me.  I particularly think that it is too confining for advanced students.  That's not to say that it's never a step up from some of the other crazy systems that many schools employ.

I personally think that the only way to improve public education for advanced students is to establish a magnet school in every city in the nation that can support one.  Homeschooling can also be a good option depending on circumstances.

I think homeschooling costs alot of money. First, parent have to stay home. Second, they have to buy text books and supplies. Third, lack of exciting school facilities (gym court, climbing wall, weight room, etc.) Fourth lack of resources (e.g. using the shcool printers to print AOPS stuff). Fifth the parent could of worked and earned millions. Sixth a parent can't be expert at everything. Seventh school offfers alot of extracuricular clubs, and stuff Eighth you get to hang out with friends. Ninth etc. etc. etc.

The only thing homeschooling is better is that no time is wasted.

I agree that homeschooling is often infeasible, but I do think that it would be better in some cases.

I personally feel like the vast majority of my classes were near or complete wastes of time and that I was almost entirely self-educated.  Being homeschooled would have at least saved time (or added time to my other activities).

The great thing about homeschooling children who are more advanced than their peers is that it lends itself to the greatest method of teaching available: give a child the appropriate materials and get out of the way while they learn.

Also, as much as a parent might be able to earn, an education might be a higher priority.  Yes, there are opportunity costs, but having costs doesn't mean that capital is wasted.

Well, not really.  I'm sure tokenadult will respond to this too, but here are my quick thoughts.  (Or maybe not so quick.)

[quote="churchilljrhigh"]I think homeschooling costs alot of money. First, parent have to stay home.[/quote]

This is generally the case, but not necessarily, especially for high school homeschoolers, who have greater autonomy (and may be at classes / activities outside the home for much of the school day anyway).  Even for younger students, sometimes grandparents or other friends or relatives help out with taking care of the kids.  At an

[quote]
Second, they have to buy text books and supplies.
[/quote]

Plenty of ways to get them cheap, with some creativity -- library (free!), buying books used... lots of resources out there.  Most homeschoolers really don't spend that much on supplies.

[quote]Third, lack of exciting school facilities (gym court, climbing wall, weight room, etc.
[/quote]

Well, there are organizations like Boys and Girls clubs, the YMCA/YWCA, JCC, many of which will give discounts for homeschooled families -- I went to a homeschool swim/gym time for several years.   At one point a homeschool mom who worked at a rock gym (yes, homeschool moms can work -- see below!) arranged for us to get reduced rates there.

[quote]
 Fourth lack of resources (e.g. using the shcool printers to print AOPS stuff).
[/quote]

Never heard that objection before... but it's not as if printing AOPS stuff is that expensive.

[quote]
 Fifth the parent could of worked and earned millions.
[/quote]

Both parents still can work, even if not after the standard model -- one can work at home, or part-time, or at different hours.  Sure, it's harder to make millions, but for my part I'm happier that my parents gave me the flexibility to learn at my own pace instead of designer clothes, expensive vacations, etc...

[quote]
Sixth a parent can't be expert at everything.
[/quote]

That's what outside resources are for: the library, classes (eg at local colleges), homeschooling groups, etc...

[quote]
 Seventh school offfers alot of extracuricular clubs, and stuff
[/quote]

Extracurriculars can be done anywhere, and homeschoolers often have greater choice in what they can do:  I attend a philosophy discussion group and my sister acts in an amateur theater troupe, neither of which we could do if we went to school because they meet on weekday mornings.  We have also done many activities with our local homeschooling group.  In addition, many schools are receptive to homeschoolers attending their extracurriculars (although in many places they can't compete on school sports teams) -- I attended math club meetings at the local high school for several weeks when a math club for our area teamed up with the school to send a team to NYSML.

[quote] Eighth you get to hang out with friends.
[/quote]

Homeschoolers have plenty of opportunity to hang out with friends (see all the activities above) without teachers getting in the way and trying to make them learn stuff.  I may not spend as much total time with friends as I would if in school, but I definitely spend more quality time with them.

[quote]
 Ninth etc. etc. etc.
[/quote]

Sorry, can't exactly refute that.

Perhaps if this becomes a debate, tokenadult should make it a new thread?

[quote]The only thing homeschooling is better is that no time is wasted.[/quote]

I think the theme here is that there's no universally "right" way of education. Different strokes work well for different folks - in different areas and different situations. Homeschooling is great for some kids, but certainly not for everyone.

Likewise, the value of the IB program really varies from place to place and person to person. First of all, some schools execute the program better than other schools. And some students benefit from the structure of the program better than other kids. It's really an individualized decision of works best for you.

They just started an IB program at a local high school here. I think that the rigidity of the program does have it's drawbacks. But it's diversity and depth also has it's advantages. In my opinion the IB Math program is much better than the typical Honors to AP program at most schools. The curiculum for the IB Calculus program covers some very good topics not covered in the AP program.

But as with everything, a curiculum is really only as good as it is taught. So it all depends on how good the teachers are at teaching the material.

The best resources in making this decision, are going to be people who've been through the IB program at that school. You should ask about their experiences, and look at how well you think it prepared them. I know that most people I know who have gone through the IB program have had positive experiences and have generally come out of it as very strong, well rounded students.

[quote="gauss202"]
The best resources in making this decision, are going to be people who've been through the IB program at that school. You should ask about their experiences, and look at how well you think it prepared them. I know that most people I know who have gone through the IB program have had positive experiences and have generally come out of it as very strong, well rounded students.[/quote]

I have a bit of an issue with this last point.  Teachers are all people who have been through educational systems, yet I only had 3 or 4 of them whose "positive experiences" had anything to do with my educational needs.

While the IB programs might be good for some students, I have serious doubts as to whether or not it could challenge the kinds of students who take the USAMO.  But, like I've said before, I have serious doubts about any non-magnet public school providing that challenge (with a very small handful of exceptions).

I don't have a whole lot of opinion on homeschooling, but it's worth pointing out that while churchill's issues with it might not be serious problems for already-middle-or-upper-class families (such as Alison's), there are plenty of obvious problems implementing it on a larger scale, particularly among poorer groups of people.

Not that I think anyone was advocating millions of students being educated at home -- just that it's a model with some very clear limitations on who it can be applied to.

And, I know absolutely nothing about the IB system.

Just to be clear about my opinion: I think it will be many years before homeschooling could possibly be effectively implemented for most students, though I do believe that day will come.

[quote="MCrawford"]
While the IB programs might be good for some students, I have serious doubts as to whether or not it could challenge the kinds of students who take the USAMO.  But, like I've said before, I have serious doubts about any non-magnet public school providing that challenge (with a very small handful of exceptions).[/quote]

Oh, I agree with you Matt. But I don't think the IB program is meant to particularly address those kinds of students. Most students, even in an IB program, aren't the kind of students who take the USAMO. And the program isn't designed specifically for them. Its strength is actually supposed to be its stress on a breadth of knowledge. So the program isn't designed to be around any particularly specialized instruction.

I'm a big advocate of magnet programs and magnet schools that group kids for level appropriate education though. At least in places where they can be practically implemented. In some places they're just not geographically feasable. In places that don't have magnet programs, or for students who choose not to go - I think the IB program can sometimes be a good option.

Good point.  I sort of went into reflexive homeschooling-defense mode there, so I should probably qualify myself there.  Although I know and know of homeschoolers from assorted backgrounds, it doesn't work for everyone.

I've never seriously wished that everyone could be homeschooled (okay, so maybe when I was an idealistic middle schooler I had some ambitious plans for subverting the public school system).  What I would like to see is homeschoolers and the public schools working together to help give all students the opportunities to develop their particular talents.

[quote="JBL"]I don't have a whole lot of opinion on homeschooling, but it's worth pointing out that while churchill's issues with it might not be serious problems for already-middle-or-upper-class families (such as Alison's), there are plenty of obvious problems implementing it on a larger scale, particularly among poorer groups of people.
[/quote]

I think that IB would probably be best for the well rounded student, so if you enjoy all subjects, then you should choose the IB program.  If you are strongest in math and science and weaker in the humanities, I would recommend taking either APs or taking classes at a local college.  I know that in my situation, I took a lot of APs in high school, which looked good on my resume, but the only ones I got credit for were BC, CS, Physics C, Chem, the rest did absolutely nothing for me.  Since they had nothing to do with my major.

I am a freshman in Shanghai American School and we offer both IB and AP here.

I have taken AP Calculus and AP Music Theory, and am enrolled in three AP subjects next year (AP Statistics, US History, and Comp Gov) but I plan to go to the UK to study, therefore I would choose IB.

Some math teachers recommend that I study IB Further level mathematics, but I don't understand the concept around that. Do you have to take IB Math HL before that or something? Someone please explain this more clearly to me  :roll:

You don't have to take Math HL beforehand, but they assume a level of knowledge equal to a student that would have taken Math HL.

just browsing through some old threads..

[quote="kentliu"]IB is crap.. it restricts ur choice of classes... i think the most important thing is taking classes you like and will succeed in.. i know for a fact IB has wasted me many periods... which is why i quit before junior year.. i know many others that quit during the jr year.. i think there are only like 15 people left of the 250 people who signed up during freshman year.. but of course most of those people were unqualified...
basically, IB forces you to be good at everything..
I think its more important to EXCEL at 1 or a few things..[/quote]

I attend an IB school (rickards in FL) and I know for a fact that IB doesnt expect you to be good at everything. there are certain requirements you have to meet, but no, they dont fail you if you're not the state champion of latin or whatever. There are some people who can't handle the IB curriculum, if that is the case, then these people shouldn't be in IB. I understand that sometimes it does become annoying with the mountains of homework assigned (internal assessments, IB exams, extended essay, world lit, CAS, etc.), but in the end, the pressure will really help you deal with college. Sometimes I wished I had more time to study math like some of my other math-oriented friends (gaku liu, for example), but thats the sacrifice you have to make.
and yes, ib schools DO have usamo qualifiers (we had Keenan Pepper this year, who qualified for black MOSP), and three others.

[quote="cutiepi"]I think the IB program in my area is excellent. Next year I plan to attend my city's IB high school, and everyone i know that goes to or has gone to it has thoroughly loved it. I disagree with people who say IB sutdents don't make USAMO because the IB school I speak of has people who make the USAMO. I keep hearing people calling the IB curriculum rigid, but I like the way it gives you the classes you need. In college I will have time to take the classes I love, anyways in high school it is hard to know for sure what you want to be and what classes you need for that, so IB lets you hvae many higher level subject classes to help you find what you really like. IT seems like everyone here wants a mathematical career, but I am undecided. I have no idea what I want to major in or do with my life. All that I know is I like math, but I like the IB because it gives a nice advanced broad curriculum that gets you into college and leaves you with a choice of majors and gives you a good high-school education.[/quote]

aww karlanna is sticking up for rickards. thats nice of you.

[quote="WindSlicer"]Instead of creating another thread, I think I will choose to propose a question here.

Ok, well here's some background information (I'll get to the question in a bit). I'm currently a sophomore in a IB floridian highschool, ranked 3rd in the state and 21st in the nation for public schools. I've been enrolled in pre-IB since my freshman year, and albeit 3 Bs in the first semester, I've adjusted well and I could make it through the rest of the program with As, or close enough. The situation is, is that recently as I've been thinking that I might want to pursue a mathematically-based major at a college, and my IB program is seriously lacking in the mathematical area.

We only offer two math classes, and they are both standard level classes (If you are familar with IB, they are IB Math Studies and IB Math Methods). And if you are familiar with IB, then you know that those classes are far from the best. Even if you look a post up at Blah Blah's curriculum you can see that he took a math Higher Level (HL) course and a SL one (Further Mathematics) which is much more advanced that Math Studies/Methods. The thing  is, is that I've spent two years in this program which has limited me from taking any AP classes I may have been able to take if I were not in this program.

So I have two choices. I can stay at my school, 3rd in the state, a good math team and overall good teachers, but with a pretty bad math program. OR I can go to another public school, no ranking, offers pretty good AP classes, and would offer a better oppurtunity for dual enrollment since I wouldn't have to take history or english courses once I had all of my classes, but they have no math team and like most public schools their teachers are not always the top of the line.

So does anyone have any opinions on this? All opinions are appreciated, and if you have any other questions to help you make an opinion, feel free to ask!

Thanks in advance![/quote]

hey windsilcer, what school do u go to? im in jacksonville, fl

yeah i almost went to Rickards for the IB program.
however, i ended up going to Chiles, mainly because it was easier on my family for tranportation reasons.
sometimes i think back on this and wish that i had pushed them harder to let me go to Rickards, but right now im relativly pleased with where i am with my education.

Hola,
I was wondering if anyone has or is going to take the Further Math Sl exam. I signed up for the exam earlier this year. I know the calculus and stats parts because we prepare for them for the Hl and AP exam. Does anyone have any resources or tips on learning the "sets, relations, groups" and discrete math parts? I'm not sure, but I heard they're not testing Euclidean Geometry anymore. Thanks in advance!

Here is what my teacher had as an outline of the topics.
Chapter 2: Sets, Relations, and Groups

2.1
Finite and Infinite Sets, Operations on sets: union, intersection; complement, De Morgan’s Laws, subsets
Pg 77-97

2.2
Ordered Pairs; the Cartesian product of two sets.
Pg 83

Relations
Pg 471-536

Equivalence Relations
507-516

2.3
Functions: injections; surjections; bijections. Composition of functions and inverse functions. Composition of functions and inverse functions
97-111

2.4
Binary Operations: definition; closure; operation tables
1-28

2.5
The associative, distributive and commutative properties of binary operations
1-28

2.6
Identity
24, 89

2.7
The Axioms of a group {S,*} Abelien Groups

2.8
Examples of groups
2.9
Finite and Infinite Groups

2.10
Cyclic Groups and Generators of a group
Proof that all cyclic groups are Abelian

2.11
Definition of a subgroup
Langrange’s theorem, without proof, and its corollary

2.12
Isomorphism and isomorphic groups: formal definition in terms of a bijection; the property that an isomorphism maps the identity of one group onto the identity of the other group; a similar property for inverses

557-577

Chapter 3 Discrete Mathematics

3.1
Natural Numbers and the well-ordering principle
78
516-531

3.2
Division and Euclidean algorithms.
158-159
177-179
The greatest common divisor of integers a and b, (a,b)
Relatively prime numbers; prime numbers and the fundamental theorem of arithmetic
155
General
153-181

3.3
Congruence modulo p as an equivalence class
Residue Class
161-168

3.4
Recurrence relations Difference equations: basic definitions and solutions of a difference equation

401-435
3.5-2 days
Simple graphs, connected graphs, complete graphs; multigraphs, directed graphs; bipartite graphs; planar graphs; subgraphs, complements of graphs, graph isomorphism

537-630

3.6
Walks;
Hamiltonian paths and cycles;
583-588
Eulerian trials and circuits,
577-583
Graph coloring and chromatic number of a graph
613-621

3.7
Networks and Trees: definitions and properties
631-700
The traveling salesman Problem
586,595,599-601
Prim
688-690
Kruskal
690-691
Dijksrtra
595-599

I took Further Math SL (well, there's no HL) in May 2004 and got a 7.

For algebra, I used Fraleigh's 'A First Course in Abstract Algebra' and for discrete math, I used Grimaldi's 'Discrete and Combinatorial Mathematics'.  Grimaldi is a good reference and contains everything you need  with copious examples.  I wasn't a huge fan of Fraleigh, and I like it even less now (admittedly I used an old version) but pretty much any introductory text would do.  Also find some past papers; they will help a lot.  If you can't find any, PM me and I may be able to *cough cough* point you in the right direction.

SL is Senior Level.

I'm in the Pre-IB programme which means im going into IB next year.  Personally i don't really like the programme.  People are more than "a bit" competitive.  Since Im from Canada, theres no advantage for me if I go to a Canadian university (because Ib marks more strictly, you will lower marks than your main-school peers).  It would only be helpful if you go to the US for university.  The Programme also conflicts with some government requirements so all the IB ppl MUST TAKE SUMMER SCHOOL  :mad:

As someone mentioned previously, IB is VERY rigid.  To go off the topic a little, our International Studies teacher told us the other day...

In a country in southeast Asia the students from a school were writing an IB exam.  In the middle of the exam, planes flew over the school and started dropping bombs.  The teacher called the IB Headquarters in Geneva and ask permission to stop the exam because of the bombs.  The HQ refused.

Poor students!  Enroll at your own risk.   :D

Literally everything in your post is factually incorrect, starting with the fact that SL is Standard Level and ending with the fact that your last anecdote is likely apocryphal.

Good call, blablahblah. :10:

To previous question,
Further Math is a math class that includes all the Math HL math options. In all subject areas in IB, you get to choose options to focus on in class which corresponds to questions you choose on your exam.

goddessxx, u dont know what ur talking about...

SL=standered level
HL=higher level (which only seniors can take)

and the anecdote was a pretty pathetic lie...

Okay, in an attempt to respond to all that I remember:
-IB does not require 2 years of arts as someone said and my school offers 3 years of Programming for IB students as well as a comprehensive art program which produces a very nice journal and portofolio that college usually really like by the end of IB. AP and IBs are fairly different. Different schools offer different courses.
-You can take SLs in your second year, just not HLs in your first.
-APs tend to be more general while IB delves deeper into the material. For instance the schools don't teach "world history" but instead choose one area and focus on its entire existance. My school begins with some BC history and moves through the rest of Europeon history up until the 1900s in its first year.
After Math HL students can do independant study or get work sent to them by a college, or at least that's what my school does.
-There's a lot of work and IB students complain about it all the time, but its usually not as bad as we make it out to be. We whine constantly about stress, but everyone I know loves it and would never leave the program. I've started essays at 4am and been fine. I went to a magnet middle school and although its math/science/compsci were great, its history and english departments were abysmal. Whether you like english or not, colleges care, and you're going to have to take it again there.
-You don't have to excel at english and become a writer, but IB makes you substantially better well-rounded than any selective magnet could. For CAS(creativity action service) action and service are fairly standard but the creativity hours force you into finding unusual tasks and services that end up really enriching and beneficial, even if only for a brag sheet.
-And if you really have a passion for math, the Extended Essay is a great way to act on it. Extensive research, 4000 words. It's more research than experimental, though.
Okay. I hope that wasn't an excessive response.

nicely put Inc, i couldnt have said it better.


(If you already know the answer, don't ruin it for those who might want to figure it out for themselves.  ;) )

Given a base $b$, consider the numbers with the property that the last digits of their multiples give all the possible last digits in base $b$.

For example, in base $10$, the multiples of $3$ give all the possible last digits in base $10$:  $0, 3, 6, 9, 12, 15, 18, 21, 24, 27$.

[b]Part One:[/b]

a)  It's pretty easy to see that this property only depends on the last digit.  Which digits in base $10$ have this property?

b)  Which digits in base $15$ have this property?  Base $16$?  Base $19$?

c)  How many digits in base $b$ have this property?  Can you prove it?

[b]Part Two:[/b]

a)  Given a number with last digit $a$ in base $b$, consider the sequence $1, a, a^{2}, a^{3}, ...$.  The last digits of this sequence repeat with some period.  What is this period for $a = 3, b = 10$?  For $a = 7, b = 10$?  For $a = 17, b = 10$?  For $a = 5, b = 13$?

b)  How does this period relate to the decimal expansion of $\frac{1}{b}$ in base $a$?  Can you prove it?

(Don't worry if you can't do the proofs.  They are not exactly obvious.  Just think some of this over for awhile.  It's very interesting, to say the least. :) )

(Edit:  If you think you can figure all of this out without doing numerical examples, think again! ;) )

[hide="Part One"]
Would it just be the phi function of b (where b is 10, 15, etc.)?

Because if a digit was not relatively prime to b, then there would be some last digits not represented.

I hope this is right.
[/hide]

Well, yes.  Can you find a formula for it?


quadratics
modular arithmetic
number theory unsolved
number theory

If $ p\equal{}2^n\plus{}1$ is a fermat prime,Show that $ 3$ is a primitive root modulo $ p$.

[quote="Albanian Eagle"][hide="hint"]Let $ M$ be the set of primitive roots of $ p$, and let $ K$ be the set of quadratic nonresidues of $ p$.
What can you say about the cardinalities of $ M,K$?
Can you say 3 is in $ K$?[/hide][/quote]
Thanks for the hint!

(I assume $ n > 1$)
My solution:
Set $ p \equal{} 2^n \plus{} 1$.
Lemma 1: $ 3$ is not a quadratic residue $ \bmod p$. Proof: We notice that $ p \equiv 1 \bmod 4$ and $ 2^n \plus{} 1 \equiv \minus{}1 \bmod 3$. (Since $ n$ must be a power of $ 2$) So $ (3|p) \equal{} (p|3) \equal{} (\minus{}1|3) \equal{} \minus{}1$. So $ 3$ is not a quadratic residue $ \bmod p$.
Let $ ord_p(3)$ denote the order of $ 3$. Obviously $ ord_p(3) \equal{} 2^k$ for some $ k$. ($ ord_p(3) \mid p\minus{}1 \equal{} 2^n$). Assume $ k < n$. Then we have $ 3^{2^{n\minus{}1}} \equiv 1 \bmod p$. But $ 3$ is not a quadratic residue $ \bmod p$. So $ (3|p) \equiv 3^{2^{n\minus{}1}} \bmod p$ and thus $ 3^{2^{n\minus{}1}} \equiv \minus{}1 \bmod p$ And we get $ p \mid 1\minus{}(\minus{}1)\equal{}2$ which is not possible.!

Every primitive root is a quadratic non-residue. And there are $ \frac{p\minus{}1}{2}\equal{}2^{n\minus{}1}$ quadratic non-residues.
On the other side there are $ \varphi(\varphi(p))\equal{}2^{n\minus{}1}$ primitive roots. So the two sets coincide.

A primitive root modulo $ n$ is any number $ g$ with the property that any number coprime to $ n$ is congruent to a power of $ g \pmod{n}$.


Can someone show me the proof of Cantor-Bernstein theorem.

Is it : If $ A,B$ are sets and $ f: A\to B,g: B\to A$ are injections, then there exists a bijection betwen $ A$ and $ B$? If so, please see here .http://www.mathpath.org/proof/Sch-Bern/proofofS-B.htm

It is more appropriately to call this [b]Schroder-Berstein[/b] theorem because Cantor proved it via Axiom of Choice  :wow: . Which isnt necessay.


geometry
3D geometry
sphere
inequalities
triangle inequality

Guessing..

[hide]
The farthest possible distance is if they are on opposite sides...

I think this is when the points are collinear with the centers..

So...using the distance formula:
$\sqrt{(12-(-2))^2+(8-(-10))^2+(-16-5)^2} \rightarrow \sqrt{196+324+441} \rightarrow \sqrt{961} \rightarrow 31$.

Uh...so $31+19+87 \rightarrow \boxed{137}$?[/hide]

[hide="Solution"]The centers are $\sqrt{(-2-12)^2+(-10-8)^2+(-16-5)^2} = 31$ units apart. The maximum distance is obtained by adding the distance between the centers to the lengths of the radii, or $19+31+87 =137$.[/hide]

[quote="nsato"]What is the largest possible distance between two points, one on the sphere of radius 19 with center $(-2, -10, 5)$ and the other on the sphere of radius 87 with center $(12, 8, -16)$?[/quote]

[hide]First, find the distance between the two points.

The distance is $\sqrt{14^2+18^2+21^2}=31$

The maximum distance is when the points are on the same line that connects the circle's centers that intersects the spheres.

so add the radiuses to 31 to get 137[/hide]

Haha, that's what I assumed, too (though it took me about 2 mins to realize this)... but is there a nice geometric proof for this fact?

[hide="Proof sketch"]Apply the triangle inequality to the two radii of the sphere and the segment joining their centers to get the maximum length being $ 137$. The equality case for the triangle inequality is when all endpoints are collinear, as described.[/hide]


How many subsets are there of the set of subsets of the set $ \{ alpha, tree, \pi^{e}\}$?

[hide="solution"] We have three elements in the original set. When we create a subset, each element has two choices; either to be in the subset or not. Thus there are a total of $ 2^3\equal{}8$ subsets of the orignal set; two ways for each element. Following the same reasoning, there are $ 2^8\equal{}256$ subsets of all the subsets, since each of the eight elements has two choices.[/hide]

Unfortunately, $ 2^8$ is not 64.

your answer is close.
And could you state why there are **** subsets of the original set?

[hide]
not sure if this is right...
There are 8 subsets of the set; 1 is null, 3 have 1 element, 3 have 2 elements, and 1 has all 3.
Thus the total number of subsets of subsets is $ 2^0 \plus{} 3 * 2^1 \plus{} 3 * 2^2 \plus{} 2^3 \equal{} 1 \plus{} 6 \plus{} 12 \plus{} 8 \equal{} 27$
[/hide]


geometry
projective geometry
geometry solved

let $A,B$ be two constant points and $C$ be a constant circle.
we draw line $L$ from $B$ such that it intercepts $C$ in $D,K$. let lines $AD, AK$ intecept $C$ in $R,S$
show that $RS$ passes throw a constant point by changing the line $L$

If $X=DS\cap KR$ and $Y=DK\cap RS$, then both of these lie on the polar of $A$ wrt the circle, so $XY\cap AB$ is a fixed point $T$. $RS$ passes through the harmonic conjugate of $B$ wrt $A,T$, which is clearly a fixed point, since $A,B,T$ are fixed.

mr.groober

did you find that $RS$ passes through the harmonic conjugate of $B$ wrt $A,T$ from this?

$XY$ is the $A$  polar wrt the $C$ so points $A ,R, Z=RD \cap XY , D$ are harmonic.

then lines $YA,YR,YZ,YD$ are haronic lines(what is their name? i dont know it) so $YA, YR, YT, YB$ are harmonic ones.
and this follows with $(ATBW)=-1$)(it means $A,T,B,W$ are harmonic).where $W=RS \cap AB$

can we prove this by some other things or not?

[quote="lomos_lupin"]can we prove this by some other things or not?[/quote]
yes we can prove it with many other ways.  but I think U can't :rotfl:
If U finished ... mayebe U can.

a you wish amir .i will finish it .

i seriosly advise everyone to start learning the advance geometry .

it is pure intelligence of humanity.

by the way dige rahehalhae man ro nanevis  :mad:  :(  ;)  :)  :lol:

hey guys
the solution i have posted  is for amir2 (amir are you satisfied? :bruce: )


geometry
power of a point
radical axis
geometry proposed

Let $C_k =C(O_k ,r_k ),\ k\in \overline {1,2}$, be two fixed circles,$\ O_1 \ne O_2$, a fixed point $A\in C_1$ and a mobile point $M\in C_2$. I note the tangent $d$ in the  point $M$  at the circle $C_2,\ N\in AM\cap C_1$, the second intersection $P$ of the circle $C_1$ with the parallel from the point $N$ to the line $d,\ R\in PM\cap C_1,\ L\in AR\cap d.$
Ascertain the geometrical locus of the point $L$ when the mobile point $M$ traverses the circle $C_2 .$

directed angles $\measuredangle{LAM} = \measuredangle{RAN} = \measuredangle{RPN} = \measuredangle{MPN} = \measuredangle{PML} = \measuredangle{RML} \Rightarrow LAM$ is (oppositely) similar to $LMR \Rightarrow LR \cdot LA = LM^2$, i.e. $L$ lies on the radical axis, i.e. it is the intersection of the radical axis and the tangent at $M$, so as $M$ moves around, it is clear that $L$ traverses the complete radical axis.


analytic geometry
geometry
3D geometry
circumcircle
national olympiad

[i]First day -- October 17th[/i]

[url=http://www.mathlinks.ro/viewtopic.php?t=307541]Problem 1[/url]
Emerald writes $ 2009^2$ integers in a $ 2009\times 2009$ table, one number in each entry of the table. She sums all the numbers in each row and in each column, obtaining $ 4018$ sums. She notices that all sums are distinct. Is it possible that all such sums are perfect squares?

[url=http://www.mathlinks.ro/viewtopic.php?t=307542]Problem 2[/url]
Let $ q \equal{} 2p\plus{}1$, $ p, q > 0$ primes. Prove that there exists a multiple of $ q$ whose digits sum in decimal base is positive and at most $ 3$.

[url=http://www.mathlinks.ro/viewtopic.php?t=307545]Problem 3[/url]
There are $ 2009$ pebbles in some points $ (x,y)$ with both coordinates integer. A operation consists in choosing a point $ (a,b)$ with four or more pebbles, removing four pebbles from $ (a,b)$ and putting one pebble in each of the points
\[ (a,b\minus{}1),\ (a,b\plus{}1),\ (a\minus{}1,b),\ (a\plus{}1,b)\]

Show that after a finite number of operations each point will necessarily have at most three pebbles. Prove that the final configuration doesn't depend on the order of the operations.

[i]Second day -- October 18th[/i]

[url=http://www.mathlinks.ro/viewtopic.php?t=307546]Problem 4[/url]
Prove that there exists a positive integer $ n_0$ with the following property: for each integer $ n \geq n_0$ it is possible to partition a cube into $ n$ smaller cubes.

[url=http://www.mathlinks.ro/viewtopic.php?t=307547]Problem 5[/url]
Let $ ABC$ be a triangle and $ O$ its circumcenter. Lines $ AB$ and $ AC$ meet the circumcircle of $ OBC$ again in $ B_1\neq B$ and $ C_1 \neq C$, respectively, lines $ BA$ amd $ BC$ meet the circumcircle of $ OAC$ again in $ A_2\neq A$ and $ C_2\neq C$, respectively, and lines $ CA$ and $ CB$ meet the circumcircle of $ OAB$ in $ A_3\neq A$ and $ B_3\neq B$, respectively. Prove that lines $ A_2A_3$, $ B_1B_3$ and $ C_1C_2$ have a common point.

[url=http://www.mathlinks.ro/viewtopic.php?t=307548]Problem 6[/url]
Let $ n > 3$ be a fixed integer and $ x_1,x_2,\ldots, x_n$ be positive real numbers. Find, in terms of $ n$, all possible real values of
\[ {x_1\over x_n\plus{}x_1\plus{}x_2} \plus{} {x_2\over x_1\plus{}x_2\plus{}x_3} \plus{} {x_3\over x_2\plus{}x_3\plus{}x_4} \plus{} \cdots \plus{} {x_{n\minus{}1}\over x_{n\minus{}2}\plus{}x_{n\minus{}1}\plus{}x_n} \plus{} {x_n\over x_{n\minus{}1}\plus{}x_n\plus{}x_1}\]

What stage of the olympiad cycle is this exactly?

I don't know if I understand your question, huang, but I can try to explain what is done the Brazilian National Math Olympiad.

There are four levels of participation in Brazilian NMO: the Levels 1,2 and 3, and the Undergrad Level.
The Level 1 is for 5th and 6th grades of Fundamental Cycle.
The Level 2 is for 7th and 8th grades of Fundamental Cycle.
The Level 3 is for 1st to 3rd grades of Medium Cycle (all Medium Cycle, by the way).
(I don´t know what is the correspondent scholarhip levels of other countries, but this can be sufficient to understand).

The Undergrad is, of course, for undergrad students :). This level is the most recent (less than 5 years old, I think).

The Levels 1,2,3 has the same format: a multiple-choice exam at June, a 6-question exam at September and a two-day-with-three-questions-each last exam at October.
The Undergrad has no multiple-choice phase, just the two last ones.

The questions cyshine post here are the Third Phase of Level 3.

Sorry my bad English...


trigonometry
AMC
AIME
geometry
trig identities
Law of Sines

1) What is the largest value that $3x+4y$ can have if $x^2+y^2=14x+6y+6$?

2)$(1+x+x^2)^8=a_0+a_1x+a_2x^2+...+a_{16}x^{16}$, Find $\sum_{k=0}^{8} a_{2k}$

3)In triangle ABC, CD is the bisector of angle C, with D on AB. If $\cos(\frac{C}{2})=\frac{1}{3}$ and CD=6, find $\frac{1}{BC}+\frac{1}{AC}$

1) [hide]I'm thinking it would be optimal if x^2=14x and y^2=6(y+1), making x 14, y 5, and 3x+4y 62. But I'm only random guessing and I doubt I'm actually right.[/hide]

Hint to #1:

[hide]Try to graph the equation x^2+y^2=14x+6y+6, it should come out to be a circle.[/hide]

Can't resist a "cool"  way to answer the question  #3.  The trick is what I call "Physicst way of guessing an answer - almost cheating but still cool" .

Let us assume BC= a and AC= b and we want to find (1/a+1/b)

Now the question is a little clever. It so happens that although given restrictions does not define a triangle (A triangel needs three values  defined to make it unique - you are given only two  - angle C and lenght CD) ) it so happens that  the value (1/a+1/b) is constant. (Although 'a' and  b'  are not unique)

Assuming the question is "clever", and assuming  (1/a+1/b)  does not depend on individual values of a and b let us assume a=b

In this case angle   BDC is right angle and  CD/BC = cos (C/2)  which gives the side  BC =18

so the answer is  (1/18+1/18) = 1/9

(The  shortcut works on the assumption that the  you would not be asked a question whose answer is not unique - If this is a AIME type question you save lot of time

(If you do have  more time you can, assume the angle BDC = T   ( Rest of The work in spoiler)

[hide]then    a/sin (T) = 6/sin (T+c/2)   <  === Law of sines triangle  DBC  ===>

 and     b/sin(T)  = 6/sin(T-c/2)     < ====Law of sines traingel ADC ===?

Now find 1/c+1/b and you see (Sin T ) cancles out neatily and you get the result  1/9) when you substitute  cos( c/2)=1/3.

[/hide]

Nice(to the second solution)!!!!  Another cool way to do #3:

[hide][ABC]=[ACD]+[CDB]

[ACD]=6bsin(c/2), [CDB]=6asin(c/2), [ABC]=absinc

so 6bsin(c/2)+6asin(c/2)=absinc

       6=absinc/((a+b)sin(c/2))

=2absin(c/2)cos(c/2)/((a+b)sin(c/2))=2abcos(c/2)/(a+b)=

sub 1/3 for cos(c/2) we get ab/(a+b)=9, 1/9=(a+b)/ab=1/a+1/b[/hide]

I also edited #2, hopefully it will make sense now...

Hint for 2:    Sum of   all cofficients   in  (1+x+x^2)^n  is nothing but  3^n  and ... (rest in spoiler) .. [hide]sum of  cofficients of even powers of x in (1+x+x^2)^n is (3^n+1)/2 ... think of  f(+1) and f(-1) ... [/hide]


This problem is not pure geometry.
in Iran ,these problems are a part of 3th level exam .

Problem.Missile $\mathcal E$ is designed such that it follows  the target with a constant speed ,which  means
the vector of the missile velocity is always toward the target.
Let the target be a $F-14$ plane running away for its life in a direct line with the speed $V`$ and let the speed of $\mathcal E$ be $V$ such that $V` \ge V$.
Find the places where the missile will hit the $F-14$?

How common are these "political" subjects in the mathematics textbooks in Iran?

As for the problem itself, it's been discussed before. Try searching through Omid Hatami's messages in the "Geometry" section.

Yes , i found the topic ,but i didnt like the solutions there .(the  one from Hatami was good)
here is mine :
Assume that the problem is wrong ,let they collide at $C$
Let $P$ denote the position of the plane at the begining and $M$  denote the first position of missile .
and $L$ be the line that plane is moving on it.
if $\angle RPM \ge 90$ then the problem is trivial ,so assume
$\angle RPM < 90$.
Let $B=\text{Bisector of PM}$ and $B \cap L=A$.
Clearly $PA \le PC$ .
So now let the plane moves till it  reaches $A$.
$\begin{cases} P \longrightarrow A ,\\[.1in] M \longrightarrow L`,\\[.1in]  L` \in \tri PMA \end{cases} \longrightarrow \angle CAL`> 90 \longrightarrow CA > CL`$ which mean clearly Plane will reach $C$ sooner than missile so $\leftrightarrow $ ABSURD.

fleeting guest:You are not trying to say something are you?
There is no POLITICAL matter here.
I bymyself have put the name $F-14$ for the plane ,cause i like its form .
In any case these things are not political ,but related to militry

Until this explanation, the phrasing with "F-14 running for its life" appeared to be part of the original problem, from an Iranian textbook or a competition.    There is some history of military or political material in mathematical textbook problems, in certain countries.  (Imagine if in a Chinese competition one had a problem of "F-16 running for its life from a Silkworm missile", or a US textbook with problems about a Stealth bomber attacking a T-72 tank.)   The above missile problem seemed like an example until your clarification; the reason for my question is that I never heard of such mathematical propaganda being used in Iran and it would definitely be interesting to hear about it.

As for trying to say something: the word "political" in my question was chosen as a nicer phrase than "militaristic propaganda" or a list of the countries where this actually was the practice.


How do I make a degree symbol for geometry in LaTeX

but what is the code command?  I use x^{\circ} which comes $x^{\circ}$

That is \circ. $x$ is just a short character, so kunny's example looks higher.


[hide="1"]
Recall that $ \cos x \equal{} \minus{} \cos(\pi \minus{} x)$ and $ \sin{2x} \equal{} 2\sin x\cos x$; rewrite as $ \minus{} \left(\cos\frac\pi7\cos\frac {2\pi}7\cos\frac {4\pi}7\right)^2$. Note that $ \sin\frac\pi7\cos\frac\pi7\cos\frac {2\pi}7\cos\frac {4\pi}7 \equal{} \frac18\sin\frac {8\pi}7 \equal{} \minus{} \frac18\sin\frac\pi7\implies\cos\frac\pi7\cos\frac {2\pi}7\cos\frac {4\pi}7$
$ \equal{} \minus{} \frac18$, so $ \minus{} \left(\cos\frac\pi7\cos\frac {2\pi}7\cos\frac {4\pi}7\right)^2 \equal{} \boxed{ \minus{} \frac1{64}}$.
[/hide]

[hide="2"]
Similar to the first part, we find that $ \cos\frac\pi9\cos\frac{2\pi}9\cos\frac{4\pi}9\equal{}\frac18$, so $ \cos\frac\pi9\cos\frac{3\pi}9\cos\frac{5\pi}9\cos\frac{7\pi}9\equal{}\cos\frac\pi3\cos\frac\pi9\cos\frac{2\pi}9\cos\frac{4\pi}9\equal{}\frac18\cos\frac\pi3\equal{}\boxed{\frac1{16}}$.
[/hide]


calculus
derivative
limit
function
trigonometry
integration
real analysis

Let $ f\colon {\bf R}\to {\bf R}$ and $ g\colon {\bf R}\to {\bf R}$ two derivatives that is $ G'\equiv g$ and $ F'\equiv f.$ Is it true that $ fg$ is also a derivative?

A counterexample: $ F \equal{} G$, with $ F$ differentiable everywhere but $ F'$ not absolutely integrable. $ F(x) \equal{} \begin{cases}0 & x \equal{} 0 \\
x^2\sin(x^{ \minus{} 10}) & x\neq 0\end{cases}$ is one version. If $ h$ is an antiderivative of $ (F')^2$ on $ (0,\infty)$, $ \lim_{x\to 0^ \plus{} }h(x) \equal{} \minus{} \infty$, and if $ h$ is an antiderivative of $ (F')^2$ on $ ( \minus{} \infty,0)$, $ \lim_{x\to 0^ \minus{} }h(x) \equal{} \infty$. There's no way to glue that together into a continuous function, so $ (F')^2$ has no antiderivative on the whole line.

This is false.

Here's my example: let $ F(x) \equal{} G(x) \equal{} \begin{cases}x^2\sin\left(\frac1{x^2}\right) & x\ne 0 \\
0 & x \equal{} 0\end{cases}.$

Then $ F'(x) \equal{} f(x) \equal{} g(x).$ $ f(0) \equal{} 0$ but for $ x\ne 0,\ f(x)$ is a function whose dominant term near zero is $ \minus{} \frac {2}{x}\cos\left(\frac1{x^2}\right).$

That means that $ f(x)g(x) \equal{} f(x)^2\ge 0.$ If $ h'(x) \equal{} f(x)^2$ on $ (0,1],$ then $ h(1) \minus{} h(a) \equal{} \int_a^1f(t)^2\,dt$ would be comparable to $ C\int_a^1\frac1{t^2}\,dt,$ which is in turn comparable to $ \frac {C}{a}$ as $ a\to 0.$ There is no way to define $ h$ so that it is continuous at $ 0,$ and hence $ f(x)^2$ is not a derivative.

EDIT: Simultaneous post. Essentially the same example.


The equation $ x^2\minus{}9\equal{}x\plus{}3$ can be written as $ x^2\minus{}x\minus{}12\equal{}0,$ which we know has two solutions. Factoring the quadratic equation, we get that $ (x\minus{}4)(x\plus{}3)\equal{}0\implies{x\equal{}\minus{}3,4}.$ Thus, the sum of all solutions to the equation $ x^2\minus{}9\equal{}x\plus{}3$ is $ 4\minus{}3\equal{}\boxed1$

[quote=GameBot]What is the sum of all solutions to the equation $ x^2 \minus{} 9 \equal{} x \plus{} 3$?[/quote]

$ x^2 \minus{} 9 \equal{} x \plus{} 3$
$x^2-x-12=0$
$(x-4)(x+3)=0$
$x=-3$ or
$x=4$
$-3+4=1$

by vieta -> sum of solutions = x-coeffient * -1

-1*-1=1


Square BCFE is inscribed in right triangle AGD, as shown below. If AB=28 units and CD=58 units, what is the area if square BCFE?
I don't have a picture right now, but it's number 29 on the 2005 State sprint.

[hide]Hint:
Try to find a way of incorporating similar triangles.

If you do, it will lead you to the answer.[/hide]

[hide="hint"]similar triangles[/hide]
[hide="hint expanding on the first one"]ABE is similar to ECB[/hide]
[hide="answer"]Label the side of the square x. Using hint 2. you get $\frac{28}{x}=\frac{x}{58}$. Cross multiplying gives you $1624=x^{2}$ We are only looking for $x^{2}$ So there is your answer.[/hide]

[quote="mathgeniuse^ln(x)"]Yes, that is the correct answer.

And fix some of your latex, that you had in your answer.[/quote]
Thanks for pointing that out...


If Kn is decomposed into n complete graphs, then no 2 of them are disjoint.

Sorry ,I know that problem may be easy, but I still can't solve it.  I'm poor in combinatorics, so I'll appreciate if anyone can post a whole solution and help me out.

 :help:  :help:

This problem is another form of the 2008 Singapore TST problem 6. Shasha said by proving a stronger statement we could solve the problem, but I'm  still puzzled. So I post this topic here. I suppose the above statement may help.


calculus
derivative
calculus computations

In your seventh line of calculations: $ \sqrt {(b \minus{} a)^2}$ equals $ (b \minus{} a)$ if $ b > a$, and $ (a \minus{} b)$ if $ a > b$.


let x,y,z \geq 0
prove that
 \sqrt (1-x) +  \sqrt (4-y+x) +  \sqrt (9-z+y) +  \sqrt (16+z)  \leq 10.
it's easy but i think it's the interesting inequality.
try it  :D

It is interesting in guessing the factor,right?But if we found it,it becomes easy.
By Cauchy,we have:
$(1-x)+1 \geq 2 \sqrt {1-x}$
$(4-y+x)/2+2 \geq 2 \sqrt {4-y+x}$
$(9-z+y)/3+3 \geq 2 \sqrt {9-z+y}$
$(16+z)/4+4 \geq 2 \sqrt {16+z}.$
Now put it together,we have
\[ \sqrt  {1-x} +   \sqrt {4-y+x} +   \sqrt {9-z+y} +   \sqrt  {16+z} \leq 10- \frac {6x+2y+z}{12} \leq 10 . \]


In a quadrilateral $ ABCD$ we have $ \angle A \equal{} \angle C$. In the interior of $ ABCD$ there is a point $ N$ such that
 the circumcircles of  triangles [b]a) [/b] $ NAB,NCD$ and [b]b) [/b]$ NBC,NDA$ are tangent.Prove that points $ B,D,N$ are collinear and that the points $ C,N,A,K$ are concyclic, where $ K$ is the common point of lines $ CB,DA$.

[quote="The QuattoMaster 6000"]Look here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1148389#1148389[/quote]

Yes :)  ! I have read your nice solution , before sending the problem. I changed the wording in order to gather some  new approach.!
 Babis


induction
number theory unsolved
number theory

279. 5 (Vũ Đùc Sơn) Find all positive integers n such that $n < t_n$, where
$t_n$ is the number of positive divisors of $n^2$.


Prove that the number that n numbers can be ordered in different manners is n!.
I'm having trouble to prove it by induction.

Without induction:

[hide] n numbers could go in the first slot, (n-1) could then go in the second slot, (n-2) in the third slot, so n! ways.[/hide]

With induction:

[hide] 1 number may be arranged in 1 way.  If k numbers can be arranged in k! ways, there are (k+1) places to put the next one, so (k+1)! ways for k+1 numbers.[/hide]


group theory
abstract algebra
superior algebra
superior algebra solved

Let $G$ be a finitely generated nilpotent group and $x \in G.$ Is it true that there exist
subgroups $G_{0}= <x>, \ G_{1}, \ ... \ , \ G_{n}= G$ such that for every $0 \leq i \leq n-1,$
$G_{i}$ is a characteristic subgroup of $G_{i+1}$?

It's not even true for abelian groups. Consider for example elementary abelian p-groups. They have no characteristic subgroups at all beside the two trivial subgroups.


advanced fields
advanced fields theorems

Global existence and blowup solutions for quasilinear parabolic equations

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6WK2-4MY6GGK-5&_user=10&_coverDate=11%2F01%2F2007&_rdoc=15&_fmt=summary&_orig=browse&_srch=doc-info(%23toc%236894%232007%23996649998%23662155%23FLP%23display%23Volume)&_cdi=6894&_sort=d&_docanchor=&_ct=55&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=6e001113acd9b6a045b4a19c6a94ff97

I would like to have this article but I can not download it, would you help me.

My address is hoc.toan@yahoo.com

Thanks.

[quote="N.T.TUAN"]Hmm, what is its title,author, Vol,...?[/quote]

Dear Prof. N.T.TUAN,

I'm in VN so I can not have it. Please, help me.
It is:

[b]Global existence and blowup solutions for quasilinear parabolic equations
Shaohua Chena, and Deming Yub,

Journal of Mathematical Analysis and Applications
Volume 335, Issue 1, 1 November 2007, Pages 151-167 [/b]

Thanks.

Học, hoc.toan@yahoo.com

[quote="hoc.toan"]

Dear Prof. N.T.TUAN,

 [/quote]
Don't said as that I am not a Prof :D

[quote]

Global existence and blowup solutions for quasilinear parabolic equations
Shaohua Chena, and Deming Yub,

Journal of Mathematical Analysis and Applications
Volume 335, Issue 1, 1 November 2007, Pages 151-167
[/quote]
I will try! :)

Học đã nhận được bài báo. Học cảm ơn anh Tuân rất nhiều.

Kính chúc anh well.

L.U.Học.

No problem :) But please use English on this from  :wink:


calculus
integration
trigonometry
real analysis
real analysis unsolved

We all know how to calculate the following Riemann integral,

$\int_{0}^{\pi/4}\ln(1+\tan(x))dx$,

since it has been discussed so many times on the forum, however this problem raises the following question:

Problem. Let $n\geq 2$ be a natural number and let

$I(n)=\int_{0}^{\pi/4}\cdots\int_{0}^{\pi/4}\ln(1+\tan(x_{1})\cdots \tan(x_{n}))dx_{1}\cdots dx_{n}$.

Calculate:

a) $I(2)$.

b) Is there a closed formula for $I(n)$, when $n\geq 2$, perhaps a linear combination of well-known constants?

i was think to use Cauchy formula for repeated integration, but we have multiple variables, not only one.

so, to solve this problem we need a generalization for cauchy formula.

so i think that if the solution exist, is very hard

I was thinking. isnt the rigth formule the formula below?
$I(n)=\int_{0}^{\pi/4}\cdots\int_{0}^{\pi/4}\ln(n+\tan(x_{1})\cdots \tan(x_{n}))dx_{1}\cdots dx_{n}$

If yes, we can use a relation.

${n+\sum_{i=1}^{n}\tan \left(x_{i}\right)=\sqrt{2}\sum_{i=1}^{n}\frac{\cos \left(\frac{\pi }{4}-x_{i}\right)}{\cos \left(x_{i}\right)}}$

yagaron,
The integral is as written.
I think that the case when $n=2$ is "solvable" via the substitutions $\tan(x_{1})=y_{1}$ , $\tan x_{2}=y_{2}$, however the general case as far as I know is unresolved.


Hi, when i was in midd school my school did not offer algebra 1 or geometry

(since it used to be for seventh graders only but changed when i entered it),

 anyways the problem is that now in high school I cant take any advanced

 classes such as Calculus(when i get to 11th). I have already asked the

 teachers and they said that nothing can be done..but what worries me is that

when i get to college there will be classes that i might have to take(such as

 Calc 1 and 2) and these classes will a waste of time since i will allready know

 calc once i get there. :(  My question is do most colleges offer a test to see if i

can start in more advanced classes, or maybe if i write it on my app the might

 understand, also has anyone been in a similar situation? :maybe:

Try skipping some of the easier classes so you can be placed in higher ones. Other than that, there's pretty much nothing you can do.

personally, that is sorta similiar to my situation....

you see, i skipped 2 grades and on top of that im doing calculus(12yrs.old)

basically for me i would take tests and back up the fact that im smart. basically you have to "prove" to othe "authorities" that you are really smart.

that could include taking tests....SAT's etc...

That used to be my situation.

Basically, try to take Calculus at a nearby community college. That's what I'm currently doing.

Then, in your junior or senior year, take the AP Calc test for BC. (the AB test is sort of useless)...

Also, an advantage of taking classes at a community college is that you get the credits and they're transferable to the college you will attend in a few years. So if you take calc, and calc is a required class at the college you are going to attend, you have that over with.


[quote="boogie master"]O is the circumcentre of triangle ABC and H is the orthocentre. If AB = c, BC = a, CA = b and T is arbitrary point, express
$ \vec{TO} \plus{} \vec{TH}$ with a, b, c and $ \vec{OA}, \vec{OB}, \vec{OC}, \vec{OH}$.[/quote]
Hmm I think its incorrect because $ \vec{TO} \plus{} \vec{TH}$ when ABC fixed not constant

$ \vec{TH}\equal{}\vec{TO}\plus{}\vec{OH}$
Use Hamilton's theorem which says $ \vec{OH}\equal{}\vec{OA}\plus{}\vec{OB}\plus{}\vec{OC}$
$ \vec{TO}$ you can find from triangles $ TOA,TOB,TOC$:
$ \vec{TO}\equal{}\vec{TA}\plus{}\vec{AO}$
$ \vec{TO}\equal{}\vec{TB}\plus{}\vec{BO}$
$ \vec{TO}\equal{}\vec{TC}\plus{}\vec{CO}$

$ \vec{TO}\plus{}\vec{TH}\equal{}\frac{1}{3}\cdot {(2(\vec{TA}\plus{}\vec{TB}\plus{}\vec{TC}\plus{}\vec{AO}\plus{}\vec{BO}\plus{}\vec{CO})\plus{}3\vec{OH})}$
It's not constant because of $ \vec{TA}\plus{}\vec{TB}\plus{}\vec{TC}$.

If G is centroid of $ \triangle{ABC}$, then: $ \overrightarrow{TA} \plus{} \overrightarrow{TB} \plus{} \overrightarrow{TB} \equal{} 3.\overrightarrow{TG}.$


Prove that there exist a double sequence $b(k,n)$ of rationnals numbers such that

For any $n>1$ integer
$\sum_{k=1}^{n}\frac{b(k,n)}{k}=1$

and for any $n>1$ integer

$\sum_{k=1}^{n}\frac{b^2(k,n)}{4^n}$ is bounded


algebra
polynomial
factorial
number theory
greatest common divisor
calculus
integration

Let $ R\equal{}\mathbb Z_{p^n}$ for some prime p and natural n. $ f(x),g(x)\in R[x]$ with $ f\equal{}\sum_{i\equal{}0}^ra_ix^i,\; g\equal{}\sum_{j\equal{}0}^sb_jx^j$ and $ f(x)\cdot g(x)\equal{}0$. Then I have to prove that $ a_ib_j\equal{}0$ for all i,j. Am I missing something trivial that I couldn't yet have done that? Or else please give the solution.

[quote="saurav"]Let $ R \equal{} \mathbb Z_{p^n}$ for some prime p and natural n. $ f(x),g(x)\in R[x]$ with $ f \equal{} \sum_{i \equal{} 0}^ra_ix^i,\; g \equal{} \sum_{j \equal{} 0}^sb_jx^j$ and $ f(x)\cdot g(x) \equal{} 0$. Then I have to prove that $ a_ib_j \equal{} 0$ for all i,j. Am I missing something trivial that I couldn't yet have done that? Or else please give the solution.[/quote]

Hello.
It amounts to proving the following statement : if $ f(x),g(x)\in \mathbb{Z}[x]$ with $ f \equal{} \sum_{i \equal{} 0}^ra_ix^i,\; g \equal{} \sum_{j \equal{} 0}^sb_jx^j$, if the polynomial $ f(x)\cdot g(x)$ is divisible by $ p^{n}$ (in other words, if each coefficient of $ f(x)\cdot g(x)$ is divisible by $ p^{n}$), then each $ a_ib_j$ is divisible by $ p^{n}$.
I think we can prove by recurrence on $ n$ that there exist integers $ r,s$ ($ 0 \leq r,s \leq n$) such that $ r \plus{} s \equal{} n$, that each  $ a_i$ is divisible by $ p^{r}$ and each $ b_j$ is divisible by $ p^{s}$.
Our statement is trivial for $ n \equal{} 0$.
Assume our statement is true for $ n$ and let us prove it for $ n \plus{} 1$.
Let $ f(x),g(x)\in \mathbb{Z}[x]$ with $ f \equal{} \sum_{i \equal{} 0}^ra_{i}x^i,\; g \equal{} \sum_{j \equal{} 0}^sb_{j}x^j$, the polynomial $ f(x)\cdot g(x)$ being divisible by $ p^{n \plus{} 1}$. ..... (1)
By hypothesis of recurrence, there exist integers $ r,s$ ($ 0 \leq r,s \leq n$) such that $ r \plus{} s \equal{} n$, that each  $ a_i$ is divisible by $ p^{r}$ and each $ b_j$ is divisible by $ p^{s}$.
Write $ a_{i} \equal{} p^{r}a'_{i}$ and $ b_{j} \equal{} p^{s}b'_{j}$.
Then (1) shows that the product of the polynomials $ \sum_{i \equal{} 0}^ra'_{i}x^i$ and $ \sum_{j \equal{} 0}^sb'_{j}x^j$ is divisible by $ p$. Since $ \mathbb{Z}/p\mathbb{Z}$ is a field (and if a product of polynomials with coefficients in a field is null, one at least of these polynomials is null) , either each $ a'_{i}$ or each $ b'_{j}$ is divisible by $ p$. In the first case, $ r \plus{} 1$ and $ s$ are what we need; in the second case, take $ r$ and $ s \plus{} 1$.

Friendly,
P.

P. S. It is a particular case of a "lemma of Gauss". If $ A$ is a factorial ring, if $ h(X)$ is a polynomial in one variable with coefficients in $ A$, a gcd of the coefficients of $ h(X)$ is called a content of $ h(X)$. If $ f(X)$ and $ g(X)$ are polynomials with coefficients in $ A$, if $ c(f)$ is a content of $ f(X)$ and $ c(g)$ a content of $ g(X)$, then $ c(f)c(g)$ is a content of  $ f(X)g(X)$. (See Zariski and Samuel, Commutative algebra, vol. I, 1958, p. 32.)
In our case, $ A\equal{} \mathbb{Z}$, we assume that the content of $ f(X)g(X)$ is divisible by $ p^{n}$, thus $ c(f)c(g)$ is divisible by $ p^{n}$. If $ p^{r}$ is the greatest power of $ p$ which divides $ c(f)$ and $ p^{s}$ the greatest power of $ p$ which divides $ c(g)$ (I assume we are not in the trivial case where a polynomial is null), we must have $ r\plus{}s \geq n$, like in our argument.

Hello panurge.
Thanks for the excellent solution. I was really taken with the step that said $ \sum a_i'x_i$ etc. are elements of $ \mathbb Z_p[x]$ which is an integral domain.

However, as you said that this can be deduced from the theorem on factorial rings, may I ask (for I don't have a copy of Zariski and Samuel's book) whether the proof of the theorem is an easy one or assuming something too involved to be easy to grasp?

Many thanks again.
Regards,
Saurav

Hello Saurav.

Let $ A$ be a factorial ring and $ f(x),g(x)\in A[x]$ , let $ h(x)$ be their product, let $ c(f)$ be a content of $ f(X)$, $ c(g)$ be a content of $ g(X)$ and $ c(h)$ a content of $ h(x)$.
Clearly, $ c(h)$ is divisible by $ c(f)c(g)$. Let us prove that $ c(h)$ divides $ c(f)c(g)$.
Let $ p$ an irreducible factor of $ c(h)$, let $ p^{n}$ the greatest power of $ p$ which divides $ c(h)$. Thus $ h(x)$, i. e. $ f(X)g(X)$, is divisible by $ p^{n}$.
As in the proof above, there exist $ r$ and $ s$ such that $ r \plus{} s \equal{} n$, $ p^{r}$ divides $ f(x)$ and $ p^{s}$ divides $ g(x)$. (The fact that $ \mathbb{Z}/p\mathbb{Z}$ is a field, in the proof above, can be replaced by the fact that $ A/(p)[x]$ is an integral domain, because $ A/(p)$ is an integral domain.) Then $ c(f)$ is divisible by $ p^{r}$ and $ c(g)$ is divisible by $ p^{s}$, thus $ c(f)c(g)$ is divisible by $ p^{r \plus{} s}$, i.e. by $ p^{n}$.
Thus, for each irreducible divisor $ p$ of $ h(x)$, $ c(f)c(g)$ is divisible by the greatest power of $ p$ which divides $ c(h)$, thus $ c(f)c(g)$ is divisible by $ c(h)$.
(And thus, each $ a_{i}b_{j}$ is divisible by $ c(h)$.)

So, $ c(f)c(g)$ is a content of $ h(x) \equal{} f(x)g(x)$.

I didn't reread Zariski and Samuel, but I presume that their proof amounts to this.

Friendly,
P.

[quote="saurav"]Many thanks, sir.
But here I find one thing that I should ask you about, that is, should we say that c(h)"equals" c(f)c(g), or that c(h) is an associate of c(f)c(g)? Because, we have shown that c(h) divides c(f)c(g), and obviously c(f)c(g) divides c(h), which together only shows that they are associates.

Regards,
Saurav[/quote]

Yes, they are only associates, because a gcd is only defined as one of (perhaps) several associates.
Thus a content of a polynomial is only defined as one of (possibly) several associates.
Friendly,
P.


induction
probability
combinatorics unsolved
combinatorics

i have this problem:
given two numbers $n,k$ with conditions $k\leq n$
how do i find the number of permutations of the set {$1,2,3,...,n$} so that
in each permutation(of these chosen ones) k is bigger than all elements on it's left.

so..
i think it will be somethin like this {$p_1,p_2,p_3,...,p_{k-1},k,p_{k+1},...,p_{n-1},p_n$}
and k with the property that it is bigger than all behind her so we have:
$k\geq p_{k-1},p_{k-2},...,p_2,p_1$
so i think it should be something like the sum of the permutations of $p_1,p_2,...,p_{k-1}$ for each position of k
so what we have is for $t$=position of k

t=1 => number of posibilities 0
so we can say:
$
t=1 => 0 posibilities\\
t=2 => k-1 posibilities\\
t=3 => (k-1)(k-2)\\
t=4 => (k-1)(k-2)(k-3)\\
.
.
.
$
in general i think i can say the number of posibilities for k on position t
$W(t)=\frac{(k-1)!}{(t-1)!}$
it would suffice an induction proof but i won't do it because i don't want to
so...continuing my probably wrong assumptions we have further:
$S=\sum_{t=2}^{t=n-1}W(t)$
i think...
i'm probably wrong...
anyway what i would need to do now is to find a fast formula to calculate this $S$ of mine

I think the answer is $(k-1)!\sum_{t=1}^{k}\frac{(n-t)!}{(k-t)!}$.

Ok, Stefan.

If $k$ stays on position $t$ ($1\leq t\leq k$), then we have $A_{k-1}^{t-1}$ possibilities to choose numbers for first $t-1$ positions, then also we have independent $(n-t)!$ posibilities to place remaining $n-t$ numbers on positions $t+1$, $t+2$, ..., $n$. It gives $\frac{(k-1)!}{(k-t)!}\cdot (n-t)!$ variants if $k$ stays on position $t$.

[code]
#include <stdio.h>
#define mare unsigned long
mare f(mare q)
{mare i,r=1;
for(i=1;i<=q;i++)r*=i;
return r;}//f(q)=q!

int main()
{
mare N,K,p=0,t;
//in p i have final result
FILE *fi=fopen("perm.in","r");
FILE *fo=fopen("perm.out","w+");
fscanf(fi,"%d%d",&N,&K);
if(K!=N)
for(t=2;t<=K;t++)

p+=(f(K-1)/f(K-t))*f(N-t);

fprintf(fo,"%d",p);
return 0;
}
[/code]

this is the code for the problem
it does not work for N=5 and K=3
it is supposed to give 35,it's getting only to 16
have i done something wrong ?

Why for(t=2;...)? It should be for(t=1;...).
It will give you 40.

consider the relationship of k with elements in A = {k+1, k+2, ... , n-1, n}. the probability that all elements in A are on the right side of k is (n-k)!/(n-k+1)! = 1/(n-k+1), hence the result n!/(n-k+1).

i really can't understand the last thing you said
that the result is n!/(n-k+1).


geometry
incenter
LaTeX
geometry proposed

Let AH1,BH2,CH3 be the altitudes of scalene triangle ABC. The incircle of triangle ABC is tangent to BC, CA, and AB at T1,T2, and T3, respectively. Line HiHi+1 (where H4=H1) intersects TiTi+1 at Pi+2. Prove that the incenter of triangle ABC is the orthocenter of triangle P1P2P3.

I found that three lines P1T1,P2T2,and P3T3 are concurrent, and their intersection (denote by T4) is on the incircle of triangle ABC (I cannot prove it),  If that is true, then four points T1,T2,T3,and T4 are concyclic, denote by I the incenter of triangle ABC, then I is their circumcenter. Diagonals T1T3,T2T4 meet at P2 (suppose that P2 is inside triangle ABC),The opposite sides T1T4,T2T3 meet at P1, T1T2,T4T3 meet at P3, then I is orthocenter of triangle P1P2P3. my problem is solved.

Are you a Chinese? where do you come from?

Dear Darij Grinberg, Cosmin Pohoata, Kostas Vittas and Han-sol Shin:
Could you check my solution carefully, if it is right,then please you latex it.
I have solved this problem.
Lemma: Let $ ABC$ be a triangle, let $ D$, $ E$ be points on sides $ CA$, $ AB$ of triangle $ ABC$, respectively. $ P$ is point on $ DE$. If $ \frac {BE}{AB} \cdot \frac {PD}{PE} \cdot \frac {AC}{CD} = 1$ (1), then $ AP$, $ BD$ and $ CE$ are concurrent.

Drawing parallel lines to $ BC$ from point $ D$, $ E$ and intersect $ AP$ at $ G$, $ F$, respectively. Let $ AP$ intersect $ BC$ at $ Q$, then
\[ \frac {EP}{DP} = \frac {EF}{DG} = \frac {EF}{BQ} \cdot \frac {CQ}{DG} \cdot \frac {BQ}{CQ} = \frac {AE}{AB}\cdot\frac {AC}{AD}\cdot\frac {BQ}{CQ.} \ \ \ (2)
\]
By multyplying (1) with (2), we have $ AE/BE \cdot BQ/CQ \cdot CD/AD = 1$, thus $ AP$, $ BD$, and $ CE$ are concurrent.
At first, we prove that $ P_{1}H_{1}$, $ P_{2}H_{2}$, $ P_{3}H_{3}$ and $ P_{1}T_{1}$, $ P_{2}T_{2}$, $ P_{3}T_{3}$ are concurrent, respectively.
By Menelaus theorem, we have $ \frac {H_{1}P_{1}}{H_{3}P_{2}} \cdot \frac {P_{1}H_{3}}{P_{1}H_{2}} \cdot \frac {H_{2}P_{3}}{H_{1}P_{3}} = \frac {H_{1}T_{1}}{H_{3}T_{3}} \cdot \frac {H_{3}T_{3}}{H_{2}T_{2}} \cdot \frac {H_{2}T_{2}}{H_{1}T_{1}} = 1$ (3). Hence by lemma, $ P_{1}H_{1}$, $ P_{2}H_{2}$, $ P_{3}H_{3}$ are concurrent.

By Menelaus theorem, similarly we have $ \frac {P_{2}T_{1}}{P_{2}T_{3}} \cdot \frac {P_{1}T_{3}}{P_{1}T_{2}} \cdot \frac {P_{3}T_{2}}{P_{3}T_{1}} = \frac {BH_{3}}{BH_{1}} \cdot \frac {H_{1}T_{1}}{H_{3}T_{3}}\cdot \frac {AH_{2}}{AH_{3}} \cdot \frac {H_{3}T_{3}}{H_{2}T_{2}} \cdot \frac {CH_{1}}{CH_{2}} \cdot \frac {H_{2}T_{2}}{H_{1}T_{1}} = 1$ (4). Likewise, by lemma, $ P_{1}T_{1}$, $ P_{2}T_{2}$, $ P_{3}T_{3}$ are concurrent.

Then, we prove that $ A$, $ P_{2}$, $ P_{3}$ and $ B$, $ P_{1}$, $ P_{3}$, respectively $ C$, $ P_{2}$, $ P_{1}$ are collinear.
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=684346#684346[/url]

Let $ P_{2}P_{3}$ intersect $ T_{2}T_{3}$, $ H_{2}H_{3}$ at $ M$, $ N$ respectively. By Menelaus theorem and (4), we have $ T_{2}M/T_{3}M = T_{1}P_{2}/T_{3}P_{2} \cdot T_{2}P_{3}/T_{1}P_{3} = P_{1}T_{2}/P_{1}T_{3}$. So the quadruple $ (T_{2},M,T_{3},P1)$ is harmonic.

Similarly, $ H_{2}N/H_{3}N = P_{1}H_{2}/P_{1}H_{3}$, $ (H_{2},N,H_{3},P_{1})$ is harmonic. Hence, $ A$, $ M$, $ N$ are collinear, thus $ A$, $ P_{2}$, $ P_{3}$ are also collinear and likewise $ B$, $ P_{1}$, $ P_{3}$, respectively $ C$, $ P_{2}$, $ P_{1}$ are collinear.

Then we prove the intersection of $ P_{3}T_{3}$ with $ P_{1}T_{1}$ is on the incircle of triangle $ ABC$.
Let $ P_{3}T_{3}$ intersect the incircle of triangle $ ABC$ at $ T_{4}$. $ T_{1}T_{4}$ intersect $ T_{2}T_{3}$ at $ P'_{1}$, since $ BT_{1}$, $ BT_{3}$ are tangent of the incircle of triangle $ ABC$, so $ B$, $ P'_{1}$, $ P_{3}$ are collinear, because $ B$, $ P_{1}$, $ P_{3}$ are collinear, so $ P'_{1}\equiv P_{1}$, thus the intersection of $ P_{1}T_{1}$, $ P_{2}T_{2}$, $ P_{3}T_{3}$ is on the incircle of triangle $ ABC$.

[color=red][EDITED by pohoatza: LATEX and English spelling within the text (but not exactly from top to buttom).][/color]

Add to two lemmas as follows:
Let k be a circle with center O, and let A,B,C,D be four points on this circle k. Let P=AC∩BD,Q=AB∩CD,R=BC∩AD.
Let MB,MD be tangent of circle k. then
(I):Three points Q,M, and R are collinear.
(II):The point O is the orthocenter of triangle PQR.

Very nice, Fang-jh! Actually this result (along with a similar proof) appears in a Russian papper by Lev Emelyanov and Tatiana Emelyanova, Semejstvo Feuerbacha, [i]Matematicheskoe Prosveshjenie[/i], 2002.

In fact, more generally, the line $ P_{i}P_{i \plus{} 1}$ is the polar of $ P_{i \plus{} 2}$ w.r.t. the incircle. I also studied this configuration and I will be back in a week or two with more details and some generalizations since I think it is better to not discuss it yet.

Thank you very much! Dear Cosmin Pohoata.


real analysis
calculus
derivative
function
vector
limit
absolute value

Define $ f(0,0) \equal{} 0$ and

$ f(x,y) \equal{} \frac {x^3}{x^2 \plus{} y^2}$ if $ (x,y) \neq (0,0)$

a) Prove that the partial derivatives of f are bounded functions in $ R^2$.

b) Let $ \mathbf{u}$ be any unit vector in $ R^2$. Show that the directional derivative $ (D_{\mathbf{u}} f)(0,0)$ exists, and its absolute value is at most 1.

c)Let $ \gamma$ be a differentiable mapping of $ R^1$ into $ R^2$, with $ \gamma(0) \equal{} (0,0)$ and $ |\gamma'(0)| > 0$. Put $ g(t) \equal{} f(\gamma(t))$ and prove that $ g$ is differentiable for every $ t \in R^1$.

I can do parts a) and b). I need help with part c) at t = 0.

[quote="benorin"]Start off with $ \gamma (t) \equal{} \gamma_1 (t)\mathbf{e_1} \plus{} \gamma_2 (t)\mathbf{e_2}$  and substitute this into f(x,y) and take a derivative (deal with t=0 separately).[/quote]

I tried that.

$ g(t) \equal{} \frac {\gamma_1(t)^3}{\gamma_1(t)^2 \plus{} \gamma_2(t)^2}$ when $ t \neq \equal{} 0$ and $ g(0) \equal{} 0$. Clearly $ g$ is differentiable whenever $ t \neq 0$. To show that $ g$ is differentiable at $ t \equal{} 0$, we need to show that
$ \lim_{t \to 0} \frac {\gamma_1(t)^3}{t\gamma_1(t)^2 \plus{} t\gamma_2(t)^2}$
exists. We can apply l'Hopital but we can only apply it once because we don't know whether $ \gamma$ is twice-differentiable or not:

$ \lim_{t \to 0} \frac {t\gamma_1(t)^3\gamma_1'(t)\gamma(t)^2 \minus{} \gamma(t)^2\gamma_1(t)^3 \minus{} 2t\gamma(t)\cdot\gamma'(t)}{t^2(\gamma_1(t)^2 \plus{} \gamma_2(t)^2)^2}$

and then I am stuck.

[quote="10000th User"]Example 9.18 has somewhat an sketchy way of showing that and then use part b...[/quote]

But Example 9.18 assumes f is differentiable everywhere.


superior algebra
superior algebra solved

Decompose in $\mathbb{Z}[i\sqrt{2}]$ the following number $35+25i\sqrt{2}.$ and in $\mathbb{Z}[i]$ the following number $18+36i$. :)


[quote="b-flat"]I noticed that the forum Getting Started is now called High School Basics...

Why was the name changed, and when was it changed?

I should edit some of the topics like "Getting Started Difficulty."[/quote]

Hmm, well, I just noticed the change as well... maybe in the last few hours?

I renamed the forum because clearly too many people were interpreting it as 'Getting Started with the Forum' and posting their first question there, no matter how hard it is.  I probably spend 10 minutes a day moving threads out of that forum that are too hard, and I'm hoping the name change reduces that to 5 minutes a day.

[quote="PenguinIntegral"]That's [i]really wierd[/i]. They should be off writing books, not renaming our forums... :maybe:[/quote]Recently, all you've been doing on the forums is:

i) Making complaints and cynical comments, as others stated.
ii) Stating that you can consistantly score 12's on the AIME, yet you can't solve most 1's and 4's on the USAMO. Seriously, sit down and work on some problems or something. You should know by now that there's no "silver bullet" when it comes to doing well in math contests. People have been telling you that for months, yet the message hasn't seemed to get across...

FYI, it's weird, not wierd. You know, spelling is an important skill in life too.  :wink:

[quote="mysmartmouth"][quote="PenguinIntegral"]That's [i]really wierd[/i]. They should be off writing books, not renaming our forums... :maybe:[/quote]

Dude what's your problem?  Lately, you seem to have a cynical comment to add to every thread you can find.  Just chill...[/quote]
How was that cynical? Must I start using smilies? :roll:

Seriously though, (most) every post I make is in good humor. Please don't interpret it as mean or cynical or anything.

[quote="ragnarok23"][quote="PenguinIntegral"]That's [i]really wierd[/i]. They should be off writing books, not renaming our forums... :maybe:[/quote]
Seriously, what is your problem? They've given you so much, and all you do is demand more? It's not like changing the name of the forum takes month. Besides, it's not your forum, it's theirs. Take it easy, chill out.[/quote]
1) I have no problem.
2) I don't demand more.
4) I know.
5) I know.
6) I wish I could.

Once again, please don't interpret my comment as somehow demaning or demanding. :huh:  I don't do that kind of thing. Much. :blush:

[quote="4everwise"]

i) Making complaints and cynical comments, as others stated.
ii) Stating that you can consistantly score 12's on the AIME, yet you can't solve most 1's and 4's on the USAMO. Seriously, sit down and work on some problems or something. You should know by now that there's no "silver bullet" when it comes to doing well in math contests. People have been telling you that for months, yet the message hasn't seemed to get across...

FYI, it's weird, not wierd. You know, spelling is an important skill in life too.  :wink:[/quote]
Once again, I'm not making a complaint, and I'm not making a cynical comment. It's just the name change was sort of random, and now I know, so I really don't care anymore.

Yeah, I realize there's no silver bullet. I've just been looking for one because I'm lazy and I really don't want to do math anymore. You can check my post history, I finally realized this.

And please though, don't streach the facts :maybe:  I never stated I can get 12s consistently ( I can get a good score, yes, but not 12s.) And I never stated I couldn't ever get 1s or 4s on USAMOs.  :roll:
Thanks for the spelling tip :P

Seriously though, you're taking sarcastic little comments and taking it way too far.  :P

Um, this discussion has gotten a bit off topic - I think I'll lock this thread, as the initial question has been answered. ;)


inequalities
geometry
3D geometry
tetrahedron

a) In a triangle $ MNP$, the lenghts of the sides are less than $ 2$. Prove that the lenght of the altitude corresponding to the side $ MN$ is less than $ \sqrt {4 \minus{} \frac {MN^2}{4}}$.

b) In a tetrahedron $ ABCD$, at least $ 5$ edges have their lenghts less than $ 2$.Prove that the volume of the tetrahedron is less than $ 1$.

a) I use Apollonius Theorem and Pythagorean Theorem and approach the solution :(
Can you give a hint?

[hide="solution"][b]a).[/b] Let $ PQ$ be the median related to the line $ MN$ . Now we have $ PQ^2\equal{}\frac{2(PM^2\plus{}PN^2)\minus{}MN^2}{4}<4\minus{}\frac{MN^2}{4}$

[b]b).[/b]  Let $ CD\equal{}a<2$ and $ AB$ isn't necesarily less than 2. Denote $ M$ the projection of the point $ B$ on the line $ CD$ .In the triangle $ BCD$ the sides are less than 2 , therefore $ BM<\sqrt{4\minus{}\frac{a^2}{4}}$ Analogously the altitude $ h$ from vertice $ A$ in the triangle $ ACD$ is less than $ \sqrt{4\minus{}\frac{a^2}{4}}$ therefore $ h$ is less then $ \sqrt{4\minus{}\frac{a^2}{4}}$.Then $ V[ABCD]\equal{}\frac{1}{3}\cdot h\cdot [BCD]<\frac{a(16\minus{}a^2)}{24}$
Remains to prove that $ a(16\minus{}a^2)\leq 24$.
But $ a^3\minus{}16a\plus{}24\equal{}(a\minus{}2)(a^2\plus{}2a\minus{}12)$ and $ a<2$ , $ a^2\plus{}2a<4\plus{}4<12$ from where the conclusion results.[/hide]

:blush: I made a stupid mistake  :blush:  :blush:


Find$ x,y$ natural number $ x^y\equal{}y^{x\minus{}y}$

First of all, this should go into the Number Theory fora.

Second of all, [hide="Solution"]
[b]Lemma.[/b] If $ x,y,a,b$ are natural numbers such that $ x^a \equal{} y^b$, then $ \min\{x,y\}\mid\max\{x,y\}$.

[b]Proof.[/b] WLOG assume that $ x\le y$; then $ a\ge b$. Let $ g \equal{} (x,y)$ and $ x \equal{} gx_1,y \equal{} gy_1$. Then our equality is equivalent to $ g^{a \minus{} b}x_1^a \equal{} y_1^b$. Therefore $ x_1\mid y_1^b$. But $ (x_1,y_1) \equal{} 1$; hence $ x_1 \equal{} 1$ and thus we conclude that $ x\mid y$. $ \blacksquare$

Now for our problem, note that $ x\ge y$. If $ x \equal{} y$ or $ y \equal{} 1$, we get the solution $ (x,y) \equal{} (1,1)$. So assume that $ x > y > 1$.

By our lemma, we deduce that $ y\mid x$. Let $ x \equal{} my$. Then our equation becomes $ m^yy^y \equal{} y^{(m \minus{} 1)y}$, which after taking the $ y$-th root on each side, becomes $ m \equal{} y^{m \minus{} 2}$. But this implies $ m\ge 2^{m \minus{} 2}$ which is false for $ m\ge 5$. Thus $ 3\le m\le 4$. Checking all possible cases we find that $ (m,y) \equal{} (3,3),(4,2)$ are the only possible pairs. Therefore we conclude that $ (x,y) \equal{} (1,1),(9,3),(8,2)$ are the only solutions.[/hide]


Would anyone want me to post some problems daily? I was thinking of selecting the better mandelbrot problems from the book I have.

Sure. I don't always get on daily, but I wouldn't mind the practice. And I'm sure no one else really minds, either. (Correct me if I'm wrong)

NO!  YOU MAY NOT EVER POST MATH PROBLEMS EVER EVER EVER!  PENALTY FOR THIS IS HANGING!!!  </sarcasm>

hai guys,
just feel free to ask me any maths problem
==================================
jackhenry
http://www.drugtreatments.com/iowa


inequalities
LaTeX
calculus
derivative
algebra
polynomial
inequalities open

Hi evevrybody,do you know Schur inequality for 4 numbers.I and arosisi are looking for an inequality similar to it,we have had some,but it is not too very nice although it is quite strong,for example:
    [tex] \sum a^3b+\sum a^2b^2+\sum abcd \geq 3 \sum a^2bc [/tex]
     {sum here is symmetric sum}.
 So we must ask for your ideas.Please tell me if you find something,thanks :P

[color=red][Admin Edit: Made Latex compliant][/color]

I am very interested in the solution of this beautiful inequality.

Can you post it, please?

Thank you very much.

I'm even more interested in a form that I could understand. Can anyone please write the full formula for me? I don't understand those symetric sums.

the inequality that Anh and Arosisi are talking about is in fact

[tex] a^3b+a^3c+a^3d+b^3a+b^3c+b^3d+c^3a+c^3b+c^3d+d^3a+d^3b+d^3c+2a^2b^2+2a^2c^2+2a^2d^2+2b^2c^2+2b^2d^2+2c^2d^2+12abcd \geq 3(a^2bc+a^2bd+a^2cd+b^2ac+b^2ad+b^2cd+c^2ab+c^2ad+c^2bd+d^2ab+d^2ac+d^2bc) [/tex] :D

I'm in a good shape tonight. It's not very hard after all. All we need is to write down Schur for any triple:
  [tex] d(a^3+b^3+c^3)+3abcd\geq a^2(db+dc)+b^2(cd+ad)+c^2(da+db) [/tex]. Write down three other inequalities like this one and add them up. You will find that [tex]\sum a^3(b+c+d)+12abcd\geq\ 2\cdot\sum a^2(bc+bd+cd) [/tex]. Now we need to show that [tex] 2\cdot\sum a^2b^2\geq\sum a^2(bc+bd+cd) [/tex]. But this is easy, since [tex]2\cdot\sum a^2b^2=\sum a^2(b^2+c^2+d^2)\geq\sum a^2(bc+bd+cd) [/tex].

Thanks Harazi,your inequality is exactly Schur inequality.I  found the general form of Schur inequality yesterday,so our inequality is really weak.But we are finding in another way,so we hope that we will find stronger inequality.
In fact our form is [tex] \left\sum \ (a-b)^2(ac+bc+ad+bd-ab-cd\right) \geq \ \left\ 0 \right[/tex]
I will post if I find something interesting.And I really grateful if you send to me something about this one too,thanks. :)

i have a problem. i has sent it in one topic but i have no reply.
let a,b,c,d is possitive number and
a 2 +  2 + c 2 + d 2 + abcd = 5
prove that a+b+c+d  <= 4
i'm not sure it's right
if yes i think it can be solved by some Schur inequalities for 4 varies as anhcuong has demained
i hope that this ineq interests you
thanks  :D

Damn, I have written a lot in Latex and after that a nice message: page has expired. And I lost everything. That is why I won't write now in Latex:
    First of all, I beg for someone to write explicitely the symetric sum so that I can understand it. Secondly, the last problem is super-hard. I think I have a solution, if my computations aren't wrong:
    Suppose we have such a 4-uple a,b,c,d for which a+b+c+d=4r, with r>1. Put a=rx, b=ry, c=rz and d=rt. We find that x+y+z+t=4 but x^2+y^2+z^2+t^2+xyzt<5. So, we need to prove that for any x,y,z,t>=0 such that x+y+z+t=4 we have x^2+y^2+z^2+t^2+xyzt>=5. This isn't easy at all. Take f(x,y,z,t)=LHS-RHS. Let us choose x>=1. First we prove that f(x,y,z,t)>=f(x,u,u,u) where u=(x+y+z)/3. This comes down to:
  x*[ (y+z+t)^3/27-yzt]<=2/3*[ sum y^2-sum yz]. Now, let us observe that
 (y+z+t)^3/27-yzt=1/27*[ (y+z+t)(sum y^2-sum yz)+3(sum yz(y+z)-6yzt)]<=1/27*4(sum y)(sum y^2-sum yz), since from Schur we have sum yz(y+z)-6yzt<=(sum y)(sum y^2-sum yz). So, it's enough to have 4/27*x*[sum y]*(sum y^2-sum yz) <=2/3*(sum y^2-sum yz) or x(y+z+t)<=9/2. Fortunately, this is true, since x(y+z+t)=x(4-x)<=4<9/2. So, it's enough to prove that f(x,u,u,u)>=0. Since x=4-3u, the inequality comes down to (4-3u)^2+3u^2+u^3(4-3u)>=5, or after lots of computations (which I hope aren't wrong) to (u-1)^2(3u^2+2u-11)<=0. But thi follows from the fact that x>=1 , thus u<=1.

i think if we write the last inequality in symetric sum, it will become easier
we need to prove that for all possitive real number x,y,z,t have sum 4, then
$x^2 +y^2+z^2+t^2+xyzt\geq5$ (*)
let $x=\frac{4a}{a+b+c+d}, y=\frac{4b}{a+b+c+d}...$
then (*) is equivalent to
$\sum (sym) 11a^4 +36a^3b+3a^2b^2+34abcd-84a^2bc\geq0$
but we have $sum (sym) 34a^2b^2+34abcd-68a^2bc=34\sum(a-b)^2(c^2+d^2-2cd)\geq0$ which is right.
so we have (*).  :D

I think (in fact I am sure) that the following stronger inequality holds:

If $x, y, z, t$ are positive numbers such that $x+y+z+t=4$, then

\[3(x^2+y^2+z^2+t^2)+4xyzt\geq16.\]

Try somebody to prove this ?

Let's try in Latex this time. Take $ f(x,y,z,t)=3(x^2+y^2+z^2+t^2)+4xyzt $. It's easy to verify that the inequality $ f(x,y,z,t)\geq f(x,u,u,u) $ where $ u=\frac{y+z+t}{3} $ is equivalent to:
    $\frac{27}{2}\cdot (y^2+z^2+t^2-yz-zt-ty)\geq x\cdot[y^3+z^3+t^3+3yz(y+z)+3zt(z+t)+3yt(y+t)-21yzt] $. Now use Schur to find that $ y^3+z^3+t^3+3yz(y+z)+3yt(y+t)+3zt(z+t)-21yzt\leq 4(y+z+t)(y^2+z^2+t^2-yz-zt-ty) $. So, for the inequality $ f(x,y,z,t)\geq f(x,u,u,u) $ to be true it is suficient to have $ 4x(y+z+t)\leq\frac{27}{2} $. But this is true under the assumption $ x\leq 1$ , which can be made. Thus it remains to prove that $ f(4-3u,u,u,u)\geq 16 $ for any $ \frac{4}{3}\geq u\geq 0$. And this is probably not so difficult. Hope my computations are right, otherwise the solution being wrong.

This one is also my solution. Congratulation!
And a good counsel:
Before press SUBMIT, press PREVIEW.

I quote this from solution to Vietnam 1996/6 in book "Problems and Solutions from All Around the World":
Define s_i as the average of the products of the i-element subsets of {x_1..x_n}...
...Schur's inequality:
3s_1^3+s_3>=4s_1s_2
...(!!)Finally, for those who have only seen Schur's inequality in three variables, note that in general ANY inequality involving s_1..s_n which holds for n>=k variables also holds for n+1 variables, by replacing x_1..x_n+1 by roots of the derivative of the polynomial (x-x_1)..(x-x_n+1)

Harazi, your proof is not only correct but also magnificient.

The final expression is:

$12(u-1)^2(u+2)(u-4/3) \leq0$  which is obvious for  $4/3 \geq u \geq0$


vector
analytic geometry
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superior algebra
superior algebra theorems

Let $ E$ be a vector space over a field $ K$ with finite dimension m. Let $ A: E \to E$ be a linear operator. The definition of the determinant depends on what basis you use. It is the unique antisymmetric multilinear form so that $ \det(b_1,...,b_m) \equal{} 1$ if you use $ b_1,...,b_m$ as your basis.

My question is this. Are the roots of $ \det(A \minus{} \lambda I) \equal{} 0$ always the same as the eigenvalues of $ A$ or does that only hold when the determinant is taken using the standard basis? Please explain why or why not.

The determinant is actually basis-independent (this should be true, as eigenvalues/vectors are also basis-independent). The most basic way to see this is that change of basis is a conjugation operation (i.e. $ M A M^{\minus{}1}$), and so it leaves the determinant invariant.

Since you seem to think in terms of multilinear forms, let me give the coordinate-free description. Let D be the mth exterior power of E, (that is, the vector space with the universal property that alternating multilinear forms of m arguments from E to a vector space V are the same as linear functions from D to V, and the correspondence results from precomposition with a natural alternating multilinear m-form from E to D). Note that D is a vector space of dimension 1. If we have a linear map L from E to itself, we can extend that to a map of D (such that $ v_1 \wedge ... \wedge v_m$ goes to $ L(v_1) \wedge ... \wedge L(v_m)$). Since D is 1-dimensional, that map must be multiplication by some scalar; this scalar is called the 'determinant' of L.


LaTeX
modular arithmetic
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relatively prime

Find all integers p with the following condition :

                         the square root of [i](p2-1)[/i]  is an integer
I found a long solution and I want know if there isn’t a simple one

Which one you want¿:
a) $ \sqrt{p^{2}-1}$ an integer.
a) $ \sqrt{p^{5}-1}$ an integer.

And by the way, see http://www.mathlinks.ro/Forum/viewtopic.php?t=5261 for how to use $ \text{\LaTeX}$.

Excuse me the right problem is
Find all integers p with the following condition :

the square root of (p^5-1) is an integer
I found a long solution and I want know if there isn’t a simple one

[hide="Something possibly helpful"]
We have $ p^5 - 1 = k^2$.

$ p^5 \equiv \pm1, 0 \pmod{11}$ (after testing...), and $ p^2 \equiv 1,4,9,5,3,0 \pmod{11}$. Therefore, $ 11|k$ and $ p^5 \equiv 1 \pmod{11}$.

Also, since $ p$ is obviously odd, (if not, we have $ p^5 - 1\equiv - 1\pmod{8}$ which in not a square modulo 8) we have $ p^5 - 1 \equiv p - 1 \pmod{8}$ by Euler's Theorem, so $ p\equiv 1,5 \pmod{8}$.
[/hide]

and btw, don't use "code"; use dollar signs to enclose $ \text{\LaTeX}$ code.

EDIT: oh, you did; just get rid of the [code] tags

my first solution is unfinished :
$ p=1$ is a trivial solution.
Now if $ p>1$ we have
   \[ p^5-1= (p-1)(p^4+p^3+p^2+p+1) =(p-1)[(p-1)(p^3+2p^2+3p^+4)+5]\]
if $ ^delta=\gcd(p-1,p^4+p^3+p^2+p+1)$ then $ \delta\mid5$
if $ \delta =1$ then $ p-1=m^2$ and $ p^4+p^3+p^2+p+1=n^2$ (*) for some integers $ m$ and $ n$.
it known that (*) have the only solution $ p=3$, $ n=11$.we obtain that $ m^2=2$, it's impossible.
if $ \delta =5$ then $ p-1=5q$ for some positif integer q
   \[ q[5^{3}q^4+5^{3}q^3 +2*5^{2}q^2 + 10q+1]\] is square
we have $ 5^{3}q^4+5^{3}q^3 +2*5^{2}q^2 + 10q+1=a^2$ with simple algebric operations we obtain
  \[ 5(2^{4}5q^2+2^{3}5q+9)^2-26(2^{4}5q^2+2^{3}5q+9) +85=(16a)^2\]
the condition for the last equation to have integers solutions is its discriminant must be a square: $ \Delta=4(5b^2-16^2)=d^2$ for some integer d and $ b=16a$
we obtain the equation $ 5a^2-t^2=1$(**)( $ 5b^2-16^2$ is square and $ b=16a$).
the solutions of (**) are
               \[ \left\{\begin\{array}{cc}x_(n+1)=9x_n+4y_ny_(n+1)=20x_n+9y_n\end{array}\right\]  (***)
  $ x_0=1$ and $ y_0=2$.
$ \Delta=4(5b^2-16^2)=d^2$ is sqaure $ \Longleftrightarrow$ $ a$ is in $ {x_n; n\in N}$.
if $ a=x_n$ then $ Delta= (2^5{y}_n)^2$.Then, $ \displaystyle 2^{4}5q^2+2^{3}5q+9= \frac{26+2^{5}y_n}{10}$ (****)
Obviously we have $ y_n=2 \bmod (5)$. With (***) we can see by a simple reccurence that
  \[ y_n=\left\{\begin\{array}{cc}2 \bmod (5), & \mbox{if} n is even\\3 \bmod (5), & \mbox{if} n is odd\].
we substitue $ q$ by $ p$ in (****) then we obtain that $ p$ is in the sequence of integers $ (p_n)$ such that
                                                          \[ 2*(p_n)^2+p_n+1=2y_2n\]
it's simple to give to $ y_2n$ a direct form (with (***)), then we can have a direst form for $ p_n$.
if someone have an idea please post it.

my second solution :
i don't like it but we can use it to  resolve the problem in  general case : find all integers n such that $ \sqrt{n^q-1}\in N$ where $ q$ is a prime.
the problem is equivalent to resolve the equation $ n^q-1=y_2$(*). obviously, if $ y$ is odd then $ y^2+1=2 \bmod(8) which is impossible for a$q$ th power with$q\geq 2.
So $ y$ is even and $ n$ is odd. On the other hand (*) dosen't have solution for $ q=2$ (only $ (0,1)$), from now let $ q$ an odd prime.
we  have $ (y+i)(y-i)=n^q$, since n is odd, then $ y+i$ and $ y-i$ are coprime in $ Q[i]$, so $ y+i= i^k*\alpha^q$ where $ 0\geqq\leq3$.But since $ p$ and $ 4$ are coprime and by BEZOUT, we can write that
$ uq+4v=1$ so $ i^k= (i^u)^q$ and $ i$ is a $ p$th power. we obtain that $ y+i= \beta^q$.
Let $ \beta= a+ib$. We have $ \beta^p= N+iM$ with
                \[ \ M=\sum{k=0)^{\frac{q-1}{2}}frac{(q)!}{(2k)!(q-2k)!}a^{2k}b^{q-2k}(-1)^{frac{q-1}{2}-k}}\]
We have $ b\mid$ and $ M=1$ so $ b=\pn1$. we have
                \[ \sum{k=0)^{\frac{q-1}{2}}frac{(q)!}{(2k)!(q-2k)!}a^{2k}(-1)^{frac{q-1}{2}-k}=\pn1}\]
since $ q\mid frac{(q)!}{(2k)!(q-2k)!}$ for $ 0\geqq\leq frac{q-1}{2}$. we have
                \[ \D=sum{k=1)^{\frac{q-1}{2}}frac{(q)!}{(2k)!(q-2k)!}a^{2k}(-1)^{k}=0}\]
it easy to see that $ a$ is even.
Let $ \displaystyle n_k=frac{(q)!}{(2k)!(q-2k)!}a^{2k}=frac{q(q-1)}{2k(2k-1)}frac{(q-2)!}{(2k-2)!(q-2)!}a^{2k}$.
we have $ \displaystyle frac{n_k}{n_1}=frac{1}{2k-1}frac{(q-2)!}{(2k-2)!(q-2)!}a^{2k-2}$. then for $ k>1$ ($ q>3$), we have
                \[ v_{2}(n_k)-v_{2}(n_1)\geq (2k-2)v_{2}(a)-v_{2}(k)\geq (2k-2))-v_{2}(k)geq 1\]
So $ v_{2}(n_k)\geq v_{2}(n_1)$ for $ k>1$. Since $ \D=sum{k=1)^{\frac{q-1}{2}}(-1)^{k}n_k=0}$ then $ v_{2}(D)=v_{2}(n_1)=v_{2}(frac{p(p-1)}{2}a^2)=0$ which is imposiible for $ a$ not null.
So $ a=0$ hence $ \beta=\pni$, $ y=0$ and $ n=1$.

i don't know if it's true for p odd

my first solution is unfinished :
$ p=1$ is a trivial solution.
Now if $ p>1$ we have
   \[ p^5-1= (p-1)(p^4+p^3+p^2+p+1) =(p-1)[(p-1)(p^3+2p^2+3p^+4)+5]\]
if $ \delta=\gcd(p-1,p^4+p^3+p^2+p+1)$ then $ \delta\mid5$
if $ \delta =1$ then $ p-1=m^2$ and $ p^4+p^3+p^2+p+1=n^2$ (*) for some integers $ m$ and $ n$.
it known that (*) have the only solution $ p=3$, $ n=11$.we obtain that $ m^2=2$, it's impossible.
if $ \delta =5$ then $ p-1=5q$ for some positif integer q
   $ q[5^{3}q^4+5^{3}q^3 +2*5^{2}q^2 + 10q+1]$ is square
we have $ 5^{3}q^4+5^{3}q^3 +2*5^{2}q^2 + 10q+1=a^2$ with simple algebric operations we obtain
  \[ 5(2^{4}5q^2+2^{3}5q+9)^2-26(2^{4}5q^2+2^{3}5q+9) +85=(16a)^2\]
the condition for the last equation to have integers solutions is its discriminant must be a square: $ \Delta=4(5b^2-16^2)=d^2$ for some integer d and $ b=16a$
we obtain the equation $ 5a^2-t^2=1$(**)( $ 5b^2-16^2$ is square and $ b=16a$).
the solutions of (**) are
               \[ \{\begin{array}{cc}x_(n+1)=9x_n+4y_ny_(n+1)=20x_n+9y_n\end{array}\]  (***)
  $ x_0=1$ and $ y_0=2$.
$ \Delta=4(5b^2-16^2)=d^2$ is sqaure $ \Longleftrightarrow$ $ a$ is in $ {x_n; n\in N}$.
if $ a=x_n$ then $ \Delta= (2^5{y}_n)^2$.Then, $ \displaystyle 2^{4}5q^2+2^{3}5q+9= \frac{26+2^{5}y_n}{10}$ (****)
Obviously we have $ y_n=2 \bmod (5)$. With (***) we can see by a simple reccurence that
  \[ y_n=\{\begin{array}{cc}2 \bmod (5), & \mbox{if} n is even\\3 \bmod (5), & \mbox{if} n is odd\end{array}\].
we substitue $ q$ by $ p$ in (****) then we obtain that $ p$ is in the sequence of integers $ (p_n)$ such that
                                                          \[ 2(p_n)^2+p_n+1=2y_2n\]
it's simple to give to $ y_2n$ a direct form (with (***)), then we can have a direst form for $ p_n$.
if someone have an idea please post it.

my second solution :
i don't like it but we can use it to  resolve the problem in  general case : find all integers n such that $ \sqrt{n^q-1}\in N$ where $ q$ is a prime.
the problem is equivalent to resolve the equation $ n^q-1=y_2$(*). obviously, if $ y$ is odd then $ y^2+1=2 \bmod(8) which is impossible for a$q$ th power with$q\geq 2.
So $ y$ is even and $ n$ is odd. On the other hand (*) dosen't have solution for $ q=2$ (only $ (0,1)$), from now let $ q$ an odd prime.
we  have $ (y+i)(y-i)=n^q$, since n is odd, then $ y+i$ and $ y-i$ are coprime in $ Q[i]$, so $ y+i= i^k*\alpha^q$ where $ 0\geqq\leq3$.But since $ p$ and $ 4$ are coprime and by BEZOUT, we can write that
$ uq+4v=1$ so $ i^k= (i^u)^q$ and $ i$ is a $ p$th power. we obtain that $ y+i= \beta^q$.
Let $ \beta= a+ib$. We have $ \beta^p= N+iM$ with
                \[ \ M=\sum_{k=0)^{\frac{q-1}{2}}{\frac{(q)!}{(2k)!(q-2k)!}a^{2k}b^{q-2k}(-1)^{frac{q-1}{2}-k}}}\]
We have $ b%Error. "midM" is a bad command.
$ and $ M=1$ so $ b=\np 1$. we have
                \[ \sum_{k=0)^{\frac{q-1}{2}}{\frac{(q)!}{(2k)!(q-2k)!}a^{2k}(-1)^{\frac{q-1}{2}-k}}=\ pn 1}\]
since $ q\mid \frac{(q)!}{(2k)!(q-2k)!}$ for $ 0\geqq\leq frac{q-1}{2}$. we have
                \[ \D=sum_{k=1)^{\frac{q-1}{2}}{\frac{(q)!}{(2k)!(q-2k)!}a^{2k}(-1)^{k}}=0}\]
it easy to see that $ a$ is even.
Let $ \displaystyle n_k=\frac{(q)!}{(2k)!(q-2k)!}a^{2k}=\frac{q(q-1)}{2k(2k-1)}\frac{(q-2)!}{(2k-2)!(q-2)!}a^{2k}$.
we have $ \displaystyle \frac{n_k}{n_1}=\frac{1}{2k-1}\frac{(q-2)!}{(2k-2)!(q-2)!}a^{2k-2}$. then for $ k>1$ ($ q>3$), we have
                \[ v_{2}(n_k)-v_{2}(n_1)\geq (2k-2)v_{2}(a)-v_{2}(k)\geq (2k-2))-v_{2}(k)geq 1\]
So $ v_{2}(n_k)\geq v_{2}(n_1)$ for $ k>1$. Since $ \D=sum_{k=1)^{\frac{q-1}{2}}{(-1)^{k}n_k}=0}$ then $ v_{2}(D)=v_{2}(n_1)=v_{2}(\frac{p(p-1)}{2}a^2)=0$ which is imposiible for $ a$ not null.
So $ a=0$ hence $ \beta=\pni$, $ y=0$ and $ n=1$.

i don't know if it's true for p odd.

[quote="NA_219"]find all positive integers p which

[code]sqrt(p^5-1)[/code] in integer[/quote]
[hide="solution using Gaussian integers"]
$ p^{5}\equal{}(y\plus{}i)(y\minus{}i)$
gcf(y+i,y-i)=2i at most.
i is a unit, so we discard it, and 2 can't be a factor of y+i, since as not trig showed, p is odd.
So y+i and y-i are relatively prime, and must both be 5th powers.
$ y\plus{}i\equal{}(a\plus{}bi)^{5}$
$ a^{5}\minus{}10a^{3}b^{2}\plus{}5ab^{4}\equal{}y$
$ 1\equal{}5a^{4}b\minus{}10a^{2}b^{3}\plus{}b^{5}$
$ b|1$, so $ b\equal{}1$.
$ 0\equal{}5a^{4}\minus{}10a^{2}$*
$ a^{2}\equal{}2$ has no integral solutions.
At * we divided by $ 5a^{2}$, so keep in mind the case where a=0. If a=0, b=1, then y=0, and $ p^{5}\equal{}1$, so our only solution is $ \boxed{p\equal{}1}$[/hide]
:D

my first solution is unfinished :
$ p=1$ is a trivial solution.
Now if $ p>1$ we have
   \[ p^5-1= (p-1)(p^4+p^3+p^2+p+1) =(p-1)[(p-1)(p^3+2p^2+3p^+4)+5]\]
if $ \delta=\gcd(p-1,p^4+p^3+p^2+p+1)$ then $ \delta\mid5$
if $ \delta =1$ then $ p-1=m^2$ and $ p^4+p^3+p^2+p+1=n^2$ (*) for some integers $ m$ and $ n$.
it known that (*) have the only solution $ p=3$, $ n=11$.we obtain that $ m^2=2$, it's impossible.
if $ \delta =5$ then $ p-1=5q$ for some positif integer q
   \[ q[5^{3}q^4+5^{3}q^3 +2*5^{2}q^2 + 10q+1]\] is square
we have $ 5^{3}q^4+5^{3}q^3 +2*5^{2}q^2 + 10q+1=a^2$ with simple algebric operations we obtain
  \[ 5(2^{4}5q^2+2^{3}5q+9)^2-26(2^{4}5q^2+2^{3}5q+9) +85=(16a)^2\]
the condition for the last equation to have integers solutions is its discriminant must be a square: $ \Delta=4(5b^2-16^2)=d^2$ for some integer d and $ b=16a$
we obtain the equation $ 5a^2-t^2=1$(**)( $ 5b^2-16^2$ is square and $ b=16a$).
the solutions of (**) are
               \[ \{\begin{array}{cc}x_{n+1}=9x_n+4y_ny_{n+1}=20x_n+9y_n\end{array}\]  (***)
  $ x_0=1$ and $ y_0=2$.
$ \Delta=4(5b^2-16^2)=d^2$ is sqaure $ \Longleftrightarrow$ $ a$ is in $ {x_n; n\in N}$.
if $ a=x_n$ then $ \Delta= (2^5{y}_n)^2$.Then, $ \displaystyle 2^{4}5q^2+2^{3}5q+9= \frac{26+2^{5}y_n}{10}$ (****)
Obviously we have $ y_n=2 \bmod (5)$. With (***) we can see by a simple reccurence that
  \[ y_n=\{\begin{array}{cc}2 \bmod (5), & \mbox{if} n is even\\3 \bmod (5), & \mbox{if} n is odd\end{array}\].
we substitue $ q$ by $ p$ in (****) then we obtain that $ p$ is in the sequence of integers $ (p_n)$ such that
                                                          \[ 2*(p_n)^2+p_n+1=2y_2n\]
it's simple to give to $ y_2n$ a direct form (with (***)), then we can have a direct form for $ p_n$.
if someone have an idea please post it.

my second solution :
i don't like it but we can use it to  resolve the problem in  general case : find all integers n such that $ \sqrt{n^q-1}\in N$ where $ q$ is a prime.
the problem is equivalent to resolve the equation $ n^q-1=y_2$(*). obviously, if $ y$ is odd then $ y^2+1=2 \bmod(8) which is impossible for a$q$ th power with$q\geq 2.
So $ y$ is even and $ n$ is odd. On the other hand (*) dosen't have solution for $ q=2$ (only $ (0,1)$), from now let $ q$ an odd prime.
we  have $ (y+i)(y-i)=n^q$, since n is odd, then $ y+i$ and $ y-i$ are coprime in $ Q[i]$, so $ y+i= i^k\alpha^q$ where $ 0\geq k \leq3$.But since $ p$ and $ 4$ are coprime and by BEZOUT, we can write that
$ uq+4v=1$ so $ i^k= (i^u)^q$ and $ i$ is a $ p$th power. we obtain that $ y+i= \beta^q$.
Let $ \beta= a+ib$. We have $ \beta^p= N+iM$ with
                \[ M=\sum_{k=0)^{\frac{q-1}{2}}\frac{(q)!}{(2k)!(q-2k)!}a^{2k}b^{q-2k}(-1)^{\frac{q-1}{2}-k}}\]
We have $ b%Error. "midM" is a bad command.
$ and $ M=1$ so $ b= \pn 1$. we have
                \[ \sum_{k=0)^{\frac{q-1}{2}}\frac{(q)!}{(2k)!(q-2k)!}a^{2k}(-1)^{\frac{q-1}{2}-k}=\pn 1}\]
since $ q\mid \frac{(q)!}{(2k)!(q-2k)!}$ for $ 0\geq k \leq \frac{q-1}{2}$. we have
                \[ D=\sum_{k=1)^{\frac{q-1}{2}}\frac{(q)!}{(2k)!(q-2k)!}a^{2k}(-1)^{k}=0}\]
it easy to see that $ a$ is even.
Let $ \displaystyle n_k=\frac{(q)!}{(2k)!(q-2k)!}a^{2k}=\frac{q(q-1)}{2k(2k-1)}\frac{(q-2)!}{(2k-2)!(q-2)!}a^{2k}$.
we have $ \displaystyle \frac{n_k}{n_1}=\frac{1}{2k-1}\frac{(q-2)!}{(2k-2)!(q-2)!}a^{2k-2}$. then for $ k>1$ ($ q>3$), we have
                \[ v_{2}(n_k)-v_{2}(n_1)\geq (2k-2)v_{2}(a)-v_{2}(k)\geq (2k-2))-v_{2}(k)geq 1\]
So $ v_{2}(n_k)\geq v_{2}(n_1)$ for $ k>1$. Since $ D=\sum_{k=1)^{\frac{q-1}{2}}(-1)^{k}n_k=0}$ then $ v_{2}(D)=v_{2}(n_1)=v_{2}(frac{p(p-1)}{2}a^2)=0$ which is imposiible for $ a$ not null.
So $ a=0$ hence $ \beta=\pn i$, $ y=0$ and $ n=1$.

i don't know if it's true for p odd.

let $ \sqrt {p^5 \minus{} 1} \equal{} q$, then, $ p^5 \minus{} q^2 \equal{} 1$. by mihailescu, catalan, or mihailescu-catalan (don't know how to state it) we know that the only solution to $ a^x \minus{} b^y \equal{} 1$, with $ a,b,x,y > 1$ is $ (a,b,x,y) \equal{} (3,2,2,3)$. so there's no other solution than $ (p,q) \equal{} (1,0)$.


calculus
integration
real analysis
real analysis theorems

let $f: [a,b]\rightarrow R$ be continuous except at an interior point c. also assume that f is bounded. show that f is integrable . could any of you give me some hints? only hints please

Take a partition that isolates the bad point in a very small interval; use the boundedness to control that bad interval and the fact that continuity implies integability for the rest.

This is easy, just split the interval $[a,b]$ into $[a,c-\epsilon),(c-\epsilon,c+\epsilon)$ and $[c+\epsilon,b]$. $f$ is continous on the first and third of these intervals, hence integrable, and letting $\epsilon$ be sufficiently small, the boundedness of $f$ ensures that the contribution from the second integral is small, and  hence $[a,c-\epsilon] \cup [c+\epsilon,b]$ constitutes a partition of $[a,b]$ where $f$ is integrable..

edit: too sloow..