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Hi,

I'm looking for a textbook for my 10yo son to use for math next year (6th grade), independently with a tutor once a week for support.  (He attends a very small private school, and there's no one else who picks things up as quickly as he does, so he will be working independently.)

This year he has covered much of what I think of as "pre-algebra" using a combination of 7th and 8th grade texts.  For a number of reasons, we don't want him to get much beyond "Algebra I" before he starts high school, so he has 3 years to fill any pre-algebra gaps, do Algebra I and spend time on problem solving and other enrichment.  (He is in agreement with this plan.)

I'm wondering (a) if he's ready for the "Introduction to Algebra" book, and (b) if it sounds like a good fit for him to work with, taking his time and perhaps more than a year to go through it, and supplementing with problem solving practice from other sources (perhaps even AoPS volume I?)

I was thinking that maybe the "Are you ready" test from the Intro Algebra class would be a good indicator of whether he is ready or not, but I couldn't find the test since the class isn't being offered right now.  So... would this be a good indicator, and if so, is the test available somewhere that we could access it?

If he's not ready, can someone recommend a good book to use before it?

Thanks in advance!

[quote="funcia"]I've seen a 6th grader(age wise) use the [url=http://videotext.com/homeschool.htm]video text algebra program[/url] though it is mostly the basic algebra concepts. If your son is ready I'd think sticking with the AoPS algebra book would be the best choice. I'm pretty sure if he has a tutor working with him  and has done well in the pre-algebra that doing it wouldn't be that hard. I think that it sounds like a "good fit" for him when he is able to do it, but it really depends on what your definition of Algebra 1 is.[/quote]

Thanks for your reply.  I am leaning toward sticking with the AoPS book.  The videos look like a more basic coverage, and I'd rather have him do a "deep" coverage such as the AoPS book will offer.  I just need to figure out if he is ready, or if there are pre-algebra topics he has to solidify first.

Unfortunately, we have not developed on online Intro Algebra course yet, so there is no pre-test available for that course.

If you son has seen most of pre-algebra, then he should be ready for our Intro Algebra book.  It does not presume any prior algebra experience, and it starts at the very beginning of the algebra curriculum.

We have put the [url=http://www.artofproblemsolving.com/Books/IntroAlgebra/toc.pdf]table of contents[/url] on the website so that you can see specifically what is in the book.

[quote="DPatrick"]Unfortunately, we have not developed on online Intro Algebra course yet, so there is no pre-test available for that course.[/quote]

That would explain why I can't find it! ;-)

I think what we'll do in that case is have him try the placement tests from Singapore math 6a and 6b.  If those reveal pre-algebra topics that he stumbles on, we'll have him firm those up, probably using the Singapore books, and then he should be well set to move onto the AoPS Algebra book.  If he seems fine on the Singapore placements, we'll go straight to the Algebra and fix any gaps we find along the way when we find them.

I would suggest that you consider, at some point, developing pre- and post-tests for your textbooks like you have for the classes.  I know I'm not the first person here to wonder about the levels of the books, though I also appreciate that getting the remaining books published is probably a higher priority.

Thanks for your help.

I'll echo DPatrick's comments: if he has worked through pre-algebra once already, he should be ready for our Algebra book.  The first few chapters will revisit variable manipulation and solving linear equations (which I'm guessing he has already done in pre-algebra), then move on to more challenging material.

After he has studied some of the Introduction to Algebra book, we would recommend he study the Introduction to Counting & Probability and Introduction to Number Theory books, as well.  This will help broaden his understanding of math, expose him to new subject areas, and prevent him from falling into the [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php]Calculus Trap[/url].

[quote="rrusczyk"]I'll echo DPatrick's comments: if he has worked through pre-algebra once already, he should be ready for our Algebra book.  The first few chapters will revisit variable manipulation and solving linear equations (which I'm guessing he has already done in pre-algebra), then move on to more challenging material.[/quote]

OK, this does sound like it will be appropriate.  It's hard for me to know if he's done "pre-algebra" or not because he attends a small private school and they don't really label their math that way.  He's working with a group of 6th to 8th graders on something that resembles pre-algebra, but I'm not really sure what is "supposed to" be in pre-algebra, and they're not following any specific text, though they use several.  They have done solving linear equations, and are doing simple factoring right now (like, factoring out a constant or an x, but not, I think, factoring a quadratic to find its roots).

[quote]
After he has studied some of the Introduction to Algebra book, we would recommend he study the Introduction to Counting & Probability and Introduction to Number Theory books, as well.  This will help broaden his understanding of math, expose him to new subject areas, and prevent him from falling into the [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php]Calculus Trap[/url].[/quote]

Those are great suggestions.  Your "Calculus Trap" article is one of the reasons I don't want him to go beyond Algebra I in middle school, but rather stick with the usual honors stream for math in our district.  He may want to do more geometry for Mathcounts though -- I guess we'll see how that goes!

Oh, one more question -- do any of your books go into how to use a graphing calculator to solve the problems, or is that something he'll have to learn another way?

Thanks for all your help!

That's a good idea.  The [url=http://www.singaporemath.com/]Singapore Math textbooks[/url] provide an excellent foundation in pre-algebra.  You might consider the 5B as well.  I thought that it was the most challenging of the series.    The main advantages of these texts for advanced pre-algebra students include:

1.  difficult problems that can be approached with both algebraic and non-algebraic techniques.  This makes for a relatively seamless transition to using symbols in algebra.

2.  thin, easy to use, and cheap texts (< 10 dollars for each book)

If you can work the tougher problems in the 5B, 6A, and 6B Singapore books, the AopS Intro Algebra book and other extracurricular math books (Counting/Probability, Number Theory) should be very accessible.

The biggest "hole" most people have in their pre-algebra preparation is lack of experience with "difficult" multi-step problems.  When these folks enter an algebra course they get hit with

1) much harder problems than they've ever seen,
2) completely foreign symbols
3) higher levels of abstraction

Even after they've taken a few courses beyond Algebra I, most people just don't get basic algebra. I think that a big part of the problem is in the pre-algebra preparation - kids need to be comfortable working on much more difficult problems than are available in typical elementary texts if they are to really understand Algebra.

[quote="gt59"]That's a good idea.  The [url=http://www.singaporemath.com/]Singapore Math textbooks[/url] provide an excellent foundation in pre-algebra.  You might consider the 5B as well.  I thought that it was the most challenging of the series.[/quote]
Thanks for that pointer.  I'll check that one out as well.  What combination of text/workbook/intensive practice/challenging word problems do you recommend?  My youngest is doing parts of the 2A workbook and intensive practice (just for fun -- he's in kindergarten, though also accelerated in math in school, so he is not dying of boredom) but I've been wondering whether the challenging word problems would have been a better choice than the intensive practice.  For what it's worth, if the workbook and text are at about the same level, he'd prefer the workbook.  Handwriting is an issue for him, and the workbook would require less copying than the text.

[quote]The biggest "hole" most people have in their pre-algebra preparation is lack of experience with "difficult" multi-step problems.  When these folks enter an algebra course they get hit with

1) much harder problems than they've ever seen,
2) completely foreign symbols
3) higher levels of abstraction
[/quote]
I work (as a volunteer) with my son's math group for 2 hours every 2 weeks on Mathcounts/Math Olympiad (MOEMS, not IMO!) type problem solving.   So he does have some experience with challenging problems already, and is quite skilled at that kind of thinking.

[quote]Even after they've taken a few courses beyond Algebra I, most people just don't get basic algebra. I think that a big part of the problem is in the pre-algebra preparation - kids need to be comfortable working on much more difficult problems than are available in typical elementary texts if they are to really understand Algebra.[/quote]

I once read somewhere that the best predictor of success in high school math was a thorough understanding of decimals, fractions, percents and ratios.  It does seem to take a long time for many kids to really "get" those concepts, which may be some of what you're observing.  Experience with hard problems as well is of course another important factor.

Thanks for taking the time to respond to my post!

[quote="mathmom"]
Oh, one more question -- do any of your books go into how to use a graphing calculator to solve the problems, or is that something he'll have to learn another way?

Thanks for all your help![/quote]

Don't give him a graphing calculator until he can prove the quadratic formula.

I strongly agree with your comment about the key to success in high school being an understanding of fractions, decimals, ratios, etc.  It's the introduction of calculators too early that has resulted in a generation of math students taking a big step back in math education.  Nearly every student I've worked with that had trouble in high school can trace their troubles back to fractions.  Almost all of them had calculators in their elementary school classes.

For a student with a good foundation in mathematics, a graphing calculator is a tool that is easy to use.  For a student with a lot of training with their graphing calculator, mathematics is often a tool that is impossible to use.  Don't worry about training your son with a calculator.  When the time comes that he needs one, it's use will be quite clear.  If your concern is about *programming* a calculator, skip the calculator and go straight to programming a computer.

From the math background you describe, your son is ready for the Introduction to Algebra book.  After he has finished the first 9 chapters of that book, he'll be  ready for the Intro Counting and Intro Number Theory books as well.  If he's very eager about math, it's likely he'll get into the Intro Geometry book before high school, which is fine.

We've heard of a lot of people using the Singapore books prior to ours with success.

[quote="rrusczyk"]Don't give him a graphing calculator until he can prove the quadratic formula.

I strongly agree with your comment about the key to success in high school being an understanding of fractions, decimals, ratios, etc.  It's the introduction of calculators too early that has resulted in a generation of math students taking a big step back in math education.  Nearly every student I've worked with that had trouble in high school can trace their troubles back to fractions.  Almost all of them had calculators in their elementary school classes.[/quote]

You're preaching to the choir here.  I believe I've posted on this site before about an experience I had while working with the middle schoolers at our kids' school.  *Several* of them tried to convert 8/100 to a percent using a calculator, and got it WRONG!  :o That was when I went on the warpath against the calculators.  I convinced the math teacher to use them much less often, including encouraging her to design problems where they should not be needed (by picking convenient numbers).  We do pull out the calculators when the numbers get ugly (say, on a Mathcounts Work-Out, which are often clearly designed to be done with calculators -- no doubt the influence of the engineers running the program).  The elementary kids use them much less now as well.  I do have to allow them occasionally, because i have a mixed group of 7- to 10-year-olds working on the 4th grade level of the [url=http://www.curriculumassociates.com/products/detail.asp?topic=T0M&title=FigureItOut]Figure It Out[/url] problem solving series, and some of the younger ones haven't learned to multiply or divide (multi-digit numbers) yet!  So for them, recognizing that they need to multiply and doing it with a calculator is reasonable.  I check with their regular math teacher regularly to make sure I'm not letting anyone use a calculator who should be practicing doing the computation by hand.

[quote] Don't worry about training your son with a calculator.  When the time comes that he needs one, it's use will be quite clear.  If your concern is about *programming* a calculator, skip the calculator and go straight to programming a computer.[/quote]

My concern is that the public school's Honors Algebra I curriculum does explicitly teach the kids how to use a graphing calculator, and to place into Honors Geometry, he'll eventually have to take the final exam from that class.  We've got lots of time though, and I'm in no hurry to get him one.  We'll know more details about this soon as my older son will be taking the exam next month.  He's been following the public school's textbook, and it explicitly teaches the use of the calculator for various things.

[quote]
From the math background you describe, your son is ready for the Introduction to Algebra book.  After he has finished the first 9 chapters of that book, he'll be  ready for the Intro Counting and Intro Number Theory books as well.  If he's very eager about math, it's likely he'll get into the Intro Geometry book before high school, which is fine.

We've heard of a lot of people using the Singapore books prior to ours with success.[/quote]

Thanks for your detailed suggestions.  We have AoPS volume I at home too, so if you have suggestions as to where that might fit in with the others, I'd be grateful to hear about them. [/url]

[quote="mathmom"]
My concern is that the public school's Honors Algebra I curriculum does explicitly teach the kids how to use a graphing calculator, and to place into Honors Geometry, he'll eventually have to take the final exam from that class.  We've got lots of time though, and I'm in no hurry to get him one.  We'll know more details about this soon as my older son will be taking the exam next month.  He's been following the public school's textbook, and it explicitly teaches the use of the calculator for various things.  [/quote]

Don't worry about it - if he masters the algebra in our textbook, it will be obvious how to use the calculator where needed.  If you can see a copy of an old test, you'll be able to see if there's anything on there that is strictly button pushing that he doesn't know how to do.

As for AoPS1, I would recommend holding off on that until he's worked through some of the other books, and I would largely use that for contest prep (if he gets interested in that) and extra challenging problems in areas where he needs or wants them.

[quote]What combination of text/workbook/intensive practice/challenging word problems do you recommend? [/quote]

My kids just worked Singapore textbook problems with emphasis on the challenging word problems.  They were supposed to work enough of the exercises to get the concept then all of the word problems.  If they couldn't get the word problems then they worked more exercises until they could solve the problems.  When they could work the word problems then it was time to move to the next section.  It's been a couple of years, but I don't remember making use of the workbooks.  I just had them write the answers (preferably "solutions") in a notebook but they didn't copy the problems.

I found it a bit difficult to explain how to solve some of the tougher Singapore problems without algebra until I read through a bit of the book and thought about the problems.  I'm fairly sure that I was never exposed to problems at that level prior to learning algebra.

Richard's right about the calculator - don't get it until it's absolutely necessary.  One of my kid's has become a calculator programming fiend.  Add "time-wasting" to it's other faults.  Educators are slowly coming to the realization that calculators are part of the problem, not part of the solution.

Good luck trying to avoid the "Calculus Trap."  It might be harder to avoid than you might imagine if your kids are really interested and good at math.  I agree with Richard that you should provide extracurricular opportunites before acceleration.  Most kids that I see accelerating in the local schools without taking on extracurricular math are making a mistake.
However, some of the brightest kids can exhaust available extracurriculars (AoPS courses....) within a couple of years.  After taking on the extracurriculars, they may be  able to absorb material at a much, much faster rate than others in typical high school math classes.  There may not be much point in holding them back if you get that far.

Many students on this forum have extensive knowledge of extracurricular math and also take Calculus early in their high school careers.  For them, relatively early exposure to Calculus may be the best of the math choices available in high school.  The Algebra II and Pre-Calculus courses offered are most high schools are about 50% "review."  If your kids can complete the AoPS Intermediate Algebra and Intermediate Trig/Complex numbers classes then there are not many reasons to avoid Calculus.

[quote="gt59"]Good luck trying to avoid the "Calculus Trap."  It might be harder to avoid than you might imagine if your kids are really interested and good at math.  I agree with Richard that you should provide extracurricular opportunites before acceleration.  Most kids that I see accelerating in the local schools without taking on extracurricular math are making a mistake.
However, some of the brightest kids can exhaust available extracurriculars (AoPS courses....) within a couple of years.  After taking on the extracurriculars, they may be  able to absorb material at a much, much faster rate than others in typical high school math classes.  There may not be much point in holding them back if you get that far.[/quote]

My boys all currently seem to be very good at math, but not really motivated to work at it outside of school hours.  So, I don't seem them running out of stuff to do too quicly.  Of course if that changes, and one or more of them becomes particularly driven, then all bets are off.  I'll still try to help them find enrichment opportunities where possible to supplement and improve their experience.

Thanks for your additional feedback on the Singapore books.

[quote="gt59"]
However, some of the brightest kids can exhaust available extracurriculars (AoPS courses....) within a couple of years.  After taking on the extracurriculars, they may be  able to absorb material at a much, much faster rate than others in typical high school math classes.  There may not be much point in holding them back if you get that far.

Many students on this forum have extensive knowledge of extracurricular math and also take Calculus early in their high school careers.  For them, relatively early exposure to Calculus may be the best of the math choices available in high school.  The Algebra II and Pre-Calculus courses offered are most high schools are about 50% "review."  If your kids can complete the AoPS Intermediate Algebra and Intermediate Trig/Complex numbers classes then there are not many reasons to avoid Calculus.[/quote]

I pretty strongly agree with this.  Once a student has a good, broad foundation in problem solving math, then acceleration is fine, and, in most cases, desirable.  That said, I'd try to get this acceleration from a suitably challenging source.  In many schools, I'm not too confident that AP Calculus fits that bill.

That said, even once a student accelerates, I would encourage the student to do more than simply plow through the standard 'calculus, multi-variable calculus, differential equations' track.  Take a modern algebra course.  Take a discrete math class.  See something besides just more and more calculus.  There's a big world of higher math out there.

[quote="rrusczyk"]Once a student has a good, broad foundation in problem solving math, then acceleration is fine, and, in most cases, desirable.  That said, I'd try to get this acceleration from a suitably challenging source.  In many schools, I'm not too confident that AP Calculus fits that bill.  [/quote]
How can you tell if a given school's AP calculus is "good" or not?  I think our HS only teaches the AB course, but they co-teach it with AP physics, which strikes me as a pretty good idea.  Beyond AP Calc AB, they let the kids co-enroll in local colleges, which offer many of the more "interesting" courses you suggest below in addition to the calculus track.

[quote]That said, even once a student accelerates, I would encourage the student to do more than simply plow through the standard 'calculus, multi-variable calculus, differential equations' track.  Take a modern algebra course.  Take a discrete math class.  See something besides just more and more calculus.  There's a big world of higher math out there.[/quote]
Good advice as always.  Calculus is nice if you want to be a physicist or an engineer.  But the other stuff is much more fun. IMO. :)

A first cut on a school's AP course is simply the test scores.  Any good treatment of calculus should result in lots of 4s and 5s on the AP test and very few 1s and 2s.  This said, test scores alone don't mean the program is great - it only means it isn't awful (of course, a class with virtually all 5's is probably a good class).

If the students' results in the course are very scattered, or many opt not to take the test, then any student in that class who is serious about math should spend a lot of time studying the material on her/his own.

If you have access to recent graduates of the high school, another indicator is finding out how they fared in their early college math classes.

Bundling calculus with physics is a fine idea - I've often thought that if I had all the time in the world, I'd try to write a book combining them.  Alas, I don't have all the time in the world.

The fact that your school has a teacher that can integrate these two is probably a good sign.

[quote="rrusczyk"]I strongly agree with your comment about the key to success in high school being an understanding of fractions, decimals, ratios, etc.  It's the introduction of calculators too early that has resulted in a generation of math students taking a big step back in math education.  Nearly every student I've worked with that had trouble in high school can trace their troubles back to fractions.  Almost all of them had calculators in their elementary school classes.[/quote]
This definitely strikes a chord with me.  I remember that in elementary school we had exercises where you were given several math exercises and asked whether you would solve them with mental math, paper and pencil, or a calculator.  I got many of the questions "wrong" because I said that I would solve almost all of the exercises as mental math.  It's so clearly an opinion question that I don't see how the teachers could grade something like that as "right" or "wrong", especially since I only selected mental math because I had performed the calculations that way (I also often got points taken off because I would subtract multiple quantities from the same number rather than doing them one step at a time).

Mathmom, I think your idea of having your son work through the various AoPS books you've proposed during middle school (rather than rushing ahead headlong towards the calculus trap) is a very good idea.

However, I noticed that you left the AoPS Geometry book off your list.

I would also suggest that the AoPS Geometry book should be a high priority during middle school.

Geometry is often the weakest part of the math curriculum in most American schools, and there's a lot to be said for gettting a solid foundation in geometry [i]before [/i]high school.  If his high school geometry class turns out to be weak (as many are), he may not have time or interest to supplement the curriculum with self-study of geometry fundamentals at that time, because of so many other competing demands on time during high school.

On the other hand, if he has developed a strong geometry foundation (through the use of the AoPS Geometry book) in middle school, you needn't worry that he will run out of geometry to study on his own in high school.  There is still more geometry available to pursue in high school (if he is so inclined--and hopefully his early exposure to AoPS Geometry will help him develop a taste for challenging geometry)--he could follow up at that time with Coxeter's [i]Geometry Revisited[/i].

[quote="sophia"]I would also suggest that the AoPS Geometry book should be a high priority during middle school.

Geometry is often the weakest part of the math curriculum in most American schools, and there's a lot to be said for gettting a solid foundation in geometry [i]before [/i]high school.  If his high school geometry class turns out to be weak (as many are), he may not have time or interest to supplement the curriculum with self-study of geometry fundamentals at that time, because of so many other competing demands on time during high school.[/quote]

Thanks, Sophia, this is a good point.  My oldest is starting high school this fall, so we'll soon have a good idea soon how good (or not) the Geometry class is (though it may in part depend on which teacher they each get).  My older son has had some limited geometry exposure while preparing for MATHCOUNTS, but nothing formal.

If the course is good, though, I'd like to stand by my plan to mostly avoid it during middle school.  He pretty much "has to" take geometry his first year in high school, and I don't want it to be totally boring/review for him.

[quote]On the other hand, if he has developed a strong geometry foundation (through the use of the AoPS Geometry book) in middle school, you needn't worry that he will run out of geometry to study on his own in high school.  There is still more geometry available to pursue in high school (if he is so inclined--and hopefully his early exposure to AoPS Geometry will help him develop a taste for challenging geometry)--he could follow up at that time with Coxeter's [i]Geometry Revisited[/i].[/quote]

But again, in high school, he probably won't take the time to study geometry on his own, but rather just take the geometry class and be bored, which is what I'm trying to avoid.

If it's a priority for [b]him [/b]to learn more geometry for MATHCOUNTS etc. while he's still in middle school, I won't stop him, and I'll try to work out some arrangement for him to do some guided independent geometry study instead of the usual geometry class when he gets to high school.  But I think pushing it when there's so much more out there that he'd never see in high school may be the wrong move, at least if the geometry class he will eventually get in HS is a good one.

But, it's definitely worth keeping in mind that if the high school class is not great, now may be the time to do it.

Thanks!

I think that there are other orders possible for teaching math, and that "rushing into calculus" is not such a terrible idea.

My son worked through the "Key to Algebra" series in 4th and 5th grade (not *my* choice of algebra books, but the school's), and learned a little calculus in 5th grade for his science fair project about bouncing balls (deriving how high the ball should bounce based on the velocity out of the bounce being r*velocity into the bounce).  He did this by seeing what happened in the discrete simulation he wrote, then recognizing the area under the curve as being the area of a triangle.  Most of his instruction in this came during long walks, in which we had nothing to write on or with, so we had to do everything mentally.  After he found out that what he was doing was calculus, he bought himself "Calculus for Dummies", which he has read through several times for fun.

This year in 6th grade he is working on the AoPS Intro to Geometry book, which is at just the right level for him.  I think that the school will also give him some chapters from one of the AoPS algebra books, since he wants to learn more about complex numbers, trigonometry, and logarithms.  (I introduced the math teachers at the school to the Singapore math books, which they have now adopted for most of the math classes, and to the Art of Problem Solving books, which they will use for those 6th graders that finish the Key to Algebra series.)

He recently read "the Pea and the Sun", which presents the Banach-Tarski paradox in a very readable way.

We also started him on set theory and mathematical logic, using "Mathematical Foundations of Linguistics" which provides a fairly rigorous treatment in a reasonably readable style.

So: calculus before trigonometry.  Banach-Tarski paradox before geometry.  *Not* an order that any rational person would pick.  But it seems to be working for him.

[quote="engineering_prof"](I introduced the math teachers at the school to the Singapore math books, which they have now adopted for most of the math classes, and to the Art of Problem Solving books, which they will use for those 6th graders that finish the Key to Algebra series.)[/quote]
Congratulations on convincing the school to use the excellent Singapore and AoPS books.  How did you manage to convince the school?  In my experience, that's not so easy.

[quote="Ravi B"]
Congratulations on convincing the school to use the excellent Singapore and AoPS books.  How did you manage to convince the school?  In my experience, that's not so easy.[/quote]

The small private school for gifted kids evaluates some aspect of their curriculum each year.  I had complained to the principal about the math books being used, and she asked me for advice on texts.  I brought in the Singapore math books that my son had been working on in previous years and she loaned them out to the teachers to evaluate.  The teachers liked them, so the series was chosen for their main math classes.  The school is a K-6 school, so this covered most of their needs, but they also have a 7th and 8th grade math group doing pre-algebra stuff.  (The 8th-grade group also does some contest prep work.)  For the 7th and 8th grade groups, I shared what I was finding on the net with the instructors.  They tried various things.

(My son was put in the 7th grade math group when he was in 4th grade, so he is having to do the 8th grade math group over this year in 6th.  He does the contest prep stuff with the other kids, then works in the AoPS geometry book on his own. I'm not sure what book the other kids are working from, but I don't think it is an AoPS text.)

The school starts algebra in 4th grade with the "Key to Algebra" series, and a few of the kids finish the series before the end of 6th grade.  They are planning to use the AoPS algebra book for continuing the algebra for those who finish Key to Algebra.  Personally, I'd like to see them switch over to the AoPS book from the beginning, since the Key to Algebra series is rather boring.  The algebra class uses very little instruction---the kids work through the workbooks at their own paces, with rather variable results in how much they retain.

Note: "algebra" and "math" are two separate subjects at this school.  K-6 all do math at the same time, so that kids can be placed by ability, not grade level.  Only 4-6 do algebra and they are self-paced so there is no need to synchronize class schedules.
I think that "math" is 3 hours a week and "algebra" is 2, but algebra may only be 1 hour a week in some grades.


function
analytic geometry
graphing lines
slope
real analysis
Functional Analysis
real analysis unsolved

Let $ E_n$ be the set of all $ f$ in  $ C([0,1])$ for which there exists $ x_0\in [0,1]$ (depending on $ f$) such that
$ |f(x)\minus{}f(x_0)|\leq n|x\minus{}x_0|$ for all $ x\in [0,1]$

a. Prove that $ E_n$  is nowhere dense in $ C([0,1])$(hint: any real $ f$ in $ C([0,1])$ can be uniformly approximated
 by a piecewise linear function whose linear pieces , finite in number have slope +2n and -2n. )

b.The set of nowhere differentiable functions is residual in $ C([0,1])$

It is a classic result of Banach, Mazurkiewicz. It may be found on many advanced books of real analysis or functional analysis (Oxtoby, Yosida, Boas, Hewitt-Stromberg, Kuratowski)


From $a_{k+1}-a_{k}=\frac{1}{a_k}$, squaring of both sides and adding both of sides  for from $k=1$ to $k=999$,yielding to ...... :D

kunny


That was a huge hint. So basically the answer would be [hide]$\frac{2007^{3}-1337^{3}-670^{3}}{2007 \cdot 1337 \cdot 670}= \frac{3 \cdot 2007 \cdot (-1337) \cdot (-670)}{2007 \cdot 1337 \cdot 670}= 3$[/hide]

Is that all :| ?

I remember this problem! [hide]Its always 3 when you have it like this.[/hide]

[quote="kunny"][hide="hint"]$a+b+c=0\Longrightarrow a^{3}+b^{3}+c^{3}=3abc.$[/hide][/quote]
[hide]
well now we just divide the abc, and it'll always be 3...[/hide]

Even if you didn't know about that nice fact, it's not so bad :)

[hide="The key is..."] To use variables.  Let $a = 1337, b = 670$.  Then

$\frac{(a+b)^{3}-a^{3}-b^{3}}{ab(a+b)}= \frac{3ab(a+b)}{ab(a+b)}= 3$

By a simple binomial expansion - nothing complicated.  Kunny's remark can be proven this way, but it is also a consequence of the identity

$a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$ [/hide]

How would you come up with ways to factor equations like this?

The fundamental observation is that $2007 = 1337+670$.  After that, I think my logic is the most "natural" way to go (without fancy identities).  Often for strange computations of this type replacing the numbers with variables helps the algebraic relationships become more apparent.

I like how 1337 was the random numer picked  :P

Yes, like an example of the sum of the terms in an infinite geometric series on Wikipedia.

They use $191$ as the first term and $\frac{6}{7}$ as the common ratio.


Find $ a;b \in Z;p \in P$ such that $ pa^5\equal{}b^4\plus{}3$

Because $ b^4\plus{}3\equal{}3,4\mod 16$ we get a is odd, b is even. If $ 3|b$ then p=3 and equation had not solution.
Therefore $ p\equal{}\frac{16x^4\plus{}3}{a^5}, \ b\equal{}2x, \ 3\not |b.$
I think, that these equation hadinfinetely many solution. For example $ x\equal{}1\to (a,b,p)\equal{}(1,2,19)$, $ x\equal{}4, (a,b,p)\equal{}(1,8,4099),...$


A circle is inscriptible in the isosceles trapezoid $ ABCD$.Let the circle meet diagonal $ AC$ at $ K$ and $ L$ (with $ K$ between $ A$ and $ L$.Find the value of $ \frac{AL*KC}{AK*LC}$.

[hide="Solution"]WLOG, suppose that $ AB\parallel CD$. Let E,F,G,H be the tangency points of the incircle with AB,BC,CD,DA respectively. It’s well know to see that AC, BD, EG, FH meet at the midpoint M of HF. We easily get $ \frac {AL}{AK} \equal{} {{\left( \frac {EL}{EK} \right)}^{2}}$, $ \frac {CK}{CL} \equal{} {{\left( \frac {FK}{FL} \right)}^{2}}$$ \Rightarrow \frac {AL\cdot CK}{AK\cdot CL} \equal{} {{\left( \frac {EL\cdot FK}{EK\cdot FL} \right)}^{2}} \equal{} {{x}^{2}}$ ($ x > 0$). We find $ x$.
$ EF\cdot KL \equal{} EL\cdot KF \minus{} EK\cdot FL\Rightarrow \frac {EF\cdot KL}{EK\cdot FL} \equal{} x \minus{} 1\,\,\,\,\,\,\,(1)$,
$ KE\cdot HL \equal{} HK\cdot EL\Rightarrow \frac {EH\cdot KL}{EK\cdot HL} \equal{} 2\,\,\,\,\,\,(2)$
Compare (1) and (2) we get $ \frac {HL}{FL} \equal{} \frac {x \minus{} 1}{2}$ (note that $ EF \equal{} EH$).
In other hand, $ \frac {HL}{KF} \equal{} \frac {MH}{MK} \equal{} \frac {MF}{MK} \equal{} \frac {FL}{HK}$
$ \Rightarrow \frac {HL}{FL} \equal{} \frac {KF}{HK} \equal{} \sqrt {\frac {HL\cdot KF}{FL\cdot HK}} \equal{} \sqrt {\frac {EL}{EK}\cdot \frac {FK}{FL}} \equal{} \sqrt {x}$
Solve $ \frac {x \minus{} 1}{2} \equal{} \sqrt {x}$, we get $ x \equal{} {{(\sqrt {2} \plus{} 1)}^{2}}$.
Answer: $ \frac {AL\cdot CK}{AK\cdot CL} \equal{} {{(\sqrt {2} \plus{} 1)}^{4}}$  :D  :D   :D [/hide]


Let p be an odd prime and gcd(a,p) = gcd(b,p) = 1. Prove that either all three of the quadratic congruences x^2= a(mod p), x^2= b(mod p), and x^2= ab(mod p) are solvable or exactly one of them admits a solution.

Consider the [url=http://en.wikipedia.org/wiki/Legendre_symbol]Legendre symbols[/url] $ \left(\frac{a}{p}\right)$ and $ \left(\frac{b}{p}\right)$; the symbol is completely multiplicative so $ \left(\frac{ab}{p}\right)\equal{}\left(\frac{a}{p}\right)\cdot\left(\frac{b}{p}\right)$. We want to prove that the number of Legendre symbols taking on value $ \plus{}1$ is odd. If it is not, however, then one side of the previous equality will be $ \plus{}1$ while the other side will be $ \minus{}1$... contradiction. We can have e.g. $ \left(\frac{a}{p}\right)\equal{}1$ and $ \left(\frac{b}{p}\right)\equal{}1$ so $ \left(\frac{ab}{p}\right)\equal{}1$, etc.

Why is the Legendre symbol multiplicative, you ask? This property follows from [url=http://en.wikipedia.org/wiki/Euler's_criterion]Euler's Criterion[/url], which states that $ \left(\frac{a}{p}\right) \equiv a^{(p\minus{}1)/2} \pmod{p}$. This is proven essentially by writing $ a\equal{}g^k$ where $ g$ is a primitive root $ \pmod{p}$ and $ k$ is even iff $ k$ is a quadradic residue $ \pmod{p}$ and using Fermat's little theorem.

I understand how you showed it was multipicative, but does that mean the equations are solvable?

Ah well, the Legendre symbol's definition is (if you didn't go to the link!):

$ \left(\frac{a}{p}\right) \equal{} \begin{cases}
0 & a \equiv 0 \pmod{p} \\
\plus{}1 & a \not\equiv 0 \pmod p \mbox{ and } \exists x \mbox{ s.t. } x^2 \equiv a \pmod{p} \\
\minus{}1 & a \not\equiv 0 \pmod p \mbox{ and } \nexists x \mbox{ s.t. } x^2 \equiv a \pmod{p}
\end{cases}$

By the given condition, the first case is not true.

If you think about it, Euler's Criterion is pretty interesting. Quadratic reciprocity is nice as well...  :lol:


There are 9 unit squares, 4 2 by 2 squares, and 1 3 by 3 square.

$ 9\plus{}4\plus{}1\equal{}\boxed{14}$ squares total.


inequalities
geometry
parallelogram
rectangle
complex numbers
geometry unsolved

Given a triangle ABC,BC=a,CA=b,AB=c.P is an arbitary point in space.Prove the ineq:
$ \frac{PA.PB}{ab}\plus{}\frac{PB.PC}{bc}\plus{}\frac{PC.PA}{ca}$≥$ 1$
I need a elementary solution for this problem(without using complex plane).Thanks.

[quote]The most easy way to do it is to work in the complex plane, letting M be the origin and assigning to the points A, B, C the affixes u, v, w. Then,

$ \frac {MB\cdot MC}{bc} \equal{} \frac {MB\cdot MC}{CA\cdot AB} \equal{} \frac {\left|v \minus{} 0\right|\cdot \left|w \minus{} 0\right|}{\left|w \minus{} u\right|\cdot\left|u \minus{} v\right|} \equal{} \frac {\left|v\right|\cdot \left|w\right|}{\left|w \minus{} u\right|\cdot\left|v \minus{} u\right|} \equal{} \left|\frac {vw}{\left(w \minus{} u\right)\left(v \minus{} u\right)}\right|$.

Similarly,

$ \frac {MC\cdot MA}{ca} \equal{} \left|\frac {wu}{\left(u \minus{} v\right)\left(w \minus{} v\right)}\right|$;
$ \frac {MA\cdot MB}{ab} \equal{} \left|\frac {uv}{\left(v \minus{} w\right)\left(u \minus{} w\right)}\right|$.

Thus, by the triangle inequality,

$ \frac {MB\cdot MC}{bc} \plus{} \frac {MC\cdot MA}{ca} \plus{} \frac {MA\cdot MB}{ab}$
$ \equal{} \left|\frac {vw}{\left(w \minus{} u\right)\left(v \minus{} u\right)}\right| \plus{} \left|\frac {wu}{\left(u \minus{} v\right)\left(w \minus{} v\right)}\right| \plus{} \left|\frac {uv}{\left(v \minus{} w\right)\left(u \minus{} w\right)}\right|$
$ \geq \left|\frac {vw}{\left(w \minus{} u\right)\left(v \minus{} u\right)} \plus{} \frac {wu}{\left(u \minus{} v\right)\left(w \minus{} v\right)} \plus{} \frac {uv}{\left(v \minus{} w\right)\left(u \minus{} w\right)}\right|$.

But since

$ \frac {vw}{\left(w \minus{} u\right)\left(v \minus{} u\right)} \plus{} \frac {wu}{\left(u \minus{} v\right)\left(w \minus{} v\right)} \plus{} \frac {uv}{\left(v \minus{} w\right)\left(u \minus{} w\right)} \equal{} 1$,

we thus get

$ \frac {MB\cdot MC}{bc} \plus{} \frac {MC\cdot MA}{ca} \plus{} \frac {MA\cdot MB}{ab} \geq \left|1\right| \equal{} 1$,[/quote]
I haven't learn complex plane yet,so I need a better way to solve this problem.The solution above is nice but I cannot use it in competitions(I am only in grade 10).If anyone have another solution, please help me.Thanks a lot.

Let $ M$ be the point in the plane of triangle $ ABC$. Then the following inequality holds:
\[ a|MB|\cdot |MC|\plus{}b|MC|\cdot |MA|\plus{}c|MA|\cdot |MB|\geq
abc.\]
Equality holds if and only if $ M$  is either orthocenter of acute-angled triangle or one of the vertices of triangle.

[b]Solution:[/b]

Let $ E$ and $ F$ be the points such that
$ BCME$ and $ BCAF$ are parallelograms. Then, $ EMAF$
is also parallelogram. So, we have $ |AF|\equal{}|ME|\equal{}a$, $ |EF|\equal{}|MA|$,
$ |EB|\equal{}|MC|$, $ |BF|\equal{}b$.

By applying Ptolomey inequality on $ AFEB$ and
$ AEBM$ we have:
\[ c|MA|\plus{}a|MC|\equal{}c|EF|\plus{}a|EB|\geq |BF|\cdot|AE| \equal{}b|AE|,\]
\[ |MB|\cdot|AE|\plus{}|MA|\cdot |MC|\equal{}|MB|\cdot|AE|\plus{}|MA|\cdot|EB|\geq
c|ME|\equal{}ac.\]
So, we have
\[ a|MB|\cdot |MC|\plus{}b|MC|\cdot |MA|\plus{}c|MA|\cdot |MB| \equal{} |MB|\left( c|MA|\plus{}a|MC| \right)\plus{}b|MC|\cdot
|MA| \geq b|MB|\cdot|AE|\plus{}b|MC|\cdot |MA|\equal{}b\left(
|MB|\cdot|AE|\plus{}|MC|\cdot |MA| \right)\geq abc.\]
Equality holds if the quadrilaterials
$ AFEB$ and $ AEBM$ are cyclic. Since $ AFEB$ is cyclic
we have $ \angle EBA\plus{}\angle AFE\equal{}180^\circ$. Further, since
 $ AEBM$ is cyclic we have $ \angle EMA \equal{}\angle EBA$. So,
$ \angle EMA\plus{}\angle AFE\equal{}180^\circ$ and $ EMAF$ is cyclic. From the other side  $ EMAF$ is parallelogram, so it is rectangle, It follows that
$ MA \perp BC$. Further, we have
$ \angle EBA\equal{}90^\circ$, so  $ EB\perp AB$, and $ MC\perp AB$.
Hence, $ M$ is orthocenter and triangle is acute angled. Also, equality holds if $ M$ is one of the vertices of triangle..[hide=Index][/hide]

[quote="SUPERMAN2"]I haven't learn complex plane yet,so I need a better way to solve this problem.The solution above is nice but I cannot use it in competitions(I am only in grade 10).If anyone have another solution, please help me.Thanks a lot.[/quote]
Why can't you use that in competitions? Aren't you allowed?
How old are 10 graders in the US? :) I'm 16, and we are allowed to use anything in our solutions - including complex numbers, matrices, etc., which really aren't a part of the highschool curriculum(in denmark at least). :)


linear algebra
matrix
linear algebra unsolved

Is it true that an automorphism in a matrix group preserves trace?

What if the group is finite?

Similarity preserves trace. $ F$-algebra automorphisms preserve trace over $ F$. A random automorphism of a (multiplicative) group of matrices doesn't.
An example: consider the $ 1\times 1$ invertible matrices over $ \mathbb{C}$. Conjugation is an automorphism, and it doesn't preserve trace.


modular arithmetic
number theory proposed
number theory

Prove that there exist infinite numbers $h=2^\omega+1$ for a certain $\omega\in\mathbb N$ so that if

$2^{h+1}+1\equiv h\pmod{2h}$

then

$2^{h+1}+1\equiv h\pmod{2(h+1)}$

I think it is something wrong in [b]Logic[/b].I don't know how to spell some word .so I can't point out what it is.

I can't really find this logic mistake you are thinking of.. I mean, the condition that $h=2^\omega+1$ is redundant maybe. I could have asked just to show that there exist infinite $h\in \mathbb N$ that satisfy the problem...

By the way, maybe with an example it can be easier

If we choose $\omega=2\Rightarrow h=5$ we get $2^{6}+1\equiv 5\pmod{10}$ and $2^{6}+1\equiv 5\pmod{12}$

EDIT: I've thought about your problem tiger.. let's change the thesis into this one

prove that there exist infinite $h$ that satisfy the two congruences above at the same time.. tell me if you have other problems.. bye  :P

No,I don't have now.
I suggest that you edit the first the post ,and change the word "then" by the word "and".

Prove that there exist infinite numbers $h=2^\omega+1$ for a certain $\omega\in\mathbb N$ that verify

$2^{h+1}+1\equiv h\pmod{2h}$

and

$2^{h+1}+1\equiv h\pmod{2(h+1)}$

at the same time

I'm sorry but I'm not able to edit the first post... i don't know why.. :blush:


function
geometry
3D geometry
sphere
symmetry
inequalities

Let $ x;y;z$ be positive real numbers satisfying $ x^2\plus{}y^2\plus{}z^2\equal{}3$
Find min of $ A\equal{}\sqrt{x\plus{}y}\plus{}\sqrt{y\plus{}z}\plus{}\sqrt{z\plus{}x}$

The following solutions are both in the same spirit.

[hide="Level Surfaces"]Note that any arbitrary level surface $ A(x,y,z) = c$ is convex (curls away from the origin) near the point $ \left(\frac {c^2}{18}, \frac {c^2}{18}, \frac {c^2}{18} \right)$ because the squareroot is a concave function: $ \sqrt {a + b} + \sqrt {a - b}$ decreases with $ b$, so along a level surface, $ a$ must increase as $ b$ increases to prevent the sum from falling.

However, $ x^2 + y^2 + z^2 = 3$ is a sphere, and thus concave.

Also note that both equations are symmetric in $ x,y,z$, and that for higher $ c$, the surface $ A(x,y,z) = c$ becomes further from the origin.

We want the value $ c_{\mathbf{min}}$ such that the level surface $ A(x,y,z) = c_{\mathbf{max}}$ is tangent to the sphere. We know there will be only one point because the sphere is concave and $ A = c$ is convex. By symmetry, the tangent point will occur at $ x = y = z$. So, $ x = y = z = 1$, and $ \boxed{c_{\mathbf{max}} = 3 \sqrt {2}}$.[/hide]
[hide="Jensens"]
Since $ \sqrt {x}$ is concave in $ x$, we have that $ \frac {A}{3} = \frac {\sqrt {x + y} + \sqrt {y + z} + \sqrt {z + x}}{3} \leq \sqrt {\frac {(x + y) + (y + z) + (z + x)}{3}} = \sqrt {\frac {2}{3}(x + y + z)}$ ${ \implies A \leq \sqrt {6(x + y + z)}}$.

We now want to minimize $ x + y + z$, for which we will use $ RMS \geq AM$:

$ 1 = \sqrt {\frac {x^2 + y^2 + z^2}{3}} \geq \frac {x + y + z}{3}$ $ \implies x + y + z \leq 3$ $ \implies A \leq \sqrt {6(3)} = 3 \sqrt {2}$

Checking, we have equality at $ (x,y,z) = (1,1,1)$, so indeed, $ \boxed{A_{\mathbf{max}} = 3 \sqrt {2}}$
[/hide]

[color=red][u]Edit:[/u] Both of these actually compute the max value of A(x,y,z)[/color]

Thats wrong, the inequality is pointing in the wrong direction. You found the maximum value...

ex. $ x\equal{}y\equal{}0$; $ z\equal{}\sqrt{3}$ gives $ 2\sqrt[4]{3}<3\sqrt{2}$.

Ok yeah ... $ A \leq 3 \sqrt {2} \implies c_{\mathbf{min}} \equal{} 3 \sqrt {2}$... wow, what was I smoking

So then, I guess you can just reduce $ c$ and push the level surface towards the origin:

[hide="Level Surfaces take 2"]Since $ A \equal{} c$ and the sphere have opposite convexity, we know that the level $ c$ which hits one of the three corners of the sphere is the closest to the origin along the line $ x \equal{} y \equal{} z$, and also has no other intersection with the sphere. These three corners are $ (0,0,\sqrt {3})$ and the symmetric cases, which gives a minimum of $ 3 \sqrt [4]{2}$[/hide]


quadratics
geometry
3D geometry
algebra
factorization
sum of cubes

[b]1[/b]  Find integers $ m > n > 100$ that satisfy $ m \minus{} n \equal{} (\sqrt {m} \minus{} \sqrt {n})^3$.

[b]2[/b]  Find integers $ m\ne n$ that satisfy $ m^3 \minus{} n^3 \equal{} (m \minus{} n)^4$.

Solution to number 1.

[hide]The subtraction of the radicals will be ugly unless m and n are perfect squares.  Let $ a^2 \equal{} m, b^2 \equal{} n$.  Rewriting the equation yields $ a^2 \minus{} b^2 \equal{} (a \minus{} b)^3$.  Dividing by $ a \minus{} b$ and rearranging yields $ a^2 \minus{} 2ab \plus{} b^2 \minus{} a \minus{} b \equal{} 0$.  This looks as if it could be somehow "factored," so we rewrite the expression as $ (a \minus{} b \plus{} 1)^2 \equal{} 3a \minus{} b \plus{} 1$.  Let $ x \equal{} a \minus{} b \plus{} 1$.  The equation is $ x^2 \equal{} x \plus{} 2a$, a quadratic, and solving yields $ x \equal{} \dfrac{1\pm\sqrt {1 \plus{} 8a}}{2}$.  Since we know $ 1 \plus{} 8a$ is a perfect square and $ a > 10$, we find a solution at $ a \equal{} 15$, but this makes $ b \equal{} 10$, which cannot happen.  Therefore, we find another solution at $ a \equal{} 21, b \equal{} 15$, making $ m \equal{} 441, n \equal{} 225$.[/hide]

For number 2, can we have $ m\equal{}1, n\equal{}0$ or $ m\equal{}0, n\equal{}\minus{}1$?

[hide="1"][u]Lemma:[/u] If $ m, n$ are positive integers, then for nonzero integers $ r,s$, $ t = r\sqrt {m} + s\sqrt {n}$ is a nonzero integer only if $ m,n$ are both perfect squares.
[u]Sketch of proof:[/u] move one radical to left side and then square both side.

Since $ 0 < m - n = (\sqrt {m} - \sqrt {n})^3 = (m + 3n)\sqrt {m} - (n + 3m)\sqrt {n}$, we conclude $ m,n$ are perfect squares by our lemma. Let $ a = \sqrt {m}, b = \sqrt {n}$, then $ a > b > 10$ are integers. Now we have
\begin{align*} a^2 - b^2 & = (a - b)^3 \\
a + b & = (a - b)^2 \end{align*}

Setting $ k = a - b$,
\begin{align*} 2b + k & = k^2 \\
\Rightarrow b & = \frac {k^2 - k}{2}, a = \frac {k^2 + k}{2} \end{align*}

Because $ \frac {k^2 - k}{2} = b > 10$, $ k > 5$. Hence, all such ordered pairs $ (m,n)$ are
$ \boxed{\left(\frac {(k^2 + k)^2}{4}, \frac {(k^2 - k)^2}{4}\right), \forall k > 5 \in \mathbb{Z}}$. $ \blacksquare$
[/hide]

[quote="timwu"][hide="1"]Because $ \frac {k^2 \minus{} k}{2} \equal{} b > 10$, $ k > 5$. Hence, all such ordered pairs $ (m,n)$ are
$ \boxed{\left(\frac {(k^2 \plus{} k)^2}{4}, \frac {(k^2 \plus{} k)^2}{4}\right), \forall k > 5 \in \mathbb{Z}}$. $ \blacksquare$
[/hide][/quote]
I think you mean $ \boxed{\left(\frac {(k^2 \plus{} k)^2}{4}, \frac {(k^2 \minus{} k)^2}{4}\right), \forall k > 5 \in \mathbb{Z}}$. And also note that the answer has to do with the sum of cubes formula... possibly having to do with the cubic thing on the RHS of the original equation?

I edited that typo before you posted... :lol:

That is an interesting observation. Indeed, since we defined $ k\equal{}\sqrt{m}\minus{}\sqrt{n}$, the original equation is just
\[ \sum_{i\equal{}1}^{k}i^3\minus{}\sum_{j\equal{}1}^{k\minus{}1}j^3\equal{}k^3
\]
Unfortunately, I didn't see a way of knowing that before finding the answer...

Let $ k\equal{}\sqrt{m}\minus{}\sqrt{n}$. Then $ k\plus{}\sqrt{n}\equal{}\sqrt{m}$, squaring yields $ k^2\plus{}2k\sqrt{n}\equal{}m\minus{}n$. By the given, $ k^3\equal{}m\minus{}n$.
Setting equal, we have $ k^2\minus{}k\minus{}2\sqrt{n}\equal{}0$. Then continue as in the other proofs.


geometry
rectangle
analytic geometry
modular arithmetic

An $ 11$x$ 12$ rectangle is given. Show that

(a) the rectangle can be tiled with $ 20$ rectangles of sizes $ 1$x$ 6$ or $ 1$x$ 7$.

(b) it cannot be tiled with $ 19$ such rectangles.

[hide="Solution"]a)   Break up the rectangle into a $ 7\times 12$ rectangle and a $ 4\times 12$ rectangle.  The first can be tiled with 12 $ 1\times 7$ rectangles, and the second can be tiled with 8 $ 1\times 6$ rectangles.

b)   Put the big rectangle in a coordinate plane with the lower left corner at the origin, so that the bottom left square is $ (1,1)$.  Consider the squares with sum of coordinates equivalent to $ 5\pmod{7}$.  There are 4 squares with sum 5, 11 with sum 12, and 5 with sum 19.  Therefore there are 20 of these squares total.  But the difference in the sum of the coordinates of two squares in the same tile is at most 6, so each tile can only cover one of these squares.  Therefore they cannot all be covered by 19 tiles.[/hide]


Let $ (X_n)_{n \in \mathbb{N}}$ be a sequence of independent identically distribuited random variables, each of which can take any value in $ \{0,1,\cdots,9\}$ with equal probability $ \frac {1}{10}$. Show that $ \sum_{n \equal{} 1}^{\infty}\frac {X_n}{10^n} < \infty$ a. s. and find its distribution.

The distribution of the sum is the uniform distribution on $ [0,1].$ You're simply building decimal expansions at random, one digit at a time. The only slight complication is the set of real numbers that have two different decimal representations - but the probability of landing in that set is zero.


If a group of things is not specified in a counting problem as being distinct or not distinct, what do we assume they are?

For example:

One person with $7 and six people with $4 pool their money to buy a $20 gift, in how many ways can they do it?
Are the people in the group of six people with 4$ distinct or not?

[hide]It depends on the question. For this one, I definitely think it's distinct, because there is a difference if you have person x pay 4 dollars instead of person y.[/hide]

Spoilered for the benefit of beginners.

General rule: people (or other animal-type creatures) are distinct, things are not.  If we have distinct red balls, the problem will explicitly tell you that the  balls are distinct.  If you have identical red balls, it might say identical, or it might just say red.

If you are taking the test in a central location (for example, a Saturday math contest, as opposed to a contest at your school like the Mandelbrot), I would recommend that you ask a proctor to resolve the ambiguity. You shouldn't lose points if a question is poorly-worded. Interestingly, on one problem from a MOP test, it was unclear whether some things (people I think) were distinct or not, so I asked, and I was told that they were. Some people who were taking the test in a different room also asked whether they were distinct or not and were told that they were not. (They ended up accepting both interpretations.) There was also a similarly ambiguous question on the ARML relay round 2003 (1-1) in which you were supposed to assume that some things (people again?) were distinct, and some people assumed that they were not distinct. We tried to appeal to get both answers accepted, but it was denied.

If this distinctness issue can be so ambiguous at times, then why don't test-writers just distinct clearly whether each object of the problem is distinct or not to avoid ambiguity. Like in your example, ARML is a pretty major competition, and for a problem on there to be ambiguous...

Yeah, in Red A, they told us people weren't distinct, which I thought was very odd.  I would've sided with the ARML people (and not just because it would have been to NYC's advantage) -- people are always supposed to be distinct.

"people are distinct" => stereotype => bad => topic should be moved to round table

I agree with Fiery.  Test writers should add a word (like "distinguishable" or "indistinguishable") to make such questions clearer.


calculus
integration
limit
calculus computations

Here is a nice problem:

Find the limit:
$ \lim\limits_{n\rightarrow \infty}\int\limits_{0}^{1}\frac{ne^{x^{2}}}{1\plus{}n^{2}x^{2}}dx$.

[hide="Is this right?"]
$ \lim\limits_{n\rightarrow \infty}\int\limits_{0}^{1}\frac {ne^{x^{2}}}{1 + n^{2}x^{2}}dx$
integration by parts leads to
${ etan^{-1}(n)-1-\int_{0}^{1}2xne^{x^{2}}}tan^{-1}(nx)dx$
${ tan^{-1}(na)(e-e^{a^2})<=\int_{0}^{1}2xne^{x^{2}}}tan^{-1}(nx)dx<=tan^{-1}(n)(e-1)$
(a>0)
plugging in these bounds to the previous equation and taking the limit, we get a lower bound of
$ e\frac{\pi}{2}-1-\frac{\pi}{2}(e-1)=\frac{\pi}{2}-1$
and an upper bound of
$ e\frac{\pi}{2}-1-\frac{\pi}{2}(e-e^{a^2})=\frac{\pi}{2}e^{a^2}-1$
since a can be made arbitarily close to 0, the answer must be
$ \frac{\pi}{2}-1$
[/hide]

The limit should be greater than or equal to $ \frac{\pi}{2}$.

I think I made a mistake in the first step
it should be
${ etan^{ - 1}(n) - \int_{0}^{1}2xne^{x^{2}}}tan^{ - 1}(nx)dx$ instead of ${ etan^{ - 1}(n) - 1 - \int_{0}^{1}2xne^{x^{2}}}tan^{ - 1}(nx)dx$

that leads to an answer of $ \frac{\pi}{2}$


Let a,b,c>0. Prove that:
$ a\sqrt{a^2\plus{}2bc}\plus{}b\sqrt{b^2\plus{}2ac}\plus{}c\sqrt{c^2\plus{}2ab} \geq \sqrt{3}(ab\plus{}bc\plus{}ca)$

[quote="mathstarofvn"]Let a,b,c>0. Prove that:
$ a\sqrt {a^2 \plus{} 2bc} \plus{} b\sqrt {b^2 \plus{} 2ac} \plus{} c\sqrt {c^2 \plus{} 2ab} \geq \sqrt {3}(ab \plus{} bc \plus{} ca)$[/quote]

Using Holder's Inequality.

 :lol:

[quote="mathstarofvn"]Let a,b,c>0. Prove that:
$ a\sqrt {a^2 \plus{} 2bc} \plus{} b\sqrt {b^2 \plus{} 2ac} \plus{} c\sqrt {c^2 \plus{} 2ab} \geq \sqrt {3}(ab \plus{} bc \plus{} ca)$[/quote]

We have:
$ \sum \frac{a}{a^2\plus{}2bc} \le \frac{a\plus{}b\plus{}c}{ab\plus{}bc\plus{}ca}$
And:
$ LHS^2(\sum \frac{a}{a^2\plus{}2bc}) \ge (a\plus{}b\plus{}c)^3$

[quote="langtuvietnam"]My solution, please check it  because I am not sure the solution is true :)
  http://www.mediafire.com/?wjtonem2ldg[/quote]
[quote="langtuvietnam"]Let a,b,c >0  satisfying $ \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}\equal{}3$. .Prove that:
\[ 8(a^2\plus{}b^2\plus{}c^2)\geq 3(a\plus{}b)(b\plus{}c)(c\plus{}a)\][/quote]
Here is my solution to your problem:

It's the best solution for it. :) . And I have known it.
I have a solution, it's different :lol:  and......... it's ugly although only using AM-GM
try
It's the same proof with the problem of mathstarofvn
http://www.mediafire.com/?wjtonem2ldg (I have posted this link above)
I will post my solution soon.

Let a,b,c >0  satisfying $ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \equal{} 3$. .Prove that:
 \[ 8(a^{2}\plus{}b^{2}\plus{}c^{2})\geq 3(a\plus{}b)(b\plus{}c)(c\plus{}a)\]

here is my solution for this problem
http://www.mediafire.com/?0fjntmeonnz

Happy new year :)


ok i'm really not sure if this goes in this part of forum but here goes

Six boys and five girls are to line up for a picture.
How many ways can they be lined up such that no girls are together?

i get 604800
by doing 6! x 7 x 6 x5 x 4
but my teacher uses a more complicated method and gets
1,814,400

sorry i duno how to use latex

I think this is how you do it.

[hide] First you line up all the boys. This gives $6!$ permutations. Picture 6 boys standing in a line _ stands for an empty space

_$B_{1}$_$B_{2}$_$B_{3}$_$B_{4}$_$B_{5}$_$B_{6}$_

We see that there are 7 empty spaces where the girls can go, so we have $P^{7}_{5}$ permutaions for placing the girls.

Therefore the total number of permutations are $6! \times P^{7}_{5}=1814400$.[/hide]


$ a_i,b_i,r > 0$
\[ \left[\sum_i(a^r_i \plus{} b^r_i)^{\frac {1}{r}}\right]\left[\sum_i(a^{ \minus{} r}_i \plus{} b^{ \minus{} r}_i)^{\frac {1}{ \minus{} r}}\right]\le \left[\sum_i a_i\right]\left[ \sum_i b_i\right]
\]
when $ r \equal{} 1$

the problem becomes
\[ \left[\sum_i(a_i \plus{} b_i)\right]\left[\sum_i\frac {a_ib_i}{a_i \plus{} b_i}\right]\le \left[\sum_i a_i\right]\left[ \sum_i b_i\right]
\]
which can be solved by Cauchy Schwarz
http://www.mathlinks.ro/Forum/viewtopic.php?t=249474

I think in this inequalities solve by Gelder's inequalitie -

$ (\sum(a_i^p)^{\frac{1}{p}})(\sum(b_i^q)^{\frac{1}{q}})\ge\sum (a_ib_i)$
$ \frac{1}{p}\plus{}\frac{1}{q}\equal{}1$


In a given pentagon $ABCDE$, triangles $ABC$, $BCD$, $CDE$, $DEA$ and $EAB$ all have the same area. The lines $AC$ and $AD$ intersect $BE$ at points $M$ and $N$. Prove that $BM = EN$.

It is well-known that in any qudrilateral $XYZT$ we have:
$YZ||XT$ $\Longleftrightarrow$ $S_\triangle{XYZ}=S_\triangle{YZT}$ (by $S$ is meant area).

Using this property we get that:
$AB||CE$,$BC||AD$, $CD||BE$,$DE||AC$,$AE||BD$.
Hence $\frac{BM}{MN}=\frac{MC}{AM}=\frac{ND}{AN}=\frac{EN}{MN}$, or $BM=EN$.

from condition :[ABC]=[AED] , BE//CD, because  area[BCD]=[CDE] and , triangle ABE, &,ABM, & ANE have same vertex, so  [ABM]=BM.h/2, [ANE]=NE.h/2,  [CBM]=BM.h'/2, [DNE]=NE.h'/2, [ABC]=[ABM]+[BCM], & [AED]=[ANE]+[DEN]  therefor  BM(h+h')/2=NE(h+h')/2 ,    BM=NE.  Q E D.

[Hide=Solution by Saucepan_man02] We know that $ [ABC] = [BCD] $. Hence, the altitude from A to CD(extended if necessary) has same length as the altitude from D to CD(extended if necessary). Hence $ AD \parallel BC $. Similarly, $ AC \parallel DE $ and $ BE \parallel CD $. Here there are many parallel lines that intersect, and we try to see some parallelograms. Notice that $ BNDC $ and $ MEDC $ are parallelograms. Hence, $ BN = CD = ME $ or $ BN = ME $. Subtract $ MN $ both sides and we get $ BM = EN $.
Hence Proved! :-D

First I will show that $\overline{BE} \parallel \overline{CD}$. This is simply because $B$ and $E$ are equidistant from $\overline{CD}$ by the equal areas condition. Thus, $\frac{AN}{DN} = \frac{AM}{CM}$ implies $[AEN] = [ABM]$, so $BM=EN$, as required.


The Fibonacci numbers are defined by F1 = F2 = 1, Fn = Fn-1 + Fn-2.
Suppose that a rational a/b belongs to the open interval (Fn/Fn-1, Fn+1/Fn).
Prove that b \geq Fn+1.

I think this is about Farey series, somehow, because we can see that one of the 2 following equalities is true (alternatively) for all n:

Fn2-1=Fn-1*Fn+1 and
Fn2+1=Fn-1*Fn+1

so in our case we have the second one. However, I don't exactly know how to put it all together to yield a solution. We should prove that the 2 fractions are consecutive in the Farey series of fractions wtih the largest denominator Fn.

It's not necesary to go so far. We have bF_(n+1) \geq aF_n+1 and aF_(n-1) \geq bF_n+1. Thus, bF_(n+1)>=[(bF_n+1)*F_n]/F_(n-1) +1, from where b(F_n-1*F_n+1-(F_n)^2)>=F_n+1. Since |F_n-1*F_n+1-(f_n)^2|=1, the conclusion follows.


Find out all functions f:Q intersected with [0,1] -->R for which we have:

f(p/(p+q))=p/(p+q)f((p-1)/(p+q-1))+q/(p+q)*f(p/(p+q-1)), where p,q are natural \ {0}.

cheers! :D


inequalities
function
integration
calculus
calculus computations

Let $ x_0$ be a real number. $ \forall x\geq x_0 | f'(x) | \leq g'(x)$. Prove, that $ \forall x\geq x_0: |f(x) \minus{} f(x_0)|\leq g(x) \minus{} g(x_0)$.

Are you sure all those inequality signs are pointing in the right directions?

I'm going to assume you were talking about real-valued functions and give a proof under that assumption. It should still be true if you allow $ f$ to be complex-valued, but I'll let someone else do that proof. There are also tacitly given hypotheses here that $ f$ and $ g$ are continuous for $ x\ge x_0$ and differentiable for $ x > x_0$ -- in other words, the standard hypotheses for the mean value theorem and Rolle's Theorem.

Define $ h(x) \equal{} [g(x) \minus{} g(x_0)] \minus{} [f(x) \minus{} f(x_0)].$ Then $ h(x_0) \equal{} 0$ and $ h'(x) \equal{} g'(x) \minus{} f'(x)\ge 0$ That makes $ h$ a nondecreasing function and hence $ h(x)\ge 0$ for $ x\ge x_0.$ We can then rearrange $ [g(x) \minus{} g(x_0)] \minus{} [f(x) \minus{} f(x_0)]\ge 0$ so that it says $ f(x) \minus{} f(x_0)\le g(x) \minus{} g(x_0).$

That's one side. We need the other side of the absolute value. This time, define $ k(x) \equal{} [g(x) \minus{} g(x_0)] \plus{} [f(x) \minus{} f(x_0)].$ We have $ k(x_0) \equal{} 0$ and $ k'(x) \equal{} g'(x) \plus{} f'(x)\ge 0,$ so again $ k(x)$ is increasing. Rearrange $ [g(x) \minus{} g(x_0)] \plus{} [f(x) \minus{} f(x_0)]\ge 0$ so that it says $ f(x) \minus{} f(x_0)\ge \minus{} (g(x) \minus{} g(x_0)).$

Put these two arguments together and you have what you want.

-----

Now, for a lesson in reading proofs: why [i]didn't[/i] I say this?

[quote="What I did not say"] By the MVT, there exists $ \xi\in(x,x_0)$ such that $ f(x) \minus{} f(x_0) \equal{} f'(\xi)(x \minus{} x_0)$ and $ g(x) \minus{} g(x_0) \equal{} g'(\xi)(x \minus{} x_0).$ Note that $ x \minus{} x_0 > 0$ and $ |f(x) \minus{} f(x_0)| \equal{} |f'(\xi)|(x \minus{} x_0).$ Since $ |f'(\xi)|\le g'(\xi),$ the inequality follows.[/quote]

That wouldn't work because the same $ \xi$ need not exist for both $ f$ and $ g$.

The weakness of that proof is that it requires stronger hypotheses. In particular, the statement that $ f$ is continuous for $ x\ge x_0$ and differentiable for $ x>x_0$ does not guarantee that $ f'$ is integrable on $ [x_0,x].$ In order to you your proof (after fixing the typo in the last integral, deleting the extra integral sign, and making it $ g'$), you would need to separately hypothesize the integrability of $ f'.$

Or use the generalized MVT: $ g'(\xi)(f(x)\minus{}f(x_0))\equal{}f'(\xi)(g(x)\minus{}g(x_0))$ (proved in the same way as the regular MVT). This gives $ |f(x)\minus{}f(x_0)|\le |g(x)\minus{}g(x_0)|$. Another application of MVT shows that $ g(x)\minus{}g(x_0)\ge 0$.


talking about cancun , imo 2005 was first going to be held there , then they changed it to merida . any mexican knows the reason?

It's said on the official site: http://www.imo2005.org
Elections will take hold in the same period at Cancun.

[quote="rokas"]I am problem coordinator and will spend two days in Cancun before going to Merida on July 8-9. It would be nice if I could find some people to hang out with  :) Anybody in Cancun those days?[/quote]You're a problem coordinator? :) That's cool. I'm one also :D
I'll be arriving in Merida on the 9th, so I guess I'll see you there.

[quote="wisemaniac"]even contestants want to see you valentin , and even take pics with you , you ll be for sure the superstar of this imo !![/quote]Hehe, superstar? :) I don't think so  :blush:


Hey guys,

If you're from Illinois and you wanna have lots of fun, let's arrange a time to all play a big game on FTW! :D  Normal ranking game, 45 seconds per problem, 10 problems.

Please post here if you're interested in participating! Let's also arrange a time to meet!

I'd prefer time instead of ranking, but sure.

Good idea! But with so many people, I would go with time instead of ranking.

Signups:

gauss1181
suma_milli
Stargirl240
whitesoxfan

Scoring system is now changed to Time.

Sorry AIME, but you may not join. This is an [b][i][u]ILLINOIS ONLY[/u][/i][/b] game.

Additionally, I'm waiting for responses from avec_une_h, her sister, RussianRocket, ManHatTaNaTiVe94, Aryth, guineapiggies, JoshD, and jgjgjmath.

[quote="AIME15"]...I'm from illinois...[/quote]
Sorry, but no. I can pretty much tell that this is a joke...Don't underestimate me. :wink: Or argue either.

[quote="suma_milli"]:rotfl:

Location: CA?[/quote]
HAHA YEAH
THAT'S THE CLEAR EVIDENCE THAT IT'S A LIE!!!! :wink:

Y'all should probably say what times your available before you say you're in... anybody could WANT to play, but it takes more than "wanting" to get the meet together.

I might play if I have time.

Once we get as many people as we can in, we'll talk about when to meet.

Signups:

gauss1181
suma_milli
Stargirl240
whitesoxfan
Aryth

gauss1181 (1352)
suma_milli (1200)
Stargirl240 (1533)
whitesoxfan (1250)
Aryth (1200)
JoshD
guineapiggies (1300)
stuck
avec_une_h (1250)
RussianRocket
helagha
enryK (1200)

I'm probably gonna have to PM those four people who haven't posted their highest ratings.

Yay we have 12 people!  That's like a perfect number for an FTW meet :D

That will be our final number. In that case, we're going to have four selected people receiving byes into the second round. :D

[quote="suma_milli"]lol that will be me! :P[/quote]
bleh oh come on suma that may or may not be you...Probably you're guessing that b/c I was so laughed up with you on Gmail voice chat... :rotfl:

[quote="suma_milli"]Huh why would I be guessing that b/c of our gmail voice chat?  :maybe:[/quote]
because ur funny and all that matters is that....
so i was originally thinking of giving you a bye...
but IDK really. :rotfl:

See my PM. I just wasn't sure who to give byes to.... :maybe:

it'd just have to depend on whatever I select.

Byes should be random. Several people, includingme, haven't played much since they changed the rating system. It used to be impossible to reach 1600, let alone the 2000 that some people have now.

Update: stuck PM'ed me. He's dropping out.

gauss1181 (1352)
suma_milli (1200)
Stargirl240 (1533)
whitesoxfan (1250)
Aryth (1200)
JoshD
guineapiggies (1300)
avec_une_h (1250)
RussianRocket
helagha
enryK (1200)

Update: (JoshD's rating is now up)

gauss1181 (1352)
suma_milli (1200)
Stargirl240 (1533)
whitesoxfan (1250)
Aryth (1200)
JoshD (1350)
guineapiggies (1300)
avec_une_h (1250)
RussianRocket
helagha
enryK (1200)

Still waiting for RussianRocket and helagha...

Update: helagha's rating is now up. If I don't get any response from RussianRocket sooner or later, we're gonna start this without him.

gauss1181 (1352)
suma_milli (1200)
Stargirl240 (1533)
whitesoxfan (1250)
Aryth (1200)
JoshD (1350)
guineapiggies (1300)
avec_une_h (1250)
RussianRocket
helagha (1200)
enryK (1200)

wait, when are you starting. Is there still time to join?

lol I think this just died a long time ago.


Can anyone suggest a link to some skill builders. ie; pages of consecutive arithmetic questions, addition subtraction multiplication and division.

123 x 456
123 / 34
32 + 216
734 - 432

and so on...

[url]http://zetamac.com/arithmetic/[/url]

the only thing is is that it's timed

There are some patterns that I learned from Scott Flasburg.

any number that ends with a 5 squared is like.

$(10n+5)^{2}=$
$100*n(n+1)+25$


[quote="10000th User"][hide]Let $x=2^{100}$. Divide $16x^4+8x^3+4x^2+2x+1$ by $x+1$. Use remainder theorem and find that remainder $r(-1)=16-8+4-2+1=11$[/hide][/quote]

Wow, nice solution.

we can also say that x is congruent to -1 mod x+1 so we subtract the sum of the coefficients of the odd powered-terms from the sum of the coefficients of the even powered-terms.... nice substitution nonetheless...

[quote="10000th User"][hide]Let $x=2^{100}$. Divide $16x^4+8x^3+4x^2+2x+1$ by $x+1$. Use remainder theorem and find that remainder $r(-1)=16-8+4-2+1=11$[/hide][/quote]

I had the idea!  :roll:  I just didn't realize that $16x^4$ was equal to $2^{404}$  :roll:


All of these can be written as $k/100$ where $k$ is a 4-digit number. However, note that $15.59=\frac{1559}{100}$ in lowest terms since $GCF[1559,100]=1$.

Your lowest common denominator for these numbers is $100$. This kind of problem generally goes into HSB  :maybe:


Let  $ x,y,z>0$ $ x^2\plus{}y^2\plus{}z^2\equal{}1$.Find the minimum  value of the expression   $ \frac{1}{xy\plus{}z}\plus{}\frac{1}{yz\plus{}x}\plus{}\frac{1}{zx\plus{}y}$    :?:  :rotfl:

Hello !! I'm a new member in this nice forum ^^ , and this is my first post.

[url=http://www.casimages.com][img]http://nsa09.casimages.com/img/2009/09/21/090921042241886249.gif[/img][/url]

 :)

[quote="Pirkuliyev Rovsen"]Let  $ x,y,z > 0$ $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$.Find the minimum  value of the expression   $ \frac {1}{xy \plus{} z} \plus{} \frac {1}{yz \plus{} x} \plus{} \frac {1}{zx \plus{} y}$    :?:  :rotfl:[/quote]

$ \ LHS\geq \frac{9}{xy\plus{}yz\plus{}zx\plus{}x\plus{}y\plus{}z}\geq \frac{9}{x^2\plus{}y^2\plus{}z^2\plus{}\sqrt{3(x^2\plus{}y^2\plus{}z^2)}}\equal{}\frac{9}{1\plus{}\sqrt{3}}\equal{}\frac{9(\sqrt{3}\minus{}1)}{2}$

Sorry for my noobiness but how do you get

$ \frac {9}{x^2 \plus{} y^2 \plus{} z^2 \plus{} \sqrt {3(x^2 \plus{} y^2 \plus{} z^2)}}$

EDIT: Nevermind, I see Cauchy-Schwarz. Nice..

i wanted to mean $ \frac{9(\sqrt{3}\minus{}1)}{2}$ $ not \frac {9(\sqrt {3} \minus{} 2)}{2}$


[hide]
let $h$ be the height of $DBC$ and $x$ be the distance from $B$ to the altitude of $DBC$

Without loss of generality, let the base = $1$

$\tan 6 = \frac{h}{x}$
$\tan{18}= \frac{h}{1-x}$

$\left(1-\frac{h}{\tan 6}\right)\tan{18}= h$

$h = \frac{\tan{6}\tan{18}}{\tan 6+\tan{18}}$

by similar logic, the height of the larger triangle, which I'll call $H$ is:
$H = \frac{\tan{54}\tan{24}}{\tan{54}+\tan{24}}$

now,
$\frac{A(ABC)}{A(DBC)}= \frac{H}{h}$

$\frac{A(ABDC)}{A(DBC)}= \frac{H-h}{h}= \frac{H}{h}-1$

$= \frac{(\tan{54}\tan{24})(\tan 6+\tan{18})}{(\tan{54}+\tan{24})(\tan{6}\tan{18})}-1$

There must be a nice way to simplify that
There also must be an elegant solution to the problem. I went the long way
[/hide]

yes, i solved with  geometric ways.It is good question.I will wait for yours different solution.

I will translate solution.
[img]http://img114.imageshack.us/img114/2012/45y45h6h5cd5.png[/img]
[img]http://img523.imageshack.us/img523/3532/f43f434su9.png[/img]


hello guys;pls help me :blush: from this problem;"If 'n' points of the plane don't lie on the same line then there exists a line passing through exactly two points" :)

It's the well-known Sylvester-Galai theorem:

http://en.wikipedia.org/wiki/Sylvester–Gallai_theorem .


A sequence $ < L_{n} >$ is defined on positive integers ,$ L_{1} \equal{} 1$ and $ L_{2} \equal{} 3$ , $ L_{n} \equal{} L_{n \minus{} 1} \plus{} L_{n \minus{} 2}$ . Show that $ L_{p} \minus{} 1$ is divisible by $ p$ if $ p$ is prime .

[hide]Easy to get that $ L_n\equal{}\frac{(1\plus{}\sqrt5)^n\plus{}(1\minus{}\sqrt5)^n}{2^n}$ (it can also be proved by induction). For $ p\equal{}2$, $ L_2\minus{}1\equal{}3\minus{}1\equal{}2$ is divisible by $ 2\equal{}p$. For odd $ p$, we have $ p|L_p\minus{}1$ iff $ p|2^p(L_p\minus{}1)$ iff $ p|(1\plus{}\sqrt5)^n\plus{}(1\minus{}\sqrt5)^n\minus{}2^p$. Note that $ (1\plus{}\sqrt5)^n\plus{}(1\minus{}\sqrt5)^n\equal{}2\left(1\plus{}\binom{p}25\plus{}\binom{p}425\plus{}\ldots\plus{}\binom{p}{p\minus{}1}5^{(p\minus{}1)/2}\right)$. It is easy to see that $ p|\binom{p}k$ for $ 0<k<p$. So $ (1\plus{}\sqrt5)^n\plus{}(1\minus{}\sqrt5)^n\equiv2\pmod{p}$. By Fermat's Theorem, $ 2^p\equiv2\pmod{p}$. So $ p|(1\plus{}\sqrt5)^n\plus{}(1\minus{}\sqrt5)^n\minus{}2^p$ as desired. Q.E.D.[/hide]

[quote="Johan Gunardi"][hide]Easy to get that $ L_n \equal{} \frac {(1 \plus{} \sqrt5)^n \plus{} (1 \minus{} \sqrt5)^n}{2^n}$ (it can also be proved by induction). For $ p \equal{} 2$, $ L_2 \minus{} 1 \equal{} 3 \minus{} 1 \equal{} 2$ is divisible by $ 2 \equal{} p$. For odd $ p$, we have $ p|L_p \minus{} 1$ iff $ p|2^p(L_p \minus{} 1)$ iff $ p|(1 \plus{} \sqrt5)^n \plus{} (1 \minus{} \sqrt5)^n \minus{} 2^p$. Note that $ (1 \plus{} \sqrt5)^n \plus{} (1 \minus{} \sqrt5)^n \equal{} 2\left(1 \plus{} \binom{p}25 \plus{} \binom{p}425 \plus{} \ldots \plus{} \binom{p}{p \minus{} 1}5^{(p \minus{} 1)/2}\right)$. It is easy to see that $ p|\binom{p}k$ for $ 0 < k < p$. So $ (1 \plus{} \sqrt5)^n \plus{} (1 \minus{} \sqrt5)^n\equiv2\pmod{p}$. By Fermat's Theorem, $ 2^p\equiv2\pmod{p}$. So $ p|(1 \plus{} \sqrt5)^n \plus{} (1 \minus{} \sqrt5)^n \minus{} 2^p$ as desired. Q.E.D.[/hide][/quote]
Nice ..Even I got the same solution but can you give me a solution without finding a general expression for $ L_{n}$


induction
superior algebra
superior algebra unsolved

suppose that $G$ is a group.and $x,y\in{G}$.
if $z=xyx^{-1}y^{-1},zx=xz,zy=yz$ show that for all $n\in{\mathbb{N}}$:
$(yx)^{n}=z^{\frac{n(n-1)}{2}}y^{n}x^{n}$

I can solve this problem by induction on $n$, i think so :) have you got other solution?

so if, please post that solution, because i can't   :(

ok N.T.THUN
I used these propositions in the solution:
1)If $m,n\in{\mathbb{Z}}$and $a,b\in{G}$ and $ab=ba$ then :
$a^{m}b^{n}=b^{n}a^{m}$
and
$(ab)^{m}=a^{m}b^{m}$
2)If $a,b,c\in{G}$ then:
$ab=ac\Leftrightarrow{b=c}$ and $ba=ca\Leftrightarrow{b=c}$
3) If$a,b\in{G}$ then: for all natural $k$ :
$(aba^{-1})^{k}=ab^{k}a^{-1}$
to solve the problem by induction we shoud prove that :
If $(yx)^{k}=z^{\frac{k(k-1)}{2}}y^{k}x^{k}$ then:$(yx)^{k+1}=z^{\frac{(k+1)k}{2}}y^{k+1}x^{k+1}$
we have:
${(yx)^{k}=z^{\frac{k(k-1)}{2}}y^{k}x^{k}}\Rightarrow(yx)^{k+1}=yxz^{\frac{k(k-1)}{2}}y^{k}x^{k}$
Therefore we shoud prove:$yxz^{\frac{k(k-1)}{2}}y^{k}x^{k}=z^{\frac{(k+1)k}{2}}y^{k+1}x^{k+1}$
we have:
$yxz^{\frac{k(k-1)}{2}}y^{k}x^{k}= z^{\frac{(k+1)k}{2}}y^{k+1}x^{k+1}\Leftrightarrow{yxz^{\frac{k(k-1)}{2}}y^{k}=z^{\frac{(k+1)k}{2}}y^{k+1}x}\Leftrightarrow\\ z^{\frac{k(k-1)}{2}}yxy^{k}=z^{\frac{(k+1)k}{2}}y^{k+1}x\Leftrightarrow{yxy^{k}=z^{k}y^{k+1}x}\Leftrightarrow{yxy^{k}=y^{k+1}z^{k}x}\Leftrightarrow\\xy^{k}=y^{k}z^{k}x$
and the last relation is true because:
$z=xyx^{-1}y^{-1}\Rightarrow{zy=xyx^{-1}}\Rightarrow{(zy)^{k}=xy^{k}x^{-1}}\Rightarrow{y^{k}z^{k}=xy^{k}x^{-1}}\Rightarrow{y^{k}z^{k}x=xy^{k}}$
  :)


Shouldn't AoPS have an official IRC Chat (preferablly LaTeX enabled)? YEah sure the private messages are something but they can be a little slow..

Search function is also good for other that kind of stuff ;)
http://www.mathlinks.ro/Forum/viewtopic.php?t=98568


Hello.  I currently feel the urge to set up a marathon in which all topics involving the intermediate and pre-olympiad level are brought up. You know how a marathon works, as a result, I'll post the first question.
Prove the inequality
\[ \\\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{9997}+\sqrt{9999}}>24 \].

Masoud Zargar

Yup.  Would you like to post the next problem? :)

A little one in "geometrical enumerative combinatorics" (title longuer than problem ;))

Consider an arrangement of $n \geq 3$ lines in the plane.
Find the minimal number of the minimal connected regions which are triangles.


geometry
geometric transformation
reflection
ratio

Hi,

recently someone informed me that we Belgians are actually among the few that actually connect their letters in one word when we write on paper, when I write 'book' the lower right part of the b is also the lower left part of the o, the o is connected to the other o,etc...

I know at least one Dutch person who doesn't do that.
When my niece went to school in the United States, she was ordered to stop doing so by the teachers there...

So, how do you write?  And why is there a difference?

Well, when I write neatly, most of the letters are unconnected.

If I write quickly (ie. taking notes), most of the letters are connected.

Are you refering to cursive? I usually don't do cursive, but sometimes my letters get somewhat connected anyways...hand is too tired to lift pencil.

When I need to write fast, I do cursive, so they are obviously connected.  When I take notes, I usually try to write print, but they sometimes end up being connected in a sort of intermediate state.

[img]http://upload.wikimedia.org/wikipedia/commons/4/48/Cassowarys_Victorian_Modern_Cursive.png[/img]

This is what I meant, so yeah I guess so : cursive.

I write like this all the time.  It's neater when I write slowly, but it is always connected like this, and so do all my fellow students and professors

I don't write in cursive,mainly because it looks horrible when i do it. :D

I have a habit of writing illegible cursive; my letters are connected but aren't immediately recognizable.

:(  I realise now that my question is not the proper one :
I meant : what are you supposed to do?  How did your teachers want you to do it.
I myself have a disgusting way of writing, but it's about the culture at your place.

Eh here it doesn't really matter...the majority of people write in print, but cursive was taught in elementary school, and some people do write in cursive.

I don't think the teachers really care, but I guess print is easier to read (to me, at least).

i never use cursive except for my signature on my paper.
even so, all my letters are connected anyway because my handwriting is scribbly lines.  in elementary school penmanship was my lowest grade. i don't know why they worry about it, because every adult i know signs their name like this:
my dad's name is richard cassidy.

[size=200]R[/size][size=75]scribbly line scribbly line scribbly line[/size][size=200]C[/size][size=75]scribbly line scribbly line scribbly line[/size][size=200]y[/size]

Every student I've ever talked to says this about handwriting:
In elementary school (2nd or 3rd grade) they start teaching you to write this new way, with your letters connected: cursive. They swear to you that you need to be really good at writing in cursive, because when you get to high school, all of your teachers will require that you write in cursive [you are now terrified of high school, because this SUCKS!]. So then you get to high school, and you don't write in cursive. All of the students that do write in cursive are told to stop turning things in that way, because the teachers can't read it. If the teachers can't read it, they can't grade it. Now, I don't even write my notes in cursive, for the same reason: I want to be able to read it later. However, cursive IS faster- some people write notes in cursive so they can get everything down. Everyone signs their name to legal documents in cursive, so it is required for something. However, to my elementary school teachers, I say: LIES!! ALL LIES!

[quote="fredbel6"]but that is what you were taught right???
because every student in Belgium I know was taught to write with the letters connected[/quote]we are taught for the reason that it's a form of writing, and we should be able to read it.  we were taught the 'print' style of writing long before we were taught cursive - so just because we were taught cursive, doesn't mean it's considered the standard or anything.  we are not supposed to write in cursive/handwriting, and i've never had a teacher that requires anything to be submitted in cursive.

printing is standard, but some like cursive and they choose to write in cursive.  that's all there is to it, i believe.  someone mentioned that cursive is faster - that's interesting.  i've never compared.

my personal opinion, i hate cursive, and i hate when people write in cursive when they can't do it neatly.  however, when a person has very nice cursive writing, very smooth and elegant, it can be a lot nicer than print.  not my thing tho, still.

I personally can't read my cursive, but if i look at it at a glance, it seems neater than print.

To make a few comments and corrections:

The fastest way to write (that is readable) is a mix between print and cursive, it is mainly a style of print but certain groups of letters can be combined together into one pen movement.

The most asthetically pleasing way to write is a copperplate calligraphic style but this requires too much training and work for it to become practical in schools, however, 70+ years ago it was taught in a large number of schools as handwriting was far more important then. Here is a nice example of a relatively practical copperplate style: http://www.anomaly.org/debbie/calligraphy/images/copperplate-sometimes-when-I-reflect-back-2004-july.jpg

I model my own handwriting on Leonardo Da Vinci's because it is attractive (to me anyway!) and relatively neat, although it does not have the grace of other hands.

i agree, i write with a mix because cursive is a little too elaborate and i'm to impatient to write in regular manuscript. so i have a mix and its also mostly legible from what everyone else says.

[quote="fredbel6"] what are you supposed to do?  How did your teachers want you to do it.
I myself have a disgusting way of writing, but it's about the culture at your place.[/quote]
Teachers don't really care how you do it, as long as they can read it. THats how it is in my school.

In Elementary we had to write cursive (3-5 mainly). I was one of the few people who actually liked it. I alternate now.

(Messy printing does end up with connected letters but I wouldnt consider it cursive..)

I write with a mix of cursive and print. I used to write mainly in cursive, but it looked like a bunch of smeared loops and was illegible.

I generally don't write in cursive (Actually, almost never). However, since my classes are all based on note-taking, I write really fast in way that it's all connected and only leigble to me. (Sometimes, it's not even legible to me). I DO have good handwriting, I just don;t have the time to write neat anymore. :D

I prefer to use print, partially because it's neater than my cursive. I only use cursive when I sign my name (my school requires us to write "I pledge my honor" and sign every piece of work to indicate that we didn't cheat, etc).

Answering the original question, I connect the letters when I write. Most of my friends don't connect them. I'm a bit untidy when I write, but well, as my teachers say, the important thing is the content, not the way you write.

I don't write in cursive. To compensate for that, I write small. My handwriting is already mediocre in print, it's just illegible in cursive.

Conjecture: All mathemticians have not a nice handwritting... any countraexample?

I'm pretty sure almost all Americans were taught cursive and [u]forced[/u] to use it throughout elementary school.

But in middle and high school, teachers started letting you print. So I guess 95% of the kids started printing, including me.

I have horrible handwriting and my letters are connected if I write too fast although I don't intend for them to be.

I write cursive. TOO cursive. Sometimes even I can't read my own writing after I take notes at school...most times...
I used to have good handwriting, but now I just don't have to time to write better, so teachers often scorn at me after I turn something in.

They never taugh us cursive, but i prefer to write cursive for my own notes that are not too importand so it will not be a great loss if i cannot read them. but at tests,etc i use print. and i have created a very nice (according me) way of writing that i use when i have time:

In India, we're taught to write in cursive, but quite a few of us switch to print........But then the cursive to print ratio is nearly 1, so we're pretty balanced in terms of handwriting...... :D


search
number theory proposed
number theory

Let $d(n)$ be number of divisors of n. Find all natural $k$ for which exists such $n$ that \[d(n^{2}) = kd(n)\] Pretty nice :)

It is an IMO problem. Search at the sercouses page

IMO 1998,problem 3!
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=124439#p124439[/url] :wink:


find integer solution of $ x^3\plus{}y^3\equal{}3xy\plus{}3$

[quote="zaya_yc"][quote="nguyensilva"]find integer solution of $ x^3 \plus{} y^3 \equal{} 3xy \plus{} 3$[/quote]

$ x \plus{} y \equal{} a$
$ xy \equal{} b$ ----> $ x^{3} \plus{} y^{3} \equal{} 3xy \plus{} 3$<-> $ a^{3} \minus{} 3ab \equal{} 3b \plus{} 3$--> $ 3b \equal{} \frac {a^{3 \minus{} 3}}{a \plus{} 1} \equal{} a^{2} \minus{} a \plus{} 1 \plus{} \frac {2}{a \plus{} 1}$ --->

$ a \plus{} 1 \equal{} \minus{} 2; \minus{} 1;2;1$ ---> $ a \equal{} \minus{} 3; \minus{} 2;1;0$
$ a \equal{} \minus{} 3$--> $ b \equal{} 5$--->$ \phi$
$ a \equal{} \minus{} 2$--> $ 3b \equal{} 11$--->$ \phi$
$ a \equal{} 1$--> $ 3b \equal{} \minus{} 1$--->$ \phi$
$ a \equal{} 0$--> $ b \equal{} \minus{} 1$-----> $ (x,y) \equal{} (1, \minus{} 1) \equal{} ( \minus{} 1,1)$[/quote]

thank you very much , but DO YOU THINK ABOUT (2;1)

@zaya_yc
$ \frac{a^3 \minus{}3}{a\plus{}1}\equal{}a^2\minus{}a\plus{}1\minus{}\frac{4}{a\plus{}1}$


induction
LaTeX
number theory solved
number theory

Using iduction it is easy to show that for all $n\geq 5$ we have $a_1^2+a_2^2+\dots+a_n^2+70-n=a_1a_2\dots a_n$.

wher do you find all these problems Iris aliaj

this is the site where i've taken the problems from:
http://www.geocities.com/CapeCanaveral/Lab/4661/

Hey, do you know that you are not forced to post two or three times the same message? :D
Just wait a bit after sending the first one....

Pierre.

Oh really,i didn't know that!! :D
Well in fact there is a problem.Many times,when i'm working on this site and i click on the back button this message appears on the screen:[i]Your webpage has expired.Click on the refresh button[/i](or sth like that).But if i click on the refresh button another message appears leaving me with two opportunities
1-Clicking on retry and this way continuing the work normally,but reposting a topic
2-clicking cancel and working offline and this way not reposting the topic
I'd really like not to annoy the moderators :)  by posting a message 2 times but...
So what should i do?(i can't work offline,can I ? )

Iris

Iris aliaj, what do you mean by (a70)2 ? ${2a_70}^2$? or $a_{70}^2$?
it's time for you to learn LaTeX !

Now Arthur, it is time for you to learn about editing your post.... :D

Pierre.

haha...l was waiting for a reply like that. this wasn't the moment for me to make a typing misatke.

Ok guys, stop quarreling like that now! :D
When i posted it i was in a hurry so i did that typing mistake.By (a70)2 i mean simply (a70) 2 .I know that it's time for me to learn about LaTex cuz not knowing  anything about it hampers me to post many problems :(
but there is only a week left before the 8th Junior Balkan,for which i am preparing now,so i don't have any time.I promise i'll learn it after a week. :)