Princeton
college
calculus
derivative
geometry
3D geometry

I'm currently in AP Chemistry and we just had our second test of the year and I finished in hysterics. I don't think it's that I'm underestimating the course because I've taken AP courses before and I know a lot of times the tests will have types of problems that you may not have seen before or problems where you need to go those extra steps. I'm very diligent with the homework readings and I will go out of my way to make sure that every calculation is correct before the day it's due.

I know alot of colleges and universities put of practice exams but I'm not sure how similar the AP Chemsitry syllabus and content is to an actual college course.

The test today was on atomic structure; mainly giving electron configurations, quantum numbers, calculating wavelengeht/frequency, De broglie wavelength, etc. I was fine for the first half but the questions got really tricky and I got wound up with all the applied concepts. When I get the test back I will post some of the problems but I think my problem solving skills could use a lot of work. Is there any site that I can find ap chem exercises and problems to help me?

The exercises at the end of each chapter in my textbook don't seem to help as they either test very basic concept and calculation or the challenge problems at the end apply concepts from other previous chapters that we haven't gone over yet (we jump around in the book a lot).

You can go to the collegeboard's website to get old free response questions, and hope to find some from atomic structure.  Otherwise, just do more problems from the book.  Read it over and over if you have to.  But post the problems, and if possible, your solutions.

There are plenty of AP books out there.

Get Princeton Review. I hear that is good.
Or if you want additional reading that delves deeper than just AP, I would recommend reading/working through Atkins'  Chemical Principles.

The one concern about AP Chemistry I have heard a lot is that it seems overwhelming. And it is, because it is required for students to know (a little of) everything in this broad subject. But in my experience, overestimating the difficulty of AP Chemistry (daunting, not difficult) can lead to some failure...at my previous high school, the first test of the year is infamous for people to get F's (which is pretty much unheard of at my school as well...besides this class). Also, the tests are curved generously, and a F-range grade = 30% raw score. However, it was just a review of the all-so-loved concept of stoichiometry. In fact, two people (out of 44 students, a record number more than double the number who normally take the class) got 99% on the first test.

Given my personal experience in chem, I would say not to freak out before or during a test. Freak out all you want afterwards (sometimes it pays off well :-D), but just don't go crazy.

Another resource you should check out is the U.S. Chemistry Olympiad tests. The problems there are the most difficult you will ever see on the AP, and even more difficult. A few people would disagree with me, but I strongly advocate studying at least for the AP with the local section tests.

If you want a review book that is harder than the actual test, get Barron's.  It is absolutely thorough and contains a lot of information and pracice questions.  Princeton Review is good, I used it and got a 5, as is Barron's.

Blkmage, I'm probably worse off than you. My teacher doesn't teach. He said water was organic, used the answer for a stoich problem in his calculation, and doesn't know how to grade.

If there are any friends in your class that are also struggling, you can have a study group meeting. It really helps when you have other people to talk to about the course material. Otherwise, make sure you're on top of the class, and you're never cramming during that last night.

I guess how you prepare for an AP chemistry course depends on what you want to get out of it. I will therefore assume you want to get an A and recieve a 5 on the AP exam. Therefore, I recommended that you follow these simple steps.
THE BASICS
1. Believe you can do it :You won't do very well if you doubt your abilities.
2. Use your head: This is a LOT easier said than done. Chemistry is kind of similar to math in this respect. For example, in algebra, we always knew the answer was correct because you could plug back your solution into the problem and you would get the solution "0=0" This carried over in calculus when we realized that if you integrated something and took its derivative, you would end up with the same function. Although it isnt apparent immediately, there are checks like that in chemistry that will indicate whether you have a plausible answer. For example, if you are doing a thermodynamics problem where you add 1 5gram cube of ice to  2000 kg water at 98 degrees C and get an answer of   some final temperature of 30 degrees celcius.... you should probably check your work again. In essence, ask yourself... does this actually happen irl?

And now... regarding the actual chemistry:
1. Chemistry exams test the same concepts in different forms: For example, during my atomic theory test our teacher placed the question "what is lanthanide contraction?" This is something from chapter 21 of the book (we were on chapter 6 or so), yet, if we thought about the word we could reason that the lanthanides contract due to f orbitals. Therefore, if ever theres a question you feel you cant answer, stop, think about what the chapter talks about, and try to apply those concepts. Sometimes you just have to state the obvious a couple times in order to make a sudden breakthrough. This brings us to the next point.
2. Know your stuff... If you do badly because you dont know what the test is testing then theres no excuse for failure.... u gotta put in the time
3. Practice on hard, multi-step problems: I used zuhmdal for ap chem and I really liked it as the problems in the back of the book were pretty challenging. When the problems are hard and multistepped, you gain a more insightful view of chemistry. The beauty of it all is, that once you solve a hard problem, it becomes so easy, that you can do the whole thing again in your head without a paper or pencil.An excellent source for difficult problems is old ap exams (although these become easy very fast....). I suggest old USNCO problems for the bold and daring.

If you follow these steps and still dont see any results........ ummm...... I dont think anyone can help u....

i agree that zuhmdal chemistry book is the best for ap chem. we used that book in my class and it was really helpful. also, i would recommend the book: fact tract to a 5. use this every other day to review stuff you have learn so that in may, you don't forget what you did in september.

Actually, my class uses the Zumdahl book too. The chapter questions are great for reinforcing concept and calculations but seriously, the test made me feel like I got hit by a truck! The quizzes are very straightforward and seem to just reinforce basic concept and I can get a great grade easily. The tests are about 20 steps above.

And 7h3.D3m0n.117, my teacher isn't bad at all so I don;t think I'm on the same boat as you in that respect. He explains everything thoroughly and I feel like I understand everything the first time around. Everything is explained thoroughly as far as concepts go. I know that obviously those tricky application questions will be something that can't be "taught."

I will definitely post up the questions when I get the test back though. Each question just got progressively harder. I realize some of the mistakes I made now but some I still have no idea how to do the problems.

I was actually going to get with a group of 3 or so friends and do the remaining problems that we haven't done for classwork/homework for each chapter and exchange answers; not the just trade answers method but we would help each other out and such and see if that makes a difference.

And the first test of the year that we took was extremely easy. The questions were definitely the "think outside the box" type but definitely do-able if you thought about it for a few minutes.

I'm also using the Zumdahl book, and I wouldn't recommend using it as your primary resource. In all honesty, you could have a better AP text book. Some explanations in the book are very decent and thorough (like hybridization) but some are too elaborate and way to complicated for you to get a simple idea (like excess reactants- I mean, he incoorporated large visuals and pictures that took up like 5 pages to explain the concept...).

The secret to master a given subject is to read different, complementary, and appropriate books.

I used the Zumdahl book as well, as our only textbook.  It was okay, but nothing special.  However, my teacher was really good.  If you have 1 or 2 good review books to supplement, Zumdahl is fine.

I'm currently taking the class as a sophomore. My class uses the book, "Chemistry - The Central Science" by Brown, Lemay and Bursten. It is very detailed and has helpful problems to work through. Despite this, it does get somewhat too complicated when describing some simple concepts. I recommend using a textbook, and other simpler, more specific books. This helps to understand each topic in depth.


What can you say if $ \inf S \equal{} \sup S$?

$ S$ is empty?


algorithm
continued fraction
Diophantine equation

Find all primes $ p$ such that $ 47p^2\plus{}1$ is a perfect square.

lets say that $ 47p^2 \plus{} 1 \equal{} a^2$.

then $ 47p^2 \equal{} a^2 \minus{} 1 \equal{} (a \plus{} 1)(a \minus{} 1)$

since 47 is a prime number, 47 is either a+1 or a-1, and $ p^2$ should be another.

if 47=a-1,a=48, $ p^2 \equal{} a \plus{} 1 \equal{} 49.$ so p=7.

answer : 7

[quote="FantasyLover"]since 47 is a prime number, 47 is either a+1 or a-1, and $ p^2$ should be another.[/quote]

Your solution doesn't address all cases. The problem is here.

Continuing on FantasyLover's solution...

If you let $ a\plus{}1\equal{}47p$ and $ a\minus{}1\equal{}p$, you get non-integer solutions, so that's not possible.

If you let $ a\plus{}1\equal{}p$ and $ a\minus{}1\equal{}47p$, you also get non-integer solutions.

If you let $ a\plus{}1\equal{}47p^2$ and $ a\minus{}1\equal{}1$, you get non-integer solutions.

If you let $ a\plus{}1\equal{}1$ and $ a\minus{}1\equal{}47p^2$, you get non-integer solutions.

These are all of the possibilities since 47 is prime...so [b]7[/b] is the only possible answer.

Note that $ \gcd(a \plus{} 1, a \minus{} 1) | 2$.  Verify that $ p \equal{} 2$ doesn't work, so $ \gcd(a \plus{} 1, a \minus{} 1) \equal{} 1$ and we only have two cases to check instead of four.

Right. It is clear that $ 47$ divides one of those factors individually, but you have to allocate the other factors as well.

$ a\minus{}1\le a\plus{}1\le 2(a\minus{}1)$ and that eliminates all of the other cases.

[quote="mathcrazed"]P could be a multiple of 47...[/quote]

if $ p\equal{}47k$, k<2 because p is a prime number.

but if k=1, 47*47+1 should be a perfect square, which is not true.

[quote="Altheman"]Find all primes $ p$ such that $ 47p^2 \plus{} 1$ is a perfect square.[/quote]
The equation can be written $ n^2 \minus{} 47p^2 \equal{} 1$ which is Pell’s equation. By calculating the convergents in the continued-fraction expansion of $ \sqrt {47}$ we find the fundamental solution to be $ n \equal{} 48,\,p \equal{} 7.$ All other solutions to the Pell equation will be of the form $ n \equal{} a,\,p \equal{} b$ where
\[ a \plus{} b\sqrt {47}\ \equal{} \ \left(48 \plus{} 7\sqrt {47}\right)^k\]
for some integer $ k.$ However it is clear that $ b$ will be a multiple of 7 and so cannot be prime. Hence the only prime solution is $ p \equal{} 7.$

[quote="itiselizabeth"]By calculating the convergents in the continued-fraction expansion of $ \sqrt {47}$ we find the fundamental solution to be $ n \equal{} 48,\,p \equal{} 7.$[/quote]

How [i]do[/i] you calculate the convergents in a continued-fraction expansion of the square root of a positive non-square integer?

[list]$ \sqrt {47}\ \equal{} \ 6 \plus{} \cfrac1{1 \plus{} \cfrac1{5 \plus{} \cfrac1{1 \plus{} \cfrac1{12 \plus{} \cdots}}}}$[/list]
Set $ q_0 \equal{} 6,\ q_1 \equal{} 1,\ q_2 \equal{} 5,\ q_3 \equal{} 1,\ q_4 \equal{} 12,\ \ldots.$

The convergents are terms in the sequence $ \left\langle\frac {a_i}{b_i}\right\rangle_{i\, \equal{} \,0}^\infty$ where the $ a_i$ and $ b_i$ are calculated as follows.

[list]$ a_0\ \equal{} \ q_0\ \equal{} \ 6$
$ a_1\ \equal{} \ q_1a_0 \plus{} 1\ \equal{} \ 7$
$ a_2\ \equal{} \ q_2a_1 \plus{} a_0\ \equal{} \ 41$
$ a_3\ \equal{} \ q_3a_2 \plus{} a_1\ \equal{} \ 48$

[list]$ \vdots$[/list]
$ a_{i \plus{} 2}\ \equal{} \ q_{i \plus{} 2}a_{i \plus{} 1} \plus{} a_i$[/list]

[list]$ b_0\ \equal{} \ 1$
$ b_1\ \equal{} \ q_1\ \equal{} \ 1$
$ b_2\ \equal{} \ q_2b_1 \plus{} b_0\ \equal{} \ 6$
$ b_3\ \equal{} \ q_3b_2 \plus{} b_1\ \equal{} \ 7$

[list]$ \vdots$[/list]
$ b_{i \plus{} 2}\ \equal{} \ q_{i \plus{} 2}b_{i \plus{} 1} \plus{} b_i$[/list]
Now you just have to try $ a_i^2 \minus{} 47b_i^2$ until you find a pair for which the expression equals 1; these will be the fundamental solution. But in this particular problem, you know that $ b_i$ has to be prime, so you know you can pick $ b_3\equal{}7$ right away. :lol:

This is a technique I only learned recently so I may not be terribly good at it yet. ;)

The algorithm is given [url=http://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Continued_fraction_expansion]here[/url].  There is a very elegant way to organize the work in a table that makes the computation extremely fast, but it's a bit hard to describe in words.


inequalities
real analysis
real analysis unsolved

$a_n >0$
$p\geq 2$ integer such that

$\sum_{n\geq 1}a_n$ converge

Prove that  $\sum_{n\geq 1}{a_n}^{\frac{n-p+1}{n}}$ converge.

--------------------------------------------------------------------

Do you consider this problem easy, medium ,hard ?
thank you

I consider it a bit strange as written because the second condition is not needed. Indeed, split all indices $n$ into 2 groups: those with $a_n<e^{-n}$ and those with $a_n>e^{-n}$. The sum of $a_n^{\frac{n-p+1}n}$ over the first group clearly converges while the sum over the second group is essentially the same (up to a factor $e^{p-1}$) as the sum of $a_n$. Probably there is some misprint somewhere   :? .

Also, compare this problem with discussion in [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35093[/url].

I don't think you should change the condition.Because the solution will be nice if there is the second condition.
Use the extension Cauchy(I don't know exactly the name) inequality we have :
  ax+by>=x^{a}.y^{b} (a,b>0 and a+b=1)
Replace x=a_{n},y=a_{n}^{1/p}
and a=(n-1)/n and b=1/n we have
(n-1)/n *a_{n}+b{n}>=c{n}
in this b{n}=a_{n}^{1/p}/n
and c{n}=a_{n}^[(n-p+1)/n]
  It show that sum c_{n} is converges


group theory
abstract algebra
modular arithmetic
limit
number theory
least common multiple
Galois Theory

Here are two interesting problems in Galois Theory, purportedly proposed by Artin. I couldn't solve them. Could anyone help?

1) Let F be a fixed algebraic closure of Q. Let K be a maximal subfield of F not containing the square root of 2. Show that any finite extension is cyclic.

2) Let K be a field such that any finite extension is cyclic. Show that there exists an automorphism of F (a fixed algebraic closure of K) over K which fixes precisely K.

Does anyone have any idea? These problems can be found in Lang's Algebra textbooks. He noted that this shows how to 'dig in algebraic closure.' How does this hint help?

[hide="Hint for 2)"]Exhaust $ F$ by a strictly increasing sequence of subfields $ K_n$ with $ K_0\equal{}K$. Then lift carefully a generator of $ Gal(K_1|K)$ step by step till heaven.[/hide]

[quote="Third Edition"]Exhaust $ F$ by a strictly increasing sequence of subfields $ K_n$ with $ K_0 \equal{} K$.[/quote]
That is not so easy.
Let $ p_1,p_2,p_3,\ldots$ be all primes and take $ K_n$ a maximal subfield of $ F$
with the degree $ |K_n: K|$ dividing $ q_n: \equal{}p_1^{n}p_2^{n\minus{}1}\cdots p_{n\minus{}1}^2p_n$ for all $ n\in\mathbb{N}$.
Then show $ K_1\subseteq K_2\subseteq \cdots$ and $ F\equal{}\bigcup_{n\in\mathbb{N}}K_n$.

[hide="Hint for 1)"]
Any nontrivial finite Galois extension $ L/K$ has $ K(r)$ as the only minimal nontrivial intermediate field.
So the Galois group $ G\equal{}Gal(L/K)$ has the only maximal subgroup $ H\equal{}Gal(L/K(r))$.
Then for $ x\in G\setminus H$ we have $ \langle x\rangle \not\le H$ and $ G\equal{}\langle x\rangle$ cyclic.
Then $ L/K$ abelian gives any intermediate field also Galois over $ K$.
And any finite extension lies in some Galois extension, and therefore is Galois.

[/hide]

2) we have to extend a generator of $ Gal(K_n/K)$ to a generator of $ Gal(K_{n\plus{}1} / K)$, why is that possible?

[quote="-oo-"]2) we have to extend a generator of $ Gal(K_n/K)$ to a generator of $ Gal(K_{n \plus{} 1} / K)$, why is that possible?[/quote]
$ G\equal{}Gal(K_{n \plus{} 1} / K)$ and $ H\equal{}Gal(K_n/K)$ are cyclic with restriction epimorphism $ f: G\to H$.
Say $ G\equal{}\langle x\rangle ,H\equal{}\langle y\rangle$. Then $ H\equal{}f(G)\equal{}\langle f(x)\rangle$ gives $ f(x)\equal{}y^k$ with some $ k\in\mathbb{Z}$ coprime to $ n\equal{}|H|$.
Let $ p_1,\ldots ,p_r$ be the prime factors of $ |G|$ coprime to $ n$.
Then using [url=http://en.wikipedia.org/wiki/Chinese_remainder_theorem]Chinese remainder theorem[/url] we find $ m\in\mathbb{Z}$ with $ m\equiv k^{\varphi (n)\minus{}1}\pmod{n}$ and $ m\equiv 1\pmod{p_i}$ for $ i\equal{}1,\ldots, r$.
Then no prime factor of $ |G|$ divides $ m$, so $ x^m$ is generator of $ G$.
And $ km\equiv k^{\varphi (n)}\equiv 1\pmod{n}$ gives $ f(x^m)\equal{}f(x)^m\equal{}y^{km}\equal{}y$.
So $ y$ extends to $ x^m$.

[quote="Enya"][quote="Third Edition"][quote="Enya"]how do you show that any finite extension of K is finite Galois?[/quote]Extend the finite extension to a Galois extension which was shown to be Abelian. This implies that every sub-extension is Galois.[/quote]

I get what you mean. but what extension is that? I cannot figure out. Is it related to the square root of 2? I know that Q(sqrt(2)) is abelian galois extension  over Q. But how to figure out an abelian galois extension of K containing L as an intermediate subfield?[/quote]All you need is given in the hint of [url]http://www.mathlinks.ro/viewtopic.php?p=1264687#1264687[/url] noting that the following fact: given a finite extension $ L|K$, then let $ L'$ be a normal hull of $ L$ in $ F$. Then $ L'|K$ is finite normal, automatically separable (characteristic is zero), hence Galois. Now apply olorin's hint.

[quote="Third Edition"][/quote][quote="Enya"][quote="Third Edition"][quote="Enya"]how do you show that any finite extension of K is finite Galois?[/quote]Extend the finite extension to a Galois extension which was shown to be Abelian. This implies that every sub-extension is Galois.[/quote]

I get what you mean. but what extension is that? I cannot figure out. Is it related to the square root of 2? I know that Q(sqrt(2)) is abelian galois extension  over Q. But how to figure out an abelian galois extension of K containing L as an intermediate subfield?[/quote][quote="Third Edition"]All you need is given in the hint of [url]http://www.mathlinks.ro/viewtopic.php?p=1264687#1264687[/url] noting that the following fact: given a finite extension $ L|K$, then let $ L'$ be a normal hull of $ L$ in $ F$. Then $ L'|K$ is finite normal, automatically separable (characteristic is zero), hence Galois. Now apply olorin's hint.[/quote]

can you explain why the normal closure L'/K is automatically separable? what do you mean by char 0?

[quote="Enya"]can you explain why the normal closure L'/K is automatically separable? what do you mean by char 0?[/quote] Every finite extension in characteristic 0 is separable.

[quote="Algebra Galois"]For Problem 2, is it clear that Kn/K (which is maximal) is a finite extension?[/quote]
well, the degree is bounded ...

Can somebody help me with this.
How can we see that the selected K_n forms an increasing chain . And also how does the union gives us the algebraic closure?

[quote="-oo-"]$ K_n / K$ is cyclic, thus has a unique subextension of every degree which divides $ |K_n : K|$.
if $ a \in \overline{K}$, the chinese remainder theorem applied to $ Gal(K(a)/K)$ shows that $ a$ is a sum of products of elements whose degree is a prime power. these prime powers divides some $ q_n$, so you're done.[/quote]

hi, i don't see how it is related to Aut(F/K).

think about it a second before you post :P.

once we have fould the fields $ K_n$ with the desired properties, we find generators $ \sigma_n$ of their galois groups such that $ \sigma_{n \plus{} 1}$ extends $ \sigma_n$ (see above). then $ \lim_{\rightarrow} \sigma_n$ is the desired element of $ Aut(F/K)$.

[quote="-oo-"]then $ \lim_{\rightarrow} \sigma_n$ is the desired element of $ Aut(F/K)$.[/quote]I think you mean $ \lim_{\leftarrow} \sigma_n\in\lim_\leftarrow Gal(K_n/K)$, don't you?

[quote="-oo-"]$ K_n / K$ is cyclic, thus has a unique subextension of every degree which divides $ |K_n : K|$.
if $ a \in \overline{K}$, the chinese remainder theorem applied to $ Gal(K(a)/K)$ shows that $ a$ is a sum of products of elements whose degree is a prime power. these prime powers divides some $ q_n$, so you're done.[/quote]
I don't follow your argument, but I think it's more tricky than that.
[quote="olorin"]
Let $ p_1,p_2,p_3,\ldots$ be all primes and take $ K_n$ a maximal subfield of $ F$
with the degree $ |K_n: K|$ dividing $ q_n: \equal{} p_1^{n}p_2^{n \minus{} 1}\cdots p_{n \minus{} 1}^2p_n$ for all $ n\in\mathbb{N}$.
Then show $ K_1\subseteq K_2\subseteq \cdots$ and $ F \equal{} \bigcup_{n\in\mathbb{N}}K_n$.
[/quote]
We need to show $ L\subseteq K_n$ for any subfield $ L$ of $ F$ with $ L/K$ finite of degree dividing $ q_n$.
Then $ K_{n \minus{} 1}\subseteq K_n$ for all $ n\in\mathbb{N}$, because $ |K_{n \minus{} 1}: K|$ divides $ q_{n \minus{} 1}|q_n$ by definition of $ K_{n \minus{} 1}$
and for all $ a\in F$ the finite degree $ |K(a): K|$ must divide some $ q_n,n\in\mathbb{N}$ by contruction of the $ q_n$, so $ a\in K(a)\subseteq K_n$.

So let $ L/K$ be a finite extension in $ F$ with $ l \equal{} |L: K|,k \equal{} |K_n: K|$ both dividing $ q_n$.
Let $ G \equal{} \langle x\rangle$ be the cyclic Galois group of the compositum $ K_nL/K$ in $ F$.
Then the Galois groups of $ K_nL/K_n$ and $ K_nL/L$ must be $ \langle x^k\rangle,\langle x^l\rangle$ the only subgroups of index $ k,l$ of $ G$.
But then the intersection $ \langle x^k\rangle\cap\langle x^l\rangle \equal{} \langle x^{{\rm lcm}(k,l)}\rangle$ must have fixed field $ M$ containing both $ K_n,L$
and with degree $ |M: K| \equal{} {\rm lcm}(k,l)$ also dividing $ q_n$.
So the maximal choice of $ K_n$ gives $ K_n \equal{} M\supseteq L$.


Ring Theory
superior algebra
superior algebra unsolved

Hello,
Can anyone help me with these problems I can't do in a textbook ?

1. Let S be the subset of Z[sqrt(m)] consisting of all elements of Z[sqrt(m)] with even norm. Prove that S is a subring of Z[sqrt(m)].

2. R is a ring with property that x^2=x for each x in R.

a) Show that each element of R equals its own negative. (Hint: Consider (x+x^2).)

b) Hence show that R is a commutative ring.

1. trivial calculation, where's the problem?
2. a) clear since $ (\minus{}x)^{2}\equal{}x^{2}$ in every ring.
b) expand $ x\plus{}y\equal{}(x\plus{}y)^{2}$ and use a).

I am still stuck, can anyone please help me ?

For Q1, the bit that said "even norms' confuses me, how exactly can you solve this problem ?

For Q2a, I got

(x+x)^2=4x^2
x+x=4x
2x=4x
0=2x

How do I get to

x=-x

Is this right ? Can someone help me ?

For Q2b, I still don't get it, can someone help me on this please ?

P.S. For the hint given in Q2a, it should say (Consider (x+x)^2) rather than (Hint: Consider (x+x^2).), sorry about the typo.


Hard problem.
Let f(x)=x4+17x3+80x2 +203x+125. Find the polynomial, g(x), of the smallest degree for which
f(3 \pm  \sqrt3)=g(3 \pm  \sqrt3) and f(5 \pm  \sqrt5)=g(5 \pm  \sqrt5).

Did anybody achieve anything? May be we have to show that f(x)=g(x) ?

Unless I have misunderstood what you are asking, the question does not seem very difficult.

Let h(x) = (x - 3 - \sqrt 3)(x - 3 + \sqrt 3)(x - 5 -  \sqrt 5)(x - 5 + \sqrt 5). g(x) has the smallest degree when f(x) - g(x) = h(x).  Now just expand out h(x) to find g(x).

Or have I missed out some important point?


A regular hexagon is inscribed in a circle.  The ratio of the length of a side of the hexagon to the length of the shorter of the arcs intercepted by the side, is:

$ \textbf{(A)}\ 1: 1 \qquad \textbf{(B)}\ 1: 6 \qquad \textbf{(C)}\ 1: \pi \qquad \textbf{(D)}\ 3: \pi \qquad \textbf{(E)}\ 6: \pi$

Since the radius of the circle is equal to the side length, the ratio is simply $ \frac s{\frac16\cdot2s\pi}\equal{}\boxed{\textbf{(D)}\ 3: \pi}$.


geometry
3D geometry
tetrahedron
trigonometry
similar triangles

Four balls of radius $1$ are mutually tangent, three resting on the floor and the fourth resting on the others. A tetrahedron, each of whose edges has length $s$ is circumscribed around the balls. Then $s$ equals:

(A) $4\sqrt{2}$
(B) $4\sqrt{3}$
(C) $2\sqrt{6}$
(D) $1 + 2\sqrt{6}$
(E) $2 + 2\sqrt{6}$

See the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=59459]Four balls of radius 1 are mutually tangent[/url].

[hide]I got E.  First, I showed that the angle between two faces is $\cos^{-1}{\frac{1}{3}}$.  Thus, by similar triangles, the distance from the center of one circle to the nearest vertex of the tetrahedron is $3$.  So then, the height of the tetrahedron is $4+2\frac{\sqrt{2}}{\sqrt{3}}$, which implies that a side length is $2+2\sqrt{6}$.[/hide]


No one help me to solve this interesting problem...
I just can prove when n=2...

This is kind of interesting. But I can't solve it. :blush:
Maybe you can use mathematical induction.

[quote="Jianxing113725"]:( Can't anyone give the solution...Is it very hard?[/quote]

Sorry. But your second inequality is wrong. Try x1=1, x2=2.

Your sign of inequality must be inversed, I think.


Choose!

I chose Challenger for patriotic reasons (plus, we are going to win!).

i honestly dont know who will win
probly challenger
but pizitz will win
wow im bipolar

lawl, tony why didnt you vote?
i think pizits will win eventhough ms chilton wont put tony on the team
XD

What?! Tony isnt on the team? (Yaaaay...)

wait, did i post that?
well tony is on the team dynamo, sorry to break the news :P  :D

:(  :noo:  :w00tb:
dont know where that last one is from...

wait, your chapter is tomorrow, right?
who is on the pizitz team?

ummm tony, me, bo, and a kid named farhan or ferhan... forgot his name spelling

hi yaofan. I'm assuming you go to berry? What place did your team/you get at chapter? What was your score?

ya

i got 2nd place individual, 3rd place team

i got 39

heheheh...
your the kid who tied with me in chapter, but i got the first one wrong....
so i got third and you got second!!!! :mad:
but oh well...
im gonna own you in state competition

[quote="weird888"]Either pushkar or bo or xinke[/quote]

wait does bo have an account that you know of?

My predictions

Challenger!
Pizitz
Baldwin
Randolph
Auburn

Pizitz
Challenger
Berry
Baldwin
Simmons
Randolph


what is and arcsin? i have always thought $ \arcsin x\equal{}\sin^{\minus{}1}(x)$

You are correct. They are the same thing.

I've always wondered how you would tell the difference between $ \sin^{\minus{}1}x \equal{} \arcsin x$ and $ \sin^{\minus{}1}x \equal{} \csc x$

The notation $ \sin^{\minus{}1} x$ [b]always[/b] refers to inverse function notation (same with the notation $ f^{\minus{}1}(x)$), so it always refers to the arcsine.  Write $ (\sin x)^{\minus{}1}$ if you want to avoid the use of the cosecant notation.

The really confusing notation is that $ \sin^2 x$ does not mean $ \sin (\sin x)$.

[quote="t0rajir0u"]The really confusing notation is that $ \sin^2 x$ does not mean $ \sin (\sin x)$.[/quote]

I never thought about that.  :rotfl:

Yeah, especially after seeing a bunch of $ \sin^{2} x \ , \ cos^{3} x \ , $ etc.

$ \sin^{\minus{}1}x$ denotes the inverse of $ \sin x$. So like since $ \sin\left(\frac{\pi}{6}\right)\equal{}\frac{1}{2}, \sin^{\minus{}1}\left(\frac{1}{2}\right)\equal{}\frac{\pi}{6}$

[quote="Poincare"][quote="7h3.D3m0n.117"]Remember, $ \sin ^{ \minus{} 1} x \neq \csc x$. The $ \minus{} 1$ denotes an [b]inverse[/b], not the reciprocal.

I think it's more conventional to write inverse trig functions as $ \text{arc}$... to minimize confusion.[/quote]

Can you plz explain that a wee bit more clearly  :blush:[/quote]
When $ x^{ \minus{} 1}$ is written, usually the reciprocal, $ \frac {1}{x}$, is meant. However, when $ \sin^{ \minus{} 1}x$ is written, it does not mean the reciprocal, $ \frac {1}{\sin x} \equal{} \csc x$, but instead the inverse function, $ \arcsin{x}$.

This tends to cause confusion sometimes because $ \sin^2x$ is still $ \left(\sin x\right)^2$. To avoid confusion, $ \arcsin x$ is often used instead of $ \sin^{\minus{}1}x$.


Let $ a,b,c>0, abc\equal{}1$
Prove that:
$ \frac{a}{b^4\plus{}2}\plus{}\frac{b}{c^4\plus{}2}\plus{}\frac{c}{a^4\plus{}2} \geq 1$

[quote="marsupilami"]Please help me :|[/quote]

It's problem 67 in this file at topic :

http://www.mathlinks.ro/viewtopic.php?t=283154

:)


Use the numbers $1$, $2$, $3$, and $4$ each once to replace the variables $a$, $b$, $c$, and $d$. What is the maximum possible value for the expression $a+b\cdot c^{d}$?

Adam4444 is right.

Extention, what is the least value that can be made?

To Quevvy's Question:
[hide] Make it $3+2*(1^{4})$, which is $3+2$, which is $5$. [/hide]

[quote="snipefreak"]5??? NO WAY! He is asking for the MAX possible value, not the least. Adam is right, because the highest exp. value that can be made is 3^4, and that practically determines the problem.[/quote]
That was for the extension to find the smallest value possible, which is 5.


Suppose that $ P(x)$ and $Q(x)$ are polynomials with real coefficients such that for all $x$, $P(x)$ is an integer if and only if $Q(x)$ is an integer. Prove that $P(x)-Q(x)$ is a constant.

$ P(x)\in Z\Leftrightarrow P(x)\equal{}a\forall x\in R[x]$
$ Q(x)\in Z\Leftrightarrow Q(x)\equal{}b\forall x\in R[x]$
So $ P(x)\minus{}Q(x) \equal{} a\minus{}b$
We put a-b=c and here is the answer
I am not sure though about the answer.

This problem haven't solved .
It is very good.
Has some problem from this problem
Find all $ P(x)\in R[x]$ such that
$ P(x\plus{}1)\in Z\leftrightarrow P(x)\in Z$

[quote="TTsphn"]Suppose  $ P(x),Q(x)\in R[x]$
$ \forall x : P(x)\in Z\Leftrightarrow Q(x)\in Z$
Prove that $ P(x)\minus{}Q(x) \equal{} c$[/quote]

I think we miss some precision :

Let $ P(x)\equal{}x$ and $ Q(x)\equal{}1\minus{}x$, we have $ \forall x : P(x)\in Z\Leftrightarrow Q(x)\in Z$ but $ P(x)\minus{}Q(x) \equal{} 2x\minus{}1\neq c$

maybe you wanted to add "monic" polynomials ?

I am sorry $ P(x)\plus{}Q(x)$ or $ P(x)\minus{}Q(x)$=c

[quote="TTsphn"]Suppose  $ P(x),Q(x)\in R[x]$
$ \forall x : P(x)\in Z\Leftrightarrow Q(x)\in Z$
Prove that $ P(x)\minus{}Q(x) \equal{} c$[/quote]

Here is a solution :

With the condition $ P(x)\minus{}Q(x) \equal{} c$ or $ P(x)\plus{}Q(x) \equal{} c$, WLOG consider  that $ P(x)$ and $ Q(x)$ have positive coefficients of highest degree (you just have to appropriatly replace $ P$ by $ \minus{}P$ or $ Q$ by $ \minus{}Q$)

Then it exists $ x_{0}$ such that $ \forall x > x_{0}$, $ P(x)$ and $ Q(x)$ are strictly increasing (and grow up to $ \plus{}\infty$).
Let then $ n_{1}\in\mathbb{Z}$ such that $ n_{1}> P(x_{0})$
$ \forall n\geq n_{1}$, $ \exists$ unique $ a_{n}\in(x_{0},\plus{}\infty)$ (edited) such that $ P(a_{n}) \equal{} n_{1}\plus{}n\minus{}1$ and $ \{a_{n}\}$ is a strictly increasing sequence.
Since $ P(a_{n}) \equal{} n_{1}\plus{}n\minus{}1\in\mathbb{Z}$, $ Q(a_{n})\in\mathbb{Z}$ too.

Since $ a_{n\plus{}1}> a_{n}> a_{1}> x_{0}$, $ Q(a_{n\plus{}1}) > Q(a_{n})$. Then if $ Q(a_{n\plus{}1}) > Q(a_{n})\plus{}1$, $ \exists y\in(a_{n},a_{n\plus{}1})$ such that $ Q(y) \equal{} Q(a_{n})\plus{}1\in\mathbb{Z}$
So $ P(y) \equal{} m\in\mathbb{Z}$ and, since $ a_{n}< y < a_{n\plus{}1}$ and $ P(x)$ strictly increasing, $ P(a_{n}) \equal{} n_{1}\plus{}n\minus{}1 < P(y) \equal{} m < a_{n\plus{}1}\equal{} n_{1}\plus{}n$, which is impossible.

So $ Q(a_{n\plus{}1}) \equal{} Q(a_{n})\plus{}1$ and, since $ P(a_{n\plus{}1}) \equal{} P(a_{n})\plus{}1$, we have $ P(a_{n\plus{}1})\minus{}Q(a_{n\plus{}1}) \equal{} P(a_{n})\minus{}Q(a_{n}) \equal{} c$ $ \forall n > 0$

So $ P(x)\minus{}Q(x)\minus{}c$ has infinitly many zeroes and so $ P(x)\minus{}Q(x) \equal{} c$ $ \forall x\in\mathbb{R}$


Given $ 12$ points in plane and all the segments with the ends among them, find the maximum number of intersection points determed by these segements(of course we don't consider the case when a segment lies on another and gives infinitely many "intersection" points  :P )

$ \binom{12}{2}$ lines, and choosing 2 of them makes an intersection, so the answer is $ \boxed{\dbinom{\binom{12}{2}}{2}}$

We are talking about segments not lines...not any 2 segemtns must intersect....

hmm, either this is a trick question, or it doesn't really matter since we're looking for maximum/

We are looking for the [i]possible[/i] maximum....you cannot obtain all the segements to intersect  :P

Chart:
Segments, intersection:
1,0
2,1
3,3
4,6
triangular number
13th triangular number is 91.

be careful, note that there are 12 [u]points[/u], not line segments

And alan, I don't think that two lines that have the same endpoint are considered to be "intersecting"

So...can anyone give a solution to this problem? :maybe:

[quote="10000th User"][quote="Fraenkel"]i have a hunch (after  doing casework) that the maximum number of intersections the line segments of $ n$ points are those that are formed by the intersections of the diagonals of a regular $ n$-$ gon$[/quote]But, consider those regular polygons where the diagonals intersect at the center like hexagon with 3 diagonals intersecting at the center.[/quote]

That does disprove my hunch.
I guess if we move it a little on a certain vertex we will get more intersections.

Do you have a hunch?

I thought about arranging the points on a circle and counting for each segment then taking care of the carrings,too....but I know that the answer is [hide]$ 495$[/hide] and I couldn't get it in this way... :maybe:

[hide]I looked for a pattern... for $ n$ points, you can have $ \left(\frac{1}{2}\right)\binom{n}{2}(n\minus{}3)$... this agrees with $ 495.$[/hide]

Sorry to bump a thread, but a correct solution is available at http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1582761869&t=212526 .  (The answer of 495 given by two posters was correct, although the formula given by jsc is wrong and doesn't match the right answer.)


I noticed that when Mr. Crawford posted the word "competition" there were a lot more people. So I decided to have an honorary "AoPS member of the Month" and a competition to win that title.
Each week I will post a question, three total each month (I know, I know, I wanted it to be an odd number so there wouldn't be any ties...not! :) just an excuse so in case I can't post for various reasons, I won't be criticized...I'm very protective...of myself :) ). The winner of the competition would be the AoPS member of the Month for the [b]next[/b] month (because the competition would usually end at the end of the month, so even if that person is made the AoPS member of that month at that point, there isn't much time left.)

Oh, and everyone's invited, s'long as that person doesn't keep on winning over and over again...but oh well, we'll worry about that when that happens. For now, come on in! :)

I plan to start off the competition tomorrow. Are you ready?

OK, here's the rules.
A.) Everyone's invited to participate, but if you think it's too easy, then I'm hoping you'll refrain yourself
B.) [b]Please post all solutions in Spoiler[/b]. Again, [b]Please post all solutions in Spoiler[/b]. :) This is important, because this is an internet competition, and people from all over the world is participating. So when you post your answer, Budapest might be asleep. Now I don't want people to say "you're cheating" or the equivalent when people post after someone already post the answer. So please, use Spoilers. Otherwise your automatically disqualified for the round. ([b]Everything[/b] should be in Spoilers under the "round" threads)
C.) There would be 3 or 4 rounds, depending on how many problems I can come up with :) If there is three rounds, then we would have a winner for each round. If there are three different winners, then we shall decide the winner by the total points. Points are given mostly for three reasons:

1.) Correct Answer: The main way to earn the points
2.) How many ways in which you got the answer: How you got the answer is just as important as the answer itself. Also, this is to make the game fair for beginners competing, because this would be mostly busy work.
3.) Special points awarded for outstanding/brilliant answers: This is to make it fair for the super-geniuss out there. If you can come up with an answer that everyone can say "Wow :shock: " to, then you get awarded points for that. But keep in mind that this'll rarely happen, because most of the questions are really easy.

D.) Keep in mind this whole things works on an "honor" system. The prize you get is the "honor" of being the AoPS member of the Month, so you need to be honorary the whole time, not just be smart :)[b] However[/b], if I find out that you cheated or asked for help or looked at others' answers, then that's an automatic disqualification.
E.) There is no "due"s for each individual rounds, but you have to turn in the answers before the end of the month because the AoPS Member of the Month would be announced at the start of each month
F.) You can say if someone's answer is correct or incorrect, but you can't help them (like tell them where it went wrong, for example) Still, [b]that's only if you're not going to participate[/b], but that's common sense :)
G.) There won't be any "penalty" for incorrect answers. [b]However[/b] (there's always a "string" attached, eh? :) ), repeated bad manners such as name-calling, split/double-posting, etc. could result you in deduction of points, disqualification, etc.
H.) Have fun!

Any questions or comments can be private messaged to me. Enjoy!

I don't get on frequently enough any more to be able to participate.  Plus, I'm sort of beyond mathcounts now.  Sigh.\

-mathfanatic

P.S. I find it interesting to note that you got mad at rep123max for double-posting (even though you posted in between those "doubles"), yet you double-posted yourself anyway.

Well the second post is [b]the[/b] "Official Rules" and the first part's just my post, so I separated them out.
Also, as I've said, you can
a.) solve them even if you are beyond MC level
b.) after the round is over

If I'm allowed online, I'll try this :)

Hmm...think there's any way you could start one for USAMO people?  If it's a weekly thing, I could help run one for upper-level math...

Again, "I'm the man with the plan!" so just wait patiently until October 1, just like tons of other jitterish people (well, 2 to be exact :) ) who wants to take this competition to their own forum. (hint, hint :) )

Not at all. You can start whenever you want, and end whenever you want, as long as it's before the end of September (the 30th).

In fact, you could even do Round 1 on the 29th and I'll be fine with that.


Let $a,x,y,z\in R^{+}$, show that

$x\frac{a+z}{a+x}+y\frac{a+x}{a+y}+z\frac{a+y}{a+z}\leq x+y+z\leq x\frac{a+y}{a+z}+y\frac{a+z}{a+x}+z\frac{a+x}{a+y}$

See http://www.mathlinks.ro/Forum/viewtopic.php?t=32551

  darij


AMC AIME USA(J)MO USAMO trigonometry geometry search

Where would be a good place to start in practicing olympiad problems? Any particular competitions' archive/old problems or books? (This is for AIME preparation, and I have no previous experience with olympiad level)

Thanks :D

Competitions: Canada MO, Ireland MO, Early USAMO

Books: ACOPS, 102 Combinatorial Problems, 104 Problems in Number Theory, 103 Problems in Trig, Geometry Revisited (only the triangles & circles chapters)

meh USAMO 2002 #1 (go search it using the "contests" button, not wiki)

here's a question: "why do proofs like the one in my blog not work?(blog link below, should be a more recent post or use categories)

And definitely do USAMTS. It not only adds a bit more competitiveness, it takes the "spend time on problems" concept to the next level.

While I think USAMTS can be a good way to improve problem-solving skills, its problems are significantly easier than USAMO problems. I would say USAMTS problems are generally AIME level.

[quote="andersonw"]I would recommend even recent USAMO #1/4s, because although they are harder than earlier ones, they are only harder in that they require more creativity, not more mathematical knowledge. If you can usually score about 5-7+ on the AIME and know some basic proof techniques, you definitely have all of the skills needed to solve 1s and 4s. One important note when solving any olympiad problem: You will get a LOT more out of any problem if you try solving it for a decent amount of time before giving up than if you look at the solution after the first 10 minutes of seeing the problem. If you can't solve a problem and you feel like you haven't been making any progress for a while, I would recommend putting it aside for a bit and trying some new approaches later (I'm pretty sure many users here can tell you of how they solved an olympiad problem in the shower or in bed).
Other than USAMO, Canada MO is pretty good, as well as Juniar Balkan MO and British MO (http://www.bmoc.maths.org/).[/quote]

This.

Though for most contests, it's typically useful to ask a person who's done the problem whether the problem requires advance knowledge/just sucks.


If $ a,b,c>0$, then

$ \frac{5a^2\plus{}2bc}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{5b^2\plus{}2ca}{c^2\plus{}ca\plus{}a^2}\plus{}\frac{5c^2\plus{}2ab}{a^2\plus{}ab\plus{}b^2}\geq7$


Find the value in function of $ n$:

$ 1\cdot1!\plus{}2\cdot2!\plus{}\cdots\plus{}n\cdot n!$

[hide="outline"] Conjecture that it is $ (n\plus{}1)!\minus{}1$, use induction, done.[/hide]

Too easy for Preolympiad...

Since it is easy would you mind deducing how you found $ (n\plus{}1)!\minus{}1$ ?

Evaluate the first few $ n$.

You can also note that $ k\cdot k!\equal{}(k\plus{}1)k!\minus{}k!\equal{}(k\plus{}1)!\minus{}k!$ to get a nice telescoping series.


Let $ ABCD$ be a quadrilateral inscribed in a circle of center $ O$. Let $ M$ and $ N$ be the midpoints of diagonals $ AC$ and $ BD$, respectively and let $ P$ be the intersection point of the diagonals $ AC$ and $ BD$ of the quadrilateral $ ABCD$.It is known that the points $ O,M,N,P$ are distinct. Prove that the points $ O,M,B,D$ are concyclic if and only if the points $ O,N,A,C$ are concyclic.

[b]The is CE.229 of MATHESIS, 1886, p.24.[/b]
view topic:  http://www.mathlinks.ro/viewtopic.php?t=163543
(page 6)

Let A,N,O,C are concyclic  NP∩OM=X
∠ONC=∠OAM=90-1/2∠AOC=90-∠ABC    =〉  ∠CNP=90-∠ONC=∠ABC
=〉 ∠BNC=∠ADC  =〉∠AND=∠BNC  =〉  ABCD are harmonic quadrilateral
=〉 ∠PNC=∠COM=∠ABC  =〉X,N,O,C,A are concyclic
XM*XO=XP*XN    but   XP*XN=XN^2-NP*XN
IF D,M,O,B are concylic  XM*XO=XD*XB
That means  XP*XN=XD*XB   〈=〉 XN^2-ND^2= XN^2-NP*XN
  〈=〉 ND^2= NP*XN   〈=〉ND^2=PN^2+AP*PC=OP^2-ON^2+R^2-OP^2=R^2-ON^2
  〈=〉ON⊥BD
but we already know ON⊥BD
so D,M,O,B are concylic
And when D,M,O,B are concylic ,  analogously  we acquire A,N,O,C are concyclic

Done!


inequalities
function
inequalities unsolved

Let $ 0\leq a,b,c \leq1$. Prove that:
$ \frac{1}{bc\plus{}1}\plus{} \frac{1}{ca\plus{}1} \plus{}\frac{1}{ab\plus{}1} \leq \frac{5}{a\plus{}b\plus{}c}$

Multiplying each side by $ a\plus{}b\plus{}c$, and considering the $ LHS$ as a function of $ a$, we have $ f''(a)\geq0$ so the function is strongly convex in each of the variables. Since a strongly convex function obtains is maximum at one of its endpoints, it remains to check the function for the 8 possibilities of $ {a,b,c}\equal{}{0,1}$.  We find that equality occurs when two of $ a,b,c$ are $ 1$ and the other is $ 0$.


symmetry
real analysis
real analysis unsolved

Let $ z$ be a complex number. Show that the series:

$ \frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\frac{z^4}{1-z^8}+\frac{z^8}{1-z^{16}}+ ...$

converges to:

${ \frac{z}{1-z}}$ if $ |z|<1$

${ \frac{1}{1-z}}$ if $ |z|>1$

and diverges if $ |z|=1$

If $ |z|<1$, expand everything as power series in $ z$, prove the required condition to show you can rearrange.

Thanks, transserries, I got that.

Any ideas about the cases $ |z|>1$ and $ |z|\equal{}1$ ?

For $ |z| \equal{} 1$, try term test.
For $ |z| > 1$, exploit the symmetry of the expression - you know how to deal with the case $ |z| < 1$ already. How can you change the variable to turn $ |z| > 1$ into $ |z| < 1$?

Thanks for answer, Soarer, your ideas helped a lot.


function
group theory
abstract algebra
superior algebra
superior algebra unsolved

The purpose of this proposal is to present the reader with a [b]novel[/b] fallacy in mathematics. The exposition is self-contained, yet, it is required that the reader keep both [i]mind[/i]'s eyes open all the way through it.
--
We begin by introducing the notion on which all of our discussion will hinge:

[b]Definition 1.[/b] Let $ \mathbb{X}$ be a set and $ \mathrm{G}$ a group (written multiplicatively). A [i]group action[/i] of $ \mathrm{G}$ on $ \mathbb{X}$ is a binary function $ \mathcal{F} : \mathrm{G} \times \mathbb{X} \to \mathbb{X}$ that satisfies the following axioms:

a) $ \mathcal{F}(1,x) \equal{} x$ for every $ x \in \mathbb{X}$.
b) If $ g, h \in \mathrm{G}$ then $ \mathcal{F}(g,\mathcal{F}(h,x)) \equal{} \mathcal{F}(g \cdot h,x)$ for every $ x \in \mathbb{X}$.

Associated to group actions we commonly encounter the construct that motivates our

[b]Definition 2.[/b] Let $ \mathbb{X}$, $ \mathrm{G}$, and $ \mathcal{F}$ be as in definition [b]1[/b]. If $ x \in \mathbb{X}$, the [i]stabilizer subgroup[/i] of $ x$, denoted by $ \mathbf{Stab}(x)$, is the set (subgroup) of all elements of $ \mathrm{G}$ that fix $ x$ under $ \mathcal{F}$, that's to say

$ \mathbf{Stab}(x) \equal{} \{g \in \mathrm{G} : \mathcal{F}(g,x) \equal{} x\}.$

From definitions [b]1[/b] and [b]2[/b], we readily derive the rather inoffensive-looking

[b]Scholium.[/b] Let $ \mathbb{X}$, $ \mathrm{G}$, and $ \mathcal{F}$ be as in definition [b]1[/b]. If $ x \in \mathbb{X}$ then $ g \cdot \mathbf{Stab}(x) \equal{} h \cdot \mathbf{Stab}(x) \Leftrightarrow \mathcal{F}(g,x) \equal{} \mathcal{F}(h,x)$.

[b]Proof.[/b]

$ \Rightarrow$) The assumption that $ g \cdot \mathbf{Stab}(x) \equal{} h \cdot \mathbf{Stab}(x)$ implies that $ g \equal{} h \cdot s$, where $ s \in \mathbf{Stab}(x)$. Hence, $ \mathcal{F}(g,x) \equal{} \mathcal{F}(h \cdot s, x) \equal{} \mathcal{F}(h,\mathcal{F}(s,x)) \equal{} \mathcal{F}(h,x)$.

$ \Leftarrow$) From the hypothesis that $ \mathcal{F}(g,x) \equal{} \mathcal{F}(h,x)$ we obtain that $ \mathcal{F}(h^{ \minus{} 1} \cdot g,x) \equal{} \mathcal{F}(h^{ \minus{} 1},\mathcal{F}(g,x)) \equal{} \mathcal{F}(h^{ \minus{} 1},\mathcal{F}(h,x)) \equal{} \mathcal{F}(h^{ \minus{} 1} \cdot h,x) \equal{} \mathcal{F}(1,x) \equal{} x.$ It follows that $ h^{ \minus{} 1}g \in \mathbf{Stab}(x)$ and we're done.

[b]QED.[/b]

The latter result has paved our way for a detour to the exciting world of bijective proofs.

[b]Lemma. [/b] If $ \mathbb{X}$, $ \mathrm{G}$, $ \mathcal{F}$, and $ x$ are as in definition [b]2[/b], then $ \vert \mathbb{X} \vert \equal{} [\mathrm{G} : \mathbf{Stab}(x)].$

[b]Proof.[/b] Let $ \mathcal{C}$ denote the collection of all left cosets of $ \mathbf{Stab}(x)$ in $ \mathrm{G}$. Since $ \vert \mathcal{C} \vert \equal{} [\mathrm{G} : \mathbf{Stab}(x)]$, we'll be done if we are able to exhibit a bijection from $ \mathcal{C}$ to $ \mathbb{X}$. Let $ f: \mathcal{C} \to \mathbb{X}$ be defined by

$ g \cdot \mathbf{Stab}(x) \mapsto \mathcal{F}(g,x).$

We claim that this function $ f$ is the bijection that we need. It is straightforward from the scholium that $ f$ is a one-to-one application from $ \mathcal{C}$ to $ \mathbb{X}$. On the other hand, if $ x \in \mathbb{X}$ the equality $ f(1 \cdot \mathbf{Stab}(x)) \equal{} \mathcal{F}(1,x) \equal{} x$ implies the surjectivity of $ f$, as desired.

[b]QED.[/b]

We are ready now for the proclaimed demonstration of the [size=150]main[/size] result of the post:

[b]Proposition.[/b] $ \vert \mathbb{Z} \vert \equal{} 1.$

[b]Proof.[/b] For $ n \in \mathbb{N}$ we denote with $ \mathrm{C}_{n}$ the cyclic group of order $ n$. Now, if we let $ \mathbb{X} \equal{} \mathbb{Z}$, $ \mathrm{G} \equal{} \mathrm{C}_{2009}$, and $ \mathcal{F}(g,x) \equal{} x$ for every $ (g,x) \in \mathrm{G} \times \mathbb{X}$, a direct application of the above lemma gives us that $ \vert \mathbb{Z} \vert \equal{} [\mathrm{C}_{2009} : \mathbf{Stab}(1)]$. Since in this case we also have that

$ \mathbf{Stab}(1) \equal{} \mathrm{C}_{2009}$

we conclude that

$ [\mathrm{C}_{2009} : \mathbf{Stab}(1)] \equal{} [\mathrm{C}_{2009} : \mathrm{C}_{2009}] \equal{} 1$.

Hence, $ \vert \mathbb{Z} \vert \equal{} 1$ and the proof terminates.

[b]QED.[/b]
--
Did I hear you utter something along the lines of [i]gosh, it can't be...[/i]? Well, believe it or not, we've just proven that [i]it[/i] can definitely be, can't it? ;)

Clever.

[hide="One word"]Transitive.[/hide]

- Thanks for taking the time to reply, dear [b]jmerry[/b].

- Who else feels like posting his/her solution? ;)

[quote]Lemma.  If $ \mathbb{X}, \mathrm{G}, \mathcal{F}$, and $ x$ are as in definition 2, then $ \vert \mathbb{X} \vert \equal{} [\mathrm{G} : \mathbf{Stab}(x)]$.
...
On the other hand, if $ x \in \mathbb{X}$ the equality $ f(1 \cdot \mathbf{Stab}(x)) \equal{} \mathcal{F}(1,x) \equal{} x$ implies the surjectivity of $ f$, as desired. [/quote]

In the second statement you're using $ x$ both as a fixed element in $ \mathbb{X}$ and a varying element. We've already fixed $ x$; we need to show that for $ \underline{y} \in \mathbb{X}$ we can find a coset (by varying $ g$) such that  $ f(g \cdot \textbf{Stab}(x)) \equal{} y$. Therefore we have not established the subjectivity of $ f$ in the second statement.

Perfect, [b]dlc122[/b]! That was so down to the nitty-gritty...

By the way, you can find the [color=orange]May proposal[/color] here:

[url=http://www.mathlinks.ro/viewtopic.php?t=275696]-_-[/url]

Regards,

José.

P.S. How did you like this [i]fallacy[/i] of mine?


this probability business if VERY CONFUSING!

[img]http://i43.tinypic.com/295v884.jpg[/img]

You know the variance and you know the mean, which allows you to recover $ \mu$ and $ \sigma$.


im interested how to get this

$ \sqrt[3]{ \sqrt{393}\plus{}67 }\plus{} \sqrt[3]{ 67 \minus{} \sqrt{393} } \plus{}4$

from

$ h^3\equal{}12h^2\plus{}6$

1. Google Cardano's formulas and follow the steps.

2. Derive Cardano's formulas yourself.
[list][*]Write the equation as $ x^3 \plus{} ax \plus{} b$ (you may have to shift $ x$ to get rid of the $ x^2$ coefficient)
[*]Note the factorization $ x^3 \plus{} u^3 \plus{} v^3 \minus{} 3xuv \equal{} (x\plus{}u\plus{}v)(x^2\plus{}u^2\plus{}v^2\minus{}xu\minus{}xv\minus{}uv)$
[*]Therefore, try to find $ u,v$ such that $ \minus{}3uv \equal{} a, u^3 \plus{} v^3 \equal{} b$. Your answer will be nasty.
[*]Plug in and use the factorization.[/list]

There is also a faster way, which involves making a largely unmotivated substitution after the first step. It's something like $ x \equal{} t \minus{} \frac{3}{t}$.


Solve the system equations in x and y:
x + y = a and x^5 + y^5 = b.


search
real analysis
real analysis unsolved

Prove that $\mathbb R^{2}$ has a dense subset such that has no three collinear points.

Tim Gowers is a master of explaining not just how to prove something, but how to discover the proof. This problem is Example 1 [url=http://gowers.wordpress.com/2008/08/16/just-do-it-proofs/]here[/url].


calculus
function
algebra
domain
topology
real analysis
real analysis unsolved

Let $E$ be an arbitrary set and, for $p, q \in E$, define $d(p, q) = 0$ if $p = q$, $d(p, q) = 1$ if $p \ne q$. This is a metric space.

So the question is: What are the open and closed balls in this metric space?

I don't understand what the question is asking. Graphically, it talks about the set of points on a 1D number line such that if $p = q$ then the distance from $p$ to $q$ is $0$ and if $p \ne q$, then the distance b/w the two points is $1$. Wouldn't this make the whole number line covered, b/c there is always a point within a distance of $1$ from any given point? And any given point is a distance of $0$ from itself? So would the open and closed balls both be the whole number line?

Any help is appreciated.

The radius matters.  Let $B(x,r)$ denote the open ball of radius $r$ around $x$.  Then, for example, if $E=\mathbb{R}$, then $B(x,1/2)=\{x\}$, $B(x,1)=\{x\}$, and $B(x,2)=\mathbb{R}$.

blahblahblah, could you please explain your equations a little further? I don't understand why $B(x, 1/2) = \{x\}$, Wouldn't it have an infinite number of "R" numbers. Also, are you saying that what I wrote is right or wrong? Thanks though for replying

I don't think questions about open sets, compact sets, metric spaces, etc. - in other words, the more abstract and general parts of a "real analysis" or "advanced calculus" course - belong in the "Calculus Computations and Tutorials" forum, even if they are (as this one is) textbook exercises. In my view, that forum should remain accessible to students taking the courses that run from first-year calculus through differential equations. So I'm moving this to a "higher" forum.

Bythecliff, your last response reveals that you have trapped yourself into trying to visualize this as the usual distance. It is not the usual distance and it has no geometric meaning at all. $B(x,1/2)=\{x\}$ because there are no elements $y$ other than $x$ itself for which $d(y,x)<1.$

This is called the "discrete metric" and it gives rise to the "discrete topology." Bythecliff, did the question actually ask what are the open and closed [b]sets[/b] in this metric space? If that's the question, then we reason as follows:

As blahblahblah has pointed out, single-element sets are open balls, hence open sets.

The arbitrary union of open sets is open. But any set whatsoever is the union of all the single-element sets contained in it. Therefore, every subset of the whole space is open.

It follows that every subset of the whole space is also closed. It's also true that any function with this metric space as its domain into any other metric space is automatically continuous.

Kent Merryfield,

I am still a bit confused over what you mean by $B(x, 1/2) = \{x\}$. I keep on thinking that 1/2 has something to do with the radius. It's really confusing. Can you shed some light on this matter?

B(p, r) means the (open) ball centered at p with radius r. In this case, B(x, 1/2) would mean the set of points that have distance less than 1/2 from x. What are these points? Well, x itself is certainly in this set, as d(x, x) = 0. Any other point is not in this set, as the distance is 1. Thus the ball is simply {x}.

[quote="bythecliff"]Oh, ok, that makes more sense, mysticterminator!

So $B(x,1) = \{x\}$ also because we have an open ball with points 1 unit from x but not including that exact distance of 1 unit (b/c of open parent), but say $B(x,3) = R$, right?[/quote]

Yes.  (Although in the original question, your set was named $E$, not $R$.)

The discrete metric is designed to shatter your basic intuition, because your basic intuition isn't very good at dealing with things like an abstract metric space.  The point, more or less, is that there are things which [i]don't[/i] look like the number line or more generally real Euclidean spaces, but which are perfectly good metric spaces.

Here's another intuition-bender:

Let $E$ be the set of 7-character strings, in which each character is either $0$ or $1.$ Note that $E$ is a finite set with $|E|=128.$ Define the distance between strings as the number of places at which the strings differ. For instance:

$d(0011001,1011001)=1$
$d(0011001,1100110)=7$
$d(1111111,1101101)=2$

Can you prove that this is indeed a metric?

Pick some $p\in E.$ How many elements are there in the closed ball centered at $p$ of radius $1?$ That is, find the cardinality of $\{x\in E: d(x,p)\le 1\}.$

Ultimately, we'd like to know how many pairwise disjoint closed balls of radius $1$ we can pack into $E.$


AIME 1983 (not exact text)

15. BC is a chord length 6 of a circle center O radius 5. A is a point on the circle closer to B than C such that there is just one chord AD which is bisected by BC. Find sin AOB.

I just copied and pasted this from my solution document. I hope you don't mind. :D

[hide="Answer"]Let A be any fixed point on circle O and let AD be a chord of the circle O. The locus of midpoints N of the chords AD is a circle P, with diameter AD. Generally, the circle P can intersect the chord BC at 2 points, touch it at one point, or it may not have a common point with the chord BC. By the problem condition, the circle P touches the chord BC at one point N. Let M be the midpoint of the chord BC, $BM = \frac{BC}{2} = 3$. From the right angle triangle $OMB$,

$OM = \sqrt{OB^2 - BM^2} = \sqrt{5^2 - 3^2}=4$, which implies that $\tan BOM = \frac{BM}{OM} = \frac 3 4$

Since circle P is tangent to the line BC at the distance OM from the point O,

$OM = PN + PO \cos AOM = r(1 + \cos AOM)$

$\cos AOM = \frac{OM}{r} - 1 = \frac{2OM}{R} - 1 = \frac 8 5 - 1 = \frac 3 5$

$\tan AOM = \frac{\sqrt{1 - \cos^2 AOM}}{\cos AOM} = \frac{\sqrt{5^2 - 3^2}}{3} = \frac 4 3$

Next, $\angle AOB = \angle AOM - \angle BOM$

$\tan AOM = \frac{\tan AOM - \tan BOM}{1 + \tan AOM \cdot \tan AOM} = \frac{\frac 4 3 - \frac 3 4}{1 + \frac 4 3 \cdot \frac 3 4} = \frac{7}{24}$

$\sin AOM = \frac{7^2}{\sqrt{7^2+24^2}} = \frac{7}{25}$

The answer is thus $7 \cdot 25=\boxed{175}$.
[/hide]


I need a step by step explanation of the Quadratic formula, but not with variables, can someone explain to me were a=2, b=3, and c=5? PLZ i really need help, i have a test tmrw.

What do you need to know about it? The formula is $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$. The general form for a quadratic is $ax^2+bx+c$, which in your case would be $2x^2+3x+5$. You would just substitute your numbers into the quadratic formula:

$\frac{3 \pm \sqrt{3^2-4(2)(5)}}{2(2)}=\frac{3 \pm \sqrt{9-40}}{4}=\frac{3 \pm \sqrt{-31}}{4}$
Because the stuff under the square root is negative, there are no (real) solutions to this problem.

The $\pm$ symbol means you have a number with a $+$ and a number with a $-$. Instead of saying "$-1\ \text{and}\ 1$", you could say "$\pm 1$". Another example: $3 \pm \sqrt{4}$ is equivalent to $3+\sqrt{4}=3+2=5$ and $3-\sqrt{4}=3-2=1$.


floor function
number theory
prime factorization

Let $2^n$ be the greatest power of $2$ which can devide $200!$ evenly. Find $n$.

[hide] i got $2^{197}$ i only concerned myself with the part of the prime factorization of $200!$ that had $2$ in it. So $\boxed{n=197}$... i think  :? [/hide]

I think you mean the other way around... or maybe not.
Just so I can understand, 200! is divisible by 2^n...

200! has 100 even factors. Each can be divided by 2, making n 100. After that, there are 50 multiples of 4 which can be divided by 2 again. n = 150.
25 are multiples of 8, so n = 175.
12 are multiples of 16, so n = 187.
6 are multiples of 32, so n = 193
3 are multiples of 64, so n = 196
1 more is a multiple of 128, so the final n is $ 197 $.

Hope that made sense.  :)

Yes Hexaditidom, that's exactly how I did it, just keep dividing by 2, and if you run into a number that doesn't divide by it, then go to the closest number that does. Good job both of y'all :10:

[quote="236factorial"]ISn't this related to finding how many trailing zeros n! has?
Divide by 5+divide by 25....[/quote]

Yeah, speaking of that.....Say you have $200!$ and you wanna know how many zeroes it has, would it just be:

$200/5=40$
$40/5=8$
$5/5=1$

So would there be $49$ zeroes at the end? Or do you gotta divide by $25$ also?

[quote="chess64"]$\lfloor 200/5 \rfloor + \lfloor 200/25 \rfloor + \lfloor 200/125 \rfloor + \cdots = 49$, so $49$ zeros.[/quote]

Okay, so either way, it's the same answer...I always thought it was easier if you just keep dividing by $5$ but essentially it's the same thing either way.

[quote="chess64"]411 is not a perfect square, Hexaditidom ;)[/quote]
Yeah, well you know what I mean...  fixed it!


Hmm ok..
Show that : 6! * 7! = 10! (without calculating 10! and without using a calculator)
:)

[size=75][color=darkblue]<split from another thread by moderator RCV>[/color][/size]

[quote="mathmanman"]Hmm ok..
Show that : 6! * 7! = 10! (without calculating 10! and without using a calculator)
:)[/quote]

$10!=10\cdot9\cdot8\cdot7!=6!7!$

No no , post different problems on different threads  :)

Never mind, ignore this post.

Alright, I didn't know that :)

btw, congrats Shyong!

(if someone wants to develop it a bit, feel free)

[hide]Since $6!=720$ and $10*9*8=720$, you can substitute each other for another.

Since $10!=10*9*8*7!$, you can also write that as $6!*7!$[/hide]


[CENTER][COLOR="Red"][b]CÁC LỚP THCS[/b][/COLOR][/CENTER]
[b]Bài T1 (lớp 6 )[/b]
Xét [b]n[/b] điểm liên tiếp $ A_1, A_2, A_3,...,A_n$ cùng thuộc một đường thẳng sao cho $ A_1A_2=A_2A_3=A_3A_4=...=A_{n-1}A_n$. Tìm [b]n[/b] biết rằng trên đường thẳng đó có tất cả [b]2025[/b] đoạn thẳng nhận một trong các điểm đã cho làm trung điểm.
[b]Bài T2(lớp 7)[/b]
Cho tam gaíc [b]ABC[/b] vuông ở [b]A[/b]. Trên nửa mặt phẳng bờ [b]BC[/b] không chứa điểm [b]A[/b] vẽ các điểm [b]D, E[/b] sao cho [b]BD[/b] vuông góc và bằng [b]BA, BE[/b] vuông góc và bằng [b]BC[/b]. Gọi [b]M[/b] là trung điểm của đoạn thẳng [b]CE[/b]. Chứng minh rằng ba điểm [b]A, D, M[/b] thẳng hàng.
[b]Bài T3[/b]
Tìm tất cả các số [b]x, y, z[/b] nguyên dương sao cho $ 2xy - 1 = z(x-1)(y-1)$
[b]Bài T4[/b]
Giải phương trình : $ 4x - x^2 = \sqrt{4-3\sqrt{10-3x}}$
[b]Bài T5[/b]
Cho tam giác [b]ABC[/b] vuông ở [b]A[/b]. Kẻ phân giác [b]AD[/b]. Gọi [b]M[/b] và [b]N[/b] lần lượt là hình chiếu của [b]D[/b] trên [b]AB[/b] và [b]AC[/b]. [b]BN[/b] cắt [b]CM[/b] tại [b]K, AK[/b] cắt [b]DM[/b] tại [b]I[/b]. Tính số đo của góc [b]BID[/b].
[CENTER][COLOR="Red"][b]CÁC LỚP THPT[/b][/COLOR][/CENTER]
[b]Bài T6[/b]
Cho đa thức $ f(x)=2009x^5-x^4-x^3-x^2-2006x+1$. Chứng minh rằng, với [b]n[/b] là số nguyên tùy ý thì các số $ f(n), f(f(n)), f(f(f(n)))$... đôi một nguyên tố cùng nhau.
[b]Bài T7[/b]
Tìm các số [b]a, b[/b]  để $ max|\sqrt{x}+ax+b|, x \in [0,16]$ đạt giá trị nhỏ nhất. Hãy tính giá trị nhỏ nhất đó.
[b]Bài T8[/b]
Đường tròn nội tiếp tam giác [b]ABC[/b] tiếp xúc với các cạnh [b]BC, CA, AB[/b] thoe thứ tự tại [b]A', B', C'[/b]. Chứng minh rằng $ AB'^2+BC'^2+CA'^2 \geq AB'.B'C'+BC'.C'A'+CA'.A'B' \geq B'C'^2+C'A'^2+A'B'^2$
Đẳng thức xảy ra khi nào ?
[CENTER][COLOR="Red"][b]TIẾN TỚI OLIMPIC TOÁN[/b][/COLOR][/CENTER]
[b]Bài T9[/b]
Với số nguyên dương [b]k[/b] bất kì, ta phân tích [b]k[/b] thành tích các thừa số nguyên tố ( không nhất thiết phân biệt, chẳng hạn [b]k = 18 = 3.3.2[/b]. Gọi [b]T(k)[/b] là tổng của tất cả các thừa số đó. Tìm hằng số [b]C[/b] lớn nhất sao cho $ T(k) \geq C lnk$, với mọi số nguyên dương [b]k[/b].
[b]Bài T10[/b]
Cho các số thực không âm [b]a, b, c[/b] có tổng bình phương bằng 3. Tìm giá trị lớn nhất của biểu thức $ P = ab^2+bc^2+ca^2-abc$
[b]Bài T11[/b]
Cho dãy số $ (x_n) (n = 1, 2,...)$ được xác định bởi :
$ {\begin{x_1=\frac{1}{2}}\\{x_{n+1}=x_n-x_n^2+x_n^3-x_n^4+...+x_n^{2007}-x_n^{2008}$
Tính $ lim nx_n$ khi $ n --> \infty$
[b]Bài T12[/b]
Giả sử [b]ABCD[/b] là một tứ diện có các đường cao đồng quy. Gọi [b]R[/b] là bán kính mặt cầu ngoại tiếp tứ diện : $ h_1, h_2, h_3, h_4$ theo thứ tự là độ dài các đường cao hạ từ các đỉnh [b]A, B, C, D[/b] của tứ diện. $ R_1, R_2, R_3, R_4$ theo thứ tự là bán kính đường tròn ngoại tiếp các mặt đối diện các đỉnh [b]A, B, C, D[/b] của tứ diện. Chứng minh bất đẳng thức $ \frac{1}{h_1+2\sqrt{2}R_1}+\frac{1}{h_2+2\sqrt{2}R_2}+\frac{1}{h_3+2\sqrt{2}R_3}+\frac{1}{h_4+2\sqrt{2}R_4} \geq \frac{1}{R}$
Đẳng thức xảy ra khi nào ?

Mãi mới mua được tờ báo, mọi người làm post lên luôn đi. Số cũ chắc các bài gửi hết rồi, để mọi người xem với.
[COLOR="Red"][b]Các lớp THCS[/b][/COLOR]
[b]Bài T1/364( Lớp 6 )[/b]
Tìm hai chữ số tận cùng của hiệu $ 3^{9999}\minus{}2^{9999}$
[b]Bài T2/364( Lớp 7 )[/b]
Xét tổng gồm [b]n+1[/b] số hạng $ S_n\equal{}\frac{2}{101\plus{}1}\plus{}\frac{2^2}{101^2\plus{}1}\plus{}...\plus{}\frac{2^{n\plus{}1}}{101^{2n}\plus{}1}$
So sánh $ S_n$ với [b]0,02[/b].
[b]Bài T3/364[/b]
Cho các số dương a, b, c thỏa mãn $ ab\plus{}bc\plus{}ca\equal{}1$. Chứng minh rằng
$ \frac{1}{ab}\plus{}\frac{1}{bc}\plus{}\frac{1}{ca} \geq 3\plus{}\sqrt{\frac{1}{a^2}\plus{}1}\plus{}\sqrt{\frac{1}{b^2}\plus{}1}\plus{}\sqrt{\frac{1}{c^2}\plus{}1}$
Dấu "=" xảy ra khi nào ?
[b]Bài T4/364[/b]
Giải phương trình : $ \sqrt{7x^2\minus{}22x\plus{}28}\plus{}\sqrt{7x^2\plus{}8x\plus{}13}\plus{}\sqrt{31x^2\plus{}14x\plus{}4} \equal{} 3.\sqrt{3}.(x\plus{}2)$
[b]Bài T5/364[/b]
Cho hình chữ nhật [b]ABCD[/b] với [b]AB < BC[/b]. Vẽ nửa đường tròn đường kính [b]AB[/b] nằm trên nửa mặt phẳng chứa [b]CD[/b] bờ là đường thẳng [b]AB[/b]. [b]M[/b] là điểm nằm trên nửa đường tròn ( [b]M[/b] khác [b]A[/b] và [b]B[/b]). Các đường thẳng MA, MB cắt đường thẳng [b]CD[/b] theo thứ tự tại [b]P, Q[/b]. Các đường thẳng [b]MC, MD[/b] cắt đường thẳng [b]AB[/b] theo thứ tự tại [b]E, F.[/b] Xác định vị trí của [b]M[/b] trên nửa đừơng tròn để tổng [b]PQ+EF[/b] đạt giá trị nhỏ nhất. Tính giá trị nhỏ nhất đó ?
[b][COLOR="Red"]Các lớp THPT[/COLOR][/b]
[b]Bài T6/364[/b].
Tìm tất cả các bộ hai số nguyên dưong [b]a, b[/b] sao cho khi chia $ a^2\plus{}b^2$ cho [b]a+b[/b] được thương là [b]q[/b] và số dư [b]r[/b] thỏa mãn $ q^2\minus{}r\equal{}2007$.
[b]Bài T7/364[/b]
Xét phương trình $ ax^3\minus{}x^2\plus{}bx\minus{}1\equal{}0$ với a,b là các số thực $ a \neq 0, b \neq 0$ sao cho các nghiệm đều là số thực dương. Tìm giá trị nhỏ nhất của $ P\equal{}\frac{5a^2\minus{}3ab\plus{}2}{a^2(b\minus{}a)}$
[b]Bài T8/364[/b]
Trên mặt cầu bán kính [b]R[/b] lây năm điểm [b]A, B, C, D, E[/b] sao cho $ \widehat{BAC}\equal{}\widehat{CAD}\equal{}\widehat{DAE}\equal{}\widehat{EAB}\equal{}\frac{2}{3}\widehat{BAD}\equal{}\frac{2}{3}\widehat{CAE}$. Chứng minh rằng : $ AB\plus{}AC\plus{}AD\plus{}AE \leq 4\sqrt{2}R$
[b][COLOR="Red"]Tiến tới Olimpic toán[/COLOR][/b]
[b]Bài T9/364[/b]
Trên các cạnh [b]AB, BC, CA[/b] của tam giác [b]ABC[/b] lấy các điểm [b]M, N, P[/b] theo thứ tự sao cho [b]AM+BN+CP=MB+NC+PA[/b]. Chứng minh rằng $ S_{MNP} \leq \frac{1}{4}S_{ABC}$
[b]Bài T10/364[/b]
Tính $ lim_{n \to \infty}(\sqrt{2\minus{}\sqrt{2}}\sqrt{2\minus{}\sqrt{2\plus{}\sqrt{2}}}...\sqrt{2\minus{}\sqrt{2\plus{}\sqrt{2\plus{}\sqrt{2\plus{}...\sqrt{2}}}}})$  . ( biểu thức cuối là [b]n[/b] dấu căn )
[b]Bài T11/364[/b]
Ki hiệu [b][x][/b] là số nguyên lớn nhất không vượt quá [b]x[/b] và kí hiệu [b]{x}=x-[x].[/b]. Hãy tìm $ lim_{n \to \infty}{(7\plus{}4\sqrt{3})^n}$.
[b]Bài T12/364[/b]
Cho [b]n[/b] là một số nguyên dương và [b]2n+2[/b] số thực $ a,b,a_1,a_2,...,a_n,b_1,b_2,...,b_n$ với $ a_i \neq 0 (i\equal{}1,2,...,n)$ sao cho hàm số $ F(x)\equal{}\sum_{i\equal{}1}^n \sqrt{a_ix\plus{}b_i}\minus{}(ax\plus{}b)$ có tính chất:
Tồn tại hai số thực phân biệt $ \alpha, \beta$ sao cho $ F(\alpha)\equal{}F(\beta)\equal{}0$. Chứng minh rằng $ \alpha, \beta$ là tất cả các nghiệm thực của phương trình $ F(x)\equal{}0$.[/


function
floor function
induction
inequalities
algebra unsolved
algebra

Let $ n$ be a natural number and let $ x > 0$ be a real number such that none of the numbers $ x,2x,...,nx,\frac {1}{x},\frac {2}{x},...,\frac {\lfloor{nx}\rfloor}{x}$ is an integer. Prove that
\[ \lfloor{x}\rfloor \plus{} \lfloor{2x}\rfloor \plus{} ... \plus{} \lfloor{nx}\rfloor \plus{} \lfloor{\frac {1}{x}}\rfloor \plus{} \lfloor{\frac {2}{x}}\rfloor \plus{} ... \plus{} \lfloor{\frac {\lfloor{nx}\rfloor}{x}}\rfloor \equal{} n\lfloor{nx}\rfloor\]

We present an induction on $ n$. Suppose that the statement is true for $ n$. Consider $ n \plus{} 1$, we need to prove the following identity:
$ \lfloor (n \plus{} 1)x\rfloor \plus{} \lfloor \frac {\lfloor nx\rfloor \plus{} 1 }{x} \rfloor \plus{} \lfloor \frac {\lfloor nx\rfloor \plus{} 2 }{x} \rfloor \plus{} ... \plus{} \lfloor \frac {\lfloor (n \plus{} 1)x\rfloor }{x} \rfloor \equal{} (n \plus{} 1)\lfloor (n \plus{} 1)x\rfloor \minus{} n\lfloor nx\rfloor$
$ \Leftrightarrow \lfloor \frac {\lfloor nx\rfloor \plus{} 1 }{x} \rfloor \plus{} \lfloor \frac {\lfloor nx\rfloor \plus{} 2 }{x} \rfloor \plus{} ... \plus{} \lfloor \frac {\lfloor (n \plus{} 1)x\rfloor }{x} \rfloor \equal{} n(\lfloor (n \plus{} 1)x\rfloor \minus{} \lfloor nx \rfloor)$
$ \Leftrightarrow \lfloor \frac {\lfloor nx\rfloor \plus{} 1 }{x} \rfloor \plus{} \lfloor \frac {\lfloor nx\rfloor \plus{} 2 }{x} \rfloor \plus{} ... \plus{} \lfloor \frac {\lfloor (n \plus{} 1)x\rfloor }{x} \rfloor \equal{} n\lfloor x \plus{} \{nx\}\rfloor$
$ \Leftrightarrow \lfloor \frac {\lfloor nx\rfloor \plus{} 1 }{x} \rfloor \plus{} \lfloor \frac {\lfloor nx\rfloor \plus{} 2 }{x} \rfloor \plus{} ... \plus{} \lfloor \frac {\lfloor (n \plus{} 1)x\rfloor }{x} \rfloor \equal{} n\lfloor x \rfloor \plus{} n\lfloor \{x\} \plus{} \{nx\}\rfloor$

Consider the two following cases:
1. $ \{x\} \plus{} \{nx\} < 1$ then $ n\lfloor \{x\} \plus{} \{nx\}\rfloor \equal{} 0$ and $ \lfloor (n \plus{} 1)x \rfloor \equal{} \lfloor nx \rfloor \plus{} \lfloor x \rfloor$. We have to prove: $ \sum_{k \equal{} 1}^{\lfloor x \rfloor} \lfloor \frac {\lfloor nx \rfloor \plus{} k}{x}\rfloor \equal{} n\lfloor x \rfloor$.
But it is easy to see that $ n\leq \frac {\lfloor nx \rfloor \plus{} k}{x},\forall 1\leq k \leq \lfloor x \rfloor$. Indeed, the inequality is equivalent to $ nx \leq \lfloor nx \rfloor \plus{} k$, or $ \{nx\} \leq k$, which is true since $ k\geq 1 > \{nx\}$.
On the other hand, we will prove that $ \frac {\lfloor nx \rfloor \plus{} k}{x} < n \plus{} 1,\forall 1\leq k \leq \lfloor x \rfloor$. This inequality is equivalent to $ k\leq \lfloor x \rfloor \plus{} \{x\} \plus{} \{nx\}$, which is true by $ k\leq \lfloor x \rfloor$.
Hence $ n\leq \frac {\lfloor nx \rfloor \plus{} k}{x} < n \plus{} 1,\forall 1\leq k \leq \lfloor x \rfloor$, or $ \lfloor \frac {\lfloor nx \rfloor \plus{} k}{x} \rfloor \equal{} n,\forall 1 \leq k \leq \lfloor x \rfloor$.
From this: $ \sum_{k \equal{} 1}^{\lfloor x \rfloor} \lfloor \frac {\lfloor nx \rfloor \plus{} k}{x}\rfloor \equal{} n\lfloor x \rfloor$
So the identity is proved in this case.

2. $ \{x\} \plus{} \{nx\} > 1$, the same argument yields what desired.

QED.


algebra
polynomial
group theory
abstract algebra
Euler
number theory
number theory solved

For any prime number p, show that there exists a natural number k such that $k^{p-1} - 1$ is divisible by p, while $k^m - 1$ is not divisible by p for any natural m with 0 < m < p.

I believe this one is very hard and I am searching for an elementary proof. I don't want to hear anything about polynomials on fields, all what I know are some beginnings of group theory (order of subgroup divides order of group, if this can be used) and a bit of number theory (Euler-Fermat, chinese remainder, inverses in modulo groups).

Any help?

  Darij

Sorry, do yo want to show there is a primitive root mod p?

I don't know what a "root" is but my question is (under another guise) to show that the multiplicative group mod p contains at least one primitive element.

  Darij

the answer to pascual's question is 'yes', and this was posted before, but I think the answers in that topic might not match your criteria. Anyway, I don't think there's a way to attack this without any field consideration whatsoever.

The proof of the existence of a primitive root modulo a prime can be found in almost any book of elemntary theory of numbers.
See for exemple 'Elementary theory of numbers' by Sierpinski (chap VI,-6), which is available for free here :
http://matwbn.icm.edu.pl/kstresc.php?tom=42&wyd=10

Pierre.


inequalities
function
symmetry
three variable inequality

if a2b+ab2 +b2c+bc2 +c2a+ca2 =6
then prove that
3(a2b2 +b2c2 +c2a2 ) >= abc[a2b2 +b2c2 +c2a2 +ab(a2 +b2 )+bc(b2 +c2 )+ca(c2 +a2) ]

I personally find this inequality nicer in the homogenized version:

[color=blue]If a, b, c are any three nonnegative reals, then prove the inequality

$\left(b^2c^2+c^2a^2+a^2b^2\right)\left(bc^2+cb^2+ca^2+ac^2+ab^2+ba^2\right)$
$\geq 2abc\left(b^2c^2+c^2a^2+a^2b^2+bc\left(b^2+c^2\right)+ca\left(c^2+a^2\right)+ab\left(a^2+b^2\right)\right)$.[/color]

My solution of this inequality is a bit non-standard, and I would like to hear some comments about it. I start with a lemma:

[color=blue][b]Lemma 1.[/b] If a, b, c, x, y, z are six nonnegative reals such that $a\geq b\geq c$ and $x\leq y\leq z$, then

$x\left(b-c\right)^2\left(b+c-a\right)+y\left(c-a\right)^2\left(c+a-b\right)+z\left(a-b\right)^2\left(a+b-c\right)\geq 0$.[/color]

[hide="Proof of Lemma 1"]
[i]Proof of Lemma 1.[/i] If $b+c-a\geq 0$, then the inequality $x\left(b-c\right)^2\left(b+c-a\right)+y\left(c-a\right)^2\left(c+a-b\right)+z\left(a-b\right)^2\left(a+b-c\right)\geq 0$ is almost trivial (in fact, $c+a-b\geq 0$ follows from $a\geq b$, and $a+b-c\geq 0$ follows from $a\geq c$). Hence, it is enough to consider the case when $b+c-a<0$. In this case, since $ \left(b-c\right)^2\geq 0$, we have $\left(b-c\right)^2\left(b+c-a\right)\leq 0$. Since $x\leq y$, this yields $x\left(b-c\right)^2\left(b+c-a\right)\geq y\left(b-c\right)^2\left(b+c-a\right)$. Also, $a+b-c\geq 0$ (this is because $a\geq c$); together with $\left(a-b\right)^2\geq 0$, this yields $\left(a-b\right)^2\left(a+b-c\right)\geq 0$. But since $y\leq z$, we have $z\geq y$, and thus $z\left(a-b\right)^2\left(a+b-c\right)\geq y\left(a-b\right)^2\left(a+b-c\right)$.

So altogether,

$x\left(b-c\right)^2\left(b+c-a\right)+y\left(c-a\right)^2\left(c+a-b\right)+z\left(a-b\right)^2\left(a+b-c\right)$
$\geq y\left(b-c\right)^2\left(b+c-a\right)+y\left(c-a\right)^2\left(c+a-b\right)+y\left(a-b\right)^2\left(a+b-c\right)$
$=2y\left(a\left(a-b\right)\left(a-c\right)+b\left(b-c\right)\left(b-a\right)+c\left(c-a\right)\left(c-b\right)\right)$.

Since $a\left(a-b\right)\left(a-c\right)+b\left(b-c\right)\left(b-a\right)+c\left(c-a\right)\left(c-b\right)\geq 0$ by Schur, we thus have

$x\left(b-c\right)^2\left(b+c-a\right)+y\left(c-a\right)^2\left(c+a-b\right)+z\left(a-b\right)^2\left(a+b-c\right)\geq 0$,

and Lemma 1 is proven.[/hide]

Now, let's prove the initial inequality:

After some algebra,

$\left(b^2c^2+c^2a^2+a^2b^2\right)\left(bc^2+cb^2+ca^2+ac^2+ab^2+ba^2\right)$
$-2abc\left(b^2c^2+c^2a^2+a^2b^2+bc\left(b^2+c^2\right)+ca\left(c^2+a^2\right)+ab\left(a^2+b^2\right)\right)$
$=a\left(ca-ab\right)^2\left(ca+ab-bc\right)+b\left(ab-bc\right)^2\left(ab+bc-ca\right)+c\left(bc-ca\right)^2\left(bc+ca-ab\right)$.

But if we WLOG assume that $a\leq b\leq c$, then $bc\geq ca\geq ab$ (in fact, $a\leq b$ yields $ca\leq bc$, so that $bc\geq ca$, and similarly $ca\geq ab$ follows from $b\leq c$). Hence, we can apply Lemma 1 to $a$, $b$, $c$, $bc$, $ca$, $ab$ instead of $x$, $y$, $z$, $a$, $b$, $c$, and get

$a\left(ca-ab\right)^2\left(ca+ab-bc\right)+b\left(ab-bc\right)^2\left(ab+bc-ca\right)+c\left(bc-ca\right)^2\left(bc+ca-ab\right)$
$\geq 0$.

Hence,

$\left(b^2c^2+c^2a^2+a^2b^2\right)\left(bc^2+cb^2+ca^2+ac^2+ab^2+ba^2\right)$
$-2abc\left(b^2c^2+c^2a^2+a^2b^2+bc\left(b^2+c^2\right)+ca\left(c^2+a^2\right)+ab\left(a^2+b^2\right)\right)\geq 0$,

so that

$\left(b^2c^2+c^2a^2+a^2b^2\right)\left(bc^2+cb^2+ca^2+ac^2+ab^2+ba^2\right)$
$\geq 2abc\left(b^2c^2+c^2a^2+a^2b^2+bc\left(b^2+c^2\right)+ca\left(c^2+a^2\right)+ab\left(a^2+b^2\right)\right)$.

QED.

Now, what I am really interested in is whether my Lemma 1 above, or some of its modifications (I see a lots of possibilities to modify it right now) can be useful in proving other inequalities. Any ideas are appreciated!

  Darij

Just want to post an ugly generalization.
If f is decreasing and g is increasing and both of them are positive, then
\[
\sum {f(a)g(|b - c|)(b + c - a)}  \geqslant 0
\]

Ugly??

As usual, I rewrite your inequality with changed notations:

[color=blue][b]Lemma 2.[/b] If a, b, c are three nonnegative reals such that $a\geq b\geq c$, if x, y, z are three nonnegative reals such that $x\leq y\leq z$, and if $g: \mathbb{R}^{+}_{0}\to\mathbb{R}^{+}_{0}$ is a monotonically increasing function, then

$xg\left(\left|b-c\right|\right)\left(b+c-a\right)+yg\left(\left|c-a\right|\right)\left(c+a-b\right)+zg\left(\left|a-b\right|\right)\left(a+b-c\right)\geq 0$.[/color]

[i]Proof of Lemma 2.[/i] First, since $a\geq b\geq c$, we have $b-c\geq 0$, $c-a\leq 0$ and $a-b\geq 0$, and thus | b - c | = b - c, | c - a | = a - c and | a - b | = a - b. Consequently, the inequality in question,

$xg\left(\left|b-c\right|\right)\left(b+c-a\right)+yg\left(\left|c-a\right|\right)\left(c+a-b\right)+zg\left(\left|a-b\right|\right)\left(a+b-c\right)\geq 0$,

rewrites as follows:

$xg\left(b-c\right)\left(b+c-a\right)+yg\left(a-c\right)\left(c+a-b\right)+zg\left(a-b\right)\left(a+b-c\right)\geq 0$.

Since $a\geq b$, we have $c+a-b\geq 0$, and since $b\geq c$, we have $a+b-c\geq 0$. Now, if $b+c-a\geq 0$, then the inequality

$xg\left(b-c\right)\left(b+c-a\right)+yg\left(a-c\right)\left(c+a-b\right)+zg\left(a-b\right)\left(a+b-c\right)\geq 0$

is trivial (a sum of nonnegative numbers is always nonnegative), and thus there is nothing to prove. So we will only consider the case when b + c - a < 0.

In this case, since $a\geq b$, we have $b\leq a$ and thus $b-c\leq a-c$; hence, since the function g is monotonically increasing, $g\left(b-c\right)\leq g\left(a-c\right)$. Since g (b - c), g (a - c), x and y are all nonnegative, and since $x\leq y$, we thus get $xg\left(b-c\right)\leq yg\left(a-c\right)$. Since b + c - a < 0, we thus have $xg\left(b-c\right)\left(b+c-a\right)\geq yg\left(a-c\right)\left(b+c-a\right)$.

Also, since $a+b-c\geq 0$ and $g\left(a-b\right)\geq 0$, we have $g\left(a-b\right)\left(a+b-c\right)\geq 0$, so that $y\leq z$ yields $zg\left(a-b\right)\left(a+b-c\right)\geq yg\left(a-b\right)\left(a+b-c\right)$. Hence,

$xg\left(b-c\right)\left(b+c-a\right)+yg\left(a-c\right)\left(c+a-b\right)+zg\left(a-b\right)\left(a+b-c\right)$
$\geq yg\left(a-c\right)\left(b+c-a\right)+yg\left(a-c\right)\left(c+a-b\right)+yg\left(a-b\right)\left(a+b-c\right)$
$=yg\left(a-c\right)\left(\left(b+c-a\right)+\left(c+a-b\right)\right)+yg\left(a-b\right)\left(a+b-c\right)$.

This is $\geq 0$, since $\left(b+c-a\right)+\left(c+a-b\right)=2a\geq 0$ and $a+b-c\geq 0$. Hence, Lemma 2 is proven.

Very nice generalization, Siuhochung! I found it quite surprising that Peter Scholze and me discovered this method of proving inequalities on two subsequent days (Peter Scholze was the first - see his post #4 in http://www.mathlinks.ro/Forum/viewtopic.php?t=23108 ; then, I followed with this topic, http://www.mathlinks.ro/Forum/viewtopic.php?t=149630 post #5 and http://www.mathlinks.ro/Forum/viewtopic.php?t=3547 post #10). Actually, when I discovered this method I was inspired by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=1166]Valentin Vornicu's extension of the Schur inequality[/url], and I would call the method the "Schur modification method".

  Darij

I get your idea of proving inequalities and in fact, it's very similar to the method many members (vietnam members i guess?) use, that they always try to put inequalities in the form $S_a (b-c)^2 + S_b (c-a)^2 + S_c (a-b)^2 \ge 0$.

In fact, when i was staying in the library today, I tried hard to generalize $b+c-a$ into a function as I think it would be very very nice to have this generalized. Unforunately I didn't get it done, and till now I can't really think of any inequalities that can be proven by my generalization. That's why I call it ugly.

I think that it would be great if you could make a new topic describing your method, Darij. (I mean how to start and what's the next step and so on) Then with the help of other forum members it would be possible to make sure, which inequalities are solvable using this method.

Thank you

Well, the problem is, I am myself still not too sure about what "my" method really is (as Siuhochung correctly pointed out, this method is similar to what some Vietnamese members - but Vasc too - have been using on this forum for a long time). I should rather have said that there is a number of tricks applying in different inequality problems which all are somewhat related. Probably the crux of the method will be more clear to me if I solve more inequalities using this method. As for now, I can just explain how I proceeded for the initial problem of this thread:

I had to prove the inequality

$\left(b^2c^2+c^2a^2+a^2b^2\right)\left(bc^2+cb^2+ca^2+ac^2+ab^2+ba^2\right)$
$-2abc\left(b^2c^2+c^2a^2+a^2b^2+bc\left(b^2+c^2\right)+ca\left(c^2+a^2\right)+ab\left(a^2+b^2\right)\right)\geq 0$.

After expanding the left hand side and fumbling around with algebra, it took the form

$a\left(ca-ab\right)^2\left(ca+ab-bc\right)+b\left(ab-bc\right)^2\left(ab+bc-ca\right)+c\left(bc-ca\right)^2\left(bc+ca-ab\right)$
$\geq 0$.

Now, this [i]almost[/i] proves the inequality: If bc, ca, ab are the sidelengths of a triangle - what is the usual, "central" case, especially since the original inequality becomes an equality only for a = b = c -, then we are done. However, we get into troubles in the "marginal" case when bc, ca, ab are not the sidelengths of a triangle. Now, the usual way is to give up the approach and to try another one (e. g. keeping on fumbling around with the left hand side until it takes a form which proves it trivially $\geq 0$ for [i]all[/i] a, b, c). I took the other way - actually, it is natural that proving the inequality in a "marginal" case is easier than proving it in the "central" case, particularly if equality is achieved in the central case only (in other words, the "central" case is the really important one). So I sought a particular argument that could prove the inequality for the "marginal" case, too. I remembered that similar troubles occur if one tries to prove the Schur inequality

$a\left(a-b\right)\left(a-c\right)+b\left(b-c\right)\left(b-a\right)+c\left(c-a\right)\left(c-b\right)\geq 0$,

by rewriting it as

$\left(b-c\right)^2\left(b+c-a\right)+\left(c-a\right)^2\left(c+a-b\right)+\left(a-b\right)^2\left(a+b-c\right)\geq 0$,

what yields the proof for the "central" case when a, b, c are the sides of a triangle but not for the "marginal" case when they are not. The actual proof of the Schur inequality for all positive a, b, c is different, it works via breaking the symmetry by WLOG setting $a\geq b\geq c$ and then applying trivial positive-negative observations (I hope you know that proof, so I needn't quote it here; else, look at http://www.mathlinks.ro/Forum/viewtopic.php?t=1166 ). So I looked out for a similar proof of

$a\left(ca-ab\right)^2\left(ca+ab-bc\right)+b\left(ab-bc\right)^2\left(ab+bc-ca\right)+c\left(bc-ca\right)^2\left(bc+ca-ab\right)$
$\geq 0$.

I noted soon that, as in the Schur inequality, only the order of the numbers is important, i. e. we can replace the a, b, c in front of the brackets by any three numbers x, y, z such that the number arrays (bc; ca; ab) and (x; y; z) are oppositely sorted. I renamed bc, ca, ab into a, b, c and came up with Lemma 1. The rest is clear. Siuhochung nicely generalized this approach so it became what one could call a method. But still it's not a method in the usual meaning of this word, since it is generally hard to say clearly how to proceed by this method in a given case.

  Darij

very clever indeed, thank you very much, Darij. If I can find any inequalities to solve with this "method", I'll be sure to let you know.

[quote="galois"]If $ a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b) \equal{} 6$, then disprove that

$ 3\left(b^2c^2 \plus{} c^2a^2 \plus{} a^2b^2\right)\geq abc\left[b^2c^2 \plus{} c^2a^2 \plus{} a^2b^2 \plus{} a^3(b \plus{} c) \plus{} b^3(c \plus{} a) \plus{} c^3(a \plus{} b)\right].$[/quote]$ a \equal{} \minus{} 3,b \equal{} 1,c \equal{} 5$ is a counterexample :

$ a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b) \equal{} 9\times6 \plus{} 1\times2 \plus{} 25\times( \minus{} 2) \equal{} 6;$

$ 3\left(b^2c^2 \plus{} c^2a^2 \plus{} a^2b^2\right) \equal{} 3\times 259 \equal{} 777;$

$ abc\left[b^2c^2 \plus{} c^2a^2 \plus{} a^2b^2 \plus{} a^3(b \plus{} c) \plus{} b^3(c \plus{} a) \plus{} c^3(a \plus{} b)\right] \equal{} 2265.$[quote="darij grinberg"][color=blue]If $ a, b, c$ are any three positive reals, then prove the inequality

$ \left(b^2c^2 \plus{} c^2a^2 \plus{} a^2b^2\right)\left[a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b)\right]$

$ \geq 2abc\left[b^2c^2 \plus{} c^2a^2 \plus{} a^2b^2 \plus{} a^3(b \plus{} c) \plus{} b^3(c \plus{} a) \plus{} c^3(a \plus{} b)\right].$[/color][/quote]After some algebra, we can see this inequality holds for [b]real[/b] numbers $ a,b,c$ with $ b \plus{} c\geq 0,c \plus{} a\geq 0,a \plus{} b\geq 0:$

$ \left(b^2c^2 \plus{} c^2a^2 \plus{} a^2b^2\right)\left[a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b)\right]$

$ \minus{} 2abc\left[b^2c^2 \plus{} c^2a^2 \plus{} a^2b^2 \plus{} a^3(b \plus{} c) \plus{} b^3(c \plus{} a) \plus{} c^3(a \plus{} b)\right]$

$ \equiv F(a,b,c) \equal{} F(r \plus{} s \plus{} t,r \minus{} s \plus{} t,r \plus{} s \minus{} t)$

$ \equal{} 8(s^2 \minus{} st \plus{} t^2)r^5 \plus{} 40(s \plus{} t)(s \minus{} t)^2r^4 \plus{} 80(s^2 \plus{} st \plus{} t^2)(s \minus{} t)^2r^3$

$ \plus{} 80(s \plus{} t)(s^2 \plus{} t^2)(s \minus{} t)^2r^2 \plus{} 8(5s^4 \plus{} 5s^3t \minus{} 4s^2t^2 \plus{} 5st^3 \plus{} 5t^4)(s \minus{} t)^2r$

$ \plus{} 8(s \plus{} t)^3(s \minus{} t)^4\geq 0,$

which is clearly true for $ a \equal{} \max\{a,b,c\}.$

[color=blue]By the way, if $ a,b,c$ are real numbers such that $ a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b) \equal{} 6\geq \minus{} 2abc,$ then

$ 3\left(b^2c^2 \plus{} c^2a^2 \plus{} a^2b^2\right)\geq abc\left[b^2c^2 \plus{} c^2a^2 \plus{} a^2b^2 \plus{} a^3(b \plus{} c) \plus{} b^3(c \plus{} a) \plus{} c^3(a \plus{} b)\right],$

with equality if and only if $ a \equal{} b \equal{} c \equal{} 1.$[/color]


calculus vector geometry Princeton college MIT Harvard

Hello, I am a grade 11 Canadian student who wants to apply to the Ivy leagues to pursue a P.H.D. in mathematics, and a career as a mathematician. I have been passionate about mathematics all of my life, and have recently gone to the IMO, winning silver. I have also done well in many of the olympiads and local contests, but other than those, my transcript is pretty minimal, especially with volunteering, although I will have volunteer hours by the end of my schooling because I am required to have 150 hours for the IB program, in which I am planning to get the IB diploma.

I would say that my grades are hovering around the 90s, my exact grades being:
--100% in math30IB(higher level), math31IB(higher level),the first being equivalent to typical grade 12 with slightly more advanced topics, the latter being calculus, vectors, and minimal statistics
--100% on my high school diploma exam for mathematics
--100% in whatever physics and french courses I am taking right now
around 94-95 in whatever higher level chemistry I'm stuck in currently, though still at a grade 11 level
--And the stain on my report card, 80% in social (social teacher can be described as less than ideal)
--Also, haven't taken english yet in grade 11, but expecting somewhere around 90%

Note that all of the above courses are IB courses, but I am guessing that Ivy league schools do not care about that kind of fluff too much though (I just took it so that the ceilings in math and the sciences wouldn't be detrimental to me, since I tend to rush tests).

My questions:

1. Which Ivy League school is best for pure, theoretical mathematics, if it matters, I appreciate much more the continuous branches rather than the finite ones such as number theory.

2. Considering the rest of my marks are pretty decent, will the ivy league schools reject me on the basis of my social mark if I emphasize that I am planning to pursue mathematics on my application?

3. If I pile on crazy amounts of volunteer hours in grade 12, will the universities consider this to be just polishing up my resume, and not take them as seriously?

4. Saying I put in ridiculous hours in grade 12, and assuming my SAT scores are decent because I will take the test every conceivable time that I can to get the highest mark possible, and assuming that any trends in my marks hold for another year, what do you think my chances are of getting into the Ivy league school most applicable to question 1?

One last comment, I really hope that universities like people who are specialized in 1 area, because under all extra-curricular activites/hobbies/sports with be the word math (and maybe girls for the second). I am in the high school choir though, but that is the extent of any activity other than math that I can say that I do.

I really appreciate any replies that I might receive.

Cheers,

Hunter

Edit: Also, I am pretty high with piano, and have done some stuff I can put on my application dealing with that as well.

-----Bump, any attempt at an answer to any one of these would be much appreciated.

Cheers,

Hunter

Hey Hunter! Long time no talk. How is your training going?

Well here are some of my opinions:

1. For pure, theoretical math I'd say Princeton is the best, although MIT, Harvard and Stanford all have extremely strong programs too.

2. Definitely not; Ivy League schools will never reject someone based on one subject's mark. I think 80 is fine, but if there is a trend that your social mark is constantly below 80 in high school then it might cause a slight problem. I think you'll prbly be fine; pretty much everyone has a hard time with Social 20IB. It gets easier in Grade 12 when you are more and more familiar with the kind of writing you're doing.

3. I can't say it's a sure yes but it's very probable. The universities care more about why you do these activities than how much you did, so it's always better to start early and show that you have an interest in them. Also, piling up volunteer hours in Grade 12 can be very stressful, especially with all the university applications and course load you'll have by that time.

4. You'll get in. Just get some decent uni app essays and you're all set; but do start volunteering early.

Good luck! And have fun in Winter Camp!

ZMOG JARNO!!!!!!!

Thx. I guess I'll try for Princeton (although you're probably just saying that since you go to Princeton, lol).

Cheers,

Hunter

Well not really. Princeton is largely known for its pure math and sciences.

And Princeton really likes Canadians for some reason...

for MIT anyway, you have an excellent chance because of your contest results.  most of their application (in terms of space) is fluffier stuff, but you'll stand out well.  do people really think volunteering is necessary??  and good luck writing a good 'essay' - it was a trying experience for everyone i know

you have a silver at IMO! I'm sure they'll overlook one bad grade if you have good essays etc.. On the part of the app where u can explain any bad grades I would mention that the teacher in social studies is less than ideal.

[quote="Rofler"]1. Which Ivy League school is best for pure, theoretical mathematics, if it matters, I appreciate much more the continuous branches rather than the finite ones such as number theory.[/quote]

If you go to a sufficiently highly ranked school (e.g. Princeton, Harvard, Stanford, MIT, Chicago, maybe one or two others) the quality of your undergraduate education will depend mostly on factors that you can't possibly anticipate ahead of time (e.g., which professors end up teaching the courses you take, who you do summer research with, etc).  This 'randomness' will likely play a much bigger role in your education than small differences in faculty quality (which is really only relevant at the graduate level) and the like.


Each square of an $m\times n$ chessboard is painted black or white in such a way that for every black square, the number of black squares adjacent to it is odd (two squares are adjacent if they share one edge). Prove that the number of black squares is even.

Something similar was given in some Iranian pre-TST this year. Consider the graph having the black squares as vertices, and in which two vertices are adjacent iff the corresponding squares are adjacent. We are told that each vertex has an odd degree, an since the number of vertices with odd degree in a graph is always even, we get the conclusion.

yes ,grobber.I think so .This problem were easy if we would use graph

another solution with graph
let $x_i,y_i$ be the numbers of column and row in which the black square is
consider a bipartitie graph such that one subset has vertices such that $x_i+y_i \equiv 0(2)$ and another with vertices such that $x_i+y_i \equiv 1 (2)$. suppose for contrary that the number of black squares is odd then one of this subsets has an odd power and one has even power, then one subset has even number of edges, and one has odd number


All right, who designed the flyer that was sent out in the mail? It's really cool! Thanks for the nice "Proofs without Words" poster. I was actually just thinking that would be neat to find a poster of those...  ;)

Glad you like it - that was almost entirely vRusczyk's doing.

Yes, the poster that came in the mail was very nice.  On the website with the animations, its even cooler. I wonder who designed the animations, more of these should be made!   ;)


inequalities
geometry
circumcircle
inradius
inequalities unsolved

Let $ ABC$ be a triangle with sides $ a,b,c$, circumradius $ R$ and inradius $ r$. Prove that:
$ \frac{R}{2r} \geq (\frac{64a^{2}b^{2}c^{2}}{(4a^{2}\minus{}(b\minus{}c)^{2})(4b^{2}\minus{}(c\minus{}a)^{2})(4c^{2}\minus{}(a\minus{}b)^{2})})^{2}$

Let $ a\equal{}y\plus{}z$ , $ b\equal{}x\plus{}z$ , $ c\equal{}x\plus{}y$ so $ R\equal{}\frac{(x\plus{}y)(y\plus{}z)(x\plus{}z)}{4\sqrt{xyz(x\plus{}y\plus{}z)}}$ and

$ r\equal{}\frac{\sqrt{xyz(x\plus{}y\plus{}z)}}{x\plus{}y\plus{}z}$ so inequality equivalents to;

$ [(3x\plus{}y)(3y\plus{}x)(3x\plus{}z)(3z\plus{}x)(3y\plus{}z)(3z\plus{}y)]^2 \geq 2^{15}xyz(x\plus{}y)^3(y\plus{}z)^3(x\plus{}z)^3$

Now let use this lemma: $ (3x\plus{}y)^2(3y\plus{}x)^2 \geq 32(x\plus{}y)^3\sqrt{xy}$

Let prove this lemma. Because of homogeneity we can take $ x\plus{}y\equal{}1$ so;

inequality equivalents to $ 4xy\plus{}3 \geq 4\sqrt{2}\sqrt[4]{xy} \Longleftrightarrow$

$ (\sqrt{2}\sqrt[4]{xy}\minus{}1)^2(2\sqrt{xy}\plus{}2\sqrt{2}\sqrt[4]{xy}\plus{}3) \geq 0$ and solution is done. :wink:


333....333  *  333...3334
     (n 3's)           (n 3's and units digit is 4)

How many digits of the product above are 1's?

do we have to give an explanation? i got :[hide]n[/hide]

yup... and yes, just say how you found the answer on this one

since at the end there is an extra 4 it adds a 1 to the next place value when you multiply giving it an extra 1?

umm... you're not being clear... can you just tell what you did? but then... I worked it out through an algebraic approach before, but its way easier with another method.


See poll.  :)

People always say y'all is a Texas thing, but I've heard it everywhere.

EDIT: AAAAAAAAHHHHHHHHH I always forget to submit the last poll choice  :( . The last poll choice was "You".

"y'all" is more generally Southern rather than specifically Texan. For some other parts of the U.S., some other possible poll choices could be "you guys" (which doesn't intend to imply gender), or even "youse guys".

Most European languages (I don't know about Indian or Iranian languages) maintain a distinction between the second person familiar and second person formal, and the second person familiar usually distinguishes between singular and plural. The problem is that English has nearly completely lost the second person familiar. "You" is the formal, and is the same in the singular and plural (hence creating a need in slang and vernacular to invent a plural, such as "y'all.") The second person familiar singular is "thou" ("thee" in the accusative); the plural would be "ye." (That's been so thoroughly lost from the language that I had to look up "ye.")

Interesting.  :)  I always wondered about "ye".

I say y'all when I'm not mad and you guys when I'm mad, but I'm not mad most of the time so I voted y'all.

I say "You All". This is kind of random, though...

you when I'm not mad/frustrated, GUYS!!!! if I'm mad/frustrated.

No offense, but this is dumber than my Mexican Food poll.

this is really random...

gahhh i would've chosen the "third choice" if it was there


[quote="zaya_yc"]$ a > 0$ and $ ax^{2} \plus{} bx \plus{} c \equal{} 0$----> $ 0 < x_{1} < x_{2} < 1$ . prove: $ a\geq5$[/quote]

Let $ f(x) \equal{} ax^2 \plus{} bx \plus{} c\ (a > 0)$.
$ 0 < x_1 < x_2 < 1\Longleftrightarrow f(0) > 0,\ f(1) > 0,\ D > 0,\ axis: 0 < \minus{} \frac {b}{2a} < 1$

$ \Longleftrightarrow c > 0,\ a \plus{} b \plus{} c > 0,\ b^2 > 4ac,\ \minus{} 2a < b < 0$