modular arithmetic
number theory
relatively prime
number theory proposed

Let $ f(m)$ the number of residues modulo $ n$ with order $ m$, prove that $ \phi{(m)}|f(m)$

let $ a$ have order $ m$ $ \pmod{n}$. then $ 1, a, a^2, ..., a^{m\minus{}1}$ are all different $ \pmod{n}$. moreover, $ a^{r}$ has order $ m$ iff $ r$ is relatively prime to $ m$. the number of such $ r$ in the range $ [0, m\minus{}1]$ is $ \phi(m)$. if $ b$ is another number with order $ m$ $ \pmod{n}$, such that $ b$ is different $ \pmod{n}$ from all of $ 1, a, a^2, ..., a^{m\minus{}1}$, then $ b^s$ has order $ m$ $ \pmod{n}$ for all $ s$ relatively prime to $ m$, and all these $ b^s$ are different from all of the $ a^r$. to see this, assume on the contrary that $ a^r \equal{} b^s$. then since $ s$ is relatively prime to $ m$, there exists $ x$ such that $ sx\equiv 1 \pmod{m}$, so that $ a^{rx}\equal{}b^{sx}\equiv b\pmod{n}$. but we've assumed that $ b$ is different $ \pmod{n}$ from all powers of $ a$.

this proves that the residues modulo $ n$ that have order $ m$ come in groups of $ \phi(m)$, and that completes the proof.


geometry
AMC
USA(J)MO
USAMO
invariant
pigeonhole principle
extremal principle

I just finished AoPS book2.

I have four books:

Problem-Solving Strategies by Engel
Art and Craft by Zeitz
Geometry Revisited by Coxeter and Greitzer
MO challenge by Andreescu and Gelca

what should be the sequence of study(as in, which one should i begin with)? are there any books i need to study concurrently?

answers about those and other commments relating will be helpful. Thanks in advance..

P.S. this is prepping for USAMO. assuming that i made the cutoff.

Can I get any of the four books online for free??
I am in China and it is not so easy to get one...

No, they're not available for free online.  Also, this thread should be in another forum..probably Other Problem Solving Topics.

Ok, so out of those 4, I only have Zeitz.  I think that you could do a few of those books concurrently.  I would do Zeitz first and then Engel.  You could work on Geometry Revisited at the same time as going through Zeitz since Zeitz does not have any geometry in his book.  Then the MO challenges is probably just all problems so you could go through these whenever.

thanks...

another question:

Which of those books are at USAMO-level?? or are there any other books that are at USAMO level??

thanks in advance.

I think they all are.

I know for sure that Geometry Revisted is. It's quite difficult; some of the problems are used in the Olympiad Geometry class.

And MO Challenges... well the title speaks for itself.

In addition to what paladin said Zeitz has a bunch of Olympiad problems and a couple chapter devoted to stuff like invariants, pigeonhole, the extremal principle and so on.

I don't know anything about Engel except that it is used in the Olympiad Independent Study here on AoPS so I would assume it is USAMO level.

lol, Engel is all USAMO level in my opinion, most of the problems come from national olympiads and IMOs.


Let $ S\equal{}\{(x,y)\mid x,y\in\mathbb Z\wedge 0\le x,y\le 3\}$ and let $ A$ be any $ 7$ element subset of $ S$. Prove that there exists three points in $ A$ which are the vertices of a right triangle.

[hide]Well pigeonhole says there must be a row with at least 2 points in A:

Exactly 2 in a row: If there is a point in either of the columns that the two points are in then they form a right triangle.  If not then five points fit into the remaining 2x3 grid, and clearly every point except a corner or a side is missing in the 2x3 grid and either way there is a right triangle.

Exactly 3 in a row: Since the four remaining points can't all be in the last column, (otherwise it wouldnt be exactly 3) one point must be in the same column as the three, forming a right triangle.

4 points in one row: clearly wherever another point goes it will make a right triangle[/hide]

[hide]From PHP, choose 7 points out of 4 columns, implies there exists a column with at (at least) two points.

Given that there exists two points in the same column, neither of the two rows that contain one of those two points may have another point, otherwise the three make a right triangle. Thus, there exists 8 places left, in two rows each with 4 places, to pick 5 points.

Considering only those 8 places, from PHP (chose 5 points in 4 columns) there exists a column such that both places have points. But since the other 3 points must be in the same two rows as those two points, thus they form a right triangle, contradiction.[/hide]

@ pianoforte: Your solution is incomplete. At first you say there exists a column with [b]at least[/b] two points. But later on you continue your solution assuming that the column contains [b]exactly[/b] two points, which is not necessarily true. But your solution can be completed easily. :wink:


[quote="Anthony2"]Find all triples of primes (p,q,r) such that

p | 2qr + r
q | 2rp + p
r | 2pq + q[/quote]

If two of them are equal, Wlog say $ p \equal{} q$, we get $ p|(2p \plus{} 1)r$ and so $ p \equal{} q \equal{} r$ and this indeed is a solution.

If the three primes are pairwise different, we get :

$ p|(2q \plus{} 1)r$ and so $ p|2q \plus{} 1$ and so $ 2q \plus{} 1 \equal{} ap$ with some odd $ a\ge 1$
$ q|(2r \plus{} 1)p$ and so $ q|2r \plus{} 1$ and so $ 2r \plus{} 1 \equal{} bq$ with some odd $ b\ge 1$
$ r|(2p \plus{} 1)q$ and so $ r|2p \plus{} 1$ and so $ 2p \plus{} 1 \equal{} cr$ with some odd $ c\ge 1$

Wlog say $ \max(a,b,c) \equal{} a$. We get from the three equations above : $ (abc \minus{} 8)p \equal{} bc \plus{} 2c \plus{} 4$ and so $ 0 < abc \minus{} 8\le bc \plus{} 2c \plus{} 4$

So $ abc > 8$ and $ ((a \minus{} 1)b \minus{} 2)c\le 12$

$ a \equal{} 1$ is impossible since then $ abc \equal{} 1$
$ a \equal{} 3$ implies $ (b,c)\in\{(3,3),(3,1),(1,3)\}$ implying $ (p,q,r)\in\{(1,1,1),(9,13,19),(13,19,9)\}$ and so no solution
$ a \equal{} 5$ implies $ (b,c)\in\{(3,1),(1,3),(1,5)\}$ and none gives natural values for $ p,q,r$
$ a \equal{} 7$ implies $ (b,c)\in\{(1,3)\}$ implying $ (p,q,r) \equal{} (1,3,1)$ and so no solution
$ a\ge 9$ implies $ b \equal{} c \equal{} 1$ and $ a\le 15$ and so only 4 possibilities for $ (a,b,c)$ :
$ (a,b,c) \equal{} (9,1,1)$ $ \implies$ $ (p,q,r) \equal{} (7,31,15)$ not all primes
$ (a,b,c) \equal{} (11,1,1)$ $ \implies$ $ (p,q,r)$ not integers
$ (a,b,c) \equal{} (13,1,1)$ $ \implies$ $ (p,q,r)$ not integers
$ (a,b,c) \equal{} (15,1,1)$ $ \implies$ $ (p,q,r) \equal{} (1,7,3)$ not all primes

Hence the unique family of solutions $ \boxed{(p,q,r) \equal{} (p,p,p)}$ for any prime $ p$


For 3 real numbers $a,b,c$ let $s_n=a^{n}+b^{n}+c^{n}$.
It is known that $s_1=2$, $s_2=6$ and $s_3=14$.
Prove that for all natural numbers $n>1$, we have $|s^2_n-s_{n-1}s_{n+1}|=8$

:wow:  beautiful solution! the problem looks so hard but after looking at pbornsztein's solution, i completely understood it. thank you!

a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab + bc +ac) then 6 = 4 - 2(ab + bc + ac) and we conclude that ab + bc + ac = -1

Besides that, (a+b+c)^3 = a^3 + b^3 + c^3 + 3(a+b+c)(ab + bc + ac) - 3abc then 8 = 14 + 3*2*(-1) - 3abc

 3abc = 0 so abc = 0.

By Newton's Sums:

sn = (a+b+c)s(n-1) - (ab+ bc + ac)s(n-2) + (abc)s(n-3)

sn = 2s(n-1) + s(n-2)

Now, you prove by induction that this sequence satisfies what we want


Find the inverse of this matrix:
A=
1 -2 2
-1 1 3
1 -1 4

Can be done either [url=http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=mathcom&s1=matrices&s2=inverse&s3=basic]here[/url] or [url=http://www.easycalculation.com/matrix/matrix-inverse.php]there[/url]. And if you want to learn to do such computations by hand, read [url=http://www.purplemath.com/modules/mtrxinvr.htm]this[/url].

If you have Matlab, this can be done by:
[code]
A = [1, -2, 2; -1, 1, 3; 1, -1, 4]
inv(A)
[/code]


calculus
integration
trigonometry
calculus computations

evaluate the indefinite integral as a power series

(integral)x/(1+x^5) dx

not even sure where to start

and do the same for this

(integral) arctan(x^2)dx

I think i have to do something with substituting x^2 in for x in the power series, but im having troubles finishing it up

I can't understand what you want.

Is your aim to obtain the series, which represents these integrals? Inthis case, it is quite easy.

First one: ${x=\tan ^{\frac{2}{5}}\theta}$

Wait for 20 minut. I have to go to dinner. :)


let a;b;c>0 such that $ ab\plus{}bc\plus{}ac\equal{}a\plus{}b\plus{}c$.Show that
                      $ \frac {a \plus{} b \plus{} 1}{a^2 \plus{} b^2 \plus{} 1}\plus{}\frac {b\plus{}c \plus{} 1}{b^2 \plus{} c^2 \plus{} 1}\plus{}\frac {a \plus{} c \plus{} 1}{a^2 \plus{} c^2 \plus{} 1}\le3$

[quote="Victory.US"][quote="rachid"][quote="Victory.US"]let a;b;c>0 such that $ ab \plus{} bc \plus{} ac \equal{} a \plus{} b \plus{} c$.Show that
                      $ \frac {a \plus{} b \plus{} 1}{a^2 \plus{} b^2 \plus{} 1} \plus{} \frac {b \plus{} c \plus{} 1}{b^2 \plus{} c^2 \plus{} 1} \plus{} \frac {a \plus{} c \plus{} 1}{a^2 \plus{} c^2 \plus{} 1}\le3$[/quote]
Using Caushy-Schwarz inequality we have :

$ \sum \frac {a \plus{} b \plus{} 1}{a^2 \plus{} b^2 \plus{} 1} \leq \sum \frac {3}{a \plus{} b \plus{} 1} \leq \sum \frac {3(a \plus{} b \plus{} c^2)}{(a \plus{} b \plus{} c)^2} \equal{} 3$[/quote]

could you explain it?[/quote]
I've only used Caushy-Schwarz inequality !

What part of the proof didn't you understand ?  :)

[quote="imedeen"]$ \frac {3}{a \plus{} b \plus{} 1}\leq \sum \frac {3(a \plus{} b \plus{} c^2)}{(a \plus{} b \plus{} c)^2} \equal{} 3$  could you explain clearly[/quote]
Because from Caushy-Scwharz inequality we have : $ (a \plus{} b \plus{} c^2)(a \plus{} b \plus{} 1) \geq (a \plus{} b \plus{} c)^2$,and $ \sum \frac {3(a \plus{} b \plus{} c^2)}{(a \plus{} b \plus{} c)^2} \equal{} \frac {3( a^2 \plus{} b^2 \plus{} c^2 \plus{} 2a \plus{} 2b \plus{} 2c )}{(a \plus{} b \plus{} c)^2} \equal{} 3$ since $ a \plus{} b \plus{} c \equal{} ab \plus{} bc \plus{} ca$
 :)


Prove the existence of a polynomical $ P(x) \in Z[x]$ satisfying the condition:
\[ P(\sqrt[n]{2-\sqrt{3}}+\sqrt[n]{2+\sqrt{3}})=0\]

I think that the polynomial which has $z_iu+z_jv$ as its roots has integer coefficients, where $z_i,\ i \in \overline{0,n-1}$ are the roots of the equation $x^n-1=0$ and $u=\sqrt[n]{2-\sqrt3},v=\sqrt[n]{2+\sqrt 3}$.

I'm not sure, but it looks like it's Ok right now :).

Finally I found the sol by proving:  $ 4= 2-\sqrt{3}+2+\sqrt{3}= Q(\sqrt[n]{2+\sqrt3}+\sqrt[n]{2-\sqrt3})$ with $ Q(x) \in Z[x]$ . So take $ P(x) = Q(x)-4$ we have $ P(\sqrt[n]{2+\sqrt3}+\sqrt[n]{2-\sqrt3})=0$.
Thanks for your help.

:D  :D  I think it's very easy problem . I can prove it by 3 lines .
 x= P1(x) = a+1/a . P2(x) = a^2 + 1/a^2 . Use inductive method , find out : Pn+1(x) = xPn(x) - Pn-1(x) = a^n +1/a^n.  P(x) is  a polynomial with interger coefficient . x is a root of equation Pn(x) =4


function
logarithms
limit
inequalities proposed
inequalities

If $f(a,b,c)$ is a continuous and symmetric function of the positive real variables $a,b,c$ and if it is true that

$f(a,b,c) \geq f(\sqrt{ab},\sqrt{ab},c)$ then you can deduce that

$f(a,b,c) \geq f(\sqrt[3]{abc}, \sqrt[3]{abc}, \sqrt[3]{abc})$

This can be generalized to n variables.

Define the sequences $u_n$, $v_n$, $w_n$ as follows:
$u_1=\sqrt{ab}$, $v_1=\sqrt{ab}$, $w_1=c$, $u_{n+1}=v_{n+1}=\sqrt{v_nw_n}$, $w_{n+1}=u_n$. Then $f(a,b,c)\geq f(u_n,v_n,w_n)\geq f(u_{n+1},v_{n+1},w_{n+1})$ and $u_nv_nw_n=abc$. Let $x_0=\ln c$, $x_1=\frac{\ln a+\ln b}2$, $x_{n+1}=\frac{x_n+x_{n-1}}2$. Then $u_n=v_n=e^{x_n}$, $w_n=e^{x_{n-1}}$.
Since $(x_{n+1}-x_n)=-\frac12(x_n-x_{n-1})$, $x_n$ converges.
Therefore $\lim_{n\rightarrow\infty}u_n=\lim_{n\rightarrow\infty}v_n=\lim_{n\rightarrow\infty}w_n=\root3\of{abc}$. Thus $f(a,b,c)=f(\root3\of{abc},\root3\of{abc},\root3\of{abc})$.

I thought it was rhetorical question... ;)

Thank you very much Sasha for your proof.

Has anybody a more intuitive and less formal proof not involving the concept of limit ?

I think Sasha gave the best solution.
You can think directly:
We have abc =1 and f(a,b,c) ≥ f(√ab,√ab,c)
Let us note for f(u,v,w) - S = u+v+w
and you can see that if two numbers aren't equal
we can get a triple which we put in f(x) and obtain
a value less then predesessor, in this time
S(a,b,c) > S(√ab,√ab,c) So S decreases.

Suppose we have min f(x) = f(u0,v0,w0) with min(S0)
S0 = u0+v0+w0 and if there are two unequal
numbers then we will get S' < S0 - a sum less then precedent
hence not minimal and f(x) won't increase - a contradiction.

Same as Sturm method.


Let n and k be positive integers, and a,b and c be positive reals. Prove that
(a^(n+k)/b^n)+(b^(n+k)/c^n)+(c^(n+k)/a^n)>=(a^k)+(b^k)+(c^k)

By AM-GM we have $k\frac{a^{n+k}}{b^{n}}+nb^{k}\geq (n+k)\sqrt[n+k]{a^{(n+k)k}}=(n+k)a^{k}$.  :wink:

Use rearrangement inequality.

You have 2 sequences

{a^(n+k),b^(n+k),c^(n+k)}
{1/a^n,1/b^n,1/c^n}

Arranged opposite ways, so therefore,

Multiplying the 2 sequences 1st terms, second terms, and trids terms, summimg gives smalles possible value of multiplying the 2 sequences term by term together by rearrangement. Therefore, this value is a^k+b^k+c^k, and multiplying 1st term by second, second by third, third by first, gives the L.H.S. which is greater than the minimum value a^k+b^k+c^k.


percent
articles
geometry
calculus
integration
geometric transformation
reflection

In Texas there is a law stating that all students who graduate within the top 10% of their class is guaranteed automatic admission to any public university in Texas.  Right now there is controversy over it.  The supporters say that it gives a chance to those attending school with less opportunity and promotes diversity on campuses, while opponents say that the law gives students in highly competitive schools a disadvantage, and has done nothing to improve the diversity/quality of students.

What do you guys think about this law?  (especially those not in Texas)

Considering that the Texas public schools are decently strong, I think this is bad, because it is difficult to judge the caliber of someone who finishes in the top 10 percent of his/her class.  However, I do believe there should be an extra incentive for doing well at school, maybe not measured by grades.

Well, I am not and have never been in Texas, even when I know they have pretty descent schools there. I like that law, I really consider that people should have an oportunity regardless their origin. I mean, isn't it enough punishment that you don't have an equal access to the education as people with more money than you, or at least who live in a better place than the one where you live.

I honestly don't think that it is a real disadvantage for the students of highly competitive schools... because there are not that many, and if it is, well, the Education Boreau of Texas could enlarge the number of acceptees on the public universities.

I was able to dig up a lot of talk about wanting the law repealed, but nothing about it actually having been repealed.  Do you have a source, Miths?

Also, this topic probably belongs in the Round Table section.

I have to say, I'm kind of surprised Texas would do anything to rectify educational inequity.  While public schools in any one state might be generally good, this does not mean they are all good, and clearly students do not have equal access to all public schools.

I'm generally in favor of such public policies, though I would not make them so broad that it would disallow student 10.1% in a really good school to go to a top state school (as often happens).  However, I do think equal access to education is a goal worth pursuing, within reason.  Ideally, all schools would be close in quality, so we wouldn't have to correct for inequity later on.  Clearly this isn't happening at the elementary or middle or high school level.  Therefore, I don't think it's absurd to give the top few percent at all schools automatic admission to college.  (I'd wager on those students at the top 1-2% of bad schools over number 15% at an above average school if given equal college education, anyway, so I'd say this isn't just good social policy, but is a good metric in general for college admissions.  I just don't think I'd make the automatic bar reach to 10%.)

[quote="MithsApprentice"]The law has been repealed last time I checked. Enforcement of this new change should start in 2006 or 2007 if it hasn't already.[/quote]I am $\sum_{k=0}^{\infty}\frac{50}{2^{k}}$ percent sure that it is still in effect. ;)

[quote="eryaman"]I am $\sum_{k=0}^{\infty}\frac{50}{2^{k}}$ percent sure that it is still in effect. ;)[/quote]
There is a proverb that says just a stubborn is 100% sure about what he says (it is a proverb, don't blame me!)

We do things a little differently here in Minnesota by putting all schools in the whole state on an equal funding footing, and then allowing open enrollment into any school district. (There are exceptions to both general principles, but that's the basic idea on how the system works here.) The state university systems don't have any such rigid rule about whom they can or must admit.

[quote="JBL"]I was able to dig up a lot of talk about wanting the law repealed, but nothing about it actually having been repealed.  Do you have a source, Miths?[/quote]

Yeah...I looked it up just now and didn't find anything on the net either. Really odd. Our district newsletter (note that I'm a homeschooler now, so it's actually just the regional school district) had an article on it being repealed a few months ago.

And also, apparently some kids are saying that UT's admissions policy has been changed so that not all top 10% kids are admitted.

I'll try to find out some more about it though. A lot of this has been muted by the method of distributing funds to public school districts being declared unconstitutional. So the state got all worked up about Robin Hood last year and the top ten thing was dropped for a bit.

Not only is the law unfair to those at competitive schools, I have a hunch it won't work out. Naturally all top 10% students, since they're guaranteed admission, would choose the best Texas public school, and I doubt they will all fit.  ;)

[quote="notsosilly"]Naturally all top 10% students, since they're guaranteed admission, would choose the best Texas public school, and I doubt they will all fit.  ;)[/quote]

Well, there's the affordability factor. Secondly, not all "top 10%" kids will go to college in-state and those that do won't all go to public schools. Furthermore, those that do go to public schools will go to different ones as there is no absolute "best public school". Note that UT at Austin has well over 45,000 students. That that's only one branch of a whole network of UT schools.

i think that this law is great... although it would give some disadvantage like someone have posted earlier... with the competitive school...

I don't think there's an issue with competitive schools at all. If you go to a competitive school and you don't hit the top 10%, the state public schools can still accept you. The inverse of this law is not true.

[quote="eryaman"][quote="probability1.01"]Considering that the Texas public schools are decently strong[/quote]
[quote="djimenez"]even when I know they have pretty descent schools there. [/quote]
I'm a bit surprised about your impression of Texas public schools, considering the average SAT scores for Texas are below 1000 (I think around 980-990), compared to the national average of over 1000.[/quote]

I think that both probability1.01 and I were talking about universities (more likely grad schools). At least in Mathematics there are  a few very good and very strong grad schools!

[quote="MithsApprentice"] The inverse of this law is not true.[/quote]

Might you mean converse? :D

[quote="MithsApprentice"][quote="probability1.01"]Might you mean converse? :D[/quote]

-.- What ever. Haha...actually I forgot which is which. Man...lol. Must review logical reasoning. :lol:[/quote]

Well actually, upon closer inspection, they are the same, but I never use inverse, ever, so I guess I overlooked that. lol.

On the original topic, I think this law is a good idea.  It's just that 10% is too large.  I think if it were something like 5%, then you could still let the kids in less fortunate areas have a chance while not reducing drastically the quality of students.  Students who do well even in poor areas could probably improve their SAT scores dramatically given the right resources.  Also, the few outliers among high schools could mess it all up a bit.

I think the legislature made some changes in the law, but stopped short of getting rid of it. Even though it did not accomplish its purpose of increasing minority enrollment (and the Supreme Court changed the ruling on affirmative action that led to this law anyway), this law has many supporters. Texas schools are BAD. Texas ranks low in spending per pupil and in student performance. There is also enormous disparity btwn wealthy and poor school districts (a result of the way schools have traditionally been funded). That means that the bottom slice of schools are not just bad, but truly AWFUL. Most of those schools are actually full of white students, not the minority students who were the intended beneficiaries of this law. People in those school districts love the 10% rule, especially in rural areas where they would rather spend school money on football than on acdemics.

In addition to providing a disincentive to schools to crank up the academics so that their students can compete on their merits, it also discourages students from challenging themselves w/in their schools. many students try to game the system by avoiding taking difficult courses which could harm their GPA. Why take an AP course and risk a B when you can take a regular course and be assured an A? The top 10% by grades is not necessarily the top 10% in terms of being prepared for college. The changes in the law are supposed to try to address this, but I don't know the details.

well my cousin lives in Austin, and she seems to enjoy UT-Austin...for mere education... and also i heard Texas A&M is a great school... and Rice, of course...

although for schools like Texas Tech, UTEP, and others, i would have to agree with you, texas137.

[quote="texas137"][quote="xxreddevilzxx"]well my cousin lives in Austin, and she seems to enjoy UT-Austin...for mere education... and also i heard Texas A&M is a great school... and Rice, of course...quote]

No, I didn't mean that the colleges are bad. I meant the high schools are bad. And if UT-Austin continues to be forced to accept top-10%ers from crappy football-obsessed rural schools who would not normally get in, it is going to harm the university.[/quote][/quote]

is UT really into sports??? i mean ive been watching college sports, but UT seem to suck in basketball or football...(look at this year's basketball debacle)

[quote="eryaman"]UT won the Rose Bowl this year...that's pretty successful.[/quote]

well even the selection to Rose Bowl was controversial... even though Vince Young(the quarterback i believe) was electrifying.

BTW, back to the Texas 10% law,

[quote="texas137"]
No, I didn't mean that the colleges are bad. I meant the high schools are bad.
[/quote]

in terms of math, some of the high schools arent that bad. there are many other high schools worse in the nation. Texas has one of the best schools.(i mean like public)

Okay, let's be very, very clear here:

Having one good school is not the same as having a good school system.

Texas is the second-most populous state -- if it didn't have any above-average schools, I would be absolutely dumbfounded.  As you'll note, texas137 specifically pointed out that Texas has a very large disparity in school quality -- much like the rest of the south, suburban schools tend to be reasonably (or even very) good and well-funded while urban and rural schools are some of the worst in the nation.

I would dispute texas137 on one thing -- she said, "Most of those schools are actually full of white students, not the minority students who were the intended beneficiaries of this law."  I think she's definitely wrong on one part of this: if the 10% law were really designed to help minority students, they would have stuck with an honest Affirmative Action program.  In any event, in terms of achieving racial integration, it's certainly better than not having such a program.  After all, most students in the really awful schools may be white, but almost all of the minority students are also in those awful schools.
The thing that annoys me about the 10% program is that it essentially takes away any burden from the state to improve the schools -- "Why should we improve your schools?  You get to go to UT Austin anyhow!"  This is, I think, just the mirror image of texas137's worries -- she phrased it in terms of individual students and school districts, while I see it as a state-level problem.

*Edited to reflect TokenAdult's comment*

Two points:

1) The state of TX went with the top 10 percent law because a more frank "affirmative action" of the kind TX used to have was known to be vulnerable to court challenge. The top 10 percent law was intended to be a back-door, legally impeccable way to boost minority ethnic enrollments in TX colleges.

2) Texas137 is a famous mathlete's MOM, and should be referred to as "she."  :)

[quote="texas137"][quote] is UT really into sports??? i mean ive been watching college sports, but UT seem to suck in basketball or football...(look at this year's basketball debacle)[/quote]

I was saying that crappy rural high schools in Texas are really into football, as opposed to academics. Is UT having a bad year for sports? I don't follow it. They invest huge resources in their teams, esp football, and are very successful some years.[/quote]

Just as an FYI, Sports Illustrated named UT one of the nation's two top sports programs a year or so ago (along with Stanford)


The negation of the proposition "For all pairs of real numbers $a$, $b$, if $a=0$, then $ab=0$" is: There are real numbers $a,b$ such that

$\text{(A)} \ a\ne0,ab\ne 0 ~~\text{(B)} \ a\ne 0, ab=0 ~~ \text{(C)} \ a=0,ab\ne 0$
$\text{(D)} \ ab\ne0,a\ne0 ~~\text{(E)} \ ab=0,a\ne0$


Let $ x$ be a real number satisfy the following condition:

For any natural number $ q$ there exists a integer $ p$ such that the inequality is true:

$ \bigg|x \minus{} \frac pq\bigg|<\frac1{3q}$

Prove that $ x$ is an integer.

this kind of reminds me of the cantor set

for each $ q$, we get a possible solution space

$ q \equal{} 1$: says $ x \mod 1$ is contained in $ \left[0,\frac13\right]\cup\left[\frac23,1\right]$
$ q \equal{} 2$: says $ x \mod 1$ is contained in $ \left[0,\frac16\right]\cup\left[\frac13, \frac23\right] \cup \left[\frac56,1\right]$
$ q \equal{} 3$: says $ x \mod 1$ is contained in $ \left[0,\frac19\right]\cup\left[\frac{2}{9},\frac{4}{9}\right] \cup \left[ \frac{5}{9}, \frac{7}{9}\right] \cup \left[ \frac{8}{9},1 \right]$
$ \vdots$
$ q \equal{} n$: says $ x \mod 1$ is contained in $ \left[ 0,\frac{1}{3n} \right]  \cup \left(\bigcup_{i\equal{}1}^{n\minus{}1} \left[ \frac{3i\minus{}1}{3n},\frac{3i\plus{}1}{3n}\right] \right)\cup \left[ \frac{3n\minus{}1}{3n}, 1\right]$

taking the intersection as $ n \to \infty$ we get $ x\mod 1 \in \{0,1/3,2/3,1\}$. (actually, I just noticed, just the first two intersections give us this set)

frankly, I think this is correct. $ x$ must be an integer or an integer $ k\pm 1/3$,  with one exception of $ k \equal{} 0$.

take $ x \equal{} 4/3$ for example. for each $ q$, we can set $ p$ to one of $ \frac{4q}{3},\frac{4q\minus{}1}{3},\frac{4q\plus{}1}{3}$. One of these is guaranteed to be an integer, making the inequality hold.

To make $ x$ an integer, I think the inequality should be strict.

[quote="shallow"]take $ x \equal{} 4/3$ for example. for each $ q$, we can set $ p$ to one of $ \frac {4q}{3},\frac {4q \minus{} 1}{3},\frac {4q \plus{} 1}{3}$. One of these is guaranteed to be an integer, making the inequality hold.

To make $ x$ an integer, I think the inequality should be strict.[/quote]

You are right. The inequality is strict. I edited the problem.


μιας και μιλαμε για γεωμετρια ,θα ηθελα να μπλεξουμε και τους μιγαδικους που σε ορισμενα γεωμετρικα θεματα βοηθουν στη λυση τους.για αρχη μπορει κανεις να λυσει το τεταρτο θεμα της ΔΜΟ 2003 με μιγαδικους ? οπως το εκανε και ενα μελος απο την Κινα. εναλακτικα να αποδειξει την ευθεια Simson με τη χρηση παλι μιγαδικων.οποιος δεν γνωριζει τι ειναι μιγαδικοι να με ρωτησει οκ? με εκτιμηση.... :)

φίλε κώστα εγώ δεν μπορώ να βοηθήσω παρόλο που ξέρω μιγαδικούς δεν ξέρω πώς χρησιμοποιούνται στην Γεωμετρία
αν θές με βοηθάς
χαιρετώ

Λοιπόν αν θέλετε έχω μια λύση στο ΙΜΟ 1993 με μυγαδικούς..... :wink:

βεβαιως Αμβροσιε.υπαρχουν αρκετα θεματα που λυνονται ετσι.εγω προσωπικα θα βαλλω τα δυο που ρωτησα.αν δεν τα βαλλει αλλος.Σιλουανε φυσικα και θελω να δω το θεμα που λες  :)

Λοιπόν ....
Προβλημα :Σημείο $D$ επιλέγεται στο εσωτερικό σκαληνού τριγώνου $ABC$ ώστε $\angle{ADB}=\angle{ACB}+90$
and $AC\cdot BD=AD\cdot BC$  .Βρείτε το λόγο
$\frac{AB\cdot CD}{AC\cdot BD}$

Λύση
Έστω $D$ η αρχή του συστήματος μας και έστω $x$ η εικόνα  του $X$ στο μυγαδικό επίπεδο .Έστω $s=\frac{AC}{AD}=\frac{BC}{BD}$  και $\angle{CAD}=k$
Εύκολα βρίσκουμε ότι $(a-b)c=sabe^{ik}(1+i)$
Άρα $\left|AB\right|\cdot\left| CD\right|=\left|a-b\right|\cdot\left|c\right|=s\left|a\right|\cdot\left|b\right|\sqrt{2}=\left|AC\right|\cdot\left|BD\right|\sqrt{2}$
άρα ο ζητούμενος λόγος είναι $\sqrt{2}$
***
Οι συμβολισμοί με κεφαλαία φανερώνουν μέτρα παντού (ας μην τα έχω σημειώσει)

Λοιπον,ας πω  ενα θεμα.Εστω ABC ενα οξειγωνιο τριγωνο.εστω S η εφαπτομενη του περιγεγραμμενου κυκλου του τριγονου ABC στο  B. απο το ορθοκεντρου του ΑΒC φερνουμε την καθετη στην S που την τεμνει στο Κ.αν L το μεσο  της AC.  να δειξετε οτι KLB ισοσκελες

η λυση στη δικη μου:θεωρουμε χωρις βλαβη της γενικοτητας να υποθεσουμε οτι ο πρειγεγραμμενος κυκλος του ABC ειναι ο μοναδιαιος κυκλος στο μιγαδικο επιπεδο.εστω  Α=α+bi, B=-i, C=c+di .( ta A,B,C ειναι εικονες),τοτε το ορθοκεντρο Η ειναι Η=A+B+C,K=a+c-i, L=(a+c)/2+(b+d)^2/2. o τυπος για το ορθοκεντρο ειναι γνωστος και οι υπολοιποι βγαινουν ευκολα(δηλ το Β,C).τοτε το μετρο του LB ειναι =ριζα του....(a+c)^2+(b+d+2)^2=KL απλα με πραξεις. και εδειχθει

Λοιπόν παιδιά,
 η πρόταση του Κώστα είναι πολύ χρήσιμη.Τον Αύγουστο συνάντησα το Σιλουανό στη Χαλκιδική και ένα μεσημέρι που καθόμασταν σε ένα τραπέζι στην παραλία του έδειχνα πώς λύνονται γεωμετρικά προβλήματα με μιγαδικούς.Συγκεκριμένα κοιτάζαμε την παρακάτω άσκηση που είχα βρει σε παλαιό τεύχος στη Gazeta Matematica.
[size=150]
  Έστω $S$ σημείο του περιγεγραμμένου κύκλου ενός τριγώνου $ABC$. Αν $K,L,M$ είναι τα κέντρα των κύκλων του Euler των τριγώνων $SAB , SBC , SCA$, να αποδειχθεί ότι το τρίγωνο $KLM$ είναι όμοιο με το τρίγωνο $ABC$ με λόγο ομοιότης $\frac{1}{2}$  [/size]
  Η άσκηση αυτή λύνεται πολύ σύντομα με μιγαδικούς.Παρακαλώ το Σιλουανό , που ξέρει τη λύση , να τη στείλει με την πρώτη ευκαιρία.Να θεωρήσει γνωστά αυτά που γράφει στη λύση ο Κώστας .
   Πάντως , το καλλίτερο εισαγωγικό βιβλίο για το αντικείμενο αυτό είναι του Andrescu: Complex Numbers , From A to Z- Birkhauser.
  Το έχει έτοιμο ο γνωστός φίλος και κοστίζει κάπου 15 ευρώ.Μπορείτε να το παραγγείλετε.Το πρωτότυπο κοστίζει 80 ευρώ!

  Μπάμπης

Είναι πράγματι τρομερή η λύση με μιγαδικούς......Δείτε
http://www.mathlinks.ro/Forum/viewtopic.php?t=108225

να δωσω αλλο  ενα :4ο θεμα ΔΜΟ 2003:Αν ΑΒCD τετραπλευρο εγγ.σε σε κυκλο.αν P,Q,R ειναι τα ιχνη των καθετων που φερνω απο το D stis BC,CA,AB αντιστιχα.να δειξετε οτι PQ= QR αν και μονο αν οι διχοτομοι των ABC ADC τεμνονται στο ΑC.  για να δωσουμε μια λυση με μιγαδικους πρεπει να επιλεξουμε το καταληλλο συστημα συντεταγμενων.καταρχην Χωρις βλαβη της γεν. μπορουμε να υποθεσουμε οτι τα A,B,C,D βρισκονται πανω στο μοναδιαιο κυκλο.θεωρουμε επισης οτι η μεσοκαθετος του AC ειναι πανω στον αξονα των πραγματικων (αξονα χ).εστω Α=cosθ+isinθ tote C=cosθ-isinθ ωστε AC=1.(αφου το C ειναι ο  συζυγης του Α). και A+C=2cosθ.αφου τωρα οι διχοτομοι των ABC,ADC γωνιων περνουν απο τα μεσα του του μεγαλου και μικρου τοξου AC μπορουμε να υποθεσουμε οτι περνουν απο το 1 και το -1 αντιστιχα.αν Ζ το σημειο τομης του AC me tin διχοτομο της ABC γωνιας τοτε Z+ACZ'=A+C =>Z=(2Bcosθ-Β-1)/Β-1 οπου Ζ' =συζυγης Ζ και σχυει Ζ+ΒΖ'=Β+1. αν Ζ'' το σημειο τομης της ΑC με τη διχοτομο της γωνιας ADC τοτε Ζ''=(2Dcosθ+D-1)/D+1 .στη συνεχεια βρισκω  τα P,Q,R χρησιμοποιοντας την ιδιοτητα τους.μετα απο το θεωρημα Simson ta P,Q,R ειναι συνευθειακα.αρα για να εχω PQ=QR πρεπει Q=(P+R)/2 . αντικαθηστω τα P,Q,R και βρισκουμε Ζ=Ζ''

λοιπον μιας και μου ζητηθηκε ας πω δυο λογια παραπανω:  το μιγαδικο επιπεδο ειναι ενα συστημα συντεταγμενων.(αξονας Χ αξονας=πραγματικος αξ,αξονας Υ=φανταστικος αξονας) καθε σημειο του εστω Μ(χ,ψ) εχει μοναδικες συντεταγμενες.και ειναι εικονα ενος μιγαδικου του χ+ψi.ετσι αυτη η αντιστιχια χρησημευει για να μεταφερω το σχημα που θελω στο μιγαδικο επιπεδο χρησιμοποιωντας συντεταγμενες.  οσον αφορα στην ασκηση της ΔΜΟ χρησιμοποιω οτι Α=cosu+sinu, δηλ την γνωστη και στα σχολικα βιβλια τριγονομετρικη μορφη μιγαδικου οπου z=p(cosu+sinu)...(εδω p=1) για καθε μιγαδικο,οπου p  ειναι το μετρο του ή η αποσταση του απο το Ο(0,0). απλα το σημειο που πρεπει να προσεξει κανεις ειναι το συστημα συντεταγμενων του .στη συγκεκριμενη ασκηση θετω αξονα των Χ ΤΗ μεσοκαθετο του AC.γενικα σε θεματα με κυκλους χωρις προβλημα πολλες φορες το θεωρουμε ως το μοναδιαιο κυκλο.ισχυουν τα βασικα θεωρηματα οπως στην αναλυτικη γεωμετρια.ομως διαφερει λιγο.αν εχω το τριγωνο ΑΒΓ στο μιγαδικο επιπ. τοτε το βαρυκεντρο ειναι (Α+Β+Γ)/3 κτλ για ορθοκεντρο πχ ειναι Α+Β+Γ δηλ το αθροισμα των εικονων.απο κει και περα εχεις τους γνωστους τυπους του κυκλου της μεσοκαθετου,συνθηκες καθετοτητας,παραλληλιας κτλ.

ελπιζω να καλυπτω καποια .αν μπορει καποιος ας πει και αλλα πανω σε αυτο το θεμα ωστε να γινει τελειως κατανοητο,εγω θα ριξω κανα θεματακι ακομα...


Let $\ a_{1},a_{2},.....,a_{n}$ be positive real numbers such that $\sum_{i=1}^{n}a_{i}=1$.    Prove that
          $\sum_{i=1}^{n}\frac{a_{i}}{S-a_{i}}$ $\geq$ $\frac{n}{n-1}$
                       with $\ S=\sum_{i=1}^{n}a_{i}$
         good luck  :D  :play_ball:

Let $f(x)=\frac{x}{1-x}$.Then $f(x)$ is convex for $x<1$.So using Jensen you can prove this inequality. :D

can you post other case ?
Thankyou !!!  :ninja:  :blush:


So Texnic Center is a windows program. I recently switched to a mac, and have no idea about any mac versions of the software. anyone know?

thanks

Have a look at [url=http://www.mecheng.adelaide.edu.au/~will/texstart/]TeX on Mac OS X: The most simple beginner’s guide[/url] which leads to the more detailed [url=http://www.cs.wright.edu/~jslater/mac-tex/mac-tex-intro/mactexintro.html]Getting Started With TEX on MacOS X[/url] both of which recommend [url=http://www.uoregon.edu/~koch/texshop/]TeXShop[/url]


[quote="inny"]$ x,y,z \ge 0$
$ x^{2} \plus{} y^{2} \plus{} z^{2} \equal{} 1$

Proof that
$ \frac {x}{1 \minus{} x^2} \plus{} \frac {y}{1 \minus{} y^2} \plus{} \frac {z}{1 \minus{} z^2} \ge \frac {3 \sqrt {3} }{2}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64294


geometry
advanced fields
advanced fields theorems

What do you think of this?  Is it a good text to use in a classroom? http://www.cacr.math.uwaterloo.ca/~ajmeneze/publications/124.pdf

To be used in a classroom for what?  At what level?

As a supplementary text on a college level class.  I have to do a presentation to suggest some stuff like that, and I was looking for some input, so....

You're still not being very clear.  That is not a text, it is a paper.

What kind of college level course are we talking about?  To understand what's going on there, you probably need some graduate-level algebra (e.g, some field theory + algebraic geometry)


group theory
abstract algebra
linear algebra
linear algebra unsolved

Let $ G$ be a subgroup of $ \mbox{GL}_n (\mathbb{F}_p )$, and $ m\equal{}|G|$.
1. If $ m\equal{}p^{n(n\minus{}1)/2}$, show that all $ M\in G$ are simultaneously trigonalizable.
2. For which values of $ m$, the following statement is true: all $ M\in G$ are simultaneously trigonalizable?

The 1) is wellknown (earlier Grobber has given a sketch of the answer).
$ card(GL_n(\mathbb{Z}/p\mathbb{Z}))\equal{}(p^n\minus{}1)(p^n\minus{}p)(p^n\minus{}p^2)\cdots(p^n\minus{}p^{n\minus{}1})\equal{}p.p^2\cdots{p^{n\minus{}1}}s\equal{}p^{n(n\minus{}1)/2}.s$ where $ gcd(p,s)\equal{}1$. Thus $ G$ is a $ p$-Sylow which is conjugate with $ S_p\equal{}\{I\plus{}N\in{GL_n(\mathbb{Z}/p\mathbb{Z}))}: N \;is\; strictly \;upper\; triangular\}$. QED
2) If $ m$ is a power of $ p$ then $ G$ is conjugate with a subgroup of $ S_p$ and we are done.
If $ gcd(m,s)>1$ then I don't know.


number theory
greatest common divisor
number theory unsolved

prove that the fraction

$\frac{21n+4}{14n+3}$

is irreducible for all $n \in \mathbb{N}$.

Thats the very first IMO problem: http://www.mathlinks.ro/Forum/viewtopic.php?t=54814

hahah that's interesting it was also Indian regional math olympiad 1986.  i guess they figured no one would've thought of using the very first IMO to study for the indian RMO...  it is very easy by today's IMO standards.

so is it useful to use it...i think it is


limit
logarithms
inequalities
real analysis
real analysis solved

Let $(a_n)_{n\geq 1}$ be a set of real numbers such that :
     $(i) a_n>0 ,\forall n \in \mathcal l N$
     $(ii) \lim_{n\to\infty} a_n=0$
     $(iii) \lim_{n\to\infty}na_n=a, a\in \Re \cup \{+\infty\}$

Compute $\lim_{n\to\infty}(1+a_1)(1+a_2)...(1+a_n)$

I think that you need $a\neq 0$ to actually can calculate the lim . Then your problem can be written as:

Calculate $\sum^\infty_{n=1} \ln (1+a_n)$

But $\sum^\infty_{n=1} \ln (1+a_n) \ge \sum^\infty_{n=N} a_n-a_n^2$ (for sufficiently large $N$, because of taylor expanding).
Now $\sum^\infty_{n=N} a_n-a_n^2 \ge \sum^\infty_{n=N} (a-\epsilon)\frac 1 n-(a+\epsilon)\frac 1 {n^2}\~ \ln n\to infty$ (for sufficiently large $N$, $\epsilon$ can be chosen such that $(a-\epsilon)\ge 0$ ).

So this limit like your limit is $\infty$.
If $a=0$ the limit can be finite by choosing $a_n=\sqrt[3]{a_n}$ but is very hard to calculate.

that`s were I got stuck too...:
if $(1+a_1)(1+a_2)...(1+a_n)$ should be bounded then it would converge ,$\lim_{n\to\infty}(1+a_1)(1+a_2)...(1+a_n)=\lim_{n\to\infty}\sqrt[n]{(1+a_1)^n(1+a_2)^n...(1+a_n)^n}$ and from the radical convergence cryteria I got
$\lim_{n\to\infty}(1+a_1)(1+a_2)...(1+a_n)=\lim_{n\to\infty}(1+a_1)(1+a_2)...(1+a_n)(1+a_{n+1})^{n+1}$
 then it would be convergent and $l=le^a$ if $a\neq 0$ then $l=0$ false because $l\geq 1$ so $(1+a_1)(1+a_2)...(1+a_n)$ is unbounded and as a consequence $\lim_{n\to\infty}(1+a_1)(1+a_2)...(1+a_n)\rightarrow \infty$. So $a=0$ is my problem.
With a different approach we can write:
$\lim_{n\to\infty}(1+a_1)(1+a_2)...(1+a_n)=\lim_{n\to\infty}(\sqrt[n]{(1+a_1)(1+a_2)...(1+a_n)})^n=\lim_{n\to\infty}[1+(\sqrt[n]{(1+a_1)(1+a_2)...(1+a_n)}-1)]^n=e^{(\sqrt[n]{(1+a_1)(1+a_2)...(1+a_n)}-1)n}$
And judging by the fact that $\sqrt[n]{(1+a_1)(1+a_2)...(1+a_n)}$ and $1+a_n$ have the same order of convergence(is this correct?) we can replace it and get: $\lim_{n\to\infty}(1+a_1)(1+a_2)...(1+a_n)=e^{(1+a_n-1)n}=e^a$..........  :? I`m confused.

You misused the radical criteria :P  It states that [b]if $\lim \frac{a_{n+1}}{a_n}$ exists[/b] and is equal to $\ell$then $\lim\sqrt[n]{a_n}$ exists too and is equal to $\ell$. You do not know if $\\lim_{n\to\infty}(1+a_1)(1+a_2)...(1+a_n)(1+a_{n+1})^{n+1}$ exists.

$x_n=\sqrt[n]{(1+a_1)(1+a_2)...(1+a_n)}$ and $y_n=1+a_n$ may not have the same convergence order. Let $a_n=\frac 1 n$. Then, $x_n=\sqrt[n]{n+1}=e^{\frac{\ln (n+1)}{n}}$ and $y_n=\frac{n+1}{n}$.
We have that $x_n-1\equiv\frac{\ln n+1}{n}$ and $y_n-1\equiv \frac 1 n$and not have the same order. From here comes your mistake. For $a=0$ the limit can take (I think) every ral value greater than 1.

thanks for the explanations I am not very much acquainted to convergence orders :blush:

No problem..  ;)  You can get a lot of resouces about convergence order on this forum .

I think that your product is not determinated in the general case, since for $(a_n)\subseteq\mathbb{R}^+$
\[ \prod_{n=1}^\infty(1+a_n) \quad converges\quad \Longleftrightarrow\quad \sum_{n=1}^\infty a_n\quad converges \]
just compare the with the series $\sum_{n=1}^\infty ln(1+a_n)$ and use $ln(1+x)-x=x^2\left(\frac{1}{2}-\frac{x}{3}+\ldots\right)$ so there exists $0<\alpha<1/2<\beta$ such that for $x$ enoughtly small $\alpha x^2<ln(1+x)-x<\beta x^2$.

Now  taking $a_n=\frac{1}{n}$, we have $a=1$ and the product diverges but for $a_n=\frac{1}{n^2}$, $a=0$ and the product converges, so the limit is not determinated by the conditions above.

Theat was our problem too: $a=0$. I claim that the set of points which can be limit of that sequence is dense in $[1,\infty]$ or it is exactly $[1,\infty]$.
Proof: Choose an interval $[a,b]$. The product $\Pi_{k=1}^\infty (1+\frac 1 {k^2})$ converge hence for any $\epsilon$  there exists $N$ such that $\Pi_{k=N}^\infty\frac{1}{k^1}\le (1+\epsilon)$.Tanking the forst $N$ terms such that $\Pi_{k=1}^n (1+a_k)=a$ and $a_n+k=\frac 1 {(n+k)^2}\forall k>0$ and $\epsilon<\frac{b-a}{a}$, we have the limit point of the product in the interval $[a,b]$ (the product is convergent since it is increasing and bounded). Since $[a,b]$ was chosen arbitrarly, the set of those points is dense in $[1,\infty]$.
[b]Problem[/b] Prove or disprove that the set of points which can be limit of a sequence of that form is $[1,\infty]$.

You have already done it, well for $\alpha\in [1,\infty)$, taking $N$ st $0<\Pi_{k>N} (1+\frac 1 {k^2})=c<1$
then $\frac{\alpha}{c}>1$ so just choose $a_1,\ldots,a_N$ st $(1+a_1)\cdots (1+a_N)=\frac{\alpha}{c}$ and for $n>N$ $a_n=\frac{1}{n^2}$ then the whole product must converge to $\alpha$.

Well it was your idea I just used it :D ....

[quote="xirti"][quote="hcast"]well for $\alpha\in [1,\infty)$, taking $N$ st $0<\Pi_{k>N} (1+\frac 1 {k^2})=c<1$
[/quote]
this inequality does not hold :!: since $c\ge 1$. But I agree that your method works. Thanks.
Does there exist a sequence of this form which tends to $\infty$ when $a=0$?
Or, in other words: Does there exists a divergent series $\sum a_n$ with $a_n=o(\frac 1 n)$?[/quote]

$a_n=(n\ln n)^{-1}$


Find the sum of all the digits in the integers from 1 to 100000.  (Note that we are asking for all the sum of the digits in all the numbers, not all the numbers themselves).

Is there any methods to figure out the answer??

Any help would be appreciated.....

From 00000~99999 notice that we count every digit 10000 times. So, 450000 for 0~99999. Add 1 to count 100000.

So, our answer is 450001

[quote="Bachukas"]From 00000-99999 we count every digit 50000 times.

$ 50000\cdot45 \plus{} 1 \equal{} 2250001$[/quote]

Where do you get the one and the 45?

First of all, consider how many times the sequence of digits; 1,2,3,4,5...9 occurs. IMagine the numbers in a table like this:

00000
00001
00002
00003
00004
.
.
.
.
10000

Consider the digits on the the right side to be called the 'first digits'. THe digits just to the left of these as the 'second digits' and so on.
The sum we're after is obviously equal to the sum of the first digits, plus the sum of the second digits etc.
The sum of the digits from 1 to 9 is 45(see arithemetic progression). by adding up the first digits from 1-100 u get 45*10, since the cycle 1,2,3,4,5,6,7,8,9 happens 10 times; once where 0 is the second number, once where 1 is the second number...
Consider the bigger cycle(the cycle 01,02,03,04...99), this is what we basically did in the previous paragraph. how many time does that repeat when we consider the numbers from 1-1000? It actually happens 10 times. so the sum of the first digits from 1-1000 is 45*10*10. so the sum from 1-10^3 is 45*10^2. This pattern continues if u go further and hence the sum of the first digits from 1 to 10^n is 45*10^(n-1).

How about the second digits?

Well they dont come in the cycle 1,2,3,4... but rather in 1,1,1,1,1,1,1,1,1,0,2,2,2,2,2,2... etc. The sum of the second digits from 1-100 is
1*10+2*10+3*10...9*10=(1+2+3+4+...+9)*10=45*10
Which is the same as the sum of the first digits from 1-100!!!!!!! (those are exclamation marks, not factorial signs).
again, this cycle happens 10 times if we consider the numbers from 1-1000(creating what i call a bigger cycle). This bigger cycle happens 10 times if we consider the number 1-10000. hence the sum of the second digits is 45*10*10. This pattern continues(the proof of which will be left as an exercise for the reader, or i'll just explain it later as i cbf explainin so many things in one post) Hence the sum of the second digits from 1-10^n is 45*10^(n-1). Which is the same as the formula for the first digit sums. If you think about it(if u dont want to think then tell me and i'll explain in a further post) then the same formula is for third, second, fourth... digits.
We want to find the sum of digits from 1 to 10^n. So we want the sum of the fisrt, second , third, fourth.. nth digits.
and so this is 45*(10^(n-1))+45*(10^(n-1)).... (an n number of times since we're considering up to n digits)
of course we then hav to add a 1 since this is only really from 1 to 9999...

45*(10^(n-1))+45*(10^(n-1)).... (an n number of times)=n*45*(10^(n-1))
since multiplication is really just repetitive addition.

now in our case n=5 since we want up to 100000(five zeros yea?). so the sum is:
5*45*10(5-1)
=5*(10^4)*45
=5000*45

so congratz Bachukas!!! we agree =]

Oh wait!! we're wrong since this guy wants from 1-10000, and we were counting to 9999. So the correct answer is 5000*45+1

Thanks kamil9876... You have made things much more clearer...


function
inequalities
triangle inequality
real analysis
real analysis unsolved

[quote="amcavoy"]Let $\left\{f_{n}\right\}$ be a sequence of continuous functions converging to $f$.  Is it true that $\left\{f_{n}\right\}$ is uniformly convergent if and only if $f$ is continuous?  If so, how do you prove this?[/quote]

[b]Theorem:[/b] If $\{f_{n}\}$ are continous on $\{\}\subset S\subseteq\mathbb{R}$ which converge uniformly to $f(x)$ then $f(x)$ is continous on $S$.

[b]Proof (Outline):[/b] We want to show $|f(x)-f(x_{0})|$ is "small" for some $x_{0}\in S$. We can write,
$|f(x)-f(x_{0})|=|f(x)-f_{n}(x)+f_{n}(x)-f_{n}(x_{0})+f_{n}(x_{0})-f(x_{0})|$
Triangle inequality,
$\leq |f(x)-f_{n}(x)|+|f_{n}(x)-f_{n}(x_{0})|+|f_{n}(x_{0})-f(x_{0})|$

You can make first inequality small because it is uniformly convergent for all $x\in S$ you can make the second small because $f_{n}$ is continous and you can make the third one small because it is the property of uniform convergence at a point. (I am write out the full prove if you like).

I dont think the converse is true.

[quote="sylow_theory"]
I dont think the converse is true.[/quote]

Of course it isnt true.

Define the "Dirichlet function":
${g(x) = \{ \begin{array}{c}1 \, ,x\in \mathbb{Q}\\ 0 \, ,x\not \in \mathbb{Q}\end{array}}$

Define,
$f_{n}(x) = \frac{g(x)}{n}$ for $x\in \mathbb{R}$

We see that $f_{n}\to 0$ uniformly on $\mathbb{R}$ because,

$|f_{n}(x)-f(x)| = | \frac{g(x)}{n}| \leq \frac{1}{n}$ for all $x\in \mathbb{R}$.

So choose $N=\frac{1}{\epsilon}$ to actually prove it.

Note the limit function $f(x)$ is continous everywhere.
Yet the Diminishing Dirichelt function is continous nowhere  :rotfl:

The converse would be: "if $f_{n}$ are continuous and their pointwise limit, $f$, is continuous as well, then $f_{n}\to f$ uniformly". This is false: the sequence $f_{n}(x)=nx^{n}(1-x)$ converges to zero on $[0,1]$, but not uniformly. (Because $f_{n}(1-1/n)=(1-1/n)^{n}\to e^{-1}$.)


I even gave it the same name ("Lucas divisibility"):

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53771[/url]


AMC
AMC 12
AMC 12 A
email
geometry
rhombus
AIME

After looking at both of them, which contest did you think was harder? AMC [u][b]12[/b][/u] A or B?

I thought 12B was somewhere between somewhat and much harder but closer to somewhat harder.  I managed to somewhat solve all of the 12A problems but got killed by #24 and #25 on B.

My school had a snow day when we were supposed to take the AMC A, so I took it after the fact.  I answered more questions on the AMC A, but missed about 5 of them, so I guess I have to say it was harder.

AMC A: 108 (unofficial)
AMC B: 123 (not official ... yet)

12B much harder. I got 15 less than on the A

[quote]My school had a snow day when we were supposed to take the AMC A, so I took it after the fact. [/quote]

Wtf, my school had a FOG day yesterday when I was suppose to take the amc b.  So i like called the school to ask if they would still allow us to take it today, but they said no.  So i asked them could I just come to school to take it (the fog cleared at like 11:00), they were like no.  I even went to the school to ask and they were like no.  but at 12:00, the person in charge of the test said we could take it, as he emailed us.  But i dont check my email every hour and I didnt know until a friend called me because he wanted to go and he had no ride.  I was so mad at the school.. since it also said in the email that the test would start at 1:00.  ANyways, I ended up going but at 2:00, everyone was already taking the test, and the principal(who was administrating this) got really mad at us, even thought like 2 hours earlier I asked another administrator if we wre going to take it and she said no.  ANyways, I ended up taking it, but I dont even know if i signed my name because the principal made us fill everything else in a rush because he wanted to do his other work.  so i was in such a rush i misread the problem with BD, as i found CD and number 17... anyways, i ended up with a 108...which is like my worst score ever on amc 12..(luckily i already passed the amc 10 so i wanst under thaqt much pressure, but still...)

And yes the amc 12b was harder (i practiced on the amc 12a, because i took the 10a, and in that practice i got 130.5)

and this was made even worse because i called like 3 other peole in my school and they said we werent taking it and even someone on aops (not mentioning name) who goes to my school told me we werent taking it and prepare for next year.. also, beacuse we were late i dont even think the principal gave me enough time (the calculator clock after i finished said only like 70 mins had passed)
and the sad thing is, our superintendent never gives snow days, but she gave a FOG DAY, when it wanst even that foggy..

All I can say is who's ever heard of #24 on 12=#25 on 10.

I think that B is harder looking at the last but I can't tell.

AMC 12B REALLY was HARD, i got at least 12 pts lower

The AMC 12B was much harder, both the middle and last set of problems were harder than the 12A. I took the 12A as practice and got a 132, but i got either a 115.5 or 109.5 (depending on if i bubbled a question in right  :( .

I think they might lower the cut off even to like 90 for the 12B. Then again, im not an expert on this, so who knows?

[quote="paladin8"][quote="beta"]lol AMC12A much much harder, I did 18 points better on B than on A.[/quote]

I did better on the B, but I felt that the B was harder. Except for #25, which was sorely disappointing after finally visualizing it. That said, #25 on the A was pretty easy too.

I think questions 20-24 were harder on the B test. A lot of number theory, in particular (21, 23, 24).[/quote]

That is true, I think I just had a worse day on AMC12A. I thought #22, 25 were hard to visualize (actually for #25 if you just guess right angles you can visualize it)

Maybe that's why... I enjoy number theory alot more than trigo. and counting.

[quote="roadnottaken"][quote="usaha"][quote]My school had a snow day when we were supposed to take the AMC A, so I took it after the fact. [/quote]
and the sad thing is, our superintendent never gives snow days, but she gave a FOG DAY, when it wanst even that foggy..[/quote]
That's funny, because my school almost never calls snow days either, but we had two days off the week of the AMC A due to the cold, and two snow days the following week :huh: !  (Although we are probably one of the few schools in Ohio that haven't used up more than our quota of snow days this year :maybe: .)[/quote]

OMG, that's exactly like me.  Except 3 cold days on the day week of the AMC.

[quote="white_horse_king88"]Whereas for the A, I didn't have time for 24 nor 25, the B I had plenty of time for both and just screwed up 24. So I voted equally hard.[/quote]

On the A I did not have time for 24 and 25 because I checked each question after doing it and such, so on the B I went faster and got to all the questions.. but could not figure out how on earth you could tape AB to DC on that rhombus question, messed up 23, couldn't get 24, and for 25.. I was trying to math it out and wasn't getting anywhere.  In history class after taking the test though I just pictured it without writing anything and was like.. that was #25??

But so I thought B was definitely harder because I got 24 and 25 on the A right after the test.

It reminded me of last year's difficulties, although many people disgreed with me if I remember correctly (that 2006 A was easy and brutable compared to B).

I'm just trying to qualify...  So, I found both about equal.  The beginning problems were obviously not that different between the tests, but the middle questions on the B did seem easier.  So, I would have to go with B being a bit easier, but not by much.

[quote="h_s_potter2002"][/quote][quote="roadnottaken"][quote="usaha"][quote]My school had a snow day when we were supposed to take the AMC A, so I took it after the fact. [/quote]
and the sad thing is, our superintendent never gives snow days, but she gave a FOG DAY, when it wanst even that foggy..[/quote]
That's funny, because my school almost never calls snow days either, but we had two days off the week of the AMC A due to the cold, and two snow days the following week :huh: !  (Although we are probably one of the few schools in Ohio that haven't used up more than our quota of snow days this year :maybe: .)[/quote][quote="h_s_potter2002"]

OMG, that's exactly like me.  Except 3 cold days on the day week of the AMC.[/quote]

Oh that's nothing... try an [b]entire week[/b] of cold days on the week of the AMC so I had to make my teacher give the test in the afternoon.

[quote="Go Beyond"]I think it totally depends on what type of questions you are good at. Maybe some people who are good at combinatories will like B better. Some people who are good at geometry will like A better... (just personal opinion)[/quote]

snap!!!!!!!!!!

im sooooooo much better in geometry than combinatories er watever

i WAS going to take 12a  but stupid school
they didnt even ORDER them. seriously
since my school doesnt offer
i like talk about taking it last fall

but still they hadnt order 12a
i was like sooooooooo mad
so i went to office like everyday
to make sure they'll order 12b -one i did so bad.

I thought the B was harder, but then again I am much better at combinatorics than I am at geometry.

[quote="mysmartmouth"][quote="Kent Merryfield"]It's possible that the 12B was harder than the 12A for the people who answer all the questions, trying for 130+ scores, while at the same time easier than the 12A for those trying for 100 who don't even look at the back page. I don't know that either half of what I just said is correct - it's all speculation. But it makes some sense to me.[/quote]

That's funny, because I was thinking the exact same thing.  The first 15 were easier on the 12B, the last 10 were easier on the 12A.[/quote]

I agree.  I actually thought the 10B was harder than the 12A ( :o )  and someone I know had his score drop 24 points from the 12A to the 12B.  The two of us were the only AIME qualifiers in my club on the A.  But we had 8 AIME qualifiers on the B.  So I think AIME qualification was easier but getting a really high score was harder on the B.

Well, I did much better on the B so I suppose I must say that it was easier...
I got a 100.5 on the A (15 miss 3) and a 132 (22 miss 3)
However, I knew that I qualified on the A and was going to go all out on the B and I happened to guess 23 right (gg) but I also made a dumb mistake on 17...

I found the 12B muchhhh harder than the 12A... I wanted to cry after the 12B was over yesterday, hahah

[quote="Ad Infinitum"]I completely agree with what Kent Merryfield said.  I never even try the 20s, but I did a lot better on the 12B than the 12A (17 correct answers to 14).[/quote]

With my limited math skills, I generally can't do them.  However, I always look at what they are asking because I may happen to know how to do one of them quickly.  A number 23 one time dealt with side lengths and medians, with Stewart's Theorem it took about twenty seconds.  It just depends on the problem.


conics
parabola
calculus
derivative
analytic geometry

My problem is similar to a homework problem, but its for a computer program I'm developing...I am not that good with physics, but I am trying..

I need to display a graph of the trajectory of an object, for instance a ball. When the user clicks a spot on the screen, my program will draw the trajectory of the ball where the "thing" that launches the ball will be facing at a given angle.

So: I have a ball that needs to be launched at a given angle, and where the trajectory of the ball will intercept the position of the mouse cursor...

Say the launcher is at (0,0) and at an angle of 60 degrees...hence, to graph the equation of the parabula to reach the mouse point (for example (5, 15))...

How would I go about in solving for the initial velocity? My problem is similar to this thread: http://www.artofproblemsolving.com/Forum/topic-36425.html

but how would I go about solving for the velocity when y is also greater than 0?

My final result will be the ball launching from the "launcher" at 60-degrees and will intercept the mouse cursor position of (5, 15), then continue to the ground..

Any help would be appreciated..thx

Call the general equation of the parabola $y=ax^2+bx+c$ and the launcher at (0,0), cursor at (x,y) and angle = $\theta$.

Then 0=c because it has to touch (0,0)
And $y=ax^2+bx$ by the second point.
For the last condition, take the derivative:
$dy/dx=2ax+b$
at x=0, $dy/dx = tan\theta$
$tan\theta=b$

So then $y=ax^2+tan\theta x$, put in x and y (the cursor's coordinates), solve for a:
$a=\frac{y-tan\theta x}{x^2}$

Then you have a final equation.

gggggg.

x = v0 cos a
y = v0 sin a - 16t^2

Assuming you're in feet per second


Let  $ {AEFG}$ be a concyclic quadrilateral  and let $ {AE \cap GF \equal{} B}$ and $ { EF \cap AG \equal{}C }$ $ {D}$ is the midpoint of segment $ {BC}$, join AD meet the circle at point $ {H}$;
(1) prove that $ {AH*BC\plus{}BH*AB\plus{}CH*AC\equal{}2AD*BC}$;
(2) let $ {AD \cap GB \equal{}M}$ and $ {CM \cap AB \equal{}N}$ GN meet the circle at point $ {L}$
     PROVE :$ { C, H, L}$ are conllinear

Hello!
(2) is true $ \left(\forall\right)D\in BC$

The points $ A,E,F,G,H,L$ lie on the circle $ \Gamma$ and $ d$ is the polar of the point $ C$ with respect to the angle $ \angle ABG$.

$ E,N\in BA$, $ F,M\in BG$, $ \left\{P\right\} \equal{} AF\cap GE$ and $ \left\{Q\right\} \equal{} AM\cap GN$ $ \Rightarrow B,P,Q\in d$

$ A,E,F,G\in\Gamma$, $ \left\{P\right\} \equal{} AF\cap GE$,$ \left\{B\right\} \equal{} AE\cap GF$ and $ B,P\in d$ $ \Rightarrow d$ is the polar of the point $ C$ with respect to the circle $ \Gamma$
$ \left\{P\right\} \equal{} AF\cap GE$, $ \left\{Q\right\} \equal{} AH\cap GL$ and $ B,P,Q\in d$ $ \Rightarrow$ the points $ C,H,L$ are collinear.

Best regards, Petrisor Neagoe :)

Hello!

The solution of (1):

The points $ A,E,F,G,H$ lie on the circle $ \Gamma$ and let $ \Sigma$ be the circumcircle of the triangle $ BFC$.
$ \left\{F,H'\right\}\equal{}\Gamma\cap\Sigma$ and $ \left\{D'\right\}\equal{}AH'\cap BC$.

$ \angle H'BD'\equal{}\angle H'BC\equal{}\angle H'FC\equal{}\angle EAH'\equal{}\angle BAD'\Rightarrow$ the triangles $ H'BD'$ and $ BAD'$ are similar $ \Rightarrow D'B^{2}\equal{}D'H'\cdot D'A$     (1)
$ \angle H'CD'\equal{}\angle H'CB\equal{}\angle GFH'\equal{}\angle GAH'\equal{}\angle CAD'\Rightarrow$ the triangles $ H'CD'$ and $ CAD'$ are similar $ \Rightarrow D'C^{2}\equal{}D'H'\cdot D'A$     (2)
(1),(2) $ \Rightarrow D'B\equal{}D'C\Rightarrow D'\equal{}D\Rightarrow H'\equal{}H$. So $ \left\{F,H\right\}\equal{}\Gamma\cap\Sigma$.

$ K\in AD$ such that $ D$ is the midpoint of $ HK$ $ \Rightarrow BHCK$ is a parallelogram.
$ \angle BKC\equal{}\angle BHC\equal{}\angle BFC\equal{}\angle EFG\equal{}180^{\circ}\minus{}\angle EAG\equal{}180^{\circ}\minus{}\angle BAC\Rightarrow$
$ \Rightarrow\angle BKC\plus{}\angle BAC\equal{}180^{\circ}\Rightarrow$ the quadrilateral $ ABKC$ is cyclic $ \Rightarrow$
$ \Rightarrow CK\cdot AB\plus{}BK\cdot AC\equal{}AK\cdot BC\Rightarrow BH\cdot AB\plus{}CH\cdot AC\equal{}\left(AD\plus{}DH\right)\cdot BC\Rightarrow$
$ \Rightarrow BH\cdot AB\plus{}CH\cdot AC\equal{}\left(2\cdot AD\minus{}AH\right)\cdot BC\Rightarrow$
$ \Rightarrow AH\cdot BC\plus{}BH\cdot AB\plus{}CH\cdot AC\equal{}2\cdot AD\cdot BC$

Best regards, Petrisor Neagoe :)


When P(x) is divided by x+1 and x-2 the remainders are 2 and -3 respectively. Find the remainder when P(x) is divided by (x-1)(x-2).
$%Error. "pmdiv" is a bad command.
\cdot \ast \star \dagger \ddagger \amalg \cap \cup \uplus \sqcap \sqcup \vee \wedge \oplus \ominus \otimes \circ \bullet \diamond \lhd \rhd \unlhd \unrhd \oslash \odot \bigcirc \triangleleft \Diamond \bigtriangleup \bigtriangledown \Box \triangleright \setminus$

Should that be $(x+1)(x-2)$?  I've seen this problem before.

[hide="Hint"] It's equivalent to the corresponding problem for integers.  You can use the polynomial version of the Chinese Remainder Theorem. [/hide]

I know the answer and solution.
But I wrote it for any other interesting solutions.


Two monoprotic acids, HX and HY, both 0.100M in conc. are titrated with 0.100M NaOH. The pH at the equivalence point for HX is 8.8 and for HY is 7.9
Which is the weaker acid and explain your answer?

-------

Calculate the mL for 0.0850 M NaOh to titrate 40.0mL of 0.0900M $ HNO_{3}$ to the equivalence point.

From both equivalence points we can conclude that both acids are weak. So let's consider the titration of an weak acid HA with NaHO: in the equivalence point we will have a solution of NaA (a weak base, and so $ pH_{eq}> 7$), and so the following equilibrium will exist in solution:

$ A^{-}(aq)+H_{2}O(l) \rightleftharpoons HA(aq)+HO^{-}(aq)$.

As in our problem the concentration of both acid and base (C) are the same, and as they react in 1:1, then the volume of equivalence (of NaHO) is equal to the volume of acid sample used: let's denote it by V. Then the "initial" concentration of A- will be

$ [A^{-}] = \frac{CV}{2V}= \frac{C}{2}$

and so

$ K_{b}= \frac{K_{w}}{K_{a}}= \frac{[HA][HO^{-}]}{[A^{-}]}= \frac{2[HO^{-}]^{2}}{C}$
$ \Rightarrow  [HO^{-}] = \frac{K_{w}}{[H_{3}O^{+}]}= \sqrt{\frac{CK_{w}}{2K_{a}}}\Rightarrow  [H_{3}O^{+}]_{eq}= \sqrt{\frac{2K_{a}K_{w}}{C}}$

and now we can see that the weakest acid (the one with the lowest Ka) will have the greatest pH of equivalence.


calculus
integration
trigonometry
function
algebra
rational function
calculus computations

I know this might have been posted before... but i didn't find it :(

Evaluate:

$I=\int\sqrt{\tan u}\ du$

First substitute $u = \arctan t$. Since $t \geq 0$, we can substitute $t = x^{2}$. You get a rational function, which can be handled by standard methods :agent: The answer won't look very nice though :noo:

ok, we get $\int\frac{x^{2}}{1+x^{4}}dx$ :D

[b]perfect_radio[/b] wrote:
[quote] which can be handled by standard methods [/quote]

now what i do is write $\frac{x^{2}}{1+x^{4}}=\frac{1}{x^{2}+\frac{1}{x^{2}}}$(maybe we can do sth with this :maybe: )

or maybe write
$\frac{x^{2}}{1+x^{4}}=\frac{A}{x^{2}-i}+\frac{B}{x^{2}+i}$ :huh:

I believe it would be:

$\int \frac{x}{x^{4}+1}dx$

Hmmmmmmmmm.....  use partial fraction decomposition knowing that:

$x^{4}+1 = (x^{2}+1)^{2}-2x^{2}= (x^{2}-\sqrt{2}x+1)(x^{2}+\sqrt{2}x+1)$


LaTeX
algebra
polynomial
algebra proposed

Find all positive real solutions to x3 + y3 + z3 = x + y + z, x2 + y2 + z2 = xyz.

Come on, it is time now to learn basic latex :wink:

Pierre.

yes i will post again
find all posive real solution :
        $x^{3}+y^{3}+z^{3}=x+y+z$ and $x^{2}+y^{2}+z^{2}=xyz$

[hide="Sorry for reviving an old thread but someone might be interested (especially if this is moved to intermediate)"]
Let $p=x+y+z, q=xy+yx+xz, r=xyz$. Then (1) is just $p^{2}-2q=-r$ and (2) is $(p^{2}-3q)(-p)+3r=-p \iff p(p^{2}-3q-1)=3r$

Substituting, $p^{3}-3pq+3p^{2}-p-6q=0$ is a cubic in either $p$ or $q$. For arbitrary $p$, for instance, one can solve for $q$ and plug in to $r$ to yield a polynomial of the form: $a^{3}-pa^{2}+qa-r=0$ and the roots can be found, giving $x,y,z$. Clearly, there are an infinity of solutions but the problem is fixed we are given $p=1$, for instance. This is expected since there are $3$ variables and two equations.[/hide]


For $ a,b,c$ positive reals

$ \sum \frac {a \plus{} c^3}{ab \plus{} c^2} \geq 3$

My friend, my inequality is \[ \sum \frac{a\plus{}c^3}{ab\plus{}c^2} \ge 3\]

Thank you very much, I have just corrected it. :blush:


abstract algebra
group theory
search
superior algebra
superior algebra unsolved

Let $ G$ be a solvable torsion group in which every Abelian subgroup is finitely generated. Prove that $ G$ is finite.

[i]J. Pelikan[/i]

It's only a slight rewording of [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1627962730&t=165615]this one[/url].

Though perhaps the solution is somewhat more confusing then it needs to be. All one needs to say is--take a maximal abelian subgroup $ A\subset{G}$. This exists by the hypotheses given. Furthermore, we see at once that $ A$ is finite. Then notice that $ A$ must be self-centralizing, because if $ x$ is in $ C_G(A)$ but not $ A$, then $ (A,x)$ is still abelian and strictly larger than $ A$. But then the map
$ G\to\operatorname{Aut}(A)$ is an injection.

[quote="sra"]Though perhaps the solution is somewhat more confusing then it needs to be. All one needs to say is--take a maximal abelian subgroup $ A\subset{G}$. This exists by the hypotheses given.[/quote]
Actually, a maximal abelian subgroup exists without any hypotheses. Of course, you do need the hypothesis of finite generation for your next step (to observe that $ A$ is finite), but not here.

[quote="sra"]Furthermore, we see at once that $ A$ is finite. Then notice that $ A$ must be self-centralizing, because if $ x$ is in $ C_G(A)$ but not $ A$, then $ (A,x)$ is still abelian and strictly larger than $ A$. But then the map
$ G\to\operatorname{Aut}(A)$ is an injection.[/quote]
How do you get a map $ G \to \mbox{Aut}(A)$? You don't know that $ A$ is a normal subgroup (or, at least, you never showed that it is).

Easy to see that this problem follows from http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1281300&sid=0f7f849e22c93e78a1d9cdcca76a483d#p1281300


geometry
circumcircle
trigonometry
projective geometry

[hide="1"]
2 exterior tangents of 2 circles with different radii, each one touches both circles in A,B,C,D.
Prove that ABCD is circumscribed if and only if 2 circles touch each other in 1 point
[/hide]
[hide="2"]
ABCD is a parallelogram. Excircle of ABD touches AD and AB in M and N. Prove that if P and Q are the points in which MN cuts BC and CD, then P and Q are on the incircle of BCD.
[/hide]
[hide="3"]
A circle with center Q which contains points B and C cuts sides AB and AC of a triangle ABC in K and L. Circumcircle of AKL cuts circumcircle of ABC in A and N. Prove $ \angle QNA\equal{}90^\circ$
[/hide]
[hide="4"]
Construct squares AKLB and ACPQ outwards on AB and AC of triangle ABC. Prove that the intersection point of BP and CL lies on the altitude from A to BC
[/hide]
Please, use only elementay geometry, no trig, no inverse and projective geometry, no complex numbers.
Thanks.

I don't think number one is correct
 [geogebra]1e99a29807a7ec43a664608556a26d47e7a6c9d1[/geogebra]

Is this what you mean by number two?

 [geogebra]29a8dcbb1913a4a19a20fc20bafc6d3de8b58c78[/geogebra]

If so note that $ \bigtriangleup ABD, \bigtriangleup DBC, \bigtriangleup AB'D'$ are all similar and have the same incenter. Therefore we know $ MN$ and $ PQ$ are parallel.

i gotta run ill come back to this tommorow..
[/hide]

1) If and only if the 2 circles touch in 1 point, quadrilateral formed by touching points of tangents is circumscribed (it has an incircle)
2) M,N,P,Q are collinear


Find all integers $x,y,z$ (or prove there's an infinite number) that satisfy the equation $n^x+(n+1)^y = (n+2)^z$
when $n = $
$(a) 3$
$(b) 5$
$(c) 7$

Is that the only solution? Could you show why?


Ok ill make it harder
 { (a, b) | a Z+, b Z+,  f(a) = f(b), where f is a homotopically equivalent function  }

The mod one is not a partial or equivalence


[quote="evarist"][color=purple] Let a,b,c are the side length of a triangle.Prove that[/color]
$\frac{5R}{16R-2r}\geq \frac{ab+bc+ca}{ (a+b+c)^{2}}\geq \frac{4r}{10r+R}$[/quote]
Here is my solution
Setting $p=\frac{a+b+c}{2}$, then we have the following identify
\[ab+bc+ca=p^{2}+r^{2}+4Rr \]
And we also have the well-known inequality (Gerretsen's Inequality):
\[4R^{2}+4Rr+3r^{2}\geq p^{2}\geq r(16R-5r) \]
the left inequality becomes
\[\frac{5R}{16R-2r}\ge \frac{p^{2}+r^{2}+4Rr}{4P^{2}}\]
Since $p^{2}\geq r(16R-5r)$, so it remains to prove that
\[\frac{5R}{16R-2r}\ge \frac{r(16R-5r)+r^{2}+4Rr}{4r(16R-5r)}\]
\[\Leftrightarrow r(R-2r) \geq 0 \]
which is true.
The right inequality becomes
\[\frac{p^{2}+r^{2}+4Rr}{4P^{2}}\geq \frac{4r}{R+10r}\]
Since $p^{2}\leq 4R^{2}+4Rr+3r^{2}$, so it suffices to prove that
\[\frac{4R^{2}+4Rr+3r^{2}+r^{2}+4Rr}{4(4R^{2}+4Rr+3r^{2})}\geq \frac{4r}{R+10r}\]
\[\Leftrightarrow (R-2r)(R-r)^{2}\geq 0 \]
which is true.
We are done. :)

[color=purple]Thanks .It is my solution posted on diendantoanhoc.net I think you should kill these problems
   [url]http://mathnfriend.net/index.php?showtopic=7250[/url]
  Why i can't post new topic in Ineq solve problem  anh Can? [/color]


function
calculus
derivative
number theory unsolved
number theory

Let $ f(x)$ be the exponential generating function of the Fibonacci sequence:

$ f(x)=F_{0}+F_{1}\frac{x}{1!}+F_{2}\frac{x^{2}}{2!}+F_{3}\frac{x^{3}}{3!}+\cdots$

(i) Find a relation satisfied by $ f(x)$ and its derivatives $ f'(x)$ and $ f''(x)$

(ii) Find a simple formula for $ f(x)$

Check the worked example B in
http://en.wikipedia.org/wiki/Examples_of_generating_functions


geometry
circumcircle
radical axis
perpendicular bisector
geometry proposed

Let H and O be the orthocenter and circumcenter of acute triangle ABC respectively, with AB distinct from AC. Let C´the circuncircle of triangle ABC. The median AM of triangle ABC, meets C´at point N and the circumference of diameter AM meets C´at points A and P. Prove that lines AP, BC and OH are concurrent if and only if AH=HN.

Let $ AM\cap OH \equal{} \{R\}, AH\cap BC \equal{} \{J\}$
$ AH \equal{} HN \Leftrightarrow AN\perp OH$ (because $ OA \equal{} ON$)
But $ \angle APH \equal{} \angle AJM \equal{} 90^o$
$ \Rightarrow APHR, HJMR, APJM$ are cyclic. Applying the concept of radical center we get $ PA, HO, BC$ are concurrent

Since $ HN \equal{} HA$, thus $ H\in$ the perpendicular bisector of $ AN\Longrightarrow OH\bot AM$. Let $ B_0,C_0$ respectively be the projections of $ B,C$ onto $ AC,AB$. Let $ P_a\equiv B_0C_0 \cap BC$. We have $ P_aH$ is the polar of $ A$ wrt $ (M,MB)\Longrightarrow MA\bot P_aH$. Therefore $ P_a,H,O$ are collinear. Now, let $ P'\equiv AP_a\cap (O)$. We have $ \overline {P_aP'}\cdot \overline {P_aA} \equal{} \overline {P_aB}\cdot \overline {P_aC}$. But $ (P_aA_0BC) \equal{} \minus{} 1$, $ M$ is the midpoint of $ BC$ $ \Longrightarrow$ $ \overline {P_aA_0}\cdot\overline {P_aM} \equal{} \overline {PB}\cdot \overline {PC}$, where $ A_0$ is the projection of $ H$ onto $ BC$. Therefore, $ \overline {P_aP'}\cdot \overline {P_aA} \equal{} \overline {P_aA_0}\cdot\overline {P_aM}\Longrightarrow P',M,A_0,A$ are concyclic $ \Longrightarrow P\equiv P'$. Thus, $ AP,BC,OH$ are concurrent.
Our proof is completed then $ \square$


Find the polynomial $f(x)=x^2+px+q$ for which $\max_{x\in[-1,1]}|f(x)|$ is minimal.

Intuition says that this polynomial must be symmetric about $x=0$. Guess and check then yields the answer, but is there a more straightforward approach? Thanks.

[quote="towersfreak2006"]Find the polynomial $p(x)=x^2+px+q$ for which max$_{x\in[-1,1]}|p(x)|$ is minimal.[/quote]
What do you mean by $p(x)=x^2+px+q$? Possibly $p(x)=x^2+p(x)+q$, i.e. $x^2+q=0$? But in this case $x$ can't be an arbitrary number in $[-1,1]$. I'm sorry but I have no idea what you are asking.

Oh. Bad notation. It has been fixed. Sorry about that.

The maximum of $|p(x)|$ occurs for one of $\{-1,1,-p/2\}$, which should be clear by convexity.

So the maximum is one of $\{|1-p+q|,|1+p+q|,|q-p^2/4|\}$.  We want to minimize these.

If we put $p=0$, then the minimum is the max of $\{|1+q|,|q|\}$, which is smallest for $q=-1/2$.

Otherwise, assume that $p>0$ (or else replace $x$ with $-x$) and $q<0$ (or else $|1+p+q|>1$)

Put $q=-r$, where $r>0$, and we get

$r+p^2/4\leq 1/2$, so that $r\leq 1/2$

and

$|1+p-r|<1/2$

But this leads to a contradiction, since then $|1+p-r|\geq |1/2+p|>1/2$

The result follows.


[quote="abacadaea"]not spam[/quote]
I beg to differ.

and i lost

EDIT: i am going to stop. feel free to continue. i dont want to get in trouble.(although i probably already am)


search
function
number theory
relatively prime

:?:
 What is the Chicken McNugget Theorem?  Please explain.

ok...this is what i understand of it....i might be wrong...you can use the chicken mcnugget theorem to find the largest number that cannot be produced by a combination of two relatively prime numbers....(legend goes that it originated when some person wondered what number of chicken nuggets they could not get by combining like 4 and 6 pieces) :P ...so the formula goes if the two numbers that are combined are m and n, the formula is m*n -(m+n)...so if you have like two numbers 3 and 7....the largest number that cannot be formed by combining them would be 3*7-(3+7)=11...yup..that's the theorem...personally, i think it should be made a job requirement for every person who works at McDonald's to know this theorem...just in case one of us comes in as a customer and asks this problem :)  :roll:

Please use the search function and AoPSWiki. This has been posted at least 10 times.

Sounds like it's pretty easy, I never saw it on the MC test this year.

It could have been in disguise. I mean it's not like they'd say, "Use the Chicken Nugget Theorem." :D


least common multiple
number theory unsolved
number theory

[quote="ociretsih"]Proof that given three integers $a>b\geq n>1$
\[ (a-b)\vert[1,2,\dots,n]\left(\binom an-\binom bn\right)  \]
where $[1,2,\dots,n]$ denotes the least common multiple of the first $n$ numbers.[/quote]
[color=red][EDIT: the proof below is incorrect!][/color]

Note that $\frac{[1,2,\dots,n]}{n}$ is an integer so
\begin{eqnarray*}[1,2,\dots,n]\left(\binom an-\binom bn\right)&=&\frac{[1,2,\dots,n]}{n}\left(a\binom {a-1}{n-1}-b\binom {b-1}{n-1}\right)\\ &\equiv&\frac{[1,2,\dots,n]}{n}\left(a\binom {b-1}{n-1}-b\binom {b-1}{n-1}\right)\\ &\equiv&\frac{[1,2,\dots,n]}{n}\binom{b-1}{n-1}(a-b)\equiv 0 \mod a-b \end{eqnarray*}
and the result follows.

There is an error in the above argument by Xixas.
How does it pass from $a\binom{a-1}{n-1}$ to $a\binom{b-1}{n-1}$ in the congruences?
If you follow the argument, it apparently proofs that $n\left(\binom an-\binom bn\right)$ is a multiple of $a-b$, but this is wrong!
The divisibility relation is strict, I mean that if $d$ is a proper divisor of $[1,2,\dots,n]$ then there are integers $a,b$ such that $(a-b)\nmid d\left(\binom an-\binom bn\right)$.


Determine all values of x for which $ tan(x)=tan(x+10)tan(x+20)tan(x+30)$

I am really baffled by this question. Through a hefty amount of algebraic manipulation I obtained:

$ (-2a^{4}+6a^{2})u^{4}-(a^{5}-12a^{3}+11a)u^{3}+2(4a^{4}-14a^{2}+6)u^{2}+(a^{5}-12a^{3}+11a)u-2a^{4}+6a^{2}= 0$

where $ u = \tan{x}$ and $ a = \tan{10}$

Clearly degrees.  And algebra is not really the way to go.

[hide="Try this..."] Rewrite as

$ \frac{\sin x \cos(x+30)}{ \cos x \sin (x+30)}= \frac{\sin (x+10) \sin (x+20) }{\cos (x+10) \cos (x+20) }$

And use product-to-sum formulas. [/hide]

[url=http://imageshack.us][img]http://img211.imageshack.us/img211/5995/tanuh7.gif[/img][/url]
Shot at 2007-07-16


function
number theory proposed
number theory

Find all positive integer solutions of the equation $k!l! = k!+l!+m!.$

Have a look here
http://www.mathlinks.ro/Forum/viewtopic.php?t=116937

Pierre.


This has been posted before too. (or am I missing something  :? )

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=72357]www.artofproblemsolving.com/Forum/viewtopic.php?t=72357[/url]

Silverfalcon, try to discuss with 4everwise what problems to post.  :P


calculus
integration
trigonometry
calculus computations

Evaluate $ \int \allowbreak \frac{x\plus{}1}{\sqrt{x\left( x\minus{}1\right) \left( \sqrt{x}\minus{}x\plus{}1\right) }}dx$

hmm.. are you sure the denominator is typed correctly ? :maybe:

hello, i can't find a primitive function.
Sonnhard.

Sorry Nora and Sonnhard, made a typo. I will repost it now :blush:

Here it goes...

Evaluate $ \frac {x \plus{} 1}{\sqrt {x^{3/2}\left( x \minus{} 1\right) \left( \sqrt {x} \minus{} x \plus{} 1\right) }}$

put   $ t\equal{}\sqrt x\minus{}\frac{1}{\sqrt x}$

Unless I'm mistaken, This would work out nicely if the $ x^{\frac {3}{2}}$ were $ x^{2}$.

We would then have $ t \equal{} \sqrt {x} \minus{} \frac {1}{\sqrt {x}} \equal{} x^{\frac { \minus{} 1}{2}}(x \minus{} 1)$

and $ dt \equal{} \frac {1}{2}x^{\frac { \minus{} 1}{2}} \plus{} \frac {1}{2}x^{\frac { \minus{} 3}{2}}\,dx \equal{} \frac {1}{2}x^{\frac { \minus{} 3}{2}}(x \plus{} 1)\,dx$

$ I \equal{} \int\frac {2\,dt}{\sqrt {t(1 \minus{} t)}}$

$ \equal{} \int\frac {2\,dt}{\sqrt {\frac {1}{4} \minus{} (t \minus{} \frac {1}{2})^{2})}}$

$ \equal{} 2\sin^{ \minus{} 1}2(\sqrt {x} \minus{} \frac {1}{\sqrt {x}} \minus{} \frac {1}{2}) \plus{} c$


trigonometry
AMC
AIME
USA(J)MO
USAMO
geometry
complex numbers

I am planning to take either intermediate counting or trig this fall, and I can't decide which one. I am taking these classes for AIME preparation. On the counting I can answer the first few questions on the AIME and usually I can't answer any questions on trig/complex numbers. However, counting seems to be a much more frequently tested subject on the AIME, which class would you recommend?

Looking at it through the narrow focus of the AIME, the Intermediate Counting class is probably more relevant -- there are usually more counting questions on the AIME than there are questions involving trig or complex numbers.

However, there usually is a trig or complex numbers question on the AIME too.

If you don't know trig/complex numbers at all, you should obviously learn it (like from a text book).  Since most applications of trig/complex numbers at the AIME level are relatively simple (usually you know where to start right away), I would also say counting is more important.  Besides, stupid mistakes occur very often in counting problems.

Apologies for the double post...

I believe I would overwhelm myself if I take two AOPS intermediate clases, especially if its during school time.

Also, I haven't gotten through all the AIME tests yet, but for those of you who have what would you say is the average amount of counting and trig problems you usually see?

Two subject courses would be manageable.  If you want to take more than one class but don't think you can handle two subject courses, try taking a subject course and a problem series.  But it also depends on how much extracurricular stuff you do.  If you don't think that you can fit in 1.5 hrs for class plus a few more to work on message board problems, midterm/final, and review content then only go for one.  Also, when they have a trig problem it [i]usually[/i] is in the last half of the test.  So you will need to have it down solid.

Last year, they only had about one problem that required trig/complex numbers on both AIME's, I believe.  There shouldn't be any more than 2 trig/complex numbers on the AIME going by past years.  Combinatorics is maybe about 2 to 5 problems

I believe taking two classes will be a little overwhelming because I still have past AIME problems to do, along with relearning some material from previous classes I have taken.

Is the trig/complex numbers in AOPS volume 2 good enough so I can answer at least any trig question in the first 10? Or can someone also name another text that is good in trig/complex numbers.

[quote="Calculus"]I also have both the AOPS series and also the art and craft of problem solving, and I will make sure I understand at least all the trig/counting material in the AOPS series.

Also, can I take a class for example trig, and if I feel I would benefit more from counting, can I switch before 10 days?[/quote]

Yes - you can switch if you do so before the drop date.

If making USAMO is your goal, Counting is probably more likely to appear more often on the AIME. Usually Trig problems won't appear in the first 8, and sometimes only appear in the form of a relatively difficult #13, 14, or 15. Trig is very powerful and useful if properly applied (see: people who brute trig a majority of Olympiad geometry problems). This decision also depends on which topic you feel you need more rigor in (Having seen some of the AoPS class finals, I can tell you they are quite rigorous)

just a question...
is the counting course better or is number theory better?
my goal is to make usamo...

I had another similar question. What is more common on the AIME: number theory or trig. Which would help me more the intro NT or intermediate trig. I am thinking to take one of those alongside of intermediate counting, but I am not certain yet.

Sorry for being impatient but I do not want to wait too long after the start date to enroll. Any advice?

If I were you, I would take Intermediate Counting/Probability and Intermediate Trig/Complex.  The things you learn will be more difficult, but if you spend your time on it, you'll learn way more.  Learn your basic number theory from a book.

I'm only doing trig.... counting seems a lot more self-studiable, plus I am awful at Trig (2 questions I could answer in do you need this) Plus, I can answer 60% of the "do you need this?" test, and I haven't even started the probability/counting chapters in AoPS Volume 2. The questions I could not answer are 2, 3, 9, and 10, and all of them except 2 have generating function/summation stuff (which frankly scares me, but I hope I can understand it when I actually to AopS v 2).  Am I making the right choice? :| Or should I still take both? I am fairly busy.

I am also doing trig for the same reasons as 1234567890 :)

I think you should take the Counting class and get a trig book from a libray and look through it. I've found that trig is easier to learn on your own than Counting.
 :roll:

I don't think so. And also, I have a few counting books. ;)


vector
analytic geometry
linear algebra
linear algebra unsolved

[quote="Moubinool"]E an finite R-vector space it is well known
dim(A+B)=dim(A)+dim(B)-dim(A inter B) for A, B sub vertor space of E

Do you some formula or inegality for dim(A(1)+...+A(k)) with k \geq 3
involving dim(A(i)), dim(A(i) inter A(j)), ...., dim(A(i) inter...inter A(q))

for A(1),...,A(k) sub vector space of E

thank's for your reply[/quote]Inclusion-exclusion principle applied to a suitable basis of the vector space sum

[quote="jmerry"]There is no such formula, and there can't be. What is the dimension of the the sum of three lines with pairwise trivial intersection?[/quote]

But only lines passing through the origin are vector spaces, right? So if $A, B, C$ are lines passing through the origin, then

$\dim(A+B+C) = \dim(A+B)+\dim(C)-\dim((A+B) \cap C) \\ = \dim(A)+\dim(B)+\dim(C)-\dim(A \cap B)-\dim(B \cap C)-\dim(A \cap C)+\dim(A \cap B \cap C)$.

Something like that? Which is just PIE.

[quote="jmerry"]In the case I mentioned, your formula gives an answer of 3 (the first three terms are each 1, and all others are zero). We could have put all three lines in $\mathbb{R}^{2}$. Alternately, we could make our lines the three coordinate axes in $\mathbb{R}^{3}$.[/quote]

Oh, I see. It's because

$\dim((A+B) \cap C) \neq \dim(A \cap C)+\dim(B \cap C)-\dim(A \cap B \cap C)$.

Right?


MATHCOUNTS function AMC AIME USA(J)MO USAMO search \/closed

The Art of Problem Solving Summer classes will begin soon.  We are offering six classes this summer.  Details about the courses are listed below.  More information, including how to enroll, can be found by [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]clicking here[/url].

If you have any questions about any of the classes, you can post them here.  More information about the instructors can be found [url=http://www.artofproblemsolving.com/AboutUs/AoPS_A_Company.php]here[/url].

[b]Introduction to Number Theory[/b] (Jun 7 - Aug 23, 7:30 PM - 9 PM ET/4:30 PM - 6 PM PT).  This course covers fundamental principles of number theory, including primes and composites, divisors and multiples, divisibility, remainders, modular arithmetic, and number bases.  The course is taught by Joshua Zucker, former member of the US Physics team, Math Olympiad Program invitee, and past MATHCOUNTS Problem Committee member.

[b]Intermediate Trigonometry/Complex Numbers[/b] (Jun 14 - Aug 30, 7:30 PM - 9 PM ET/4:30 PM - 6 PM PT). This course covers an introduction and evaluation of trigonometric functions, trigonometric identities, complex numbers, exponential form of complex numbers, De Moivre's Theorem, geometric representation of complex numbers, roots of unity.  Taught by Art of Problem Solving founder Richard Rusczyk, who scored a perfect score on the 1989 AIME and was a 1989 USAMO winner.

[b]MATHCOUNTS Problem Series[/b] (Jun 11 - Aug 27, 7:30 PM - 9 PM ET/4:30 PM - 6 PM PT). This course is designed for students preparing for MATHCOUNTS, the premier middle school mathematics contest in the US. The class will cover much of the material needed for success at State and National MATHCOUNTS.  The class is taught by Ashley Reiter Ahlin, a past winner of the Westinghouse Science Talent Search, MATHCOUNTS problem committee member, and top-3 finisher in National MATHCOUNTS (1987).

[b]AMC 10 Problem Series[/b] (Jun 6 - Aug 29, 7:30 PM - 9 PM ET/4:30 PM - 6 PM PT).  This class is designed for students preparing for the AMC 10. The class will consist of discussion of problems from past exams, as well as strategies for taking the test. The AMC 10 is an exam for students in 10th grade and below that serves as a first test in the series of contests used to determine the US IMO team.  The class is taught by 1988 AHSME perfect scorer David Patrick, who also was a USAMO winner in 1988 and holds a PhD in Mathematics from MIT.

[b]AMC 12 Problem Series[/b] (Jun 8 - Aug 24, 7:30 PM - 9 PM ET/4:30 PM - 6 PM PT). This course is designed for students preparing for the AMC 12. The class will consist of discussion of problems from past exams, as well as strategies for taking the test. The AMC 12 is the first exam in the series of contests used to determine the US IMO team.  The class is taught by MathLinks founder and International Math Olympiad medalist Valentin Vornicu.

[b]AIME Problem Series A[/b] (Jun 12 - Aug 28, 7:30 PM - 9 PM ET/4:30 PM - 6 PM PT).  The class will consist of discussion of problems from past AIME exams, as well as strategies for taking the test. The AIME is the second exam in the series of contests used to determine the US IMO team.  The class is taught by 2-time IMO medalist Naoki Sato, who has also served as the deputy director of the Canadian IMO team 3 times.

i'm probably taking the MATHCOUNTS problem series. I got the books too!  :lol:

[quote="daermon"]will there be a trig class in the fall?[/quote]

Almost certainly not.  We may offer it next spring, but that's not certain.

These look good. I'll probably end up taking the NT one.

[quote="Temperal"]Hmmm... ET is the same as EST, right?[/quote]

Since these courses are offered in the summer it'd be EDT.

What do I do if I'm on vacation during the time of the courses? Do major problems arise, or can I just catch up to where the class is at?

And is it possible to take two (or more) courses concurrently?

[quote="stuck"]What do I do if I'm on vacation during the time of the courses? Do major problems arise, or can I just catch up to where the class is at?[/quote]

Not a problem.  Full transcripts are made of each class, so you can catch up on your own time.  Many of our students will miss a few weeks during the summer and catch up when they are back.

[quote="stuck"]
And is it possible to take two (or more) courses concurrently?[/quote]

Yes.  Many of our students do so.

[quote="jhredsox"]just wondering, but if i knew a really good math teacher, could he/she volunteer to help out?[/quote]

If you mean, help with teaching the course, we have plenty of assistants already.  But we're always happy to have more good teachers contribute to the forum in general.

AIME Problem Series A (Jun 12 - Aug 28, 7:30 PM - 9 PM ET/4:30 PM - 6 PM ET).

What does this mean? Does it mean Jun12 -Aug 28 at 7:30 PM-9PM Eastern and at 4:30 PM -6PM Eastern? Or was the second ET supposed to be California time?

Yes, the second time (4:30-6:00) was supposed to be Pacific Time (PT).  It has been corrected.

Are there any books that are recommended to go along with the Mathcounts or the Intermediate Trig/Complex Number course?
Also, when will the next Intro. to Number Theory class be?

For the MATHCOUNTS class: No text required, though we generally recommend our Introduction series of texts and AoPS Volume 1 for students preparing for MATHCOUNTS.

For the Trig/Complex Numbers class: No text required, though select chapters of AoPS Volume 2 would be useful.

The next offering of the Number Theory class will be Winter/Spring 2008 or Summer 2008.  We'll determine that in the next few weeks.

[quote="Phelpedo"]I guess the Olympiad classes still haven't come back yet  :([/quote]

Nope - they've essentially been replaced by WOOT.  We don't plan to bring them back anytime soon.

Hmmm, I might enroll in the Trigonometry/Complex class.

The classes are generally on different days of the week (except for Number Theory and Trig).  So, there's not a time conflict.

The email reminder that AoPS sent stated that,

"The next Art of Problem Solving AIME Problem Series A class is on Tuesday at 7:00 PM ET (4:00 PM PT)."

However, on the first post, it said,

"AIME Problem Series A (Jun 12 - Aug 28, 7:30 PM - 9 PM ET/4:30 PM - 6 PM PT)"

These two times are a half hour apart. Which one is the correct one?

We start the first class of each course a half hour early to give new students a chance to get used to the classroom and make sure everything is working properly.  The actual math instruction will start at the usual time (7:30 PM ET/4:30 PM PT).

[quote="mintybear"]could you please explain how we can take two classes concurrently? aren't the classes all at the same time?[/quote]

no, theyre on different days at different times.  There is only one classroom

[quote="mintybear"]could you please explain how we can take two classes concurrently? aren't the classes all at the same time?[/quote]

Possibly, But they won't be on the same day. (i.e. Mathcounts on Monday, Trig on Tuesday,AMC 10 on Wendesday, Number Theory and Thurday, AMC 12 on Friday)

Too bad that there are no olympiad level classes this summer. I surely would have taken one of them. The olympiad geometry class played a [i]fundamental[/i] role in my problem solving development. Furthermore, I still find myself looking back at problems that I never solved in my olympiad geometry binder [yes I actually printed everything out :P].

[quote="archimedes1"]But Altheman, isn't WOOT basically the same thing?[/quote]

Altheman's looking for a summer class for Olympiad level.

It's not likely we'll bring those back any time soon, but as we write the Intermediate books, some of the Intermediate classes will be expanded to include olympiad material.  For example when the Intermediate Counting & Probability class is expanded, it will include a significant chunk of the material in the Olympiad Problem Solving class.  This will likely occur this fall.

In any case, I think an Olympiad Geometry class would probably cover more than the two or three Geometry weeks in WOOT. Unfortunately I can't say for sure, as I found this site a little bit too late.