What is the nearest integer to $ (5\plus{}2\sqrt7)^4$?

[quote="pinkmuskrat"]The trick is to add $ (5 \minus{} 2\sqrt7)^4$. $ 5 \minus{} 2\sqrt7$ has a value of about 0.2915, and taking this to the fourth power makes it extremely small, about .00722, so it doesn't affect our final value.

We do this addition so our square roots clear away. Now through the Binomial Theorem,

$ (5 \plus{} 2\sqrt7)^4 \plus{} (5 \minus{} 2\sqrt7)^4$

$ 625 \plus{} 1000\sqrt7 \plus{} 4200 \plus{} 1120\sqrt7 \plus{} 784 \plus{} 625 \minus{} 1000\sqrt7 \plus{} 4200 \minus{} 1120\sqrt7 \plus{} 784$

$ 11218$[/quote]
How do you recognize this manipulation?

@droid347 first we notice that $2\sqrt{7}$ is pretty close to $5$  (so $5-2\sqrt{7}$ is close to $0$) and that simplifying $(a+\sqrt{b})^4+(a-\sqrt{b})^4$ results in no radical signs.

[quote=ChrisalonaLiverspur]But...How do you just think of it? I mean, it doesn't seem like you would just think of it, right? You would take a while. Even then, I couldn't think of it.[/quote]

You automatically try to use this method if you have solved a lot of similar problems before.

Ok. That makes sense. I just should try to think about a lot, maybe I will remember.

[quote=RyanWang]Couldn't you just use the calculator?[/quote]

I mean you can but what’s the fun in that lul

[quote=ChrisalonaLiverspur]But...How do you just think of it? I mean, it doesn't seem like you would just think of it, right? You would take a while. Even then, I couldn't think of it.[/quote]

$(5+2\sqrt7)^4$
=$[(5+2\sqrt7)^2]^2$
=$(53+20\sqrt7)^2$


Euler
number theory proposed
number theory

Prove that if g is a primitive root of p, g is a nonquadratic residue of p.

Products of squares are squares again. So a square can never be a primitive root because all it's powers are squares again. Thus it will never reach the nonsquares (which exist).
Another way is using Eulers criterion: if $g$ is a square, $g^\frac{p-1}{2}\equiv 1 \mod p$, thus it isn't  primitive root.

A longer but interesting proof is; Let r be a residue mod p and g be a primitive root. By Wilson's Theorm,

-1 = (p - 1)! = r_1r_2...r_(p - 1) = g^1g^2...g^(p - 1) = g^(1 + 2 + ...+ p - 1) = ( g^( p - 1)/2)^p = -1 iff g^(p - 1)/2 = -1 which, by Euler's Criterion, means g is a nonquadratic residue of p.


How do I solve the following equation, finding a and b ?

$ \begin{bmatrix}1&\minus{}2\\2&\minus{}3\end{bmatrix} \begin{bmatrix}a\\b\end{bmatrix} \equal{} \begin{bmatrix}\minus{}a\\\minus{}b\end{bmatrix} \plus{}\begin{bmatrix}1\\1\end{bmatrix}$

Thanks.

Hi !
multipling at left, addition at right, we get

$ a\minus{}2b\equal{}\minus{}a\plus{}1$ and $ 2a\minus{}3b\equal{}\minus{}b\plus{}1$ and so on...sorry for my english !

Note that $ \begin{bmatrix}\minus{}a\\\minus{}b\end{bmatrix}\equal{} \begin{bmatrix}\minus{}1&0\\0&\minus{}1\end{bmatrix} \begin{bmatrix}a\\b\end{bmatrix}.$

Write the equation as

$ \begin{bmatrix}1&\minus{}2\\2&\minus{}3\end{bmatrix} \begin{bmatrix}a\\b\end{bmatrix} \equal{}\begin{bmatrix}\minus{}1&0\\0&\minus{}1\end{bmatrix} \begin{bmatrix}a\\b\end{bmatrix} \plus{}\begin{bmatrix}1\\1\end{bmatrix}$

$ \left(\begin{bmatrix}1&\minus{}2\\2&\minus{}3\end{bmatrix}\plus{} \begin{bmatrix}1&0\\0&1\end{bmatrix}\right) \begin{bmatrix}a\\b\end{bmatrix}\equal{} \begin{bmatrix}1\\1\end{bmatrix}$

$ \begin{bmatrix}2&\minus{}2\\2&\minus{}2\end{bmatrix} \begin{bmatrix}a\\b\end{bmatrix}\equal{} \begin{bmatrix}1\\1\end{bmatrix}$

You should be able to finish from there. Note that the answer is not unique. In most of the circumstances in which this problem would arise, I think that a single answer from among the many possibilities would usually suffice.

Underlying this: $ \begin{bmatrix}1&\minus{}2\\2&\minus{}3\end{bmatrix}$ has $ \minus{}1$ as its only eigenvalue and is not diagonalizable. This equation is related to the work needed to put the matrix into Jordan canonical form; alternatively, this work would arise in solving a linear constant coefficient system of differential equations with this matrix as coefficients.

Thanks.

I tried at first exactly what mstoenescu offered but it didnt work (The equations cancelled each other).
Why is that?

you have found $ 2a\equal{}2b\plus{}1$ isn't it ?

[quote="Kent Merryfield"]Note that $ \begin{bmatrix} \minus{} a \\
\minus{} b\end{bmatrix} \equal{} \begin{bmatrix} \minus{} 1 & 0 \\
0 & \minus{} 1\end{bmatrix} \begin{bmatrix}a \\
b\end{bmatrix}.$

Write the equation as

$ \begin{bmatrix}1 & \minus{} 2 \\
2 & \minus{} 3\end{bmatrix} \begin{bmatrix}a \\
b\end{bmatrix} \equal{} \begin{bmatrix} \minus{} 1 & 0 \\
0 & \minus{} 1\end{bmatrix} \begin{bmatrix}a \\
b\end{bmatrix} \plus{} \begin{bmatrix}1 \\
1\end{bmatrix}$

$ \left(\begin{bmatrix}1 & \minus{} 2 \\
2 & \minus{} 3\end{bmatrix} \plus{} \begin{bmatrix}1 & 0 \\
0 & 1\end{bmatrix}\right) \begin{bmatrix}a \\
b\end{bmatrix} \equal{} \begin{bmatrix}1 \\
1\end{bmatrix}$

$ \begin{bmatrix}2 & \minus{} 2 \\
2 & \minus{} 2\end{bmatrix} \begin{bmatrix}a \\
b\end{bmatrix} \equal{} \begin{bmatrix}1 \\
1\end{bmatrix}$

You should be able to finish from there. Note that the answer is not unique. In most of the circumstances in which this problem would arise, I think that a single answer from among the many possibilities would usually suffice.

Underlying this: $ \begin{bmatrix}1 & \minus{} 2 \\
2 & \minus{} 3\end{bmatrix}$ has $ \minus{} 1$ as its only eigenvalue and is not diagonalizable. This equation is related to the work needed to put the matrix into Jordan canonical form; alternatively, this work would arise in solving a linear constant coefficient system of differential equations with this matrix as coefficients.[/quote]

You can proceed furthe by calling
\[ X \equal{} \begin{bmatrix}2 & \minus{} 2 \\
2 & \minus{} 2\end{bmatrix}
\]
and finding its pseudoinverse $ X^{\dagger} \equal{} \begin{bmatrix}0.1250 &  0.1250 \\
\minus{}0.1250 & \minus{} 0.1250\end{bmatrix}$. Thus
\[ \begin{bmatrix}a \\
b\end{bmatrix} \equal{} \begin{bmatrix}0.1250 & 0.1250 \\
\minus{}0.1250 & \minus{} 0.1250\end{bmatrix}\begin{bmatrix}1 \\
1\end{bmatrix} \equal{} \begin{bmatrix}0.2500 \\
\minus{} 0.2500\end{bmatrix}
\]


I study in second grade of high school.I studied Holiday1 for olympiad.Can you interduce me some books for studying physics olympiad?

Dude, wtf.  Please don't triple post the same topic.  Also, I think this should be in the advanced physics forum.

I just reaed Halliday-Resnick-Walker last year, and that seemed to be enough.

THanks for your guidance and I beg your pordon because I really didnt want to triple the same topic 3 times but
abruptly It got like that any way I am so sorry


Prove that for any positive real numbers $a,b$ and $c$ such that $a+b+c=1$,
\[\sqrt[4]{9+\frac{1}{abc}}\ge\frac{4}{3\sqrt{(1-a)(1-b)(1-c)}}\]

[quote="X-men"]I think this ineq can solve by using the well - known ineq: (a + b)(b + c)(c + a) >= (8/9)(a + b + c)(ab + bc + ca).[/quote]
You haven't tried that, have you?

[quote="manlio"]Has anyone tried to solve it? I cannot :blush:

Thank you very much.[/quote]
Try ABC method,i'm sure. :wink:

I would like to see a direct proof. :wink:

[quote="manlio"][quote="ductrung"]Prove that for any positive real numbers $ a,b$ and $ c$ such that $ a\plus{}b\plus{}c \equal{} 1$,
\[ \sqrt[4]{9\plus{}\frac{1}{abc}}\ge\frac{4}{3\sqrt{(1\minus{}a)(1\minus{}b)(1\minus{}c)}}\]
[/quote]

Following X-men hint I found this expression

$ (\sum ab)^{2}(9abc\plus{}(\sum a)^{3})\geq 4abc(\sum a)^{4}$

Then I used Shur $ (\sum a)^{3}\plus{}9abc\geq 4(\sum a)(\sum ab)$ to get

$ (\sum ab)^{3}\geq abc(\sum a)^{3}$

but I think this last is false.

So I don't think X-men hint works. Has anyone tried to solve it or has any hint to propose?

Thank you very much.[/quote]

To continue Manlio'solution  :)

No, Manlio . By ABC-Theorem , we have two cases :

Case 1. $ a \equal{} 0$ ( or $ b \equal{} 0$ or $ c \equal{} 0$ ).
The inequality is equivalent to $ (bc)^{3}\ge 0$ .It is true because $ b,c > 0$.

Case 2. $ a \equal{} b$.
The inequlity is equivalent to $ a\minus{}c)^{2}(a\plus{}c)\ge 0$.
We're done.Equality occurs only if $ a \equal{} b \equal{} c \equal{}\frac{1}{3}$

What is the 'ABC-Theorem'?

$ \left(9\plus{}\frac{1}{xyz}\right)*(x\plus{}y)^{2}(y\plus{}z)^{2}(z\plus{}x)^{2}\minus{}\frac{256}{81}\ge 0$

Let $ a\equal{}y\plus{}z$, etc. Then $ a\plus{}b\plus{}c\equal{}2$; $ s\equal{}1$. Our inequality is then equivalent to

$ \left(9\plus{}\frac{1}{sr^{2}}\right)*(abc)^{2}\equal{}9(4srR)^{2}\plus{}(4R)^{2}\ge\frac{256}{81}$.

$ 9(rR)^{2}\plus{}s^{2}R^{2}\minus{}\frac{16}{81}s^{4}\ge 0$

This should be easy...I cannot finish it though...

[quote="Altheman"]What is the 'ABC-Theorem'?

$ \left(9\plus{}\frac{1}{xyz}\right)*(x\plus{}y)^{2}(y\plus{}z)^{2}(z\plus{}x)^{2}\minus{}\frac{256}{81}\ge 0$

Let $ a \equal{} y\plus{}z$, etc. Then $ a\plus{}b\plus{}c \equal{} 2$; $ s \equal{} 1$. Our inequality is then equivalent to

$ \left(9\plus{}\frac{1}{sr^{2}}\right)*(abc)^{2}\equal{} 9(4srR)^{2}\plus{}(4R)^{2}\ge\frac{256}{81}$.

$ 9(rR)^{2}\plus{}s^{2}R^{2}\minus{}\frac{16}{81}s^{4}\ge 0$

This should be easy...I cannot finish it though...[/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?t=165132

You can read soem things about ABC-Theorem on here  :wink:

I think this is false $ (\sum ab)^{3}\geq abc(\sum a)^{3}$

Prove with $ a\equal{}b\equal{}1/2; c\equal{}1$ and you have $ 125/64\geq 2$ which is false. :wink:

[quote="manlio"]I think this is false $ (\sum ab)^{3}\geq abc(\sum a)^{3}$

Prove with $ a \equal{} b \equal{} 1/2; c \equal{} 1$ and you have $ 125/64\geq 2$ which is false. :wink:[/quote]
Yes maybe the proof that was given by quangpbc is wrong.He can't use ABC theorem.But i mean to use ABC theorem not in your inequality(which is maybe wrong)you should use it in ductrung one :wink:

[quote="manlio"]I think this is false $ (\sum ab)^{3}\geq abc(\sum a)^{3}$

Prove with $ a \equal{} b \equal{} 1/2; c \equal{} 1$ and you have $ 125/64\geq 2$ which is false. :wink:[/quote]

$ a\plus{}b\plus{}c\not \equal{} 1$  :(

Here is my solution
Let $ ab\plus{}bc\plus{}ca \equal{}\frac{1\minus{}q^{2}}{3}$  with $ 0\leq q\leq 1$ .By Vo Quoc Ba Can method (it was first introduced to this forum by Vasc) we have that $ abc\leq\frac{(1\minus{}q)^{2}(1\plus{}2q)}{27}\equal{} K$

So it suffices to prove that $ 9\plus{}\frac{1}{K}\geq\frac{256}{81(\frac{1\minus{}q^{2}}{3}\minus{}K)^{2}}$
This is equivalent to
$ \frac{9q^{2}(48\minus{}32q\minus{}4q^{2}\plus{}24q^{3}\plus{}13q^{4}\plus{}2q^{5})}{(q\minus{}1)^{2}(2\plus{}q)^{4}(1\plus{}2q)}\geq 0$

which is true since $ 48\geq 32q\plus{}4q^{2}$  due to the fact that $ 0\leq q\leq 1$
The equality holds for q=0 so $ ab\plus{}bc\plus{}ca \equal{}\frac{1}{3}$

This is a complete and good solution.
Thank you very much, Silouan. :)


If you has a sequence ${a_1}, {a_2},  {a_1}{a_2},  {a_1}{a_2}^2,  {a_1}^2{a_2}^3,  {a_1}^3{a_2}^5\ldots$ then what would be the 10th term of the sequence?  What is the pattern?

looks like fibonacci sequence but delayed.  Its not started at the same time. You just have to identify where it starts.
[hide]solution:
it starts with the last 1 at (a_1)(a_2) which is the 3rd number in this sequence so the 10th number has (a_2)^34
now for the a_1 the last 1 starts at (a_1)(a_2)^2 which is the 4th number in this sequence so the (a_1)^21

answer:
((a_1)^21)((a_2)^34)
[/hide]


analytic geometry
graphing lines
slope
Support
calculus
integration
Gauss

[quote="JBL"]Incidentally, since I see you're from Michigan, you will get a great deal on your education if you go to UMich, which has a quite good honors program in mathematics.  Although the massive-state-school format is not for everyone.[/quote]
Nor is an admissions system riddled with the markings of affirmative action something to which everyone wants to subject themselves.

Well, I guess there aren't any moderators on this forum, so would one of the administrators please delete confuted's message on this post (and this one, as well), since
a) I am mildly offended by it
b) It is irrelevant to the question posed by Hikari
c) It is irrelevant to a discussion of this sort in general since every major university in the country, with a small handful of exceptions, uses Affirmative Action.

JBL, I certainly didn't mean to offend you, but I LIVE in Michigan. A bunch of my friends have gone and visited the campus and stuff at U of M. A few days ago, one of them (a white male) was telling me that down there, there were so many minority students that whites were a minority themselves on campus. I don't have anything against minorities, but you know that those kids weren't all admitted because of talent/grades/such - a lot of them got in based on skin color/whatever. That's not right.

If you look at how their admissions system was structured before the court ruling, a minority student got 20 points on the basis of their skin, and a perfect essay got somebody something like 2 points. I believe you needed a total of 80 for admission. Sorry, but some people prefer to go to a college where every admitted student is there because of something they've done, not because of the way they were born. Hopefully you understand why I made the comment.

According to the University of Michigan web site, here is the racial/ethnic breakdown of freshman who entered the Ann Arbor campus in 2003:

Caucasian             64.9%
Asian American    13.7%
Other/unknown        8.2%
African American    7.7%
Hispanic American 4.8%
Native American     0.7%

Because this is a math site, I'll leave out my thoughts on Affirmative Action.

[quote="gauss202"]Futhermore the state of Michigan itself is 14.2% African-American.[/quote]
Almost all of which is in Detroit, Flint, and Lansing.

This was originally part of a thread about good colleges for math.  I moved it here as it's not really part of that discussion, but is appropriate for discussion in this forum.

[quote="gauss202"]That should tell you something about the quality of the schools and the access to educational resources in those city schools. No matter where they live, they are still citizens of Michigan.[/quote]
Schools in Detroit get more money per student than schools in most other cities.

I'm not from Detroit, or even Michigan for that matter, so I can't confirm or deny whether that is true (though it would be very unusual for inner city schools if it were).

But the real point here is much deeper than just money in and of itself. Affirmative Action is a deep and complex issue that can be approached from a lot of different angles - both from the perspective of pro and con.

But arguing against racial Affirmative Action for [i]moral[/i] reasons leads to a very slippery slope, and it is tenuous at best. For one thing, most such arguments never address other types of "affirmative action" policies such as athletics, legacies, geographical diversity considerations - which in most cases account for an even greater percentage of admissions than racial affirmative action policies. And secondly, they rarely, if ever, face the essential question of what is the [i]purpose[/i] of a public (and even private) University is supposed to be, and who is it supposed to serve.

Most disturbing of all though, is when these arguments try to recharacterize Civil Rights equal protection laws to fit their agenda (in many cases by people who didn't support Civil Rights laws in the first place). It's analogous to a basketball game where one team has been cheating for three and a half quarters. Then when they're finally called on it, they say, "Yeah, you're right. Lets all just play fair now." But the score is already 100 to 50 and the other team is demoralized. And when the losing team asks to reset the score, they're accused of cheating.

Will affirmative action fix everything that's wrong with this country? Absolutely not. But does it make a difference? Absolutely. Because lets face it - people who are not from inner-city Detroit, or Flint, or Lansing are not going to go there and integrate themselves into families or communities, or act as leaders or mentors there. Having at least some degree of plurality among your educated population is essential to a healthy functioning democracy.

[quote="confuted"]If you look at how their admissions system was structured before the court ruling, a minority student got 20 points on the basis of their skin, and a perfect essay got somebody something like 2 points. I believe you needed a total of 80 for admission.[/quote]

I know how the admissions system was structured.  The essay comment is frankly absurd -- they didn't care about the essay at all, they didn't have staff to read real essays and award them real points.  I don't even understand why they had it on the application.  The essay was worthless compared to every single aspect on the application -- to point out that it was worthless compared to the affirmative action segment is just decieving.   Also, you actually needed a total of 100 for admission.

Furthermore, you haven't bothered to consider [i]why[/i] a school might [i]want[/i] Affirrmative Action.  Obviously, they are making some sort of judgement.   For whatever reason, they have found that there are fewer minority applicants who meet their other standards for admission.  Now, you can ascribe this to whatever you want.  Personally, I think it points to a serious problem in this country, namely that our society has created barriers to achievement and never bothered to do something to fix them.  So should college administrators just continue to affirm those barriers, or should they recognize that there are a lot of minority applicants who [i]would have gotten in[/i] if those barriers hadn't been there?

And, finally, I visited UMich too.  It didn't seem that minority-heavy to me.  Now, maybe if I lived in a place where "almost all" the Black population lived in three cities and I, along with almost all the rest of the white population, lived as far away from them as possible, I would have noticed the integration a little bit more.  I might even have complained about it.

My two cents on this topic: If people really believed all people were equal they wouldn't have "special" programs just because of race or ethnicity and some people are just using this as an opportunity to make things look good (as a campaign sort of) which can cause reverse-discrimination (which is sometimes just as bad as discrimination, well they're basically same thing except the positions are opposite and such)

Also believe it or not people tend to (not meaning you guys :wink:) look at English deficient people as "stupid" people so people also set up special program which lets people who can't speak English well into schools and other places and that's clearly different from the first case so yeah, just pointing that out.

JBL - OF COURSE there are fewer minority students that meet the admissions standards. That's because they make up a smaller percentage of the population! Gasp.

I agree with Tare in that if people are equal, they don't need special programs. If somebody wants to get into a certain college, they should earn their way in by merit of their accomplishments. Life much more closely resembles an oligarchy than a democracy. If you let kids into a school that don't have the educational background necessary to be there, and then expect them to succeed in said school, you'll be presented with two options.
[tab]1) Lower the educational standards of the school so that the kids without the background can handle the curriculum
[tab]2) Provide these students with some form of a way to "coast by," much like they likely did in high school if they don't have the test scores and such. That statement isn't bigotry - it's fact.

The consequences of action 1 are obvious. You'll produce classes full of young adults that don't have the skills to cut it in real life. They won't assume CEO positions and such - they'll be stuck in more mundane jobs, and there will be fewer (no?) qualified (by the original standards) applicants for high ranking jobs.

The consequences of action 2 are a bit different. Instead of the entire class being uneducated, the students admitted on the basis of their ethnicity will be the only ones that struggle in life. I don't know if you've ever taken a look at it, or ever studied any history, but life isn't kind to those without the knowledge to succeed.

If the minority groups want to succeed in being equals, special treatment is not the way to achieve it. That's simply discrimination against a different group. The way to succeed in life, the way to get the future you want, is to reach out and TAKE IT.

Regardless, my original statement was that for some, a college in which a large percentage of the people were admitted almost soley on the basis of their ethnicity, is not a place they want to be. Some of us want to be surrounded with others that can cut the cake. If I apply to a college and I don't get in, I'm not going to whine that it was because I was being discriminated against. I'll know that I wasn't good enough and didn't deserve to be there. That is all. Similarly, when I go to college, I expect to be surrounded by others that got in the same way I did. As for those that couldn't get in because they weren't smart enough - I couldn't care about them less. I don't care that I'm not politically correct when I say that if somebody wants to succeed, a little thing like skin color won't stop them.

If there wasn't a box to check for race, how would the Asians and Africans and Indians and South Pacific Islanders and [insert obscure ethnic group (no offense)]-ers be discriminated against in the admissions? Perhaps we should eliminate this silly concept of race and give people what they earn and only what they earn. I'd be up for that.[tab]

First of all, I meant percentagewise fewer.  Gasp.  Now go re-read what I wrote with this deeper understanding and see if it makes sense to you.  Then you can criticize it.

Ready:
'equality'
versus
'equal chances'
Think about that.  Think particularly about what Gauss wrote, because he is much more eloquent than I.

Also:
Accomplishments.  Interesting.  One of the accomplishments of a girl in my class was that she went all over the world and carried our a study that involved comparing views of highschool students in Tibet, New York, and France.  That's quite an accomplishment.  Personally, I couldn't have afforded the plane fares and hotel expenses to make those trips.  Arguably, I could have done a better study than she did.  But, obviously, I couldn't carry out anything comparable.  So, is she more deserving of a place in college?  Accomplishments are hardly a measure of abilities.  When you can have all your math awards in a school where they don't offer the competitions, where you have no way to know about the competitions, and where the math teacher doesn't know what he is doing, then, then you can tell me accomplishments are important.  Accomplishments out of context are irrelevant.  Race is context.

Also: I challenge you to show me one ounce of evidence that affirmative action lowers standards.  At the scale that universities accept applicants, the "ability gap" or whatever you want to call it is far to small to make any meaningful difference.

Tare -- how does it make sense to try to "look good" to a minority group at the expense of the majority group?  Hardly seems like good campaigning to me -- I'd much rather have the majority of people want to vote for me than the minority.

Well not for political campaigns but "campaign" as in commercial to invite more people, you know slogans like "Diversity and Mixture" "Equal opportunities here!" and so on. Still I'm not criticizing the ads themselves I'm just criticizing people who makes good use of it and do it just so their thing has a better public image. I'd like to back away now...

Matthew -- how do you imagine that the private school system can possibly accomodate all the students wishing to spend their vouchers?  What about the (large) number of private schools in which a voucher would not come remotely close to covering tuition?  How do you suggest that a school system will be more accountable by transfering students into a less accountable system?  What about costs to school systems that are independant of the number of students?  Will students only get the amount of money that is dependant on their showing up (which isn't so much, especially in over-crowded schools) or will they get their share of the money for general administrative needs and so on that are independant of the number of students (which will drag money out of the public school systems)?

You complain AA has too many flaws, but then you suggest a program without even considering the difficulties, imbalances, or major impracticalities it would encounter.  That hardly seems fair to me.

Also, in response to your first post:  you just justified racism by making an appeal to the racism of older generations.  That's good.  We shouldn't have AA because there are racist Americans who will use it to validate their viewpoints.  That argument is totally morally invalid -- it's the same one that has always been used against integration.

Let me start out by asking two devil's advocate questions for clarification: 1) Why should we even have AA for the economically disadvantaged? All of the arguments that have been made for racial AA lowering standards and leaving skewed perceptions of achievement should apply here as well. and 2) Where do you stand on the concept of reparations - which is something that has amazingly not been brought up yet in this discussion. Are economic and racial AA mutually exclusive?

As for the sociological studies, I have some disagreements with your analysis. First of all, I don't think it's right to discredit a study or an argument based on the fact that they have a motive to reach a certain conclusion. Certainly it means that you should examine their methodology very closely; but if we discredited every claim made by someone with a motive to reach a conclusion then even this discussion on AA could not go very far. Because any white person who argued against AA, would be discredited by saying they were reaching their conclusions to protect their own self-interest; and likewise for any minority arguing in favor of AA.

I do agree with you though that [i]some[/i] explanation of the results of those studies could be attributed, not to personal prejudice on the part of the employer, but on other factors such as their analysis of general racial conditions within the society or how it would affect their business. But I don't see how that makes the conclusions of those studies any less significant!! It's the same thing as if a white restaurant owner in the 60s said, "I myself am not racist, but I can't allow black people in my restaurant because it will scare off my other customers." How does that make any material difference to the outcome? It still results in less access to opportunities for the minority candidate.

Furthermore, I think you're failing to consider the momentum of history. Suppose this same study had been done in 1953, 1963,.. and every decade up until 2003. There isn't going to be some magic moment when everything all of a sudden became even. Sociological changes happen over long periods of time - perhaps even several generations. So it's understandable why the biases these studies show would still be there, even now. The fact that there are so many studies all showing the same thing should be an indication that the results are not part of some error in methodology.

Then as for skewed perceptions of achievement, I don't think this is a valid argument at all against AA. Because for the most part, the same people who have those perceptions are the same ones who feel that minorities are generally less qualified anyway. Now they can just use AA as their reason for feeling that way. The stigma of being less qualified never seems to hinder white athletes when they apply for jobs after graduation - in fact I suspect that they tend to get hired at a higher rate than non-athletes. So I think the solution there is to reverse the false perception that minorities are less qualified than their counterparts (because no one ever knows who was and who wasn't admitted because of AA, or because they were an athlete, or for any other reason).

My opinion on vouchers is that they have to be looked at on a case by case basis. But the claim that vouchers would help poor people the most is completely untrue. The people who would benefit the most from vouchers are middle class families who either already have children in private schools, or who allllmost make enough to send their children, but not quite. The poorest families still wouldn't be able to afford to send their children to private schools even under such a system - especially the ones with multiple kids.

Meanwhile, the exodus of some students (mainly the ones with influential parents and involved parents) from the public schools would leave them even further drained of resources - making the problems there even worse for the students remaining. (You'd never get teachers to teach there).

I some parts of the country private schools are not even the best schools. This is especially true in the south. The best schools down here are usually public schools in wealthy (suburban) neighborhoods. In many rural areas (which often have the worst schools) there aren't any private schools at all - and likely wouldn't be even with vouchers.

I feel that vouchers should be a local issue though. In some areas they may have more of a usefulness and justification than others. And the funds for those vouchers should come from locally. In some areas, charter schools might be a better option. In others the problems could best be fixed by busing students from bad public schools to better public schools. But I'm absolutely certain that the people who would most benefit from general issue vouchers are rich and middle class families.

Also a side note: In almost any public school desperate enough for good teachers they will waive certification requirements (which I also think are ridiculous). This is especially true in the south. Unfortunately you may not make as much money as the certified teachers - because pay is usually related to the "amount" of certification you have. But you could definitely be hired.

[quote="gauss202"]My opinion on vouchers is that they have to be looked at on a case by case basis. But the claim that vouchers would help poor people the most is completely untrue. The people who would benefit the most from vouchers are middle class families who either already have children in private schools, or who allllmost make enough to send their children, but not quite. The poorest families still wouldn't be able to afford to send their children to private schools even under such a system - especially the ones with multiple kids.[/quote]

I don't understand your argument that poor people still couldn't afford private schools.  Average private school tuition is $2k less than the state spends on public schools.  Vouchers would accomodate 100% of the tuition of the majority of private schools.

[quote="gauss202"]
Meanwhile, the exodus of some students (mainly the ones with influential parents and involved parents) from the public schools would leave them even further drained of resources - making the problems there even worse for the students remaining. (You'd never get teachers to teach there).

I some parts of the country private schools are not even the best schools. This is especially true in the south. The best schools down here are usually public schools in wealthy (suburban) neighborhoods. In many rural areas (which often have the worst schools) there aren't any private schools at all - and likely wouldn't be even with vouchers.
[\quote]

The fact that suburban schools are often fine is a reason not to worry about a mass exodus from schools that are doing a good job.  So what if there is a mass exodus from the bad schools.  So long as a voucher gets every child into school and most into better schools makes them a net benefit to the current system.

Also, the few good teachers at the worst schools are often there due to Teachers for America and similar programs.  Those teachers reach out to those in need in even the worst schools.  And with vouchers available, why would we believe schools would get any worse.  Honestly, it's hard for me to imagine this is a big drawback to the current system.  Private schools are more stable in the inner city and are more likely to attract teachers concerned for their safety.  I see no reason why they couldn't attract teachers to accomodate the influx of students -- especially with the best teachers from those failing schools looking for jobs.

[quote="gauss202"]
I feel that vouchers should be a local issue though. In some areas they may have more of a usefulness and justification than others. And the funds for those vouchers should come from locally. In some areas, charter schools might be a better option. In others the problems could best be fixed by busing students from bad public schools to better public schools. But I'm absolutely certain that the people who would most benefit from general issue vouchers are rich and middle class families.

Also a side note: In almost any public school desperate enough for good teachers they will waive certification requirements (which I also think are ridiculous). This is especially true in the south. Unfortunately you may not make as much money as the certified teachers - because pay is usually related to the "amount" of certification you have. But you could definitely be hired.[/quote]

Vouchers are a local issue, but nearly every locale pays more for public schooling than it would cost to fund private school.  I think this may be your misunderstanding of vouchers -- they cost LESS than public schooling.  The average public school in the US spends nearly $7k per student per year.  The average private school tuition is under $5k.  The middle and upper classes will benefit a little, but the greatest option will be given to the poor who will go from having no options to having a very solid one.

I will agree that charter and other kinds of schools are often solutions, but the nice thing about vouchers is that they are not mutually exclusive.

Also, most states will not let public schools hire uncertified teachers.  This is not a problem at private schools.

[quote="JBL"]Matthew -- how do you imagine that the private school system can possibly accomodate all the students wishing to spend their vouchers?  What about the (large) number of private schools in which a voucher would not come remotely close to covering tuition?  How do you suggest that a school system will be more accountable by transfering students into a less accountable system?  What about costs to school systems that are independant of the number of students?  Will students only get the amount of money that is dependant on their showing up (which isn't so much, especially in over-crowded schools) or will they get their share of the money for general administrative needs and so on that are independant of the number of students (which will drag money out of the public school systems)?

Also, in response to your first post:  you just justified racism by making an appeal to the racism of older generations.  That's good.  We shouldn't have AA because there are racist Americans who will use it to validate their viewpoints.  That argument is totally morally invalid -- it's the same one that has always been used against integration.[/quote]

First, I find this last remark a bit offensive.  I did not justify racism in any sense.  I simply provided a plausible explanation for behavior by companies who may be incentivized by a small number of racists to make more blacketly racist decisions.  I didn't say "Rah rah!  Go negative externality!"

Second, morals don't exist.  They are just constructs that are at best redundant of the heuristics by which we evaluate positive and negative actions.

In regards to your first paragraph: Private schools cost less on average than public schools.  All students, especially those in poorer economic conditions would be able to afford private schooling at a par level, that is, at a level significantly above the worst schools.  Vouchers don't drain money from the system either -- they save it by purchasing a less expensive education.  What they really save however is in future spending -- it takes less government to take care of an educated populace.

Also, I can't fathom calling private schools less accountable than public schools.  Private schools can fire bad teachers and aren't nearly as influenced by politics -- at all its levels.  What private schools already have to face is proving that they are worth the expense -- and they do.  Their students, even by neighborhood cross-sections, outperform those of public schools consistently.  If they didn't, why would people pay them?

When your livelihood depends on your performance -- [i]that's[/i] accountability.

[quote="gauss202"]
As for the sociological studies, I have some disagreements with your analysis. First of all, I don't think it's right to discredit a study or an argument based on the fact that they have a motive to reach a certain conclusion. Certainly it means that you should examine their methodology very closely; but if we discredited every claim made by someone with a motive to reach a conclusion then even this discussion on AA could not go very far. Because any white person who argued against AA, would be discredited by saying they were reaching their conclusions to protect their own self-interest; and likewise for any minority arguing in favor of AA.

I do agree with you though that [i]some[/i] explanation of the results of those studies could be attributed, not to personal prejudice on the part of the employer, but on other factors such as their analysis of general racial conditions within the society or how it would affect their business. But I don't see how that makes the conclusions of those studies any less significant!! It's the same thing as if a white restaurant owner in the 60s said, "I myself am not racist, but I can't allow black people in my restaurant because it will scare off my other customers." How does that make any material difference to the outcome? It still results in less access to opportunities for the minority candidate.
[/quote]

I didn't completely discredit the study, I simply talked about various skews and moral hazards.

Also, I absolutely have a self interest in this discussion.  I believe that I will live in a safer, cleaner, healthier, more educated world if we create a better system for educating those who can advance my surroundings.  Of course, that doesn't discount the effect of moral hazard.

I recognize that there is still racism in hiring.  I was not seeking to dismiss that notion.  I did however want to assert that politics, including the legal costs of business, are also a factor.

I think you're ignoring the effects of supply and demand here. If you doled out general issue vouchers to everyone for the full cost of private school tuitions, the private schools would just raise their tuitions and the poorest people would still be priced out. There certainly aren't enough spaces in private schools for all public school students. Which students do you imagine would be left out? Meanwhile, giving vouchers to people who already have children in private schools is like giving them a [i]huge[/i] tax break with funds coming from the current public education budget and no means of replacing them!!

In your analysis that private schools generally provide better educations (while that's true in some places), don't forget that some of them are also self selecting of their students. Some don't have to deal with learning disabled students, special needs students, or general remedial students. They have the ability to discriminate or reject students in ways that public education cannot. Not to mention that a lot of them are religiously affiliated, which could lead to a lot of issues about separation of church and state if they receive public funds.

I think that public education could certainly learn some things from private education. But the whole-sale privatization of education (which is basically what we're talking about if we give vouchers worth the full cost of private school tuition), is not a good idea. The economics of it will either price the poorest people out of the system, or make the whole system even more expensive than it is now.

Gauss, I think some of your concerns are valid, but fortunately we have test cases that seem to allay many of these fears.

Vouchers have not seemed to raise private school tuitions.  In fact, they have seemed to encouraged the building and expansion of private schools that cater to students who can afford no more than voucher tuition.

Additionally, supply/demand theory suggests just the opposite of the idea of pricing out.  If the voucher amount is capped that the price of the service should only rise so high.  After that private schools would have no incentive to raise tuition because it would mean a reduction, not an increase, in revenue.  Even so, many private schools (perhaps most) do not work for real profit.  Their profit is often in raising educational standards.  Religious schools have very little [overt] profit motive and are subsidized by churches.

Additionally, voucher trials have indicated that the reduction in educational costs has outweighed the tax break given to wealthier families.  In other words -- a better job was done to educate students with a lower cost to everyone.  It's okay to give a tax break to the wealthy if it doesn't translate to a reduction in services.

I am also not personally concerned with the religious affiliation of some schools.  Vouchers are about [i]choices[/i].  It's not a matter of the state encouraging students to be religious.  It's about parents deciding that, whether or not they subscribe to a particular religion, that such schools are preferable to public schools.  Besides, the axiom of public schooling is itself as religious to me as having no alternative.  It's a particular form of social indoctrination with all the trimmings of a localized culture, traditions, holidays, ceremonies, and an institutionally selected hierarchy.  I can't think of many ideas more religious than tenure.

Also, the Seton Hall study indicated an improvement in the public schools in areas where vouchers were available.  That sounds like a win-win.

[quote="JBL"]I'm skeptical of your claim that an average private school education costs less than what public schools spend on their students.  I don't know any facts about this other than what I've personally observed, however, which includes private schools costing in excess of $20,000 a year for high school, which is vastly more than any public school in the country spends on its students.

As for less acountable -- they are less regulated, so what goes on in them is more out-of-control.  If you're telling everyone to abandon the public school system, then you still have a marked advantage in the private schools over their students -- the students can't leave because they can't get a better (or equal) education anywhere else.  Furthermore, private schools have an entirely different mission than public schools -- as private institutions, they are run to provide a living to their owners.  Public schools are tasked with educating the populace.  This runs into the selectivity problem that gauss brought up -- you're going to end up with the least-equipped students and the students with the least active, envolved parents stuck in the public school system -- this is exactly the group of people who cost the most to educate and who are least likely to cause the improvement of the public school system.[/quote]

http://www.cato.org/dailys/09-08-03.html

I am not sure why you would be skeptical about public versus private school costs.  Public school teachers are paid far more than private school teachers.  They also have enormous pensions.  In my former state of Illinois teachers have such large pensions that many collect 7 digits [i]after retirement[/i].  That's pretty nice for a profession that is granted 3 months of vacation per year.  It is in fact more than the median worker in Illinois makes in a lifetime.  And that's just the pension.

Also, you characterize regulation as the only thing that stands between order and anarchy or a matter of quality.  Regulation has costs and it is important to weigh those costs.  Regulation requires regulators that are often highly paid -- soaking up the resources of the system.  Regulation also brings politics into the mix with all its costs and a reduction of individual choice.  Regulation also keeps quality teachers out of the system.  Nearly all public school teachers are required to be certified.  This limits out teachers like me who have no desire or intention (or in some cases resources) to spend three years in school paying an institution for a rubber stamp.

The most interesting metrics are performance versus costs.  Private schools seem to do a better job on average with safer conditions and lower costs.  And safer conditions are not a small factor to the educational health of the poorest students.

Also, I am not telling anyone to abandon the public school system (yet anyway).  I am suggesting that people have options in regards to educating their children.

And as for the theoretical selectivity problem -- I would be concerned if voucher trials had not indicated an increase in the quality of local public schools.  But since they have, I won't lose too much sleep about it.

Finally, as a graduate of a high school that was often regarded as the best in my home state, I cannot fathom ever sending any children of mine to a public school.  I would sooner have my big toes amputated with no anesthetic.

I was very interested in finding out more about this Seton Hall study that you were refering to Matt, because I've never actually seen a voucher program implemented on a large scale. I agree that it's one thing to speculate about what the effects of such a program would be, and another to have actual real results where it's been implemented.

It took me a little while to find some information about the study, but when I did was surprised by how unrigorous it was. If you were skeptical of the motives and methods of the studies on hiring discrimination, then you should be even more skeptical of this one!

Here goes some brief information about it: http://www.evangelist.org/archive/htm2/0824giff.htm

First of all, the study looked at one single elementary school where the students were offered a $2000 dollar voucher to attend a school of their choice. Two-thirds of them went to a Catholic school that was located about a mile away.

But surprise! Seton Hall is a Catholic School!! And they published their results in the Journal of Catholic Education - http://soeap.udayton.edu/support/pub2/catholic_ed/Issues/vol3no4.htm Talk about questionable motives!

But even just looking at it closely, it was hardly a scientific or statistically significant survey. The number of schools and the number of students involved was far too small to extrapolate those results to a larger scale; and the money to fund those vouchers didn't come from the government, they came from a private organization. It gives no indication of what would happen if vouchers were implemented by the government on a large scale, or in different circumstances.

So I still have all of my concerns about what would happen if voucher programs were implemented on large scales. I still believe that if vouchers are given to everyone, most private schools will generally raise their tuitions- still pricing the poorest people out and mainly benefitting those who already have children in private school; that the money used on vouchers will leave the public schools drastically crippled and underfunded; and that many private schools will be able to unfairly discriminate against certain kinds of students. I can't see any mechanism whereby general issue vouchers make public schools better in the long run!!

Fine, so there is moral hazard in the Seton Hall study.  There is however also the undeniable fact that the public school reacted to competition introduced to its environment.  It hired more and better faculty.  Frankly, that's a far better argument for vouchers than I was making.

Also, you keep making this supply/demand argument for private schools pricing out the voucher students.  If private schools currently cost less than vouchers (and the vast majority do), then upward pressure cannot exceed the cost of the voucher.  The supply/demand curve is necessarily in equilibrium within the range formed by [i]status quo[/i] and [i]voucher inhanced[/i] circumstances.  And that assumes 100% cash profit motive which is clearly a bit on the high side.

I've done some research, and I will grant that the available data show that the average per-student expenditures for private schools are somewhat less than those for public schools.  This does not, however, necessarily imply that private schools are more efficient at educating students than are public schools.  Public schools need to spend money on a great deal of things that most private schools do not need to worry about, a few being:
-Education of disabled/special needs students.  Public schools are required to provide an appropriate education for [i]all[/i] students until age 18.  This requires a lot more resources to do for special needs students than it does for most others.  Since private schools can choose whom to enroll, they need not worry about this problem.
-Complying with state and national education "standards", such as the "No Child Left Behind" requirements and (for example) the so-called "Standards of Learning" (SOL) tests in my home state of Virginia.  I know for a fact that my school system spends a great deal of resources trying to teach its slower students how to pass these SOL tests for fear of losing accreditation.  They even have special remedial classes (required, I think, by the state) for students who fail SOL tests.  (These tests, like so many others similar to them, measure test-taking ability more than they do knowledge of the subject.)  Private schools do not have to do this.
-Busing.  Most private schools do not need to bus their students to school (Do you see fleets of private-school buses on the roads?  Didn't think so.).  The parents enrolling their kids there now have the motivation, time, and money to provide this transportation, which eliminates the need for such busing.  Busing, required by law for public schools, is another large expenditure.

The comment about retirement pensions for teachers in Illinois is misleading at best.  According to its wesite, the annual pension for a public school teacher in Chicago with 30 years of service is 66% of the average salary during the highest-paid four years of the last 10 years of service.  The average public school teacher's annual salary for teachers in Illinois in 2001-2002 was under $50,000 (Illinois is one of the top 10 states in teacher salaries, by the way: the national average (all of these are means, not medians)  was under $45,000, and the average for South Dakota, the state with lowest teacher salaries, was just a hair over $31,000 - pretty low pay for a job that requires a total of 4+ years of college).  Of course, pay in a teacher's final years is slightly more than the average pay; even taking this into account, we find that our hypothetical Chicago teacher would need 25-30 years of retirement to acheive a total of "seven figures" (and this is ignoring taxes!).  Also, it's not surprising that teacher salaries in Illinois are greater than the median - according to the 2000 census, only 26.1% of Illinois residents had at least a bachelor's degree.  If teachers earn "several times" what the median salary is in Illinois, this says more about what we pay our blue-collar workers than it does about the proportion of teachers' salaries to what is appropriate.  Compare these numbers to those for other [i]professional[/i] jobs, and I think you'd find a different story.

Concerning the claim that private schools pay their teachers less than do public schools:  This may be either true or false, for all I know, but operating under the assumption that it is true, we run into a problem.  A great deal of the problem in public education is due to the lack of quality teachers.  In Virginia, as in many other states, schools are desperate to find good (or even [i]any[/i]) teachers to replace those that are retiring.  Why?  People do not want to teach because they can [i]make much more money elsewhere[/i].  If these private schools were to take over and pay teachers [i]less[/i] than their public-school counterparts, the mass exodus from education would be exacerbated, lending more truth to the old jibe, "Those who can, do; those who can't, teach."  How driving the potentially good teachers [i]away[/i] from teaching (remember, the best teachers are primarily also the ones skilled enough to do very well in other jobs) will improve the quality of education is beyond me.


probability
expected value
probability and stats

Given a standard deck of 52 cards (26 red, 26 black), shuffle the deck, randomly distributing the cards by color.
This deck is an ordered collection of colors which will contain a largest "clump", or region of the deck where a block of $ n$ cards is an $ n$-clump if all $ n$ consecutive cards are the same color.

Find the expected value of the largest clump size in a deck.

(Yes, I realize that this will be a lot of casework. I tried simulating it in a Java program, but my programming skills are limited enough to the point where I couldn't figure out a way to track the clumps.)

Note that this problem is a bit harder than if we were using, say, coins. This is because each card is dependent on how many cards have been dealt before it.
So, we can't just use a straight recursion, I think.

I'm fairly sure I can solve this explicitly in the case of using coins (52 throws, heads, tails, Ex(clumps) should be find-able with a recursion) but the card problem has eluded me.

I also think that this sort of problem has been posted before, but I don't want to see a solution, I just want hints as to how to attack it.

If anyone has any ideas or partials solutions, it would be great to post em here :)

For the coin problem, the answer (up to a division by $ 2^n$) is sequence [url=http://www.research.att.com/~njas/sequences/A102712]A102712[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url], but they don't have the analogous sequence for your problem.

Thanks for the link. Sadly, that's pretty much all I expected. I know it's a lot easier to do when each color/coin is independent :*(


How much more is 25$ \%$ of $ \$$5 than 20$ \%$ of $ \$$2? Express your
answer in terms of cents.


Express $ 0.003 \times 0.00033$ in scientific notation.

convert both to scientific first then multiply. because i said so.
$ .003\equal{}3*10^{\minus{}3}$
$ .00033\equal{}3.3*10^{\minus{}4}$
$ 3*10^{\minus{}3}*3.3*10^{\minus{}4}\equal{}9.9*10^{\minus{}7}$

and by the way, you have to write the word times because i did 9.9*10^-7 and i got it wrong  :fool:

Wait so you have to write 9.9times10^-7?


[color=darkblue]Given are $\left\{\ m\ ,\ n\ ,\ p\ \right\}\subset N^{*}$ so that $m> np$ and $k>0 .$

Ascertain [b][u]without derivatives[/u][/b] the ratio $\frac{x}{y}$ so that the expression $x^{p}\cdot (x+y)$ is minimum for the all positive real numbers $x$ , $y$ for which $x^{m}\cdot y^{n}=k\ .$

[b]Remark.[/b] The case $m=2\ ,\ n=p=1$ ( http://www.mathlinks.ro/Forum/viewtopic.php?t=117322 ) was given at a districtual contest from Romania.[/color]

[hide="Answer."]$\frac{x}{m-np}=\frac{y}{n(p+1)}\ .$[/hide]

[hide]
We want to use an inequality to reduce $x^{p}(x+y)$ to a constant with an equality condition. Basically, some multiple/power of $x^{m}y^{n}$. We can use AM-GM to get

$(m+n)(x+y) = \sum_{i=1}^{m-np}\frac{m+n}{m-np}x+\sum_{i=1}^{n(p+1)}\frac{m+n}{n(p+1)}y \ge Cx^{\frac{m-np}{m+n}}y^{\frac{n(p+1)}{m+n}}$

$x+y \ge \frac{C}{m+n}x^{\frac{m-np}{m+n}}y^{\frac{n(p+1)}{m+n}}$

for some constant $C$. Then our expression is

$x^{p}(x+y) \ge \frac{C}{m+n}x^{\frac{m(p+1)}{m+n}}y^{\frac{n(p+1)}{m+n}}= \frac{C}{m+n}k^{\frac{p+1}{m+n}}$,

which is constant. Equality holds when

$\frac{m+n}{m-np}x = \frac{m+n}{n(p+1)}y \Rightarrow \frac{x}{m-np}= \frac{y}{n(p+1)}$

as desired.[/hide]


$ a,b,c \in R_\plus{}$.Νδο
$ \sum_{cyc}(2\plus{}\frac{a}{b})^2 \ge \frac{9(a\plus{}b\plus{}c)^2}{ab\plus{}bc\plus{}ca}$

Θαυμάσια ανισότητα :!:
[hide="solution-Pham Kim Hung"]
Είναι της μορφής:
$ \sum\frac {a^2}{b^2} \plus{} 4\sum\frac {a}{b} \ge \frac {9(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca} \plus{} 6$
Λαμβάνοντας υπ'όψιν τις ταυτότητες:
$ \sum\frac {a}{b} \minus{} 3 \equal{} \frac {(a \minus{} b)^2}{ab} \plus{} \frac {(c \minus{} a)(c \minus{} b)}{ca}$
and $ a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca \equal{} (a \minus{} b)^2 \plus{} (c \minus{} a)(c \minus{} b)$
Η ανισότητα πέρνει τη μορφή:
$ (a \minus{} b)^2M \plus{} (c \minus{} a)(c \minus{} b)N \ge 0$
Όπου $ M \equal{} \frac {4}{ab} \plus{} \frac {(a \plus{} b)^2}{a^2b^2} \minus{} \frac {9}{ab \plus{} bc \plus{} ca}$
Εαν $ a \ge b \ge c$:
$ \sum\frac {a}{b} \le \sum\frac {b}{a},\sum\frac {a^2}{b^2} \le \sum\frac {b^2}{a^2}$ $ (*)$
So we only need to consider the case $ a \ge b \ge c$
$ N \ge \frac {5}{ac} \plus{} \frac {b}{ac^2} \minus{} \frac {9}{ab \plus{} bc \plus{} ca} > 0$
$ M \plus{} N \ge \frac {8}{ab} \plus{} \frac {5}{ac} \minus{} \frac {18}{ab \plus{} bc \plus{} ca} > 0$
Let $ k \equal{} \frac {1 \plus{} \sqrt {5}}{2}$,Consider the following subcases:
$ (i)$ The first case $ a \minus{} b \le k(b \minus{} c).$ Then we have $ (a \minus{} b)^2 \le (a \minus{} c)(b \minus{} c)$,so
$ (a \minus{} b)^2M \plus{} (c \minus{} a)(c \minus{} b)N \ge (a \minus{} b)^2(M \plus{} N) \ge 0.$
$ (ii).$ The second case $ a \minus{} b \ge k(b \minus{} c).$ It suffices to prove that $ M \ge 0$ or
$ (a^2 \plus{} b^2 \plus{} 6ab)(ab \plus{} bc \plus{} ca) \ge 9a^2b^2.$
Since $ (k \plus{} 1)b \minus{} a \le kc$ and $ k \plus{} 1 \equal{} \frac {3 \plus{} \sqrt {5}}{2}$ we deduce that:

$ ab \minus{} (a \minus{} b)^2 \equal{} \left(\frac {(3 \plus{} \sqrt {5})b}{2} \minus{} a \right) \left( a \minus{} \frac {(3 \minus{} \sqrt {5})b}{2} \right) \le kc\left( a \plus{} \frac {(3 \minus{} \sqrt {5})b}{2}\right) \le 2c(a \plus{} b).$

Therefore
$ (a^2 \plus{} b^2 \plus{} 6ab)(ab \plus{} bc \plus{} ca) \minus{} 9a^2b^2 \ge ab[(a \minus{} b)^2 \minus{} ab] \plus{} c(a \plus{} b)^3 > c(a \plus{} b)^3 \minus{} 4abc(a \plus{} b) \equal{} c(a \plus{} b)(a \minus{} b)^2 \ge 0.$
Done!!!The equality holds when $ a \equal{} b \equal{} c$ [/hide]

[hide="P.S 1"]Sorry που το μισό το έγραψα στα αγγλικά αλλά θα ξημέρωνα...[/hide]
[hide="P.S 2"]Μία από τις δυσκολότερες ανισότητες που έχω δει(ίσως και πάνω από επίπεδο ΙΜΟ).....Μήπως έχει κάποιος από εσάς κάποια καλήτερη λύση???[/hide]


geometry
3D geometry
sphere
parameterization
function
calculus
derivative

Hello! I have a quick question. Say I have two parameterized lines in 2D space that intersect. Determining their intersection point is easy, but how can I tell at what point they are a certain distance apart? The reason I need to know this is because  I'm actually trying to determine the intersection point between a line and a moving circle.

Thanks in advance!

Is that a moving circle or a moving sphere? You have two parameters to work with, and simply looking at the distance as a function of both parameters would suffice for one version.

if the path was linear then the sphere would be the same problem.. but for a non linear path its ugly. lets stick with the circle..

so.. its not to bad..

a circleof radius r's center will be represented over time by $ \gamma : [0,T] \to R^2, \gamma (t) \equal{} (x(t),y(t))$

and a fixed line in the plane will be $ y\equal{}mx\plus{}b$ now they intersect at a simple equation

$ (x\minus{}x(t))^2\plus{}(y\minus{}y(t))^2\equal{}r^2 \to (x\minus{}x(t))^2\plus{}(mx\plus{}b\minus{}y(t))^2\equal{}r^2$

lets fix t and get $ x_t, y_t$ these arent derivatives but i just dont want to write so many parenthesis:

$ x^2\minus{}2xx_t\plus{}x^2_t\plus{}m^2x^2\plus{}2m(b\minus{}y_t)x\plus{}(b\minus{}y_t)^2\equal{}r^2$ or

$ (m^2 \plus{}1)x^2 \plus{}2(b\minus{}x_t\minus{}y_t)x \plus{} [x^2_t\plus{}b^2\minus{}2by_t\plus{}y^2_t\minus{}r^2]\equal{}0$

so quadratic gives us our solution: $ x \equal{} \frac{2(y_t\plus{}x_t\minus{}b) \pm \sqrt{4((y_t\plus{}x_t)^2\minus{}2b(y_t\plus{}x_t)\plus{}b^2)\minus{}4(m^2\plus{}1)(x^2_t\plus{}b^2\minus{}2by_t\plus{}y^2_t\minus{}r^2)}} {2(m^2\plus{}1)}$

$ x\equal{}\frac{(y_t\plus{}x_t\minus{}b) \pm \sqrt{2x_ty_t\minus{} 2bx_t\minus{} m^2x^2_t\minus{} m^2b^2\plus{} 2bm^2y_t\minus{}m^2y^2_t\plus{}m^2r^2\plus{}r^2}} {(m^2\plus{}1)}$

$ x\equal{}\frac{(y_t\plus{}x_t\minus{}b)}{m^2\plus{}1} \pm \frac{\sqrt{2x_ty_t\minus{}2bx_t\plus{}2bm^2y_t \plus{}[(m^2\plus{}1)(r^2\plus{}b^2\minus{}(x^2_t\plus{}y^2_t))\minus{}b^2\plus{}x^2_t\plus{}y^2_t]}} {(m^2\plus{}1)}$

$ x\equal{}\frac{1}{m^2\plus{}1}\left[(y_t\plus{}x_t\minus{}b) \pm\sqrt{2x_ty_t \plus{}2b(m^2y_t\minus{}x_t)\plus{}m^2b^2\plus{}r^2}\right]$

$ y\equal{}mx\plus{}b\equal{}\frac{m}{m^2\plus{}1}\left[(y_t\plus{}x_t\plus{}bm^2) \pm \sqrt{2x_ty_t \plus{}2b(m^2y_t\minus{}x_t)\plus{}m^2b^2\plus{}r^2}\right]$

when the solutions are real, the intersection exists..

Ah, I didn't see the 2D bit. I was thinking of a three-dimensional problem, in which there's a real difference between circles and spheres.


geometry
trigonometry
trig identities
Law of Cosines

$ ABC$ is triangle whose area is $ L$. Given $ a^2\plus{}b^2\plus{}c^2\equal{}16L$, determine $ \cot A\plus{}\cot B\plus{}\cot C$.

[hide="solution"]Applying the law of cosines three times we get:

$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2a^2 \plus{} 2b^2 \plus{} 2c^2 \minus{} 2ab \cos C \minus{} 2ac \cos B \minus{} 2bc \cos A$

Thus
$ 2 (ab \cos C \plus{} ac \cos B \plus{} bc \cos A) \equal{} a^2 \plus{} b^2 \plus{} c^2$
$ ab \cos C \plus{} ac \cos B \plus{} bc \cos A \equal{} 8L$

We devide by $ 2L\equal{}ab \sin C\equal{}ac \sin B\equal{}bc \sin A$:

$ \frac{ab \cos C}{ab \sin C} \plus{} \frac{ac \cos B}{ac \sin B} \plus{} \frac{bc  \cos a}{bc \sin A} \equal{} 4$

$ \cot C \plus{} \cot B \plus{} \cot A \equal{} 4$[/hide]


calculus
integration
logarithms
calculus computations

show :D

$ \sum_{i=1}^\infty\;\frac{\textbf H_{i}^{2}}{i^{2}}\;=\;\sum_{i=1}^\infty\;\frac{1}{i^{2}}\left( 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{i}\right)^{2}\;=\;\boxed{\frac{17}{4}\cdot\zeta{(4)}}$

[ PS: sorry if i asked this before, but somehow i don't think i did... :| ]

This series has been fully discussed in the following paper:

On an intriguing integral and some series related to $ \zeta(4)$, David Borwein and Jonathan Borwein,

Prodeedings of the American Mathematical Society, Volume 123, Number 4, April 1995.

It seems that this sum has been verified empirically by a student, Yeung, and proved later on by the two authors mentioned above.

It is a famous result, involving the square of the $ n^{th}$ harmonic number $ H_{n}$, obtained back in 1995.

OK, i will check it out.  :coolspeak:

meanwhile, can you also [b][u]show[/u][/b] :D

$ \sum_{i=1}^\infty\;\frac{\textbf H_{i}}{i^{2}}\;=\;\sum_{i=1}^\infty\;\frac{1}{i^{2}}\left( 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{i}\right)\;=\;\boxed{2\;\zeta{(3)}}$

Use this equality:

$ \sum_{n=1}^{\infty}\frac{z^{n}}{n^{2}}+\frac{1}{2}\log^{2}(1-z)=\sum_{n=1}^{\infty}\frac{H_{n}}{n}z^{n}$, valid for $ |z|<1$.

Now, divide by z and integrate from 0 to 1 and all you should do is to calculate the following integral:

$ \int_{0}^{1}\frac{\ln^{2}(1-x)}{x}dx=\int_{0}^{1}\frac{\ln^{2}x}{1-x}dx$,

and the later integral follows simply by power series expansion and integration by parts.

I'm having déjà vu: http://www.mathlinks.ro/Forum/viewtopic.php?t=155523 :D


no fair..

start at 15

boys outnumber the girlss....
[b]

15[/b]

19: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1591379#1591379

STOP MAKING 10 THREADS WHEN 1 IS ENOUGH!!!

$ 50$  :first:  :w00t:  :thumbup:

So boyz won..

Its $ boyz: 5\minus{}girlz: 4$

That was short! :D I'm gone for 2 hours and you guyz all done.

wow the last 2 rounds i didnt participate and the boys still won both of them! gj!


geometry
trigonometry
inequalities
function
complex numbers
number theory

So what is your favorite competition subject out of these? Why?
Geometry, Algebra, Trignometry, Complex Numbers, Combinatorics/Probability, Inequalities, Number Theory

I would say Geometry. I've understood most of the tools needed in solving problems.

I love combinatorics but when it comes to competitions I kinda hate combinatorics :P .  My fav is trig.

Geometry. Especially love the problems without using trig and analytics.

[quote="nr1337"]It's a 7-way tie between:
        Geometry
  	Algebra
  	Trignometry
  	Complex Numbers
  	Combinatorics/Probability
  	Inequalities
  	Number Theory

I'm going to say Geometry, though.[/quote]
I agree, except for the last line. I really cannot pick a favorite.

if you're good at geometry; go to your physics lessons.
however number theory is the real math

i don't see any big relations between geometry and high school physics.
I am taking olympiad physics for more than 7 months and i didn't see so many physics problems solved by using geometry.

Inequalities for me. Just because I have been recently doing it for a longtime... :D

inequality seems interesting....but i haven't started doing it.
is there any good website to learn basic knowledge of inqe?

Try [url=http://mathdb.org/notes_download/e_algebra_el.htm]this[/url] site and/or Kiran Kedlaya's notes [url=http://www.unl.edu/amc/a-activities/a4-for-students/s-index.html]here[/url].  I've gone through the former a little but haven't done much with the latter.

Yeah, both of those are pretty good.  You might also want to try [url=http://my.netian.com/~ideahitme/tin.html]Hojoo Lee's Topics in Inequalities[/url], or take the AoPS class  :D .

There should be one that's basic arithmetic, that is my favorite. ;)

I swear my favourite would have been "anything simple"

Combinatorics, which amazingly enough, is also the subject i'm best at.

Least favorite: Geometry

In general, number theory is my favorite, combinatorics is my least favorite, I'm best at algebra, and I'm worst at geometry... Strange, aren't I?

I like all math though, and am currently reading through Geometry Revisited. I solved my first olympiad geometry problem last week :) . Also, when it comes to the olympiad, I'm best at identities (deriving the function from them). I almost always get those problems.

i'm not too bad at algebra, though certainly not at the olympiad level. in geometry, let's just say i have serious proving flaws.... :oops: i haven't really spent too much time doing number theory, but its not one of my favorites either


Hello, I`m new user in this forum

$ x^2\plus{}y^2\plus{}z^2\equal{}\frac{1}{3}$
$ x^2y^2\plus{}x^2z^2\plus{}y^2z^2\equal{}xyz(x\plus{}y\plus{}z)^3$

How to solve this problem?

Why none of $ x, y, z, or (x\plus{}y\plus{}z)$ can equal $ 0$?

In fact they can.
1. case: $ x\plus{}y\plus{}z\equal{}0$.
Then $ (xy)^2\plus{}(yz)^2\plus{}zx)^2\equal{}0$, thus all the squares on the left side are $ 0$, thus one of $ x,y,z$ is $ 0$, so we reduced it to the second case.
2. case: $ x\equal{}0$ (similar for $ y\equal{}0$ or $ z\equal{}0$).
Then we have $ (yz)^2\equal{}0$, so $ yz\equal{}0$. W.l.o.g. let $ y\equal{}0$, then the first equation gives $ z \equal{} \pm \frac 1{\sqrt{3}}$.

Why are you so keen on the right answer, but not willing to read and think¿

I`m sorry, but I really confused, because many people wrote that $ (\frac {1}{3},\frac {1}{3},\frac {1}{3}),( \minus{} \frac {1}{3}, \minus{} \frac {1}{3}, \minus{} \frac {1}{3})$ is the right answer, but the other claim that
$ (\frac {1}{3},\frac {1}{3},\frac {1}{3}),( \minus{} \frac {1}{3}, \minus{} \frac {1}{3}, \minus{} \frac {1}{3}),(\pm{\frac {1}{\sqrt3}},0,0),(0,\pm{\frac {1}{\sqrt3}},0),(0,0,\pm{\frac {1}{\sqrt3}})$ is the right answer.

Help me...

I think ... the second answer is the right one. First sentence of solution I posted is wrong... For example ... suppose x=0. From the second equation follows that: yz=0 it means one of the variables y, z=0 ... noth both simultaneously. It is the reason that there are "two" answers. I'm too lazy sometimes ... and this solution is taken from a book "as is".


Prove that there is no retraction from the Mobius band to its boundary circle.

Let $ M$ denote the Mobius band and $ S^1$ the circle. Then the inclusion $ i : S^1 \to M$ onto the boundary circle gives an induced map $ i_* : \pi_1(S^1) \equal{} \mathbb{Z}\to \pi_1(M) \equal{} \mathbb{Z}$. This map is multiplication by 2. If there were a retract $ r : M \to S^1$ such that $ ri \equal{} id$, then $ 2r_* \equal{} r_*i_* \equal{} id : \mathbb{Z} \to \mathbb{Z}$. But there is no homomorphism $ r_*$ such that $ 2r_* \equal{} id$.

(NOTE: I'm trying to find a nice rigorous way of showing that $ i_*$ is multiplication by 2. If anyone has anything there that doesn't suck I'd appreciate it.)

Nice solution!

its multiplication by 2, consider shrinking M to its central circle


I was just wondering if there is a closed form of

\[\sum_{k=0}^{n}\binom{k}{r}.\]
If so, how do you find it?

It's equal to $\binom{n+1}{r+1}$ (this is pretty well-known). Just show that $\binom{k}{r}=\binom{k+1}{r+1}-\binom{k}{r+1}$ and then it telescopes.


calculus
integration
LaTeX
trigonometry
function
calculus computations

Show that (cosx)/(2+sinx)^2 dx = - {(1/(2+sinx)} +c

Any help or guidance would be great

$ \text{\LaTeX}$: $ \int \frac {\cos x}{(2 + \sin x)^2} dx = - \frac {1}{2 + \sin x} + C$

Simply make a substitution: $ 2 + \sin x = t$. Notice that the numerator is $ dt$, and voila... You've solved the problem.

(edited a typo)

$ P \equal{} \int\frac {\cos x}{(2 \plus{} \sin x)^2}\,dx \equal{} \int (2 \plus{} \sin x)' (2 \plus{} \sin x)^{ \minus{} 2}\,dx$

$ \equal{} \frac {(2 \plus{} \sin x)^{ \minus{} 1}}{ \minus{} 1} \equal{} \minus{} \frac {1}{2 \plus{} \sin x} \plus{} C$.

Thanks loads, I was nearly there but you helped me see the error of my ways :)

Now try the following:

1) $ \int\frac{\cos x}{2 \plus{} \sin^2x}\,dx$

2) $ \int\frac{1}{1 \plus{} 3 \cos^2x}\,dx$

[hide]$ \int \frac{\cos{x}}{2\plus{}\sin^2{x}}\text{ d}x\equal{}\int \frac{1}{2\plus{}u^2}\text{ d}u\equal{}\frac{1}{\sqrt{2}}\arctan{\bigg(\frac{\sin{x}}{\sqrt{2}}\bigg)}\plus{}C$[/hide]

For first question, you can translate  sinx=u.Then you can solve it.

[hide="#2"]$ \equal{}\int \frac{1}{1\plus{}\frac{3}{\sec^2{x}}}dx\equal{}\int \frac{\sec^2{x}}{2^2\plus{}\tan^2{x}}dx\equal{}\frac{1}{2}\arctan{\bigg(\frac{\tan{x}}{2}\bigg)}\plus{}C$
[hide="Equivalently"]$ x\equal{}\tan{y}\Rightarrow \cot{y}\equal{}\frac{1}{x}$ \[ \Leftrightarrow \int \frac{1}{1\plus{}3\cos^2{x}}dx\equal{}\frac{1}{2}\text{arccot}(2\cot{x})\plus{}C\][/hide]
[/hide]

In your second solution there is something missing.

My bad i was hoping you could follow the hop.

anyways $ \arctan{x}\equal{}y\equal{}\text{arccot}\frac{1}{x}$ and we define $ x\equal{}\frac{\tan{x}}{2}$ from the original solution to rewrite in terms of the cotangent inverse. This form of the integrated function can be reached by multiplying by $ 1\equal{}\frac{\csc^2{x}}{\csc^2{x}}$.


geometry
circumcircle
search
geometry proposed

Let $A_1$ and $B_1$ be two points on the base $AB$ of an isosceles $\triangle ABC$, where $\angle BCA > \pi /3$ and $\angle A_1CB_1 = \angle ABC$.  A circle externally tangent to the circumcircle of $\triangle A_1B_1C$ is tangent also to the rays $\overrightarrow{CA}$ and $\overrightarrow{CB}$ at points $A_2$ and $B_2$, respectively.  Prove that $A_2B_2 = 2AB$.

Nice, but it has been posted before. See the thread [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=23118]Isosceles triangle and tangent circles[/url].

Ok, thanks.  I guess I couldn't find it by a search since it was given in another Olympiad!


linear algebra
matrix
graph theory
combinatorics unsolved
combinatorics

Some pieces are placed on an 8x8 chessboard. There are exactly 4 pieces in each row and each coloumn of the board. Show that there are 8 pieces among those pieces such that no two of them are in the same row or coloumn.

People.. What's with all the beautiful problems posted today? :)

The general version holds: on an $n\times n$ board we have $m\le n$ pieces on each row and each column. Prove that there are $n$ pieces, one on each row and column.

Consider a bipartite graph $G$ having bipartition $A,B$, where $A$ has $n$ elements and represents the rows of the board, while $B$ has $n$ elements and represents the columns. $a\in A,b\in B$ are connected iff there is a piece at the intersection of row $a$ and column $b$.

$G$ is $m$-regular, and we want to prove that it has a perfect matching: $n$ edges no two of which have a common endpoint. This follows from the fact that it satisfies Hall's condition: choose a subset $T\subseteq A$ with $t\le n$ elements. Assume the set $T'\subseteq B$ of neighbours of the elements in $T$ has $t'<t$ elements. In the subgraph $G'$ of $G$ generated by $T,T'$ there are $tm$ edges, and all of them have an endpoint in $T'$. Since $|T'|=t'<t$, one of the vertices in $T'$ must have degree $>m$ in $G'$, and thus also in $G$, which is absurd, since all the vertices of $G$ had degree $m$. The assumption that the graph does not satisfy Halls condition has thus led to a contradiction.

Here's a somewhat stronger version:

Given an $n\times n$ matrix with non-negative entries s.t. the sum of the entries is the same on all rows and all columns, there are $n$ positive entries, one on each row and column.


function
number theory
totient function
relatively prime

The totient function phi(n), also called Euler's totient function, is defined as the number of positive integers <=n that are relatively prime to (i.e., do not contain any factor in common with) n,

Φ(30)=Φ(2.3.5)=Φ(2).Φ(3).Φ(5)
Φ(2.3.5)=1.2.4=8

Φ(15)=Φ(3.5)=Φ(3).Φ(5)
Φ(3.5)=Φ(3).Φ(5)=2.4=8

by the above reasoning i can say that there are 8 numbers less than equal to 30 which are co prime to 30
similarly there are 8 numbers <= 15 which are co prime to 15..
how is this possible??
between 15 and 30 there are primes 17,19,23 which are surely co prime to 30 hence the no of numbers which are co prime to 30 should be more than that of 15..
plz explain where m i wrong in my reasoning...

$\phi(30) = 8$ and $\phi(15) = 8$.  They are both correct, remember that just because $15|30$ doesn't mean they have [i]all[/i] the same factors.  $30$ is divisible by $2$ while $15$ is not - that will make a difference when finding $\phi$.

Or, slightly more simply: some of the 8 numbers less than 15 and relatively prime to 15 are not relatively prime to 30.  (For example, 2 is such a number.)

Now, if $\phi(45) = \phi(15)$, we would indeed have a problem.


Is there a prime between $n^2$ and $(n+1)^2$ for every $n > 0$ :?:

I think that's correct,but I don't know how to prove it.That's a really hard question.

Well, actually, it's an open problem...
http://primes.utm.edu/notes/conjectures/

[quote="Rust"]Best proved result about it $p_{n+1}<p_n+cp_n^a, a=0.54...$. Your problem equavalent to prove a<0.5.[/quote]

what is $c$?

[quote="Xixas"]A nice joke  :lol:[/quote]

why?

Because it is open...

And, Rust, do you know where this result come from? :)

can Rust solve :$p_{n+1}<p_n+cp_{n}^{a}, a<0,5$
 how many c=?

Baker R.C.; Harman G. The difference between consecuitive primes.
Proc. London Math. Soc. (3) 72 (1996), no 2, 261-280.
In this article a=0.535 (better result). In 2001 they proved a=0.525.


geometry
3D geometry
sphere
calculus
Putnam
integration

You are given a hollow container composed of a
cylinder of a 3 ft radius, 9 feet in length
two "ends" which go on the cylinder.  Each one of the ends is a piece of a sphere with a 6 ft radius (meaning the ends are not hemispheres).  Find the volume of the container.

(My explanation sounds pretty confusing.  I could probably draw a picture in Paint if people don't get it)

So the volume of a cylinder is less than a normal cylinder? (Can you describe the "ends" in details?)

Okay, [hide]this'll[/hide] be [hide]a calculus[/hide] problem. I [hide]can't think[/hide] of an [hide]elementary way to do it.  I could calculate the volume with calculus, but I don't really want to.[/hide]

No, the volume of the cylinder is more than the volume of a normal cylinder.  My dad claims that he did this problem once, and he doesn't know calculus, so it's either possible using other techniques or he didn't do it ;)  I don't see the solution, though.

Here, let's try this for a picture
[code]
 ___
(___)
[/code]
Note: the curved ends are NOT full hemispheres, because of the difference in radii.[/code]

[quote="gauss202"]Here's an interesting problem similar to this one:

Suppose you have a sphere of radius r with it's top and bottom caps chopped off by parallel planes. Let h be the height of the remaining truncated solid. Find the volume of the remaining solid.

Notice anything interesting about the answer?[/quote]

If we could solve that, the solution to my problem becomes trivial.  Now... how do we solve that?

if its a college question (looks familiar to a Putnam), perhaps it was meant to be solved by integration

Then shouldn't it be in Advanced? Or is calculus Intermediate at AoPS and I haven't realized that?

[quote="gauss202"]Actually, I mistated the original problem. Here's what I meant to say:

A hole is drilled through a solid sphere and the centerline of the
hole lies exactly on the diameter of the sphere. The drilled sphere is
then placed on a table and its 'truncated' height is h. Find the volume of the remaining solid.

I think this problem really is equivalent to your problem. It's a somewhat famous one. I saw it originally in college and I did it using integration. But the "niceness" of the answer makes me think that maybe there is an elementary way to solve it.[/quote]

Well, the answer to that can be approximated (somewhat roughly) by just removing the area of a cylinder with the diameter of the hole and the height of the diameter of the sphere.  However, that isn't going to be exact... and when you start adding cylinders of different heights to make it more accurate, you start doing the calculus...  So, I guess perhaps the only way is with calc.  Hmm...

Hmm...that problem seems very very familiar. I either say it on an old AHSME or in a calculus textbook in the integration/revolution of areas section. In any case, I forget how to do it now.


Numbers $ (1,2,...,n$) have been randomly put in sequence $ (a_{1},a_{2},...,a_{n})$ (each number from 1 to n occurs only once in sequence $ (a_{1},a_{2},...,a_{n})$ ) What is propability that the sequence has following feature:
\[ \forall_{k>2} \ \ \ a_{k}>a_{k\minus{}1}  \ \ \ or \ \ \  a_{k}>a_{k\minus{}2}\]

This is asking for the number of permutations that avoid the (dashless) generalized patterns 321 and 231.  With or without that characterizaton, it's not hard to describe these permutations: 1 must occur either in the first or second position.  Then we may insert 2 in either of the two leftmost unoccupied spaces.  Then we may insert 3 in either of the two leftmost unoccupied spaces.  And so on.  We have $ n \minus{} 1$ such choices, so $ 2^{n \minus{} 1}$ permutations.


Hi, I am trying to deal with a strange bug which I have not seen before.

For some reason, my program behaves as I would expect, with one problem. I have to print some output in a loop (even if its just printf ( "\n" );)

I cannot see any obvious reason for this.

Your help would be very much appriciated

Thanks

EDIT:

For the record this is for a USACO problem which my program solves (without the offending printf ("\n" ); on their computer)

[quote="t0rajir0u"]It would help if you showed us your source code...[/quote]

I was hoping no one would ask, seeing as it is a valid solution for any cheats, but here is the "offending" section

[code]		if ( (sum2 < sum1) || (depth < Deepest) )
		{
			for ( i = 0; i < G; ++i )
			{
				Best[i] = Used[i];
			}
			//printf ( "\n" );
			Deepest = depth;
		}[/code]

You still aren't providing us very much information.  Is it a compile or a runtime error?

As mentioned, theres really nothing we can do without seeing your source code, as you haven't even described what type of error you're getting :P The code you showed us is meaningless, and doesn't have any mistakes in it.

I'll take a random stab in the dark though. Are you doing any comparisons with floats in a 'bad' way? Bad as in, not allowing a small epsilon difference in comparisons, eg testing if (a==0.001) rather than abs(a-0.001)<1e-7, or if (a<b) rather than a<b+1e-7, etc?

The way compilers optimise things sometimes means if you don't use safe methods of comparing floats, adding a single cout or printf statement can cause tiny variations in numbers. Dealing with floating point values correctly means you shouldn't have to worry about those sort of problems.

That was my random guess anyway.

It is almost certainly not a complier error, since it runs on both computers,

Here is the full script

[code]/*
ID: simon.m1
LANG: C
TASK: holstein
*/

#include <stdio.h>

int Required[30];
int Current[30];
int Feed[20][30];

int Used[20];
int Best[20];

int V, G;
int Deepest;

void search ( int depth, int start )
{

	int i, j, T, sum1, sum2;

	if ( depth > Deepest ) return;

	T = 1;
	for ( i = 0; i < V && T; ++i )
		if ( Current[i] - Required[i] < 0 ) T = 0;

	if ( T )
	{
		for ( i = 0; i < G; ++i )
		{
			if ( Best[i] )
				sum1 += i;
			if ( Used[i] )
				sum2 += i;
		}
		if ( (sum2 < sum1) || (depth < Deepest) )
		{
			for ( i = 0; i < G; ++i )
			{
				Best[i] = Used[i];
			}
			//printf ( "\n" );
			Deepest = depth;
		}
		return;
	}

	for ( i = start; i < G; ++i )
	{
		if ( !Used[i] )
		{
			Used[i] = 1;
			for ( j = 0; j < V; ++j )
				Current[j] += Feed[i][j];
			search ( depth + 1, i );
			for ( j = 0; j < V; ++j )
				Current[j] -= Feed[i][j];
			Used[i] = 0;
		}
	}
}

int main ( )
{

	FILE *fin	= fopen ( "holstein.in", "r" );
	FILE *fout	= fopen ( "holstein.out", "w" );

	int i, j;

	fscanf ( fin, "%d", &V );

	for ( i = 0; i < V; ++i )
		fscanf ( fin, "%d", &Required[i] );

	fscanf ( fin, "%d", &G );

	for ( i = 0; i < G; ++i )
		Best[i] = 1;
	Deepest = G;

	for ( i = 0; i < G; ++i )
		for ( j = 0; j < V; ++j )
			fscanf ( fin, "%d", &Feed[i][j] );

	search ( 0, 0 );

	fprintf ( fout, "%d", Deepest );
	for ( i = 0; i < G; ++i )
		if ( Best[i] )
			fprintf ( fout, " %d", i+1 );
	fprintf ( fout, "\n" );

	return 0;
}
[/code]

Thanks[/code]

So its not a compile error, but you still haven't told us what the problem is :P

Exactly. If your variables are not initialized, they may contain any value (sometimes 0 and sometimes not). Calling a random function such as printf may force the compiler to generate different code and therefore put your variables sum1, sum2 at different places in memory which will contain different random data. If you're lucky enough, they will be always zero at the beginning of a call to your search() function.

EDIT: Actually Ravi B is correct, the variables are not moved around (although they could be) but instead the stack is changed and since the solve() function is called recursively, the following call to solve() may have it's stack memory zeroed out by the printf function.

In particular, calling (and returning from) printf may change the stack, which is where local variables such as sum1 and sum2 are typically stored.

[quote="Ravi B"]In your search function, you never initialized sum1 and sum2.[/quote]

Ah.. I didn't notice that.

Thanks


For each positive integer $n$ find the largest subset $M(n)$ of real numbers possessing the property: \[n+\sum_{i=1}^{n}x_{i}^{n+1}\geq n \prod_{i=1}^{n}x_{i}+\sum_{i=1}^{n}x_{i}\quad \text{for all}\; x_{1},x_{2},\cdots,x_{n}\in M(n)\] When does the inequality become an equality ?

First, we will prove the inequality for nonnegative $x_{i}$.  Define

$M_{k}\{x_{i}\}=\sqrt[k]{\frac{\sum_{i=1}^{n}x_{i}^{k}}{n}}$ for $k \in \mathbb{R}\backslash 0$
$M_{0}=\sqrt[n]{\prod_{i=1}^{n}x_{i}}$.

Then the inequality becomes

$n+n M_{n+1}^{n+1}\geq n M_{0}^{n}+n M_{1}$.

The Power Mean Inequality states that $M_{i}\geq M_{j}$ whenever $i \geq j$, so

$1+M_{n+1}^{n+1}\geq 1+M_{n+1}^{n+1}-(M_{n+1}-1) (M_{n+1}^{n}-1) = M_{n+1}^{n}+M_{n+1}\geq M_{0}^{n}+M_{1}$.

For $n$ odd, we can replace $x_{i}$ by $|x_{i}|$ for all $i = 0,1,\ldots n$.  This keeps the left hand side constant and can only increase the right hand side, and the above inequality now holds for the new numbers.  So the inequality holds for all reals.

Given $n$ even;  for any $x <-1$, we cannot have $x \in M(n)$, for then, letting $x_{i}= x$ for all $i = 0,1,\ldots,n$,

$n+n x^{n+1}-n x^{n}-n x = n (x-1) (x^{n}-1) < 0$.

We claim that $M(n) ={x | x \geq-1}$ satisfies the given condition.  Divide $x_{i}$ into $y_{i}, i = 0,1,\ldots, r$ and $z_{i}, i = 0,1, \ldots s$, where $y_{i}\geq 0$ for all $i$ and $z_{i}< 0$ for all $i$.  Replacing all $z_{i}$ with $1$, we have

$n+\sum_{i=1}^{r}y_{i}^{n+1}+s \geq n \prod_{i=1}^{r}y_{i}+\sum_{i=1}^{r}x_{i}+s.$

Also, $z_{i}^{n+1}\geq z_{i}$ for all $i = 0,1,\ldots,s$, and $\prod_{i=1}^{r}y_{i}\prod_{i=1}^{s}z_{i}\leq \prod_{i=1}^{r}y_{i}$, so

$n+\sum_{i=1}^{n}x_{i}^{n+1}\geq n \prod_{i=1}^{n}x_{i}+\sum_{i=1}^{n}x_{i}$

[hide=Solution]
We have two cases: $n$ odd, and $n$ even.

For $n$ odd, let $y_i = |x_i|$, so that $y_i \ge 0$, and we see that $(y_i^n-1)(y_i-1) \ge 0$, since the two terms are both nonnegative when $y_i \ge 0$, and both negative otherwise.

Hence we see that $1+y_i^{n+1} \ge y_i+y_i^n$, and so we see that, since $x_i^{n+1} = y_i^{n+1}$,  $$n+\sum x_i^{n+1} = \sum \left(1+y_i^{n+1}\right) \ge \sum y_i + \sum y_i^n$$

But by AM-GM, $$\sum y_i^n \ge n \prod y_i$$Hence $$\sum y_i + \sum y_i^n \ge \sum y_i +n \prod y_i \ge \sum x_i + n\prod x_i$$And so we see that $$n+\sum x_i^{n+1} \ge \sum y_i + \sum y_i^n \ge \sum x_i + \prod x_i$$Hence $M(n)$ is the set of all reals.

For $n$ even, again let $y_i = |x_i|$, so that $y_i\ge 0$. We see that if $x_i \in M(n)$, then by setting all of them equal to $x_i$, we get $1+x_i^{n+1} \ge x_i^n+x_i$. Hence, $(x_i^{n}-1)(x_i-1) \ge 0$. Checking each of the intervals $(-\infty, -1)$, $[-1,1)$, and $[1,\infty)$, we see that this means $x_i \in [-1, \infty)$.

Now, we see that this means that $$n+\sum x_i^{n+1} = \sum\left(1+x_i^{n+1}\right) \ge \sum x_i + \sum x_i^n$$

But $$\sum x_i^n = \sum y_i^n \ge n\prod y_i \ge n \prod x_i$$ by AM-GM, so we see that $$n+\sum x_i^{n+1} \ge \sum x_i + \sum x_i^n \ge \sum x_i +n \prod x_i$$Hence $M(n)$ is the set of reals at least -1.

So, our answer is $$2\mid n: \mathbb{R}_{\ge -1}, 2\nmid n: \mathbb R$$
[/hide]


algorithm
ceiling function
combinatorics unsolved
combinatorics

There are two groups of tennis players, one of $ 1000$ players and the other of$ 1001$ players. The players can ranked according to their ability. A higher ranking player always beats a lower ranking player (and the ranking never changes). We know the ranking within each group. Show how it is possible in $ 11$games to find the player who is $ 1001$st out of $ 2001$.

Let $ a_1,a_2,\ldots,a_{1000}$ be the players of the first group with the ordered ranking ie., $ a_1<a_2\ldots<a_{1000}$. Similarly there is ordered ranking among the other group $ b_1,b_2,\ldots,b_{1001}$.

 $ comp(a_i,b_j)\equal{}a_i$ if $ a_i>b_j$
                       $ \equal{}b_j$ if $ a_i<b_j$

Consider the algorithm,
if ($ comp(a_i,b_j)\equal{}a_i$)
$ comp(a_{\lceil (i/2)\rceil},b_j)$

else if ($ comp(a_i,b_j)\equal{}b_j$)
$ comp(a_i,b_{\lceil (j/2)\rceil})$

Starting with $ i\equal{}1000\equal{}j$, atmost by 11 operations we can get the player in combined 1001th spot. observe that  if $ f(n)\equal{}\lceil{n/2}\rceil$, $ f^{11}\equal{}(1000)\equal{}1$ and $ f^{11}\equal{}(1001)\equal{}2$.

In general, for two teams with n people each, we need atmost r games to determine the person with combined (x+1)th rank($ x\leq n$) iff $ f^r(x)\equal{}1$.

[quote="srikanth"]

Consider the algorithm,
if ($ comp(a_i,b_j) \equal{} a_i$)
$ comp(a_{\lceil (i/2)\rceil},b_j)$

else if ($ comp(a_i,b_j) \equal{} b_j$)
$ comp(a_i,b_{\lceil (j/2)\rceil})$

[/quote]

What do you mean? Please explain clearly...

There is a slight mistake in the algorithm, anyways I would explain it verbally as you have asked for.
compare $ a_{1000}$ with $ b_{1000}$. Let $ a_{1000}>b_{1000}$ wlog. Now compare $ b_{1000}$ with $ a_{500}$,
if $ b_{1000}>a_{500}$
then compare $ b_{1000}$ with $ a_{750}$, else compare $ b_{1000}$ with $ a_{250}$ .... and so on. Basically, it is going half way everytime. Now you can refer to my earlier post.

I think the following method works. :wink:

For game 1, play $ a_{500}$ against $ b_{501}$. If $ a_{500}$ wins, we will know that the rank of $ a_{500}$ is at most 1000, since $ a_{500}$, who is better than 500 players in group A, will have been found to be better than at least 501 players in group B. We will then have narrowed down our possible candidates to 1001 players: $ a_{501}$–$ a_{1000}$ and $ b_{1}$–$ b_{501}$. (It can’t be, e.g. $ b_{502}$ because $ b_{502}$ is better than only 499 players in his own group and at most 500 players in group A – whereas the 1001st-ranked player must be better than 1000 players in total.) Similarly, if it is $ b_{501}$ who wins, then $ \textrm{rank}(b_{501})\leq1000$ and our target player will be among ($ b_{502}$–$ b_{1001}$ and $ a_{1}$–$ a_{500}$), the possible candidates this time narrowing down to 1000 players.

For game 2, play $ a_{750}$ against $ b_{251}$ (if $ a_{500}$ won) or $ a_{250}$ against $ b_{751}$ (if $ b_{501}$ won). As before, whoever wins will turn out to be better than at least 1001 players and so will be ranked $ \leq1000$. The number of possible candidates now becomes 501 or 500. (For example, if $ a_{750}$ beats $ b_{251}$, the 1001st-ranked must be among $ a_{751}$–$ a_{1000}$ and $ b_{1}$–$ b_{251}$; if $ b_{251}$ beats $ a_{750}$, he’ll be among $ b_{252}$–$ b_{501}$ and $ a_{501}$–$ a_{750}$.)

And so on. For example, game 3 will – depending on who won game 2 – be $ a_{875}$ v $ b_{126}$, $ a_{625}$ v $ b_{376}$, $ a_{375}$ v $ b_{626}$, or $ a_{125}$ v $ b_{879}$. In every game, we:

(i) play the “middle” player in the list of possible candidates in each group against the “middle” player in the corresponding list in the other group, and
(ii) ensure that the total number of players ranked below both players is 1000.

The second condition ensures that the winner of any game will turn out to be ranked at most 1000 (because he’ll be better than 1000 players plus the player he just beat). After each game, the number of possible candidates for the 1001st-ranked player should be halved. The minimum number of games required to suss him out should then be $ n$ where $ \frac {2001}{2^n}\leq1$, i.e. $ n\geq11$. :D

Well, there could still more ways of working it out.. it is all basically about halving.


a problem from dingzhou:

All two-digit numbers have the property that if you reverse the digits and add that result to the original number, reverse the digits of the sum and add that to the sum, and continue doing this, you will eventually arrive at a palindrome. For example,

57+75=132   132+231=363

amd 363 is the palindrome at which you arrive in 2 steps (reversals) What is the smallest two-digit number which requires the greatest number of reversals to reach a palindrome.

89 requires a whole lot of reversals.
But I'm too lazy to look for more.
$ \boxed{89}$

[hide]
Let $ x$ be the tens digit and $ y$ the units digit.
We need $ 11x\plus{}11y$ to be two-digit and not a multiple of 11 or three-digit and not one of 121. The first option is ruled out, so we are left with $ 9 < x\plus{}y$ and $ x \plus{} y \not\equal{} 11k$, where $ k$ is a positive integer.

After one addition, all numbers with the same digit sum will turn out the same, so we use guess and check:
Sum 10--2 iterations
Sum 11--0 iterations
Sum 12--2 iterations
Sum 13--2 iterations
Sum 14--3 iterations
Sum 15--4 iterations
Sum 16--6 iterations
Sum 17--9+ iterations
Sum 18--0 iterations

So $ \fbox{89}$. I think.
[/hide]

[quote="distracted523"][hide]
Let $ x$ be the tens digit and $ y$ the units digit.
We need $ 11x \plus{} 11y$ to be two-digit and not a multiple of 11 or three-digit and not one of 121. The first option is ruled out, so we are left with $ 9 < x \plus{} y$ and $ x \plus{} y \not \equal{} 11k$, where $ k$ is a positive integer.

After one addition, all numbers with the same digit sum will turn out the same, so we use guess and check:
Sum 10--2 iterations
Sum 11--0 iterations
Sum 12--2 iterations
Sum 13--2 iterations
Sum 14--3 iterations
Sum 15--4 iterations
Sum 16--6 iterations
Sum 17--9+ iterations
Sum 18--0 iterations

So $ \fbox{89}$. I think.
[/hide][/quote]
lolz. i guessed correct. ok, maybe i didnt, i used common sense.


Let $a$ be a positive real number and $\{x_n\}_{n\geq 1}$ a sequence of real numbers such that $x_1=a$ and
\[ x_{n+1} \geq (n+2)x_n - \sum^{n-1}_{k=1}kx_k, \ \forall \ n\geq 1. \]
Prove that there exists a positive integer $n$ such that $x_n > 1999!$.

[i]Ciprian Manolescu[/i]

Let $S_n=\sum_{k=1}^nkx_k$. We want to show that $x_n\ge S_{n-1}$ for all $n$. This is trivially true for $n=1$. Assuming $x_n\ge S_{n-1}$, we have $x_{n+1}\ge nx_n+2S_{n-1}-S_{n-1}=nx_n+S_{n-1}=S_n$.

In particular, $x_n\ge (n-1)x_{n-1}$ and hence $x_n>(n-1)!a$, so $x_n$ is unbounded and we are done.

[b]Official Solution.[/b]

We will prove by induction that \[ x_{n+1}>\sum_{k=1}^{n}kx_k.\] For $ n=1$, the definition gives $ x_2\geq 3x_1>x_1$. Suppose $ x_{n+1}>\sum_{k=1}^{n}kx_k$ and we shall prove the corresponding formula for $ x_{n+2}$, \begin{eqnarray*}
x_{n+2}\geq (n+3)x_{n+1}-\sum_{k=1}^{n}kx_k&=&(n+1)x_{n+1}+2x_{n+1}-\sum_{k=1}^{n}kx_k\\
&>&(n+1)x_{n+1}+2\sum_{k=1}^{n}kx_k-\sum_{k=1}^{n}kx_k=(n+1)x_{n+1}+\sum_{k=1}^{n}kx_k=\sum_{k=1}^{n+1}kx_k.
\end{eqnarray*}
From this we deduce, \[ x_{n+1}>\sum_{k=1}^{n}kx_k>nx_n.\] By induction, \[ x_{n+1}>nx_n>n(n-1)x_{n-1}>\cdots >n!a.\] Obviously, for sufficiently large $ n$, we have $ n!a>1999!$.

Here my solution:
We will prove that $ x_{n+1} \ge 3 x_n $ by induction on $ n $.Suppose that $ x_{j+1} \ge 3 x_j $ is true for $ j \le n-1 $.Then ( by condtion of the problem)
$ \frac{x_{n+1}}{x_n} \ge (n+2) - (\frac{x_1}{x_n}+\frac{2x_2}{x_n} + ... + \frac{(n-1)x_{n-1}}{x_n}) \ge (n+2)-(n-1)(\frac{1}{3}+\frac{1}{9}+...)=\frac{n+5}{2} \ge 3 $, so claim is proven.Hence we can take $ n $ such that $ n>1+log_3 (\frac{1999!}{a}) $ :
$ x_n \ge 3^{n-1}x_1=3^{n-1}a>1999! $.DONE !


geometry
3D geometry
sphere
combinatorics unsolved
combinatorics

The following problem is related with an electrostatic problem, in the sense to find the minimum of
\[V(x_1,\dots,x_n)=\sum_{1\le i<j\le n}\frac 1 {|x_i-x_j|}\]
So, the problem is the following:

What is the distribution of $n$ points over an sphere such that the quantity of difference distances of the points are minimal.

In the case $n=3$ clearly solution is 1, taking 3 points  in a equilateral triangule. And same in $n=4$ taking the points ina regular polygon of 4 vertex (clearny inscribed into the sphere).

[quote="fleeting_guest"][url]http://arxiv.org/abs/hep-th/0105179[/url]

See also Sloane's tables of spherical packings.
This is a well known and messy optimization problem,
but the Atiyah and Sutcliffe paper linked above is very nice.[/quote]

hahaha, I know this paper, so, I put the problem waiting someone help me in some sense. But thx a lot for your help.


parameterization
combinatorics unsolved
combinatorics

In a square cartoon we have $15$ cubical blocks with numbers inscribed on them. We are allowed to move the blocks up and down, when there is a free place, we cannot jump over other blocks. Prove, that starting from the left configuration we cannot get to the left one

Sam Loyd's 14-15 puzzle

[hide]If we define a parameter D of each rearrangement of the initial position as the number of [i]pairs[/i] of numbers in the wrong order, then it can be shown that, as long as the spare square is in the bottom right corner, then D is always even. This is not true of the target position.[/hide]

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Sam+Loyd&t=75854


[color=darkblue]Let $ABCD$ be a parallelogram. Denote the point $M\in [CD]$ for which $\widehat{MAC}\equiv\widehat{MAD}$ and the point $N\in [BC]$ for which $\widehat{NAB}\equiv\widehat{NAC}\ .$
Define the points $X\in AB\cap MN$ and $Y\in AD\cap MN\ .$ Prove that the area [XAY] is equally to the area [ABCD] if and only if $ABCD$ becomes a rectangle.[/color]

See the attached diagram for additional notation.

$AM$ is the bisector of $\angle DAC$, which gives ${DM\over MC}={b\over d}$. Similarly, ${BN\over NC}={a\over d}$

$\triangle YDM\sim\triangle NCM\implies{[YDM]\over[NCM]}=\left({DM\over MC}\right)^{2}={b^{2}\over d^{2}}$. Similarly, ${[XBN]\over[NCM]}=\left({BN\over NC}\right)^{2}={a^{2}\over d^{2}}$

Now

\begin{eqnarray*}[XBN]+[YDM]=[NCM]\iff \frac{a^{2}+b^{2}}{d^{2}}[NCM]=[NCM]\iff a^{2}+b^{2}=d^{2}\end{eqnarray*}

which is fulfilled if and only if $ABCD$ is a rectangle.


does anyoe formula for the hexagon area?

For a regular hexagon:

$ \frac{3 \sqrt{3}}{2} s^2$

where $ s$ is the side length.

[hide="hexagon"]$ \frac {3s^{2}\sqrt {3}}{2}$[/hide]
Try figuring out why...it might help that the area of an equilateral triangle is
[hide="triangle"]$ \frac {s^{2}\sqrt {3}}{4}$[/hide]

More help:

[img]http://mathcentral.uregina.ca/QQ/database/QQ.02.06/trevor1.1.gif[/img]


An elevator m=800kg has a maximum load of 8 people or 600kg. The elevator goes up 10 stories = 30m at a constant speed of 4m/s. What is the average power output of the elevator motor if the elevator is fully loaded with its maximum weight? (neglect friction)

Hint $P = w /t=fv$

A) 34.6 B) 79.7 C) 60.999 D) 54.9

My answer seems closest to 54.9......I mean, you havent specified the units, so I'll post my solution anyway (I did things a little differently) --

Potential Energy of the body when 10 storeys high = $mgh$
$= (800+600) X 9.8 X 30$
$= 411600 J$

Since the elevator goes at a speed of 4m/s, the time that it takes is $30 \div 4 = 7.5 sec$

Power of the motor = $Work \div Time$ $=$ $411600 \div 7.5$

which is 54880 W, which is approximately 54.9 Kilowatts -- I'm assuming your unit of power is kW....do tell me if I'm wrong.

[color=blue]If I solve the problem in this way

$\large P=\frac{w}{t}=\frac{fs}{t}=fv=(800+600)\times 9.8\times 4=...$

It seems that the value of 10 stories is neddless[/color]

Oh yeah.....but then whenever that value is given, I tend to solve it the way I did.....and one more major factor is that our textbooks here generally have some confusing hints....so it's my general tendency to overlook the hints....