calculus geometry trigonometry AMC AIME USA(J)MO USAMO

...math is taught, what would you change?  (Don't put stupid things in here like give everybody an A++++)

I would agree with Steve Olson in Countdown for one thing...

Also, children should be encouraged to do logic puzzles and stuff at a young age.  That way you can expect them (sort of) to learn faster later.

[quote="BHorseMath"]I would change it by like not holding people back like they do here. There were several in my school this year that could have taken algebra and it was possible to do it but the school just wouldn't let it. Instead they let us get frustrated working on stuff that was way too easy and let us make like 113's and stuff for grades in the class. They should let a kid move as fast as they want to move. If a 7th grader has the potential to take algebra 1 and wants to let them don't hold them back at like pre-algebra.[/quote]

I agree with that, 7th grade is quite ready for Algebra 1... but I know quite a few people who have taken Calculus and remember almost nothing about geometry, algebra, or even basic arithmetic. It's funny... yet sad. (I have no idea what to do about that)

mainly though, they should really stop letting the little kids use calculators - I mean, come on, if the kid can't add 5 + 7 in his head, why give him a calculator? The concept is beyond me. they should only be allowed calculators in prealgebra at the EARLIEST - I think you can make it through without one until you start learning basic trig though.

Ehh somethings in prealgebra are very trying without calculators though.  Cut and try especially.  I mean I can do it without a calc but it is so much easier with one and doesn't get on my nerves nearly as much.  Here where I am the regular 8th grade class is prealgebra advanced 8th grade is algebra 1.  They let advanced 7th graders take prealgebra but not algebra one.  I sat there the whole yearand aced the class and payed attention maybe 3 times in class.  Otherwise I had another book and was working algebra instead.  Even my math teacher said that several in my class should have been able to go into algebra and still ace it but the counsler wouldn't stand for the idea.  My teacher being good let me and a few others just work other things as long as we could answer any questions that she asked.  That was better but I still think that schools shouldn't hold kids back who have the abilitie to go farther sooner.

Even if you do get advanced, most of the classes are still pathetic (This is quite funny: I learned school-level algebra II, geometry, and trig all in one year minus the first month maybe just studying for about 45 minutes a day on my own.  Of course, I wasn't satisfied so I studied them more deeply the next year).  Again, I think it's the problem solving ability that needs to be developed.  It's not going to help no matter how much you cram into 6-year-old kids.  It matters to learn to develop deeper understanding.  And also, of course, we should pay teachers a ton more.

I agree.  Class placement shouldn't be given on basis of age, but rather on basis of skill.  But in the classroom, I dislike the "pres-set" boring days.

(Some day: Lecture, 4.1 with proofs, problems, examples.  Classwork. Brain usage: 20%
Next day: Lecture, 4.2 with proofs, problems, examples. Classwork.  Brain usage: 20%
Next day: Lecture, 4.2 with proofs, problems, examples.  Classwork.  Brain usage: 20%
Next day: Quiz.  Brain usage: 60%
Next day: Lecture, 5.1 with proofs, problems, examples.  Claswork  Brain usage: 20%)

There are much better ways to go about teaching.

Those things would work because I know that I got really bored this year sitting there trying not to roll my eyes when the teacher asks what the pythagorean theroem is and I am the only one that knows the fomula even.  That gets frustrating after awhile.

they should have even more accelerated classes that move faster like what JMI was saying.  So that people can move as fast as they need or want to.

[quote="JMI"]I think the school system should have three different paces:
Slow- For kids who need extra help or don't want to learn
Normal-For average kids
Fast-For smart kids who want to learn more
and you might even want:
Extra Fast-For kids who want to take the course in a month instead of entire school year[/quote]

This would not be nice to the "slow" kids.  And there are other issues (some say that there would be racial discrimination, which is quite possible).  As I have said before, it is much more important to develop learning skills (problem solving and creativity help) at a young age.  School should be much more solving stuff as opposed to brute memorization.

First: As a junior in high school I have far too many friends who (despite the fact that they are in Alg II or Trig) still think that math is arithmetic.  This is not to terribly surprising considering the way math is taught in elemetary school.  I'm not sure how, but they need to find a way in the younger grades to make math more than just arithmetic.  And no, I am not ignoring the importance of learning the basic math facts, but I think that so much time is spent on timed tests that it is amazing any of us made it through and still see the beauty in math.  There has to be other aspects of math that could be understood at the elementary level.

Second:  They need to get rid of the "crutches".  No more drawing dots on the numbers for addition, no more "tool kits" or note cards with fomulas for tests, and no calculators at the lower levels of math.  I just finished AP Calc and we were only allowed to use our calculators about half the time, and then the majority of that all I used it for was the basic addition, multiplication, etc...  I have friends that made it through Geometry and don't know how to write a proof, all they had to do was copy it off of their tool kits(our school uses CPM (typed it wrong the first time :) ) curriculum for non-AP math...I hate it).

Third:  The schools need to teach us to understand math, not just memorize it.  For example, we need to understand why x^2 +y^2 = r^2 makes a circle and we need to take the time to derive the trig identites, not just have them given too us.

and as a final note...Do not listen to the counselor if they tell you that you can not take a class.  I was told that I could not take Alg in 7th grade, that it was not allowed.  After much (and I mean a ton of) persistance my parents and I convinced them to give me the Alg readiness test normally given to the 7th graders at the end of the year to place them for 8th grade.  They never told me what I scored but apparently it was high enough that I found myself in Alg 7th grade year...the first 7th grader in my district.  By the time my family was moving at the end of my 9th grade year they had 7 or 8 7th graders in Alg.  Persistance pays off!

I did take the test.  I aced it in half the time given to take it.  They still wouldn't let me.  They said it would be too hard to scedual it.  Oh brother.

I do see the beauty of math and I would love to be able to do AIME but as it stands I have been able to solve only one AIME problem.  That is parcially because though I haven't been able to study more math to be able to figure them out.  Now that I have aops to work from I have been able to solve one AIME problem.  My school doesn't even do the AMC's just one small state math contest that is patheticly easy if you can take tests.

[quote="Dark_Shadow"]Instead of blowing through the curriculum at lightspeed, you should slow down and learn to connect different parts of math until you can dissect AIME's with ease.[/quote]

I don't think you have to go [i]that[/i] far.  Some AIMEs are easy USAMO level.  But probably be able to do 7 or 8 at least.

Banish electronic calculators and use slide rules instead, or abacuses. At least they require some thought about how they work. Or, give kids an HP calculator and make them learn RPN, just for the hell of it.

Don't stick too closely to the book or standards. The standards should be turned into recommended guidelines instead.

Use the inch system as little as possible. Converting in the inch system can be a real pain.

I PMed you a big hint.  I have forgotten the more elegant method that I first used, but I have another just as good, but less elegant.

Hint: [hide]show that f(x+y) = f(x) + f(y) for positive x and y[/hide]

[quote="Nukular"]I wrote a similar proof when I took this USAMO as a freshman, and had never done proofs before. It got one point. The problem isn't quite that easy :).[/quote]

How did you find out your score on that problem?

I also answered #1 to the extent that I would expect partial credit of some sort.

I had two points in total.

First they should do more math in elementary school. I mean, all they basically teach you in those 5 (or 6) years is how to add, subtract, multiply, and divide integers, decimals, and fractions, and maybe some other stuff that is taught again anyway. Students can learn a lot more than than in those years. The problem is the teachers probably don't know any more math. (lol my 4th grade teacher didn't know order of operations). Elementary school should end earlier so that more advanced subjects can be taught by specialized more qualified (hopefully) teachers.

Another problem is that just going by the school curriculum, it would be nearly impossible for a person to be qualified enough to become a mathematician. Math learned in school comprises only a small part of my mathematical knowledge.

I think the concept that x is a variable, not a number, is explained well with functions. I don't like the fact that expressions are too often evaluated to decimal approximations. It may make sense to do this for the end result, especially in a "real world" problem, but things like ln(3) should not be approximated while solving the problem.

Counselors who hold people back are evil. Homeschooling, though, does not work for all people, for many reasons. One of the reasons I do not want to be homeschooled is that most of my day-to-day social interactions are either at school or school-related.

I wholeheartedly agree they should raise teachers' wages. This would make teaching more competitive and would result in better teachers. Currently most people who know a subject well go to a higher paying job. My math teacher once told us that when he applied to the school, they barely glanced at his college transcript.

Separating people into slower and faster classes is supposed to be done through honors and AP classes. I don't think this system is working well, though. You don't really have to be that smart to get into an honors class. There aren't enough divisions to make a noticeable difference. Unfortunately, an official separation would generate a lot of complaints. My old school, which was a 7-12 private school, had an unofficial separation into "smarter" and "dumber" classes. It was barely mentioned, and I only put it all together to figure out the system actually existed in the middle of the year after I had left the school.

Teaching why something works in addition to how it works is also important. Unfortunately many teachers do not have an adequate understanding of the subject to be able to do this.

The math textbooks should also be better. My precalculus textbook stated that when proving trig identities (like sin^4(x)-cos^4(x)=sin^2(x)-cos^2(x)), you had to transform the right side and the left side separately and that you could not do operations on both sides since you didn't know that the two sides were equal. This seems to be a very popular misconception in textbooks. In reality, as long as you don't multiply or divide both sides by anything that could be 0 it's OK.

Also, many math topics are taught over and over again. Not only is this a waste of time, but it also leads to another problem: The first time something's taught, the teacher thinks that it doesn't matter if the students don't understand it since they'll learn it again the next year. The second time it is taught the student believes that it is impossible to understand the topic. The teacher doesn't spend too much time on it since the topic was covered the previous year and will be covered again the next year.  The third time the topic is taught the student has lost all hope of ever understanding the topic and the teacher believes that everyone should have a sound understanding of the topic since it was covered twice before.

[quote="probability1.01"]And also, of course, we should pay teachers a ton more.[/quote]

Being a teacher, I like this idea  :-D

I also agree with most of you that the system should be more flexible. Placement in mathematics classes should be based on preparation rather than strictly by grade. Some students are prepared to take Algebra I in the 7th grade.

However, teachers have to be very careful about that. You have to remember that teachers have a lot more experience than you do. They have seen examples of kids who tried to accelerate through the curriculum too quickly and got themselves overwhelmed. So the only children who should be allowed to accelerate to the next class are the ones who have shown, not just the mathematical ability, but also the complete mastery of the previous material and the maturity and study skills to be successful in the next level class.

Some students have the mathematical ability to take a math class 3 of 4 years more advanced than their grade - but they have skipped or skimmed a lot of things in the intermediate geometry or precalculus curriculum. Not that there's anything wrong with that, because math topics don't have to be learned in a completely rigid sequential manner - as long as they eventually pick up on all of those things they missed before they graduate.

The biggest problem with education is the thing that no one ever seems to address, and that is discipline. Both behavioral and academic. Most of you probably go to very good schools, or are in advanced classes, and come from very good homes where this is not as much of an issue; but in many classrooms and many homes across the country this is a big issue. If you do not have discipline you cannot get down to the process of learning. Academic discipline really has to come from the parents, and behavioral discipline also has to come from both the parents and the school administration.

I agree with Yasha that repetition is a big problem in secondary school mathematics. Part of that is done because students move from school to school and don't always have the same background; but at some point spending the first month of a class reviewing topics that they should already know is a bit excessive.

I disagree with him however that elementary school mathematics should be more advanced. A lot of elementary school mathematics is repetition, and that's important developmentally at that age. They do not have the ability for abstraction at that age that older students do. This is the primary reason why algebra is not taught universally until middle school. It's not that elementary school students haven't learned the basics required for algebra; developmentally they just aren't ready for the concepts of mathematical abstraction. This is why teaching algebra even to 7th graders has to be done very carefully. If they don't really get it at that age and try to move on then you run the risk of them left behind for the rest of their careers.

Instead of the standard middle school curriculum for middle school students I advocate the MEGSSS Program [url]http://www.megsss.org/[/url] for gifted students. It's a program that uses the Elements of Mathematics books and curriculum. The books of the EM curriculum are:

Book 0: Intuitive Background
Chapter 1: Operational Systems
Chapter 2: The Integers
Chapter 3: Sets, Subsets & Operations with Sets
Chapter 4: Ordered n-Tuples
Chapter 5: Mappings
Chapter 6: The Rational Numbers
Chapter 7: Decimals and an Application of the Rational Numbers
Chapter 8: An Introduction to Probability
Chapter 9: An Introduction to Number Theory
Chapter 10: Algebra in Operational Systems
Chapter 11: Geometry: Incidence and Isometries
Chapter 12: Topics in Probability and Statistics
Chapter 13: Algebra of Real Functions
Chapter 14: Geometry: Similitudes, Co-ordinates, & Trigonometry
Chapter 15: Topics in Number Theory
Chapter 16: Introduction to Computer Programming

Book 1: Introductory Logic

Book 2: Logic and Sets

When I was in the program I believe we did Book 0, Chapters 1-9 in 6th and 7th grade, and Books 1 and 2 in 8th grade. Integrated with these books were traditional lessons in algebra and geometry in the 7th and 8th grades respectively. Mostly through problem solving books, Mathcounts exercises, and SAT prep books.

In the 9th grade we went back to using solely traditional textbooks for Algebra II, and then a tradition textbook (Brown and Robbins - an exceptional book) for Pre-Calculus in the 10th grade. Followed by 2 years of calculus in 11th and 12th grade.

The program has shown great success in the schools where it has been used. The Chapters of Book 0 are actually individual books. I have no idea of the quality of Chapters 10-16, but Chapters 1-9 and Books 1 and 2 are exceptional for gifted middle school students, and provide a solid foundation. Students not only learn proofs, but learn formal logic in books 1 and 2. All books are written in a style easy to read for gifted middle school students.[/url]

[quote="philomath"](our school uses CMP curriculum for non-AP math...I hate it).[/quote]

Actually, your District uses the CPM (College Preparatory Mathematics) series.  "CMP" is used by the slightly larger School District to your north.  [These are easy to confuse.  Just want to be sure you are dogging the right curriculum.  :)]

[quote="gauss202"]I disagree with him however that elementary school mathematics should be more advanced. A lot of elementary school mathematics is repetition, and that's important developmentally at that age. They do not have the ability for abstraction at that age that older students do. This is the primary reason why algebra is not taught universally until middle school. It's not that elementary school students haven't learned the basics required for algebra; developmentally they just aren't ready for the concepts of mathematical abstraction. This is why teaching algebra even to 7th graders has to be done very carefully. If they don't really get it at that age and try to move on then you run the risk of them left behind for the rest of their careers.[/quote]

Just to clarify, I don't think that graphing rational functions should be a standard part of a fourth grade curriculum. Most elementary school students would get bored with such in-depth discussion of a topic. But I do think more of the basic math concepts should be introduced earlier. Coming out of elementary school, students should know what a variable is and what an exponent is. They should be able to solve problems like "Bob is 5 years older than Bill and Fred is 3 years older than Bob. The sum of their ages is 28. How old is Bill?" They should know what distributivity is.

Looking at a Russian 3rd grade textbook, there are the following problems:

[quote]Solve the equation 7*(627-x)=490.

5 small boxes and 2 big boxes contain 54 colored pencils. 3 small boxes and 2 big boxes contain 42 colored pencils. How many pencils are there in one small box? How many pencils are there in one big box?

Bilbo's home has 28 rooms with windows, which is two times less than the number of windowless rooms. How many rooms are there in Bilbo's home?[/quote]

If the average Russian third grade student can learn how to solve these problems then so can the average American third-grader.

[quote="probability1.01"]And so [i]should[/i] the American third-grader.  But it actually takes an honors math 8th or 9th grader to do that.[/quote]

Surprisingly, there are some people in 6th or 7th grade that do that in MathCounts or even AMCs.  Sadly, there aren't many.

They can't go much faster in math in elementry because I know that I nearly lost my respect for math and hated it when I changed to a much faster school.  It was basicly like skipping a few grades without haveing gone over any of the material that was previously covered.  I was doing things in 4th grade that I can still show to my fellow math honors classmates who just finished prealgebra with me and it confuses them.  I have found since then that what we were doing in 4th grade and I believe this was totally wrong was quadratic formulas and problems of the sort.  How could they expect us to learn it that early???  It has taken me 3 months to straighten out what I messed up so bad in learning it that early.  I wholeheartedly believe that I saved myself switching back out of that school into public because I know love math and it comes easily to me now and I can now move faster that I have the basics.  It would be a disastor to teach faster and more advanced curriculum like my previous school did in the average 4th grade classroom.The school I went to was one of the best.  We had entrance tests to get in and the people in it were probally like a lot of people here.  They could go faster but that school went too fast.  One of my best friends that started there with me loved math going in but as she contiued after I left she started hating it.  She hates math to this day.  I can't help thinking that thats what would happen if the taught more advanced earlier.  Though it is true that some 7th graders could do algebra one.  I took the test for algebra early and aced it perfectly.  They still wouldn't let me in the class though so I did algebra the hard way.  I worked through it with the help of the algebra teacher who thought I was good enough to be in her class.  I thought my school might let me just take the summer class and get it over with but they are requiring me to take the full class next year even though they know that I have worked through it and completly understand the concepts.

[quote="Yasha"]What I was saying wasn't that elementaary schools should offer more depth or go "faster". Rather, they should offer a greater breadth of mathematical topics. I think most elementary school students would understand the concept of a contrapositive or truth tables. They could most likely understand what a logarithm is. All the properties of logarithms would, however, be too much depth.[/quote]

I would have understood (or did understand) that, but although I can't speak for everyone, I think the general population couldn't understand all of it.  What should be covered in elementary school is basic logic (reasoning skills, or at least encourage that type of thinking).  I know that I like math because I was intrigued by questions about logic I was asked when I was 3 or 4.  Only once you have mastered basic logic can you hope to deeply understand other things.

I have a few questions though. First, are Russian grades the same as American grades - meaning do they start 1st grade at the same age that we do? How do you know the book is a textbook for Russian 3rd graders? And how do you know that it's used by typical 3rd grade students there?

I really don't believe that many 3rd grade students (8 and 9 years olds) are prepared to learn algebra. There are always exceptions by the exceptionally gifted - but I don't believe there are many students like that at an average elementary school from year to year.

You have to be extremely careful with young children when it comes to accelerating them. Bad experiences with things they feel are too difficult can be almost as damaging as understimulating them with things that are too easy.

Besides, most American schools do not start tracking students into advanced and regular classes until at least the 2nd, 3rd, or sometimes 4th grade. There are important reasons for doing that.

One thing that could maybe be improved about elementary school education is to have specialized teachers for the different subjects. A lot of elementary schools have teachers who teach all, or many, subjects. They usually have degrees in elementary education - which is a lot like child psychology, but not in specific subjects like math or history, etc. I don't know enough about child psychology to know what are the pros and cons of such a change, but it may be that having specialized teachers for each subject at least by the 3rd or 4th grade could be beneficial.

Dealing with young children though, you have to be very very careful. Young children's' psychology and mental processes are very different from even middle school aged children.

[quote="gauss202"]I have a few questions though. First, are Russian grades the same as American grades - meaning do they start 1st grade at the same age that we do? How do you know the book is a textbook for Russian 3rd graders? And how do you know that it's used by typical 3rd grade students there?[/quote]

The book is in fact for 3rd graders; it says so on the cover. I'm pretty sure it's a typical textbook; I don't see anything that would say otherwise, but I can't be completely sure. The book is fairly recent: 1998, so I think the third grade there would actually correspond to fourth grade in the US. I should have mentioned this, but I don't think this fact changes the validity of my point much.

The colored pencil problem I quoted was one of the more difficult ones in the book, but the others were fairly standard.

I know that the grades thing can be very confusing from country to country though. Like in some schools in Mexico 1st grade, 2nd grade, and 3rd grade actually refer to the 3 grades of high school.

That would be very interesting though if it were a standard 3rd, or even 4th, grade book used in Russia. I have a few friends who were educated there. I will ask them what their elementary school education was like.


What is the value of the sum \[ (37037\cdot 3)\plus{}(37037\cdot 6)\plus{}(37037\cdot 9)?\]

Instead of multiplying and adding all the terms, we notice each one has a common factor: $ 37037.$ When this is factored out, we are left with $ 37037(3\plus{}6\plus{}9)\equal{}37037\times18\equal{}\boxed{666666}.$


Could anyone provide me with some online resources for practising logic games questions? I have an exam that involves this very soon and I was just told that there might be a big emphasis on these type of questions. My preparation course didn't cover enough of these and I'm really terrible at these questions especially in time constraints. Any help would be appreciated!

What I mean by logic games are questions like this:

[hide]A particular seafood restaurant serves dinner Tuesday through Sunday. The restaurant is closed on Monday. Five entrees—snapper, halibut, lobster, mahi mahi, and tuna—are served each week according to the following restrictions:

    * Halibut is served on three days each week, but never on Friday.
    * Lobster is served on one day each week.
    * Mahi mahi is served on three days each week, but never on consecutive days.
    * Halibut and snapper are both served on Saturday and Sunday.
    * Tuna is served five days each week.
    * No more than three different entrees are served on any given day.

On which of the following pairs of days could the restaurant's menu of entrees be identical?
.
(A) Friday and Sunday
(B) Tuesday and Wednesday
(C) Saturday and Sunday
(D) Wednesday and Friday
(E) Thursday and Friday

#
Which of the following is a complete and accurate list of the days on which halibut and lobster may both be served?
.
(A) Tuesday, Thursday
(B) Tuesday, Wednesday, Thursday
(C) Monday, Tuesday, Wednesday
(D) Tuesday, Wednesday, Thursday, Friday
(E) Tuesday, Wednesday, Thursday, Saturday

#
If mahi mahi is served on Saturday, it could be true that
.
(A) snapper and mahi mahi are both served on Sunday
(B) snapper and halibut are both served on Tuesday
(C) lobster and halibut are both served on Thursday
(D) tuna and snapper are both served on Saturday
(E) lobster and snapper are both served on Friday

#
Which of the following statements provides sufficient information to determine on which three days halibut is served?

(A) Mahi mahi and lobster are served on the same day.
(B) Lobster and snapper are both served on Tuesday.
(C) Tuna is served on Saturday, and lobster is served on Tuesday.
(D) Mahi mahi is served on Saturday, and snapper is served on all but one of the six days.
(E) Tuna is served on Sunday, and snapper is served on Tuesday and Thursday.[/hide]

here are some good sites:
[url]http://www.eduplace.com/math/brain/[/url]
[url]http://mathforum.org/library/problems/sets/dm_logic.html[/url]
[url]http://perplexus.info/category/2/[/url]
hope this helps  :lol:


geometry
incenter
ratio
geometry proposed

Proof, that  in scalene (non-isosceles) triangle $ IK\parallel NJ$,  where:
$ I$ – incenter $ (a: b: c)$;
$ K$ – Lemuan point $ (a^2 : b^2 : c^2 )$;
$ N$ - Nagel point $ \left( {1/( \minus{} a \plus{} b \plus{} c): 1/(a \minus{} b \plus{} c): 1/(a \plus{} b \minus{} c)} \right)$;
$ J$ - Jergonne point $ (( \minus{} a \plus{} b \plus{} c): (a \minus{} b \plus{} c): (a \plus{} b \minus{} c))$.

Call $ M$ the Mittenpunkt point of ABC and $ G$ the centroid. So, as well know, $ M$, $ I$ and $ K$ are collinear. But $ M$ is the Gergonne point of the medial triangle and $ I$ is the Nagel point of the medial triangle. So with a dilatation with center $ G$ and ratio $ \minus{} \frac{1}{2}$, $ J$ go in $ M$ and $ L$ go in $ I$.
So $ IM \parallel NJ$ and $ IK \parallel NJ$.

Let $ a$, $ b$, $ c$ be the side lengths of the given triangle and $ s$ be its semi-perimeter. We have $ I$, $ K$, $ J$, and $ N$ are the centers of mass of system of points $ A$, $ B$, $ C$ with masses $ \{a, b, c\}$, $ \{ a^2, b^2, c^2\}$, $ \{(s - b)(s - c), (s - c)(s - a), (s - a)(s - b)\}$, $ \{s - a, s - b, s - c\}$, respectively. Setting $ x = s - a$, $ y = s - b$ and $ z = s - c$. Consider an arbitrary point $ O$ on the plane. We have
\begin{eqnarray*}\overrightarrow{IK} = \overrightarrow{OK} - \overrightarrow{OI} & = & \frac {1}{\sum a^2}\sum a^2\overrightarrow{OA} + \frac {1}{\sum a}\sum a\overrightarrow{OA} \\
& = & \frac {1}{\left(\sum a^2\right)\left(\sum a\right)}\sum a\left[a\left(b + c\right) - \left(b^2 + c^2\right)\right]\overrightarrow{OA} \\
\overrightarrow{NJ} = \overrightarrow{OJ} - \overrightarrow{ON} & = & \frac {1}{\sum yz}\sum yz\overrightarrow{OA} - \frac {1}{\sum x}\sum x\overrightarrow{OA} \\
& = & \frac {1}{\left(\sum yz\right)\left(\sum x\right)}\sum\left(y + z\right)\left(yz - x^2\right)\overrightarrow{OA}\end{eqnarray*}
On the other hand, notice that $ a=y+z$, $ b=z+x$ and $ c=x+y$, so we can easily to get that $ a\left[a\left(b+c\right)-\left(b^2+c^2\right)\right]=2\left(y+z\right)\left(yz-x^2\right)$. Hence, we conclude that $ IK$ is parallel to $ NJ$, as desired.


calculus
derivative
integration
function
real analysis
real analysis unsolved

How will u show that the Radon Nikodym derivative is unique modulo set of measure zero?

Thanks.

Proof sketch: Let $ f_1,f_2$ be two Radon Nikodym derivatives of some measure $ \nu$ with respect to $ \mu$, then for each $ \mu$-measurable set $ A$ we have $ \nu(A) \equal{} \int_A{f_1d\mu} \equal{} \int_A{f_2 d\mu}$ and therefore $ \int_A{f_1 \minus{} f_2d\mu} \equal{} 0$.  This implies, in particular, that $ \mu \left(\{f_1 \minus{} f_2 > \epsilon \}\right) \equal{} \mu \left(\{f_1 \minus{} f_2 < \minus{} \epsilon \}\right) \equal{} 0$ (recall that the difference of measurable functions is measurable) for each $ \epsilon>0$ and hence $ \mu \left( \{f_1\neq f_2\} \right) \equal{} 0$.


hi,

 Find the largest power of $ 2$ that divides $ \lfloor(3 \plus{} \sqrt{11})^{2n\plus{}1}\rfloor$.

[hide="Solution"]
Since $ |(3\minus{}\sqrt{11})|^k<1$ for positive $ k$, and $ (3\plus{}\sqrt{11})^{2n\plus{}1}\plus{}(3\minus{}\sqrt{11})^{2n\plus{}1}$ is an integer for all nonnegative $ n$, we conclude that $ \lfloor(3\plus{}\sqrt{11})^{2n\plus{}1}\rfloor\equal{}(3\plus{}\sqrt{11})^{2n\plus{}1}\plus{}(3\minus{}\sqrt{11})^{2n\plus{}1}\equal{}x_n$. Note that $ ((3\plus{}\sqrt{11})^2\plus{}(3\minus{}\sqrt{11})^2)x^{n\minus{}1}\equal{}40x_{n\minus{}1}\equal{}x_n\plus{}4x_{n\minus{}2}\implies x_n\equal{}40x_{n\minus{}1}\minus{}4x_{n\minus{}2}$. With this, we can use induction to prove that the largest power of $ 2$ that divides $ x_n$ is $ \boxed{n\plus{}1}$.
[/hide]


A certain circle's area is $ x$ square units, and its circumference is $ y$ units. The value of $ x \plus{} y$ is $ 80\pi$. What is the radius of the circle, in units?


[url=http://www.youtube.com/watch?v=Ee18vXyLBMM&search=crazy%20german%20guy]This is what happens...[/url]

lol doesnt really have anything to do with anything, just some kid who goes crazy cuz his unreal tournament is loading too slow :P

:rotfl:  :rotfl:
a bit of bad language though

Sorry I can't watch videos at home. I'll go over to the broadband cyber cafe later and try it.

that was the funniest thing since..well...ever..

[quote="pianoforte"]oh my...that was, uh,...uh

gosh, I feel sorry for that keyboard[/quote]

So do all of us.

Wow...
It's people like that you really want to beat up...like ACTUALLY want to beat up...
I first would want to meet that person though... :D
waAAAAY too much black black...and sugar and caffeine and stimulants.

[quote="furious"]i sure hope he never gets a headset  :o[/quote]

don't we all?
if that guy gets a headset, ...
just wow

I have seen it finally.
Noticed the boy in the middle of it all? He didn't strike me as the epitome of patience and serenity.

That just happened to me.... I was tearing apart my cable thingy.
Then, i ran adaware 4 times and rebooted.


since i am not getting a response from the intermediate board, then i am posting my question here, sorry if it is too easy
given a line and 2 points on the same side of the line, construct a circle that goes through both points and is tangent to the line

A not-so-pretty way to do it...

Let the points be A, B; let the line be l.  Let AB intersect l at C (if AB || l, then it is trivial).  Let the midpoint of AB be D.  Construct somewhere a right triangle with hypotenuse |CD| and one leg |AD|.  Take the length of the other leg, and mark a point that length away from C (either direction).  That is your point of tangency.  Then it's easy to construct a circumcircle.


Putnam
probability
counting
distinguishability
factorial
function
binomial coefficients

Let $ k$ be a positive integer. Suppose that the integers $ 1,2,3,\dots,3k \plus{} 1$ are written down in random order. What is the probability that at no time during this process, the sum of the integers that have been written up to that time is a positive integer divisible by $ 3$ ? Your answer should be in closed form, but may include factorials.

We divide the numbers up into equivalence classes mod 3. There are k in the equivalence class with 0, there are k+1 in the equivalence class with 1, and k in the equivalence class with 2.

Let a_n be the mod 3 sum of the numbers already written after n numbers have been written. Thus, $ a_n$ is a sequence of 1's and 2's. So $ a_n$ alternates between 1s and 2s. We know that $ a_{3k\plus{}1}$ is equal to 1.

If we start with $ a_1$ = 2, then a_n goes from 1 to 2 k+1 times and from 2 to 1 k-1 times, which is impossible since you will run out numbers in the equivalence class of 2.
If we start with $ a_1$ =  1, then a_n goes from 1 to 2 k times and it goes from 2 to 1 k times, which works.

We can specify each sequence $ {a_n}$ by the number of moves between a change from 1 to 2 or from 2 to 1. These 2k numbers must add up to less than or equal to k since we have only k numbers in the equivalence class of 0.

If the sum of these numbers is 0 then there is one possibility. If the sum is 1, then there are 2k possibilities. If the sum is 2, then there are $ \binom{2k}{2} \plus{}\binom{2k}{1}$ possibilities. If the sum is 3, then there are $ \binom{2k}{3} \plus{}2\binom{2k}{2} \plus{} \binom{2k}{1}$.
If the sum is four there are
$ \binom{2k}{4} \plus{}3\binom{2k}{3} \plus{} 3\binom{2k}{2}\plus{}\binom{2k}{1}$ possibilities.

Noticing the pattern, we can write that the total number of $ {a_n}$ sequences that works is:

$ 1\plus{}\sum_{i\equal{}1}^{2k}\sum_{j\equal{}0}^{i\minus{}1}\binom{2k}{j\plus{}1}\binom{i\minus{}1}{j}$

So, we multiply this by k!k!(k+1)! to account for the sizes of the equivalence factors and divide by the total number of ways that you can write down that numbers to get:

$ \left(1\plus{}\sum_{i\equal{}1}^{2k}\sum_{j\equal{}0}^{i\minus{}1}\binom{2k}{j\plus{}1}\binom{i\minus{}1}{j}\right)\frac{k!k!(k\plus{}1)!}{(3k\plus{}1)!}$
Anyone know how to simplify that?

Here is a faster solution.

We first find the number of ways to arrange $ k\plus{}1$ 1’s, $ k$ 0’s, and $ k$ -1’s so that the sum of the first $ i$ terms, $ 1\le i\le k\plus{}1$, is not divisible by 3. Note that the 0’s have no effect on the modulo 3 residue of the sum, so they can go anywhere except at the beginning. Thus, we disregard them for now, and put them in later.

Say our sequence starts with a -1. Then, we will go -1,-1,1,-1,1,-1,1, etc., but we will always end up with more -1’s than 1’s, and we need one more 1 than a -1, so this will not work. Thus, we will need to start with a 1. The sequence will go 1,1,-1,1,-1,1, etc., so we will end with a -1.

Now, we have a string of $ 2k\plus{}1$ 1’s and -1’s, and since the 0’s can go in any ‘slot’ other than at the beginning, we have $ 2k\plus{}1$ urns and $ k$ balls, and there are $ \dbinom{3k}{k}$ ways to put in the zeroes in. Since the sequence of 1’s and -1’s is forced, there are $ \dbinom{3k}{k}$ ways to arrange the entire sequence.

Now, we must put $ 1,4,7,\ldots,3k\plus{}1$ in the places of the 1’s, $ 2,5,8,\ldots,3k\minus{}1$ in the places of the -1’s, and $ 3,6,9,\ldots,3k$ in the places of the 0’s. There are clearly $ (k\plus{}1)!k!k!$ ways to do this. Also, there are $ (3k\plus{}1)!$ possible arrangements of the original $ 3k\plus{}1$ numbers.

Thus, the desired probability is $ \dbinom{3k}{k}\dfrac{(k\plus{}1)!k!k!}{(3k\plus{}1)!}\equal{}\dfrac{3k!(k\plus{}1)!k!k!}{k!(2k)!(3k\plus{}1)!}\equal{}\boxed{\dfrac{(k\plus{}1)!k!}{(2k)!(3k\plus{}1)}}$.

So your summation is equal to just $ \dbinom{3k}{k}$.

Think of all of the numbers as being $ 1,0,$ or $ \minus{} 1.$ We have $ k \plus{} 1$ copies of $ 1$ and $ k$ copies each of $ 0$ and $ \minus{} 1.$

The final sum is going to be $ 1.$ To get there, not changing the sum by more than one each time, and avoiding multiples of $ 3,$ then all intermediate sums must be $ 1$ or $ 2.$

That means that the very first number chosen must be $ 1.$ The probability of this happening is $ \frac {k \plus{} 1}{3k \plus{} 1}.$

After that, it doesn't matter when the $ 0$s are scheduled, but the $ 1$s and $ \minus{} 1$s must go in a strict $ 1, \minus{} 1,1, \minus{} 1,\dots,1, \minus{} 1$ order.

The probability that you get a $ 1$ before a $ \minus{} 1$ is exactly $ \frac12.$

After that, the probability you get a $ \minus{} 1$ before a $ 1$ is $ \frac {k}{2k \minus{} 1}.$

The overall probability now looks like:

$ \frac {k\plus{}1}{3k \plus{} 1}\cdot\frac12\cdot\frac {k}{2k \minus{} 1}\cdot\frac12 \frac {k \minus{} 1}{2k \minus{} 3}\cdots\frac12\frac11$

We can write this as $ \frac {(k \plus{} 1)!(2^kk!)}{(3k \plus{} 1)2^k(2k)!} \equal{} \frac {(k \plus{} 1)!k!}{(3k \plus{} 1)(2k)!}.$

[quote="CatalystOfNostalgia"]
Now, we have a string of $ 2k \plus{} 1$ 1’s and -1’s, and since the 0’s can go in any ‘slot’ other than at the beginning, we have $ 2k \plus{} 1$ urns and $ k$ balls, and there are $ \dbinom{3k}{k}$ ways to put in the zeroes in. [/quote]

Can you explain this? What do the urns and the balls mean? I've never seen those terms before in this context?

Basically, the problem is to put $ b$ balls into $ u$ urns. The balls are indistinguishable, and the urns are distinguishable.

Consider a string of $ b$ *'s, and $ u\minus{}1$ dividers, |, such as ||****|***|, which will correspond to 0,0,4,3,0 balls in 5 different urns. It is not hard to see that we have a bijection between this and the original problem. Now, there are $ u\plus{}b\minus{}1$ objects, and we choose which $ b$ are *'s, so the total number of ways to do this is $ \dbinom{u\plus{}b\minus{}1}{b}$.

If my explanation is bad (which it surely is), you can google it.

[quote="Kent Merryfield"]
That means that the very first number chosen must be $ 1.$
[/quote]

Darn. Through trial and error I came to the realization that you couldn't start with some numbers. Why didn't I realize that it was precisely the ones that were congruent to $ 2 \mod 3$? Ah!

[quote="cats..."]is an answer containing binomial coefficients considered "closed form" here?  i have the correct expression, i just didnt bother simplifying it into strictly terms containing factorials[/quote]

one never wants to speak for the (seemingly capricious) putnam graders, but I think that this should be fine.

I answered it like this:

Divide the list of integers into three sets
{1, 4, 7,..., 3k+1} (denoted as (1) (I used a 1 inside of a circle in the answer, but cannot type it here))
{2, 5, 8,..., 3k-1} (denoted as (2))
{3, 6, 9,..., 3k} (denoted as (3))

The first number written down cannot be a member of (3), otherwise it would be a multiple of 3.
Since adding a (3) would not change a number that was not divisible by 3 to a number that is, let's ignore (3) for the moment.

If the first number is from (2), the next number is forced to be a (2) also, since it would be divisible by 3 otherwise (I made a footnote stating that I am still ignoring (3)). The next number would then be from (1), and the next is (2). The list would go on like this.

(2)(2)(1)(2)(1)(2)(1)(2)...

However, at the end of the list, when all k terms in (2) are used, there would still be elements from (1) left. Therefore the first number cannot be a (2).

The only possibility left is if the first number is a (1). the number followed would be also a (1), because it would be a multiple of 3 if it is a (2) (made another footnote stating that I am still ignoring (3)). The number after that can only be a (2), then (1), and so on so forth. The list of number is as followed:

(1)(1)(2)(1)(2)(1)(2)(1)...

This time, the list uses all of (1) and (2).

The number of ways to construct the list as above is k!(k+1)!.

Now let's put the (3) in the list. The first element of 3 that gets written down can go into any slot except before the first number. Thus, there are 2k+1 ways to write it into the list. After that, next number to be put into the list would have 2k+2 slot it can fit into. Therefore, there are (3k)!/(2k)!ways to fit (3) into the list, and the number of ways to construct the list is k!(k+1)!(3k)!/(2k)!

The possibility is
k!(k+1)!(3k)!/(2k)!(3k+1)!
=k!(k+1)!/(2k)!(3k+1)

does anyone know how to generalize this? That is, at no time is the sum a multiple of $ a$ for $ a>3$

The proofs posted above relied on a great deal of rigidity to the problem. As soon as you put in an $ a>3,$ you lose most of that rigidity and render those proofs useless. To me that looks like not so much a generalization as a completely new (and likely much harder) problem.

Does anyone know a solution involving generating functions? :D

The answer is $\frac{k!(k+1)!}{(3k+1)(2k)!}$.

(Note the notion of success is used to mean meeting the condition of the problem statement.)

Observe that $k$ terms of the sequences are $0$ modulo $3$, $k+1$ terms of the sequence are $1$ modulo $3$, and $k$ terms are $2$ modulo $3$. Suppose we are given the residues of an arbitrary sequence modulo $3$. Call such a sequence its [i]threequence[/i]. Any threequence admits $(k!)^2(k+1)!$ possible sequences. Since each sequence is considered equally likely, each threequence must also be equally likely.  Also, if a given sequence succeeds, then all other sequences with the same threequence succeeds. There are $\tbinom{3k+1}{k,k,k+1}$ threequences, so it remains to calculate how many threequences succeed.

Take an arbitrary threequence. Now, take its subsequence obtained by deleting the $0$ terms. From here on, when we say partial sum, we mean partial sum modulo $3$. If the threequence succeeds, none of the partial sums of this subsequence can be $0$, so these partial sums must all be $1$ or $2$. But none of the terms of the subsequence are $0$, so the partial sums must alternate between $1$ and $2$. Say, for sake of contradiction, the subsequence starts with $2$. Then the first partial sum is $2$. We know the last partial sum, the sum itself, is $1$. Given the alternation condition, there must be an even number of terms in the subsequence. But we know there are $k+(k+1)=2k+1$ terms in the subsequence, so this is impossible. So the subsequence must start with $1$. If a partial sum is $1$, since the partial sums alternate, the next term must be $1$ so that the next partial sum is $2$. Similarly, if a partial sum is $2$, the next term must be $2$. This subsequence has $k+1$ partial sums that are $1$, and $k$ partial sums that are $2$, so it needs $k$ terms that are $2$, $k$ terms that are $1$, and an additional $1$ to start the subsequence, so $k+1$ terms that are $1$ in total. We have exactly the right number of terms for this, and the construction of the subsequence is fixed by these rules. Following this finite list of steps, we construct this subsequence. As long as the $0$ terms of the threequence don't lead the threequence, the original threequence succeeds with this subsequence. The number of ways to construct a threequence with such a subsequence and no leading $0$ terms is $\tbinom{3k}{k}$ by the well-known stars and bars bijection.

So the probability is $$\frac{\binom{3k}{k}}{\binom{3k+1}{k,k,k+1}}=\frac{k!(k+1)!}{(3k+1)(2k)!},$$ as desired.

[hide=Remark]
Here are the first handful of probabilities, approximated to three decimal places.
\begin{tabular}{c|c|c|c|c|c|c}
$k$ & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
$P(k)$ & 0.250 & 0.071 & 0.020 & 0.005 & 0.001 & 0.000
\end{tabular}


On each tile in a set of dominoes, each integer 0 through 14 is paired exactly once with each integer 0 through 14 . One number of the pair is written on one half of the domino, and the other number is written on the other half. If one domino is selected at random from all the dominoes in this set, what is the probability that the same number will be on both halves? Express your answer as a common fraction in lowest terms.

I have no idea what to do with this..Help.

so find favorable over total. The favorable is obviously 15, or 0/0,1/1,2/2.....14/14.

The total is a bit trickier. So Choose the first number, 14. It can be grouped with 15 numbers, including itself. Choose the next number, 13. It can be grouped with 14 numbers, all but 14. Choose the next number, 12. It can be grouped with 13 numbers, all but 14 and 15. See a pattern? it is 15+14+13......+1.

so (15*16)/2=120

so 15/120 or 1/8


algebra
polynomial
number theory
prime numbers
greatest common divisor
geometric sequence

Find all prime numbers $p$ such that the equation $x^3 - 24x + p = 0$ has an integer solution.

Find the number $a$ so that the polynomials $x^3 - 2x^2 - x + 2$ and $x^3 + ax - 2$ have a non-trivial greatest common divisor.

I *think* this is pre-olympiad. If this is too easy and I'm just too dumb, then put them in intermediate I suppose. :)

edit: I know I am posting a lot of problems now, but here is another one that now bothers me the most:
Let $n$ be a natural number. Prove that  $(x^{n-1} + x^{n-2} + ... + x + 1) | (x^{3n} - x^{2n} + x^n - 1)$
Both sides are geometric sequences, but I don't know if that is the right track... I mean, they are supposed to divide each other as polynomials and once you start involving geometric sums you're not dealing with polynomial expressions anymore.

[hide="Sol to the easy no.1 if it is right"]
$x^3-24x+p=0$
$x(x^2-24)=-p$
If p is prime and positive then $x$ or $(x^2-24)$ should be negative or $1$
Solving,
$x$ can be $-1$ or $1$ so substituting,
So,
$p=23$
Nothing works for the 2nd pair so,
$p=23$
[/hide]
Correct?

For the 2nd one what do you mean by "non-trivial" roots?

[quote="seamusoboyle"]i would imagine for the geometric series you would use roots of unity[/quote]
Yes, it becomes pretty trivial when you think of this. I feel dumb. Thanks!

riddler:
For no 1 you didn't get all solutions, but I am too tired to look at your solution right now. I'll do it tomorrow.

Non-trivial greatest common divisor means a divisor that isn't constant, I would suppose.

[quote="riddler"]$x^3-24x+p=0$
$x(x^2-24)=-p$
If p is prime and positive then $x$ or $(x^2-24)$ should be negative or $1$
Solving,
$x$ can be $-1$ or $1$ so substituting,
So,
$p=23$
Nothing works for the 2nd pair so,
$p=23$[/quote]
Another possibility:

$x^2-24=1\Longleftrightarrow x=5,-5$
$x=-5$ gives $-p=-5\Longleftrightarrow p=\boxed{5}$

So I think the answer would be: $p=5, 23$

(EDITED :oops: )

and for the last one, doing the sum of a geometric sequence, and factoring the second one, you get your answer since the denominator cancels out

[quote="Altheman"]1 is not a prime number,

a prime number is defined as a number n with only 2 divisors 1 and n

but with 1, there is only 1 divisor, that being 1[/quote]
He only made a mistake at the end. He surely realizes that $-p = -5$ has the solution $p = 5$. So it is the right solution.

I do see that the denominator cancels out but how does that show anything about divisibility? We have only defined it for polynomials, not other expressions... So please explain.

The solutions I see for the last one is either what seamusoboyle suggested, or something I thought of today:
The RHS can be factored as $(x^{2n} + 1)(x^n - 1) =$/using geometric sequence formula in reverse/ $= (x^{2n} + 1)(x - 1)(1 + x + x^2 + \cdots + x^{n-1})$. So it is obvious. Maybe this is what you meant?

Now for the second one :)

[quote="Kalle"][quote="Altheman"]1 is not a prime number,

a prime number is defined as a number n with only 2 divisors 1 and n

but with 1, there is only 1 divisor, that being 1[/quote]
He only made a mistake at the end. He surely realizes that $-p = -5$ has the solution $p = 5$. So it is the right solution.[/quote]
Yeah, I was kinda rushing. What a stupid mistake... :blush:

Now edited.


Given a collection of convex sets on the plane. The area of their union equals S. Does there exist a constant C such that it is always possible to find a subcollection $A_{1},\,A_{2},\,\dots,A_{n}$ such that

(i) sum of area of $A_{i}\, (i=1..n)$ does not exceed $C\cdot S$

(ii) there exist sets $B_{i}$ so that each $B_{i}$ is homothetic to corresponding $A_{i}$ with coefficient $C$ and some centre inside $A_{i}$ and $\cup B_{i}$ contains a union of the sets of the initial collection

?

The answer is "No". To see it, run the classical Besicovich construction of the set of small measure containing a unit interval in every direction. It'll give you a family of narrow triangles of height $1$ with bases on the line whose total area is not much greater than $(1-a)^{2n}+na^{2}$ but whose bases cover an interval of length about $(1-a)^{n}$. Now choose an integer $n>0$ and a real number $a>0$ such that the first of these numbers is much less than the second. The rest should be clear.

Thank you! But where is it written about Besicovich construction?

By the way, in the case, which I actually need, when these sets are cups of given volume of some given convex body, the statement holds.

I don't know where on the web is a good place where you can [b]read[/b] about it, but here is a web page where you can [b]see[/b] it ($\alpha=1-a$): [url]http://www.math.ucla.edu/~tao/java/Besicovitch.html[/url]


calculus
integration
function
limit
trigonometry
logarithms
calculus computations

how do I show that the integral from 0 to infinity of 1/sqrt(1+x^3) dx exists? do I find an upper bound function and use the "integral comparison test"?

 I'm not really sure how to use the integral comparison test, does it have to be an upper bound for the function in a specific way? I don't think I can just use any kind of function I want..

thanks

The integral is $ \int_0^{\infty}\frac1{\sqrt{1\plus{}x^3}}\,dx.$

Let's step through this carefully, asking several questions as we go along. One thing we notice is that it's highly unlikely that this function has a closed-form antiderivative, so we forget about doing things by writing exact formulas.

1. For how many reasons is this integral improper? The answer is one: it's improper because it's an integral to $ \infty,$ but there are no other problems anywhere. In particular, for any $ 0<a<\infty,$ the integral of this function from $ 0$ to $ a$ is defined - it's the integral of a continuous function over a closed bounded interval.

2. Is the integrand always positive, always negative, or sometimes positive and sometimes negative? The answer here is always positive. What that does for us is activate the positive/monotone alternative: either this integral converges or it equals $ \plus{}\infty,$ and there are no other choices. (The original version of this theorem is the "monotone sequence alternative": a non-decreasing sequence either converges or tends to $ \plus{}\infty.$ This is equivalent to the completeness of the real numbers.) It's this alternative that justifies comparison tests.

3. What does the integrand "look like" for large $ x?$ In $ 1\plus{}x^3,$ the $ x^3$ matters and the $ 1$ doesn't matter, so we say it looks like $ \frac1{x^{3/2}}.$

At this point, I already know the answer: this integral converges by comparison to a "$ p$ integral" with $ p\equal{}\frac32>1.$ Now to say this a little more formally:

For $ x\ge 1,$ we have $ \frac1{\sqrt{1\plus{}x^3}}<\frac1{x^{3/2}}.$ Hence,

$ \int_1^{\infty}\frac1{\sqrt{1\plus{}x^3}}\,dx\le\int_1^{\infty}x^{\minus{}3/2}\,dx\equal{}\left.\minus{}\frac2{\sqrt{x}}\right|_1^{\infty}\equal{}2,$ which converges.

But wasn't the original integral from $ 0$ to $ \infty,$ not $ 1$ to $ \infty?$ That's no problem, because the integral from $ 0$ to $ 1$ is, as stated above, just the integral of a continuous function on a bounded interval.

If you want a cheap estimate (we don't really need this to answer the original question - it's an extra flourish) we can say that

$ \int_0^1\frac1{\sqrt{1\plus{}x^3}}\,dx<\int_0^11\,dx\equal{}1,$ so $ \int_0^{\infty}\frac1{\sqrt{1\plus{}x^3}}\,dx<1\plus{}2\equal{}3.$

One thing I always see from my own Calculus II students on this problem, and one thing that I always chide them for: they will compare the original integral to $ \int_0^{\infty}x^{\minus{}3/2}\,dx$ (and claim that the latter integral converges, as a $ p$ integral). But the latter integral actually diverges. It's improper for two reasons, not one, and it diverges because of the misbehavior near $ 0.$ But that singularity at $ 0$ is not a feature of our original problem - it's an artifact of the comparison. So don't make that comparison. $ x^{\minus{}3/2}$ is not a good comparison near zero for our integrand. Don't create a problem that wasn't there in the original. Pick a convenient number between $ 0$ and $ \infty$ and break the problem apart (which is exactly what I did above).

I'll add that this is all a very pedantic exercise aimed at a beginning calculus student. The more you know, the less you'll write down, and I would assume that every one of the long-experienced analysts and calculus teachers who read this forum simply looked at your statement and immediately saw it as a convergent integral - not by doing anything essentially different than what I did above, but simply by doing all of that immediately and informally without writing anything down.

[quote="mche"]I don't think I can just use any kind of function I want.[/quote]
Of course, you don't use just "any kind of function" - you use a function that closely resembles the original function in size. You can see why $ \frac1{x^{3/2}}$ (for large $ x$) is what occurs to us in this example, right?

Wow. that was more than I expected, thank you so much :)

   I see how  \frac1{x^{3/2}}     resembles our original function, since   \frac1{x^{3/2}} is always >= than our original function. But to use the integral comparison test, what are the exact conditions of our comparison function? does it have something to do with supremium and lowest upper bounds?

 thanks a lot !

Don't make things more complicated than they are!

This is just like series, and, just like series, we have both a one-sided comparison test and a limit comparison test.

Suppose we're dealing with $ \int_a^{\infty}f(x)\,dx,$ where $ f(x)\ge 0$ and $ \int_a^bf(x)\,dx$ exists for all finite $ b.$

Here's the one-sided comparison test:

If $ f(x)\le g(x)$ for all $ x>b$ for some $ b,$ and $ \int_{\text{somewhere}}^{\infty}g(x)\,dx$ converges, then $ \int_a^{\infty}f(x)\,dx$ converges, and

if $ f(x)\ge g(x)$ for all $ x>b$ for some $ b,$ and $ \int_{\text{somewhere}}^{\infty}g(x)\,dx\equal{}\infty,$ then $ \int_a^{\infty}f(x)\,dx\equal{}\infty.$

Here's the limit comparison test:

Suppose $ \lim_{x\to\infty}\frac{f(x)}{g(x)}\equal{}L,$ where $ L\ne 0$ and $ L\ne\infty.$ Then $ \int_a^{\infty}f(x)\,dx$ and $ \int_{\text{somewhere}}^{\infty}g(x)\,dx$ either both converge or both diverge.

I actually argued the example above as a one-sided comparison test, but I could have argued it as a limit comparison test, with the same comparison function.

What properties do you want for $ g(x)?$ You want it to reasonably resemble $ f(x)$ is size (for the limit comparison test, it has to be the "same size" as $ f(x)$ ), you want to be able to compute with $ g(x)$ so you can tell whether its integral is positive or negative (which practically limits $ g$ to being something fairly simple), and you'd like it to settle the question. (That last point is something you can't always have - this is a set of ideas, not an automatic machine.)

Try this one: does $ \int_0^{\infty}\frac{1\minus{}\cos x}{x^2}\,dx$ converge?

[color=green][Unnecessary quote removed by moderator.][/color]

 Thanks, I will try that problem.

 I found this proof:

http://www.proofwiki.org/wiki/Comparison_Test_for_Improper_Integral

 but it goes over series, limits of series..et c and I haven't gone over that stuff yet. I'm just not sure how the comparison test would be true.. Thanks!  Also. "closely resemble" is still a bit vague to me.. How do I know if 1-cos(x) is "too different" from 1-cos(x)/x^2? :S Do you just know from experience or something?

Comparison is about [i]size[/i]. It's about how fast something grows, or how fast it shrinks. Do you really think $ 1\minus{}\cos x$ qualifies as a "size"?

hello, the value of your integral
$ \int_{0}^{\infty}\frac{1\minus{}\cos(x)}{x^2}\,dx$ is $ \frac{\pi}{2}$.
Sonnhard.

Way to miss the point, Sonnhard. Yes, I know what the value of this integral is - that's not what it's doing in this topic. And I didn't ask mche for the value, just for an argument about its convergence.

so, the steps I would try to take are:

 1) find a function that is always larger or equal than our original function
 2) make sure that this function is easily computable, but it cannot be too different from the original function? ( I'm still not sure about this..) do they grow at the same rate? or they're like each other  for large x?

 I need a hint with the 1-cos(x)/x^2  integral.. I'm not sure what function to use.. thanks

[quote="Kent Merryfield"]I've already given my hint, and it's one word: size.[/quote]

 ah okay, this should work, 1/x^2 from 1 to infinity.

 Then 1-cos(x)/x^2 should converge from 1 to infinity, and thus 0 to infinity should converge as well. From assumption, we already know that the integral 1-cos(x)/x^2 exists for finite intervals [1, N], so it exists from [0,1] and we only need to be concerned with comparing 1 to infinite

 how's that?
thanks :)

hello, where is the post with the two other integrals?
$ \int_{2e}^{\infty}\frac{\sin(x)}{\ln(x)\plus{}\sin(x)}\,dx$
$ \int_{0}^{\infty}\ln\left(1\plus{}\frac{\sin(x)}{\sqrt{x}}\right)\,dx$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, where is the post with the two other integrals?
$ \int_{2e}^{\infty}\frac {\sin(x)}{\ln(x) \plus{} \sin(x)}\,dx$
$ \int_{0}^{\infty}\ln\left(1 \plus{} \frac {\sin(x)}{\sqrt {x}}\right)\,dx$
Sonnhard.[/quote]

 it was moved by a moderator to another thread, since apparently it is an exercise concerning more than what's being discussed here

[quote="Dr Sonnhard Graubner"]hello, thank you for your information, but where can i find this thread?
Sonnhard.[/quote]

here: http://www.mathlinks.ro/viewtopic.php?t=326516

:)

[quote="Kent Merryfield"]The one quibble: in the integral $ \int_0^{\infty}\frac {1 \minus{} \cos x}{x^2}\,dx,$ why is there not a problem as $ x\to 0?$

(In fact there is no difficulty there - but can you see why?)[/quote]

   Because 1-cos(x)/x^2 is still bounded around 0? but 1/x^2 is not, so wouldn't we need to compare them both on a n interval valid for both integrals?

 thanks


A fair coin is flipped 7 times. What is the probability that at least 5 consecutive flips come up heads?

There are $ 2^7 \equal{} 128$ total cases, and we find which ones work.

7 Heads - 1 way

6 Heads- 4 ways: put the tail in space 1, 2, 6, or 7

5 Heads- 3 ways: put the heads in slots 1-5, 2-6, or 3-7

$ \frac {8}{128} \equal{} \frac {1}{16}$

[hide]Here's another method:
Case 1: the run of five heads starts at the beginning, so the sequence goes HHHHH??
Case 2: the run of five heads starts on the second flip. It therefore doesn't start on the first flip, so the sequence goes THHHHH?
Case 3: the run of five heads starts on the third flip. It therefore doesn't start on the second flip, so the sequence goes ?THHHHH

P(Case 1)= (1/2)*(1/2)*(1/2)*(1/2)*(1/2)*(2/2)*(2/2)=1/2^5
P(Case 2)= (1/2)*(1/2)*(1/2)*(1/2)*(1/2)*(1/2)*(2/2)=1/2^6
P(Case 3)= (2/2)*(1/2)*(1/2)*(1/2)*(1/2)*(1/2)*(1/2)=1/2^6

So adding, we get 1/2^5+1/2^6+1/2^6=(2+1+1)/2^6=4/2^6=1/16[/hide]

XD I had 3/64 then I guessed 1/16 on a whim and was like WOA.

[quote=paperplates][quote=TheMaskedMagician]Dang it! I forgot about the 5 heads with 1 head different case![/quote]

Same same UGH
And there goes my really low streak D:[/quote]

My mistake also. :P

[font=Times new roman][hide=Solution]We can break this problem up into several cases. Since each case is composed of only Heads and Tails, then each case has a $(\tfrac12)^7$ chance of occuring.

[size=75]H=Heads, T=Tails[/size]

[b]Case I - 5 Consecutive Heads:[/b] [indent][b]Case I.I - THHHHHT:[/b] We have a $(\tfrac12)^7$ chance of this happening.
[b]Case I.II - TTHHHHH:[/b] We have a $(\tfrac12)^7$ chance of this happening.
[b]Case I.III - HHHHHTT:[/b] We have a $(\tfrac12)^7$ chance of this happening.
[b]Case I.IV - HTHHHHH:[/b] We have a $(\tfrac12)^7$ chance of this happening.
[b]Case I.V - HHHHHTH:[/b] We have a $(\tfrac12)^7$ chance of this happening.[/indent] Adding up these sub-cases, we get $5(\tfrac12)^7=\tfrac5{128}.$

[b]Case II - 6 Consecutive Heads:[/b] [indent][b]Case II.I - THHHHHH:[/b] We have a $(\tfrac12)^7$ chance of this happening.
[b]Case II.II - HHHHHHT:[/b] We have a $(\tfrac12)^7$ chance of this happening.[/indent] Adding up these sub-cases, we get $2(\tfrac12)^7=\tfrac1{64}.$

[b]Case III - 7 Consecutive Heads:[/b] [indent][b]Case I.I - HHHHHH:[/b] We have a $(\tfrac12)^7$ chance of this happening.[/indent] So for this case, the probability is obviously $(\tfrac12)^7=\tfrac1{128}.$ Adding up these cases, we have $$\dfrac5{128}+\dfrac1{64}+\dfrac1{128}=\boxed{\dfrac1{16}}.$$ $$\definecolor{color}{rgb}{0.4, 0.5, 0.6} \scriptsize{\colorbox{color}{\color{white}{A pifinity solution}}}$$

:noo: I forgot about Case 3... :(

. . . and I see I'm not the only one to do so . . . :rotfl:

[quote=mathapple101]Lol had the same mistake :)[/quote]

[quote=asdf334]this question had terrible wording[/quote]

Lol we all had the same mistake :rotfl:

[quote=A1001][quote=OlympusHero]frick this question ruined my 1022 problem streak and 99.8 cp rating

so close to 1024 :wallbash_red:

on the practical side this question was pretty tricky. at first misread it but what i'm really annoyed about was that i put 16/128 = 1/8 instead of 8/128 when i literally had figured out it was 8 outcomes

ah well[/quote]

dude rip 1022 is v good steak :o[/quote]

But not as good as my 137 streak.

dang it why. I didn't think of the case 6 heads but not in a row :facepalm:

[hide=bad alternative solution]
We can have $5$ heads in a row with anything else, for $3\cdot2^2=12$ total possibilities, but we double counted all cases where there are $6$ heads in a row with anything else, for $12-2\cdot2=8$ possibilities, and this also makes the case where there are $7$ heads in a row be counted exactly once, so the answer is $\tfrac{8}{2^7}=\boxed{\tfrac{1}{16}}$.
[/hide]

[quote=TheMaskedMagician]Dang it! I forgot about the 5 heads with 1 head different case![/quote]

[hide]we have 8 different possibilities, and 2^7=128 ways so 8/128=1/16[/hide]

This is such a silliable problem, the 6 heads but not in a row is so trolly...

[quote name="mathmax12" url="/community/p26960029"]
[quote=TheMaskedMagician]Dang it! I forgot about the 5 heads with 1 head different case![/quote]

[hide]1/16[/hide]
[/quote]

Bruh
This is more like postfarming.
Nobody cares if you just give the answer.
They want a explanation on why.
also how did you even get 8 possibilities?

is there a better sol for this prob i just have that feeling

i forgot case 3
 :oops_sign:  :wallbash_red:  :censored:  :starwars:


Here is a proof from an old thread:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11610[/url].

[hide][b]Lemma.[/b]  If $a$ is irrational and $x$ is rational, then $a+x$ is irrational.

[i]Proof.[/i]  Consider $a+x$ rational.  Then we must have $a+x-x$ rational as well.  However, $a+x-x=a$ is irrational by the hypothesis.

Take $\sqrt{1000^{2}+1}$ and $p,q\in\mathbb{N}$ with $\gcd\left(p,q\right)=1$.  We then have \[\sqrt{1000^{2}+1}=\frac{p}{q}\implies p^{2}=\left(1000q\right)^{2}+1^{2},\] which is a contradiction since this would mean that $p\vee q\notin\mathbb{N}$.[/hide]

That's the proof I got, amcavoy.  :lol: One more thing. Can this be generalized to this statement:
Let $m<n$. Prove that the following sum is irrational:
$\sqrt{m^{2}}+\sqrt{m^{2}+1}+\sqrt{m^{2}+2}+...+\sqrt{n^{2}}$

[quote="Ravi B"][quote="amcavoy"][b]Lemma.[/b]  If $a$ is irrational and $x$ is irrational or rational, then $a+x$ is irrational as well.[/quote]

Um, that lemma is incorrect.  Consider $a = \sqrt{2}$ and $x =-\sqrt{2}$.[/quote]
Right, nevermind.

[quote="vishalarul"]To make the lemma true, wouldn't we require $a,x>0$?[/quote]
No because then $x=b-a$ with $b>a$ and $b\in\mathbb{Q}$ could mean that $a+x=b\in\mathbb{Q}$.  See the edited Lemma.

[quote="vishalarul"]To make the lemma true, wouldn't we require $a,x>0$?[/quote]

$\sqrt2+(2-\sqrt2) = 2$

Random question that this raises:  how big can one make a subset of the irrational numbers and have it be closed under addition?  (Or other operations or combinations of operations, I suppose.)

One realization is that we can get a dense, countable subset by taking, for instance, the set $\{\pi \cdot q\ |\ q\in \bf Q\}$.  How much better can we do?

[quote="JBL"]how big can one make a subset of the irrational numbers and have it be closed under addition?[/quote]

Take an arbitrarily large (I think it can be made uncountable) linearly independent subset (over $\mathbb{Q}$) of the irrational numbers and consider the union of the set of their rational multiples.  (Edit:  Well, adding in all the elements necessary for closure.)

Edit:  A countable such subset definitely exists, since the dimension of $\mathbb{R}$ over $\mathbb{Q}$ is infinite.  I lack the rigorous training in the reals to give a convincing argument why an uncountable subset must also exist.


Let [i]a,b,c[/i] be [i]real[/i] numbers such that [i]abc=1[/i]. Prove that at most two of the numbers $2a-\frac{1}{b},2b-\frac{1}{c},2c-\frac{1}{a}$ are greater than [i]1[/i].

assume:
$a=\frac{x}{y}$ and b,c similar we get:
\[ 2a-\frac{1}{b}=\frac{2x-z}{y},2b-\frac{1}{c}=\frac{2y-x}{z},,2c-\frac{1}{a}=\frac{2z-y}{x} \]
then it is obvious to see that one of:
\[ y \ge 2x-z,z \ge 2y-x,x \ge 2z-y \]
will be ture.
so we proved :lol:

this is to easy...it is cleat that either all the numbers are positive,either two of them are smaller then 0 and one is bigger then 0.let as pressume that all those  numbers in question are bigger then 1
I a>0 b,c<0 let  b=-d,c=-e a,d,e>0
then 2a+1/d>1
-2d+1/e>1 and
-2e-1/a>1 and this is clearly false because the lhs is smaller then 0
IIa,b,c>0 rewrite 2a-1/b>1 as 2ab>b+1 or 2/c>b+1<=> 2>bc+c similar we get that 2>ac+a and 2>ab+b.summing up these 3 we get that ab+bc+ca+a+b+c<6 and this is false because ab+bc+ca>=3(am-gm)
a+b+c>=3(am-gm)and we have the contradiction so at most 2 of those numbers are bigger then 1

i don't know what happened.i posted 2 times the same thing and i did not see that zhaobin already posted :?

Well i didn`t say that the problem i s so hard. But i think that the idea of solving it is nice.. just like the problem :)


Find all function f:Z_Z sastify f(x^3+y^3+z^3)=f^3(x)+f^3(y)+f^3(z)

This is probably the fourth time this problem is being posted .
First :was post by Arne, second: nmtruong1986,third:pigfly :D ,fourth:Pascual2005 and was solved by Pierre.Look at:
http://www.animath.fr/tutorat/dossier_03044sol.pdf
or http://www.artofproblemsolving.com/Forum/topic-26140.html
http://www.artofproblemsolving.com/Forum/topic-14908.html
... :D
@K09:Are you the member of HaNoi team (since you post this problem)?

I find this problem in France wesbsite.,later I know it is hn PR.


Let $ a, b, c$ be positive reals such that $ abc \equal{} 1$. Prove that

$ 5(a^2\plus{}b^2\plus{}c^2)\plus{}2(a\plus{}b\plus{}c)\geq (ab\plus{}bc\plus{}ca)\plus{}2(\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})\plus{}12$

[quote="Ligouras"]Let $ a, b, c$ be positive reals such that $ abc \equal{} 1$. Prove that

$ 5(a^2 \plus{} b^2 \plus{} c^2) \plus{} 2(a \plus{} b \plus{} c)\geq (ab \plus{} bc \plus{} ca) \plus{} 2(\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}) \plus{} 12$[/quote]

1)$ a\plus{}b\plus{}c\ge \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}$

2)$ a^2\plus{}b^2\plus{}c^2\ge ab\plus{}bc\plus{}ca$

3)$ a^2\plus{}b^2\plus{}c^2\ge 3$

[quote="Pain rinnegan"][quote="Ligouras"]Let $ a, b, c$ be positive reals such that $ abc \equal{} 1$. Prove that

$ 5(a^2 \plus{} b^2 \plus{} c^2) \plus{} 2(a \plus{} b \plus{} c)\geq (ab \plus{} bc \plus{} ca) \plus{} 2(\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}) \plus{} 12$[/quote]

1)$ a \plus{} b \plus{} c\ge \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}$

2)$ a^2 \plus{} b^2 \plus{} c^2\ge ab \plus{} bc \plus{} ca$

3)$ a^2 \plus{} b^2 \plus{} c^2\ge 3$[/quote]

Compliments!!! it is a beautiful surprise for me...

[quote="pablo221092"]Very good your solution, [b]Pain rinnegan[/b].

[b]Ligouras[/b], Could you post your solution please?

Many thanks[/quote]

It belongs to my new book... you have to wait 2010 November. Sorry  :(   :(   :(  ( mine is a beautiful solution... in my opinion...)


show that for all prime numbers p,$ Q(p) \equal{}\sum^{p\minus{}1}_{k\equal{}1}k^{2k\minus{}p\minus{}1}$is an integer.

[quote="yildobekir"]show that for all prime numbers p,$ Q(p) \equal{} \sum^{p \minus{} 1}_{k \equal{} 1}k^{2k \minus{} p \minus{} 1}$is an integer.[/quote]

This must be false, because:
[hide]
Suppose p is the prime number $ 5$.

Then it becomes

$ \sum^{5 \minus{} 1}_{k \equal{} 1}k^{2k \minus{} 5 \minus{} 1} \equal{}$

$ \sum^{4}_{k \equal{} 1}k^{2k \minus{} 6} \equal{}$

$ 1^{\minus{}4} \plus{} 2^{\minus{}2} \plus{} 3^0 \plus{} 4^2 \equal{}$

$ 1 \plus{} \frac14 \plus{} 1 \plus{} 16 \equal{}$

$ 18\frac14$,

which is not an integer.

[/hide]


AMC
AMC 12
AIME
USA(J)MO
USAMO
geometry
algebra

I don't think you will need that many to qualify for the USAMO, but it's always better to be ambitious and to shoot for a higher score than what you may need.

Topics to study: Geometry, Geometry, and more Geometry. And then - all of the usual properties of polynomials, trig functions, complex numbers, and the basics of number theory.

And finally some advanced techniques of counting and probability - i.e. investigating small cases, looking for paterns, looking for symmetry, and looking for recursions.

You can study to brush up on the basics of these topics, but nothing beats working on actual problems to learn how to apply them. Most difficult AIME problems require a certain amount of "investigation" to solve them. So don't be afraid to get your hands dirty in a problem.

One year I studied hard for the AIME.  Generally, the way that I studied any type of problem that I was reasonably good at (your strengths) was to keep taking a crack at any problem that I couldn't solve until I found a way to solve it.  For those problems I nearly never (maybe never) looked at the solution before I thought I had an answer.  For my weaker subjects (for me geometry) I would try to get as many as I could, but would peek at the solutions every now and then.  When I say "peek" I really mean "peek".  I would sometimes take a 2 second glance at the diagram and see if I was simply attacking the problem poorly.  When this wasn't enough I would then go and read through the full solution.

When reading the full solution for a problem that you did not solve on your own, walk through the solution as if you plan on teaching other students how to solve it [i]tomorrow[/i].  This will ensure that you ask the important questions such as "how would I have thought to do [i]that[/i]?"  It can sometimes be frustrating to spend most of your time on the problems that are hardest to you, but that's really the only way to understand...the problems that are hardest to you.

If you are looking for sources for problems, when I ran out of AIMEs in high school I sometimes worked ARML problems.  They rarely get to be as hard as the harder AIME problems on the individual round, but the team problems are sometimes that difficult.

Good luck with your goal.  I do however doubt you'll need as high as an 11, but setting a high goal is a good way to get where you want to go.

I just thought of this, and other people may pan it, but you might look at questions that are harder than AIME problems -- maybe Putnam or Olympiad questions (but only those with small numbers, or from countries who have easy olympiads).  If you want to get double-digits on the AIME, you'll almost certainly end up working some of the last three or four questions, and they can be as hard as olympiad questions (except that you don't actually have to prove them).

The numbers 5, 6 and 8 on ARML exams tend to be on the same level as some AIME questions, except that you only have 10 minutes for a pair instead of 24.  (Although, actually, number 8 is almost always an 8 minute problem, not a 5 minute problem -- number 7 is usually as easy or easier than the first 3.)
Some ARML exams are available at ARML.com.  I dunno where else you can find them, though.

I have nothing else original to add.  Gauss and Mathew give good advice :)

I agree with JBL that working on harder problems is a great way to go.  To me this means that

If you can get a 3-4 on the AIME, keep studying AIME problems.

If you can get a 5-8 on the AIME, studying hard AIME and some easier Olympiad level problems (including Putnam).

If you can get a 9-12 on the AIME, study the hardest AIME problems and a lot of Olympiad level problems.

If you can get a 13+ on the AIME, either you are already working on Olympiad level problems or else you're probably really bored.

Very few students score higher than a 10 on the AIME without ever having studied Olympiad level problems.  Olympiad level problems require students to make numerous problem solving leaps (compared to about 1-2 for most AMC problems and about 1-4 for AIME problems) and often combine several areas of mathematics (i.e., geometry with probability or number theory with inequality).

The harder AIME problems are typically Olympiad level problems without the proof writing and using specific quantities (as opposed to finding a general answer as in many Olympiad level problems).

[quote="MCrawford"]If you can get a 13+ on the AIME, either you are already working on Olympiad level problems or else you're probably really bored.[/quote]

Hehe.

Whew, this is turning out to be a beefy thread.  :mrgreen:  Thanks for all of the responses. A couple things though:

1) When you folk suggest working on olympiad problems, do you think it is helpful to work on olympiad problems about topics that are not really on the AIME. For example, with inequalities, the AIME only requires knowledge of the very basic inequalities (AM-GM, Trivial, Triangle) and none of the more *advanced* ones that are found on olympiads (Power Mean, Rearrangement, Minkowski, Chebyshev, Jensen, etc..). Similarly, the numerous standard functional equations olympiad problems that ask something like "Find all functions f(x) such that blah blah blah..<some expressions involving f(stuff)>" cannot really be formulated as a short answer question (as on the AIME). So do you recommend also studying functional equations, inequalities, etc...?

2) Why does everyone not think that I would need a 11-12 on the AIME to make the USAMO cutoff? With my score of 126.5, I would need 3 more than the AIME floor, and since the floor was 8 last year, I would have needed about 11 on the AIME (well, not quite, since the cutoff last year was 226, but if it had been 227 or any other number that is greater than or equal to 7 mod 10).

Well, last year's tests were just laughably easy; they've increased the difficulty this year.  The AIME floor might be 6ish or something this year.

[quote="cats..."]When is the actual date for the First AIME?[/quote]

All official information like this can be found at http://www.unl.edu/amc somewhere.

[quote="cats..."]well, that's why i wasnt sure... http://www.unl.edu/amc/whatswhen.html there are 2 separate dates, though the one on the right is just tentative.  I'm guessing that the ones on the LHS are accurate, since those are the correct dates for the AMC's.[/quote]

Yep, RHS ones are for next year.

The heading on the right side is supposed to be "2004-2005 Tentative Dates", not "2003-2004 Tentative Dates".  Yeah, it's confusing.

cats..., check your private messages box.

What is a floor score?

[quote="Fierytycoon"]I need some advice...

Well, after not doing so well on the AMC12s, my higher AMC12 score is a 126.5, so I will have to get (at worse) 3 more on the AIME than the floor AIME score. Assuming that the floor score is 8 like last year (not that this will be necessarily true, but with the more difficult AMC12's, I don't think that it will be higher), I will need a 11 or 12 to comfortably make the USAMO. I am taking the AIME2 in April, and I will have a 2-week school vacation right before that date.

So I need some wise advice on what I should study during that 2-week vacation so that I can comfortably make the USAMO. I know I can work on all fo the past AIME's, which will help, but just doing that won't enable me to easily get a 11-12 on the AIME.

So...suggestions? I am willing to work almost all day during vacation, so..[/quote]

my thought is that it's good to take at least SOME full-length tests in 3 hours, since it's pretty tough to get used to thinking for 3 hours straight, at least compared to the 1.25 hrs for the AMC.  And those you dont get in the 3 hrs you can still work on afterwards.

You'll definitely want to give the 3 hour sitting a shot a few times before the real thing.  There aren't many marathoners training by just doing 1-mile sprints.  As for 'working until you get an answer', you should put a time limit on yourself - whenever you're just doing problems without a test time limit, don't spend 2 hours on a single AIME problem.  The goal is to train yourself, and while you'll learn a lot in the first 20-30 minutes banging your head against a problem, you're much less likely to learn much in the next 90 minutes (obviously, for Olympiad problems, you'll want to give yourself a longer time limit).  After some period of time, give up and look at the solutions or ask for a hint on the board.  Most importantly, make sure you understand how you could have thought of the solution.  If you don't see how someone would have thought of the given solution, post here and someone will help you out.

[quote="mathkid"]Wow. Would you happen to have anything like that for probability and counting? Or would you happen to of any sites that do?[/quote]

I compiled a similar set of probability problems for my ARML team to practice on. Most of them came from this board, so if you do a search for "probability" in the Intermediate or AMC Forums you can find them. Otherwise, the full set will be up on our website by the end of the week and I'll post a link to it.

In the meantime, here are a few problems to get you started:

1. A standard, fair, 6-sided die is rolled 8 times. Given that the number 3 appears exactly 3 times, what is the probability that no two 3's appear on consecutive rolls?

2. A fair coin is tossed 10 times. Compute the probability that heads do not occur on consecutive tosses.

3. Hubert flips through a standard deck of 52 cards. After flipping over how many cards from the top is the probability greater than  that he has already flipped over all 13 hearts?

Hmm, I'll have to tell Hubert that he's made it into the SC ARML team practice.  I wonder what he'll think ...

Since I've already worked through all the geometry sections in AOPS, are there any suggestions for which chapters in Geometry Revisited to go through? I'm definitely going to (try to) do the first two on triangles on circles, but should I go further for the AIME? (I'll do the rest eventually =))

I would say that the parts of Geometry Revisited on angles and circles would be great for the harder AIME problems.  I would also say that reading that whole book is one of the best things you could ever do for developing your geometry skills.  That and taking Richard's Olympiad Geometry course... ;)

For somebody who is stronger in geometry than algebra, Is there an algebra book similar to geometry revisited so that I can sharpen my algebra skills?

[quote="rrusczyk"]Work through AoPS, Art and Craft of Problem Solving, and Engel.  (While the latter two are pretty soft on geometry, they have tons of algebra.)[/quote]
Will Engel be useful in AIME prep?

Working Engel problems could definitely help you with the AIME, but really, it's more of an Olympiad level book.  AoPS 2 or Art and Craft of Problem Solving are much more likely to help you directly with the AIME.

That and taking AoPS courses which definitely helped me  :D  I'm a bit scared that I'll do poorly on the AIME because I haven't had the same amount of practice as last year, or at least not practice on that level  ;)  I need to get back to work or else my Olympiad preparation will go down the drain as far as USAMO is concerned. Back to the subject at hand, I took my timed AIME's about a month before the actual competition and before that I just worked problems from the earlier years and worked through some chapters of AoPS Volume 2.

[quote="WarpedKlown1335"]Does reading I[u]ntroduction To Geometry[/u] by Coxeter help?[/quote]

I'm not familiar with that book, so I can't say...

I just encountered this thread and I have a few questions regarding my first AMC and hopefully AIME (or even USAMO and IMO :P).
1.  What chapters in Mathematical Olympiad Challenges woudl you recommend reading for AIME, even though it is more of an Olympiad book?

2.  What chapters in Problem Solving strategies would you recommend?

3.  In general, what branches of mathematics are best to study for AIME and AMC?

Masoud Zargar

Do NOT revive the old threads. This is *very* bad.

Also, I began to see very large amount of people asking for advice. I think people should stop doing this because other people can't help you to get better. YOU have to do the problems to get better and this is ONLY way.

Locked.


function
integration
real analysis
real analysis solved

$f:[0,1]\rightarrow R+$ continous suppose $f$ is concave

Prove that

1/  $\int_{0}^{1}xf(x)dx\leq \frac{2}{3}\int_{0}^{1}f(x)dx$

2/   $\int_{0}^{1}x^2f(x)dx \leq \frac{1}{2}\int_{0}^{1}f(x)dx$

I've been thinking about the first one, and it looks like it's enough to prove it for linear functions (I bet there's a nice, quick way to prove it, but I couldn't find it :(). The fact that it's enough to prove it for linear functions is clear if you draw a picture. We write it as $I=\int^1_0\left(x-\frac 23\right)f(x)dx\le 0$, so if we prove it when $I$ is as large as possible, we are done.

To this end, we can use the line passing through $(0,f(0)),\ \left(\frac 23,f\left(\frac 23\right)\right)$, because its graph is below the graph of the function if $x<\frac 23$ (and we want $f$ as small as possible on this interval, so that $\left(x-\frac 23\right)f$ is as large as possible, since $x-\frac 23<0$), and it's above the graph of $f$ if $x>\frac 23$, and in this case we want $f$ as large as possible, because $x-\frac 23>0$.

Proving it for linear functions is, of course, a piece of cake :).

And the same approach seems to work for the second one: we just prove it for the linear graph passing through $(0,f(0)),(\sqrt {\frac 12},f(\sqrt {\frac 12}))$.


A car traveling at a constant speed of 115 km/hr passes a trooper hidden behind a billboard. One second after the speeding car passes the billboard, the trooper sets in a chase after the car with a constant acceleration of 3.0 m/s2. How far does the trooper travel before he overtakes the speeding car?

I know that you have to write down the equations for the position of the police car and the speeder as function of time and make sure that they are equal at the same time and then solve for "T"..but I keep getting the wrong answer.....

We have that $v_\textrm{speeder}= 115 \, \frac{\textrm{km}}{\textrm h}= 115 \cdot \frac{1000}{3600}\, \frac{\textrm m}{\textrm s}= \frac{115 \cdot 5}{18}\, \frac{\textrm m}{\textrm s}= \frac{575}{18}\, \frac{\textrm m}{\textrm s}$.

The speeder moves at constant speed, so $ x_\textrm{speeder}(t) = v_\textrm{speeder}\cdot t$.
The trooper moves with constant acceleration, so $ v_\textrm{trooper}(t) = a_\textrm{trooper}\cdot t$. Thus, \[ x_\textrm{trooper}(t) = \int_{0}^{t}v_\textrm{trooper}(u) \, du = \left. \frac{a_\textrm{trooper}\cdot u^{2}}2 \right|_{0}^{t}= \frac{a_\textrm{trooper}\cdot t^{2}}2 .\] The trooper reaches the speeder exactly when $ x_\textrm{speeder}= x_\textrm{trooper}$, i.e. $ v_\textrm{speeder}\cdot t = \frac{a_\textrm{trooper}\cdot t^{2}}2 \, \, \Longleftrightarrow \, \, t \left( v_\textrm{speeder}-t \cdot \frac{a_\textrm{trooper}}2 \right) = 0$.
Hence, $ t = \frac{2 v_\textrm{speeder}}{a_\textrm{trooper}}= 2 \cdot \frac{\frac{575}{18}\, \frac{\textrm m}{\textrm s}}{3 \, \frac{\textrm m}{\textrm s^{2}}}= \frac{575}{27}\, \textrm s \approx 21.296 \, \textrm s$.

I hope I haven't made any mistake.

yeah...i got the same time, but i get the wrong answer when i try to get the distance the police car travels.

[hide]$x_{\text{trooper}}= \frac{1}{2}a\left(t-1\right)^{2}$

$x_{\text{speeder}}= vt$

The trooper overtakes the speeder when $x_{\text{trooper}}= x_{\text{speeder}}$

$\Rightarrow \frac{1}{2}a\left(t-1\right)^{2}= vt$

$\Rightarrow t^{2}-2t+1 = \frac{2v}{a}t$

$\Rightarrow t^{2}-2(1+\frac{v}{a})t+1 = 0$

$\Rightarrow t =\frac{2(1+\frac{v}{a})\pm \sqrt{4(1+2\frac{v}{a}+\left(\frac{v}{a}\right)^{2})-4}}{2}$

$v = 115\text{ kph}= \frac{287.5}{9}\text{ m/s}$

$\frac{v}{a}= \frac{287.5}{27}$

$\Rightarrow t =\frac{2(1+\frac{287.5}{27})\pm \sqrt{4(1+2\frac{287.5}{27}+\left(\frac{287.5}{27}\right)^{2})-4}}{2}$

$\Rightarrow t = 23.25\text{ s}$. (The other solution is absurdly small.)

$\boxed{\therefore x_{\text{trooper}}= vt = (\frac{287.5}{9})(23.25) = 742.8\text{ m}}$[/hide]

:oops_sign: I have not noticed that the trooper starts chasing the speeder with a second delay. Actually, I thought it pretty weird that the trooper would start his chase exactly when the speeder passes him, but I still didn't notice the one second mentioned in the problem :oops:


pigeonhole principle
complex numbers
algebra proposed
algebra

Given $\varepsilon > 0$, consider an infinite sequence $\{z_n\}_{n \geq 1}$ of complex numbers such that, for any $i \neq j$,
\[{
|z_i - z_j| > \frac{1}{|i-j|^{\frac{1}{2} - \varepsilon}}}.
\]
Prove that $\{|z_n|\}_{n \geq 1}$ is not bounded.

--Vesselin

proof of a friend:
if the sequence were bounded then it would be included in a square of side $c$.
Cut this square into $n^2$ small squares and use pigeonholes principle to find a contradiction.

But is it possible for $\frac{1}{|i-j|^{\frac{1}{2} + e}}$ ?


projective geometry
geometry proposed
geometry

Problem: ABCD is a convex quadrilateral. intersection of the diagonals is point M. N is any point on AB. intersection of line NM and line DC is L. intersections of {AL, ND} and {BL, NC} are P, R respectively. Prove that P, M, R are collinear.

It is true for all points $ N,L$,if they are not collinear with $ M$,then the problem is true too.
It follows from Pappus theorem. :wink:

yes, this is a clear resault of pappus theorem :wink:


soale asoonie vali azash khosham oomad.ye morabba e 1*1 darim ke dar an n mostatil mikeshm ke ba ham tadakhol nadarand(magar dar marzhayeshan) sabet konid dar soorati ke in mostatil ha ro beborrim hadde aksar n+1 ghete ye dige ijad mishe.

soal e ghashangiye.
albate fekr konam maasale bayad migoft ke: mostatil ha bayad azla e movazi ba azla e moraba ro daashte bashan, chon agar nabashe angah teedad e navahiye ijad shode kheyli ziyad tar e ,$\rightarrow \infty$
dar kol bayad Induction estefadeh koni, amma
baraye shooroo:
[hide]
farz konim ke azla e in mostatil ha movaziye moraba bashe, pas baad as keshidan e yek mostatil ma 2 nahiye ijad kardeim.  baad az keshidan e $n$ mostatil, ya: nahiyeye beyn e mostatil ha  ya $i$  mostatil khahad bood, ya yek nahiyeye peyvashte. agar mostatil bashad ma dar kon $n+i=j$ mostatil sakhte im, va agar peyvaste bashad, dar kol $j+1$ nayihe khahim dasht.

QED[/hide]


A Quadrilateral with sides of length $3,6,2,\;and\;7$ is inscribed in a circle. What is the area of this quadrliateral?

this is a direct application of the formula $A^{2}=(s-a)(s-b)(s-c)(s-d)$ where $s=\frac{1}{2}(a+b+c+d)$


logarithms
function
inequalities
inequalities unsolved

If $x,y,z>0$ and $xyz=1$

Find the minimal of

$\sum_{cyc}\frac{x^x}{x^2y}$

 :)

[quote="shyong"]If $x,y,z>0$ and $xyz=1$

Find the minimal of

$\sum_{cyc}\frac{x^x}{x^2y}$[/quote]

In fact, the minimum of $\frac{x^x}{x^2y}+\frac{y^y}{y^2z}+\frac{z^z}{z^2x}$ is 3. This minimum is achieved for x = y = z = 1, and in order to show that it is really the minimum, we have to prove that if x, y, z are three positive numbers satisfying xyz = 1, then $\frac{x^x}{x^2y}+\frac{y^y}{y^2z}+\frac{z^z}{z^2x}\geq 3$.

This is quite easy: By AM-GM, we have

$\frac{x^x}{x^2y}+\frac{y^y}{y^2z}+\frac{z^z}{z^2x}\geq 3\sqrt[3]{\frac{x^x}{x^2y}\cdot\frac{y^y}{y^2z}\cdot\frac{z^z}{z^2x}}=3\sqrt[3]{\frac{x^xy^yz^z}{\left(xyz\right)^3}}=3\sqrt[3]{x^xy^yz^z}$

(since xyz = 1). Now, in order to show that this is $\geq 3$, it remains to prove that $x^xy^yz^z\geq 1$. This can be done, e. g., by noting that, since the logarithm function is monotonically increasing, the number arrays (x; y; z) and (ln x; ln y; ln z) are equally sorted, so that, by Chebyshev,

$x\ln x+y\ln y+z\ln z\geq\frac13\left(x+y+z\right)\left(\ln x+\ln y+\ln z\right)$,

but xyz = 1 yields $\ln x+\ln y+\ln z=\ln\left(xyz\right)=\ln 1=0$, so that $x\ln x+y\ln y+z\ln z\geq 0$, and after exponentiating (is there such a verb?), this becomes $x^xy^yz^z\geq 1$, and we are done.

  darij

you have $\frac{x^x}{x^2y}+\frac{y^y}{y^2z}+\frac{z^z}{z^2x}\geq 3\sqrt[3]{x^x\cdot y^y\cdot z^z}$(i used $xyz=1$) now suppose $x\geq y\geq z$ then $x^x\cdot y^y\cdot z^z=(xyz)^zx^{x-z}\cdot y^{y-z}=1^z(x^{x-y}\cdot x^{y-z}\cdot y^{y-z}\cdot z^{y-z}\cdot z^{z-y})=x^{x-y}\cdot z^{z-y}=
x^{x-y}\cdot (\frac{1}{z})^{y-z} \geq 1$
because $x\geq 1 , \frac{1}{z}\geq 1\ \ x-y\geq 0$ and $y-z\geq 0$ so we have that  $\frac{x^x}{x^2y}+\frac{y^y}{y^2z}+\frac{z^z}{z^2x}\geq 3$ but for $x=y=z=1$ the expresion is 3 so that is the minimum :D

:oops_sign: i didn't notice that while i was writing my solution [b]darij[/b]  posted his own  :lol:

Bravo ! [b]Darij[/b] and [b]Tellah[/b] !

[hide="Ignore this"]
$\sum_{cyc}\frac{x^x}{x^2y}=\sum_{cyc}x^{x-1}z$

Using Chebyshev ,
$\sum_{cyc}x^{x-1}z\geq \frac{1}{3}\sum_{cyc}x^{x-1}\sum_{cyc}z$

However , AM-GM gives you $\sum x\geq 3$ ;

weight AM-GM gives you $\sum_{cyc}x^{x-1}=\sum_{cyc}\frac{x^x}{x}\geq 3$

Combine all of them gives you

$\sum_{cyc}x^{x-1}z\geq \frac{1}{3}\cdot 3\cdot 3 = \boxed{3}$    [/hide]

[quote="siuhochung"][quote="shyong"]
Using Chebyshev ,
$\sum_{cyc}x^{x-1}z\geq \frac{1}{3}\sum_{cyc}x^{x-1}\sum_{cyc}z$
[/quote]
Sorry, I don't understand this part, why is $x^{x-1}$ and $z$ similarly sorted?[/quote]

Ya , you have a good point there siuhochong . The sequence is not similarly sorted so
I cant use Chebyshev  :blush:

[quote="peeta"][quote="siuhochung"][quote="shyong"]
Using Chebyshev ,
$\sum_{cyc}x^{x-1}z\geq \frac{1}{3}\sum_{cyc}x^{x-1}\sum_{cyc}z$
[/quote]
Sorry, I don't understand this part, why is $x^{x-1}$ and $z$ similarly sorted?[/quote]

dont understnad ur point siuhochung.

$\frac{x^x}{x^2y}+\frac{y^y}{y^2z}+\frac{z^z}{z^2x}=x^{x-1}z+y^{y-1}x+z^{z-1}y\ge\frac13 (x^{x-1}+y^{y-1}+z^{z-1})(z+x+y)$

can sopmebody explain why this should be wrong?[/quote]

Because in order to use Chebyshev , $x^{x-1}\geq y^{y-1}\geq z^{z-1}$ and $z\geq x\geq y$ must both holds . :)

Just put $z=2$ , $x=1$ , $y=\frac{1}{2}$ and you will see both of them cant hold at the same time .


trigonometry
logarithms
limit
real analysis
real analysis unsolved

Consider a sequence of reals $ (a_n)_{n \ge 0}$, $ q \in ( \minus{} 1, 1) \backslash \{0\}$ and $ x_n \equal{} \sum_{k \equal{} 0}^n a_kq^k$, $ \forall n \ge 1$.
a) Prove that if $ (a_n)_{n \ge 0}$ is bounded, than $ (x_n)_{n \ge 0}$ converges.
b) Exhibit an unbounded sequence $ (a_n)_{n \ge 0}$ for which $ (x_n)_{n \ge 0}$ converges.

first is simple say $ |a_{k}| < C$ because $ \{a_{k}\}_{k \equal{} 0}^{\infty}$ is bounded
hence $ |\sum a_{k}q^{k}| < |\sum C q^{k}| < B$ ( because it's geometrical progression)
more interestin is your second question:
let $ a_{k} \equal{} \frac {1}{1 \plus{} \sin k}$ it's unbounded (obviusly)
let's prove that  it converge
$ \sum \frac {q^{k}}{1 \plus{} \sin k }$ really because :
there exist sequance such as $ \sin k$ is like $ \minus{} \frac {1}{k}$ and such subsequance $ \{k_{j}\}_{j \equal{} o}^{\infty}$ is like geometrical progression
hence
$ \sum \frac {q^{k}}{1 \plus{} \sin k } < \sum \frac {q^{k}}{1 \minus{} \frac {1}{k^{2}}} \equal{} \sum \frac {k^{2}q^{k}}{k^{2} \minus{} 1} < C$ for some $ C$
because if we consider each terms $ k^{2}q^{k}$ we see that for big $ k$ we have (say $ k^{2}q^{k} < \frac {1}{k}$
or more generally $ k^{2}q^{k} < \frac {1}{k^{n}}$ for all $ n$) because
$ n\ln k \plus{} k\ln q < 1$ for some $ k < K\in N$ (q<1)

It's not sufficient to show that $ x_{n}$ is bounded to say that it converges (ex: $ a_{n}: \equal{} (\minus{}1/q)^{n}$ and then $ x_{2n}\equal{}0$ , $ x_{2n\plus{}1}\equal{}1$, which gives $ x_{n}$ bounded, but not convergent). The $ a_{n}$ 's are not necessarily positive, neither is q.

  You can verify $ x_{n}$ is Cauchy if $ a_{n}$'s are bounded, which proves convergence.

  A simpler example for 2 is to choose t such that $ |q|<t<1$ and $ t>0$. Then $ a_{n} \equal{} (t/q)^{n}$ is unbounded (remember $ q \neq 0$and thus $ |t/q|>1$ since $ |q|<t$). Then you can easily show that $ \lim_{n \minus{}> \infty} x_{n} \equal{} 1/(1\minus{}t)$ with this selection.

I think he proved absolute convergence ;)

Anyway, this is a strange problem to give to 11th graders. Some basic facts about series will solve both (a) and (b) in no time.