[quote="nonie"]Let a and b be positive real numbers such that a>b. Which of the following is true? Explain why.

a)$a^{a}b^{b}>a^{b}b^{a}$
b)$a^{a}b^{b}<a^{b}b^{a}$
c)$a^{a}b^{b}>a^{b}b^{a}$ if and only if a<2b
d)$a^{a}b^{b}<a^{b}b^{a}$ if and only if a<2b
e)none of these[/quote]
[hide] a) $a^{a}b^{b}>a^{b}b^{a}\Longrightarrow\frac{a^{a}}{a^{b}}>\frac{b^{a}}{b^{b}}\Longrightarrow a^{a-b}>b^{a-b}\Longrightarrow a>b$[/hide]

[hide="explination"]a is the only true one.

$a^{a}b^{b}> a^{b}b^{a}\rightarrow \frac{a^{a-b}}{b^{a-b}}> 1$[/hide] :D

[quote="Farenhajt"][quote="lotrgreengrapes7926"]$a^{a-b}>b^{a-b}\Longrightarrow a>b$[/quote]

This implication is wrong: $a^{a-b}>b^{a-b}\Longrightarrow (a>b>1\lor a<b<1)$[/quote]
But $a$ is still the correct answer? Since there are no positive integers $a<b<1$.

[quote="Farenhajt"]$a,b$ are reals, not integers.[/quote]

Perfectly true. I agree 100%. (a) is only correct if a and b are both positive integers.


Is it possible to pair off the positive integers 1,2,3....,50 in such a manner that the sum of each pair of numbers is a different prime number?

[hide="My solution"]

No, it's not possible. Let $p$ be the prime number. All $p$ must be $3 \leq p \leq 99$ because 3 is the smallest number that can be formed by adding two numbers (1, and 2). Similarly, 99 is the largest, formed by 49 and 50. If numbers 1 through 50 are paired, there will be 25 pairs, or distint primes. But in the interval $3 \leq p \leq 99$, there are only 23 prime numbers. So, we can't pair numbers so that each pair is a different prime number.[/hide]

This is too obvious.. Maybe I'm missing something? :D

YOu're right (at least a middle-schooler thinks  :D ), there are only 24 primes from [3,99], and you need 25 pairs, so it's not possible.

What kind of contest is this supposed to be?

IMTS = International Math Talent Search

Some of the problems correspond to current USAMO problems. This problem was from earlier years and I think that's why it was bit easy.

I still don't know if it's THIS easy though.. This seems too simple.  :?

Counting prime numbers might be hard  :D

[quote="Ravi B"]Nice solution, Silverfalcon.  I wouldn't say the problem is [i]that[/i] easy.  Thinking of the idea of counting primes is not obvious to those who haven't seen such problems before.[/quote]

Thanks.  :)


Find all real numbers $ a,b$ such that $ a^2\minus{}b^2,a^3\minus{}b^3,a^5\minus{}b^5$ are rational numbers.

Obviosly if $ a=b=x$ all of them rational for any x. If $ a,b$ rational, then all of them rational.
If $ b=-a=x$ then all of them rational only if $ x$ rational. So consider $ a^2-b^2\not =0$.
Then $ \frac{a^2+ab+b^2}{a+b}=a+b-\frac{ab}{a+b}=r,$ $ \frac{a^4+a^3b+a^2b^2+ab^3+b^4}{a+b}=r_1$.
Let $ a+b=x,a^2-b^2=r_0$ then $ r_1=(a+b)^3-3\frac{ab}{a+b} (a+b)^2-2(a+b)(\frac{ab}{a+b})^2=x^3-3x^2(x-r).$
It give $ x(r^2+rx-x^2)=r_1$ and $ x^4-r_0^2=4x^3(x-r)$.
From these 2 equation for x, we get $ 3r_1x^2-2x(r_0^2+2r_1)+3rr_0^2=0$ and linear equation. It mean $ x=a+b$ rational and $ a-b=\frac{r_0|}{x|}$ rational, or $ a,b$ - rational.


Solve the following equation for x:
$ \frac{\cos2x}{\sin3x}\plus{}\frac{\cos6x}{\sin9x}\plus{}\frac{\cos18x}{\sin27x}\equal{}0$

hello, applying the additions formulas i have found the following equivalent equation
${ 33554432\, \left( \cos \left( x \right)  \right) ^{26}-209715200\,
 \left( \cos \left( x \right)  \right) ^{24}+578813952\, \left( \cos
 \left( x \right)  \right) ^{22}-928514048\, \left( \cos \left( x
 \right)  \right) ^{20}+958791680\, \left( \cos \left( x \right)
 \right) ^{18}-666796032\, \left( \cos \left( x \right)  \right) ^{16}
+317521920\, \left( \cos \left( x \right)  \right) ^{14}-103194624\,
 \left( \cos \left( x \right)  \right) ^{12}+22404096\, \left( \cos
 \left( x \right)  \right) ^{10}-3111680\, \left( \cos \left( x
 \right)  \right) ^{8}+256256\, \left( \cos \left( x \right)  \right)
^{6}-10920\, \left( \cos \left( x \right)  \right) ^{4}+182\, \left(
\cos \left( x \right)  \right) ^{2}-1}=0$
Sonnhard.

wow...
the easier way would be to find x so that 2x, 6x, and 18x are equal to 2kpi +- pi/2
x=pi/4 works

hello, and how would you prove that you can find all real solutions by your method?
Sonnhard.

oh yeah, youre supposed to find all real solutions
kpi +- pi/4

hello, my TI 89 Titanium does also find your solutions, but have you read the warning "more solutions may exist"?
Sonnhard.


ARML
trigonometry
Putnam
symmetry
analytic geometry
geometric series
geometry

May I ask where you found this problem? I don't that many people took the contest I took that contained this problem.

Here's one thing you can do: The problem is awfully hard as stated, so why don't we consider cos(pi/7) - cos(2pi/7) + cos(3pi/7) - cos(4pi/7) + cos(5pi)/7 - cos(6pi/7). This will be twice the value you want. (Do you see why?) Then you can sum the geometric series e^(n*i*pi/7)*(-1)^(n+1) and take real parts.

Also, I have no idea whether or not this will work, but |1-cis(theta)|=2*sin(theta/2). That can be used for so many problems, so it's definitely one to learn.

Thanks for the suggestion.  This problem is from last year's UCSB School of Creative Studies mathematics qualifying exam.

Then you're asking the right guy - look at the winners of the advanced exam.

Yeah, that's where I thought it was from. And the first solution I gave you was the one that Prof. Ryavec told me when I emailed him asking how to do it. It's still really messy though. On the actual contest I tried my trick of approximating everything, but I was being stupid and thought that cos(3pi/7) was < 0 for some reason, and so I got it wrong.

Simon, do you remember the ARML problem with the heptagon from a couple years back?  The trig method led to a summation of sevenths of angles -- not particularly the same, but reasonably similar.  I bet you could reformulate the problem starting from the trig part to make it really unpleasant . . .

Ah yes, of course. Someone else on my team was making fun of me after the contest, saying that I failed to solve the problem because I knew too many theorems. (For example, did you know that if ABCDEFG is a regular heptagon, then 1/AB=1/AC+1+AD? I tried to use this property and got nowhere.) As for that problem, I can see sines coming in after using the half-angle sine identity, so maybe someone extremely clever could figure out how to use it to get cosines. If I remember, I'll ask Prof. Ryavec tomorrow in Putnam seminar how he came up with this problem geometrically (if at all).

I did the ARML question by trigonometry, and then estimating two of the ugly 7ths angles as 60 and 30 degrees, and it ended up working out to the same answer.  That was entertaining.  Do you remember exactly what the question was?

ABCDEFG is a regular heptagon inscribed with sidelength 2002. Find AB*DG-AC^2. I think that was it. Anyway, it was equivalent to that.

actually, i believe this problem is originally from IMO1963, and consequently one solution can be found

http://www.kalva.demon.co.uk/imo/isoln/isoln635.html

Wow. So he casually throws in an IMO problem into a contest with 30 questions to be solved in 50 minutes? That's rather harsh.

Individually or as a team?  Besides, old IMO's aren't [i]that[/i] hard :).

It's an individual contest. I agree that old IMOs are much easier, but they're still too hard for that kind of time limit.

This is sth else that might work:

If z=cisx then cos x=(z^2+1)/2z (as I recall; it might be z^2-1, I'm not sure, but whoever is interested could check it out). Then we denote z=cisPi/7 and use the fact that z^7=-1 etc.etc. I think I've solved this prb before using this, but it was really tough for me in 9'th grade, when I didn't know complex numbers.

heptagons... **shudder**

thank god for ptolemy's theorem... too bad I still added wrong!

JBL's suggestion of using a heptagon leads to a simple graphical solution.

Draw a regular heptagon, with side-length 1.  Orient the heptagon such that vertex A is at the origin.  The next vertex, B, counterclockwise, would be at (cos(pi/7), sin(pi/7)).  The remaining vertices are labelled C through G.

Draw a vertical line through the origin.  By symmetry, this line will be the perpendicular bisector of side DE.

Since side DE is of length 1, we know the x-coordinate of vertex D is 0.5 and the x-coordinate of the symmetric vertex E is -0.5.

Beginning at the origin, the x-coordinate of point A is 0.  The x-coordinate of point B is cos(pi/7).  The x-coordinate of point C is cos(pi/7)+cos(3pi/7).  The x-coordinate of point D is cos(pi/7)+cos(3pi/7)+cos(5pi/7) = 0.5.

Since cos(5pi/7) = -cos(2pi/7), the problem is proven.

[quote="timhsieh"]Without using a calculator or the Taylor series, how do you prove that cos (pi/7) - cos (2pi/7) + cos (3pi/7) = 0.5?[/quote]

[tex]\displaystyle 2\sin{\pi\over7}\left(\cos{\pi\over7} - \cos{2\pi\over7} + \cos{3\pi\over7}\right) [/tex]
[tex]\displaystyle = \sin{2\pi\over7} - \left(\sin{3\pi\over7}-\sin{\pi\over7}\right) + \left(\sin{4\pi\over7} - \sin{2\pi\over7}\right)[/tex]
[tex]\displaystyle =\sin{\pi\over7}[/tex]

or

[tex]\displaystyle 2\sin{2\pi\over7}\left(\cos{\pi\over7} - \cos{2\pi\over7} + \cos{3\pi\over7}\right) [/tex]
[tex]\displaystyle = \left(\sin{3\pi\over7}+\sin{\pi\over7}\right) - \sin{4\pi\over7} + \left(\sin{5\pi\over7} - \sin{\pi\over7}\right)[/tex]
[tex]\displaystyle =\sin{2\pi\over7}[/tex]

Alternatively,

[tex]
\displaystyle{\cos\left(\frac{\pi}{7}\right) - \cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{3\pi}{7}\right)}[/tex] [tex]\displaystyle{= \frac{\sin\left(\frac{\pi}{7}\right)}{\sin\left(\frac{\pi}{7}\right)} \cdot \left(\cos\left(\frac{\pi}{7}\right) + \cos\left(\frac{3\pi}{7}\right) + \cos\left(\frac{5\pi}{7}\right)\right)}
[/tex] [tex]\displaystyle{= \frac{\sin\left(\frac{6\pi}{7}\right)}{2\sin\left(\frac{\pi}{7}\right)} = \frac{1}{2}
}[/tex]

since telescoping occurs when we apply

[tex]
\displaystyle{2\sin(\alpha)\cos(\beta) = \sin(\alpha + \beta) - \sin(\beta - \alpha)
}[/tex]

One of my favorite trig tricks.


abstract algebra
geometry
Online Math Open
number theory

Ena apo ta pio omorfa theorimata tis algebrikis theorias arithmon einai to epomeno :
Kathe abeliani epektasi tou $ \mathbb{Q}$ periexete se mia kiklotomiki epektasi tou $ \mathbb{Q}$ i.e. iparxei riza tis monadas $ t$ (esto $ t^n\equal{}1$) oste i epektasi periexete sto $ \mathbb{Q}(t)/\mathbb{Q}$.
To parapano theorima einai kentrikis simasias, kathos odigise stin anaptiksi tis abelian class field theory. I class field theory einai sto epikentro tis mathimatikis ereunas stin theoria arithmon, kai exei polles diasindeseis me to programma tou Langland- einai sigkekrimena i Langlands correspondance gia to $ GL_1$- to opoio sindeei representation theory me number theory, algrebraic geometry akomi kai mathematical physics. Polloi mathimatikoi titanes tou aiona mas exoun strepsei tin prosoxi tou ekei p.x. to theorima tou Fermat basizete se autes tis idees.

Esto n fisikos arithmos. Na deiksete oti to $ \mathbb{Q}(\sqrt{n})$ einai iposoma tou $ \mathbb{Q}(t)$ me $ t^{4n}\equal{}1$. (Eidiki periptosi tou Kronecker-Weber)

Stergios

Einai eukolo na doume pos arkei na apodeixoume tin askisi gia $ n$ square free. Exoume $ \sqrt{-1}=\zeta_4$ kai $ 2=(\zeta_8+\zeta_8^{-1})^2$ ara $ \sqrt{2}\in\mathbb{Q}(\zeta_8)$.
Tora gia p=prime $ \neq2$ theoro $ \tau=\displaystyle\sum_{x\in\mathbb{Z}/p\mathbb{Z}}\left(\frac{x}{p}\right)\zeta_p^x$ ara $ \tau^2=\displaystyle\sum_{x,y\in\mathbb{Z}/p\mathbb{Z}}\left(\frac{xy}{p}\right)\zeta_p^{x+y}$. Omos to $ y=xt$ gia kapoio ${ t\in\mathbb{Z}/p\mathbb{Z}}$ ara
$ \tau^2=\displaystyle\sum_{x,t\in\mathbb{Z}/p\mathbb{Z}}\left(\frac{t}{p}\right)\zeta_p^{x(1+t)}=\displaystyle\sum_{t\in\mathbb{Z}/p\mathbb{Z}}\left(\frac{t}{p}\right)[\displaystyle\sum_{x,t\in\mathbb{Z}/p\mathbb{Z}}\zeta_p^{x(1+t)}]=\left(\frac{-1}{p}\right)(p-1)+\displaystyle\sum_{t\ne-1}\left(\frac{t}{p}\right)(-1)=\left(\frac{-1}{p}\right)p-\displaystyle\sum_{t\in\mathbb{Z}/p\mathbb{Z}}\left(\frac{t}{p}\right)=\left(\frac{-1}{p}\right)p=\pm p$
ara $ \sqrt{\pm p}\in\mathbb{Q}(\zeta_p)$ to opoio dinei $ \sqrt{p}\in\mathbb{Q}(\zeta_p)$ i $ \sqrt{p}\in\mathbb{Q}(\sqrt{-1},\zeta_p)\subseteq\mathbb{Q}(\zeta_4,\zeta_p)$.
Proti periptosi :  $ n=p_1p_2...p_k$ oste gia kathe $ i$ kai $ j$ me $ i\neq j$ na isxuei $ p_i>2$ kai $ p_i\neq p_j$ tote
$ \mathbb{Q}(\sqrt{n})\subseteq\mathbb{Q}(\sqrt{p_1},...\sqrt{p_k})\subseteq\mathbb{Q}(\zeta_4,\zeta_{p_1},...\zeta_{p_k})\subseteq\mathbb{Q}(\zeta_{4p_1...p_k})\equiv\mathbb{Q}(\zeta_{4n})$
Deuteri periptosi :  $ n=2p_1p_2...p_k$ oste gia kathe $ i$ kai $ j$ me $ i\neq j$ na isxuei $ p_i>2$ kai $ p_i\neq p_j$ tote
$ \mathbb{Q}(\sqrt{n})\subseteq\mathbb{Q}(\sqrt{2},\sqrt{p_1},...\sqrt{p_k})\subseteq\mathbb{Q}(\zeta_8,\zeta_4,\zeta_{p_1},...\zeta_{p_k})\subseteq\mathbb{Q}(\zeta_8,\zeta_{p_1},...\zeta_{p_k})\subseteq\mathbb{Q}(\zeta_{8p_1...p_k})\equiv\mathbb{Q}(\zeta_{4n})$

Poli oraia kataskeuastiki apodeiksi. :)

I idea auton ton athroismaton ofeilete sto Gauss. Nomizo oti exo diavasei autin tin apodeiksi sto Problems in Algebraic Number Theory ton Murty,*

Enas allos tropos tha itan na theorisoume tin epektasi Q(sqrt(d))/Q opou d i diakrinousa tis x^p-1 ( Aksiki : |d|=p^p ).

Stergios

Auta ta athroismata dinoun pragmati polla apotelesmata grigora. :)
I parapano askisi mou fanike eukoli giati ixa dei apodeiksi gia tin parakato
An p=prime$ \equiv1\bmod{4}$ tote ndo $ \sqrt{p}\in\mathbb{Q}(\zeta_p)$
kai apla eprepe na thimitho pos itan to athroisma.


find all fuctions $ f: \mathbb{N}\rightarrow \mathbb{N}$
such that :

$ f(a\plus{}ab\plus{}b)\equal{}f(a)\plus{}f(b)$

;)

Define $ f$ arbitrarily on any number that is a prime minus one; this gives a solution. If we let $ g(a) \equal{} f(a\minus{}1)$ for $ a \geq 2$, then the equation becomes $ g((a\plus{}1)(b\plus{}1)) \equal{} g(a\plus{}1) \plus{} g(b\plus{}1)$. Let $ a\plus{}1 \equal{} x, b\plus{}1 \equal{} y$, then $ g(xy) \equal{} g(x) \plus{} g(y)$ for $ x,y \geq 2$. It follows fairly easily that $ g(p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}) \equal{} e_1 g(p_1) \plus{} e_2 g(p_2) \plus{} \ldots \plus{} e_k g(p_k)$. So we define $ g$ on primes arbitrarily.

Let $ g(n)\equal{}f(n\minus{}1)$, then $ g(nm)\equal{}g(n)\plus{}g(m)$. For all $ g(p)\equal{}k_p$ we get suth function:
$ n\equal{}\prod_i p_i^{m_i}\to g(n)\equal{}\sum_i m_ik_{p_i}$.


Consider $ ABCD$ a parallelogram and $ X$ a mobile point on the side $ (BC)$. Denote by $ M$, $ N$ the intersection points of $ XA$, respectively $ XD$, with $ BD$ and $ AC$. Find the locus of the midpoint $ L$ of the segment $ (MN)$.

Are you sure that this has a nice result?

If you set $ t=BX: XC$, it is trivial to use vectors to determine $ L$. However it does not seem to have any simple geometric meaning.

For the sake of symmetry, let $ Y$ be the intersection of the diagonals. We can show that:

$ L=\frac{t^{2}+4t+1}{(t+2)(2t+1)}Y+\frac{1}{(2)(2t+1)}B+\frac{t}{(2)(t+2)}C$


AMC
USA(J)MO
USAMO
AIME
trigonometry
MATHCOUNTS
AMC 10

[hide]You can't find statistical significance with n=1.[/hide]

[quote="calc rulz"][quote="pieterminate"]THE CLASSIC:
[hide]There's always next year.[/hide][/quote]

Sorry to be negative but...no chance for freshman MOP next year  :([/quote]
Those were the kind of negative words that I used to make NERVOUS BARGE cry in Mathcounts four years ago, but they found a grading error in Target round and she ended up going to nationals anyway.

I love finding anagrams of people's names.

[quote="calc rulz"][quote="pieterminate"]THE CLASSIC:
[hide]There's always next year.[/hide][/quote]

Sorry to be negative but...no chance for freshman MOP next year  :([/quote]

But there's always regular MOP.

[quote="calc rulz"][quote="pieterminate"]THE CLASSIC:
[hide]There's always next year.[/hide][/quote]

Sorry to be negative but...no chance for freshman MOP next year  :([/quote]

Ah, same for me too.  I got 126.5 on AMC 12 and 8 on Aime.  If only they forgot the 10ers.... *sigh*

Oh wait, how 'bout this: [b]There's sometimes next year.[/b]

146.5 in the AMC10; 7 on the AIME.
A 132/8 would have made it.  :(

Keep on working. Next year is always an option.

[quote="drunner2007"]I barely missed the USAMO with an AIME score of 7. I know that I am kind of dissapointed, because I should have had a 9 or 10 (read #1 wrong!, found identity for trig question but rewrote equation wrong), but I guess there is one more year for me, and I should start working towards USAMO again.
I know that lots of other people feel the same, so if you want to post anything positive or encouraging, here's your chance.[/quote]

Although it can be disappointing to come so close, to those of us who teach all of you students, it is a phenomenal accomplishment to do what you have done.  You may be in the top 400 or 500 in the country instead of the top 268, but either way, what you've done is remarkable.

Look at is this way.  About 230,000 students took the AMC-10 and 12 exams.  You are probably in roughly the top 500.  That puts you in the top two tenths of one percent in the country, and probably similarly in the world.  It's truly an achievement to be proud of.  We your teachers certainly are proud of you.

Best,
CA

[quote="Xantos C. Guin"][quote="nat mc"]143 AMC 12A / 7 AIME  :([/quote]

If only you got a perfect on the AMC12. Then I would have made it.[/quote]
and what suck is that i got a 10 unofficially on the aime 1

[quote="calc rulz"][quote="pieterminate"]THE CLASSIC:
[hide]There's always next year.[/hide][/quote]

Sorry to be negative but...no chance for freshman MOP next year  :([/quote]

Same scenario here, except a year forward...and I made it this time around...just study as much as possible between now and next year's USAMO.

Start [i]now[/i]!

[quote="CatherineAsaro"]Although it can be disappointing to come so close, to those of us who teach all of you students, it is a phenomenal accomplishment to do what you have done.[/quote]

Yep, that's one perspective to have on this issue. I'm answering in reply to your reply mostly as a shout-out to you, a math coach whom I heard a lot of good things about when I was at MathPath last summer.

What I say to the kids I coach, whom I coach much more lousily than you coach, is that the thing to look for is year-on-year growth. There are some random ups and downs in everyone's scores on a particular day, but over time participation in AMC math tests can help a teen learner develop a much deeper understanding of precalculus math then even most math majors ever develop. And that's great, even if if never leads to any particular awards.

So to all of you AoPSers who came close this year, keep up your spirits, and keep trying. To travel hopefully is better than to arrive, as Robert Louis Stevenson wrote, and there is plenty of interesting math still to learn.

Ack; Personal experience time:

So. In 8th grade (2004) I miraculously got a 8 on the AIME after finding out about the AMC 12B the Sunday before the test. I got into USAMO and actually did considerably well (7 pts)

Then in 9th grade, I stopped working on math for quite a while, probably got somewhat cocky. I get screwed on the AIME, by...
Yes you guessed it. I got a 8 last year

So this year, I started working hard again and got a 150 on 10A; 137 on 12B; 10 on AIME I;

The moral of this story is: [b]Don't stay on the boundary of the qualifiers; its too risky; always keep working and trying to boost that future monster index of yours.[/b]

[quote="chess64"]Just study really hard for next year so that you can PWN AIME/USAMO next year. :D That's what I'm doing.

And start studying right now. ;)[/quote]

precisely what i'm doing as well. although i started studying two days ago.  :P
I think it's possible to boost my dismal AIME score (1) to a qualifying score next year, no? (Please tell me if I'm being overly optimistic :? )
Any suggestions on prep materials?

[quote="davidyko"][quote="chess64"]Just study really hard for next year so that you can PWN AIME/USAMO next year. :D That's what I'm doing.

And start studying right now. ;)[/quote]

precisely what i'm doing as well. although i started studying two days ago.  :P
I think it's possible to boost my dismal AIME score (1) to a qualifying score next year, no?
Any suggestions on prep materials?[/quote]

Read the aops books, and do a lot of practice tests.

I'm too slow and I never have time to check my work on speed contests.

[quote="K81o7"][quote="chess64"]Just study really hard for next year so that you can PWN AIME/USAMO next year. :D That's what I'm doing.

And start studying right now. ;)[/quote]

I missed it by one point last year, with 218/8, so that's what I wanted to do this year.  And then I get 208/7...
It's just those careless mistakes.  Were it not for those, we'd have all done much better.  I wouldn't have done much better last year on AIME, maybe 1 point, but if I had made none this year, it would be 12 for me  :D

Well here are my encouraging words...
1. There's generally a next year.
2. It doesn't really matter.  The actual gains in the competition are not much unless you make MOP (I suppose usamo for college).
Just have fun with it.[/quote]

What? It doesn't matter?  No, that would make history(at my school anyways)!  But yeah, have fun, that's the point of math right? :lol:

Ah, yes, I remember being a part of this group back in freshman year. All I can say is don't give up! If anything can stop you from qualifying for the USAMO again, it's that. Your math ability has gotten you this far; it's not just going to die out on you. So you shouldn't die out on it. Good luck next year to everyone who barely missed qualifying. You'll qualify next year, as long as you believe in yourself (and work hard, of course).

[quote="E^(pi*i)=-1"]I'm in the same position as a lot of people - 9th grade, AIME 7.  Here is what I always tell myself:

In 7th grade, I almost took the AMC (found out about it too late).

In 8th grade, I almost made the AIME (99.5 on AMC 12)

In 9th grade, I almost made the USAMO

So. . .

In 10th grade, I will almost make MOP

In 11th grade, I will almost make IMO

In 12th grade, I will almost win a gold medal at IMO

And a silver medal at IMO is not bad at all.  :lol:[/quote]

Or work hard and skip a few grades! :lol:

[quote="E^(pi*i)=-1"]
In 8th grade, I almost made the AIME (99.5 on AMC 12)

In 9th grade, I almost made the USAMO
[/quote]

If you go up as much between 9th and 10th grade and between 8th and 9th, you will have a USAMO index > 300! :-)

Best,
CA


geometry
circumcircle
geometric transformation
reflection
geometry proposed

Let $ ABC$ be a triangle with circumcircle $ (O)$($ O$ is center)
Suppose that $ M$ is a point in this triangle.$ AM;BM;CM$ cut $ (O)$ at $ A_1;B_1;C_1$,respectively.
$ A_2$ is the point of the match $ A_1$ through $ BC$
$ B_2$ is the point of the match $ B_1$ through $ AC$
$ C_2$ is the point of the match $ C_1$ through $ BA$
Prove that triangle $ A_1B_1C_1$ and triangle $ A_2B_2C_2$ are similar.

[hide="Solution"]
[b]Lemma: [/b]Letting $ H$ be the orthocenter of $ \triangle ABC$, we have that $ HB_2A_2C_2$ is cyclic.

[i]Proof: [/i]Notice that the reflections of $ H$ into any one of the sides of $ \triangle ABC$ lies on the circumcircle of $ \triangle ABC$, meaning that $ AC_2HB$ is cyclic and similarly for the rest.  Thus, inversion around $ H$ which maps point $ X$ to $ X'$ gives that $ A'$, $ B'$, and $ C_2'$ are collinear and similarly for the others.  From the inversions, we have that $ \frac {A'C_2'}{AC_2} \equal{} \frac {HC_2'}{HA}$ and $ \frac {B'C_2'}{BC_2} \equal{} \frac {HC_2'}{HB}$.  Dividing yields that $ \frac {A'C_2'}{B'C_2'}\cdot \frac {BC_2}{AC_2} \equal{} \frac {HB}{HA}$.  Doing the same for the rest and multiplying gives that\[ \frac {A'C_2'}{B'C_2'}\cdot \frac {B'A_2'}{A_2'C'}\cdot \frac {C'B_2'}{A'B_2'} \equal{} \frac {AC_2}{BC_2}\cdot \frac {A_2B}{A_2C}\cdot \frac {CB_2}{AB_2} \equal{} \frac {AC_1}{BC_1}\cdot \frac {A_1B}{A_1C}\cdot \frac {CB_1}{AB_1} \equal{} 1\]from the concurrency of $ AA_1$, $ BB_1$, and $ CC_1$, so by Menalaus, $ C_2'$, $ B_2'$, and $ A_2'$ are collinear, yielding that $ HC_2A_2B_2$ is cyclic.

Now, for the problem.  With this in mind, we have that\[ \angle C_2A_2B_2 \equal{} 180 \minus{} \angle C_2HB_2 \equal{} 180 \minus{} (\angle AHB_2 \minus{} \angle AHC_2) \equal{}\]\[ 180 \minus{} (180 \minus{} \angle B_2CA \minus{} \angle ABC_2) \equal{} \angle ACB_1 \plus{} \angle C_1BA \equal{} \angle C_1A_1B_1\]and similarly for the other angles, yielding that $ \triangle A_1B_1C_1\sim \triangle A_2B_2C_2$. [/hide]


superior algebra
superior algebra solved

what does it mean a factor set? but a factor group?
10x

If & is an equivalence relation on the set M then the set M/& is the set of all the equivalence classes of M with respect to &, and M/& is said to be the factor set of M with respect to &.


analytic geometry
geometry
geometric transformation

The graph of the equation $ y \equal{} |x| \minus{} 3$ is translated two units to the left and three units down. What are the coordinates of the
minimum point of the new graph?

Originally the minimum point is (0,-3). Translation leads to (-2,-6).


Find all +ve integers n such that n divides 3^n-2^n.

Here's Solution.

Let $p$ be the smallest prime that divides $n$.

then $3^n=2^n (mod p)$

If $p \not{=} 2$, $gcd(2,p)=1$, So there exist $2^*$ s.t $2\times 2^* =1 (mod p)$

so we get $(2^* 3)^n =1 (mod p)$ and here, $2^* 3 \not{=} 1 (mod p)$

Let $ord_{p} (2^* 3) = t$. Then $t|n$

cuz $t<p$. Contradiction for assume that p is smallest prime.

So, $n=1$ or $n=2^k$ form.

But cuz $3^n - 2^n $ is odd. $n$ can't be even.

Therefore, Possible value of $n$ is only $1$.

$Q.E.D.$

Another solution (it starts in the exact same way):

$p$ is the smallest prime which divides $n$. Clearly, $p\ne 2,3$. Then $p|(3^n-2^n,3^{p-1}-2^{p-1})=3^{(n,p-1)}-2^{(n,p-1)}=3-2=1$.


Hello,

[quote="pohoatza"]Consider the numbers from $1$ to $16$. The "solitar" game consists in the arbitrary grouping of the numbers in pairs and replacing each pair with the great prime divisor of the sum of the two numbers (i.e from $(1,2); (3,4); (5,6);...;(15,16)$ the numbers which result are $3,7,11,5,19,23,3,31$). The next step follows from the same procedure and the games continues untill we obtain only one number. Which is the maximum numbers with which the game ends.[/quote]

Nice problem. The answer is $19$ (and not $17$, as it seems at first glance).
Demo :

1) $19$ is possible :
$(1,8),(2,15),(3,14),(4,13),(5,12),(6,11),(7,10),(9,16)$ $\Rightarrow $ $3,17,17,17,17,17,17,5$
$(3,17),(5,17),(17,17),(17,17)$ $\Rightarrow $ $5,11,17,17$
$(5,11),(17,17)$ $\Rightarrow $ $2,17$
$(2,17)$ $\Rightarrow $ $19$

2) nothing above 19 is possible
2.1) Every number in the process is $\leq 31$ :
It is easy to see that maximum p at a given step is $\leq$ maximum q at the previous step + 2 (equality iff q and $q+2$ are primes).
And since maximum after first step is $31=16+15$ and $(31,2)\Rightarrow 11<31$, Point 2.1 is achieved.
As a consequence, maximum at end of game may only be $31,29,23$ or $19$

2.2) $31$ may only be obtained with $(31,31)$ or $(29,2)$ obvious
2.3) $29$ may only be obtained with $(29,29)$ obvious
2.4) $23$ may only be obtained with $(23,23)$ or $(29,17)$ obvious

2.5) $31$ can not be obtained
From 2.2 and 2.3 above, it is clear that $31$ at end of step 4 (the last one) demands either at least 2 numbers 31 at end of step 1, either at least 4 numbers 29 at end of step 1. And this is cleary impossible.

2.6) $29$ can not be obtain
From 2.3 above, $29$ at end of step 4 demands 8 numbers 29 at end of step 1, and this is impossible.

2.7) $23$ can not be obtained
From 2.3 and 2.4 above, $23$ at end of step 4 demands either at least 3 numbers 29, either 2*29 and 5*23, either 8*23 at end of step 1. And this is impossible.

Hence the result : $19$

--
Patrick


Prove that $sin\sqrt{x}< \sqrt{sin x}$ for $0 < x <\frac{\pi}{2}$.


for example

$\left(\frac{2x}{3}|^{3}_{2}\right)$

how do u make the | line as big as the fraction?

Do you mean $\left(\left.\frac{2x}{3}\right|^{3}_{2}\right)$ which is [code]$\left(\left.\frac{2x}{3}\right|^{3}_{2}\right)$ [/code]When you want to size only one thing use . for the other, so in this case you get \left[color=red][b][size=150].[/size][/b][/color]   \right[color=red][size=150]|[/size][/color]


this is a very easy problem from the NW math championship. i was so pissed because i made a mistake and missed it.... :(
500 pigs and chickens are standing in a field, of there are a total of 1684 legs, how many pigs are there.

Haha this problem is very easy what did you do say that chickens have 3 legs or something?

no, i wrote the number of chickens, instead of pigs.... :(


encarta@botmetro.net

This address seems to be an automatic replying machine. Try adding to your messenger contact list!

I have added it and it is yet to come online in two days, at least it never came when I was online. :(

EDIT: Sorry it does work. I had made a typo which I just corrected. It is great! It gets you information from one of four encartas!


[hide]$1512=2^3 \cdot 3^3 \cdot 7$.

Perfect squares have even exponents.

$2^2 \cdot 3^2 = 36$.[/hide]

[hide]$1512=2^3\cdot3^3\cdot7$.
We see a $2^2\cdot3^2 =36$ (remember that perfect squares have even exponents in prime factorizations). [/hide]

[quote="hwenterprise"]What is the largest perfect square factor of 1512?[/quote]
1512=2^3*3^3*7. 2^2*3^2=36.


I thought people might want to know where the unofficial, and, I suppose I am required to say, completely unaffiliated with USACO itself, USACO rankings are.

http://www.tjhsst.edu/~pwang/usaco.php

Don't know if this should be stickied or not, so I'll just let a mod decide. Unfortunately only USA Gold rankings are calculated (although on a certain level this makes sense because it would give a rough indication of who has a shot at camp).

[quote="JSteinhardt"][quote="domino"]It would be nice if we had similar rankings, but with all participants, including international.[/quote]

I will refer your request to the appropriate authorities.[/quote]

Your request has been granted. Please note the "See international competitors" link.

[quote="quantum leap"]How many ppl even compete in the series of tests for the IOI in America?  I know in terms of ap exams, more than 10 times as many students took the Calculus BC exam than the AP Computer Science one.  Plus, I'd expect (and do think) more people take tests leading to the IMO than for any other ISO, perhaps because everyone has at least some math education.  So, how competitive are these programming competitions and about how many even participate in them? :huh:[/quote]

Here's the official site for the USACO: http://oldweb.uwp.edu/academic/mathematics/usaco/

It's definitely a lot less competitive than math because the programming skill set isn't popular in this country. I'm not sure exactly how many people participate, but it's generally pretty competitive because even if there are only 50 people who can program they're still fighting for the top 4 spots...

[quote="quantum leap"]so how much practice would it take to get good at competitive programming?  On the USACO site, it says the novice practice exercises are for first year programmers...So what about the bronze, silver or gold rankings?[/quote]

Well bronze is pretty easy, still relatively basic programming. Silver you need to know how to code some algorithms but they are mostly direct applications (a little ingenuity maybe). Gold is basically adapting algorithms and techniques to fit specific problem conditions. I'm a very low gold, so maybe someone else can provide more input; you can always go to the USACO forum too: http://ace.delos.com/bb/

Green is basicly for people who are learning programming...
Bronze is for people who are learning "competition" programming...
Silver is for people who already know the most common algorithms...
Gold... well... To get good results there, you have to get a lot of practice first, I think. :)

[quote="quantum leap"]So if i start learning to program now, how far can i get by next january? (this isn't a riddle :lol: )...[/quote]

Well by learning to program do you mean you don't even know a language, or start studying competitive CS?

Just as an example, I took the Intro to CS course at CTY when I was a rising 8th grader, learning enough C to write functioning programs (at that point I could write a fairly poor chess program [just barely]). Then, during my 9th grade summer I learned a little bit of JAVA. I then took Accelerated Computer Science as a Freshman, when I learned JAVA in earnest. JAVA became my programming language for USACO. I took the USACO Qualification exam and just barely missed qualifying for silver. Then I did a few contests and managed to do well enough on 1 to qualify for silver.

At this point I read my school's USACO lectures (http://activities.tjhsst.edu/sct/). Then I took a silver contest and missed gold qualification by about 200 points (which is better than you think, btw). I took the open and got like 5th in the U.S. for silver.

I was very frustrated because I was fairly confident that if I had made gold on the one silver contest I had taken, I would've most likely gone to USAICO by virtue of being the only freshman in the Gold division from the U.S. With that in mind, I resolved that I would make Gold next year. I pretty much spent a good portion of the summer doing the first 4 chapters of the training pages and some of chapter 5. I also participated a lot in TopCoder and quickly progressed there, as well.

So then this year's USACO qualification round came, and it happened to be a [i]lot[/i] harder than last year's. So I freaked out a bit but still managed to solve one of the two problems while getting enough partial credit on the other problem to slip by into gold (probably by less than 50 points).

I got 13th in the US on the first gold contest. Then on the next 2, I got something like 28th and 24th (out of 30, so not too great). I again resolved to do better, and practiced actually coding complex problems accurately in a short amount of time. I then jumped to 4th and 9th on the next two contests, so I would say that was considerable improvement.

The other Sophomore who is in Gold, Kevin Li, I believe did a similar thing, although I don't think he did USACO last year.

So I went from knowing how to code, and being okay in Bronze, to being near the top of silver, to being in the upper half of gold in one year. Most of it is due to lots of practice, mostly in the form of just doing all of the Training Pages problems without help (and also try to do them without guessing; it's best to try to get them on your first submission, because on actual USACO contests you don't know whether it's right or wrong until it gets graded officially).

So by next January I think it is very possible that you could be in Gold and be doing pretty well if you actually set your mind to it (just be willing to spend a lot of the summer doing compsci).

Oh yeah, in response to the earlier comment that there's not that many people who do compsci, I'd like to point out that most of the people who [i]do[/i] do USACO are people who are already near the top in other competitions (namely the AMC competitions), so I'd say it's about the same level of difficulty. (For example, I believe most of the people in Gold Division for USACO also made USAMO, with Tom Morgan, Tim Loffredo, and Justin Solomon (ironically all TJ'ers) being the only exceptions I know of, unless someone wants to correct me).

EDIT: To add on to Paladin's comment, you basically need to be really comfortable with all the major Dynamic Programming and Network Flow Techniques, and be able to integrate lots of algorithms together. Usually there will be one problem that is just a direct adaptation of a standard technique, another that requires some degree of cleverness, and the last one will require you to integrate multiple algorithms together in some clever way.


How come so many topics are getting locked???? Is it a server error? I'm sorry if I'm spamming or posting in the wrong section, but I don't get it!  :o

Threads are usually locked because they get excessively saturated of spam or are straying significantly off topic. Some such topics are "recycled," but...I don't know what that means.  :blush:


integration calculus function trigonometry inequalities algebra polynomial

$ \int cos^6(x)\,dx$

How do you solve this?

Wrong forum... this should be in calculus computations.

Anyway, Mathematica says $ \frac{5 x}{16}\plus{}\frac{15}{64} \text{Sin}[2 x]\plus{}\frac{3}{64} \text{Sin}[4 x]\plus{}\frac{1}{192} \text{Sin}[6 x]$, so I don't know if there's a good way to do it.

Moved.

Any integral of a function like $ \cos^n x$ can be solved by writing $ \cos^n x$ as a linear combination of $ \cos(nx), \cos((n\minus{}1)x), \ldots \cos x, 1$ and then integrating these. The guarantee that this linear combination exists can be shown by [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=246555]Chebyshev polynomials[/url].

Try the reduction formula. For $ n\ge 2,$

$ \int \cos^nx\,dx\equal{}\frac{1}{n}\cos^{n\minus{}1}x\sin x\plus{}\frac{n\minus{}1}{n}\int \cos^{n\minus{}2}x\,dx$

This formula can be proved in a three step process:

1. Integrate by parts. Do this by peeling off one factor, so $ dv\equal{}\cos x\,dx$ and $ u\equal{}\cos^{n\minus{}1}x.$

2. Use a Pythagorean identity. That is, replace $ \sin^2x$ by $ 1\minus{}\cos^2x.$

3. Upon finding the desired unknown integral on both sides of this equations, solve for that unknown integral as you would solve a linear equation.

You can find similar reduction formulas for powers of all the trig functions using essentially the same plan.

$ \cos ^ 6 x \equal{} (\cos ^ 3 x)^2 \equal{} \left(\frac {3\cos x \plus{} \cos 3 x}{4}\right)^2$

$ \equal{} \frac {1}{16}\left(9\cos ^ 2 x \plus{} 6\cos x\cos 3 x \plus{} \cos ^ 2 3x\right)$

$ \equal{} \frac {1}{16}\left(9\cdot \frac {1}{2}(1 \plus{} \cos 2x) \plus{} 6\cdot\frac {1}{2} (\cos 2x \plus{} \cos x ) \plus{} \frac {1 \plus{} \cos 6x}{2}\right)$

Since all other ways have been discussed, here's one more. This is easier when the power of $ \cos$ or $ \sin$ is odd.

$ \int \cos^6 x dx \equal{} \int \sqrt {1 \minus{} \sin^2 x}\left(1 \minus{} \sin^2x\right)^2 \{\sin x\}'dx \\
 \\
\text{Now, set }\sin x \equal{} t$

We can also use Euler's formula: $ e^{ix}\equal{}\cos x\plus{}i\sin x\Longrightarrow \cos x\equal{}\frac{e^{ix}\plus{}e^{\minus{}ix}}{2}$.

hello, this identity is very useful
$ \cos(x)^6\equal{}\frac{1}{32}(10\plus{}15\cos(2x)\plus{}6\cos(4x)\plus{}\cos(6x))$
Sonnhard.

apply the cosine reduction formula a number of times.

or a better thing would be to do:

$ \int \cos^6 x dx$

= $ \int (\frac{1\plus{}\cos 2x}{2})^3 dx$

= $ \frac {1}{8} \int 1 \plus{} 3\cos^2 (2x) \plus{} 3\cos (2x) \plus{} cos^3 (2x) dx$

=  $ \frac {1}{8} \int 1 \plus{} 3\cos (2x) \plus{} 3\frac {1\plus{}\cos 4x}{2} \plus{} \cos^2 (2x)\cos (2x) dx$

= $ \frac {1}{8} \int 1 \plus{} 3\cos (2x) \plus{} 3\frac {1\plus{}\cos 4x}{2} \plus{} \frac {1\plus{}\cos 4x}{2} \cos (2x) dx$

= $ \frac {1}{8} \int 1 \plus{} 3\cos (2x) \plus{} 3\frac {1\plus{}\cos 4x}{2} \plus{} \frac {\cos (2x)}{2} \plus{} \frac {\cos (2x)\cos (4x)}{2}dx$

= $ \frac {1}{8} \int 1 \plus{} \frac {7\cos (2x)}{2} \plus{} 3\frac {1\plus{}\cos 4x}{2} \plus{} \frac {\cos (6x) \plus{} \cos (2x)}{4}dx$

= $ \frac {1}{8} \int 1 \plus{} \frac {15\cos (2x)}{4} \plus{} 3\frac {1\plus{}\cos 4x}{2} \plus{} \frac {\cos (6x)}{4} dx$

= $ \frac {1}{8} \int \frac {5}{2} \plus{} \frac {15\cos (2x)}{4} \plus{} 3\frac {\cos 4x}{2} \plus{} \frac {\cos (6x)}{4} dx$

now its easier.  :)


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be a finite group, which has a cyclic subgroup $A$ such that $|A|=p^kq^l$ ($k,l$ are positive integers and $p,q$ are primes).

Prove that if $|A|>\sqrt{|G|}$ then $A$ includes a subgroup which is normal in $G$.

ps. in the same words: we have to prove that there is a subgroup of $A$ which is normal in $G$.

quite interesting one...

What I should do? showing that $A$ is normal?

I corrected problem's text replacing "consists" (which does not exist in English  :blush: ) with "includes" and now it's correct...

I'm so unorganized.  :D
I didn't see 'includes'. sorry.
maybe, I consider except the trivial normal subgroup $\{e\}$?

If you ignore the trivial subgroup, this is false. Let $G$ be the simple group $A_5$ of order 60, and let $A$ be a subgroup isomorphic to $A_4$, with order 12.

You actually need $\frac{|G|}{|A|}<kp+lq$ to make this work.

shux, correction - $A$ is cyclic ! That indeed changes a lot, sorry for that.

Again: $A$ is cyclic.

[Edit] of course I don't mention trivial subgroups ;)


Hi everyone, I have got inspirations from reading Shakespeare’s poetry and Sir Arthur Conan Doyle works and have some poems that I have written. There are many platforms on which I could place my poems but I want some suggestions from you people to help me know the places to present my poems.

Er... this is the MOEMS forum, a math competition, not the poems forum...

And, sometimes, a school might have a Literary Magazine, so try that.

[quote="Dojo"]lol MOEMS, poems they are all the same...
I'm not sure if there is a forum for this though...[/quote]

under the arts and sciences topic, there is a music and arts forum
try posting this there

You could also present them in FF, but I would rather you did it in Arts and Sciences instead.


[quote="nguoivn"]Given $ a, b, c$ are the real numbers satisfy $ a \plus{} b \plus{} c \equal{} 0$.
Prove that: $ (ab \plus{} bc \plus{} ca)^6 \plus{} 4abc(a^3 \plus{} b^3 \plus{} c^3)(a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3) \ge 0$
 :)[/quote]
With $ a \plus{} b \plus{} c \equal{} 0$,we have the following identities ( easy to prove ! ):

$ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3abc$ and $ a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3 \equal{} (ab \plus{} bc \plus{} ca)^3 \plus{} 3a^2b^2c^2$;

Hence,the inequality is equivalent to:

$ (a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3 \minus{} 3a^2b^2c^2)^2 \plus{} 12a^2b^2c^2(a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3) \geq 0$

Which,clearly,can be reduced to:

$ (a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3 \plus{} 3a^2b^2c^2)^2 \geq 0$

Done !


Source: 2005 AMC-10B #15

An envelope contains eight bills: $2$ ones, $2$ fives, $2$ tens, and $2$ twenties. Two bills are drawn at random without replacement. What is the probability that their sum is $\$20$ or more?

$\text{(A)}\ \frac14 \qquad\text{(B)}\ \frac25 \qquad\text{(C)}\ \frac37 \qquad\text{(D)}\ \frac12 \qquad\text{(E)}\ \frac23$

[quote="catcurio"]An envelope contains eight bills: $2$ ones, $2$ fives, $2$ tens, and $2$ twenties. Two bills are drawn at random without replacement. What is the probability that their sum is $\$20$ or more?

$\text{(A)}\ \frac14 \qquad\text{(B)}\ \frac25 \qquad\text{(C)}\ \frac37 \qquad\text{(D)}\ \frac12 \qquad\text{(E)}\ \frac23$[/quote][hide="Answer?"]To draw at least 20, look at drawing a single bill first and then the number of combinations that allow the sum to be greater than 20. So drawing a 1, the only other bill drawn can be 20, so there are 2 combinations (2/7) if a 1 is drawn (1/4). Is a 5 is drawn, a 20 must also be drawn, so the same as for the ones. If a 10 is drawn, either a 10 or 20 can be drawn, so 3/7. If a 20 is drawn, then the probability that the next bill makes the sum >20 is 1. So summing the probabilities times their respective chances gives $\frac{1}{4}(\frac{2}{7}+\frac{2}{7}+\frac{3}{7}+1) = \frac{1}{4}(2) = \frac{1}{2}$ so $\boxed{D}$.[/hide]


inequalities
probability
inequalities unsolved

Let $m$ and $n$ be positive integers and let $x\in \left[0,1\right]$. Prove that $\left(1-x^{n}\right)^{m}+\left(1-\left(1-x\right)^{m}\right)^{n}\geq 1$.

Well, it is not a piece of cake, but it is a very well-known inequality as a special case of the following problem :
http://www.mathlinks.ro/viewtopic.php?t=1566&highlight=probability

[Moderator edit: See also http://www.mathlinks.ro/Forum/viewtopic.php?t=85072 .]

Pierre.

how many strange solution I'm encounter these days  :D very nice but according to existing so many master in this forum I call it easy.

It's true I'm not a master, but I really don't find this inequality easy.


This was meant to be a linear algebra question:

No matter what basis $ B$ is, verify if the number 101010101 is prime.

(I have no idea how to do this.)

Linear algebra? I fail to see the connection.
In basis $ b$, this number is $ 1 \plus{} b^2 \plus{} b^4 \plus{} b^6 \plus{} b^8 \equal{} \frac {b^{10} \minus{} 1}{b^2 \minus{} 1} \equal{} \frac {b^5 \minus{} 1}{b \minus{} 1}\frac {b^5 \plus{} 1}{b \plus{} 1} \equal{}$ $ (1 \plus{} b \plus{} b^2 \plus{} b^3 \plus{} b^4)(b^4 \minus{} b^3 \plus{} b^2 \minus{} b \plus{} 1)$
For $ b > 1$, $ \frac {b^5 \minus{} 1}{b \minus{} 1},\frac {b^5 \plus{} 1}{b \plus{} 1} > 1$, so it follows that the number is always composite.

Your number is $ \frac {10^{10} \minus{} 1}{10^{2} \minus{} 1}$, or $ \frac {(10^{5} \plus{} 1)(10^{5} \minus{} 1)}{10^{2} \minus{} 1}$. Note that greatest common divisor of numbers $ (10^{2} \minus{} 1;10^{5} \minus{} 1)$ is $ 9$ and $ g.c.d.(10^{2} \minus{} 1;10^{5} \plus{} 1) \equal{} 11$. It means that $ C \equal{} \frac {10^{5} \minus{} 1}{9}$ and $ D \equal{} \frac {10^{5} \plus{} 1}{11}$ are positive integers, greater than $ 1$. So, $ B \equal{} CD$ and hence, $ B$ is composite. Lasha Lakirbaia. ;)


In a soccer game, the Mathletes defeated the Sci-fi's 6 to 3. The score at half-time was 4 to 2. The score progression is a chain of successive goals made by either team such as M-M-S-M-M-S-S-M-M. In how many different ways could the final score have developed?

Draw a graph, with the X axis as the Mathletes score and the Y axis as the sci-fi's score.  We want the number of ways to go to the point (6,3) such that it passes through (4,2).  This is simply 6C2 * 3C1=45.


inequalities
function
inequalities proposed

Let $ a,\ b,\ c,\ d,\ e$ be real numbers such that $ |a|\leq 1,\ |b|\leq 1,\ |c|\leq 1, |d|\leq 1$. Prove that

$ a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2}\plus{}\sqrt{1\minus{}a^{2}}\plus{}\sqrt{1\minus{}b^{2}}\plus{}\sqrt{1\minus{}c^{2}}\plus{}\sqrt{1\minus{}d^{2}}\leq 5$.

[hide="an idea"]
If $ |a|\leq 1$, $ |b|\leq 1$ etc ..$ \exists$ $ A,B,C,D\in\left[\minus{}\frac{\pi}{2},\frac{\pi}{2}\right]$ such that $ a \equal{} sinA$ ,$ b \equal{} sin B$ ..etc.
Using this substitutions our inequality is:
$ \sum\left( sin^{2}(A)\plus{}cos(A)\minus{}\frac{5}{4}\right)\leq 0$
Studying the function $ f:\left[\minus{}\frac{\pi}{2},\frac{\pi}{2}\right]\to\mathbb R$ ,$ f(x) \equal{} sin^{2}(x)\plus{}cos(x)\minus{}\frac{5}{4}$ we get that $ f(x)\leq 0$ $ \forall x\in\left[\minus{}\frac{\pi}{2},\frac{\pi}{2}\right]$[/hide]


There are three capacitors($C$) with charges $q_{1}$,$q_{2}$ and $q_{3}$.The are connect formed a circuit.What will  be the new charges of capacitors?

It's quite simple. Let's assume that the charges are $q'_{1}$, $q'_{2}$ and $q'_{3}$, with the same signs as the initial charges on your picture. By conservation of charge, \[q'_{2}-q_{3}= q_{2}-q_{3}\] and \[q'_{1}-q_{2}= q_{1}-q_{2},\] and applying the second Kirchoff's law to the circuit we get \[q'_{1}+q'_{2}+q'_{3}= 0.\] Solving this we get \[q'_{1}=-\frac{q_{2}+q_{3}-2 q_{1}}{3},\] and just rotate indices for the other two charges.


vector
geometry
parallelogram
rhombus
linear algebra

1. Use vector methods to prove that the diagonals of a parallelogram are orthogonal if and only if the parallelogram is a rhombus (i.e., has all sides of equal length).

2. Use vector methods to prove that a triangle inscribed in circle and having a diameter as one of its sides must be a right triangle.

More generally, given two points A and B in the plane, what is the locus of points X so that ∠ AXB has a fixed measure, and what is the fixed measure? Prove your claim.

(The trouble for me is to prove that the only possible fixed measure of ∠ AXB is 90 degrees, which should be obvious.)

1. if you consider the parallelogram generated by vectors $ \vec{v_1}, \vec{v_2}$
then the diagonals are $ \vec d_1\equal{}\vec v_1\plus{}\vec v_2, \vec d_2\equal{}\vec v_2\minus{}\vec v_1$
so, $ <\vec d_1,\vec d_2>\equal{}<\vec v_1\plus{}\vec v_2,\vec v_2\minus{}\vec v_1>\equal{}$
$ <\vec v_1,\vec v_2>\minus{}<\vec v_1,\vec v_1>\plus{}<\vec v_2,\vec v_2>\minus{}<\vec v_2,\vec v_1>\equal{}\|\vec v_2 \|^2 \minus{}\|\vec v_1 \|^2$

so $ \vec d_1 \perp \vec d_2 \iff \|\vec v_2 \|^2 \equal{}\|\vec v_1 \|^2$

2. similar, the two sides $ \vec v_1, \vec v_2$ are of the form
$ v_1\equal{}\vec a \plus{}\frac{1}{2}\vec d$, $ v_2\equal{}\vec a \minus{}\frac{1}{2}\vec d$
with $ \|\vec a\|^2\equal{}\|\frac{1}{2}\vec d\|^2$
here $ \vec d$ is the diameter that is the side of the triangle

that locus thing, though... i don't know..


In a bag, there are three red marbles, five white marbles, and four blue marbles. What is the fewest number of marbles that must be pulled from the bag to ensure that there is at least one of each color?

The worst scenario is that we get five white marbles and four blue marbles, then a red marble, for a total of $ 5\plus{}4\plus{}1\equal{}\boxed{10}$ drawings.


geometry
geometric transformation
reflection
circumcircle
rhombus
congruent triangles
perpendicular bisector

Let $ABCDEF$ be a convex hexagon satisfying $AC=DF$, $CE=FB$ and $EA=BD$. Prove that the lines connecting the midpoints of opposite sides of the hexagon $ABCDEF$ intersect in one point.

[quote="Megus"]$ABCDEF$ is a convex hexagon in which $AC=DF$,$CE=FB$ and $EA=BD$. Prove that lines connecting midpoints of opposite sides intersect in one point.[/quote]

i have very simple solution for this problem.
suppose M, N, P, Q, R, S are midpoints of AB, BC, CD, DE, EF, FA.
we have $MN = \frac{1}{2}AC = \frac{1}{2}DF = QR$, then MQ, NR intersect each other at their midpoints.
similar MQ, PS intersect each other at their midpoints, too; then 3 segments MQ, NR, PS concurent.

[quote="xirti"]Why two equal segments intersect each other at their midpoints?[/quote]

i'm so sorry, i've done a mistake. i think they're parallel!

Zibi's link from post #2 immediately solves the problem, but here is a more olympiad solution:

[quote="Megus"]Let $ ABCDEF$ be a convex hexagon satisfying $ AC \equal{} DF$, $ CE \equal{} FB$ and $ EA \equal{} BD$. Prove that the lines connecting the midpoints of opposite sides of the hexagon $ ABCDEF$ intersect in one point.[/quote]

[i]Solution.[/i] We start with a lemma:

[color=blue][b]Lemma 1.[/b] If UVW and U'V'W' are two oppositely congruent triangles, then the midpoints of the segments UU', VV' and WW' are collinear.[/color]

[i]First proof of Lemma 1.[/i] Since the triangles UVW and U'V'W' are oppositely congruent, there exists a glide reflection mapping the points U, V, W to the points U', V', W'. Then, the midpoints of the segments UU', VV' and WW' are collinear, because these midpoints all lie on the axis of this glide reflection (in fact, if a glide reflection sends a point P to a point P', then the midpoint of the segment PP' lies on the axis of this glide reflection). Lemma 1 is proven.

[i]Second proof of Lemma 1.[/i] Let X, Y, Z be the midpoints of the segments UU', VV', WW'. We want to show that the points X, Y, Z are collinear.

We use vectors. Let K be the point such that $ \overrightarrow{YK}\equal{}\overrightarrow{VU}$, let L be the point such that $ \overrightarrow{YL}\equal{}\overrightarrow{V^{\prime}U^{\prime}}$, let M be the point such that $ \overrightarrow{YM}\equal{}\overrightarrow{V^{\prime}W^{\prime}}$, and let N be the point such that $ \overrightarrow{YN}\equal{}\overrightarrow{VW}$.

Since Y is the midpoint of the segment VV', we have $ \overrightarrow{YV}\equal{}\minus{}\overrightarrow{YV^{\prime}}$. Since X is the midpoint of the segment UU', we have

$ \overrightarrow{YX}\equal{}\frac{\overrightarrow{YU}\plus{}\overrightarrow{YU^{\prime}}}{2}\equal{}\frac{\left(\overrightarrow{YV}\plus{}\overrightarrow{VU}\right)\plus{}\left(\overrightarrow{YV^{\prime}}\plus{}\overrightarrow{V^{\prime}U^{\prime}}\right)}{2}\equal{}\frac{\left(\minus{}\overrightarrow{YV^{\prime}}\plus{}\overrightarrow{YK}\right)\plus{}\left(\overrightarrow{YV^{\prime}}\plus{}\overrightarrow{YL}\right)}{2}$
$ \equal{}\frac{\overrightarrow{YK}\plus{}\overrightarrow{YL}}{2}$,

so the point X is the midpoint of the segment KL. Now, since $ \overrightarrow{YK}\equal{}\overrightarrow{VU}$ and $ \overrightarrow{YN}\equal{}\overrightarrow{VW}$, the triangle KYN is directly congruent to the triangle UVW. Similarly, the triangle LYM is directly congruent to the triangle U'V'W'. Thus, the opposite congruence of triangles UVW and U'V'W' yields that the triangles KYN and LYM are oppositely congruent. Thus, LY = KY, so that the triangle KYL is isosceles, and thus, since X is the midpoint of its base KL, we have < XYL = < XYK by symmetry. Similarly, < ZYM = < ZYN; also, < LYM = < KYN because of the congruence of triangles KYN and LYM. Hence, < XYL + < LYM + < ZYM = < XYK + < KYN + < ZYN. But (< XYL + < LYM + < ZYM) + (< XYK + < KYN + < ZYN) = 360°; hence, $ \measuredangle XYL\plus{}\measuredangle LYM\plus{}\measuredangle ZYM \equal{}\frac{360^{\circ}}{2}\equal{} 180^{\circ}$, so the points X, Y, Z are collinear, and Lemma 1 is proven once again.

Now let's solve the problem: Since AC = DF, CE = FB and EA = BD, the triangles ACE and DFB are congruent. Since the hexagon ABCDEF is convex, these two triangles have the same orientation, so they are directly congruent. Hence, if O and O' are the circumcenters of these triangles ACE and DFB, then the triangles AOE and DO'B are directly congruent, too (in fact, the points O and O' are corresponding points in the directly congruent triangles ACE and DFB, being the circumcenters of these two triangles, and corresponding points in directly congruent triangles form directly congruent triangles themselves). Since O' is the circumcenter of triangle DFB, we have O'B = O'D. Thus, triangle DO'B is isosceles, so we can say that triangle DO'B is oppositely congruent to triangle BO'D. Combined with the direct congruence of triangles AOE and DO'B, this yields that the triangles AOE and BO'D are oppositely congruent. Thus, according to Lemma 1, the midpoints of the segments AB, OO' and ED are collinear. Hence, the line through the midpoints of the segments AB and ED passes through the midpoint of the segment OO'. Similarly, the line through the midpoints of the segments BC and FE and the line through the midpoints of the segments CD and AF also pass through the midpoint of the segment OO'. Thus, these three lines, which join midpoints of opposite sides of the hexagon ABCDEF, intersect in one point (namely, in the midpoint of the segment OO'). Problem solved.

  Darij

Two triangles $\triangle ABC, \, \triangle DEF$, such that $BC = EF, \, CA = FD, \, AB = DE$ are called oppositely congruent if, for example, the vertices of $ABC$ are in clockwise direction and the vertices of $DEF$ are in counterclockwise direction.
That means that, when transforming a triangle into the other, one must use a reflection at some point.

[hide="sketch"]Draw diagonals AD, BE, CF. Consider the midpoints of AD, BE, AB, and DE. These points form a rhombus, so the line connecting the midpoints of AB and DE is the perpendicular bisector of the line connecting the midpoints of AD and BE. Considering all pairs of opposite sides, the line connecting the midpoints will intersect at the circumcenter of the triangle formed by the midpoints of AD, BE, and CF. [/hide]


superior algebra
superior algebra unsolved

Let $A$ be a ring with $0\neq 1$. Prove that if for every $a\neq 1$ one can find $b$ s.t. $ab=a+b$ then $A$ is a division ring.

[hide="not sure about"] my aproach:
Write the relation for $1-a$.
$(1-a)b=(1-a)+b\Rightarrow a\cdot (1-b)=1$  hence $a$ is invertible.
Am I wrong? What is the correct solution.
[/hide]

I thought I need to prove that $a$ have left inverse also to be invertible, or at least that $1-b$ have right inverse.

I think you need to prove that $1-b$ has right inverse. Or if it is known, I'd like someone to post the demonstration of this fact.

Repeat your argument at $b$; $1-b$ has a right inverse, which must be equal to $a$.

This still doesn't make the ring into a field; there's no way you can get commutativity unless you assume it. You've just got a division ring.

That leaves you two choices: You can assume commutativity, and your original argument works unmodified, or you can leave commutativity out, and the target statement isn't true.

I didn't kno how to say to this structure: a ring in which every element have inverse (inroamnian it is called "corp" which can be translated as "body" in english). Thank you!


geometry
geometric transformation
reflection
algorithm
combinatorics unsolved
combinatorics

There are n fleas on an infinite sheet of triangulated paper. Initially the fleas are in different small triangles, all of which are inside some equilateral triangle consisting of n^2 small triangles. Once a second each flea jumps from its original triangle to one of the three small triangles having a common vertex but no common side with it. For which natural numbers n does there exist an initial configuration such that after a finite number of jumps all the n fleas can meet in a single small triangle?

It's possible for all $n$ except $n=2$ or $n=4$.

Your definition of which small triangles a flea could jump to was a bit ambiguous, but since there are 3 legal neighboring triangles, I assumed that, for a triangle ABC, they're the ones obtained by reflection: by the line parallel to AB through C, by the line parallel to AC through B, and the line parallel to BC through A.

We say that two triangles are neighbors if a flea can jump from one to the other. Take any small triangle and draw segments joining its center with the centers of the neighboring ones. Now take the 3 neighbors and join their centers with their neighbors' centers, and so on. Then you get a web of joined triangles. Suppose WLOG that there's a flea in the first small triangle. Then it can get to any triangle belonging to the web, and it can't get to any triangle outside the web.

We see that the edges of the web form a web of hexagons. Now we'll color the triangle centers. Take the first triangle and color its center white; now for any white point, if some other point is joined to it color it black and vice-versa. It's clear this can be done. Also notice a flea on a white-centered triangle jumps to a black-centered one on each turn and vice-versa. So all fleas must (in the original position) be in triangles of the same color to eventually end up on the same triangle.

For any two fleas on triangle centers A, B, you can draw a broken line on the web (taking the hexagon sides as segments) joining A and B. If they're of the same color the broken line has $2k$ segments; they both move towards each other on the broken line, so after a move they're $2k-2$ segments apart, and after $k$ moves they're in the same triangle. If any two fleas are on the same triangle, from then on, they will always jump to the same triangle.

Given the $n$ fleas, you apply the following algorithm. If there are $m$ unoccupied fleas, you pair them up (if there's an odd number one will be left out) and make the ones in a pair join each other as above. Once two fleas join each other they count as one, and the new flea is unoccupied. If a flea is unoccupied and there's no other to pair up with, it just moves around; when someone else is unoccupied they pair up, and so on. Eventually, this algorithm will lead all fleas to one triangle.

Now it's a matter of fitting fleas into the $n^2$ equilateral triangle, in triangles of the same web with the same color. I can't draw a diagram here, but it's clear that for $n=2k$, you can fit $k(k+1)/2$ triangles satisfying those conditions, and $k(k+1)/2 \geq 2k$ for $n \geq 6$. For $n=2k+1$, you can fit $(k+1)(k+2)/2 \geq 2k+1$ for all $n$. Then the only impossible cases are $n=2$ and $n=4$.


algebra
polynomial
modular arithmetic
quadratics
complex numbers
Galois Theory
number theory unsolved

Let $ g(x)$, $ h(x)$ be two non-constant monic polynomials with degrees less than 4 (coefficients in $ \mathbb{Z}$).  For any prime $ p$, the congruence $ x^{4}\plus{}1\equiv g(x)h(x)\pmod{p}$ holds for some $ g$, $ h$.  Is there a general technique or equation for finding the $ f$ and $ g$?

http://www.mathlinks.ro/viewtopic.php?t=251121 or http://www.mathlinks.ro/viewtopic.php?t=180635 .

The general technique is factoring it in complex numbers and by Chebotarev to find that you can find a factorisation mod any prime. See http://www.mathlinks.ro/viewtopic.php?t=34935 .

I don't know any algebra. :(   Could you explain the technique?

For a real understanding, you will need algebra. But even without it, you can check the factorisations after the next paragraph.

The roots of $ x^4\plus{}1$ are the primitive $ 8$-th roots of unity. Let $ \zeta$ be one of them. Then $ \zeta^2 \equal{} \pm i$, $ \zeta\plus{}\zeta^3 \equal{} \pm \sqrt{\minus{}2}$, $ \zeta\plus{}\zeta^{\minus{}1} \equal{} \pm \sqrt{2}$. Note that each of these three numbers is quadratic over $ \mathbb Q$.
In fact, the Galois group of $ x^4\plus{}1$ is $ (\mathbb Z / 8)^\times$, where the fixed fields of $ 3 \mod 8, \ 5\mod 8 , \ 7 \mod 8$ are $ \mathbb Q[\sqrt{\minus{}2}], \ \mathbb Q[\sqrt{\minus{}1}], \ \mathbb Q[\sqrt{2}]$, respectively.
So this tells is that over a field $ k$ containing a square root of at least one of $ \minus{}1,2,\minus{}2$, this polynomial must split. In fact, it tells us that a splitting is archieved by taking the factors of type $ (x\minus{}\zeta^x)(x\minus{}\zeta^{ax})$, where $ x$ is arbitrary and $ a$ is the corresponding element of the Galois group. In this case, this gives (where we now let $ \zeta \equal{} e^{\frac{2\pi i}{8}}$):
If $ \sqrt{\minus{}1} \in k$, then $ a\equal{}5 \mod 8$ and the factors are $ (x\minus{}\zeta)(x\minus{}\zeta^5) \equal{} x^2\minus{}x(\zeta\plus{}\zeta^5)\plus{}\zeta^6 \equal{} x^2\minus{}\sqrt{\minus{}1}$ and $ (x\minus{}\zeta^3)(x\minus{}\zeta^{3 \cdot 5}) \equal{} x^2\minus{}x(\zeta^3\plus{}\zeta^7)\plus{}\zeta^2 \equal{} x^2\plus{}\sqrt{\minus{}1}$.
If $ \sqrt{2} \in k$, then $ a\equal{}7 \mod 8$ and the factors are $ (x\minus{}\zeta)(x\minus{}\zeta^7) \equal{} x^2\minus{}x(\zeta\plus{}\zeta^7)\plus{}\zeta^0 \equal{} x^2\minus{}\sqrt 2 x\plus{}1$ and $ (x\minus{}\zeta^3)(x\minus{}\zeta^{3 \cdot 7}) \equal{} x^2\minus{}x(\zeta^3\plus{}\zeta^5)\plus{}\zeta^0 \equal{} x^2\plus{}\sqrt 2 x\plus{} 1$.
If $ \sqrt{\minus{}2} \in k$, then $ a\equal{}3 \mod 8$ and the factors are $ (x\minus{}\zeta)(x\minus{}\zeta^3) \equal{} x^2\minus{}x(\zeta\plus{}\zeta^3)\plus{}\zeta^4 \equal{} x^2\minus{}\sqrt{\minus{}2}x\minus{}1$ and $ (x\minus{}\zeta^5)(x\minus{}\zeta^{5 \cdot 3}) \equal{} x^2\minus{}x(\zeta^5\plus{}\zeta^7)\plus{}\zeta^4 \equal{} x^2\plus{}\sqrt{\minus{}2} x \minus{}1$.

Conclusion:
If $ \minus{}1$ is a square, then $ x^4\plus{}1 \equal{} (x^2\minus{}\sqrt{\minus{}1})(x^2\plus{}\sqrt{\minus{}1})$ is a factorisation.
If $ 2$ is a square, then $ x^4\plus{}1 \equal{} (x^2\minus{}\sqrt 2 x\plus{}1)(x^2\plus{}\sqrt 2 x\plus{} 1)$ is a factorisation.
If $ \minus{}2$ is a square, then $ x^4\plus{}1 \equal{} (x^2\minus{}\sqrt{\minus{}2}x\minus{}1)(x^2\plus{}\sqrt{\minus{}2} x \minus{}1)$ is a factorisation.

Now for a finite field, at least one of $ \minus{}1,2,\minus{}2$ is a quadratic residue, so we can always factor it; even more, the above tells us the factorisation itself. So if $ p\equal{}11$, we have that $ 3^2 \equiv \minus{}2 \mod 11$, so w.l.o.g. $ \sqrt{\minus{}2}\equiv 3 \mod 11$, giving the factors $ x^2\minus{}\sqrt{\minus{}2}x\minus{}1 \equal{} x^2\minus{}3x\minus{}1$ and $ x^2\plus{}\sqrt{\minus{}2}x\minus{}1\equal{}x^2\plus{}3x\minus{}1$.

The idea behind this is the fact that Galois groups of extensions of number fields contain Galois groups of residue field extensions of primes. Or alternatively, that Galois groups of unramified local extensions give an action on corresponding global extensions.

I appreciate the response.  Although I don't understand any of the algebra, I will look back on this once I have acquired the necessary algebra.


AMC
AIME
ARML
function
USA(J)MO
USAMO
binomial coefficients

Has anyone here read this book and would like to recommend/discourage it?

[quote="ComplexZeta"]I think the more recent book [i]A Path to Combinatorics for Undergraduates[/i] is better.[/quote]

Hmm, how so? Is one more difficult than the other?

[i]102 Combinatorial Problems[/i] is essentially just problems with no exposition. [i]A Path to Combinatorics for Undergraduates[/i] can be used as a course in combinatorics starting from the basics (the first chapter is about whether you should add or multiply, if I recall correctly) and going up to quite advanced problems. Not all the problems in [i]A Path to Combinatorics for Undergraduates[/i] are olympiad problems though. Many problems are from AIME and the later questions of the AMC12/AHSME as well as ARML and probably some other contests too. But there are definitely olympiad problems as well.

Hmm, so [i]A Path to Combinatorics for Undergraduates [/i]is structured in chapters with instructions and problems at the end of each chapter?

Hmm, does anyone know where I can find a table of contents for [i]A Path to Combinatorics for Undergraduates[/i]? amazon.com doesn't show one. :(

Otherwise, does anyone know how difficult this book gets (ie some of the topics in the later chapters)?

I think the last chapter is on generating functions.

How in depth about generating functions?

That is, I'm just wondering whether most of the stuff in this book will be useful even if I understand all of the countings/combinatorics stuff in AoPS.

Well, it has all the standard stuff with generating functions as applied to combinatorics problems. You won't need generating functions very often before USAMO, although I do recall one non-olympiad question from my sophomore year that asked for the coefficient of x2000 in (1+x)(1+x2)(1+x4)..., but that doesn't require using generating functions. It just is one.

[quote="Fierytycoon"]Hmm, does anyone know where I can find a table of contents for [i]A Path to Combinatorics for Undergraduates[/i]? amazon.com doesn't show one. :([/quote]

i was also looking for one... and luckily there was one at b&n  :D

Table of Contents
   Preface
 Introduction
 Acknowledgements
 Abbreviations and Notation
1 Addition or Multiplication? 1
2 Combinations 25
3 Properties of Binomial Coefficients 43
4 Bijections 69
5 Recursions 91
6 Inclusion and Exclusion 117
7 Calculating in Two Ways: Fubini's Principle 143
8 Generating Functions 165
9 Review Exercises 195
 Glossary 213
 Index 217
 Further Reading 221
 Afterword 227

the link: [url=http://btobsearch.barnesandnoble.com/booksearch/isbninquiry.asp?userid=33SQRX7ST0&sourceid=00395996645644787198&btob=Y&ean=9780817642884&displayonly=TOC#TOC]here[/url]

by the way... for a course in combinatorics, which would be better? this book or "Principles and Techniques in Combinatorics" by Chen Chuan-Chong, Koh Khee-Meng?

Ah, thanks, I must have missed that..somehow...even though the table of contents takes up the entire browser screen...  :mrgreen:

By the way, does anyone know how to do Simon's problem, heh?

Um, unless I misunderstood it, the answers just

[hide]

1..

 (1+x)(1+x^2)(1+x^4)(1+x^8)... = 1 + x + x^2 + x^3 + x^4...

Think either binary, or just factor out 1+x from RHS, then 1+x^2, etc..

[/hide]

Yup. I thought it was too easy, but I couldn't find anything wrong with it.

Oh, I thought the exponents were the even numbers, not the powers of 2 (Simon's original problem didn't specify). Sorry.

Well, it starts with 1+x. Anyway, I remembered another problem that looks like (well, and is) a generating function problem.

(ARML 2001 #8) Let f(x)=(1+x)(1+x4)(1+x16)(1+x64)... Evaluate f-1(8/(5f(3/8))).


When expanded under decimal form, the fraction $ \dfrac{1}{97}$ has a period with $ 96$ digits. If the last three digits are $ A67$, find $ A$.

Let $ p$ be the period.
$ \frac{1}{97}.10^{96}\minus{}p\equal{}\frac{1}{97}\Leftrightarrow p\equal{}\frac{10^{96}\minus{}1}{97}$

Let the last digit of $ x\equal{}L(x)$

We need to find the last digits of the integer $ \frac{10^{96}\minus{}1}{97}$. I'll find five of them.
$ \frac{10^{96}\minus{}1}{97}\equal{}\cdots ABCDE$.
$ 9\equal{}L(97*E)\equal{}L(7*E)\implies E\equal{}7$
$ 9\equal{}L(97*D\plus{}(97*E\minus{}9)/10)\equal{}L(97*D\plus{}67)\equal{}L(7*D\plus{}7)\implies D\equal{}6$
$ 9\equal{}L(97*C\plus{}(97*DE\minus{}99)/100)\equal{}L(97*C\plus{}64)\equal{}L(7*C\plus{}4)\implies C\equal{}5$
$ 9\equal{}L(97*B\plus{}(97*CDE\minus{}999)/1000)\equal{}L(7*B\plus{}54)\equal{}L(7*B\plus{}4)\implies B\equal{}5$
$ 9\equal{}L(97*A\plus{}(97*BCDE\minus{}9999)/1000)\equal{}L(7*A\plus{}53)\equal{}L(7*A\plus{}3)\implies A\equal{}8$

I'm trying to understand the first part of your solution.

I would like to know where are my mistake with the similar problem:

$ \dfrac{7}{198}\equal{}0,03535\ldots$

If i'm not wrong, period of $ 0,03535\ldots$ has two digits: $ 3$ and $ 5$.

Hence

$ \dfrac{7}{198}\cdot 10^2 \minus{} p\equal{}\dfrac{7}{198}\Longleftrightarrow p\equal{}\dfrac{7\cdot(10^2\minus{}1)}{198}$

But $ p$ isn't an integer.

So what i'm doing wrong? Your approach can't be used for the example that i used?

Thanks in advance.

[quote="LoboSolitario"]So what i'm doing wrong? Your approach can't be used for the example that i used?[/quote]  The denominator must be relatively prime to 10 to begin so there is no nonrepeating part at the beginning.  Otherwise the difference will be nonintegral.


algebra
polynomial
superior algebra
superior algebra unsolved

List the conjugates of 2^(1/4)
List the monomorphisms from K=Q(2^(1/4))-->C

The minimal polynomial is $ x^4\minus{}2$ (do you see why¿). The conjugates are the other roots of that polynomial; show that if $ \alpha$ is a root, then $ \alpha i$ is another root.
By this you get all roots, and every root gives another monomorphism.

for the record: it's highly uncommon to write $ 0.25$ in algebra. Just write $ \frac1{4}$ ;)


Prove that the equation $x^2+y^2+z^2 = x+y+z+1$ has no solution in rational numbers.

We write this equation in the form $(2x-1)^2+(2y-1)^2+(2z-1)^2=7$. If there is any rational solution then the equation $a^2+b^2+c^2=7m^2$ must have any solution in positive integer. Let $m$ be the minimal such solution. If $m$ is even then $a^2+b^2+c^2\equiv0(mod 4)$ which implies that $a,b,c$ are even and then $m/2$ is solution -contradiction with the minimality of $m$($x=y=z=1/2$ is impossible). If $m$ is odd then $m^2\equiv 1(mod 8)$ therefore
$a^2+b^2+c^2\equiv7(mod 8)$ which is impossible ;)


LaTeX
algebra
polynomial
inequalities unsolved
inequalities

Let $a_1$, $a_2$, $\cdots$, $a_n$ be positive real numbers, and let $S_k$ be the sum of the products of $a_1$, $a_2$, $\cdots$, $a_n$ taken $k$ at a time.  Show that
\[ S_k S_{n-k} \geq {n \choose k}^2 a_1 a_2 \cdots a_n \]
for $k = 1$, $2$, $\cdots$, $n - 1$.

This is just AM-GM.
\[ S_k=\sum_{i_1<...<i_k}a_{i_1}\cdots a_{i_k}\ge{n\choose k}\sqrt[{n\choose k}]{\prod_{i_1<...<i_k}a_{i_1}\cdots a_{i_k}}. \][I think Latex reaches its limits here :)]
The RHS of this has degree $k$ and is symmetrical in the $a_i$, so it must be $={n\choose k}(a_1\cdots a_n)^\frac kn$. Likewise $S_{n-k}\ge{n\choose n-k}(a_1\cdots a_n)^\frac{n-k}n$. The result follows.

[hide]This can also be solved by Muirhead's Inequality. The LHS expands to a variety of symmetric polynomials of degree $n$ whose coefficients add up to $\binom{n}{k}^2$[/hide]

By Cauchy-Schwarz inequality

\[ S_{k} S_{n-k}=\sum_{i_1<...<i_k} a_{i_1}\cdots a_{i_k} \sum_{i_1<...<i_k}\frac{S_n}{a_{i_1}\cdots a_{i_k}}\ge({n\choose k}\sqrt{S_{n}})^2={n\choose k}^2 S_{n} \]

Note that their are $n\choose k$ terms in $S_k$ and ${n\choose n-k}$ in $S_{n-k}$. By AM-GM we see that
$$S_k=\sum_{i_1<\ldots<i_k}{a_{i_1}\cdot\ldots\cdot a_{i_k}}\ge {n\choose k}{\sqrt[n\choose k]{\prod_{i_1<\ldots<i_k}{a_{i_1}\cdot\ldots\cdot a_{i_k}}}}$$
Now, fix any $a_i,i=1,\ldots,k$. There are now ${n-1\choose k-1}$ ways to combine $a_i$ with another $k-1$ factors, so $a_i$ appears ${n-1\choose k-1}$ terms in $S_k$ ant thus appears with power $n-1\choose k-1$ in the product on the right-hand-side. Therefore
$${\sqrt[n\choose k]{\prod_{i_1<\ldots<i_k}{a_{i_1}\cdot\ldots\cdot a_{i_k}}}}=\left(a_1^{n-1\choose k-1}\cdot\ldots\cdot a_k^{n-1\choose k-1}\right)^\frac{1}{{n\choose k}}$$
Now since
$$\frac{{n-1\choose k-1}}{{n\choose k}}=\frac{k}{n}$$
we get
$$S_k\ge {n\choose k}\left(a_1\cdot\ldots\cdot a_k\right)^{\frac{k}{n}}$$
In a similar fashion, we get
$$S_{k-n}\ge \left(a_1\cdot\ldots\cdot a_k\right)^{\frac{n-k}{n}}$$
and thus
$$S_k\cdot S_{n-k}\ge {n\choose k}^2\cdot a_1 a_2 \cdots a_n$$

$\color{red}\boxed{\textbf{SOLUTION}}$

By $\textbf{AM-GM Inequality,}$
$$S_k=\sum_{1 \le i_1<\ldots<i_k \le n}{a_{i_1}\cdot\ldots\cdot a_{i_k}}\ge {n\choose k}{\sqrt[n\choose k]{\prod_{1 \le i_1<\ldots<i_k \le n}{a_{i_1}\cdot\ldots\cdot a_{i_k}}}}$$

And,
$$S_{n-k}=\sum_{1\le i_{k+1}<\ldots<i_{k+j}\le n}{a_{i_{k+1}}\cdot\ldots\cdot a_{i_{k+j}}}\ge {n\choose k}{\sqrt[n\choose k]{\prod_{i_{k+1}<\ldots<i_{k+j}}{a_{i_{k+1}}\cdot\ldots\cdot a_{i_{k+j}}}}}$$

$\textbf{Note :}$
$\prod_{x \in (1,2...k,k+1,...k+j)}{a_{i_x}}=\prod_{1\le i \le n}{a_i}$

So, Multiplying Gives,
$$S_k\cdot S_{n-k}\ge {n\choose k}^2\cdot a_1 a_2 \cdots a_n\blacksquare$$


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Hi all,

I would like to have some feedback about a topic that is very important to me now: what is an interesting problem in mathematics. This is part of my doctoral research, so answers are welcomed.
I would like to know personal opinions, in case illustrated with some examples, but most of all I want to see if there is any possibility to get closer to a formalized view of "interestingness" in mathematics.
Should we treat "serious" (like IMO) problems treat separatelly when discussing about the subject or they are "interesting" in the same sense as "recreational" math problems are.

Your contribution is welcomed.
Thanks,

IP

[quote="fedja"][hide]Well, in my humble opinion "interesting" in mathematics means pretty much the same as "adding something new to the existing knowledge" though I realize that the latter words are not exactly formal. This may be a new viewpoint on the things that are already known, a connection between previously unrelated things, a new question, an unexpected twist in a proof, and so on and so forth.

This would be an "objective" meaning. As to "subjective" points of view, they vary tremendously and I usually cannot even come to an agreement with my co-authors about what exactly is the "really interesting" part in our joint papers. Roughly speaking, every mathematician has his own scale of values determined both by the general mathematical culture of his time and by his personal taste. So, like many other evaluation ajectives, the word "interesting" requires a lot of specifiers like "when", "for whom", "for what purpose", etc., without which you will, probably, never get two people agree on whether something is "interesting" or not. Personally, I find everything that I do not currently know "interesting" though I understand that I will have no chance to learn everything or to think of everything. Thus, I set my priorities according to my personal taste and that's about it.

One more thing I would like to say here is that almost any problem in mathematics can be developed into something "interesting". Just start with something as boring as multiplication of two numbers. Of course, the task to multiply $12345$ by $6789$ is very non-interesting but to multiply two $1000000$-digit numbers even if you are given a computer is
something quite interesting: you cannot do it by brute force anymore in a reasonable time, so you should find a better way.
Of course, not every problem has such a simple "difficulty parameter" as the length of the numbers in this example but you  can almost always change the formulation step by step to add one difficulty a time and to get something that nobody would know how to approach quite fast.

I should also say that quite many people would consider "interesting" as a synonym of "useful" or "applicable". I personally prefer to separate these two notions completely but, again, this merely reflects my taste and the culture I grew in.

At last mathematics is no better than jewelry, clothes, movies, etc. in terms of how people are swept by fashions. I would say that truly independent minds in this respect are not too many and I'm very far from pretending to be one of them :). So, for quite many people "interesting" would mean "fashionable".

There are many more possible meanings (see below) but overall, the question is very similar to "What is mathematics?". The only answer I know to that is "Mathematics is what mathematicians do". Of course, it raises the question "Who are mathematicians?" and the only answer I know to that is "They are people who do mathematics". At least the moral is the same: show that you excel in doing what is considered interesting today and then there is a good chance that you'll be able to tell others what is interesting tomorrow. Then their turn will come and so on.

As to recreational or olimpiad mathematics in particular, you should understand that it is directed at people who are not professional mathematicians (yet or already) but still love to think about mathematics just for the pleasure of it. These are subcultures of the bigger culture with their own standards, fashions, etc. that are not always close to the standards of the entire culture or its other subcultures. All I said above applies here as well plus one more thing: here "interesting" may also be used as a synonym for "beautiful" (in professional mathematics it is also so but to a much lesser extent as far as I can judge).

So, just forget about formalizing the meaning of the word "interesting" in mathematics. If you are still thinking of formalizing evaluation adjectives, start with some innocently looking nouns like "chair"; the chances are $10$ against $1$ that you'll not be able to come up with a description that fits everything that is a chair and nothing that isn't one even if you spend a year thinking of nothing but this. This is not to say that you cannot write an extensive $1000$ page treatise on the concept of chair with highly nontrivial content. This is just to say that, though each of us has his/her own concept of "interesting" and there is some concept like this in the culture, there is no short way to describe it and there seems to be no way at all to pass the description to another person, especially a layman, without distorting the information beyond recognition.

I understand that you asked not about "interesting" but about an "interesting problem", but there is not too much difference. The only easy thing is to say what problem is non-interesting: it is a problem that can be solved in a routine way by an average mathematician in a fairly short time, has no applied value for other consumers of mathematics, no internal beauty, no educational value, no place in a bigger theory, no interesting corollaries, no interesting generalizations, no interesting simplifications of generalizations, no interesting generalizations of simplifications of corollaries, and so on. I believe (though still not 100% sure) that such a problem would be deemed non-interesting by everybody but now I'd like to challenge you to find a single non-interesting problem in this sense. Or, if you want to go outside mathematics, find something unimportant for life on Earth. In a sense, every particular small thing can be easily disposed of, but if you dispose of everything, you'll get nothing left but the feeling that there was something important you threw out . This is just to illustrate that sometimes you can have a few things (say problems), each of which in isolation would be considered of no particular interest but they may form something (say, a theory) that is extremely interesting.

Well, this post is, probably too long already, so I'll stop here. My main point is that research directed at finding universal formal meanings of evaluation adjectives is, most likely, just a waste of time but an attempt to take a look at a sophisticated culture by looking at what the meaning of one word is for different people at different times may be "interesting" enough (at least, I know of some much crazier angles of view that were used). Just be very careful when drawing any conclusions about the culture from what you see at this (or any other such) angle.[/hide][/quote]

Wow that was lengthy  :lol:

I believe that you are trying to say what facinates you in mathematics if i didn't interpret the question wrong?

What makes math interesting is the way it connects to the real world and how it challenges the mind. Also, I believe it is interesting because it has one solid answer. For example,

$1+1=2$ in base 10 and it ALWAYS equal 2 in base 10. These "rules" are far more easier to understand then something from other subjects, like history, where a historian might have to sift through information that say either this happened or it didn't. (way too many possibilities and multiple correct answers for a single question)

I don't know I didn't have time to write such an essay like fedja but I hope my 3 second answer helps  :) .

interesting means to me that it doesn't make me bored, but instead makes me more motivated and more willing to advance in math

[quote="7h3.D3m0n.117"]Wow that was lengthy  :lol:[/quote]
Which is usually a good reason not to quote the entire post.

An interesting problem for me is something that I feel I should be able to solve, but I can't.  :wallbash:

Any problem I've never seen before is interesting.  When I was in Elementary School and learned to divide, I was fascinated by that.  Any problem that I've never seen before intrigues me, at least for a while.

The things that I find interesting long after I've learned about them are the results which seem too elegant to be true.  For example, I find the number $e$ to be the most interesting thing in mathematics, because of its omnipresence.  Or any connection between different branches of mathematics- like the connection between Pascal's Triangle, combinations, and the binomial expansion theorem.

What I find beautiful and interesting is connection between differenet fields in math. renatzu explained it very nice.
The other intriguing thing are the problems that have to be solved and you know you can but you need just a little more time, concentration, luck or idea. At last when you solve it you don't know what's the time and why it's dark outside.

[quote="13375P34K43V312"]An interesting problem to me is something that really takes alot of thinking to solve, but it turns out that there is such a short, solution. But it's only fun if I actually think of it, and the more creative it is, the better it is for me.[/quote]

What is also interesting about the problem if it isn't tedious, which in some cases, most math problems are.

The most interesting things in mathematics are those things that tie different fields together, especially ones that don't seem to be very related.

Currently, what is interesting me a lot are multiplicative functions and the Mobius inversion theorem. I think it is really cool because of the parallels between multiplicative functions on the integers and linear functions on the reals (for example, linear function <-> multiplicative function, basis <-> prime, both form an abelian group, matrix inverses versus extended Mobius inversion, etc.).


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Hey guys, I'm a freshmen at Bronx Science High School, and an incoming sophomore at Stuyvesant High School. I am COMPLETELY lacking motivation. I don't know why, I just don't feel like doing anything. I have this horrible 80 average, with a failed course one marking period {With a 60...}. I tell myself EVERYmarking period I'm gonna do homework, but my words aren't enough to convince myself. I've been like this since 5th grade. I'm the typical failed student. :( Any advice?  :|

PS: My college admissions have went down the drain haven't they? x.x

[quote="7h3.D3m0n.117"]I know the feeling.
My GPA is approximately 3.6 at the end of freshman year (I'm currently a sophmore) and I've renounced hope for any high-end college. [/quote]

That's a joke, right?

I had a 3.4 freshman year, we'll see how I do with college decisions in a week or two. The important thing, it seems, is to have an upward trend, if you didn't have good grades from the beginning.

[quote="7h3.D3m0n.117"]I know the feeling.

Personally, I have random periods of time where I try. In my opinion, my lack of motivation is partly caused by relentless work without many periods of relaxation. I've been playing tennis with my school team recently and it's actually done a lot in helping me relax. Even though it's a time killer (I sit through maybe 2-3 hours of practice instead of doing homework), I end up staying up very late at night (on some instances, midnight or even up til 4 AM) completing homework and studying. Usually, without the practice, I end up not doing my homework anyway, and it all gets pushed back. So find something to do to take your mind off of school!

Honestly, I can't say I see bright futures with your current grade set. My GPA is approximately 3.6 at the end of freshman year (I'm currently a sophmore) and I've renounced hope for any high-end college. But keep trying; you can still improve your current situation![/quote]

Thing is, my parents won't let me stay awake till like past 10!! >.< So I'm very limited to what I can do in a day. D:

Tell them you need more time to do your homework.

do you think you have a disease like mono or such? otherwise all i can say is some people are workers others aren't. My best friend is exactly like you. try harder??

[quote="dragon96"]Maybe your lack of motivation is due to the limited time you have. I'm not sure how much homework they give in high school, but you might keep thinking to yourself [i]I'm not going to finish all this anyways. Why bother?[/i] As 7h3 mentioned, it might be a good idea to spend some time (especially before long writing assignments) doing something you enjoy.[/quote]

Ah, I'll be sure to give that a shot. :) Any other words of advice? :o

What do you do with your time when you're awake? If you're slacking off and doing stupid things, then your parents are less likely to be lenient. If you actually sit there and do your homework, your parents will probably understand (at least, eventually they will).

Where do you lose most of your points in class?

You're only in freshman year, if you turn yourself around and do a lot better the next three years, colleges will notice that. You haven't thrown your life away yet. Actually, even if you keep slacking off, you're not quite throwing your life away, but the longer you wait, the further you're diverging from where you'd want to be and the harder it will be to get back on track.

[color=green]Moderator says: there's never a good reason to quote the entire post immediately preceding yours.[/color]

I lost most of my points from homework. :x

OK, so do you not [i]want[/i] to do it properly, or do you not [i]have enough time[/i] to do it properly?

I assume it's the first, based on the title of this thread. If you're really worried about how it may affect your future, then that should be motivation enough. If you don't care, then that's that.

If you get in the habit of avoiding distractions and getting your homework done effectively, you'll hate it less. If you've ever heard the time management analogy with rocks, take it to heart! Also, consider waking up half an hour earlier each day and doing your homework in the morning.

[quote="Forte"]Thing is, my parents won't let me stay awake till like past 10!! >.< So I'm very limited to what I can do in a day. D:[/quote]

Haha yeah I had the same problem a couple years ago - I wasn't allowed to stay past 9 and if I did because of homework, I got yelled at for "fooling around" even if I was consistently working during the day.

Now, I usually go to bed around 10, but on certain nights with piles of homework, I usually just bring all my books into my room and study there. If your parents are seriously against you staying up, even with you consistently working, then you need to talk to them. Show them that you're not goofing off. If they're not listening, you can always pretend to go to bed and then when they're all settled just whip out a small reading light and your books.

[quote="TZF"]What do you do with your time when you're awake? If you're slacking off and doing stupid things...[/quote]

Ugh darn Facebook!

[quote="sponge008"] I had a 3.4 freshman year, we'll see how I do with college decisions in a week or two. The important thing, it seems, is to have an upward trend, if you didn't have good grades from the beginning.[/quote]

You are absolutely right. Colleges to want to see improvement, and obviously, by getting better grades each year (leading up to the most important, junior year) will definately help.

For example here's my projection

[hide]
Freshman year - 3.6
Sophmore year (hopefully - everything is hinging on this week-long Spanish exam I finished today) - 3.8
Junior year (hopefully) 4.0

Overall GPA (excluding senior year) - 3.8
Projected number of APs by the end of senior year - 12-13
Extracurriculars - 6
Community service hours - 100+

Personally, in my opinion, a 3.8 overall without senior year included is very nice considering the 3.6 I had freshman year. Combined with many extracurriculars, community service, good SAT scores, and a nice number of challenging AP courses, then you are very well off. [b]But it's the hardwork and uphill climb that you must COMMIT to before you can achieve your goals.[/b]
[/hide]

[quote]
My GPA is approximately 3.6 at the end of freshman year (I'm currently a sophmore) and I've renounced hope for any high-end college. [/quote]

3.6 is not very bad for unweighted.

3.6 isn't a very bad GPA, but when applying to top-tiered colleges, it's almost shameful.

I recall an article that getting into Harvard was like the lottery. All the applicants have 4.0s, 2400 SATs, and mounds of sports/music extracurriculars fueled by the money their parents invest. From these hundreds of thousands of applicants, only a few thousand or so will get in. How do college admissions determine the most qualified if everyone has nearly the same application? I have a 3.6, haven't taken the SATs yet but my PSATs looked disappointing, and all the extracurriculars I do are based on my natural skills and aren't funded by parents.

One of my senior friends ended his high school with a 3.75, and is getting rejected and waitlisted from his top choices.

[quote="7h3.D3m0n.117"]All the applicants have ... 2400 SATs..... From these hundreds of thousands of applicants, ....[/quote]  A little reality, please: only a few hundred people in the country each year get perfect SAT scores.  (See the first link [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1007312#1007312]here[/url].)  And Harvard accepts about 10% of applicants usually, meaning it has no more than about 18,000 applicants.  The entire Ivy League probably gets a few more than 100,000 applications, but there is massive redundancy among these (applicants who apply to 3 or more are quite common, for example).

Your broader argument (that the college application process is stochastic) should, I think, be self-evident.  At least, I've [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=79026#79026]saying it[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=906841#906841]for years[/url].

Oh my this is all so discouraging. :( And also, one thing I've noticed is, my Math average increases at the expense of my other classes. For example, my report card is cut down the middle of good student and bad student. I have a 97 in Math, and a 65 in French. This is beginning to be a real problem for me...

Right. I actually finished all my homework today. Thanks for the advice everyone.

[quote="JBL"]Perhaps (and I mean this as actual advice, not purely as a quip) you should stop wasting time expressing self-pity on the internet and instead devote yourself in some more serious way to making sure you do the work you need to do.[/quote]

This advice really makes sense in a school-centered environment, where working harder is bound to yield good results.

Actually, there are a lot of activities where you can get surprisingly far just by not giving up and working harder. School, fortunately, is one of them, along with more "socially" oriented activites However, there are many activities where that's not the case. Activities that involve hard thinking such as mathematics are one of them. Surprisingly, there are many people who no matter how far they work hard never substantially improve in mathematics.

Not only that, but such stumbling blocks could be purely biological, as we now have [url=http://www.newsroom.ucla.edu/portal/ucla/more-proof-that-intelligence-is-85134.aspx]solid, empirical evidence that most of your intelligence is largely inherited.[/url] Basically meaning that working hard could be just as good as doing nothing.

So whenever I have a moment of "self-pity", it's because I face a dilemma: am I not good in math because I am applying myself in the wrong way, or is it because I am biologically limited?

The biggest problem is our technology isn't advanced enough to truly quantify intelligence (IQ tests are pseudoscience). So how do I solve this dilemma?

[quote="ZzZzZzZzZzZz"][quote="PKLove"]
Not only that, but such stumbling blocks could be purely biological, as we now have [url=http://www.newsroom.ucla.edu/portal/ucla/more-proof-that-intelligence-is-85134.aspx]solid, empirical evidence that most of your intelligence is largely inherited.[/url] Basically meaning that working hard could be just as good as doing nothing.[/quote]

Sounding a lot like James Watson here...[/quote]

I'm not proposing that there exists an intelligence difference between races (there doesn't), but individually, it is 100% futile to pursue certain interests if your brain lacks the necessary tools to do it.

[quote="PKLove"]Surprisingly, there are many people who no matter how far they work hard never substantially improve in mathematics.[/quote]  I have yet to meet a person who can't learn more mathematics then they already know (though certainly some people need to devote more effort than others).  What is your basis for this statement?

The UCLA study you cite, while interesting, has no obvious relationship to anything else you said.  In particular, your interpretation of it is ridiculous and has nothing to do with the conclusions of the study.

[quote="PKLove"]However, there are many activities where that's not the case. Activities that involve hard thinking such as mathematics are one of them. Surprisingly, there are many people who no matter how far they work hard never substantially improve in mathematics.

Not only that, but such stumbling blocks could be purely biological, as we now have [url=http://www.newsroom.ucla.edu/portal/ucla/more-proof-that-intelligence-is-85134.aspx]solid, empirical evidence that most of your intelligence is largely inherited.[/url] Basically meaning that working hard could be just as good as doing nothing.

So whenever I have a moment of "self-pity", it's because I face a dilemma: am I not good in math because I am applying myself in the wrong way, or is it because I am biologically limited?
[/quote]You're right, albeit to a certain degree. I am sure that not everybody has the intelligence to truly understand, say - quantum mechanics. But I'm sure that at least 99.99% of us have the intelligence to get a perfect score on the amc10 or to just finish all of their homework at school.

To motivate myself, I use to play a game as I was an avid sports fan. I would create a sheet with all kinds of statistics of how I was doing in each class listing all the test scores, etc. Then I had 'standings' section in which I gave myself a 'W' for every test higher than a 93 and a 'L' otherwise. Make it a game.

Other than that, you've been two of the best public high schools in the world. You'll find it easier to connect once you get going. Some say if you can work hard for 21 days straight it becomes habit.

Just rebumping this thread,
I'm having lack of motivation as well, and I'm a freshman.
In the first semester, I had a 4.0, but then I lost motivation to well, "do the work", because I didn't really learn anything out of it, ending up with 2 B's in the second semester, lowering it to a bad 3.6 GPA. (Maybe this was from worrying to much about 88.5 AMC 10 to USAJMO qualification in 6 months?). I've procrastinated alot during the second semester, and I don't understand why. My parent used to care, but now is sort of losing hope in the school system, since I'm not getting much out of it, and this lack of hope progressed very far. (In fact, my dad now doesn't care about my grades, next year's courses that I'm going to take, or really, just small things like driving me to school; he mainly watches TV and goes to online dating sites). After reading the statements, it does seem that I have the lack of motivation to work this time, but how do I get out of it? It's sort of getting to my mind that I'm one of the only ones living in a small apartment, and that I have to walk 3 miles of vehicle infested road to school, but still not learning much from it. It does sort of seem that my life is becoming more and more...imprisoned. I can't get out of this mindset just by playing with friends, since it doesn't work on me, and going the "stress" mindset only limits my mental capacity to school, losing USAMO experience. Any suggestions?
I mean, I can stay for long hours doing USAMO, because it's become a habit, but doing other school-related homework makes my mind feel trapped and limited.

It really depends on what your goals are. If you want to focus on USAJMO, then maybe you should go with that, but you need to keep in mind that colleges look at your GPA first. I think one reason why you lost motivation to do well after the first semester was because you already got a 4.0, and you probably thought that it's easy to get a 4.0, so you didn't try that hard in second semester. That happens to me sometimes. My advice is to still focus on schoolwork first and then work on USAJMO or other non-school-related stuff. Tell yourself that you can only do the USAMO stuff after you finish schoolwork. It will help you to not procrastinate as well.

Hope this helps!

Self-pity parties are terribly boring and they obviously fail the "will anyone else care" criterion of the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=138&t=214779]RT posting guidelines[/url].  The fact that you've decided to have yours on a thread on which discussion died more than two years ago is particularly odd.  (Note that your post does not directly relate to any of the 20-odd posts that precede it on this thread.)  Please don't do this.