symmetry
analytic geometry
special factorizations

Consider the graphs of $ y\equal{}x^2\minus{}3x\plus{}4$ and $ y\equal{}x^2\plus{}3x\plus{}4$. For the first one, the $ x$-coordinate of the vertex is $ 3/2$, while the $ x$-coordinate of the vertex of the second one is $ \minus{}3/2$. (These can be obtained by completing the square easily.)

So for the graph of $ y\equal{}ax^2\plus{}bx\plus{}c$, if $ a$ and $ b$ have the same sign the $ x$-coordinate of the vertex is negative; if $ a$ and $ b$ have opposite signs the $ x$-coordinate of the vertex is positive. (We all know what happens when $ a\equal{}b\equal{}0$.)

So changing $ b$ changes the $ x$-coordinate of vertex; surely there must be a transformation. What is this transformation?

Write it in this form:

$ y \equal{} a \left( x \plus{} \frac {b}{2a} \right)^2 \plus{} \left( c \minus{} \frac {b^2}{4a} \right)$

Now, you can more easily see the plane transformations in play. For example, it is now clear that the axis of symmetry is the line $ x \plus{} \frac{b}{2a} \equal{} 0$


geometry
calculus
derivative
function
graph theory
number theory
prime numbers
\/closed

A member of the College Board and two past winners of the Siemens Westinghouse Competition in Math, Science & Technology will be hosting a Math Jam on August 18th from 5:30PM to 7PM ET (2:30PM to 4PM PT).

Janet DeCarlo is the Manager, Siemens Westinghouse Scholarship Program at The College Board.

The two students who will be involved are Hannah Chung and Linda Westrick. Hannah Chung was part of the 2003 Southwestern Regional Silver Medal winning team and National Gold Medal winning team. She and her teammates produced "Eccentric Graphs of Block Graphs and Trees," a project that contributed innovative conceptual ideas to graph theory, an area of mathematics with a wide range of applications in computer science, architecture, economics, encryption, and networking.

Linda Westrick was the individual 2003 Southern Regional Silver Medal winner and National Gold Medal winner for her project "Investigations of the Number Derivative," which investigated a mathematical function, which, by definition, is closely linked with the prime numbers and calculus.  During her research, she found that the number derivative produces many surprising and beautiful patterns.

Students who are considering participating in the Siemens Westinghouse Competition should consider this a highly unique opportunity to find out about the application process, how two past participants created projects to research and found mentors and other resources, and any other questions students might have about the Siemens Westinghouse Competition.

I would also like students to take the time and ask some of the questions they are interested in hearing answers to in this thread.  The hosts of the Math Jam will be using questions asked in this thread to guide the information they prepare for the Math Jam.

Please be advised that the hosts of the discussion will not be allowed to discuss possible project ideas, "do's and don'ts", or judges perspectives or advice.

Ok,  here are some of my questions:

How much time did they spend on their projects?  How much time on average did they spend on it for the week?  Finally how long before the competition did they start working on the project?

How did they come up with their idea for the project?

What did they do when they were stuck and how did they stay motivated?

My last question is what do they think that the hardest part of the whole process was?

Do you need a mentor such as a professor at a local university?

How practical is it to do a project completely on your own?

thanks

If interested in the competition in the future, (I'm a rising sophomore) what are the best ways to prepare for the competition? Working on competition math problems? Reading science magazines frequently? Improving writing skills substantially? Of course, trying to qualifyfor the Research Science Institute program is the best way to prepare; but I believe that my skills have developed far too late for me to even have a chance at there (one has to be REALLY badass).

Also, how many of the students do not have parents as scientists? (Thus, I lack connections and a convenient person to work on the project with).

If I don't have connections, then will I absolutely must overcome any shyness? (since I would have to call professors, right)?

Is competition regional at first? Would similarly talented students in Alaska or Wyoming have an advantage over students in California or New York? And just how intense is the competition?

Will preparing for Olympiads hurt my chances at this competition? (since they are different in form and I need all the time I can get).

Finally, are you people who don't enjoy your sleeping hours?

Wow...all my questions have pretty much been asked above...I just have some more:

Exactly how did you get inspired and pick your topic? Did a teacher inspire you or did you encounter it while working on something and becoming curious of something?

Also, is the complexity of your research topic more important than the way you present it or vice-versa?

Thanks

Also...

The fields are chemistry, biology, physics, and mathematics, right? How much specialized knowledge do you need to know in order to be competitive? (among the various fields). Is taking an AP course in the particular subject virtually necessary for becoming competitive? (since many schools do not offer AP courses for all of the fields).

This thread is giving me the erroneous unread thread message.

it didn't to me.

probably the software dislikes you :D

I just have one more question:
Does the Oct 3 deadline mean you have to have your project finished by oct 3 or just register by the 3rd?

I think they don't mean this year's october :? you probably had to dig very deep for this topic.


[hide="My Method"]
$\frac{\ln e^{\ln 100}}{\ln e^{\frac{9}{2}}+\ln e^{\frac{11}{2}}}= \frac{\ln 100}{\frac{9+11}{2}}= \frac{\ln 100}{10}= 0.4605...$[/hide]

Basically, you use the fact that because e is the base of the ln, $\ln e = 1$ Thus, only the exponents of e need to be evaluated, as the expression $\ln e^{k}= k$

- Nawahd Werdna

[quote="AstroPhys"][hide="Continuing"]
$\frac{\ln 100}{10}= \frac{\frac{ \log 100}{\log e}}{10}= \frac{\frac{2}{\log e}}{10}= \frac{1}{5 \log e}$
For the simplified answer.
[/hide][/quote]

You don't need to convert to $\log$; $\ln$ is fine also for simplifying.

You could also have:

$\frac{\ln100}{10}=\frac{2\ln10}{10}=\frac{\ln10}5$

There really isn't much point in changing it any more.

Ok,
[hide]
The top is now $\ln{e^{e^{\ln{100}}}}$
Because $e^{\ln{x}}=x$,
it becomes $\ln{e^{100}}$
or $100$
The bottom hasn't changed, it's still 10.
So the answer is $\frac{100}{10}=10$

[/hide]

EDIT: Edited.

Yea I thought so...

[hide]Sorry no latex
Hey guys you're only solving half of the problem. Remember ther is a natural logarithm next to that fraction so the answer is ln 10/ln10 or 1. I know this is obvious but just putting 10 is kinda confusing.[/hide]

emjay285, miyomiyo seems to have changes the problem again.   :huh:
[hide]
$\frac{\ln e^{e^{\ln 100}}}{\ln e^{\frac{9}{2}}+\ln e^{\frac{11}{2}}}= \frac{e^{\ln 100}}{10}= \frac{100}{10}= 10$
Then $\frac{\ln 10}{\ln 10}= 1$
[/hide]

I'm not sure why, but I keep ending up with (ln 50 / ln 10). What am I doing wrong?

Oh, I get it...stupid mistake. I'm getting 1 now.


Let $a,b,c$ be real numbers with $0 < a< 1$, $0 < b < 1$, $0 < c < 1$, and $a+b + c = 2$.
Prove that $\dfrac{a}{1-a} \cdot \dfrac{b}{1-b} \cdot \dfrac{c}{1-c} \geq 8$.

Replace a , b with $\sqrt{ab}$.

Then $(1-a)(1-b)$ becomes $(1-\sqrt{ab})^2$.  If we determine the latter to be larger we are done.

We compare $-(a+b)$ to $-2\sqrt{ab}$.  The latter is clearly larger.

A less inspired solution can be :

$1-a = (b+c-a)/2$ to homogeneize

We have to prove $abc \geq \left (b+c-a\right )\left (c+a-b\right )\left (a+b-c \right )$

Or ${a}^{3} +{c}^{3} +{b}^{3} +3\,abc \geq {a}^{2}b +{a}^{2}c +a{b}^{2} +a{c}^{2} +{b}^{2}c +b{c}^{2}$ which is Schur

Put $x=1-a$ and so on, so that $x+y+z=1$

Then the inequality becomes

$\left(\dfrac{y+z}{x}\right)\left(\dfrac{z+x}{y}\right)\left(\dfrac{x+y}{z}\right)\geq 8$, which is the same as

$(x+y)(y+z)(z+x)\geq 8xyz$

which follows immediately from AM-GM.

indeed, your solution is much simpler blahblahblah  :blush:

Substitute $x=1-a, y=1-b, z=1-c$. Now obviously if you subtract  each number from 1 then it it's um will get reduced by 3 so $x+y+z=3$. Now,
\[\dfrac{a}{1-a}\cdot \dfrac{b}{1-b} \cdot \dfrac{c}{1-c} \geq 8\ \Rightarrow \dfrac{1-x}{x}\cdot \dfrac{1-y}{y} \cdot \dfrac{1-z}{z} \geq 8\ \]
After substituting $1=x+y+z$, we obtain:
\[\dfrac{x+y}{x}\cdot \dfrac{x+z}{z} \cdot \dfrac{y+z}{z} \geq 8\ \] And now we need to prove this. So, by $AM-GM$ we have:
\[\dfrac{x+y}{2} \geq \sqrt{xy}\]
\[\dfrac{y+z}{2} \geq \sqrt{yz}\]
\[\dfrac{x+z}{2} \geq \sqrt{xz}\]
Now, multiplying these we obtain
\[\dfrac{x+y}{2} \cdot \dfrac{y+z}{2} \cdot \dfrac{x+z}{2} \geq \sqrt{x^2y^2z^2} \Rightarrow (x+y)(x+z)(y+z) \geq 8xyz\]


Let

m integer > 0
a = (m+1)^(2m-1)
b = ((m-1)!)^2
c = 2m+1

(a+b) mod(c) = 0 only if c is composite

First : Is this statement true?
Second : If true can we prove it?

Thank you for any comment.

PS : sorry for my writing mistake (a+b) not (a+c)

1.  What's the role of $ b$ here?

2.  Why don't you recognize that this statement is equivalent to $ a \equiv 0 \bmod c$ (even if you intended to put $ b$ in there somewhere)?  You don't actually seem to know a lot of number theory for someone who's so interested in primality testing - have you considered taking a course or buying a book to learn more number theory so you can answer questions like these for yourself?

[quote="t0rajir0u"]1.  What's the role of $ b$ here?

2.  Why don't you recognize that this statement is equivalent to $ a \equiv 0 \bmod c$?  You don't actually seem to know a lot of number theory for someone who's so interested in primality testing - have you considered taking a course or buying a book to learn more number theory so you can answer questions like these for yourself?

3.  Yes (actually, this is "if and only if").  If $ c$ is composite, then $ c \equal{} pq$ where $ p, q < m \plus{} 1$, so both appear in $ a$ if $ m > 1$.  If $ c$ is prime, it obviously can't divide $ a$.[/quote]

Ok.
Sorry for my little knowledge in number theory. I'm just trying to understand and to learn more.
If the statement (a+b) mod c = 0  and a mod(c) = 0 (too) what about b mod(c)? b mod(c) has to be equal to 0

How can you explain me that?
Thank you for your comment.

[quote="Matharith"]a mod(c) = 0 [/quote]
This can never occur; $ \gcd(m \plus{} 1, 2m \plus{} 1) \equal{} 1$.  If you want to actually learn some number theory, there are [url=http://www.maths.mq.edu.au/~wchen/lnentfolder/lnent.html]lots of resources on the web[/url] that can help you.

Anyway, your claim is equivalent to the following:  if $ c$ is prime, then $ a \plus{} b \not\equiv 0 \bmod c$.  Now, note that $ 2(m \plus{} 1) \equiv 1 \bmod c$, so

$ a \equiv \frac {1}{2^{2m \minus{} 1}} \equiv 2 \bmod c$

by Fermat's little theorem.  On the other hand,

$ m^2 b \equiv m!^2 \equiv m! ( \minus{} 2m)( \minus{} 2m \plus{} 1)...( \minus{} m \minus{} 1) \equiv ( \minus{} 1)^m (2m)! \equiv ( \minus{} 1)^{m \plus{} 1}$

by Wilson's theorem, hence $ b \equiv ( \minus{} 1)^{m \plus{} 1} \frac {1}{m^2} \bmod c$.  Now, $ m( \minus{} 2) \equiv 1 \bmod c$, so $ \frac {1}{m} \equiv \minus{} 2 \bmod c$, hence

$ b \equiv 4( \minus{} 1)^{m \plus{} 1} \bmod c$.

And now it's clear that $ a \plus{} b$ is congruent to either $ 6$ or $ \minus{} 2 \bmod c$.  The only possible counterexample is when $ m \equal{} 1$, which is actually a valid counterexample (here $ a \equal{} 2, b \equal{} 1, c \equal{} 3$), but for all $ m > 1$ this works.

Note that $ a$ is pretty superfluous; since we already know what the result of the computation is, there's no point in doing it.  Everything depends on $ b$, so this isn't any better than the Wilson's theorem test.

Now I'm convinced by your explanations.

It is like Wilson's theorem because I used the theorem by expanding a factorial.

Thank you for all your comments and sorry my little knowledge.
I will try to do better.

[i]Note that  is pretty superfluous; since we already know what the result of the computation is, there's no point in doing it. Everything depends on , so this isn't any better than the Wilson's theorem test.[/i]

It is slightly different than Wilson theorem because when [u]m is odd [/u] we could factorize a+b
m=9 for example

a= 10^17
b= (8!)^2

More than that we could find a general factorization of the sum a + b under some conditions.
Maybe for some big numbers the solution could be easier.
There is another advantage to my test : all Carmichael numbers pass the test.
I think that there is way to improve it.

[quote="Matharith"]Maybe for some big numbers the solution could be easier.[/quote]
Pretty unlikely.  Computing $ a$ just takes too much more effort to really be worth it; in the time it takes to compute $ a$ [url=http://en.wikipedia.org/wiki/Miller%E2%80%93Rabin_primality_teste]Miller-Rabin[/url] can already decide the answer, and all of this is not taking into account how prohibitely difficult it is to compute $ b$.

[quote="Matharith"]There is another advantage to my test : all Carmichael numbers pass the test.[/quote]
This isn't an advantage over the Wilson's test, and nobody uses Fermat's test because of Miller-Rabin, as well as [url=http://en.wikipedia.org/wiki/Solovay-Strassen_primality_test]Solovay-Strassen[/url].

People use Wilson's theorem to test primality?

Of course not!  That's part of what I've been trying to tell Matharith across multiple threads  :|


Prove that there is a Borel set $E\sub\R$ such that for every non trivial interval $I$
\[ 0<m(E\cap I)<m(I) \]
(where $m$ is the lebesgue measure)

I was wonderig if for any $\alpha\in[0,1]$ there is a borel set $E_\alpha$ such
\[ m(E\cap I)=\alpha\cdot m(I) \]
For $\alpha=0,1$ is easy to see that $E_0=\mathbb{Q}$ and $E_1=\mathbb{R}\setminus\mathbb{Q}$ works. Maybe they involve positive measure cantor sets, what do you think?

I posted the first question in this section a long time ago. Try searching for it. I remember Leva1980 and fedja had replies there. As for your second question, the answer is no: there are no measurable subsets $E$ of the real line such that for every interval $I,\ m(E\cap I)=\alpha m(I)$, for fixed $\alpha\in(0,1)$. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=poorly&t=68528]this[/url].

I found it  :), thanks

the topic was [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=rudin&t=32313]cool measurable subset[/url]. The concept of density response negatively my second question, also Grobber's argument of outer measure does ;).


Please dont give me the full solution just some hints and a answer for me to check my work would also be nice, thx!
1) Charlie is at the end of a 1.5 km caravan moving a uniform speed across the Mohave
desert when he is asked to deliver a message to the leader, Mirna, at the head of the
caravan. Moving at constant speed, he delivers the message to Mirna and returns at the
same uniform speed. By the time he returns the caravan has advanced 2 km. How far did Charlie travel?

Write two separate equations - when the messenger goes forward and when he returns. Also write the equation for the distance he traveled. From all of these you'll be able to eliminate ${v_{c}\over v_{m}}$, where $v_{c}$ is the caravan's speed and $v_{m}$ the messenger's speed. Answer: $d=L+\sqrt{L^{2}+s^{2}}=4\,\mathrm{km}$, where $L$ is the length of the caravan and $s$ is the distance traveled by the caravan.

Should i find equations of the distance traveled by the messenger as he goes forward and returns? cause with the equations i made up,, I seem to have insufficent infomation.can u give a bit more hints.. i'm newb at this..

Use the concept of relative motion. Charlie is moving relative to the train.

Well, I don't seem to be able to give more hints without solving the problem, so here we go:

Let $t_{1}$ be the time of his forward travel. Then $v_{m}t_{1}=L+v_{c}t_{1}$ (he travels both the length of the caravan AND the distance traveled by the caravan itself during that time). Similarly, on the way back we have $v_{m}t_{2}=L-v_{c}t_{2}$ (his traveled distance is now diminished, because he runs in the opposite direction of the caravan).

From those two equations, $t_{1}=\frac{L}{v_{m}-v_{c}}, t_{2}=\frac{L}{v_{m}+v_{c}}$. The total time is $t_{1}+t_{2}=\frac{2v_{m}L}{v_{m}^{2}-v_{c}^{2}}$. We know that the caravan, during that time, traveled $s$:

$s=v_{c}\cdot\frac{2v_{m}L}{v_{m}^{2}-v_{c}^{2}}\implies \frac{s}{2L}=\frac{v_{m}v_{c}}{v_{m}^{2}-v_{c}^{2}}\qquad(*)$

During the same time, the messenger traveled

$d=v_{m}\cdot\frac{2v_{m}L}{v_{m}^{2}-v_{c}^{2}}=v_{m}\cdot{s\over v_{c}}={v_{m}\over v_{c}}s\qquad(**)$

Therefore, the only information we now need is ${v_{m}\over v_{c}}$, and that can be calculated from $(*)$. Put $x={v_{m}\over v_{c}}\implies v_{m}=xv_{c}$:

${s\over 2L}=\frac{xv_{c}\cdot v_{c}}{x^{2}v_{c}^{2}-v_{c}^{2}}=\frac{x}{x^{2}-1}$

That gives $x^{2}-1={2L\over s}x\iff x^{2}-{2L\over s}x-1=0\implies x=\frac{{2L\over s}+\sqrt{{4L^{2}\over s^{2}}+4}}{2}$ (the negative solution is discarded)

Simplifying gives $x={L\over s}+\sqrt{{L^{2}\over s^{2}}+1}$, and now $(**)$ gives $d=xs=L+\sqrt{L^{2}+s^{2}}$


Fat Man and Little man goes out shopping. Fat Man goes to the restaurant bar. He can order a $1,$3, or $5 dollar dish. He can buy as many dishes as he wants. Little man goes to the game center. He can buy tickets to basketball, baseball, or soccer games that cost 3, 5, and 7 dollars respectively. He can buy as many tickets as he wants to each game. How many possible ways can each of them enjoy themselves if their total budget for combined spendings is$50?

Is it the coeff of $ x^{50}$ in the expansion of
$ (x^1 \plus{} x^2 \plus{} \cdots x^{50}) \cdot (x^3 \plus{} x^5 \plus{} x^6 \plus{} x^7 \plus{} x^9 \plus{} x^{10} \cdots x^{49} \plus{} x^{50})$ ?  :)

Fat Man getting a $ 1$ and $ 5$ dollar dish or getting two $ 3$ dollar dishes ought to be counted differently in the problem, but your generating functions considers them the same.

I'm getting that it's the coefficient of $ x^{50}$ in
$ \frac1{1\minus{}x\minus{}x^3\minus{}x^5} \cdot \frac1{1\minus{}x^3\minus{}x^5\minus{}x^7}$

well i th ink it would be

$ (x\plus{}x^3\plus{}x^5)^{infinity}(x^3\plus{}x^5\plus{}x^7)^{infinity}$

and then the coefficient of $ x^{50}$

That would be correct if Fat and Little were both purchasing an infinite amount of the available commodities, and then the amount of ways they'd spend $ 50$ dollars is $ 0$ (they'd always spend an infinite amount). What you probably meant was

$ ((x\plus{}x^3\plus{}x^5) \plus{} (x\plus{}x^3\plus{}x^5)^2 \plus{} \ldots)((x^3\plus{}x^5\plus{}x^7)\plus{}(x^3\plus{}x^5\plus{}x^7)^2\plus{} \ldots )$

which simplifies to the quantity in my post.

[hide="Actually..."] The correct generating function is

$ \frac{1}{(1 \minus{} x)(1 \minus{} x^3)(1 \minus{} x^5)(1 \minus{} x^3)(1 \minus{} x^5)(1 \minus{} x^7)}$.

There's no reason to group Fat Man's purchases and Little Man's purchases together.  The only thing that's important is the particular combination of six commodities. [/hide]

[quote="#H34N1"]Yeah, I got that. But how do you now find teh coefficient?[/quote]

The general method is by partial fractions.   :(

[quote="SCHNOGGINVOGGINSCHWITZERW"]how do you guys all get these random functions?[/quote]

For more information on the topic of generating functions, I highly recommend [url=http://www.math.upenn.edu/~wilf/DownldGF.html]generatingfunctionology[/url].

[quote="SCHNOGGINVOGGINSCHWITZERW"]how do you guys all get these random functions?[/quote]
For the record, t0rajir0u's post is the correct one, so only pay attention to that.

[hide="The simple explanation as to how he got it:"]
A generating function is a function of the form:

$ \sum_{n \equal{} 0}^{\infty}a_{n}x^n \equal{} a_{0} \plus{} a_{1}x \plus{} a_{2}x^2 \plus{} \cdots$

Remember that the main problem we're doing is finding the number of solutions for:

 $ j_{1} \plus{} j_{2} \plus{} j_{3} \plus{} j_{4} \plus{} j_{5} \plus{} j_{6} \le 50$

When

 $ j_{1} \in \{0,1,2,3,4\cdots\}$
 $ j_{2}, j_{4} \in \{0,3,6,9,12\cdots\}$
 $ j_{3}, j_{5} \in \{0,5,10,15,20\cdots\}$
 $ j_{6} \in \{0,5,10,15\cdots\}$

Let's just focus on the generating function for $ j_{2}$ as an example. The coefficients ($ a_n$) have to be:

$ 1,0,0,1,0,0,1,0,0,1\cdots$

Explaining why we need to find that instead of 0,3,6,9... would be tricky for me. Just think of it as a series of booleans for the corresponding $ n$.

The generating function for those coefficients is clearly:

$ 1 \plus{} x^3 \plus{} x^6 \plus{} x^9 \plus{} \cdots$

Now let's use basic algebra and play around with this function:

$ (1 \minus{} x^3)(1 \plus{} x^3 \plus{} x^6 \plus{} x^9 \plus{} \cdots) \equal{}$
$ (1 \minus{} x^3) \plus{} (x^3 \minus{} x^6) \plus{} (x^6 \minus{} x^9) \plus{} (x^9 \minus{} x^{12}) \plus{} \cdots$

Notice how all of those $ x^n$'s cancel out except for $ x^0 \equal{} 1$. That means:

$ (1 \minus{} x^3)(1 \plus{} x^3 \plus{} x^6 \plus{} x^9 \plus{} \cdots) \equal{} 1$
$ 1 \plus{} x^3 \plus{} x^6 \plus{} x^9 \plus{} \cdots \equal{} \frac {1}{1 \minus{} x^3}$

We repeat that for all of them, and we end up getting

$ \frac {1}{(1 \minus{} x)(1 \minus{} x^3)(1 \minus{} x^5)(1 \minus{} x^3)(1 \minus{} x^5)(1 \minus{} x^7)}$
[/hide]

AFAIK this problem takes a lot of time. From there, there's a way to figure out the coefficients of $ x^1, x^2, x^3, \cdots x^{50}$ in the generating function we receive and then add them. Not only am I unsure as to how, but it would also take awhile, especially since it is an inequality (perhaps by accidentally poor wording?).

If they're spending fifty dollars [i]or less[/i], all you have to do is consider "spending on nothing" as a seventh commodity; in other words, multiply by $ \frac{1}{1 \minus{} x}$.

What's partial fractions? How do you use them? Can you please provide an example? Sorry for my noobishness. :(

See http://en.wikipedia.org/wiki/Partial_fraction .  The basic idea is that partial fraction decomposition will allow us to write the generating function as a sum of generating functions.  You'll probably want to do it on a smaller example, though; try $ \frac{1}{(1 \minus{} x)(1 \minus{} x^2)}$ first once you've read a little about it.


linear algebra
matrix
superior algebra
superior algebra solved

Let A be in M_n(R) s.t. A=A^t, where A^t is the transpose of A.
Prove that if exists U in M_n(R) s.t. A^2=UAU^(-1)-A then A=O_n.

Very easy

tr(A^2)
= tr(UAU^(-1)) - trA
= tr(U.U^(-1)A) - trA
=trA - trA = 0

A is symetric
0 = trA^2 = tr(A.tA) where tA is the transpose of A

 \sum (a(i,j))^2 = 0 => a(i,j)= 0 so A = 0_n


For every real number $x,y(y\ge 0)$ we denote $S(x,y)=max(a-2x+y,1+2x+y,x^2-y^2)$
Find the minimum value of $S(x,y)$

Oh I'm sorry, the right problem is:
For every real number $x,y(y\ge 0)$ we denote $S(x,y)=max(1-2x+y,1+2x+y,x^2-y^2)$
Find the minimum value of $S(x,y)$


geometry
3D geometry
tetrahedron
geometry unsolved

$ABCD$ is a tetrahedron with $\angle BAC = \angle ACD$ and $\angle ABD = \angle BDC$. Show that $AB = CD$.

[quote=soryn]Any ideas, pls??[/quote]

just open the tetrahedron and it make it 2D . then its really easy.


trigonometry
function
real analysis
Euler
complex analysis
algebra
rational function

I saw a nice problem given in oral examination which asked to compute $\sum_{n\in Z}{\frac{1}{(n-x)^2}}$ where $x$ is not an integer number. Of course, the problem had many questions that had as purpose to show what is the value. I won't say the answer, but instead I ask for someone to explain a little bit how we can compute it without exteriour help. Are there specific tricks or is it just a particular problem? I really doubt...

Clearly  $\frac{\pi^2}{\sin^2 (\pi x)}$  :lol:

Thank you Moubi, you really helped me!!! :P  :P  ;) . If you hadn't known it before, would you say again clearly?

[quote="harazi"] If you hadn't known it before, would you say again clearly?[/quote]
Yes, harazi, for people who learned advanced complex analysis, it is clear :P . Seems like it's time for you to become one of them too  :) .

Ok, Fedja, but if you learn all the other things I have to learn now, I promiss to learn advanced complex analysis. Of course I will try to learn it, but I still have so many things to learn before that.  :( . Anyway, some natural way to find it?

[quote="harazi"] Anyway, some natural way to find it?[/quote]
I don't know if there is any pure real-variable approach, but, from the complex analysis point of view, you are dealing with an even meromorphic function $f(z)=\sum_n \frac 1{(n-z)^2}$ that is $1$-periodic, bounded if you are at positive distance from the poles and tending to $0$ as $\text{Im}\,z\to\infty$. Other such meromorphic functions are $\frac 1{\sin^k \pi z}$ ($k=2,4,6\dots)$. So, it becomes natural to look for a linear combination that would cancel the poles at integer points. Then all we'll need to do is to apply the Liouville theorem to the difference to conclude that the difference is identically $0$. A direct computation shows that $\frac{\pi^2}{\sin^2\pi z}$ is the right one (all you need to do is to check that the pole at $0$ gets cancelled, which is a simple exercise in Taylor series).

Now, can you figure out what $\sum_n\frac 1{(n-x)^3}$ is?

[quote="harazi"]if you learn all the other things I have to learn now, I promise to learn advanced complex analysis.[/quote]

I've been through that ( +  History of the Communist Party and a few other equally exciting subjects :lol:), so I know what you mean...

[quote] Of course I will try to learn it, but I still have so many things to learn before that. [/quote]

In principle, it seems that your background in real analysis is sufficient for reading introductory books in complex analysis (though my judgement here may be erratic because all I saw is the problems you posted and solved on this forum). And it is a whole (sub)world of mathematics you are missing now. I do not know how much free time you have but if you have some, my advice would be to try to teach yourself something beyond multiplication of complex numbers. My overall experience is that the earlier you learn something, the better you remember and understand it for the rest of your life (at least in mathematics), though, of course, it is just my humble opinion.

[quote="fedja"]
Now, can you figure out what $\sum_n\frac 1{(n-x)^3}$ is?
[/quote]
It is $f(z)=-\frac{8\pi^3}{sin^3(2\pi z)}+\frac{4\pi^3}{sin(2\pi z)}$ Am i right?[/code]

[quote="fedja"] complex analysis[/quote]

What do you think a about the book
Rudin
Real & Complex analysis

[quote="fedja"]You definitely got the idea but the expression you proposed has extra poles at half-integers. Probably that is partially my fault because I just noticed a stupid misprint in my post (I wrote $k=1,2,\dots$ where I meant $k=2,4,\dots$; I edited it by now). If you want odd powers, the best building blocks to use are negative powers of $\tan\pi z$: they are $1$-periodic and have poles exactly at the integer points. They can be used for even powers too but the identity $1+\frac 1{\tan^2\pi z}=\frac 1{\sin^2\pi z}$ makes inverse even powers of $\sin\pi z$ just as good.[/quote]

I haven't check the details:I got $f(z)=-\frac{\pi^3}{tan^3(\pi z)}-\frac{3}{4tan(\pi z)}$

P.S.Thank you fedja for the explanations

[quote="eugene"]
I haven't check the details:I got $f(z)=-\frac{\pi^3}{tan^3(\pi z)}-\frac{3}{4tan(\pi z)}$
[/quote]
The first term is right, but the second is not. The correct answer is $-\pi^3\left(\frac 1{\tan^3 \pi z}+\frac 1{\tan \pi z}\right)$. Indeed, to get $-\frac 1{z^3}$, one needs to use $-\frac{\pi^3}{\tan^3 \pi z}=-\frac{\pi^3}{\left(\pi z+\frac {\pi^3 z^3}3+\dots\right)^3}=-\frac 1{z^3}+\frac {\pi^2} z+ \dots$. Now it remains to cancel $\frac {\pi^2}z$, which can be done by subtracting $\frac {\pi^3}{\tan \pi z}$. Another way to check yourself here is to note that the function must tend to $0$ as $\text {Im}\,z\to\infty$ while $\tan \pi z$ tends to $\pm i$ depending on whether $z$ is in the upper or in the lower half-plane. So, to cancel $i^3$ and $i$, one should have the same coefficient at both.

[quote="Moubinool"]What do you think a about the book Rudin, Real & Complex analysis[/quote]
In general, I like Rudin's books a lot, this one being no exception. I do not remember right now how far it goes, so if you want a more constructive opinion, wait until I go to the library next time.

[quote="harazi"] $\sum_{n\in Z}{\frac{1}{(n-x)^2}}$ where $x$ is not an integer number[/quote]

The source is probably Euler, who knew how to guess the correct answer to this type of problem, before complex analysis.  One can think of it as a partial fraction expansion for a rational function of infinite degree (something like $f(z)/g(z)$ with $f,g$ power series converging on the entire complex plane) and then use $sin(z)$ to find a candidate with the right local behavior.  I think it can ultimately be proved without complex analysis other than $exp(2 \pi i) = 1$ and (limits of) finite trigonometric formulae.

now that the pb is solved, here are the hints my friend got at the oral examination:
1/ what can you say about a continuous function $f:R->R$ such that there exists a constant $c>2$ satisfying:
  $f(\frac{x}{2})+f(\frac{x+1}{2})=cf(x)$ for all $x \in R$
2/ conclude that the sum is equal to $\frac{\pi^2}{{sin(\pi x)}^2}$

that's true that it is not at all natural  :D  but far more easy  :)


search
number theory solved
number theory

[quote="TomciO"][quote="Ilthigore"]Why bother? Why not just make the obvious point that $x=y$ is the only solution, since $x(x+1)$ is strictly increasing, so $x>y \Rightarrow x(x+1)>y(y+1)$?  :roll:[/quote]

Because that's wrong.[/quote]

Nothing wrong with Ilthigore's solution. Just consider, that for positive values is indeed $f(x)=x^{2}+x$ strictly increasing. And because we are looking for all positive integers satisfying that equation, his solution is OK.

Note that if the problem will be about  finding all such integers, his solution will not work. But this is not this case..

conceive

Euhm...$x^{2}+x=y^{2}+y\ \Longleftrightarrow\ \left(x^{2}-y^{2}\right)+\left(x-y\right) = (x-y)(x+y+1) = 0$...

[quote="conceive"]Solve in positive integers:

$x^{2}+x=y^{2}+y$[/quote]

Rewrite as $x(x+1)=y(y+1)$.
This problem was discussed before, so search for the link.
Solutions are $x=y$ or $x+y=-1$ I think.

This problem is indeed dry. Try its generalization here:

http://www.mathlinks.ro/Forum/viewtopic.php?t=103725 :lol:

conceive

I think this is obvius,and the best solution is ZetaX's reply.


[quote="Elio (n)"]$ x^{9} \minus{} 29x^{8} \plus{} 29x^{7} \minus{} 29x^{6} \plus{} 29x^{5} \minus{} 29x^{4} \plus{} 29x^{3} \minus{} 29x^{2} \plus{} 29x \plus{} 1 \equal{} ?$

Pour $ x \equal{} 28$[/quote]

Je pense que tu sais utiliser une calculette ou excel :huh: Sans doute il y a une version generale de cette expression, mais quand-mème personne ne veut faire tes devoirs...

Also, speak english on mathlinks. :P

You don't need a calculator/computer, and this probably isn't homework.

[hide="Solution"]The expression is equal to $ (x\minus{}28)(x^8 \minus{} x^7 \plus{} x^6 \minus{} x^5 \plus{} x^4 \minus{} x^3 \plus{} x^2 \minus{} x) \plus{} x \plus{} 1$. Putting in $ x \equal{} 28$ gives $ 0 \plus{} 28 \plus{} 1 \equal{} \boxed{29}$.[/hide]

[quote="Elio (n)"]THis are not mes devoir . This kind of sum are not created to solve with calculatrice.  :wink:[/quote]

Yeah, the problem is indeed not-calculator-needing (you can also solve it using $ x^9\plus{}1\equal{}(x\plus{}1)(x^8\minus{}x^7\plus{}x^6\minus{}x^5\plus{}x^4\minus{}x^3\plus{}x^2\minus{}x^1\plus{}1)$), so I'm sorry for being harsh to you Elio :( Et tu m'as donné une possibilité pour exercer mon francais... je ne peut plus l'apprendre à l'école :(


A purse contains 4 pennies, 2 nickels, 1 dime, and 1 quarter. Different values can be obtained by using one or more coins in the purse. How many different values can be obtained?

[hide="Solution"]I am pretty sure every number 1 through 49 can be obtained:

0 can't be obtained because it doesn't use any coins.
1-4 can be obtained with just pennies
5 can be obtained using just a nickel
6-9 can be obtained by adding a nickel to every case for 1-4
10 can be obtained using just a dime
11-14 can be obtained by adding two nickels to every case for 1-1
15 can be obtained using a nickel and dime
16-19 can be obtained by adding a nickel and a dime to every case for 1-4
20 can be obtained using two nickels and dimes
21-24 can be obtained by adding two nickels and a dime to every case  for 1-4
25 can be obtained using just a quarter
26-49 can be obtained by adding a quarter to every case for 1-24
50 or greater can't be obtained because 49 is the greatest possible value, which uses all the coins

So the answer is $ \boxed{49}$.
[/hide]

PowerofPi. That is the same answer I got, in the same way!!!!!!!
Anyway, thanks for saving me the work of posting it up :P


let $u=\frac{log x}{log y}$ , where $x \geq y \geq 2$. if $u \longrightarrow \infty  $ as $x \longrightarrow \infty $ $u \leq y^{\frac{1}{2}}$   then the following estimates holds:

                $\psi(x,y)=xu^{-u+o(u)}$

[quote="grodij"] $\psi(x,y)=xu^{-u+o(u)}$[/quote]

could you explain the notation a bit?
what is $ \psi $ and what is $o$?

[tex]o[/tex] is probably a notation for asymptotics saying that:
[tex]f ( x ) = o ( g ( x ) ) \equiv f ( x ) \succ g ( x ) \equiv \lim_{x
\rightarrow \infty} \frac{g ( x )}{f ( x )} = 0[/tex]
On the other hand I've got no idea on what he meant by [tex]\psi[/tex]

for every  positive integer $n$,let $\rho (n)$ denote the largest squarefree
divisor $r$ of $n$ for which $gcd(r,\frac{r}{n})=1$.It is easy to see that $s=\frac{n}{\rho (n)}$ is the largest
powerful divisor of $n$ (recall that a positive integer $m$ is said to be $powerful$ if $p^2|m$ for everyprime $p$ that divides $m$).


Given are equaliteral $\triangle ABC$ with any $M$ on plane find max value of
$\frac{MA+MB+MC}{MA^{3}+MB^{3}+MC^{3}}$   :)


Steve's empty swimming pool will hold $ 24,000$ gallons of water when full. It will be filled by $ 4$ hoses, each of which supplies $ 2.5$ gallons of water per minute. How many hours will it take to fill Steve's pool?

$ \textbf{(A)}\  40  \qquad
\textbf{(B)}\   42  \qquad
\textbf{(C)}\  44   \qquad
\textbf{(D)}\  46   \qquad
\textbf{(E)}\   48$

[hide="Solution"]Since $ 4$ hoses supply $ 2.5$ gallons a minute each, the pool is being filled at a rate of $ 4 \cdot 2.5\equal{}10$ gallons per minute. Thus, it takes $ \frac{24000}{10}\equal{}2400$ minutes to fill the pool. Converting this to hours, it takes $ \frac{2400}{60}\equal{}40$ hours, or answer choice $ \boxed{\mathbf{(A)}}$.[/hide]


algebra
polynomial
function
quadratics
superior algebra
superior algebra unsolved

Suppose $ F(X_{1},\ldots,X_{n}) \in \mathbb{Q}[X_{1},\ldots,X_{n}]$ is a polynomial over the rational field $ \mathbb{Q}$. Suppose that $ F$ has a zero over every number field $ K \neq \mathbb{Q}$. Must then $ F$ have a zero over $ \mathbb{Q}$?

The question, as it stands, has a trivial affirmative answer: take $ K : = \mathbb{Q}(2^{1/m})$ for $ m > \deg{F}$, which successfully imitates the field of rational functions $ \mathbb{Q}(t)$ in our situation. Such argument shows, more generally: if $ X$ is an algebraic variety over a field $ K$ and $ X(L) \neq \emptyset$ for all finite extensions $ L/K$, then as well $ X(K) \neq \emptyset$. [More generally, if $ X$ is a variety over a scheme $ S$ such that $ X(T) \neq \emptyset$ for all etale coverings $ T \to S$ that are not isomorphisms, then  as well $ X(S) \neq \emptyset$. ]

The interesting question is this: if $ F$ has a zero over every [b]quadratic [/b] extension of $ \mathbb{Q}$, must $ F$ have a zero over $ \mathbb{Q}$?


Δίνω ακόμη ένα πρόβλημα στερεομετρίας.. Είναι γενικά ωραίος τομέας , αλλά δυστυχώς δεν διδάσκεται στο λύκειο :(

Θεωρούμε μια κυρτή $ n$-γωνική πυραμίδα . Να αποδειχθεί ότι το άθροισμα των δίεδρων γωνιών της πυραμίδας βρίσκεται μεταξύ των $ 2(2n\minus{}3)$ και $ 2(n\minus{}1)$ oρθών γωνιών .

Επειδή ακριβώς είναι ενδιαφέρουσα η στερεωμετρία και δεν ξέρω τίποτα  :blush:, μπορείς μήπως mostel να γράψεις μια λύση;


quadratics
algebra
SFFT
special factorizations
\/closed

Hi all,

I am about to start learning Algebra.What should I keep as a goal in this subject?..like solving complex equations..

What goals you have in other subjects?

Regards,
ram

I would not say there is anything specific you should learn.  Algebra is the foundation for all high school and college math, so your goal should certainly be to learn everything you can.  If you are ignorant in algebra, you will find it impossible to pass more difficult math classes (and yes, they will get much more difficult).

Like Stank said, there aren't any specific goals for Algebra. I just completed Algebra I, so well...
an example you can strive for is to reach a certain point. Like to know how to extract a basic quadratic equation.
Or maybe you want to be able to learn the foundation for upper class math.
Whatever you choose well, know there's no right answer.

Also, this belongs in "Middle School Classroom Math".

[quote="meewhee009"]Like Stank said, there aren't any specific goals for Algebra. I just completed Algebra I, so well...
an example [i][b]you can strive for is to reach a certain point. Like to know how to extract a basic quadratic equation. [/b][/i]
Or maybe you want to be able to learn the foundation for upper class math.
Whatever you choose well, know there's no right answer.[/quote]

you can strive to reach few points....could you kindly set me few points to reach...I mean they should be very toughf
Kindly do this for me..

[b]Like to know how to extract a basic quadratic equation[/b]

could you elaboarte on this?
Any books in this direction?

many thanks for the reply,
~ram[/u][/i]

I can give you a couple I found quite difficult,
[b]Learning to work with rationale denominators [/b]
[b]Learn about Simon's Favorite Factoring Trick[/b]
and yes,
[b]Working with basic quadratic equations, using the formula  [/b]
The book all of this is in,
Introduction to Algebra. By Richard Rusczyk
That's what I took for Algebra I.
They sell it on AoPS,
and if you take the class you get a discount... in the book it's split into parts when it comes to quadratic formula.

[quote="meewhee009"]
The book all of this is in,
Introduction to Algebra. By Richard Rusczyk
That's what I took for Algebra I.
They sell it on AoPS,
and if you take the class you get a discount... in the book it's split into parts when it comes to quadratic formula.[/quote]
If you are beginning to learn Algebra, a great course to take is the Algebra 1 Class. [url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=41]Click here for more info[/url] You can see what each class will be about, and you can set goals based on that. :) Good luck!


inequalities
function
inequalities proposed

I remember that it is easy to show by Jensen inequality.

You only check convexity of graph of fuction $ log_{x}a$ for $ 0\prec x \prec1$ and decreasing part of function.
And I remember that We use AM-GM inequality in the part that function is decreasing

P.S.
$ \sum_{i \equal{} 1}^{n}{log_{a_{i}}(t_{n})} \ge n log_{\frac {\sum_{i \equal{} 1}^{n}a_{i}}{n}}t_{n} \ge n log_{\frac {\sum_{i \equal{} 1}^{n}a_{i}}{n}}(\frac {\sum_{i \equal{} 1}^{n}a_{i}}{n})^{n \minus{} 1} \equal{} n(n \minus{} 1).$

First, we use jensen and, second because this function is decreasing function, By AM-GM oR AM-HM inequality, we can get result.


The log tells you that the error is on line 7 and it is in $ \$$\textbf{531440=888888_{9}}$ \$$. What has happened is that you are using _ in text mode which gives an error. Solution: use \mathbf instead $ \$$\mathbf{531440=888888_{9}}$ \$$.

In fact, the dollar sign changes things from text to maths mode and then \textbf puts it back to text mode so they cancel each other out!


advanced fields
advanced fields unsolved

Let $ X$ the set of representatives for the equivalence in $ P(R)$
$ A\equiv B$ if and only if $ (A\setminus B)\cup(B\setminus A)$ is finite.

Evaluate $ |X|$.

in other words, we're interested in the index $ d$ of $ 2^{(R)}$ in $ 2^R$, where $ R$ is an infinite set. every equivalence class is bounded by

$ \left| \bigcup_{E \subseteq R \text{ finite}} 2^E \right| \leq \left| \bigcup_{E \subseteq R \text{ finite}} \aleph_0 \right| \leq |E(R)| \aleph_0 \equal{} |R| \aleph_0 \equal{} |R|$.

thus $ 2^{|R|} \equal{} \sum_{g \in 2^R / 2^{(R)}} |g| \leq d |R|$. this implies $ d > |R|$, and then $ 2^{|R|} \equal{} d$.

The answer is correct, but I cannot understand your notation (and so your proof)

Who is E? Who is the index d?
I cannot understand how can you find this bound for the cardinality of the equivalence classes.

Thank you very much...

$ E$ and $ d$ are defined. $ E(R)$ is the set of finite subsets of $ R$. elements which are equivalent to $ f \in 2^R$ are determined by changing a finite number of entries, i.e. by elements of $ 2^E$ where $ E$ is a finite subset of $ R$.


inequalities
AMC
USA(J)MO
USAMO
trigonometry
geometry
angle bisector

If x^2 + 3xy + y^2 =60, where x and y are real, determine the maximum possible value of xy.

[hide]
$x^2+y^2+3(xy-20)=0$

$x^2+y^2$ will always be possible, so $xy<20$?
[/hide]

err...nvm

[hide]
$x^2+2xy+y^2+xy=60$

$(x+y)^2+xy=60$

From AM-GM, we have $x+y \geq 2\sqrt{xy}$. To maximize the value of $xy$, we set $x+y=2\sqrt{xy}$. Putting this into the equation:

$4xy+xy=60 \Longrightarrow xy=12$

The maximum value of $xy$ is going to occur when both $x$ and $y$ are positive, or negative, so we know we can apply AM-GM in the first place.
[/hide]

[quote="Altheman"][quote="cincodemayo5590"][hide]
I'm not sure if this is correct:
By AM-GM we have $x+y \geq 2\sqrt{xy} \iff x^2 + y^2 \geq 2xy$
So then $5xy \leq x^2+3xy+y^2=60 \iff \boxed{xy \leq 12}$. I think.[/hide][/quote]

i believe that this is correct, but you have to take into account when x and or y are negative (or explain why those cases don't matter)

also you need to show that this exists...it is not that hard...what if there was another inequality that said that...basically find x,y such that maximum occurs...i know this is pretty trivial but you would loose points on say the usamo for leaving that out...[/quote]

I guess I'll expand on cincodemayo5590's solution (since I fail)...
[hide]
Since $(-x)^2+(-y)^2=x^2+y^2, -(xy) \le xy$, and  $(-x)\cdot(-y)=xy$, the sign of $x$ and $y$ does not affect the maximum value of $xy$.

The maximum is acheived at $x=2\sqrt{3}, y=2\sqrt{3}$ and $x=-2\sqrt{3}, y=-2\sqrt{3}$?[/hide]

[quote="Altheman"][quote="cincodemayo5590"][hide]
I'm not sure if this is correct:
By AM-GM we have $x+y \geq 2\sqrt{xy} \iff x^2 + y^2 \geq 2xy$
So then $5xy \leq x^2+3xy+y^2=60 \iff \boxed{xy \leq 12}$. I think.[/hide][/quote]

i believe that this is correct, but you have to take into account when x and or y are negative (or explain why those cases don't matter)

also you need to show that this exists...it is not that hard...what if there was another inequality that said that...basically find x,y such that maximum occurs...i know this is pretty trivial but you would loose points on say the usamo for leaving that out...[/quote]

lol ok alex, this is what i would include in a usamo or something proof:
1) If $x<0<y$, then $xy$ is negative and would be less than any of the cases in which $x \geq y \geq 0$.
2) If $x \leq y \leq 0$ then all negatives are cancelled out and $xy$ is still $\leq 12$.

weeeee

ok...here is a problem that you need to be careful with these kinds of things

let $p$ be a positive integer, take a triangle $ABC$ with $AB=BC=p$...the angle bisector of angle B hits AC at D, find the maximum integer length of BD...

obviously, we get that

$BD=p*\cos\frac{C}{2}$

we have that cos in this problem ranges from 0 to 1...so not being careful and assuming that a certain maximum exists, it seems that

max BD=p

...but this is not correct...explain why...

ok i see what you're saying, you need to make sure that the maximum that you found exists (becuase if $\cos{\frac{C}{2}}=1$ then its a line, not a triangle..lol).

ok so does that mean that i need to find $x,y$ that have the maximum $12$? If so, its easy ot show that the equality case of AM-GM is $x=y$ so $5x^2=5y^2=60 \Longrightarrow (\pm 2\sqrt3 , \pm 2\sqrt 3)$

(to reiterate, weeeeeeeeeee) :D


[color=cyan]Take the poll.  Leave a comment.  Yay.[/color]

I think you posted way too many questions at once.... maybe over a 3 week span would have been better... your posts took up too much space.

Hmm, I personally like a bunch of questions...though maybe not that many. But a good 5-7 is nice. And I think that the difficulty ranged from easy-intermediate. A couple were on the too easy side for AMC purposes, but most were ok.

You could post easier of the questions in MATHCOUNTS or Getting Started, but they were at my level (meaning AMC 9) so I loved them!

and yes, AMC 9 meaning level of difficulty in between AMC 8 and AMC 10.

i would prefer more algebra/number theory and geometry questions...

It would be more appropriate if you had collected them all in one single topic.

I wish we can have sub-forums or groups where we can organize our problems. I tried to use stickies but it didn't work.

[quote]No, that's too many questions for one post. It's better to have them separated to make it less confusing.[/quote]

Agreed.

Yea guys.

That's why I spent 10min breaking up Joel's 15question post into 15 different posts....

go look for it in this forum

its called "stuff" and author is "JBL"

it's not [b]that [/b]long...
JBL could've broken it up by subject, though.

[color=cyan]Generally speaking, I am lazy.  The problems, as I currently have them, are not arranged by topic and only very generally by difficulty.  Thus, I've just been posting what I thought were the interesting ones in the order that they are in my files.  In the future, I'll post 2 or 3 of similar topics in each message.  The 15-question post wasn't particularly long, but just imagine what would have happened as people started answering them . . .[/color]

[i]Mmm... :twisted:
Good idea... (not in here, of course. but there :twisted: )[/i]

I apologize on behalf of myself.
...is that possible?

[color=cyan]Possible, if not really necessary.[/color]


The incircle $C(I,r)$ of the triangle $ABC$ touches its sides in the points $D\in BC$, $E\in CA$, $F\in AB$. I note the intersection $L\in EF\cap AD$ and the points $M\in AB$, $N\in AC$ so that $I\in MN$ and $MN\perp AD$. Prove that:
$1^{\circ}.\ ML\cap DN\cap BE\ne \emptyset;$
$2^{\circ}.\ NL\cap DM\cap CF\ne \emptyset.$

The idea to solve this problem is using pole and polar.
   Suppose $A', B'$ be the intersections of $EF$ and $BC$, $DF$ and $AC$. Then $A', B'$ are respectively the poles of $AD, BE$ wrt the incircle $(I)$. Hence, $A', M, N$ are collinear and $AA'$ is the polar of $L$.
   Notice that the polars of $M, N$ are respectively the lines $f, e$ through $F, E$ that are parallel to $AD$. Thus, if we call $U, V$ respectively the intersections of $f$ and $AA'$, $e$ and $BC$, then $U$ and $V$ are the poles of $ML$ and $DN$. Therefore, it is sufficient now to prove that $B', U, V$ are collinear.
   Extend $VE$ so that it intersects the linte $AA'$ at $W$. Then $(AA'WU) = (LA'EF) = -1$. We have
   $\frac{C'A}{C'C}. \frac{VC}{VA'}. \frac{UA'}{UA} = (\frac{C'A}{C'C}.\frac{VC}{VD}).(\frac{VD}{VA'}. \frac{UA'}{UA}) = (\frac{C'A}{C'C}.\frac{EC}{EA}). (\frac{WA}{WA'}. \frac{UA'}{UA}) = (-1).(-1) = 1$.
  According to Menelaus' theorem, we imply that $B', U, V$ are collinear. And the problem is solved.


Does there exist a set $ M$ of points in space such that every plane intersects $ M$ at a finite but nonzero number of points?

For example $M=\{(x,y,z) \in M |x=t^5; y=t^3; z=t; t \in \mathbb{R}\}$. This problem comes from some old shortlist

$(x,y,z)=(atanv,sinv,cosv)\forall v \in R$
atanv assures the graph contains all x values
sinv and cosv assure that there are not too much points of intersection but
enough of them

just choose all points like (t,t^3,t^5)then it's done.


Solve in the set of the real numbers this equation :

$ 3 \lfloor x \rfloor \plus{}2\equal{}5x$

[hide="My Solution"]
We can look at it in terms of the fractional part rather than the floor value. We know that $ \{x\} \equal{} x \minus{} \lfloor x \rfloor \implies \lfloor x\rfloor \equal{} x \minus{} \{x\}$. Applying the substitution, we have

$ 3(x \minus{} \{x\}) \plus{} 2 \equal{} 5x \implies 3x \minus{} 3\{x\} \plus{} 2 \equal{} 5x \implies \{x\} \equal{} \minus{}\frac{2}{3}x \plus{} \frac{2}{3}.$

We know that $ 0 \le \{x\} < 1$, so we have

$ 0 \le \minus{}\frac{2}{3}x \plus{} \frac{2}{3} < 1 \implies \minus{}\frac{2}{3} \le \minus{}\frac{2}{3}x < \frac{1}{3} \implies \minus{}\frac{1}{2} < x \le 1.$

Thus, $ \boxed{x \in \left(\minus{}\frac{1}{2},1 \right]}$.
[/hide]

That isn't right. You just showed that no $ x$ outside that interval can ever be a solution, but try plugging in e.g. $ x \equal{} \minus{} {1\over 3}$ and see it doesn't satisfy the equation.

[hide="Correct solution"]Put $ x \equal{} n \plus{} a$, where $ n \equal{} [x], a \equal{} \{x\}$.

Then $ 3n \plus{} 2 \equal{} 5n \plus{} 5a\implies 5a \equal{} 2 \minus{} 2n$. Hence $ 5a\in\mathbb{Z}\implies 5a \equal{} k$ for some integer $ k$. Since $ k \equal{} 2(1 \minus{} n)$, we get that $ k \equal{} 2m$, hence $ a \equal{} {2m\over 5}$. Since $ 0\leqslant a < 1$ (by definition), the only possible values for $ m$ are $ m \equal{} 0, 1, 2$.

$ m \equal{} 0\implies a \equal{} 0\implies 0 \equal{} 2 \minus{} 2n\implies n \equal{} 1\implies x \equal{} n \plus{} a \equal{} 1$

$ m \equal{} 1\implies a \equal{} {2\over 5}\implies 2 \equal{} 2 \minus{} 2n\implies n \equal{} 0\implies x \equal{} n \plus{} a \equal{} {2\over 5}$

$ m \equal{} 2\implies a \equal{} {4\over 5}\implies 4 \equal{} 2 \minus{} 2n\implies n \equal{} \minus{} 1\implies x \equal{} n \plus{} a \equal{} \minus{} {1\over 5}$

Therefore $ x\in\left\{ \minus{} {1\over 5},{2\over 5},1\right\}$ and checking shows that all work.[/hide]


$a,b,c$ are prime numbers such that $a^b+b^a=c$, find all ($a,b,c$).

mmmm....I'm not good with these kind of problems but I found a couple:

1)a=1
   b=2
   c=3

2)a=3
   b=2
   c=17

3)a=2
   b=3
   c=17

4)a=2
   b=1
   c=3
I don't know exactly how to do this without trial and error

1) and 4) are not right because 1 is not a prime number.

[quote="jelly119"]$1$ is NOT a prime number. ;)[/quote]

Oops....I forgot that  :blush: Allright, let me try again then....

[quote="xxreddevilzxx"]well there is one [hide="hint"]one of $a$ and $b$ has to be odd number, or else it will make c an even nunmber, which is never prime number[/hide][/quote]

Yeah that does help...hmmm....let me figure this out

Clearly we can see a can't equal b.  Then it would be $2a^a=c$, which is impossible if c is prime.  Further, if a and b are both odd, then we have $odd^{odd} + odd^{odd} = c = odd + odd = even$, which is impossible if c is prime.  So a or b must be 2.  Two easy solutions are $(2,3,17)$ and $(3,2,17)$.  Now it's easy to see that if one is 2, and the other is greater than 3, the result is always divisible by 3.  But I don't know how to prove that.

[quote="Palytoxin"]well certainly one of them has to be an even number because if both $a$ and $b$ are odd primes, $c$ would be even as explained earlier by [b]xxreddevilzxx[/b]. we can choose $b=2$ since the equation on left side is very symmetrical if we switch $a,b$. then new equation looks like $a^2+2^a=c$, with $a,c$ odd primes.[/quote]

Yeah, but I think he meant to look more at odd numbers than even I guess.

[quote="Iversonfan2005"]Can anyone give another method instead of just guessing and checking...i mean, i think I got another one: a=2 b=7 c=177, but that will take forever!  :read:[/quote]
sorry but 177 is not prime, 177=3*59
ill try to prove that it must be divisible by 3....

ok we reduced problem to $c=a^2+2^a$ with $a,c$ odd primes. let $a=2t+1>3$
then $c=a^2+2^{2t+1}=a^2+2(2^t)^2=a^2+(2^t)^2+(2^t)^2$, so $c$ is a sum of 3 particular squares.
using the claim proven below, $c$ is a multiple of 3, i.e. if $a^2=3x+1$ and $(2^t)^2=3y+1$, then $c=3x+6y+3=3(x+2y+1)$. therefore only solutions are $(a,b,c)=(2,3,17),(3,2,17)$.

claim: every perfect square of an integer not divisible by 3 is of the form $3z+1$
proof: every integer not divisible by 3 is either one of the forms $n=3r\pm 1$. then $n^2=9r^2\pm 6r+1=3(3r^2\pm 2r)+1$ which is of the form $3z+1$


[size=150][b]Rules:[/b][/size]

In this game, each of the 50 states starts out with 25 points. Each time you post, you get 3 points to spend(help your state or attack other states). You may post at most one time each hour. The only exception is that you may only add points to the state live in (please don't cheat). But you can form alliances which allow you to help other states(explained later).You may post at most one time each hour.

Points: Each state starts out with 25 points. If your state reaches 0 points, then it is out of the game and the state that destroyed it gets 2 extra points. When your state has enough points, you can choose to spend some of those points on special items(these items are listed below).

Alliances: During the game, 2 or more states can choose to form an alliance. The maximum number of states there can be in an alliance is 5(name it whatever you want). In order to form one, one person from each state must first agree to it and then each state that wants to be in the alliance must give up 10 points.

No Refugees: Once your state of pledged allegiance is destroyed you may NOT continue playing.

Advantages of Alliances:

1) States that are in the same alliance can help each other(add points to each other)

2) When one of your allies destroys another state, your state gets 1 extra point(the state that destroyed it gets 2 points)

3) You can split the cost of buying special items

4) If you form an alliance with 3, 4, or 5 states you get that many points to spend(for example, if Hawaii, Alaska, and California form an alliance, then each person from those states gets 3 points to spend each time they post instead of 2. If 4 states form an alliance then each gets 4 points to spend)

You may not attack your allies unless they agree. And any state in an alliance can choose to leave at any time (you dont get the 10 points back).

[b][size=150]Special Items:[/size][/b]

[b]Shields:[/b]

Normal Shield (5), lasts 3 hours
Strong Shield (10), lasts 8 hours
Reinforced Shield (25), lasts 1 day
Defense System (45), lasts 2 days
{These only shield you from conventional attacks (no missiles/nukes)}

[b]Nuke-Rocket Defense Systems:[/b]

Nuclear Defense System (50), protects state from 1 nuke (only direct damage)
SAM Rocket-Intercept (50), protects state from 3 Rockets
Biohazard suits (25), protects you from radioactive fallout (if the state next to you is nuked)

[b]Missiles/Nukes:[/b]

V0 Rocket (5), Damage = 6 points
V1 Rocket (10), Damage = 12 points
V2 Rocket (20), Damage = 25 points
Nuke (50), Instant Destruction, Each Bordering State = 5 points damage

And PLEASE hide the score so it doesn't take forever to scroll.

[hide="States"]Alabama-25 pts
Alaska-25 pts
Arizona-25 pts
Arkansas-25 pts
California-25 pts
Colorado-25 pts
Connecticut-28 pts
Delaware-25 pts
Florida-25 pts
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Illinois-25 pts
Indiana-25 pts
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Massachusetts-25 pts
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
Missouri-25 pts
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Jersey-25 pts
New Mexico-25 pts
New York-25 pts
North Carolina-25 pts
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
Oregon-25 pts
Pennsylvania-25 pts
Rhode Island-25 pts
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
Texas-25 pts
Utah-25 pts
Vermont-25 pts
Virginia-25 pts
Washington-25 pts
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts [/hide]

Put your action in the hide tag so it is obvious what occurred.

[hide="+3 MO"]Alabama-25 pts
Alaska-25 pts
Arizona-25 pts
Arkansas-25 pts
California-25 pts
Colorado-25 pts
Connecticut-28 pts
Delaware-25 pts
Florida-25 pts
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Illinois-25 pts
Indiana-25 pts
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Massachusetts-25 pts
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
Missouri-28 pts
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Jersey-25 pts
New Mexico-25 pts
New York-25 pts
North Carolina-25 pts
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
Oregon-25 pts
Pennsylvania-25 pts
Rhode Island-25 pts
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
Texas-25 pts
Utah-25 pts
Vermont-25 pts
Virginia-25 pts
Washington-25 pts
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts [/hide][/quote]

[hide="+3 CT, Offer Alliance Treaty to Missouri"]Alabama-25 pts
Alaska-25 pts
Arizona-25 pts
Arkansas-25 pts
California-25 pts
Colorado-25 pts
Connecticut-31 pts
Delaware-25 pts
Florida-25 pts
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Illinois-25 pts
Indiana-25 pts
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Massachusetts-25 pts
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
Missouri-28 pts
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Jersey-25 pts
New Mexico-25 pts
New York-25 pts
North Carolina-25 pts
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
Oregon-25 pts
Pennsylvania-25 pts
Rhode Island-25 pts
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
Texas-25 pts
Utah-25 pts
Vermont-25 pts
Virginia-25 pts
Washington-25 pts
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts [/hide]

[hide="+3 Connecticut"]Alabama-25 pts
Alaska-25 pts
Arizona-25 pts
Arkansas-25 pts
California-25 pts
Colorado-25 pts
Connecticut-34 pts
Delaware-25 pts
Florida-25 pts
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Illinois-25 pts
Indiana-25 pts
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Massachusetts-25 pts
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
Missouri-28 pts
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Jersey-25 pts
New Mexico-25 pts
New York-25 pts
North Carolina-25 pts
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
Oregon-25 pts
Pennsylvania-25 pts
Rhode Island-25 pts
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
Texas-25 pts
Utah-25 pts
Vermont-25 pts
Virginia-25 pts
Washington-25 pts
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts [/hide]

[hide="+3 Missouri, Treaty Accepted"]Alabama-25 pts
Alaska-25 pts
Arizona-25 pts
Arkansas-25 pts
California-25 pts
Colorado-25 pts
Connecticut-24 pts
Delaware-25 pts
Florida-25 pts
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Illinois-25 pts
Indiana-25 pts
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Massachusetts-25 pts
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
Missouri-21 pts
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Jersey-25 pts
New Mexico-25 pts
New York-25 pts
North Carolina-25 pts
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
Oregon-25 pts
Pennsylvania-25 pts
Rhode Island-25 pts
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
Texas-25 pts
Utah-25 pts
Vermont-25 pts
Virginia-25 pts
Washington-25 pts
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts [/hide][/quote]

I forgot to add this:

2 State Alliances = 4 points
3 State Alliances = 5 points
4 State Alliances = 7 points
5 State Alliances = 9 points

[hide="+2 CA propose alliance to MOCT alliance"]Alabama-25 pts
Alaska-25 pts
Arizona-25 pts
Arkansas-25 pts
California-27 pts
Colorado-25 pts
Connecticut-24 pts
Delaware-25 pts
Florida-25 pts
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Illinois-25 pts
Indiana-25 pts
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Massachusetts-25 pts
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
Missouri-21 pts
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Jersey-25 pts
New Mexico-25 pts
New York-25 pts
North Carolina-25 pts
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
Oregon-25 pts
Pennsylvania-25 pts
Rhode Island-25 pts
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
Texas-25 pts
Utah-25 pts
Vermont-25 pts
Virginia-25 pts
Washington-25 pts
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts [/hide]

[hide="Alliances"]
MO/CT - Proposed, CA
[/hide]

CT accepts.  Waiting on MO.

[hide="+3 CT"]Alabama-25 pts
Alaska-25 pts
Arizona-25 pts
Arkansas-25 pts
California-27 pts
Colorado-25 pts
[color=blue]Connecticut-27 pts[/color]
Delaware-25 pts
Florida-25 pts
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Illinois-25 pts
Indiana-25 pts
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Massachusetts-25 pts
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
[color=blue]Missouri-21 pts[/color]
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Jersey-25 pts
New Mexico-25 pts
New York-25 pts
North Carolina-25 pts
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
Oregon-25 pts
Pennsylvania-25 pts
Rhode Island-25 pts
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
Texas-25 pts
Utah-25 pts
Vermont-25 pts
Virginia-25 pts
Washington-25 pts
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts[/hide]

[hide="+3 PA request to join MOCTCA(if it is made)"]Alabama-25 pts
Alaska-25 pts
Arizona-25 pts
Arkansas-25 pts
California-27 pts
Colorado-25 pts
Connecticut-27 pts
Delaware-25 pts
Florida-25 pts
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Illinois-25 pts
Indiana-25 pts
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Massachusetts-25 pts
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
Missouri-21 pts
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Jersey-25 pts
New Mexico-25 pts
New York-25 pts
North Carolina-25 pts
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
Oregon-25 pts
Pennsylvania-28 pts
Rhode Island-25 pts
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
Texas-25 pts
Utah-25 pts
Vermont-25 pts
Virginia-25 pts
Washington-25 pts
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts[/hide]

[hide="+3 CT, accepts Pennsylvania, waiting on Missouri"]Alabama-25 pts
Alaska-25 pts
Arizona-25 pts
Arkansas-25 pts
California-27 pts
Colorado-25 pts
[color=blue]Connecticut-30 pts [/color]
Delaware-25 pts
Florida-25 pts
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Illinois-25 pts
Indiana-25 pts
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Massachusetts-25 pts
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
[color=blue]Missouri-21 pts [/color]
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Jersey-25 pts
New Mexico-25 pts
New York-25 pts
North Carolina-25 pts
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
Oregon-25 pts
Pennsylvania-28 pts
Rhode Island-25 pts
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
Texas-25 pts
Utah-25 pts
Vermont-25 pts
Virginia-25 pts
Washington-25 pts
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts[/hide]

[hide="+3 CA"]
Alabama-25 pts
Alaska-25 pts
Arizona-25 pts
Arkansas-25 pts
California-30 pts
Colorado-25 pts
[color=blue]Connecticut-30 pts [/color]
Delaware-25 pts
Florida-25 pts
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Illinois-25 pts
Indiana-25 pts
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Massachusetts-25 pts
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
[color=blue]Missouri-21 pts [/color]
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Jersey-25 pts
New Mexico-25 pts
New York-25 pts
North Carolina-25 pts
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
Oregon-25 pts
Pennsylvania-28 pts
Rhode Island-25 pts
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
Texas-25 pts
Utah-25 pts
Vermont-25 pts
Virginia-25 pts
Washington-25 pts
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts[/hide]

[hide="+3 CT"]
Alabama-25 pts
Alaska-25 pts
Arizona-25 pts
Arkansas-25 pts
California-30 pts
Colorado-25 pts
Connecticut-33 pts
Delaware-25 pts
Florida-25 pts
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Illinois-25 pts
Indiana-25 pts
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Massachusetts-25 pts
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
Missouri-21 pts
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Jersey-25 pts
New Mexico-25 pts
New York-25 pts
North Carolina-25 pts
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
Oregon-25 pts
Pennsylvania-28 pts
Rhode Island-25 pts
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
Texas-25 pts
Utah-25 pts
Vermont-25 pts
Virginia-25 pts
Washington-25 pts
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts[/hide]

edit: whoops, didn't add the three points

[hide="+3 CT"]
Alabama-25 pts
Alaska-25 pts
Arizona-25 pts
Arkansas-25 pts
California-30 pts
Colorado-25 pts
[color=blue]Connecticut-36 pts[/color]
Delaware-25 pts
Florida-25 pts
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Illinois-25 pts
Indiana-25 pts
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Massachusetts-25 pts
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
[color=blue]Missouri-21 pts[/color]
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Jersey-25 pts
New Mexico-25 pts
New York-25 pts
North Carolina-25 pts
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
Oregon-25 pts
Pennsylvania-28 pts
Rhode Island-25 pts
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
Texas-25 pts
Utah-25 pts
Vermont-25 pts
Virginia-25 pts
Washington-25 pts
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts[/hide]

[hide="+3PA"]
Alabama-25 pts
Alaska-25 pts
Arizona-25 pts
Arkansas-25 pts
California-30 pts
Colorado-25 pts
Connecticut-36 pts
Delaware-25 pts
Florida-25 pts
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Illinois-25 pts
Indiana-25 pts
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Massachusetts-25 pts
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
Missouri-21 pts
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Jersey-25 pts
New Mexico-25 pts
New York-25 pts
North Carolina-25 pts
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
Oregon-25 pts
Pennsylvania-31 pts
Rhode Island-25 pts
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
Texas-25 pts
Utah-25 pts
Vermont-25 pts
Virginia-25 pts
Washington-25 pts
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts
[/hide]

[hide="+4 CT"]
Alabama-25 pts
Alaska-25 pts
Arizona-25 pts
Arkansas-25 pts
California-30 pts
Colorado-25 pts
[color=blue]Connecticut-40 pts[/color]
Delaware-25 pts
Florida-25 pts
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Illinois-25 pts
Indiana-25 pts
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Massachusetts-25 pts
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
[color=blue]Missouri-21 pts[/color]
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Jersey-25 pts
New Mexico-25 pts
New York-25 pts
North Carolina-25 pts
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
Oregon-25 pts
Pennsylvania-31 pts
Rhode Island-25 pts
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
Texas-25 pts
Utah-25 pts
Vermont-25 pts
Virginia-25 pts
Washington-25 pts
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts[/hide]

Forgot CT has a 2 state alliance. :P

[hide="+7 CA, Missile IN, WA, AZ, VA, OR (40 from CA, 20 from other 3 states in alliance)"]Alabama-25 pts
Alaska-25 pts
Arizona-PWNT
Arkansas-25 pts
[color=blue]California-15 pts- Protected From Nukes [/color]
Colorado-25 pts
[color=blue]Connecticut- 52 pts- Protected From Nukes, 3 Rockets
[/color]Delaware-25 pts
Florida-PWNT
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Illinois-PWNT
Indiana-PWNT
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Massachusetts-PWNT
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
Missouri-PWNT
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Jersey-PWNT
New Mexico-25 pts
New York-PWNT
[color=blue]North Carolina-23 pts [/color]
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
Oregon-PWNT
Pennsylvania-PWNT
Rhode Island-PWNT
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
[color=blue]Texas-38 pts [/color]
Utah-25 pts
Vermont-25 pts
Virginia-PWNT
Washington-PWNT
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts
[/hide]

Yay systematic bombing.

[hide="+7 CT, THEN Leave the Blue Alliance, Buy V0 and V1 Rockets, Pwn CA, Buy V2 Rocket, Pwn NC, +4 from kills"]Alabama-25 pts
Alaska-25 pts
Arizona-PWNT
Arkansas-25 pts
California- PWNT
Colorado-25 pts
Connecticut- 28 pts- Protected From Nukes, 3 Rockets
Delaware-25 pts
Florida-PWNT
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Illinois-PWNT
Indiana-PWNT
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Massachusetts-PWNT
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
Missouri-PWNT
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Jersey-PWNT
New Mexico-25 pts
New York-PWNT
North Carolina- PWNT
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
Oregon-PWNT
Pennsylvania-PWNT
Rhode Island-PWNT
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
[color=blue]Texas-38 pts [/color]
Utah-25 pts
Vermont-25 pts
Virginia-PWNT
Washington-PWNT
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts
[/hide]

Sorry.  But nice one-state alliance, Texas.

[hide=" +3 TX, Buy v2 and 4 v0 rockets, PWN CT's defence system and then CT"]]Alabama-25 pts
Alaska-25 pts
Arizona-PWNT
Arkansas-25 pts
California- PWNT
Colorado-25 pts
Connecticut- PWNT
Delaware-25 pts
Florida-PWNT
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Illinois-PWNT
Indiana-PWNT
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Massachusetts-PWNT
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
Missouri-PWNT
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Jersey-PWNT
New Mexico-25 pts
New York-PWNT
North Carolina- PWNT
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
Oregon-PWNT
Pennsylvania-PWNT
Rhode Island-PWNT
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
[color=blue]Texas-1 pt [/color]
Utah-25 pts
Vermont-25 pts
Virginia-PWNT
Washington-PWNT
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts
[/hide][/hide]
OOOWNED[/quote]

Screw you.  Too bad this topic's gonna be pwnt too now.

[quote="lotsofmath"][hide=" +3 TX"]Alabama-25 pts
Alaska-25 pts
Arizona-PWNT
Arkansas-25 pts
California- PWNT
Colorado-25 pts
Connecticut- PWNT
Delaware-25 pts
Florida-PWNT
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Illinois-PWNT
Indiana-PWNT
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Massachusetts-PWNT
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
Missouri-PWNT
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Jersey-PWNT
New Mexico-25 pts
New York-PWNT
North Carolina- PWNT
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
Oregon-PWNT
Pennsylvania-PWNT
Rhode Island-PWNT
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
[color=blue]Texas-4 pts [/color]
Utah-25 pts
Vermont-25 pts
Virginia-PWNT
Washington-PWNT
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts
[/hide][/hide]
OOOWNED[/quote][/quote]

[quote="lotsofmath"][hide=" +3 TX"]Alabama-25 pts
Alaska-25 pts
Arizona-PWNT
Arkansas-25 pts
California- PWNT
Colorado-25 pts
Connecticut- PWNT
Delaware-25 pts
Florida-PWNT
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Illinois-PWNT
Indiana-PWNT
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Massachusetts-PWNT
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
Missouri-PWNT
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Jersey-PWNT
New Mexico-25 pts
New York-PWNT
North Carolina- PWNT
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
Oregon-PWNT
Pennsylvania-PWNT
Rhode Island-PWNT
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
[color=blue]Texas-7 pts [/color]
Utah-25 pts
Vermont-25 pts
Virginia-PWNT
Washington-PWNT
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts
[/hide][/hide]
OOOWNED[/quote][/quote][/quote]

Also, you can't double post.  FAIL.

And go TX, I was born there. :P

[quote="lotsofmath"][hide=" +3 TX"]Alabama-25 pts
Alaska-25 pts
Arizona-PWNT
Arkansas-25 pts
California- PWNT
Colorado-25 pts
Connecticut- PWNT
Delaware-25 pts
Florida-PWNT
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Illinois-PWNT
Indiana-PWNT
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Massachusetts-PWNT
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
Missouri-PWNT
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Jersey-PWNT
New Mexico-25 pts
New York-PWNT
North Carolina- PWNT
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
Oregon-PWNT
Pennsylvania-PWNT
Rhode Island-PWNT
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
[color=blue]Texas-10 pts [/color]
Utah-25 pts
Vermont-25 pts
Virginia-PWNT
Washington-PWNT
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts
[/hide][/hide][/quote]
you put two /hides
here is how it should be:
[hide=" +3 TX"]Alabama-25 pts
Alaska-25 pts
Arizona-PWNT
Arkansas-25 pts
California- PWNT
Colorado-25 pts
Connecticut- PWNT
Delaware-25 pts
Florida-PWNT
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Illinois-PWNT
Indiana-PWNT
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Massachusetts-PWNT
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
Missouri-PWNT
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Jersey-PWNT
New Mexico-25 pts
New York-PWNT
North Carolina- PWNT
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
Oregon-PWNT
Pennsylvania-PWNT
Rhode Island-PWNT
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
[color=blue]Texas-10 pts [/color]
Utah-25 pts
Vermont-25 pts
Virginia-PWNT
Washington-PWNT
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts
[/hide]

[hide=" +3 TX"]Alabama-25 pts
Alaska-25 pts
Arizona-PWNT
Arkansas-25 pts
California- PWNT
Colorado-25 pts
Connecticut- PWNT
Delaware-25 pts
Florida-PWNT
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Illinois-PWNT
Indiana-PWNT
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Massachusetts-PWNT
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
Missouri-PWNT
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Jersey-PWNT
New Mexico-25 pts
New York-PWNT
North Carolina- PWNT
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
Oregon-PWNT
Pennsylvania-PWNT
Rhode Island-PWNT
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
Texas-13 pts
Utah-25 pts
Vermont-25 pts
Virginia-PWNT
Washington-PWNT
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts
[/hide]

[hide="+3 TX"]Alabama-25 pts
Alaska-25 pts
Arizona-PWNT
Arkansas-25 pts
California- PWNT
Colorado-25 pts
Connecticut- PWNT
Delaware-25 pts
Florida-PWNT
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Illinois-PWNT
Indiana-PWNT
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Massachusetts-PWNT
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
Missouri-PWNT
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Jersey-PWNT
New Mexico-25 pts
New York-PWNT
North Carolina- PWNT
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
Oregon-PWNT
Pennsylvania-PWNT
Rhode Island-PWNT
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
Texas-16 pts
Utah-25 pts
Vermont-25 pts
Virginia-PWNT
Washington-PWNT
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts [/hide]

[hide="+3 TX"]Alabama-25 pts
Alaska-25 pts
Arizona-PWNT
Arkansas-25 pts
California- PWNT
Colorado-25 pts
Connecticut- PWNT
Delaware-25 pts
Florida-PWNT
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Illinois-PWNT
Indiana-PWNT
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Massachusetts-PWNT
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
Missouri-PWNT
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Jersey-PWNT
New Mexico-25 pts
New York-PWNT
North Carolina- PWNT
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
Oregon-PWNT
Pennsylvania-PWNT
Rhode Island-PWNT
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
Texas-19 pts
Utah-25 pts
Vermont-25 pts
Virginia-PWNT
Washington-PWNT
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts [/hide]

cockroaches have evolved into intelligent beings from the wreckage of Massachusetts, taking over the government. now can MA become un-pwnt?

[hide="+3 TX"]Alabama-25 pts
Alaska-25 pts
Arkansas-25 pts
Colorado-25 pts
Delaware-25 pts
Georgia-25 pts
Hawaii-25 pts
Idaho-25 pts
Iowa-25 pts
Kansas-25 pts
Kentucky-25 pts
Louisiana-25 pts
Maine-25 pts
Maryland-25 pts
Michigan-25 pts
Minnesota-25 pts
Mississippi-25 pts
Montana-25 pts
Nebraska-25 pts
Nevada-25 pts
New Hampshire-25 pts
New Mexico-25 pts
North Dakota-25 pts
Ohio-25 pts
Oklahoma-25 pts
South Carolina-25 pts
South Dakota-25 pts
Tennessee-25 pts
Texas-22 pts
Utah-25 pts
Vermont-25 pts
West Virginia-25 pts
Wisconsin-25 pts
Wyoming-25 pts [/hide]


The sequence $a_1,a_2,...$ of natural numbers satisfies $\gcd(a_i,a_j)=\gcd(i,j)$ for all $i$ not equal to $j$. Prove that $a_i=i$ for all $i$

Put $n = \prod_{i=1}^k p_i^{e_i}$, the prime factorization of $n$.

For any $i$, set $r = p_i^{e_i}$.  Then $r = (r,n) = (a_r, a_n)$ implying $a_n$ has a factor $r$.  Thus, $a_n$ is a multiple of $n$.  So there exists a sequence $(b_1, b_2, \dots )$ with $a_n = nb_n$ for all natural n.

Now $k = (k, kb_k) = (a_k, a_{kb_k}) = (kb_k, kb_kb_{kb_k}) = kb_k$ implying $b_k = 1$ for all $k$.  The result follows.

$\square$


Let $ (X_1,T_1)$ and $ (X_2,T_2)$ be topological spaces and let $ X\equal{}X_1 \times X_2$ prove that
a) If $ X_1$ and $ X_2$ are separable then $ X$ is also separable
b) If $ X_1$ and $ X_2$ are first countable then $ X$ is also first countable.

part a) maybe routine, btw. what is first countable?   :blush:

Part a is pretty easy in the finite case:  let $ A_1$ and $ A_2$ be countable dense subsets of $ X_1$ and $ X_2$ respectively.
You should be able to show that $ A_1 \times A_2$ is countable and dense in $ X_1 \times X_2$.

(Just as a note extending this to countable products isn't quite as nice-you cannot just take the product of dense subsets in
each coordinate...)

As for what does "first countable mean"?  First countable means that each point has a countable basis about it (i.e.
for each point $ x$ there exists a sequence, $ U_1$, $ U_2, \hdots$ of open neighborhoods of $ x$ such that for any open neighborhood, $ V$, of $ x$, there exists an integer, $ i$, with $ U_i$ contained in $ V$).


I sent my solutions in augaust and I haven't heard back.  I noticed this today and resent them. Will they end up being counted?

Check the My USAMTS page - there will be a note there if we have received them months ago.  (A little check mark.)  Check that your email is correct in your profile by clicking the profile button above.

I checked for the check mark and it wasn't there, and I couldin't find any email address in my profile. What should I do now?

Click on the profile button above, here in the forum. That's where your email address should be.

Did you email or mail your solutions?  Where did you send them?

hmm... I sent mine in through the e-mail and I dont have a checkmark either... Is that just because you havent gotten it yet?

If you sent near the deadline, then that's the issue - we're crawling through all the mail-ins first since those have the most problems with not following directions.

I checked and the email adress is right. I emailed my solutions back in agaust then I sent another email today with my solutions. Both went to solutions@usamts.org. If the first one went to the wrong adress or something (by typo maybe) is there anything I can do now?

If you don't have a green check mark next to round 1 on the scores page by the end of Monday, let us know.  Don't worry about it until then - I'm pretty sure we have your entry.

What if we sent the copy instead of the original? And, the checkmark's already there

If the checkmark is there, then you're ok.  (And if it's not, don't worry yet - we still haven't received and processed every single paper.)

I sent my solutions by mail, but i still dont see a check mark next to my round one. what should i do?

I also have no check-mark yet.  I emailed it about a week ago.

We'll be doing a very thorough checking of everything tomorrow - if you don't have a checkmark by the end of tomorrow (Wednesday), write solutions@usamts.org and let us know.  Sorry it's taking us a while to parse the papers in the first round; it's our first time!   Future rounds will be parsed more quickly, as we won't have Entry and Permission forms to deal with.


ratio
ARML
geometry
3D geometry
AwesomeMath
summer program
Mathcamp

This is my first year applying to Mathcamp... I'm sooo nervous about it! I got my application in 30 minutes before 2 A.M. eastern time >_<

Does anyone know what the boy to girl ratio is there?

I answered 1-7 and 10a.

I ran out of time to give partial solutions to the other problems, but I realized I made really small mistakes in 1 and 7 that cost me the answer. They will still look at my solutions, right?

OMG. I'm just so psyched.

** I meant central time, sorry!

I solved the same number of questions as you: 1-6, 9, and 10a.
According to the directions on the quiz, answers are not nearly as important as solutions, so I wouldn't worry about small mistakes.
I didn't realize that the time limit was pacific time, so I had everything done by midnight eastern time. I decided to apply about a week before the application was due, so I didn't have too much time to solve the problems; I wish I had an extra week, but it feels good to be done I guess.
Good luck on getting accepted!

The website says it's usually between 25 and 35%. Last year it was 21.5% (I went last year and calculated that from my yearbook), but it's sure to be different this year.

That's pretty good! I just didn't want to get accepted and feel left out of stuff. I heard that some of the students who go to the camp are godly cubers. (My best time is 1:15... I kinda suck =o)

Well. My name is Sara, and I'm a rising junior in Illinois =) My friend Alina and I applied.... We're both really excited!!! =D

I started about a week ago aswell, and good luck to you too!

Anyway...

I wouldn't worry about getting accepted - 7.5 problems should be more than enough if the rest of your application is good.  I was accepted in '06 with 2 correct solutions.

[quote="ionave"]I heard that some of the students who go to the camp are godly cubers. [/quote]

Yeah, apparently Macky is coming back this year.

So, I hope to see you at camp this year!

[quote="ionave"]I just didn't want to get accepted and feel left out of stuff. I heard that some of the students who go to the camp are godly cubers.[/quote]
I know just how you feel. When I was accepted to my state's ARML team, I learned how to solve a cube just because I thought that everyone else would be able to. :lol: Cubing ended up not being a big deal at ARML or at AwesomeMath later in the summer. I don't think I've ever broken two minutes cubing, and I haven't touched a cube in about a year.

I applied, but didn't have much time to work on the test. I only got 1-5, 6a, 6c, 7a, 10a. :(

I think I started a little late on the test -_- but I completed 1-5, 7, 8.  On the last day, I tried to put together some partial solutions to 6, 9, and 10, but I really didn't get much for those.

[quote="ionave"]That's pretty good! I just didn't want to get accepted and feel left out of stuff. I heard that some of the students who go to the camp are godly cubers. (My best time is 1:15... I kinda suck =o)

Well. My name is Sara, and I'm a rising junior in Illinois =) My friend Alina and I applied.... We're both really excited!!! =D[/quote]

If you were worried about feeling left out because you're a girl, I don't think any girl at MC has had a problem with that. At least I didn't notice. Or if it's cubing, I didn't know a single thing about cubing (beyond seeing other people cubing) before I went and learned there. I'd probably go really slowly if I tried to solve a cube now... haven't tried in a while.

Oh cool, unless there are two people named Alina in IL applying to MC, I know your friend :). She and I went to MathPath in '06... good luck for getting in to both of you! And all of you others applying. I know about six people of those applying this year.... don't know if that's good or bad since there's obviously a chance they won't all get in :maybe: . But I hope they all will :D.

Also, when I did the QQ last year, I wrote up solutions for 1-4, 6, and 10, plus partial answers for 5a, 8, and 9. Afterwards I figured out that I got 1 totally wrong (heh, it's always good to check your work, even #1 :blush:) and had almost the right idea on 6 but messed up at the end. So don't worry too much about finding small mistakes; as long as you have good ideas/solutions for the QQ it's fine :).

I applied this year. I'm Taylor from NYC.
I got questions 1-4, 6 7, 9, 10a. Not entirely sure of my solution to 9 though, as I figured it out and wrote it all up late the day it was due. I think I got off-by-one on #3 but I'm pretty sure of the ideas in my solution. I also wrote up a whole bunch of random thoughts I had for 10b, none of which I think actually lead anywhere, but you never know.

How long do you think it'll be before we can discuss the quiz? I'm eager to know how bad/wrong all my solutions are.

Hope to see all of you there?

Also, unless there's two people named Alina then I know your friend too. And hi Laura.

[quote="Darmani"]I also applied to MathCamp for the first time this summer. I am a rising senior (sophomore-age, however) from St. Louis named James.

I started working on the problems on and off the moment soon after they were released online several months ago, and solved all of them.

As for cubing, I briefly apprenticed myself to a friend freshman year who could solve it in 30 seconds, but never got very far. I can memorize Shakespeare and Pi with ease (memorized to Feynman point for Pi day this year), but I'm not so good at things that do not come in a sequence, such as mappings between relative images and finger movements. It also doesn't help that my cube has an unusual coloring scheme and very high internal friction.[/quote]

O_O 30 seconds... Wow.... There are some kids at my high school who can do that!!! I'm 14 right now, and I have a summer b-day... You've got cube friction? I actually used vaseline in my cube, but everyone started yelling at me about how bad it is... I've had it for a year now, and it works just fine =P I really suck at memorizing stuff, though...

Yeah. You guys have got the right Alina!!! I go to IMSA with her. We're going to be roommates next year =)

I don't know how much longer I can wait for the results...

According to a bunch of other topic replies, they get acceptances out by the 2nd or 3rd weekend of May...

As far as I know, acceptances, rejections, and waitlists all go out at the same time.  I'd imagine they have a few hundred emails to send out so it'll take awhile to get them all out.  You'll probably get yours today.

Edit: Alex replied while I was posting.

[quote="E^(pi*i)=-1"]It's probably in alphabetical order. solafidefarms would be pretty early if that was the case.[/quote]

The decisions might also go out in order of when applications were received.

OMG Guess what I got!!!?!?!

I got mine today!  It said around 30% of people were accepted.
So, the first round of admissions have been completed.

They will let us know of scholarships by next week.

Yaaayyyy I got in!!!!  :D  :D

btw, check out the poll I'm gonna make in like 10 seconds  :) .

[quote="ionave"]Arg.. So If I didn't get an e-mail, does that means I'm rejected?

>o<[/quote]

No it doesn't, CHILL OUT.

ALINA! No. Lol. Now I'm going to fail all my classes in worry >_<<

We have soooo much hw to do, and if I don't know the result of my application... AHHh. I don't know what I'm going to do. Wow. I sound really immature.

Anyway, congrats to everyone who made it in =o

Edit:

Also, Alina. DO THE BC HW! its more than a WEEK LATE!

arg. I'm going to fail math too. damn

[quote="ionave"]ALINA! No. Lol. Now I'm going to fail all my classes in worry >_<<

We have soooo much hw to do, and if I don't know the result of my application... AHHh. I don't know what I'm going to do. Wow. I sound really immature.

Anyway, congrats to everyone who made it in =o[/quote]

Dude, calm down.

[quote="ionave"]
Edit:

Also, Alina. DO THE BC HW! its more than a WEEK LATE!

arg. I'm going to fail math too. -[/quote]

No. :P

Meh.. Looks like I have to apply next year D=
Anyway, I have class now so I g2g.. Bye everyone..

Those months paid off. See you lucky others in July. :)

I sure am happy that I was accepted, but unfortunately I can't go. Well, I can spend the four weeks in which I will be doing nothing teaching myself calc + linear algebra hopefully so I can have a broader range of classes to take next year.

I can't go because of a conveniently placed family trip.  If you're waitlisted, you may get my spot.

Lots of Mathpath '06 people in here. This is Leon. Word games, ping-pong, etc., remember?

Anyway, I got accepted, yay.