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I have noticed that the level of Mathematics in America is getting lower and lower. The textbooks are getting more colorful, and much of the curriculum, in my opinion, is useless.

This is discussed in the article, A Mathematician's Lament, written by Paul Lockhart.

As students and as budding mathematicians, what do you all think?

+1 to your opinion.
They probably didn't "teach" the Pythagorean Theorem over and over again in the 1960's.  :rotfl:

You may want to focus your question a bit more.

Rather, there is a lack of insightful and qualified teachers at the high school level, and the public education system is more geared towards helping those who are underachieving than those who are overachieving.

On the other hand, I bet competition mathematics is having a larger and larger outreach, in part because of the internet (and sites such as this). The difficulty, too, is going up -- it's not hard to see that today's olympiad problems are much harder than those from a few decades ago.

Also, has the college education standard gone down?

I am not sure we have any proof that it is deteriorating, although it may be.
What Paul Lockhart is saying (I did read the whole 24 pages) is that it is being "taught" the wrong way. It probably was always taught wrong.
I am not sure how much I agree with Lockhart. I think math could be taught better iff teachers realized that respect should be a two way street and were ready to learn from their students. If a teacher expects to be always right and always wiser than the students, he or she has already failed.  That's my never humble opinion, never thrown on the face of one of my teachers, but always held, anyway.
A "good teacher" could introduce the problems; let students think about it, and
listen to their ideas, if any (maybe learn a better way to solve it from one of them). Then, the teacher could show the historical solution or a "good" solution in the most jargonless possible way. In that way, a teacher may get half of their students to learn the required material, keep the most capable from terminal boredom, and not stunt anyone's creativity.
What a teacher should not do is require that students follow a special pattern, like write "ASA" on the upper left hand corner of their triangle solution, before applying a prescribed solution recipe. Or require that they memorize the long division algorithm or the times tables, without caring for their meaning.

Aside from that, I see other problems with American math education. Let me count the ways until I get bored.
1_ The way teaching is defined includes lots of unnecessary memorization and lots of unnecessary jargon.
2_ Teachers do not understand the math they are supposed to teach. Many believe that you need to work hard, do a lot of repetitive exercises, ape they way they solve problems, and stick to a prescribed format (you'll loose points if you forget to label one of the axes in your plot with "x" or "y").
3_ The curriculum goes atround in circles with little new ideas introduced, especially in middle school. I have seen elementary schools change to a more problem-solving oriented approach, only to dump their graduates into that endless repetition and drudgery that is middle school. It does not get much better in high school, either (see if they let you skip pre-calculus).
4_ Counselors and teachers try to convince students to take less challenging courses. A student who thinks he/she can manage Algebra 1, will not be allowed to take it and will have to take Algebra 1A first, then Algebra 1B the next year.
5_ High school is a rat race. Rats should keep their eyes on the scoreboard, and do whatever it takes to get the best possible GPA. Argue your test grade. Ask for extra credit. Try to get on the good side of the teacher. Bring in the senselss homework, whether done using yout own head or someone else's. Grades are used as punishment and a way to classify students, rather than as a way of communicating what the students "got" and what they did not understand.
6_ There is no incentive to ensure that material is mastered for the short term, let alone the long term. If the student did not properly learn logarithms, apply a poor grade and let the student forget logarithms and proceed with the next chapter. Regardless, the student should not bother to remember logarithms, since they will be only a small part of the midterm (or final) that accounts for only 10% of the total grade, anyway.
7_ There is a bunch of unnecessary rules and requirements (often tied to grades) like prescribed size and style binders, notebooks, and other materials, bringing the book to class every day (whether it will be used or not), having a bunch of papers (class rules contract, interim progrss reports, for example) signed by parent and/or student, and so on.
8_I'll stop the list here, but I'm sure I could go on.

If there is any "deterioration", it is probably that class rules are getting more rigid and books are getting heavier, glossier, and more colorful, without any real improvement. There is unnecessary classification of possible problems, and step by step instructions on the one right way to solve each type of problem. Pictures of buildings, mountains and volcanoes appear, which are only dimly related to some problem or example in the text. Problems refer to young people apparently of the students' ages, with names that are very diverse sounding. There is a lot of unnecessary weight added to students' backpacks also holding the mandatory 2 inch binder.

[quote="mewto55555"]Yes I have noticed in many of the different schools I have gone to, they focus more on the lower end of the spectrum usually at the expense of the upper end, by forcing kids who learn faster to be in classes with those who learn a lot slower than them.[/quote]

is this bad?

like, is this bad in general? i don't mean "do you personally find this annoying"

i find it ridiculously ironic how many mathematically gifted kids support taxing the rich, yet vehemently oppose it when mathematical resources are diverted from stronger students towards weaker ones :\

you guys gotta understand that "level of mathematics" is generally not defined by the type of education that the very top-level kids receive.

[url=http://www.maa.org/devlin/LockhartsLament.pdf]For anyone who hasn't read the essay[/url]

@KMST: I think they teach math that way because it's easier to grade (that's why they make you do two-column proofs, memorize random jargon, and "show your work"). It's easy to take points off for silly technicalities like forgetting to bring your binder to class and whatnot; it's harder for them to actually evaluate the quality of student's solutions, their reasoning, their thought processes. Many teachers take the easy way out.

I think part of the reason they do this is because of standardized testing and No Child Left Behind. Seriously, what they're really teaching is how to do well on the test, not how to actually do math.

I once heard of someone who encountered a problem like this on a test: "Add 249 and 156 in your head. Show your work." (Wait, what?!)

Obligatory:

[quote]Teaching Math In 1950’s
A logger sells a truckload of lumber for $ \$$100. His cost of production is 4/5 of the price. What is his profit ?
Teaching Math In 1960’s
A logger sells a truckload of lumber for $ \$$100 His cost of production is 4/5 of the price, or $ \$$80. What is his profit?
Teaching Math In 1970’s
A logger sells a truckload of lumber for $ \$$100. His cost of production is $ \$$80. Did he make a profit?
Teaching Math In 1980’s
A logger sells a truckload of lumber for $ \$$100. His cost of production is $ \$$80 and his profit is $ \$$20. Your assignment: Underline the number 20.
Teaching Math In 1990’s
A logger cuts down a beautiful forest because he is selfish and inconsiderate and cares nothing for the habitat of animals or the preservation of our woodlands. He does this so he can make a profit of $ \$$20. What do you think of this way of making a living? Topic for class participation after answering the question: How did the birds and squirrels feel as the logger cut down their homes? (There are no wrong answers, and if you feel like crying, it’s ok.)[/quote]

Hmmm...well one thing I noticed on my own part was the entire curriculum of the "advanced" 10th grade math (trig+algebra 2) takes about 2 hours to learn, and all of ab calc, which is taught in the second semester of the "advanced" 11th grade course at my school took me about 1 hour to learn. By state law these classes require like 70 something hours total per semester. Typical overachievers who aren't good at math (I'm not saying I am, I am significantly better than the vast majority of people at school though, since this is AoPS) would take slightly longer, maybe a few days with 2 hours a day.

I'm not sure if it's gone down as so much as there is now a huge gap. While obviously the average student is worse, good math competitions have really spread and gotten decently harder. There are "math circles" and "math camps" that are quite popular. I'd say the top is getting a lot better and bigger. While we'll be stuck with a nation of mostly innumerates, our own intellectual microcosms are improving in terms of quality and size. Whether or not that's a good thing is up to you.

EDIT: Also, I really agree about that AP note. It'd be better if mathlike people just took the calculus classes at colleges where they'd learn it the real way, though as a whole I think it's achieved a lot in teaching basic calculus so that people who need it in college will be not be so lost.

I should clarify; I don't mean that it's bad to read ahead -- it's a great idea to try to get a feel of advanced material early on, and this gives you more time to get a feel for it.

I mean that someone should not blow off the "slow" stuff just because they've seen it before, because that person would be missing out on a lot.

I totally agree with that, by the way. When I was in 6th grade I did a lot of precalc/single variable calc from my grandpa, but only a year ago when I started doing the contests and proof based stuff did I really appreciate the beauty in mathematics and the entire scheme of things. The Finding Nemo analogy really is good.

hole in the wall!

oh hey, so guys, you know nobody is actually stopping you from studying math

PS

if you really want to study math hardcore, you could try IMing me

I may or may not reply

chances are yes but it may take an hour or so depending on when I notice

I doubt anyone will actually do this though

Hmm...... a joke i heard from Dr. Andreescu (former US IMO coach)

they went to a convention in Bulgaria for math teachers. After 2 weeks, one of the American scholars said:

"Bulgarian mathematics in middle school: draw a triangle. Draw equilateral triangles externally to its sides. Prove that the lines through opposite vertices of the triangle and equilateral triangles are concurrent."

"In American mathematics in middle school: [hide="if you bother to know"][hide="seriously"][hide="you will get pwned"]Draw a triangle[/hide][/hide][/hide]

This joke tells a lot about the differences

Also, I am starting linear algebra this year, and the textbook is HORRIBLE. It starts with solving linear equations in different methods, which were exclusively dicussed in Algebra i and ii (texas curriculum)

I was pissed off that i have to waste 2 weeks of 'advanced' mathematics in college (fyi, i am going to [hide="something"]9th[/hide]grade)

I really think that it's the matter that advanced people in mathematics are not willing to teach due to low income compared to how much they studied (yes, i am also afraid of this)

As I was coaching my middle school math team (yeah;) i was able to teach what the previous coaches couldn't teach, and that made a great improvement from 20th in state to 8th for a second year math team

[quote="TZF"]You know, there's plenty to be learned by going at a slower pace. Don't think that you've "learned" Calc AB just because you can get a 5 on the AP test; the AP test is a load of BS. The HMMT is a much better indicator of basic calculus understanding and intuition for your purposes.

[hide="A story if you care to read"][color=darkblue]I was steamrolling through the curriculum while I was in Connecticut - my middle school had a cooperative program with JHU-CTY by which we could enroll in a distance math class in place of a school math class, and I had up to Algebra 2 in the bag by the end of 6th grade. I came to Massachusetts, and they threw me into Pre-Algebra.

Honestly, that was the best thing that happened to me. I obviously didn't put up with the teacher's explanations, but by being forced to revisit the material/think about it more, a lot of things "clicked" -- that is, I was able to synthesize and more deeply understand the implications of basic concepts which I had previously taken de bene.

I ended up getting to Calculus in 10th grade -- 2.5 years after I would have had I stayed in Connecticut. But this time, I was ready to understand and better appreciate the subtleties.[/color][/hide]
High school math is like a Disney movie. Little kids can watch Alladin or Finding Nemo or whatever, be like oooh fishies, ooohh princess, oooh pretty colors, and be very entertained. But they likely won't be able to appreciate the clever dialogues or life messages until they revisit them later in life.

If you don't like what your teacher is doing, you don't have to listen. But since you're being forced to revisit the material, at least think about it more. Let it roll through your head. DO NOT read calculus in freshman geometry class, that's one hell of a waste of time.[/quote]
What I mean is I can learn what I need to know to ace the class easily in that much time. I never said I was good at calculus. But school doesn't require you to be. I don't mean that I'm just going to say, "Oh I know calc :D" I know there's a lot more to it. Honestly, school doesn't teach crap. I've found out for myself that self studying is the only way to improve.

How can you say that the past's education system was better? Until you have lived through it, you don't know what it was really like.

Is there a place where students don't forget the material [i]en masse[/i] after a test? Or subjects?

Actually your correct. I have a Trigonometry book dated 1898 and another dated 2003 and frankly the 2003 version is watered down. The 2003 version is more a pound and grind book, with little care as to if the student understands what is going on. The 1898 book makes you think and has been quite challenging for me.

The 1898 book is certainly not representative of American education at the time. At that time, 17 states did not mandate education, and over a decade later, 30% of American children were not receiving public education. American education has come a [i]long [/i]way from there.

The problem with American mathematics education fails to build interest in mathematics in the average youth, and it fails to dispel stereotypes associated with enjoying math. That doesn't mean it sucks. There is no lack of opportunity for those who do develop interest. I would count math contests, summer programs, and other enrichment opportunities that are driven by universities and the academic world as part of American education, since it's such a well-established and accessible system. It's just not mandatory, and not driven by town education budgets (though several opportunities are driven by federal and sometimes state dollars).

The lament is not as much about what [i]opportunities[/i] are available for the interested students but about what [i]requirements[/i] are imposed on all students and their teachers. Those seem to go down and down. Nowdays you can be a math major and graduate from an American university without knowing how to multiply complex numbers (I'm not joking: a colleague of mine just had to explain that to a [b]graduate[/b] student taking a course in complex analysis). Learning hardcore mathematics becomes more and more "optional".

Ok, there was one guy that happened to slip through the cracks. That doesn't make a nation.

And I always notice that it's never you. It's always "I have a friend who once knew a guy, or I had a colleague who had a friend who's brother didn't know 1+1"

No, one guy doesn't make a nation and he was an extreme example, of course. As to something I saw myself within the last month, I sent my daughter to a very decent private school, and still she recently was asked to compute the area of a triangle with sides 5,6, and 8 and the altitude (to the side of length 8) equal to 3. Again, one stupid lapse of one elementary school teacher doesn't make a general rule but I'd prefer to see (and hear of) fewer of those. Also, the necessity to teach elementary algebra (I mean the skill of opening parentheses and transferring terms to the other side with minus) in college is a shame and I've seen enough business calculus students who couldn't correctly add $ \frac 12$ and $ \frac 23$ too.

Yikes, this thread is actually like 10 threads now, but...

1)  As has been pointed out several times, most of the people commenting seem to take the deterioration of mathematics in America as a given, despite the fact that no one has offered a SHRED of evidence for the assertion.

2)  As has been implicitly suggested earlier, it's entirely possible for the general public to be learning *more* while the contents of particular *courses* deteriorate.  This can happen, for example, if more students take more higher level courses (which has, in fact, happened).

3)  I heartily enjoyed Mystic's post in which he says, roughly:  If you really want to learn some hardcore math, you should IM me, but I might just ignore you.

4)  The "math is useless" idea comes up from time to time, but I don't buy it.  Just because people don't use their education directly does not mean that the education is "useless."  The fact is that higher levels of education are associated with higher levels of productivity (as argued in The Race Between Education and Technology, by Goldin and Katz, which I admit I have not read), and therefore I think we can reasonably infer that ordinary people do gain something useful from non-vocational education.

And if you can argue that "most people" don't "use" high school math in their daily lives, then you can also argue that "most people" don't use *any* high school knowledge in their daily lives.  As useless as math is, it's at least as practical and relevant to most people's lives than history, literature, science, and foreign language.  (I guess the most useful classes must be health, PE, and home ec?)

5)  I heartily agree that teaching high school is a difficult and thankless and unrewarding job. That said, you can't blame students for disliking their teachers who are incompetent; it's not the students' fault that the teachers are bad.

6)  Regarding Singapore mathematics, I once heard that although Singapore dominates in most international measures of math education, they don't produce first-rate mathematicians.  (Um, no cite.)  I will refrain from theorizing why this might be so.

[quote="yenlee"]1)  As has been pointed out several times, most of the people commenting seem to take the deterioration of mathematics in America as a given, despite the fact that no one has offered a SHRED of evidence for the assertion.[/quote]

[insert snarky comment on how the majority of commenters who take the deterioration of mathematics in America as being self-evident are in middle school or high school and have roughly the perspective one would expect of a middle schooler or high schooler here]

e: apparently jbl has made a similar comment on the first page.  how silly to try and compete with him in matters of snarkiness :(

Just my comment.

[url]http://xkcd.com/285/[/url]

(I knew I was going to use this somewhere!)

[quote="yenlee"]
3)  I heartily enjoyed Mystic's post in which he says, roughly:  If you really want to learn some hardcore math, you should IM me, but I might just ignore you.[/quote]

Thank you. Of course, I say "hardcore math" slightly sardonically, as I'm well aware that the amount of math I currently know still pales in comparison to the totality, but in the context of this thread, I think I probably have some grasp of a reasonably healthy amount.

[quote="blahblahblah"]apparently jbl has made a similar comment on the first page.  how silly to try and compete with him in matters of snarkiness :([/quote]  Well, you're in the Round Table, so I have home field advantage ;)  In [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1206461#1206461]other parts of the forum[/url] I think you may be the undisputed king of snark.  (And in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=158612&start=120]this thread[/url] your 123 faced off directly with my 121 and 128, and I'd be hard-pressed to declare one of them to be snarkier than the other.)

I really feel like I should have something on-topic to add to the thread, but I can't think of anything interesting to add that I didn't already say in the comment blahblahblah mentioned :(

It could also be that the US may not improve as quickly as other countries in math as it could be a real deterioration...

You should try to be nice to the middle schoolers on AoPS.  You were in middle school once, too.  Unless they get uppity--then you can bring the hammer.


[quote="AceOfDiamonds"]use associative property to regroup:
$ (2\plus{}8)\plus{}(12\plus{}18)\plus{}(22\plus{}28)\plus{}(32\plus{}38)$
$ 10\plus{}30\plus{}50\plus{}70$
$ 160$[/quote]

Sum of first 4 odd positive integers:

1,3,5,7

4^2

Multiply that by 10 and you get 160.


Given is a disk D of radius 1.
What is the least number of disks of radius $\frac{1}{2}$ wich cover D?

Prove that any ten points in the plane can be covered by non-overlapping unit disks.

Consider a configuration of an infinitely long stretch of tangent unit disks:
[asy]
size(300);
for (int i=0;i<5; ++i)
{
draw(circle((2i,0),1));
draw(circle((2i-1,sqrt(3)),1));
draw(circle((2i,2sqrt(3)),1));
}
draw(circle((9,sqrt(3)),1));
[/asy]
Choose any ten points in the plane. We claim that it is possible to shift all of the disks in a certain direction so that all ten points are covered. The probability that a point will be covered after such a shift is $\frac{\sqrt{3} - \frac{1}{2}\pi}{\sqrt{3}}$, so the probability that all ten points are covered is $p = 1- 10 \cdot \frac{\sqrt{3}-\frac{1}{2}\pi}{\sqrt{3}} > 0.068$, implying that there does exist such a shift.


Rounded to the nearest integer, on average, five cars passed through the intersection of Q and 16th per minute on Tuesday. On that day, what is the largest number of cars that could have passed through the intersection?

i don't quite understand the problem  :blush:

Ehhh it was phrased kinda badly. Let me rephrase it:

On average, $ x$ cars passed through the intersection of Q and 16th per minute on Tuesday. Given that $ x$ rounded to the nearest integer is 5, find the largest number of cars that could have passed through the intersection that day.

[hide]Would it be 7,919, because there are 1,440 minutes in a day, and 5.5 cars is the lowest number that would round to 6 cars. $ 5.5\times 1,440 = 7,920$. The highest number below that is 7,919.[/hide]

[hide]

$ \text{There are}$, $ 60\cdot24= 1440$ $ \text{minutes in a day. The lowest number that can be rounded to}$ $ 5$ $ \text{is}$ $ 5.4999999\dots$
$ \text{Thus, gives the answer: }$

$ 1440\cdot5.49999999999999\dots$
$ =7919.9999\dots856$

$ \text{Because, the average}, x ,\text{was rounded down, we have to round this down too.}$

$ \text{Therefore, we get}$ $ 7919$

[u][b]Answer to the Traffic Problem:[/b][/u]$ \boxed{7919\text{ cars}}$

$ \text{So yeah, I got the same answer as vamathletes.}$
[/hide]


$ ABC$   is an isosceles triangle in $ A$. $ D$ is the midpoint of [$ AC$] and $ E$ is the  orthogonal projection of $ D$ on [$ BC$] and $ F$ is the midpoint  of [$ DE$].
prove that  $ (BF)$ and  $ (AE)$ are  perpendicular $ <\equal{}>$ $ ABC$ is  equilateral.

Call M the midpoint of BC and K the projection of B on AC.

Then $ AE \perp BF \ \Longleftrightarrow \ \tan \angle FBE \equal{} \tan \angle EAM \equal{} \frac{1}{2} \tan \frac{A}{2} \ \Longleftrightarrow$ $ \tan \angle EBD \equal{} \tan \frac{A}{2} \ \Longleftrightarrow \ \ D \equiv K \ \Longleftrightarrow \ \ a\equal{}c$.


graph theory
number theory unsolved
number theory

Hello everyone.

I have been working on a math problem for some time now, but I can not finish the proof. In other words, I'm stuck. I hope someone here can solve it for me, or perhaps we could make it into a joint effort. The problem is taken from a monthly math magazine, which features 3 math problems each month.

I don't think it belongs in this forum, but I am not sure where else to post it. So admins, feel free to move it about.

Here's the problem:

----------------------------------------------------------------------------------
Consider a set $ S \subset\mathbb{Z}$ with $ |S| \equal{} 15$ and the following property:

$ \forall s \in S : \exists a,b \in S$ such that $ s \equal{} a \plus{} b$.

Prove that for every such $ S$, there exists a non-empty subset $ T \subset S$  with $ |T| \leq 7$ such that the sum of the elements of $ T$ equals zero.
----------------------------------------------------------------------------------

I have come up with a few observations that might be useful, but I think it's best to wait and see how things go. If noone has an idea on how to approach the problem or if someone asks me for the info, then I will give it.

Until then, good luck!

- NightBreeze

Nice to see an overlap with xkcd, but I think this is probably a combinatorics question

That might be true. Maybe the good people who browse that part of the forum have a take on this : )

It seems there's not a lot of interest into the problem here.. so I figure I'll just spout some of my thoughts.

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

Let S be a set of integers with |S| = 15. We may assume that S contains neither zero nor an inverse of another member of s, for it would make the choice for our subset T rather trivial. Furthermore, if S contains neither, it can be shown that it must contain [i]at least[/i] three positive [i]and[/i] three negative numbers. The proof is rather simple, just some logic steps. So take my word for it for now.
Consider now A to be the positive members of S and B to be the negative members of S. Without loss of generality, we may assume that |A| < |B|. So 3 <= |A| <= 7. Now define a graph G on A as follows: Let x, y be members of A. There exists an edge x -> y if there exists a z in S such that x = y + z. By this definition, every member of A is adjacent to at least one other member of A, because every positive number can not be written as the sum of two negative numbers. Because A is finite, there must exist a cycle X = x1, ... xk. Let k be the smallest k with this property. Then there exist Z = z1, ... zk such that

x1 = x2 + z1 = x3 + z2 + z1 = ... = x1 + zk + ... + z1.

The the sequence z1, ..., zk will sum to zero.

The way I see it, the main problem in this proof has to do with the doubles that occur in such a sequence Z. If there are no doubles in the sequence, then we have found our T. So let's assume that there are x[sub]i[/sub] and xj with i =/= j such that xi = xi+1 + z and xj = xj+1 + z. It is easy to show that z can not be a member of X. If it were, then there would exist a smaller cycle through xi, xj and z. That is not allowed since we chose X to be the smallest cycle in A. So we are left with the case that z is not a member of X. We need to proof that either we can replace z and still get a T with |T| <= 7 or that there must exist a different cycle (possibly in B) with the required properties.

I experimented some with assuming that |X| < 7. Then it's possible for z to be positive. There are some implications there, but nothing useful. We can 'expand' z a number of times till we reach a member of X. A new cycle might be possible in that way. Remember that our upper limit on the cycle size is equal to the upper limit on the amount of positive numbers in S. The other possibility is that z is in B. Maybe it's worth looking into the smallest cycle in B then? We have no guarantee that its length will be smaller than 7, but maybe we can proof that the biggest cycle in B can be no larger than the size of A. This certainly sounds plausible.. since negative numbers are absolutely required to make a cycle in the positive part of S.

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

The set I've been looking into most recently is S = {-8, -4, -3, -1, 2, 5, 7, 9}. It's of size 8, naturally, but the cycle 9 -> 7 -> 5 -> 9 has a complement Z which features the number 2 twice. I'm trying to find a structural way to avoid this. My work so far:

-------------
Let X be the smallest cycle x1 -> ... -> x1 in A (the positive members of S). Then 3 <= |X| <= 7, because 3 <= |A| <= 7 and |X| = 1 -> 0 in S, |X| = 2 -> S has an inverse.

Now let xi and xj be such that xi = xi+1 + z and xj = xj+1 + z and suppose that i < j. Now there are a few cases I was thinking of addressing.

- Suppose z = xk in X. Then there exists a cycle inside of X which is smaller than X itself. This is not possible since I chose X to be the smallest cycle in A.
- Suppose |X| < 7. Then there can possibly be m = |A| - |X| members of A which are not in X. So suppose that z in A \ X. If we expand z m times, we must arrive at a member of X (since there are only m members of A left which are not in X). Then we can construct a new cycle which includes z and might possibly be larger than X but must be smaller or equal than 7.

The final case is when z is in B. I don't know how to approach that one.

Does this make sense?

Edit: I reread it and concluded, once again, that graph theory proof are totally unreadable without some drawing or example. So I'll just give you one.

Consider S = {-8, -4, -3, -1, 2, 5, 7, 9}. Then |S| = 8 and believe me that it satisfies the important property. Then A = {2, 5, 7, 9} and we found the cycle 9 -> 7 -> 5 -> 9 with complement {2, 2, -4}. In this case, xi = 9, xj = 7 and z = 2. Note that |X| = 3 < 4 = |A|, so m = 1. Here, z is in amongst the remaining members of A which are not in X, so z in X \ A. Note that if we expand z, we get z = 5 + -3. 5 is a member of X, which is to be expected after m expansions. Now we can construct the cycle z -> 5 -> 9 -> z (because 9 = xi) which satisfies our criterium.

The trick here is that the length of such a newly found cycle can not exceed |A|, which is what we wanted.

-------

There is no guarantee that this new cycle will not conflict, but I'm fairly confident we can repeat this process till we find a cycle in A which either doesn't have a conflict, or the conflicting z is not a member of A.

Hope this helps!


calculus
integration
trigonometry
calculus computations

1/72 \[ImaginaryI] (48 \[ImaginaryI] Catalan +
   3 \[Pi]^2 + \[ImaginaryI] \[Pi] (8 ArcCsch[Sqrt[2]] + Log[512] -
      6 (Log[(1 - 2 \[ImaginaryI]) + \[ImaginaryI] Sqrt[3]] -
         Log[2 + Sqrt[3]] +
         Log[-\[ImaginaryI] ((2 + \[ImaginaryI]) + Sqrt[3])])) -
   12 (2 ArcCsch[Sqrt[
        2]] (Log[(1 - 2 \[ImaginaryI]) + \[ImaginaryI] Sqrt[3]] -
         Log[-1 + Sqrt[3]] + Log[1 + Sqrt[3]] -
         Log[-\[ImaginaryI] ((2 + \[ImaginaryI]) + Sqrt[3])]) -
      PolyLog[2, 2 - Sqrt[3]] +
      PolyLog[2, -\[ImaginaryI] (-2 + Sqrt[3])] -
      PolyLog[2, -(\[ImaginaryI]/(2 + Sqrt[3]))] +
      PolyLog[2, 1/(2 + Sqrt[3])]))

Whatever this is. :D

Hey JRav . . .  whenever you want to post Mathematica code, you need to first convert it to Raw Input Form:  Select the cell, then choose Cell/Convert To/Raw Input form.  For the record, I realize that's not what Misan is asking for; he wants the manual calculations -- just trying to help JRav guys:

[code]
In[154]:=
Integrate[ArcTan[x]/(x*(1 + x^3)), {x, 0, 1}]

Out[154]=
(-(1/72))*I*(48*I*Catalan + 3*Pi^2 +
   I*Pi*(8*ArcCsch[Sqrt[2]] + Log[512] -
     6*(Log[(1 - 2*I) + I*Sqrt[3]] -
       Log[2 + Sqrt[3]] +
       Log[(-I)*((2 + I) + Sqrt[3])])) -
   12*(2*ArcCsch[Sqrt[2]]*
      (Log[(1 - 2*I) + I*Sqrt[3]] -
       Log[-1 + Sqrt[3]] +
       Log[1 + Sqrt[3]] -
       Log[(-I)*((2 + I) + Sqrt[3])]) -
     PolyLog[2, 2 - Sqrt[3]] +
     PolyLog[2, (-I)*(-2 + Sqrt[3])] -
     PolyLog[2, -(I/(2 + Sqrt[3]))] +
     PolyLog[2, 1/(2 + Sqrt[3])]))

[/code]


Prove that the numbers $ a^2,b^2,c^2,d^2$ are in arithmetic progression if and only if the numbers $ \sqrt[3]{\frac{a^5}{bcd}},\sqrt[3]{\frac{b^5}{cda}},\sqrt[3]{\frac{c^5}{dab}},\sqrt[3]{\frac{d^5}{abc}}$ are in arithmetic progression .

[hide]Note that $ \sqrt [3] {\frac {a^5}{bcd}} \equal{} \sqrt [3] {\frac {a^6}{abcd}} \equal{} \frac {a^2}{\sqrt [3] {abcd}}$. Let $ \frac {1}{\sqrt [3] {abcd}} \equal{} x$, then since $ a^2$, $ b^2$, $ c^2$, $ d^2$ are in arith progession, so are $ a^2x$, $ b^2x$, $ c^2x$, $ d^2x$[/hide]

What interesting is there are no positive integer solutions for a,b,c,d [url]http://www.mathpages.com/HOME/kmath044/kmath044.htm[/url].


Let $n$ be a fixed positive integer. Let $S$ be any finite collection of at least n positive reals (not necessarily all distinct). Let $f(S) =$ $(\sum_{a\in S}a)^n  $, and let $g(S)$ = the sum of all $n$-fold products of the elements of $S$ (in other words, the nth symmetric function). Find $sup_S \frac{g(S)}{f(S)}.$

For any $n$ elements $a, b, ... , w$ of $S$, the coefficient of $a^1b^1...w^1$ in the multinomial expansion of $f(s)$ is just $\frac{n!}{(1! 1! ... 1!)} = n!$. In other words, $f(S) = n! g(S) $+ other terms. But the other terms are all positive, so $f(S) > n! g(S)$. Hence $\frac{g(S)}{f(S)} < 1/n!$ . This establishes that $\frac{1}{n!}$ is an upper bound.

Take $S$ to be $m$ elements all $1$. Then $f(S) = mn$ and $g(S) = mCn$. Hence $\frac{g(S)}{f(S)} = (\frac{m}{m}) (\frac {(m-1)}{m}) (\frac {(m-2)}{m}) ... (\frac {(m-n)}{m}) \frac{1}{n!}$ . As m tends to infinity, each of the $n$ terms $\frac{(m-r)}{m}$ tends to $1$ and hence$\frac {g(S)}{f(S)}$ tends to $\frac{1}{n!}$ . Hence $\frac{1}{n!}$ is the least upper bound.
$T_{h^{i^{s}}}$  $W^{a}s$ $PUTNAM \\P_{r}O_{b}L_{em}$


Putnam
calculus
abstract algebra
linear algebra
Ring Theory
college contests

can someone please tell me what topics does the putnam contest cover???

thanks

Because the Putnam is a college-level competition, some of the restrictions on high school competitions do not apply. Calculus may be required. (Problems that involve or potentially involve calculus are often among the easier ones - see 2006 A1, which turned out to be the volume of a solid of revolution.) A good working knowledge of freshman/sophomore college math material - calculus, linear algebra, and so on - helps. But many Putnam questions could potentially be offered in high school competitions. You don't generally need highly particular or specialized advanced knowledge - the distinction between a Noetherian ring and an Artinian ring is just not going to come up. (In fact, the "abstract algebra" problems may well involve unfamiliar structures studied from first principles.)

I would suggest looking at old Putnam problems to get the idea. There's a substantial archive of them linked to the AMC website.


Let $a_{n}>0$ be a decreasing sequence, such that $\Sigma^{\infty}_{n=1}a_{n}$ converges. Let $b_{n}>0$ be a bounded sequence. Show that

$\Sigma^{\infty}_{n=1}(b_{1}+......+b_{n})(a_{n}-a_{n-1})$ converges.

[hide]
Switch $a_{n-1}$ and $a_{n}$ in the last summation so the sum is positive. We want to show that

$\sum_{n=1}^{\infty}(b_{1}+\cdots+b_{n})(a_{n-1}-a_{n})$

converges. Since we can bound $b_{1}+\cdots+b_{n}$ by $Mn$, where $b_{i}\le M$ is the upper bound of $\{b_{n}\}$, it remains to show that

$\sum_{n=1}^{\infty}n(a_{n-1}-a_{n})$

converges. However, this is simply

$\sum_{n=1}^{\infty}[-na_{n}+(n-1)a_{n-1}+a_{n-1}]$,

which telescopes to

$\sum_{n=1}^{\infty}a_{n-1}$

which converges by hypothesis.[/hide]


analytic geometry
floor function
advanced fields
advanced fields open

How many integer points lies in circle with radious $ r$? It's center is at point $ (0,0)$.

It's the number of integer solutions to $ a^2\plus{}b^2 \leq r^2$, and I don't think you're going to find a closed form for it.  Clearly it's asymptotic to $ \pi r^2$.

Denote the desired number by $ f(r)$. $ f(r)$ is equal to the sum over all possible $ x$ coordinates of the number of points with that $ x$ coordinate. The possible $ x$ coordinates are clearly all $ k$ such that $ \minus{}\lfloor r\rfloor\le k\le\lfloor r\rfloor$, so
\[ f(r)\equal{}\sum_{k\equal{}\minus{}\lfloor r\rfloor}^{k\equal{}\lfloor r\rfloor}n_r(k),\]
Where $ n_r(k)$ is the number of solutions with x coordinate $ k$. It is not hard to see that $ n_r(k)\equal{}1\plus{}2\left\lfloor\sqrt{r^2\minus{}k^2}\right\rfloor$, so

\[ f(r)\equal{}\sum_{k\equal{}\minus{}\lfloor r\rfloor}^{k\equal{}\lfloor r\rfloor}\left(1\plus{}2\left\lfloor\sqrt{r^2\minus{}k^2}\right\rfloor\right),\]

which can be simplified to
\[ 1\plus{}4\lfloor r\rfloor\plus{}4\sum_{k\equal{}1}^{\lfloor r\rfloor}\left\lfloor\sqrt{r^2\minus{}k^2}\right\rfloor\]

using the method of stationary phase, you can get that the number of lattice points is $ \pi r^2\plus{}O(r^{2/3})$, which is an improvement over $ \pi r^2\plus{}O(r)$


inequalities
calculus
inequalities proposed

$a,b,c\ge0$. For real $r\ge0$, prove
$\sum_{cyc}b^rc^r(b-c)^2(a-b)(a-c)\ge 0$.

OK. But do you have an elegant way for 'interpolating' this for [i]non-integer[/i] $r$? Without too much calculus :)  I am still looking for one ;)


Could someone please explain what the following means?

$ \text{rate} \sim  \frac{\text{decrease in reactant concentration}}{\text{time}} \sim  \frac{\text{increase in product concentration}}{\text{time}}$

Thanks in advance!

[quote="elcric"]Could someone please explain what the following means?

$ \text{rate} \sim \frac {\text{decrease in reactant concentration}}{\text{time}} \sim \frac {\text{increase in product concentration}}{\text{time}}$

Thanks in advance![/quote]With "=" instead of "~" it wouold look like a definition of rate of reaction.

Why would
\[ \frac{\text{amt. of decrease in reactant concentration}}{\text{time}} \equal{} \frac{\text{amt. of increase in product concentration}}{\text{time}}
\]

Thanks!

Well why don't we consider a generic reaction:

$ \ce{aA} \plus{} \ce{bB} \rightarrow \ce{cC} \plus{} \ce{dD}$

What does it mean to measure the rate of the reaction? It really just means to measure how fast a reaction is going and since we know that during a chemical reaction the reactants are converted to products wouldn't it logically follow that measuring the rate at which the reactants are used up (or the products are produced) will give an indication of the rate of reaction?

Try and see if you can deduce why the following holds:

$ \frac{\minus{}1}{a} \frac{d[A]}{dt} \equal{} \frac{\minus{}1}{b} \frac{d[B]}{dt} \equal{} \frac{1}{c} \frac{d[C}{dt} \equal{} \frac{1}{d} \frac{d[D]}{dt}$.


Don't give me the answer, give me a few hints for both of these.

1. In a certain suburb, the power supply is interrupted during peak hours on average once every 7 days, and the power supply is interrupted during off-peak hours on average once every 17 days. On some days, the power supply is interrupted both during peak hours and off-peak hours. Peak hours are between 6am and 9am and again between 5pm and 9pm. What is the probability that the suburb's power supply is interrupted on a given day?

2. In the sequence of numbers 1,2,3....,2008, it is possible to choose two (different) numbers whose sum is divisible by 11. A new sequence is formed by excluding some numbers from the original sequence in such a way that it is impossible to choose two numbers from the new sequence whose sum is divisible by 11. What is the maximum number of numbers in this sequence?

1.  Hint I - Casework.

Hint II - Any given time - P(Peak hours) - 1/7
Any given time - P(off-peak hours) - 1/17

2.  Divisibility rule for 11 -

Alternately add and subtract the digits from left to right. (You can think of the first digit as being 'added' to zero.)
If the result (including 0) is divisible by 11, the number is also.
Example: to see whether 365167484 is divisible by 11, start by subtracting:
[0+]3-6+5-1+6-7+4-8+4 = 0; therefore 365167484 is divisible by 11.

^ (Dr. Math)

Good luck!

For your Hint I:
More subtle hints please; thanks :)

For Hint 2:
???

Easier ways:
#1:
P(a or b)=p(a)+p(b)-p(a and b)
Basically the time is irrelevant. Our problem is now:
P(off peak or peak)=P(off peak)+P(peak)-P(we have both off peak and on peak)
You should get [hide]23/119[/hide]

#2 can be solved like so:

Consider all numbers in mod 11. Write them all down mod 11. We see there are _____ numbers that are 1,2,3,4,5,6 mod 11 and ________ numbers that are 7,8,9,10, 11 mod 11. We know that if we have a 1 mod 11, we cant have any 10 mod 11, if we have a 2 mod 11, we cant have any 9 mod 11, etc. So since there are more 1,2,3,4,5 mod 11 than 7,8,9,10,11 mod 11, we want to use all of the 1,2,3,4,5 mod 11 numbers, for a total of _______ numbers. And we can also use ONLY one 0 mod 11.
You should get [hide]916[/hide]

D: I don't know modular arithmetic, so I couldn't solve it that way.  Nice solutions thunder.

Adihtya, how would you have solved it (w/o modular arithmetic)?


The other day, my acquaintance went to China, He bought me a mathematics book in Chinese ""高等数学, 上巻, 下巻”　in attached.
For what　grades students are those books used?

They are designed for college-level students (implied by the title which literally means "advanced mathematics").


Let be $ x,y,z>0$ such that $ x\plus{}y\plus{}z\equal{}3$ . Then show that :

$ \frac{x^2y}{x^2(x\plus{}1)\plus{}y^2}\plus{}\frac{y^2z}{y^2(y\plus{}1)\plus{}z^2}\plus{}\frac{z^2x}{z^2(z\plus{}1)\plus{}x^2}\le 1$

\[ %Error. "eqalign" is a bad command.
{
  \& \sum {\frac{{{x^2}y}}
{{{x^2}(x + 1) + {y^2}}}}  = \sum {\frac{{y(x + 1 + 1)}}
{{\left( {x + 1 + {{\left( {\frac{y}
{x}} \right)}^2}} \right)\left( {x + 1 + 1} \right)}} \leqslant \sum {\frac{{y(x + 2)}}
{{{{(x + 1 + \frac{y}
{x})}^2}}}} }  \leqslant \sum {\frac{{y(x + 2)}}
{{9%Error. "root" is a bad command.
3 \of {{y^2}} }}}   \cr
  \&  = \frac{1}
{9}\sum {%Error. "root" is a bad command.
3 \of y } (x + 2). \cr}\]
We have:
\[ \sum 2 %Error. "root" is a bad command.
3 \of y  \leqslant 6\]
Let
\[ f(x) = %Error. "root" is a bad command.
3 \of x \,\,\,\]
f is concave and by jensen theorem:
\[ \sum {\frac{x}
{3}f(y)}  \leqslant f(\sum {\frac{{xy}}
{3}) = %Error. "root" is a bad command.
3 \of {\sum {\frac{{xy}}
{3}} } }  \leqslant 1 \Rightarrow \sum {x%Error. "root" is a bad command.
3 \of y  \leqslant 3}\]
So finally,
\[ LHS \leqslant \frac{1}
{9}\sum {%Error. "root" is a bad command.
3 \of y } (x + 2) \leqslant \frac{1}
{9}(3 + 6) = 1\]


What does tg mean and LaTexing it would help greatly.

I believe this is what he is asking:

Prove that:

\[\frac{1}{\tan x \sin x (\cos x -\sin x)}>8 \ \ \text{for} \ \ 0<x<\frac{\pi}{4}.\]

[quote="Stas"]Prove that 1/(tgx*sinx*(cosx-sinx))>8
0<x<pi/4[/quote]
Latexing it:
Prove that $\frac{1}{\tan x \cdot \sin x \cdot (\cos x - \sin x)}>8$
$0<x<\frac{\pi}{4}$

edit: sorry, didn't see your post

Sorry for the form of my task. It's  my first post in this Forum. Is anybody know how prove the inequality?


Is there any compiled solutions to 2007 AIME's?

I got these problems from the Contests > USA > AIME 2007 and there wasn't any solutions to the problems. And i want to compare my answers to text book solutions

[[2007 AIME I Problems]] (Solutions as well)
[[2007 AIME II Problems]] (Solutions as well)

On our own [[AoPSWiki]].

[quote="gauss1181"]They're also available for order on the AIME 2007 Solutions Pamphlets, as well as on the 21st Century Contests CD.[/quote]

yeah, but why would we order if we have them for free on the web? :wink:


I'm trying to come up with a mathematically feasible definition of a straight line. Here's what I've come up with so far.

1) The locus of points where the ratio between the vertical and horizontal expansion factor is constant.

2) A line segment (defined as the shortest distance between two points) extended infinitely in both directions.

Are either of these authentic definitions? If not, how would you define a straight line?

The first one is much better. A definition shouldn't depend almost entirely on another definition.

if i were you id look it up on dictionary.com :D

I did that but I didn't get very many "mathematical" definitions. For instance, I found:

"a length (straight or curved) without breadth or thickness; the trace of a moving point"

"That which lies evenly between its extreme points." Euclid. "The shortest line between two points." Chauvenet. "A line which has the same direction through its whole length." Newcomb.
- Euclid.

and my personal favorite

"Lying in a straight line."

:wallbash: :wallbash:

This is the best one I could find:

"a line traced by a point traveling in a constant direction; a line of zero curvature"

Perhaps the definition I'm looking for is closer to that of a [b]linear function[/b]

A linear could be defined as "A function that, when graphed in Euclidean space, gives a line."


If $\sin 2x \sin 3x = \cos 2x \cos 3x$, then one value for $x$ is

A. $18^\circ$
B. $30^\circ$
C. $36^\circ$
D. $45^\circ$
E. $60^\circ$

[hide]
similar to finding $\tan 2x\tan 3x=1$

$\implies \tan 3x=\cot 2x$

$\implies \tan 3x=\tan(90^0-2x)$

$\implies x=18^0$

[/hide]

[hide="Answer"]$\sin$ and $\cos$ are reflected across $45^\circ$, so $|45-2x|=|45-3x|\Rightarrow x=18^\circ\Rightarrow \boxed{A}$.[/hide]

hello, why not a more complicated way, since i don't seem to understand the "easy" way. Use (sinu)(sinv) and (cosu)(cosv) formulas to find [cos(-x)-cos(5x)]= [cos(-x)+cos(5x)] then solve... 0=2cos(5x)  which seems rather straight forward to me. Then it becomes easy since the coefficient 2 is negligible. "what times 5 lands on the y axis on the unit circle". Sorry for the elementary explanation but i'm very new; just started ready AOPS V1 as a senior in HS, literally went from bad to decent in math in a few months(which has encouraged me to want to pursue math as a major in college); it sure would be nice to do well on the amc atleast.

[hide="Yet another way"]$ \cos 2x\cos 3x\minus{}\sin 2x\sin 3x\equal{}0$

$ \cos (2x\plus{}3x)\equal{}0$

$ 5x\equal{}90^{\circ}$

$ x\equal{}18^{\circ}$[/hide]

[hide="Solution"]
\begin{eqnarray*}\tan (2x)+\tan (3x) &=& 1\\ \tan (3x) &=&\frac{1}{\tan (2x)}\\ &=&\cot (2x)\\ &=&\tan(\pi/2-2x)\end{eqnarray*}

Now its not too hard to see that $ x=\pi/10\implies\boxed{18^{\circ}}$.[/hide]


I do not know why the other related topic was locked and if there is something illegal with this book. But I have seen the book in many sites. I would like Arne to inform me.

   Anyway  , if somebody wants any book , please write to

   stergiu@otenet.gr

 There are a lot of nice books I can give. But I'll will send you only those which are  legal and out of print.For exaple I can not give you [b]IMO compendium[/b] or  Andreescu's books.

 Babis

[quote="stergiu"] I can not give you [b]IMO compendium[/b] or  Andreescu's books.

 Babis[/quote]
I very like Andreescu's books :D

[quote="kunny"][quote="stergiu"]There are a lot of nice books I can give. Babis[/quote]

Could you give us the list of books?[/quote]
plz

[quote="Arne"][quote="stergiu"]I do not know why the other related topic was locked and if there is something illegal with this book. But I have seen the book in many sites. I would like Arne to inform me.[/quote]

I didn't lock any topic...[/quote]

 Thank you Arne !
 I thought that you locked the topic , probably because somebody demanded it from the forum , since the download was illegal.
 Any way , I know that some books are very expensive and it is difficult to buy them. In Athens , there ia a friend of mine who haw bought all the olympiads books and gives them very cheap(photocopies). So in Greece it is easy to heve them. But in other countries we can not send them, because the post is very expensive.
  We have some books in pdf form.Some books by Andreescu ,Feng , Englel , Cuzcma , Prasolov's geometry , Coxeter's geometry revisited and some other(Honsberger ect).
 I hope that in a few weeks I will have the greek site(link) to download some of them. Otherwise , I will send  some books with e mail. You can ask me what you want. Some day I will write a list with what I have.
 These days I'm very busy because I must correct the exam papers of my high school students.Sorry. But to
 [b]otman4u[/b] I have sent Prasolov's Geometry ( he asked me that , in my e mail).

[b] Babis [/b]

1)Why you can found those books?
2) Where I can find those books?

The most of these books can be found at

    http://www.amazon.com

 searching for olympiad books or problem solving books or Andreescu ect.
 But you can not find free e - pdf forms on internet , except

   [b]a) Geometry by Prasolov
   b)  Problem solving strategies by A.Engel  [/b]
For some other books I will give you links after some days, or send e mail.

[u] Babis [/u]

[quote="stergiu"][quote="N.T.TUAN"]Please send ONLY list to me because I have a lot of e-books! :P[/quote]

 Well ,
 it seems , it is easier for you to make also a list. Then we can change some of the books to complete our lists  :) .

  If I remember well I have(between other books) in pdf :

   a) Engel  : strategies
   b) prasolov: geometry
   c) Andreescu : olympiads around the world 2000 - 2001
   d) Andreescu : 102 compinatorial - 102 trigonometric problems

 [u]Babis[/u][/quote]
Hmm...  :) I had all  :P

In Greece we had an unofficial site with many e books(all the above and many others). It seems that some editors usinf spywares found the site and closed it. So , we have problems with the links. I think now that the only way to have the books is via persolal e mails. I'm afraid that editors read this forum and destroy all these ways of comunications.
 NT.TUAN has also all these books , so you can ask him too, to send you your desired pdf's.

  [u]Babis[/u]


How often does cheating in the AMC's occur? I mean, its by a school and the answers are sent along with the problems!!

Most schools have enough integrity to not cheat :wink:

lets not talk about this...  :ninja:
Most people who even take these exams are not cheaters...

They are honorable people.


The value of $ X$ is between $ \minus{}20$ and $ 10$, and the value of $ Y$ is between $ \minus{}10$ and $ 20$. Find the sum of the upper and lower bounds of $ X \times Y$.

The lower bound is the the opposite of the upper bound, thus our answer is $ \boxed{0}$.


I'll explain the general method:
(1) Let $u$ denote this long, scary expression
(2) Compute $u^{3}$. Note that many things cancel out.
(3) Now it is easy to solve for $u$.

How about you don't do that, and you just see that $(4+\sqrt{2})^{2}= 18+\sqrt{128}$.

Realizing that, we all know that the answer is [hide]2+rt(2)[/hide][/hide]

[quote="vishalarul"]Oh, I thought I could use the general method.  :fool: I ended up with a nasty cube root.

What would you do if the problem had a $129$ in the square root instead of the $128$?[/quote]

Maybe you could use the fact that $\sqrt[3]{a}+\sqrt[3]{b}= \frac{a+b}{\sqrt[3]{a^{2}}-\sqrt[3]{ab}+\sqrt[3]{b^{2}}}$.

That might help, or it might make things worse.

[quote="vishalarul"]Oh, I thought I could use the general method.  :fool: I ended up with a nasty cube root.

What would you do if the problem had a $129$ in the square root instead of the $128$?[/quote]

you either

A)write WTF as your answer
B)leave the contest hall because you dont want to put up with the hassle of such a terrible problem
C)be a man and beat it to death until you get some insane answer that has almost no chance of being correct
D)skip it and do the other ones

btw...

another method.

set the sum equal to $x$.

we now have $a+b+x=0$,

$\therefore a^{3}+b^{3}+x^{3}=3abx$

How did you get $a^{3}+b^{3}+x^{3}=3abx$ using the fact that $a+b+x=0$? Did you factor the first equation? If so, tell me what you got.

By the way,
If the $128$ really was $129$, then the number would be a solution to the equation:
$u^{3}= 2\sqrt{18+\sqrt{129}}+3u\sqrt[3]{2\sqrt{129}}$
Have fun trying to solve it!

a common factorization:

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(\sum_{\sigma}a^{2}-\sum_{\sigma}ab)$

and the result follows.

What do the "little sigmas" under the sums mean?  :?: Is this some kind of convenient notation or something, because I don't get it. Can someone explain?

cyclic sum...  i think you can compare the two identical posts to figure out their meaning.

[quote="vishalarul"]
By the way,
If the $128$ really was $129$, then the number would be a solution to the equation:
$u^{3}= 2\sqrt{18+\sqrt{129}}+3u\sqrt[3]{2\sqrt{129}}$
Have fun trying to solve it![/quote]

you still realize that even w/ those numbers that method would prove barren.


Find $\lim_{x\to3}\frac{2x^{4}-3x^{3}-81}{x^{5}-10x^{3}+27}$ Apply l’Hopital’s rule as many times as necessary, verifying your results after each application.

I wouldn't use L'Hôpital here, of course. Are the numerator and denominator both equal to zero at $x=3?$ If not, then this isn't an indeterminate case and we can just settle it on the spot. If so, then the numerator and denominator both have $x-3$ as a factor. Pull out that factor. (Knowing for sure what one factor is is a great help in factorization, of course). Then reexamine what you have.

use of l hopitals rule must be a little as possible
l hopital is just a short cut for solving
real understanding will come when it is solved in other ways
just like here using l hopital will be disaster!!!!!!!! :D  :mad:

It's true, it does make this one easier.

[quote="Dvagner"]what is wrong with l'hopital's rule when it can make ur life much easier, especially when you know how to do it the actual way[/quote]

As to Japanese high school students, they don't study l'Hopital's theorem.
So they are not allowed to use the theorem in Japanese university entrance examination.


MIT college AMC USA(J)MO USAMO Harvard Duke

post any hopes/plans for college. or basically anything

believe it or not, college prep is actually playing a really major part in my life (even though i'm still in 8th grade). Part of the reason is that i want to go to Harvard. i'm not sure what i'm going to major in, but it might be bio-med. idk, i like the field...

[quote="mwpl11"][quote="undefined117"]yeah, you should take the purdue summer research program next year if you're in bio honors.. i am, probably with some chemistry professor..

in addition to summer pe and possibly summer strings?

heck my SAT score is coming out hopefully tomorrow.. it was supposed to be today. if not, it's next wednesday

EDIT: oh yeah, imma go to MIT :D[/quote]

ya, i think i'm going to do the summer research thing next year. and i hope u know i have an extreme aversion (big word!) to orchestra. hey, maybe that's why i quit in FIFTH grade!

P.S. is it possible to start the summer research thing this year? (8th grade).[/quote]

no, it's for honors bio only

which means you already have research and paper writing experience

etc

just get a gender change

[quote="undefined117"]yeah, you should take the purdue summer research program next year if you're in bio honors.. i am, probably with some chemistry professor..

in addition to summer pe and possibly summer strings?

heck my SAT score is coming out hopefully tomorrow.. it was supposed to be today. if not, it's next wednesday

EDIT: oh yeah, imma go to MIT :D[/quote]

what did you get on your SATs?  Just wana compare...
But like I said, MIT is unintentionally being biased by not trying to be biased, but since you might make MOSP this year, it shouldnt matter.

[quote="the future"]umm
dont really care...

My PSAT scores were 210+ (i think this is NMSF for 11th graders)...but that sucked compared to the other sophomores in my school (if they werent bsing me)

and i think I made USAMO this year...

other than that, my grades suck, i suck at orchestra, i suck at sports, i suck at volunteering, and i suck at writing essays

sorry i used sucked too many times...
ahh dont care if I only go to IU if they give me full scholarship
hopefully, MIT, Cal Tech, Harvard, Duke, or Stanford  :)[/quote]

Wow typical ____ (i dont wana get in trouble)-fill in that blank.  3.95+ GPA UW doesnt constitute as grades suck (yes i stalk you and calculate your GPA)

[quote="stupidkid"]Having a crappy GPA and SAT score with only USAMO won't get you very far in the college admissions. USAMO + 3.8 + 2200 probably won't get you into MIT, Caltech, Stanford, etc.

And, don't be surprised if you get rejected from MIT with 2x USAMO, 2300, and 4.0.[/quote]

You have to write a good essay.  Its the most important part because everyones so close to eachother but the essay distinguishes and makes each person sound unique to the colleges.

Why exactly are we reviving year old topics?

[quote="firecricket91"]You have to write a good essay.  Its the most important part because everyones so close to eachother but the essay distinguishes and makes each person sound unique to the colleges.[/quote]
I think that's a common misconception. Essays aren't as important as people make them out to be. As long as it reflects who you are, there isn't really much else you can do with them. Don't spend too much time trying to make them unique. Chances are, someone has already thought of that and it isn't unique anymore. Embellishments, big words, etc don't help at all. I personally thought my essays were junk. I didn't even have any of my friends proof read them.

firecricket - my SAT scores were:
790 math (my brain failed me)
650 reading
580 writing
for a total of 2020 in 8th grade. I really need to work on writing essays (SAT, not college app) quickly.. and to remember to put a title on the page..

stupidkid - hehe what about 4x USAMO, 2400, and a 4.0?

hah 21^2th post

lol you beat my 8th grade SAT score

and i only got a 75 on the PSAT this year in math (stupid curve, missed one and it brought me down 5 points) when you couldve missed 3 on the reading and got a 75ish

hmm im not expecting anything on the USAMO (maybe 6-7) but I think if you can seperate yourself from that 0-15 USAMO score range and make MOP, that would be almost a lock for MIT (like with 3.8, 2200...)

No, there's no "lock" for MIT, unless its IMO or #1 in Siemens/IntelSTS. Blue MOPers have been rejected from MIT.

yea i should do Siemenns..but I cant find a math project...

[quote="ghjk"]Which grade are you in, the future? Since I also intend to join in Siemens as well as Intel(if possible), I saw the rule and Siemens said I must be in grade 12 to join in :([/quote]

wait really?  i thought group you could do 11th?

considering i know stupidkid^^ did it in 11th and i know a couple of others that did it

me in 10th grade.

i actually probalby can find a really smart partner (way smarter and more dedicated than me), but I cant really find  a "sponsor" so I dont know if i can do it...

Math SAT just leaves me... speechless.  I took a real test (although not officially) the other day, and I picked the Math Subject Test Level 2 (since it's the most challenging).  It took me 33 minutes out of the 60 I had, and I used a horrible calculator (I wasn't at home when I took the test).  And I got a perfect without even trying...
I mean, seriously, the MathCounts School test is harder to get perfect on than the Math SAT... It's true that the SAT has more advanced [i]subjects[/i], but even programs such as MathCounts require much, much, much more advanced [i]thought[/i].
I remember that when I was in 4th grade, I knew this guy who was in college who got a perfect on the Math SAT, and I was impressed.  But 4-5 years later, which is now, it is not impressive.  It actually makes me [i]angry[/i] the way the Math SAT is designed.
Someone who wants to get a perfect doesn't need to know math very well.  They can just plug in answer choices or eliminate the wrong ones.  They can just memorize formulas the week before and use them all on the day of the test.  They can use their calculators on a majority of the problems, and for calculations that AoPSers can do in their heads faster.
I'll give an example of one of the many disappointing problems that I see.  As I said earlier, the SAT Math Level 2 Subject Test is the hardest math test they offer.  And (for those of you who don't know) the problems go from easy to hard based on their number, and they are numbered from 1 to 50.  So a problem like #39, especially on the Math Level 2 Test, is supposed to be hard.  I'll give a REAL example of a #39 (this problem is on the May 2002 Test):
"The Fibonacci sequence can be defined recursively as
$ a_1 \equal{} 1$
$ a_2 \equal{} 1$
$ a_n \equal{} a_{n \minus{} 1} \plus{} a_{n \minus{} 2}$ for $ n\ge3$.
What is the $ 10th$ term of this sequence?"
Yes, this was one of the most challenging problems on the test.  And, yes, only 52% of the nation's best math students (because only the best would take the hardest test) got this question correct.  Pitiful.

Sorry for the rant...  When I take an official SAT Math Level 2 Subject Test this October, I'm probably going to get something like this wrong on accident...  Oh well, I can just use the remaining time (which will be about 27 out of 60 minutes) to double check!

comparing that to normal SAT math.. :rotfl:

The hardest problem on the SAT I math on 2007 was like
what is the product of
(a-x)(b-x)(c-X)......(z-x)
I mean what is that? (Sighs...)
I guess because there are so many people who have trouble w/ math, and this is just to say that they exist.

um.....that was a riddle i gave to you at school once. i don't think that was on a SAT, the answer is 0.

Uh, it was both a riddle from you and an SaT question.
Yes I know the answer is 0 (duh)

@cognos:have you checked your PM box yet?

Come on, the SAT is easy once you study it.
Writing - just learn the grammar rules. You only need a 9 on the essay, which is not too difficult.
C.R. - Learn the vocab. Easy 700 there afterward.
Math - No tips needed.

I got my scores from 1600 in 8th to 2280 average now (going to 11th) and up 200 points since last year.

A better forum to ask these types of questions would be collegeconfidential.com.
I don't know much about that place, but I went there once and it's pretty informative.

Um C.R. isn't all vocab. I would consider myself as being able to pwn all SAT vocab but yeah my reading score isn't exactly that high, while my writing score pwns my reading

sentence completiton isnt even all vocab, its all logic...

i guess you have to know some vocab for the one line blanks but for the two lines, you really dont, just need to know the definition relative to the other word....

theres about 1/3 things that tests "vocab" in CR, 19 from SC and like 4-5 from reading passages and words in context or how author portrayed something...

yea writing is pretty easy to score well on because if you get like a 12 on essay, you can miss like 5 and still get 800 (i think...)

umm but werid, i have never gotten perfect on math in any test i have taken (ACT back in middle school, SAT, or PSAT...) but i always get perfect in practice tests (of course on the real ones i miss perfect by like 1-2 questions but thats weird cause i have time to check every question 3 times but for some reason i make same mistake...lol)

mooraga....dang nice improvement (way better than me), i only like imporved 200 or so points from th grade

for some reason my reading sucks...i dont know why but i can get like 750-800M/750-800W(no essay since i would be biased in giving myself a 12)/650-700CR and this always happens on practice tests...its kinda hard for me to do the stupid reading comprehension especially the "implied" questions...and i ONLY study CR too...

but if you really want a good score take the ACT.  the english and math for the ACT are pretty easy, math and science arent that bad but the time limit sucks...if you can get a 36E/36M/32R/32S its pretty good...science is all interpretation of material...no background knowledge required...

hey all you West lafayette kids, how good is your school...like how many people did you guys send to top schools this year (ivy league, standford, duke, northwestern...)

two years ago, one went to princeton and like 5 got admission to caltech (some turned it down). this year, i'm not so sure. i think one went to princeton again?

this year, we had someone going to princeton.....

that person in question actually is going to Caltech
(idk, better engineering program?)
(Just making sure, the person in question has the initals S.M. right?)

yeah, but he/she got admission to princeton.... (yes his/her initials are S.M.).

You guys announce so many names in public here anyway. Posting someone's initials is basically the same.

@MustIt: Not really, if I gave you an initial and gave you a few hints, there is a high probability that you wouldn't know that person.

[quote="moogra's signature"]There are 10 types of people in the world: those who understand binary, and those who don't.
[/quote]

There are 10 types of people in the world: those who understand binary, and those who get laid.


Prove that if $a,b,c,d$ are non-negative reals such that $a^2+b^2+c^2+d^2=4$, then:

$\frac{1}{7-3abc}+\frac{1}{7-3bcd}+\frac{1}{7-3cda}+\frac{1}{7-3dab} \leq 1$.

Very nice, VASC.

To Thuan. You should affirm any problems of you if it true or false before posting here.

[quote="pvthuan"]I prefer the following form:
Let $a,b,c,d\geq0$ and $a^2+b^2+c^2+d^2=1$, determine the maximum of
\[ p=\frac1{1-abc}+\frac1{1-bcd}+\frac1{1-cda}+\frac1{1-dab}.  \][/quote]
Your proposed inequality is equivalent to
$\sum \frac 1{8-abc} \leq \frac 4{7}$ for $\sum{a^2}=4$.
It is a weak inequality because we see that $8>\frac 7{3}$.

Yes, of course weaker. I prefer this weak form.

[quote="Albanian Eagle"]Hi !
I wonder about this solution...
we know that $xyzt\geq1 \rightarrow \sum \frac{1}{3+x}\le 1$ so we need to prove $\prod (4-3abc)\geq 1$
but from bernoulli :
$\prod (4-3abc)\geq 1+3(4-\sum abc)$
but $\sqrt{\frac{\sum a^2}{4}}\geq \sqrt[3]{\frac{\sum abc}{4}}$ and we are done...[/quote]

\[ x^2+y^2+z^2+t^2=4 \implies xyzt \le 1  \]

[quote="Albanian Eagle"]but from bernoulli :
$\prod (4-3abc)\geq 1+3(4-\sum abc)$
...[/quote]
Nice idea.
But I think you can't use bernoulli here.
Because  bernoulli is $\prod(1+x_i) \ge 1+\sum x_i$
it call for $x_i>-1$  and all $x_i$ be positive and negitive...

[quote="Albanian Eagle"]Hi !
so we need to prove $\prod (4-3abc)\geq 1$ [/quote] This inequality is not true: see $a=b=1$ and $c=4/3$.


Prove that for any positive integers a_1,...,a_n
the set of primes who divide a_1^k+...+a_n^k for a certain k (integer)
is finite

You lose somethings . Set of prime divisor of $ f(k) \equal{} \sum_{i \equal{} 1}^na_i^k$ is finite if and only if $ a_1 \equal{} .. \equal{} a_n$

Sorry.
For $ a_1,\cdots,a_n$ with not all the same,
Prove that the set of primes who divide$ a_1^k\plus{}...\plus{}a_n^k$ for a certain $ k$  (integer)
is infinite.
Could anyone prove it?

Could anyone help please?
This, I think is a hard problem and I have tried many times to solve it.


Yes it is easy indeed! Your solution resembles that of mine.

It can also be proved using the following sharp inequality,

$ \frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{2n\minus{}1}{2n}\leq{\frac{1}{\sqrt{3n\plus{}1}}},$        $ n\geq{1}$

But I failed to prove this sharp one. Will anybody help me, please!


In 1998, a runner jogged 709.7 miles from January 1 through the end of May. If the runner continued to average the same number of miles per day, how many miles did the runner jog total in all of 1998? Express your answer as a decimal to the nearest tenth.

The runner jogs 709.7 miles in 31+28+31+30+31=151 days.

$ \frac{709.7 \cdot 365}{151}$ which yields when rounded $ \boxed{1716.0}$

It should be 1715.5 becuz it says round to the nearest tenth

i thought 1998 was a leap year so i put
over 152 instead of 151 am i correct?

It's not a leap year.

Is there any fast way to do this?

@matsumoto 1998 is not a leap year since it is not divisible by 4


The perfect squares 1,4,9,16,25,... are put into a string of digits 1491625... What is the 40th digit in  the string?

[hide]
There are 3 single-digit perfect squares (1, 4, and 9).

The 2-digit perfect squares are from 4 to 9, which is 6, which is a total of 12 digits.

There are 25 digits left, so we need to find the first digit of the 9th 3-digit perfect square. The first 8 are from 10-17, so the ninth is $18^{2}$, and the first digit of it is $3$.[/hide]


I want to figure out the age that is the most common for people to join AoPS.

I expect everyone will answer around the same grade, since it seems new people of a grade only show up after a certain year.

No bumping  :wink: .

And I joined in 8th grade.

Well, let's see... I first learned of AOPS by meeting Richard Rusczyk at ARML 2004, shortly after graduating from college. I was posting here within two weeks after that.

I joined in the winter of my Sophomore year. I ran across it randomly on the internet... I think I searched for some math words on Google and found an AoPS thread with those words in it. And I was like... "Art of Problem Solving"... that's cute, that sounds like my romanticized view of competition mathematics, I'll check it out.

I was in 6th grade, I went on the FCMS site (to see if they had any MC resources  :D ) and found this.

March 19, in sixth grade. Within hours of Temperal.

In this thread I joined earlier than all but jmerry(TZF joined 6 days after me(!!!)) when I was in 6th grade.

i joined as an 8th grader. cuz our math team told us about it. after mc..............................

I joined between 8th and 9th grade. I voted for the 9th-10th grade on the poll.

I joined mostly to play FTW, but I don't really do that anymore...

Hmm...lets see now, I joined when a friend introduced me to it.

I joined when I was in 7th grade. My coach told me about it, but I didn't really listen. I forgot the reason why I joined...now you see how forgetful I am. :(

[quote="isabella2296"]Last March, in 6th grade. My brother found an AoPS book and lent it to me. I thought it was really cool and decided to check out the site. And the rest is history.  :P[/quote]

Same.

Our join dates are like 10 days apart...however, i didn't really start using the forum until i signed up for classes.

I joined because my Nationals coach wanted to get me a username to do Alcumus.

But then the rest is history (Obscure reference).

the summer before 6th grade. this website was well short of incredible and it openeda new door to math for me.  too bad i didnt care for that too much last summer.

I first found out about this site in 7th grade, and occasionally read the MATHCOUNTS forum, but I didn't actually join until 9th grade.

joined in 7th grade...

i think i joined aops mostly to play ftw  :D

Yeah, same here. When I first came on to AoPS, I glanced at the forums, then saw "For The Win". It sounded interesting so I clicked on it and was amazed. I played a bunch of games as guests, decided I liked it, made an account, and that's basically it.

My original account joined back in the past, before I entered highschool.
But this account joined obviously this year.
And i'll probably stuff this account away after 8-12 more months.

Back (in the day), forums didn't really interest me, only FTW did.  I joined mainly because of FTW.

I joined at the beginning of 7th grade, on my birthday.

For some reason, I have a really bad MC coach, so I decided to self-teach myself. I googled some websites to help me with math, and after somehow googling "math forum", I happened to find AoPS and it seems pretty cool. It really takes me around math in a different direction than normal people in school approach it.


algebra
polynomial
geometry
3D geometry
quadratics
Galois Theory
superior algebra

$K$ splitting field over $Q$ of $A(x)=x^5-x^4+x^3-x^2+x-1$

1/ What are the possible values for the minimum degree among the irreducible
factors of polynomial of degree 5 ?
2/ Write $A$ as product of factors irreducible over $R$
3/ Same question as 2/ over $Q$
4/What is the degree of $K$ over $Q$
5/What is the Galois group of $K$ over $Q$

$A(x)=(x-1)(x^4+x^2+1)=(x-1)(x^2-x+1)(x^2+x+1)$
The primitive sixth roots of unity and the primitive cube roots of unity generate the same field, which has degree 2 whether over $\mathbb{R}$ or $\mathbb{Q}$ or any field in which these quadratic polynomials are irreducible. The only automorphisms are the identity and conjugation.

Any Galois questions about cyclotomic polynomials are easy.

Sorry, I have problems understanding 1)....

$(x^5-x^4+x^3-x^2+x-1)(x+1)=x^6-1$, so the roots are the 6th roots of unity different from $-1$
I will call $\zeta = e^{\frac{ \pi i}{3}} = \frac{1}{2} + \frac{\sqrt 3}{2} i$. Then:

2), 3): $(x^5-x^4+x^3-x^2+x-1) = (x-1) ((x-\zeta)(x-\zeta^{-1})) ((x-\zeta^2)(x-\zeta^{-2})) = (x-1)(x^2-x+1)(x^2+x+1)$

4) the degree is $2$, since every root of $(x^2-x+1)$ generates the whole splitting field
5) since 4) its trivially $\mathbb{Z}_2$


[quote="funmath"]When I try to use the commands [code]\binom{}{}[/code],[code]\boxed{}[/code], or [code]\nmid[/code], the output thingy displays Undefined Control Sequence.  Does anyone have any suggestions to make this stop?[/quote]

Where do you use for, TeXinc, or the forum?

Questions concerning LaTeX should go in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=123]part of the forum specifically designated for LaTeX questions[/url].

Hunter, there is never a good reason to quote in its entirety the post immediately preceding yours, especially when it's the only post in a thread.

The probable answer: you need to include more packages. In particular, I think \binom is in amsmath.

sweet! the amsmath package made everything work!
THANKS!!!


induction
combinatorics unsolved
combinatorics

In the cabinet 2004 telephones are located; each two of these telephones are connected by a cable, which is colored in one of four colors. From each color there is one cable at least. Can one always select several telephones in such a way that among their pairwise cable connections exactly 3 different colors occur?

:? I'd say no, but I'm not exactly sure if I understood the question

the cable 1-2 is red, the cable 3-4 is green, the cable 5-6 is yellow, and all the other cables are blue. No possibilities having less than 2 blue cables...

But in the group of telephones 3,4,5,6,7,...,n only appear exactly 3 colors.

Pierre.

I wonder that nobody can solve this problem! It's rather easy one, I suppose.

Answer: yes.

We will consider not telephones but respective graph $G$. Suppose there exist a vertex $u$ s.t. the edges of the graph $G \setminus \{u\}$ are colored in 4 colors. Then we will delete this vertex. And then we will continue deleting vertices with preserving 4 colors. We will do it until we can't do it. Then it is easy to see that deleting any vertex implies disapearing at least 2 colors (if for some vertex only one color disapears then we are done). It means that there exist a vertex $u$ s.t. all blue edges and all red edges incident to the vertex $u$ (for some colors "blue" and "red"). Let $uv$ be blue edge and $uw$ be red edge. But then the edge $vw$ can't be blue or red, since it doesn't incident to $u$. Hence the set $(u,v,w)$ is desired. The problem is completely solved.

Induction is really easy too. The cases with 4,5,6 vertices are easy.

Then, in an arbitrary graph with more than 4 vertices, choose a red, and green, and a blue edge. Consider all edges amongst those vertices. At least three colors appear. If exactly three, then we're done. If 4, then we're also done by induction (since at least 1 vertex was not selected).

Similar to #4.

Assume the contrary. Let $S$ be a subgraph of our original graph. At first lets put an arbitrary triangle in $S$ so it is colored in one or two colors. Then we add the vertices one by one (with their edges).

Since the graph is colored in 4 colors. There will be a moment when $S$ is colored in less than 3 colors before the operation and in more then 3 colors after it. So at least 2 of the colors of the edges of the vertex we are adding $v$ didn’t appear in $S$ before. Therefore there is a triangle with vertex $v$ colored in 3 colors.


USAMTS Round 1 is a few months away, and school's almost over, so I'm thinking of getting together 5 problems of varying difficulty and passing it off as a mock USAMTS round. Solutions would be sent to me, and after a week or two, I'll post results and winners. There will be some kind of prize. I will put up problems tommorow. Anyone interested?

Sure.  I probably won't win but I always love doing math problems(and contests).

I could supply a problem or two if you were interested.  Anyhow, it sounds like a good idea to me.

sounds like a good idea... and if you want, I'll tex up the 5 problems into a pdf so people can download and print out the problems like they would in a typical USAMTS round.

can we PM you our solutions, and can we PM them one at a time?

PM me solutions. If you must, it can be a problem at a time, altough I prefer if you could send all your work at once.

tetra--would you like me to grade number 5?

Yeah, that would be great. I will send you the solutions in a couple weeks.


[quote="Vasc"]

If $a,b,c$ are non-negative real numbers such that $a+b+c=3$, then
$\frac{a^2}{b^2+2}+\frac{b^2}{c^2+2}+\frac{c^2}{a^2+2}\geq 1$.[/quote]

Also,
If $a,b,c,d$ are non-negative real numbers such that $a+b+c+d=4$, then
${\frac{a^2}{b^2+3}+\frac{b^2}{c^2+3}+\frac{c^2}{d^2+3}}+\frac{d^2}{a^2+3}\geq 1$.

this last Vasc ineq is so hard that it is better to deserve it own topic ;)

Then, for this topic I propose the following nice inequality:
For $a,b,c$ non-negative numbers such that $a^2+b^2+c^2=3$, the inequality holds

$\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+a} \ge \frac{3}{2}$.

It neither is hard, nor easy.

Cauchy it.

\[\sum{\frac{a}{b + 1}} \sum{a(b+1)} \geq (\sum{a})^2.\]

It suffices to prove \[2(\sum a)^2 \geq 3\sum a + 3\sum{ab},\] or

\[6 + \sum{ab} \geq 3\sum a.\] Square to get

\[9 + (\sum ab)^2 \geq 6\sum{ab},\] which is true because

\[(3 - \sum{ab})^2 \geq 0.\]

But what do you say about this very nice complementary inequality:
For $a,b,c$ non-negative numbers such that $a^2+b^2+c^2=3$, the inequality holds

$\frac{a}{b+2}+\frac{b}{c+2}+\frac{c}{a+2} \le 1$.

Is it hard ?

Multiply by denominators to obtain equivalent inequality:

\[2 + abc \geq \sum a^2c.\]

But from here I am stuck. Any easy solution to this?

[quote="StRyKeR"]
\[2 + abc \geq \sum a^2c.\]
But from here I am stuck. Any easy solution to this?[/quote]
Yes! there is a very easy one line solution! You can see it [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=549851&p=3191142#p3191142]here[/url]
But can it be proved easily without expanding   :maybe: ?


[quote="PKMathew"]Prove that

$ \tan 10^{\circ} \plus{} \sec 50^{\circ} \equal{} \sqrt 3$[/quote]

It's equivalent to proving $ \tan 60^{\circ}\minus{}\tan 10^{\circ}\equal{}\sec 50^{\circ}$

$ \Leftrightarrow \frac{ \sin 50^{\circ}}{ \cos 50^{\circ}}\equal{}\frac{\tan 60^{\circ}\minus{}\tan 10^{\circ}}{1\plus{}\tan 60^{\circ} \tan 10^{\circ}}\equal{}\frac{1}{ \cos 50^{\circ} (1\plus{}\tan 60^{\circ} \tan 10^{\circ})}$

$ \Leftrightarrow \sin 50^{\circ} (\frac{ \cos 60^{\circ} \cos 10^{\circ}\plus{}\sin 60^{\circ} \sin 10^{\circ}}{\cos 60^{\circ} \cos 10^{\circ}})\equal{}1$

$ \Leftrightarrow 2 \sin 50^{\circ} \cos 50^{\circ} \equal{}2 \cos 60^{\circ} \cos 10^{\circ}$

$ \Leftrightarrow \sin 100^{\circ}\equal{} \cos 10^{\circ}$, which is obviously true.


calculus
integration
geometry
analytic geometry
AMC
AIME
Putnam

Find the area of the figure  surrounded by the locus of the point $ P$ inside the square with side length $ a$ such that the distance to the center of the square is equal to the minimum distance to the side of the square.

AIME would never ask you to calculate area under a parabola.  Putnam 1989: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106690

In that thread, why does he have something with $ \sqrt {2}$ for his probability? If my solution did not have any errors, the probability should just be $ \frac {5}{12}$, right?

Even if you use calculus;
$ P \equal{} \frac {8}{a^2} \int_{0}^{a/4} \left[ \left(\frac {a}{2} \minus{} x \right) \minus{} \left( \frac {a}{8} \plus{} \frac {2}{a} x^2 \right) \right] dx \equal{} \frac {8}{a^2} \cdot \frac {5 a^2}{96} \equal{} \frac {5}{12}$  :maybe:

[quote="The Zuton Force"][hide="Solution"]Place the square with coordinates $ (\pm a/4, \; \{0,a\} )$, and break it into four pieces along the diagonals.

Now, consider a point P(x,y) on the curve. It satisfies: $ y \equal{} \sqrt {x^2 \plus{} \left( y \minus{} \frac {a}{2}\right)^2} \iff y \equal{} \frac {a}{8} \plus{} \frac {2}{a} x^2$

At $ x \equal{} \pm \frac {a}{4}, \; y \equal{} \frac {a}{4}$ and at $ x \equal{} 0, \; y \equal{} \frac {a}{8}$, so the height of each parabolic section is $ \frac {a}{8}$. The width is clearly $ \frac {a}{2}$. So, the area of each parabolic section is $ \frac {2}{3} \frac {a}{2} \cdot \frac {a}{8} \equal{} \frac {a^2}{24}$.

We multiply this by 4 to get $ \frac {a^2}{6}$.

Finally, there is a square in the middle with area $ \left( \frac {a}{2} \right)^2 \equal{} \frac {a^2}{4}$.

Hence, the total area is $ \frac {a^2}{6} \plus{} \frac {a^2}{4} \equal{} \frac {5}{12} a^2$.[/hide]

I'm fairly certain this is essentially the same as a problem I've seen before...

[b]Edit:[/b] [i]Wow, what was I smoking... yeah, the problem cited in the post below is what I was talking about![/i][/quote]

Regrettably your answer is incorrect, The Zuton Force.


inequalities
quadratics
inequalities proposed

[u][b]The author of this posting is : harazi[/b][/u]
____________________________________________________________________

Here is a little bit more: find the best k such that for any $x_1, ..., x_n$ real numbers we have

$\left(1+x_1^2\right)...\left(1+x_n^2\right) \geq k\left(x_1+...+x_n\right)^2$.

And there are even more complicated extensions, though I think this is the nicest.

[u][b]The author of this posting is : Myth[/b][/u]
____________________________________________________________________

[quote="manlio"][quote="Vasc"]Find $k$ such that
$\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right) \geq k\left(a+b+c\right)^2$
hold for all $a$, $b$, $c$ real numbers.[/quote][/quote]

My answer is $k=\frac34$.

[u][b]The author of this posting is : Myth[/b][/u]
____________________________________________________________________

[quote="harazi"]Here is a little bit more: find the best k such that for any $x_1, ..., x_n$ real numbers we have

$\left(1+x_1^2\right)...\left(1+x_n^2\right) \geq k\left(x_1+...+x_n\right)^2$.[/quote]

$k=\frac{n^{n-2}}{\left(n-1\right)^{n-1}}$.

[u][b]The author of this posting is : StRyKeR[/b][/u]
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I can solve for n = 3, but the general case seems out of reach.

\[\frac34 \left(a^2+1\right)\left(1+\left(b+c\right)^2\right) \geq \frac34 \left(a+b+c\right)^2\]

by Cauchy.

So it suffices to prove

$4\left(b^2+1\right)\left(c^2+1\right) \geq 3\left(1+\left(b+c\right)^2\right)$,

or

$4b^2c^2+4b^2+4c^2+4 \geq 3+3b^2+6bc+3c^2$,

or

$4b^2c^2+b^2+c^2+1 \geq 6bc$.

But this is evident because

$4b^2c^2+1 \geq 4bc$ and $b^2+c^2 \geq 2bc$, both by AM-GM.

Equality holds iff $a=b=c=\frac{1}{\sqrt2}$.

[u][b]The author of this posting is : Megus[/b][/u]
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Myth: can you now show your solution for generalization, please ?  :D

[u][b]The author of this posting is : Myth[/b][/u]
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[quote="Megus"]Myth: can you now show your solution for generalization, please ?  :D[/quote]

I can, but I want Harazi did it first.  ;)

[u][b]The author of this posting is : harazi[/b][/u]
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Ok, I hope I will post it this week-end, it's not short and I do not have now the time. If you know what I mean...  ;) . Hint for the others, hoping that someone will post a solution: use Bernoully inequality and Cauchy.

[u][b]The author of this posting is : Myth[/b][/u]
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I have nice solution using IPT and properties of quadratic trinomials.

[u][b]The author of this posting is : harazi[/b][/u]
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What is IPT? Ok, now I post my solution  :D  and I would like to post yours, Myth. I am very interested in the method. Here goes my solution. Using the tranfomation $a_i=\frac{x_i}{\sqrt{n-1}}$ we can reduce the problem to proving that $\prod_{i=1,n}{\left(1+\frac{a_i^2-1}{n}\right)} \geq \left(\frac{\sum_{i=1,n}{a_i}}{n}\right)^2$. Now, consider that $a_1$, ..., $a_k$ are greater than 1 and the others are smaller than 1. Then by Bernoully inequality we have $\prod_{i=1,n}{\left(1+\frac{a_i^2-1}{n}\right)} \geq \left(\frac{a_1^2+...+a_k^2+n-k}{n}\right) \left(\frac{a_{k+1}^2+...+a_n^2+k}{n}\right)$. And now Cauchy finishes the job.

[u][b]The author of this posting is : Myth[/b][/u]
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Umm... All harazi's proofs are very nice, but almost all his proofs are specified for particular problems. On the other hand, my approaches are general, and they allow to prove many inequalities, but due to their generality they are losing attractiveness.
So after harazi's proof I don't want to tire you.

@harazi. IPT is Inflection Point Theorem (see "Super-Hard" topic).

[u][b]The author of this posting is : harazi[/b][/u]
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Please, Myth, tire me, I really want to see how it works your approach. If possible.

Here is my problem O:752 in Gazeta Matematica 6/1994.
If $a_1,a_2,...,a_n$ a real numbers, then
(1)	$(a_1^2+n-1)(a_2^2+n-1)...(a_n^2+n-1)\geq n^{n-2}(a_1+a_2+...+a_n)^2$;
(2)	$(a_1^2+2n-1)(a_2^2+2n-1)...(a_n^2+2n-1)\geq 2(2n)^{n-1}(a_1+a_2+...+a_n)$.
The Harazis proof of (1) is very nice. It remains now to show (2). If Harazis method does not work, maybe Myths general proof (?!) works.

(2) is trivial (my favorite word today) ;)  :D  :D  :D

I solved it earlier for $n=3$ (it was one of "Vasc_3 delightful ineq" or "Vasc_1 delightful ineq"). And, of course, method works fine in the general case too, especially if you pointed to exact value of optimal constant, though it was very simple to find it.

Of course, (2) follows from (1) using the inequality $a_1^2+2n-1 \geq 2\sqrt{n(a_1^2+n-1)}$.
But Myth, how does YGM (your general method) work for the following problem:
 Find the minimum of the product
$P=(x_1^2+1)(x_2^2+1)...(x_n^2+1)$
for  $x_i$ positive numbers such that $x_1+x_2+...+x_n=n$.

All I can say is that IPT or RCF work perfect here, so answer is $\frac{((n-1)^2+1)^n}{(n-1)^{2(n-1)}}$ if I didn't miscalculated.  ;)

[quote="Vasc"]You are right, Myth.  :)
P is minimum for $x_1=...=x_{n-1}=\frac{1}{n-1}$ and $x_n=n-1$. ;)[/quote]
I know. ;)


This problem is in fact more difficult than it can be deduced from its solution:
  The solution is based on the following remark:
  (x+y+z)^2>=(2x+y)(y+2z). Thus we can write prod (a_i-1+a_i+a_i+1)^2>=(2a_n+a_1)(a_1+2a_2)(2a_1+a_2)(a_2+2a_3)...=prod (a_i-1+2a_i)(a_i+2a_i-1)>=prod 2(a_i-1+a_i)^2. We used the inequality (x+2y)(y+2x)>=2(x+y)^2.

Beautiful, surprising  and very hard solution, Harazi.

[quote="manlio"]Let n \geq 3 be positive integer. For all positive numbers a_1,..,a_n show that
 ((a_1+a_2)/2)*((a_2+a_3)/2)*..*((a_n+a_1)/2) \leq ((a_1+a_2+a_3)/2 \sqrt 2))*((a_2+a_3+a_4)/2 \sqrt 2)*..*((a_n+a_1+a_2)/2 \sqrt 2)
Source: RUSSIA 2000 mathematical olympiads[/quote]
Let $ n \geq 3$ be positive integer. For all positive numbers$ a_1,\cdots a_n $show that\[\frac{a_1+a_2}{2}\cdot\frac{a_2+a_3}{2}\cdots \frac{a_n+a_1}{2}\leq \frac{a_1+a_2+a_3}{2 \sqrt 2}\cdot \frac{a_2+a_3+a_4}{2 \sqrt 2}\cdots \frac{a_n+a_1+a_2}{2 \sqrt 2}.\]
                                                          (RUSSIA 2000 mathematical olympiads)


$ a,b,c$ are positive reals with $ a\plus{}b\plus{}c \equal{} ab \plus{} bc \plus{} ca$

Prove the following inequality: $ a\plus{}b\plus{}c\plus{}1 \ge 4abc$.

[quote="Agr_94_Math"]$ a,b,c$ are positive reals with $ a \plus{} b \plus{} c \equal{} ab \plus{} bc \plus{} ca$

Prove the following inequality: $ a \plus{} b \plus{} c \plus{} 1 \ge 4abc$.[/quote]

$ \iff (a\plus{}b\plus{}c)^2(ab\plus{}bc\plus{}ca)^2\plus{}(ab\plus{}bc\plus{}ca)^3 \ge 4abc(a\plus{}b\plus{}c)^3$

$ \iff \sum_{cyc}{(c^4\plus{}c^3(a\plus{}b)\plus{}\frac{1}{2}c^2(ab\plus{}bc\plus{}ca))(a\minus{}b)^2} \ge 0,$

clear true.

Hey anyone for an alternate answer....like there seems to be a possible (but hard) approach using ELEMENTARY SYMMETRIC SUMS....


Are calculus-based solutions allowed on the application exam?

As long as the solution is correct, it is fine. You shouldn't need calculus though. If you plan to Lagrange multiplier bash problem 5, you need to be pretty careful with assumptions, etc.

In general, when you are doing contest problems, you should assume that there is a relatively simple solution. (and there is).

Can we rotate the figure of #1? If not, I don't think there exist a square being constructed by that figure.I just do it for practice  and maybe application.Anyway, the $ \$$50 FEE IS TOO EXPENSIVE for me to apply! Does anybody know any ways to submit application without paying this fee?

Yes you can rotate the block in problem 1.

If you cannot rotate the block, then it is impossible to make a square. As a little add-on, you could prove that statement. (It isn't that hard).

As for the application fee question, e-mail Mr. Andreescu about it.

I emailed to Titu and he didn't reply me. Well, this is the second time he hasn't replied me when I email to him :( .Maybe he doesn't mind seeing what I say


Where can I find a free calculator to solve questions like these:

1/x^2+5    +     x+6/x      =      x+2/x

Sure i know how to do them, but I want to be able to get the answer quicker. Thanks.

[url]http://www.wolframalpha.com[/url]

Answer to that specific problem: [url]http://www58.wolframalpha.com/input/?i=1%2Fx%5E2%2B5+%2B+x%2B6%2Fx+%3D+x%2B2%2Fx+[/url]


It is currently impossible to choose a page to go to by its number, as it used to be. I have to jump from page 1 by 2 and 1 every time I need to go to page 20, say, which is quite boring. Together with poorly functioning search it makes finding old posts a very unpleasant task even if you know approximately where they are and what is there. I'm surprised that nobody has reported it yet. Perhaps it's affecting only my computer? Or is it just too obvious to mention like the thing with missing Favorites? :?

I'm using Mozilla 1.7 under Windows XP.

That is annoying.

A workaround for now: edit the address line. The address for page 2 should end with "&start=50" (in the default 50 posts/page setting). If you want page 20, change this 50 to 950.


[hide]$ F(x)\plus{}F(1\minus{}x)\equal{}1$, so the sum is 1004.[/hide]

[quote="Brut3Forc3"][hide]$ F(x) \plus{} F(1 \minus{} x) \equal{} 1$, so the sum is 1004.[/hide][/quote]

right.

So you find it so fast. :roll:


is there a really really really fast way to do this (countdown round of 2004 states)

solve for $x$ : $x = \frac{x}{6} + \frac{x}{12} + \frac{x}{7} + 5 + \frac{x}{2} + 4$

I think someone did it in 4 seconds or so.

Aha you're right, but how do you solve it that quickly.

I think that the person who got it in 4 seconds did this problem before, and just happened to see it again on some tv show. Everyone, even the hosts, were very surprised.

[quote="nat mc"]I got 84[/quote]

How long did it take you?


parameterization
calculus
integration
function
LaTeX
real analysis
real analysis unsolved

find  $ a$ such that these functions defined for $ x \in(0, \plus{} \infty)$ are integrable :
$ f(x) \equal{} \frac {e^{ \minus{} x}}{x^a}$. and $ g(x) \equal{} \frac {arctan(x)}{x^a}.$

$ \int _0^\infty {e^{ \minus{} x} x^{ \minus{} a} dx} \equal{} \Gamma(1 \minus{} a)$ if $ a < 1$, and is clearly divergent if $ a\geq 1$ (look in a neighborhood of $ 0$).

$ \int_0^\infty {\frac {\arctan x}{x^a}dx}$ is divergent for $ a\geq 2$ since $ \arctan (x)\geq x \minus{} \frac {x^2}{3}$ for $ x\geq 0$.  It's also clearly divergent for $ a\leq 1$ since $ \arctan x$ is monotonically increasing.  Therefore the integral converges if and only if $ 1 < a < 2$, since $ \arctan x \leq x$ for all $ x\geq 0$.

for the second question other approch is to sse that :near that $ 0$ we have $ \frac {\arctan x}{x^a} \sim \frac {x}{x^a} \equal{} \frac {1}{x^{a \minus{} 1}}$so $ a > 1$ and where $ x$ tend to $ \plus{} \infty$ whe have :$ \frac {\arctan x}{x^a} \sim \frac {\pi}{2x^a}$ so $ a < 2.$

tchebytchev: use \infty in your $ \text{\LaTeX}$ expressions, not $ 00.$ I have edited your posts to make this change.


[quote="VinCHESTER"][/quote]
Kuat Yessenov
Azamat Ukpegaziyev (me)
Daniyar Kelbetov
Damir Yelliusizov
Andrey Kim
Nurzhas Makishev

I hope two of these students will take part in next IMO in mexico.
They are Daniyar Kelbetov and Andrey Kim

[Rashyd Unbayev has very big chance to take part on next IMO.... Wish them good luck!!!!! :lol:

here are the  best IMO participicants from Kazakhstan:
  Kuat Yessenov: silver (Scotland)
                          gold (Japan)
                          gold (Greece)
Now In MIT
  Baurzhan Bektemirov: gold (USA)
                                  silver (Scotland)
                                  silver (Japan)
Now in Harvard

[quote="VinCHESTER"]Our country took apart on West China math olimpiad and got 2 silver and 1 bronze for 4 participicants. last year it was much better Kuat-gold, me-silver, Sarin, Marat-bronze[/quote]

Oh well.
Is Kuat the perfect scorer in CWMO last yr?
I joined in cwmo last yr too, hk team leader said ur country has a perfect scorer, followed by a 117/120 score by a chinese.

[quote="VinCHESTER"]Our country took apart on West China math olimpiad and got 2 silver and 1 bronze for 4 participicants. last year it was much better Kuat-gold, me-silver, Sarin, Marat-bronze[/quote]Congratulations to you guys! :)


calculus
integration
geometry
function
real analysis
real analysis theorems

Let D be a compact & connected set having area. Let f be continuous on D. Then show that there exists a point p in D such that

[b]double integral of over D[/b] = f(p)A(D) , where A(D) denotes the area of D.

Assume $ A(D)>0;$ else the statement is vacuous.

$ f$ achieves both a maximum and a minimum on the set $ D$. Call the maximum value $ M$ and the minimum value $ m.$

A straightforward estimate shows that $ m\cdot A(D)\le \int_{D}f\le M\cdot A(D).$

Hence $ \frac1{A(D)}\int_{D}f\in [m,M].$

But the continuous image of a connected set is connected, so $ f(D)$ is a connected subset of $ [m,M].$ The only connected subset of $ [m,M]$ that contains both $ m$ and $ M$ is the whole interval $ [m,M].$ The average value of the function (as defined above) lies within $ [m,M],$ hence it equals $ f(p)$ for some $ p.$

Suppose $ D$ has a connected interior. Can you show that the point $ p$ is not unique?