linear algebra
matrix
number theory unsolved
number theory

Let $ a$ be a fixed integer. Find all integer solutions $ x,y,z$ of the system:

$ 5x\plus{}(a\plus{}2)y\plus{}(a\plus{}2)z\equal{}a,$
$ (2a\plus{}4)x\plus{}(a^2\plus{}3)y\plus{}(2a\plus{}2)z\equal{}3a\minus{}1,$
$ (2a\plus{}4)x\plus{}(2a\plus{}2)y\plus{}(a^2\plus{}3)z\equal{}a\plus{}1.$

[hide]Take equation #2 and subtract equation #3 to get:
$ (a^2\minus{}2a\plus{}1)(y\minus{}z) \equal{} 2a\minus{}2 \Rightarrow (a\minus{}1)^2(y\minus{}z) \equal{} 2(a\minus{}1)$

Thus, either $ a \equal{} 1$ OR $ y\minus{}z \equal{} \dfrac{2}{a\minus{}1}$ which must be an integer,

So, $ a\minus{}1 \equal{} \pm1 , \pm2 \Rightarrow a \equal{} \minus{}1,0,2,3$.

Therefore, integer solutions $ (x,y,z)$ can only occur when $ a \in \{\minus{}1,0,1,2,3\}$.

$ a \equal{} \minus{}1$ yields:
$ \begin{pmatrix} 5 & 1 & 1 \\ 2 & 4 & 0 \\ 2 & 0 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} \equal{} \begin{pmatrix} \minus{}1 \\ \minus{}4 \\ 0 \end{pmatrix}$ $ \Longrightarrow (x,y,z) \equal{} (0,\minus{}1,0)$

$ a \equal{} 0$ yields:
$ \begin{pmatrix} 5 & 2 & 2 \\ 4 & 3 & 2 \\ 4 & 2 & 3 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} \equal{} \begin{pmatrix} 0 \\ \minus{}1 \\ 1 \end{pmatrix}$ $ \Longrightarrow (x,y,z) \equal{} (0,\minus{}1,1)$

$ a \equal{} 1$ yields:
$ \begin{pmatrix} 5 & 3 & 3 \\ 6 & 4 & 4 \\ 6 & 4 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} \equal{} \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$ $ \Longrightarrow (x,y,z) \equal{} (\minus{}1,2\minus{}t,t)$ (Which is an integer solution iff $ t \in \mathbb{Z}$

$ a \equal{} 2$ yields:
$ \begin{pmatrix} 5 & 4 & 4 \\ 8 & 7 & 6 \\ 8 & 6 & 7 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} \equal{} \begin{pmatrix} 2 \\ 5 \\ 3 \end{pmatrix}$ $ \Longrightarrow (x,y,z) \equal{} (\minus{}6,5,3)$

$ a \equal{} 3$ yields:
$ \begin{pmatrix} 5 & 5 & 5 \\ 10 & 12 & 8 \\ 10 & 8 & 12 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} \equal{} \begin{pmatrix} 3 \\ 8 \\ 4 \end{pmatrix}$ $ \Longrightarrow (x,y,z) \equal{} (\minus{}\frac{2}{5}\minus{}2t,t\plus{}1,t)$ (Which is not an integer solution for any $ t \in \mathbb{R}$.)

Therefore, the solutions are:
$ a\equal{}\minus{}1$: $ (x,y,z) \equal{} (0,\minus{}1,0)$
$ a\equal{}0$: $ (x,y,z) \equal{} (0,\minus{}1,1)$
$ a\equal{}1$: $ (x,y,z) \equal{} (\minus{}1,2\minus{}t,t)$ for all $ t \in \mathbb{Z}$
$ a\equal{}2$: $ (x,y,z) \equal{} (\minus{}6,5,3)$ [/hide]


Let the shaded square's side length be $ x$. By isosceles right triangles, the side length of the large square is $ 2\sqrt{2}x$. Ratio of side length of small square to side length of large square is $ \frac{1}{2\sqrt{2}}$, so the ratio of their areas is the square of that, or $ boxed{\frac{1}{8}}$.


AMC
USA(J)MO
USAMO
AIME
USAJMO
AMC 10
AMC 12

This message is to announce changes in the structure and selection
process for the 2010 USAMO.

On May 18, 2009, the MAA Executive COmmittee formally approved the attached proposal from the Committee on the American Mathematics Competitions.

In short, the changes are:

The current USA Mathematical Olympiad administered to about 500
students in all grades will be split into two flights:

1. The USA Mathematical Olympiad administered to about 270 students
who qualify from the AMC 12 with a top USAMO index based on their AMC
12 score. (along with a few students who took only the AMC 10
and scored 11 or more on the AIME.)

2. The USA Junior Mathematical Olympiad administered to about 230
students who qualify from the AMC 10 with a top USAMO index based on
their AMC 10 score.

3.  The current "floor score"  system will no longer be used.

In practical terms this means:
1.  For 11th and 12th graders, there is essentially no change.

2.  Students who want to qualify for the USAMO should take the AMC 12,
since the AMC 12 is the gateway to the USAMO.  (The exception is
students who take the AMC 10 only but score 11 or more on the AIME.
Evidence from the last few years indicates that this is typically
about 5 students.  These students later excelled at all levels of the
AMC program, and so based on this evidence we have built in this
exception.)

3.  Students in 10th grade and below who want a chance at the USAMO
should take the AMC 12, but may also want to hedge their chances by
taking the AMC 10.  Students in 10th grade and below who take the AMC
12 will have their AMC 12-based USAMO index considered without
consideration of age or grade or AIME score.  Of course this means
they are considered with 11th and 12th graders and compete for the
approximately 250-270 USAMO spots on AMC 12 index alone.  Students in
10th grade and below who qualify for the USAMO are eliminated from the
pool of AMC 10 takers competing for the 230 invitations to the USA Junior
Math Olympiad.

4.  The gateway for the USA Junior Math Olympiad is the AMC 10.  This
automatically limits participants in the Junior Math Olympiad to
grades 10 and below.

5.  The USA Junior Mathematical Olympiad would have 6 problems and be
administered over 2 days, the same as the USAMO.  Problems J1,J2 on
Day 1, and Problems J4, J5 would be different from the USAMO, and
would be close in level and content to problems 13-15 on the AIME.
Problem J3 would be the same as Problem 1 (Day 1) on the USAMO,
Problem J6 would be the same as Problem 4 (Day 2) on the USAMO.  All
six problems on the USA Junior Mathematical Olympiad would require
written answers, perhaps a detailed algebraic, number theoretic,
combinatoric or geometric solution rather than a proof.  Problems
would be graded on the same 7 point scale (42 points total) as the
USAMO, and have the same rigorous grading.

6. Selection to "Red MOSP" will work in the following way (assuming
funding for Red MOSP continues.)  Among all 9th graders taking the
USAMO and JMO, the scores on problems J3 = 1 and J6 = 4 will be
totalled and ordered.  Then we will select the students with the top
25 scores on these two problems.  Ties among the top scores in each
category will be broken by using the scores from the other 4 problems
on each Olympiad.

So if someone hypothetically qualified for both JMO and USAMO, can they choose which one to take or is it left to the guidelines?

After discussing that issue, we decided that: No, if someone qualifies for both USAMO and USAJMO, they will be invited to the USAMO only.  See the last sentence of point 3:
[quote]Students in
10th grade and below who qualify for the USAMO are eliminated from the
pool of AMC 10 takers competing for the 230 invitations to the USA Junior
Math Olympiad. [/quote]
We want students working at the best of their abilities.

Steve Dunbar
MAA Director, American Mathematics Competitions

Out of curiosity (and to get a vague general idea), what would it have take on the AMC 10/12 and AIME to qualify for the USAJMO and USAMO respectively this year (had this new system been in place this year)?

EDIT: Also, a question.

Clearly, if 1/4 are too hard on USAMO, they will be on USAJMO too. If students can solve problems that are of the approximate difficulty of 13-15 on the AIME, don't you think they will be able to score an 11 or higher? That is, won't the problems on the USAJMO be either too easy or too hard to everyone who qualifies?

A quick skim of the data suggests that it would have taken an index of approximately 210 on both the
AMC10+10*AIME -> USAJMO and AMC12+10*AIME -> USAMO
this 2009 contest year.

Steve Dunbar
MAA Director, American Mathematics Competitions

So a tenth grader who wants to try to make USAMO and MOSP most take the AMC12?

[quote]Clearly, if 1/4 are too hard on USAMO, they will be on USAJMO too. If students can solve problems that are of the approximate difficulty of 13-15 on the AIME, don't you think they will be able to score an 11 or higher? That is, won't the problems on the USAJMO be either too easy or too hard to everyone who qualifies?[/quote]

The item difficulty on the both the AIME and the USAMO as well as analysis of scores by grade indicates that neither of those concerns is valid.  Students solving some or all of problems 13-15 on the AIME should be scoring 11 or higher on the AIME and will be taking the USAMO, in any case.

Finally, don't you trust the USAMO Committee,  all of them highly qualified mathematicians, most of them MO veterans and IMO veterans, and with years of experience in creating problems, tests, and contests including the USAMO and AIME to carefully estimate and calibrate the difficulty of the contests?  My intention is to indicate the relative difficulty level in relation to something most readers on this forum will recognize.  I am sure that the USAJMO will be a satisfactory experience for those who qualify.

Steve Dunbar
MAA Director, American Mathematics Competitions

Since I'm going to be a senior next year, the same number of spots are still here, so I guess I'm really not affected by this. Just gotta hit the books now and qualify :]

I think this is a very thoughtful and creative approach.

The USAMO was originally designed for roughly 100+ students, of whom 8 would be designated "winners" and another 20 or so of the most promising students in 11th grade and below would be invited to the Math Olympiad Summer Program.  Moreover, all qualifiers were selected from a single qualifying path (originally just the AHSME, later just a single version of the AHSME and a single version of the AIME.)

Expecting a single test with six questions to be appropriate for 500 students, some as young as 4th grade (!), selected by a combinatorial explosion of paths (10A/12A/10B/12B/USAMTS with multiplicity possible and AIME I or AIME II) is really a lot to ask!  The goal of identifying this year's team and very strong prospects for teams in the next two years (i.e., black/blue MOP) require different questions than the goals of encouraging and developing the talents of younger students with much less experience.

There are no perfect solutions, but I like this idea a lot, in principle.  The junior math olympiad strikes me as having great possibilities as a "stepping stone" for younger students with less experience.

Stepping stones yes, but if students are able to do that well that young, they shouldn't be worried about having "too hard" of a USAMO. The USAMO is the most prominent competition in America (for high schoolers) for a reason, by adding a newer, easier level, I am of the opinion we give some of these younger kids a disservice. Rather than push them to work harder and master more difficult problems, we are giving them easier, AIME level problems, which they could get from elsewhere (USAMTS perhaps?)

Also, now qualification for red MOP has been reduced from 6 problems to 2. Is that a good thing?

Also, is there still the state best rule? That is, the highest index in the state? So now will there be [i]two[/i] of those? That is, will there be a qualifier for both USAJMO and USAMO solely based on the fact they live in a less competitive state?

How will CGMO selection work? Will it be done similarly to the Red cutoff? It doesn't affect me personally, but it hasn't been addressed yet.

[quote="MellowMelon"][quote="CircleSquared"]Could you please enlighten me about these "qualification problems"?  The way I see it, narrowing down the selection criteria from 6 problems to 2 makes the system more based on luck and unstable, and would result in Red MOP qualifiers who potentially are farther from the top 25 9th graders than the qualifiers by the old system.[/quote]
Qualification problems means the issue of taking the AMC10 being beneficial to qualifying for USAMO, which is the topic that has been hotly debated on this forum.

[/quote]

Yes, we have solved the problem of taking the 10 or not if you want to take the Junior, but now 10th graders who want to make MOP, the real goal of the whole process, now must take the AMC 12.

I am hijacking D2 (daughter #2)'s account to post here.

[quote]but now 10th graders who want to make MOP, the real goal of the whole process, now must take the AMC 12.[/quote]

That is not true. A 10th grader can still take AMC10 but score 11 and above on AIME will still qualify for USAMO and have a short at MOP.

No system is absolutely right for every occasion. If an underclassman wants to qualify for USAMO (which definitely sounds better than qualifying for USAJMO) should choose to take AMC12.

Like bookaholic, D1 (just finish her freshman year in college), always did AMC12 in high school. She did AMC10 incidentally in 8th grade (first time knew there was such thing like AMC) and never took AMC10 again. D2 is about finishing her 7th grade and took AMC10 this year. This is the first time that her middle school ever participated in AMC10 (at my suggestion and support). I will suggest to her middle school to do both AMC10 and AMC12 next year.

One solution is: students should push their schools to participate in both AMC/A and AMC/B, then an underclassman would have opportunity to do both 10 and 12 and have chance to qualify USAMO (if made the index for USAMO) or USAJMO (if made the index for USAJMO).

[quote="CatalystOfNostalgia"][quote="AMCDirector"](assuming funding for Red MOSP continues.)[/quote]

On a tangent, this is pretty ominous. Is this likely not to happen?[/quote]
This was probably more of a reminder that the funding is a source outside of the MAA, so it's not something they can guarantee. They probably don't know or have an idea if the money is going to get pulled from them or not, but it's probably safe to count on it continuing.

I'm about to do some splitting on this thread.

[quote="Kent Merryfield"][quote="mathemonster"]this basically makes the amc10 useless[/quote]
I can't let this comment pass unchallenged.[/quote]

Lol.  Troll fed.

This is a very interesting decision.  I do not like the possible departure from proof though.  I think that all the problems should be proofs, and if you want an easier contest, you should just ask easier proof questions.

^ They are proof question, but at 13-15 AIME [i]level[/i] Is this definite?[/i]

Actually,
[quote="AMCDirector"]All
six problems on the USA Junior Mathematical Olympiad would require
written answers, perhaps a detailed algebraic, number theoretic,
combinatoric or geometric solution rather than a proof.
[/quote]

1) Can 10th graders qualify for blue MOP through USAJMO?
2) If so, how will these 10th graders be compared to those (10th graders) who took the USAMO in terms of MOP qualification?

Questions about the USAJMO question format have been addressed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=280817]here[/url] by AMC officials.

mathking: I have not seen this answered for sure. But if USAJMO contestants could qualify for blue MOP the whole idea would be laughably stupid. Based on what the USAJMO difficulty sounds like, most blue MOP qualifiers should be able to get a perfect on it, and people not at blue level would probably be able to as well.

[quote]1) Can 10th graders qualify for blue MOP through USAJMO? [/quote]

No.

[quote] If so, how will these 10th graders be compared to those (10th graders) who took the USAMO in terms of MOP qualification?[/quote]

By the answer to 1), not applicable.

Steve Dunbar
MAA Director, American Mathematics Competitions

As someone who essentially qualified for the USAMO in 9th grade [i]because[/i] of the expansion and who met the blue mop cutoff (although was placed in red mop, which I feel was better for me), I am somewhat biased against this change. Even as a ninth grader, I felt like the AMC10 was uninteresting, having gotten a 146.5 on it the year before (that was before the 2.5->1.5 score change). Therefore, I only took the AMC12 my freshman year, even though I realized that it may hurt my chances of making USAMO. I ended up with a 144/6, for a 204 index. The cutoff that year was 197.5 (If I remember correctly), but if the expansion were not in place, I feel like I would not have qualified. What worries me is that I took the AMC12 because I find easy problems uninteresting, not because it would help qualifying for any future exam, so I am not sure that I would have taken the AMC10 at all even with this new rule. I would strongly object to having no way to qualify for USAJMO having only taken the AMC12. Actually, I am worried that you might accidentally miss some 10th graders who are worthy of blue mop (as would have been the case if I had been a year older), whereas with the expansion this is pretty much impossible.

I do realize that I am by far the exception, and you won't miss more than one or two such people, but I definitely feel that it's doing a disservice to that minority of underclassmen who would want to take the AMC12 rather than the AMC10 for the interesting problems and are worthy of making USAJMO but not necessarily USAMO, or those ninth graders who are ready for red mop but miss the USAMO cutoff and didn't try for the USAJMO cutoff.

I took 2 AMC12s both 9th and 10th grade for the same reason.  I didn't qualify in 9th though I got an 8 on the AIME.

I think it should be at least part of the objective of the selection processes to have people taking the most challenging tests they will do reasonably well on.  I see no reason for someone who will get a 0 on the USAMO to take it instead of an easier test, but that isn't true on AMC10/12 for anyone with a remote chance of making either Olympiad.  We can all do the AMC12, and the selection process shouldn't make us waste our time on the AMC10.

(For the record, I'll be an 11th grader next year and be mostly unaffected by the exact process.)

There is no possible solution that will leave everybody happy. The AMC committee or whatever attempts to find the solution that satisfies the most people. This is what they came up with. Obviously, there will always be some people who don't like their decisions. Personally, I think it is a good move. I will be in 7th grade and will take the AMC 12 next year based on this change. This is what the AMC committee is looking for. They are encouraging taking the AMC 12.

Does this mean that AMC 12 is gonna get much harder?(Since we will only see AMC 12 index)

Because 10th graders in USAJMO can't qualify for blue MOP, doesn't that mean that USAJMO is essentially a dead end for 10th graders?  :|

[quote="AMCDirector"]As of this writing, there are now 50 posts in this thread filled with conjecture, speculation, questions, opinions and counter-opinions.  This is too many, too diverse, and too cacophonous to respond to intelligently in this thread.

If you have a question about either a fact or a procedure related to the 2010 USAMO or the 2010 USA Junior Math Olympiad, please begin a new thread and ask your question in that new thread. (One question per thread for simplicity and clarity, please).   Either I or AMCFinance (E. Claassen) will answer your question in that thread for all to see.  Thank you.

Steve Dunbar
MAA Director, American Mathematics Competitions[/quote]

After Mr. Dunbar made this comment, there have been entirely one another page about it. I think that the discussion will continue, so I'm going to lock the thread and ask everyone to ask questions separately. This is, after all, an announcement thread.


function
complex analysis
calculus
calculus computations

How to prove that Taylor's expansion is valid to expand a function?

Thank you for helping. :P

You can prove for example that it's Lagrange rest tends to zero.

[quote="sylow_theory"]Here is a less elementary way.

Say you have $ f(x)$ a real function on a set $ (-R,R)$ for $ R>0$.

If you can find a holomorphic complex function $ g(z)$ in the open disk $ \{|z|<R : z\in \mathbb{C}\}$ such that $ f(y) = g(y)$ for all $ y \in (-R,R)$ then it turn out amazingly that $ f(x)$ [b]must[/b] be infinitely differentiable and furthermore is analytic.

So for example we can to show that the Taylor series for $ f(x)=e^{x}$ works for $ (-1,1)$ then consider $ g(z)=e^{z}$ and we notice that $ f(x) = g(z)$ when the variable is real. And since $ g(z)$ is holomorphic in the open disk or radius 1 centered on the origin it means $ e^{x}$ is equal to its Taylor series.[/quote]

Very interesting indeed, thank you sylow_theory. :P

[quote="IrinaSharova"]
Very interesting indeed, thank you sylow_theory. :P[/quote]
Oops... "holomorphic" means "differentiable" just with complex numbers.
(I should have said that, since maybe you never learned complex analysis).

Indeed it is, it is one of the the two most amazing thing I found in Complex Analysis.
1)Cauchy's Theorem: Which states integrating in a closed loop a differentiable functions is always zero.
2)And this. That holomorphic functions are analytic functions (analytic means "having a Taylor series").

The real amazing think about #2 is that if a complex function is differentiable once, then it is infinitely differentiable! And furthermore is analytic. I just love that so much.


Each term of the sequence X_1, X_2,      X_n is either +1 or -1

Prove that the equality (X_1)(X_2) +(X_2)(X_3) +...........(X_(n-1)(X_n) +
(X_n)(X_1)    = 0  is possible only when n is divisible by 4.

This really is classical :).

clearly, $n=2k$ because that sum is $0$, so there are equal numbers of $1$'s and $-1$'s among $x_ix_{i+1}$ (consider $x_{n+1}=x_1$). Now look at the product of those $2k$ numbers. On the one hand, we get $(-1)^k$, and on the other hand we get $1$, because it's the product of the squares of $x_i$, so $(-1)^k=1$, meaning that $2|k\Rightarrow 4|n$.

I'm not sure if I completely understand grobber's solution.  I think that this might be exactly what he was saying, but here's my solution:

We must have $n = 2p$ as he said.

Now, consider the number of pairs of $1\cdot 1$ and $-1\cdot -1$, which is clearly equal to the number of pairs with $1\cdot -1$ (otherwise the sum would not be $0$).  There are clearly $p$ pairs of the form $1\cdot -1$

Now take $\displaystyle \prod a_ia_{i+1}$.  On one hand, by the above logic, they are equal to ${-1}^p$.  On the other hand, they are equal to ${-1}^{2p}$ because each number is counted twice.  This implies that $p\equiv 2p \bmod 2$, or that $p=2q$, so that $n=4q$ as desired.

And a generalization.

Prove that if $a_i=\pm 1$, $0\leq i\leq n$ (with indices modulo n), and $\displaystyle \sum^{n}_{i=1} a_ia_{i+1}a_{i+2}a_{i+3}=0$, then $4\vert n$.

And if you've seen it in Engel, please refrain from posting that solution  :cool:


[quote="mathwizarddude"]How did you get that common factor of $ 11$ (by hand)?[/quote]

To answer this question I am asking you a question: Do you remember divisibility test for 11 ?

clue: add up the 1st, 3rd, 5th, etc. digits in a number, and then compare with the even set (2nd, 4th, 6th, etc.)

The divisibility rule for 11:

Add up the 1st, 3rd, 5th, etc. places.

Then add up all the others (2nd, 4th, 6th, etc.)

Find the difference. If it is divisible by 11, then the whole number is divisible by 11.

[hide="my favorite divisibility rule for 11"]
Where's my TI-84? :lol:  :P  :rotfl: [/hide]

is so-called 'Extended Euclidean Algorithm' faster than normal disivibility rule of 11?

I think the divisibility rule is faster, but if it isn't 11, then you should use the Extended Euclidean Algorithm.

izzy is right how do you know you did not miss a factor?


superior algebra
superior algebra solved

The topological space of prime ideals in $\mathbb{Z}[X]$.

Ask more definite questions please.

I am sorry.what is the structure of spec(Z[X])?It is the 16th execrise in chapter 1 of Atiyah and Macdonald Introduction to Commutative Algebra.

jiulai $Spec(\mathbb{Z}[X])$ was defined on exercise 15 of the same book Atiyah and Macdonald Introduction to Commutative Algebra.

Take a look!

You will find that

$Spec(\mathbb{Z})=\{(p), p\in\mathbb{Z}\ \ \text{prime}\}$
$Spec(\mathbb{R})=\varnothing$
$Spec(\mathbb{C}[X])=\{(f(x)), f(x)\in \mathbb{C}[X] \ \ \text{of degree}\ \ 1\}$
$Spec(\mathbb{R}[X])=\{(f(x)), f(x)\in \mathbb{R}[X] \ \ \text{irreducible}\}$
$Spec(\mathbb{Z}[X])=\{(f(x)), f(x)\in \mathbb{Z}[X] \ \ \text{irreducible}\}$

You always forgot the prime ideal $\{0\}$!

And the fifth is wrong, that would imply that $\mathbb{Z}[x]$ is a PID.


The simplest form of $ 1 \minus{} \frac{1}{1 \plus{} \frac{a}{1 \minus{} a}}$ is:

$ \textbf{(A)}\ {a}\text{ if }{a\not\equal{} 0} \qquad
\textbf{(B)}\ 1\qquad
\textbf{(C)}\ {a}\text{ if }{a\not\equal{} \minus{}1}\qquad
\textbf{(D)}\ {1 \minus{} a}\text{ with not restriction on }{a}\qquad
\textbf{(E)}\ {a}\text{ if }{a\not\equal{} 1}$

[hide]
Time to abuse multiple choice:
The only restriction on is that a != 1 (or else we get a divide by 0 error). Hence,
$ \boxed{\text{E}}$[/hide]

[hide="Solution w/o using MC"]
$ 1\minus{}\frac{1}{1\plus{}\frac{a}{1\minus{}a}} \equal{} 1 \minus{} \frac{1}{\frac{1 \minus{} a \plus{} a}{1 \minus{} a}} \equal{} 1 \minus{} \frac{1}{\frac{1}{1 \minus{} a}} \equal{} 1 \minus{} (1 \minus{} a) \equal{} a$. We can't have $ a \equal{} 1$ in the original expression, so our final answer is $ \fbox{(E)}$.
[/hide]

of course answer is a. if a isn't 1. because 1+a/1-a=1/1-a, then this is equal to 1-1+a=a!

It looks there is a typo for answer choice $ \mathbf{(D)}$. Should it read $ 1 \minus{} a \text{ with no restriction on } a$?


Mac OS 10.3.9
Firefox 2.0.0.2
Mathlinks board

I was viewing the MOEMs forum and got a very odd bold.  It doesn't look like mz94 did it himself.  Is this odd?

I've seen this myself (10.4.n, Firefox 2.0.n). It's a bug in either Mac OS X or Firefox for Mac- not something that Valentin can do anything about.

You can get rid of it by forcing the system to redraw the page; scrolling is usually the simplest way to do it.


I have been reading the following page http://en.wikipedia.org/wiki/Bilinear_interpolation about bilinear interpolation. The method is very simple. One interpolates linearly in x, after in y (or vice-versa), according to proportions given by the distances in x and y from the point to be interpolated to its neighbour points. What I am in doubt is if it is a restriction of this method that should always exist four points "around" the point to be interpolated. For example, if just two points were neighbours at left the other two points at right would have to be extrapolated and then used in the desired point to be interpolated?

Bilinear interpolation creates an approximation of the form:

$ f(x,y) \approx c(x\plus{}h)(y\plus{}k)\plus{}b \equal{} a_1 x \plus{} a_2 y \plus{} a_3 xy \plus{} a_4$

There are thus 4 variables to be determined, and four points will generally be able to determine such an interpolation.

Generally as in, as long as no three are collinear. If three or four are collinear, the points offer redundant information, and cannot uniquely determine such a function.

In this case, hope that (x,y) of the point you wish to estimate $ f$ is collinear with the (x,y)'s of the sample points that are collinear; if so, you can just use normal linear interpolation and you're home free. If not, you're sunk.

There will be no collinear points as I am working with a regular grid. The problem is when I want to interpolate values in the grid that are in its border when changing the grid resolution, as in the border the number neighbours are not the same as for cells "inside" the grid.


Sorry,I want to know where is solution of this problem:a/sqrt(a+b)+b/sqrt(b+c)+c/sqrt(c+a)<=5sqrt(a+b+c)/4.
Sorry because i did'n remember Tex

See http://www.mathlinks.ro/Forum/viewtopic.php?t=28276 .

  darij


probability
linear algebra
matrix
ratio
absolute value
combinatorics unsolved
combinatorics

Consider the system in $x,y$:

$ax + by = e$

$cx + dy = f$

with $a,b,c,d,e,f \in Z^+$.

a) Prove the system has only 1 solution (not necessary integer) $\forall e,f$
if and only if: $ad \neq bc$.

b) Suppose to choose at random $a,b,c,d,e,f \in Z$, with absolute value $\leq n$,
with $n \in Z^+$ choosen.
Prove that the probability $p$ that the system has exactily 1 solution (not necessary integer) is:
$1 - \frac{1}{2n} < p < 1 - \frac{1}{3n^2}$

 :read:

for a) I got x = (ed - bf) / (ad - bc) and y = (af - ce) / (ad - bc), therefore x and y are undefined if ad = bc.

Do you know how to solve b)?

Ok, so (a) is really easy, either just solve for $x$ and $y$, write it as a 2x3 matrix and show that cramer's rule gives a denominator that would be zero.

b) is tougher, but here's an attempt at a solution
[hide]
the choice of $e$ and $f$ are irrelevant, because they are not used in the restriction for the uniquness of the system. The restriction seems very complicated until we come up with another way to right it:
$\displaystyle\frac{a}{c}=\frac{b}{d}=k$
where $k$ is a constant

If $\mid a\mid,\mid b\mid<n$, then $\mid a-b\mid<2n$
The same follows for $c$ and $d$, that $\mid c-d\mid<2n$,
By the laws of ratios, $k=\displaystyle\frac{a-b}{c-d}\\$ Without a loss of generality, let$\mid a-b\mid\ge\mid c-d\mid\\$
Now we see that $k$ must be greater than 1.

$p=1-q$, where $q$ is the probability that the solution set will not be unique.
$q=\displaystyle\frac{\alpha}{n^\beta}$, where $\alpha$ is the number of ways that the solution cannot be unique, and where $\beta$ is some integer between 1 and 4.
[/hide]
Unfortunately i do not know enough about probability to take this any further. See if someone can go from here.

[quote="gopherhole112"]
If $\mid a\mid,\mid b\mid<n$, then $\mid a-b\mid<2n$
The same follows for $c$ and $d$, that $\mid c-d\mid<2n$,
By the laws of ratios, $k=\displaystyle\frac{a-b}{c-d}\\$ Without a loss of generality, let$\mid a-b\mid\ge\mid c-d\mid\\$

[/quote]

there's a problem in the TEX code...

I have a solution, I'll post it into some days... Bye


Prove that in any way we choose $ 504$ distinct natural numbers less or equal than $ 2007$ we can always find $ 2$ of them
which have their difference equal to $ 2$,$ 3$,$ 5$ or $ 8$.

[hide="hint"]Let the set in question be $ S$.  Go about it through contradiction and define $ S_k\equal{}\{s\minus{}k|s\in S\}$.  What is $ S_3,S_5,S_8$?[/hide]


inequalities
function
logarithms
calculus
derivative
inequalities proposed

$ \bigstar$ Problem:
Let $ a,b,c,d,e,f$ be non-negative real numbers. Prove that
\[ \sqrt {\frac a{a \plus{} b}} \plus{} \sqrt {\frac b{b \plus{} c}} \plus{} \sqrt {\frac c{c \plus{} d}} \plus{} \sqrt {\frac d{d \plus{} e}} \plus{} \sqrt {\frac e{e \plus{} f}} \plus{} \sqrt {\frac f{f \plus{} a}}\leq{3\sqrt2}.
\]
Hope that I have not made mistake in created it  :P

[hide="Hint"]By Jensen's Inequality for the concave function $ f(x_i)\equal{}\dfrac{1}{\sqrt{1\plus{}x_i}}$
$ \left( \ Let x_1\equal{}\dfrac{b}{a}, \ x_2\equal{}\dfrac{c}{b}, \ x_3\equal{}\dfrac{d}{c}, \ x_4\equal{}\dfrac{e}{d}, \ x_5\equal{}\dfrac{f}{e}, \ x_6\equal{}\dfrac{a}{f} \right)$[/hide]

Zaizai-hoang, I think there's no place for hope in here :)
Anyone can notice that for $ \frac{a}{b} \equal{} \frac{b}{c} \equal{} ... \equal{} \frac{e}{f} \equal{} k$, the left-hand side's limit as $ k$ tends to infinity is $ 5 > 3\sqrt{2}$.
Now here's the rigorous explanation:
[hide="My approach"]As Kouichi Nakagawa noticed, we may generalize this to:
$ \sum \frac{1}{\sqrt{1 \plus{} x_k}} \le \frac{n\sqrt{2}}{2}$,
for any positive $ x_k$ with product $ 1$ and $ n \ge 2$ a positive integer.
Consider the function $ f : \mathbb{R} \rightarrow \mathbb{R}_\plus{}^*$, given by the law
$ f(t) \equal{} \frac{1}{\sqrt{1 \plus{} e^t}}$.
Now we notice that $ f$ is concave on $ (\minus{}\infty, \mathrm{ln} 2]$ and convex on $ [\mathrm{ln} 2, \infty)$. Now let $ a_i \equal{} \mathrm{ln}(x_i)$. The sum of all $ a_k$'s is constant (equal to $ 0$). However, by LCRCF theorem we know that the maximum of the left-hand side, which is $ \sum f(a_k)$, is attained for $ n \minus{} 1$ equal $ a_k$'s, let them be $ a_1 \equal{} ... \equal{} a_{n\minus{}1} \equal{} a \le b \equal{} a_n$, with $ (n\minus{}1)a \plus{} b \equal{} 0$.
Now consider $ x \equal{} e^a$ and $ y \equal{} e^b \equal{} \frac{1}{x^{n\minus{}1}}$, with $ 0 < x \le 1 \le y$. It remains to be shown
$ \frac{n \minus{} 1}{\sqrt{1 \plus{} x}} \plus{} \sqrt\frac{x^{n\minus{}1}}{1 \plus{} x^{n\minus{}1}} \le \frac{n\sqrt{2}}{2}$. However this is false because the left-hand side is greater than $ \frac{n \minus{} 1}{\sqrt{1 \plus{} x}} > \frac{n}{\sqrt{2}}$, which for $ n \equal{} 6$ (the topic problem's case) can be checked as true for $ x < \frac{7}{18}$.
Consequently the problem is not correct.
However, there is a related result, namely Vasc's well-known
$ \sum \sqrt{\frac{2a}{a \plus{} b}} \le 3$.[/hide]

Thanks for your help    :)  Here is wrong solution, but I can not find out my mistake. Hope someone will help me  :wink:
By substitution $ b/a\equal{}x$, $ c/b\equal{}y$, $ d/c\equal{}z$, $ e/d\equal{}t$, $ f/e\equal{}u$, $ a/f\equal{}v$, we have $ xyztuv\equal{}1$ and the inequality takes the form
\[ \frac1{\sqrt{1\plus{}x}}\plus{}\frac1{\sqrt{1\plus{}y}}\plus{}\frac1{\sqrt{1\plus{}z}}\plus{}\frac1{\sqrt{1\plus{}t}}\plus{}\frac1{\sqrt{1\plus{}u}}\plus{}\frac1{\sqrt{1\plus{}v}}\leq3{\sqrt2}.\]
Consider the function
\[ f(w)\equal{}\frac{\sqrt2}2\minus{}\frac1{\sqrt{1\plus{}w}}\minus{}\frac{\sqrt2}8\ln w.\]
Calculating the derivative of the function, we obtain
\[ f'(w)\equal{}\frac{4w\minus{}\sqrt2(w\plus{}1)^{3/2}}	{8w(w\plus{}1)^{3/2}}.\]
The equation $ f'(w)\equal{}0$ has two roots $ w\equal{}1$, and $ w\equal{}2\plus{}\sqrt5$. Then we have
\[ f(w)\geq \min\left[f(1), f\left(\frac{96}{100}\right)\right]\equal{}0.\]
It follows that $ f(w)\geq 0$ for all $ 0\leq w\leq \frac{96}{100}$. This implies that the desired inequality is true for all $ 0\leq x,y,z,t,u,v\leq\frac{96}{100}$. Otherwise, for $ x,y,z,t,u,v\geq\frac{96}{100}$, we have
\[ f(x)\plus{}f(y)\plus{}f(z)\plus{}f(t)\plus{}f(u)\plus{}f(v)\leq \frac6{\sqrt{1\plus{}\frac{96}{100}}}\leq 3\sqrt2.\]
The proof is complete. But I am really strange because of your comment... Where is the mistake in my proof  :)

[quote="zaizai-hoang"][...] This implies that the desired inequality is true for all $ 0\leq x,y,z,t,u,v\leq\frac {96}{100}$. Otherwise, for $ x,y,z,t,u,v\geq\frac {96}{100}$ [...] [/quote]
And then, what happens when not all the variables are on one side of $ \frac{96}{100}$ ?


An artist has 50 inches of oak trim to frame a painting.  The frame is to have a border 3 inches wide surrounding the painting.

If the painting is rectangular with a length twice its width, what are the dimensions of the painting?  What are the dimensions of the frame?

[quote="Concentric"][quote="chickendude"]Can you please stop creating so many different topics?

If you have several questions, compile them into a list and make one big topic with 5-10 questions.
That would be much better than having the entire topic list be populated by your questions

thank you

[hide]
the dimensions are
W x 2W

the perimeter is 6W
because the border is 3 inches wide, we have to add 3 inches for each corner, or 12 inches

6W + 12 = 50
6W = 38
W = 6.33333
[/hide][/quote]

Wait, wouldn't it be 6x+24 because of the 3 inches on each side? so wouldnt x equal 4 and 1/3?
or did i screw up[/quote]
I think that you can just put one 2x+6 in. piece of wood on the top and bottom so the two sides would only need x in. wood each (the width of 3 in. on top and bottom takes off 6 in. from left and right sides). 2*(2x+6)+2*x=6x+12


modular arithmetic
number theory proposed
number theory

[color=Red]Prove that for no $ a,b,c \in \mathbb{N}$

$ \frac{a^2\plus{}b^2\plus{}c^2}{3(ab\plus{}bc\plus{}ca)}$

is an integer.
(sorry if it has been posted before :oops:)[/color]

it is a nice application of jacobi symbol.I will post my solution later because there are some others who are still trying.. :wink:

For $ a\in\mathbb{N}^*$ and $ p$ is an odd prime, let $ \left( \frac{a}{p} \right)$ denote the Legendre's symbol.

Assume the contradiction that there exists $ k\in\mathbb{Z}$ such that $ \frac{a^2 \plus{} b^2 \plus{} c^2}{3(ab\plus{}bc\plus{}ca)} \equal{} k$ then $ (a\plus{}b\plus{}c)^2 \equal{} (3k\plus{}2)(ab\plus{}bc\plus{}ca)$. Let $ p$ be a prime such that $ p \equiv 2 \pmod{3}$ and $ p^{2s\plus{}1} \parallel 3k\plus{}2$, we have $ p^{2s\plus{}1} | (a\plus{}b\plus{}c)^2$ then $ p^{2s\plus{}2} | (a\plus{}b\plus{}c)^2$ and hence $ p | a\plus{}b\plus{}c$ and $ p | ab\plus{}bc\plus{}ca$. Therefore, $ ab \plus{} c(a\plus{}b) \equiv ab \minus{} (a\plus{}b)^2 \equiv 0 \pmod{p}$, meanwhile $ p | a^2 \plus{} ab \plus{} b^2$. From this we can deduce that $ p | (2a\plus{}b)^2 \plus{} 3b^2$ and so $ \left( \frac{\minus{}3}{p} \right)\equal{} 1$. Moreover, since $ p \equiv 2 \pmod{3}$ then $ \left( \frac{p}{3} \right) \equal{} \minus{}1$ and the Gauss's laws yields $ \left( \frac{\minus{}1}{p} \right) \cdot \left( \frac{3}{p} \right) \cdot \left( \frac{p}{3} \right) \equal{} \left( \frac{p}{3} \right) \equal{} \minus{}1$ or $ (\minus{}1)^{\frac{p\minus{}1}{2}\plus{}\frac{p\minus{}1}{2}\cdot\frac{3\minus{}1}{2}} \equal{} (\minus{}1)^{p\minus{}1} \equal{} \minus{}1$, contradiction. QED.

There is a small mistake in Mashimaru's solution . I really want to complete that nice one :lol:

   Frist , You should assume WLOG that $ (a ; b ;c ) \equal{} 1 \ \ (*)$

 For $ a\in\mathbb{N}^*$ and $ p$ is an odd prime, let $ \left( \frac {a}{p} \right)$ denote the Legendre's symbol.

Assume the contradiction that there exists $ k\in\mathbb{Z}^{ \plus{} }$ such that $ \frac {a^2 \plus{} b^2 \plus{} c^2}{3(ab \plus{} bc \plus{} ca)} \equal{} k$ then $ (a \plus{} b \plus{} c)^2 \equal{} (3k \plus{} 2)(ab \plus{} bc \plus{} ca)$. Let $ p$ be a prime such that $ p \equiv 2 \pmod{3}$ and $ p^{2s \plus{} 1} \parallel 3k \plus{} 2$, we have $ p^{2s \plus{} 1} | (a \plus{} b \plus{} c)^2$ then $ p^{2s \plus{} 2} | (a \plus{} b \plus{} c)^2$ and hence $ p | a \plus{} b \plus{} c$ and $ p | ab \plus{} bc \plus{} ca$.

We can prove that $ p \geq 5$ , Indeed , if $ p \equal{} 2$ . Due to the fact that $ p | a \plus{} b \plus{} c$ , we can deduce that in three numbers $ a ; b ; c$ , Exist at least one even number , assume $ 2 |b$ then $ a;c$ have the same parity , but $ 2 | ac$  so $ a ; c$ are also even integer
      $ 2 | gcd(a;b;c )$ , Contradiction with $ (*)$

Therefore, $ ab \plus{} c(a \plus{} b) \equiv ab \minus{} (a \plus{} b)^2 \equiv 0 \pmod{p}$, meanwhile $ p | a^2 \plus{} ab \plus{} b^2$. From this we can deduce that $ p | (2a \plus{} b)^2 \plus{} 3b^2 \rightarrow \left( \frac { \minus{} 3}{p} \right) \ \equal{} \ 1$ .

  (We have $ p | (2a \plus{} b)^2 \plus{} 3b^2$ .Then if $ p|b \rightarrow p| (2a \plus{} b) \rightarrow p|2a \rightarrow p|a$ , $ p|b$

   But $ p | a \plus{} b \plus{} c \rightarrow p|c$ , Contracdition with $ (*)$

  So , we must have $ (b ; p) \equal{} 1$ . On the other hand ,  $ \minus{} 3b^2 \equiv (2a \plus{} b)^2 \ \ (mod \ \ p)$
 $ \minus{} 3 (b .b^{ \minus{} 1} )^2 \equiv (b^{ \minus{} 1}(2a \plus{} b))^2 \equiv \minus{} 3 \ \ (mod \ \ p) \rightarrow \left( \frac { \minus{} 3}{p} \right) \ \equal{} \ 1$ )

. Moreover, since $ p \equiv 2 \pmod{3}$ then $ \left( \frac {p}{3} \right) \equal{} \minus{} 1$ and the Gauss's laws yields

$ \left( \frac { \minus{} 3}{p} \right) \equal{} \left( \frac { \minus{} 1}{p} \right) \cdot \left( \frac {3}{p} \right)$

$ \equal{} \left( \frac { \minus{} 1}{p} \right) \cdot \left( \frac {p}{3} \right) ( \minus{} 1)^{\frac {p \minus{} 1}{2} \cdot \frac {3 \minus{} 1}{2} } \equal{} \minus{} ( \minus{} 1)^{\frac {p \minus{} 1}{2} \plus{} \frac {p \minus{} 1}{2} \cdot \frac {3 \minus{} 1}{2} } \ \equal{} \ \minus{} ( \minus{} 1)^{p \minus{} 1} \equal{} \minus{} 1$, Contradiction

Foolish posts!
Mashimaru (pham hy hieu) and nguyenvuthanhha shouldn't repeat the solution in Dusan Djukic's article, rigidly. This problem is only an application of the Fermat little theorem.  :wink:

[quote="N.N.Trung"]This problem is only an application of the Fermat little theorem.  :wink:[/quote]can you post your solution?

You can see the solution [hide="in this file"]http://www.imomath.com/tekstkut/quadcong_ddj.pdf[/hide]
The problem appeared in [hide="Mathlinks contest edition 7_round 3"]http://www.mathlinks.ro/contest.php?edition=7&round=3[/hide]

[quote="Mashimaru"]For $ a\in\mathbb{N}^*$ and $ p$ is an odd prime, let $ \left( \frac {a}{p} \right)$ denote the Legendre's symbol.

Assume the contradiction that there exists $ k\in\mathbb{Z}$ such that $ \frac {a^2 \plus{} b^2 \plus{} c^2}{3(ab \plus{} bc \plus{} ca)} \equal{} k$ then $ (a \plus{} b \plus{} c)^2 \equal{} (3k \plus{} 2)(ab \plus{} bc \plus{} ca)$. Let $ p$ be a prime such that $ p \equiv 2 \pmod{3}$ and $ p^{2s \plus{} 1} \parallel 3k \plus{} 2$, we have $ p^{2s \plus{} 1} | (a \plus{} b \plus{} c)^2$ then $ p^{2s \plus{} 2} | (a \plus{} b \plus{} c)^2$ and hence $ p | a \plus{} b \plus{} c$ and $ p | ab \plus{} bc \plus{} ca$. Therefore, $ ab \plus{} c(a \plus{} b) \equiv ab \minus{} (a \plus{} b)^2 \equiv 0 \pmod{p}$, meanwhile $ p | a^2 \plus{} ab \plus{} b^2$. From this we can deduce that $ p | (2a \plus{} b)^2 \plus{} 3b^2$ and so $ \left( \frac { \minus{} 3}{p} \right) \equal{} 1$. Moreover, since $ p \equiv 2 \pmod{3}$ then $ \left( \frac {p}{3} \right) \equal{} \minus{} 1$ and the Gauss's laws yields $ \left( \frac { \minus{} 1}{p} \right) \cdot \left( \frac {3}{p} \right) \cdot \left( \frac {p}{3} \right) \equal{} \left( \frac {p}{3} \right) \equal{} \minus{} 1$ or $ ( \minus{} 1)^{\frac {p \minus{} 1}{2} \plus{} \frac {p \minus{} 1}{2}\cdot\frac {3 \minus{} 1}{2}} \equal{} ( \minus{} 1)^{p \minus{} 1} \equal{} \minus{} 1$, contradiction. QED.[/quote]
My solution only have a bit of difference from Mashimaru's  :P :
$ p | a^2 \plus{} ab \plus{} b^2|a^3 \minus{} b^3 \Rightarrow a^3\equiv b^3 \pmod p$.
Let $ p \equal{} 3k \plus{} 2$. The Little Fermath theorem shows that $ a^{3k \plus{} 1}\equiv b^{3k \plus{} 1}\pmod p$.
So, $ a \equiv b \pmod p \Rightarrow p|(3a^2) \Rightarrow p|a \Rightarrow p|b, p|c$ (contradiction).
No Legendre's symbol in solution!  :roll:


real analysis
inequalities
triangle inequality
real analysis unsolved

Studying for my analysis qual, came across this problem:

$ \{E_n\}$ Lebesgue measurable subsets of $ [0,1]$, and $ \sum_{n\equal{}1}^\infty m(E_n) < \infty$.  Define $ f \equal{} \sum_{n\equal{}1}^\infty \frac{\chi_{E_n}}{n}$ and show that $ f \in L^p[0,1]$ for $ 1 \leq p < \infty$.

I got the case $ p\equal{}1$, which was rather trivial.  However when you stick the $ p$ in there, it messes everything up.  It seems like you need to say something about the convergence rate of the measures of E_n and compare this to the convergence rate of $ \frac{1}{n}$ somehow, but I have no idea.

It looks like you're using Holder's to say:

$ \sum_{n \equal{} 1}^m \left(\frac {1}{n}\right)\left(\chi_{E_n}(x)\right) \leq \left(\sum_{n\geq 1}^{m}\frac {1}{n^{p}}\right)^\frac {1}{p}\left(\sum_{n\geq 1}^{m}[\chi_{E_{n}}(x)]^{p}\right)^\frac {1}{p}$

Is that correct?  But you can only use Holder's when $ \frac {1}{p} \plus{} \frac {1}{q} \equal{} 1$.

Edit: I suppose you could just have it

$ \left(\sum_{n\equal{}1}^m\frac {1}{n^{q}}\right)^\frac {p}{q}$

And it seems like the rest should work... maybe?

It's a series problem in disguise. Apply the triangle inequality; if $ f_n$ is the $ n$th term of that series, we want to show that $ \sum_n \|f_n\|_p<\infty$. What is $ \|f_n\|_p$? It's $ \frac1n(m(E_n))^{1/p}$; let $ a_n\equal{}m(E_n)$.
Now, Holder's inequality: $ \sum_n \frac1n\cdot a_n^{1/p}\le \|\frac1n\|_q\cdot \|a_n^{1/p}\|_p$. The latter term, which is $ \left(\sum a_n\right)^{1/p}$, is always finite for $ 1\le p<\infty$. The former is finite for $ 1<q\le\infty$, or equivalently $ 1\le p<\infty$.

In the $ \infty$-norm, this is of course false; just nest the $ E_n$, and it's easy to construct an unbounded sum.


Euler
number theory
least common multiple
number theory unsolved

Hello..

Got one more problem Im having difficulties with.

Prove using fermat-euler thoerem that theres no primitive root mod $ 2009 ( \equal{} 7^2*41)$.

quick calculation shows that $ \varphi(2009) \equal{} 1680$ so for every $ a$ coprime to $ 2009$
we have $ a^{1680} \equal{} 1 (mod 2009)$.

I tried everything I know to continue and reach some conclusion with no luck.

Thanks.

Use the totient theorem $ \bmod 49$ and $ \bmod 41$ and then use CRT.

this problem is to show that there is no g mod 2009 with period 2008, or period 1680?

$ 1680$.  That's the order of the multiplicative group $ \bmod 2009$.

ok, so we have:

$ a^{40}\equal{}1 (mod 41)$  for some a coprime to 41.

and

$ b^{36}\equal{}1(mod 49)$ for some b coprime to 49.

then according to the CRT:

$ X\equal{}a(mod 41)$ and $ X^{40}\equal{}a^{40}\equal{}1(mod 41)$
$ X\equal{}b(mod 49)$ and $ X^{36}\equal{}b^{36}\equal{}1(mod 49)$

How can I continue from here?

Thanks..

[quote="t0rajir0u"]If $ x^{40} \equiv 1 \bmod 41$ and $ x^{36} \equiv 1 \bmod 49$ then what can you say about $ x^{\text{lcm}(40, 36)}$?[/quote]

$ x^{360}\equal{}1(mod 2009)$ ?

One last thing:

Why are we allowed to consider the solution as the power to the LCM of $ 40$ and $ 36$ ?

Thanks.

It's CRT, which you haven't yet applied.  If $ x^{40} \equiv 1 \bmod 41$ then $ x^{360} \equiv 1 \bmod 41$; simlarly, if $ x^{36} \equiv 1 \bmod 49$ then $ x^{360} \equiv 1 \bmod 49$.  By CRT it follows that $ x^{360} \equiv 1 \bmod 2009$.


function
integration
limit
real analysis
inequalities
complex analysis
complex analysis theorems

Hi, I began to study subharmonic functions and I learned some theorems on the regularization of those functions, but there's one theorem I can't understand the proof of, to be exact there's one mysterious moment in the proof. Here's the theorem:

"If $u \in SH(\Omega) \cap L^{1}(\Omega, loc)$, then $\Delta u \geq 0$ on $\Omega$ in distribution sense, i.e.
\[\int_\Omega u\Delta \varphi dL^{2}\geq 0 \]
for $\varphi \in C^\infty_{0}(\Omega, \mathbb{R}_{+})$. Conversely, if $u \in L^{1}(\Omega, loc)$ is a function such that $\Delta u\geq 0$ on $\Omega$ in distribution sense, then there exists a function $v\in SH(\Omega)$ such that $u=v$ $L^{2}$-almost everywhere in $\Omega$."

That's it. And the problem is in "$\Leftarrow$", from the word "Conversely". The author of the book I read uses regularization functions of $u\in L^{1}(\Omega, loc)$ which are:
\[u_\epsilon (z) : = \int_\Omega u(w) \Psi_\epsilon (z-w) dL^{2}(w) = \int_{E}u(z+\epsilon w) \Psi(w) dL^{2}(w) \]
for $z \in \Omega_\epsilon : = \{z\in \Omega: d_\Omega (z) > \epsilon\}$, where
\[\Psi_\epsilon (z) : = \frac{1}{\epsilon^{2}}\Psi (\frac{z}{\epsilon}) \]
for $z\in \mathbb{C}, \epsilon > 0$ and $\Psi \in C^\infty_{0}(\mathbb{C}, \mathbb{R}_{+})$ is such that $supp \Psi = \overline{E}$ ($E = \{z\in \mathbb{C}: |z|<1\}$), $\Psi(z) = \Psi(|z|), z \in \mathbb{C}$ and $\int_\mathbb{C}\Psi dL^{2}= 1$; $d_\Omega (z) : = dist (z, Bd \Omega)$. The proof of "$\Leftarrow$" consists of few steps; first it is shown, that $u_\epsilon\in SH(\Omega_\epsilon)$, then that the sequence $u_\epsilon$ decreases when $\epsilon \to 0$. Finally it is introduced:
\[v : = \lim_{\epsilon \to 0}u_\epsilon \]
which is subharmonic, of course, on $\Omega$. And there's the moment when the author writes: "On the other hand, as you know, $u_\epsilon \to u$ $L^{2}$-almost everywhere in $\Omega$, whence $u=v$ $L^{2}$-al. e. in $\Omega$." These words end the proof. And here's my question: what is it: "as you konw"?? I don't know, why $u_\epsilon \to u$ $L^{2}$-al. e. in $\Omega$. Please, help me to understand, if you can.

Thanks for your help. ;)

It is because almost every point is a [url=http://en.wikipedia.org/wiki/Lebesgue_point]Lebesgue point[/url] of $u$ (this is a non-trivial theorem) and the convergence takes place at every Lebesgue point of $u$ (this you, probably, should be able to prove yourself).

fedja, THANK YOU  :lol:  This (theorem) was the one, that made me so tired! Unfortunately, I'm not too good at measure theory (as can be seen) and your advice seems to be some basic in this theory. Am I right? I must say, that while trying to find an argument for the "phrase" in the book, I thought it would be fine if such inequality hold, but I wasn't able to explain why it had to work. Now I konw it's non-trivial. Of course the rest if obvious.

Greetings! ;)


For $a,b,c $ sides of a triangle  such that

$2( \frac{1}{a} +\frac{1}{b} + \frac{1}{c} ) = 4 + \frac{a}{bc} + \frac{b}{ca} + \frac{c}{ab}  $

prove the inequality

$ 64(1-a)(1-b)(1-c) \leq1 $

After some computations, the given condition becomes [tex] 4R+r=4Rp[/tex]. Homogenize the inequality as follows [tex]64\prod [4Rp-a(4R+r)]\leq (4Rp)^3 \Leftrightarrow 64\prod [4Rrp-ar(4R+r)]\leq (4Rrp)^3 \Leftrightarrow 64\prod (bc-4Rr-r^2)\leq a^2b^2c^2 \Leftrightarrow 64\prod (p^2-ab-ca)\leq a^2b^2c^2 \Leftrightarrow \prod (-a+b+c)^2\leq a^2b^2c^2 [/tex], which is well known, and the proof is complete.


How would you solve this problem?

  :arrow: A cube's faces are painted green, then divided into unit cubes. If there are 64 unit cubes that are not painted, how many unit cubes are there in total?

I don't understand how to solve this. Please help.  :maybe:

Thank you.  :roll:

If 64 cubes are not painted, then on the inside there's a cube with side length $ \sqrt[3]{64}\equal{}4$.  Thus, the number of unit cubes is $ (4\plus{}2)^3\equal{}216$.


Prove that there exists an irrational real number $ x > 0$ such that $ x^{\sqrt{2}}$ is rational.

[hide="Explanation"]We don't even have to bother trying to see if the number in the previous post is rational or not.

[b][u]Case 1:[/u][/b] $ \sqrt {2}^{\sqrt {2}}$ [i]is rational [/i]
Then, $ x \equal{} \sqrt {2}$ works.

[b][u]Case 2:[/u][/b] $ \sqrt {2}^{\sqrt {2}}$[i] is irrational [/i]
Then, $ x \equal{} \sqrt {2}^{\sqrt {2}}$ works.[/hide]


Here is another sum.I think I did it, so please check the answer:

 Find the value 0f $ x^3 \minus{} 6x^2 \plus{} 18x \plus{} 11$ when $ x \equal{} 2 \minus{} 2^{\frac {1}{3}} \plus{} 2^{\frac {2}{3}}$.My answer is 34.Is it correct?The book gives 33.

Thank you in advance.

Oh!I forgot to mention that I brought 2 to the left and cubed the expression.

[quote="muks"]Here is another sum.I think I did it, so please check the answer:

 Find the value 0f $ x^3 \minus{} 6x^2 \plus{} 18x \plus{} 11$ when $ x \equal{} 2 \minus{} 2^{\frac {1}{3}} \plus{} 2^{\frac {2}{3}}$.My answer is 34.Is it correct?The book gives 33.

Thank you in advance.[/quote]

You must have made some mistakes on calculation.

$ x^3 \minus{} 6x^2 \plus{} 18x \plus{} 11\equal{}(x\minus{}2)^3\plus{}6x\plus{}19\equal{}2(\sqrt[3]{2}\minus{}1)^3\plus{}6x\plus{}19$

$ \equal{}(2\minus{}6\sqrt[3]{4}\plus{}6 \sqrt[3]{2})\plus{}6(2\plus{}\sqrt[3]{4}\minus{}\sqrt[3]{2})\plus{}19\equal{}33$


Let \[x = .123456789101112\ldots998999,\] where the digits are obtained by writing the integers 1 through 999 in order. The 1983rd digit to the right of the decimal point is

$ \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 $

[hide]There are $9$ numbers with one digit. They produce 9 digits.
There are $90$ numbers with two digits. They produce 180 digits.
There are $900$ numbers with three digits. They produce 2700 digits.

That's probably too much, don't you think? :D We're looking for the 1983rd one.

Let's say there are $x$ three digit numbers. Then

$1983 = 9 + 180 + 3x\\ x=598$

The digit right before that is D: 7.[/hide]

[hide="Answer"]We use 9 digits for the 1-digit numbers and 180 digits for the two-digit numbers, so we have $1983-189=1794$ digits left to use. $1794\equiv 0\bmod{3}$, so we are looking at the final digit of a number. $\frac{1794}{3}=598\Rightarrow 598+99=697$, so our answer is $7\Rightarrow \boxed{D}$.[/hide]


[quote="José"]I can't jump to another forums... why?[/quote]

What do you mean by "jump"?  Can you not see any part of our forums other than this one?  Or do you mean that where you see links, you cannot access other forums than this one?  Or do you mean something else?

I'm having this problem too.

Near the bottom right hand corner there is a place where it says "Jump to" and "select a forum" and it used to also have a list of all the forums under the "select a forum". It just says "none" now under "select a forum".

[quote="Valentin Vornicu"]Clear the cookies (there's a clear cookies set by this board link just above the jumbpox in the forum view) and try again.[/quote]

Sorry, but where is it? I can't find it

[quote="joml88"]Go to the front Index and look for it right above the "Jump to..." pull down. (This is right above where it lists all the people viewing the forums).[/quote]

Fixed. Thanks.

[quote="José"]Without touching anything, the same is happening again!!!!!!!!!!!![/quote]

What can I do????

Well, this is really strange... without doing anything, it fixed again!!! why is it?


calculus
integration
function
calculus computations

So what stinks about this problem is I have 1 shot to get them all right and no partial credit! I could figure it if it were just converge or diverge but using only the integral test is the trick because I never use the integral test ha...Can anyone help?

Test each of the following series for convergence by the Integral Test. If the Integral Test can be applied to the series, enter CONV if it converges or DIV if it diverges. If the integral test cannot be applied to the series, enter NA. (Note: this means that even if you know a given series converges by some other test, but the Integral Test cannot be applied to it, then you must enter NA rather than CONV.)

1.) {summation 1 to infinity} ne^(6n)
2.) {summation 1 to infinity} ne^(-6n)
3.) {summation 1 to infinity} (n+8)/(-6^n)
4.) {summation 1 to infinity} 8/(n*ln(3n)
5.) {summation 1 to infinity} 8/(n(ln(3n))^3)

I think it's:
1) NA It's easy to see this series diverges, it's term goes to infinity.

2) CONV. the function is positive and ultimately deacresing. And it's infinite integral is finite.

3) Same as 2.

4) CONV. or NA b/c this easy to see comparing with the p series.

5) Same as 4.

Keep in mind that this problem doesn't care whether you can apply another test or not; the only question is whether you can apply the Integral Test, so don't talk about the other tests.

Does 3) mean $ \sum_{k \equal{} 1}^{\infty} \frac {n \plus{} 8}{( \minus{} 6)^n}$?  In this case, the Integral Test can't be applied because the integrand isn't continuous.

I will tell you that 4) and 5) do not have the same answers.  Keep working :)

[quote="MegPie"]So what stinks about this problem is I have 1 shot to get them all right and no partial credit! [/quote]
I think that's exactly what you should expect in every real life situation. It would be nice if a neuro-surgeon would be allowed to try to operate his patient several times and got the partial credit for just holding the scalpel right but, quite surprisingly, the patient will probably run screaming into the night if he is told about such rules :roll:

On the other hand, in this particular set I would be quite perplexed about 1): should I write "diverges" or "NA"? The answer depends heavily on how exactly the theorem was stated in your course.

As to the rest, the integral test application is very simple. On a cookbook level, it reduces to checking that $ a_n\equal{}f(n)$ where $ f(x)$ preserves sign and is monotone eventually (if not, the test does not apply) and then just checking if $ \int^\infty f(x)\,dx$ converges.


[quote="jedaihan"]$ a,b,c$ are the sides of a triangle. $ a \plus{} b \plus{} c \equal{} 3$, $ x,y > 0$. Prove that
$ \sum_{cyc}\sqrt {(x \plus{} y)ab \plus{} xbc \plus{} xca}\geq3\sqrt {(x \plus{} y)(ab \plus{} bc \plus{} ca) \minus{} 2y}$[/quote]
The inequality is true for all $ a,b,c \ge 0$, not necessary to be the side lenghts of a triangle.
By sqaring both side we got
\[ \Leftrightarrow \left( {3x \plus{} y} \right)\sum {ab} \plus{} 2\sum {\sqrt {\left( {x\sum {ab} \plus{} yab} \right)\left( {x\sum {ab} \plus{} bc} \right)} } \ge 9\left( {x \plus{} y} \right)\sum {ab} \minus{} 18y
\]
\[ \Leftrightarrow 9y \plus{} \sum {\sqrt {\left( {x\sum {ab}  \plus{} yab} \right)\left( {x\sum {ab}  \plus{} bc} \right)} }  \ge \left( {3x \plus{} 4y} \right)\sum {ab}
\]
By the Cauchy Schwarz inequality we have
\[ \sum {\sqrt {\left( {x\sum {ab} \plus{} yab} \right)\left( {x\sum {ab} \plus{} bc} \right)} } \ge 3x\sum {ab} \plus{} y\sqrt {abc} \left( {\sum {\sqrt a } } \right)
\]
Thus it suffices to show that
\[ 9 \plus{} \sqrt {abc} \left( {\sum {\sqrt a } } \right) \ge 4\sum {ab}
\]
which is true by Schur for third degree, :).

Nice proof, Honey_S!
[quote="Honey_S"][quote="jedaihan"]$ a,b,c$ are the sides of a triangle. $ a + b + c = 3$, $ x,y > 0$. Prove that
$ \sum_{cyc}\sqrt {(x + y)ab + xbc + xca}\geq3\sqrt {(x + y)(ab + bc + ca) - 2y}$[/quote]
The inequality is true for all $ a,b,c \ge 0$, not necessary to be the side lenghts of a triangle.
[/quote]
If so, why ${ (x + y)(ab + bc + ca) - 2y}\geq0$ $ ?$ :wink:


inequalities
trigonometry
inequalities unsolved

$a_1, ..., a_n$ are non-negative and $a_p+a_q \leq \pi$ for $p \neq q$
Prove that $\frac{1}{n}\sum\cos{a_i} \leq \cos(\frac{1}{n}\sum a_i)$

[quote="blahblahblah"]Not quite; we can have $a_i>\pi/2$[/quote]
Hmm... does the inequality still hold in that case?


Given a positive prime $p$ and two positive integers $a,b$, prove that

$\binom{pa}{pb} \equiv \binom{a}{b} \mod{p^3}$.

Please do not post the same problem in two sub-forums, you have posted this problem in:

http://www.mathlinks.ro/Forum/viewtopic.php?t=37337

Well, in that thread blahblahblah said that the problem is more appropriate for Olympiad NT.....

[quote="zscool"]edit: I think it was solved by Grobber.[/quote]

Then it shouldn't be hard to find, just look through his posts!
On second thought... :lol:

[quote="zscool"]this problem has appeared here before, try a search.

edit: I think it was solved by Grobber.[/quote]

I tried and didn't find anything, but I'll try some more.

EDIT: ok, i found it, but i dont fully understand the solution... so could anyone either clarify or post another one? thanks.

[url=http://artofproblemsolving.com/Forum/viewtopic.php?highlight=mod+p%5E3&t=21509]http://artofproblemsolving.com/Forum/viewtopic.php?highlight=mod+p%5E3&t=21509[/url]

I think I understand grobber's part in the link, but I don't quite get pbornsztein's.


$ \log_3 5 \plus{} \log _3 6 \minus{} \log _3 10 \equal{} \log_3 \left(\frac {5 \times 6}{10}\right) \equal{} \log_3 3 \equal{} 1$.
Thus, $ \boxed{D}$.


For example, would taking intro to geometry cover everything you would learn in a school year of geometry?

It depends on your school. But in most cases you would learn more than a school class, have a better understanding. Also a great group of classmates in most cases.

If your school will let you skip yes!
Its totally possible, as long as you put the work in.

If you're trying to use the classes as credit, make sure you talk it over with your school counselor or whoever is in charge of scheduling before you sign up.

[quote="Raynee"]So would it be possible to skip one grade of math in school?  Like taking geometry eighth grade, taking algebra 2 on AOPS during the summer, and skipping to pre-calc in 9th grade?[/quote]

Eh, I tried to do that with geometry, it didn't work; so it depends on your school.

@Raynee: The AoPS stuff is way harder than regular school math. I'm getting really board in Geometry so I do Precalc aops during that time.

[quote="Raynee"]So would it be possible to skip one grade of math in school?  Like taking geometry eighth grade, taking algebra 2 on AOPS during the summer, and skipping to pre-calc in 9th grade?[/quote]

You would have to make sure your school will allow you to do this; don't simply assume your school will accept our classes as substitutes.

Our classes are considerably more challenging than most regular school classes.

Yeah some schools like mine are really inflexible, so I'm stuck in algebra II for the year.   We just got to learning what $ i$ was  :roll: and it's the beginning of 2nd semester.  You tend to learn much more in an aops class than a typical school class, also the time you need to devote to the class is much more than you'll probably be used to.  But it's definitely worth it! :D

AIME15, you have quite some lofty goals lol, at least for the average person. But good for you, and good luck achieving those goals :)

I do not think AIME15 is an average person.  In fact, I don not think anyone on AoPS is an average person.

I kinda hope there is some sarcasm in AIM15's siggy.  Cause I'm annoyed.  But in my state, the is an extra-cric option (with a test) that allows you to do 2 years of math in one year, and is generally accepted throughout the state.  So I did Algebra 1 and 2 in 6th grade, and geo+PC in 7th grade and am doing Calc I now.  Woot.  But I don't think I learned it as well as I could have if it was AoPS, so...meh.


induction
number theory unsolved
number theory

Prove that for that for every positive integer $ n$, the smallest integer that is greater than $ (\sqrt {3} \plus{} 1)^{2n}$ is divisible by $ 2^{n \plus{} 1}$.

in fact $ 2^{n\plus{}1} | [(\sqrt{3}\plus{}1)^{2n}]\plus{}1$ and also $ 2^{n\plus{}1} | [(\sqrt{3}\plus{}1)^{2n\plus{}1}]$
just use the fact that the sequence :
$ x_1\equal{}2,x_2\equal{}8,x_n\equal{}2(x_{n\minus{}1}\plus{}x_{n\minus{}2})$
satisfies $ x_n\equal{}(\sqrt{3}\plus{}1)^n\plus{}(1\minus{}\sqrt{3})^n$
now use induction and the fact that $ x_n\equal{}[(\sqrt{3}\plus{}1)^n]$ for odd $ n$ and $ x_n\equal{}[(\sqrt{3}\plus{}1)^n]\plus{}1$ for even $ n$.

[quote="PIRISH"]\text<ясно что > \{sqrt(3)+1}^n-{sqrt(3)-1}^n\  \text<целое число>
\{sqrt(3)-1}^n \text<дробное  число ,тогда целое част данной вырожений будет верхное вырожение >
\text<ясно что верхное вырожение делится на >  \2^(n+1)
\text<то можно доказат по методом математической индукций>
 :rotfl:[/quote][code][/code][code][/code]

The smallest integer greater than it is (sqrt(3) +1)^2n + (sqrt(3) -1)^2n = 2^n * [ (2+sqrt(3))^n + (2-sqrt(3))^n ] . Binomial expansion of the expression in the square braces will reveal that it is a multiple of 2. Hence problem solved[/quote][/code]

if $ x = (\sqrt {3} + 1)^2n$ is a fraction then,its fraction which has to summed up to make an integer is $ y = (\sqrt {3} - 1)^2n )$

so, we know that x and y are both divisible by $ 2^{n}$[expand the square] .and we know that $ y < 1$.so, x+y is the least  integer.so,add x and y.we get it as x+y=$ 2(%Error. "comb" is a bad command.
{2n}c{2n}3^{n} + %Error. "comb" is a bad command.
{2n}c{2n - 2} . ........... + %Error. "comb" is a bad command.
{2n}c{2n} 1)$.but we know that the whole expression is divisible by again $ 2$.so, it is divisible by $ 2^{n + 1}$.

In fact, we can also prove that $ (\sqrt{3}\plus{}1)^{2n}$ is not divisible by $ 2^{n\plus{}2}$.

[quote="Stephen"]In fact, we can also prove that $ (\sqrt {3} \plus{} 1)^{2n}$ is not divisible by $ 2^{n \plus{} 2}$.[/quote]
Are you sure about that?
Let $ \alpha \equal{} 2\plus{}\sqrt{3}, \beta \equal{} 2\minus{}\sqrt{3}$. Then the smallest integer greater than $ (\sqrt{3}\plus{}1)^{2n}$ is $ (\sqrt{3}\plus{}1)^{2n} \plus{} (\sqrt{3}\minus{}1)^{2n} \equal{} (2\alpha)^n \plus{} (2\beta)^n \equal{} 2^n( \alpha^n \plus{} \beta^n)$.
Let $ a_n \equal{} \alpha^n \plus{} \beta^n$. Then $ a_1 \equal{} 4, a_2\equal{}14$ and $ a_{n\plus{}2} \equal{} 4a_{n\plus{}1}\minus{}a_n$. Hence $ 2 \mid a_n$ and $ 4 \mid a_n$ when $ 2 \nmid n$ and $ 4 \nmid a_n$ when $ 2 \mid n$.
So $ 2^{n\plus{}2}$ divides the smallest integer greater than $ (\sqrt{3}\plus{}1)^{2n}$ iff $ 2 \nmid n$.

[quote="Moonmathpi496"][quote]The smallest integer greater than $ (\sqrt {3} \plus{} 1)^{2n}$ is $ (\sqrt {3} \plus{} 1)^{2n} \plus{} (\sqrt {3} \minus{} 1)^{2n}$ [/quote]
Actually I don't understand how you people have gotten this result. What was the motivation of taking conjugate? Can anyone explain?[/quote]
If $ \alpha, \beta$ is roots of $ x^2\minus{}nx\minus{}m\equal{}0$ and we define $ a_n \equal{} x \alpha^n \plus{} y \beta^n$ for some $ x,y \in \mathbb{C}$ then we have $ a_{n\plus{}2} \equal{} na_{n\plus{}1} \plus{} ma_n$.

If $ |\alpha| < 1$ we can approximate $ a_n$ to be $ y \beta^n$ for large $ n$. Don't know how to explain the "motivation". It's just a standard thing to do :P

thanks mathias:dk...nice!!!
but plz can u say how can i conclude with the part i mean how can i form the recursion??? :blush:
can u plz say this??? :oops:


we have a quadrilateral. we can change its angles, without changeing its sides. which quadrilateral will have the maximum area? there is an easy sol with trigo and calculus.

[hide="The relevant formula is..."]
[url=http://mathworld.wolfram.com/BretschneidersFormula.html]Bretschneider's Formula[/url] [/hide]


For n natural denote by f(n) the aruthmetic mean of positive divisors of n.
Show that  \sqrt n <=f(n) <=(n+1)/2

Observe that if d_1,...,d_k are the divisors then n/d_1,...,n/d_k are excatly the same divisors ( in other order). Thus, n*(1/d_1+...+1/d_k)=d_1+...+d_k. Then d_1+...+d_k \geq n*k^2/(d_1+...+d_k) from where we get the left inequality. For the second one, all we need to ptove is that if x_1<...<x_k are natural numbers then 1/x_1+...+1/x_k \leq k/2. This is easy, since 1/x_1+...+1/x_k \leq 1+1/2+...+1/k \leq k/2.


geometry
3D geometry
number theory proposed
number theory

Show that any integer can be expressed as a sum of two squares and a cube. :)

There is a proof by identities showing:
- for all odd $u$, $u$ can be represented.
- for all odd $u$, $2u$ can be represented.
- for all odd $u$, $4u$ can be represented.
- for all $n$ that can be represented, also $8n$ can be.

Could you explain more clearly.I am not able to understand your solution.

For example $2n+1 = (n^{3}-3n^{2}+n)^{2}+(n^{2}-n-1)^{2}+(2n-n^{2})^{3}$, thus every odd integer can be represented.
By a similar way we can show that $2\cdot (2n+1)$ and $4\cdot (2n+1)$ can be represented. Last but not least we show that when $u$ can be represented, then $8u$ can be represented, too (same way again).
This shows (inductively) that all integers can be represented.

But I don't want to spoil the fun of finding those identities ;)

Thank you very much,Zetax.I indeed try to find the solution myself but I failed ,So I guessed I moght not understand your post .


geometry
inequalities
geometric transformation
rotation
Vieta
factorial
reflection

It's the Friday after the test, so according to http://www.artofproblemsolving.com/Forum/viewtopic.php?t=141364 , we may discuss the test now. Can someone smart please post what they think are the correct answers?

I applied, so here's what I got...
[hide]
1. 25 feet
2. n=139
3. 20 triangles
5. 4 030 056
7. 190???
8. 360 => pwned this!
9. (3,18)(18,3)
10. Use Cyclic Quads, then use common chords, and you're done!
[/hide]
By the way, do you think I will get in???
Please share your answers!
- Nawahd Werdna

I got
[hide]
1. 25
2. 139 (double checked with my calculator)
3. 30 (double checked by brute forcing)
4. 98 (not sure)
5. 4,030,056
6. 197 (fun problem!)
7. 1020 (not sure, seems kind of high)
8. 360 (fun problem!) (double checked with my calculator)
9. Didn't try
10. Didn't try
[/hide]
EDIT: how did you go from not having applied to having applied in a couple of minutes, Nawahd Werdna?

isn't number 10 the outer napolean triangle?
It's a very very famous problem, I'm surprsed they used it.
It's mentioned in Intro to Geometry, so it must be provable using extremely basic techniques.

[quote="Twiz"]EDIT: how did you go from not having applied to having applied in a couple of minutes, Nawahd Werdna?[/quote]

Sorry, that was a typo... I thought the application wouldn't reach in time, but it did :)... although, I guess that i did make quite a few errors.... by the way, could you please post your solution to numer 3? I want to see where I went wrong...

For Number 3, you use the triangle inequality theorem (sum of 2 sides is greater then the third side). Then you list:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
Sum of the other sides, other than 13, must be at least 14.
If one other side is 12, there are 10 choices,
if one side is 11, there are 8 choices.
If one side is 10, there are 6 choices
If one side is 9, there are 4 choices,
if one side is 8, there are 2 choices
If one side is 7, there are no more choices since 6+7=13, and
all the greater choices, B= 7+ x have been counted. Thus,  you have
10 + 8 + 6 + 4 + 2 = 30 choices

can somebody please post their solution to number 4 and number 9

number 10 is EASILY brute forceable using complex numbers
lol

[quote="Twiz"]Could someone please tell me whether or not my answers to numbers 4 and 7 are correct?[/quote]

I can't be very sure, but I think your #7 is very high compared to mine...

Meh, I already knew the complex number one, so I was like "guys, sorry, no cheating here", and had a REALLY BAD one by rotation, which I think reduced to "here guys, these triangles are similar...AND OH LOOK THESE LOOK EQUILATERAL I BET THEY ARE QED"

[quote="lena_siz"]I thought that number 10 didn't involve any concepts with complex numbers, I solved it by rotation and similar triangles. What do the complex numbers have to do with it?[/quote]
you can brute force it with complex numbers =D

[quote="kohjhsd"][quote="lena_siz"]Isn't number 10 the Napoleons theorem?[/quote]

It is :o
[url]http://www.cut-the-knot.org/proofs/napoleon_intro.shtml[/url][/quote]

lol...i used the same exact website

for numba nine!

$xy+\frac{x^{3}+y^{3}}{3}=2007$

Let $p=xy$ and $s=x+y$

We know $x^{3}+y^{3}=s^{2}-3sp$

So $p+\frac{s^{3}-3sp}{3}=3^{2}\cdot 223$
$\Rightarrow 3p+s^{3}-3sp=3^{3}\cdot 223$

We know that $s^{3}$ must be divisible by 3 so $s$ is also divisible by 3.  Now let $3a=s$

$3p+3^{3}a^{3}-3^{2}ap=3^{3}\cdot 223$
$\Rightarrow p+3^{2}a^{3}-3ap=3^{2}\cdot 223$
$\Rightarrow 9a^{3}-p(3a-1)=9\cdot 223$

$3a-1$ is not divisible by 9 so $p$ must be divisible by 9.  Let $9b=a$.

$9a^{3}-9b(3a-1)=9\cdot 223$
$\Rightarrow a^{3}-b(3a-1)=223$
$\Rightarrow b=\frac{a^{3}-223}{3a-1}$

---

By vieta's theorem, the solutions to $m^{2}-sm+p=m^{2}-3am+9b=0$ are $x$ and $y$.  The discriminant is:

$9a^{2}-4\cdot 9b=9a^{2}-36b$.

This must be greater than 0 so $a^{2}\ge 4b$

$\Rightarrow a^{2}\ge 4\frac{a^{3}-223}{3a-1}$Solving for this equation, we see that $a$ is between 0 and 9 inclusive. I used calculator for this by the way. Now test them out and there is only one pair of $(a,b)$ that gives integers. Then we can find $s$ and $p$ thus giving us $x$ and $y$.  The answer is $\boxed{(3,18) (18,3)}$

I probably did it the worst, most tedious way but whatever.  I think I got an 8 or something

:o

I was going to do that

but then I was like

nah its too long

will never work

AMP writers are evil grr

[hide="I got"]
1. 25
2. 139
3. 30
4. 100
5. 4030056
6. 197
7. 215 (rather brute-force-ish :( )
8. Didn't do--found the answer to be 360 using my calculator :roll:
9. (18,3) and (3,18)--very brut-force-ish :mad:
10. Didn't do--Isn't looking it up somewhat cheap?
[/hide]

I wouldn't be surprised if I either over- or under-counted in 7, and am not at all proud of 9--I basically limited the range and then checked each possibility individually.

Could you post or PM me your solution to number 4? I'm curious to see whether you overcounted by two or I undercounted by two (or we were both wrong). Reviewing my solution, I feel that I may have overcounted by two, but still can't find where I may have undercounted.

My solution 4 is attached.  If anyone spots a potential error, please feel free to shoot holes in it.

Roadnottaken: Okay, your solution is far more convincing (and neat) than mine. Anyways I found a possible error in my solution (which uses a completely different method than yours does) and an error in your solution that lead to the answer being $96$. In case three, you subtract $4$ cases when the squares contain horizantal pieces and $4$ additional cases when the squares contain vertical pieces; however, when the squares are against the corners, both the vertical pieces and the horizantal pieces are contained in square; therefore, I believe that subtracting $4$ only one time would have accounted for the overcounting. This possible mistake took me about ten minutes to notice, and again, I could be wrong, but I guess we'll see when the answers come out.
EDIT: It turns out that your answer was right.

[quote="lena_siz"]you guy all got 139 for number 3. I got 169, and i used the fact that 2*5 = 10 and 10 * 1 = 10 , could someone please post their solution to this question? :huh:[/quote]
It looks like you may have forgotton to consider that a factor of 25 contains two fives and thus adds two zeroes to the product and a factor of 125 contains three fives and thus adds three zeroes to the product (I did this the first time I did one of these types of problems as well).

They've posted solutions to the admissions test, but unlike test a, none of the solutions were student-submitted.  I wonder why?

well, I got 6 problems perfect (ladder, factorial, triangles, 2007 rt. triangle, the fraction sum, primes (loved that one)), had a fairly shoddy proof for 10 (used reflections and concurrency, made one HUGE assupmtion that I didn't prove), and got 1944 for the cubes simply because it was 5:30 on the due date, I didnt want to do case work, and so I said there are $6^{6}$ ways to color the cube, $24$ ways to orient the cube, so the answer is $\frac{6^{6}}{24}=1944$, which seemed decently reasonable from the small amount of casework I had done previous to writing the solution. I totally didn't read the problem right to see that each face had to be adjacent to faces of different colors.

I got a call today from Dr. A, and I'm in.

Sapphyre, what grade are you in??? I just want to know... because I think that I also got 6 or 7 problems perfect.... So, just curious...

- Nawahd Werdna

9
[I'm typing this because my message is too short]

[quote="Aryth"]you can use analytic geometry for #10 if you're inclined to brute force your way through it.[/quote]

Using Complex Numbers is much cleaner, and essentially the same thing. It's also a well-known theorem, so you can just google outer napoleonic triangle and BAM

I came up with this proof:

http://reflections.awesomemath.org/2007_2/Napoleons_Theorem.pdf

that's exactly how I did it (with the inner point and everything) but I pretty much assumed everything rather than proving it, which I knew I couldn't do... :roll:


trigonometry
calculus
derivative
real analysis
real analysis unsolved

Let $F_n(x)=\sum_{m=1}^n \frac{1}{2^m} |\sin (\pi m!x)|.$ For rational numbers $r=\frac{q}{p}\ (p: natural\ number,\ q: integer),$
if $n\geq p$ then prove that $F_n(x)$ isn't differentiable at $x=r.$

Hmmm... I think you mean "[b]rational[/b] numbers". But isn't it obvious that the right derivative exceeds the left derivative at each rational point with small denominator? :?


hello every body :) !!!

find all fuctions $ g: \mathbb{R}_\plus{}^* \longrightarrow \mathbb{R}_\plus{}^*$

$ g \left( \frac{a}{x} \right) \equal{} g(x)$ with $ a \in \mathbb{R}_\plus{}^*$

$ good \ luck$

mathema, is $ a$ represent a given positive real? If so, then there are infinitely many $ g(x)$ because $ g(\dfrac{a}{\dfrac{a}{x}})\equal{}g(x)$.

If $ a$ represent arbitrary positive real, then put $ a\equal{}x$ and we get $ g(x)$ is a constant.

Hi stephen :) !

for exemple the functions that wa have $ g(\frac{1}{x} ) \equal{}g(x)$

or also find all functions that: $ g( \frac{\pi}{x})\equal{} g(x)$ ....

PS: that's obvious that may be exists infinity solutin :)

Let $ A$ be the set of all real numbers between 0 and $ \sqrt{a}$, where $ \sqrt{a}$ belongs to $ A$, and $ B$ be the set of all real numbers between $ \sqrt{a}, \inf$, where $ \sqrt{a}$ does not belong to $ B$.

Define G arbitrarily on $ A$, then for every number in $ B$ define $ f(x)\equal{}f(\frac{a}{x})$ where $ \frac{a}{x}$ must obviously belong in $ A$.


hi,
let $a,b,c,d \in \mathbb{R}$ and $0 < |a| < |c|$ and $f : \mathbb{R}\longrightarrow \mathbb{R}$ with the property $f(ax+b)=f(cx+d)$, $\forall x \in \mathbb{R}$. Prove that there $\exists x_{0}\in \mathbb{R}$ s.t.
if $f$ is continous in $x_{0}$ than $f$ is constant.

Let $x_{0},y_{0}$ defined by $ax_{0}+b=cx_{0}+d=y_{0}$ and $s=a/c, \ g(y)=f(y_{0}+y/c)$.
Then $g(y)=g(sy), \ |s|<1$, therefore if $g(y)$ is continiosly in 0, then $g(y)=g(0) \ \forall y$ or
if $f(x)$ continiosly in $y_{0}$, then $f(x)=f(y_{0})$.


Let $ ABC$ be a right triangle with $ \angle A \equal{} 90^\circ$. Let $ M$ and $ H$ be the mid-point of the side $ BC$ and the foot of the perpendicular from $ A$ to $ BC$, respectively. Extend segment $ AM$ through $ A$ to $ P$. The lines through $ H$ and perpendicular to $ AB$, $ AC$, intersect $ PB$, respectively $ PC$ at $ Q$ and $ R$. Prove that $ A$ is the orthocentre of triangle $ PQR$.

Let $ V_C^{\frac{CP}{CR}}: R\rightarrow P,H\rightarrow I,A\rightarrow J$ we have:
$ \angle{JAP}\equal{}90^0\plus{}\angle{BAP}\equal{}90^0\plus{}\angle{BIP}\equal{}\angle{JIP}$
Then $ AJPI$ is inscribe quadrangle, it mean $ \angle{IAP}\equal{}\angle{IJP}$
Now $ \angle{IBP}\equal{}\angle{IAP}$, then $ \angle{IBP}\equal{}\angle{IJP}$.
Hence $ \angle{BPJ}\equal{}\angle{BIJ}\equal{}90^0$, it mean $ BP,PJ$ are perpendicular.
But $ PJ,AR$ are parallel then $ BP,AR$ are perpendicular.
Similarly we have solution!
[img]http://i227.photobucket.com/albums/dd72/TPFC/april.png[/img]


Given the graph of $ y\equal{}f(x)$, how would you sketch the graph of $ y\equal{}f(\sqrt{x})$ and $ y\equal{}\sqrt{f(x)}$?

As an example, consider $ f(x)\equal{}e^x$, $ x \in \mathbb{R}$.

Hello
First of all we need to understand what parameters r going to change and then we can proceed easily.
$ f(x) = e^x$
All d three func $ y = f(x),\sqrt {f(x)},f(\sqrt{x})$ all are going to meet at$ (0,1)$
For $ y = \sqrt {f(x)}$
As for $ x < 0$ $ f(x) < 1$ so sqrt of this func will give larger value than this function.
But for $ x > 0$ $ f(x) > 1$ so sqrt of this value will give smaller value than this function.
So the graph $ y = \sqrt {f(x)}$  will be above the graph $ y = f(x)$ before it reach $ x = 0$.At $ x = 0$, both the graph will meet and after that $ f(x)$ will come above to $ \sqrt {f(x)}$.
For $ y = f(\sqrt x)$
no need to think about $ - ve x - axis$.For $ + ve$ x-axis ,like $ y = \sqrt {x}$ it opens concave
at x=0 $ f(x)$ and $ f(\sqrt {x})$ meets .in the interval $ (0,1)$ $ f(\sqrt {x})$ will be above $ f(x)$ both will meet again at $ x = 1$.And after that $ f(\sqrt {x})$ will come below the $ f(x)$.
Try with other functions.Have fun with graphs.
Thank u.

[quote="kabi"]
As for $ x < 0$ $ f(x) < 0$ so sqrt of this func will give larger value than this function.
Thank u.[/quote]

I don't understand your grammar, sorry. It is not possible for any real value of $ x$ for $ e^x<0$.

Hello
Sorry but by mistake I wrote that.I have edited that.


calculus
absolute value
calculus computations

(a)
Prove that if f'(x)>=M whenever a<=x<=b, then f(b)>=f(a) + M(b-a)

(b)
Prove that if f'(x)<=m whenever a<=x<=b, then f(b)<=f(a) + m(b-a)

(c)
Formulate a similar theorem when absolute value of f'(x)<=M for all x in [a,b]

By mean value theorem we have $f(b)-f(a)=f'(\zeta)(b-a)$ for some $\zeta\in (a,b)$. Since $f'(\zeta)\geq M$, then $f(b)-f(a)\geq M(b-a)$. The rest problems are similar.

thanx,

How do I do  part (c) of the question?

I think it the same  :)  write $|f'(x)| \leq M$ as $-M \leq f'(x) \leq M$ and use similar argument.

$|f(b)-f(a)|\leq M|b-a|$

This is what's called a Lipschitz condition.


function
linear algebra
linear algebra unsolved

les E and F two $ K$vectorial space of finit dimension p,n respectively,and let $ f \in L(E,F)$.
we set $ G = \{u \in L(F,E)\ fuf = 0_{L(E,F) \}}$
prove that G is a subspace of $ L(E,F)$ and find  his dimension in function of $ p,n, rank(f)$

I cannot decipher this statement, nor can I guess what it should have said. $ G$ is the set of $ u$ such that .. what? The condition that follows doesn't mention $ u.$ And $ fgf$ doesn't make sense as written.

Maybe $ G\equal{}\{g\in L(F,E)\mid fgf\equal{}0\}?$

No, it's not OK now. If $ u\in L(E,F)$ and $ E$ is not the same space as $ F,$ then $ fuf$ is not defined. Perhaps you want $ u\in L(F,E)?$

G is a subspace of L(F,E) :).for the dimension i've just find some inéqualitise beetwen p,n;rg(f).it is also easy to see that G is the Ker of a linear application.
if you have a nice solution dear Driss post it  :D

$ Im(f)\oplus{H}=F,dim(Im(f))=r$. $ u|_{Im(f)}$ is any linear function$ : Im(f)\rightarrow{ker(f)}$. $ u|_{H}$ is any linear function $ : H\rightarrow{E}$.
Then $ dim(G)=r(p-r)+(n-r)p=np-r^2$.


toi mừng Việt Nam thay cho tuyệt IMO.
toi thich  Việt Nam lam.

Cam on ban rat nhieu, dat nuoc cua chung toi dep nhi? :D Ban thay Ha Long the nao? Toi o do day! :P

Viet Nam that dep voi nhung con nguoi that tuyet voi.Ha Long cung vay,that tuyet! Toi hi vong se co dip quay tro lai Viet nam !
(sorry for my bad vietnamese)

Oa anh nao ma moi tieng Viet Nam chuan vay . Cho em dia chi email di .Co gi em giup nang cao von tu Viet :D .[hide]Anh Tuan co thi cho em nha[/hide]

[quote="N.T.TUAN"][quote="Hong Quy"]ne thiet !  ban hoc tieng viet lau chua ma` noi rat chuan.
    ban wa VN tuan toi lam j vay ?[/quote]
Cậu ấy nói: Tuân, tôi ước ao gặp bạn khi đó! :D[/quote]
 anh ay noi :
[quote="dashmiz"]Cam on ban rat nhieu'day la email cua minh,ban lien lac voi minh nhe. (shayandashmiz@yahoo.com)
tuan toi ước ao  gap ban khi do![/quote]
LA Y NOI GAP EM MA  :P
:spider:
Anh ay noi tieng Viet cung tot ma khi chat voi em toan chat bang tieng Anh  :starwars:
Lam em toan dung tu dien :read:   :wallbash:

Khong ngo co nguoi nuoc ngoai tren ML noi tieng Viet gioi nhu vay! Khong biet co phai dashmiz di thi IMO dot vua roi khong nhi?


Please if it possible someone to send me on private message or via e-mail the problems from Australian Math Olympiad 2008.
Without solutions.


Where can I find The problems of AHSME 1990-1994? I don't find it.
Can anyone help me?

Check the AOPS wiki section [url=http://www.artofproblemsolving.com/Wiki/index.php/AHSME_Problems_and_Solutions]here[/url]

Some ASHME problems are missing I believe so the only way to obtain them is to buy one of the contest books the MAA sells.

All the ones from 1990 to 1994 can also be found [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44]here[/url].


geometry
circumcircle
angle bisector
geometry proposed

In a trinagle with $ h,t,m,R$ as altitude, internal angle bisector, median(all three corresponding tot he same vertex) and circumradius  respectively, prove that
$ 4R^2h^2(t^2\minus{}h^2)\equal{}t^4(m^2\minus{}h^2)$.
Try to avoid a lot of computaion  while proceeding by algebra or trigonometry.

Note that $ t\equal{}\frac{2\sqrt{bc(p\minus{}a)}}{b\plus{}c}, m^2\equal{}\frac{2(b^2\plus{}c^2)\minus{}a^2}{4}$ and $ 4R^2.h^2\equal{}4\frac{a^2b^2c^2}{16S^2}.\frac{4S^2}{a^2}\equal{}b^2c^2$
Is it enough to calculate?  :)


calculus geometry probability MATHCOUNTS AMC AIME USA(J)MO

I thought it would be cool for all of U.S. students to give their math profile (good for future reference). Here is the format, (and my profile) and please, don't me modest with your achievements; be very specific as well.

[b]Grade:[/b] 10
[b]Age:[/b] 15
[b]Math Class Currently in:[/b] Calculus BC
[b]Weakest Math Area:[/b] Geometry
[b]Favorite Math Area:[/b] Probability
[b]Chapter Mathcounts (tell ranking, score): [/b]too low to be ranked, 30
[b]State Mathcounts(same as above): [/b]didn't participate
[b]National Mathcounts(same as above):[/b] didn't participate
[b]AMC 10 [i]Highest [/i]Score:[/b] 118
[b]AMC 12 [i]Highest [/i]Score:[/b] 105
[b]AIME [i]Highest [/i]Score:[/b] 2
[b]USAMO [i]Highest [/i]Score: [/b]didn't participate
[b]Other Achievements (list): [/b]qualified for MMPC (Michigan Math Prize Competition) Part 2

Grade: 9
Age: 14
Math Class Currently in: Trig/Analysis Honors
Weakest Math Area: dunno
Favorite Math Area: Probability or Algebra
Chapter Mathcounts (tell ranking, score): 3, don't remember
State Mathcounts(same as above): 2, don't remember
National Mathcounts(same as above): 3 written, 40 or 41
AMC 10 Highest Score: 13_ don't remember last digit
AMC 12 Highest Score: 11_ same as above
AIME Highest Score: 7
USAMO Highest Score: 4

[b]Grade:[/b] 11
[b]Age:[/b] 15
[b]Math Class Currently in:[/b] Calculus BC and Statistics (in school); Intermediate Independent Study, Intermediate Counting and Intermediate Trig (on AoPS)
[b]Weakest Math Area:[/b] Number Theory
[b]Favorite Math Area:[/b] Geometry (as of now)
[b]Chapter Mathcounts: [/b]NA (never even did MC)
[b]State Mathcounts: [/b]NA
[b]National Mathcounts:[/b] NA
[b]AMC 10 [i]Highest [/i]Score:[/b] can't remember 80-90 something
[b]AMC 12 [i]Highest [/i]Score:[/b] 106.5
[b]AIME [i]Highest [/i]Score:[/b] 1  :blush:
[b]USAMO [i]Highest [/i]Score: [/b]NA
[b]Other Achievements (list): [/b]SC ARML team(All State Team); placed 2nd and 3rd at a bunch of statewide competitions

[b]Grade:[/b] 9
[b]Age:[/b] turned 14 today :)
[b]Math Class Currently In:[/b] Pre-AP Alg. II
[b]Weakest Math Area:[/b] don't know
[b]Favorite Math Area:[/b] don't know
[b]Chapter Mathcounts:[/b] 7th Grade: 33, rank: 23
                                             8th Grade: 31? (guess) rank: don't know
[b]State Mathcounts:[/b] 7th Grade: 37-38, Rank: 31
                                        8th Grade: 33, Rank: 51 or 55, don't remember
[b]National Mathcounts:[/b]N/A
I have not competed in AMC
[b]Other Achievements:[/b]can't think of any

Grade: 7
Age: 12
Current math class: Algebra 1
Weakest Math Area: Geometry and Number Theory
Favorite Math Area: I like everything, even my weak spots, I really can't decide
Chapter Mathcounts (tell ranking, score): 24th place, 34 points
State Mathcounts(same as above): N/A
National Mathcounts(same as above): N/A
AMC 10 Highest Score: N/A
AMC 12 Highest Score: N/A
AIME Highest Score: N/A
USAMO Highest Score: N/A
Other Achievements (list): none that I can think of
[EDIT] Whoops...I didn't see the 'if you are a [b]high school[/b] student...'

Grade: 9
Age: 14
Current math class: Algebra II
Weakest Math Area: Number Theory
Favorite Math Area: Counting and Probability
Chapter Mathcounts (tell ranking, score): 7th-3rd(42 or 43)
State Mathcounts(same as above): N/A 7th-6th(40 or something)
National Mathcounts(same as above): N/A
AMC 10 Highest Score: N/A
AMC 12 Highest Score: N/A
AIME Highest Score: N/A
USAMO Highest Score: N/A

[quote="LynnelleYe"][EDIT] Whoops...I didn't see the 'if you are a [b]high school[/b] student...'[/quote]

Oh no! You've violated a huge rule! Now we will have to forever brand you as an outcast on AoPS and make a giant cyber-bird peck your brains out every daaaaaay!!!!!  :D  ;)

Grade-9
Age-14 and some even months
Math Class Currently in: Honors Trig
Weakest Math Area: Your guess is as good as mine
Favorite Math Area: Easy problems that I can do in less than a minute
Chapter Mathcounts (tell ranking, score): 7th grade- 3, 40 something. 8th grade-1,44.
State Mathcounts(same as above): 7th-3, 39. 8th- 3, 40.
National Mathcounts(same as above): 7th- 69, 33. 8th- 38, 33?
AMC 10 Highest Score: 7th grade- 120.5 (B)
AMC 12 Highest Score: Nope
AIME Highest Score: 7th grade-3
USAMO Highest Score: guess
Other Achievements: I won a no-talking contest with someone on my 2003 Mathcounts  team. Ah, good times.  :D

[b]Grade:[/b] 10
[b]Age:[/b] 15 (16 on the 19th)
[b]Math Class Currently in:[/b] Math Analysis (basically Algebra II/Trig)
[b]Weakest Math Area:[/b] Probably Probability
[b]Favorite Math Area:[/b] Geometry
[b]Chapter Mathcounts (tell ranking, score): [/b]7th Grade: 2nd (forgot my score, I'm afraid; that'll be something of a theme); 8th Grade: 1st
[b]State Mathcounts(same as above): [/b]7th Grade: 18th or something like that; 8th Grade: 2nd, 44
[b]National Mathcounts(same as above):[/b] 8th Grade: 138th
[b]AMC 10 [i]Highest [/i]Score:[/b] N/A
[b]AMC 12 [i]Highest [/i]Score:[/b] N/A
[b]AIME [i]Highest [/i]Score:[/b] N/A
[b]USAMO [i]Highest [/i]Score: [/b]N/A
[b]Other Achievements (list): [/b]5th Place in the 2004 National Mathematics High School Championship

Grade:  11
Age:  16
Math Class Currently In:  teaching myself mutivariable calc and linear algebra, without much success, though
Weakest Math:  proving stuff
Favourite Math Thingy:  Probably calculus, for some reason
Mathcounts:  none
AMC 12:  116
AIME:  5
USAMO:  not yet :-/
Other:  Nope.

How fun...does that imply my brains grow back? Like, you know, that Greek god person's liver? LOLOLOL

Anyway, what scores did you guys get for [i]6th grade[/i] MathCounts? Just curious...

You mean Prometheus?

Wasn't in MathCounts in 6th Grade, because I moved halfway through the year and my new school had already started the MathCounts season.

I think you are talking about Tityus for the greek myth reference...but I m sure that's besides the point...

[b]Grade:[/b] 9
[b]Age:[/b] 14
[b]Math Class Currently in:[/b] Algebra II Honors(next semester...due to block scheduling, i have no maths this semester. *tearz*)
[b]Weakest Math Area:[/b] i dunno
[b]Favorite Math Area:[/b] algebra i guess
[b]Chapter Mathcounts (tell ranking, score):[/b] 1st, 37 i believe
[b]State Mathcounts(same as above):[/b] 15th, 36 or 37 ish
[b]National Mathcounts(same as above):[/b] n/a
[b]AMC 10 Highest Score:[/b] n/a
[b]AMC 12 Highest Score:[/b] n/a
[b]AIME Highest Score:[/b] n/a
[b]USAMO Highest Score:[/b] n/a
[b]Other Achievements (list):[/b] can't think of any atm

[b]Grade:[/b] 10
[b]Age:[/b] 15
[b]Math Class Currently in:[/b] H Math Analysis
[b]Weakest Math Area: [/b]Geometry
[b]Favorite Math Area:[/b] Number Theory
[b]Chapter Mathcounts (tell ranking, score):[/b] 22nd, 38
[b]State Mathcounts(same as above):[/b] didn't participate
[b]National Mathcounts(same as above): [/b]didn't participate
[b]AMC 10 Highest Score:[/b] N/A
[b]AMC 12 Highest Score:[/b] 103
[b]AIME Highest Score:[/b] 2
[b]USAMO Highest Score: [/b]didn't participate [/b]

Yeah, I was kinda talking about Prometheus. Probably because my English class read the story and went really in-depth about it.

Grade: 11
Age: 16
Math Class Currently in: Calculus BC/IB Calc HL (school), Modern Algebra (local U.)
Weakest Math Area: Geometry
Favorite Math Area: Analytic Number Theory
Chapter Mathcounts: 1st place (8th grade)
State Mathcounts: 4th place (8th grade)
National Mathcounts: 63rd place (8th grade):?
AMC 10 Highest Score: 150 (9th)
AMC 12 Highest Score: 124 (10th):?
AIME Highest Score: 10 (10th)
USAMO Highest Score: 17 (10th grade)
Other Achievements (list): Made MOP

Yeh, there are just way too many of em. Half of the Princeton staff does that. Seriously, when I visited and went to the Math Building, every single person I met was an analytic number theorist.

Grade: 12
Age: 17
Math Class Currently in: Indpt Study through University of Illinois-Chicago in group theory.
Weakest Math Area: Geometry
Favorite Math Area: Number Theory.
Chapter Mathcounts (tell ranking, score): never did
State Mathcounts(same as above): never did
National Mathcounts(same as above): never did
AMC 10 Highest Score: Don't remember.  Not good.
AMC 12 Highest Score: 100.5
AIME Highest Score: 4
USAMO Highest Score: never did
Other Achievements (list): Solved the Riemann Hypothesis at the age of 6.  Ok, maybe not.  But it'd be pretty cool if I did.

As some of our peers might say:

"Analytic number theory is the new black..."

Who does analytic number theory at Princeton? I guess Sarnak does, but who else?

The people I spoke to included only two that I actually remember distinctly. One was Jordan Ellenburg, who professed to being one. The other was a graduate student who was about to present his thesis in a week. He said that his area was anayltic number theory and that most of the people there are the same (He showed me on the board with all of the professors what some people did).

I see. Around here, practically everyone is either a ring theorist or a topologist. (Well, okay, there are some people at this university who aren't mathematicians, but we'll ignore them.)

[b]Grade:[/b] 12
[b]Age:[/b] 17
[b]Math Class Currently in:[/b] Self learning real analysis
[b]Weakest Math Area:[/b] geometry (I don't think I'm alone here)
[b]Favorite Math Area:[/b] Number Theory
[b]Chapter Mathcounts (tell ranking, score):[/b] 5th, too long ago to remember...
[b]State Mathcounts(same as above):[/b] 13th, ditto
[b]AMC 12 [i]Highest[/i] Score:[/b] 133
[b]AIME [i]Highest[/i] Score:[/b] 5
[b]Other Achievements (list):[/b] MN ARML Gold and Math Bowl! w00t!

Grade 9
Age 14  (15 next month)
Current Math Course:  Algebra II
Weakest Area: Counting/Probablility
Favorite Area: Geometry
Highest AMC 12 score: 97
never did mathcount, AIME, and USAMO

No other achievements

Grade: 12
Age: 16
Math Class Currently in: Linear Alg/ Multivar Calc / other topics
Weakest Math Area: hopefully I've eliminated them by now
Favorite Math Area: Geometry
Chapter Mathcounts (tell ranking, score): didn't participate
State Mathcounts(same as above): didn't participate
National Mathcounts(same as above): didn't participate
AMC 10 Highest Score: don't remember haven't taken it since 9th gr
AMC 12 Highest Score: 132.5 (2004)
AIME Highest Score: 8 (2004)
USAMO Highest Score: 22 (in 2004)

Grade: 11
Age: 16
Math Class Currently in: Geometry (going through Geometry Revisited by Coxeter and Greitzer)
Weakest Math Area: Geometry
Favorite Math Area: Number Theory, Algebra, Combinatorics, Discrete are all tied
Chapter Mathcounts: 7th-1st, 39 pts; 8th-1st, 43
State Mathcounts: 7th-74th, 16 pts; 8th- 24th, 35 pts
National Mathcounts: didn't participate
AMC 10 Highest Score: 131
AMC 12 Highest Score: 132.5
AIME Highest Score: 8
USAMO Highest Score: 9
Other Achievements (list): MOP '04

Grade: 11
Age: 16
Math Class Currently in: College Course combining Linear Algebra, Multi-D Calc and Differential Equations
Weakest Math Area: don't know
Favorite Math Area: Topology/Knot theory
Chapter Mathcounts (tell ranking, score): 7th-9th, don't remember the score, 8th-2nd, 43
State Mathcounts(same as above): 7th-too low, 8th-13th, don't remember the score.
National Mathcounts(same as above): didn't participate
AMC 10 Highest Score: 150
AMC 12 Highest Score: 130
AIME Highest Score: 5
USAMO Highest Score: didn't participate
Other Achievements (list): Qualified for the APMO

Age: 17, almost 18
senior
math studies (self studied, and self studying):
self studied calculus, passed with 5 (11th)
multivariate calculus (11th)
number theory, fractal stuff, complex analysis (12th)

Note these are practice tests, I've never taken them in a testing environment, because my school doesnt offer these (i'll take them somewhere else though, this year)

amc12 highest: around 22 or 23 correct
aime highest: 10-11 correct
usamo: never taken this as a practice test

Grade: 12
Age: barely 18
Math Class Currently in: none
Weakest Math Area:
Favorite Math Area: Discrete Math
Chapter Mathcounts (tell ranking, score): did not participate
State Mathcounts(same as above): did not participate
National Mathcounts(same as above): did not participate
AMC 10 Highest Score: never taken
AMC 12 Highest Score: 124?
AIME Highest Score: 7?
USAMO Highest Score: did not participate
Other Achievements (list): learned virtually all I know by 8th grade (intense algebra, trig, geometry, calculus, logic, games, etc... mostly precalculus stuff though.)

Grade: 11
Age: 15
Math Class Currently in: AP Calculus AB (BC is not offered at my school, how sad...-_-')
Weakest Math Area: Geometry
Favorite Math Area: interesting problems, weird ones even
Chapter Mathcounts: 2
State Mathcounts: 4
National Mathcounts: 42
AMC 10 Highest Score: 146.5 (guess what I forgot: how to calculate the area of a parallelogram given base and height, yep)
AIME Highest Score: 6 or 7
Other Achievements (list): top assassin in the Verbal Assassin game  :cool:

Grade: 12
Age: 17
Math Class Currently in: None (will take Enriched Math Topics a.k.a. Calc BC next semester, though)
Weakest Math Area: Number Theory
Favorite Math Area: Number Theory
Chapter Mathcounts: Didn't Participate
State Mathcounts: didn't participate
National Mathcounts: Didn't participate
AMC 10 Highest Score: Didn't participate (did the AMC 12 in 10th grade)
AMC 12 Highest Score: 99.5 (Aargh! So close!)
AIME Highest Score: Didn't participate
USAMO Highest Score: Didn't participate
Other Achievements: 9th in state at 2003-4 Minnesota H.S. Math League Team Competition

What if you're not in high school :P ?

Grade: not telling
Age: not telling
Math Class Currently in: Calculus BC
Weakest Math Area: Vectors
Favorite Math Area: Probably counting, or maybe adding single-digit numbers
Chapter Mathcounts: 1st, 45
State Mathcounts: 4th, 41
National Mathcounts: 36th, unknown
AMC 10 Highest Score: didn't take it
AMC 12 Highest Score: didn't take it
AIME Highest Score: didn't take it
USAMO Highest Score: didn't take it
Other Achievements (list): Probably three or four years younger than most of you, placed 1st in NJ math league, placed 2nd in NJ math league, and I can count to ten on my fingers :D !

Grade: 10
Age: 16
Math Class Currently in: Alg II/Trig AL
Weakest Math Area: Probability
Favorite Math Area: Number Theory
Chapter Mathcounts (tell ranking, score): didn't participate
State Mathcounts(same as above): didn't participate
National Mathcounts(same as above): didn't participate
AMC 10 Highest Score: 125
AMC 12 Highest Score: didn't take it in 9th grade
AIME Highest Score: 4 (Totally would have gotten five if I had remembered about that one single cube left to make it a complete cube on AIME 2004)...
USAMO Highest Score: didn't participate
Other Achievements (list): None that I can think of...I got a 6-7 trophies at MAO as a freshmen, which was pretty cool.

Grade: 9
Age: 13
Math Class Currently in: Calculus BC
Weakest Math Area: Dunno
Favorite Math Area: Dunno
Chapter Mathcounts: 7th?
State Mathcounts: 4th.
National Mathcounts: 6th
AMC 10 Highest Score: 146.5
AMC 12 Highest Score: N/A
AIME Highest Score: 11
USAMO Highest Score: 7
Other Achievements (list): I'll come up with them later.

Grade: 10
Age: 15
Math Class Currently in: Calculus AB
Weakest Math Area: State MathCounts and AMC tests
Favorite Math Area: I like prime numbers and really hard stuff
Chapter Mathcounts: 2nd (4 times...)
State Mathcounts: 36th, 40th  :twisted: I was angry at myself for a while..
National Mathcounts: N.A.
AMC 10 Highest Score: 119 (in 8th grade)  (so close...)
AMC 12 Highest Score: 99.5 (in 9th grade)  (so close...)
AIME Highest Score: N.A.
USAMO Highest Score: N.A.
Other Achievements (list):
-I have a row of MathCounts trophies in my bedroom
-17th in MN Math League last year
-ARML!  :lol:
-25 w/ 1 commended solution in USAMTS Year 16 Round 1  :lol:

[quote]
National Mathcounts: 6th
[/quote]

Was that this year's competition? If it was... *looks you up in my MathCounts paper thingy showing everybody who competed*

Grade: 12
Age: 17
Math Class Currently In: Honor's Mathematics I 295, Problem Solving Seminar 285 (both at  U of M)
Weakest Math: advanced geometry
Favourite Math Thingy: number theory
Mathcounts: 3rd in region, choked at states
AMC 10: 124 freshmen year
AIME: 1 freshmen year
My school stopped doing AMC stuff after that, hopefully we'll do it for my senior year
Other: MMPC; 9th grade, 94thish. 10th grade, 46thish, 11th grade 2nd. Two year member of Michigan ARML team. Honorable Mention Lawrence Tech High School Math Competition.