so guys, what do yo think this year's usamo cutoff and floor values are going to be?

I would start off by saying 215 and 7.

eh you just copied last year's cutoff. I think the usamo expansion will make a diff:

225/8 or 9.

mid 220's/8 (seems to be a common guess)

probably 215/7 because of the expansion and I don't think it was as easy as last year and we are better this year than last year and last year's seniors are gone so..

I'd say 216.5 and 7, though that might be wishful thinking since I got a 7.

i don't see how some people think that the Expansion will affect the index/floor that much!
i mean think about it, the expansion of 150 students means increase of 2~3 students for each state on average!
now consider this, last year there were 9/10 graders in each state whom had the floor value, but did  not qualify because of tie-brakers, also in each state there were   [i]way more than 2~3[/i] students whom were within one point of the floor so they the expansion does not mean a drop of one point.
now that is only for 10- graders, as for the 11/12 graders, there are again people who were very close to the cutoff and did not qualify, because somebody else in their state was closer to the cutoff than them, and had qualified through state index!

anyways, my guess:
230/8+ (by 8+ i mean that not all 10- graders will qualify and they will have tie-brakers)

People should keep in mine that Index-Floor*10 <= 150... some of them are pretty borderline...

In any case, I think 226.5/8 (for those silly 146.5'ers).

Looking at last year's and they year before's statistics, about half the people who got the floor value or more got one less than the floor.  So the expansion should drop the floor by about one.

The floor's for previous years has been 6, 8, 7, 9  in that order.  There's a pattern there, so without the expansion, this year's floor should have been an 8.  But because of the expansion, it's a 7.

And I think that this AIME was a bit harder than last year's, for the people who are borderline.

So looking at all that, the USAMO cutoff should be 216/7.

Gahh, I feel like I'm just making excuzes for me to get in.

the cutoff is definitely 6, there's no question about it. :harhar:

You automatically qualify for USAMO if you guess :D

I would say 7 or 8 floor, seeing as it was slightly harder than last year (which had a 9 floor), and the expansion will probably drop it by a point.

Let's say the floor is 8, but there are more than 120 10- with 8. They compare the AMC+AIME scores to determine which ones get it. Question though, would they just compare among the people who scored 8 or will an 8er with 10 points higher than a 9er on the AMC get preference?

[quote="scorpius119"]Hmm... I'll say a 242 index, yay i'll barely qualify  :rotfl:

No, seriously, my guess is 221/8, possibly a few 7's with tiebreakers.[/quote]
How do they do tiebraekers?

explained in post #21 ;)

based on AMC score

Umm 250/12.

Just kdding.

I agree with 223/ 8 ish. 220 lets some 7's in so a little over 220 makes sense.

[quote="paladin8"][quote="Fermatprime"]Hmm...gonna go with 555.5/5.

Seriously though, 222.5/8. Same thing really.

If we get it right, what do we win? :P[/quote]

How do you get a 142.5 on the AMC?[/quote]

The person who sets the floor doesn't have to have the lowest index, it's possible that someone scores a 143/8.

Ah, good point. That's why you get the really low floors. I still think one of those 146.5 scorers got an 8. Just to spite them for leaving a single question blank.

300/15.   Duh!

But seriously, I'm thinking 220ish/hopefully 7 or 8.

The floor is definitely going to be 9.
The index is going to be 233.0 give or take 1.5

The floor is definitely going to be 9.
The index is going to be at most 233, the 13th Fibonacci Number.

Neal, just for that, the floor will be a 10.  :rotfl:

[quote="h_s_potter2002"][quote="h_s_potter2002"]So looking at all that, the USAMO cutoff should be 216/7.[/quote]

Kind of funny that I was off by one for each thing.[/quote]

I guessed 216.5/8 the day before yesterday.
See, I missed only 0.5!!!


What's the easiest way to find how to say/write something in a different language?

for example: how do you ask "what supplies do we need for Chinese 2" in Chinese?

It might be possible to find individual words easily through a dictionary but I'm not sure how to piece the words together unless I assume that the grammar is the same as English's

If the phrase you want to figure out is pretty standard, then a phrasebook or an introductory textbook would be your best bet.  Otherwise, a grammar of the language will tell you the syntactic and inflectional rules, but you may have to spend a while figuring out how to correctly apply those rules.

Finding a native speaker to help you is always great, but you have to trust that they didn't misunderstand what you want or are pulling a prank on you (cf. "My Big Fat Greek Wedding").


trigonometry
inequalities solved
inequalities

Let $a,b,c$ be positive reals satisfying $a+b+c=abc$
Show that
\[ \sum a(1-b^2)(1-c^2) \leq \frac{4}{27}(a+b+c)^3 \]

[quote="darkmaster"]By using trigonometry ( puting a = tanA ,b= tanB, c=tanC ...) we can show that:
    \[ \sum a(1-b^2)(1-c^2) = 4abc  \]
After that, apply AM-GM and get the answer.    :lol:[/quote]

 \[ \sum a(1-b^2)(1-c^2) = 4abc  \]

WHY??


A gum ball machine contains 9 red, 7 white and 8 blue gum balls. What is the least number of gum balls a person must buy to ensure getting four balls of each color?

If we take 17 balls, it could be 9 red, 8 blue. So the answer is 18.

Unfortuantely, the problem asks for 4 balls of each color. Let's take the worst case scenario. We get 9 red and 8 blue. We still need 4 white so its 9+8+4=21.

Oooh I see! I thought 4 balls with balls in every color. Sorry all!


algebra
polynomial
system of equations
algebra unsolved

Hopefully I am not totally off-base on how to use these forums (if I am, I'm sorry), but I have a couple of problems that I would really like to see the solutions to:

1: First show that if
p(z) = z^4 + 3z - 2 = (z^2 - az + b)(z^2 + az + c)
then a^6 + 8a^2 - 9 = 0. Then solve the equation p(z) = 0

2: Solve the equation z^10 + z^8 + 2z^6 + 2z^4 + 2z^2 + 2 = 0

If you expand (z^2 - az + b)(z^2 + az + c) and compare coefficients you get a non-linear system of equations:
c - a^2 + b = 0
a(b-c) = 3
bc = -2

I would guess that you're supposed to show that a^6 + 8a^2 - 9 = 0 from this, but I simply can't do it. I can show some other things that let you solve the equation, like
(b^2 - 4/b^2)*(b + 2/b) = 9
but I still can't show what's asked for.

For the second problem I'm pretty much clueless.

All help appreciated (these are from past tests, http://www.matematik.lu.se/matematiklu/kurser/MAT131/material/ttu.pdf 73 and 74)!

From $b+c=a^2$ and $b-c=\frac{3}{a}$, you can get $2b=a^2+\frac{3}{a}$ and $2c=a^2-\frac{3}{a}$.

Plugging these into the last one gives $\left(a^2+\frac{3}{a}\right)\left(a^2-\frac{3}{a}\right)=a^4-\frac{9}{a^2}=-8$, which rearranges to the equation in $a$ you're looking for.

Thanks!

The other one can be solved like this: Let w = z^2.
w^5 + w^4 + 2w^3 + 2w^2 + 2w + 2 = 0
w^4(w + 1) + 2w^2(w + 1) + 2(w + 1) = 0
(w + 1)(w^4 + 2w^2 + 2) = 0

Then you factorize and solve it.


我不是上海人...在上海上学...由于我们班上没什么上海人...我还是不会上海话

[quote="shobber"]This is from: [url=http://www.icoool.com/index.asp]http://www.icoool.com/index.asp[/url]

Maybe this is for liyi!  :D

一、选择题(每题10分):

1.上海话里通常管“今天”叫
A.今日 B.今夜 C.今朝 D.今天

2.上海话里通常管“昨天”叫
A.昨日 B.昨夜 C.昨朝 D.昨天

3.上海话里“你”的读音为
（[]中为国际音标）
A.nong B.ou C.yi D.[ne]

4.上海话里“二十二”的读音和上海话中 ____ 相同
A.尼石而 B.而尼 C.难石而 D.难尼

5.现代上海话中，下列哪个字的发音和其他不同？
A.鹿 B.六 C.陆 D.露

6.同上,下列哪个字的发音和其他不同？
A.王 B.望 C.黄 D.横

二、选择填空题(每题5分):

1.“侬只小赤佬！”其中“赤佬”可以用 ___ 代替。
A.垃三 B.瘪三 C.猪头三 D.肮三

2.“阿拉夜饭吃格是油焖落苏。”落苏就是 ___ 。
A.番茄 B.西红柿 C.茄子 D.萝卜

3.“真倒霉，又被黄牛斩了一刀。”这里，黄牛可能是 ___ 。
A.屠夫 B.黄色的牛 C.票贩子 D.穿着黄色牛仔服的人

4.“辰光不多了，叫部叉头吧！”叉头是指 ___ 。
A.出租车 B.面包车 C.直升飞机 D.摩托车

5.“伊老坍板格。”坍板可以用 ____ 代替。
A.丘 B.洋盘 C.十三点 D.贼忒兮兮

6.哪句是错误的？
A.侬阿是搭错了？ B.侬搭错是伐？ C.阿搭错啦侬？ D.侬是搭错伐？

7.哪种说法不合乎通常用法？
A.侬当我洋盘啊？ B.侬当我冲头啊？ C.侬当我屈死啊？ D.侬当我戆大啊？

8.“伊拉专门斩洋葱头。”这句话的意思可能是 ___
A.他们是加工洋葱的专家。 B.他们敲了一个叫洋葱头的人一笔竹杠。 C.他们既是洋盘也是冲头。 D.他们专骗外国人的钱。[/quote]

Ou Xuo Sang Zei Wu, Gu Jiae Dong Sang Hae Wu Cu Fe Dou!
我  说   常   州 话,  估  计  和   上    海  话   差  不 多!
选几个注音碰碰运气 1.C 2.C 3.A 4.A 5.D 6.B

最后一题应该是 D 吧! (洋葱的那个).


inequalities trigonometry quadratics algebra polynomial AMC Elmo

So... I'm going to shamelessly take advantage of the mathematical excitement generated by IMO season and release the 2009 Entirely Legitimate Junior Mathematical Olympiad for your mid-summer delight!

Traditionally, the veterans at MOP divide themselves into team leaders and coordinators, prepare an ELMO for the rookies, and then mark the exam in IMO fashion.  The most agonizing decision of the ELMO Grand Jury, perhaps even more so than the choice of the problems themselves, is what the acronym "ELMO" shall stand for.  In past years, ELMO has been short for "Extremely Loquacious Mathematical Olympiad" and "Eric Larson Mathematical Olympiad."  In honor of the new USA Jr. Math Olympiad, however, we opted to pose an ELMO for Krishanu Roy Sankar and an ELJMO for everyone else, and we selected a title for the exam that we thought would underscore its legitimacy.  Not being exactly masters of subtlety, we wrote what was indeed an entirely legitimate junior math olympiad.

The six problems below represent the best of the 2009 ELJMO shortlist, as voted by the team leaders, who comprised the enfranchised portion of the jury.  Being one of the seniors in black who did not secure a position on the IMO team, I faithfully agreed to serve as Supreme Grand Ayatollah and facilitate the problem selection and coordination processes.  Afterward, Sherry Gong and I worked to compile a LaTeX solution manual, which will be posted after everyone here has had the opportunity to wrestle with the problems.  All of us at MOP invested a great deal of effort into this contest, whether we helped put together the exam or took it as competitors, so I hope you will all enjoy the fruits of that labor as best you can.  If you would like to take the ELJMO as a mock test, you may feel free to send your solutions to me via private message, and I will grade them in accordance with the official rubrics (also voted by the team leaders).  Happy problem-solving!

--DBR

DAY I

1. (Evan O'Dorney) Let $ a,b,c$ be positive integers such that $ a^2 \minus{} bc$ is a square. Prove
that $ 2a \plus{} b \plus{} c$ is not prime.

2. (David B Rush) Let $ ABC$ be a triangle such that $ AB < AC$. Let $ P$ lie on a line through $ A$ parallel to line $ BC$ such that $ C$ and $ P$ are on the same side of line $ AB$. Let $ M$ be the midpoint of segment $ BC$.  Define $ D$ on segment $ BC$ such that $ \angle BAD \equal{} \angle CAM$, and define $ T$ on the extension of ray $ CB$ beyond $ B$ so that $ \angle BAT \equal{} \angle CAP$.  Given that lines $ PC$ and $ AD$ intersect at $ Q$, that lines $ PD$ and $ AB$ intersect at $ R$, and that $ S$ is the midpoint of segment $ DT$, prove that if $ A,P,Q,$ and $ R$ lie on a circle, then $ Q,R,$ and $ S$ are collinear.

3. (Wenyu Cao) Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ a(a \minus{} b)(a \minus{} 2b) \plus{} b(b \minus{} c)(b \minus{} 2c) \plus{} c(c \minus{} a)(c \minus{} 2a) \geq 0.\]
DAY II

4. (Allen Yuan) Let $ n$ be a positive integer. Given $ n^2$ points in a unit square, prove that there exists a broken line of length $ 2n \plus{} 1$ that passes through all the points.

5. (Evan O'Dorney) Let $ ABCDEFG$ be a regular heptagon with center $ O$. Let $ M$ be the centroid of $ \triangle ABD$. Prove that $ \cos^2(\angle GOM)$ is rational and determine its value.

6. (John Berman) Let $ p$ be an odd prime and $ x$ be an integer such that $ p \mid x^3 \minus{} 1$ but $ p \nmid x \minus{} 1$. Prove that
\[ p \mid (p \minus{} 1)!\left(x \minus{} \frac {x^2}{2} \plus{} \frac {x^3}{3} \minus{} \cdots \minus{} \frac {x^{p \minus{} 1}}{p \minus{} 1}\right).\]

[quote="xpmath"]Is it necessary to send in a specific way, i.e. do I have to send in everything at the same time, or can I send a problem at a time?[/quote]

Whatever you like.  :)

My dear Grand Toyota (did I get that right?),

Could the ELMO problems also be released?  I have only seen the problems 1 and 4 on the ELMO as they were also on the ELJMO.

Hmm...is there a way to somehow get hold of ELMO problems from past years?

[quote=Phelpedo][hide]1. Consider the quadratic $ p(x)\equal{}bx^2\plus{}2ax\plus{}c$. Note that this is a polynomial over $ \mathbb{Z}$. Since its discriminant is a perfect square, it has rational roots, i.e. it can be factored over the rationals. Therefore it can be factored over the integers. Now just note that $ p(1)\equal{}2a\plus{}b\plus{}c$. [/hide][/quote]

Really great solution here is my idea.
Let $x^2=a^2-bc$ and $t=2a+b+c$
So $tb=(a+b-x)(a+b+x)$
Let's take the contrary. $t$ is prime. So we have two cases $t|a+b-x$  or $t|a+b+x$
$1)$ $t|a+b-x$ which means $a+b-x\ge 2a+b+c$ $\Longrightarrow $  $a+c+x\le 0$
$2)$ $t|a+b+x$ which means $\sqrt{a^2-bc}\geq a+c$ $\Longrightarrow $ $t\le 0$.

[quote=gighiuhui]
3. (Wenyu Cao) Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ a(a \minus{} b)(a \minus{} 2b) \plus{} b(b \minus{} c)(b \minus{} 2c) \plus{} c(c \minus{} a)(c \minus{} 2a) \geq 0.\]
[/quote]
Let  $a,b,c$ are positive real numbers . [url=https://artofproblemsolving.com/community/c6h1660340p10538050]Prove that [/url]$$a^3+b^3+c^3+2(ab^2+bc^2+ca^2)\geq3(a^2b+b^2c+c^2a).$$

For [hide=1, you get get]
$bc=a^2 -k^2
=> c= \frac{a^2 -k^2}{b}
=> 2a +b +c = 2a + \frac{b^2 +a^2 -k^2}{b}= \frac{a^2 +b^2 +2ab -k^2}{b}= \frac{(a+b)^2 -k^2}{b}= \frac{(a+b+k)(a+b-k)}{b}$, which we know is an integer.
As $a > k$, we are done.


Hello everyone,

I am working on my political stats homework,  and I am having trouble interpreting the data.  I understand all the computations and I believe I have done them all correctly and they are shown in the PDF.  However, I don't know how to interpret the data, I don't know exactly what the data is telling me.  If someone could help in explaining what the data means and what its telling me that would be great.  According to the instructor, the exam will be very similar to the following problems I have in the PDF. a I understand the math, but I don't understand exactly what he wants me to interpret and see in the data. There are two problems and everything I don't understand is marked in Red text in the PDF.  I have checked my answers a couple of times and they seem to be all correct so its not so much the math, its the interpretation of the data.

here is the link to my PDF.

http://www.geocities.com/terry.green6/StatHW.pdf

if someone has enough time to check my math that would be great!! Thank you all!!!!

I can help you only with the mathematical part:

1a) Imagine that you have just 2 voting districts $ A$ and $ B$ with populations $ 10$ and $ 90$ respectively. Imagine that in $ A$ 0% voted for the candidate and in $ B$ 100% voted for him. Then the mean of 0% and 100% is 50% but the portion of the total population voting for the candidate is 90%

2) You set up the equation for $ Z$ correctly but $ \frac{0.52}{1}$ is meaningless. Instead you need to determine the value of $ Z$ from the table looking at where the probability becomes $ 0.02$.

As to the interpretation part, you can restate the definitions, of course, but that's all that comes into my head. Ask your instructor what he wants there.

Thanks you, I think I got all thw stuff down too after massive amounts of googeling.  Apreciate the help.


superior algebra
superior algebra solved

Let R=F[x] with F field. Describe the field of fractions of R.

Thanks


Let $a,b,c$ are real number such that $a+b+c=0$ and $a^{2}+b^{2}+c^{2}=6$. Find the maximum of following expression:
$N=a^{2}b+b^{2}c+c^{2}a$

Nobody solve it  :roll: welcome and waiting a proof to it ... General case is true .


function
analytic geometry
calculus
derivative
Asymptote
limit
calculus computations

Hi,

I need to analyze this function $ y\equal{}\frac{1\minus{}x^3}{x^2}$ by these steps:
1. Find function's definition of area.
2. Find crosspoints with the coordinate axes.
3. Find 1st derivative, determine  intervals where function is ascending/decreasing. Find extreme points.
4. Find 2nd derivative, determine  intervals where function is concave upward /concave downward. Find points of inflection
5. Find asymptotes.

This is my work so far:

1. $ Dom y\equal{}\mathbb{R}$ \ $ {0}$
[b][color=red]2. [/color][/b]No  crosspoints  with 0y axis.
    with 0x axis: $ \frac{1\minus{}x^3}{x^2}\equal{}0$

[b][color=red]3.[/color] [/b]

$ y'\equal{}\frac{(1\minus{}x^3)'x^2\minus{}(1\minus{}x^3)(x^2)'}{(x^4}\equal{}\frac{\minus{}3x^2(x^2)\minus{}(1\minus{}x^3)2x}{x^4}\equal{}\frac{\minus{}3x^4\minus{}2x\plus{}2x^4}{x^4}\equal{}\frac{\minus{}2\minus{}x^3}{x^3}\equal{}\frac{\minus{}2}{x^3}\minus{}1$

Now, I have to find critical points, but I trapped here.

[b][color=red]4. [/color][/b]

$ y''\equal{} (\frac{\minus{}2}{x^3}\minus{}1)'\equal{}(\frac{\minus{}2}{x^3})'\minus{}(1)'\equal{}\frac{(\minus{}2)'(x^3)\plus{}(\minus{}2)(x^3)'}{x^6}\equal{}\frac{\minus{}6}{x^4}$

$ \frac{\minus{}6}{x^4}\equal{}0$ There is no inflection points.

5. Vertical asymptote is $ x\equal{}0$ because $ \lim_{x\to 0}\frac{1\minus{}x^3}{x^2}\equal{}\infty$

There is no horizontal asymptotes because because the degree of the numerator is larger than the denominator.

Slant asymptote is $ y\equal{}\minus{}x$ because $ (1\minus{}x^3): x^2\equal{}\minus{}x\plus{}\frac{1}{x^2}$

Coud you check me and help with missing points?

To find the critical points, you just set the derivative equal to 0: $ \minus{}\frac{2}{x^{3}}\minus{}1\equal{}0\Rightarrow \minus{}\frac{2}{x^{3}}\equal{}1\Rightarrow \minus{}2\equal{}x^{3}\Rightarrow x\equal{}\sqrt[3]{\minus{}2}$ (only one of these is real).

hello, the only real solution of the equation $ x^3\equal{}1$ is $ 1$ so you have only one intersection point with the $ x$ axes.
Sonnhard.


[hide]
If Alex does $\frac{1}{t}$ jobs per day, and he works for a total of 4.8 days, he does $\frac{4.8}{t}$ jobs in the time he works. Bob works for 7.8 days at a rate of $\frac{1}{t+5}$ jobs per day, so he does $\frac{7.8}{t+5}$ jobs in the time he works.

Together, they complete 1 job.
$\frac{7.8}{t+5} + \frac{4.8}{t} = 1$
$t = 10 \textbf{ days}$

[/hide]

[quote="robinhe"][quote="aidan"]Alex takes $t$ days to complete a job and Bob takes $(t + 5)$ days to complete the same job. Suppose Bob works alone on the job for $3$ days. After this, Alex works together with Bob on the job and they take a further $4.8$ days to complete the job.
What is the value of $t$?

Source: Singapore ASEAN Scholarship Math Entrance Examination[/quote]

[quote="aidan"]C'mon, this isn't that hard.. application of rates.[/quote]

Maybe you should wait a couple of days before saying that ;)

[hide="Solution"]Bob can do $\frac{1}{t+5}$ in one day, so after working for 3 days he did $\frac{3}{t+5}$ of the job.

Alex can do $\frac{1}{t}$ of the job in one day.  So he working together with Alex can do $\frac{1}{t+5}+\frac{1}{t}=\frac{2t+5}{t(t+5)}$ of the job in one day.

So, our equation is $\frac{4.8(2t+5)}{t(t+5)}+\frac{3}{t+5}=1$ and solving gives
$t=\boxed{10}$.[/hide][/quote]

Yeah, instead of a couple of minutes at least wait a couple of hours.

People can't always be online when you are.

I also got the same and hopefully correct answer of

[hide]$10$[/hide]


geometry
parallelogram
perpendicular bisector
geometry proposed

[url=http://mathworld.wolfram.com/vanLamoenCircle.html]vanLamoen circle[/url] can be rewrite as

Three medians $ AA',BB',CC'$ of triangle $ ABC$ are concurrent at $ G$, then perpendicular bisector of $ GA,GA',GB,GB',GC,GC'$ intersect base a cyclic hexagon.

Two my extensions

Three medians $ AA',BB',CC'$ of triangle $ ABC$ are concurrent at $ G$

1. Take $ A_1,A_2\in AA'$ such that $ \overrightarrow{A_1A_2} \equal{} \frac {1}{2}\overrightarrow{AA'}$ similarly we have $ B_1,B_2,C_1,C_2$. Draw two lines through $ A_1,A_2$ and perpendicular to $ AA'$, similarly for cyclically $ B,C$, we get six lines. Prove that six lines intersect, respectively (cyclically $ A,B,C$), base a cyclic hexagon.

2. Let $ A_1,A_2$ be midpoint of $ GA,GA'$ similarly we have $ B_1,B_2,C_1,C_2$. Draw through $ A_1,A_2$ two parallel lines, similarly for cyclically $ B,C$, we get six lines. Prove that six lines intersect, respectively (cyclically $ A,B,C$), base a hexagon inscribed a conic.

See the figure. $ A'$ is the midpoint of $ SV$
$ TMUQ$ is a parallelogram $ \Rightarrow S_{TMUQ}\equal{}B_1B_2.TQ\equal{}A_1A_2.TM$
$ \Rightarrow \frac{TM}{TQ}\equal{}\frac{B_1B_2}{A_1A_2}\equal{}\frac{BB'}{AA'}$
$ \angle TNR\equal{}\angle ASC'\equal{}\angle CSV, \angle TRN\equal{}\angle C'SB\equal{}\angle SCV$
Therefore $ \Delta TRN\sim \Delta VCS \Rightarrow \frac{TR}{TN}\equal{}\frac{VC}{VS}\equal{}\frac{BB'}{AA'}$
$ \Rightarrow \frac{TM}{TQ}\equal{}\frac{TR}{TN} \Rightarrow TM.TN\equal{}TR.TQ$
$ \Rightarrow M,N,Q,R$ are concylic. Similarly we get the result (1)


OK so this isn't reaaaally maths, its more psychology or something, no idea. But has anyone does any sort of study on how people can't pick random numbers? Like just picking numbers from 1 to 10, most people would pick 3 or 7, and in the game rock/paper/scissors or janken or whatever its called, I think people have worked out strategies based on the fact that humans cannot pick a random option of out three.. has anyone thought about this much? I thought it could make a good project or something.

Well holmes, in theory you are correct no one can pick random numbers. There is a specific area of computer science which deals with this for there are many uses in programming that have to do with picking random numbers. Gah, if only I could find that windows 2k magazine article...

Google it, you'll find loads of stuff..

[quote="TripleM"] Like just picking numbers from 1 to 10, most people would pick 3 or 7[/quote]

Or they might NOT pick 3 or 7 just because most people DO.  Or they might because some people wouldn't pick them.  Or not.  Or maybe.  Or not...I think this would be WAY too complicated, since you'd have to study everything that could possibly influence a human's decision (for instance, my birthday might be on the 3rd, or my best friend's dog's mom's birthday might be on the 7th, or...

I think it's easier to just say they're close enough to be random.  Or maybe "random with slightly different probabilities."  But nothing is truly random.  Except for my e-mails.

Yeah...they're so random that forget to send them.  :wink:

-interesting_move

If you're interested in the learning aspects of this, I would suggest Fudenberg and Levine's unpublished paper "Conditional Universal Consistency"

Where would I get an unpublished paper from :)

You can go to the Mathematics ArXiV (pronounced Archive I guess) at [url]http://front.math.ucdavis.edu/[/url].

Oooh... this is enough material to keep me busy for a lifetime.  Especially considering that I don't understand a lot of the math at this point.

It's even more definitely enough to keep you busy for a lifetime considering that it gets updated frequently.


function
algebra
domain
geometry
geometric transformation
reflection
trigonometry

Suppose you tell a student that $ f(x) \equal{} \frac{1}{x}$ is not continuous at $ x \equal{} 0$ because it is not defined at $ x \equal{}0$. Is $ g(x) \equal{} \frac{x^2 \minus{} 1}{x \minus{} 1}$ continuous at $ x \equal{} 1$?

So, this obviously goes back to the issue raised here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=304721

My personal opinion is that we may consider both of these functions as mapping from $ \bf R$ to $ \bf R$ plus a new value "is not defined here," and that in this case both functions are discontinuous.  Both functions are continuous on the set of values at which they don't take the value "is not defined here," (which in a more serious mathematical context we would just call "their domains"), but the latter function has the convenient property that it can be easily extended to a continuous function from $ \bf R$ to $ \bf R$.

The other way to answer this question (which, on further reflection, may be both more honest and more useful) is to say, "the function isn't defined at $ x \equal{} 1$, so it surely can't be continuous there."  Again, this particular function is interesting because there's an obvious way to extend it to fill the gap, but this extension is not actually the same function.

For a significant fraction of the time, I tend to take the attitude that "removable singularities deserve to be removed." If you ask me what the value of $ \frac{\sin x}x$ is at $ x\equal{}0,$ I'll tell you $ 1.$ If we're talking about $ \int_0^1x\ln x\,dx,$ I'll say that that's not even an improper integral - it's the integral of a continuous function on $ [0,1].$

As a Calculus I issue: we say that the function is not defined at a certain point, but it's just that point. (In other words, an isolated singularity.) But if I want to have any understanding of how the function behave, I really need to know not just that the function may not be defined there but also how it behaves as $ x$ tends to the troublesome value. And it's in that - the "how it behaves in the neighborhood" - that distinguishes $ \frac1x$ from $ \frac{x^2\minus{}1}{x\minus{}1}.$ Or, for that matter, $ \frac{\sin x}{x}.$

Yes in a lecture, both explanations given by Kent and JBL will set things straight. But if in addition the answer in the back of the book says that g(x) is continuous at the point in question, then offering a correct view as noted by JBL: "the function isn't defined at x = 1, so it surely can't be continuous there" to a student who has already looked up the solution in the back of the book will create a confusion.  I am convinced that Kent's suggestion that "removable singularities deserve to be removed" before going to the question of continuity will help reduce contradictions in the mind of a beginner.

I disagree: in particular, I think people should be introduced to the notion that textbooks can be wrong as early as possible :)


geometry
3D geometry
symmetry
real analysis
real analysis unsolved

Three ants sit in the three centers of adjacent faces of unit cube 1x1x1.
At time t=0 all three ants begin to walk towards
each other 1->2->3->1 at equal constant speeds v=1.

How much time before the ants meet?

The time seems to be exactly $ 1$ -- neat problem!

Oh, a proof, you say?

Let $ C$ be the corner point that the ants are converging towards. (by symmetry, they must meet there)  Let $ A$ be one of the ants, and $ B$ another.

We claim that at any time $ t$ prior to the collision, $ \angle CAB \equal{} \frac{\pi}{4}$.

Proof: Let $ \theta$ be the angle between line $ CA$ and the adjoining side, and $ D$ be the projection of $ B$ onto the other edge of its face.  Then angle $ \angle BCD \equal{} \theta$, by symmetry, so $ \angle ACB$ is a right angle.  Since $ \triangle ACB$ is isoceles, we're done.

Note that we do not need the assumption that each ant remains on their own face.

Finally, we conclude that each ant has a constant radial velocity towards the corner of $ \frac{\sqrt{2}}{2}$, and the starting distance is $ \frac{\sqrt{2}}{2}$.

Ah, anyone reading this should note that I am obviously working on the unfolded cube, and take that into account when I am talking about line segments and such.


if there is a square with side length of r, and there are four circles, each with a radius of r, centered at the four vertices of the square, what is the area of the intersection of the four circles?

Let me give you an overview of the solution:
Find the area of the region enclosed by 2 circles. Let this be y
Find the area of the region that looks like an equilateral triangle (the area is actually a sector). Let this be x
After you've done that, the area is y-2x.

If you are still helpless, download the solution.

in your solution, the funny section that looks like a triangle with arcs for sides, is the sector you defined as x, right?
"notice that x is just an ordinary sector with an angle of 60 degrees."
how do you know its 60 degrees? and isn't a sector a slice of a circle (and must include the circle's center, from which the radii are measured)?


The primeter of an equalateral triangle exceeds the perimeter of a square by 1989 cm. The length of each side of the triangle exceeds the length of each side of the square by d centemeters. How many integers are not possible values of d?

Are you sure you didn't mean how many integers d there are?  :?

Please [b]show your solutions[/b]! Posts without solutions have been deleted. :mad:

Heyyyyyyy u deleted my post!!!1I was askin him whether i was goin right or not...only then would i know whether my solution was goin right or not  and then i would post it.....not fair.

[hide="Solution"]
Let the sides of the triangle be $x+d$,
So the sides of the square is $x$,
$3x+3d-4x=1989$
$3d-1989=x$
$3(d-663)=x$
Now,
$d$ has to be a positive no. so all integral values of $d$ lesser than $663$ will make $x$ a negative no. or $0$ which is not possible.
So $663$ values of $d$ are not possible.
Answer=$\boxed{663}$
[/hide]
Correct?


complex analysis
complex analysis unsolved

Find all Möbius transformation $ \tau(z) \equal{} \frac {az \plus{} b}{cz \plus{} d}$, with $ ad \minus{} bc \equal{} 1$ such that $ \tau(D) \equal{} D$ where $ D$ is the unit disk $ D \equal{} \{z\in\mathbb{C}: |z| < 1\}$.


Would someone either give me a hint or explain how to solve one/both of the problems below?

"How many integers less than 500 have an odd number of distinct, positive, integer factors? Note: One is a always a factor of a positive integer."

"Arrange the digits 0 through 9, using each exactly once, to form five two-digit numbers whose product is as large as possible. For example, $10*23*45*67*89=61717050$."

Thanks for any help provided.

[hide]For first given any $n$, if $a$ is a divisor, so would be $\dfrac n a$ making number of divisors even...unless  it so happens that $a=\dfrac n a$ ... hmm.. hint ..[b] big hint .[/b].... when is that foing to happen?  :)  [/hide]

Thanks for the help Gyan & SnowStorm. Snow, would it be accurate if I said that you're trying to say we want each term to be as big as possible?

[hide="hint"]if a number is a perfect square, then it has an odd number of factors...[/hide]

Alright, thanks guys:

[hide="Number 1"]
Since perfect squares are the only numbers that have an odd number of distinct postive integral factors, we want the number of squares less 500. $\fbox{22}$
[/hide]

For #2,
[hide]Wouldn't you try to make the numbers as close possible to each other? :?[/hide]

[quote="SnowStorm"]2
[hide]
This probably isn't the greatest explanation, but you want the largest amount of 10s in the product possible and to create the greatest numbers possible at each step which should occur when you have
$90 * 81 * 72 * 63 * 54 = 1785641760$

I think this is right, but not 100% sure.[/hide][/quote]
Maybe this resolves your doubt:
Consider the following mini-problem: Which one is larger 42x31 or 41x32 without using calculator? The main thing is not to solve this mini-problem but to learn how numbers are compared.

So let's think about the numbers 31 and 41. If we rewrite them, 42x31=(41+1)x31 and 41x32=41x(31+1). Now lets expand a bit

42x31=(41+1)x31=41x31+31
41x32=41x(31+1)=41x31+41

So, is your solution giving the greatest possible product? :D

So we compare every two numbers and such like that? that's a pain....

No, I think he was just using an example to prove something. :)

ok, so what's the answer and solution?  :? I've seen this many times...

SnowStorm said it best when he said we want the greatest number of tens in the product as possible. To achieve this we want the bigger digits in the tens place of each term. So the solution is $(90)(81)(72)(63)(54)$


Rosalee plans to open a savings account and a checking account. She has decided to
deposit a total of $ \$45$ per week, such that $ \$20$ goes into the checking account each
week and the remaining money goes into the savings account. When she has deposited
a total of $ \$200$ into her savings account, how much will she have deposited into her
checking account?

if $ \$20$ goes in the checking account, $ \$25$ goes in the savings account. so after 200/25=8 deposits, the savings will have $ \$200$, and the checking account will ahve $ \$(20*8)\equal{}\$160$


geometry
perpendicular bisector
angle bisector

I am looking for someone to solve the questions below using steps as you go through the lesson.  I would like to learn how to solve similar questions on my own.  Ready?

1) How many points are equidistant from points A and B and also 6 inches from line segment AB?

2) How many points are equidistant from two intersecting lines and also 4 inches from the point of intersection of the lines?

#1 Both points must lie on the perpendicular bisector of $AB$. On each side of the line $AB$ there must be exactly one point 6 inches from the line $AB$. (As a point slides along the perpendicular bisector, its distance from $AB$ increases continuously, at some point it passes 6 inches.)

#2 The locus of points equidistant from two lines is the [url=http://www.artofproblemsolving.com/Wiki/index.php/Angle_bisector]angle bisector[/url] of the two lines: this is a line and passes through the point where they meet. By the same logic as before, there is one point either side of their intersection which is exactly 4 inches from their intersection.

So, for question 2, the answer is one point?

Are not intersecting lines the same as vertical angles?

Vertical angles form 4 angles, right?

For question 2 there are [b]2[/b] points: both on the same line that passes through the point of intersection, but one either side of it.

The fact that this line is the angle bisector is not really important, so there is no need to worry about vertical angles, we just have a point (where the lines meet) and a light that goes through it.


function
limit
inequalities
algebra proposed
algebra

Find all functions $ f: \mathbb{R^\plus{}} \rightarrow \mathbb{R^\plus{}}$ satisfying $ f(f(x)\minus{}x)\equal{}2x$ for all $ x>0.$

The equation immediately yields $ f(x)>x$ and $ f$ is surjective. But this requires $ 2x>f(x)\minus{}x\Longrightarrow f(x)<3x$. Define two positive sequences via $ a_0\equal{}1,b_0\equal{}3$ and $ a_{n\plus{}1}\equal{}\frac{2\plus{}b_n}{b_n},b_{n\plus{}1}\equal{}\frac{2\plus{}a_n}{a_n}$ and assume we have shown $ a_nx<f(x)<b_nx$ for some $ n\equal{}0,1,\ldots$. Then $ f(f(x)\minus{}x)\equal{}2x$ implies $ a_n(f(x)\minus{}x)<2x<b_n(f(x)\minus{}x)$ which is equivalent to $ a_{n\plus{}1}x<f(x)<b_{n\plus{}1}x$, i.e. we can summarize $ \forall n,x: \,a_nx<f(x)<b_nx$.

[b]Claim[/b]: $ \lim_{n\rightarrow\infty}a_n\equal{}\lim_{n\rightarrow\infty}b_n\equal{}2$ and $ f(x)\equal{}2x$.

[i]Proof[/i]: Only the assertion about the sequences has to be verified.

(1) We verify $ 1\leq a_n<2<b_n\leq 3$ directly from the initial values and the recursive definition.
This yields $ a_{n\plus{}1}b_{n\plus{}1}\equal{}\left(1\plus{}\frac{2}{a_n}\right)\left(1\plus{}\frac{2}{b_n}\right)$ $ \geq 2\cdot\frac{5}{3}>3$.
(2) We observe for $ n>1$ the inequalities $ a_{n\plus{}1}\minus{}a_n\equal{}(b_{n\minus{}1}\minus{}b_n)\frac{2}{b_nb_{n\minus{}1}}<b_{n\minus{}1}\minus{}b_n$ $ \equal{}(a_{n\minus{}1}\minus{}a_{n\minus{}2})\frac{2}{a_{n\minus{}1}a_{n\minus{}2}}$. Taking into account the initial values we conclude $ a_n$ is monotone increasing and $ b_n$ is monotone decreasing.
(3) For $ n>1$ we note the inequality $ b_{n\plus{}1}\minus{}a_{n\plus{}1}\equal{}(b_n\minus{}a_n)\frac{2}{a_nb_n}<\frac{2}{3}(b_n\minus{}a_n)$.

(1),(2) imply that $ a_n,b_n$ converge to limits $ a\leq 2\leq b$. From (3) we conclude $ a\equal{}b\equal{}2$.


If you play on FICS what's your handle??

mine is [b]andonehotsauce[/b]. Maybe we can play each other

Handle: davidyko
Warning: I absolutely suck at anything except for standard. I can't think right in faster games.

Are are you supposed to acquire an account? I have no non-free emails.

Meh, chess64 gave me an email address on his site so that's the only way I got my handle.


parameterization
function
topology
geometry
algebra
domain
real analysis

1) Show that if S is a (piecewise) smooth k-dimensional surface in $ R^n$ then up to an at most countable number of piecewise smooth surfaces, S can be written as the union of at most countably many parametrized k-dimensional surfaces.

2) Show that the partition in 1) can be made to be locally finite.

Definitions?

My guess is that you are asked to eliminate the parameters locally, using the implicit function theorem. For example, if $ x \equal{} x(t)$, $ y \equal{} y(t)$ is a parametrized curve, then $ y$ can be written as a function of $ x$ at a neighborhood of any point where $ x'(t)\ne 0$.

Here are the definitions:

We call a point a zero-dimensional surface of any smoothness class.
A [i]piecewise smooth one-dimensional surface[/i] is a curve in $ R^n$ which breaks into smooth one-dimensional surfaces when a finite or countable number of zero-dimensional surfaces are removed from it.
A surface S in $ R^n$ of dimension k is [i]piecewise smooth[/i] if a finite or countable number of piecewise smooth surfaces of dimension at most k-1 can be removed from it in such a way that the remainder decomposes into smooth k-dimensional surfaces $ S_i$ (with boundary or without).

As you can see, the way the definition is given makes it almost impossible to do any useful mathematics. In the statement of the problem it's not clear exactly what "partition" means since it seems likely that the sets in the partition might have some (negligible) overlap. I'm inferring from comments the author makes elsewhere that the intersection of any $ S_i$ and $ S_j$ is a piecewise smooth surface of dimension at most k-1 contained in the intersection of their (nontopological) boundaries.  :|

---

The partition of the surface S is locally finite if any compact set K in S can intersect only a finite number of the surfaces $ S_i$.

I considered something similar to what you said with local charts but I don't see how to show the collection $ S_i$ is countable. In what I interpret as a hint, the author made some offhand comment about each of the pieces $ S_i$ being contained in the range of some local chart of the surface S. Hopefully you see something I don't.

Since the surface is embedded into a Euclidean space, its topology has a countable basis. I think this should imply countability of s_i

But we don't know that the countable basis has the negligible overlap property, for one. The point of the exercise is to show that these Si can always be constructed and the area of S can be defined invariantly as the sum of the areas of the Si.

On further thinking, I think you may be right about the first part. By a similar reasoning, S has a countable atlas and the domains of action of the local charts of that atlas might give the partition we are looking for. However it's still hard to see how to show that they abut on piecewise smooth surfaces of lower dimension. It seems more likely that we should use the connected components of S but is it true that a connected smooth manifold is parametrizable?

If it's any help, the author makes an offhand comment which I interpret as a hint: he says that each of the smooth pieces Si is contained in the range of some local chart of the surface S.

And the local finiteness condition?

I made made a few errors in my explanation so I'll start over:

We call a point a zero-dimensional surface of any smoothness class.
A [i]piecewise smooth one-dimensional surface[/i] is a curve in $ R^n$ which breaks into smooth one-dimensional surfaces when a finite or countable number of zero-dimensional surfaces are removed from it.
A surface S in $ R^n$ of dimension k is [i]piecewise smooth[/i] if a finite or countable number of piecewise smooth surfaces of dimension at most k-1 can be removed from it in such a way that the remainder decomposes into smooth k-dimensional surfaces $ S_i$ (with boundary or without).

I presume the union is supposed to be pairwise disjoint.

We say $ S \subset R^n$ is a k-dimensional parametrized surface if there is an [i]open connected set[/i] $ D \subset R^k$and a $ C^1$
homeomorphism (basically a local chart)$ r: D \to S$.

Let's break the problem into smaller steps.

If M is a smooth k-dimensional surface in $ R^n$, can we always remove at most countably many piecewise smooth surfaces of dimension at most k-1 so that M breaks up into a union of smooth [i]parametrized[/i] surfaces?

My idea was to consider the connected components of M and show that once an atlas for a connected component A of M is fixed, M can be represented as a smooth parametrized surface. This seems plausible but I don't see how to do it.

[url=http://books.google.com/books?id=q118Hw_7wKgC&pg=PA189&lpg=PA189&dq=%22the+partition+be+locally+finite%22&source=web&ots=Ui4wmQyxMZ&sig=0WXM-Q4YoQOjpDjlvA6NGCnKAtw&hl=en]See here also[/url]


If $ A \equal{} \left[\begin{array}{cc} 7 & 4 \\ \minus{}9 & \minus{}5 \end{array}\right]$ compute $ A^n$.

Quite easy

Well known that for n = 2 matrices $ \ A^2 \minus{} TraceA*A \plus{} det(A)I_2 \equal{} 0$
$ \ A^2 \equal{} 2A \minus{} I_2$ by induction $ \ \minus{} >$  $ \ A^n \equal{} nA \minus{} (n \minus{} 1)I_2$


Yes, you got it.

I needed to square both sides because it is a radical equation.
Good job.

[quote="symmetry"]I have worked on this all day.  I need to solve for L, which is in the radicand.

2 = 2PI(sqrt{L/32})[/quote]

Another way to solve it:

[hide]
$2=2\pi\sqrt{\frac{L}{32}}$

$1=\pi\frac{\sqrt{L}}{\sqrt{32}}$

$1=\frac{\pi4\sqrt{2L}}{32}$

$1=\frac{\pi\sqrt{2L}}{8}$

$\frac{8}{\pi}=\sqrt{2L}$

$\frac{64}{\pi^2}=2L$

$L=\frac{32}{\pi^2}$

It may be a little bit longer, but...
[/hide]

 :?


As the subject description already says, What do you think your rank at nats will be?

as i can not vouch for myself, i will do so for my team member, dan.

53rd

that's how we do it!

jorian

Probably (hopefully) in the top 100 written... not in countdown for sure. I'm pretty bad.  :blush:

most likely in the top 228 spots... unless i get 229 ... which would be kinda sad -_-...

[quote="13375P34K43V312"]According to my blog 6th written, 1st countdown.  :rotfl:
But I really don't know, I'm hoping I'll get into countdown, but I probably won't crack the top 4 written to get a bye. Anyhow, I consider myself to be pretty strong at countdown, so I have a good shot at going far as long as I don't get matched up against someone really good. But I've countdown maybe 100 times this year, I'm 99-1.[/quote]

note that i almost won, but he got the win, 3-5

who won against you? ni?

jorian

[quote="13375P34K43V312"]According to my blog 6th written, 1st countdown.  :rotfl:
But I really don't know, I'm hoping I'll get into countdown, but I probably won't crack the top 4 written to get a bye. Anyhow, I consider myself to be pretty strong at countdown, so I have a good shot at going far as long as I don't get matched up against someone really good. But I've countdown maybe 100 times this year, I'm 99-1.[/quote]

What did you get at states?

he got 1st the last 2 years, and i'm randomly guessing he got 112th at nats last year...

jorian

its just a guess, but based on my not so awesome past nats scores, id say between 120-140, but dqmot  (dont quote me on that)

EDIT:  As a score, i think anywhere between 20-25 maybe, if im lucky, a 27 or 28

a 28 gets you around 70th :wink:

jorian

ok then, maybe not that high... im gonna get owned at nats.  but my goal is to get in the top 100.  if not, oh well...

This year I predict the Countdown cutoff to be around 30, and the Top 57 cutoff to be around 25.

It should be harder this year

Isn't the cutoff countdown a little low, even if the test is hard?

scores at nationals -_-... the cutoff for countdown will be high this year, considering the fact that most of us can get like 35+ on old nats test... and 30+ on recent ones.  or maybe im just assuming things ...

i think the cutoff's gonna be like 32 or 33. 30 sounds a bit low.

uh,.. i think its gonna be like 35 -_-... really high

I agree with Mathcrazed, I think its going to be high, too.

Winner: 43
Top four cutoff: 41
Top 12: 37

My predictions, even if it's a "hard" year.

I can't say for me but here is some of the others:

Kevin Chen: 1st (or 2nd if by countdown he messes up...  :( )
Bobby Shen: In the top 15
Other people: In the top 30

Team: 1st! w00t

Um: I think that people are getting smarter as the years go by so I think that even if the contest is hard, everyone is just going to own it and so the cutoff is 42 to make it to countdown and 45 is the winner.

you can't really translate scores

i know everyone hates my dt examples, but here's one

'06: dt got 3rd at school, dt got 12th at chapter, dt got 2nd at state, dt got 92nd at national
'07: dt got 1st at school (should be 2nd, but w/e), 3rd at chapter, dt got 1st at state, dt got x at national

im not even going to try to guess any of dan's placings...

jorian

For nationals I want a 15+ on the sprint and a double digit on target (harder than you think)

Last year I got 105.
This year, I'm hoping for the top 50+~. ( :o That's a jump of more than 50 places!)

I want at least 14 on target and at least 20 on sprint

I wat at least a 36.

I'm hoping for top 25.

[quote="zserf"]I got 33 at States, what does that normally translate to at nationals?

Probably not accurate though considering I got 29 at Chapters

Chapter:21/8
State:19/14

Nationals?:17/20.....[/quote]
Translations usually don't work.
Last year I went from a 35 at states to a 33 at nats.
Neal Wu stayed constant with a 42.
Someone I know went from a 40 to a 24.
Someone else I know went from a 41 to a 25.
Someone else I know went from a 39 to a 28.
So it all depends on the person.

Someone I knew went from 30 to 34 :maybe:

But yeah, it is weird.

[quote="diophantient"]Someone else I know went from a 41 to a 25.
[/quote]

Yeah that would have been me.
There are, as I discovered the hard way last year, much more factors than a first-time student my think there are. This is where experience helps, whether you are a coach or a competitor.

Basically, sleep, quality/quantity of food, type of problems, bathroom issues, and even stuff like clothing and size of paper (they give you like 6*9 sheets) play a role. And the pressure that proctors put on you...my proctor last year thought I was cheating and threatened she would disqualify me so yeah.

I hated that moment Carl, it got me distracted.

JKLOL
But yeah, I was like, who would cheat at a Math Competition.

my prediction: top score = 46 (k.chen i wonder who that could be?)

and the 40 to 24 was me.
Carl and I had horrid days, we easily could've been much much higher, so I don't think you should try translating scores.

i dont know what ill get but ill predict ill make it into countdown.  My parents will probably get mad if i dont make top 5, cuz they've "invested so much money in mathcounts".  jeez, i wonder whatll happen if i drop.  i go t 23rd last yr.  i think high score will be 45(dunno who probably kevin.  Do i even need to say a last name?).  cd cutoff :35, top 4:40


Our free AIME Math Jam will be on Thursday, March 19, starting at 7 PM Eastern / 4 PM Pacific (please note early start time).  We will discuss solutions to all 15 problems on the contest.  Click on the "Math Jam" tab at the top of the page to get more information.

As long as it takes to get through all 15 problems.  Probably about 2 hours, give or take a half-hour.  Of course, you don't have to be there for the whole thing, and a complete transcript will be posted afterwords.

Where will be able to find the transcript?

if you hover your mouse over the "math jams" option above, you will find a drop down menu that gives two options
one says transcripts

click on that and you should find it

*note* you may have a to wait a while for the transcript to be uploaded onto the site

If you missed the Math Jam, or want to review some of the solutions, the [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=251]transcript[/url] is now posted.

The AIME II Math Jam is on Friday, April 3 at 7 PM Eastern / 4 PM Pacific.  We will discuss the solutions to all 15 problems.


[quote="toanIneq"]Let $ a,b\in R$ and $ f: R\rightarrow R$ is an increasing function.Find the functions $ g: [a;b]\rightarrow [f(a);f(b)]$ such that:
$ g(x) \minus{} g(y)\geq|f(x) \minus{} f(y)|$ for any $ x,y\in [a;b]$[/quote]
I think you have a typo .
$ g(x)\minus{}g(y)\geq |f(x)\minus{}f(y)|$
Let $ x\to y,y\to x$ then
$ g(y)\minus{}g(x)\geq |f(x)\minus{}f(y)|$
Therefore $ |f(x)\minus{}f(y)|\leq 0$ so $ f(x)\equal{}c$


$n$ and $l$ are positive integers.

Find all possible pairs of $n$ and $l$ such that

\[{n\choose 0}+{n\choose 1}+{n\choose 2}+{n\choose 3}=2^{l}.\]

Well the one solution I know immediately is [hide] $(n,l)=(3,3)$.[/hide] But I don't know if there exist more solutions. If this is useful, I'll save everyone a bit of arithmetic and tell you that it reduces to $n^{3}+5n+6=6 \cdot 2^{l}$. The usual mods don't give much, other than $4 \not | \ n$.

The polynomial on the left factors as $(n+1)(n^{2}-n+6)$.  Thus, you have to have that $n+1 | 3\cdot 2^{l+1}$, so either $n = 2^{k}-1$ or $n = 3\cdot 2^{k}-1$.  From there, some calculations will give you bounds on the possible size of $k$, from which you can find all values of $n$.  (There are 6 of them, I believe.)  Unfortunately, this is kind of ugly; there may be a nicer way.

[quote="mao_zai"]jbl, there are only three. but your approach sounds sound. any elaboration on how u obtain your bounds?[/quote]

No, there are 5.  (I counted 0 in my previous response, which made 6.)

${1\choose 0}+{1\choose 1}+{1\choose 2}+{1\choose 3}=2 = 2^{1}$, ${2\choose 0}+{2\choose 1}+{2\choose 2}+{2\choose 3}=4 = 2^{2}$, ${3\choose 0}+{3\choose 1}+{3\choose 2}+{3\choose 3}=8 = 2^{3}$, ${7\choose 0}+{7\choose 1}+{7\choose 2}+{7\choose 3}=64 = 2^{6}$, and ${23\choose 0}+{23\choose 1}+{23\choose 2}+{23\choose 3}=2048 = 2^{11}$.

As for the actual process, it's fairly ugly.  Let me outline the calculations necessary for the case $n = 2^{k}-1$.

If $n = 2^{k}-1$, we have $2^{k}\cdot (2^{2k}-3\cdot2^{k}+8) = 6\cdot 2^{l}$.  The left hand side is dominated by the $2^{3k}$ term of that product, so one can show that for sufficiently large $k$, it gets bounded strictly between $6 \cdot 2^{3k-3}$ and $6 \cdot 2^{3k-2}$.  (Basically, you subtract and look at the differences: one term will grow much faster than the others, so eventually the difference will have the same sign as that term.)  When that happens, we know there can't be any solutions, so we only need to check the values below the bound.  Something similar happens in the case $n = 3\cdot 2^{k}-1$.

thanks for the solution. i omitted the first two cases n=1,2.


limit
trigonometry
real analysis
real analysis unsolved

Let $a,b,A,B$ are positive real numbers. Let sequence $x_{1},x_{2},x_{3},...$ by $x_{1}=a,x_{2}=b,x_{n+1}=A\sqrt[3]{x_{n}^{2}}+B\sqrt[3]{x_{n-1}^{2}}(n=2,3,...)$. Prove that exists $lim_{n\to\infty}x_{n}$ and find it.

[quote="HilbertThm90"]Tried the standard trick, but something doesn't seem quite right. Suppose $\lim_{n\to\infty}x_{n}=L$, then $\lim_{n\to\infty}x_{n+1}=L$ and $\lim_{n\to\infty}x_{n-1}=L$ as well. Letting both sides of the equation go to infinity (and using the standard limit laws) we get: $L=A\cdot L^{2/3}+B\cdot L^{2/3}\Rightarrow L=L^{2/3}(A+B)\Rightarrow L^{5/2}=A+B\Rightarrow L=(A+B)^{2/5}$. As for proving the limit exists, I'll come back later today, don't have much time right now.[/quote]

I think you want $L = L^{\frac{2}{3}}(A+B) \Rightarrow L = (A+B)^{3}$.

Ah yes. I knew something wasn't right when I did a quick check.

And $L=0$???????
We show that $\lim_{n\to\infty}u_{n}=(A+B)^{3}$.
Remark: if $a<1<b$ then $a<a^\frac{2}{3}<1<b^\frac{2}{3}<b$.
We may assume that $A+B=1$ ($v_{n}=\frac{u_{n}}{(A+B)^{3}}$).
$u_{n}\in[u_{n-1}^\frac{2}{3},u_{n-2}^\frac{2}{3}]$ up to order. Let $z_{n}=\sup(1,u_{n},u_{n-1})$; by the remark $z_{n}$ is a decreasing sequence and $\lim{z_{n}}=\alpha\ge1$.
Suppose $\alpha>1$; then there exists $\epsilon$ such that $(\alpha+\epsilon)^\frac{2}{3}<\alpha$. For all sufficiently large $n$ $u_{n-1}\le\alpha+\epsilon$ and  $u_{n-2}\le\alpha+\epsilon$; thus $u_{n}\le (\alpha+\epsilon)^\frac{2}{3}$ , $u_{n+1}\le (\alpha+\epsilon)^\frac{2}{3}$ and $z_{n+1}\le (\alpha+\epsilon)^\frac{2}{3}$; contradiction and $\alpha=1$ therefore $\limsup{z_{n}}\le1$. We show by same method that $\liminf{z_{n}}\ge1$.    QED

First , it is to be prove that the limit exist.

First we have to prove $x_n$ do not oscillate. We rewrite the eq. in this way:

$\frac{A\sqrt[3]{x^2_n}}{x_{n+1}}+\frac{B\sqrt[3]{x^2_{n-1}}}{x_{n+1}}=1$,   let $\sin^2 \theta_n=\frac{A\sqrt[3]{x^2_n}}{x_{n+1}}$ and  $\cos^2 \theta_n=\frac{B\sqrt[3]{x^2_{n-1}}}{x_{n+1}}$.

We know $sin^2 y$ & $\cos^2 y$ both are periodic. So we consider $\frac{A\sqrt[3]{x^2_2}}{x_3}=\frac{A\sqrt[3]{x^2_m}}{x_{m+1}}\Rightarrow{x^\frac{2}{3}_{m}}=c_{(2,3)}x_{m+1}$, where $c_{(2,3)}$ be a constant. Again $\frac{B\sqrt[3]{x^2_1}}{x_3}=\frac{B\sqrt[3]{x^2_{m-1}}{x_{m+1}}\Rightarrow{x^\frac{2}{3}_{m-1}}=c_{(1,3)}x_{m+1}$, where $c_{(1,3)}$ be a constant. Then $(\frac{x_m}{x{m-1}})^{\frac{2}{3}}=\frac{c_{(2,3)}}{c_{(1,3)}}=d_m$ ,or, $x_m=d_mx_{m-1}$ .
  Since $d_m\neq{1}$ for any value of $m$, then $x_n$ do not oscillates.

Now we have two cases.
[b]case 1:[/b]$x_n>x_{n+1}$, then it is need not to prove that $x_n$ has a limit as $x_n>0$.
[b]case 2:[/b]$x_n<x_{n+1}$. If we consider that $A\sqrt[3]{x^2_n}>B\sqrt[3]{x^2_{n-1}}$, then
we know that $x_{n+1}<2A\sqrt[3]{x^2_n}\Rightarrow{x^3_{n+1}<8A^3x^2_n}\Rightarrow{x_{n+1}<8A^3}$, this gives $x_n$ is bounded as $A$ is finite.

  If we consider that $A\sqrt[3]{x^2_n}<B\sqrt[3]{x^2_{n-1}}$, then also  $x_n$ is bounded as $B$ is finite.
At last we proved $x_n$ has a limit as $n\to\infty$(let the limit is $L$)
      We get $x_n=x_{n+1}=x_{n-1}=L$ as $n\to\infty$ , then $L = L^{\frac{2}{3}}(A+B) \Rightarrow L = (A+B)^{3}$

Hence  $ \lim_{n\to\infty}x_{n}=(A+B)^{3}$ (ans.)


Prove that if a body has a force pointing to a fixed point acting on it, then the area swept out by the body in a certain length of time is always the same.

[quote="dangvh"]isnt this one of kepler's laws?[/quote]
Yes. The Kepler's Law is a special case.
But i think the coolest thing about Kepler is that he got the conclution by observation and analyzing the notes, rather than proving by Angular Momentum!

[quote="probability1.01"]Isn't the force of gravity not constant though?[/quote]
Do you mean $mg$ or $G\frac{Mm}{R^2}$?

Hint:
[hide]compare the expression for area swept out in a small interval of time  to the expression for angular momentum[/hide]


Prove that

$0<\frac{\ln{x}}{x}<\frac{2}{\sqrt{x}}$.

for all $x>1$

(the fact that $0<\ln{x}<x$ and $\ln{x}=2\ln{\sqrt{x}}$ might be helpful).
If this is too easy, bump it to Getting started.

Umm, if $x < 1$, the left inequality is not true... in fact $\frac{\ln{x}}{x} \to -\infty$ as $x \to 0$.

for all $x>1$

How's this?
[hide]$0<\frac{\ln{x}}{x} = \frac{2\ln{\sqrt{x}}}{x}$

Since $x>1$, $\sqrt{x}>1$ so ${\frac{\sqrt{x}}{\ln{\sqrt{x}}}}>1$

Then multiplying the upper bound by this factor will increase it, so

${0<\frac{\ln{x}}{x} < \frac{2\ln{\sqrt{x}}}{x} ( \frac{\sqrt{x}}{\ln{\sqrt{x}}}} ) = \frac{2}{\sqrt{x}}$[/hide]

Well, I'm assuming its $0\leq\frac{\ln x}{x}<\frac{2}{\sqrt{x}}\ \forall x\geq 1$.  Now, multiplying both sides by $x$, we have ${0\leq\ln x<2\sqrt{x}}$.  Raising $e$ to all of them, we have $1\leq x<e^{2\sqrt{x}}\ x\neq 0$. $x=e^{2\ln\sqrt{x}}$, thus, we have $1\leq e^{2\ln\sqrt{x}}<e^{2\sqrt{x}}\ x\neq 0\implies 2\ln\sqrt{x}<2\sqrt{x}\implies \ln x<x$, which is true.

Masoud Zargar


Please tell me whether the posted problem is adequate for the given level or should I post such problems in other forum. Also suggest which forum.

[quote="varunn97"]Prove (k^7)/7 + (k^5)/5 + (2*k^3)/3 - k/105
is an integer for every positive integer k.[/quote]

[hide="using Number theory"]

It is also equivalent to $\frac{k(15k^6+21k^4+70k^2-1)}{105}$

If $gcd(k,3,5,7)=1$ then it is true by FLT that the denominator divides the numerator.

If $k$ has the factor of at least one of $3,5,7$ , then $k|105$ while $15k^6+21k^4+70k^2-1 | 105 $ by FLT . So we are done . [/hide]


calculus
derivative
algebra
polynomial
vector
induction
superior algebra

I have had problems to prove the following theorem. Can someone say if the proof is correct?

Theorem. Let $ \mathbb{F}_q$ be a finite field with $ q$ element. Let $ n\geq 1$ be a fixed integer. Then there exist a unique extension of $ \mathbb{F}$ of degree $ n$, and this extension is $ \mathbb{F}_{q^n}$.

Proof. Let us show first that there exist an extension of $ \mathbb{F}_q$ of degree $ n$. Let $ q_1 \equal{} q^n$ and let $ K$ be the splitting field of $ f(x) \equal{} x^{q_1} \minus{} x$. Now the derivative of $ f$ is $ q_1x^{q_1 \minus{} 1} \minus{} 1 \equal{} \minus{} 1$ as $ q_1$ is divisible by the characteristic of $ \mathbb{F}_q$. Now a splitting field of given polynomial is unique.

Let us show that every extension of $ \mathbb{F}_q$ of degree $ n$ contains $ q_1$ elements. The number of elements in the prime field $ \mathbb{F}_{q_0}$ of $ \mathbb{F}_q$ is some prime $ p$. Now $ \mathbb{F}_{q_0}$ is isomorphic to $ \mathbb{Z}/p\mathbb{Z}$.Then the degree of $ \mathbb{F}_q$ over $ \mathbb{F}_{q_0}$ is $ n$. Let $ \{x_1,\ldots ,x_n\}$ be a base of $ \mathbb{F}_q$ over $ \mathbb{F}_{q_0}$. Then every $ x\in \mathbb{F}_q$ can be expressed of the form
\[ x \equal{} a_1x_1 \plus{} \ldots \plus{} a_nx_n,
\]
where $ a_1,\ldots,a_n\in \mathbb{F}_{q_0}$ and this expression is unique. As one can choose every coefficient $ a_i$, $ i \equal{} 1,\ldots,n$ in $ q_0$ ways, $ \mathbb{F}_q$ has $ q_0^n$ elements.

If $ E$ is an arbitrary field containing the roots of $ f$. Then $ K$ is a subfield of $ E$. If $ K\ne E$, the extension $ E$ of $ \mathbb{F}_q$ has degree greater than $ n$.

I have a few comments.

First, the problem is to show that some extension of order $ q^n$ exists, and also that [i]any[/i] two extensions of that order are isomorphic.  You have definitely not shown the latter, and it is questionable whether you have shown the former.  Your proof should contain the following two lines: "We conclude that the extension field $ K$ has order $ q^n$." and, later, "Therefore, the two extension fields $ K$ and $ L$ are isomorphic."

Second, you use the variable $ n$ to refer to two different things.  First you say that $ q_1 \equal{} q^n$, then you say that $ q \equal{} q_0^n$.

Third, you assert the following: "the degree of $ \mathbb{F}_q$ over $ \mathbb{F}_{q_0}$ is $ n$".  Probably what you mean to say was: "the degree of the splitting field over $ \mathbb{F}_q$ is $ n$".  But you never proved this!  You only stated it.

Fourth, your last paragraph serves no purpose.

In short, I would start at the end of your first paragraph and ask yourself where you are really trying to go with this proof.  You say that the derivative is $ \minus{} 1$.  What can you conclude from this?  What can you say about the relationship between the roots of $ x^{q_1} \minus{} x$ and the splitting field?

Thanks for your comment!
[quote="LydianRain"]You say that the derivative is $ \minus{} 1$.  What can you conclude from this?  What can you say about the relationship between the roots of $ x^{q_1} \minus{} x$ and the splitting field?[/quote]
Because the derivative is $ \minus{}1$, $ x^{q_1}\minus{}x$ has no multiple roots so $ K$ is the field of $ q_1$ elements. What is the next step?

[quote="MeKnowsNothing"]
Because the derivative is $ \minus{} 1$, $ x^{q_1} \minus{} x$ has no multiple roots so $ K$ is the field of $ q_1$ elements. What is the next step?[/quote]

You've already skipped a step!  It's still thinkable that there are elements of $ K$ which are not roots of that polynomial, which would make the size of $ K$ bigger than $ q_1$.  You have to make a convincing argument that this case is not possible.

Also, to be clear: why do you say that the derivative being $ \minus{}1$ implies that there are not multiple roots?

After you've shown this, you're done with the first part of the problem.  For the second part, my advice is to pick an arbitrary extension and ask, "Is this a splitting field?"

Thanks for helping again!
[quote="LydianRain"][quote="MeKnowsNothing"]
Because the derivative is $ \minus{} 1$, $ x^{q_1} \minus{} x$ has no multiple roots so $ K$ is the field of $ q_1$ elements. What is the next step?[/quote]

You've already skipped a step!  It's still thinkable that there are elements of $ K$ which are not roots of that polynomial, which would make the size of $ K$ bigger than $ q_1$.  You have to make a convincing argument that this case is not possible.

Also, to be clear: why do you say that the derivative being $ \minus{} 1$ implies that there are not multiple roots?
[/quote]
It is proven in Lang's Algebra that if $ F$ is a field and $ f\in F[x]$, $ f\ne 0$ identically, then $ a\in F$ is a multiple root of $ F$ if and only if $ f(a)\equal{}0\equal{}f^\prime (a)$.

[quote="LydianRain"]It's still thinkable that there are elements of $ K$ which are not roots of that polynomial, which would make the size of $ K$ bigger than $ q_1$.  You have to make a convincing argument that this case is not possible.
[/quote]
I see. Does it follows from the fact that $ x^{q_1} \minus{} x\equal{}0$ has no multiple roots?

[quote="MeKnowsNothing"][quote="LydianRain"]It's still thinkable that there are elements of $ K$ which are not roots of that polynomial, which would make the size of $ K$ bigger than $ q_1$.  You have to make a convincing argument that this case is not possible.
[/quote]
I see. Does it follows from the fact that $ x^{q_1} \minus{} x \equal{} 0$ has no multiple roots?[/quote]

Yes.  But let me be perfectly clear: you do not have a proof in your hands until you can say, in plain words, [i]why[/i].

Here is some help: Assuming that the polynomial has no multiple roots (which I am not yet convinced that you know how to prove :) ) we see that the polynomial has $ q_1$ distinct roots in the splitting field.  What we want to show is that the splitting field has exactly $ q_1$ elements.  That means that the $ q_1$ rooots would have to form a field, right?

So, if you can show that the set of roots is a field, then it [i]must[/i] be the splitting field. (splitting fields are defined to be "as small as possible", so the splitting field can be no larger)

Now how can you show that the set of roots forms a field?

[quote="LydianRain"]Now how can you show that the set of roots forms a field?[/quote]
Because if $ a\in\mathbb{F}_{p^n}$ then $ a^{p^n}\minus{}a\equal{}0$ and the mapping $ f: \mathbb{F}_{p^n}\to \mathbb{F}_{p^n}$ $ a\mapsto a^{p^n}$ is an automorphism.

[quote="MeKnowsNothing"][quote="LydianRain"]Now how can you show that the set of roots forms a field?[/quote]
Because if $ a\in\mathbb{F}_{p^n}$ then $ a^{p^n} \minus{} a \equal{} 0$ and the mapping $ f: \mathbb{F}_{p^n}\to \mathbb{F}_{p^n}$ $ a\mapsto a^{p^n}$ is an automorphism.[/quote]

Good! (but you should make sure that you know how to prove the last thing you said)

Writing good proofs is tricky, because even if you've solved the problem completely, if you fail to convey that in your writing, it will look like you didn't understand it at all.  Just remember that proofs are for other people to read!

The only piece missing from the answer now (if we treat the assembled totality of the posts thus far as constituting a proof) is that if $ F_q\subset K$ and $ F_q\subset L$ are two extensions of degree $ n$, then $ K$ is isomorphic to $ L$.  Here is the proof: Since $ K$ and $ L$ are splitting fields for $ x^{q_1}\minus{}x$ over $ F$, they are isomorphic by the uniqueness of splitting fields.

This is very important!  This proof relies in a crucial way on another proof, that shows that splitting fields are isomorphic.

I'm still working on the problem.
[quote="LydianRain"]
The only piece missing from the answer now (if we treat the assembled totality of the posts thus far as constituting a proof) is that if $ F_q\subset K$ and $ F_q\subset L$ are two extensions of degree $ n$, then $ K$ is isomorphic to $ L$.  Here is the proof: Since $ K$ and $ L$ are splitting fields for $ x^{q_1} \minus{} x$ over $ F$, they are isomorphic by the uniqueness of splitting fields.[/quote]
If $ K$ is an extension of degree $ n$ of $ \mathbb{F}_q$, why do $ K$ is the splitting field for $ x^{q_1} \minus{} x$ over $ F$?
[quote="LydianRain"]So, if you can show that the set of roots is a field, then it must be the splitting field.[/quote]
What is exactly the set of roots? The polynomial $ x^{q_1}\minus{}x$ may have roots in any field.

[quote="MeKnowsNothing"]If $ K$ is an extension of degree $ n$ of $ \mathbb{F}_q$, why do $ K$ is the splitting field for $ x^{q_1} \minus{} x$ over $ F$?[/quote]
Do you know the theorem that says that $ x^q \equal{} x$ for any $ x$ in a finite field of order $ q$?  It follows from that.

[quote="MeKnowsNothing"]What is exactly the set of roots? The polynomial $ x^{q_1} \minus{} x$ may have roots in any field.[/quote]
At the stage of the proof to which you are referring, we are working explicitly in a splitting field for this polynomial, so it makes sense to talk about the set of roots.

I have still some problems with this exercise and professor won't accept my proof.

1. Why the splitting field of $ x^{q_1} \minus{} x$ is the same as the field formed by the set of roots of $ x^{q_1} \minus{} x$? So why the field formed by the set of roots of $ x^{q_1} \minus{} x$ has no proper subfield in where $ x^{q_1} \minus{} x$ splits completely? And this should be done without using the fact that $ x^{q_1} \minus{} x$ has no multiple roots.

2. If a field $ A$ has $ q^n$ elements and a field $ B$ has $ q$ elements, why $ A$ is an extension of $ B$ with degree $ n$?

[quote="MeKnowsNothing"]I have still some problems with this exercise and professor won't accept my proof.[/quote]
As I said, you should try to concentrate on writing out the details very fully.  It's the only way that somebody else can understand your proof, but it also makes it easier to find holes in your reasoning, and fix them.  You should also keep going back to the definitions, as I get the sense that you don't know what many of the terms that you use, actually mean...
[quote="MeKnowsNothing"]1. Why the splitting field of $ x^{q_1} \minus{} x$ is the same as the field formed by the set of roots of $ x^{q_1} \minus{} x$? So why the field formed by the set of roots of $ x^{q_1} \minus{} x$ has no proper subfield in where $ x^{q_1} \minus{} x$ splits completely? And this should be done without using the fact that $ x^{q_1} \minus{} x$ has no multiple roots.
[/quote]
I don't know how your textbook defines splitting field.  Mine defines it as a minimal field that contains all the roots.  In other words, if you have a field, you have to check two things to figure out that it's a splitting field.  1) It has to have all the roots! and 2) No smaller field can contain all the roots.

The set of roots is your field; it can't get any smaller than that.  Think about it--how could anything smaller than the set of all the roots, contain all the roots? :)

[quote="MeKnowsNothing"]2. If a field $ A$ has $ q^n$ elements and a field $ B$ has $ q$ elements, why $ A$ is an extension of $ B$ with degree $ n$?[/quote]
What is your definition of degree?  My definition is the dimension of the vector space of $ A$ over $ B$.  A counting argument shows you that a vector space of dimension $ l$ will have $ q^l$ elements.

Let me show what I know already. I have used some lemmas I have proven already.
Let $ \mathbb{F}_q$ be a finite field with $ q$ element and $ n\geq 1$ a fixed integer. Then there exists at least on extension of $ \mathbb{F}_q$ whose degree is $ n$. That extension is $ \mathbb{F}_{q^n}$

proof.
Let us show first that there exists at least one extension of $ \mathbb{F}_q$ with degree $ n$. Let $ q_1 \equal{} q^n$ and $ K$ be the set of roots of $ f(x) \equal{} x^{q_1} \minus{} x$ in the splitting field $ L$ of $ f$.

Let us show that $ K$ is a field be showing that it is a subfield of $ L$. As $ 0^{q_1} \minus{} 0 \equal{} 1^{q_1} \minus{} 1 \equal{} 0$, $ K$ contains at least two elements. Let $ x,y,z\in K,z\ne 0$. By the subfield criterion, one has to show that $ x \minus{} y\in K$ and $ xz^{ \minus{} 1}\in K$.

Let us show that every $ x\in \mathbb{F}_{q}$ is a root of $ f$. As the order of the multiplicative group of $ \mathbb{F}_q$ is $ q \minus{} 1$, we have $ x^{q \minus{} 1} \equal{} 1$ for all $ x\in \mathbb{F}_q\setminus \{0\}$. Furthermore $ 0^{q \minus{} 1} \equal{} 0$. Now by induction with respect to $ n$ we get
\[ x^{q^{n \plus{} 1} \minus{} 1} \equal{} x^{q^nq \minus{} 1} \equal{} x^{ \minus{} 1}(x^{q^n})^q \equal{} x^{ \minus{} 1}x^q \equal{} x^{q \minus{} 1} \equal{} 1
\]
for every $ x\ne 0$. Of course, $ 0^{q^{n \plus{} 1} \minus{} 1}$. Thus every $ x\in \mathbb{F}_{q}$ is a root of $ f$.

Now
\[ (x \minus{} y)^{q_1} \equal{} (x \plus{} ( \minus{} y))^{q_1} \equal{} x^{q_1} \plus{} ( \minus{} y)^{q_1}.
\]
If the characteristic of $ K$ is odd we have that $ ( \minus{} y)^{q_1} \equal{} \minus{} y^{q_1}$ so $ (x \minus{} y)^{q_1} \equal{} x^{q_1} \minus{} y^{q_1}$. If the characteristic of $ K$ is even, it is two so $ ( \minus{} y)^{q_1} \equal{} y^{q_1}$ as $ \minus{} 1 \equal{} 1$. In that case we have also $ (x \minus{} y)^{q_1} \equal{} x^{q_1} \minus{} y^{q_1}$. If $ z^{q_1} \equal{} z$ and $ z\not \equal{} 0$ we have $ (z^{ \minus{} 1})^{q_1} \equal{} z^{ \minus{} q_1} \equal{} z^{ \minus{} 1}$. Then $ (xz^{ \minus{} 1})^{q_1} \equal{} x^{q_1}(z^{ \minus{} 1})^{q_1} \equal{} xz^{ \minus{} 1}$. This shows that $ K$ is a field.

As $ K$ is a field and every $ x\in \mathbb{F}_{q}$ is a root of $ f$, we have shown that $ K$ contains $ \mathbb{F}_q$ as a subfield. Thus $ K$ is an extension of $ \mathbb{F}_q$.

Now I have to show that $ f$ has $ q_1$ roots to prove that $ K$ has $ q_1$ elements and thus is an extension of $ \mathbb{F}_q$ of degree $ n$. The derivative of $ f$ is $ q_1x^{q_1 \minus{} 1} \minus{} 1 \equal{} \minus{} 1$ as the characteristic of $ \mathbb{F}_q$ divides $ q_1$. Thus $ f$ has no multiple roots. Now $ f$ has at least $ q$ roots (all $ x\in \mathbb{F}_q$). But why it has $ q_1$ roots? Note that I don't know yet that $ \mathbb{F}_{q_1}$ exists so I can't say that every $ x\in \mathbb{F}_{q_1}$ is a root of $ f$. What I know is that for every prime number $ p$ there exists a field with $ p$ elemens, namely $ \mathbb{Z}/p\mathbb{Z}$ and, of course, the assumption that $ \mathbb{F}_q$ exists. I also know that $ f$ has at most $ q_1$ roots since $ K$ is a field and $ \operatorname{deg}(f) \equal{} q_1$.

I also know the following result:

Let $ F$ be a finite field and $ F_q$ a subfield of $ F$ with $ q$ elements. Let $ n: \equal{} [F: F_q]$. Then the number of elements of $ F$ is $ q^n$. But I don't know if it is helpful as I don't know how to compute $ [F: F_{q}]$ without knowing that $ F$ has $ q^n$ elements.


induction
combinatorics proposed
combinatorics

Prove the following identity:
$\sum_{k=0}^n(-1)^kC_{2n-k}^k2^{2n-2k}=2n+1$.
Try to find a combinatorics using solution.

This is not the combinatorial solution you wanted, but it's the best I managed in $15$ minutes:
Denote $a_m=\sum_{k\in Z} (-1)^kC_{m-k}^k 2^{m-2k}$.
Using the identities $C_a^b=0$ for $0\leq a<b$ and $C_a^b=0$ for $b<0\leq a$ we deduce your sum to find is $a_{2m}$.
Now we have
$a_m+a_{m-2}=\sum (-1)^k C_{m-k}^k 2^{m-2k}+(-1)^{k-1}C_{m-k-1}^{k-1} 2^{m-2k}=
\sum (-1)^k C_{m-k-1}^k 2^{m-2k}=2a_{m-1}$.
(Recall that $C_a^b+C_a^{b-1}=C_{a+1}^b$)
Hence we deduce $a_m-2a_{m-1}+a_{m-2}=0$ and a banal induction to show that $a_m=m+1$ solves your problem

But no one tried to find computational solution. Killer sent two similar problems before, and we managed to find combinatorical solution.
So I was waiting combinatorical approach again...


Find the natural n if the number A=(2^17)+17*(2^12)+(2^n) is a perfect square of an integer

Can you explain your solution analitically? :o

[quote="heartwork"]$n=16$ and $n=20$.
Easily check for $n>20$ by bounding the number between two consequently squares (little bit attention for $n$ - odd)and then check similar for $n < 12$ and finally for $ 12 \leq n \leq 19$ and get also $n=16$.[/quote]

For $n=16$ , $2^{12}(2^5+17+2^4)=2^{12}(65)$ which is not a perfect square
For $n=20$ , $2^{12}(2^5+17+2^8)=2^{12}(305)$ also not a perfect square .... :?

In my opinion , theres not solution

observe that $2^{17}+17(2^{12})+2^n=2^{12}(2^5+17+2^{n-12})=2^{12}(49+2^{n-12})$

If $n> 12$ ,then letting $49+2^{n-12}=(2k+1)^2$  :arrow:  $12+2^{n-14}=k(k+1)$ :arrow: $2^{n-14}=(k+4)(k-3)$ . Since one of $k+1$ or $k-3$ must be odd , it follows that $k-3=1$  :arrow: $k=4$ . When $k=4$ it gives us $n=17$ .

For $n\leq 12$ since any even prefect square modulo 8 $\equiv 0$ . So $n$ must be multiple of $3$ . By trivial for $n=3,6,9,12$ , there are no solution also .
So the only answer is $n=17$  :P

EDIT : Thanks for pointing out where is wrong  :D

I asked someone and he told me that there exist solution, but i didnot find it. Please help me :?

There is one solution: $n=17$.

$2^{17}+2^{16}+2^{12}+2^{17}=2^{12}+2\cdot2^{15}+2^{18}=(2^6+2^9)^2$

For $n>12$, shyong's reasoning works except that the $k=4$ case should not have been ruled out. For $n\le12$, we can directly check $n=0,2,4,6,8,10,12$ and verify that none are solutions.

A question from an oral round of a local competition:
Find a positive integer value of $n$ such that $2^8+2^{12}+2^n$ is a perfect square.

[quote="shyong"]

If $n\geq 12$ ,then letting $49+2^{n-12}=(2k+1)^2$  :arrow:  $12+2^{n-16}=k(k+1)$ [/quote]
You were wrong this step shyong  :D
It's $12+2^{n-14}=k(k+1)$
$\rightarrow 2^{n-14}=(k-3)(k+4)$
Then $k-3=1\rightarrow k=4\rightarrow k+4=8\rightarrow n=17$ and we've done now  :D

Opps , I didnt notice i made a mistake there  :blush:  I will edit it .Thanks for pointing it out badboy :P

I was rushing and i skipped the "17" 1st time... definitely this is not hard. :(


If

$1$ , $x$ , $y$ , $\frac{8}{27}$ , $\frac{16}{81}$

is a geometric series, find $x$ and $y$

[hide]x= 2/3 y= 4/9 multiply by 2/3[/hide]


A square of side $n$ is formed from $n^2$ unit squares, each colored in red, yellow or green. Find minimal $n$, such that for each coloring, there exists a line and a column with at least 3 unit squares of the same color (on the same line or column).

[hide]Is it just $n=7$?  $n=7$ works because of the pigeonhole principle.  For any $n$ smaller than this you can just make each row go RYG and cycle through that.[/hide]

What if it's the minimal n for which you will always have either 3 consecutive of the same color in a row or column or 3 consecutive of different colors?

I haven't actually thought about this one (is it even finite?) but seems like an interesting question.

Edit--OK, it doesn't work... but what if you have the condition that every 2x2 subsquare has all 3 colors?

[img]http://www.jia-han.net/images/math.png[/img]

6 doesn't work, 7 does through pigeonhole principle.


Why is standard deviation more commonly used than mean deviation, when in fact, the formula for mean deviation is easier to use than standard deviation?

This is essentially because while mean deviation has a natural intuitive definition as the "mean deviation from the mean," the introduction of the absolute value makes analytical calculations using this statistic much more complicated than the standard deviation.

This is wrong.  The real reason is the appearance of the standard deviation (and not the mean deviation) in the [url=http://en.wikipedia.org/wiki/Central_limit_theorem]central limit theorem[/url], which is one of the two most important theorems in probability theory, and also that the standard deviation is better when sampling from normally distributed populations.  There is no corresponding important result which uses the mean absolute deviation.


function
calculus
integration
algebra
domain
real analysis
real analysis unsolved

I need some examples of functions which are differentiable but not integrable..

If it's differentiable, it's continuous. If it's continuous, it's locally integrable. The only sense in which integrability can fail is that the integral on an infinite domain might not converge.

In other words, $ f(x)\equal{}1$ is the example you're looking for. If that isn't it, you're looking for omething that doesn't exist.

[quote="sos440"]If you are searching for a differentiable function whose 'derivative' is not (Riemann) integrable, then refer to [url=http://en.wikipedia.org/wiki/Volterra_function]here[/url].[/quote]

Thank u very much//SOS 440// Volterra's function is what is was searching for!!!


Hola amigos de este foro, les propongo el siguiente problema de geo,..

Sean P y C sobre el semicirculo con diametro AB, tal que los arcos BC y CD son iguales.
Si AC y BP se intersectan en E y AD y CP se intersectan en F, pruebe que EF es perpendicular a AD, haciendo uso de numeros complejos. :)
Bueno, suerte y espero sus respuestas...

D es el punto que hace BC y CD igual y por eso C es mas cerca de B.  P es variable y creo que necesitamos considerar 3 casos: si los puntos en orden son BCDPA, BCPDA, y BPCDA porque en el primer caso, E y F son adentros del semicirculo, en caso 2 E es adentro, y en caso 3, E y F son afueras.

No estoy seguro, pero creo que mi interpretacion es correcto porque parece verdadero en mis dibujos.  Pero ya no lo he probado  :oops:

EDIT: ay, lo siento, no es necesario que C es mas cerca de B.  Si es mas cerca de A, necesitamos extender el circulo, pero todavia es cierto.

[quote="nicoopm"]Sean P y C sobre el semicirculo con diametro AB, tal que los arcos BC y CD son iguales. Si AC y BP se intersectan en E y AD y CP se intersectan en F, pruebe que EF es perpendicular a AD.[/quote]
Solo basta ver que como los arcos $ BC$ y $ DC$ son iguales, entonces los ángulos inscritos $ \angle CPB$ y $ \angle DAC$ son iguales, esto es $ \angle EPF \equal{} \angle EAF$ $ \Longrightarrow$ el cuadrilátero $ PFEA$ es cíclico $ \Longrightarrow$ $ \angle AFE \equal{} \angle APE \equal{} 90^{\circ},$ i.e. $ EF \perp AD.$


Given p>5 is a prime, prove:
there exists no less than two positive integers $ 1\leq a,b\leq p-1$ such that
$ a^{p-1}\equiv 1(\mod p^{2}),b^{p-1}\equiv 1(\mod p^{2})$

It fails for p=7, where the only possibility is 1.


I have Mathematica at home and can use a school graphing calculator at school. Should I still buy a graphing calculator? If so, which model?

Depends what it's for.  If you're permitted to use a graphing calculator in your classes, you might as well get one since the teacher will probably design exams that utilize a graphing calculator.  If it is strictly optional, it's your call.  In high school, we were all required to get a graphing calculator but in college, almost no classes allow calculators on exams, and they are usually not much help.  I currently use a graphing calculator only to perform quick arithmetic, graph functions, and check work-I don't use it as a problem solving technique.

So in short, it's up to you.  I like the TI series  I think they start somewhere around $ \$$80-90, although I could be wrong.  The 89 and the 92 are extremely powerful, but their costs are quite high.  If you just need simple graphing and algebra capabilities, an 80, 83+, or 84+ will all work fine (I think those are the model numbers).

How often would you be away from your computer and need to graph something with greater accuracy than a sketch by hand?

I feel like graphing calculators are only useful inside the classroom, for tests, class exercises, and the AP exams. It's your call -- if you can get by at school with one and can find one to use for the AP exam, don't bother! Just get a cheap scientific calculator for a few bucks.

If you must buy one, peers and eBay sell TI-83+'s cheap. Some people like the HP graphers that use Reverse Polish Notation.

[quote="JRav"]The 89 and the 92 are extremely powerful, but their costs are quite high.[/quote]

Are you [i]shunning[/i] the $ n$spire??? :P

Graphing calculators are useful for, well, graphing. I have no idea if they can program or not. I have a TI $ n$spire (scientific and graphing, even though I only use the TI-84 mode) so I wouldn't know.
Now tat the AMC 10/12 don't allow calculators I have a sinking feeling that my calculator is going to collect a lot of dust. :P

I don't think you should get one, you already have Mathematica.

Most graphing calculators are $ \$$100 bricks with 10-30 year old technology and public domain code, imo.

But they're $ \$$100 bricks that you can use on tests! Yay!

 :|

As someone has said, Ebay is great. I bought an 84 silver a long time ago, when I didn't know anything. Then I lost it, so I went on Ebay and ordered the cheapest calculator I could find, a TI-86, 35 dollars shipped. Works great for annoying computation in physics and chem. That is the only use I have found for my calculator, as well as when the teacher tells you to do something with the calculator in class.

Sidenote, a pitch for the 86: I believe it's the ideal calculator for Mathcounts because functions can be accessed with one button press from menus.


If a number $ X$ consists of 66 digits of threes, and a number $ Y$ consists of 66 digits of sixes, name all the digits that will appear in the product of $ X$ and $ Y$.

[hide]3*6 = 18
33*66 = 2178
333*666 = 221788
We can accurately predict that for an equal number of 3's and 6's above 1, the only digits that appear will be 1,2,7, and 8.[/hide]


Let $ g: \mathbb{R}\rightarrow \mathbb{R}_ \plus{}$, nondecreasing on $ \mathbb{R}_\plus{}$, $ g(x) \equal{} g( \minus{} x)$. Let $ X$ be a random variable such that $ |X(\omega)|\leq C,\forall \omega \in \Omega, C > 0$. Then for $ \varepsilon > 0$

$ \frac {\mathbb{E}[g(X)] \minus{} g(\varepsilon)}{g(C)}\leq \mathbb{P}(|X \minus{} \mathbb{E}[X]|\geq \varepsilon)\leq \frac {\mathbb{E}[g(X \minus{} \mathbb{E}[X])]}{g(\varepsilon)}$

I have proved right inequality and I would be grateful for some tips who to prove the left one.

[quote="Mr.Doe"]Let $ g: \mathbb{R}\rightarrow \mathbb{R}_ \plus{}$, nondecreasing, $ g(x) \equal{} g( \minus{} x)$ .[/quote]
strange isn't it :maybe:

And what is $ C$ ? And what if $ P(|X \minus{} E[X]| > \epsilon) \equal{} 0$ ?

Thank you for pointing out some of the errors, I have corrected them. The problem doesn't say anything in case X is a degenerate variable, and I didn't even thought about it  :blush: . I hope everything is ok now.


after i registered for SATII test, can i change my mind, and take another additional test on the test day?

Yes.  The College Board website is a good place to go for questions like this.


Find the ten-thousandths digit of the following series:

$ 1,4,8,\frac{32}{3},...$

(A) 8
(B) 9
(C) 0
(D) 1
(E) 2

I assume this is asking to find the ten-thousands digit of the [b]sum[/b] of the given series.  I have two questions:  1.  Is this a divergent series (because the ratios are greater than 1) that wouldn't have a sum? 2.  How do I find the next terms of this series (what is the pattern)?

The question is just awful -- poorly conceived and poorly implemented.  It is, however, possible to guess the pattern that the sequence follows: look at the ratios between consecutive terms.


A math teacher requires Noelle to do one homework assignment for each of the first five homework points she wants to earn; for each of the next five homework points, she needs to do two homework assignments; and so on, so that to earn the $ n^{\text{th}}$ homework point, she has to do $ n\div5$ (rounded up) homework assignments. For example, when she has 11 points, it will take $ 12\div5\equal{}2.4\rightarrow3$ homework assignments to earn her $ 12^{\text{th}}$ point. What is the smallest number of homework assignments necessary to earn a total of 25 homework points?


Well , I just had a doubt with the IUPAC nomenclature....Is the name of the compound in the slide that I've attached
$4,4-diethyl-3-methyl-5-Heptyne$  or $4,4-Diethyl-5-methyl-2-Heptyne$ ??

I can't see any compound... the first name you listed is incorrect.  The proper name for that compound is 3,3-diethyl-4-methyl-1-heptyne (or, using the revised IUPAC rules, 3,3-diethyl-4-methyl-hept-1-yne, although that correct form is less common in most current textbooks and classes).

Since this gives a completely different compound than the other name, I can't really tell which is correct.  Here are the chemical formulas for each one:

3,3-diethyl-4-methyl-1-heptyne: $\text{CH}\equiv\text{C}-\text{C(C}_2\text{H}_5\text{)}_2-\text{CH(CH}_3\text{)}-\text{CH}_2-\text{CH}_3$

4,4-diethyl-5-methyl-2-heptyne: $\text{CH}_3-\text{C}\equiv\text{C}-\text{C(C}_2\text{H}_5\text{)}_2-\text{CH(CH}_3\text{)}-\text{CH}_3$


algebra
polynomial
Vieta
geometric sequence

Source: math olympiad treasure
Find $m$ and solve the following equation, knowing that its roots form a geometric sequence:
$x^{4}-15x^{3}+70x^{2}-120x+m=0$.

Well, 64 only has only so many factors, and they form a geometric sequence... I was thinking of just synthetic-dividing the whole thing

[quote="srulikbd"]nice! \begin{eqnarray*}x^{4}-15x^{3}+70x^{2}-120x+64=0 &\iff& x^{4}-3x^{3}+2x^{2}-12x^{3}+36x^{2}-24x+32x^{2}-96x+64=0\\ &\iff& x^{2}(x^{2}-3x+2)-12x(x^{2}-3x+2)+32(x^{2}-3x+2)=0\\ &\iff& (x^{2}-12x+32)(x^{2}-3x+2)=0\end{eqnarray*}
how did you factor it? is there any method or just trial and error?[/quote]

Basically, it was kind of obvious what the roots were: $a\cdot aq^{3}=aq\cdot aq^{2}$, and $a\cdot aq\cdot aq^{2}\cdot aq^{3}=64$. Therefore the product of the smallest and the largest one is $8$, as well as the product of the other two. From there you very easily see that the roots must be $1,2,4,8$, and the "factorization" is just writing that out.


function
modular arithmetic
algorithm
number theory
greatest common divisor
floor function

Prove that there is no solution to $72x+23y=845$ in $\mathbb{N}_+^2$.

Notice that $845=21^2\cdot 2+23$
y must be odd
solve the equation in function of x

Or with some modular trickery:

$845 \equiv 53 \pmod {72}$

So it must be true that $23y \equiv 53 \pmod {72}$

Using the euclidian algoritm, we find the modular inverse: $47$.
So: \[ y \equiv 43 \pmod {72} \]

Since we are dealing with $\mathbb{N}^2_+$, we don't have to consider negative $y$.
The smallest value for $y$ is $43$, giving \[ 72x+989=845 \]
Since $x \geq 0$, this equation has no solutions.

The same method as Jan, except take modulo 23.  You get 3x  = 17 mod 23, where it is easier to find $x\ge 21$.  Contradiction.

here is a method without mod but i think the concept are still the same ...

rewrite it as

$23y=845-72x$

$\implies y=36-3x+\frac{17-3x}{23}$

Since $y$ is integer , $\frac{17-3x}{23}$ must also be an integer . By inspection we know that $x=-2$ gives an integer value . And for this value of $x$ , we can get $y=43$

Now we know that one of the solution is $(x,y)=(-2,43)$

and we also know that for $ax+by=c$ , if $x_0,y_0$ is the solution , then for any integer $k$ , $x_0+bk,y_0-ak$ is also the solution .

From here you can see that $43-72=-29<0$ or $-2-23=-25<0$ . Both give you negative value . Hence there are no solution for potive integer $x,y$

[quote="gdeskos"]We  know  that  a  diofantic  equation ax+by = c   have  intigers  solutons  if  and  only  if  the  c  divides    the       gcd(a,b)  .  The  proof   simple  . Let  d= gcd ( a, b)  then  there  are   p,q in N  with  gcd (p,q ) = 1  for  which  it holds  that     a = p*d   and  b = q * d  so  dxp + bqd =c  so  c /d   and  if  c/d  then  the  eqyation  has  intiger solution  . For  the  eqyation  72 x + 23 y = 845  we  obtain that  gcd(72,23)=1  so  because  845 dont  divide  1  the  equation  has  no  solution  in  N ^2[/quote]

Not true ..example

$3x+4y=7$ . Although $(3,4)=1$ and $7$ doesnt divide $1$ , we still have solution $(x,y)=(1,1)$

In the other hand, given the equation $ax+by=c$, where $a, b$ and $c$ are positive integers and $\gcd(a,b)=1$, the number of positive integer solutions of the equation is $\left\lfloor\frac{c}{ab}\right\rfloor$ (Prove that).

Hence, $\left\lfloor\frac{845}{72\cdot23}\right\rfloor=0$ and we are donne.

[quote="probability1.01"]Err what about 3x + 4y = 7 again?[/quote]

Forget my last post. :blush:

:blush: Forgot  my  last  reply .  The  theorem  is  like  that  .The  equation  ax + by = c  has  solutions  in  Z  if  and  only  if  gcd(a,b)  / c . But  this  situation  holds  because  gcd( 72 , 23) =1  and  c = 845  then  1/  845 . So  the  equation  has  intiger   solutions  and  not  solution  in  N^2 .

Let  ( p,q )  is  a  solution  of  the  equation  ax + by = c  . Then  the  solution  of  the  equation  is  in  the  form  of  x = p + (b/d)* t  and  y = q - (a/d)* t  where  d=gcd(a,b)  and  t is in  Z .

@Jan:

[quote]Using the euclidian algoritm, we find the modular inverse: 47.
So:[/quote]

What's the Euclidian Algorithm  :?


My Algebra II/Trig teacher gave us a problem set. Here is the "example" problem:

A factory manufactures chairs and tables, each requiring the use of three operations: Cutting, Assembly, and Finishing. The first operation can be used at most $40$ hours, the second at most $42$ hours, and the third at most $25$ hours. A chair requires $1$ hr. of cutting, $2$ hrs. of assembly, and $1$ hr. of finishing. A table needs $2$ hrs. of cutting, $1$ hr. of assembly, and $1$ hr. of finishing. If the profit is $\$20$ per unit for a chair and $\$30$ for a table, how many units of each should be manufactured to maximize revenue?

Here was the objective equation ($C$=chair, $T$=table): $20C+30T$.

Here are the constraints:
$C+2T\leq{40}$
$2C+T\leq{42}$
$C+T\leq{25}$
$C\geq{0}$
$T\geq{0}$

The teacher told us to graph the region, locate the vertices of the feasibility region, and figure out which vertex maximizes profit.

I tried "solving" it as a system of equations, but my method did not work. Why?

Well I think your main issue is that given $A\leq B$ and $C\leq D$, you [b]can't[/b] say that $A-C\leq B-D$. All you can say is that $A+C\leq B+D \implies A-B\leq D-C$

What I'm trying to say is that you probably saw that:
$2C+T\leq 40$
$C+T\leq 25$ and then tried to say:
$C\leq 15$
but that isn't true; $C=16$ works perfectly...

This is the problem of Linear Programming Method.

[quote="vishalarul"]I tried "solving" it as a system of [b]equations[/b], but my method did not work. Why?[/quote]

Simply because they are not [b]equations![/b]  That is, $\le, \ge$ are not equivalence relations.

Hi !

We teach that in France in 11th class. Do you want some good courses ? :maybe:

I got the vertices as (0,0) (21,0) (0,20) nd (44/3,38/3)..Hence the maximum value gets obtained at C=0 T=20 which 600..Was that the answer???

See this one :
[url]http://people.hofstra.edu/faculty/Stefan_waner/RealWorld/Summary4.html[/url]

I learned the linear programming method in algebra 2... which is about 9th-10th grade for the U.S.

Did you learn the inner product of vector?

The answer is $650$ units when $(C,\ T)=(10,\ 15).$

[quote="me@home"]inner product of vector? i'm not exactly sure what you mean...
anyway thats a great idea to use vectors in this kind of problem.. i haven't thought of that before... let me work it out now![/quote]

It was dot product or scalar product. :)


complex numbers
algebra unsolved
algebra

P is an odd prime. The integers a, b, c, d are not multiples of p and for any integer n not a multiple of p, we have {na/p} + {nb/p} + {nc/p} + {nd/p} = 2, where { } denotes the fractional part. Show that we can find at least two pairs from a, b, c, d whose sum is divisible by p.

But Don't use complex to solve

It has a solution with complex numbers? Show it!

I never do it ;) This a problem solving forum, but NOT "read it on kalva" forum!

It's your problem. I find it completely senseless to ask for a solution using complex numbers and to refuse to look at some other source. Anyway, you are right, this is your forum, so why am I bothering?

You will never understand my pedagogical reasons. Sadly.

I would rather say fortunately. It means I would spend my life and energy on math forums, which is the least thing I want. Anyway, please delete this message after that, I start spaming.

[quote="harazi"]It means I would spend my life and energy on math forums, which is the least thing I want.[/quote]

Dont say that harazi , many people have learned countless things from you , .this

http://www.mathlinks.ro/Forum/viewtopic.php?t=22364&highlight=integrate&start=20 is a way good enough example of your contributions.

[quote]I never do it  This a problem solving forum, but NOT "read it on kalva" forum

 you are right, this is your forum, so why am I bothering?

You will never understand my pedagogical reasons. Sadly[/quote]

Cut it guys . you are just speaking about REFERING TO ANOTHER SOLUTION and see how you are reacting .
Anyone who likes , go look at the other source and who doesnt like, dont . OK now start agian ,Say  " YOU just cant  understand My reasons"


Given any positive integer $n$ , show that there are two positive rational numbers $a$ and $b$ , $a \not= b$, which are not integers and which are such that $a - b, a^2 - b^2 , \ldots a^n - b^n$ are all integers.

[hide]$a - b | a^n - b^n$ for every positive integer $n$.

So we can put $a = \frac{x}{z}$ and $b = \frac{y}{z}$.

Hence $a^n - b^n = \frac{x^n - y^n}{z^n}$.

Now we can find a number $z^n$ that divides $x - y$.

An example can be $x = 2005^n + 1$, $y = 1$ and $z^n = 2005$.

So $a = \frac{2005^n + 1}{2005}$ and $b = \frac{1}{2005}$.[/hide]

Shouldnt it be something like:
[hide]$x=2^n+1,y=1,z=2$, or anything else in the place of 2 (3,4,5,...)?[/hide]

Take $a=\frac{m}{n}$ and $b=\frac{1}{n}$.

We have to find suitable $m$ such that $\begin{cases} n\mid m-1\\ n^2\mid m^2-1\\ \cdots \\ n^n\mid m^n-1\end{cases}$

Taking $m=n^n+1$ suffices as $n^n\mid m-1\mid m^i-1$ for any $1\le i\le n$.

So, $(a,b)=\left(\frac{n^n+1}{n}, \frac{1}{n}\right)$.

taking $a=2^n+\frac{1}{2}$ and $b=1/2$ does the job! :D

a=(x^n+1)/x and b=1/x does the job in fact (x is an integer)
Bye


[quote="yangml"]Can you send your test in English![/quote]
Who are you asking? What are you asking for? :huh:


[quote="Lovasz"]\[ a,b,c \in [0,1] \Rightarrow a^2+b^2+c^2 \ge 1+a^2b+b^2c+c^2a  \][/quote]
Maybe $a^2+b^2+c^2 \le 1+a^2b+b^2c+c^2a$ $?$ ;)
This is a very old inequality.

What about these ones:
$a,b,c \in [0,1]$
\[ \Rightarrow 2(a^3+b^3+c^3) \le 3+a^2b+b^2c+c^2a  \]
\[ \Rightarrow \frac{a}{bc+1}+\frac{b}{ca+1}+\frac{c}{ab+1} \le 2  \]

Can you post solution for these ones, i don't know what to do with the problems like these.
By the way, could you show me some methods to kill them?
Merci.

[quote="Lovasz"]\[ \Rightarrow \frac{a}{bc+1}+\frac{b}{ca+1}+\frac{c}{ab+1} \le 2  \][/quote]

This one is also well-known.

$LHS\leq \frac{a+b+c}{abc+1}$, so it remains to prove:
$a+b+c-2abc\leq 2$, and it is enough to check $(0,0,0), (0,0,1),(0,1,1),(1,1,1)$ and all its permutations (why?)

The other one I don't know about...

[quote="Lovasz"]\[ a,b,c \in [0,1] \Rightarrow a^2+b^2+c^2 \le 1+a^2b+b^2c+c^2a  \][/quote]

$f(a)=a^2(b-1)+ac^2+(b^2c+1-b^2-c^2)$

because b-1<0 We have minimum of $f(a)$ on 0 or 1. By the analogy $b,c=0 or 1$ We check all the combination and we will see that this inequality is true :).


To solve the equations
\begin{align*}
2a+3b+c &= 11\\
6ab+2ac+3bc &= 24\\
abc &= -6
\end{align*}
Volume 2 says that I can create a polynomial and solve for its roots.  However, I don't get how to force these equations to be symmetric roots.  Could someone please explain this part?

Let $ 2a \equal{} k$ and $ 3b \equal{} m$
Then you have
$ k\plus{}m\plus{}c \equal{} 11$
$ km \plus{} kc \plus{} mc \equal{} 24$
$ kmc \equal{} \minus{}1$


geometry
3D geometry
calculus
calculus computations

A cone is made for a circular sheet of paper having radius r, by cutting out a sector and gluing the cut edges of the remaining piece together. What is the maximum volume attainable for the cone.


inequalities
logarithms
inequalities unsolved

For $x,y,z$ positive real numbers such that $x+y+z=3$ prove that

$\displaystyle
x^ry^r + y^rz^r + z^rx^r \leq \max(3, 3^{2r} / 2^{2r})$

On CRUX Vasc proved this one:

For $x,y,z$ positive real numbers and $m=\frac{ln9-ln4}{ln3}$ we have

$(\frac{x^m+y^m+z^m}{3})^2 \geq (\frac{xy+yz+zx}{3})^m$

and he said this one is the same as our ineq.

I really don't see how the two inequalities are related :?  :?

take $a=x^r,b=y^4,c=z^r$. then $x^ry^r+x^rz^r+y^rz^r=ab+ac+bc\leq 3\left(\frac{a^{1/r}+b^{1/r}+c^{1/r}}{3}\right)^{2r}=3$ if $r\leq \frac{\ln 3}{\ln 9-\ln 4}$ by the other one(by power means, we see that it works for $m\geq \frac{\ln 9-\ln 4}{\ln 3}$. i don't get the time to post what happens the other way round, wait a minute :D

Peter, are you running at the speed of light ?

I say that as your minute is very dilated in time according to Einstein Relativity Theory, in fact it is again passing.

Naturally, I am joking  ;)

uhm yeah, the other way round is just power means as well. to see why we explicitly calculate for $r=\frac{\ln 3}{\ln 9 - \ln 4}$ observe that for this we obtain equality $3=\frac{3^{2r}}{2^{2r}}$.

Dear Peter, can you please detail better your post.

I think you mean that:

[quote]If we put $a=x^r$,$b=y^r$and$c=z^r$ we have two cases.[/quote]

[quote]First case: $mr \leq 1$

$\displaystyle x^ry^r+x^rz^r+y^rz^r=ab+ac+bc\leq 3\left(\frac{a^{rm}+b^{mr}+c^{mr}}{3}\right)^{2/m} \leq 3\left(\frac{a+b+c}{3}\right)^{2r} = 3$ [/quote]

[quote]Second case: $mr > 1$

 $\displaystyle x^ry^r+x^rz^r+y^rz^r=ab+ac+bc\leq 3\left(\frac{a^{rm}+b^{mr}+c^{mr}}{3}\right)^{2/m}  \leq = ?????$[/quote]

Thank you very much. :)

[quote="chakding"]is the original ineq. correct? it doesnt seem to work for x=y=z=1;

maybe the RHS should read MIN(3,3^2r/2^2r)?[/quote]

it works for x=y=z=1 - try to check it once more

for those who didn't understand my previous posts :lol:

it is easily seen that vasc's inequality is equivalent to the case of $r=\frac 1m$ of our inequality - and the case $r=\frac 1m$ is the case where $3=\frac{3^{2r}}{2^{2r}}$.

now, for arbritary $r$ one can use one the one hand that if $r\leq \frac 1m$, then

$x^ry^r+x^rz^r+y^rz^r\leq 3(\frac{x^{\frac 1m}+y^{\frac 1m}+z^{\frac 1m}}{3})^{rm}\leq 3$

and for $r\geq \frac 1m$ we have

$x^ry^r+x^rz^r+y^rz^r\leq (x^{\frac 1m}+y^{\frac 1m}+z^{\frac 1m})^{rm}\leq frac{3^{2r}}{2^{2r}}$.

each by the case of $r=\frac 1m$. therefore it suffices to prove vasc's inequality.