A manufacturer of cotton pins knows that 5% of the product he is manufacturing is defective. If he sells the pins in boxes of 100 and guarantees that no more than 4 pins will be defective, what is the approximate probability that the box will fail to meet the guaranteed quality?

When I fit this into the Poisson model with $ \lambda\equal{}5$,  I get the required probability is  $ P(X>4) \approx 0.5595$, where $ X$ is the no. of defective ones. Isn't the answer unbelievable to our intuitive reasoning?

My manual computation also lead to the same answer, but it looks something odd as you said.
As a contrast a defect probability is just $ 0.05$, but the probability that the box will have an extra defective grows to $ 0.5$..  :!:

I get $ 56.4019\ldots \%$, personally ([url=http://tinyurl.com/yz62u9v]via[/url]).  I have no idea what's surprising to you: the average box has too many bad pins (five instead of four), so why should it be surprising that most boxes have too many bad pins?

On average 5% are defective. His claim is that in a box containing 100,no more than 4 are defective. 4 is just nearer to 5. So one would naturally think that his claim is almost true. Isn't it? Am I missing something here?

I don't understand what idea or intuition "4 is just nearer to 5" is meant to express here.

My intuition is simple: this is a reasonably nice randomized process and so the median number of bad pins in a box should not be too far from the mean number of bad pins in the box.  The mean number of bad pins in a box is five, so I would expect the median number of bad pins in a box to be close to five.  Thus, I would expect that somewhat more than half the time, one would get at least five bad pins in the box.  And indeed, this is what we've calculated (although I don't understand how you ended up with a different number than I did, largely because the words "Poisson process" don't mean very much to me).  In fact, I wouldn't have been surprised if the answer had turned out to be as large as 60% or a little more.

What do you think the probability would be if we replaced "100" with "1000" and "4" with "49"?  What about if we replaced "100" with "1000" and "4" with "40"?

[hide="Links to my answers"]
Clicking won't work, you'll have to copy-paste:
http://www.wolframalpha.com/input/?i=1-(Sum[(.95)^(1000-i)+(.05)^i+Binomial[1000%2Ci]%2C{i%2C+0%2C+49}])
http://www.wolframalpha.com/input/?i=1-(Sum[(.95)^(1000-i)+(.05)^i+Binomial[1000%2Ci]%2C{i%2C+0%2C+40}])
[/hide]

I believe that while you're using the binomial distribution, he's using the Poisson one which is a good approximation for large $ n$ (i.e. large box size).

The quality of the word "good" might account for the difference in your answers.

Ya you are right. If we replaced "100" with "1000" and "4" with "40" , we would expect his claim is most likely to fail; and the result is also affirmative, because the actual prob. is more than 0.9..In the [b]if we replaced "100" with "1000" and "4" with "49"[/b] case,  the actual prob. 0.520259 which should not be surprising to one because, since mean is 50, I think, one should expect $ P(X<49)\approx P(X>49)$.
I don't really understand your median argument!

[quote="limsup"]In the [b]if we replaced "100" with "1000" and "4" with "49"[/b] case,  the actual prob. 0.520259 which should not be surprising to one because, since mean is 50, I think, one should expect $ P(X < 49)\approx P(X > 49)$.
I don't really understand your median argument![/quote]
You just restated my median argument and then said that you didn't understand it :huh:.   In the original problem, we should expect $ P(X < 5) \approx P(X > 5)$, but also $ P(X \equal{} 5)$ should be nontrivial.


What is the equation in polar coordinates of the equianglar spiral which starts at the origin and goes clockwise?

Also called logarithmic spiral, the equation is
$ r \equal{} ae^{b\theta}$ where $ a$ and $ b$ are arbitrary constants.

Using complex numbers, it's $ z(t) \equal{} z_0^t$ for an arbitrary complex number $ z_0$.


yo this was like the hardest ashme of life..
corrections?
help on the blanks?

13. C
14. A
15. E
16. C
17. C
18. E
19. C
20. E
21. B
22. C
23. E
24. B
25.
26. E
27. A
28. E
29. C
30.

what do the letters mean? I'm slightly confused. Was this a multi-choice test?


Please don't double-post ( http://www.mathlinks.ro/Forum/viewtopic.php?p=367970#p367970 )


N is the set of nonnegative integers. For any subset S of N, let P(S) be the set of all unordered pairs of members of S. (A unordered pair is a set with two distinct members.) Partition P(N), arbitrarily, into two sets (of pairs) P1 and P2. Prove that N must contain an infinite subset S such that either P(S) is contained in P1 or P(S) is contained in P2.

this is just Ramsey's Theorem for infinite graphs. a proof appears here for the case of two colors (which is the case for this problem): http://en.wikipedia.org/wiki/Ramsey's_theorem


I haven't yet recieved any conformation that my solutions for Round 2 were recieved.  I am pretty sure that I recieved conformation for Round 1.

By the way, I e-mailed my solutions (in RTF) for Round 2.

Is there still the possiblity that they have not yet had a chance to send me conformation?  What should I do?


How many latice points are in a circle that is centered at the origin and has a radius of $ r$?

Is there a generalization of that formula to make it simpler?

If there were, don't you think it would be in that article?  Mathworld tends to be pretty thorough.  :)

If you're looking at a competition problem for a particular $ r$, it's easiest just to draw it out.  In particular, for $ r \equal{} 5$ (which is a MAO problem, I think) the only tricky part is thinking about $ 3 \minus{} 4 \minus{} 5$ triangles (in other words, noticing that the points $ (\pm 3, \pm 4), (\pm 4, \pm 3)$ are on the circle).

are you guys here just to memorize formulas?

That's what alot of mathcounts kids do.  Understanding the concept is better than memorizing the formulas.

well I didn't know anything about that formula until he asked. I googled it up, and found it. I memorize some formulas that are crucial to the competition, and then I try to understand why they work. I've actually come up with a few myself. (and then later found that others had made them before me)


calculus
integration
trigonometry
function
calculus computations

evaluate the following

$\int\frac{1}{2+\cos x}dx$

[hide="hint"]Denote $\tan{\frac{x}{2}}=t$ and we need  integration rational functions of $t$.[/hide]

Another (equivalent) way, with substitution $x=2u$

$\int \frac{dx}{a+cos x}= \int \frac{2 \, du}{a \left( \sin^{2}u+\cos^{2}u \right)+\left( \cos^{2}u-\sin^{2}u \right)}=$

$= 2 \int \frac{du}{(a+1)\cos^{2}u+(a-1)\sin^{2}u}= \frac{2}{a-1}\int \frac{\sec^{2}u \, du}{\frac{a+1}{a-1}+\tan^{2}u}=$

$= \frac{2}{a-1}\int \frac{d(\tan u)}{\frac{a+1}{a-1}+\tan^{2}u}= \dots$


Hi people! Here's good biology forum
http://www.biology-online.org/biology-forum/
and for bio news visit this site
 http://www.biologynews.net/
Enjoy(know u will)!!! :lol:

The aforementioned biology forum is pretty cool.


[quote="anirudh"]If the sum of two numbers is $12$ and their product is $-4$, then determine the sum of their reciprocals.[/quote]
[hide]
$x+y=12$
$xy=-4$
$\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=-3$[/hide]

[quote="Wonsuk"]you need to make the questions answerable to humans. lol. :lol:  :help:[/quote]
What?
I answered it, read my solution it isn't that complicated.
It is important to realize this:
$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$

Use least common denominator, so

$\frac{1}{a}+\frac{1}{b}=\frac{b}{ab}+\frac{a}{ab}=\frac{a+b}{ab}$.

Look at the LCD used.  :wink:

[quote="jhredsox"]so the sum of 2 numbers' reciprocals is equal to their sum/product?

-jorian hoover[/quote]
It is a very common and very useful manipulation. My method is more general, and I suggest you learn it (though it won't give you the fraction in simplest form).

oh ya...i was too lazy too look at the ab fraction thing to figure out it was actually really simple :P

-jorian


geometry
inradius
circumcircle
incenter
parallelogram
geometry unsolved

Let $ r$ be the inradius of triangle $ ABC$.  The circle $ \omega_{1}$ passes though $ B$ and $ C$ and is externally tangent to the incircle of triangle $ ABC$.  The circle $ \omega_{1}'$ is tangent to $ AB$ and $ AC$ and is externally tangent to $ \omega_{1}$.  Circles $ \omega_{2}$, $ \omega_{3}$, $ \omega_{2}'$, and $ \omega_{3}'$ are defined similarly.

Let $ r_{i}'$ denote the radius of circle $ \omega_{i}'$.  Show that $ r_{1}'+r_{2}'+r_{3}' = r$.

[img]http://www.artofproblemsolving.com/Admin/latexrender/pictures/5ad3765b0cf883fb4aae37cee7aceab5.png[/img]

Very nice problem Naoki,thank you! :)

 The basic idea in my solution is to prove that the common internal tangent to $ {\omega}_{1}'$ and the incircle is parallel to $ BC$. From here you can easily conclude that $ \frac{r_{1}'}{r}=\frac{p-a}{p}$ and similarly for the other circles.  :wink:

This was my idea too when I first saw the problem.

Actually the configuration says more:

First, consider the mixtrilinear incircles (the circles tangent to two sides and internally to the circumcircle) $ \omega_{a}$, $ \omega_{b}$, $ \omega_{c}$, such that $ \omega_{a}$ is tangent to $ AB$, $ AC$ in $ X_{3}$, respective $ X_{2}$, $ \omega_{b}$ tangent to $ BA$, $ BC$ in $ Y_{3}$, $ Y_{1}$ and $ \omega_{3}$ to $ CA$, $ CB$ in $ Z_{2}$, $ Z_{1}$.

Now, it is well-known, that the incenter is the midpoint of each segment $ X_{3}X_{2}$, $ Y_{3}Y_{1}$, $ Z_{2}Z_{1}$, thus one obtains that $ Y_{3}Z_{2}$, resulting that the $ Y_{3}Z_{2}Y_{1}Z_{1}$-type quadrilaterals are parallelograms, i.e the $ Y_{3}Z_{2}$-type lines are parallel to $ BC$ and the distances $ d(I;Y_{3}Z_{2})$ and $ d(I;Y_{1}Z_{1})=d(I;BC)=r$, are equal, i.e the $ Y_{3}Z_{2}$-type lines are tangent to the incircle.

In the proposed problems, $ Y_{3}$, $ Z_{2}$ are the points of intersection of the common internal tangent of $ \omega_{1}'$ and the incircle, with $ AB$, respective $ AC$.

[color=darkblue]1. Please, [b]Nsato[/b], receive and the my contents for this problem !

2. Very nice the your proposed problem and the its [b]Pohoatza's proof[/b].

3. Please [b]Yetti[/b], I hope to see and the [b]your proof[/b] !

4. [b]Remark.[/b] If $ w_{1}=C(O_{1},r_{1})$ a.s.o., then $ r_{1}=\frac{a^{2}+r_{a}^{2}}{4r_{a}}$ a.s.o. I think that it is interestingly.[/color]

Pohoatza: I fixed your typos as requested.

  darij

Okay I will write more details.

We draw the tangent line $ t$ to the incircle of the triangle which is parallel to $ BC$ (and of course doesn’t coincide with it). As I said above our mission is to prove that the incircle $ \omega_{a}$ of the triangle formed by the sides $ AB,AC$ and line $ t$ is tangent to the circle $ \omega_{1}$(from here obviously follws that $ \omega_{1}'\equiv \omega_{a}$). Now we turn to a very useful but undeservedly infamous generalization of Ptolemy’s theorem.

[color=blue]Theorem: Let $ A$,$ B$,$ C$,$ D$ be four points on the circle $ \Omega$ and $ \omega_{A},\omega_{B},\omega_{C},\omega_{D}$ are four circles which are touching $ \Omega$ at the points $ A,B,C,D$, respectively. Then

\[ t_{AB}t_{CD}+t_{BC}t_{AD}=t_{AC}t_{BD}\]

where $ t_{XY}$ is the length of the common tangent to the circles $ \omega_{X}$ and $ \omega_{Y}$(the common tangent is external if $ \omega_{X}$ and $ \omega_{Y}$ touch the circle similarly(both internally or both externally) and internal otherwise ). [/color]

Note that the reversed statement also holds. That is if \[ t_{AB}t_{CD}+t_{BC}t_{AD}=t_{AC}t_{BD}\]  for some four circles  then there is a circle which touches all those four circles. In our problem we will be using this reversed statement.

So our four circles will be $ \omega_{a}$,points $ B$ and $ C$(note that points also can be viewed as circles) and incircle of $ ABC$. According to the

theorem, to prove that there is a circle which touches $ \omega_{a}$ externally and the incircle internally and also passes trough $ B$ and $ C$(clearly if

there is such a circle then it is $ \omega_{1}$) it is sufficient to show that

\[ (p-b)*CC_{1}+a*[t]=BB_{1}*(p-c)\]

where $ C_{1}$ and $ B_{1}$ are the touching

points of $ \omega_{a}$ and sides $ AC$ and $ AB$, respectively, and $ [t]$ is the length of the common internal tangent to the incircle and circle $ \omega_{a}$.

Now it is not difficult to show that $ BB_{1}=c-\frac{(p-a)^{2}}{p},CC_{1}=b-\frac{(p-a)^{2}}{p}$ and $ [t]=(b-c)(\frac{p-a}{p})$(with assumption that $ b\geq c$ ).

Tiks, thank you for your nice solution!  I didn't think of using Casey's theorem, but it is quite natural.


On the AMC order form, I saw a USAMO CD with all the problems from 1972-2008. I am wondering whether it comes with solutions. If not, how can I obtain them?

Thanks,

EP

I don't own it, but I am almost certain it does. Evidence in that an AMC staff member posted a thread here awhile back asking about solutions for some old USAMOs to put in that CD.

EDIT: Here is that thread: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=236430

Yes, the CD does include solutions (I have it).


Let a triangle  $ABC$  with $AB < AC$ , the  bisector $AD$ of the triangle  $ABC$  and the midpoints $M, N$ of sides $AC , BC$ respectively. If line $AD$ meets line $MN$ at $E$ , prove that $AE \perp EC$

[u] Babis[/u]

Let $CE\cap AB=F$. Then $CE=EF$ by Thales ($CM=MA$ and $CN=NB$). With $\angle CAE=\angle FAE$ and $AE=AE$, we get $\triangle AEC\cong\triangle AEF$, giving $AC=AF$, which means that $AE$ is the height of the isosceles triangle $ACF$. Hence the result.

Hi ! here's my solution:
[hide]$MN$ parallel to $AB$ then angle $BAE$ equal angle $AEM$ (angles correspondants). But angle $BAE$ equal angle $EAM$ and so we have angle $AEM$ equal angle $EAM$. So the triangle $AME$ isoceles with $AM=ME$.But $ME=MC$ and so the median $EM$ in the triangle $AEC$ is half of side $AC$.
That means the triangle $AEC$ is rectangle on $E$.[/hide] :roll:

[hide="my solution"]
Line parallel to $AC$ and containing $E$ intersects with $AB$ at
$K$. $AE$ is diagonal in $AKEM$ what bisects its angles so sides of $AKEM$ are equal. $MC=AM=EM$ so $M$ is circumcenter of $AEC$ what lies on middle of $AC$, what shows $AEC$ has right angle $E$.[/hide]


calculus Harvard college ratio quadratics geometry integration

A new [url=http://www.centerforpubliceducation.org/site/lookup.asp?c=kjJXJ5MPIwE&b=2466963]education report[/url] addresses the question: does homework help or hinder student learning? It turned out that "[t]here is no conclusive evidence that homework increases student achievement across the board". However, homework seems to be beneficial under certain circumstances:
[quote]Studies that have delved more deeply into this topic suggest, however, that the amount of homework [b]assigned[/b] by teachers is unrelated to student achievement, while the amount of homework actually [b]completed[/b] by students is associated with higher achievement[/quote]
(emphasis mine). There is an even more controversial finding: homework has a disparate impact on the ground of race:
[quote]A national study of the influence of homework on student grades across five ethnic groups found that homework had a stronger impact on Asian American students than on students of other ethnicities[/quote]
Given that the impact of homework is somewhat narrow and uneven, do you think that schools should get rid of it? Or replace it with something race-neutral?

getting rid of homework wouldn't help most people, as they need the extra reinforcement. The [i]amount[/i] of homework is another matter.

(btw I'm an Asian-American :wink: )

[quote="not_trig"](btw I'm an Asian-American :wink: )[/quote]

same here  :D

[quote="not_trig"]getting rid of homework wouldn't help most people, as they need the extra reinforcement. The [i]amount[/i] of homework is another matter.[/quote]

well it really matters what the student is doing instead of homework doesn't it?

lets take three cases:

1) hangs at mall instead - the students does not learn anything but doesn't get stressed by the homwork

2) goes ahead in their textbook and learns more stuff

3) homework is assigned, like usual, and reinforces what the students are learning

the thing is, they can't ban homework because people aren't going to go ahead in their textbook. some might, but the vast majority will not and therefore it is not benefitial.

[quote="mlok"]... homework has a disparate impact on the ground of race...[/quote]
This almost certainly follows from the point which precedes it, that only homework completed is of value (color me shocked at that finding, by the way), combined with disparate attitudes between racial groups on the importance of doing assigned homework.

Homework is important. Strip out all the exercises from Apostol and give it to a student to learn calculus from, and then give a complete copy to another student of comparable ability. I believe the latter student will do better.

That example wasn't very good, I admit. If we take homework to mean intellectually stimulating problems, ones that by completing a student will gain or reinforce knowledge with, then homework is essential. However, what passes for homework in the public education system is just scary.

Mind numbing, repetitive exercises have their place, but only because absolute mastery of some things are required before progressing on to more stimulating endeavors(multiplication tables, spelling, grammar, geography, etc), but public schools use them way too much. When 14 year-olds are still coloring in pictures with colored pencils for a grade, something is wrong.

[quote="PenguinIntegral"]Mind numbing, repetitive exercises have their place, but only because absolute mastery of some things are required before progressing on to more stimulating endeavors(multiplication tables, spelling, grammar, geography, etc), but public schools use them way too much. When 14 year-olds are still coloring in pictures with colored pencils for a grade, something is wrong.[/quote]

Yeah, that's "busy work".

Ideally, the teachers would know how much homework is needed for each individual student. However, since it's impossible to achieve this, homework is unavoidable.

[quote="7h3.D3m0n.117"]the thing is, they can't ban homework because people aren't going to go ahead in their textbook.[/quote]

Huh?

[quote="PenguinIntegral"]
Mind numbing, repetitive exercises have their place, but only because absolute mastery of some things are required before progressing on to more stimulating endeavors(multiplication tables, spelling, grammar, geography, etc), but public schools use them way too much. When 14 year-olds are still coloring in pictures with colored pencils for a grade, something is wrong.[/quote]

I completely agree. There's nothing worse than being forced to follow a long complicated boring assignment that requires little or no thinking. It's completely unproductive. A critical thinking worksheet will help gain a much better understanding of something that a simple but tedious task (such as multiply 4x4 matricies or something. that homework assignment took forever.)

And the coloring part is the most annoying.

[quote="plucesiar"][quote="tjhance"]And the coloring part is the most annoying.[/quote]

Absolutely. That's why some teachers give out bonus marks for doing that

 :rotfl:[/quote]
I'd like to elaborate on this. Certainly, how you present your work makes a difference. If I were a teacher, I certainly would have a category on projects for general neatness. Making sure your text is straight and there isn't dried glue all over the place is part of communicating effectively. I would certainly give points for presenting work in a way that is neat and easy to read.

I would not, however, give points for dotting i's with hearts or putting pink squares behind every paragraph.

Would anyone be interested in getting in a discussion about the standard curriculum today versus the Latin Language/Rhetoric/Logic trivium thing? I find that method of teaching very interesting.

[quote="mathnerd314"]Ideally, the teachers would know how much homework is needed for each individual student. However, since it's impossible to achieve this, homework is unavoidable.[/quote]

But a good teacher can gague how much homework is needed for a class.  Last year I took two math classes at the same time.  One teacher assigned problems 1-40 every night.  Most kids in the class didn't do their homework.  The other teacher would give a quiz (not for a personal grade) at the end of each class/lecture, and depending on the overall grade of the class he would decide how much/what problems of homework to assign.  People payed attention in class and worked hard because it meant less homework.  So it is possible, but it takes a special teacher to do that.

And this is at a public high school.  So although there are teachers who make you color pictures for homework (yes, I have those teachers too) there are some that recognize homework as a tool not to be overused.  Everything is good in moderation...

[quote="PenguinIntegral"]That example wasn't very good, I admit. If we take homework to mean intellectually stimulating problems, ones that by completing a student will gain or reinforce knowledge with, then homework is essential. However, what passes for homework in the public education system is just scary.[/quote]

You're basically saying "If we define homework as beneficial, then homework is beneficial."

[quote]Studies that have delved more deeply into this topic suggest, however, that the amount of homework [b]assigned[/b] by teachers is unrelated to student achievement, while the amount of homework actually [b]completed[/b] by students is associated with higher achievement[/quote]

Well well well, who would have thought of that? So assigning homework doesn't actually help unless the students do it. Must have taken a huge study to figure out, eh? I think I'll take out a big government grant for studying cholesterol, take the money for myself, and submit

[quote] Studies that have delved more deeply into this topic suggest, however, that the amount of exercise and diet [b] assigned [/b] by doctors is unrelated to lowering cholesterol level, while the amount of diet and exercise actually [b] completed [/b] is associated with lowering cholesterol. [/quote]

Well maybe they were trying to show causation, as opposed to simply having better schools giving more homework.

I do see what you're saying, but common sense could probably have arrived at that result.

The finding doesn't seem so obvious to me.  It seems to divide into two parts:

(1) The amount of homework [b]assigned[/b] by teachers is unrelated to student achievement.

(2) The actual homework [b]completed[/b] by students is associated with higher achievement.

Part 2 makes sense to me.  Part 1 surprises me, because I figured that if teachers assigned more homework, then students would do more homework, which would lead to higher achievement.  I guess my chain of reasoning was wrong.

[quote="Ignite168"][quote="footballrocks41237"]I don't get whats the deal with reading assignments.  I don't read the books anyways.  I get all the information either from a friend or I get it from SparkNotes.  And in History were doing WWII now, so reading it is pointless...  I've done well on the pop quizzes without reading  :D[/quote]

You're an idiot.

How do you plan to write a paper or understand the themes and central ideas of the book without reading it? The best way to expand your vocabulary (and SAT Verbals as well) is to read. Your not getting that when you're not reading the book. Also, the "Analysis" on SparkNotes is pretty much crap. It'll be enough to scrape a C- usually but nothing even close to the B range.[/quote]

As far as I know I've gotten an A+ in C.A. for the past 3 quarters...

[quote="footballrocks41237"]I don't get whats the deal with reading assignments.  I don't read the books anyways.  I get all the information either from a friend or I get it from SparkNotes.  And in History were doing WWII now, so reading it is pointless...  I've done well on the pop quizzes without reading  :D[/quote]
...
not reading the book totally defeats the purpose of the class.
The purpose isn't to just simply analyze a book, it's to teach you how to think(in terms of literature and whatnto).

ignite, please don't flame: you can explain to someone what about his comments is idiotic without calling him an idiot.  This is a useful skill, in fact.

[quote]How would this conflict with my absolute hatred of mysticism? The state, in my opinion, should for the most part not be involved in determining how people live their life. (eg basic health class good, abstinence education bad).

I also really think parents and family should be making the decision about education and lifestyle, certainly not the child(until he/she is capable of making those decisions).[/quote]
I didn't say it conflicted with your [i]dislike[/i] of creationism; I want to know how you would reconcile it with your belief that creationism shouldn't be taught in school.  Because the latter implies, "shouldn't be taught in school, even if the students (or their families) want it to be taught in school."  Similarly, abstinence education is popular among families in some communities this country (its complete uselessness notwithstanding*), while comprehensive health education is very unpopular in those same places.  Why do you think [i]those[/i] students/families shouldn't have a say in what they're taught?

* If you (not "you PenguinIntegral," "you the generic reader") disagree with this statement, start a new thread complaining about it, and I or others will come explain to you why you're wrong.

[quote="PenguinIntegral"]I would like to know how this wrong. It's how I learned it in school: some of the framers were very free market and stuff.
But anyway, my point is that I've been told, by books that aren't that biased, that America rose to power precisely because they had more freedom and the free enterprise system was working well.[/quote]
So, the answer to the first question obviously is that there is nothing whatsoever about capitalism in the constitution.  (I mean, Adam Smith had only been dead a year when the damn thing was getting ratified.)  Amendments III, IV and V can be construed to provide a right to property, but their purpose was certainly to protect against monarchy/despotism (i.e., to provide political freedoms), not to protect a free-market system (economic freedoms).  (Of course, the words "communism" and "socialism" as applied to economic systems wouldn't be coined for decades after this all got finished with.)  And I hold your claim about some of the Framers in deep skepticism until you provide some names and some evidence.

For the latter point, I would argue that any book that told you that would by definition be biased, although perhaps biased in a way that represents a majority viewpoint.  And, of course, that's a totally separate question from what the framers intended by the writing of the constitution.

I'm not saying we should be forcing even basic health classes on children. If the parents object to factual, unbiased information about the human body, then that is their problem. But most abstinence education I have seen draws "conclusions" for the students using some shoddy facts and reasoning, and is extremely biased.

So, given that we will have a public school system in place, basic education about one's body is simply basic biological science, factual and scientific and part of the canon of basic knowledge. Abstinence is not.

Whether or not the founders intended to give economic freedoms, the country certainly did have them, and I believe that was a contribution to the rise of the nation.

I do not know much about the founders other than suspect information I can parrot back form my class, I give you that, but I know Thomas Jefferson and the Democratic-Republicans wanted a free economy and favored a small government. A quick wikiquote brings up:
[quote]
To compel a man to furnish contributions of money for the propagation of opinions which he disbelieves and abhors is sinful and tyrannical.
[/quote]
[quote]I agree with you that there is a natural aristocracy among men. The grounds of this are virtue and talents.[/quote]
[quote]
I know no safe depository of the ultimate powers of the society but the people themselves; and if we think them not enlightened enough to exercise their control with wholesome discretion, the remedy is not to take it from them, but to inform their discretion.[/quote]
[quote]I think myself that we have more machinery of government than is necessary, too many parasites living on the labor of the industrious.[/quote]
[quote]The legitimate powers of government extend to such acts only as are injurious to others. But it does me no injury for my neighbor to say there are twenty gods or no God. It neither picks my pocket nor breaks my leg.[/quote]
[quote]Sometimes it is said that man cannot be trusted with the government of himself. Can he, then, be trusted with the government of others? Or have we found angels in the forms of kings to govern him? Let history answer this question.[/quote]

[quote="JBL"]Why not this?[quote="PenguinIntegral"]So, given that we will have a public school system in place, basic education about [Shakespeare] is simply basic [cultural/historical/linguistic knowledge], factual and [analytic] and part of the canon of basic knowledge.[/quote][/quote]
Then one can not take that course, or drop it. I never advocated not teaching Shakespeare, he's a great playwright.

But it's a cheap way to get a skeletal summary of the book. You miss the rhetoric and literary devices that have made the book a classic for hundrds of years. You said it yourself: Reading a book is more meaningful and you actually get more of an impact.

Sorry if I sound trite.

By the way, what does this have to do with the original topic? :lol:

Not everyone cares for meaning; I certainly don't read every single book we are assigned in class (I certainly didn't feel like reading Heart of Darkness...too dense). They miss out on it, but not everyone cares about that, and so it is not a loss for them to just do the Sparknotes route.

If you're responding to the full post directly preceding yours, DON'T USE THE QUOTE BUTTON!  (I've edited the posts in question.)

[quote="Elemennop"][quote="Ignite168"][quote="footballrocks41237"]I don't get whats the deal with reading assignments.  I don't read the books anyways.  I get all the information either from a friend or I get it from SparkNotes.  And in History were doing WWII now, so reading it is pointless...  I've done well on the pop quizzes without reading  :D[/quote]

You're an idiot.

How do you plan to write a paper or understand the themes and central ideas of the book without reading it? The best way to expand your vocabulary (and SAT Verbals as well) is to read. Your not getting that when you're not reading the book. Also, the "Analysis" on SparkNotes is pretty much crap. It'll be enough to scrape a C- usually but nothing even close to the B range.[/quote]

Not that I disagree that reading a book is more meaningful and you actually get more of an impact, but SparkNotes is MORE than enough in order to get a high grade on a paper or essay. Unless you don't ever have any in class discussion in English about the book (and in that case, what the heck do you do in there?), then Sparknotes lays down the plot and main themes and the class discussion will give you all the filler. That's more than enough to get A's on both tests and papers. I don't see where you're getting this "not even B range" stuff.[/quote]

Umm.. our tests have no factual really at all (those are only to pad your grade) It's 8 quotations from the book, and you have to ID 6 of them (where they occured, who said to who about who, significance, themes, etc.) I'm pretty sure using SparkNotes will make it impossible to get anything higher than a F on that, much less an A.

LOL, SparkNotes an A on a paper? The anaylsis on SparkNotes is crap. It usually falls into one of two categories:

1) Some obscure analysis that doesn't really make sense at all and is from seriously stretching the book, obscure does not equal better, ex. of this: (not necessarily on SparkNotes) trying to diagnose characters with mental conditions, etc.

2) Obvious anaylsis. A paper should have incisive analyzation, the SparkNotes stuff only scratches the surface.

Our teacher says he checks SparkNotes sometimes and lets students use it. Why? Because the papers that students hand in from SparkNotes get Ds and Fs and they it's pretty much impossible to ID quotes reading SparkNotes.

Side Note: I still haven't gotten an A- on any paper this year and I believe my analysis is much better than SparkNotes. If anyone is interested in reading them:
[hide="Far From the Madding Crowd"]In Far from the Madding Crowd, by Thomas Hardy, many character relationships are explored while the personality of the protagonist, Bathsheba Everdene evolves. She starts out as an independent carefree woman and through three courts, her life is nearly wrecked but she finds contentment in the end. In her search for love, Bathsheba’s inexperience and vanity almost causes her to lose her identity but she eventually finds a realistic desire of contentment in her destiny.
	Bathsheba was originally vain and indifferent toward how other people felt, and searched for an unrealistic true love rather than being content with what could already be attained. After Oak asked her hand in marriage, "'People would talk about me, and think I had won my battle, and I should feel triumphant, and all that, But a husband...'Why won't you have me?'...Because I don't love you'" (36-37), Bathsheba possessed a free spirit and did not want to settle down. However, she valued finding someone she loved more than being loved. In addition, Bathsheba did not care about Oak's feelings at all as she made the decision. After she finally married Troy, “In those earlier days she had always nourished a secret contempt for girls who were the slaves of the first goodlooking young fellow who should choose to salute them” (271). Bathsheba did not just want a good-looking young fellow as evidenced by her contempt. Her vanity forced her to search for an unrealistic person and hurt Oak's feelings in the process rather than being content with him and his farm. Her pursuit led her to Troy, who would ultimately be her downfall.
	Bathsheba's was inexperienced in terms of love yet she did not regard her inexperience as important. Bathsheba toyed with Boldwood's emotions, “'What fun it would be to send it to the stupid old Boldwood, and how he would wonder!'...Of love as a spectacle Bathsheba had a fair knowledge; but of love subjectively she knew nothing” (100-101). Bathsheba knew of how great and obsessive love could become but she did not know how such love could be formed. Yet, she decided to experiment in the trifling of love anyway. Bathsheba knew the potential consequences of her actions and decided to proceed anyway. After her experience with Troy, “Heaven knows I say it in a spirit the very reverse of vain, for I am grieved and troubled to my soul about it…I hold that man’s future in my hand…I tremble at my responsibility, for it is terrible” (348), her demeanor changed. She had experienced the consequences of love first-hand, and now cared about the fate of other people rather than about her pride. Now that she was experienced, her vanity and indifference toward other people disappeared. Her vanity and indifference toward other people were caused by inexperience.
	Bathsheba's downfall from inexperience almost caused her to lose her identity; however, it was a necessary step for her to reach her destiny of contentment. When she found out about Fanny, “Bathsheba became at this moment so terrified at her own state of mind that she looked around for some sort of refuge from herself” (293). Bathsheba does not know what emotions she is feeling anymore and has completely lost sense of who she is. She was even shocked at herself and not at all the free spirit she was. After Troy affirmed he never loved her, “There was something so abnormal and startling in the childlike pain and simplicity of this appeal from a woman of Bathsheba's calibre and independence” (295). Even under the one-of-a-kind headstrongness, vanity, naiveness, and all the emotions Bathsheba displayed, underneath it all she was searching for love. The emotion and personality that made up her identity did not matter and threatened to disappear during this event which was all provoked by her inexperience.
	Bathsheba finds her destiny of contentment but not before she learns a few more lessons about herself and learns to let go of all her arrogance. It is only after she lets go of all of her arrogance does her only personality begin to ebb back into her. After Oak threatened to leave, “You've been with me so long—through bright times and dark times—such old friends that as we are—that it seems unkind almost” (384). Even through all the anxiety and trauma she experienced, she still considered Oak someone to fall back on. From the beginning of the book, she had always considered Oak as a fall back option that was always there. When the topic of marriage was brought up, “Surely, surely you be the last person in the world I think of marrying” (388), Bathsheba is still being unrealistic and lying to herself. It took the threat of Oak leaving for Bathsheba to finally realize how much depended on and appreciated him. When they finally wed, “Yet, though so plainly dressed there was a certain rejuvenated appearance about her: As though a rose should shut and be a bud again” (393). Bathsheba finally let go of all her vanity and was slowly regaining the spirit of Bathsheba during the days she first met Oak without the vanity and most importantly, without inexperience. However, “Bathsheba smiled (for she never laughed readily now)” (395), whether Bathsheba will find with Oak the level of love she was looking for in Troy is uncertain.
	Bathsheba Everdene undergoes a significant transformation throughout this book. She starts out as a carefree and headstrong but inexperienced and vain. After her first relations with Oak and Boldwood, she learned nothing about herself but was shown to be indifferent to the feelings of others and searching for an unreasonable mate. Her interaction with Troy built her up for a crash back to reality when she realized she was being played and that Troy's love for her was never real. Everything that defined Bathsheba then and what Oak fell in love with, was put in jeopardy yet even afterwards, she still held on to some of her vanity. It was only after she risked losing the last thing that had always stood by her, Gabriel Oak, did she realize she could not take things for granted and found contentment with Oak. Inexperience and vanity was the root of all of Bathsheba's downfalls. Bathsheba and Oak's destinies were to be married and Far from the Madding Crowd is the journey of Bathsheba to realize her destiny.[/hide]
[hide="Lord of the Flies"]In William Golding's Lord of the Flies, there are many different objects throughout the book which represent masks. In addition, the boys on the island range from young and impressionable to emerging leaders, but they are all still children. The boys hide behind masks throughout the book for different reasons to escape from reality and the taboos of society which eventually causes their downfalls. Each boy has their own motivation for hiding behind masks. Piggy wants to escape the limitations of his normal personality, Roger wants to be able to escape taboos, and Jack wants power and control over the other boys.
There are symbols throughout the book which the boys use to escape reality for different goals. When they were looking around the island for the beast, Ralph thought, “They would reach the castle some time; and the chief would have to go forward” (104). Ralph uses the hunt for the beast to fantasize and relate it to civilization. Right before Jack had started his own tribe, Ralph saw, “He noticed, without understanding, how the flames were visible now against the dull light. Evening was come, not with calm beauty but with the threat of violence” (150). Jack used the darkness to escape from the rules of the society that the boys had set up. While Ralph was talking to the twins, “If it were light shame would burn them at admitting these things. But the night was dark” (188). The darkness functions as a mask to allow the twins to escape reality and shame.
	Piggy hides behind masks to escape the reality of what he is. The day after the boys killed Simon, 'We left early,' said Piggy quickly, 'because we were tired'” (158). Even Piggy, the most mature of the boys, refuses to acknowledge reality. After Piggy lost his glasses, “They blinded me” (169). Piggy is now blind which causes everything to be masked to him. Shortly after, he says, “Young Simon was murdered. And there was that other kid what had a mark on his face. Who’s seen him since we first came here” (171). While he and the other boys refused to speak about the deaths earlier as it would acknowledge that they had killed Simon, Piggy now being masked allows him to speak freely. Without the blindness, he could not acknowledge reality, but with it, he can. Piggy’s blindness is his mask and it allows him to escape from qualities he usually possesses.
	Roger used the pigs as masks to escape reality which allowed him to do anything he wanted to. As a sadist, he chose to hurt the other boys for his pleasure. While they were hunting for the pigs, “The pigs lay, bloated bags of fat” (134). They killed this pig and later, when Roger is watching Piggy walk towards Castle Rock, “Below him...Piggy a bag of fat” (180). Seeing Piggy as a pig allows Roger to murder him yet a short while before, “Roger took up a small stone and flung it between the twins, aiming to miss” (175), the imprints of society rendered him unable to even attempt to harm the twins. However, once Piggy became a pig to him, he was able to escape the taboos and murder Piggy. This shows that the pigs were used as masks to allow Roger to escape the taboo of murder in society. When Robert acted like a pig, “The desire to squeeze and hurt was over-mastering” (115), the savage desire inside him overpowered the ones from society and it was the idea of a pig as a mask that allowed him to become that savage.
Masks affect all the boys, whether they are conscious of it or not. When Jack had a mask, he used it to assert power over the boys.  As Simon was crawling out of the forest, “A thing was crawling out of the forest...The shrill screaming that rose before the beast was like a pain” (152). Simon became a pig to each of the boys and they all had the urge and impulse to hurt him. They could not have hurt him if society’s teachings remained, but Simon being “a thing” and “the beast” to the boys allowed them to become savages. Regardless of what the beast is, it had a power over all the boys and gave each of them a mask like the one Jack literally had. The mask that Jack painted over his face had a similar effect, “The mask compelled them” (64). Jack used this mask for his own goals and to get meat. He having access to meat gave him power and even furthered his desire for more which spawned a cycle that Jack could not escape from. It eventually led to him hunting Ralph and completely losing touch with society. The mask became a permanent influence on him.
Masks are an important underlying of everything that occurs on the island. They reveal more about each of the boys that can not be found in their actions alone. The beast, night, and the pigs all function as masks for the boys. More of each character’s inner wills and personality is revealed through masks as well as let us trace the evolution of each boy losing touch with society. The boys all use masks to cover themselves to try and change something about themselves or the island about them. But this change causes Piggy too bold which gets him killed, Roger to become a savage, and Jack to be so power hungry that he cannot focus on anything else. The mask each character uses leads to their downfall from escaping reality.[/hide]

Always looking for ways to improve my writing. First one is slightly weak on specifics of quote analysis and incisive analysis, second one isn't deep or specific enough. B+ on both =/

I'm sorry I don't have any specific facts right now but earlier when I was researching John Dewey for a term paper on educational progressive reform I came across something like a "contract" plan that is still used in some colleges, but I would have to find more info about that before making a cohesive post... I can't find any links to it right now :(

[quote="Aznwithabrain"]Back on topic (idk if what I'm saying is correct. I've only skimmed through the topic)

Homework is necessary for these classes:
Math
Foreign Language

Light homework load for these classes:
English
Science
Social Studies (or some ppl call History)
Electives (non-core classes)

No Homework:
Music Classes (not including practice)[/quote]

Pretty sure Music Theory, English, and History should have some heavy homework and "electives" is way too broad a term. I see no reason to have more foreign language than say science.


trigonometry
inequalities unsolved
inequalities


[quote="stevengoh911"]Given
x,y ∈ [0,2c]
f(x),g(y) ∈ [0,2c]
Show
| xy-f(x)+g(y)|≥ c²
exist

I really don't get it.
Please help me. thanks[/quote]

See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=155063[/url] which seems nearly the same.


inequalities
function
inequalities unsolved

No one?? I've been trying to solve this for a while but I can't. I've used all strategies I know for solving this kind of problems.

Yeah, nice solution  :) , even easier. Likewise for my second one, as $[2,2,2]+[4,1,1] \ge 2[3,2,1]$ is plain Schur, and $[4,2,0]\succ [3,3,0]$.

So that means we can combine both by writing
\[[4,2,0]-2[3,2,1]+[2,2,2] \ge |\ [4,1,1]-[3,3,0]\ | \]
which is interesting: The LHS of this is trivially $\ge0$ by AM-GM, and this is intersparsed by the (absolute) difference of two partition functions not comparable by $\succ$.

I suppose that more generally for each pair of partitions $\alpha, \beta \vdash n$ such that $\alpha\not\succ \beta$ and $\beta\not\succ\alpha$, we can find partitions $\gamma, \delta\vdash n$ such that $[\gamma]+[\delta]-2[\frac{\gamma+\delta}2]\ge |\ [\alpha]-[\beta]\ |$. How to find the [i]best [/i]$\gamma$ and $\delta$, ('best' meaning here that the LHS is smallest possible)? Is the best pair $(\gamma, \delta)$ uniquely determined by $\alpha, \beta$?  :lol:

[quote="rem"][quote="jasamper88"]Prove or disprove.
$[2,2,2]+[3,3,0]+[4,2,0] \ge 2[3,2,1]+[4,1,1]$
I feel stupid, because I can't prove this.[/quote]
Well $[4,2,0]\ge [4,1,1]$ by Muirhead;
$[2,2,2]+[3,3,0]\ge 2[3,2,1]$ by Schur(in degree $3$) for $ab,bc,ca$
So this ineq is true[/quote]

What is Schur in degree 3?  :blush: Could you explain please? I tried schur but did not get anything.


I am writing a mock-MATHCOUNTS using $ \text{\LaTeX}$. I don't care about the cover page, but I want the problems to look like the MATHCOUNTS papers, so I want to know how to make two colums with the answer blanks in one and answer choices in another. If you don't know what I am talking about, please look at the previous tests at http://www.mathcounts.org. I also need a border. Thanks so much!

Use twocolumn option in \documentclass and see section 9 of the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=fancybox]fancybox manual[/url] for a frame for the whole page eg \fancypage{\fbox}{}

I don't understand. The manual doesn't show me how to add just a plain square border and I don't even know where to download fancybox.sty. And the twocolumn thing won't make my text 12pt and I want my problems to be like 2/3 the page and my answer blank to be about 1/3.

I've just told you how: \fancypage{\fbox}{}. fancybox.sty is included in all LaTeX distributions unless you have MiKTeX Basic in which case it will fetch and install it for you.

What has twocolumn got to do with font size; they are independent options? \documentclass[twocolumn,12pt]{article} will give two equal size columns and 12pt font size. If you want unequal columns then use a table with p{[i]width[/i]} to give the size of each column.


geometry
circumcircle
incenter
search
geometric transformation
angle bisector

Let be given a triangle $ABC.$ Let $O_{1}$ be the center of the circle $C'$ which touches $AB,\,AC$ at $E,\, F$ and internal tangent to the circumcircle $C(O)$ of $ABC$ at $T,$ respectively. Let $M$ be the second intersection of $AO'$ and $C(O).$
Prove that $BC,\,EF,\,MT$ are concurrent.

It's known that the incenter $I$ of $\triangle ABC$ is the midpoint of $EF.$ For some proofs, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1164894600&t=31739]Incenter I is the mid point of DE [mixtilinear incircle][/url] and elsewhere.

Let $TA,TC$ cut  $(O_1)$ again at $A',C'.$ Since $T$ is the exsimilicenter of $(O) \sim(O_1),$ then $AC \parallel A'C'.$ Thus, the arcs $FA'$ and $FC'$ of $(O_1)$ are equal, i.e. $TF$ is angle bisector of $\angle ATC$ $\Longrightarrow$ $BI,TF$ meet at the midpoint $N$ of the arc $AC$ of $(O).$ By Pascal theorem for the cyclic non-convex hexagon $TMACBN,$  the intersections $P \equiv MT \cap BC,$ $ I \equiv AM \cap BN$ and $F \equiv NT \cap CA$ are collinear $\Longrightarrow$ $EF,BC,MT$ concur at $P.$

Dear Mathlinkers,
you can see also
http://jl.ayme.pagesperso-orange.fr/  vol. 4 A new mixtilinear incircle adventure I, p. 31-33.

Sincerely
Jean-Louis


Is there any quick way to jump from binary to base 10 or base 10 to binary? I understand how binary works with the first digit being $ 2^0$, the next digit being $ 2^1$ and so on, but how can this be done quickly?

There isn't necessarily a "fast" way to do it.
I normally just do it manually.

Make a program. :wink: Unless if of course, it's a sprint round problem, in which case, do it manually.

If you need a good program to convert from base to base (for Target), just go to http:\\www.ticalc.org and search base converter.lol

[quote="hm29168"]Ok, but say there's a semi-big number such as 496 that you need to convert on a Sprint Round to binary. What would be a good strategy? Do you just subtract 256 and so on til you get an answer?[/quote]

Funny you chose 496: $ 496\equal{}16\times 31$, so $ 496\equal{}32\times 16\minus{}16$, which in binary is $ 10000000\minus{}10000\equal{}1110000$, I think.

Wow. I just chose that number randomly.

That's sorta what I meant by quick way. So you just factor the number and hope you get some $ 2^n$ values?

[quote="alanchou"][quote="hm29168"]Ok, but say there's a semi-big number such as 496 that you need to convert on a Sprint Round to binary. What would be a good strategy? Do you just subtract 256 and so on til you get an answer?[/quote]

Funny you chose 496: $ 496 \equal{} 16\times 31$, so $ 496 \equal{} 32\times 16 \minus{} 16$, which in binary is $ 10000000 \minus{} 10000 \equal{} 1110000$, I think.[/quote]

Clever!

If there is no trick like this that will work for the number you are given, manually is probably the way you should do the problem. There ARE ways to quickly jump from base 8 or base 16 to base 2 and vice versa, however, not to base 10 (or back) that I know of.

hmm, oops, it looks like I thought $ 2^5\cdot 2^4\equal{}2^7$, but it's the idea that counts

the correct answer is 111110000, btw

[quote="Nexmus"]If you have a titanium, just type 0b (binary number) and click enter. (I think...been some time)[/quote]

that's to convert from binary to decimal.  The other way around is to do some other stuff

uhhh...the other way around is u use binary mode and type in the decimal without 0b

Actually the other way around is to type $ x \blacktriangleright \text{Bin}$ where $ x$ is an integer in the range $ (\minus{}\infty, \infty)$.


limit
logarithms
real analysis
real analysis unsolved

Find
\[ \lim_{n\to\infty}\left(1 + \frac {1}{n}\right)^{n}\left(1 + \frac {2}{n}\right)^{\frac {n}{2^{3}}}\ldots\left(1 + \frac {n}{n}\right)^{\frac {n}{n^{3}}.
}\]

Denoting by $ L$ the limit we seek we have :
$ \ln(L) \equal{}\lim_{n\to\infty} \sum _{k\equal{}1}^n \frac{n}{k^3}\ln(1\plus{}\frac{k}{n})$ .
And now using $ x\minus{}\frac{x^2}{2}<ln(x)<x$  $ \forall x>0$ we get that :

$ \sum_{k\equal{}1}^n \frac{1}{k^2}\minus{} \frac{1}{2} \sum _{k\equal{}1}^n \frac{1}{kn}<\ln(L)<\sum _{k\equal{}1}^n \frac{1}{k^2}$
Because (from Stolz's lemma) $ \lim_{n\to\infty} \frac{1}{n} \sum_{k\equal{}1}^n \frac{1}{k} \equal{} \lim _{n\to\infty} \frac{1}{n\plus{}1}\equal{}0$  we can conclude that
$ \ln(L)\equal{}\lim_{n\to\infty}\sum _{k\equal{}1}^n \frac{1}{k^2}\equal{}\frac{\pi^2}{6}$ .
So $ \boxed{L\equal{}e^{\frac{\pi^2}{6}}}$

There was a very similar problem, if not the same, posted some time back. Unfortunately I have no idea what to look for.

Incidentally:
[quote="Svejk"]Denoting by $ L$ the limit we seek we have :
$ \ln(L) \equal{} \lim_{n\to\infty} \sum _{k \equal{} 1}^n \frac {n}{k^3}\ln(1 \plus{} \frac {k}{n})$ .
[/quote]

ln (1+x) ~ x as $ x \to 0$
$ \sum\limits_{n \equal{} 1}^\infty {\frac {1}{{n^2 }}} \equal{} ?$

 :P

[quote="§outh§tar"]There was a very similar problem, if not the same, posted some time back. Unfortunately I have no idea what to look for.

Incidentally:
[quote="Svejk"]Denoting by $ L$ the limit we seek we have :
$ \ln(L) = \lim_{n\to\infty} \sum _{k = 1}^n \frac {n}{k^3}\ln(1 + \frac {k}{n})$ .
[/quote]

ln (1+x) ~ x as $ x \to 0$
$ \sum\limits_{n = 1}^\infty {\frac {1}{{n^2 }}} = ?$

 :P[/quote]

idea is true I agree but you also need to check changing of $ \ln (1+x) - x$ when $ 0<x\leq 1$
or I mean :
$ n \sum^{n} \frac{1}{i^{3}} \ln (1+\frac{i}{n})= n \sum\frac{1}{i^{3}}  (\frac{i}{n}-\frac{i^{2}}{2n^{2}}+o())=(\zeta(2)-\frac{\ln n}{n}+o() )\limits_{n \to \infty} \to \zeta(2)$
you are done
.
and here is generalization:
$ n\sum^{n}\frac{1}{i^{m}}\ln (1+\frac{i}{n})$
answer will be $ \zeta (m-1)$


linear algebra
matrix
limit
induction
Functional Analysis
absolute value
superior algebra

Let $ A$ and $ B$ be $ n\times n$ complex matrices. Let $ C \equal{} AB\minus{}BA$.

If $ C$ commutes with $ A$, then prove that $ C$ is nilpotent.

wow i dont even know what a spectral radius is!

Could you explain to me its importance, and how to compute it?

the spectral radius $ r(A)$ of a bounded operator $ A$ on a complex hilbert space $ H$ is the "size" of the the spectrum $ \sigma(A) \equal{} \{\lambda \in \mathbb{C} : A \minus{} \lambda \text{~is invertible}\}$, precicely $ r(A) \equal{} \sup_{\lambda \in \sigma(A)} |\lambda|$. It is well-known (so you can find it in every introduction to functional analysis) that
\[ r(A) \equal{} \lim_{n \to \infty} {\left\| A^n \right\|}^{1/n}.
\]
If $ H \equal{} \mathbb{C}^n$ is finite-dimensional, i.e. $ A$ is just a matrix, the definition says that $ r(A)$ is the largest absolute value of an eigenvalue of $ A$.

in your exercise, we have to show $ r(C) \equal{} 0$, because then $ C$ must be nilpotent. define $ \delta(X) \equal{} AX \minus{} XA$. then $ \delta$ is easy to be seen a derivation and our assumption $ CA \equal{} AC$ tells us $ \delta^2(B) \equal{} 0$. now an easy induction yields $ \delta^n(B^n) \equal{} n! (\delta(B))^n$. then we have $ \left\|C^n\right\| \leq 2^n/n! \left\|A\right\|^n \left\|B\right\|^n$ and the claim follows by taking limits. many gaps are left here, but I think that you can work them out.

well "my" approach works in every normed algebra: if ab-ba commutes with a, then r(ab-ba)=0 (where r is defined by the limit above).

what do you mean by traces of $ C^k$?

[quote="harazi"]I have to admit that it is very difficult to get a more complicate solution to this simple problem. However, I do like your approach. The trivial way is to prove that the traces of $ C^k$ are all zero. This is immediate from hypothesis and the fact that the trace of a commutator is $ 0$.[/quote]

De gustibus non est disputandum.

To answer -oo-'s question, $ tr(C^k) \equal{} tr([A,BC^{k \minus{} 1}]) \equal{} 0$, so the sum of the kth powers of the eigenvalues is always zero, so the eigenvalues themselves must be zero.  So the matrix is nilpotent. (obviously there are a couple details to fill in for this argument, but they're all "elementary")

Just for curiosity: is there a way to see this via the structure theory of lie algebras? If, for example, $ [A,B]\equal{}A$, then one can look at the solvable subalgebra spanne by $ A$ and $ B$, and then $ A$ is nilpotent as it lies in the commutator subalgebra. Is there a similar way to deal with $ [A,[A,B]]\equal{}0$?

I think the statement is also true for a field K with n not in (p),p=char(K)
Patrick


geometry
trigonometry
rhombus
combinatorics unsolved
combinatorics

Let $ \mathcal{F}$ be a family of hexagons $ H$ satisfying the following properties:

i) $ H$ has parallel opposite sides.

ii) Any 3 vertices of $ H$ can be covered with a strip of width 1.

Determine the least $ \ell\in\mathbb{R}$ such that every hexagon belonging to $ \mathcal{F}$ can be covered with a strip of width $ \ell$.

Note: A strip is the area bounded by two parallel lines separated by a distance $ \ell$. The lines belong to the strip, too.

This is a very nice problem!

The answer is quite a surprise: $ \sqrt 2$. To see that $ \ell\geq\sqrt 2$ consider a square $ ABCD$ and translate the triangle $ ABC$ by $ \overrightarrow{v}\equal{}\epsilon\overrightarrow{BD}$, obtaining hexagon $ A'BB'CC'D$. Any strip covering this hexagon must be of width close to $ \sqrt 2$, and by varying $ \epsilon$ one can make such width close as we want to $ \sqrt 2$.

Now to see that a strip of width $ \sqrt 2$ cover any hexagon $ UVWXYZ$ from $ \cal F$ consider the triangle $ UWY$ obtained by connecting alternating vertices from the hexagon. If one side of $ UWY$, say $ UW$, has length not exceeding $ \sqrt 2$ we are done, since the distance between opposite hexagon sides $ ZU$ and $ WX$ does not exceed $ UW$ and we can cover the hexagon with a strip of width $ UW\leq\sqrt 2$. So suppose that all sides of $ UWY$ are larger than $ \sqrt 2$. Since its smallest altitude must not exceed $ 1$, one can see that $ UWY$ is obtusangle: if the smallest altitude $ YT$ is perpendicular to $ UW$, then $ \cos\angle TYU <{1\over\sqrt 2}\Longrightarrow\angle TYU > 45^{\circ}$ and, analogously, $ \angle TYW > 45^{\circ}$ and, finally, $ \angle UYW > 90^{\circ}$.

Let $ r$ and $ s$ lines parallel to $ UY$ and $ UW$, respectively, with $ d(r,UY) \equal{} d(s,UW) \equal{} 1$ . It's not hard to see (but it's not so easy either!) that $ X$ must belong to the rhombus determined by the four lines $ UY$, $ UW$, $ r$ and $ s$; otherwise, either the hexagon is not convex or one of the triangles $ XUW$ and $ XUW$ have all altitudes larger than $ 1$. Since $ \angle UYW > 90^{\circ}$, $ UX <\sqrt 2$ and the opposite hexagon sides $ UV$ and $ XY$ are less than $ \sqrt 2$ apart and we can cover the hexagon with a strip of width $ \sqrt 2$ again.

The hexagons must be convex hexagons!

you can check it in the official web http://www.oei.es/oim/prova2es.pdf

You need to consider the hexagon D'A'ABCC' to obtain that \ell\geq\sqrt 2

this seems very similar to this japanese problem [url]http://www.imomath.com/othercomp/Jap/JapMO07.pdf[/url] .... quite curious, huh?


conics
parabola
quadratics
algebra
polynomial
geometry
geometric transformation

A standard parabola is the graph of a quadratic polynomial $y = x^2 + ax + b$ with leading co\"efficient 1. Three standard parabolas with vertices $V1, V2, V3$ intersect pairwise at points $A1, A2, A3$. Let $A \mapsto s(A)$ be the reflection of the plane with respect to the $x$-axis.

Prove that standard parabolas with vertices $s (A1), s (A2), s (A3)$ intersect pairwise at the points $s (V1), s (V2), s (V3)$.

Denote by $P(Q)$ the standard parabola with vertex at point $Q$.

[b]Lemma.[/b] If $B\in P(A)$, then $s(A)\in P(s(B))$.
[i]Proof.[/i] Let $A=(a_x,a_y),B=(b_x,b_y)$, so that $s(A)=(a_x,-a_y),s(B)=(a_x,-b_y)$. Then
\begin{align*}
B\in P(A)&\iff b_y=(a_y-a_x)^2+a_y\\\iff -a_y=(a_x-a_y)^2-b_y&\iff s(A)\in P(s(B)).
\end{align*}

Denote now $A_1=P(A_2)\cap P(A_3),A_2=P(A_3)\cap P(A_1),A_3=P(A_1)\cap P(A_2)$. By the lemma, we know that $A_2\in P(V_1)\implies s(V_1)\in P(s(A_2))$. Similarly, $A_3\in P(V_1)\implies s(V_1)\in P(s(A_3))$.

Since two different standard parabolas intersect in at most one point, it follows that $P(s(A_2))\cap P(s(A_3))=s(V_1)$. Analogously, it follows that $P(s(A_3))\cap P(s(A_1))=s(V_2)$ and $P(s(A_1))\cap P(s(A_2))=s(V_3)$, as desired.


The average of 12, 21 and $ x$ is 18. What is the value of $ x$?


We have to solve simple inequalities, too. Here's one example:
  Prove that if a,b,c>0 have product 1, then
  27(a^2+bc)(b^2+ca)(c^2+ab)>=(a+b+c+1/a+1/b+1/c)^3.

I don't really like this solution (I'm pretty sure there are simpler ones out there  :D ):

First of all, I used the inequality (x+y)^3 <= 4(x^3+y^3), so (a+b+c+1/a+1/b+1/c)^3 <= 4(a+b+c)^3+4(1/a+1/b+1/c)^3. Then we can see that 27(a^2+bc)(b^2+ca)(c^2+ab)=27(a^3+1)(b^3+1)(c^3+1)=
27(abc+a^3+b^3+c^3+1/abc+1/a^3+1/b^3+1/c^3), so it would be enough to prove that, for positive a,b,c we have 27(abc+a^3+b^3+c^3)>= 4(a+b+c)^3. This reduces after all the calculations to 3abc+23(a^3+b^3+c^3)>=12(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b). We know from Schur 3abc+a^3+b^3+c^3>=a^2b+a^2c+b^2a+b^2c+c^2a+c^2b and a^3+b^3+c^3>=3abc from AM-GM, so

3abc+23(a^3+b^3+c^3)>=36abc+12(a^3+b^3+c^3)>=12(a^2b+a^2c+..), so the inequality we wanted is proved.

Yes, there are. I have found two solutions: an ugly one and a nice one. I will present only the nice one. Clearly, we have prod( x^2+yz)=prod (x^3+1). Let us denote A=(prod(x^3+1)^1/3. Then x^3/(x^3+1)+y^3/(y^3+1)+1/(z^3+1)>=3xy/A, x^3/(x^3+1)+1/(y^3+1)+z^3/(z^3+1)>=3xz/A, 1/(x^3+1)+y^3/(y^3+1)+z^3/(z^3+1)>=3yz/A, x^3/(x^3+1)+1/(y^3+1)+1/(z^3+1)>=3x/A and two other inequalities of this type. Sum up all these inequalities.


geometry
geometric transformation
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Support
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Something I'm writing:

CHIMERA

chimera - chimaera - a foolish, idle fancy.
[L. chimaera, <Gk. khimaira, she-goat, akin to Gk. kheimon, winter].

MONSTERS, in the language of mythology, were beings of unnatural proportions or parts, usually regarded with terror, as possessing immense strength and ferocity, which they employed for the injury and annoyance of men. Some of them were supposed to combine the members of different animals; such were the Sphinx and Chimaera and to these all the terrible qualities of wild beasts were attributed, together with human sagacity and faculties.

THE CHIMAERA was a fearful monster, breathing fire. The fore part of its body was a compound of the lion and the goat, and the hind part a dragon’s.

- Bulfinch's Mythology, Ch. XVI.

Translator's Note to the Terran Edition:

Not long ago a curious object was recovered from deep space.  Its trajectory suggests that the object originated in the vicinity of a small star in the Sirius Supercluster.  The star is middle-aged, and is graced with several planets.

The object weighs about 350 grams, and is composed primarily of cellulose.  It is formed in a series of flexible rectangular sheets, 20x28x.01cm in size, folded in groups of six along half of their longest dimension, where they are bound together along the 20cm axis, producing a sort of many-leaved, rectangular fan.

Across the surfaces of these sheets of matted cellulose, at right angles to the axis along which they are bound, we find 43 regular rows of marks. The marks themselves are regular: 36 different marks occur in short strings, ranging from one to 28 marks in length, each string separated by a small gap.

As yet no theory of natural origin has survived serious scrutiny.

Fantastic as it might seem, the possibility that this object is the result of intelligent design cannot, therefore, be entirely discounted.

Object or artifact?  While public opinion is predictably divided, debate rages among the experts.

This is an account of the author's attempt to find and describe any evidence of intelligent design to be found in the sequences of these marks.

I have transmitted  this translation of my analysis for the edification of any, should such creatures actually exist, who might actually use these marks as a means of communication.

In the admittedly unlikely event that a copy of this translation should find its way into the hands of someone who can actually make some kind of sense out of it, the author wishes them well, and hopes to reward their attention with a deeper, broader, and more coherent understanding of what it is that we are actually doing when we use language.

The methods employed in this study may be fruitfully applied to the analysis of many languages, be they spoken, telepathed, or secreted, among the inhabited regions of the Orion Arm.

Dedication

For whom burns the midnight oil?  For you, for one.  For the author, it's like sluicing down Sliding Rock in springtime when the stream flashes down the chute in an icy sheet and you come hurtling over the ledge and sock wham into the ice-cold pool at the bottom, whoosh.
Sometimes, though, it's more like pulling teeth.

Unh.

And then there are those memorable moments when the limb breaks and you fall clean out of the  tree.

Thud.

Gasps, bruises, and dental work have all been rendered under the constant supervision of the reader in the imagination of the author.  Running the rapids down this mountainside of data to plunge into this book would have been inconceivable without your constant, if unwitting, participation.

And so it is with the utmost humility, delight, and candor, that the author offers this token of his gratitude for the reader's steadfast support.

Preface

To the cultivator of literary sensibilities, and to the investigator of linguistic intricacies, the author pleads guilt with premeditation for the crimes that he has committed to paper, in the words that are to follow.  For what it's worth, he tried.  The author has endeavored to contrive a single vehicle to bear a variety of readers to a diversity of destinations, and if the mainsail gets mangled in the tank treads, or if in high winds the pterodactylerons suspending the gondola yaw so precipitously athwart that the fruit bats can no longer navigate, send one bat back, and we'll dispatch a search and rescue team who'll talk you down to where you can resume tracking the bear, on foot.

Don't worry if none of this makes much sense right now.  The vehicle is self propelled, and if from time to time it seems to wander off the track, strap yourself in, and trust the bats to find the way.

So let us pray, prostrate, and perambulate, as we prepare to peregrinate.

The author has shepherded this wild herd of words through a mighty wilderness of sin and pain, and prays they will bear the reader without mishap through canyon and crevasse .  Still, even if their alphabetical meanderings are reflected in the Jeweled Net of Indra, hitched to a magic carpet and led down the Yellow Brick Road to the gates of the Emerald City by Pythagoras, Tinkerbell, and Yogi Berra, this monstrous throng that has kept half the county up into the wee hours through the frenzied corralling of, may ultimately fail to bear anyone, anywhere.

Avalokitesvara* hums, twinkling in her tiara, as the author opens the gate.

Neighbors dive for cover.

Some rare collections of words reveal the majesty of language in all her beauty, magnificence, and power.

This is not one of them.

The prospect of luring the reader away from other, more profound pursuits, like, say, Thucydides, has dissuaded the author from becoming one, until now.

Desperation and madness have driven him to this.

What this is, is a sort of linguistic fashion show, where we model a little of language's lingerie.

Flattening it out to the thickness of a sheet of paper does strange things to a language.  Not to mention what it does to what we do with it, or what it does with us.  Reducing the number of dimensions in the channel of communication generates certain paradoxes in the translation of the transmission.

Certain very useful paradoxes.

For this moment is, in fact, two.  As these words appear in the author's mind, they also appear in yours.  Right here, right now, the two of us meet in a strangely timeless place.  It's an odd form of communion, arising out of our mutual willingness to engage each other in this way.  As our fingers tap the keys, our fingers turn the pages.  The keyboard is the rudder, the paper is the sail.  We set out for lands yet undiscovered, across a sea of meaning, the Great Semantic Ocean.  So lay in provisions, say goodbye to your loved ones, and let us

Cast Off

Pursuing our relationship this way, instead of face to face, live and in person, takes a special effort, and grants us special powers.  Powers to consider, gossip,  evaluate, speculate, and revise our interpretations and consequent responses, which amounts to inventing new ideas.

HOW DO YOU SPELL POWER?

WITH THE ALPHABET.

Think about it.

See what I mean?

Amazing.

Isn't it?

Why are we not shocked into silence by the audacity of the whole idea of using words, in the first place?  How does it happen that we perform this miracle without even blinking, much less reveling in amazement at the feat?

WHAT IS GOING ON HERE?

Concentrating our attention on teasing meanings from these marks builds bridges that span the gaps that lie between us in space and time.  Ironically, when we erase time and space, we also become more independent.  More free.  I write down whatever comes into my head, and you read whatever you want.  I take a break to contemplate, you answer the door.  And in a strange, timeless way that I don't pretend to understand, as I encounter these words for the first time, you do, too.  This moment of discovery is one that we both share.  You miss out on all the things that I experience along the way to encountering these words, like the wind blowing in the trees, and all the words that came and went before bearing these.  I miss out on all the things that you've experienced along the way to meeting them, too.  So in this sense, we're both in the same boat.  For as long as we care to ship together.

What seems like temporal teleportation from our immediate perspective is actually only virtual temporal teleportation.  It is achieved and maintained only within certain strict constraints.  The circular paradoxes encountered in the event of actual time travel do not arise, since the reader can see, but cannot touch, while the author can touch, but cannot see.

Temporal teleportation by dimensional compression is achieved by precluding precognition in one direction, and by tying history's hands, in the other.

So we both give up something, in order to be here, together.

Seasoned explorers of these waters, 'distributional analysts', 'computational linguists', and 'corpus linguists', are wandering these seas without the wherewithall to navigate them.  These intrepid linguistinauts do not suffer from any deficiency of subtlety or sophistication among the instruments that they devise, but they have yet to form a concept of the realm that enables them to find their way.

Every languageographic expedition that's returned so far has failed to find this ocean's limit.  Some think it might be infinite.  Others who set out to find one, have yet to find an edge.
No knotted line is long enough to span it.  No mountain high enough to see across it.

Some think that even though it's limitless, it may not have an edge.  If this world of ours is round, and if space itself is curved, then maybe the latitudes and longitudes of language bend around some as yet undiscovered dimensions, too.

The tack the author takes starts from a different understanding of where we stand in the universe from the one he absorbed as a child, and proceeds perpendicularly, from there.  Are we sandwiched between heaven above and hell below, or are we pinned to the surface of a big wet rock, twirling as it spins around a nearby star?

Or are we somewhere else, entirely?

The author is searching for an answer.  A means by which to navigate this Great Semantic Ocean.

Curiosity, as it turns out, proves to be a manifestation of the only truly irresistible force in the entire universe, the only force that in the end, is capable of moving even the most immovable object, which clearly, must be apathy.

Are these ideas that seem so clearly revolutionary to the author, to others, already anachronistic?  Is he really, in his heart of hearts, just an unreconstructed structuralist?

Or just plain crazy?

Foreword

This is a preliminary investigation into the phenomenon of human language.

If the commonsensical ideas that we share about what it is that we are doing when we use language are correct, then those of us who are led to question these ideas, insofar as we actually discover the truth, will discover that the rest of us who already thought that we knew what we were doing when we use language, were right all along.

Sigh.

The author started losing his grip on common sense many lunacies ago.

It's hard to pinpoint just when he lost it, completely.

Actually it all started with Socrates.  The authority that Socrates led the author to question, is the authority of his language, over his mind.

Whether these common sense ideas broke down because the  author abused them, or because they never actually worked very well in the first place, is anything but academic, but impossible for him to determine without your help.

And so, out of the aboriginal collision of the author's immovably jaundiced view of his own intellectual powers, and the irresistible temptation to find out, once and for all, just where he went wrong, the words that are to come, follow.

These selfsame words that the author is at this very moment groping to arrange in the order that they were meant to be, these very words that the reader is somehow simultaneously, yet consequently, groping to find a reason for, drizzle, fall, and sometimes pour in cataracts across the page, and out of  the inundation, ideas grow.

Might this particular concatenation constitute a kind of insertion mutation in our conceptual genome, a particular permutation that might promote the transformation of the metabolism of our thought-processes?

Whatever is about to happen, it would seem, is entirely up to us.

Is it possible for us to discover something new?  Can two people who have never met, discover something, together?

If so, how?  With words that we both already know?

How?

By arranging them in some special way?

How?

Does our desire to learn something new have anything to do with whether or not we actually do?  How does it happen that we learn something new, when we're hoping to?  What about when we're not expecting to?   Do our hopes and fears, our expectations and predispositions, play some role in what we learn?

a.) If they play no part at all, then we have no worries.    Life, sometimes, can be painful.  Getting stubbed is just part of being a toe.  Get used to it.

b.) If, perversely, hoping to discover something new actually eliminates any chance of actually doing so, then the only hope there is, is for those who don't have any.  Our first mistake is trying to learn from our mistakes.  We lose our balance when we look down, and we whack our throbbing toes, yet again.

c.) If, conversely, hoping to discover something can somehow engender its objective, how might those of us who hope to, do so?

As the author inundates, let the reader demystify.

I did not invent these words, nor did I contrive their arrangement.  How these words wound up appearing here is a complete mystery to me.  I decide to write, the words come out of nowhere.  I learned to do this from other people, kind people, a whole forestful of people who have been caring for me and looking out for me, for my whole life.

This may go some way to explaining what drives these digits to such dervish frenzy.  I want to thank you.  I want to thank you, and these words wind up landing here.

But if gratitude is the thing that these digits are turning into characters, gratitude in no wise explains HOW these words decide to arrange themselves in the order that they do.  Words come, I spell them.  It's like collecting laundry in a windstorm, or catching fish in a whirlpool.  All of these words are blowing and wriggling around, and I spin around in a lot of confusion, trying to catch them.  Why these,words, in this order?

Why these words, and not others, in this order, and not another, and where they all come from, to begin with, and where they all go, afterwards, I have no idea.  Maybe these words arise from the same nothingness into which they consequently vanish.  Maybe it doesn't make any sense to think of these words as arising from anything at all, even nothingness, but are really just transient, illusory features of an ever-changing present as it inexplicably elides into something else.

All I know is, before these words appeared and eventually settled into the order that we now find them, others arose and vanished, many others, and a whole lot more others, never even occurred to me at all.  Where did these words come from, in the first place?  And of all of the words blowing around in this windstorm, swimming around in this whirlpool, why are these the ones that have wound up getting caught?

While I can't seem to come to grips with where they're all coming from, once they land here, it's another story.  Now we've got something to work with.  Now we've got some data to sink our analytical teeth into.  Mysticism gives birth to logic.

If we really understood the results of whatever it is that tempts us to try using language in the first place, if we understood the things that we actually say, it might help us understand how we wind up saying them.  Maybe we could reverse-engineer the process by analyzing its products.

An adequate description of the products of linguistic creativity might shed some light on the process that creates them.  Maybe this is how we could define 'adequacy' in the first place.  Before Copernicus, several dozen generations of good Christians imprinted their offspring with the belief that the sun went around the earth.  Jesuit astronomers refined the  geocentric cosmology of Ptolemy for hundreds of years to  predict the motions of the planets through the heavens  with ever greater accuracy.  Was Ptolemy's ancient method of predicting planetary motions 'adequate'?  A large body of evidence supported its predictions.  But other evidence, that did not, seemed irrelevant at the time.  So how do we select what evidence is relevant, and what evidence, is not?  Do comets, and eclipses, count?

It depends on whether the evidence supports our theory, or not.

If, just for the sake of argument, we decide to question whether our understanding of what it is that we are doing when we use language is correct, then we might consider reevaluating our notions of adequacy.  What evidence might a better theory more adequately describe?  What aspects of language lie outside of our common sense ideas of adequacy that we use to circumscribe the aspects of language that we consider to be relevant to our understanding of it?

Grammatical comets?  Semantic eclipses?

Some of the ideas that have been blowing through this salty brain tend to drift a bit from the course of common sense.  If in fact the author's dinghy has truly lost its bearings, then the delusional state that buoys his boat is a truly wondrous, immeasurably immense, daedal chimera.

Something is extracting these words from this brain in this particular order.

Is it really magic, or just a clever trick?

Next: Chapter One

:)

...This seems interesting.

And no, you're not alone. I've thought thoughts like this--as others have, undoubtedly.

No worries. I assure you--you are just as insane as the rest of us. Coming from me, though, may dispute the authority of the statement. :lol:

May the words arrange themselves skillfully. (Best wishes.)


[quote="nmtvn"]Algebraic Inequality[/quote]
To Mods: Please lock this topic, this is a problem from M&Y, issue 9, and we can't discuss it now.


real analysis
advanced fields
advanced fields theorems

I've been thinking about the chromatic number of a plane. Basically, I assumed it was 4 and tried to find a contradiction, or some other interesting result. I've found one or two cool things; for example, if you have three collinear points of the same color a certain distance from each other, then a specific point on that line a certain distance further down is also the same color. I was wondering if anyone has seen similar results around.

That depends on what the edges are.  The standard version of this problem is that two points are connected if they differ by a distance of $ 1$, and this is the [url=http://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem]Hadwiger-Nelson problem[/url], which is open.

You misunderstood.

I am talking about the "The standard version". But I'm not asking about the Hadwiger-Nelson problem; I know it is open. I simply want to know what characteristics a 4-coloring of the plane (if it exists) has.

You can find some in the Wikipedia article.  For example, the color classes cannot all be Lebesgue measurable.  This suggests, at least morally, that the axiom of choice is necessary in some form to construct a 4-coloring, if it exists.


Let $ z_1$ , $ z_2$ , $ z_3$ be distinct complex no.s such that

$ |z_1| \equal{} |z_2| \equal{} |z_3| >0$

If $ z_1 \plus{} z_2z_3$ , $ z_2 \plus{} z_3z_1$ , $ z_3 \plus{} z_1z_2$ are real no.s , prove that $ z_1z_2z_3 \equal{} 1$

I see no way to do this as the first condition follows from the second.

the second condition \implies

$ z_1 \equal{} \bar{z_2}\bar{z_3}..............(1)$

$ z_2 \equal{} \bar{z_1}\bar{z_3}..................(2)$

$ z_3 \equal{} \bar{z_2}\bar{z_1}....................(3)$.

from (1) and (2),

$ z_1\bar{z_1}\bar{z_3} \equal{} z_2\bar{z_2}\bar{z_3}$

$ \implies$ $ z_1\bar{z_1} \equal{} z_2\bar{z_2}$.

similarly, from (2) and (3), we can get

$ z_2\bar{z_2} \equal{} z_3\bar{z_3}$

from which we get the first condition  :huh:


Evaluate the infinite sum $ \sum_{n\equal{}1}^{\infty}\frac{n}{n^4\plus{}4}$.

[hide="Solution"]$ n^4 \plus{} 4 \equal{} n^4 \plus{} 4n^2 \plus{} 4 \minus{} 4n^2 \equal{} (n^2 \plus{} 2)^2 \minus{} (2n)^2 \equal{} (n^2 \plus{} 2n \plus{} 2)(n^2 \minus{} 2n \plus{} 2)$

Now, by partial fractions, we get that this is equal to $ \frac {1}{4} \sum_{n \equal{} 1}^{\infty} \frac {1}{n^2 \minus{} 2n \plus{} 2} \minus{} \frac {1}{n^2 \plus{} 2n \plus{} 2}$ $ \equal{} \frac {1}{4} \sum_{n \equal{} 1}^{\infty} \frac {1}{(n \minus{} 1)^2} \minus{} \frac {1}{(n \plus{} 1)^2 \plus{} 1}$

This telescopes, and all terms cancel but the first two, so $ S \equal{} \frac {1}{4} \left( \frac {1}{0^2 \plus{} 1} \plus{} \frac {1}{1^2 \plus{} 1} \right) \equal{} \boxed{\frac {3}{8}}$[/hide]

[hide]
Sophie Germaine anyone?
\[ \displaystyle\sum_{n \equal{} 1}^{\infty}\frac {n}{n^{2} \plus{} 4} \equal{} \frac {1}{4}\sum_{n \equal{} 1}^{\infty} \minus{} \frac {1}{n^{2} \plus{} 2n \plus{} 2} \plus{} \frac {1}{n^{2} \minus{} 2n \plus{} 2}
\]
Writing out the first few terms, note that everything telescopes except for $ \frac {1}{4}\left(1 \plus{} \frac {1}{2}\right) \equal{} \boxed{\frac {3}{8}}$. [/hide]

Edit: Hm, I could've sworn I posted before Zuton Force  lol :huh:


geometry
perimeter
trigonometry
geometry unsolved

Let us consider a convex hexagon ABCDEF. Let $A_1, B_1,C_1, D_1, E_1, F_1$ be midpoints of the sides $AB, BC, CD, DE, EF,FA$ respectively. Denote by $p$ and $p_1$, respectively, the perimeter of the hexagon $ A B C D E F $ and hexagon $ A_1B_1C_1D_1E_1F_1 $. Suppose that all inner angles of hexagon $ A_1B_1C_1D_1E_1F_1 $ are equal. Prove that \[ p \geq \frac{2 \cdot \sqrt{3}}{3} \cdot p_1 .\] When does equality hold ?

Let us note that $p_1$ is the sum of the projections of the sides of ABCDEF to corresponding sides of $A_1B_1C_1D_1E_1F_1$. For example,

  projection of AB = projection of $AA_1$ on $F_1A_1$ + projection of $A_1B$ on $A_1B_1$
  = $\frac{AB}{2}\cdot\left(\cos x+\cos\left(60^{\circ}-x\right)\right) \leq AB\cdot\frac{\sqrt3}{2}$.

Thus, $p_1\leq p\cdot\frac{\sqrt3}{2}$.

Here is my solution for this problem
[b]Solution[/b]
Denote $AA_1 = a$, $BB_1 = b$, $CC_1 = c$, $DD_1 = d$, $EE_1 = e$, $FF_1 = f$, $A_1B_1 = x$, $B_1C_1 = y$, $C_1D_1 = z$, $D_1E_1 = t$, $E_1F_1 = u$, $F_1A_1 = v$
Let $I$, $J$ be reflections of $B$, $C$ through $A_1B_1$, $B_1C_1$, respectively
Since: $\angle{A_1B_1C_1} = 120^o$ then: $\angle{IB_1J} = 60^o$
But: $IB_1 = BB_1 = CB_1 = JB_1$ so: $\triangle IJB_1$ is equilateral triangle
Hence: $IJ = IB_1 = BB_1 = b$
Then: $a + b + c = A_1I + IJ + JC_1 \ge A_1C_1 = \sqrt{A_1B^2_1 + A_1C^2_1 + A_1B_1 . A_1C_1} \ge \dfrac{(A_1B_1 + A_1C_1)\sqrt{3}}{2} = \dfrac{(x + y)\sqrt{3}}{2}$
Similarly: $b + c + d \ge \dfrac{(y + z)\sqrt{3}}{2}$, $c + d + e \ge \dfrac{(z + t)\sqrt{3}}{2}$, $d + e + f \ge \dfrac{(t + u)\sqrt{3}}{2}$, $e + f + a \ge \dfrac{(u + v)\sqrt{3}}{2}$, $f + a + b \ge \dfrac{(v + x)\sqrt{3}}{2}$
So: $3(a + b + c + d + e + f) \ge (x + y + z + t + u + v)\sqrt{3}$ or $a + b + c + d + e + f \ge \dfrac{x + y + z + t + u + v}{\sqrt{3}}$
Hence: $p \ge \dfrac{2p_1}{\sqrt{3}}$
Equality holds when: $ABCDEF$ is regular hexagon


:?:  :?:

I find myself not wanting to do any work unless I really like it; so I usually find myself doing a large project at the last minute (USAMTS and Olympiad PS midterm), homework at 2 AM, and sometimes not doing important things at all (AoPS Volume 2 and AoPS Class HW).  As a result, I've become slightly nocturnal and very tired, and not working to my maximum potential in math.

How do you get yourself to do something beneficial even if you don't feel like it?

STEP ONE: Get off this forum. No offense, AoPS guys, I don't mean it like that. I'm talking to the habitual "Fun and Games" posters, the ones who get wrapped up in circular Round Table (pun!) arguments, and the people who ask (for the MILLIONTH TIME) "how do I get into Prestigious College?"

STEP TWO: Make a list of the things you need to do. I find that if I write down EVERY LITTLE THING I need to do that day, I get great satisfaction from looking at the list and crossing things off, and the bigger things don't seem as daunting.

STEP THREE: Get 'er done. Make a resolution. "I will not [watch TV, play video games, get online] until I get [my homework, this project] done." Then stick to it, and when you're done, play that video game or whatever.

STEP FOUR: [b]Do NOT[/b] make plans with your friends if you know you've got things you need to do. There's a really easy solution for this: "Oh, I'm really sorry. I have to [do my homework, work on this project]. Maybe next weekend?" Stick to your guns.

STEP FIVE: Pay for your AoPS classes yourself. You'll find that when it's your money going towards something like that, you are more likely to appreciate it, and thus your homework seems a little more necessary.


In a school of 500 students, 65 are in the school piano club, 255 students are in the choir and 40 students participate in BOTH activities.  What percent of the number of students in the school participate in EITHER the school piano club or in the choir BUT NOT on BOTH?

My answer, after spending 20 minutes trying to find the correct percentage, is 50 percent.

The book's answer is 48 percent.

Who's right?

Why?

[hide]
It looks pretty simple actually - Just take the (65 + 255 - 80)/500=48%

I bet I got it wrong though - especially since the book was 48% AND your was 50% ....
[/hide]

EDIT: Oh thanks usaha - i wasn't thinking when i read this (this happens TOO much) - i changed it

[quote="Anaxerzia"]
It looks pretty simple actually - Just take the (65 + 255 - 40)/500=56%

I bet I got it wrong though - especially since the book was 48% AND your was 50% ....
[/quote]

Too easy but you misread the question (or I hope you did because then...yea)  NOT BOTH..when you do that, you still count BOTH once.  So 65+255-80=240.  So 240/500=48 percent


graph theory
combinatorics unsolved
combinatorics

On a circle there are $ n$ red and $ n$ blue arcs given in such a way that each red arc intersects each blue one. Prove that some point is contained by at least $ n$ of the given coloured arcs.

[hide]Letting the arcs constitute as vertices in a graph and letting edges signify an intersection, it is clear that at the least, we have a complete bipartite graph with $ n$ vertices on each side.  Now it is obvious that there must exist a cycle of length at least $ n$.[/hide]


invariant
combinatorics proposed
combinatorics

A deck of $ 2n\plus{}1$ cards consists of a joker and, for each number between 1 and $ n$ inclusive, two cards marked with that number. The $ 2n\plus{}1$ cards are placed in a row, with the joker in the middle. For each $ k$ with $ 1 \leq k \leq n,$ the two cards numbered $ k$ have exactly $ k\minus{}1$ cards between them. Determine all the values of $ n$ not exceeding 10 for which this arrangement is possible. For which values of $ n$ is it impossible?

I'm not even sure how to approach this question (which is bit unusual since Canada Math Olympiad problems usually [i]aren't[/i] that bad).

[hide="My Questions"]
1. Is $ n$ an integer or can include ones like $ n \equal{} \frac{5}{2}$? I'm just saying this because the question says $ 2n\plus{}1$ (it does say $ n$ later but still)
2. The first sentence ends as "...two cards marked with THAT number" What is "that number"? $ 2n\plus{}1$?
3. $ k$ is an integer, right (or integers)? [/hide]

1 does not work
2 does not work
3 works: 232J311
4 works: 2423J4311
7 and higher don't work because there are $ (n\minus{}2)(n\minus{}1)/2$ spaces inbetween the numbers that have to be filled by 2n-1 spaces. ( the 2n+1 spaces - the endpoints). Which doesn't work for $ n \ge 7$

Now bash 5 and 6.

[quote="Silverfalcon"]1. Is $ n$ an integer or can include ones like $ n \equal{} \frac {5}{2}$? I'm just saying this because the question says $ 2n \plus{} 1$ (it does say $ n$ later but still)
2. The first sentence ends as "...two cards marked with THAT number" What is "that number"? $ 2n \plus{} 1$?
3. $ k$ is an integer, right (or integers)?[/quote]

1. $ n$ is an integer.
2. "for each number between 1 and $ n$ inclusive."  That is, there are two cards marked 1, two cards marked 2, etc.
3. yes.

rofler, your argument for $ n \geq 7$ is unconvincing: what quantities, exactly, are you comparing?

Two related questions:
(1) It's easy to see that if the joker can be placed anywhere (not just in the middle), there's always a way to do what we want.  How many ways are there to do it in this case?
(2) When is it possible when there is no joker?

Hi,

Just wondering if someone could post a solution for this problem (hopefully a non-bashy one). Also, if there is another  place to get Canadian math olympiad problems, could someone post a link?

Thanks

Here's the invariant part of the proof:

Given an ordering on the deck of cards (not necessarily of the desired form), define the "total skip" of that arrangement to be the sum over $i$ of the number of cards between the two cards numbered $i$.  Thus, the total skip of an ordering meeting the desired conditions is $0 + 1 + \ldots + (n - 1) = \binom{n}{2}$.

Notice that every arrangement of the $2n + 1$ cards such that the joker is in the middle can be arrived at from any other such arrangement by repeated application of two operations: (1) switching two adjacent, non-J cards, and (2) switching the card to the left and the right of the J.  Observe that both of these operations may change the total skip, but they do so always by an even number.  The arrangement $12\cdots n J 12\cdots n$ has total skip $n^2$.  Thus, if there exists an arrangement meeting the desired conditions, we must have that $\binom{n}{2}$ and $n^2$ have the same parity.  It's easy to see that this implies that $n$ is equivalent to 0 or -1 modulo 4.  What remains is to construct solutions for these values of $n$.

Several past years of CMO are available in the resources section http://www.artofproblemsolving.com/Forum/resources.php and there are also some (a proper subset?) in the wiki http://www.artofproblemsolving.com/Wiki/index.php/Canadian_MO_Problems_and_Solutions

Wow! What a solution! Tried this for 1.5 hrs, and thought was dumb not to get it. I give you one up-vote.

Let the places have integer coordinates from $1$ to $2n+1$, where the leftmost card has the coordinate $1$, and the joker has coordinate $n+1$. Then, consider a number $i$ and let $y_i$ be a coordinate of the leftmost card marked with $i$. Then, by the definition of the sequence the second card marked with $i$ will have coordinate $i+y_i$. Let $a_i$ be the sum of $i$ and the number on the card with coordinate $i$, and let $S=\sum\limits_{\substack{i\ne n+1\\1\le i\le2n+1}}a_i$. On the one hand, it's the sum over all the numbers on cards and all the coordinates except of $n+1$:

$$
S=2\sum_{i=1}^ni+\sum_{\substack{i\ne n+1\\1\le i\le2n+1}}i=n(n+1)+\frac{(2n+1)(2n+2)}{2}-(n+1)=3n(n+1).\qquad (1)
$$

On the other hand, if we pair up the cards with the same numbers on them, then in a pair with the number $i$ on cards we will have sum $(i+y_i)+(i+y_i+i)=3i+2y_i$, and we obtain

$$
S=\sum_{i=1}^n(3i+2y_i)=\frac{3n(n+1)}{2}+2\sum_{i=1}^ny_i. \qquad (2)
$$

Combining $(1)$ with $(2)$, we finally get

$$
S=3n(n+1)=\frac{3n(n+1)}{2}+2\sum_{i=1}^ny_i\Rightarrow \frac{3n(n+1)}{2}=2\sum_{i=1}^ny_i.
$$

This implies $4\mid n(n+1)$, which is possible only for $n=4k$ and $n=4k-1$. Thus, $3$, $4$, $7$, and $8$ are the only answers for $n\le 10$, and it's impossible to construct such a sequence for the numbers of the form $n=4k+1$ and $n=4k+2$.


This problem is dedicated to my Vietnamese Friends who sent me a lot of nice inequalities these day. Other Mathlinkers are very well-come

Let $a,b,c,d$ be non-negative real numbers with sum $3$. Prove that
\[ab(b+c)+bc(c+d)+cd(d+a)+da(a+b) \le 4.\]

I have a very nice solution to this one,

[quote="Little Gauss"]WLOG $b+d\le a+c$.

$ab(b+c)+bc(c+d)+cd(d+a)+da(a+b)=(b+d)(ab+cd)+(a+c)(bc+da)$

$=(b+d)(ab+cd)+(a+c)((a+c)(b+d)-(ab+cd))$

$=(a+c)^{2}(b+d)-(ab+cd)(a+c-b-d)\le (a+c)^{2}(b+d) \le 4$.[/quote]

This is must be one of the most clever solution I see here. Thank you very much, Little Gauss.

Dear my Vietnamese friend! What is your solution?

Ok. Here is two generalization for this problem

$\bigstar$
If $a_{1},a_{2},...,a_{n}$ be non-negative real numbers with sum $3$ then
\[a_{1}a_{2}(a_{2}+a_{3})+a_{2}a_{3}(a_{3}+a_{4})+...+a_{n-1}a_{n}(a_{n}+a_{1})+a_{n}a_{1}(a_{1}+a_{2})\le 4. \]
$\bigstar$ (one generalization of a recent VASC's one)
If $a_{1},a_{2},...,a_{n}$ be non-negative real numbers with sum $2$ and $a_{1}\ge a_{2}\ge...\ge a_{n}$ then
\[a_{1}a_{2}(a_{2}+a_{3})+a_{2}a_{3}(a_{3}+a_{4})+...+a_{n-1}a_{n}(a_{n}+a_{1})+a_{n}a_{1}(a_{1}+a_{2})\le 1. \]
The common method I use is induction.

[quote="hungkhtn"]This problem is dedicated to my Vietnamese Friends who sent me a lot of nice inequalities these day. Other Mathlinkers are very well-come

Let $a,b,c,d$ be non-negative real numbers with sum $3$. Prove that
\[ab(b+c)+bc(c+d)+cd(d+a)+da(a+b) \le 4. \]
I have a very nice solution to this one,[/quote]
Very nice, Hung. Let I post my solution
WLOG, suppose $b=\max\{a,b,c,d\}$, then we have
\[ab(b+c)+bc(c+d)+cd(d+a)+da(a+b) \le ab^{2}+b(c+d)^{2}+(c+d)a^{2}+ab(c+d) \]
Indeed, it's equivalent to
\[cd(b-a)+d^{2}(b-c)+a^{2}c \ge 0 \]
which is obvious. Now, it suffices to prove that
\[xy^{2}+yz^{2}+zx^{2}+xyz \le 4 \]
where $x,y,z \ge 0,x+y+z=3$.
WLOG, suppose $y$ is the value between $x$ and $z$, that means
\[x(y-z)(y-x) \le 0 \]
Hence
\[xy^{2}+zx^{2}\le xyz+x^{2}y \]
Therefore
\[xy^{2}+yz^{2}+zx^{2}+xyz \le y(z+x)^{2}\le 4\left(\frac{y+(z+x)}{3}\right)^{3}=4 \]
The inequality is proved. :)

Yes, it is exactly my solution. Thank you, toanhocmuonmau. I wonder if this is your way to solve me general inequalities? :)