geometry
geometric transformation
CMU
college

Do you have to take a foreign language when you are a math major?  I am not 100% positive I want to be a math major, but would you suggest sticking w/French (I have taken 6 years of it), or starting a new language?  I am not really interested in French, or really good at it (my grammar is really lacking), but I suppose that after 6 years I can't be that bad.

Most bachelor's degree programs in mathematics don't have a specific language requirement, although there are exceptions. Check the catalogs of the schools you're interested in. (On-line for nearly everyone.) Requirements aside, should you study a language? Two answers:

If you're planning to teach, be it middle school, high school, or even possibly community college or college, it could be quite useful to have a speaking knowledge of a language that many of your students and their parents speak. In most parts of the United States, that's an argument for Spanish, but there are many other languages that this could apply to, including such languages as Vietnamese or Tagalog.

If you're planning to get a Ph.D. and be a research mathematician, you may need to be able to read mathematical papers in other languages. It's traditional for Ph.D. programs to have a language proficiency requirement in two languages, although I don't know how things may have changed recently. The proficiency exam often simply involves producing a translation of a technical paper, something that is quite a bit easier than translating literature, as long as you are familiar with the subject of the paper and its terminology. The three main languages for this requirement are French, German, and Russian. Of these, French is usually considered the easiest - the "big words" of a mathematical paper will be the Latin/Greek words you're familiar with. German is a little more likely to create new words out of German roots, and you have to get used to German word order. Russian is written in a different alphabet. You probably already know more than enough French. Perhaps you could consider one of the other languages.

colleges that require a foreign language for graduation require it for everyone, regardless of what they are majoring in. Frequently you can fulfill this requirement before you start college by getting a particular score on the SAT II or AP tests. Go to the websites of colleges you are interested in and see what the foreign language requirement is. Unless you have a burning interest in a particular language other than French, I would stick with the language you've already studied until you have the requirement met.

Thanks for the replies!  I don't think CMU has a language requirement, but I could be mistaken.

[quote="texas137"]colleges that require a foreign language for graduation require it for everyone, regardless of what they are majoring in. [/quote]

While you are technically correct, your statement is ambiguous and in some cases in incorrect.

First, let's clarify that a university and a college are different things... many colleges exist within a university and different majors are given by different colleges. At most universities, the Mathematics degree is given through the Liberal Arts College. Almost all liberall arts colleges I've heard of require foreign language for all majors. This is not for all, but it's for most.

Most of the time colleges will have general requirements for all majors within the college and foreign language is usually one of them. HOWEVER, there are exceptions... for example, at the University of Florida, the business college has a general foreign language requirement for all majors with the sole exception of finance. A business finance major doesn't have to take foreign language while every other business major does. So not all majors within a college have the same requirements for foreign language. I trust this trend is prolific among other universities.

With 6 years of language, I would be astounded if you were required to continue taking a foreign language at college.  If you think you might go to grad school, as Kent said, you will be expected to be able to read math in some non-empty subset of {French, German, Russian}.  You may be able to get away with only one of those nowadays, rather than 2.

Reading math is actually pretty easy -- if you pick up a textbook in a foreign language, you can recognize a lot of words right off the bat.  This is true of French, German, and even Russian textbooks once you've learned Cyrillic.  (I sort-of know the Cyrillic alphabet, although I don't speak any language that uses it, and I can figure out the titles of many of the Russian-language math textbooks I've picked up in our library.)

If you're not sure you're going to major in math, of course, talking about math graduate school is a bit premature.  There are plenty of other reasons to take language classes, though -- even as a mediocre French student (and I mean that seriously -- in the last 4 semesters of French classes that I have taken, I have amassed an unimpressive B- average), I find learning languages is a lot of fun, and you might want to do it just for that reason.  Also, you might think about doing a study-abroad some semester or over a summer, in which case it would be nice to know something of the language in advance.


$ \{a_i\}_{i \equal{} 1}^{\infty}$ is an infinite sequence of natural numbers; $ a_1 \equal{} 1$.

 Let $ d$ is the largest odd divisor of $ n$. If $ d \equal{} 4k \plus{} 1$, $ \Rightarrow a_n \equal{} a_{n \minus{} 1} \plus{} 1$.
If $ d \equal{} 4k \plus{} 3$, $ \Rightarrow a_n \equal{} a_{n \minus{} 1} \minus{} 1$. Prove that in the sequence is written every natural number infinitely many times.

Your sequence is not correct...

Proof idea:
- Show that $ a_{2^n \minus{} 1} \equal{} 1$, $ a_{2^n} \equal{} 2$.
- Show that $ a_{2^n \plus{} i} \equal{} 2 \plus{} a_{i}$ for $ 1 \leq i \leq 2^{n \minus{} 1} \minus{} 1$.
- Conclude.

alterrnatively you recognise that your sequence counts the number of 1's when n is written in Gray code. And then conclude by saying that with n 1's we can write infinitely many numbers in Gray code.  :)


linear algebra
matrix
linear algebra unsolved

Find a Jordan canonical form for an 8x8 matrix A such that (A-I) has nullity 2, (A-I)^2 has nullity 4, (A-I)^k has nullity 5 for k ≥ 3 and (A-2I)^j has nullity 3 for j ≥ 1. I is the identity matrix.

thank u in advance

Can you explain how did u find it ? thank u in Advance

If a matrix $A$ has an $m\times m$ Jordan block with eigenvalue $\lambda,$ then this particular block contributes $1$ to the nullity of $A-\lambda I,$ $2$ to the nullity of $(A-\lambda)^2,$ and so on, until you get to $m.$ All in all, this block contributes $\min(k,m)$ to the nullity of $(A-\lambda)^k.$

The rest is just following the consequences of this. It's clear that we need more than one block for the eigenvalue $1,$ and so forth.


Let $ \triangle ABC$, let $ BF \perp AC$ and $ CE \perp AB$ as picture. Let $ \angle BAC \equal{} \alpha$ and area of $ \triangle ABC \equal{} a^2$. Find area of $ \triangle AEF$.

[hide="solution"]
Let the foot of the perpendicular from $ F$ to $ AB$ be $ G$.

Triangles $ AFP$ and $ ACE$ are similar for having equal angles. So
$ \frac{FP}{CE}\equal{}\frac{AF}{AC}\equal{}\frac{AB\cos(\alpha)}{AC}\Rightarrow FP\equal{}(\frac{AB\cos(\alpha)}{AC})CE$

On the other side
$ \frac{AE}{AB}\equal{}\frac{AC\cos(\alpha)}{AB}\Rightarrow AE\equal{}(\frac{AC\cos(\alpha)}{AB})AB$

We have
$ (AFE)\equal{}\frac{1}{2}AE*FP\equal{}\frac{1}{2}\cdot (\frac{AB\cos(\alpha)}{AC})CE\cdot (\frac{AC\cos(\alpha)}{AB})AB\equal{}\frac{1}{2}\cdot\cos^2(\alpha)\cdot AB*CE\equal{}\cos^2(\alpha)(ABC)\equal{}(a\cos(\alpha))^2$
[/hide]

 :)

This is my solution:

$ \bullet$ We have $ AE \equal{} AC.cos\alpha$ and $ AF \equal{} AB.cos\alpha$.

$ \bullet$ $ S_{\triangle AEF} \equal{} \frac {1}{2}AE.AF.sin\alpha \equal{} \frac {1}{2}AC.cos\alpha.AB.cos\alpha.sin\alpha \equal{}$ $ \cos^2\alpha.\frac {1}{2}AB.AC.sin\alpha \equal{}$ $ \boxed {\cos^2\alpha.S}$  .

[quote="thanhnam2902"] $ \boxed {\cos^2\alpha.S}$  .[/quote]

what is $ S$ ??

If $ S\equal{}a$
then u r totally wrong because Area have two dimensions.

my answer is $ a^2\cos^2\alpha$


what is the value of the three digit number HEE if

$ .$  A H
_[u]+   A[/u]
H E E

ignore the dot

[hide]100. Basically since the E is in the 10s digit, it must be one more than A. Then you look at how H must be one since it's impossible for HEE to be 200+.[/hide]

From $ AH\plus{}A\equal{}HEE$  we can find that$ A\equal{}9$ and  $ 10> H>1$ or $ H\equal{}1$
Than is easy  to conclude that ($ 9H\plus{}9\equal{}HEE$ from this $ 91\plus{}9\equal{}100$

$ A\equal{}9$

$ H\equal{}1$

$ E\equal{}0$

[hide="Solution"]A = 9, because that is the only way for a 2 digit number plus a one digit number to equal a three digit number. H = 1 because it is impossible that a 2 digit number plus a one digit number $ \geq  200$. So we have 91 + 9 = 1EE. Of course, we just calculate 91 + 9, and we find that HEE = 100.[/hide]


[quote="toetoe"]hi,

[u]problem:[/u]

let a and b be two positiv integers such that $gcd(a,b)=1$

$S ={0,1,\dots,ab-a-b}$

prove that only the half of S can be written in the following

form: $ax+by$ such that $x,y$ are intergers.[/quote]
This may be an old problem.
First we know that $ab-a-b$can not be written in the form of $ax+by$(otherwise ,suppose $ax+by=ab-a-b$ then $x\equiv-1(\mod b),y\equiv-1(\mod a)$,then $ax+by>ab-a-b$)
secondly,if $t$can be written in the form of $ax+by$,then $ab-a-b$can not be written in the form of $ax+by$(otherwise $ab-a-b=(ab-a-b-t)+t$ can be written
thirdly,if $t$cannot be written in the form of $ax+by$
then suppose $t=am+bn$where $0\leq m<a-1,n<0$ then $ab-a-b-t=a(b-1-m)+(-n-1)b$ can be written in the form of$ax+by$

We know that if $\ gcd(a,b)=d$
then there are two positive integers $\ (x,y)$ such that:

$\ ax+by=d$

so in this case $\ d=1$ and we have:

$\ ax+by=1$ so we can define any integer in this form:

just let $\ x_{n}=nx$ and $\ y_{n}=ny$ and we have all the integers that we want regardless of $\ S$

Sorry,I didn't read the toetoe's post carefully,But I think it's a typo.
It shoud be $x$ and $y$are postive integers. then all the thing is right.


logarithms
integration
calculus
calculus computations

Define $ \displaystyle{S(n)\equal{}\sum_{k\equal{}1}^{n}\frac{1}{k(n\minus{}k\plus{}1)}}$ ,Evaluate the series $ \displaystyle{\sum_{n\equal{}1}^{\infty}\frac{S(n)}{n\cdot 2^n}}$ :D

Using that

$ \frac{1}{k(n\minus{}k\plus{}1)}\equal{}\left(\frac{1}{k}\plus{}\frac{1}{1\plus{}n\minus{}k}\right)\frac{1}{n\plus{}1}$

you get that

$ S_{n}\equal{}\frac{2H_{n}}{n\plus{}1}$, where $ H_{n}\equal{}1\plus{}\cdots\plus{}\frac{1}{n}$ is the $ n^{th}$ harmonic number.

Your sum reads:

$ 2\sum\limits_{n\equal{}1}^{\infty}\frac{H_{n}}{n(n\plus{}1)2^{n}}$.

For calculating this sum use the following expansion

$ \frac{1}{2}\ln^{2}(1\minus{}z)\equal{}\sum\limits_{n\equal{}1}^{\infty}\frac{H_{n}z^{n\plus{}1}}{n\plus{}1}$, valid for $ \minus{}1\leq z<1$.

If you use this formula you should get, after some computations, that your sum equals

 $ \int\limits_{0}^{1/2}\frac{\ln^{2}(1\minus{}x)}{x^{2}}dx\equal{}\frac{\pi^{2}}{6}\minus{}2\ln^{2}2$.

 (Assuming that  I made no computational mistakes).

Didi, Your answer is correct. :D
My Solution...
Observe that $ \displaystyle{\sum_{n\equal{}1}^{\infty}\frac{S(n)}{n\cdot 2^n}\equal{}\int_{0}^{0.5}\frac{\ln^2 (1\minus{}x)}{x^{2}}dx}$ and substitute $ u\equal{}\minus{}\ln(1\minus{}x)$ then we get
$ \displaystyle{\int_{0}^{0.5}\frac{\ln^{2} (1\minus{}x)}{x^{2}}dx\equal{}\int_{0}^{\ln 2}\frac{u^2e^{\minus{}u}}{(1\minus{}e^{\minus{}u})^{2}}du\equal{}\sum_{k\equal{}1}^{\infty}k\int_{0}^{0.5}u^{2}e^{\minus{}ku}du}$
$ \displaystyle{\equal{}\sum_{k\equal{}1}^{\infty}k\left[\left[\minus{}\frac{1}{k}u^{2}e^{\minus{}ku}\right]_{0}^{\ln 2}\plus{}\frac{2}{k}\int_{0}^{\ln 2}ue^{\minus{}ku}du\right]}$
$ \displaystyle{\equal{}2\sum_{k\equal{}1}^{\infty}k\left[\left[\minus{}\frac{1}{k^{2}}ue^{\minus{}ku}\right]_{0}^{\ln 2}\plus{}\frac{1}{k^{2}}\int_{0}^{\ln 2}e^{\minus{}ku}du\right]\minus{}\ln^{2} 2}$
$ \displaystyle{\equal{}\frac{\pi^{2}}{3}\minus{}3\ln^{2} 2\minus{}2\sum_{k\equal{}1}^{\infty}\frac{1}{k\cdot 2^{k}}}$
Now let $ \displaystyle{I(a)\equal{}\int_{0}^{a}\minus{}\frac{\ln(1\minus{}x)}{x}dx}$,where $ a\in (0,1)$, we also know that $ \displaystyle{I(a)\equal{}\sum_{n\equal{}1}^{\infty}\frac{a^n}{n^2}}$
$ \displaystyle{I(a)\equal{}\int_{0}^{a}\minus{}\frac{\ln(1\minus{}x)}{x}dx\equal{}\int_{0}^{\minus{}\ln(1\minus{}a)}\frac{ue^{\minus{}u}}{1\minus{}e^{\minus{}u}}du\equal{}\sum_{n\equal{}1}^{\infty}\int_{0}^{\minus{}\ln(1\minus{}a)}e^{\minus{}nu}du}$
$ \displaystyle{\equal{}\sum_{n\equal{}1}^{\infty}\left[\left[\minus{}\frac{1}{k}ue^{\minus{}nu}\right]_{0}^{\minus{}\ln(1\minus{}a)}\plus{}\frac{1}{n}\int_{0}^{\minus{}\ln(1\minus{}a)}e^{\minus{}nu}\right]}$
$ \displaystyle{\equal{}\frac{\pi^{2}}{6}\minus{}\ln 1\ln(1\minus{}a)\minus{}\sum_{n\equal{}1}^{\infty}\frac{(1\minus{}a)^{n}}{n^2}\equal{}\frac{\pi^{2}}{6}\minus{}\ln a\ln(1\minus{}a)\minus{}I(1\minus{}a)}$
$ \displaystyle{\therefore I(a)\plus{}I(1\minus{}a)\equal{}\sum_{n\equal{}1}^{\infty}\frac{a^{n}}{n^{2}}\plus{}\sum_{n\equal{}1}^{\infty}\frac{(1\minus{}a)^{n}}{n^{2}}\equal{}\frac{\pi^{2}}{6}\minus{}\ln 1\ln(1\minus{}a)}$
$ \displaystyle{\therefore\sum_{n\equal{}1}^{\infty}\frac{S(n)}{n\cdot 2^n}\equal{}\frac{\pi^{2}}{3}\minus{}3\ln^{2} 2\minus{}2I(0.5)\equal{}\frac{\pi^{2}}{6}\minus{}2\ln^{2} 2}$
With this method we can also get $ \displaystyle{\sum_{n\equal{}1}^{\infty}\frac{H(n)}{n\cdot 2^{n}}\equal{}\frac{1}{2}\sum_{n\equal{}1}^{\infty}\frac{1}{n^{2}}}$ too. :)


real analysis
topology
real analysis unsolved

There exist compact subsets of $\mathbb{R}^2$ which is not Lebesgue-measurable.
I think it has been settled somewhere.Could you please give some hints? :(

That's strange: every compact set is Borel and, therefore, Lebesgue measurable. Probably, something else was meant but it is hard to guess what.  :?

I'm sorry  :blush: .
How do you find about Jordan-measurable($B$ compact and$\partial{B}$ has Jordan measure $0$).I heard that we must use choice-axiom or some peculiar things like that,but I don't have the idea.)

For Jordan measurable sets, the construction is fairly simple and does not require anything like axiom of choice. Just take the unit square and delete a dense open set of Lebesgue measure strictly less than $1$. The resulting compact $K$ is the same as its boundary $\partial K$ but has positive Lebesgue measure, which makes it not Jordan measurable.


The result is posted in AoPS MATHCOUNTS Forum.  Washington State won 2nd place team (only 1 point away from National Champion), terrific result and congratulations!
Samath is 6th place in written and Quarterfinalist in Countdown.

Congratulations guys! That's awesome! :D

One point away!? It'd be so cool if they'd won. Oh well, it's a lot better than we've ever had, even when adam hesterberg was there.

I stand corrected, when result was posted officially on Saturday, WA is only 0.25 away from VA, but all 4 students got into top 50 lists (the only state that achieved this)

I got 6th
Joy got 15th.
Jonathan got 29th.
Lawrence got 49th.


inequalities
number theory
algebra unsolved
algebra

Hi all, just asking

How do I go about solving the question:

Find all integer solutions of the equation

1+x+x^2+x^3+x^4=y^4

The solution that I have came up on my own was (0,1), (0,-1), (-1,1), (-1, -1) but I don't know if there are any other solutions. I don't have the standard answers with me so I am kind of stuck.

Thanks in advance!

When $ |x|$ is very large, $ x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1$ lies between $ x^4$ and $ (x\plus{}1)^4$, two consecutive fourth powers. Therefore, $ y$ can not be an integer. It just remains to figure out how big $ |x|$ needs to be to guarantee no integer $ y$ and to check the equation for small values of $ |x|$.

[hide="Solution"]an old shortlist(83) states that $ 1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} x^4$ is a perfect square iff $ x \equal{} \minus{} 1 , 0, 3$
[hide="HINT"]$ (x^2 \plus{} \frac{x}{2})^2 < 1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} x^4 < (x^2 \plus{} \frac {x}{2} \plus{} 1)^2$[/hide][/hide]

enoughsaid05, this problem should be posted in Number theory section  :wink:

Hi Fourrier,

I understand the solution, but I don't get the hint! XP

Mind explaining the hint??

Thanks a lot!

since:
$ (x^2 \plus{} \frac {x}{2})^2 < 1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} x^4 < (x^2 \plus{} \frac {x}{2} \plus{} 1)^2$
we have 2 cases:

$ n$ is even $ \implies 1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} x^4$ cannot be a perfect square since its between 2 perfect squares.

$ n$ is odd then we have the only perfect square between $ (x^2 \plus{} \frac {x}{2})^2$ and $ (x^2 \plus{} \frac {x}{2} \plus{} 1)^2$ is $ (x^2 \plus{} \frac {x}{2} \plus{} \frac {1}{2})^2$ so $ 1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} x^4 \equal{} (x^2 \plus{} \frac {x}{2} \plus{} \frac {1}{2})^2$ this leads to the solutions I gave above.

WOW! That's incredible! Thanks a lot! Anyway how do you manage to come up with the inequality? I am curious to know. Don't mind if I ask ;)

you're welcome and dont hesitate to ask  :D
well, Im not the creator of this inequality, but It's not very hard to come up with it: Just try, Trust me you'll get it after 3 or 4 atempts, I've already seen some examples of that kind, here are some:( and don't see quickly the solution, first try to solve :wink: )

can $ n^2 \minus{} n \plus{} 1$ be a perfect square for $ n > 1$?

same question for
$ y^4 \minus{} y^3 \plus{} y^2 \minus{} y \plus{} 1$ but this time $ y$ is an arbitary integer

[hide="HINT"]
$ (2y^2 \minus{} y)^2 < 4(y^4 \minus{} y^3 \plus{} y^2 \minus{} y \plus{} 1) < (2y^2 \minus{} y \plus{} 2)^2$
[/hide]

Hm...The questions are tricky though...

Hopefully I am correct. I have to see the hint for the second part though.

1) For $ n > 1$, $ \left(n \minus{} 1\right)^2 < n^2 \minus{} n \plus{} 1 < \left(n \plus{} 1\right)^2$
This means that the only possible integer that the middle expression can get is $ n^2 \minus{} n \plus{} 1 \equal{} n^2 \Rightarrow n \equal{} 1$

2) The inequality given is true. So I take it as that.

As the difference between the lower bound and the upper bound is 2, so $ 4y^4 \minus{} 4y^3 \plus{} 4y^2 \minus{} 4y \plus{} 4 \equal{} \left(2y^2 \minus{} y \plus{} 1\right)^2 \Rightarrow y^2 \plus{} 2y \minus{} 3 \equal{} 0\Rightarrow \left(y \plus{} 3\right)\left(y \minus{} 1\right) \equal{} 0$

Hence y = -3 and 1

OK! :) Now I know that if I have any further math problems I can turn to this forum. This is a great place to learn!


I am also a heavy email user. I use gmail for most. Still I find many of the people are in need of gmail invitations. So, I planned a website that distribute gmail invitations. Please help to run this service smooth by providing some gmail invitations at
http://www.ibight.com

There is a lot of such places. I also recomend [url=http://www.googlemania.com]GoogleMania[/url]. The site contains a lot of information about Google, but it is written in spanish. Nevertheless, if you do not speak spanish, there is a text field in the mail page and it has a button in the right that says "Enviar" (Send), You can just write your current email account and they will send you an invitation for Gmail to that address.

Best regards,

[url]http://www.ibight.com[/url] doesn't guarantee a gmail for you. It is just for my satisfaction. I had invites that are wasted. So, I thought of giving these invitations to those who are really in need. So, after putting couple of validations, they are distributed one per email.
I just tried entering my email ID at googlemania. But there is no email sent to me. May be they send that later when they find invitations. But I wish not to hide my invitations remaining. And the gmail invitation is sent within two minutes or more depending on your mail service provider.


I'm interested in [url=http://www.selectsmart.com/president/]this[/url] for a couple of reasons--to see people's results and to see what math people think of the tool's accuracy.

To receive as much information as possible, uncheck the three candidate viability boxes.

Here are my results:
1. 	Your ideal theoretical candidate.   (100%)
2. 	Cobb, David - Green Party   (99%)
3. 	Nader, Ralph - Independent   (92%)
4. 	Brown, Walt - Socialist Party   (85%)
5. 	Kucinich, Rep. Dennis, OH - Democrat   (79%)
6. 	Moseley-Braun, Former Senator Carol, IL - Democrat   (78%)
7. 	Dean, Gov. Howard, VT - Democrat   (74%)
8. 	Sharpton, Reverend Al - Democrat   (72%)
9. 	Kerry, Senator John, MA - Democrat   (72%)
10. 	Clark, Retired General Wesley K., AR - Democrat   (71%)
11. 	Gephardt, Rep. Dick, MO - Democrat   (71%)
12. 	Edwards, Senator John, NC - Democrat   (69%)
13. 	Lieberman, Senator Joe, CT - Democrat   (40%)
14. 	LaRouche, Lyndon H. Jr. - Democrat   (36%)
15. 	Badnarik, Michael - Libertarian   (12%)
16. 	Hagelin, Dr. John - Natural Law   (9%)
17. 	Peroutka, Michael - Constitution Party   (8%)
18. 	Bush, President George W. - Republican   (5%)

I must say, I don't understand the point of "Your ideal theoretical candidate."
[/url]

Well, My one on one is accurate:
1.  Your ideal theoretical candidate.   (100%)
2.  Kerry, Senator John, MA - Democrat   (62%)
3.  Bush, President George W. - Republican   (10%)

But what is up with Nader and Cobb being up on my scale?
On the other hand, Howard Dean was my first choice, so it is accurate there and my ideal theoretical candidate is in the right place  :cool:
1.  Your ideal theoretical candidate.   (100%)
2.  Nader, Ralph - Independent   (86%)
3.  Cobb, David - Green Party   (86%)
4.  Dean, Gov. Howard, VT - Democrat   (82%)
5.  Moseley-Braun, Former Senator Carol, IL - Democrat   (74%)
6.  Sharpton, Reverend Al - Democrat   (73%)
7.  Clark, Retired General Wesley K., AR - Democrat   (72%)
8.  Brown, Walt - Socialist Party   (72%)
9.  Kucinich, Rep. Dennis, OH - Democrat   (69%)
10.  Kerry, Senator John, MA - Democrat   (62%)
11.  Edwards, Senator John, NC - Democrat   (60%)
12.  Gephardt, Rep. Dick, MO - Democrat   (60%)
13.  LaRouche, Lyndon H. Jr. - Democrat   (47%)
14.  Lieberman, Senator Joe, CT - Democrat   (42%)
15.  Badnarik, Michael - Libertarian   (40%)
16.  Peroutka, Michael - Constitution Party   (20%)
17.  Hagelin, Dr. John - Natural Law   (17%)
18.  Bush, President George W. - Republican   (10%)

Truthfully, I don't think the thing is all that accurate. I've been looking into positions, and I'm seriously considering voting for the Constitution Party candidate (11) over Bush (1), and I would CERTAINLY vote for ANY Constitution Party candidate before I voted for Kerry or ANY other candidate that supported abortion (I mean genocide). Heck, I'd vote for Nader, as a man who sticks to his views, even though I blatantly disagree with everything the man says, before I would vote for John "Waffler" Kerry, even though I disagree with what Kerry says some small degree less.

1. 	Your ideal theoretical candidate.   (100%)
2. 	Bush, President George W. - Republican   (73%)
3. 	Badnarik, Michael - Libertarian   (56%)
4. 	Kucinich, Rep. Dennis, OH - Democrat   (45%)
5. 	Kerry, Senator John, MA - Democrat   (45%)
6. 	Edwards, Senator John, NC - Democrat   (42%)
7. 	Dean, Gov. Howard, VT - Democrat   (38%)
8. 	LaRouche, Lyndon H. Jr. - Democrat   (37%)
9. 	Lieberman, Senator Joe, CT - Democrat   (35%)
10. 	Sharpton, Reverend Al - Democrat   (32%)
11. 	Peroutka, Michael - Constitution Party   (32%)
12. 	Gephardt, Rep. Dick, MO - Democrat   (30%)
13. 	Nader, Ralph - Independent   (23%)
14. 	Cobb, David - Green Party   (23%)
15. 	Clark, Retired General Wesley K., AR - Democrat   (22%)
16. 	Moseley-Braun, Former Senator Carol, IL - Democrat   (18%)
17. 	Brown, Walt - Socialist Party   (15%)
18. 	Hagelin, Dr. John - Natural Law   (15%)

1. 	Your ideal theoretical candidate.   (100%)
2. 	Nader, Ralph - Independent   (81%)
3. 	Cobb, David - Green Party   (81%)
4. 	Moseley-Braun, Former Senator Carol, IL - Democrat   (77%)
5. 	Sharpton, Reverend Al - Democrat   (75%)
6. 	Brown, Walt - Socialist Party   (71%)
7. 	Clark, Retired General Wesley K., AR - Democrat   (70%)
8. 	Dean, Gov. Howard, VT - Democrat   (70%)
9. 	Kerry, Senator John, MA - Democrat   (68%)
10. 	Edwards, Senator John, NC - Democrat   (67%)
11. 	Kucinich, Rep. Dennis, OH - Democrat   (66%)
12. 	Gephardt, Rep. Dick, MO - Democrat   (61%)
13. 	LaRouche, Lyndon H. Jr. - Democrat   (44%)
14. 	Lieberman, Senator Joe, CT - Democrat   (39%)
15. 	Badnarik, Michael - Libertarian   (30%)
16. 	Peroutka, Michael - Constitution Party   (14%)
17. 	Hagelin, Dr. John - Natural Law   (5%)
18. 	Bush, President George W. - Republican   (1%)

This is what I got.

I was pleased to see that Lyndon LaRouche the nut got 0 percent from me. Other than that, I'll just say that the two candidates I'm most strongly considering voting for in November were the two that were on the top of my list, so I must have already checked them out.

[quote="confuted"]...ANY other candidate that supported abortion (I mean genocide). [/quote]

Alright, I've said it before, and I'll say it again. Abortion is NOT genocide, no matter how you look at it. "Genocide" is the extermination of a group of people based on race, religion, etc. For example, the killing of Jews in the Holocaust was genocide. However, a serial killer who goes out and shoots people completely at random is not guilty of genocide. Because abortion is not killing a group of people based on their race or religion or hair color or any other characteristic, it's not genocide.

BTW, my list went:

1.  Your ideal theoretical candidate.   (100%)
2.  Nader, Ralph - Independent   (81%)
3.  Cobb, David - Green Party   (81%)
4.  Brown, Walt - Socialist Party   (76%)
5.  Sharpton, Reverend Al - Democrat   (72%)
6.  Kerry, Senator John, MA - Democrat   (64%)
7.  Kucinich, Rep. Dennis, OH - Democrat   (62%)
8.  LaRouche, Lyndon H. Jr. - Democrat   (49%)
9.  Badnarik, Michael - Libertarian   (35%)
10.  Peroutka, Michael - Constitution Party   (15%)
11.  Bush, President George W. - Republican   (14%)

...And I have no idea why Bush got 14%.

Your Results:
Exciting results! I had trouble doing the test because I had no clue what those organizations were.
1.  Your ideal theoretical candidate.   (100%)  Click here for info
2.  Dean, Gov. Howard, VT - Democrat   (78%)  Click here for info
3.  Clark, Retired General Wesley K., AR - Democrat   (77%)  Click here for info
4.  Sharpton, Reverend Al - Democrat   (75%)  Click here for info
5.  Kucinich, Rep. Dennis, OH - Democrat   (75%)  Click here for info
6.  Kerry, Senator John, MA - Democrat   (74%)  Click here for info
7.  Nader, Ralph - Independent   (72%)  Click here for info
8.  Cobb, David - Green Party   (72%)  Click here for info
9.  Brown, Walt - Socialist Party   (72%)  Click here for info
10.  Moseley-Braun, Former Senator Carol, IL - Democrat   (69%)  Click here for info
11.  Edwards, Senator John, NC - Democrat   (66%)  Click here for info
12.  Gephardt, Rep. Dick, MO - Democrat   (62%)  Click here for info
13.  Lieberman, Senator Joe, CT - Democrat   (50%)  Click here for info
14.  Badnarik, Michael - Libertarian   (41%)  Click here for info
15.  LaRouche, Lyndon H. Jr. - Democrat   (35%)  Click here for info
16.  Hagelin, Dr. John - Natural Law   (12%)  Click here for info
17.  Bush, President George W. - Republican   (11%)  Click here for info
18.  Peroutka, Michael - Constitution Party   (9%)  Click here for info

very puzzled at some of mine...

1.  My ideal Theoretical candidate
2. Bush, President George W.- republican (95%)
3. Kerry, Senator John, MA - Democrat  (80%)
4. Kucinich, Rep. Dennis, OH - Democrat (69%)
5. Edwards, Senator John, NC - Democrat (65%)
6. Lieberman, Senator Joe, CT - Democrat (63%)
7.  Gephart, REp. Dick, MO - Democrat (61%)
8. Dean, Gov. Howard, VT - Democrat (49%)
9. Nader, Ralph - Independent (49%)
10.  Cobb, David- Green Party (46%)
11. Sharpton, REverand Al - Democrat (43%)
12. Brown, Walt - Socialist Party (39%)
13. Moseley-Braun, Former Senator Carol, IL - Democrat (37%)
14. LaRouche, Lyndon H. Jr. -Democrat (28%)
16 Badnarik, Michael - Libertarian (14%)
17. Hagelin, Dr. John - Natural Law (11%)
18. Peroutka, Michael - Constitution Party (7%)

1.  Nader, Ralph - Independent   (82%)
2.  Brown, Walt - Socialist Party   (78%)
3.  Kerry, Senator John, MA - Democrat   (67%)
4.  Edwards, Senator John, NC - Democrat   (64%)
5.  Badnarik, Michael - Libertarian   (32%)
6.  Peroutka, Michael - Constitution Party   (13%)
7.  Bush, President George W. - Republican   (9%)

I was very surprised to get Nader #1. Not that I'd ever vote for him (I'm too young anyway), but I never expected to have him ahead of Kerry.

Even more surprising was that Bush got 9%! I was expecting 3 or less.

[quote]1.  Your ideal theoretical candidate.   (100%)
2.  Dean, Gov. Howard, VT - Democrat   (80%)
3.  Nader, Ralph - Independent   (75%)
4.  Cobb, David - Green Party   (75%)
5.  Sharpton, Reverend Al - Democrat   (72%)
6.  Clark, Retired General Wesley K., AR - Democrat   (70%)
7.  Brown, Walt - Socialist Party   (70%)
8.  Kerry, Senator John, MA - Democrat   (70%)
9.  Kucinich, Rep. Dennis, OH - Democrat   (70%)
10.  Moseley-Braun, Former Senator Carol, IL - Democrat   (67%)
11.  Edwards, Senator John, NC - Democrat   (67%)
12.  Gephardt, Rep. Dick, MO - Democrat   (65%)
13.  Lieberman, Senator Joe, CT - Democrat   (45%)
14.  Badnarik, Michael - Libertarian   (36%)
15.  LaRouche, Lyndon H. Jr. - Democrat   (34%)
16.  Peroutka, Michael - Constitution Party   (14%)
17.  Hagelin, Dr. John - Natural Law   (12%)
18.  Bush, President George W. - Republican   (10%)
[/quote]

Wow, "your ideal theoretical candidate" is really popular!  I bet he/she will win the election!  ;)  lol

This is a little odd seeing that I can actually vote this next election.  I was hoping Kerry would be first on my list - that would make it easy.  I don't think which political candidate you choose can really be quantified with a quiz though...

[quote="confuted"][quote="Fermatprime"][quote="confuted"]...ANY other candidate that supported abortion (I mean genocide). [/quote]

Alright, I've said it before, and I'll say it again. Abortion is NOT genocide, no matter how you look at it. "Genocide" is the extermination of a group of people based on race, religion, etc. For example, the killing of Jews in the Holocaust was genocide. However, a serial killer who goes out and shoots people completely at random is not guilty of genocide. Because abortion is not killing a group of people based on their race or religion or hair color or [b]any other characteristic[/b], it's not genocide.[/quote]
[i](emphasis mine)[/i]

I suppose that in Fermatprime's world, age is not a characteristic.

If abortion is to be allowed for unborn babies, I want to see the right extended to normal, crying, smelly babies, toddlers, preteens, teens, heck, I want the right to kill my offspring at any time and say "it was an abortion." No, don't bother to tell me I'm being sarcastic. I'm serious about this. There is no difference.[/quote]

Alright, fine. I suppose I have to hand it to you that abortion is characterized by the killing of someone based on age. But since you're really not killing a group of people (unless you deliberately conceive fetuses in order to abort them, in which case, yes, you're a sicko), so it can't be genocide. I quote from Merriam-Webster, which defines genocide as:

"[T]he deliberate and systematic destruction of a racial, political, or cultural [b]group[/b]."

(emphasis mine)

Keep in mind that I never said otherwise. If you satisfactorily explain how abortion is systematically exterminating a racial, political, cultural, or even age group, I'll concede that it's genocide.

(P.S. I'm not really for abortions either; I oppose the complete elimination of abortion because sometimes childbirth can be seriously dangerous for the woman, in which case it can be justified IMO.)

I would argue that not supporting universal healthcare is legislating death based on social class. The fact is, confuted, that both left and right tend to support positions that can be considered a form of mass killing.

By the way, confuted, would you at least support abortion in cases  in which neither the mother nor the child is expected to survive childbirth?

1. 	Your ideal theoretical candidate.   (100%)
2. 	Kucinich, Rep. Dennis, OH - Democrat   (80%)
3. 	Nader, Ralph - Independent   (74%)
4. 	Cobb, David - Green Party   (74%)
5. 	Brown, Walt - Socialist Party   (72%)
6. 	Kerry, Senator John, MA - Democrat   (67%)
7. 	Sharpton, Reverend Al - Democrat   (67%)
8. 	Dean, Gov. Howard, VT - Democrat   (65%)
9. 	Clark, Retired General Wesley K., AR - Democrat   (63%)
10. 	Moseley-Braun, Former Senator Carol, IL - Democrat   (61%)
11. 	Edwards, Senator John, NC - Democrat   (58%)
12. 	Gephardt, Rep. Dick, MO - Democrat   (47%)
13. 	LaRouche, Lyndon H. Jr. - Democrat   (43%)
14. 	Lieberman, Senator Joe, CT - Democrat   (43%)
15. 	Bush, President George W. - Republican   (31%)
16. 	Badnarik, Michael - Libertarian   (30%)
17. 	Hagelin, Dr. John - Natural Law   (18%)
18. 	Peroutka, Michael - Constitution Party   (16%)

How is it tha one democrat gets 43% and another one gets twice as much as that?

Lieberman is very conservative for a Democrat, and Kucinich is extremely liberal. So it makes sense that you could get 80% for Kucinich and only 43% for Lieberman.

Your Results:

1.  Your ideal theoretical candidate.   (100%)
2.  Kucinich, Rep. Dennis, OH - Democrat   (56%)
3.  Sharpton, Reverend Al - Democrat   (54%)
4.  Cobb, David - Green Party   (54%)
5.  Kerry, Senator John, MA - Democrat   (50%)
6.  Dean, Gov. Howard, VT - Democrat   (49%)
7.  Moseley-Braun, Former Senator Carol, IL - Democrat   (48%)
8.  Nader, Ralph - Independent   (47%)
9.  Brown, Walt - Socialist Party   (47%)
10.  Clark, Retired General Wesley K., AR - Democrat   (46%)
11.  Badnarik, Michael - Libertarian   (45%)  12.  Gephardt, Rep. Dick, MO - Democrat   (44%)
13.  Edwards, Senator John, NC - Democrat   (43%)
14.  Lieberman, Senator Joe, CT - Democrat   (35%)
15.  Bush, President George W. - Republican   (30%)
16.  LaRouche, Lyndon H. Jr. - Democrat   (29%)
17.  Peroutka, Michael - Constitution Party   (22%)
18.  Hagelin, Dr. John - Natural Law   (14%)

I'm rather amazed. I guess my feelings are undevoted and rather ambiguous too. However, I'm happy Kerry was placed above Bush, though theoretically, I support neither of them.

My thoughts are dramatically influenced by outside concerns - Kerry plans to cut down the pharmaceutics profession (both parents work as chemists / pharmaceutical researchers), and Bush is just horrible as a leader - no more to be said.

According to Rosen, I'm a radical supporter of liberalism - wuhahahaha. :twisted:

1. 	Your ideal theoretical candidate.   (100%)
2. 	Cobb, David - Green Party   (99%)
3. 	Nader, Ralph - Independent   (93%)
4. 	Brown, Walt - Socialist Party   (84%)
5. 	Dean, Gov. Howard, VT - Democrat   (78%)
6. 	Kucinich, Rep. Dennis, OH - Democrat   (78%)
7. 	Moseley-Braun, Former Senator Carol, IL - Democrat   (77%)
8. 	Clark, Retired General Wesley K., AR - Democrat   (77%)
9. 	Sharpton, Reverend Al - Democrat   (75%)
10. 	Kerry, Senator John, MA - Democrat   (72%)
11. 	Edwards, Senator John, NC - Democrat   (69%)
12. 	Gephardt, Rep. Dick, MO - Democrat   (67%)
13. 	Lieberman, Senator Joe, CT - Democrat   (41%)
14. 	LaRouche, Lyndon H. Jr. - Democrat   (38%)
15. 	Badnarik, Michael - Libertarian   (19%)
16. 	Peroutka, Michael - Constitution Party   (10%)
17. 	Hagelin, Dr. John - Natural Law   (10%)
18. 	Bush, President George W. - Republican   (5%)

I didn't like several of their questions ... in particular, I felt the drug laws question did not allow enough freedom.  One should have been able to choose "support education and counseling programs" in addition to checking one's views on whether laws should be tightened or loosened.  Also, I wonder the extent to which saying I agree with organizations affected my views.  No energy to go back and do it again, though.

[quote="white_horse_king88"]According to Rosen, I'm a radical supporter of liberalism - wuhahahaha. :twisted:[/quote]

I am too!  :lol:

The way they worded some of those questions was definely tricky.  It was obvious that they leaned towards the right.

1. 	Your ideal theoretical candidate.   (100%)
2. 	Bush, President George W. - Republican   (84%)
3. 	Badnarik, Michael - Libertarian   (57%)
4. 	Kerry, Senator John, MA - Democrat   (41%)
5. 	Edwards, Senator John, NC - Democrat   (38%)
6. 	Gephardt, Rep. Dick, MO - Democrat   (37%)
7. 	Lieberman, Senator Joe, CT - Democrat   (34%)
8. 	Peroutka, Michael - Constitution Party   (33%)
9. 	Dean, Gov. Howard, VT - Democrat   (28%)
10. 	Kucinich, Rep. Dennis, OH - Democrat   (27%)
11. 	Clark, Retired General Wesley K., AR - Democrat   (23%)
12. 	Sharpton, Reverend Al - Democrat   (21%)
13. 	LaRouche, Lyndon H. Jr. - Democrat   (21%)
14. 	Hagelin, Dr. John - Natural Law   (16%)
15. 	Moseley-Braun, Former Senator Carol, IL - Democrat   (11%)
16. 	Brown, Walt - Socialist Party   (11%)
17. 	Nader, Ralph - Independent   (6%)
18. 	Cobb, David - Green Party   (6%)

1.  Your ideal theoretical candidate.   (100%)
2.  Cobb, David - Green Party   (74%)
3.  Nader, Ralph - Independent   (71%)
4.  Dean, Gov. Howard, VT - Democrat   (69%)
5.  Moseley-Braun, Former Senator Carol, IL - Democrat   (62%)
6.  Kucinich, Rep. Dennis, OH - Democrat   (62%)
7.  Brown, Walt - Socialist Party   (61%)
8.  Clark, Retired General Wesley K., AR - Democrat   (59%)
9.  Sharpton, Reverend Al - Democrat   (59%)
10.  Gephardt, Rep. Dick, MO - Democrat   (58%)
11.  Kerry, Senator John, MA - Democrat   (55%)
12.  Edwards, Senator John, NC - Democrat   (51%)
13.  Lieberman, Senator Joe, CT - Democrat   (41%)
14.  Badnarik, Michael - Libertarian   (36%)
15.  LaRouche, Lyndon H. Jr. - Democrat   (35%)
16.  Peroutka, Michael - Constitution Party   (17%)
17.  Hagelin, Dr. John - Natural Law   (16%)
18.  Bush, President George W. - Republican   (11%)

I'm not surprised by my results at all.

[quote="farmertim"][quote="Fermatprime"][quote="JBL"]1. 	Your ideal theoretical candidate.   (100%)
[b]2. 	Cobb, David - Green Party   (99%)[/b]
[/quote]

 :shock: That's absolutely insane. JBL could virtually run as a Green Party candidate someday.[/quote]

i think the conclusion to be drawn here is that he could run as David Cobb someday.  It's a fair guess that neither agrees with the official Green Party platform as much as they agree with each other.[/quote]

Yeah, that's probably a better conclusion, but it'd be sort of scary for anyone but David Cobb to run as David Cobb.

[quote="Fermatprime"]
Yeah, that's probably a better conclusion, but it'd be sort of scary for anyone but David Cobb to run as David Cobb.[/quote]

Maybe David Cobb could run for David Cobb and lose.  Kind of like that enduring legend about how Charlie Chaplin placed third in a Charlie Chaplin look-alike contest..

Hey, I was the second person to be 99% Cobb.  Without any knowledge to support this claim, I would imagine that Cobb's beliefs look more like the Green Party's beliefs than Nader's did.

Although I don't live in the USA, I do have an interest in the politics of USA.  Anyways, here are my results.

1. 	Your ideal theoretical candidate.   (100%)
2. 	Cobb, David - Green Party   (92%)
3. 	Nader, Ralph - Independent   (92%)
4. 	Brown, Walt - Socialist Party   (90%)
5. 	Sharpton, Reverend Al - Democrat   (74%)
6. 	Kucinich, Rep. Dennis, OH - Democrat   (69%)
7. 	Kerry, Senator John, MA - Democrat   (66%)
8. 	Edwards, Senator John, NC - Democrat   (63%)
9. 	Moseley-Braun, Former Senator Carol, IL - Democrat   (62%)
10. 	Dean, Gov. Howard, VT - Democrat   (60%)
11. 	Clark, Retired General Wesley K., AR - Democrat   (59%)
12. 	Gephardt, Rep. Dick, MO - Democrat   (54%)
13. 	LaRouche, Lyndon H. Jr. - Democrat   (44%)
14. 	Badnarik, Michael - Libertarian   (39%)
15. 	Peroutka, Michael - Constitution Party   (15%)
16. 	Lieberman, Senator Joe, CT - Democrat   (12%)
17. 	Bush, President George W. - Republican   (10%)
18. 	Hagelin, Dr. John - Natural Law   (2%)

Your Results:

1.  Your ideal theoretical candidate.   (100%)  Click here for info
2.  Cobb, David - Green Party   (86%)  Click here for info
3.  Nader, Ralph - Independent   (82%)  Click here for info
4.  Brown, Walt - Socialist Party   (76%)  Click here for info
5.  Dean, Gov. Howard, VT - Democrat   (74%)  Click here for info
6.  Clark, Retired General Wesley K., AR - Democrat   (73%)  Click here for info
7.  Sharpton, Reverend Al - Democrat   (72%)  Click here for info
8.  Moseley-Braun, Former Senator Carol, IL - Democrat   (70%)  Click here for info
9.  Kucinich, Rep. Dennis, OH - Democrat   (66%)  Click here for info
10.  Kerry, Senator John, MA - Democrat   (64%)  Click here for info
11.  Gephardt, Rep. Dick, MO - Democrat   (62%)  Click here for info
12.  Edwards, Senator John, NC - Democrat   (60%)  Click here for info
13.  Badnarik, Michael - Libertarian   (39%)  Click here for info
14.  LaRouche, Lyndon H. Jr. - Democrat   (38%)  Click here for info
15.  Lieberman, Senator Joe, CT - Democrat   (37%)  Click here for info
16.  Peroutka, Michael - Constitution Party   (16%)  Click here for info
17.  Hagelin, Dr. John - Natural Law   (9%)  Click here for info
18.  Bush, President George W. - Republican   (5%)  Click here for info

Not bad.  Not bad at all.

Donald Trump - 100%
Jo Jorgensen - 98%
Michael Bloomberg - 20%
Joe Biden - 10%
Bernie Sander - 9%
Howie Hawkins - 6%
Hillary Clinton - 0%
It seems that the spacehead effect is real.

[quote name="hannahptl" url="/community/p18065010"]
Donald Trump - 100%
Jo Jorgensen - 89%
Joe Biden - 10%
Howie Hawkins - 6%

It seems that the spacehead effect is real.
[/quote]

don't bump 16 year old posts?

It's cool though. Here are mine:

Joe Biden: 100%
Jo Jorgensen: 94%
Donald Trump: 60%

As an election is coming up, I think it could still be relevant. Besides, Richard Rusczyk himself posted here.

[quote=rrusczyk]I'd be very curious to see the results of some of our foreign community members, if they have the time to give this a whirl.  We've always heard how much farther left on the political spectrum Europeans are, I'm interested to see how they'd come out on this scale.[/quote]

Wow, he did!

Here's mine
Joe Biden 100% (best match)
Howie Hawkins 80%
Jo Jorgensen 54%
Donald Trump 10%
How am I not surprised

Joe Biden 100% (I am against the democratic party, so I would say no)
Howie Hawkins 91% (Definitely!)
Jo Jorgensen 50% (Great candidate, but I support Rhine capitalism, not laissez-faires )
Donald Trump: 8%

1. Joe Biden 100%
2. Howie Hawkins 90%
3. Bernie Sanders 89%
4. Jo Jorgensen 77% (I feel personally insulted by this)

1. Donald Trump 100% (fair enough)
2. Jo Jorgenson 91% (agreed, I like her candidacy)
[some other people]
6. Joe Biden 51%
[some withdrawn Democrats + other people]
24. Kamala Harris 31%
-----------------------------
All in all, not terribly surprising.

[quote=Supernova283][quote=IAmTheHazard]It's worth noting that, based on what I'm seeing form the site, it's not the most impartial when it comes to politics...[/quote]

Also, why does TRT have a lot of USA politics but no foreign ones? Is TRT just for politics?[/quote]

It's just because most people on AoPS are American.


geometry
complex numbers
geometry proposed

For △ABC,AB>CA and the incircle of △ABC touches BC at E.The second intersection point of the incircle and AE is D.
A point F is on AE such that CE=CF and the intersection of CF and BD is G.Prove that CF=FG.
Is this problem easy ? Why can I hardly solve geometric problems ? ToT
Please  :help:
I don't want to use complex numbers for this problem.
Can anyone give me so beautiful Euclidean geometric proof ? please~ I want to know the solution.


\[ \begin{array}{l}
 a,b,c > 0,ab \plus{} ac \plus{} bc \equal{} 1 \\
 prove: \sum {\left( {\sqrt[3]{{\left( {6b \plus{} \frac{1}{a}} \right)}}} \right)}  \le \frac{1}{{abc}} \\
 \end{array}
\]
 :!:

[quote="Nguyen The Ky Cuong"]
\[ \begin{array}{l} a,b,c > 0,ab \plus{} ac \plus{} bc \equal{} 1 \\
prove: \sum {\left( {\sqrt [3]{{\left( {6b \plus{} \frac {1}{a}} \right)}}} \right)} \le \frac {1}{{abc}} \\
\end{array}
\]
:!:[/quote]
You should post in inequality section :wink:
Cuong hả? bài này tự chế hay chép sách vậy? :lol:

[quote="Nguyen The Ky Cuong"]
\[ \begin{array}{l} a,b,c > 0,ab \plus{} ac \plus{} bc \equal{} 1 \\
prove: \sum {\left( {\sqrt [3]{{\left( {6b \plus{} \frac {1}{a}} \right)}}} \right)} \le \frac {1}{{abc}} \\
\end{array}
\]
:!:[/quote]
See here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30865&search_id=1943420323
This is not your inequality :(


If $ x$ is greater than zero, then the correct relationship is:
$ \textbf{(A)}\ \log (1\plus{}x) \equal{} \frac{x}{1\plus{}x} \qquad\textbf{(B)}\ \log (1\plus{}x) < \frac{x}{1\plus{}x} \\
\textbf{(C)}\ \log(1\plus{}x) > x \qquad\textbf{(D)}\ \log (1\plus{}x) < x \qquad\textbf{(E)}\ \text{none of these}$

[hide="Cheap"]Raising 10 to both sides of each given inequality/equation, we get:

A: $ 1+x=10^{\frac{x}{1+x}}$
B: $ 1+x<10^{\frac{x}{1+x}}$
C: $ 1+x>10^x$
D: $ 1+x<10^x$
E: $ 10^\text{none of these}$  :lol:

We can see that the LHS of D grows linearly, while the RHS grows exponentially. When $ x=0$, we have equality, so for $ x>0$ it should hold. Thus, the answer is $ \boxed{\textbf{(D)}}$.

We can also disprove the rest of them:

For A, letting $ x=1$ gives $ 2=10^{\frac{1}{2}}$, which is... nonsense.
For B, letting $ x=1000$ gives $ 1001<10^{\frac{1000}{1001}}$, which is closeish to 9, a contradiction.
For C, letting $ x=1000$ gives $ 1001>10^{1000}$, which is just plain wrong.[/hide]

hello
U can draw all the 4 graphs.
only y=x always lie above y=log(1+x)
Answer -D
thank u

[quote="mathwizarddude"]Just note that $ d(10^x)/dx \equal{} \ln10 \cdot 10^x > 1$... here's one example where calculus becomes useful.[/quote]Hah! Calculus is overkill on an old AHSME question :P
Anyways, consider $ 1\plus{}x<e^x<10^x$. That leads to $ \bf (D)$ quickly (if you know this inequality, that is :D)


2.Compute lim{n->oo}(nln(n))/([n/1]+[n/2]+[n/3]+...[n/n]).
i denote with [x] the integer part of x.

this one is simple.. it's 1.. I don't feel like posting a full solution now! anyone interested, I will post it later! :D

cheers! :D :D

Oh, but this a "hard" problem. It is allmost impossible to see that it is enough to see that [x]<=x<[x]+1. Very hard problem! Why do they give such problems in contests?

sincerly i don't know.... i expeted to be harder but... it seems this contest was only a fake (hope this means what i' thinking about :) ).


Let $ m$ and $ n$ be positive integers such that $ mn$ is a perfect square. Show that

[list](i) if $ \gcd(m,n)\equal{}1$, then $ m$ and $ n$ are perfect squares.

(ii) if $ mn>1$, then $ \gcd(m\plus{}\sqrt{mn},\,n\plus{}\sqrt{mn})>1$.[/list]

[hide="Hint"]
Write $ m\equal{}a^2 k$ and $ n\equal{}b^2 l$ where $ a,b,k,l$ are all positive integers and $ k$, $ l$ are squarefree (i.e. not divisible by any square). Any positive integer can be written in this way (how can you read $ a$ and $ k$ off from a prime factorization of $ m$?). Now, given that $ mn$ is a perfect square, what can you say about $ k$ and $ l$? This should let you quickly solve both parts of the problem.
[/hide]

Thanks. By the way, it don’t think the condition $ mn>1$ is necessary in part (ii).

[hide="(i)"]Let $ mn\equal{}P_1^{2a_1}P_2^{2a_2}\cdots P_N^{2a_N}$, where $ P_i$ are primes. Since $ P_i$ is a prime, it must divide at least one of $ m$ or $ n$. But since $ \gcd (m,n)\equal{}1$, it can't divide both, so one of the variables must be a multiple of $ P_i^{2a_i}$. Since that takes care of all the primes, $ m$ and $ n$ are the products of squares, which means that $ m$ and $ n$ are squares.[/hide]

I am assuming for (ii) that you meant to say $ \gcd(m,n)>1$.

[hide="(ii)"]Let's say that $ p_a|m,n$, where $ p_a$ is a prime. Thus $ p_a^2|mn$. Since $ mn$ is a perfect square, $ p_a|\sqrt{mn}$. Since $ p_a|m,n$, $ p_a|m\plus{}\sqrt{mn},n\plus{}\sqrt{mn}$, and $ \gcd(m\plus{}\sqrt{mn},n\plus{}\sqrt{mn})>1$.[/hide]

[quote="1=2"]
I am assuming for (ii) that you meant to say $ \gcd(m,n) > 1$.
[/quote]

(ii) is true even for $ \gcd(m,n)\equal{}1$. Use (i): $ m\equal{}a^2,n\equal{}b^2,\sqrt{mn}\equal{}ab$...

[quote="1=2"]I am assuming for (ii) that you meant to say $ \gcd(m,n) > 1$.[/quote]
No! I am not assuming that.

In any case, I know how to prove (ii) now. [hide]There are two cases: (1) $ m$ is a perfect square, (2) $ m$ is not a perfect square.

(1) If $ m$ is a perfect square, then $ n$ must be a perfect square as well (since $ mn$ is a perfect square). Say $ m \equal{} a^2$, $ n \equal{} b^2$. Then

$ m \plus{} \sqrt {mn} \equal{} a^2 \plus{} ab \equal{} a(a \plus{} b)$
$ n \plus{} \sqrt {mn} \equal{} b^2 \plus{} ab \equal{} b(b \plus{} a)$

Hence $ a \plus{} b$ divides both $ m \plus{} \sqrt {mn}$ and $ n \plus{} \sqrt {mn}$; $ \therefore\ \gcd(m \plus{} \sqrt {mn},n \plus{} \sqrt {mn})\ge a \plus{} b > 1$.

(2) If $ m$ is not a perfect square, then there must be a prime $ p$ which divides $ m$ an odd number of times. Since $ p$ divides $ mn$ an even number of times, it follows that $ p$ must divide $ n$ an odd number of times as well. Hence $ p$ divides both $ m$ and $ n$. And obviously $ p$ divides $ \sqrt {mn}$. $ \therefore\ \gcd(m \plus{} \sqrt {mn},n \plus{} \sqrt {mn})\ge p > 1$.

(Incidentally, case (2) also proves part (i) by proving the contrapositive statement.)[/hide]


Over the past few years, Youtube has been taking small steps that have effectively ended "viral video."  By messing with view counts, making star ratings of videos less visible, hiding the "most viewed" and "highest rated" video pages, censoring videos that are completely appropriate, and more, Youtube has made it much more difficult for those random hilarious videos to become popular, favoring their "featured videos" instead so that they can make more money.  Plus, Youtube has another big change planned that will completely restructure the site.  Essentially, Youtube is removing "You" from their site.

Check out this video to get the details:

http://www.youtube.com/watch?v=V025E_b5I4g

well, youtube [url=http://www.internetevolution.com/author.asp?section_id=715&doc_id=175123&]clearly needs to do something different, from a business standpoint[/url].

All I really took out of the video was:

1. Google messes with ratings and rankings. I have proof that they do, but I won't show you it.

2. Youtube is being reformatted and some things now take 1/2 a second longer to get to.

3. Youtube is making deals with big companies to get their videos.

The first is just a claim, not verified.

The second is just a reformation of the tabs. While people might be angry for a while (like on facebook,) after a while people get used to it and stop caring.

The third is expected. They need money, so they're paying to put certain videos on their site.

So what's wrong with that?

http://www.youtube.com/watch?v=FCTAqdDfL58

A satirical view on copyright.  Perhaps not suitable for very young children.

Interesting link blahblahblah, I'm glad I saw the other side of this (if you can call it that).

By the way, my OP is not my personal opinion; it's just a summary of what the video says.

thats why fred is so popular even though hes not even that good...just some kid who speeds up a video.
youtube cant decide the first most subscribed ppl...thats just not fair...lol i sound so serious... :)

Can't stand google and subsidaries anymore. Even though I do use gmail, I almost never go on the real thing- instead, I use something called an offline client- Windows Live Mail. Anyways, Google is getting more and more obnoxious. That's just my own opinion, and it angers one of my friends, whose Dad works for Google. Heck, Google's search techniques are not unique, in fact, here

[hide]

Results for John Sutter- Google Search

en.wikipedia.org/wiki/John_Sutter
http://www.sfmuseum.com/hist2/gold.html
http://www.pbs.org/weta/thewest/people/s_z/sutter.htm
http://www.syix.com/YUBACITY/johnsutter.html
http://www.sfmuseum.org/bio/sutter.html

Results for John Sutter- Windows Live Search

en.wikipedia.org/wiki/John_Sutter
http://www.syix.com/yubacity/johnsutter.html
http://www.sfmuseum.org/hist2/gold.html
http://www.sfmuseum.org/bio/sutter.html
http://www.pbs.org/weta/thewest/people/s_z/sutter.htm

Besides, Microsoft's translator is available through IM and you learn something new every day through the search engine!

[/hide]

Another search test
[hide]

Maplestory on Windows Live Video

5 Myspace
3 Youtube

On Google

5 Youtube
2 Google Videos
1 Guba.com

[/hide]

Conclusion of argument- Google is relying on itself to make money. Not a good business model is it?

I'm personally not a big fan of viral videos, some things that float around the internet are pretty stupid.

But it does anger me when videos are [i]wrongfully[/i] removed/flagged, or when stupid copyright claims are made, or when view counts are frozen. A frozen view count isn't a view count anymore. If they must freeze a view count, they should mark, next to the counter, the date on which the counter was frozen, just like locked threads on forums have a locked sign next to them.

[b]@ Quickstone:[/b] Why do you think Google is obnoxious?

In any case, other search engines produce similar results to Google because they, at some point in time, acknowledged its superiority and tried to replicate it. For example, Yahoo search used to blow Yahoo out of the water, until Yahoo announced its "new and improved" search engine. Now, their results are almost the same.

I also don't understand how Google showing Youtube videos in its results is at all a problem, or how the strength of Google's business model has anything to do with it being obnoxious.

[quote="matt276eagles"]Essentially, Youtube is removing "You" from their site.

Check out this video to get the details:

http://www.youtube.com/watch?v=V025E_b5I4g[/quote]

I seriously thought this was a rickroll when I saw these lines  :lol:
lol, but seriously, it looks just like a rickroll set up.

@quickstone84: Google uses their PageRank technology. Results you're seeing are there because of innumerable factors: pages linking to those pages, number of hits within a recent timeframe, meta-tag/keyword analysis, etc, etc. The fact that YouTube dominates the search is that YouTube shows up in a lot of searches is because it's ridiculously popular—and yes, Google tries to promote it a little by giving it a little section at the top of search results in some cases, but it's typically relevant.

In any case, [url=http://www.google.com/search?q=maplestory]I'm not seeing any of the results you mentioned.[/url]

@TZF: You've clearly not seen [hide="M. Bison Yes Yes!"][youtube]p0j0lO7uQBo[/youtube][/hide] arguably the most exciting of all the viral videos.  :rotfl:


It is called "Norwegian", whatever it has to do with the problem (which is not quite pre-olympiad by the way...).

[b]EDIT:[/b] Okay, it's preolympiad, if one knows what ln is. ;)

  darij

[quote="lalalala"]Let $ k$ be a positive integer, we write k as a product of prime numbers (not necessarily distinct, par example 18=2.3.3). Let $ T(k)$ be the sum of all factors in the factorization above. Find the largest constant $ C$ such that $ T(k) \geq C.lnk$ for all positive integer k.[/quote]

This is an [i]open[/i] problem? I found it straightforward... You need to find the largest constant $ C$ such that $ p\geq C\cdot\ln p$ for every prime $ p$.

  darij

[quote="darij grinberg"]You need to find the largest constant $ C$ such that $ p\geq C\cdot\ln p  (1)$ for every prime $ p$.[/quote]
Darij, i dơn't understand! Why must the largest constant $ C$ satisfy (1)? Please answer my question?

Sorry, but knowing the history of your previous posts, I am not going to answer any of your questions.
Mods: please lock this topic because of
http://www.mathlinks.ro/Forum/viewtopic.php?t=188838
(and you are free to delete it afterwards, since we don't need 2 topics on one problem).

  darij


$ \boxed {\triangle\ ABC}\ \ \wedge\ \ b\ge c\implies \frac ba\plus{}\frac {a}{b\plus{}c}\ge \sqrt {2\plus{}\frac {b^2\minus{}c^2}{a^2}}$ (with equality iff $ A\equal{}2C$).

[quote="Virgil Nicula"]$ \boxed {\triangle\ ABC}\ \ \wedge\ \ b\ge c\implies \frac ba \plus{} \frac {a}{b \plus{} c}\ge \sqrt {2 \plus{} \frac {b^2 \minus{} c^2}{a^2}}$ (with equality iff $ A \equal{} 2C$).[/quote]
Here is my solution:
$ \frac ba \plus{} \frac {a}{b \plus{} c}\ge \sqrt {2 \plus{} \frac {b^2 \minus{} c^2}{a^2}}$
<=>$ \frac{b^2}{a^2}\plus{}\frac{2b}{b\plus{}c}\plus{}\frac{a^2}{(b\plus{}c)^2} \geq 2\plus{}\frac{b^2\minus{}c^2}{a^2}$
<=> $ \frac{c^2}{a^2}\plus{}\frac{2b}{b\plus{}c}\plus{}\frac{a^2}{(b\plus{}c)^2} \geq 2$
By AM-GM we have:
$ \frac{c^2}{a^2}\plus{}\frac{a^2}{(b\plus{}c)^2} \geq \frac{2c}{b\plus{}c}$
So $ \frac{c^2}{a^2}\plus{}\frac{2b}{b\plus{}c}\plus{}\frac{a^2}{(b\plus{}c)^2} \geq \frac{2b}{b\plus{}c}\plus{}\frac{2c}{b\plus{}c}\equal{}\frac{2(b\plus{}c)}{b\plus{}c}\equal{}2$
Done!


algebra
polynomial
quadratics
algebra open

I seem to have got a result on my own (though I later came across it in an olympiad handout written by an IMO medallist)-maybe it is well known:
Any polynomial $ p(x)$ with real coefficients , there exist  reals $ \alpha_k, \beta_k, gamma_k$   such that
$ p(x) \equal{} (x\minus{}\alpha_1)(x\minus{}\alpha_2)(x\minus{}\alpha_3)....(x\minus{}\alpha_k)((x\minus{}\beta_1)^2 \plus{} \gamma^2_1)).........((x\minus{}\beta_k)^2 \plus{} \gamma^2_k))$
Well, it is very nice but I do not have any brains to think of a suitable application now, maybe I am not that talented or a genius. So, can anyone  please tell me how to apply this result somewhere or give an example of an application of this before?

Thanks.

Well, it is only true if $ p(x)$ is monic (i.e. its leading coefficient is $ 1$), and is a trivial corollary of The Fundamental Theorem of Algebra, viz. a polynomial has as many (complex) roots as its degree. Just group the conjugate roots to get the quadratic factors.

So any application of it will also be an application of the F*n*a*e*t*l T*e*r*m.


I accidentally submitted the problems instead of the solutions...Is there some way to resubmit before the deadline?

Thanks!

It would be best to email them at usamts@usamts.org. Also, you are only allowed to submit solutions once, but there are probably exceptions.

EDIT: Make sure you include your USAMTS ID# and name in the email.

Thanks for responding. I've emailed them...I just hope it all gets sorted out in time!

*returns to beating self up into an barely discrenible pulp*

[quote="timwu"]You can still mail your solutions, though.[/quote]

It says somewhere on USAMTS website that you can only submit them via one way, so e-mailing the solutions and explaining the problem would be the best thing to do.


[quote="anirudh"]What is the product of the real roots of the equation $x^{2}+18x+30 = 2 \sqrt{x^{2}+18x+45}$?[/quote]

[hide]Hint, make a substitution of

$x^{2}+18x+30=a$[/hide]

[quote="mysmartmouth"][hide="Solution"]Let $d=x^{2}+18x+30$, and substitute into the original equation:

$d=2\sqrt{d+15}$
$d^{2}= 4(d+15)$
$d^{2}= 4d+60$
$d^{2}-4d-60 = 0$
$(d-10)(d+6)=0$
$d=10$ or $d=-6$

Substituting back in the original substitution:

$d=x^{2}+18x+30=10$
$d=x^{2}+18x+30=-6$

$x^{2}+18x+20=0$
$x^{2}+18x+36=0$

By Vieta, these have products of 20 and 36, respectively.  Thus, we multiply those products together, and get a product of $\boxed{720}$.[/hide]

This was really easy for an AIME problem.  What year was it?

EDIT: Whoa, looking at the Ohio forum, I see where anirudh's post count comes from.   :rotfl:[/quote]

Well, once again, if you say they are easy, think again.  This was 1984.

$d \ge 0$, so $d \neq-6$, so $d=10$.

Hence the product is only 20.

$020$

Oh, wow, stupid mistake.  It was still an easy problem..., I just didn't read it and see that it was asking for [b]real [/b]solutions.


in the texniccenter, when i use eqnarray, the aligned equations move to the center.  is there a way to keep them left-aligned?  thanks.

Use the fleqn option in the document class at the beginning of the document. For example, [code]\documentclass[fleqn]{article}[/code]


Να εξετάσετε αν υπάρχει συνάρτηση ώστε$f(x+y+z)\leq 3(xy+yz+zx)$ (1) και να υπάρχει φυσικός αριθμός n ώστε
$g(g(x))=x^{2n+1}$ (2)  and  $f(g(x))=(g(x))^2$.  (3)

Η λύση που βρήκα :
H (1) για  $y=-z$ γίνεται  $f(x)\leq -3z^2\leq 0$.
Τώρα επειδή $g(R)=R$  η (3) δίνει
$g^2(x)\leq 0$ άρα $g(x)=0$  άρα  $x^{2n+1}\leq 0$ για κάθε χ ανήκει R  , άτοπο

Τι θέματα είναι αυτά ? Δεν ντρέπονται λιγάκι???
 :mad:  :mad:

Που να ντραπούν μας δουλεύεις ?Αντίθετα καθηγητής από τα Τρίκαλα ετοιμαζόταν να βάλλει διαγώνισμα στα παιδιά στο φροντηστήριο και του έφαγε ένα μήνα για να βρει πρωτότυπα θέματα και αυτοί που ασχολούνται και με την ΕΘΝΙΚΗ ΜΑΘΗΜΑΤΙΚΗ ΟΛΥΜΠΙΑΔΑ που δεν είναι δα και μικρό πράγμα προσπαθούν να καροϊδέψουν άτομα που ασχολούνται με το αντικείμενο. Είναι βαθιά μα πολύ βαθιά νυχτωμένοι :rotfl:

Σιλουανέ ,

    μη το παίρνεις κατακαρδα , ας έχεις δίκιο. Είμαστε μια μικρή χώρα αλλά μεγάλη σε παρακμή . Πόσοι νομίζεις ασχολήθηκαν με τη μαθηματική ολυμπιάδα; Ενώ με το νέο famestory εκατομμύρια !!!!!!!!!!! Λοιπόν , οι έλληνες μαθηματικοί - κατα κανόνα - δεν αγαπούν τους διαγωνισμούς και τις ολυμπιάδες , έχουν ίσως και κάποια ελαφρυντικά , αλλά αυτό είναι άλλο πράγμα.Ποιος λοιπόν θα φτιάξει θέματα ; Εμείς αναγκαζόμαστε να αντιγράφουμε - πλην ελάχιστων περιπτώσεων. Μη νομίσεις ότι δεν υπάρχουν πολύ καλοί μαθηματικοί ή άλλοι επιστήμονες για να φτιάξουν θέματα! και μαθητές ακόμα μπορούν να φτιάξουν υπέροχα θέματα,. Το πρόβλημα είναι ότι υπάρχει ΑΔΙΑΦΟΡΙΑ , όπως και για όλα τα σημαντικά πράγματα. Εσεις όμως τα νέα παιδιά συνεχίστε , παλέψτε , μιλήστε , συγκρουστείτε με το άδικο και το στραβό. Μόνο έτσι μπορεί να αλλάξει κάτι !

 Μπάμπης

Συμφωνώ με τον Μπάμπη.

παιδιά μην το βάζετε κάτω!


conics
ellipse
geometry
perimeter
vector
linear algebra
geometry unsolved

Given an ellipse of area $A$ on the Cartesian plane. It is known that the semiaxes of ellipse are all greater than 1. Show that the number of lattice points lying on the perimeter of the ellipse is less than $6A^{1/3}$.

(I remember seeing this problem on [url=http://problems.math.umr.edu/]20 000 Problems Under the Sea[/url] a long time ago. But the site has been down ever since for some reason.)

Just a comment: one of the oldest problems on the board (still unsolved :)), [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=56]this one[/url], is a particular case of the problem you have posted.

[color=red][Moderator edit: And [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=3440]the same question in the Geometry forum[/url] is also unsolved.][/color]

lol, this might be of use: (scroll all the way to the bottom)

http://www.cms.math.ca/Competitions/MOCP/2004/sol_jan.html

Now certainly the ellipse case is more troublesome... :lol:

[quote="al.M.V."]lol, this might be of use: (scroll all the way to the bottom)

http://www.cms.math.ca/Competitions/MOCP/2004/sol_jan.html[/quote]

Wow, thanks for the link, that solution is really fascinating. Still no idea how one can come up with something like this. I have edited the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=3440]unanswered post in the Geometry Unsolved Section[/url] to contain the link, and moved it into the Solved Section.

BTW, "20000 problems under the sea" seems to be down for already more than a year, I think. Did they get copyright troubles like Kalva?  :?

  Darij

[quote]
http://www.cms.math.ca/Competitions/MOCP/2004/sol_jan.html

Wow, thanks for the link, that solution is really fascinating. Still no idea how one can come up with something like this. [/quote]

Well, when my friend solved that problem (in the exact same way) and presented the solution to me, I was quite puzzled as well. But he said it's always better to work with a regular n-gon inscribed in a circle than just a circle itself, and since he was taking a linear algebra course at university, all he was encountering were vectors and their various products... :D So I guess the solution just "naturally" occurred to him... :lol:


Define non-descending number as a positive integer with digits $a_1a_2a_3 \ldots a_i$, such that $a_1$ <= $a_2$ <= $a_3$ <= $\ldots$ <= $a_i$.

How many 4 digit non-descending numbers are there?

If I randomly select a 5 digit number, what is the probability that it is non-descending? Express as common fraction in lowest terms.

[quote="mna851"][quote]Define non-descending number as a positive integer with digits , such that         .

How many 4 digit non-descending numbers are there?[/quote]

Is the answer to this one 900?[/quote]

How did you do it?

[hide]That's not the answer I got, but maybe you are right. Depends on the work you show.  ;) [/hide]

Do you know of a shortcut to this? I can't find one.

Again, I'm not good at counting problems, but I'm giving it a try :D I hope it's right lol. If there's a better way than mine, please tell me.

1) [hide]First, we'll count the non-descending 4 digit numbers that have all different digits. It is obvious that no matter which four different numbers you choose, you can order the digits in increasing order. It is also obvious that 0 cannot be one of the digits chosen, because 0 would then have to be the first digit, and cannot be. Therefore, there are $\binom{9}{4}=126$ such numbers.

Now, we'll count the number of non-descending 4 digit numbers that have two similar digits. The two similar digits [i]must[/i] be right next to each other, otherwise the number wouldn't be non-descending. There are three different digits, so for each combination of those three, exactly one non-descending number can be formed. There are $\binom{9}{3}=84$ such numbers.

Now, we'll count the number of non-descending 4 digit numbers that have three similar digits. Following the same reasoning, we arrive at the conclusion that there are $\binom{9}{2}=36$ such numbers.

Finally, we count the amount of numbers that have 4 of the exact same digits--there are obviously $\binom{9}{1}=9$ of them, for each of the digits 1 through 9.

I forgot the first time, but now, consider the numbers that have 2 pairs of 2 similar digits. Clearly, the number of such non-descending numbers is also $\binom{9}{2}=36$

Our final total is $\binom{9}{4}+\binom{9}{3}+2(\binom{9}{2})+\binom{9}{1}=291$ possible non-descending numbers.
[/hide]

2) [hide]Following the same logic as Part 1, the amount of 5 digit non-descending numbers is:

$\binom{9}{5}+\binom{9}{4}+2(\binom{9}{3})+2(\binom{9}{2})+\binom{9}{1}=501$ numbers.

There are a total of 90000 five digit numbers, so the probability that a randomly chosen 5 digit integer is non-descending is $\frac{501}{90000}=\frac{167}{30000}$
[/hide]

[quote="ln(dx/dy)"][quote="white_horse_king88"]That's not the answer I got, but maybe you are right. Depends on the work you show.  [/quote]

The answer doesn't depend on the work you show. There can only be one solution to this.[/quote]

Haha. Yeah. I was just trying to get him to show some work, and maybe show that my answer is wrong, considering this is a problem I made up.

[quote="Zyloch"]Again, I'm not good at counting problems, but I'm giving it a try :D I hope it's right lol. If there's a better way than mine, please tell me.

1) [hide]First, we'll count the non-descending 4 digit numbers that have all different digits. It is obvious that no matter which four different numbers you choose, you can order the digits in increasing order. It is also obvious that 0 cannot be one of the digits chosen, because 0 would then have to be the first digit, and cannot be. Therefore, there are $\binom{9}{4}=126$ such numbers.

Now, we'll count the number of non-descending 4 digit numbers that have two similar digits. The two similar digits [i]must[/i] be right next to each other, otherwise the number wouldn't be non-descending. There are three different digits, so for each combination of those three, exactly one non-descending number can be formed. There are $\binom{9}{3}=84$ such numbers.

Now, we'll count the number of non-descending 4 digit numbers that have three similar digits. Following the same reasoning, we arrive at the conclusion that there are $\binom{9}{2}=36$ such numbers.

Finally, we count the amount of numbers that have 4 of the exact same digits--there are obviously $\binom{9}{1}=9$ of them, for each of the digits 1 through 9.

Our final total is $\binom{9}{4}+\binom{9}{3}+\binom{9}{2}+\binom{9}{1}=255$ possible non-descending numbers.
[/hide]

2) [hide]Following the same logic as Part 1, the amount of 5 digit non-descending numbers is:

$\binom{9}{5}+\binom{9}{4}+\binom{9}{3}+\binom{9}{2}+\binom{9}{1}=381$ numbers.

There are a total of 90000 five digit numbers, so the probability that a randomly chosen 5 digit integer is non-descending is $\frac{381}{90000}=\frac{127}{30000}$
[/hide][/quote]

1) Very close!

How about a number, that can be written in the unique forms, where a, b, c, and d are unique digits:

abcd, aabc, abbc, abcc, aabb, abbb, aaab, aaaa

2) Same thing.

Here's what I got:

[hide="1"]There are $114$ 4-digit non-desending integers.[/hide]
[hide="2"]If you randomly select a 5-digit integer, the probability that it will be non-desending is $\frac{7}{5000}$[/hide]

I don't have time to explain how I got this.

Any more ideas?

Zyloch - try the hint I gave.

ln(dx/dy), what did you do. Whenever you have time, if you have time. Because numer 1 is wrong.


Sea $P_{n}(x)=(...(((x-2)^{2}-2)^{2}-...)^{2}-2)^{2}$, donde $2n$ indica la cantidad de parentesis que hay en total, hallar el coeficiente de $x^{2}$.
--
ahi les va ese, lo pongo asi, ya que si pongo demostrar que el coeficiente es cierto valor, aplicas induccion y ya estaria, espero que lo resuelvan, chau.

[quote="aev5peru"]Sea $P_{n}(x)=(...(((x-2)^{2}-2)^{2}-...)^{2}-2)^{2}$, donde $2n$ indica la cantidad de parentesis que hay en total[/quote]

[color=blue]A cual $2n$ te refieres?[/color]

Según lo que entiendo hay $n$ parejas de parentesis es decir asi $n$ asi ().
Sean $f(k)$ el coeficiente de $x^{2}$ y $g(k)$ el coeficiente de de $x$ en $P_{k}(x)$. Primero veamos que el termino constante de $\sqrt{P_{k}(x)}$ es $2$ para todos los valores de $k > 1$ puesto que $P_{k}(x) = (P_{k-1}(x)-2)^{2}$. Entonces tenemos que $g(n+1) = 4g(n)$. Como $P_{1}(x) = x^{2}-2x+4$ tenemos que $g(n) =-2^{2n-1}$.

Además tenemos que:
$f(n) = 4f(n-1)+g^{2}(n-1) = 4f(n-1)+2^{4n-6}$ puesto que $2$ solo se puede escribir como suma de dos enteros no negativos como $2+0$ o $1+1$.

Entonces tenemos que:
$f(n+1) = 4f(n)+2^{4n-2}$ y
$16f(n) = 64f(n-1)+2^{4n-2}$
Restando las dos obtenemos la recurrencia:
$f(n+1) = 20f(n)-64f(n-1)$

Solucionamos la ecuación característica $x^{2}-20x+64 = 0$ para obtener que:
$f(n)= A\cdot 4^{n}+B\cdot 16^{n}$. Pero podemos calcular fácilmente $f(1)=1$ y $f(2)=8$ y basta solucionar el sistema:
$4A+16B = 1$
$16A+256B = 8$
que nos deja $A= \frac{1}6$ y $B=\frac{1}{48}$ es decir:
$f(n) = \frac{4^{n}}{6}+\frac{16^{n}}{48}= \frac{2^{2n-1}(2^{2n-3}+1)}{3}$

QED

¿Alguien tiene otra solución?


Find the maximum value and the minimum value of $(1+\sin{\theta}) (1+\cos{\theta})$ if $0\le \theta <2\pi$.

Expanding, we get $(1+\sin x)(1+\cos x)=1+\sin x+\cos x+\sin x\cos x$. Let $u=\sin x\cos x$, then $\sin x+\cos x=\sqrt{1+2u}$. Now we need the minimum and maximum of $1+u+\sqrt{1+2u}$. I don't really know how to proceed from here without using calculus.

Hint

[hide]$(1+\sin x)(1+\cos x)=\frac{2+2\sin x+2\cos x+2\sin x\cos x}{2}$[/hide]

 :)

We are to deal with a quadratic function by setting certain replacing.

We'll use a few trig transformations:

$1+\sin\theta=1+\cos({\pi\over 2}-\theta)=2\cos^2({\pi\over 4}-{\theta\over 2})$

$1+\cos\theta=2\cos^2{\theta\over 2}$

Hence

\begin{eqnarray*}f(\theta) &=& (1+\sin\theta)(1+\cos\theta)\\ &=& 4\left(\cos({\pi\over 4}-{\theta\over 2})\cos{\theta\over 2}\right)^2\\ &=& 4\left[{1\over 2}\left(\cos({\pi\over 4}-\theta)+\cos{\pi\over 4}\right)\right]^2\\ &=& \left({1\over\sqrt{2}}+\cos(\theta-{\pi\over 4})\right)^2\end{eqnarray*}

Minimum value of $f(\theta)$ is $0$, for $\cos(\theta-{\pi\over 4})=-{1\over\sqrt{2}}$ or $\theta_1=\pi,\theta_2={3\pi\over 2}$

Maximum value of $f(\theta)$ is $(1+{1\over\sqrt{2}})^2=\frac{3+2\sqrt{2}}{2}$, for $\cos(\theta-{\pi\over 4})=1$ or $\theta={\pi\over 4}$

By $(\sin \theta +\cos \theta)^2+(\sin \theta -\cos \theta)^2=2,$ we have $-\sqrt{2}\leq \sin \theta +\cos \theta \leq \sqrt{2}.$

Thus $(1+\sin \theta)(1+\cos \theta)=\frac{1}{2}(1+\sin \theta +\cos \theta)^2,$

Maximum value is $\frac{3+2\sqrt{2}}{2}.$

minimum value is $0.$


algebra
polynomial
number theory unsolved
number theory

[quote="harazi"]Is it true that if P is a polynomial with integer coefficients such that for any natural number n P(n) is the sum of two squares, then P is the sum of the squares of two polynomials with rational coefficients?[/quote]

dig, dig, dig :P
Anybody, any solution?


superior algebra
superior algebra unsolved

In any finite group $ G$,

if $ S\equal{}\{x\ne e: x^5\equal{}e\}$ , then $ |S|$ is odd.

 :)

No, it's always even. They come in fours.

O.K.

That's what I thought about it.

So ,$ |S|$ is a multiple of $ 4$. ??? :maybe:

I think my thought about order to be a multiple of $ 4$ is true , since we can prove $ x,x^2,x^3,x^4$ are distinct from identity. Using this argument we could prove it has an order which is multiple of $ 4$.

Is this true ?? :)

You also have to prove that if $ y\not\in \{x,x^2,x^3,x^4\}$, then $ \{x,x^2,x^3,x^4\}\cap\{y,y^2,y^3,y^4\}\equal{}\emptyset$.

Yes , I knew that and I did it like this manner :

if $ b$ is a solution which is not a power of $ x$ , then if possible :

assume $ b^2 \equal{} x^{i}, i\in \mathbb Z$ , thus $ (b^2)^3 \equal{} x^{3i} \equal{} b$ .

Hence , $ b \equal{} a^{3i}$ which is contradiction .

Similarly for other powers .

Note that : $ b,b^2 , b^3 , b^4$ are different from identity .

 :) Thx


function
inequalities
vector
inequalities proposed

Suppose that $x_1,x_2,..,x_n$ are positive reals and:
$\dfrac{1}{x_1}+\dfrac{1}{x_2}+..+\dfrac{1}{x_n}=1$

Find the minimum of:

$P=\dfrac{1}{\sqrt{1+x_1^2}}+\dfrac{1}{\sqrt{1+x_2^2}}+...+\dfrac{1}{\sqrt{1+x_n^2}}$

________________________

I guest that we can find the maximum, but I don't solve it at once.

An important Idea for it, the minimum is $\dfrac{1}{\sqrt{2}}$

The function $f(x)=x/\sqrt{1+x^2}$ is concave. Thus, the inequality follows
by Karamata's Inequality:
$f(x_1)+...+f(x_n) \ge (n-1)f(0)+f(1)$.

Vasc you based at $(x_1,x_2,.....,x_n)\succ (0,0,0,.....,1)$. Why this majorization holds?

[quote="Vasc"]The function $f$ is concave. Hence, the function $-f$ is convex.
The vector $(1,0,0,.....,0)$ majorizes the vector $(x_1,x_2,.....,x_n)$:
$(1,0,0,.....,0)\succ (x_1,x_2,.....,x_n)$.[/quote]

Good but the sums of the vectors $A=1+0+0+0+...+0$ and $B=x_1+x_2+...+x_n$ are equal?

Edit: Sorry for the annoying and thanks for the help

OK! I don't know the function is concave.

I use Cauchy-Swacter:

$(\sum\dfrac{a_i}{\sqrt{1+a_i^2}})^2(\sum{a_i(1+a_i^2)} \ge (\sum{a_i})^3$

And $\sum{{a_i}^3} \le (\sum {a_i})^3$

[quote="Vasc"]Siluan, first we replace all $x_i$ by $1/x_i$. Is it clear now?[/quote]

Yes thnk you very much for the information

[quote="silouan"][quote="Vasc"]Siluan, first we replace all $x_i$ by $1/x_i$. Is it clear now?[/quote][/quote]
But there is something else. If you replace all $x_i$ by $1/x_i$ you cannot tell about $f(0)$ I am right?

Here's another approach.  We know that $x_i > 1$ (for $n \ge 2$).  Thus
\[ \dfrac{1}{\sqrt{1 + x_i^2}} > \dfrac{1}{\sqrt{x_i^2 + x_i^2}} = \dfrac{1}{x_i \sqrt{2}} \, .  \]
Summing over $i$, we get
\[ \sum_{i=1}^n \dfrac{1}{\sqrt{1+x_i^2}} > \sum_{i=1}^n \dfrac{1}{x_i \sqrt{2}} = \dfrac{1}{\sqrt{2}} \, .  \]
Thus the infimum (not really minimum) is $\dfrac{1}{\sqrt{2}}$.

Beautiful approach Ravi B. So the equality holds when $n=1$.

[quote="silouan"][But there is something else. If you replace all $x_i$ by $1/x_i$ you cannot tell about $f(0)$ I am right?[/quote] $f(x)=x/\sqrt{1+x^2}$.
Hence $f(0)$ exists: $f(0)=0$.

Vasc's solution  for  Silouan :

Let  $y_i = \dfrac{1}{x_i}$.Then we  have  $y_1+y_2+...+y_n=1$  and  we  want to find the minimum  value  of the  $P=\sum\dfrac{y_i}{\sqrt{1+y_i^2}}$

The  function  $f(y)=\dfrac{y}{\sqrt{1+y^2}}$  is  strictly  concave  for  $y>0$ , and  $(1,0,0,...,0)\succ(y_1,y_2,...,y_n)$. By Karamata's  inequality  we  have :

$f(y_1)+f(y_2)+...+f(y_n)\geq f(1)+f(0)+...+f(0)$  or  $\sum\dfrac{y_i}{\sqrt{1+y_i^2}}\geq f(1)+(n-1)f(0)$

or  $P\geq\dfrac{\sqrt{2}}{2}$

Equality  holds  if-f  one  of  numbers  $x_i$  is  equal  to  1  and the  rest  $x_i\rightarrow\propto$
Thats  why  Ravi  B  called  it  ''infimum'' and not  ''minimum''


The positive real numbers $x,y,z$  satisfy $x+y+z=1$. Show that
\[\sqrt{3xyz}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{1-x}+\frac{1}{1-y}+\frac{1}{1-z}\right)\geq4+
\frac{4xyz}{(1-x)(1-y)(1-z)}.\]

I can't solve it.
Could I see your proof,sbos?

Eh, here we go...

By weighted AM-GM, we have

$\sum_{\text{cyc}} \left(3\cdot xy+\frac{1}{9}\right) \geq 12\sqrt[12]{\frac{x^6y^6z^6}{9^3}} \iff xy+yz+zx+\frac{1}{9} \geq \frac{4}{\sqrt{3}}\sqrt{xyz}.$

Now, note that

$\sqrt{3xyz}\left(\sum_{\text{cyc}}\frac{1}{x}+\sum_{\text{cyc}}\frac{1}{1-x}\right) = \sqrt{3xyz}\left(\frac{xy+yz+zx+1}{(1-x)(1-y)(1-z)}+\frac{xy+yz+zx}{xyz}\right)$

$\geq \frac{4xyz}{(1-x)(1-y)(1-z)}+\sqrt{3xyz}\left(\frac{8/9}{(1-x)(1-y)(1-z)}+\frac{xy+yz+zx}{xyz}\right),$

so it suffices to prove that

$\sqrt{3xyz}\left(\frac{8/9}{(1-x)(1-y)(1-z)}+\frac{xy+yz+zx}{xyz}\right) \geq 4.$

By weighted AM-GM again, we have

$\sqrt{3xyz}\left(\frac{8/9}{(1-x)(1-y)(1-z)}+3\cdot \frac{xy+yz+zx}{3xyz}\right) \geq 4\cdot \sqrt{3xyz}\cdot \sqrt[4]{\frac{8(xy+yz+zx)^3}{9\cdot 3^3x^3y^3z^3 (1-x)(1-y)(1-z)}},$

and this reduces our claim to the following:

$9x^2y^2z^2\cdot 8(xy+yz+zx)^3 \geq 9\cdot 3^3x^3y^3z^3 (1-x)(1-y)(1-z),$ or equivalently,

$8(xy+yz+zx)^3 \geq 27xyz(x+y)(y+z)(z+x) = 27(xy+yz)(yz+zx)(zx+xy).$

Now set $a=xy, b=yz, c=zx$, so we have to prove that

$8(a+b+c)^3 \geq 27(a+b)(b+c)(c+a).$

But, this is trivial:

$8(a+b+c)^3 - 27(a+b)(b+c)(c+a) = 8(a^3+b^3+c^3)-3\sum_{\text{sym}} a^2b -6abc,$

so we may apply $2a^3+b^3 \geq 3a^2b$ repeatedly and $a^3+b^3+c^3 \geq 3abc$ (by AM-GM).

Whew...

[quote]Now set $a=xy, b=yz, c=zx$, so we have to prove that

$8(a+b+c)^3 \geq 27(a+b)(b+c)(c+a).$

But, this is trivial:

$8(a+b+c)^3 - 27(a+b)(b+c)(c+a) = 8(a^3+b^3+c^3)-3\sum_{\text{sym}} a^2b -6abc,$

so we may apply $2a^3+b^3 \geq 3a^2b$ repeatedly and $a^3+b^3+c^3 \geq 3abc$ (by AM-GM).[/quote]

If you put $a+b =A$ and similar it is AM-GM for three numbers ;)

Heh, I guess that works too... anyways, it was a trivial last step, but thanks for pointing that out!  :D

what does [b]weighted AM-GM[/b] state? (i never heard of it)

[b]Weighted AM-GM:[/b]

If $x_1,...,x_n$ are nonnegative real numbers and $w_1,...,w_n$ are positive real numbers satisfying $w_1+\cdots +w_n=1$, then

\[\prod_{i=1}^n x_i^{w_i} \leq \sum_{i=1}^n w_i x_i.\]


Hi,

I'm trying to make that package (http://cbr.jic.ac.uk/dicks/software/newicktree/) work, but it doesn't. I think I have to compile my tex file using LaTeX => PS => PDF, but I don't have this option in TeXniC Center, how can I get it ?

Newicktree does work with TeXniC Center to generate ps files, but you need to use ghostview or some other thing to create the pdfs.

George

Ok, it works, thanks !

However I'm loosing all my "hyperlinks" when going from ps to pdf, it's normal ?


Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc\equal{}2.$ Prove that:
\[ a^3\plus{}b^3\plus{}c^3\geq a\sqrt{a\plus{}b}\plus{}b\sqrt{b\plus{}c}\plus{}c\sqrt{c\plus{}a}\]


Please write down here your favourite quote...
well, probably, I should elaborate on what I meant.
A quote might be not associated with math, moreover, your quote can also be just a little story that touched you...

Old age isn't so bad when you consider the alternative
- Maurice Chevalier

"I never let my schooling get in the way of my education" ~M. Twain
"People demand freedom of speech as compensation for freedom of thought, which they rarely use." ~S. Kierkegaard
"When you categorize me, you negate me." ~S. Kierkegaard
"A witty saying proves nothing." ~Voltaire

Don't quote me on those, they might deviate from the originals.

"The world is a dangerous place, not because of those who do evil, but because of those who look on and do nothing."-Albert Einstein

just the first one i thought of :wink:

Wow, last year our Social Studies teacher had us describe what that quote meant on the WWII test (in reference to the Holocaust).

"Success is 1% inspiration and 99% perspiration"-Thomas Edison

Because it describes me so well. I'm not all that smart :)

See the signature. I have THREE
||
||
||
||
||
||
||
\/


Let $S=\{a,b,c,a+b-c,b+c-a,c+a-b,a+b+c\}$ be a set of seven distinct Primes and let $d=\max S-\min S$. if $800\in\{a+b,b+c,c+a\}$, find the maximum value of $d$.please dont just sit around giving a link, try to get your own solution. :D

[b]lemma.[/b] Let $a,d$ be positive integers such that $a>3$ and $a,a+d,a+2d$ are primes, then $6|d$.
[i]Proof.[/i] Because $a$ and $a+d$ are odd therefore $d=(a+d)-a$ is even. No has element of set $A=\{a,a+d,a+2d\}$ is divisble $3$, so there at least two elements $m,n$ of $A$ such that $3|m-n$, therefore $3|d$. Also $6|d$.

Now, for your problem. If $a+b-c,b+c-a,c+a-b$ all great than $3$ then by $2b=(a+b-c)+(b+c-a)$,etc... and lemma we have $a\equiv b\equiv c(\mod{6})\Rightarrow 3|a+b+c$ , contradiction for $a+b+c$ is prime. So if $a+b-c$ is minimum number in them then $a+b-c=2$ or $3$, if $a+b-c=2$ then $a+b=c+2$ is odd, contradiction. Final $a+b-c=3$.

At this time, that is all what i know on your problem :D

will i post my soluiton? ah, here's my soluiton:

i also got here. then proceeded like this(tell me if i'm wrong):

WLOG $a>b>c$.since $\max S=a+b+c$ and $\min S=b+c-a=3$, we have $d=2a$ hence to find max d we find max a. also since $800\in \{a+b,a+3,a+c\}$, and $b>c>3$, we find max a when a=797(note that this is prime). hence max d=1594.


draw the picture and attach it........... don't just descrbe it, your describing suxs.

G                   G is a right angle
       /\
      /   \
     /     \
E  /____\F
   / l       l \
A/_l___ l_\D
   B      C

i said draw it and attach it as a seprate file............ nt draw it w/ the buttons on a keyboard...........

STOP MAKING FUN OF ME  IM NOT SLOW!!!!!! zzzzzzzzzzz :rotfl:

ill show u tommorow at schoo'        mesa a bit sleepy

chao


Show that if :$a^m-b^n=1$ where $(a^m,b^n)$ is diffirent from pair $(9,8)$ then $(m,n)$ is Wayerferich pair.
  Denote $(m,n)$ is called Wayerferich pair iff $n^2\equiv 1(\mod m^{n-1})$ and $m^2\equiv 1(\mod n^{m-1})$
  This result was found out by Preda Mihailescu,Romanian,work in Swedden,when he proved Catalan's conjecture.I know it's very hard but anyway it's too very nice and
 well worth talking about.


ratio
geometry
circumcircle
geometric transformation
reflection
perpendicular bisector
similar triangles

In a triangle $ ABC$ , let $ \overline{BM}$ be the median, $ \overline{BD}$ the internal bisector and $ \overline{BN}$ the median simetric over the bisector.

Show that $ \frac {AN}{CN} \equal{} \frac {c^2}{a^2}$

There is a typo... the relation should be AN/CN = (c/a)^2, classic; use Steiner to prove, trivial.

Best regards,
sunken rock

Except I asked to use triangle proportion, so I don't think it's that trivial.

Thank you for identifying the typo, I fixed it.

Does anybody know how to solve it this way?

Let $ P$ be the second intersection of $AM$ with the circumcircle of  $\triangle ABC.$ We have $ \triangle ACN \sim \triangle ABP$ and $ \triangle ABN \sim \triangle ACP.$ Therefore

$ \frac {CN}{PB} \equal{} \frac {CA}{AP} \ , \ \frac {PC}{BN} \equal{} \frac {AP}{AB} \Longrightarrow \frac {CN}{BN} \equal{} \frac {CA \cdot PB}{AB \cdot PC}$

Since P lies in the reflection of the A-Apollonius circle about the perpendicular bisector of BC, we have then

$\frac {PB}{PC} \equal{} \frac {CA}{AB}$ $\Longrightarrow \frac {BN}{CN} \equal{} \frac {AB^2}{CA^2}.$

Thank you once again for your solution though, I have a doubt, I would appreciate it if you could clarify how can I show that $ P$ belongs to the apolonious circumference?


Is it possible for a knight to land on every square of the chess board starting from one corner and end on the opposite (diagonally across) corner?

I know that this is one of the first problems is the ussr problem book. Also I saw that a computer checked all the combinations or something. It says that on Wolfram.

314159265358979323846 wrote:Correct me if I did anything wrong. (spoiler)

[hide]Every time the knight moves, it goes to a square of a different color (if it was on white, it is now on black). So when the total number of moves which have occured is 1 (mod 2), the knight is on a square of a different color from the one it started on. It would take 63 moves for the knight to land on all 64 squares. 63= 1 (mod 2). But the diagonally opposite corners are of the same color. Therefore, this case is impossible.
[/hide]

I believe that is correct, correct me if I'm wrong, I'm too tired.

The knight problem was actually headline news on MathWorld ([url]mathworld.wolfram.com[/url]).  There are combinations like this:

<img src=http://mathworld.wolfram.com/news/2003-08-06/magictours/KnightsTours.gif>

(These are 6 out of the 140 total)

However, none of them meet the requirements, as Jennifer proved.

BTW, the headline news was that there was no [url=http://mathworld.wolfram.com/news/2003-08-06/magictours/][i]magic tour[/i][/url].

The problem you asked was much simpler than the computer checked one on the MathWorld site. The problem on the MathWorld site asked if it was possible to have a magic knight's tour on an 8*8 board. That is, if we mark off the number of moves it takes for the knight to get to each square and write that number on the square, can the board be an 8*8 magic square?

The proposed solution is correct, and I thought I made this problem up...but apparently not... :)

Presumably you did make it up. It is likely that other people came up with this problem before. (That certainly has happened to me before.) After your problem was solved, people started talking about another related problem.

Here's another problem that's similar to the one you proposed. If you have seen it before, please don't post a solution. Suppose we take an 8*8 chessboard and cut off two opposite corners. Thus we have 62 squares left. If we take 31 dominos, each of which can cover two consecutive squares, can we cover the mutilated chessbaord?

The answer is no.

Consider the outermost squares.  There are 26 of them (7*4 - 2).  The gaps at the corners split them into 13 on one side and 13 on the other.  In order to cover this perimeter, we can put 6 dominoes on completely, and have one overlap inward, like so:

<img src=http://userpic.livejournal.com/6678973/1236582>

But then we have another chessboard missing two corners, this time 6x6.  Applying the same process, we have a 4x4 missing two corners.  Then a 2x2 missing one corner, which looks like:

._....
|_| _
....|_|

Which can't be covered.  Thus, we conclude that we cannot cover the mutilated chessboard with dominoes.

You don't necessarily have to outline the entire perimeter before putting any dominoes over the next layer inside; you could, but it's not necessary, so I don't believe your proof is rigorous.

Spoiler:

[hide] take a look at the color of the squares being removed. [/hide]

That's true.  [hide]Adjacent (non-diagonal) squares have to have opposite colors, when there are really two too many of one color...[/hide].


group theory
abstract algebra
superior algebra
superior algebra solved

$p,q$ are different primes. $|G|=pq$. Prove that there exists a subgroup with $q$ elements in $G$.

Do not use Sylow's Theorem.

How about Cauchy's Theorem? :) There is an element of order $q$, and it generates a subgroup of order $q$.

[quote="grobber"]There is an element of order $q$[/quote]
Can you explain it please?
I know the abelian case only...

Look for Cauchy's Theorem with Google:

If a prime $p$ divides the order of a finite group (it doesn't have to be abelian), then that group has an element of order $p$. It really is well-known, man :).


AMC
USA(J)MO
USAMO
number theory
least common multiple
modular arithmetic

Prove that no integer solutions $(x,y)$ exist to $19^{19}=x^3+y^4$

I have no idea how to start the casework..... :(

[hide]consider the equation modulo 13.  By Fermat's theorem we find the right side to be congruent to 7 mod 13.  Also $\displaystyle x^{12}\equiv 0,1 \mod 13$.  This implies that $\displaystyle x^3\equiv -1,0,1 \mod 13$ because $x^{12}=(x^3)^4$.  Additionally we find this implies $y^4\equiv 0,1,3,9 \mod 13$ because $y^{12}=(y^4)^3$.  However none of these combinations produce 7 therefore there are no integer solutions to the given equation. [/hide]

[quote="amd2357"] This implies that $\displaystyle x^3\equiv -1,0,1 \mod 13$ because $x^{12}=(x^3)^4$.  [/quote]

Why? For example:   $2^3 \equiv 8 \mod 13 $

you're right Gyan I was a little tired at the time.  It made sense then.  Anyway, for a more brute force method checking different residues of $x^3 \mod 13$ we have $x^3 \equiv -5,-1,0,1,5$ upon which we still find no possible combination with the set $0,1,3,9$ produces a residue of $7 \mod 13$

This problem becomes very similar to USAMO #2 from this year. I wish I had seen it sooner so I could have thought of modulo 13.

why did you choose mod 13? is it b/c that LCM(3,4) = 12? could you chose other mods as well besides 13?

Fermat's Little Theorem and the fact that the LCM of 3,4=12.

If you don'y mind me asking, how does taking modulo get you anywhere. Could you describe each step in detail?

1234567890, do you know fermat's little theorem?

http://mathworld.wolfram.com/FermatsLittleTheorem.html

mod 13 was chosen because when you use this theorem, you are reduced to a power of 12, into which both x^3 and y^4 can go into.
do you follow it somewhat now?

[quote]1234567890, do you know fermat's little theorem? [/quote] Yes.
5^2 conguent to 1 mod 3, 3^7 congruent to 1 mod 8, etc...

[quote]mod 13 was chosen because when you use this theorem, you are reduced to a power of 12, into which both x^3 and y^4 can go into.
do you follow it somewhat now?[/quote]
Would you mind showing me the intermediate steps.? I don't understand what you mean by taking mods.

1234567890, you have some errors in your statements. ;)

Have you tried finding the possible equivalences of $x^3, x^4$ in modulo 13?
Also notice that $a^{12}\equiv1\pmod{13}$.

1234...., you know that FLT only applies to primes, right?  So $3^7\equiv 1\bmod 8$ isn't true.


quadratics limit function inequalities algebra topology calculus

Depends on the starting value.

If $ u_0\equal{}0,$ then $ u_n\to0$ since $ u_n\equal{}0$ for all $ n.$

If $ u_0\equal{}1,$ then $ u_n\to 0$ since $ u_n\equal{}0$ for $ n\ge 1.$

If $ u_0<0$ or $ u_0>1,$ then $ u_n\to\minus{}\infty.$

If $ 0<u_0<1,$ then $ u_n\to\frac35.$

Any limit this sequence can have must be a root of the equation $ x\equal{}f(x),$ where $ f(x)\equal{}\frac52x(1\minus{}x).$ This quadratic equation has two roots, $ x\equal{}0$ and $ x\equal{}\frac35.$

$ f'(x)\equal{}\frac52\minus{}5x.$

$ |f'(0)|\equal{}\frac52>1,$ so the fixed point at $ 0$ is unstable. We can only converge to it if at some point we hit it exactly.

$ \left|f'\left(\frac35\right)\right|\equal{}\left|\minus{}\frac12\right|\equal{}\frac12<1.$ so the fixed point at $ \frac35$ is stable; every sequence which comes sufficiently close to it will converge to it.

The entertaining part of the iteration $ u_{n\plus{}1}\equal{}Cu_n(1\minus{}u_n)$ takes place with $ 3<C<4.$

I forgot that:

$ u_0 \equal{} \frac {1}{2}$  :mad:  :mad:  :mad:

So in this case $ u_n \rightarrow \frac {3}{5}$

Why?

I have found: $ \frac {1}{2}\leq u_n \leq \frac {5}{8}$ but I couldnt prove that $ u_n$ is monotonic

To say that the limit of $ u_n$ is the root of $ f(x)\equal{}x$ we must prove that $ u_n$ is monotonic  :)

[quote="FOURRIER"]I have found: $ \frac {1}{2}\leq u_n \leq \frac {5}{8}$ but I couldnt prove that $ u_n$ is monotonic[/quote]
Once you have that $ u_n\in\left[\frac12,\frac58\right]\equal{}I$ and that $ u_{n\plus{}1}\equal{}f(u_n),$ note also that you have shown that $ f: I\mapsto I$ and that on $ I,\ \minus{}\frac58\le f'(x)\le 0,$ from which we can show that $ |f(x)\minus{}f(y)|\le\frac58|x\minus{}y|.$

Now quote the Contraction Mapping Lemma (also known as the Banach Fixed Point Theorem.)

Easy to show that $ \frac{1}{2}\le x_n\le\frac{5}{8},\forall n\in\mathbb{N}$

Let $ f(x)\equal{}\frac{5}{2}x(1\minus{}x),\forall x\in\; [\frac{1}{2},\frac{5}{8}]$

$ \minus{}\frac{5}{8}\le f'(x)\equal{}\frac{5}{2}\minus{}5x\le 0$ then $ |f(x)|\le\frac{5}{8},forall x\in\;[\frac{1}{2},\frac{5}{8}]$

We also have :

the equation $ f(x)\equal{}x$ has only a root $ \frac{3}{5}\in\;[\frac{1}{2},\frac{5}{8}]$

By Lagrange theorem exists $ c\in\;[\frac{1}{2},\frac{5}{8}]$ so that :

$ \frac{f(x_{n\plus{}1})\minus{}\frac{3}{5}}{x_{n\plus{}1}\minus{}\frac{3}{5}}\equal{}f'(c)\le \frac{5}{8}$ or $ x_{n\plus{}2}\minus{}\frac{3}{5}\le\frac{5}{8}(x_{n\plus{}1}\minus{}\frac{3}{5})\le\left(\frac{5}{8}\right)^n(x_2\minus{}\frac{3}{5})$ then $ \lim_{n\to\infty}x_n\equal{}\frac{3}{5}$

[quote="Kent Merryfield"]
Now quote the Contraction Mapping Lemma (also known as the Banach Fixed Point Theorem.)[/quote]

Is this a well-known theorem?

[quote="jmerry"]It's not monotonic. It alternates between the two sides of the fixed point.[/quote]
 :o
Are you sure? how do you know?

As for the alternation - why don't you try just running some values of this sequence numerically? That could be done quite quickly using a calculator and even more quickly using a spreadsheet program. Here's a short list:

$ 0.5$
$ 0.625$
$ 0.5859375$
$ 0.606536865$
$ 0.596624741$
$ 0.601659149$
$ 0.599163544$
$ 0.600416479$
$ 0.599791327$
$ 0.600104228$
$ 0.599947859$
$ 0.600026064$
$ 0.599986966$
$ 0.600006516$
$ 0.599996742$
$ 0.600001629$
$ 0.599999185$
$ 0.600000407$

The alternation (as well as the convergence) is obvious.

For more of a visual explanation, try drawing the "spiderweb diagram." Sketch on the same axes a graph of $ y\equal{}x$ and of $ y\equal{}f(x).$ In this particular case, $ f(x)\equal{}\frac52x(1\minus{}x).$ Start with the point $ (u_0,u_0)$ on the line $ y\minus{}x$. Draw a vertical line segment from there (whichever of up or down you need) until you intersect the curve $ y\equal{}f(x).$ Then draw a horizontal line segment (whichever of left or right you need) until you intersect $ y\equal{}x$ again. That is now the point $ (u_1,u_1).$ Repeat this process.

In this particular example, because $ y\equal{}f(x)$ is a decreasing function in a neighborhood of the intersection point at $ x\equal{}\frac35,$ your sequence of line segments will make a curve that spirals around the point of intersection. Hence the resemblance to a spider web. (Because $ |f'(x)|<1$ in that neighborhood, the spiral goes inward toward the intersection; you'll note that both quangpbc and I seemed to be concerned with the size of $ |f'(x)|$ - that's why.)

-----

Yes, the Banach Fixed Point Theorem (aka Contraction Mapping Lemma) is a well-known theorem. Here's the statement:

Suppose $ X$ is a complete metric space. (For your elementary purposes, you may use that $ X$ is a closed subset of $ \mathbb{R}.$) Suppose that $ f$ maps $ X$ into $ X$ and is a [i]contraction[/i], which is defined as follows: there is a positive number $ K<1$ such that for all $ x$ and $ y$ in $ X,$ $ d(f(x),f(y))\le Kd(x,y).$ (For subsets of $ \mathbb{R},$ read that last inequality as $ |f(x)\minus{}f(y)|\le K|x\minus{}y|.$) Then $ f$ has a unique fixed point in $ X.$ That is, there is precisely one solution in $ X$ to the equation $ x\equal{}f(x).$

The proof goes like this: choose an arbitrary $ u_0\in X.$ Recursively define the sequence by $ u_{n\plus{}1}\equal{}f(u_n).$ Then the contraction condition and an inductive argument shows that $ |u_{n\plus{}1}\minus{}u_n|\le K^n|u_1\minus{}u_0|\equal{}CK^n.$ Hence for $ m>n,$

$ |u_m\minus{}u_n|\le C\sum_{j\equal{}n}^{m\minus{}1}K^j\equal{}\frac{C(K^n\minus{}K^m)}{1\minus{}K}\le\frac{CK^n}{1\minus{}K}.$

This can be made arbitrarily small by making $ n$ sufficiently large; hence the sequence $ (u_n)$ is a Cauchy sequence. In a complete metric space, a Cauchy sequence must converge to a limit $ x.$ Noting that $ f$ is a continuous function (the contraction condition implies that), we take the limit in $ u_{n\plus{}1}\equal{}f(u_n)$ to get $ x\equal{}f(x).$ Hence we have a fixed point.

I leave to you the proof that this fixed point is unique, other than to offer the hint that the contraction condition is also needed for this.

(Some of you may notice that I lapsed back into saying $ |x\minus{}y|$ rather than $ d(x,y).$ I did that for familiarity, but the proof still works in an arbitrary complete metric space.)


calculus
integration
function
algorithm
calculus computations

Calculate: $ \int\sqrt {\frac {x^3}{A^3 \minus{} x^3}}\,\text dx$. Here $ A$ is a constant.

According to mathematica, the antiderivative is not elementary.

hello, the antiderivative contains the ellitpic function.
Sonnhard.

Thank you for your help.

Could you please tell me how to know that some antiderivative is not elementary or contains elliptic function?

hello, i hope this is somthing for you
http://www.cs.berkeley.edu/~fateman/282/lects/14.ppt
Sonnhard.

Integration is really sensitive.  The [url=http://en.wikipedia.org/wiki/Risch_algorithm]Risch algorithm[/url] can help, but it's not necessarily short and simple.  As seen on that page, change one term, even by 1, and an elementary antiderivative may not exist!