Here is very difficult log question I'm having trouble with:

[tex]1.  Log_8 (n-3) + Log_8 (n+4) = 1[/tex]

[tex]2.  Log_2 (2x+8) - Log_2 (2x^2+21x+61) = -3[/tex]

I got 4 for number one and I stuck at [tex]0 = 8x^2+85x+246[/tex]in number 2.

I really appreciated if someone helps me out here.

Thanks!

4 is correct for the first one.  But you're totally off on the second one.  Just try the laws of logs. What do you do when you subtract logs of the same base?  Then try getting the log out of the picture.

I got whre I screwed up.

2^(-3) is 1/8, not -8..

Ok, good then you should be able to get the right answer.


modular arithmetic
quadratics
number theory
greatest common divisor

[quote="frt"]Prove that $n^7-n$ is divisible by 42.[/quote]
[hide]
for $n \ne 0$ we can factor $n(n-1)(n^2 +n + 1)(n + 1)(n^2 - n + 1) \equiv 0 \pmod{42}$

We need to show that we can always find the primes $2,3,7$ in the terms.
we know we can find $2$ and $3$ because we have 3 consequtive terms $n - 1, n, n +1$
Now if we look at quadratic residues of 7. We see $1,4,2,2,4,1$ and if we factor $n^2 + n + 1 = (n + 1)^2 - n$ and $n^2 - n + 1 = (n - 1)^2 + n$, we see that we always get a 7 in either of the factored forms. That is, either $(n - 1)^2 + n \equiv 0 \pmod{7}$ or $(n + 1)^2 -n \equiv 0 \pmod{7}$ This is sufficient to prove for all N because of the modulus cyclesw.
[/hide]

There's a whole bunch of problems like this in the USSR Olympiad Problem Book.

Note that Fermat's little theorem immediately tells you that $n^7 - n$ is divisibly by 7.

[quote="tarquin"]Well $n^7 \equiv n \pmod n \Rightarrow n^6 \equiv 1 \pmod n$ iff $GCD(n,7)=1$[/quote]Yeah but that GCD condition will always be true unless $7|n$, which is what we were trying to show.  :P


[quote="chess64"]What is the maximum value of $n$ for which there is a set of distinct positive integers $k_1, k_2, \ldots, k_n$ for which

\[k_1^2+k_2^2+\ldots+k_n^2 = 2002.\]

(A) 14       (B) 15       (C) 16       (D) 17       (E) 18[/quote]

[hide]I know the answer is D. I have the practice AMC 10 sheet too.

I don't really understand the proof to the answer... it seems like it was purely guess and check...  :( [/hide]

They're trying to say that you know that $1^2+2^2+3^2...+18^2$ is more than 2002, so you know you can't have 18 numbers, since we just proved that the smallest possible sum is over 2002. Then out of no where, they noted that $1^2+2^2+3^2+...+19^2-18^2-12^2 = 2002$ that's 19-2=17 numbers.

...
well
...
we already know that
...
I posted this here b/c somebody might have a better solution
...
:huuh:


[quote="alex2008"]Let be $ a,b,c > 0$ . Prove that :

$ \frac {a^2b}{c^2(a^2 \plus{} bc)} \plus{} \frac {b^2c}{a^2(b^2 \plus{} ca)} \plus{} \frac {c^2a}{b^2(c^2 \plus{} ab)}\ge \frac {a^2}{a^2 \plus{} bc} \plus{} \frac {b^2}{b^2 \plus{} ca} \plus{} \frac {c^2}{c^2 \plus{} ab}$[/quote]

Try $ a\equal{}b\equal{}c\equal{}10$  :wink:

Sorry for my mistake , dduclam . Now i edited :)

[quote="alex2008"]Let be $ a,b,c > 0$ . Prove that :

$ \frac {a^3b}{c^2(a^2 \plus{} bc)} \plus{} \frac {b^3c}{a^2(b^2 \plus{} ca)} \plus{} \frac {c^3a}{b^2(c^2 \plus{} ab)}\ge \frac {a^2}{a^2 \plus{} bc} \plus{} \frac {b^2}{b^2 \plus{} ca} \plus{} \frac {c^2}{c^2 \plus{} ab}$[/quote]
$ \Leftrightarrow \sum \frac{a^2}{a^2\plus{}bc}(\frac{ab}{c^2}\minus{}1) \ge 0$

$ \Leftrightarrow \sum\frac{1}{1\plus{}\frac{bc}{a^2}}(\frac{ab}{c^2}\minus{}1)\ge 0 (*)$

setting $ x\equal{}\frac{ab}{c^2}$; $ y\equal{}\frac{bc}{a^2}$; $ z\equal{}\frac{ca}{b^2}$

$ \Rightarrow xyz\equal{}1$

$ (*) \Leftrightarrow \frac{x\minus{}1}{y\plus{}1}\plus{}\frac{y\minus{}1}{z\plus{}1}\plus{}\frac{z\minus{}1}{x\plus{}1}\ge 0$

$ \Leftrightarrow \frac{x\plus{}y}{y\plus{}1}\plus{}\frac{y\plus{}z}{z\plus{}1}\plus{}\frac{z\plus{}x}{x\plus{}1}\ge 3$

$ \Leftrightarrow \sum xy^2 \plus{}\sum x^2\ge \sum x \plus{}3$

Apply $ Am\minus{}Gm$ inequality, we have:

$ \sum xy^2\ge 3xyz\equal{}3$

$ \sum x^2\ge 2(x\plus{}y\plus{}z)\minus{}3\equal{}x\plus{}y\plus{}z\plus{}(x\plus{}y\plus{}z\minus{}3)\ge x\plus{}y\plus{}z\plus{}(\sqrt[3]{xyz} \minus{}3)\equal{}x\plus{}y\plus{}z$

DoNe.

[quote="alex2008"]Let be $ a,b,c > 0$ . Prove that :

$ \frac {a^3b}{c^2(a^2 \plus{} bc)} \plus{} \frac {b^3c}{a^2(b^2 \plus{} ca)} \plus{} \frac {c^3a}{b^2(c^2 \plus{} ab)}\ge \frac {a^2}{a^2 \plus{} bc} \plus{} \frac {b^2}{b^2 \plus{} ca} \plus{} \frac {c^2}{c^2 \plus{} ab}$[/quote]
Here is my solution:
$ \sum \frac {a^3b}{c^2(a^2 \plus{} bc)} \equal{} \sum \frac {1}{\frac {c^2}{ab}(1 \plus{} \frac {bc}{a^2})}$

Denote $ (x,y,z) \equal{} (\frac {bc}{a^2},\frac {ca}{b^2},\frac {ab}{c^2})\Longrightarrow xyz \equal{} 1$
Hence the inequality is equivalent to proving:
$ \frac {x}{1 \plus{} y} \plus{} \frac {y}{1 \plus{} z} \plus{} \frac {z}{1 \plus{} x}\ge \frac {1}{1 \plus{} x} \plus{} \frac {1}{1 \plus{} y} \plus{} \frac {1}{1 \plus{} z}$, $ (1)$, where $ xyz \equal{} 1$.
$ (1)\Leftrightarrow (x \plus{} y \plus{} z) \plus{} (xy \plus{} yz \plus{} zx) \plus{} (x^2 \plus{} y^2 \plus{} z^2) \plus{} z^2y \plus{} x^2z \plus{} y^2x\ge 3 \plus{} 2(x \plus{} y \plus{} z) \plus{} (xy \plus{} yz \plus{} zx)$
$ \Leftrightarrow x^2 \plus{} y^2 \plus{} z^2 \plus{} x^2z \plus{} z^2y \plus{} y^2x\ge 3 \plus{} (x \plus{} y \plus{} z)$, $ (i)$.
By [b]AM-GM[/b], we obtain
$ x^2 \plus{} y^2 \plus{} z^2\ge 3\sqrt [3] {x^2y^2z^2} \equal{} 3$
$ x^2z \plus{} z^2y \plus{} y^2 \equal{} \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x}\ge \frac {x \plus{} y \plus{} z}{\sqrt [3] {xyz}}\equal{}x\plus{}y\plus{}z$
Which implies that $ (i)$ is true obviously.
The result is lead as follow.

[quote]$ x^2z \plus{} z^2y \plus{} y^2 \equal{} \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x}\ge \frac {x \plus{} y \plus{} z}{\sqrt [3] {xyz}}\equal{}x\plus{}y\plus{}z$
Proof of this:
By $ AM\minus{}GM$, we obtain:

$ \frac {x}{y}\plus{}\frac {x}{y}\plus{}\frac {y}{z}\ge 3 \sqrt [3] {\frac {x^2}{yz}}\equal{}\frac {3x}{\sqrt [3] {xyz}}$

$ \Longrightarrow 3LHS\ge \frac {3(x\plus{}y\plus{}z)}{\sqrt [3] {xyz}}$. The result is lead as follow.[/quote]

[quote="caubetoanhoc94"][quote="mathVNpro"][quote="alex2008"]Let be $ a,b,c > 0$ . Prove that :

$ \frac {a^3b}{c^2(a^2 \plus{} bc)} \plus{} \frac {b^3c}{a^2(b^2 \plus{} ca)} \plus{} \frac {c^3a}{b^2(c^2 \plus{} ab)}\ge \frac {a^2}{a^2 \plus{} bc} \plus{} \frac {b^2}{b^2 \plus{} ca} \plus{} \frac {c^2}{c^2 \plus{} ab}$[/quote]
Here is my solution:
$ \sum \frac {a^3b}{c^2(a^2 \plus{} bc)} \equal{} \sum \frac {1}{\frac {c^2}{ab}(1 \plus{} \frac {bc}{a^2})}$

Denote $ (x,y,z) \equal{} (\frac {bc}{a^2},\frac {ca}{b^2},\frac {ab}{c^2})\Longrightarrow xyz \equal{} 1$
Hence the inequality is equivalent to proving:
$ \frac {x}{1 \plus{} y} \plus{} \frac {y}{1 \plus{} z} \plus{} \frac {z}{1 \plus{} x}\ge \frac {1}{1 \plus{} x} \plus{} \frac {1}{1 \plus{} y} \plus{} \frac {1}{1 \plus{} z}$, $ (1)$, where $ xyz \equal{} 1$.
$ (1)\Leftrightarrow (x \plus{} y \plus{} z) \plus{} (xy \plus{} yz \plus{} zx) \plus{} (x^2 \plus{} y^2 \plus{} z^2) \plus{} z^2y \plus{} x^2z \plus{} y^2x\ge 3 \plus{} 2(x \plus{} y \plus{} z) \plus{} (xy \plus{} yz \plus{} zx)$
$ \Leftrightarrow x^2 \plus{} y^2 \plus{} z^2 \plus{} x^2z \plus{} z^2y \plus{} y^2x\ge 3 \plus{} (x \plus{} y \plus{} z)$, $ (i)$.
By [b]AM-GM[/b], we obtain
$ x^2 \plus{} y^2 \plus{} z^2\ge 3\sqrt [3] {x^2y^2z^2} \equal{} 3$
$ x^2z \plus{} z^2y \plus{} y^2 \equal{} \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x}\ge \frac {x \plus{} y \plus{} z}{\sqrt [3] {xyz}}$
Which implies that $ (i)$ is true obviously.

The result is lead as follow.[/quote]

Your solution same bunhiacovski' solution  :lol:[/quote]
Actually, I am editing my solution so that it can be cleared and easy to understand.But they are not exactly the same :lol:. I think that I am still the one who post at first. :D


calculus
derivative
geometry
geometric transformation
integration
real analysis
abstract algebra

Thank you very much for your answers on my last question!!

Here is another and final question:

Suppose f`(x) is bounded on an open interval (lets say (- :inf:, + :inf: ).
Is it then [i]always[/i] true that f(x) is uniform continuos on that interval?

Again, I am quite sure the answer is yes, but need to be 100 % sure :-)

You can use Largrange's theorem here: $f(x)-f(y)=f'(z)(x-y)$, where $z\in[x,y]$, so $|f(x)-f(y)|\leq M\cdot |x-y|$ ($M=\sup |f'(x)|$), and if $\delta=\varepsilon/M$, then $|x-y|<\delta$ implies $|f(x)-f(y)|<\varepsilon$.

[quote]You can use Lagrange's theorem[/quote]
American calculus and analysis textbooks would call that the Mean Value Theorem. It's the only argument, of course.

We have the following implications:

(has a bounded derivative) $\implies$ (Lipschitz) $\implies$ (uniformly continuous) $\implies$ (continuous).

A good exercise in knowing your standard counterexamples is to supply counterexamples to the converses of each of these implications.

Mean Value Theorem in Russia is about mean values of continuos functions. (just a remark)

The translation to English of the book by Spivak says that the Mean Value Theorem is exactly that one you mention ($f(a)-f(b)=(a-b)f'(c)$ for $c\in(a,b)$, and $f$ differentiable), but it also mention the [b]Intermediate[/b] Value Theorem as the one that says that, if $f$ is continuous in $[a,b]$, $f(a)=A$, $f(b)=B$, and $C$ is an intermediate value between $A$ and $B$. then $\exists c\in(a,b)$ such that $f(c)=C$.

Best,

[quote]The translation to English of the book by Spivak[/quote]
Translation "to" English? Is that what you meant to say? Spivak is an American author who writes in English. His terminology in this particular instance (Mean Value Theorem, Intermediate Value Theorem) is standard among American authors.

[quote="Kent Merryfield"][quote]You can use Lagrange's theorem[/quote]
American calculus and analysis textbooks would call that the Mean Value Theorem. It's the only argument, of course. [/quote]We also call it Lagrange's Theorem in Romania, where as the mean value theorem has a similar form, that of there exists a $c \in (a,b)$ such that
$ \int_a^b f(x) dx =(b-a) f(c) $ if $f$ is continous, whereas in Lagrange's theorem $f'$ need not be continous.

Most American authors would call the theorem Valentin just quoted the Mean Value Theorem for Integrals. (And some would go on to a Second Mean Value Theorem for Integrals using a variable weight in the integral.)

However, the particular textbook I'm using this semester, Kenneth Ross's [i]Elementary Analysis: the Theory of Calculus[/i] somewhat confusingly calls that the Intermediate Value Theoerm for Integrals.

[quote="Kent Merryfield"][quote]The translation to English of the book by Spivak[/quote]
Translation "to" English? Is that what you meant to say? Spivak is an American author who writes in English.[/quote]

Ups!! :oops: You know, when I took my first undergrad calculus we used the Apostol and the Spivak Textbooks (actually, the version we used is quite different to the third edtion that they use currently), and by some reason, my professor (who did his Ph.D in Russia) had the Spivak Text book in russian. Then I said, well, Spivak is a Russian last name (I don't know more than two or three words in Russian, but I believe it means "drunk man") and the text in Russian, I thought probably he was Russian, but never checked  :blush:

Sorry,


calculus
integration
function
calculus computations

Hi everybody!!!!
I need your help :( I've stack on this problem and I can't get rid of it anymore... I've tryed again and again... just looking at it I feel it has to end up with the Gamma function but I don't menage to show it properly

I've got this definte integral

${\int_{-\infty}^{\infty}x^{2m+1}e^{-x^{2}/2}}dx$   with  $m \geq 1$

What can I do?

If first I think to the indefinite integral... if I tried by part I find

 ${\int x^{2m+1}e^{-x^{2}/2}}dx = \\ \\ \frac{x^{2(m+1)}}{2(m+1)}e^{-x^{2}/2}+\frac{x^{2(m+2)}}{4(m+1)(m+2)}e^{-x^{2}/2}+\frac{x^{2(m+3)}}{2^{3}(m+1)(m+2)(m+3)}e^{-x^{2}/2}+\int \frac{x^{2(m+3)}}{2^{3}(m+1)(m+2)(m+3)}e^{-x^{2}/2}dx$

So iterating I should have
 ${\int x^{2m+1}e^{-x^{2}/2}}dx = e^{-x^{2}/2}\sum_{i=1}^\infty \frac{x^{2(m+i)}}{2^{i}\prod_{j=1}^{i}(m+j)}$

And now??????? :(

Please help me

many many thanks in advance
vale

Use the substitution $x=-y$ and you get that the integral is 0.

:maybe:  :(  :(
Sorry mazbe I'm too tired or dull but I don't get it ...

with the substitution I will get

$x =-y \: \: dx =-dy$

$\int_{-\infty}^{\infty}y^{2m+1}e^{-y^{2}/2}dy$

isn't it the same??? :(

Hi,

well I can't say you why , but if you plug this integral in Mathematics it gives to you

Integrate[$x^{2m+1}e^{-x^{2}/2}}dx$,x,$-\infty$,$+\infty$, Assumption -> $m\geq 1$] $=$

$-2^{m-1}(-1+(-1)^{2m}) \Gamma(m)$

and not zero
but I don't know how to help...

sorry
kiko

notice that
$(-1+(-1)^{2m})$ is equal to $0$ for all $m$ so the answers are the same.

:blush:  ooops you're right

by the way I think vale is right in claiming there is a gamma function and I think the algebra she did is right but I don't know how the gamma comes out?
It would be intresting to figure it out... maybe from the series convergence she write???

and moreover I don't understand when Didi write
[quote="didilica"]let $I=\int_{-\infty}^{\infty}x^{2m+1}e^{-x^{2}/2}dx$.
Now, let $x=-y$ then

$I=\int_{\infty}^{-\infty}(-y)^{2m+1}e^{-y^{2}/2}(-dy)=\int_{-\infty}^{\infty}-y^{2m+1}e^{-y^{2}/2}dy=-I$ hence[/quote]

 :maybe: but $-dy$ times $(-y)^{2m+1}$ isn't $+y^{2m+1}$

sorry but I'm quite curius now ...

kiko

Is $m$ an integer?  If it is, then the integrand is odd since $2m+1$ is odd for all integers $m \geq 1$.

The integral of an odd integrand over an interval with endpoints that are opposites is zero (if the integrand is defined for all values in that interval, of course).

Let ${J=\int_{-\infty}^{\infty}x^{2m+1}e^{-\frac1{2}x^{2}}}dx$   with  $m \geq 1$. Break-up the integral as

\[{{J=\int_{-\infty}^{0}x^{2m+1}e^{-\frac1{2}x^{2}}}dx+\int_{0}^{\infty}x^{2m+1}e^{-\frac1{2}x^{2}}}dx\]

Substitute $y=-x\Rightarrow dy=-dx$ so that $-\infty < x\leq 0\Rightarrow \infty < y\leq 0$ in the first integral to get

\[{{{J=(-1)^{2m+1}\int_{0}^{\infty}y^{2m+1}e^{-\frac1{2}y^{2}}}dy+\int_{0}^{\infty}x^{2m+1}e^{-\frac1{2}x^{2}}}dx = \left(1-(-1)^{2m}\right)\int_{0}^{\infty}x^{2m+1}e^{-\frac1{2}x^{2}}}dx\]

Now substitute $u=\frac1{2}x^{2}\Rightarrow du=xdx$ so that $0\leq x<\infty\Rightarrow 0\leq u<\infty$ and the integral becomes

\[{J=2^{2m}\left(1-(-1)^{2m}\right)\int_{0}^{\infty}u^{m}e^{-u}}du = 2^{2m}\left(1-(-1)^{2m}\right)\Gamma (m+1)\]

[color=red][size=200] Thank you so much  [/size][/color]

benorin you've been so clear and many thanks to everybody

 :jump:

bye bye,
happy vale


Putnam calculus AMC USA(J)MO USAMO integration Harvard

I'm a senior and will be going to college next september, which seems very close right now.  Anyway, I want to be a math major, so I was hoping for advice from people who are in college now or just know about it.  I always get off to a slow start, but this time I want to hit the ground running.

Anway here's a list of basic worries/questions that I have:
(1) What classes should I take?  Any I should not take?  What are the common freshman math courses?  If you are in college, what classes have you taken?
(2) Math team/math club is a big part of my HS math career.  Are there organizations like this in college that I should join?
(3) Im a little afraid of the Putnam exam...any advice here?
(4) In HS most of my math classes tests and quizzes are the only grades that we have.  In college is it true that there are no grades except one big final test?
(5) Is there anything I should know before going to college?  I talked to a math professor at the college I am going to and he said that I should have calculus experience (I sortof had a problem w/my HS and I am in Adv. precalc now), so I am taking Calc I and Calc II at a local university this summer.  Is there anything else I should def. know?
(6) Off the topic of math, is it hard to live in the same room as someone else 24/7?  Do people party all the time and sleep during class, etc. (general college student stereotypes)?
(7) Is it hard adjusting to the workload of college classes from HS?  Is it difficult going from being one of the smartest kids in your HS to being average (or worse, yikes)?
(8) General advice that dosent fall under the questions I have already asked? (both about (a)being a math major and (b) being a college student).

Thanks!  Sorry for the long list of questions... Some of the other seniors on AoPS are probably thinking the same things.

Have you visited the Web site for the math department of the school you are going to next year? That should have answers for a few of those questions. Some of the answers to your questions are definitely specific to particular schools. If you could kindly name the school, I could take a look at the math department Web site myself and think about some of your questions.

Tokenadult, thanks for responding.  I thought that these would be general sortof questions for any school or about other people's experiences.  The school I am going to next year is carnegie mellon.  I have looked at the website and I have even met w/a math professor from there (which was very helpful).  I did ask him a lot of questions, a lot of these things (ie question 6) are not stuff I would have asked.  I meant for these questions to be about being a math major in general, not anything specific to a certain college.  Thanks for the response!

Wherever you go, whatever you do, the main piece of advice I can give for college is to ditch the slow start habit.  In college, I always spent the first 3 weeks immersing myself in my course - working way ahead where I could, etc.  Then the rest of the semester was a cakewalk.  I'd spend much of the latter part of the semester doing other things while everyone else was stressing out, trying to catch up, cramming.  Don't fall behind - it's very easy to do in college, as there are many classes with no homework, or with homework that is easy to get deferred.

Get on top of the game early.  You'll work less than other students, learn more, and do better.  Complicated concepts take a while to sit your in your brain before they become yours.  If you fight with them early, by the time finals roll around, they'll have been in your head long enough for you to fully understand them.  Midnight cram sessions won't have that effect, and won't get you through math at the high level you'd like at any decent college.

As for the Putnam - it's not the end of the world if you don't take it.  I'm probably on the very short list of USAMO winners who never took the Putnam.  Doesn't seem to have hurt me too much.

I agree completely with Mr. Rusczyk.  Heed what he says, not what I did.

Math had always come very easy for me -- easier than falling off a stegosaurus.  Never studied in HS.  Always aced the tests.  Always one of the highest in the class.  Etc.  Perhaps a lot like you.

My first semester at Bedrock A&M, I took Calc 3.  This was one of the courses they used to "weed out" marginal students.  I aced the first exam.  Didn't turn in any homework.  I got a low A/high B on the second exam.  Didn't turn in any homework.  I got a low C on the third exam.  Didn't turn in any homework.  I failed the fourth exam.  Didn't turn in any homework.  [I had better things to do!  This was University!]

It was time for the final.  I knew I was in deep dino-doo.  Just prior to the Final Exam, I spent about 3 days in the library -- just me and my calculus textbook -- cramming.  Cramming.  CRAMMING!

It was time for the final.  Adrenaline was running higher than the last time I was chased by a sabertooth tiger. I knew I was in trouble.

About fifteen minutes after the exam began, one of the hundreds of students taking the exam stood up, shouted some obscenities at no one in particular, and stormed out of the examination room.  ["That could have been me!", I thought to myself.]

To my surprise, I finished the 2 hour exam with time to spare.  I even had time to check my work.  My exam was done.  I had checked my work.  There were still some minutes remaining.  Hardly anyone had left the room.  There was nothing more I could do.  As I got up to turn in my paper, my Graduate Assistant recitation teacher caught my eye and shook his head in sorrow.  I turned in my paper, and left.

Three days later, the scores were out.  I had scored a perfect 198 out of 198 on the Final.  Among hundreds of students, the next higher score was more than 20 points below mine.  I passed the course with a "B".  It's the only "B" I have ever gotten in a math class.  I always felt lucky that I didn't fail the class.  At the same time, I resented not getting an "A".  And I also knew that if I had just kept up, I could easily have earned the "A".

Follow Richard's advice.  Get ahead at the start.  Stay ahead if you can.  Don't allow yourself to fall behind.

First of all, you probably won't want to take any advice I might give you on (1).

(2) It depends on the university. We have an undergraduate math club here that meets every Friday at 3. Last quarter we had two guest speakers (Joe Gallian and Isidore Singer) who gave interesting talks. Every few weeks, one of the professors gives us a problem sheet with challenging problems (some of which he doesn't know how to solve), and we work on those. I'll try to post some of the problems. The Putnam is the only math contest for undergraduates that I know of though, and I find that a shame.

(3) The Putnam is a pretty scary exam. I think many universities have a Putnam seminar where you go through a bunch of problems and techniques. That might help you. I'm still terrified of Putnam integrals, even though we spent about half the fall quarter learning how to do them.

(4) In the US, that doesn't tend to be true. In some classes, the grade is something like 40% midterm and 60% final, but more often, there are homework assignments. I had one class last quarter that was 30% homework and 70% research project or final (it was up to you which one you wanted to do; I did the research project). Most classes I have taken have had homework assignments.

(5) I don't know. I wasn't in the least bit prepared for most of the math classes I was taking in the first quarter.

(6) Yes, it is hard to live in the same room as someone else, especially if that someone else goes to sleep six hours later than you do and has his computer right next to your bed. Well, most people seem to get on with their roommates much better than I do, but next year I get a roommate of my choice who will almost certainly be much easier to live with. Some people party all the time, and some people (including myself) stay as far away as possible from anything resembling a party. Plenty of people sleep through class, but plenty of people sleep through class in high school as well, so that isn't anything new for me.

I don't think I can answer (7) or (8) properly. It just depends on your constitution.

Well, I think I'll take number 7 that Simon passed on, because as far as I can tell other people have covered everything else pretty well.  Also, I think number 5 is probably covered by number 7, since there isn't really a "base of knowledge" that they expect you to have learned or anything, beyond what was necessary to get accepted.

As for number 7, I will compare my experience to that of some of my roommates.  I came to college and said, "Wow, this seems really different -- I'd better take some easy classes my first semester in order to make sure I can get adjusted to the system and whatnot."  My roommate came to college and said, "Well, I've taken all of the hardest classes my highschool offered and aced all of them, so let me just continue to take all of the hardest classes offered."  (Note: not direct quotations.)  I will leave it up to the reader to guess which one of us has had more leasure time and relaxation.  Now, if you happen to like living in libraries and working all of the time, then my roommate's option isn't necessarily a bad one -- it's just one that (if you are like me) you might want to avoid to preserve your mental health.  Don't take 6 classes first semester.  Don't take graduate level classes first semester.  These are my pieces of mental health advice.  They might also help your GPA :).

Oh, as for grading, I've been in one class where the final grade was essentially determined by one paper.  My math and science courses, and those of my friends, have been graded on a final+midterm+problem-set basis.  It varies from school to school and from professor to professor.  Weekly problem sets seem pretty standard for Harvard and MIT.

It's so much fun taking graduate classes as soon as you get to college though. But I already told you not to listen to my advice for (1) for exactly that reason.

Thanks for taking the time to give me advice.  I really appreciate it.  As for (1) I don't think I would be able to handle graduate classes my first year in college.  I actually know what classes I will be taking next year.  There is a calculus class, another class that teaches how to write proofs, and some classes that teach you how to use math software.  Maybe I shoulden't have asked that if I already knew the answer (sorry!).  If there are any classes after the basics that are really cool...  FredFlinstone -- lol.  That was actually really good advice. "easier than falling off a stegosaurus" lol. =)

Another question I had was about the structure of classes.  Are most classes really big and people are not allowed to ask questions?  I keep myself awake in most of my classes by participating/asking questions, so this might be an adjustment for me too.

[quote="JS1527"]There is a calculus class, another class that teaches how to write proofs[/quote]
Excellent choices if you want to keep doing math.

[quote="JS1527"]Another question I had was about the structure of classes.  Are most classes really big and people are not allowed to ask questions?  I keep myself awake in most of my classes by participating/asking questions, so this might be an adjustment for me too.[/quote]

Generally math classes are not taught in a large lecture format, although this is not always the case.  More usual is probably a 20-30 student size, taught by a graduate student (at least for your calculus class).  You certainly could e-mail someone at the math department there and ask them.  At most schools, there's someone whose job it is to answer questions like that for incoming freshmen (perhaps the professor you've talked to already), so that's a good way to get a lot of your questions answered.  They'll probably also answer many of them for you when you get there :)

I think the calculus classes here are in large lecture format with a few hundred students (hopefully) listening and a professor lecturing. Then there are small sections with maybe 25 students and a graduate student discussing the material the professor went over in greater detail.


In right triangle $ ABC$,$ \angle CBA \equal{} 90^{\circ}$. Point $ D$ is constructed so that $ \angle DCA \equal{} 90^{\circ}$ and $ \angle CAD \equal{} \angle BAC.$ Point $ E$ lies on $ AC$ beyond $ C$ such that $ \angle ADE \equal{} 90^{\circ}$. If $ AB \equal{} 4$ and $ BC \equal{} 3$, find$ DE$.

(Pardon my drawing..)

[asy]size(200); pair A,B,C,D,E; A=(0,4); B=(0,0); C=(-3,0); D=(-8,3); E=(-4.5,-2); draw(A--B--cycle,blue); draw(B--C--cycle,blue); draw(C--D--cycle,blue); draw(C--A--cycle,blue); draw(D--A--cycle,blue); draw(C--E--cycle,blue); draw(D--E--cycle,blue); label("A",(0.5,4)); dot(A--cycle,red); label("B",(0.5,0)); dot(B--cycle,red); label("C",(-3,-0.5)); dot(C--cycle,red); label("D",(-8.5,3)); dot(D--cycle,red); label("E",(-5,-2)); dot(E--cycle,red); label("3",(-1.5,-0.5)); label("4",(0.5,2));[/asy]

[hide="hint"]Similar Triangles.[/hide]
Can't post solution, I gotta go right now...

[quote="spiders_hate_chestnut"][hide]25/4[/hide]
let me guess...[/quote]
Check your work. I think you messed up in the process.

Here's a solution:[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170125[/url]

[quote="Eragon"]I think the answer is 75/16. I didn't cheat, though!(I was tempted!)
Send me a pm showing how to work it if I'm wrong,okay?[/quote]

You are correct! Good Job.


Show that $ f(x)\equal{}x^2\plus{}2x$ is continuous at $ x\equal{}1$ with $ \epsilon \minus{} \delta$.

Won't work if $ \varepsilon\equal{}1000$ ;)

How about writing:
$ |x^{2} \plus{} 2x \minus{} 3|\equal{}
|x\minus{}1\parallel{}x\plus{}3|$
$ \leq |x\minus{}1|(|x| \plus{}3)$
$ <5|x\minus{}1|  if |x\minus{}1| < 1$
$ <5*\frac{\epsilon}{5}$    if   $ |x\minus{}1|< \frac{\epsilon}{5}$
$ <\epsilon$
Hence, pick
$ \delta\equal{}$ min($ {1, \frac{\epsilon}{5}}$).


Consider $ M\equal{}\{ z \in \mathbb{C}\minus{}\mathbb{R} \mid z^2\minus{}z\plus{}1 \not\equal{} 0$ and $ z^4\plus{}5z^2\plus{}1 \not\equal{}0 \}$ and the function $ f: M \rightarrow \mathbb{C}$ such that:

$ f(z)\equal{}\frac{(z^2\plus{}z\plus{}1)^2\plus{}(z^2\minus{}z\plus{}1)^2\plus{}z^4\plus{}z^2\plus{}1}{(z^2\plus{}z\plus{}1)^2\plus{}(z^2\minus{}z\plus{}1)^2\minus{}z^4\minus{}z^2\minus{}1}$

Prove that if $ f(z) \in \mathbb{R}$ then $ Re (z)\equal{}0$ or $ \mid z \mid\equal{}1$

[hide]
$ f(z)\equal{}\frac{3z^4\plus{}7z^2\plus{}3}{z^4\plus{}5z^2\plus{}1}\equal{}3\minus{}\frac{8z^2}{z^4\plus{}5z^2\plus{}1}\in \mathbb{R}$

$ \Longrightarrow \frac{8z^2}{z^4\plus{}5z^2\plus{}1}\in \mathbb{R}$

$ \Longrightarrow \frac{z^4\plus{}5z^2\plus{}1}{z^2}\equal{}z^2\plus{}\frac{1}{z^2}\plus{}5\in \mathbb{R}$

$ \Longrightarrow z^2\plus{}\frac{1}{z^2}\in \mathbb{R}$.

which easily leads us to the result.[/hide]


1.) show that $ 5^{n \minus{} 1} \plus{} 1$ is divisible by $ 6$ for all $ n>\equal{}1$
2.) show that $ 9^{n} \minus{} 1$ is divisible by $ 9$ for all $ n>\equal{}1$

Those are both false. The first one is only true for even $ n$, while the second one is never true.

[quote="guile"]1.) show that $ 5^{n \minus{} 1} \plus{} 1$ is divisible by $ 6$ for all $ n > \equal{} 1$
2.) show that $ 9^{n} \minus{} 1$ is divisible by $ 9$ for all $ n > \equal{} 1$[/quote]
Proof for why 1 is only true for even $ n$.
[hide="1"]
We have that \[ 5\equiv \minus{} 1\bmod 6\implies (5)^{n \minus{} 1} \plus{} 1\equiv ( \minus{} 1)^{n \minus{} 1} \plus{} 1\equiv \minus{} 1 \plus{} 1\bmod 6\equiv 0\bmod 6\implies 6|5^{n \minus{} 1} \plus{} 1\]
[/hide]

Well, #2 is divisible by 8 ... but that's pretty easy to prove :\


USAMTS
vector
calculus
function
trigonometry
geometry
similar triangles

Is there an error in this solution? How could have I made it right? I am not good with AM-GM and don't know most other stuff on the official solution (other than similar triangles and other basics. Any help would be appreciated.

(Sorry about the bad picture quality.)

i think you are mistaken in assuming that PR is nonnegative, because PR has a direction
you can't just substitute PQ+PR for QR unless you are certain that P is between Q and R (which it's not, think if BP< 2.5).
you have to make PQ,PR, and QR into vectors (put arrows over them i think?).
plus, using calculus, you can find that the maximum is 2.53 or something like that.

HEre are two ways i can think of to bash it easily:
in both ways, you have to find PQ in terms of some value x (i used x=PM, where M is the midpoint of BC and PM has a direction). This can be done easily with POP+Pythagorean Theorem. let f(x) equal this expression.
nuke #1: differentiate f(x) (bunch of chain rules and division rules). set f'(x) equal to 0, find the critical point,show that it's an absolute maximum, blah blah...
nuke #2 (with mathematica): set f(x) equal to 25/(4sqrt(6)), use mathematica to show that there are no real solutions, and since the function is continuous, you get the point, blah blah....
EDIT:oh wow, nuke #1 could've been a lot easier for me if i had solved using a different definition for x.

Yes, near the maximum value of $ PQ$ we have $ P$ between $ A$ and $ R$ instead of $ R$ between $ A$ and $ P$. When you have complicated diagrams you often have to check for stuff like this... but what you really should've noticed is that your solution seemed too easy.

I still have difficulty doing circle geometry. I only used law of sine in my solution. and the fact that AQ<5, and -1<sin <1

and I proved that PQ<\frac{125}{49}<\frac{25}{4\sqrt{6}}

Yes I just used Stewart's theorem and POP to get a function for PQ. Then, I just found the maximum. Here it is (actually these are all my solutions):

Yeah, I did the same thing as modularmarc.  It turns out the maximum is significantly less than $ \frac{25}{4\sqrt{6}}$, right?

But I used calc and then mathematica to get the exact value for the CV, so my solution's not as pretty as the official solution.

Math Geek, I liked your solution (you posted in the other thread). I see where I went wrong.

look at my solution for 5. It is actually a good solution without circle geometry.


inequalities
vector
inequalities proposed

[quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that $ abc = 1$.Prove that:
$ (a^3 + b^3 + c^3 + 1)(a + b - c)(b + c - a)(c + a - b) \leq4$[/quote]
Let $ a + b + c = 3u,$ $ ab + ac + bc = 3v^2,$ $ abc = w^3$ and $ u^2 = tv^2.$
$ (a^3 + b^3 + c^3 + 1)(a + b - c)(b + c - a)(c + a - b) \leq4\Leftrightarrow$
$ \Leftrightarrow(a^3 + b^3 + c^3 - 3abc + 4w^3)(3u - 2a)(3u - 2b)(3u - 2c)\leq4w^6\Leftrightarrow\Leftrightarrow$
$ \Leftrightarrow4w^6 + 4u(9u^2 - 10v^2)w^3 + 27u^2(3u^4 - 7u^2v^2 + 4v^4)\geq0.$
If $ t\geq\frac {4}{3}$ then, your inequality is true since,
$ 3u^4 - 7u^2v^2 + 4v^4 = (u^2 - v^2)(3u^2 - 4v^2)\geq0.$
While for $ 1\leq t\leq \frac {4}{3}$ we need to prove that
$ w^3\geq\frac {10uv^2 - 9u^3 + \sqrt {(9u^2 - 8v^2)u^2v^2}}{2}.$
$ (a - b)^2(a - c)^2(b - c)^2\geq0$ gives $ w^3\geq3uv^2 - 2u^3 - 2\sqrt {(u^2 - v^2)^3}.$
Id est, it remains to prove that
$ 3uv^2 - 2u^3 - 2\sqrt {(u^2 - v^2)^3}\geq\frac {10uv^2 - 9u^3 + \sqrt {(9u^2 - 8v^2)u^2v^2}}{2}.$
But $ 3uv^2 - 2u^3 - 2\sqrt {(u^2 - v^2)^3}\geq\frac {10uv^2 - 9u^3 + \sqrt {(9u^2 - 8v^2)u^2v^2}}{2}\Leftrightarrow$
$ \Leftrightarrow5u^3 - 4uv^2 - uv\sqrt {9u^2 - 8v^2}\geq4\sqrt {(u^2 - v^2)^3}\Leftrightarrow$
$ \Leftrightarrow5t - 4 - \sqrt {9t - 8}\geq4\sqrt {\frac {(t - 1)^3}{t}}\Leftrightarrow$
$ \Leftrightarrow(t - 1)^2(3t - 4)^4\geq0.$ :)

[quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that $ abc \equal{} 1$.Prove that:
$ (a^3 \plus{} b^3 \plus{} c^3 \plus{} 1)(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b) \leq4$[/quote]

The inequality can be rewriten as
\[ \frac{{\left( { \minus{} a \plus{} b \plus{} c} \right)\left( {a \minus{} b \plus{} c} \right)\left( {a \plus{} b \minus{} c} \right)}}{{abc}} \le \frac{{4abc}}{{a^3  \plus{} b^3  \plus{} c^3  \plus{} abc}}
\]

\[ 1 \minus{} \frac{{\left( { \minus{} a \plus{} b \plus{} c} \right)\left( {a \minus{} b \plus{} c} \right)\left( {a \plus{} b \minus{} c} \right)}}{{abc}} \ge 1 \minus{} \frac{{4abc}}{{a^3  \plus{} b^3  \plus{} c^3  \plus{} abc}}
\]
\[ \frac{{\sum\limits_{cyc} {a\left( {a \minus{} b} \right)\left( {a \minus{} c} \right)} }}{{abc}} \ge \frac{{\sum\limits_{cyc} {a\left( {a \minus{} b} \right)\left( {a \minus{} c} \right)} }}{{a^3  \plus{} b^3  \plus{} c^3  \plus{} abc}}
\]
the last one is obvious. :)

Nice proof, Loc. And Cauchy Schwart also kill it easily  :)

[quote="Honey_S"]
\[ 1 \minus{} \frac {{\left( { \minus{} a \plus{} b \plus{} c} \right)\left( {a \minus{} b \plus{} c} \right)\left( {a \plus{} b \minus{} c} \right)}}{{abc}} \ge 1 \minus{} \frac {{4abc}}{{a^3 \plus{} b^3 \plus{} c^3 \plus{} abc}}
\]

\[ \frac {{\sum\limits_{cyc} {a\left( {a \minus{} b} \right)\left( {a \minus{} c} \right)} }}{{abc}} \ge \frac {{\sum\limits_{cyc} {a\left( {a \minus{} b} \right)\left( {a \minus{} c} \right)} }}{{a^3 \plus{} b^3 \plus{} c^3 \plus{} abc}}
\]
[/quote]
This step is wrong. :wink:

[quote="nguoivn"] ...Cauchy Schwart ... kill it easily  :)[/quote]
I wish to see your proof, nguoivn. Thank you!

By the way, after $ (a\plus{}b)^2(a\plus{}c)^2(b\plus{}c)^2\geq2abc\left(\prod_{cyc}(a\plus{}b)\plus{}\sum_{cyc}(a\plus{}b)^3\right)$ an expanding gives:
$ \sum_{sym}(a^4b^2\plus{}a^3b^3\minus{}a^4bc\minus{}2a^3b^2c\plus{}a^2b^2c^2)\geq0,$ which is obvious.

[quote]Let a,b,c be positive numbers such that abc=1.Prove that:
(a^3+b^3+c^3+1)(a+b-c)(b+c-a)(c+a-b) ≤4[/quote]

It is so amazing!I created this problem 1 month ago and I have a very nice solution for it(I created this ineq when I solved a geometric problem in an Olympiad contest in my country(which was held in 2003).
In fact,the ineq is equivalent to $ 4a^2 b^2 c^2$≥$ (a^3\plus{}b^3\plus{}c^3\plus{}abc)(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)$
We can easy to prove this when a,b,c are not 3 sides of triangle.
If a,b,c are 3 sides of triangle,then this ineq is equivalent to
$ \frac{a^3\plus{}b^3\plus{}c^3\plus{}abc}{a\plus{}b\plus{}c}$≤$ 4R^2$,which is very easy to prove by using vector
It a really nice problem.
 Another nice problem we can apply:
Let a,b,c>0 satisfying (a+b-c)(b+c-a)(c+a-b)=1.Prove that $ a^3\plus{}b^3\plus{}c^3$≤$ 4a^2 b^2 c^2\minus{}abc$

[quote="SUPERMAN2"][quote]Let a,b,c be positive numbers such that abc=1.Prove that:
(a^3+b^3+c^3+1)(a+b-c)(b+c-a)(c+a-b) ≤4[/quote]

It is so amazing!I created this problem 1 month ago and I have a very nice solution for it(I created this ineq when I solved a geometric problem in an Olympiad contest in my country(which was held in 2003).
In fact,the ineq is equivalent to $ 4a^2 b^2 c^2$≥$ (a^3 \plus{} b^3 \plus{} c^3 \plus{} abc)(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)$
We can easy to prove this when a,b,c are not 3 sides of triangle.
If a,b,c are 3 sides of triangle,then this ineq is equivalent to
$ \frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} abc}{a \plus{} b \plus{} c}$≤$ 4R^2$,which is very easy to prove by using vector
It a really nice problem.
 Another nice problem we can apply:
Let a,b,c>0 satisfying (a+b-c)(b+c-a)(c+a-b)=1.Prove that $ a^3 \plus{} b^3 \plus{} c^3$≤$ 4a^2 b^2 c^2 \minus{} abc$[/quote]
this is also opinion  when i post this inequaliti :)


No  ab,ac,bc  are not under Sqrt  :ab/sqrt(ab+bc)         and so on.  thanks

I am sorry .  :oops:
     Let $\ a,b,c$ be positive real num bers such that  $\ a+b+c=1$
prove that  $\frac{ab}{\sqrt{ab+bc}}+\frac{bc}{\sqrt{bc+ca}}+\frac{ca}{\sqrt{ca+ab}}\leq\frac{1}{\sqrt{2}}$.
Which is true ?   :D


articles
blogs
USAMTS
linear algebra
matrix

I really like Wikipedia, I use it mainly for history and recent news (it is extremely up to date, usually when something happens in a country that country's page is changed within 24 hours) and math only comes second.

From a theoretical point of view, it may seem strange to rely on something everyone can change.

However, experience learns that remarkably enough, it is quite decent

Read the disclaimer on the website. I only use Wikipedia when I can verify the information by comparing it with an independent source.

the problem is that wikipedia is like any other website and easy to spread false information, in reality if you want reliable information you should only use print sources.  Wikipedia is a little better than a random website but it shouldn't be expected to be more reliable.  Personally i use wikipedia for random research (that it just reading abut random topics) and as an introduction to topics i have to research for school.  Any internet source may have false information that you should check for (except for sites that are attached to print sources of reliability)

Just use it like any other private website- if you really need something to be right, look it up in a few places to be sure.

[quote="catcurio"][quote="mel"][quote="catcurio"]I like Wikipedia, simply because it's the only free online encyclopedia that I know of. It's been pretty reliable for me so far. Aren't there restrictions on how you can write things for it though?[/quote]

You can access Wikipedia from China?  :huh:[/quote]I live in the US. :) But I think you should be able to access Wikipedia in China, since they have it in Chinese as well as many other languages.[/quote]

It's not a matter of whether or not you [i]should[/i] be able to view it in China.  Neither does having Wikipedia in Chinese mean that you can access it in China.  China's government has been known to censor stuff very quickly if they don't like something about it.  Google, for example, was censored for a little while a couple years back.  And, unless something has changed since the writing of [url=http://my.opera.com/leolee/blog/show.dml/47770]this[/url] article, Wikipedia has indeed been blocked by China.

[quote="joml88"]
It's not a matter of whether or not you [i]should[/i] be able to view it in China.  Neither does having Wikipedia in Chinese mean that you can access it in China.  China's government has been known to censor stuff very quickly if they don't like something about it.  Google, for example, was censored for a little while a couple years back.  And, unless something has changed since the writing of [url=http://my.opera.com/leolee/blog/show.dml/47770]this[/url] article, Wikipedia has indeed been blocked by China.[/quote]

Sharp stuff, joml88. (Although, I wouldn't believe everything that is said on that blog.)

[quote="catcurio"][quote="joml88"]It's not a matter of whether or not you [i]should[/i] be able to view it in China.  Neither does having Wikipedia in Chinese mean that you can access it in China.  China's government has been known to censor stuff very quickly if they don't like something about it.  Google, for example, was censored for a little while a couple years back.  And, unless something has changed since the writing of [url=http://my.opera.com/leolee/blog/show.dml/47770]this[/url] article, Wikipedia has indeed been blocked by China.[/quote]Hmm. I didn't know that. But what's wrong with Wikipedia, or [i]Google [/i]for that matter? And what would be the point of having Wikipedia in Chinese if you can't even view it in China?  :huh:[/quote]

Well, for one, you don't have to live in China to speak Chinese.  But, of course, the percentage of Chinese speaking people who would use this service would live in China.  You have to understand that Wikipedia isn't restricting people in China from viewing their site but rather China is censoring what the Chinese people can view.

I don't really know if Wikipedia is being censored by China, if it's only certain articles, or whatnot.  But it still remains that China stringently censors their people's access to media.  [url=http://www.post-gazette.com/pg/05350/623619.stm]This[/url] article appeared on the front page of the Wall Street Journal.  It talks about Google and their relations regarding China.

[quote="joml88"]You have to understand that Wikipedia isn't restricting people in China from viewing their site but rather China is censoring what the Chinese people can view.[/quote]Sorry if I implied that [i]Wikipedia [/i]was doing the restricting - I didn't mean to.

[quote="joml88"]But it still remains that China stringently censors their people's access to media.[/quote]I believe that's why my parents came to the US. :)

By the way, much better site reference this time. :D

[quote="ccy"]Well, I was in China this summer, and I could access Wikipedia.  Granted it was in English, but it still was Wikipedia (Didn't check Chinese version).[/quote]

I think it was blocked around August/September 2005. I can't remember when.

[quote="joml88"]I don't really know if Wikipedia is being censored by China, if it's only certain articles, or whatnot.  [/quote]

The entire site is blocked.

[quote="joml88"]But it still remains that China stringently censors their people's access to media.  [url=http://www.post-gazette.com/pg/05350/623619.stm]This[/url] article appeared on the front page of the Wall Street Journal.  It talks about Google and their relations regarding China.[/quote]

[b]That[/b] was an extremely slanted article.

But what most people don't realize is that intellectual freedom is a very elusive concept. Yes, it's true that media supposedly isn't blocked in the U.S., but that doesn't necessarily mean that there is true freedom in press.

For instance, when the Bush administration came to office, several progressive (which shall remain unnamed) sex education resources were asked to take down their site.

More recently, a University of Colorado professor, Ward Churchill, got into quite a bit of trouble for an alternative view of the 9-11 attacks.

If you look at what is published in mainstream media, there is one voice, one view. This one voice may be disguished to take on a spectrum of perspectives, but the band is quite narrow. Alternative views are usually placed in the Editorials section so that the newspaper doesn't look like it was published by the Ministry of Truth.

The point of the matter is that media is never truly free, no matter what laws are in place, media will never be able to fulfill the idealistic purpose (if it has one) of disseminating the bare naked "truth," whatever that may be.

[quote="mel"]The point of the matter is that media is never truly free, no matter what laws are in place, media will never be able to fulfill the idealistic purpose (if it has one) of disseminating the bare naked "truth," whatever that may be.[/quote]Agreed. The media often slants the facts to make it into what they [i]want[/i] people to hear, not what may actually be the truth.

I cannot read that one article (that blog) strange

I am really interested in how exactly the Chinese regime works (although slighty off topic here) , in fact i am interested in all communist regimes

What I heard in media here is that :

they block CNN
they block Penthouse and stuff like that
they give death penalty to people spreading porn  :|  on the web (I do not know if they mean 'just porn' or  porn that would be consider very illegal in Belgium too, like child porn or anything)

I also heard that they execute more people than all other countries combined (a dazzling 3400 in 2004) and that they come second in executions per capita (so their big number is no excuse)

I also heard that they have killed/unrepairably wounded falun gong members

I do stress the 'i have heard' because that is the only thing I know for sure, however, I have witnessed protests by falun gong at the UN in new york
that is no proof either however

[quote="fredbel6"]they block CNN[/quote]

This is true, more often in the past than now. CNN has a long history with China, and most of the details are unknown to the public. When the Chinese government blocks CNN, it is more of a political statement than a matter of limiting what the people can view. (The locals don't normally watch CNN anyway, and the government does not really care what foreigners watch.)

Moreover, media increasingly becoming an entertainment outlet than an information source. When CNN broadcasts controversial material, it is more likely that the purpose of the broadcast is to make a political statement or to provide entertainment that sells than to provide "truth."

[quote="fredbel6"] they block Penthouse and stuff like that [/quote]
That's true. In fact, it used to be impossible for people in China to access the USAMTS website through local internet providers.

[quote="fredbel6"]they give death penalty to people spreading porn [/quote]
While pornography (in any form) is illegal in China, I don't think they give the death penalty for that.

[quote="fredbel6"]I also heard that they execute more people than all other countries combined (a dazzling 3400 in 2004) and that they come second in executions per capita (so their big number is no excuse)[/quote]
Never heard of that.

[quote="fredbel6"]I also heard that they have killed/unrepairably wounded falun gong members.[/quote]
Falun Gong was a media circus. A lot of the practitioners didn't realize it, but they were actually pawns in a political struggle (between the U.S. and China). There are still Falun Gong practitioners in the U.S. today that are being used (without their knowledge) for political purposes.

Moreover, if people are setting themselves and buildings on fire (in any country), there are bound to be casualties when the police comes to attempt to control the situation.

The thing is, no matter how hard we try, there is no way for us to know what is true and what is not. We can read all the sources that exist, and all we get are different perspectives. The approximation of the truth gets better with each additional source read, but the diversity of the views is most likely still quite narrow.

You can attempt to learn about different sides of the story by learning different languages and reading the local newspapers of different countries, and it is only then can we see the absurdity and preposterousness of certain media releases.

http://en.wikipedia.org/wiki/Death_penalty

There is a matrix showing in 2004 the number of executions for certain countries per million
China has a dazzling number, although rivaled in per capita by Kuwait and Singapore

When you say China is not communist, do you mean it hasn't got the ideology of Karl Marx, because as I see it that has never really been accomplished.  When you speak of similarities with for instance Stalinism and Maoism, it seems North Korea it definitely still communist

China is probably more capitalist than U.S. today(economically).

[quote="mel"]Also, China is not communist. In fact, there are no truly communist regimes remaining today.[/quote]
Some people don't think there has ever been a trult communist nation.  (Even is there was, they sure didn't last long.)

Exactly what I meant in my last post, when I ask is it still communist , I actually mean is it still like Stalinism.

Seriously I don't want to bash USA or China government, it would be nice though to know more.  China has the biggest population in the world, and only India is a rival in that field, and their economy is changing.  Already is it showing global consequences (social unrest in my country as factories are closed either because of losing the competition against chinese firms or simply because they are moving over there themselves)  Not being informed is certainly never good.

[quote="ccy"]Recently, a person posted an article on wikipedia with false information, and noone has corrected the issue for quite a while.  Since then retired from the job he was working at at the time.

[[url=http://www.guardian.co.uk/usa/story/0,12271,1665974,00.html]http://www.guardian.co.uk/usa/story/0,12271,1665974,00.html[/url]]

Is Wikipedia credible enough to be used as a reliable source of information?[/quote]
(back on topic here)

i would only trust them if they had citations to a credible source, since otherwise, regular wiki moderaters are unable to detect whether they have been hoaxed if they were given a piece of information in the article that they did not know beforehand

and I would definitely trust it (whether it has a citation or not) if it was an article that a lot of people feel strongly about (such as religion, government, etc)

First off, i would like to point out this article:  http://www.msnbc.msn.com/id/10478207/?GT1=7516

Second, I think some people underestimate how well handled wikipedia is. I don't know this for a fact, but from personal experience and chekcing the histories of several pages has led me to believe that every change is screened by some moderator. Furthermore, this moderator will not only remove nonesense pranks but will also, among other things, put question marks for uncited material which seems to need a citation.


I'm looking for some good books in Physical and Inorganic Chemistry, to help me develop an interest in the subject.
I'm not looking for books that contain shortcuts for IITJEE prep. or anything; I just want an interesting read on the topic, that I can understand [i](I'm starting class XII)[/i], and will make me really like the subject.
([i]Morrison & Boyd[/i] did it for me with Organic Chemistry; So, I'm looking for something along those lines - clear explanations,etc.)
Could someone help me with this?

Peter Atkins, University Chemistry and J.D. Lee maybe?

Chemical Principles - Atkins & Jones . ebook available

You'll find ebooks for nearly every foreign author. Just need to know where and how to look :P

Could you guys give me links to the e-books or something?

Which ones exactly do you want?
First check up esnips.com and then tell me which ones you want. Only 5$/book :)

i think u get Paula Atkins from 4shared

and also try  http://www.freebookspot.in/


calculus
integration
trigonometry
limit
logarithms
calculus computations

Evaluate the following definite integrals.

(1) $\int^{\frac{\pi}{4}}_0 \frac{dx}{\cos x}$

(2)$\int^{\frac{\pi}{4}}_0 \frac{dx}{\cos ^3x}$

(3)$\int^1_0 \sqrt{x^2+1} dx$

Note:You should use the result of (1)oｒ(2)

I'd have no worries about putting any or all of these on an ordinary midterm exam for a Calculus II course. (Same course as AP Caclulus BC.) The appropriate procedures are in all of the textbooks that would be used for such a course.

I'm just answering these because I'm a bit rusty with integration.

1.  $\int \sec x = \int \frac {\cos x}{\cos^2 x} = \frac {1}{2} \int \frac {\cos x}{1-\sin x}+\frac {1}{2} \int \frac {\cos x}{1+\sin x} $.  The rest is straightforward

2.  Erm ... integrate by parts and then use (1).

3.  Substitute $x = \tan u$.  However, I would much rather use a hyperbolic substitution for (3).

Using a solution without trig, I've gotten number 3 (I think) to be about (though it's not quite complete)

$\lim_{h\to0}((x\sqrt{x^2+1}+x^2\cdot\ln{| x-1|}-2x\ln{|x-1|}+3x^2-4x-2\cdot\int(x-1)\ln{|x-1|}\text{d}x)|_{x=h}^1)$

I could easily be wrong.  I know I should have used trig.  I just wanted to see what happened without trig.

\[ \int_0^1 \sqrt{x^2 + 1} dx = \int_0^{\frac{\pi}{4}} \sec^3u \.\, du \] So the second and third are the same.


logarithms
floor function
algebra
binomial theorem

how would you find the number of digits the number $12^{16}$

If you take log of a number, and round up, you will find the number  of digits it has.

$\log 12^{16}= 16 \log 12$.  But you still need a calculator to find $\log 12$...

Suing a calculator  :wink:

Why don't you use the binomial theorem? $(10+2)^{1}6=10^{16}+\binom{16}{1}10^{15}\cdot2+\binom{16}{2}10^{14}\cdot 2^{2}+\binom{16}{3}10^{13}\cdot 2^{3}+\cdots$ so it has approximately 17 digits.

How would you bring suit against a calculator? I'd like to sue a calculator, but I don't know how. It gave me the wrong answer once. ;)

just to give another easier way to solve it,

for 12.. as each power goes up the number of digits increase by 1..

Don't know where this is derived from but it is true.

so since 12^2 = 144 which is 3 digits . 12^3=1728 and so on. We see that the power+1 is the number of digits The answer would be 17 since its power is 16.

( I have noticed this works only until a certain power and does work for some other numbers until a certain point. It seems to work well for 12 in this case until about the 26th power and for 13, the 17th power and so on...)

this seems to be similar to the log solving technique... but this one is unreliable i guess.

The log technique always works.  For a concrete example, consider $10^{3}= 1000$.  $\log 10^{3}= 3$, so 10000 must have $3+1 = 4$ digits.

(We don't actually round the number up, we round down, and then add 1.)

Looking at the bounds $10^{2}$ and $10^{3}$, the numbers in between, and their logs, it is easy to see why this works.

[quote="8parks11"]It seems to work well for 12 in this case until about the 26th power...[/quote]

You're wrong. $12^{12}$ has $13$ digits and $12^{13}$ has $15$ digits.

Probably not, seeing as he factored it.

Memorizing logs is always good; then you can interpolate! Woot!

[quote="Jeremic"]Farenhajt, would your method work for a prime number? Such as 17?[/quote]

It can work: $17^{2}=289\approx 300=3\cdot 10^{2}$ and $17^{3}=4913\approx 5000=5\cdot 10^{3}$, hence you can find workable numbers.

[quote="Meghrabi"]Hi

I think the best way to find the answer is, to see that:

10^16 < 12^16 < 20^16
Its easy to see that 12^16 has 17 digits. :)[/quote]

But you're wrong :) as all the others who claim that there are 17 digits :)

And your estimate is way too rough: $20^{16}$ has 21 digits.

Oops, you're wright 20^16 is not the same is 2*10^16. :)

Remembering logs is not hard .(its not hard to remember them, for example log 2 is about  (a little more than) 0.3  because $2^{10}$ is about $1024$ ,  a little more than $10^{3}$, and log of  3 is about .5 (a little less  than 0.5, as 3^2 is  little less than 10) so log  of 12 (=log 4+ log 3) is about  0.5+.3+.3 is about 1.1 )

Rest is simple.  (All you have to check 1.1*16 = 17.6)

Also, taking approximation is simpler than one might think.
For example you can easily say $30000>12^{4}> 20000$
[hide="because"]$12^{4}= 4^{4}\cdot 3^{4}= 256*81 > 250*80 = 20000$
 and of course $256*81 < 300*90= 27000 < 30000$ [/hide]

In either case,  $12^{16}> 20000^{4}= 16 \cdot 10^{16}$
and   $12^{16}< 30000^{4}= 81 \cdot 10^{16}$

You get a  18 digit number in either case .
(Typo edited above)

Yes, I am fairly certain that everyone is forgetting that $10^{x}$ has $x+1$ digits, not $x$ digits, so there are in fact 18 digits in $12^{16}$.

but your base is only 10...the question's base is 12...does that make a difference?

and the exponent is 16 in the question...so how does $16+1 = 18$???

No, becaues we are taking $\log(12^{16})$, to find the number of digits in base 10, not $\log_{12}12^{16}$.

what about the exponent $x+1$ rule that you said...?

When multiplying a number by 10, you are adding a 0 to the end.  Thus, each time you multiply by 10, you add 1 digit.  Note that $10^{0}=1$, $10^{1}=10$, etc, so $10^{x}$ has $x+1$ digits (iff $x$ is an integer).

More generally,

$10^{\left\lfloor x\right\rfloor}$ has $\left\lfloor x\right\rfloor+1$ digits.

[quote="Gyan"]In either case,  $12^{16}> 20000^{4}= 16 \cdot 10^{16}$
and   $12^{16}< 30000^{4}= 81 \cdot 10^{16}$

You get a  17digit number in either case.[/quote]

Ahem... Number $16$ plus 16 zeros - how many digits is that?
The same question for $81$ plus 16 zeros?  :lol:

yeah - typo above, has been edited. Thanks.

Wow just found a very good method on Dr Math.
here it is [url]http://mathforum.org/library/drmath/view/62942.html[/url]


calculus
integration
trigonometry
real analysis
real analysis unsolved

$\int_{0}^{\frac{\pi}{2}}\frac{\sin (2m-1)x}{\sin x}dx $

When $m=1$ ,its simply
$\int_{0}^{\frac{\pi}{2}}\frac{\sin (1)x}{\sin x}dx  = \frac{\pi}{2}$

using some trigonometric identities,
we can get
$\int_{0}^{\frac{\pi}{2}}\frac{\sin (2m+1)x}{\sin x}dx - \int_{0}^{\frac{\pi}{2}}\frac{\sin (2m-1)x}{\sin x}dx = 0 $

So, $\int_{0}^{\frac{\pi}{2}}\frac{\sin (2m-1)x}{\sin x}dx = \frac{\pi}{2}$ for all positive integers $m$


geometry
inradius
circumcircle
trigonometry
trig identities
Law of Cosines

[quote="LawOfSigns"]A circle of center  O  of radius  42  is inscribed in an equilateral triangle  ABC  . Another smaller circle of center  P  is tangent to the circle of center  O  and to  AB  and  BC  . There are three such circles. Let us suppose that  P  is closer to  B  than  A  or  C  . Find the radius of the circle of center  P  .

Can you prove your answer?[/quote]
Um, could a mod move this to another forum? I don't think these types of questions should be here. And I don't think that they have questions like this on the Sprint.

ok idk if this got moved or if it did, where it got moved, but here's my answer:

[hide]
14
[/hide]

i'll provide the solution in a few mins/hours/days? heh. soon.

If you use 30,60,90 triangles, the problem becomes so easy..and this can easily be a sprint round problem..

but i was using inradius area formula.. and circumradius, extended laws of sines..so it look me a while..even though this should only take like 30 secs max if you realize somehting..

and i agree with undefined

haha

First I did it with the Law of Cosines

Then saw the 30-60-90

I need to realize things faster. Or ahead of time.

Which is probably what failed me during sprint rounds :)

I don't understand what exactly you are supposed to do with the 30 60 90 triangles... :blush:

[quote="LawOfSigns"]I found that BO is 84 with the 30-60-90 triangle.  Now I'm not sure how to find the radius from circle P to the triangle.[/quote]

[hide]Now you know that $BO=84$, let me label the points like the picture attached below (see picture).

Then $OD$ is the radius, which is $42$.

Therefore, $DB=42$.

In triangle $PKB$, it is also a 30-60-90 triangle, so if the radius of circle $P$ is $r$, $PB=2PK=2r$.

Also, $DP=r$.

Therefore, $PB+DP=2r+r=3r$.

We also know that $PB+DP=DB=BO-OD=42$.

In other words, $3r=42$.

So $r=\boxed{14}$.[/hide]

thanks, that is helping me clear up some stuff...I just hope I will be ready for states.


ratio
geometry
3D geometry
algebra
polynomial
AMC

Here are the first five problems from the first AMC Mock Contest.  Please post your solutions here so that we can hear different methods of doing each problem.

1. The ratio (3^9)^3/(9^3)^9 is equal to
(A) 1/(9^8)
(B) 3^(-9)
(C) 6^27/(6^54*8^9)
(D) 9^16/3^59
(E) (1^3)^1/(3^1)^3

Source: mathfanatic

2. Queen Rippledances palace features a big lawn. The two gardeners, Tare and Mith, hate this lawn because it takes a long time to mow. Tare can mow the lawn in 10 hours. Mith can mow the lawn in 8 hours. How long will it take the two to mow the lawn if they both work at the same time using two lawnmowers?
(A) 40/9 hours
(B) 9/2 hours
(C) 9 hours
(D) 79/18 hours
(E) none of these

Source: mathfanatic

3. The sum of 27 consecutive integers is 2646. Find the sum of the digits in the lowest integer of these integers.
(A) 8
(B) 10
(C) 11
(D) 12
(E) 13

Source: mathfanatic

4. Find the sum of the squares of the roots of the equation x^4 - 3x:^2: + 2 = 0.
(A) 6
(B) 8
(C) 9
(D) 12
(E) 15

Source: mathfanatic

5. The volume of a cube is 216 cm^3. Find the sum of the lengths of all the edges.
(A) 56
(B) 64
(C) 72
(D) 84
(E) 92

Source: mathfanatic

- mathfanatic

([quote="zscool"]4) let y = x :^2: , then the polynomial is
y :^2: - 3y + 2, and the sum of the roots of this polynomial is 3; it remains to note that the roots of this polynomial are squares of the roots of the original polynomial, and if we want the sum of the squares of the roots of the original polynomial, we double the sum (to account for the negative roots squared) and get 6.[/quote]

^
|
like your idea better than the previous one

It means a failed attempt to use BBCode :).  It meant to have :^2: I think.


Find the moment of inertia of a segment of height h of a sphere of radius a about the axis of the segment.

I'm having a hard time visualizing this.  Do you have a picture, or could you describe it better?

Remember that finding moment of inertia is just plugging something into the formula $ I\equal{}\int r^{2}\,dm$.

I have the answer only,

$ \frac{Mh(20a^2\minus{}15ah\plus{}3h^2)}{10(3a\minus{}h)}$.


I was wondering if the MC competitions are biased to one subject every year like they were this year (c and p). If so, can someone please list the subjects MC was biased towards for the last 3 or 4years.

It's not intentional, I don't think...they were some time ago...

Usually you can get an idea of what is going to be stressed the most by looking at the school handbook.  They have a problem index in the back that is very useful.  All topics can be covered on any test but usually what is in the handbook is stressed a little more than the other topics.

You judt need to make sure that you are reasonably good at everything rather than concenctrating on just a few areas.  As soon as you do that , then the topics that you did not study will show up.  Every MS in the country gets a copy of the handbook.

Good luck preparing for this year's competitions.


I can't figure out this geometry problem, some help would be appreciated.

See diagram: degree measures of some arcs and angles of circle C are shown. Can you determine the degree measure of minor arc AB, and if so, what is it?

I don't see any obvious reason that you can't move the whole thing around, shrinking or increasing AB at will.  This is just a first examination, however.  Is C meant to be the center of the circle?

C is the center.

and that may well be the case.

Actually, I take it back -- the 96 and 38 do fix our angle in a particular position.  I'm too tired for this -- I'll look at it again tomorrow and see what I can do.

You were right the first time JBL. The problem, as it stands now, is unconstrained. Arc AB can be as small as 0:^o:, or as large as 190 :^o: (try drawing a few pictures to convince yourself).

I'm not sure but I set up a ratio.  If 66 degrees gives you 36 degrees of the circle 180-66 or 114 degrees would give you  62 2/11 or 62.18 with repeating 18.  x / 114 = 36 / 66
I'm not sure about my logic though.

I'm pretty sure x/114=36/66 ratio does not work. If the 66 degree angle were closer to the circle's edge it would surely not work.

yeah, the only way that works is if one of the intersecting chords is a diameter.

I drew a few diagrams and I'm pretty sure you were right, gauss202. Still, if anyone has more input...

[quote="bthings_2000"]Still, if anyone has more input...[/quote]

Just a bunch of angle chasing and applications of the inscribed angle theorem.  [For simplicity, I will not deal with the corner cases.]

We'll start with an arbitray angle, x, and we'll show that all other parameters of the problem are self-consistent.

Points B, D, and E lie on a circle, such that BCD=36 and DCE=38.  Choose a point A on the circle, such that ACB=x.

Extend AD.  Choose point Q on line AD, such that DQE=29 degrees.  Extend QE to intersect the circle at point F.

Draw line segment BF.  Label the intersection of BF and AQ as point, P.

By IAT, BFE=37.  [In the original problem, this is extraneous information.]

Triangle PQF has two known angles,  29 and 37.  So angle FPQ must be 114.  By supplementary angles, APF must be 66.  [In the original problem, this is extraneous information.]

Triangle ACD is isoceles.  ACD=x+36.  So CAD=CDA=72-x/2.

By IAT, AFB=ACB/2=x/2.

Triangle APF has two known angles.  APF=66.  AFP=x/2.  So PAF=114-x/2.

PAF-CAD=CAF=42.

Triangle ACF is isoceles.  CAF=CFA=42.  So ACF=96.  [In the original problem, this is extraneous information.]

Since we started with the measure of arc AB as unknown x, and we found consistent values for all other parameters of the problem, we cannot uniquely determine x.


rotation
trigonometry
linear algebra
matrix
geometry
geometric transformation
linear algebra unsolved

If det $ A$ > 0, show that $ A$ can be connected to $ I$ by a continuous chain of matrices $ A(t)$ all with positive determinants. (The straight path $ A(t) \equal{} A\plus{}t(I\minus{}A)$ does go from $ A(0) \equal{}A$ to $ A(I)\equal{}I$, but in between $ A(t)$ might be singular.)

Thanks,

$ GL^+(n,\mathbb{R})$ is pathwise connected.
With $ t\rightarrow{tA}$ we may assume that $ det(A)=1$.
Let $ \mathcal{B}=(E_{ij})_{i\not=j}$ be a part of the canonical basis of $ \mathcal{M}_n(\mathbb{R})$. There exist $ B_1,\cdots{B_k}$ in $ \mathcal{B}$ and real $ (\alpha_i)_i$ s.t. $ A=(I+\alpha_1{B_1})\cdots(I+\alpha_k{B_k})$. We conclude with $ t\rightarrow {(I+t\alpha_1{B_1})\cdots(I+t\alpha_k{B_k})}$.     QED

wow....is there any easier solution? i'm kindof new to linear algebra....

Try a version of Gaussian elimination: the operation of adding $ t$ times one row to another doesn't change determinants and can be made continuous, and similarly scaling a row by a positive factor is continuous. Row interchanges are trickier, but we can rotate: replacing $ r_1$ by $ \cos t\cdot r_1+\sin t\cdot r_2$ and $ r_2$ by $ -\sin t\cdot r_1+\cos t\cdot r_2$ doesn't change determinants, and it interchanges rows (with a sign change) if we go to $ \frac{\pi}{2}$.
By applying these three operations, we can reach a matrix with all nonzero entries on the diagonal, and all of those entries either $ 1$ or $ -1$. The negative entries come in pairs, of course. To get rid of those negative entries, we use the same rotation as the row interchanges (to $ \pi$). Equivalently, repeating that interchange twice on the same pair of rows reverses both signs.


Can anyone tell me why is that true when triangle ABC is orthologic with triangle PQR and PQR is orthologic with ABC and the orthologic center coincide at one point ,then we can say that ABC and PQR is perspective.

Dear mathlinkers,
for a proof, see
http://perso.orange.fr/jl.ayme   vol.1 Le théorème de Sondat (in French)
or
http://www.oei.es/oim/revistaoim/numero29.htm   el theorema de Sondat (in Spanich)
Sincerely
Jean-Louis


Let K be a commutative ring with identity and let V be a K-module. If L is an r-tensor over V we associate L with another r-tensor $ \pi_{r}L$ by

$ \pi_{r}L \equal{}\sum_{\sigma}{({\mathop{\rm sgn}}\sigma )}L_{\sigma}$, where $ L_{\sigma}(a_{1},...,a_{r}) \equal{} L(a_{\sigma 1},...,a_{\sigma r})$

We say an r-tensor L on V is alternating if $ L(a_{1},...,a_{i},...,a_{j},..,a_{r}) \equal{} 0$ whenever $ a_{i}\equal{} a_{j}$and $ i\ne j$

Is $ \pi_{r}L$ alternating?

I say no and I think the proof given by my book is false since $ L_{\sigma}\equal{} ({\mathop{\rm sgn}}\sigma )L$ does not imply L is alternating.

EDITED to correct errors

[quote="§outh§tar"]$ L_{\sigma}\equal{} ({\mathop{\rm sgn}}\sigma )L$ does not imply L is alternating.[/quote]
I think it does. Suppose $ a_{i}\equal{} a_{j}$ where $ i\ne j$. Let $ \sigma$ be the permutation that interchanges $ i$ and $ j$ and leaves other elements alone. As any other transposition, $ \sigma$ has $ \mathrm{sgn}\,\sigma\equal{}\minus{}1$. On the other hand, $ L_{\sigma}(a_{1},\dots,a_{r})\equal{}L(a_{1},\dots,a_{r})$. Hence $ L(a_{1},\dots,a_{r})\equal{}0$.

How does your proof work when K is a field of characteristic two, for example?

$ \sum_{\sigma\in S_{n}}\equal{}\sum_{\sigma\in A_{n}}\plus{}\sum_{\sigma\in A_{n}\circ(ij)}$, hence
$ \sum_{\sigma\in S_{n}}\mathop{\mathrm{sgn}}\sigma\cdot L_{\sigma}(a)\equal{}\sum_{\sigma\in A_{n}}(L_{\sigma}(a)\minus{}L_{\sigma\circ(ij)}(a))$. Now if $ a_{i}\equal{}a_{j}$, this sum vanishes term by term. The real problem in characteristic 2 is $ \pi_{r}^{2}\equal{}0$ for $ r>1$. owk

[quote="owk"]$ \sum_{\sigma\in S_{n}}\equal{}\sum_{\sigma\in A_{n}}\plus{}\sum_{\sigma\in A_{n}\circ(ij)}$, hence
$ \sum_{\sigma\in S_{n}}\mathop{\mathrm{sgn}}\sigma\cdot L_{\sigma}(a) \equal{}\sum_{\sigma\in A_{n}}(L_{\sigma}(a)\minus{}L_{\sigma\circ(ij)}(a))$. Now if $ a_{i}\equal{} a_{j}$, this sum vanishes term by term. The real problem in characteristic 2 is $ \pi_{r}^{2}\equal{} 0$ for $ r > 1$. owk[/quote]

How do you know that $ \pi_{r}\equal{} 0$ for $ r > 1$.?

If $ L$ is already alternating, then $ \pi_{r}L\equal{}r!\,L\equal{}0$, therefore $ \pi_{r}^{2}\equal{}0$. owk


for a real number "a" five quantities p,q,r,s,t are defined as p=4-6a, q=3a-5, r=2-5a, s=min(p,q,r), t=max(p,q,r),
find the values of a for which s is maximum and t is minimum

im not entirely sure what you mean by s is maximum and t is minimum
but u can start by trying to set as many of them equal as possible

He means that the a that will make one of  p,q,r maximum and one of them minimal.(I think so at least...)

Not sure if I got your question correctly, but let me try to explain my view :

Sketch graphs of $ y \equal{} 4 \minus{} 6x, \ y \equal{} 3x \minus{} 5$  and $ y \equal{} 2 \minus{} 5x$. Then observe $ s$ achieves its maximum at $ x \equal{} 7/8$ and $ t$ achieves its minimum at $ x \equal{} 1$


Euler
function
calculus
integration
calculus computations

Prove that $u_n=\sum_{k=1}^{n}\frac{1}{k\sqrt{k}}$ converge
$\mathrm{FERMAT -EULER}$

Of course this is well known and extensively taught in the classroom the first time the convergence of series is presented. It's a "p-series." What is well-known is that

$\sum_{k=1}^{\infty}\frac1{k^p}$ converges if $p>1$ and diverges if $p\le 1.$

Since $\frac1{k^p}$ is a decreasing sequence that tends to zero (at least for $p>0$) and it's the values of a well-behaved and easily manipulated function, we have several techniques for proving the convergence.

The usual classroom demonstration is based on the integral test: if $f(x)$ is eventually decreasing to zero, then $\sum_{k=1}^{\infty}f(k)$ converges if and only if $\int_a^{\infty}f(x)dx$ converges.

I like the Cauchy condensation test, which also works well for this case: if $a_k$ decreases to zero, then $\sum_{k=1}^{\infty}a_k$ converges if and only if $\sum_{j=1}^{\infty}2^ja_{2^j}$ converges.


$A$ is a positive definite matrix, and $R$ is its Cholesky factor so that $A=R^TR$
Could u help me to show that $\Vert x \Vert_A = \Vert Rx \Vert_2$ for all $X \in R^n$
Thx :)

Write out the definitions of $\|x\|_A^2$ and $\|Rx\|_2^2,$ using matrix multiplication rather than any special symbols for inner products.

What do you see?

Hi, do U mean by plugging $A=R^TR$ into the definiton of $\Vert x \Vert_A = (X^TAx)^{1/2}$, then get the result of $\Vert Rx \Vert_2$?? :)
Thanks!!

$\Vert x \Vert_A^2= ((x^TAx)^{1/2})^2 =x^TAx$

and $\Vert Rx \Vert_2^2= (\sum_{i=1}^n \vert Rx_i \vert^2)^{1/2}$

and then prove both are equal, am I in the right track?
Please explain. Thx :oops:

Better yet - leave the summation notation out of it and make everything matrix multiplcation if  possible.

$\|y\|^2=y^Ty$ so $\|Rx\|^2=(Rx)^T(Rx)=x^TR^TRx.$

Now, what's $R^TR$ ?


Show that the matrix $ M$ below does not represent a 1-1 mappping by at least finding one pair of vector $ u \neq v$ but $ Mu \equal{} Mv$.

Martrix M

4 -12
-2 6

Can someone pls explain what the question means?

[quote="t0rajir0u"]We want to show that the function $ v \mapsto Mv$ is not one-to-one by finding an element of the range mapped to by two elements in the domain.[/quote]

thanks.


function
logarithms
algebra solved
algebra

Prove that $\forall n \geq 1$ the function $f$

\[ f(x) = \sum^n_{k=0} \frac{x^k}{k!}\]

has at most one real root.

By induction.
If $f_n(x)=0$ and $n$ is even, then equaiton $f'_n(x)=0$ have real root $x'$ and $f_n(x')<0$, but $f_n(x')=f_{n-1}(x')+\frac{x'^n}{n!}=\frac{x'^n}{n!}>0$ for all even $n$ (it is clear $x'\neq 0$) -- contradiction. For odd $n$ we have $f'_{n}(x)=f_{n-1}(x)>0$, i.e. $f_n$ increasing.

I think, your question was discussed here before. I remember, that I solved it, but I am not sure  :?

harazi, use the \ln or \log commands for the logarithm :)

[quote="harazi"]I ask for an equivalent of the form  $ x_n= an+bn\ln n+c\ln n+d$ and this is not easy at all. I did not see this problem till now on the forum.[/quote]
Are you sure that term $bn\ln n$ is valid? Because I think $x_n=-n/e+o(n)$ or something like that. :?

Using approximate estimates I have obtained that
\[x_n=-\frac{n}{k}+\frac{\ln n}{2(k+1)}+o(\ln n),\]
where $k$ satisfies $\ln k=1+\frac{1}{k}$.
I must say that MathCad doesn't agree with me :(

[quote="harazi"]Great problem: find an equivalent for the unique solution of the equation in case n is odd.[/quote]

It is treated in Polya & Szegö
Problems Theorems in Analysis I

Pt.II,Chap.5, n°215

$x_n=an+b\ln n + c +o(1)$

$a$ is real root of $e^{1+a}a=1$

$b=1/2.(a/(1+a))$

$c=a/(1+a)\ln (\sqrt{2\pi}(1+a)/a)$


LaTeX
modular arithmetic
quadratics
number theory

This problem has been bugging me for a while now and I just can not seem to solve it. The problem is:
If a and b are integers prove that [img]http://www.efnet-math.org/wiki/images/math/cbb19147df2c225897ddba1b12646b2d.png[/img] is not an integer. I know that it has a solution however I have been unable to work it out myself.
Source: Efnet Math Problem of The Day, 4th Nov 05.

This is probably overkill, but (13) from

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

basically kills it... There should be a simpler solution though!

No, $\frac{a^2}{2b^2+3}$ must be equal to $n+\frac{2}{2b^2+3}$, where $n$ is an integer.

[quote="Andreas"]Here there's a hint  :)
Suppose that $2b^2 + 3 | a^2 - 2$.
$2b^2 + 3 | -2b^2(a^2 - 2)$
$2b^2 + 3 | 2b^2(2 - a^2)$. But...[/quote]

I'm not sure if this is going to work... are you trying to prove that $2b^2+3\nmid 2b^2(2-a^2)$? Because that's not true, take for example $a=3, b=3$, then we get $18|{-126}$.

[quote="Extra"]$(a^2-2)/(2b^2+3)=a^2/(2b^2+3) - 2/(2b^2+3)$

since b is an integer, the second part cannot be an integer.
therefore, for this to be an integer, ${(a^2)/(2b^2+3)} must equal to 2/(2b^2+3)$

hope this helps[/quote]
Are you saying that the result of the bottom part of the fraction (2b^2+3) has to be greater than two and because of this the second part (-2/(2b^2+3) is dividing a number large than two by two it will always have a decimal part?

[quote="joml88"][quote="Extra"]right sorry about that, somehow, i couldn't make the {} appear in LATEX[/quote]

You have to put \{ and \} instead of just the curly braces.  Curly braces by themself are used when specifying what the argument of a command is.  E.g. we use {n+2} for the exponent if we want to write $2^{n+2}.$[/quote]

Thanks for the tip.

about the question:

$\{\frac{a^2}{2b^2+3}\}$ must be equal to $\frac{2}{2b^2+3}$
therefore $a^2 = 2 (mod 4)$ , this is not true, therefore this cannot be an integer

Why is the mod 4 thing true?

You get $a^2 \equiv 2 \pmod {2b^2+3}$ but I don't see an easy way to show this is false besides using the stuff from the link I posted above...

No, that wasn't what I was asking -- I mean why does the fractional part thing imply they are congruent mod 4? (Which isn't necessarily true, btw)

[quote="t0rajir0u"][quote="ThAzN1"]Why is the mod 4 thing true?[/quote]

We have that $a \equiv 0, 1, 2, 3 \bmod 4$, yes?  Then square both sides.  $a^2 \equiv 0, 1, 0, 1 \bmod 4$.  For no number can $a^2 \equiv 2 \bmod 4$, so contradiction.[/quote]

I think the problem lies in why $\frac{a^2}{2b^2+3}$ "must" be equal to $\frac{2}{2b^2+3}$. You only proved that the fraction cannot equal 0.

[quote="ThAzN1"]Yeah...

(and to fill in the gaps for the solution using quadratic reciprocity stuff)

It suffices to show that $\frac{(2b^2+3)^2-1}{8}$ is odd, or that $(2b^2+3)^2 \equiv 9 \pmod {16}$ but this equals

$4b^4+12b^2+9 \equiv 4b^2(b^2-1) +9 \equiv 9 \pmod {16}$ so yeah...[/quote]

That's pretty clever. I forgot most of my number theory, but this looks good to me.

Okay, I get the stuff about why 2 / 2b^2+3 has to be a non-integer and that the only way that a^2 / 2b^2+3 - 2 / 2b^2+3 could be an integer is if a^2 / 2b^2+3 has the same fractional part as 2 / 2b^2+3, could someone explain to me why a^2 / 2b^2+3 could not have the same fractional part as 2 / 2b^2+3?
Thank you very much for all of your help.

We show $a^2 \equiv 2 \pmod{2b^2+3}$ has no solutions.  But this follows from the quadratic reciprocity argument above... see posts #2, #15


Find all integers n which are divisible by all integers less than square root of n.
Prove these numbers, and prove there are no more.

I've found the numbers to be 1, 2, 3, 4, 6, 8 , 12, 24

This could be a way to prove there are no more...
numbers above 25 need a factor 5 (As well as 1,2,3,4). But it already needs a factor 12 (coming from 3 and 4), So it needs a factor 60. So n is at least 60. But 7<sqrt(60), so it also needs a factor 7. So n is at least 420. And so on.
Is there a way to prove it from this angle. Perhaps by a sequence or limit or something?

This is quite number theory based but the proof might be from this angle which could be sequences etc, so I guessed this forum might know.
It could be from another method though.

simple.
n is divisible by a=[ :sqrt: n] and b=[ :sqrt:n]-1 and thus by their product.
hence either n=ab or n>=2ab>2( :sqrt: n -1)( :sqrt: n -2)>n so long as n - 6 :sqrt: n +4 >0 which is satified once n>36.
Now if n=ab. then [ :sqrt: n -2] needs to be either one or two. (why?) hence implying a set of answers which are easy to obtain.

well i just realized that  :sqrt: n is excluded from divisors but that doesn't make any big difference. the whole argument can be applied to
[ :sqrt: n -1] , [ :sqrt: n -2] and [ :sqrt: n -3]


A rectangle's perimeter is 22. Its length is 5 less than 3 times its width. How many square units are in its area?


[quote="suma_milli"]
My initials are SM = Smiley Monster :P
[/quote]

LOL suma can i use you as one of my Problem of the Day characters, but refer you as Smiley Monster? :D

lol xD I wonder how ManHatTaNaTiVe94, Aryth, avec_une_h, and other IL peeps (INCLUDING MYSELF!!!!!!!!) should get a nickname using their actual initials..

MC = More Cake...?

MC = Many Congrats

MC = Marco Carlo

Yeah I like Marco Carlo.   :gleam:

you could just call me Mr. Who. that's one of the more common ones.

Marco Carlo... :maybe:

Or Mister Censored :wallbash_red: since I bash my head on the wall...but it sounds uncreative...lol


Game[b] Theory[/b] is very much familiar to most economists[u],political[/u] scientists[u],and even sociologists[/u].[u] Its result[b] is very much tested during the 1[/b]9[b]5[/b]0[b]'[/b]s[b].Why is it that game theorists[/b][/u] hardly apply[b] their mathematical craftmanship to predicting[/b] future[b] terrorists[/b] assault[b]?[/b] Why such theory limited[b] in its applicability?

                                                       Respectfully,
                                                        A.Eddington[/b]

I think that the whole idea of terrorism is that it's unpredictable.... I mean it's like reading thoughts...
As far as I know there are objectives that are marked as possible terorist targets but I think Statistics has more to do with this thing than Game Theory.

How can they possibly predict terrorist attacks?

I'm not sure exactly what you mean when you say that game theory's results were very much tested during the 1950's.

The fact is that game theory constructs very simplified models of reality, abstracting away many of the complications of real life.

For example, a typical game theory model has two players, each of whom has a clear and well-specified objective function, and each of whom has a limited number of decision options.  Furthermore, game theory typically assumes that each player knows the other player's objective function and decision options.

There are clear limitations in applying game theory to the real world--there are generally far more than two players, the players don't always know their opponents' objective functions or even the set of choices available to them (e.g., do they have weapons of mass destruction or not), the players don't always even know how many players there are (e.g., because they might not be fully aware of the number of warring internal factions within the other side.)

While game theory is certainly not capable of generating precise forecasts of terrorist behavior, it can sometimes yield some useful insights.

Nobel Laureate Tom Schelling wrote a number of highly readable and interesting books with many insightful observations on the real world, often based on game theory.  [i]Strategy and Conflict[/i] and [i]Micromotives and Macrobehavior[/i] are good places to start reading.


linear algebra
matrix
superior algebra
superior algebra solved

$A,B \in M_n(R)$ invertible such that
$\frac{Com(A)}{det(A)}+\frac{Com(B)}{det(B)}=I_n$

Prove $A,B$ commute

Comatrice of A or $A^{*}$

$A.A^{*}=det(A)I_n$

But why didn't you write $A^{-1}$ instead of $\frac{Com(A)}{\det A}$? We get $A^{-1}+B^{-1}=I_n$, meaning that $A^{-1},B^{-1}$ commute, so $A,B$ commute too.

What am I missing here?


if $\alpha, \beta, \gamma$ are the roots of $x^3-x-1=0$, then find:

$\frac{1-\alpha}{1+\alpha}+\frac{1-\beta}{1+\beta}+\frac{1-\gamma}{1+\gamma}$

[hide]
Put under common denominator to get (I think):

$\frac{3+(\alpha+\beta+\gamma)-(\alpha\beta+\beta\gamma+\gamma\alpha)-3\alpha\beta\gamma}{1+(\alpha+\beta+\gamma)+(\alpha\beta+\beta\gamma+\gamma\alpha)+\alpha\beta\gamma}$

By Viete's Relations, we have

$\alpha+\beta+\gamma = 0$
$\alpha\beta+\beta\gamma+\gamma\alpha = -1$
$\alpha\beta\gamma = 1$

So the above expression becomes $\frac{3+0-(-1)-3(1)}{1+0+(-1)+1} = 1$.

Hopefully, I didn't make any silly algebraic mistakes.[/hide]

The given expression can also be expressed as:

$2\left( \frac{1}{\alpha+1}+\frac{1}{\beta+1}+\frac{1}{\gamma+1} \right) - 3$

now we go back to the formula $x^3-x-1=0$ if we denote it as $P(x)=x^3-x-1$ having roots of $\alpha, \beta, \gamma$, then the function $P(x-1)=(x-1)^3-(x-1)-1=x^3-3x^2+2x-1$ has roots of $\alpha+1, \beta+1, \gamma+1$. By Viete's, sum of the reciprocal of the roots of $P(x+1)$ is 2. so then the whole expression becomes $2(2)-3=1$

EDIT: I forgot to edit after copying and pasting around... :oops:

could someone plz post a link that explains viete's relations?? I've never heard of it and most of what i got while searching was just a biography or a representation of pi...thanks

All it is is the description of the coefficients of a polynomial in terms of its roots that you get from expanding the factored form.


[size=150]Any recurrence equation of the form
$ T(n) \equal{} T(n \minus{} 1) \plus{} T(n \minus{} 2)$

has a general solution

$ T(n) \equal{} A\left(\frac {1 \plus{} \sqrt {5}}{2}\right)^n \plus{} B\left(\frac {1 \minus{} \sqrt {5}}{2}\right)^n$[/size]

Prove the above statement.

Edit: I haven't learnt solving recurrence equations yet, but how do we know that we can use the characteristic equations to solve recurrence equations?  Proof?

I don't see what should be the problem with this (ordinary member of Fibonacci family): the characteristic equation for this relation is

$ r^n\equal{}r^{n\minus{}1}\plus{}r^{n\minus{}2}$

$ n\minus{}2$ roots of it are null, and two of the rest are $ \frac{1 \plus{} \sqrt {5}}{2},\frac{1 \minus{} \sqrt {5}}{2}$.

Hence, $ T(n) \equal{} A(\frac{1 \plus{} \sqrt {5}}{2})^n \plus{} B(\frac{1 \minus{} \sqrt {5}}{2})^n$ qed.


In a class of 20 students, 4 students forgot to put their names on a typed assignment. If the teacher randomly guesses, what is the probability that she correctly guesses which paper belongs to each of the four students? Express your answer as a common fraction.

There is one way to guess the papers correctly. There are only 4!=24 ways to guess the papers total. Thus our desired fraction is $ \frac{1}{24}$. I think...

Yeah, it's just $\frac{1}{4}\cdot\frac{1}{3}\cdot\frac{1}{2}\cdot\frac{1}{1}=\boxed{\frac{1}{24}}$

1/24 SHOULD be the correct answer. No Grantees.  :wacko:  :what?:  :whistling:


In the boxes below, fill in suitable numbers from the set $\{ 0,1,2,...,9 \}$:
$\frac 1{\Box \Box \Box \Box \Box }+\frac 1{1991}=\frac 1{\Box \Box \Box \Box}$.


Let $x,y,z\in Z^+$such that$x+xyz$,$y+xyz$and$z+xyz$are square perfect.Prove that
two squares perfect in ${x,y,z}$
generalization? ;)
Merry Xmas :) and happy new year  :lol:


I know this has been brought up before but I don't want to dig through and find it.
How are moderators chosen?
What do you need to do to become a moderator?

This has been discussed a LOT :D :
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=moderators&t=46679]click here[/url]

[quote="samath"]I know this has been brought up before but I don't want to dig through and find it.[/quote]Bad idea. That's usually what most people do (not search). However by searching you find your answers quicker ;)

[quote="samath"]How are moderators chosen?[/quote]
Randomly after their merits and the trust us the Administrators have in them :D

[quote="samath"]What do you need to do to become a moderator?[/quote]
A clean sheet first of all (no past spamming activities, etc.), some age on the site, time to come on the site once every couple of days atleast, etc. Of course exceptions have been known to be made, but only in special cases ;)


x and y are positive and real.
Prove that if:
$x^2+y^3 \geqslant x^3+y^4$
than:
$x^3+y^3 \leqslant 2$

It is from ARO 1999 and can be found here http://www.mathlinks.ro/Forum/viewtopic.php?t=1012


[quote="paladin8"][quote="Jan"]For $x,y,z > 0$ and $xyz=1$, prove that
\[\frac{x^{9}+y^{9}}{x^{6}+x^{3}y^{3}+y^{6}}+\frac{x^{9}+z^{9}}{x^{6}+x^{3}z^{3}+z^{6}}+\frac{y^{9}+z^{9}}{y^{6}+y^{3}z^{3}+z^{6}}\geq 2 \]
[/quote]

What is the point of all the high powers? We can just substitute $a = x^{3}$, $b = y^{3}$, $c = z^{3}$ to get the same inequality with each power divided by $3$.[/quote]

That's what I did at first, but I like to work in integer powers and there would be fractional ones when homogenized. It's just one of my preferences.

Nice solutions!  :lol:  I like the idea of using $(x^{3}-y^{3})+2y^{3}$.

Here's my attempt. I hope it's correct

In the first inequality, we use $a^{2}+b^{2}\geq 2ab$
In the second inequality, we use rearrangement: $a^{3}+b^{3}\geq \sqrt{ab}(a^{2}+b^{2})$
In the third inequality, we use am-gm.

We prove that
\[\sum_{cyc}{\frac{a^{3}+b^{3}}{a^{2}+ab+b^{2}}}\geq \sum_{cyc}\frac{a^{3}+b^{3}}{\frac{3}{2}(a^{2}+b^{2})}\geq \frac{2}{3}(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}) \geq 2(\sqrt[3]{abc})=2\]


analytic geometry
graphing lines
slope
calculus
calculus computations

Find the equation of the normal to the curve $ x^{3}\plus{}2x^{2}y\equal{}y^{3}\plus{}15$ at the point $ (2, 1)$.

[quote="AndrewTom"]Find the equation of the normal to the curve $ x^{3} \plus{} 2x^{2}y \equal{} y^{3} \plus{} 15$ at the point $ (2, 1)$.[/quote]
[hide]$ 3x^2 \plus{} 4xy \plus{} 2x^{2}y' \equal{} 3y^{2}y'$
$ 12 \plus{} 8 \plus{} 8y' \equal{} 3y'$
$ 20 \equal{} \minus{} 5y'$
$ y' \equal{} \minus{} 4$, so the normal slope would be $ \frac {1}{4}$.
$ y \minus{} 1 \equal{} \frac {1}{4}(x \minus{} 2)$
$ y \equal{} \frac {1}{4}x \plus{} \frac {1}{2}$[/hide]

Edit: Assuming you meant the equation of the normal line :D

Thanks ckck: in Britain and Ireland it's just called the normal.


Hi! I was thinking about this function. It seems quite nice, but I had some problems about this question... Can you help me please?

Do there exist inifinite positive integers x such that

[pi^x]*10^x

is a perfect square?

Thank you very much

C-Vane-G

Oh I forgot: [n] = floor(n)

I think I lost something here :blush:

what does the [] means?...

TaUrUs*
----------------------------------------
Life is GREEN! :D

ahhh floor.. ok  I know i have missread something, jeje :blush:

looks interesting funtion... maybe working a litlle whit log may work...

TaUrUs*
-------------------------
Life is GREEN! :D

$ M_x \equal{} \lfloor \pi^x \rfloor . 10^x$


limit
trigonometry
calculus
calculus computations

[quote="frt"]How do you show that $\lim_{x\to 0} \frac{x}{\sqrt{\tan x}}=0$?[/quote]
$\lim_{x\to 0} \frac{x}{\sqrt{\tan x}}=\lim_{x\to 0} \sqrt{\frac{x}{\tan x}}\cdot \sqrt{x}=0$
Note that $\lim_{x \to 0} \frac{\tan x}{x} = 1$.

Yes, this is simple solution without using l'Hopital or Landau sign.

kunny, I have a [b]question:[/b]

Must we classify two cases $x>0$ and $x<0$?


[quote="hxtung"]Find all pairs of positive integers $ x$ and $ y$ satisfying $ 7^x \equal{} 3^y \plus{} 4$.[/quote]

pf) If y>=2 ⇒ 7^x ≡ 4 (mod 9) ⇒ x= 6k +2 .....*

7^x ≡ 3^x ≡ 3^y (mod 4) so x and y have same parity! ....*'

3^y +4 ≡ 0 (mod 7) ⇒ y = 6m+1 .... *''

by *,*',*'' If y >=2  contradiction!!

so (x,y)=(1,1) is a unique solution!

^^ (Sorry for my broken English)


geometry
circumcircle
similar triangles
geometry solved

Given three points $P$, $A$, $B$ and a circle such that the lines $PA$ and $PB$ are tangent to the circle at the points $A$ and $B$, respectively. A line through the point $P$ intersects that circle at two points $C$ and $D$. Through the point $B$, draw a line parallel to $PA$; let this line intersect the lines $AC$ and $AD$ at the points $E$ and $F$, respectively. Prove that $BE = BF$.

easy with projetive geometry.

let $CD\cap{AB} = M$.

we know that $P,C,M,D$ are one harmonic quadruple, so, $AP,AE,AB,AF$ are one harmonic beam.

$AP||EF$ implies that $B$ is the midpoint of $EF$.  :)

Since the tangets to the circumscribed circle of $ACBD$ intersect on the diagonal of it we conclude that $ACBD$ is a harmonic quadrilateral. But then $AB$ is the symedian of $\triangle{ACD}$. Moreover $\angle{BEA}=\angle{EAP}=\angle{PDA}=\angle{CDA}$ and hence $CD$ and $EF$ are antiparallel with respect to $\angle{EAF}$. Consequently a symedian in $\triangle{ACD}$ is a median in $\triangle{AEF}$. Q.E.D.

[u]Remark.[/u] $N\in (CD),\ NC=ND;\ S\in AN\cap \overline {EBF}\Longrightarrow$ the point $S$ belongs to the given circle.

I have a little bit long proof using Euclidean Geometry.  :D

$pf$) Let $T \in circumcircle$ s.t $BT \parallel PD$, $S=AT \bot CD$.

Here, $\angle PBA = \angle BTA = \angle TSD = \angle ASC$, or $ASBP$ is cyclic.

And cuz $ABPO$ is cyclic, we get $ABPOS$ is cyclic. ($O$ is the circumcenter)

therefore, $\angle OSP = \angle OAP = \frac{\pi}{2}$. or $DS = CS$.

and cuz $CD \parallel BT$, $BC = DT$. or $\angle BAC = \angle TAD$......$(*)$.

And, $\angle FEA = \angle CAP = \angle CDA$ or $\triangle AEF$ resembles $\triangle ADC$.

cuz $AS$ is median of $\triangle ADC$, we get $AB$ is the median of $\triangle AEF$ from $(*)$.

therefore, we get $BE=BF$.

$Q.E.D.$

the problem is fromchina(westen mathematical olympiad)chendu,sichuan $2005$.geometry of first day. problem $2$ .

ok lets solve the problem:

suppose $(C)$ is a circle and $P$ is a point in the exterior of it.each time we draw a variable line and let it.

intersect the circle in points $C,D$.

the locus of points like $K$,inwhich we have:$(PKCD)=-1$, is a line which we name it polar.

name the intersection $PD$ and $AB,K$.

its clear that $(PKCF)=-1$

theorem:we have,$(ABCD)=-1$.and $O$ is a arbitrary point in plane.

for each line which intersects $PA,PB,PC,PD$, in points $A',B',C',D'$,we have:$(A'B'C'D')=-1$

the we have:$A(PKCF)=-1$

then the points $F,B,E$,and teh intersection of $AP$ and $FE$ are harmonic.

but $AP$ and $EF$ are paralell.

then: $FB=BE$

Easy to get that : $ {D,K,C,P}$ are harmonic points group  ;draw parallel line to $ {PD}$ through $ {B}$ $ \Longrightarrow$ $ \frac{PC}{CK}\equal{}\frac{PD}{DK}$$ \Longrightarrow$
$ {AT}\frac{YB}{YT}\equal{}{AT}\frac{XB}{XT}$;By similar triangles : we get :  $ {BE}\equal{}{BF}$...$ {Q}$.$ {E}$.$ {D}$ :)

[img]https://i.imgur.com/InhyNv2.png[/img]

First of all we note that the quad $ACBD$ is harmonic,thus meaning that $\frac{AC}{BC} = \frac{AD}{BD}$
So now first of all note that $\triangle CAB \sim \triangle BAE$.This is because $\angle BEA = \angle PAC = \angle CDA = \angle CBA$ and becaus e $\angle BAC = \angle EAB$,Thus from here we get that $EB= \frac{CB}{AB} EA$.From the power of point we have that $EB.EF=EC.EA$,thus we have the following:
$$\frac{CB}{AB}.EF = EC \iff \frac{CB}{AB} = \frac{CE}{EF}$$
With the conditions of a same angle $\angle CEF = \angle CBA $,we have that $\triangle CEF \sim \triangle CBA $, which implies $\angle ECF = \angle ACB$.
Now we do a simple angle-chase:
$$\angle ECF = \angle ACB \implies \angle ECB +\angle BCF = \angle ACF + \angle FCB \implies \angle ECB = \angle ACF$$
Because the quad $ACBD$ is cyclic we have that $\angle ECB = \angle BDA = \angle BAP = \angle ABF$,thus we have $\triangle ABF \sim \triangle ADB$.
Now from this similarity we get that $BF = \frac{AB}{AD}.BD$.So now let's see if that equals $\frac{CB}{AB}.EA$
$$\frac{AB}{AD} BD = \frac{CB}{AB} EA \implies \frac{AB}{AD} \frac{BD}{AD} = \frac{CB}{AB} \implies \frac{AC}{AB} \frac{BD}{AD} = \frac{CB}{AB}$$
Now this implies that $\frac{BD}{AD} = \frac{BC}{AC} $
Which is true,because we have that the quad $ACBD$ is harmonic,which means $(A,B;C,D)=-1$,and the ratio directly follows from this.
Thus $BE=BF$..... :D


number theory
relatively prime
combinatorics proposed
combinatorics

This is also quite number - theoretic combinatorics:

Let a, b, c be pairwise relatively prime positive integers. The positive integer n is said to be stubborn if n cannot be written in the form bcx + cay + abz for positive integers x, y, z. Count the number of stubborn integers.

do we choose ABC for given XYZ or do we choose XYZ for given ABC?

Well, a, b, c are fixed (pairwise relatively prime) integers and x, y, z may vary.

oh i see

we gotta prove it generally, regardless of what a,b,c is, right? :)

Yup :) It's not easy at all :D

Btw, you have to count, not to prove ;)

yaya, ok, we gotta [b]count[/b] it generally, regardless of what a,b,c is...


Prove that, for all $x,y\in\mathbb{R}$ \[3(x+y+1)^{2}+1\geq 3xy\]
[hide="Solution"]
First case: $xy<0$
- Then, \[3(x+y+1)^{2}+1\geq 1>0>3xy\]We don't have equality
Second case: $xy\geq 0$
- Then $x,y$ have the same sign. By a trivial application of Cauchy Schwarz, we have the two inequalities for [b]positive[/b] $x,y$:\[(x+y+1)^{2}\geq (2\sqrt{xy}+1)^{2}\qquad (x+y-1)^{2}\geq (2\sqrt{xy}+1)^{2}\]
Thus, $(x+y+1)^{2}\geq (2\sqrt{xy}+1)^{2}$ for all [b]reals[/b] with $xy\geq 0$
Now notice that \[(2\sqrt{xy}+1)^{2}+\frac13\geq xy\iff 3(\sqrt{xy}+\frac23)^{2}\geq 0\] hence the inequality is proven.
[/hide]
The reason I'm posting is, does anyone have a nice solution without cases and/or only factoring?

I know that method also... but can you show nice calculations etc.?

Let $x+y=s,\ xy=t$,

$\left\{ \begin{array}{ll}x\in{\mathbb{R}}&\quad \\ y\in{\mathbb{R}}&\quad \end{array}\right. \Longleftrightarrow \left\{ \begin{array}{ll}x+y\in{\mathbb{R}}&\quad \\ x-y\in{\mathbb{R}}&\quad \end{array}\right. \Longleftrightarrow \left\{ \begin{array}{ll}(x+y)^{2}\geq 0 &\quad \\ (x-y)^{2}\geq 0 &\quad \end{array}\right.$

$\Longleftrightarrow s^{2}\geq 4t$.

$\therefore 3t\leq \frac{3}{4}s^{2}\leq 3(s+1)^{2}+1$.
$\because \frac{3}{4}s^{2}\leq 3(s+1)^{2}+1\Longleftrightarrow (3s+4)^{2}\geq 0$. Q.E.D.

[quote="bibobeo"]Let $F=3(x+y+1)^{2}+1-3xy$
One can see that $4F= 3(2x+y+2)^{2}+(3y+2)^{2}$
The rest is easy I think.[/quote]

Your solution is correct.
This was my first solution.
$3(x+y+1)^{2}+1-3xy=3x^{2}+3(y+2)x+3y^{2}+6x+6y+4$
$=3\left(x+\frac{y+2}{2}\right)^{2}+\frac{9}{4}y^{2}+3y+1$
$=3\left(x+\frac{y+2}{2}\right)^{2}+\frac{9}{4}\left(y+\frac{2}{3}\right)^{2}\geq 0$, which is same as yours.

Nice solutions guys, also I'm embarassed but my solution is wrong in the case $x,y<0$ because $(x+y-1)^{2}\geq (2\sqrt{xy}+1)^{2}$ is false for $x=y=1$ (my mistake in the cauchy, $-1<0$) so I would have to expand in that case which I was trying to avoid.