Who went to the JR thing at Google?  What were some of your likes and dislikes?  What prizes sis you get?  How was it compared to last year?  What were your favorite problems?

I heard that Ccy owned a person there, and won a book for it.

[quote="serialk11r"]Problems were sorta meh, but I liked the free food and all :D[/quote]

that's not what I heard

Basically I think there were way too many things going on. Some stations were pretty nice, a lot were too elementary. I didn't really like how they just had so many tables everywhere. The descriptions on the sheet they gave told basically nothing about what each table was about.

Some of the upstairs tables were nice, except the Pythagoras ones could be easily pwned using some Trig...

I never got to Folding Fractals (my school people were obsessing over the Its-It, so I had to defend mine...), does anyone have a scanned copy of that? I heard it was interesting...

I spent[wasted] all my time at the Zome place.

I was there with a few other AoPS kids at one table for lunch.. I'll try and summarize the CCY story.

So anyways, this lady (fibonacci combinatorist, I think) was doing interesting proofs with Fibonacci identities by actually using combinatorics (i.e. there was a REALLY cool determinant proof done by counting non-intersecting lattice paths and showing a bijection to something. I didn't really get the whole thing, but it was cool)

Anyways, to start off her talk, she did a "cool math trick" which was pretty obvious if you've ever done anything with the Fibonacci #s.. anyways the idea was that each student generated a lucas sequence, and she would sum the first ten terms. (or something like that)

So she called on three people from the audience to do the trick on.
Naturally, CCY raised his hand and screamed like a mad man. So he was selected to go up.
The first two kids dutifully gave here their nice, integer lucas sequences.

Then, the best part of the whole event happened.
Ccy gave her his list, and she tells him Point-Blank to go sit down and not be such a nuisance.
Why?
He had started with $ \pi$ and $ \phi$, and her mental trick relied upon a quick multiplication of the sum $ 5\pi \plus{} 8\phi$ by 11.
Well, needless to say, he RickRolled her SOOO hard. The look of incredulity on her face was worth the drive.
I was thinking to myself
* You just got Pwnt, ms. Fibonacci lady. Welcome to sparta.*

I was watching that from upstairs...it was hilarious.

What is a Fibonacci combinatorist? Are they related to Pascal's trianglers?

She messes around with Fibonacci Numbers and uses Combinatorics to write proofs

wait... could this lady be by any chance
jennifer quinn who wrote proofs that really count: the art of combinatorical proof?
or was she a different lady

um does that mean "yar" to teh former or "yar" to the latter
and how the * can you post such short a message

[quote="ccy"]yar[/quote]

kudos to you. i would have loved to have been there.

Dang congyue is pro. Oh well I guess the trick was pretty obvious then

Yea it was Jennifer Quinn. How come I didn't get a free book  :|

[quote="ccy"]yar[/quote]
hmm it's been 2 days dude

man this thread is still on-topic

that is ... kinda boring

Haha Jennifer Quinn came to Mathcamp and tried to explain to us why C(n,k)=n!/[(n-k)!k!]. We were like, "Yeah, uh, you're kind of underestimating us. Just a little bit."

That's so pro. I never knew anything about combinations!

Well, that may be the case, but:
I talked to her after the event (after everyone had cleared out) and she said that she was torn.
She knew the audience had a number of really smart high schoolers, but she also knew that there were a bunch of newbie middle schoolers who didn't have that strong of a background in combinatorics.

So she had difficulty deciding upon the content of the talk, and she went more towards the lower end of the spectrum. She said that she'd rather bore some older kids than confuse the younger ones and possibly drive them away from problem solving. Which is an entirely admirable viewpoint, in my opinion.
So yes, I guess I'm defending her.

[quote="junggi"][quote="serialk11r"]Hmmm...yea I thought that was the case too.
But doing that at Mathcamp?  :huh:[/quote]
wait till you get a week of dr kevin's classes[/quote]
ooh, please elaborate .  And isn't he from Mathzoom or something?

I wasn't at Julia Robinson, so I'm not arguing with whatever she did there. But there are no newbie middle schoolers at Mathcamp. Well, there are middle schoolers. But definitely no newbies.

For the guys who went to Google, I won the Google tour but I haven't been contacted yet. I lost the contact card that the lady gave out. Does anyone know the email of the person to contact?

Does anybody know how to contact with Mr. Joshua Zucker besides throughout email? I tried to email him 3 times but he hasn't replied me anything! :(  I'm curious about the result since I couldn't come! Does anybody know the specific result of this event? Pleased show me the practical way to contact to him if you know(as phone, through someone else,...).He also promised that he would let me know the result after the events and I wait for a long time since then.

[quote="serialk11r"][quote="ghjk"][quote="junggi"]there is no "result"; it's not a competition
the only thing you could win was from the raffle and i doubt they kept a record who won what in a convenient place[/quote]
How about those recipients for the Problem Solving Fermat-Pythagore, which is known as the 2nd round? Nobody got anything about it in that event? :o[/quote]
What? What's that? There were [b]no[/b] competitive events. Period.[/quote]
Here's what I meant! Prizes for the Problem solving challenge Round 2:
http://www.msri.org/specials/festival/index_html
I seem to be lost! Really confused about the prizes for this Problem Solving Challenge Round 2.Maybe they don't talk about it in the event, so is it faked? I put greatly effort on this Problem Solving. :(

Hmmm...I seem to vaguely remember a bunch of middle schoolers going up to the stage and getting awards for this...


find the inverse function for the following equation:

f(x)= 3x+2f(x)= 3x+2


Prove that, if $ F_0\equal{}F_1\equal{}1$, and $ F_n\equal{}F_{n\minus{}1}\plus{}F_{n\minus{}2}$, that $ (F_n)! \geq n(F_0)!(F_1)!...(F_{n\minus{}1})!$

Edit: IT SHOULD BE $ F_{n\minus{}2}$ instead of $ F_{n\minus{}1}$ :blush:

[hide="hint"] first prove that if $ a\plus{}b\plus{}c\plus{}...\plus{}x\plus{}y \leq z$, then $ z! \geq a!b!c!...x!y!$[/hide]


real analysis
search
real analysis theorems

let M be a lebesgue measurable subset of the real line with positive measure =a. let 0<b<a, then show that there exists a subset of M which has measure b.

Since you placed this on the line, we have access to a rather cheap argument - an argument that would go away if we tried to generalize.

Let $ E_x \equal{} (\minus{} \infty,x]\cap M$ and let $ g(x) \equal{} m(E_x).$ We have that $ g(x)$ is nondecreasing, $ g(x)\to0$ as $ x\to \minus{} \infty,$ and $ g(x)\to a$ as $ x\to\infty.$ Furthermore, we have that $ g(x)$ is continuous. It's even Lipschitz, since for $ x < y,$ $ g(y) \minus{} g(x) \equal{} m(E_y\setminus E_x) \equal{} m(M\cap (x,y])\le y \minus{} x.$

Hence we can quote the Intermediate Value Theorem to claim that there exists $ x$ such that $ g(x) \equal{} b.$ Then that $ E_x$ has measure $ b.$

yes this was the exact thing that i also thought of. how do we show it for higher dimensions? what kinds of measure space have this property? surely we cant expect an arbitrary measure space to have this property.

In $ \mathbb{R}^n$, we could still use essentially the same argument: Let $ E_x\equal{}\{(x_1,x_2,\dots x_n)\in M: x_1\le x\}.$ We'll have to work a tiny bit harder to show that $ g(x)\equal{}m(E_x)$ is continuous, but we can still make that work.

On the other hand, consider this example: Take $ \{p_k\}$ to be some sequence of positive numbers. Consider $ \mathbb{N}$ to be a measure space, with all subsets measurable and with $ m(\{k\})\equal{}p_k.$ Then this theorem will be false.

For a condition to consider, try looking up the word "nonatomic" as applied to measure spaces.


For a set $E$ we can find a $G_\delta$ set $G$ such that $m^\ast(E) = m(G)$ (where $m^\ast$ and $m$ denote Lebesgue outer measure and Lebesgue measure, respectively)

What is this $G$ called in English? In both Chinese and Japanese, it is called '等測包', which literally means equi-measure-superset. I'd like to know what it is called in English.

I doubt that there is a standard term for such $G$. I'd rather use a descriptive phrase like "$G_{\delta}$ superset of equal measure".


calculus
integration
logarithms
trigonometry
parameterization
analytic geometry
search

Can someone provide (not too hard) examples where the evaluation of a simple integral is made easier by converting it first to a double integral?

Thanks in advance.

how do you mean ? r u talking about $ IuIs$ ?  :maybe:

e.g.

$ \int_0^1\frac{x\minus{}1}{\ln\, x} \; dx \equal{}  \int_{x\equal{}0}^1\int_{y\equal{}0}^1x^y\; dy\;dx\equal{}   \int_{y\equal{}0}^1\int_{x\equal{}0}^1x^y\; dx\;dy\equal{}\int_{0}^1\frac{1}{y\plus{}1}\;dy\equal{}\ln\;2$

Remember those integrals you post and for which you say something like "try to convert it first to a suitable double integral"? I am referring to such examples, but not too hard, like the one you posted. Thanks.

My all-time favorite:

Compute $ \int_0^{\infty}\frac{1\minus{}\cos x}{x^2}\,dx$ by noting that

$ \frac1{x^2}\equal{}\int_0^{\infty}te^{\minus{}tx}\,dt.$

And, by integration by parts, this is equivalent to $ \int_0^{\infty}\frac{\sin x}{x}\,dx.$

A simple one, consider $ \int_{1}^{2}{\ln x\,dx}.$ Avoid partial integration and recall that $ \ln x \equal{} \int_1^x\frac {dy}y,$ hence, the integral becomes $ \int_{1}^{2}\!{\int_{1}^{x}{\frac {dy\,dx}{y}}} \equal{} \int_{1}^{2}\!{\underbrace{\int_{y}^{2}{dx}}_{2\, \minus{} \,y}\,\frac {dy}{y}} \equal{} \int_{1}^{2}{\frac {2 \minus{} y}{y}\,dy}.$

Another as $ \int_{0}^{\infty }{\frac {2 \minus{} 2\cos x}{xe^{x}}\,dx}.$ Consider $ \frac {1 \minus{} \cos x}{x} \equal{} \int_{0}^{1}{\sin (xy)\,dy}.$

Also $ \int_{0}^{\infty }{\frac {e^{ \minus{} \alpha x} \minus{} e^{ \minus{} \beta x}}{x}\,dx}.$ Since $ \frac {e^{ \minus{} \alpha x} \minus{} e^{ \minus{} \beta x}}{x} \equal{} \int_{\alpha }^{\beta }{e^{ \minus{} xy}\,dy}.$ The rest follows.

(Well, there're lots of single integrals which can be tackled by using parameters to construct double ones, all depends of your "first ideas" and what motivates you to it.)

A famous integral over the entire real axis,
\[ I \equal{} \int_{ \minus{} \infty}^{\infty} e^{ \minus{} x^2} dx
\]
is not defined using elementary functions.
We express as follows to evaluate,
\[ I^2 \equal{} \left(\int_{ \minus{} \infty}^{\infty} e^{ \minus{} x^2}dx\right) \left( \int_{ \minus{} \infty}^{\infty} e^{ \minus{} y^2}dy\right).
\]
Or,
\[ \int_{ \minus{} \infty}^{\infty}\int_{ \minus{} \infty}^{\infty} e^{ \minus{} (x^2 \plus{} y^2)} dxdy
\]
Now, using a polar coordinate transformation we have the new bounds for the region we integrate over, \[ R \equal{} \left\{ 0 \leq r \leq \infty ;
0 \leq \theta \leq 2 \pi \right\}\] so the integral becomes
\[ I^2 \equal{} \int_{0}^{\infty} \int_{0}^{2\pi} r e^{ \minus{} r^2} d\theta dr
\]
Evaluating we have
\[ I^2 \equal{} \pi \Rightarrow I \equal{} \sqrt {\pi}
\]

The first line of |-.-|'s post is a reminder that all cases of integration by parts can be rewritten using double integrals.

For instance:

$ \int_0^axf(x)\,dx \equal{} \int_0^a\int_0^xf(x)\,dt\,dx \equal{} \int_0^a\int_t^af(x)\,dx\,dt$

Use that to compute:

$ \int_0^{\pi}x\sin nx\,dx \equal{} \int_0^{\pi}\int_t^{\pi}\sin nx\,dx\,dt$

$ \equal{} \frac1n\int_0^{\pi}( \minus{} 1)^{n \plus{} 1} \plus{} \cos nt\,dt \equal{} \frac {( \minus{} 1)^{n \plus{} 1}\pi}{n}.$

Of course we can get the same thing by integrating by parts - but perhaps one makes fewer sign errors this way.

Here's one if anyone's interested:

Evaluate

$ \int_{0}^{\infty} x(\exp(3x)\minus{}1)^{\minus{}1/3} dx$

[quote="Carcul"]Remember those integrals you post and for which you say something like "try to convert it first to a suitable double integral"?[/quote]

 :rotfl:

[quote="Carcul"]
 I am referring to such examples, but not too hard, like the one you posted. Thanks.[/quote]

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=381803483&t=194082

 :D

also

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=197640


already posted http://www.mathlinks.ro/viewtopic.php?p=1040#p1040
and zaya_yc why did you assume n even?


seeing as how there is "induction" in your title, I will try to write an induction solution.
[hide="induction solution"]
Our base case is $ n\equal{}1$ which is just $ 1\minus{}x_1 \ge 1\minus{}x_1$ which is true.

ok so suppose that it is true for n=a-1. Now we want to prove that the statement is true for n=a.

So we have our a numbers $ x_1, x_2, \ldots x_a$
and it is true that $ \prod_{k\equal{}1}^{a\minus{}1} 1\minus{}x_k \ge 1\minus{}\sum_{k\equal{}1}^{a\minus{}1} x_k$ by our induction hypothesis. We want to prove that
$ \left(1\minus{}x_a\right)\prod_{k\equal{}1}^{a\minus{}1} 1\minus{}x_k \ge 1\minus{}\sum_{k\equal{}1}^{a\minus{}1} x_k\minus{}x_a (1)$
Notice that both sides (especially the LHS) are less than or equal to 1. This is true for the LHS because each product is positive and less than or equal to 1.
Therefore, $ x_a \ge x_a \cdot \prod_{k\equal{}1}^{a\minus{}1} 1\minus{}x_k \implies \minus{}x_a \cdot \prod_{k\equal{}1}^{a\minus{}1} 1\minus{}x_k \ge \minus{}x_a (2)$
adding our two inequalities (1) and (2) gives the desired result.

hmm
equality is achieved when n or n-1 of the $ x_i$ are equal to 0.
This can be seen from setting both (1) and (2) as equalities.
[/hide]
[/hide]


inequalities
linear algebra
matrix
algebra
polynomial

Let $ A \in \mathcal{M}_n \left( \mathbb{R} \right)$ with $ \det{A} \equal{} 0$ and $ X \in \mathcal{M}_{n} \mathbb{R}$ the matrix with all elements $ 1$, $ x_{ij} \equal{} 1 , \forall 1 \leq i,j \leq n$.

Prove that:  $ \det{(A \plus{} X)} \cdot \det{(A \minus{} X)} \leq 0$.

$ X$ is similar over $ \mathbb{R}$ to $ diag(0,\cdots,0,n)$. Then we may assume $ X\equal{}diag(0,\cdots,0,n)$. If $ A\equal{}[a_{ij}]$ then $ det(A\plus{}X)\equal{}det(A)\plus{}n\alpha\equal{}n\alpha$ where $ \alpha$ is the cofactor of $ a_{nn}$. By the same way $ det(A\minus{}X)\equal{}\minus{}n\alpha$ and $ det(A\plus{}X)det(A\minus{}X)\equal{}\minus{}n^2\alpha^2$.

hi, i am not good at matrix so can you show me that why X is similar to diag(0,0,...,n) loup blanc, please? thank you very much

$ X$ is a real symmetric then it is diagonalizable over $ \mathbb{R}$. $ rank(X)\equal{}1$ then $ X$ has at least $ n\minus{}1$ eigenvalues equal to $ 0$. The last one is $ trace(X)\equal{}n$.

Maybe this will sound stupid, but I`m not sure if we have $ A,X \in \mathcal{M}_n \left( \mathbb{R} \right)$, and $ X$ is similar over $ \mathbb{R}$ with another matrix $ Y \in \mathcal{M}_n \left( \mathbb{R}\right)$ then:

\[ \det(A\plus{}X)\equal{} \det(A\plus{}Y)\]

If this is true, please post a proof. Thank you
And yes I know that $ X$ and $ Y$ share the same characteristic polynomial, but this gives only that:
$ X\equal{}PYP^{\minus{}1}$ where $ P$ is invertible, and:
$ \det(A\plus{}X)\equal{} \det(P^{\minus{}1}AP\plus{}Y)$, and this is not sufficient..

Hi linker,
"if we have $ A,X \in \mathcal{M}_n \left( \mathbb{R} \right)$, and  is similar over  $ \mathbb{R}$ with another matrix $ Y \in \mathcal{M}_n \left( \mathbb{R}\right)$ then:
$ det(A\plus{}X)\equal{}det(A\plus{}Y)$."
I didn't say that. When I write "we may assume $ X\equal{}Y$", then I hide the following:
$ X\equal{}PYP^{\minus{}1},A\equal{}PBP^{\minus{}1}$ where $ P$ is a real invertible matrix. Then $ B$ is a real matrix s.t. $ det(B)\equal{}0$. Thus we replace $ X,A$ with $ Y,B$.

we have one more general problem:
" $ A,B \in \mathcal{M}_n(R)$ with rank(B)=1. prove that $ det((A\minus{}B)(A\plus{}B)) \leq (detA)^2$ " :)

Use the same method but now there exist 2 cases:
case 1: we may assume $ B\equal{}diag(0,\cdots,0,\lambda)$ where $ \lambda$ is a non zero real.
case 2: we may assume $ B\equal{}diag(0_{n\minus{}2},\begin{pmatrix}0&1\\0&0\end{pmatrix})$.


You have a 17 x 17 chess table, and on every square there is a coin. Every coin has 2 sides: the face and the number. Innitially, all the coins ar with their faces up, but at each step, you take 5 consecutive coins from any line, column or one of the 2 diagonals and you turn them upside down. Is it possible that after a limited number of steps, all the coins will be with their number sides up?

Color the grid as follows:

[code]
ABBAAABBAAABBAAAB
AABBAAABBAAABBAAA
AAABBAAABBAAABBAA
BAAABBAAABBAAABBA
BBAAABBAAABBAAABB
ABBAAABBAAABBAAAB
AABBAAABBAAABBAAA
AAABBAAABBAAABBAA
BAAABBAAABBAAABBA
BBAAABBAAABBAAABB
ABBAAABBAAABBAAAB
AABBAAABBAAABBAAA
AAABBAAABBAAABBAA
BAAABBAAABBAAABBA
BBAAABBAAABBAAABB
ABBAAABBAAABBAAAB
AABBAAABBAAABBAAA
[/code]

Every strip contains either 2 or 0 $ B $ squares.  Therefore after every flip there will be an even number of $ B $ squares number side up.  But by my count there are 115 $ B $ squares in the coloring, so one is not number side up


Let $ a,$ $ b,$ $ c$ be positive real numbers such that $ a\plus{}b\plus{}c\equal{}ab\plus{}bc\plus{}ca.$ Prove that
$ a^{\frac{a}{b}}b^{\frac{b}{c}} c^c \ge 1.$

[quote="can_hang2007"]Let $ a,$ $ b,$ $ c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} ab \plus{} bc \plus{} ca.$ Prove that
$ a^{\frac {a}{b}}b^{\frac {b}{c}} c^c \ge 1.$[/quote]
It looks like sadism.  :D


quadratics
number theory unsolved
number theory

I'll assume that you wish to find the number of quadratic residues $ mod 2^n$.

Let the number of quadratic residues $ mod 2^n$ be $ f(n)$.

It is easy to verify that $ f(2)=2, f(3)=3$.

Claim: The number of odd quadratic residues are $ 2^{n-3}$.

Proof) We can verify that for all odd k, $ k^2 == (2^{n-1} + k)^2$  $ (mod 2^n )$.

Therefore, we only need to consider the integers $ 1,3,...,2^{n-1} - 1$.

We have that $ k^2 == (2^{n-1} - k)^2$  $ (mod 2^n )$ : we need only to consider the integers $ 1,3, ... , 2^{n-2}-1$.

However, if $ 0<x,y<2^{n-2}$, it is easy to verify that $ x^2 =\= y^2$ $ (mod 2^n)$.

Therefore, the quadratic residues are $ 1^2 , 3^2 , .. , (2^{n-2}-1)^2$.

Also, any odd quadratic residue is a multiple of 4: if we let this quadratic residue be $ 4x$, then x is a quadratic residue of $ 2^{n-2}$.

From this, we have that $ f(n) = 2^{n-3} + f(n-2)$.

Inductively, we get that
$ f(n) = 2^{n-3} + 2^{n-5} + .. + 2 + 2$ (n even)
$ f(n) = 2^{n-3} + .. + 4 + 3$ (n odd)


modular arithmetic
algebra proposed
algebra

[quote="Jean-Louis"]Determine a positive constant c such that the equation
x*y^2 − y^2 − x + y = c
has precisely three solutions (x, y) in positive integers.[/quote]

Choose $ c\equal{}1570$.
Since $ y\minus{}1$ divides $ c\equal{}2\times 5\times 157$, the only values available for $ y\minus{}1$ are $ 1,2,5,10,157,314,785,1570$ and just $ 3$ give an integer value for $ x$ : $ 1,2,10$

And so the three solurions for $ (x,y)$ : $ (14,11)$, $ (197,3)$ and $ (524,2)$

$ xy^2 \minus{} y^2 \minus{} x \plus{} y \equal{} c \Leftrightarrow (y \minus{} 1)(x(y \plus{} 1) \minus{} y) \equal{} c$

If $ y \equal{} 1$ then $ c \equal{} 0$ contracdiction .

If $ y\ge 2$ then $ c\;\vdots\; y \minus{} 1$  :wink:

[quote="Jean-Louis"]Determine a positive constant c such that the equation
x*y^2 − y^2 − x + y = c
has precisely three solutions (x, y) in positive integers.[/quote]

Here is a method to find such contants $ c$ :

In fact, we have, as quangpbc said : $ (y-1)(x(y+1)-y)=c$ and so :
$ c=u(y-1)$ and $ x=1+\frac{u-1}{y+1}$

So we need $ (y-1)|c$ and $ (y+1)|(u-1)$. Notice that we also have $ y+1\leqq-1$ and so $ c\geq y^2+y+1$
In order to be sure to have just $ 3$ solutions (and no more), one way is to have very few divisors of $ c$.

I suggest to study the case where $ c=p_1p_2p_3$ where $ p_1<p_2<p_1p_2<p_3$ are three primes.
$ c$ has $ 8$ divisors but the constraint $ p-1p_2<p_3$ implies that divisors containing $ p_3$ can not be considered (since we must have $ c\geq y^2+y+1$).

So only $ 4$ divisors are to be considered : $ 1, p_1, p_2, p_1p_2$ and so $ 4$ values for $ y$ : $ 2, p_1+1, p_2+1, p_1p_2+1$
Then the rule $ (y+1)|(u-1)$ gives :
$ 3|p_1p_2p_3-1$, $ p_1+2|p_2p_3-1$, $ p_2+2|p_1p_3-1$ and $ p_1p_2+2|p_3-1$

So, we have to find 3 primes $ p_1<p_2<p_1p_2<p_3$ such that exactly 3 or the 4 following equations are true :
$ p_1p_2p_3=1\pmod{3}$
$ p_2p_3=1\pmod{p_1+2}$
$ p_1p_3=1\pmod{p_2+2}$
$ p_3=1\pmod{p_1p_2+2}$

We can take $ p_1=2$ and we have :
$ 2<p_2<2p_2<p_3$ such that exactly 3 or the 4 following equations are true :
$ p_2p_3=2\pmod{3}$
$ p_2p_3=1\pmod{4}$
$ 2p_3=1\pmod{p_2+2}$
$ p_3=1\pmod{2p_2+2}$

Taking then $ p_2=3$, we have :
$ 2<3<6<p_3$ such that exactly 3 or the 4 following equations are true :
$ 3p_3=2\pmod{3}$ (obviously wrong)
$ p_3=3\pmod{4}$
$ p_3=3\pmod{5}$
$ p_3=1\pmod{8}$, which is in contradiction with $ p_3=3\pmod{4}$.
So we can't have $ p_2=2$ if $ p_1=2$

Taking then $ p_2=5$, we have :
$ 2<5<10<p_3$ such that exactly 3 or the 4 following equations are true :
$ p_3=1\pmod{3}$
$ p_3=1\pmod{4}$
$ p_3=4\pmod{7}$
$ p_3=1\pmod{12}$

which gives a family of solutions with $ p_3=1\pmod{12}$ and $ p_3\neq 4\pmod{7}$

And so $ p_3$ may be any prime $ >10$ and belonging to any sequence $ \{84k+1\}$, $ \{84k+13\}$, $ \{84k+37\}$, $ \{84k+61\}$ or $ \{84k+73\}$.

For example : $ 13,37,61,73,97,157,181,...$

Which gives for $ c$ : $ 130,370,610,730,970,1570,1810,...$

For example, with $ c=130$, we have the three solutions $ (2,11)$, $ (17,3)$ and $ (44,2)$


For $ a,b,c$ positive reals prove that

$ \sum \frac{ab}{3a\plus{}4b\plus{}2c}\leq \frac{a\plus{}b\plus{}c}{9}$

It's obviously trues by CS because:
$ \sum\ \frac{ab}{a\plus{}b\plus{}2c} \leq\ \frac{3}{4} and \sum\ \frac{ab}{2a\plus{}3b} \leq\ \frac{3}{5}$
PS: $ 3a\plus{}4b\plus{}2c\equal{}a\plus{}b\plus{}2c\plus{}2a\plus{}3b$  :)


function
integration
trigonometry
calculus
calculus computations

I am reading a text called "A guide to distribution theory and Fourier transforms" by Strichartz and one of the problems asks one to use the Fourier inversion formula to compute $ \int_{ \minus{} \infty}^{\infty}\frac {\sin x}{x}\,dx.$ I know how to do this integral via elementary real variable approaches and contour integration, but does anyone know the solution suggested by the text?

Also, is it permissible to treat $ \int_{ \minus{} \infty}^{\infty}\frac {\sin{\alpha\cdot x}}{x}\,dx$ as a generalized function and then differentiate under the integral with respect to $ \alpha$ (I am wondering since rewriting $ \sin{\alpha\cdot x}$ as an exponential seems to be the only way I can get Fourier phases to appear in the integrand)?

I catch a whiff of possible circularity here, in that I prefer to have this specific integral already computed so that I can use it in a proof of Fourier inversion.

But let's say that somehow we already have Fourier inversion. I'm guessing that Strichartz puts his $ 2\pi$s in a different place than I'm used to; let me go back to my favorite notation.

Given appropriate $ f,$ define $ \widehat{f}(\xi) \equal{} \int_{ \minus{} \infty}^{\infty}f(x)e^{ \minus{} 2\pi i\xi x}\,dx.$

Fourier inversion would be the statement that under certain fairly mild conditions on $ f$, we can recover $ f(x) \equal{} \int_{ \minus{} \infty}^{\infty}\widehat{f}(\xi)e^{2\pi i \xi x}\,d\xi,$ where that integral may be interpreted as an improper integral or even as a principal value (that is, the limit of the integral from $ \minus{} N$ to $ N.$)

We define the "box function": $ \sqcap(x) \equal{} \begin{cases}1 & \text{if }|x| < \frac12 \\
0 & \text{otherwise}\end{cases}.$ (Under some circumstances, I may even want to specify the value of the box function at $ \pm\frac12$ as $ \frac12$ rather than as either $ 0$ or $ 1$ - but that's not particularly important right now.)

By direct computation, $ \widehat{\sqcap}(\xi) \equal{} \frac {\sin\pi \xi}{\pi\xi} \equal{} \text{sinc}\,(\xi)$

That was the definition of the function $ ``\text{sinc}"$ except that we also have $ \text{sinc}\,(0) \equal{} 1,$ making $ \text{sinc}$ a smooth function.

Now if we believe Fourier inversion, we have $ \sqcap(x) \equal{} \int_{ \minus{} \infty}^{\infty}\text{sinc}\,(\xi)e^{2\pi i\xi x}\,d\xi$ or in particular,

$ \sqcap(0) \equal{} 1 \equal{} \int_{ \minus{} \infty}^{\infty}\text{sinc}\,(\xi)\,d\xi \equal{} \int_{ \minus{} \infty}^{\infty}\frac {\sin\pi x}{\pi x}\,dx$

 which is (essentially) the integral you're looking for.

You could make the same argument under different "where are the $ 2\pi$s" conventions - just make sure to have those in the right places, and to use the appropriate version of the box function.


The following question is from the 2006 Mathcounts chapter competiton:

Five balls are numbered 1 through 5 and placed in a bowl. Josh will randomly choose a ball from the bowl, look at its number and then put it back into the bowl. Then Josh will again randomly choose a ball from the bowl and look at its number. What is the probability that the prodcut of the two numbers will be even and greater than 10? Express your answer as a common fraction.

Can I also have an explanation on how you came to your answer, please.

[hide="hint"]
Look at products of numbers between 1,5 and then see which ones are even
[/hide]

You can make a table. That would simplify this problem a lot.

[hide] There aren't many cases so we can count them.

Consider the first one drawn.

For example, if the first ball is:

1, the successful outcomes for the second ball are 1-5

So all the successful outcomes are

1, (none), but 1, 3, and 5 are not even so
0 successful outcomes

2, (none)
0

3, (4 through 5)
1

4, (3 though 5)
3

5, (3 though 5)
1

0+0+1+3+1=5

5*5=25 possible outcomes

5/25=[b]1/5[/b]
[/hide]


If $ r$, $ s$, and $ t$, are the complex roots of the equation:
$ x^3 \plus{} 2x^2 \plus{}5x \plus{}7$,
what is $ r^2 \plus{}s^2 \plus{}t^2$?

\[ \left\{\begin{array}{lcl}
r\plus{}s\plus{}t & \equal{} & \minus{}2, \\
rs\plus{}st\plus{}tr & \equal{} & 5, \\
rst & \equal{} & \minus{}7.
\end{array}\right.\]
And
\[ r^2\plus{}s^2\plus{}t^2 \equal{} (r\plus{}s\plus{}t)^2\minus{}2(rs\plus{}st\plus{}tr).\]

Thus
\[ r^2\plus{}s^2\plus{}t^2 \equal{} (\minus{}2)^2 \minus{} 2 \cdot 5 \equal{} 4 \minus{} 10 \equal{} \boxed{\minus{}6}.\]

A general way of doing this is [url=http://www.artofproblemsolving.com/Wiki/index.php/Newton%27s_sums]Newton's Sums[/url].


Find all function $f:Z\longrightarrow Z$ such that $f(-1)=f(1)$ and for any two integer number like $m,n$ we have:
$f(m)+f(n)=f(m+2mn)+f(n-2mn)$

substituting m=1, gives: f(1)+f(n)=f(2n+1)+f(-n)
substituting m=n, n=-1, gives: f(n)+f(-1)=f(-n)+f(2n-1)
So f(2n-1)=f(2n+1) for all n, or f(n) = c for all odd integers n, where c is a constant.
It follows that f(n)=f(-n) for all n.
can't finish for some reason...


:(

[hide]Why is this message considered too small?[/hide]

i would have answered 7 hours ago if i had arrived at the answer. Try i little more right now!
and dude, calm down!

[quote="rnwang2"]No one solved this in the Pre-Olympiad forum, so I've decided to post it here too.

$f(x)+f\left(\frac{1-x}{x}\right) = 1-x$.[/quote]

for $y=\frac{1-x}{x}$ we have $f(x)+f(y)=xy$.

posted before. and there is a generalisation.you can use  surch. :ninja:


Prove that
$a^3b+a^3c+b^3a+b^3c+c^3a+c^3b\geq 2(a^2bc+b^2ac+c^2ab)$

$ab(a^2 + b^2) + ac(a^2 + c^2) + bc(b^2 + c^2) \geq 2abc(a + b + c)$

$\sqrt{\frac{ab(a^2 + b^2)}{2}} + \sqrt{\frac{ac(a^2 + c^2)}{2}} + \sqrt{\frac{bc(b^2 + c^2)}{2}} \geq \sqrt{abc(a+ b+ c)}$

For AM-GM it is

$\sqrt{\frac{ab(a^2 + b^2)}{2}} + \sqrt{\frac{ac(a^2 + c^2)}{2}} + \sqrt{\frac{bc(b^2 + c^2)}{2}} \geq ab + ac + bc$

Also it is $\sqrt{abc(a + b+ c)} \leq ab + ac +bc$ because
$a^2bc + b^2ac + c^2ab \leq a^2b^2 + a^2c^2 + b^2c^2 + 2a^2bc + 2b^2ac + 2c^2ab$
So the inequality is proved.

From Muirhead imediatelly we have that ${3,1,0}\succ{2,11}$
So $a^3b\geq a^2bc$
We add them and we are done.

Yes, I know it's easy using Muirhead but I wanted to see how other solutions would go  :)

[quote="Andreas"]$ab(a^2 + b^2) + ac(a^2 + c^2) + bc(b^2 + c^2) \geq 2abc(a + b + c)$

$\sqrt{\frac{ab(a^2 + b^2)}{2}} + \sqrt{\frac{ac(a^2 + c^2)}{2}} + \sqrt{\frac{bc(b^2 + c^2)}{2}} \geq \sqrt{abc(a+ b+ c)}$
[/quote]
How do you get from the first step to the second?

[quote="peeta"]well, first of all i think its rearangment as well, but those can be done by am-gm in most cases:
$\frac{a^3b+a^3c+ab^3+ac^3}{2}\ge2\cdot\sqrt[4]{a^8b^4c^4}=2a^2bc$

do that cyclic with $a,b,c$ and add them up u get the needed inequality[/quote]

Shall i assume a,b,c to be +ve reals???

yeah it should be a,b,c positive reals. otherwise muirhead and am-gm doesnt work and also try $a=-100;b=1;c=0$ ...


topology
linear algebra
matrix
advanced fields
advanced fields unsolved

Let $ k$ to be an odd integer. Let $ M$ to be a $ 2k$-dimensional compact, oriented, smooth manifold without boundary. Prove that the k-th Betti number, i.e. the dimension of the k-th de Rham cohomology $ H^k (M)$, is even.

Use Poincare duality; the bilinear pairing defined in it on $ H^k (M)$ is antisymmetric, so the matrix of the pairing has purely  imaginary eigenvalues in pairs with their complex conjugate.


points $A,B,C$are given on circle, find $D$ on circle,inwhich:

$\frac{1}{AD}=\frac{1}{AB}+\frac{1}{AC}$

[quote="draperp"]$AD=\frac{AB\dot{AC}}{AB+AC}$

Multiplication, addition, and division can all be done with a compass and a straightedge, if that's what your looking for.[/quote]

im looking for a cleverly solution ;) .

[quote="Ashegh"][quote="Altheman"]let $I$ be an inversion about a circle radius 1, centered at $A$

$I(B)=B'$ and $I(C)=C'$, have $D'$ on $AC'$, with $C'$ between $A$ and $D'$ such that $C'D'=B'A$, then $I(D')=D$ using inversion[/quote]

i designed this problem by inversion.

i really enjoyed, ure solution.because i love inversive solutions.

u are very clever. ;)[/quote]

ha, when i saw you post a problem, i knew immediately to think inversion since inversion is your favorite technique  :)


floor function
number theory
prime factorization

[hide="hint"]
What primes in the prime factorization of 2006! would add zeros to the end of the number, and how many of them are there?
[/hide]

[quote="Franz Joseph"]How many $0$'s does $2006!$ end with?[/quote]
[hide="solution"]
counting the #of zeros is same as counting the # of 5's in $2006!$.
so $2006!$ has 401 5's, 80 of 25's, 16 of 125's, and 3 of 625's.
Total $401+80+16+3=500$ zeros.
[/hide]

[quote="lotrgreengrapes7926"][hide="Generalization"] The number of zeroes at the end of $n!$:

$\sum_{i=1}^{\infty}\left\lfloor\frac{n}{5^{i}}\right\rfloor$[/hide][/quote]
 :coolspeak:  Thanks a lot.


2006 Nats sprint #26 & #30

26. What is the product of the two smallest prime factors of $ 2^{1024} \minus{} 1$ ?

[hide="my answer"] 21 [/hide]

[hide="their answer"] 15 [/hide]

30. The number $ \sqrt (17 \plus{} 12\sqrt {2})$ is an example of a number expressed in "embedded radical form" because of the $ \sqrt {2}$ under the radical. What is the equivalent value of this number expressed in simplest radical form $ a \plus{} b\sqrt {c}$ where $ a$ and $ b$ are integers and $ c$  is a positive integer containing no perfect square factors greater than one?

[hide="my answer"] guh....[/hide]

[hide="their answer"] $ 3 \plus{} 2\sqrt {2}$ [/hide]

Notes: The first square root goes over all that junk. That was a doozy for me Latex-wise. It took me 8 minutes to type! ( :rotfl: )

solution for 26

[hide]
for number 26 factor as the difference of perfect squares.
so it is (2^512+1)(2^512-1)
you can factor the (2^512-1) as the difference of perfect squares
you can continue until as (2^512+1)(2^256+1)(2^128+1).........(2^2+1)(2^1+1)(2^1-1)
the last 3 terms are 5, 3, and 1
so 5*3 is 15
[/hide]

[quote="dynamo729"]2006 Nats sprint #26 & #30

26. What is the product of the two smallest prime factors of $ 2^{1024} - 1$ ?

[hide="my answer"] 21 [/hide]

[hide="their answer"] 15 [/hide]

30. The number $ \sqrt (17 + 12\sqrt {2})$ is an example of a number expressed in "embedded radical form" because of the $ \sqrt {2}$ under the radical. What is the equivalent value of this number expressed in simplest radical form $ a + b\sqrt {c}$ where $ a$ and $ b$ are integers and $ c$  is a positive integer containing no perfect square factors greater than one?

[hide="my answer"] guh....[/hide]

[hide="their answer"] $ 3 + 2\sqrt {2}$ [/hide]

Notes: The first square root goes over all that junk. That was a doozy for me Latex-wise. It took me 8 minutes to type! ( :rotfl: )[/quote]
[hide="26"]$ 2^{1024}-1=(2^{512}-1)(2^{512+1)}$. If you see why this works, you can keep breaking it down until you get your answer.
$ (2^{256}-1)(2^{256}+1)(2^{512}+1)$
$ (2^{128}-1)(2^{128}+1)(2^{256}+1)(2^{512}+1)$
$ (2^{64}-1)(2^{64}+1)(2^{128}+1)(2^{256}+1)(2^{512}+1)$
$ (2^{32}-1)(2^{32}+1)(2^{64}+1)(2^{128}+1)(2^{256}+1)(2^{512}+1)$
$ (2^{16}-1)(2^{16}+1)(2^{32}+1)(2^{64}+1)(2^{128}+1)(2^{256}+1)(2^{512}+1)$
$ (2^{8}-1)(2^{8}+1)(2^{16}+1)(2^{32}+1)(2^{64}+1)(2^{128}+1)(2^{256}+1)(2^{512}+1)$
$ (2^{4}-1)(2^{4}+1)(2^{8}+1)(2^{16}+1)(2^{32}+1)(2^{64}+1)(2^{128}+1)(2^{256}+1)(2^{512}+1)$
$ (2^{2}-1)(2^2+1)=15$[/hide]

[hide="30"]First some reasoning. There is only one possibility for $ c$, which is 2. So now we have $ \sqrt{17+12\sqrt{2}}=a+b\sqrt{2}$. Square both sides and match up.
$ 17+12\sqrt{2}=a^{2}+2ab\sqrt{2}+2b^{2}$
So now we have two equations, $ a^{2}+2b^{2}=17$ and $ ab=6$. Quick guessing results in $ a=3$ and $ b=2$. So our answer is $ 3+2\sqrt{2}$[/hide]

Some are trickier, such as $ \sqrt{5\plus{}2\sqrt{6}}\equal{}\frac{\sqrt{10\plus{}4\sqrt{6}}}{\sqrt{2}}\equal{}\frac{\sqrt{6}\plus{}2}{2}$.
Try this one for practice:

$ \sqrt{8\plus{}4\sqrt{3}}\equal{}?$

how do you know to divide by $ \sqrt{2}$

You multiply by $ \sqrt{2}/\sqrt{2}$.  It's intuitive. Or you could set up the equation as $ a\plus{}b\sqrt{6}$
$ a^2\plus{}6b^2\equal{}5$ and $ ab\equal{}1$.

[quote="mathcrazed"]Some are trickier, such as $ \sqrt {5 \plus{} 2\sqrt {6}} \equal{} \frac {\sqrt {10 \plus{} 4\sqrt {6}}}{\sqrt {2}} \equal{} \frac {\sqrt {6} \plus{} 2}{2}$.
Try this one for practice:

$ \sqrt {8 \plus{} 4\sqrt {3}} \equal{} ?$[/quote]

I got.... $ \sqrt {5} \plus{} \sqrt {3}$, but I don't know how I got it.....

[quote="mathcrazed"]Some are trickier, such as $ \sqrt {5 + 2\sqrt {6}} = \frac {\sqrt {10 + 4\sqrt {6}}}{\sqrt {2}} = \frac {\sqrt {6} + 2}{2}$.
Try this one for practice:

$ \sqrt {8 + 4\sqrt {3}} = ?$[/quote]

is the answer $ \sqrt{2} +\sqrt{6}$?

srry if i didnt use $ \text{\LaTeX}$ correctly....

[quote="mathking123"][quote="mathcrazed"]Some are trickier, such as $ \sqrt {5 + 2\sqrt {6}} = \frac {\sqrt {10 + 4\sqrt {6}}}{\sqrt {2}} = \frac {\sqrt {6} + 2}{2}$.
Try this one for practice:

$ \sqrt {8 + 4\sqrt {3}} = ?$[/quote]

is the answer $ \sqrt {2} + \sqrt {6}$?

srry if i didnt use $ \text{\LaTeX}$ correctly....[/quote]

yep

[hide="method 1"]
Plug it into the TI-89 and it will spit out the simplified answer
[/hide]
[hide="method 2"]
$ \sqrt{8+4\sqrt{3}}=a+b\sqrt{3}$
$ 8+4\sqrt{3}=a^2+3b^2+2ab\sqrt{3}$
Thus, $ a^2+3b^2=8$ and $ ab=2$.
Solve and you find that $ a=b=\sqrt{2}$.
Note that you will also find extraneous solutions, so make sure your answer is positive, since that's what the square root sign implies
[/hide]


Charlene considers herself to be a great bargain shopper because she found a prom dress that cost her only 22 dollars before tax. The dress was on a rack labeled “50% off lowest marked price,” and the lowest marked price was already a 75% reduction from the original price. What was the original price of the dress?

[hide] If the dress was already $ 50\%$ off the marked off price was $ 22(1+\frac12)==33\$$. Then, since it was reduced by $ 75\%$, we have $ x(1-\frac34)=33\implies x=\boxed{132\$}$.[/hide]

[hide]Once again, my solution is quite similar to gaussintraining's.  So the dress was already 50% off the marked off pricece was 33 dollars. Since it was reduced by 75%, we have x(1/4)=33 so the answer is 132 dollars.[/hide]


function
search
algebra proposed
algebra

Suppose  $ f : R^\plus{}\rightarrow R^\plus{}$  is a decreasing continuous function such that for all $ x; y\in R^\plus{}$ , $ f(x \plus{} y) \plus{} f(f(x) \plus{} f(y)) \equal{} f(f(x \plus{} f(y))) \plus{} f(y \plus{} f(x))$. Prove that $ f(f(x)) \equal{} x$.


Simplify $ (x\minus{}a)(y\minus{}b) \equal{} ab$. Hence solve $ \frac{2}{x} \plus{} \frac{3}{y} \equal{} 1$ for integer values of $ x$ and $ y$.

this simplifies to a/x+b/y=1 and second  results in same diophentine equation

[hide="Solution"]
$ (x\minus{}a)(y\minus{}b) \equal{} ab \Rightarrow ay\plus{}bx\equal{}xy$

$ \frac{2}{x}\plus{}\frac{3}{y}\equal{} 1 \Rightarrow 2y\plus{}3x\equal{}xy \Rightarrow a\equal{}2,b\equal{}3$

$ (x\minus{}2)(y\minus{}3)\equal{}6\equal{}6*1\equal{}3*2 \Rightarrow (a,b): (8,4),(3,9),(5,5),(4,6)$[/hide]


Let $ a/b$ be a fraction with $ a,b$ naturals and $ b>\equal{}2 gcd(a,b)\equal{}1$ and $ gcd(b,3)\equal{}1$ prove that there exists a fraction of the form $ n/3^k\minus{}1$ equal to the first one

is that supposed to be $ \frac{n}{3^k\minus{}1}$?

$ k \equal{} \text{ord}_b(3), n \equal{} \frac {3^{\text{ord}_b(3)} \minus{} 1}{b} \cdot a$ which is an integer by definition.

yeah, that's too easy a problem to be russian :), that's why i ask


A fair 6-sided die is rolled.  If I roll $ n$, then I win $ n^2$ dollars.  What is the expected value of my win?  Express your answer as a dollar value rounded to the nearest cent.

if only I could win n^2 dollars every time I rolled a die...


I HEARD THAT MCRAWFORD IS NOT AN ADMIN ANYMORE AND IS MOVING AWAY FROM AOPS!

WHAT AM I GOING TO DO NOW?

OH MY GOD! s

[hide]As arnav would say...  :P [/hide]

Sorry, it's only funny when it's Arnav. :D

[quote="Karth"][quote="Pakman2012"]you're:

a) so not funny

b) late

c) hes still teaching woot

d) hes got other things he wants to pursue, which is good for him, and we wish him good luck on his way.[/quote]

 :|

Part a)

That was not really the purpose of this post. I added the z at the end because post titles have to contain at least one lower case letter.

Part b)

I came to know of this only recently.

Part c)

I didn't know this

Part d)

I completely agree with.[/quote]

karth you're in woot aren't you : |

yeah Mr. Crawford taught the first class

remember

building bridges...

:ninja:

seriously guys

come on now

let's stay ON TOP OF THE BALL ok


[hide]admission=3
food=3
soveniers=half the money he had left
multiply all that by 3 to get what he had at the begining.
3*3=9
3*3=9

18
if theres 30 dollars originally, then the 18 for admission and soveniers then $12 left half that is 6 half that is 3 and then you leave with nothing.  There for the answer is he spent $30 at the festival.
[/hide]

[quote="MCrawford"]John attended a three-country multicultural festival.  At the first exhibit, John paid a 3 dollar admission fee.  He then spent half the money he had left on souvenirs.  Finally, he spent 3 dollars on food and left.  At each of the second and third exhibits, he again paid a 3 dollar admission fee, then spent half of the money he had left for souvenirs, then spent 3 dollars on food, and left.  When he left the third exhibit, he had no money remaining.  How many dollars did he spend at the festival?[/quote]

[hide="working backwards"]
At the beginning of the third exhibit, he had 9 dollars.
At the beginning of the second exhibit, he had 27 dollars.
At the beginning of the first exhibit, he had [b]63 dollars[/b].
[/hide]

[quote="MCrawford"]John attended a three-country multicultural festival.  At the first exhibit, John paid a 3 dollar admission fee.  He then spent half the money he had left on souvenirs.  Finally, he spent 3 dollars on food and left.  At each of the second and third exhibits, he again paid a 3 dollar admission fee, then spent half of the money he had left for souvenirs, then spent 3 dollars on food, and left.  When he left the third exhibit, he had no money remaining.  How many dollars did he spend at the festival?[/quote]

[hide="how i did it"]3*2+3=9
12*2+3=27
30*2+3=[hide="the answer"]63[/hide][/hide]


geometry
cyclic quadrilateral
geometry solved

Read up something about pole and polar/reciprocation or inversion and I'm sure this problem will become a easy task for you.

[quote="Saul"]Let $ABCD$ be a cyclic quadrilateral with circumcentre $O$. Let $AC$ and $BD$ meet at $X$, $AB$ and $CD$ meet at $Y$ and $BC$ and $AD$ meet at $Z$. Show that $OY$ and $XZ$ are perpendicular.[/quote]

This is just http://www.mathlinks.ro/Forum/viewtopic.php?t=5326 , applied to the cyclic quadrilateral ACBD instead of ABCD.

  darij


Let $ a,b>0$ and $ x\in [0,\frac{1}{2}]$. Prove or disprove $ (a\minus{}b)^2\le 2(a^xb^{1\minus{}x})^2(lna\minus{}lnb)^2$.

The inequality does not hold when x=0. Simply choose x=0, a=2b, LHS = $ b^{2}$,  RHS =$ 2b^{2}(ln2)^{2} < b^{2}$


When writing all the integers from 1 to 2007 inclusive, how many 0's are used?

[hide]

In the ones place, a zero appears $ \lfloor{\frac{2007}{10}}= 200$ times
In the tens place, a zero appears 10 times in every hundred, or $ 190+8 = 198$
In the hundreds place, a zero appears 100 times in every thousand or $ 100+8 = 108$

add it up and you get $ 506$
[/hide]


[quote="Inspired By Nature"]You given 25 different positive numbers. Prove that You can choose two of them such that none of the other numbers equals to the sum or to the difference of the chosen numbers.[/quote]

What about choosing $1, 2, 3, ... , 24, 25$?

[quote="SplashD"]You can choose the 24,12 for example.[/quote]

Ohh obvious. Thanks

you can always choose the greaters and none of the other will have the same sum..

sorry for my english :oops:

[quote="SplashD"]The question uses or, not and, so you can't just choose the greater ones[/quote]
exactly. :)  Its a very nice problem.

[quote="ragnarok23"]Do you mean positive difference?

You can do 29 and 30, 30+29=59 and  29-30=-1  :huh:[/quote]

obviously because all of the integers are positive

i can see the delete button now, only for my latest post though. :huh:


I am making a personality test about patience there are four possible choices and 9 questions.

One question gives 2 points

One gives 1

One gives 1

One gives 0

On every page except the last one, becuase I want to be able to tell if they lied. I am wondering how to make it so there are 4 possibilities

Liar

Patient

Impatient

Very Impatient

The test purposely lies to them at the end and says they wont get a score and I wasted their time, but they do. So if they answer this question incorrectly it means they lied. But I can't figure out how to do that becuase I suck at math lol.

and the score goes from 1 to 100 points but I can't even figure out the proper range between score types..

I think you need to be more specific... I'm not quite sure what you are doing.

Anyway, from what I can tell, you have 8 question which count from 0-2 each for a max of 16 points.  To make the score out of 100, compute 100x/16 where x is the number of points earned.

As the lie detector, I don't know what you have in mind.  Perhaps if their score lie in the range of the answer they choose, then the test should decide they lied.

For the score ranges, my best suggestion is just to experiment or set them arbitrarily - perhaps 0-25, 25-50, 50-75, and 75-100.  But be sure to specify the exact boundaries (that is, include every possible score in one of the ranges), because it will be possible to get a score of 25, 50 or 75, and these cannot be between the categories - they have to belong to one or the other.  I'd say just count it in the higher category or the lower for every time.

Note that if you do the scoring as you described, there will only be 17 possible scores - not a very specific scale.


n cirles are given in the plane. They divide the plane into parts. Show that you can color the plane with two colors so that no colors with a common boundary line are covered the same way. Such a coloring is called a proper coloring

Induction. For n=0 this is obvious. Let we have a plane which is divided by n circles and was 2-colored by +1 and -1. Put n+1-th plane circle colored by +1 inside and by -1 outside and then multiply signs.


geometry
incenter
trigonometry
circumcircle
parallelogram
angle bisector
geometry proposed

Let ABC be a rectangular triangle with <C = 90°, and let R and S be the points where the incircle of ABC touch  AC and  BC respectively. If K = RS^AI where I is the incenter of ABC and l the parallel through K to BI, and L = l^CB, then prove that LI bisects AB.

[quote]K = RS^AI[/quote]
What do you mean by this notation?

We have to prove from Menalaus’s theorem that:
$ \frac{BL}{LD}=\frac{AI}{DI}$. But $\frac{AI}{ID}=\frac{b+c}{a}$.
So we shall prove that: $ \frac{BL}{LD}=\frac{b+c}{a}$.  (1)
Since $\frac{BL}{LD}=\frac{BD+LD}{LD}=\frac{BD}{LD}-{1}$.
(1)	becomes :  $\frac{BD}{LD}=\frac{2(p-a)}{a}$.
But $\frac{BD}{LD}=\frac{ID}{DK}=\frac{a \cdot{l_a}}{2p\cdot{DK}}$. (2)
In $\triangle{SDK}$ ,$\angle{DSK}=45^\cdot$, $\angle{DKS}=(45-A/2)^\cdot$, and $SD=CD-CS=\frac{ab}{b+c}-(p-c)$.
Appling sine theorem is this triangle we find $DK$and then from  (2)we get the result.

Here D is the foot of bisector of $\angle{A}$  p is the semiperimeter of triangle.and $l_{a}$ the bisector from A.

to jpark : XY^UV denote the intersection point of the lines XY and UV.

to sailor:  Could you explain   what do you would mean  BD/LD = (BD+LD)/LD = BD/LD - 1, ... there is a typo probably.

to all: in any case there is a completelky synthetic proof.

Yes of course, I was in a hurry  :D
I'll correct it!

I'll sketch a proof for the following generalization:

Given triangle $ABC$, all the notations stay the same as in the original problem, except for $K$, which is the point where $AI$ cuts the circumcircle of $ABC$ the second time (in the case when $\angle C=\frac\pi 2$, this definition coincides with the one given by sprmnt for his $K$). We have to prove the same thing, namely that $LI$ bisects $AB$.

Let $M$ be the midpoint of $AB$. It's easy to see that $KL$ is, in fact, the external angle bisector of $\angle BKA$, or, in other words, $KL$ cuts the circle $(ABC)$ again in the midpoint of the arc $AB$ which contains $C$. Let $T$ be the point where the internal bisector of $\angle AKB$ cuts $BC$. We have $(KT,KL;KB,KA)=-1$, so if we show that $(IT,IM;IB,IA)=-1\ (*)$, then we will have shown that $IM$ also cuts $BC$ in $L$, which is what we want.

Since $M$ is the segment of $AB$, $(*)$ amounts to showing that $IT\|AB$, which is rather easy, so I won't do it :).

Only a sketch.

Let N = KL^c(ABC), one can prove that LCK~CNK, Then NK/CK = CK/KL. One can prove also that CK = BK = IK, then IK^2 = KN*KL, wich implies that  <NIK = <ILK.
After some angle chasing and considering that JIKN [where J=BI^c(ABC)] is a parallelogram, one can prove that <NIK = <OIB. Then <ILK = <OIB wich togheter the fact that LK//BI, implies the thesis.


we have a square with side length $ a$,we draw 4 circles with radius $ a$ from each of the 4 vertices of the square,calculate the area of the intersection of these 4 circles.

P.S:I'm not sure if this is the right place to post this problem,so sorry if it isn't...

Denote by $A,B,C,D$ the vertices of the square with side length $R.$ Circles $(C),(D)$ centered at $C,D$ with radii $R$ meet inside $ABCD$ at $P.$ Likewise, $Q \equiv (D) \cap (A),$ $R \equiv (A) \cap (B)$ and $S \equiv (B) \cap (C).$ Clearly, $\triangle ABR$ is equilateral with side length $R,$ thus $\angle RAB=60^{\circ}$ $\Longrightarrow$ $\angle DAR=30^{\circ}$ and due to obvious symmetry, $\angle BAQ=30^{\circ}.$ Thus, the region $S$ overlapped by $(A),(B),(C),(D)$ is the sum of the square $PQRS$ and 4 circular segments with radii $R$ and central angle equal to $30^{\circ}.$

$S=R^2\left (2-2 \cos \frac{\pi}{6} \right)+4 \left [ \frac{R^2}{2} \left (\frac{\pi}{6}-\sin \frac{\pi}{6} \right) \right]=R^2 \left ( 1-\sqrt{3}+ \frac{\pi}{3} \right).$


What is the value of $ x$ in the equation $ \frac{1}{x} \plus{} \frac{2}{x} \div \frac{4}{x} \equal{} 0.75$?

If we simplify the equation we find that x is 4.


If f(x) = |(x^2-12)(x^2+4)|, how many numbers in the interval -2 <= x <= 3 satisfy the conclusion of the Mean Value Theorem?

A. None
B. One
C. Two
D. Three
E. Four

Please don't double post.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=172088

Also, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26438]PLEASE do not post calculus problems in this forum![/url]


A right triangle ABC with legs 3, 4, and hypoteneuse 5 is drawn. AB is 3 AC is 4, and BC is 5. A line segment is drawn down from A to D to form two right angles. What is the area of triangle ABD?

He probably means with the hypotenuse of the triangle -- that is, an altitude is drawn.

Si si.

Hint: [hide]Similar Triangles[/hide]

[quote="Leeyuhang"]i don't get it. can't you use any number for BD and AD?[/quote]
Draw a diagram, see if you can figure it out...

I believe that is incorrect but quite frankly I'm not sure either right now. Gotta go!

[quote="Leeyuhang"]i don't get it. can't you use any number for BD and AD?[/quote]
D is meant to be the base of the altitude from A to BC.. ie on BC somewhere so AD is perpendicular to BC. I think thats what is meant.

After coming from a three-day field trip, I'm probably quite rusty but I got 54/25.  Just use similar triangles like Tare said.

well, for one, you dont even know where d is on...

D is the foot of the altitude.  It doesn't explicitly state that, but that's the only thing that makes sense.   Then it's you're job to figure out where that is, distance-wise.

since the extended line ad is the alt of the triangle it is also the height. you already have the base so 5x=12 so x=2 2/5.
therefore 3^2-2.4^2=BD and BD=1.8.
1/2(1.8 * 2.4) = 2.16 or 54/25


prove that 5^99 + 11^99 + 17^99 is divisible by 33

The classic method (study the power of 5, 11, 17 modulo 11 and 3) is long.
5+11+17=33 strange ...
I search

[quote]The classic method (study the power of 5, 11, 17 modulo 11 and 3) is long. [/quote]
Not really (it is realy not long):
Actually it is true :[b] Not only for 99 but for ANY odd power:
[/b]

(With mod 3 you have:  ("==" used for conguance)
(-1)^99+(-1)^99+(-1)^99 == 1+1+1  == 0
And mod 11 you have
(-6)^99+0+(6)^99 == 6^99* (-1+1) == 0
QED.

How about using a little Fermat's Little Theorem.

We know for primes, any element of U_m has property a^(phi(m))=1 (mod p) so a^p=a (mod p)

Therefore, splitting divisibility, we look at the three expressions in mod 11 and mod 3.

therefore, a^100=1 (mod 11)
so dividing by a in Z_11 gets 5^99=9 (mod 11) 11^99=0 (mod 11) 17^99=2 (mod 11) which sums to 11 (mod 11) =0 (mod 11)

Similarly, the process holds in mod 3, so therefore congruency to 0 in both mod 3 and 11 yields congruency to 0 mod 33.

Therefore, the expression is divisible by 33

yep. and the same can be done with euler phi $\phi$ function.


Let $S=\{x_0, x_1, \cdots, x_n\} \subset [0,1]$ be a finite set of real numbers with $x_{0}=0$ and $x_{1}=1$, such that every distance between pairs of elements occurs at least twice, except for the distance $1$. Prove that all of the $x_i$ are rational.

Solution in first reply at http://www.mathlinks.ro/viewtopic.php?p=15303 .

We consider the basis of $x_i$. Let that be $\mathbb{Q}[r_1,r_2,\ldots,r_m]$. Then choose any $r$ let us say $r_1$. Since $r_1$ is in the basis thus there must exist two $x_i$ such that there difference contains $r_1$ (This ensures that the component of $r_1$ in the difference we'll define isn't $0$) . Let

$X_1$ = [i]Those $x_i$ who have the maximum component of $r_1$[/i] and
$X_2$ = [i]Those $x_i$ having minimum component of $r_1$.[/i]

Then consider the difference of "rightmost" (i.e. [b]greatest[/b]) member of $X_1$ and "leftmost" ( i.e. [b]smallest[/b] ) member of $X_2$. If this has to have a friend then one vertex must be in $X_1$ and other in $X_2$. But if you choose any such pair (other than this one) then its length will be strictly less than the earlier chosen pair, the most preposterous statement! So all lengths must be rational (as we must dustbin any $r_i's$ if they pop up just like that [size=50]quantum fluctuations anyone?)[/size] $\square$.
[size=150][u]IMPORTANT:[/u]This stunt has been performed under the supervision of [b]biomathematics.[/b] Do not try at home.[/size]
[size=75][b]Patreon supporters[/b]: 1. (to be disclosed) 2. (to be disclosed)[/size]


$a_1, a_2, ... , a_{121}$ is a sequence of positive integers not exceeding $1000$. The value $n$ occurs more frequently than any other, and $m$ is the arithmetic mean of the terms of the sequence. What is the largest possible value of $[m - n]$?

[hide="Answer"]Essentially, we want the lowest possible value for $n$ and the highest possible value for the sum of all of the rest of the terms. We must have $n=1$, and we can possibly have two $n$'s and one each of $1000+999+998+...+881$; three $n$'s and two each of $1000+999+998+...+941$; four $n$'s and three each of $1000+999+998+...+961$; etc. Upon testing, we find that having four $n$'s and three of each of the others gives the greatest possible value, and we have $\lfloor m-n\rfloor=\boxed{947}$.[/hide]


function
limit
inequalities
real analysis
real analysis unsolved

Is there a function $ f$ which is discontinuous at every point, and which has only removable discontinuities?

Def:If $ \lim_{x \to a} f(x)$ exists, but is $ \# f(a)$, then $ f$ is said to have a removable discontinuity.

I'll assume there is such a function $ f$ defined on some closed interval $ I \equal{} [a, b]$ and derive a contradiction.

Define $ g(x) \equal{} \lim_{t\to x} f(t)$. Then $ g(x)$ is a well-defined function $ I$. Choose $ c \in I$. For $ \epsilon > 0$, there is $ \delta > 0$ such that $ 0 < |x \minus{} c| < \delta$ implies $ |f(x) \minus{} g(c)| < \epsilon$. Also, choose any $ d$ such that $ 0 < |d \minus{} c| < \delta$. Taking limit $ x \to d$, we have $ |g(d) \minus{} g(c)| \leq \epsilon$. This shows that $ g$ is continuous.

For each $ n$, define $ A_n \equal{} \{ x \in I \, | \, |f(x) \minus{} g(x)| \geq 1/n \}$. Since $ \bigcup_{n\equal{}1}^{\infty} A_n \equal{} I$, some of $ A_n$ must be uncountable. Then $ A_n$ must have at least one limit point, so there is a convergent sequence $ (x_k)_{k\equal{}1}^{\infty} \subset A_n$. Let $ x \equal{} \lim_{k\to\infty} x_k$. Then $ g(x) \equal{} \lim_{k\to\infty} f(x_k)$. But taking $ k \to \infty$ to the inequality $ |g(x_k) \minus{} f(x_k)| \geq 1/n$, we have $ 0 \equal{} |g(x) \minus{} g(x)| \geq 1/n > 0$, a contradiction.

Therefore, there is no such $ f$.

Slightly modifying the proof, we find that if $ f : [a, b] \to \mathbb{R}$ has only removable discontinuity, then $ f$ has at most countable discontinuity.


we have a insulating ring of mass $ M$ and radius $ r$ and with very light spokes.The spokes join the outer region of the ring to an inner circle of radius $ a$ concentric with the  the ring's circumference.we have a magnetic field  $ \vec B$ in this region(circle of radius $ a$ which contains no material) only perpendicular to the plane of the ring,a charge density $ \lambda$ is given to the circumference of the ring.the magnetic field was instantaneously switched off.find the final angular speed of the ring

By Faraday's law, $ E*2 \pi r = \pi a^{2}B / t$.

Thus, $ F*t = E*q * t = E* 2 \pi r * \lambda * t = \pi a^{2}B \lambda$.

$ \Delta v = r \Delta \omega = \frac{F*t}{M}$.

Therefore, $ \omega = \frac{F*t}{M*r}= \frac{\pi a^{2}B \lambda}{M r}$.

u r right :)
well the direction is of course $ -\vec k$

If the magnetic field was instantaneously switched off, what is t ?

yes well that $ t$ should actually have been $ dt$.just like impulse

Faraday's Law : the electromotive force (in this case could be $ \; E.2\pi.r\;$) is given by

\[ \varepsilon\,=\;-{{\partial \Phi}\over{\partial t}\;}\]  that in this case is infinity ?!
And what about the self-inductance ?  It's negligible?

instantaneously means fastly and not in an 'instant'

We need to know " how fast".  That is; we need the law that govern the flux behavior.

Sorry, but I think that in order to aplly  Faraday's Law, you must know how the magnetic field varies with time ( see above), an d not the opposite

well here u know in a given short time $ \Delta t$ the magnetic field was reduced to 0 and $ \frac{d \phi}{dt}\approx \frac{\Delta \phi}{\Delta t}$


limit
trigonometry
function
calculus
calculus computations

[quote="bgbgbgbg"]find the limit:
lim (sinx)/x  as x goes to 0 if x in degrees[/quote]
Isn't it $ 1$?

[quote="bgbgbgbg"]notice that x in degrees not in radian[/quote]

Make the substitution u = x*2Pi/360.  x tends to 0+ iff u tends to 0+.  Replace all x's with u.

it doesnt even count much in hyperreals.

$ \lim_{x \to 0} \frac {\sin x} {x} \equal{} 1$. It does not matter the unit of measure for $ x$, be it radians, degrees (or maybe $ \textrm{Fahrenheit}^{\circ}$  :) ).

it would makes sense if you could tell what is the meaning of a real number divided by a degree ?

Given the level of understanding this question was asked at, the last few posts really don't fit.

There is a perfectly sensible $ \sin$ function defined on angles, which can be measured in degrees. Before calculus, degrees are as good as any angle measurement system, and cive us nice numbers for nice angles.
So, now we want to develop the calculus of trigonometric functions. The first crucial limit that comes up is this one: what is $ \lim_{x\to 0}\frac{\sin x}{x}$. Since we don't know any better, that angle is marked in degrees for now. What is the value of this limit? Some ugly number; if we look closer, it turns out to be $ \frac{\pi}{180}$.
We don't like that, so [i]we invent an entirely new angle measurement[/i], called the radian. Scaling the angle measurement changes the limit, and for the radian's arclength scaling it turns out to be $ 1$.
That is [b]the[/b] point of radians. From there on, we never use degrees in a calculus setting again.


vector
linear algebra
linear algebra unsolved

Let V and W be finite-dimensional vector spaces, and let T:V-->W be a linear transformation.  Suppose that B is a basis for V.  Prove that T is an isomorphism if and only if  T(B) is a basis for W.

Please help!

It suffices to prove that isomorphism preserves linear independence. Let $A={\alpha_{1}, \dots, \alpha_{n}}$ be a basis of V. Suppose that $T(A)=T(\alpha_{1}), \dots, T(\alpha_{n})$ is linearly dependent, i.e. $a_{1}T(\alpha_{1})+\dots+a_{n}T(\alpha_{n})=0$ for some nonzero $a_{1}, \dots, a_{n}$. Then $0=a_{1}T(\alpha_{1})+\dots+a_{n}T(\alpha_{n})=T(a_{1}\alpha_{1}+\dots+a_{n}\alpha_{n})$ and since T is an isomorphism, $T(x)=0$ implies $x=0$ (kerT={0}) - a contradiction, since A is linearly independent.

The inverse implication is similar.

I am with you so far in that direction of the proof. I am still having trouble with the reverse direction.  Would you mind elaborating please?

Thanks for your help!

Let $A=\alpha_{1}, \dots, \alpha_{n}$ be a basis of V and $T(A)=\beta_{1}, \dots, \beta_{n}$ $(\beta_{i}= T(\alpha_{i}))$ be a basis of W and suppose that there is some $\alpha \neq 0, \alpha \in kerT$. Let $\alpha = a_{1}\alpha_{1}+\dots+a_{n}\alpha_{n}$ for some $a_{1}, \dots, a_{n}$. $T(\alpha)=0 \rightarrow T(a_{1}\alpha_{1}+\dots+a_{n}\alpha_{n})=0 \rightarrow a_{1}T(\alpha_{1})+\dots+a_{n}T(\alpha_{n}) =0 \rightarrow a_{1}\beta_{1}+\dots+a_{n}\beta_{n}=0$ - a contradiction, since $\beta_{1}, \dots, \beta_{n}$ were supposed to be linearly independent.


AMC USA(J)MO USAMO inequalities geometry USAMTS algebra

With the USAMO's coming up in less than two weeks, and me never having had a proof formally graded, I was curious as to what USAMO proofs earned high scores, and which did not.

If anyone has links to a site with proofs given for particular problems, and their respective scores, I'd appreciate it a lot. (The USAMO invitation letter said that people taking the test often overestimate their scores; I'm erring on the side of caution to make sure I write clear, concise proofs.) If anyone's kind enough to write their own proofs for problems from previous USAMO's and the like and tell what the score was, I'd much appreciate that too.

On that note, I've got a few questions:
1. What does it mean for a solution to be well-written?
2. When are 7s awared? Are they awarded when the solution is completely correct? Do they have to be well-written? Can they be brute force?
3. When and how is partial credit awarded?
4. How is brute force graded? (E.g., massive casework, Muirhead, analytic bashing, and worse...)
5. In geometry problems, is it mandatory for a diagram to be included? Preferred?
6. Which facts can we use in the proof without citation? Which facts can we use with citation? How precise must the citations be? (I once recall solving a problem by using another USAMO solution as a solution; I clearly cannot recall this off the top of my head. Am I simply not allowed to cite this, or can I just say "on a previous USAMO, [...]"?)

Sorry if pieces of this post seem incoherent/jumbled up/out of order. I'm quite tired now. Thanks for any responses.

1.  "Well-written" means "easy to understand."  I think a good way to explain this is that the grader shouldn't have to think too hard to understand your solution, and the grader shouldn't have to pull out a piece of paper to check any of your steps.

2.  7's are given for any completely correct solution.  Score doesn't depend on elegance at all, so an extreme brute-force solution will get a 7 if it is correct.  You can sometimes get a 7 even with a small arithmetic mistake.

3.  In the documentary "Hard Problems" about the 2006 IMO team, the concept of "grading up" vs. "grading down" is discussed.  If your solution is basically right but with a few minor errors, it will be graded down, meaning you start with a 7 and lose points for errors.  In this case you'll probably end up with a 5-7 depending on the errors.  If your solution has a big hole or is just a list of facts or ideas, it will be graded up, meaning you start with a 0 and get points for good stuff.  In this case you'll probably get a 0-2 depending again on how useful the stuff you wrote actually is.  On some problems it's possible to get a 3-4, but those scores are more unusual.  I think the graders have a rubric that tells them how many points to award for each problem.

4.  It's graded the same way as any other solution.  If the problem is fundamentally not brute-forceable, like if you tried to reduce a problem with infinitely many cases to finite casework, you'll get nothing.  Many people have said that graders aren't as lenient with brute force solutions, but I'm not sure if this is really true.

5.  It's not mandatory as far as I know, but not including a diagram is a good way to get your grader annoyed from the start.

6.  This has been asked a million times.  Don't cite a problem from another contest.  You don't need to cite really obvious stuff like the associative property of addition, but you need to cite most things.  Use common sense.

It's extremely true that people overestimate their scores.  The first reason for this is because people don't understand the scoring system.  If you have a half-complete solution, you don't get a 3 or 4, you get a 1.  A significant hole in your solution doesn't cost you 1 point, it costs you 6.  That's a big difference.  The second reason is because people don't know what constitutes a rigorous proof.  You need to consider every detail, and even if your proof ignores just one aspect of the problem or one small case, it's not complete.  A lot of people will think they turned in a complete solution and end up losing anywhere from 1 to 6 points.

Ironically, the official USAMO solutions are generally not a good resource for the quality of solution that the USAMO requires.  The reason for this is that the official USAMO solutions seem to often gloss over significant details (it's taken me 30 minutes to figure out a single line of an official solution), and that the official solutions will use a method that's easy to rigorize.  Often the method you will come up with will not be so easy to rigorize and will require a much longer proof.  My USAMO solutions average about 2-3 pages.  The official USAMTS solutions might be a better resource for good solutions.

Can you cite any theorem that has a name?

(I tried to post this message for 10 minutes but each time it said "No post mode specified.")

If you cite a theorem which trivializes the problem, don't expect points. However, as a general rule, if you can name it then you can cite it.

If a well known theorem kills the problem, I don't think they would put the problem on the test in the first place. But if that was the case, I think it is only fair that citing that theorem would earn you full credit, since it becomes the contest writers' fault for not putting up a good problem. I think that is what Richard Rusczyk said in the WOOT USAMO session.

Um, no.  the powers of 2 mod 9 go $ 2, 4, 8, 7, 5, 1$ with period 6.  This is not a factor of 9.  I think the correct statement, instead of integers, is primes: this problem is pwned by Fermat's little theorem.   :)

[quote="sdkudrgn88"][quote="MellowMelon"][quote="serialk11r"]But if that was the case, I think it is only fair that citing that theorem would earn you full credit, since it becomes the contest writers' fault for not putting up a good problem.[/quote]
Regardless of whether it's fair or not, you're probably going to get hammered in practice. As an example, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=875055#875055]this ridiculously easy IMO shortlist problem[/url] was on a MOP test once. During the test review session, the graders said that you could not cite that $ 2, 2^2, 2^3, \ldots$ was periodic modulo $ k$ for any $ k$ because it trivialized the problem; you had to prove it yourself. (much groaning ensued) I would presume there is little difference between how MOP tests and USAMOs are graded.[/quote]

If you didn't prove the periodicity by yourself, would you still get 2/7 because the proof was correct otherwise?

And to prove the $ {2^n}$ is periodic thing: note that each modulus, after being multiplied by two, has exactly one modulus that can result. Since there are a finite number of moduli (moduluses) that can result (to be exact, at most k-1 because 0 x 2 = 0), the sequence $ {2^n}$ is eventually periodic.

Here's another problem you might like if you're practising to write proofs.

Prove that, for any odd integer $ k \ge 3$, the sequence $ 2^0, 2^1, 2^2, 2^3, \ldots (\text{mod } k)$ has a period that is a factor of $ k \minus{} 1$.[/quote]

Its not k-1; its phi(k), which is k-1 if k is prime.

If we prove the theorem as a lemma though, would we get full points?

I think so, since that problem is pretty much the lemma.

You guys were going a bit off-topic, but ah well.

What qualifies as "a theorem with a name?" Would, say, obscure theorems like Ravi's substitution work?

To all of you who responded to my question, thank you.

EDIT: One more thing. How much work do we have to show? For instance, if we're expanding a polynomial expression (perhaps even an inequality =P), do we have to show the work for the expansion?

[quote="Zhero"]What qualifies as "a theorem with a name?" Would, say, obscure theorems like Ravi's substitution work?[/quote]

If this is what I think you're talking about, with something so simple you might as well prove it (you are speaking of the a=x+y,b=y+z,c=z+x thing with triangles, correct?). In this case, you would just be like, construct the incircle, and by equal tangents AE=AF=x, etc. or whatever. That takes two lines.

Also if you feel like you're spitting out the name of an obscure theorem, you should state it. That being said, make sure you actually have the statement correct. If you're using a wrong statement...you will probably not get more than negative zero.

I think a lot of people giving advice about USAMO grading are overstating the deductions that can result from small mistakes. If you really only made a calculation error, if your method is what they had in mind, I'm pretty sure you will still get full credit. If you missed an easy case, you're still pretty safe on getting at least a 6. At the same time, though, it is important to understand what constitutes a "small mistake." Just because you think a case you missed looks easy to deal with doesn't mean it is. You can really only judge partial work in the context of a complete solution that relates to it. No matter how interesting and non-trivial your work is, it's only worth points if it can be used to solve the problem.

[quote]If you're using a wrong statement...you will probably not get more than negative zero.[/quote]
Well, I suppose I could raise the bounds to positive zero to make me feel slightly better.

And yeah, thanks for the rest of that post too. :P And yes, that was the Ravi's substitution I was referring to.

[quote]You can get a 7 without showing work, but it will piss off the grader, so you should be sure that it's right or else you run a huge risk.[/quote]
Thanks. I'll keep that in mind.

probability 1.01, that makes me feel slightly better. =D

what about geometric transformations like dilation, inversion, etc? do you need to prove there propertises as lemmas (like what implies dilation and what dilation implies etc)...

Eh, I think those properties are sufficiently well-known, but in certain cases, you might want to remind the reader of certain facts (e.g., "Since inversion maps circles to lines, [...]".)

On that note, I've had some discrepancies when informed of the necessary amount of work. If, say, we are doing a Muirhead expansion, what would be optimal:

1. We actually show our expanding it on the proof sheet.
2. We just show the final expanded sum of symmetric sums.

In the case we do have to show the expansion, would you suppose that they would be knowledgeable of the technique I described in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=272496]this thread[/url] and just show the work described in that? Or should I describe that shortcut first? (I'm still not clear on how well-know it is eaxctly, so a response to that thread would be appreciated as well.)

Similarly, with an analytic bash, should we show the algebra?

Also, I take it that the official solutions are model solutions? In my opinion, they could be a bit clearer at certain areas, as they assume some facts that aren't ENTIRELY obvious (though still somewhat obvious), which I would feel slightly uncomfortable doing in my proofs...

Uhh yeah as I said, you can get a 7 without showing work. The example I have is during a MOP test. I did a muirhead bash and my solution went like this.

We want <insert inequality here>.
By clearing denominators, expanding, and collecting like terms, we get <insert simplified inequality>.

Yeah, Yi was pretty pissed at me for that, but I got 7/0.6!

MOP grading in that respect may be slightly more lenient though? I still think, to be safe: show all work.

[quote="not_trig"]I think you should show it, and NOT skip steps. I doubt graders would appreciate solutions by "magic." (Also, it is probably a way for them to discourage brute force, because they don't want to read your long super-computational equations...)[/quote]

Then would the method in my topic be sufficiently well-known for that to suffice as the work shown for inequality expansion? Or would I have to give a rigorous explanation as to why that is true?

And thanks, Hamster1800, but I'd rather err on the side of caution. Unless you mean to tell me that MOP grading is identical to USAMO grading, in which case I'd be... very pleased.

I heard the mop test had a style score between 0 and 1, which means the probability that you'd get the full solution score on a real competition. So that means Hamster1800's bash solution has 60% chance of earning a perfect 7 on the usamo, which is pretty risky.

Thank you, timwu.

Just in case a bashable inequality shows up: Does anyone know exactly how well-known [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=272496]this technique[/url] is for expanding them? (Just tell if you've seen it before or not.) I do intend to show my work, and this would be far more efficient (and safe, for that matter) than just expanding it out. I just want to know if I have to justify this reasoning in this proof (since if it's sufficiently well-known, I don't.)

And thanks to everyone for all the responses. It helped a lot.  :lol:

[quote="timwu"][quote="worthawholebean"]Show enough work that the grader can easily follow your solution without pencil and paper.[/quote]

I'm just quoting this.

Using symmetric notation (correctly!) could probably simplify your technique, but I'm not very familiar with inequalities. I highly doubt usamo would put on such bashable problems anyways.[/quote]

lol there was a certain #2 last year that elicited quite... a lot of bashing. Coordinates and trig lolol

Do we need to give proof before we use Pythagorean Theorem ?

I would assume that the Pythagorean theorem can be used without proof.

You can use theorems a lot more heavy than the Pythagorean Theorem without proof. Even stuff like Dirichlet's Theorem with no known elementary proof can be used. In general named theorems are okay, although a lot of caveats have already been given in the topic (e.g. write the statement of the theorem if it is obscure).

I still don't understand why Muirhead isn't allowed.

[quote="worthawholebean"]It's allowed.[/quote]

I still think it is [i]preferable[/i] to write the AM-GMs though? I think all this stems from Mildorf's inequality packet which says something to the effect of "graders frown upon Muirhead."

I don't understand why. Muirhead is a perfectly valid mathematical tool.

Do you have to prove these on the USAMO, or are they "well-known"?

1. |x| + |1-x| [u]>[/u] 1 for all real x.

2. The sum of the first k multiples of n is k(k+1)n/2.

3. The sum of the first k squares is n(n+1)(2n+1)/6.

4a. For a 2-D figure with fixed area, the one with the smallest perimeter is the circle.

And its converse,

 4b. For a 2-D figure with fixed perimeter the one with the max area is the circle.

1. A specific case of the triangle inequality, which states that $ |x|\plus{}|y|\ge |x\plus{}y|$ for $ x,y\in \mathbb{C}$.

2. Well-known, don't think it has a name.

3. Well-known, don't think it has a name.

4. Isoperimetric principle.

[quote="mathemonster"][quote="Temperal"]

4. Isoperimetric principle.[/quote]
or calculus?[/quote]
uh, what? calculus of variations is what the bernoullis used to prove the general result, but the name of the theorem is the isoperimetric principle.


i know that the round 1 due date is october 14th.

but when does the round 1 scores come out? is it october 14th?

It certainly takes some time for the graders to grade the papers, upload the grades, etc. However, due to the change in the USAMTS official solutions format, official solutions will be released earlier, so you'll be able to predict your own score sooner.

The Round 1 scores and comments will hopefully be released during the first week of November.

We are hoping to release both Round 1 solutions and the Round 2 problems within a week of the Round 1 submission deadline.

should i submit if i only got 3.5 problems right???

[quote="gauss1181"][quote="Lawrence Wu"]are the official solutions 5 solutions?[/quote]
ummm..yeah, just by common sense?[/quote]

i think he means, are there going to be 1 solution for each problem, or will there be more than 1 solutionfor some problems.

and i dont know the answer to this

varies by problem.  There is always at least one solution; occasionally a second or third solution will be given too.

Most problems will have one solution.  If there's a interesting second (or third) solution method we might present that too.

FYI -- they should be posted later today (hopefully).

the solutions will be posted today?

YAY!!!!

Probably if the grader thinks its outstanding....or unique.

There are no more commended solutions anymore though, I believe.

[quote="USAMTS site"]

We will no longer be issuing "Commended" scores. Some issues with doing this in the past is that there were no clear criteria for the graders to follow in determining which solutions would be commended, and also commended solutions were not "worth anything" in terms of the point total. Removing commended scores should make the grading process a bit more streamlined.
[/quote]
QED :D


Compute the positive integer $x$ such that
\[4x^3 - 41x^2 + 10x = 1989.\]

[hide]$1989 = 3^2*13*17$

use synthetic division to test the factors

$13$ works[/hide]

Yep.  This question shows the value of knowing the prime factorization of the current year.  Does everyone know that the prime factorization of 2005 is $5 \cdot 401$?


Im unsure of where this goes, so I just put it here.

My question is about a proof that he gives (chapter 8) for the intermediate value theorem, defined as following: If $f(x)$ is continuous on $[a,b]$ and $f(a) < 0 < f(b)$, then there is some $x$ in $[a,b]$ such that $f(x) = 0$.

"$A = ${$x: a \le x \le b$, and $f(x) < 0$ on the interval $[a,x]$}

Clearly $A \neq \emptyset$, since $a \supset A$. In fact, there is some  $\epsilon > 0$ such that $A$ contains all points x satisfying $a \le x < a +  \epsilon$ . Similarly, b is an upper bound for A and, in fact, there is an  $\epsilon >0$ such that all x satisfying $b -  \epsilon < x \le b$ are upper bounds for A."

This is my first gripe. Consider a function that monotonically increases from $b - \epsilon $ to $b$. Then the only upper bound for the set is b, not all of those other points from $b -  \epsilon $ to $b$, because in the set, there is a value larger than them and bounding.

The next part goes on to say:

"From these remarks, it follows that A has a least upper bound $\alpha$ and that $a < \alpha < b$. We now wish to show that $f(\alpha) = 0$...."

How can this be, though? How can the least upper bound always be between a and b and not necessarily be b? Plus, why does this least upper bound have to have value = 0? Why cant we then consider negative values to be least upper bounds?

The definition provided for least upper bounds at the beginning of the chapter was:

"A set A of realy numbers is bounded above if there is a number x such that $x \ge a$ for ever a in A. Such a number x is called an upper bound for A."

By this definition, I dont see how his claims are founded.

[quote="jelyman"]

This is my first gripe. Consider a function that monotonically increases from $b - \epsilon $ to $b$. Then the only upper bound for the set is b, not all of those other points from $b -  \epsilon $ to $b$, because in the set, there is a value larger than them and bounding.[/quote]

About your first question:

Pick any point x from $b -  \epsilon $ to $b$.
If x were not an upper bound to A that would mean
that f must be negative for all points between point $a$
and point $x$. In particular, f(x) would be negative
which contradicts the right continuity of f at b...
that is to say it would contradict the fact that f(x) is positive.

It follows then, that b can not be a least upper
bound to A, because any point from $b -  \epsilon $ to $b$
is an upper bound for the set A.

I'm going to move this to Other Problem Solving Topics. In general, calculus questions should be posted there (unless you think they belong in one of the College forums).

BUMP

(someone pls answer it completely. I really need to understand this)

But..let me ask you something...

Did you understand why the least upper bound of $A$
must be strictly less than $b$?

Notice that any value $x$ for which $f(x)>0$
must be an upper bound of $A$.
Another way to look at it, would be:
Let's say $f(x)>0$ for some $x>a$.
This means that $x$ can not belong to $A$,
and since $x>a$ then $x$ must be an upper bound of $A$.
(This is due to the fact that $a$ belongs to $A$.)
Now, by the right continuity of $f$ at $b$ there must
exist some interval $(b-\epsilon,b]$ where
$f$ is stricly positive. Thus, any point belonging
to this interval will be an upper bound to $A$.

This shows that $\alpha<b$. A somewhat similar
reasoning applies to $a<\alpha$. (But in this case
you need to use the left continuity of $f$ at $a$)

Finally, to complete the proof Spivak considers
two cases $f(\alpha)>0$ and $f(\alpha)<0$.
Both cases lead to contradiction so $f(\alpha)=0$

Note that Spivak isn't claiming that $\sup\{x:a\le x\le b\}$ isn't $b$. Instead he's claiming that $\sup\{x:a\le x\le b, f(x)<0\}$ isn't $b$. I think that's what you're misunderstanding.

In fact, Simon, the set Spivak is taking the least upper bound of isn't the set where $f(x)<0$ but a more complicated (and smaller set): the set of all $x$ for which $f(y)<0\amp\forall y\in [a,x]$.

I don't particularly like this proof, partly because the description of the set $A$ is so complicated that it's too easy for students like jelyman to get tied into knots trying to cope with the definition. When I teach the course, I tend to rely on a different and more direct proof of the IVT: the successive bisection algorithm.

Assume that we have been introduced to the completeness of the real numbers, and that we have already proceded from that completeness to the all-important Monotone Sequence Theorem: a monotone bounded sequence converges.

The plan: Let $a_0=a, b_0=b$. Set $c=(a_0+b_0)/2$. Evaluate $f(c)$. If $f(c)=0$ we are done. If $f(c)<0$ let $a_1 = a_0$ and $b_1=c$. If $f(c)>0$ let $a_1=c$ and $b_1=b_0$. Repeat this process recursively, at each step setting $c=(a_n+b_n)/2$ and evaluating $f(c)$. At every stage of the computation, unless the process terminates, $a_n<b_n, f(a_n)<0$ and $f(b_n)>0$

If the process doesn't terminate, we have $(a_n)$ as a bounded nondecreasing sequence and $(b_n)$ as a bounded nonincreasing sequence. By the Monotone Sequence Theorem, both sequences converge. Since $b_n-a_n = (b-a)/2^n \to 0$ as $n\to\infty$, both sequences converge to the same number, which we'll call $x$.
On the one hand, $f(x)=f(\lim a_n)=\lim f(a_n) \le 0$.
On the other hand, $f(x)=f(\lim b_n)=\lim f(b_n) \ge 0$.

It's both a complete proof and a moderately useful numerical algorithm.

The essence of anything that is equivalent to the completeness of the real numbers (the existence of least upper bounds, the Monotone Sequence Theorem, the statement that Cauchy sequences converge, the Bolzano-Weierstrass Theorem) is this: if you have a process that ought to produce a number, then [b]there is[/b] such a number. This allows us to talk about real numbers that we have no conventional names for, there being more real numbers than such names. A good example of such a real number with no conventional name (since we are talking about the IVT): that number $x$ such that $\cos x = x$.

[quote="Kent Merryfield"]I don't particularly like this proof, partly because the description of the set $A$ is so complicated that it's too easy for students like jelyman to get tied into knots trying to cope with the definition. When I teach the course, I tend to rely on a different and more direct proof of the IVT: the successive bisection algorithm.[/quote]

The main virtue of Spivak's proof is that it only  relies on
the completeness of the real numbers. The bisection method
requires many concepts about sequences, convergence, etc..

The bisection method is in fact easier to understand
but I think Spivak's proof is much more elementary
since it doesn't rely on any previous results (which
would require proofs on their own)

This is of course, only a matter of opinions.

Ok I think I understand it now. b can be an upper bound, but the point is that it cant be the least upper bound. Somehow I skipped over that when analysing it.

[quote="jelyman"]b can be an upper bound, but the point is that it cant be the least upper bound. [/quote]

Exactly....You got it now!!!!  :cool:


superior algebra
superior algebra solved

Suppose [tex]x[/tex] is an algebraic number on the unit circle. Does this imply, that [tex]x[/tex] is a root of unity?

I believe there are counterexamples in the "Algebra" section posted by vess. I think there's a general construction there, and a particular counterexample.

The number is a root of unity if all its conjugates also have modulus $1$ though, if I remember correctly.

I think every nontrivial rational point on the unit circle will do:
Let be $\left( \frac{a}{c} \right)^2 + \left( \frac{b}{c} \right)^2 = 1$ with integers $a,b,c$.
Then assume there is some $n$ for that you have $\left( \frac{a}{c} + \frac{b}{c} i \right)^n = 1$ or $ (a+bi)^n = c^n$. By unique prime factorisation in gaussian integers (I think there is a simpler argument here, but I don't see it) you  get $a+bi = i^k \cdot c $ or equivalently $\pm a=c, b=0$ or $a=0, \pm b=c$

[quote="grobber"]
The number is a root of unity if all its conjugates also have modulus $1$ though, if I remember correctly.[/quote]

Sorry, when I said this I was thinking about algebraic integers, not merely algebraic numbers.


integration
trigonometry
function
limit
calculus
calculus computations

suppose we have $ \int_0^1\frac{\sin x}x\,dx.$

what does say to me that the integrand has limit as $ x\to0$ ? can I guarantee that converges?

----------

i know a theorem that it says when having a bounded function on a compact interval and if it has a finite number of discontinuities is integrable.

$ \lim_{x\to 0}\frac{\sin x}{x}$ is one of those limits you should just [i]know[/i]. You need it for all of the calculus of trigonometric functions anyway.
So it's a removable discontinuity, and this is basically just the integral of a continuous function on a bounded interval. It can't be calculated exactly, so use numerical methods if you want something quantitative.

I do know that limit, and yes, you're recalling the theorem I said before, right?


What is the least value of n such that every subset of {1, 2, ..., 100} with n elements has the property that we can choose 4 distinct elements a, b, c, d so that a+b=c+d ?

We will prove that for any subset of at least $ 14$ numbers, the condition holds :
Supposed the subset is {$ a_1,a_2,\cdots,{a}_{14}$} and the condition doesn't holds, then : ${ ({a}_{14}-{a}_{13})+(a}_{13}-{a}_{12})+\cdots+({a}_{2}
-{a}_{1})\leq (1+2+3+\cdots+14)=101$ contradiction
We will show that there exist a subset of less than $ 13$ numbers that are not statisfying :
{$ 1,2,3,4,5,6,7,8,9,10,11,12,13$}.

It seems as if you had misuderstood the problem.

I think the answer is $ 16$ and I'm quite sure it was posted before..

This is a proof for n is no greater than 16, but I'm not sure a set with 15 elements that doesn't fulfil the criteria can be constructed.

Since $ a \minus{} c \equal{} d \minus{} b$ we wish to add as many elements into our set as possible without repeating a difference between two elements.  Note that an element cannot appear on both sides of this equation, therefore each element that we add is allowed to have a difference between itself and one element only the same as a diffference already included. Hence the first element $ k_1$ does not create any difference, $ k_2$ creates 1 difference, and any other $ k_m$ creates a further $ m \minus{} 2$ differences. If we use the smallest possible difference, we have $ 1 \plus{} 2 \plus{} 3\dots \plus{} 13 \equal{} 91$,  Therefore the maximum number of different differences is 13.  Therefore the maximum number of elements without having too many differences is that of $ k_{15}$, that is $ n \equal{} 15$.  I do not think that this optimum can be constructed though.

[quote="surshu"]I think it is 21.[/quote]

Any justification?  Have you found a counterexample set for n=20?


Determine the highest power of $1980$ which divides \[\frac{(1980n)!}{(n!)^{1980}}.\]

[quote="Peter"][quote="baz"]I think I can show that this tends to infinity.  Does this sound correct to anyone else?  See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=153926
for more information.  Thanks[/quote]It certainly tends to infinity. But I think the problem wants an explicit formula in function of $ n$ (and hopefully a more elegant one than the immediate form t0rajir0u gives). At least I think that, I don't have a solution myself, but from the statement I think that is what the problem authors intended.

Welcome to this forum by the way. :)[/quote]

It definitely does not tend to infinity.

The largest number k such that 1980^k is a factor of ((1980n)!)/((n!)^1980) is no larger than the largest number k such that 11^k is a factor of ((1980n)!)/((n!)^1980). Denote this number by k(n)

Now, let f(n)=(11n)! / (11^n) (n!). I claim that f(n) is an integer, but is not divisible by 11. For f(0)=1 and f(n+1)/f(n)=(11n+11)(11n+10)(11n+9)(11n+8)...(11n+1)/(11(n+1))=(11n+10)...(11n+1).

Thus the largest power of 11 that divides into (11n)!/n! is n.

Let T(n)=((1980n)!)/((n!)^1980). Then T(11n)/T(n)=f(1980n)/(f(n)^1980). The right hand side is expressed as a fraction where neither top nor bottom is divisible by 11. Thus T(11n) and T(n) are divisible by exactly the same powers of 11. Thus k(11n)=k(n), so k(n) does not tend to infinity.

Interesting. Baz proved there is a subsequence $ (n_k)$ that converges to infinity, and you proved that not every sequence converges to infinity.

Both do, however, not answer the original question, do they?

[quote="Peter"]Determine the highest power of $ 1980$ which divides
\[ \frac {(1980n)!}{(n!)^{1980}}.
\]
[/quote]
\[ \frac {(1980n)!}{(n!)^{1980}}\equal{}\prod_{i\equal{}1}^{1980}{in \choose n}
\]
Trying to extend this to something more useful but it just keeps going back into what t0r said


Let $a,b,c$ be positive real numbers and $abc=1$. Prove that the inequality
\[\frac{1}{1+a^{2}}+\frac{1}{1+b^{2}}+\frac{1}{1+c^{2}}\ge \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\]
[b]can not be true.[/b]

What do u mean? Just give an counter example? :maybe:
Counterexample: just substitute x,y,z with 0.5, 0.5 and 4

Don't check any detailed examples? Could you prove that it is false?

Sorry, I didn't understand and I don't think you can because for some a,b,c the inequality holds whereas for the others it fails


AMC
USA(J)MO
USAJMO
ARML
geometry
geometric transformation
dilation

I am curious, what are your opinions of the USAJMO. As for me, I am pretty skeptical currently, but am withholding judgement for a bit. Please discuss here (although I don't expect much serious discussion until the ARML people get back).

EDIT: If a mod could fix the poll I would appreciate it.

I agree, the USAJMO will simply dilute the overall quality of the participants associated with USAJMO/USAMO, as top underclassmen will continue to attempt to qualify for USAMO by taking the AMC 12, whereas, the remaining qualifiers for USAJMO will be significantly less talented.  Furthermore, this will adversely affect upperclassmen wishing to secure USAMO qualification.

I believe if you look at the results and statistics, you will notice that the number of upperclassmen (11-12) vs underclassmen (10 and below) is about what the split for the qualification for the new USAMO/USAJMO.  Therefore there will be little to no dilution, nor disadvantage to anyone.  You will also notice that a great majority of the younger students score almost nothing on the USAMO making the possibility of scoring positively on the USAJMO a better use of their time and skills.

(probably one of the first of said 'ARML people'; I can't sleep)

About the MOP selection process I have no comment.

A bit troubled by the 270 qualifiers, though. There were 60 more than that this current year, only counting those that came from AMC12, and my score qualified by a margin of 'only' ~30. Getting rid of the 60 might reduce that to around 20-15, and bringing in AMC10 people narrows it even further.

I also have a feeling that this is a more dire concern for a lot of people who perform much better than I ever could on USAMO.

I don't like the idea of USAJMO at all for multiple reasons. I feel like they're pretty much saying that people in 10th grade and below do not score high on USAMO and this discourages them so we should give them an easier test so they don't get discouraged. Well i personally find failure more as a bigger incentive for me to study then I find from success. When i came into seventh grade i couldn't even divide fractions but i made the first cut for the mathcounts team (8/30 was cutoff and i got 9/30) and when i went to the meetings i was like the dumbest person there. Since i was the dumbest person there i worked hard so i could get better and not be the dumbest person there whereas no one else on the team really worked that hard because they were already the top kids in the school (or maybe they had some other reason i don't know about). Due to my hard work the teacher let me past the next cutoff even though my score wasn't too high (10/30 I think it was). However i continued to work hard and i became the best in my middle school at the end of seventh grade. Now in eighth grade I have gotten a perfect on amc 10b a 4 on AIME (should've been 6) and went to various other math competitions. You can see how my performance in math has developed in the past two years. I think that if i was already the best in seventh grade when i first walked in i would not have the motivation to continue to work as hard as i did and so i wouldn't have improved as much. The moral of the story is that failure should motivate students more than success and if it doesn't motivate them then do you really think that being successful will? When you fail you should be thinking that i should work harder to improve my score but when you succeed are you always thinking that?

I completely agree with peregrinefalcon, but the smaller number of seats for USAMO might encourage some of the younger students to work even harder. (At least that's what it's making me do.)

W8 what is USAJMO. Is it a typo error for USAMO?

You should try to...read the stickies, and the many other threads around the AMC forum related to USAJMO?

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=279691]here[/url].

[quote="peregrinefalcon88"]I read that whole thread before i made a post so i could make an educated opinion.[/quote]
I dislike the USAJMO idea because it makes everything more complicated.


[quote="spring_sotana"]Let a,b,c be positive numbers such that a+b+c=ab+bc+ca. Prove that: a+b+c+abc \ge 4[/quote]
$a+b+c+abc\geq4\Leftrightarrow$
$\Leftrightarrow(a+b+c)^2(ab+ac+bc)^2+abc(a+b+c)^3\geq4(ab+ac+bc)^3\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(a^4b^2+b^4a^2-2a^3b^3+3a^4bc-a^3b^2c-a^3c^2b-a^2b^2c^2)\geq0.$
The last inequality is obviously true.

[quote="jin"]Why the last one is true?[/quote]
I haven't check the last step into the last one, but if it was true, you can see from the highest power.
$a^4b^2$ (or $b^4a^2$)can beat all negative term, while $a^4bc$ can beat all negative but $a^3b^3$

Oh,I see,thank you. I don't care $\sum(2a^3-a^2b-a^2c)=\sum(a^3+b^3-a^2b-b^2a)$

[quote=spring_sotana]Let $a,b,c$ be positive numbers such that $a+b+c=ab+bc+ca$. Prove that: $$a+b+c+abc \ge 4$$[/quote]
[b]Strengthening[/b]
Let $a,b,c$ be positive numbers such that $a+b+c=ab+bc+ca$. Prove that: $$a+b+c+\frac{3abc}{a+b+c} \ge 4$$


are they supposed to send the scores to your coach, or do you not get them?

In my state (MA), the coach gets the scores during the ceremony... but it's different for every state. Your coach should get the scores, however, within one month.

Some states get their scores at the competition, but some states (like mine) don't give you scores, and just mail them to your coaches. Also, in some states, you have an option to protest and they give you the answers.

yeah we got to do that. i wanted to write one about something i cant discuss b/c it would give away diffuculty