geometry
calculus
ARML
analytic geometry
geometric transformation
rotation
reflection

OK guys so Wake Tech is coming up for all the people in the Triangle area... so I have some questions:

1. How much time is it? It seems to fluctuate randomly from year to year; sometimes they give us more, sometimes less... (also for the contest finals, especially comprehensive)

2. Scoring on the comprehensive: is it 4 for right multiple choice, -1 for wrong, and 10 for right short answer?

Thanks.

Also apparently 8 people get to go from Wake Tech according to the website (?) (for comprehensive)

EDIT: why did my post count randomly shoot up? Are they counting regional posts again?

I'm not sure. I think the test is around 1 hr, but this is my first year.

:furious:  :furious:  :furious:
I think the test is 75 min.

[quote="bisbmard9"]Yeah apparently they are. Hooray. This is pretty much the only place I post, so now I have a chance at 1000. Is Wake Tech the place where really good people get 9th and don't go to the state contest?[/quote]

If you consider Damien to be good, then yes.

Similarly, I also might reach 2500 posts (though not very soon)...

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=141149]Actually, it is a bit more extreme than that...[/url]

oh man I remember back in the day when charlotte used to be the hard place to be

and then I think elon became the hard place to be?

and now it's wake tech, huh

PRETTY CRAZY STUFF, GUYS

although you know what's not a really hard place to be

NCSSM

seriously what is with sending like 78% of your student population

what is up with that, guys

[quote="MysticTerminator"]haha I was going to correct him but whatever?

last year only 6 people made it  :o[/quote]

999 posts.Not bad.

I'm going to fail at elon  :o O well. I'm only in 8th grade  :). I bet I'll get like 3% correct  :mad: .

Well, good luck to all of you guys on Tuesday (I hope I make it into state Geometry!!!)

I have to make state algebra. Good luck to all.

Test just ended.

70 minutes by the way.

9 Algebra I students go on, 10 geometry, i think 7 algebra 2, and like 7 comprehensive or something like that.

I did horrible. I should have gotten like 3rd or 4th if I wasn't so stupid. We had scantrons (pretty annoying). I put the answer to question 68 into the answer for question 67. Funny thing was that I originally had the answer to 67 right... I feel awful. That stupid mistake pushed me to tenth.

Algebra 1:
3rd: Carnage MS
2nd: I think Salem MS.
1st: (not Ligon yay!!!) Smith MS.

Geometry:
3rd: Martin MS (I think)
2nd: West Cary MS
1st: Ligon MS (noooooooooo!!!!!!)

Algebra 2:
3rd: Can't remember
2nd: Can't remember
1st: Ligon MS (noooooooooo again!!!!!!)

Comprehensive:
I wasn't paying attention to these scores. I don't have time to memorize all the calculus nerds.

The food there wasn't all that good. The pizza was bad but they had twinkies for dessert (YAY!!!).

Individual Scores were:
Algebra 1:
10th: Can't remember
9th: Can't remember
8th: Can't remember
7th: Can't remember
6th: 7th grader from Carnage MS
5th: Can't remember
4th: 7th grader from Carnage MS
3rd: Can't remember
2nd: Can't remember
1st: Jeffery An (West Cary)

(BTW don't blame me for a bad memory)
Oh and the top 10 for Algebra 1 were all made up from Salem MS, Smith MS, Carnage MS, Ligon MS, and one person (Jeffery) from West Cary.

Geometry:
10th: Me (Apparently they put my school as some kind of imaginary school. They couldn't read their own had writing and thought I was from Carrington [Wherever that is] instead of Carnage.)
9th: Can't remember
8th: Can't remember
7th: Can't remember
6th: Can't remember
5th: Can't remember
4th: Can't remember (I should be here!!! :mad: )
3rd: Can't remember (I might have been here too!!! :mad: )
2nd: Alan Yang (Ligon MS)
1st: Calvin Deng :mad:  (Ligon MS)

Algebra 2:
10th: Can't remember
9th: Can't remember
8th: I think it was Tejas Sundarsun (that's how you spell it right?) (Ligon MS)
7th: I think it was Ivan somereallylonglastnamethaticantspellorremember (Ligon MS)
6th: Can't remember
5th: Can't remember
4th: Can't remember
3rd: Can't remember
2nd: Nick Tobey (Ligon MS) (WOOHOO!!! GO Nick!!! :D )
1st: Can't Remember

Comprehensive:
Can't remember any of these though I think I remember that Daniel Vitek (Enloe HS) got like third and Vivek somethinglong (Enloe HS) got first (Obviously. He got first just when he walked into the door. He's like the smartest person ever. I think he got like 8th in National MC or something...

Anyways... The state competition is going to be fun. Its at UNC at Greensboro on May 1. We get to miss the day of school and we get to tour the campus. They even have a hotdog and icecream shop (mmmmmm!!!). Good luck to everyone that made state!!! :D

Oh and another side note, there was a fun round in which we competed for money (cha-ching!!!). Alan Yang's group got first though... I feel so bad. My group was (not trying to be offensive or anything) really dumb. I was with a bunch of clueless Algebra 1 kids and they kept turning the paper in before I could tell them the right answer. I had to fight them for the sheet of paper on number one so I could write down the answer. It was almost as if they purposely were trying to lose!!! On one of the questions, they said it looked to hard so they simply put pi as the answer without even reading the question :mad: !!! They didn't even know how to add percentages. They though 20% of something plus 5% of the new number is 30% of the old number :fool: . Anyways, Every time they went up to turn in a wrong answer, I got the right answer almost two second after they left!!! I feel so mad!!! My group might have actually gotten like second or third if they weren't so uneager to get free cash!!! Anyways, all I am saying is that I feel really bad about the events of today...

[quote="MysticTerminator"]umm dude

why would you post all of that when there is little or no substance in your post[/quote]

dude

I agree wholeheartedly

[quote="aspiringchesslete"]I should have gotten like 3rd or 4th if I wasn't so stupid.[/quote]

if I [i][b]weren't[/b][/i] so stupid.

so what site was he blabbering about? was this wake tech?

[quote="MysticTerminator"]Umm...that is the only place in the U.S. where four states meet at one point, and those states occur in a definite order.  So once we define axes, there is only one right answer.  Sorry.

ON A COMPLETELY DIFFERENT NOTE

REAL RESULTS
----------------------------------------------------------------------
INDIVIDUAL
Place	Name	School	Score (if known)
1	Vivek Bhattacharya Enloe High School	                 140
2	Damien Jiang	      Cary Academy	                   136
3	Daniel Vitek	        Enloe High School	              135
4	William Schlieper     Green Hope High School	         135
5	Jeremy Hahn   	     Michaux Academy*	                125 (?)
6	Hyun Ji Bae	       East Chapel Hill High School
7	Jessie Duan	       Green Hope High School
8	Haoru Liu	        East Chapel Hill High School
9	Peitong Duan	      Green Hope High School
10	Patrick Yang	      Enloe High School	                     78
11	James Choi	      Raleigh Charter High School
12	Katherine Rettig      H.O.M.E.
13	Sean Reynolds	     Fuquay-Varina High School
14	Eason Lee	      Green Hope High School
15	Claire Zeng	      Enloe High School	                      69
16	Akhil Jariwala	       Raleigh Charter High School
17	Deanna Wung	   Green Hope High School
18	John Hanna	     Raleigh Charter High School
19	Brian Simel	      Jordan High School
20	Saumil Jariwala	     Raleigh Charter High School
21	Julie Ma	       Enloe High School	                55
22	Kevin Munger	    Raleigh Charter High School
23	Shayon Saleh	    Green Hope High School
24	David Gu	      Enloe High School	                       50

*This is what Jeremy calls his home, which is where he gets schooled.

TEAM
1. Enloe High School
2. Green Hope High School
3. East Chapel Hill High School
4. Raleigh Charter High School
5. Cary Academy

Nothing too surprising.  After the first 6 I would say that both scores and ARML experience drop rapidly.  :lol:[/quote]

Hey, number 23, that's me! Our number 1, William Schlieper, was a beast (Ranked #4) that got into AIME and USAMO.

Still, the ranked #1 is a pretty crazy guy. Valedictorian, Siemens Competition Semifinalist... Sheesh. I thought mathematicians were lazy.

Actually, I felt lost for a majority of that test... *sigh* I was quite surprised by my placement.

... why would you think mathematicians are lazy

that's got to be the randomest stereotype ever

[quote="b-flat"]Actually, I have heard that one before; I'm pretty sure the logic is that mathematicians are lazy because they always try to make things simpler and reuse things when possible (i.e. use theorems, etc)...[/quote]

Basketball players use the same tactics forever and noone is calling them lazy

[quote="PI-Dimension"]Basketball players use the same tactics forever and noone is calling them lazy[/quote]

I did not say that that was correct logic; I only said that that was the logic (I happen to not agree with it, but that is irrelevant).  There was also something about them always trying to do the least amount of work (aka summation notation, silly other things like that)...

[quote="MysticTerminator"]oh

if that's what you call lazy, then yeah ok

that's totally unrelated to what the dude was applying it to though[/quote]

Agreed.

[quote="dnkywin"][quote="shayonsaleh"]Sheesh. I thought mathematicians were lazy.

[/quote]

*points to PI-Dimension's sig*[/quote]

Lolz

[quote="Teki-Teki"]Whatever happened to minimum posting word limits?  Such a shame.  :lol:[/quote]

A quote is ALOT more than 1 word XD

[quote=dnkywin][quote="shayonsaleh"]Sheesh. I thought mathematicians were lazy.

[/quote]

*points to PI-Dimension's sig*[/quote]

A wild mod has appeared  :w00t:

$                   $


[quote="Rep123max"]Hey, this problem is annoying me. I can't figure out how to get it into a determinable form. Any suggestions?

$\displaystyle\lim_{x\rightarrow0}\left(\frac{1}{x}\right)\left(\frac{1}{\sqrt{1+x}}-1\right)$.

THANKS[/quote]

This isn't a particularly easy question, it took me about 10 minutes to figure out how to get the answer to fall out.  And when I did find it, I felt really stupid.

We use conjugates, of course.

Let j = 1/sqrt(1+x)-1

Then j*conj = 1/(x+1)-1 = -x/(x+1).  From here, it's pretty straightforward.

[quote="zoso"]I just finished a unit on limits (i'm taking my first calc course) but this seemed right to me:
Let

$\sqrt{1+x} = r$

then

$\lim_{x\to\(0}\displaystyle \left( \frac{1}{x}\right)\displaystyle \left(\frac{1}{\sqrt{1+x}}-1 \right) = \lim_{x\to\(0}\displaystyle \left( \frac{1}{r^2 - 1}\right)\displaystyle \left(\frac{1}{r}-1 \right)$

$= \lim_{x\to\(0}\displaystyle \left( \frac{1}{r^2 - 1}\right)\displaystyle \left(\frac{1-r}{r}\right) = \lim_{x\to\(0} \frac{-(r-1)}{r(r-1)(r+1)}
$

$= \lim_{x\to\(0} \frac{-1}{r(r+1)} = \lim_{x\to\(0} \frac{-1}{\sqrt{1+x} (\sqrt{1+x}+1)}
$

$= -\frac{1}{2}$

Is this okay? I'm unsure...[/quote]

It's correct, and a very nice way of doing the question.

Can you use l'Hopital?

Edit:  I got -1/2 with l'Hopital.

$\displaystyle\lim_{x\rightarrow0}\left(\frac{1}{x}\right)\left(\frac{1}{\sqrt{1+x}}-1\right)$ is the definition for the derivative $\displaystyle\frac {d}{dt}\left(\frac {1}{\sqrt{t}}\right)$ at $t=1$.  Just another way to get the same answer.


quadratics
inequalities
function
algebra
algebra solved
JBMO
inequalities solved

The determinant of the quadratic is $a^{2}-4(a^{2}-6)=3(8-a^{2})$ which must be $\ge 0$ if the solutions are real. but then we must have $a\le 2\sqrt 2$. so we have $6=a(a^{2}-6)\le 2\sqrt 2(8-6)=4\sqrt 2$, contradiction.

[quote="kunny"]We can also rewrite the given equation as $ax(x+a)=-6.$ This doesn't 0 or positive roots, now if this has negative roots, .....[/quote]

Shouldn't it be $ax(a+x)=-1$?
And I still don't see your point... :oops:

[quote="behemont"][quote="kunny"]We can rewrite the given quadratic equation as $x^{2}+ax+\frac{6}{a}=0$.
$D=a^{2}-\frac{24}{a}=\frac{a^{3}-24}{a}=\frac{6(a-3)}{a}$.[/quote]

Why are we done here?[/quote]

well, from what i understood, since $D\ge 0$, we have $a\ge 3$. but this implies  $a^{3}>6(a+1)$.

[quote="kunny"]No, you are wrong. :)[/quote]

God bless my multiplying..  $6*1=1$ :)

Note that the equation $a^{3}-6a-6=0$ has root the $a_{1}=\sqrt[3]{2}+\sqrt[3]{4}$

 :wink:

I just saw the topic.

My solution:

If there were real roots, then the discriminant should be positive or equal to 0. That is, according to the equation given in the problem:

$ a^{2}-4(a^{2}-6)\geq0\implies-3a^{2}+24\geq0\implies a^{2}\leq8\implies a\leq2\sqrt{2}\ (1)$ (since a is positive)

Since a is positive, we can multiply both sides of the inequality $ a^{2}\leq 8$, without changing anything. As a result, we would have:

$ a^{2}\leq 8\implies a^{3}\leq 8a\stackrel{hyp}{\implies}6(a+1)\leq 8a\implies 6\leq 2a\implies a\geq 3\ (2)$

From (1), (2) we have:

$ 3\leq a\leq 2\sqrt{2}$

But $ 2\sqrt{2}<3$, so we are forced to conclude that there are no real roots of the above equation and we are done.

Excuse me if I have missed something.

My solution for problem 1 :
 We have :        α3=6(α+1)α3-9α+3α-9+3=0
α(α-3)(α+3)+3(α-3)+3=0 (α-3)(α2+3α+3)=-3  and we have  α-3<0.  And α3=6(α+1) α3-8α=-2α+6α(8-α2)=2(α-3) then 8-α2<0 and the equation has no real solutions .

Let, the equality x^2+ax+a^2-6=0 has real root. Then, D=3(8-a^2)>=0.So,a^2<=8,but a is positive integer
So,a=1 or a=2. If a=1, a^3=6(a+1) doesn't take place.
If a=2, a^2+6(a+1) isn't correct.
So,the roots of equality  x^2+ax+a^2-6=0, aren't real.

lasha , your solution is very simple if  a is positive integer. But  a
is a real positive number.

[color=darkblue][b]Proof I (generally).[/b] Show easily that the function $ \{\begin{array}{c}f\ : \ (0,\infty )\rightarrow\mathcal R\\\ f(x)=x^{2}(x+1)\end{array}$ is strict increasing.
Therefore, the equation $ f(\frac{1}{a})=\frac{1}{6}$, i.e. the equation $ a^{3}=6(a+1)$ has at most one positive root.
Observe that for the function $ \{\begin{array}{c}g\ : \ (0,\infty )\rightarrow \mathcal R\\\ g(a)=a^{3}-6(a+1)\end{array}$ there is the relation $ g(2\sqrt 2)=2(2\sqrt 2-3)<0<3=g(3)$, i.e. $ g(2\sqrt 2)<0<g(3)$.
In conclusion, the single root $ a\in (2\sqrt 2, 3)$. The equation $ x^{2}+ax+a^{2}-6=0$ has no real roots iff $ a^{2}> 8$, what is truly.

[b]Proof II[/b] (similarly to [b]Mostel[/b]). $ f(a)=a^{3}-6(a+1)$ $ \implies$ $ f(0)\cdot f(3)<0$ $ \implies$  $ (\exists)\ a>0$ so that $ a^{3}=6(a+1)$. But
$ 0<a\le 2\sqrt 2$ $ \implies$ $ a^{2}\le 8\implies a^{3}\le 8a$ $ \implies 6(a+1)\le 8a\implies 3\le a$, what is falsely. Thus $ a>2\sqrt 2$.
$ a\ge 3\implies$ $ a^{3}\ge 3a^{2}\implies$ $ 6(a+1)\ge 3a^{2}\implies$ $ 3a\le a^{2}\le 2(a+1)\implies$ $ a\le 2$, what is falsely. Thus $ a<3$.
In conclusion, $ a\in (2\sqrt 2,3)$ and the equation $ x^{2}+ax+a^{2}-6=0$ has no real roots iff $ a^{2}> 8$, what is truly.

[b]Remark.[/b] $ a^{3}-8=6a+6-8=2(3a-1)\implies (a-2\sqrt 2)(3a-1)>0$ $ \implies a\in (0,\frac{1}{3})\cup (2\sqrt 2,\infty )$.
If $ a\in (0,\frac{1}{3})$ then $ 6<6(a+1)=a^{3}<\frac{1}{27}$ $ \implies$ $ 6<\frac{1}{27}$, what is falsely. In conclusion, $ a>2\sqrt 2$ a.s.o.[/color]

[quote="pohoatza"]Let $ a$ be positive real number such that $ a^{3}\equal{} 6(a\plus{}1)$. Prove that the equation $ x^{2}\plus{}ax\plus{}a^{2}\minus{}6 \equal{} 0$ has no real solution.[/quote]

D=$ 24\minus{}3a^{2}$

If we suppose that $ 24\minus{}3a^{2}\geq 0$ we have a^2=<8. But from  $ a^{3}\equal{} 6(a\plus{}1)$ we have $ a^{3}\minus{}6a\equal{}6$, so must $ a^{2}\geq 6$.

Now we have contradictions in $ a(a^{2}\minus{}6)\equal{}6$, because no have solutions this equation if $ a$ is positive number and $ 6\equal{}<a^{2}\equal{}<8$! (is very easy to proof)

$ \delta \equal{} 24\minus{}3a^{2}$.

We have $ a^{3}\equal{} 6(a\plus{}1)$

Let $ f(x) \equal{} a^{3}\minus{}6a\minus{}6$.

$ f'(x) \equal{} 3x^{2}\minus{}6$. $ f'(x) \equal{}\pm\sqrt{2}$ then $ f(x) < 0,\forall x\in (\minus{}\infty,\sqrt{2})$ and $ ,f(x)\nearrow ,\forall x\in [\sqrt{2},\plus{}\infty)$

But $ f(2\sqrt{2}) < 0$ then if $ f(x) \equal{} 0$ so $ x > 2\sqrt{2}$

Hence $ a > 2\sqrt{2}$ so $ \delta < 0$ ,we're done  :)

[quote="pohoatza"]Let $a$ be positive real number such that $a^{3}=6(a+1)$. Prove that the equation $x^{2}+ax+a^{2}-6=0$ has no real solution.[/quote]
Sorry for taking back this topic
Here is my solution
from this equation $a^{3}=6(a+1)$ from kardanos formula we have
$a=2^\frac{1}{3} +4^\frac{1}{3} $
and $a^{2}-6=\frac{6}{a}$
and we have the equation
$ax^{2}+a^{2}x+6=0$
$D=a^{4}-24a\ge0$ ==>    $a<0$ or $a>2\cdot3^\frac{1}{3}$
but from this inequality
$(n-1)^{\frac{1}{3}}+(n+1)^{\frac{1}{3}}<2(n)^\frac{1}{3}$
from this we get this equatin has no real solution

[hide="YAY"]
So, first we look at the discriminant of the quadratic which is just $24-3a^2$. For the roots to be complex $8-a^2<0$
This should mean that $|a|>2\sqrt{2}$. But we know that the root must be positive so we have that $a>2\sqrt{2}$ and also since $a^3-6a-6$ is increasing when $a>\sqrt{2}$ we know that $a$ must be greater than $2\sqrt{2}$. So we are done.
[/hide]
.Are JBMO problems supposed to be easy?

For $a>0$ obtain that $a^3=6(a+1)$ $\implies$ $a\left(a-2\sqrt 2\right)\left(a+2\sqrt 2\right)=2(3-a)$ $\implies$ $\left(a-2\sqrt 2\right)(3-a)>0$ $\implies$ $a\in\left(2\sqrt 2,3\right)$ . Observe that $\Delta\equiv a^2-4\left(a^2-6\right)=-3\left(a^2-8\right)<0$ . In conclusion, the mentioned equation $x^{2}+ax+a^{2}-6=0$ has no real solution.

OK, i will try with elementary solution

Assume that, $x^{2}+ax+a^{2}-6=0 $ have real solution. So,

$a^2 - 4(1)(a^2 - 6) \ge 0$
$a^2 \le 8$

So, $8a \ge a^3 = 6(a+1) \implies a \ge 3 \implies a^2 \ge 9$. Contradiction with $a^2 \le 8$. CMIIW

[quote="youarebad"]$a^2 \le 8\implies 8a \ge a^3$ [/quote]Only for $a\ge 0$ ....

[quote="pohoatza"]Let $a$ be positive real number such that $a^{3}=6(a+1)$. Prove that the equation $x^{2}+ax+a^{2}-6=0$ has no real solution.[/quote]

Proof 1 :

$a\geq 3\Longrightarrow a^3-6a-6=(a-3)(a^2+3a+3)+3>0\ \therefore 0<a<3$.

$x^2+ax+a^2-6=0\Longleftrightarrow (a+1)x^2+a(a+1)x+(a+1)(a^2-6)=0$

$\Longleftrightarrow (a+1)x^2+a(a+1)x+a^3-6(a+1)+a^2=0$

$\Longleftrightarrow (a+1)x^2+a(a+1)x+a^2=0$

$\therefore D=a^2(a+1)^2-4a^2(a+1)=a^2(a+1)(a-3)<0\ Q.E.D.$

Proof 2 :

$a\geq 3\Longrightarrow a^3-6a-6=(a-3)(a^2+3a+3)+3>0\ \therefore 0<a<3$.

$x^2+ax+a^2-6=0\Longleftrightarrow x^{2}+ax+\frac{6}{a}=0$.

$\therefore D=a^{2}-\frac{24}{a}=\frac{a^{3}-24}{a}=\frac{6(a-3)}{a}<0\ Q.E.D.$

Proof 3 :

$0<a\leq 2\sqrt{2}\Longrightarrow a^3-6a-6=(a-2\sqrt{2})(a^2+2\sqrt{a}+2)+2(2\sqrt{2}-3)<0$

$\therefore a>2\sqrt{2}$.

$\therefore D=3(8-a^2)=3(2\sqrt{2}+a)(2\sqrt{2}-a)<0\ Q.E.D.$

By the way, for what grades ,or ages, is the problem posed in JBMO?

[quote="behemont"][quote="kunny"]We can rewrite the given quadratic equation as $x^{2}+ax+\frac{6}{a}=0$.
$D=a^{2}-\frac{24}{a}=\frac{a^{3}-24}{a}=\frac{6(a-3)}{a}$.[/quote]

Why are we done here?[/quote]

because we can see that to have a real solution , we just prove that if D > 0 no solutions exist

:)

[hide="Solution"]The discriminant of the given quadratic is $a^2-4(a^2-6)=24-3a^2=3(8-a^2)$. Note that $a$ must be less than $3$ because $a^3-6(a+1)=(a-3)(a^2+3a+3)+3$; if $a$ were greater than $3$, then the RHS and thus the LHS would always be positive, making it impossible to have a positive real root that satisfies the equation. Thus, $$2a<6\Longrightarrow8a-6a<6$$$$\Longrightarrow6a+6>8a$$$$\Longrightarrow a^3>8a$$$$\Longrightarrow a^2>8$$ because $a$ is positive. Hence, $8-a^2<0$ and the discriminant is always negative. So there are no real roots.[/hide]

Solving the given equality by Tartaglia, we get "a" about 2.847, but the equation only has solution if a <= 2.sqrt(2) <= 2.83, so the equation hasn't solution. Q.E.D.

[quote=Devastator][hide=my solution]Proof by contradiction
Assume it has roots
Then the discriminant is nonnegative
So a^2-4(a^2-6)>=0
Which implies a^2=<8
Since a is positive, we can multiply both sides by a
So a^3=<8a
Implies 6a+6=<8a
Implies a>=3
Implies a^2>=9
Contradiction with a^2=<8
Therefore the discriminant is not nonnegative, or it is negative
So it has no real roots[/hide]
Is this right? Please check cause I have found holes in my proofs every now and then.[/quote]

Nice proof, yes i think that's right.

[hide]My solution:
First,let's take control on "that" equation.From the condition,we have to show that the equation has no real roots.

We have: $\Delta=3(8-a^{2})\Leftrightarrow a>\sqrt{8}$
So we have to show,that when $a^{3}=6(a+1)\Rightarrow a>\sqrt{8}$
Assume contrary:suppose $a\leq\sqrt{8}$
Then,by the condition $6(\sqrt{8}+1)$ should be equal to $\sqrt{8^{3}}$,contradiction.And the conclusion follows.[/hide]

In order to have no real solution, we must have $a>2\sqrt2$. By AM-GM, \[a^3=3a+3a+6\geq 9\sqrt[3]{2a^2}\implies a\geq \sqrt[7]{9^3\cdot 3}\] so it is suffices to check that \[\sqrt[7]{9^3\cdot 3}> 2\sqrt{2} \iff 3^{12}>2^{19}\iff 531441>524288\]

First we claim that there are no negative solutions to $a^3 = 6a+6$. If $-1<a<0$, then $a^3$ is negative and $6a+6$ is positive, so this is not possible. If $a$ is $-1$, then the LHS is greater than the RHS, and increases much faster. Now note that for the equal to have no real solution, we need $a^2-4(a^2-6) \leq 0$, or $a^2 \geq 8$. This means it suffices to show that $a \geq 2\sqrt2$. We can see that starting at $0$, the RHS is greater than the LHS, and it will continue to be greater until both sides are equal, creating the first root. We simply have to check whether $2\sqrt2$ makes the RHS greater than the LHS. It does, so the value of $a$ is greater than $2\sqrt2$, as desired.

Because the determinant of the given equation is $$a^2 - 4 \cdot 1 \cdot (a^2 - 6) = 24 - 3a^2$$ it suffices to show $a^2 > 8$.

Let $x \ge y \ge z$ denote the roots of $a^3 - 6a - 6 = 0$, i.e. the solutions to $a^3 = 6(a + 1)$. Vieta's yields $x + y + z = 0$ and $xyz = 6$. Inspection gives $x, y, z \ne 0$, so the sum condition implies either $$x \ge y > 0 > z$$ or $$x > 0 > y \ge z$$ is true. But $xyz$ is positive, so we conclude that the second condition must hold.

Hence, if $k$ is a positive real between $0$ and $x$, we have $$6 = xyz = x \cdot (-y) \cdot (-z) = x \cdot k \cdot (x - k) \le x \cdot \left(\frac{x}{2} \right)^2 = \frac{x^3}{4}$$ which implies $$x^2 \ge 4 \sqrt[3]{9} > 4 \sqrt[3]{8} = 8.$$ Because $a$ is positive, we know it cannot equal $y$ or $z$, so we're done. $\blacksquare$

Let me supply a proof which covers the scenario when $a$ is a (not necessarily positive) real number with $a^3 = 6(a+1)$.

It suffices to show $D = 24 - 3a^2 < 0$, i.e. $a^2 > 8$. If we suppose $a^2 \leq 8$, then since $a^3 - 6a \leq 4\sqrt{2}$ (equivalent to $(a-2\sqrt{2})(a+\sqrt{2})^2 \leq 0$, which works for $a^2 \leq 8$) and $4\sqrt{2} = \sqrt{32} < 6$, we have obtained a contradiction.


[quote="livetolovemath030894"]Let $ ABC$ be a isoceles triangle. $ J$ be incenter of $ ABC$. $ E$ be a point move $ (JBC)$. the parallel line from $ E$ to $ BC$ intersect $ AB$,$ AC$ at $ F$,$ G$ respectively. The parallel line from $ E$ to $ AB$,$ AC$ intersect $ BC$ at $ I$,$ H$ respectively. Let $ D$ be the intersection $ GH$ with $ FI$. prove that $ ABCD$ is cyclic[/quote]
You can see your problem and its solution at this topic: http://www.mathlinks.ro/viewtopic.php?p=1630947#1630947


$0\times3\neq0\times5$
I'm korean. nice to meet yo man.

gj i'm korean too

there aren't that many korean people here i don't think, most of them are chinese or indian lol


factorial
number theory
prime factorization

The prime factors of a positive integer $n$ can be multiplied in $60$ different ways. If $n<1000$, how many such integers are there?

I'm not quite sure I understand the question
Are you saying that n has 60 factors?

No. The factors of $n$ can be multiplied in $60$ different ways.

For example, if you have the number $24$, whose prime factorization is $2^{3}\cdot3$, you can do $2\cdot2\cdot2\cdot3$, or $2\cdot2\cdot3\cdot2$, etc.

Partial Solution

The number of ways to rearange n objects, with p,q,r,... non-distinct objets, you can arrange them in n!/(p!q!r!...) ways.

Let the objects be the factors of the number, and same factors be the non-distinct objects.

Then, n must be between 5, and 9, since 10 factors would make the number over 1000, and 4! is only 24, and dividing by anything won't change that it's less than 60.

Table
5! divided by 2
6! divided by 2*6
7! divided by 2*6*7
8! divided by 2*6*7*8
9! divided by 2*6*7*8*9

Now, we just need to find all possible products of factorials, which multiply to the numbers in the divided by column, and take every possible combination of factors (eughh).

I don't know where to approach from here besides brute force.


calculus
integration
geometry
similar triangles

One of my friends is on the marking committee for the Western region. He told me some very funny things.

For Hypathia #2, most people were able to use similar triangles to deduce the answer 2/3, but he can't even remember how many of them wrote 2/3=0.7.

Most people can't do proofs, and a lot of the Hypatia scores were 10-10-10-2.

How much does 4 c worth approximately?  Although I got it, I didn't have time to write it well enough.

Anyone know when the results will be released?

Doesn't surprise me.

A lot of people probably tried 'case' arguments for 4 (I saw it because I write up solutions for younger kids at my school) and avoided the nice, neat, algebraic solution which basically says something like 3z=2 has no integral solution.

No one from my school was close to perfect, which surprised me (I was expecting at least two scores to be close to perfect).

Last year, I got 39, which kinda pissed me off because I was so sure everything was correct.

i got 32 last year, i suck... i did bad on comc, and euclid....
i got 52 on comc
and euclid probably 7*....
=*(


Can someone give me a little hint for this homewok problem?

Prove the statement using the $\epsilon,\delta$ definition of limit:
\[
\lim_{x \to -2} (x^2-1) = 3
\]

Well, I'll just rephrase what you have to do.  Given $\epsilon > 0$, find $\delta > 0$ such that
\[|x + 2| < \delta \implies |x^2 - 4| < \epsilon.\]

In other words, if x is near -2, then $x^2$ is near 4.  Can you prove that?

Another hint: $|x^2-4|<\epsilon\Leftrightarrow-\epsilon<x^2-4<\epsilon$.

[hide](edit: deleted because my solution contained many many errors)[/hide]

I'll play off of Ravi's hint.

We want to make $(x-2)(x+2)$ very small.  By choosing a small $\epsilon$, we can make $x+2$ as small as we want.  Now, if we make $x+2$ small, we are actually bounding the size of $x-2$; say, that if $\vert x+2\vert<1$, then certainly $-3<x-2<-1$.  Can you finish it off from here?

Yeah, I was able to finish the problem - thanks - but now I'm stuck on another problem:

Verify, by a geometric argument, that the largest possible choice of $\delta$ for showing that $\lim_{x \to 3} x^2 = 9$ is $\delta = \sqrt{9+\epsilon} - 3$.

It'd be nice if a problem somewhat related to this was mentioned in the reading...

Hmmm... there are two possible things I can think of that they are getting at:

1) The graph of the function
2) A square (shape)

I don't see anything, however, that is any more intuitive than the algebraic way.  Sorry, I think I'm not being very helpful :(

It's a square.  On the base we have $x-\epsilon, x, x+\epsilon$, on the height we have $f(x)-\delta,f(x),f(x)+\delta$.


Have anyone solved it? The one w/ Triangle ABC has area 1. D is on BC, E on CA, F on AB and AFDE is cyclic. Prove that (area of DEF) is smaller or equal to EF^2/(4AD^2)

Discussion of those challenge problems are at the Advanced Problems section. The thread for the one that you mentioned is at:

http://www.artofproblemsolving.com/wforum/viewtopic.php?t=577


Hello, I have been solving this problem and got result, but I am not sure if it's correct.

Problem: You have 10x10 quadratic table. You are in the field in bottom left corner. You need to reach the upper right corner field (diagonal one). You can move up and right as many times as you want. In how much ways you can do it? :)

First I have observed smaller cases, like 2x2, 3x3, 4x4 and 5x5, found a sample and continued to 10x10.

My answer is 48620 ways.

[hide="Solution"]This is simply $ \dbinom{20}{10}\equal{}\boxed{184756}$.[/hide]

So your answer is incorrect.

[quote="pytheagle"][hide="Solution"]This is simply $ \dbinom{20}{10} \equal{} \boxed{184756}$.[/hide]

So your answer is incorrect.[/quote]

I cant seem to understand why did you use that, but can you tell me instead, how much ways there are for 5x5 quad. table, using your calculation?  :)

You have to go right 10 times, and up ten times, so out of 20 moves, 10 of them have to be right.  So, our answer is just the number of ways to choose 10 of our 20 steps to be the right, or $ 20 \choose 10$.

So in a 5 x 5 table it would just be $ \binom{10}{5}$

Well, In 5x5 result is 70, so it's not 10! / (5!)^2.... You can feel free to count them, it will take 2-3 mins only.

In 2x2 table there are 2 ways.

In 3x3 table there are 6 ways.

In 4x4 table there are 20 ways.

In 5x5 table there are 70 ways.

In 6x6 table there are 252 ways.

In 7x7 table there are 924 ways.

In 8x8 table there are 3432 ways.

In 9x9 table there are 12870 ways.

In 10x10 table there are 48620 ways.

Well I don't know how you're counting them but I'm pretty sure you're missing some.

This topic is about the same thing >> http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1586568#1586568

[quote="Tomekk"]In 5x5 result is 70, so it's not 10! / (5!)^2[/quote]  $ \frac {8!}{4!4!} \equal{} 70$ and also $ \frac {18!}{9!9!} \equal{} 48620$ -- you all just don't agree on what a 10 by 10 table is, apparently. :|  :roll:

Ohhh wow lol.  So I guess that's it then.


Consider the sequence of numbers defined recursively by $t_{1}=1$ and for $n>1$ by $t_{n}=1+t_{(n/2)}$ when $n$ is even and by $t_{n}=1/t_{(n-1)}$ when $n$ is odd. Given that $t_{n}=19/87$, the sum of the digits of $n$ is

(A) 15
(B) 17
(C) 19
(D) 21
(E) 23

[hide]
Construct it backwards, if $t_{n}> 1$, then $t_{(n/2)}= t_{n}-1$, and if $t_{n}< 1$, then $t_{(n-1)}= \frac{1}{t_{n}}$. So every time we subtract $1$ the index doubles and every time we take the reciprocal the index increases by $1$. The sequence is then

$\frac{19}{87}, \frac{87}{19}, \frac{11}{19}, \frac{19}{11}, \frac{8}{11}, \frac{11}{8}, \frac{3}{8}, \frac{8}{3}, \frac{2}{3}, \frac{3}{2}, \frac{1}{2}, 2, 1$

where I subtracted $1$ multiple times in some steps. Then the index of each term (in reverse) is

$1, 2, 3, 6, 7, 28, 29, 58, 59, 118, 119, 1904, 1905$

so the answer is $1+9+0+5 = 15$.
[/hide]


\[\begin{gathered}\text{Fie }f: M_{3}(R) \to M_{3}(R),f(X) = AX \hfill \\ A = \left({\begin{array}{*{20}c}{11}& p &{13}\\{121}&{p^{2}}&{169}\\{1331}&{p^{3}}&{2197}\\ \end{array}}\right). \hfill \\ \text{Pentru ce valori ale lui p ,f este bijectiva?}\hfill \\ \end{gathered}\]

Daca $A$ nu este inversabila atunci $f(X)$ nu este inversabila deci nesurjectiva.Deci $A$ trebuie sa fie inversabila.
Daca $A$ e inversabila atunci $\forall B\in M_{3}(R)$ putem alege $X=A^{-1}B$ pentru surjectivitate, injectivitatea fiind evidenta.
Conditia este $\det(A)\neq0$ adica $11\cdot 13\cdot p\cdot (p-11)\cdot (13-11) \cdot (13-p)\neq 0$adica $p\neq0,11,13$.

Cand ai ales matricea $B$ e ca si cum ai spune ca ai ales o functie $g$ astfel incat $g(X)=A^{-1}X$ .Tu presupui insa ca $g$ este surjectiva(mai exact din $X=A^{-1}B$ spui ca obtii toate matricile $X$ din multimea matricilor, pentru $B$ variabil).Sa zicem ca injectivitatea este evidenta.$g$ este inversa lui $f$.daca presupui ca $g$ este surjectiva atunci este si bijectiva deci si inversa, $f$, este bijectiva.Nu sunt convins ca este o abordare corecta.

Spui ca injectivitatea este evidenta din urmatoarea idee:din $f(X1)=f(X2)$ rezulta ca $X1=X2$ din $A(X1-X2)=O3 ,\forall X1,X2\in M_{3}(R)$?

Matricea $B$ este fixata.Si vreau sa arat ca ecuatia $f(X)=B$ are soltuti $\forall B$ , ceea ce este definitia surjectivitatii. Si am gasit solutia $A^{-1}\cdot B$, $B$ este fixata, nu variabila.

Poate nu am fost eu destul de clar.
Sa studiem echivalenta $A$ este inversabila daca si numai daca $f$ este bijectiva.
Intr-un sens:
  Daca $f$ este bijectiva atunci $\exists X_{A}\in M_{n}(R)$ astfel incat $f(X_{A})=I_{3}=A\cdot X_{A}$. Deci $A$ este inversabila.
In celalalt sens:
  Daca $A$ este inversabila atunci din $f(X)=f(Y)$ rezulta ca $AX=AY$ si inmultind cu $A^{-1}$ din stanga rezulta $X=Y$,deci $f$ este injectiva. Apoi, $f$ este surjectiva daca $\forall B\in M_{3}(R),\exists X\in M_{3}(R)$ astfel incat $f(X)=B$ adica $X=A^{-1}\cdot B$ de unde $f$ este surjectiva, deci bijectiva.
Mai departe cred ca e clar.


quadratics
advanced fields
advanced fields unsolved

For $n,$ how many distinct equivalence relations are there on the set $\{1,2,\dots,n\}?$

For $n=1,2,3,4,5,$ the answers are $1,2,5,15,52.$

Find a useful and tractable recursive method of computing this number, and demonstrate it by computing the number for $n=12.$

---

No, as I discovered [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63469]here[/url], this isn't the same as the number of topologies. Unlike that one, this question does have an answer.

The sequence is well, known, since it's just the number of partitions of the set into subsets.
So one method would be to add Stirling numbers of second kind (should be a quadratic complexity).
Another one would be:
Let's write $s_n$ for that numbers and let them already been calculated for smaller values as $n$.
Take any subset of $\{1,2,3...,n-1\}$, add ${n}$ to that set and partition the rest. Now there are $\binom{n-1}{k}$ ways to choose $k$ numbers and $s_{n-k-1}$ possibilities to partition the rest. It's also easy to see that every partition is reached by that and that none occure twice.
So define $s_0=s_1=1$ and get
$s_n = \sum_{k=0}^{n-1} \binom{n-1}{k} s_{n-k-1} = \sum_{k=0}^{n-1} \binom{n-1}{n-k-1} s_{n-k-1} = \sum_{k=0}^{n-1} \binom{n-1}{k} s_{k}$.
But that has also quadratic complexity...
And sorry, but I was to foolish to calculate.

ZetaX: I was computing using the Stirling numbers of the second kind. I did go and try your scheme, and the results do agree, completely. In a procedural programming language, they'd probably be about the same, but the Stirling numbers are a little easier to implement in a spreadsheet.

As a reference, the value for $n=12$ is [hide="this."]$4213597$[/hide]


In an open book, the product of the two facing page numbers is 1190. What is the sum of the page numbers on the two facing pages?

The two numbers are clearly one apart.

1190=10x7x17=5x2x7x17=34x35 so 34+35=69.


Lilly has a brown bag, a red bag, a green bag, a yellow bag, and two identical blue bags. If she wants to distribute 8 identical chocolate chip cookies into the 6 bags, how many distinct ways are there for her to do it?

Can someone please post a clarification on the dividers idea used in the solution?
Thanks a lot.

[quote="lena_siz"]Can someone please post a clarification on the dividers idea used in the solution?[/quote]
Here is how the dividers idea works:
[hide] Consider a simpler case where you have 4 identical objects to be placed into 3 distinct bags.  If we use a notation where $ o$ represents an object and $ |$ represents a "divider", then a valid distribution of the objects will be a permutation of the string $ oooo\parallel{}$, where the number of $ o$'s to the left of the first $ |$ represents the number of objects in the first bag, the number of $ o$'s between the $ |$'s represents the number of objects in the middle bag, and the number of $ o$ to the right of the last $ |$ represents the number of objects in the last bag.  Because there is a bijection (the correspondence between permutation and valid distributions is one-to-one AND all valid distributions are accounted for), the number of ways to distribute the 4 identical objects into 3 distinct bags will be equal to the number of permutations, which is $ {{6}\choose{2}}$.

You can generalize the $ m$ identical objects into $ n$ distinct containers as $ {{m\plus{}n\minus{}1}\choose{n\minus{}1}}$.

NB: the objects must be identical (since two objects in different containers could be swapped to yield a different distribution but the same string) and the bags must be distinct (or $ oooo\parallel{}$ would be the same distribution as $ |oooo|$ and $ \parallel{}oooo$ which are different strings.[/hide]

FYI, I actually did the problem in a different way:
[hide]
which was to consider the cases where the number of cookies in the identical bags were the same vs. different.  Basically, if we could ensure that the number of cookies in blue bag 1 was different from the number of cookies in blue bag 2, then we know that if we did the computation with 6 distinct bags, we would double-count each configuration.  However, there are cases where the number of cookies is the same in blue bag 1 and blue bag 2, and you can figure out how many of these cases there are by distributing the leftover cookies to the remaining $ 4$ distinct bags.

The number of configurations for $ 8$ identical cookies in $ 6$ distinct bags is $ {{13}\choose{5}} \equal{} 1287$.  The number of configurations where the number of cookies is the same in blue bag 1 and blue bag 2 is $ {{11}\choose{3}} \plus{} {{9}\choose{3}} \plus{} {{7}\choose{3}} \plus{} {{5}\choose{3}} \plus{} {{3}\choose{3}} \equal{} 165\plus{}84\plus{}35\plus{}10\plus{}1 \equal{} 295$.  So the number of configurations that were double-counted is $ 1287\minus{}295 \equal{} 992$, meaning the number of unique configurations is $ 992/2 \plus{} 295 \equal{} 496\plus{}295 \equal{}$ $ \fbox{791}$

Or in a more simplified format $ (1287\plus{}295)/2 \equal{} 791$[/hide]

[quote="Jessicatian"]I dont get when you said

The number of configurations where the number of cookies is the same in blue bag 1 and blue bag 2 is  $ {{11}\choose{3}}+{{9}\choose{3}}+{{7}\choose{3}}+{{5}\choose{3}}+{{3}\choose{3}}= 165+84+35+10+1 = 295 $  So the number of configurations that were double-counted is $ 1287-295 = 992 $, meaning the number of unique configurations is $ 992/2+295 = 496+295 = $$ \fbox{791} $[/quote]

I'm not sure I do either, it was a long time ago! :)

To clarify, if both blue bags contain $0$ cookies, the number of ways to distribute the remaining cookies among the remaining bags is ${{11}\choose{3}}$.
If both blue bags each contain $1$ cookie, the number of ways is ${{9}\choose{3}}$.
and so on.


This is the thread where you'll discuss about AMC-10 B. However, this thread will be locked until AMC-10 B is done.

[quote="Zellex"]A 2 X 3 rectangle and a 3 x 4 rectangle are contained within a square without overlapping at any interior point, and the sides of the square are parallel to the sides of the two given rectangles. What is the smallest possible area of the square?

(a) 16
(b) 25
(c) 36
(d) 49
(e) 64

I have a question about the wording.

When it says "any interior point", does it mean "any interior point in the square?", since it was the last one mentioned in the sentence. Therefore, if we had a square with area 25 (side of 5), wouldn't there be overlapping interior points between the two rectangles, since the   sides of the rectangle would overlap, thus, creating an overlap at an interior point?[/quote]

"Any interior point" refers to the points inside the rectangles. The sides of the rectangles are not interior points, so the rectangles do not overlap.


Prove that if $ x$ is a perfect number in $ \sqrt{x \plus{} \sqrt{x \plus{} \sqrt{ x \plus{} \dots}}}\equal{}y$ then $ y$ is a integer power of $ 2$.

Hmm?  $ \sqrt{6 \plus{} \sqrt{6 \plus{} \ldots}} \equal{} 3$.  (Even perfect numbers are of the form $ y(2y \minus{} 1)$ for $ y$ a power of 2, not $ y(y \minus{} 1)$.)

:| I can not believe did not remember the definition of a perfect number correctly.
Even though JBL kind of gave a big hint, two possible revisions

Prove that if $ x$ is a perfect number in $ \sqrt {2x \plus{} \sqrt {2x \plus{} \sqrt {2x \plus{} \dots}}} \equal{} y$ then $ y$ is a integer power of $ 2$.

Prove that if $ x$ is a Mersenne prime in $ \sqrt {x\left(x\plus{}1 \right) \plus{} \sqrt {x\left(x\plus{}1 \right) \plus{} \sqrt {x\left(x\plus{}1 \right) \plus{} \dots}}} \equal{} y$ then $ y$ is a integer power of $ 2$.

 :wink: Hope those are plausible now.

The first one still need not be true -- what if $ x$ were odd?  :P

But if there [i]is[/i] one, then the statement isn't true, so a proof of the truth of the statement requires a proof of the nonexistence of odd perfect numbers :)

Yes, then, for that proof, let $ x$ be an even perfect number.  :)


Soon, aops must start celebrating because the 10,000th user will soon join the AOPS-ML website, a great accomplishment!

yeah AoPS is pretty big..does anyone know how many new members exactly we get per day?

[quote="Myself"][quote="stupidkid"]They should give a prize to the 10000th user.  :D[/quote]That wouldn't quite work.  I remember back in the day where there was some guy made a second account just to be the 1000th user.  It would have been so significant if he hadn't done anything.  :([/quote]

Wasn't that Tare? Back in the old days of AoPS. Not quite sure, though.

The 10,000th user should become the head administrator.  just kidding. :)

[quote="NoSoupForYou"]Maybe the admins should install the Statistics Mod for phpBB...[/quote]Maybe not. There are around 20 new users / day registering in average.

PS There will be absolutely no special prize for the 10,000th user :)

[quote="nr1337"][quote="A+MATH"]At Mathpath me and my friends have promoted AoPS like crazy. I'm soly responsable for about three new people thank you.[/quote]

Yes, and a few people who haven't posted for a while have started posting again.

You know, 10,000 people isn't much if only a few of them post....

(Many people have 0 or 1 posts...)[/quote]

That's how it works on all forums. Many people like to just read through the topics, others messup the e-mail address reply thing, and others just signup one day and never come back. Can't help it.

But to even have 10,000 people signing up on a math forum is definitely an accomplishment.

Exactly its an accomplishment. Also, the growth rate is supposed to increase because as more people join the site, there are more people that can tell others, if that is so on this site, the number of members should soon multiply (at least mathematically it seems true).

[quote="noneoftheabove"]Actually after a while the growth rate will slowly decline because the amount of members left who are willing to join become less and less[/quote]If you are estimating that only ~10.000 people in the world like mathematics then I'm afraid you are (very) wrong ;) Yet the site is fairly unknown to the general public :) (just think of how many people take the AMC's in US only per year! :) ).

I think the growth rate is still increasing, I'm sure Valentin could provide a statistic on that :P I don't think it will ever curb off and slow down because old users still exist and as more kids enter the age of competitive math more will come. And from local competitions I can say that more and more kids are beginning to do competitive math; it's becoming more difficult to be at the top.

[quote="hello"]In genereal or in aops? I dont think it is illegal though.,[/quote]

its under aops policy

[quote="shobber"]there is a user with name: 10000th user.[/quote]

Thanks, Captain Obvious.

[quote="furious"]i'm just wondering how the 10,000th user knew (s)he was the 10,000th user.
[b]in my opinion[/b], i wouldn't expect a new user to find the number of registered users, see that they were at 9999 and then call him/her-self the 10000th user. i think it could possibly be someone who was already a member who created the 10000th account to gain whatever sort of "fame" or "prize" that goes along w/ that.[/quote]

  Yes, thats what I thought too.  Someone who is new to the AOPS website wouldn't probably notice the number of users who have already registered (at least I didn't) and then see his account would change the tally to 10,000.  The only thing that might stop one from doing this is that his or her other account will probably go to waste, and if he has something like 4 or 5 stars on it, it might not be very easy to let it go.

???:furious:  :furious:  :furious:  :bomb:  :bomb:  :bomb:  :cursing:  :cursing:  :cursing:  :evilgrin:

LOL, I was wondering when the 10000th user would actually post in this thread.  :D

[quote="hello"]....I ghouth so beofre...[/quote]

Does that mean "I thought so before"???

I think it's a bad idea to name yourself 10000th user. Then everybody talks about you and stuff.  :D I wouldn't want that.

[quote="chess64"][quote="hello"]....I ghouth so beofre...[/quote]

Does that mean "I thought so before"???

[/quote]

Of course it means "I thought so before"...thats one of his easier-to-figure-out typos.....

Ameem, come down, obviously that's what it means. Ok, lets play trivia!!!! I know this is not the right forum but hez has gone crazy so we must play.

[quote="hello"]The 10,000th user speaks funny. Yeah I agree with that guy, I ghouth so beofre, I forgot what his name is, the one with the picture in which he smiles as his avatar, 10,000th user already has a different account, dont u? HAHAHAHa. I wont mind my own business. The business of business is business. LOL[/quote]

he speaks "funny" because he's not from the US...

OK, what aops should do is dominate the world!!! We have to make specialists in all areas of knowledge and be the best!!!! I know that was random.

i think we should expand our forums from math to other things like sports, english, etc.  Oh, i just realized we do.  Nevermind :)

15Yearoldbump also I am the 542404th user. Nice repetition, wright?


calculus
integration
calculus computations

Since the definite integral is constant, so let $ \int_{ \minus{} 1}^1 g(t)\ dt \equal{} A,\ \int_{ \minus{} 1}^1 tg(t)\ dt \equal{} B\ (A,\ B\ \text{are constants})$,

we have $ g(x) \equal{} Ax^3 \plus{} (B \plus{} 1)x^2$. Thus $ A \equal{} \int_{ \minus{} 1}^1 g(t)\ dt \equal{} 2\int_0^1 (B \plus{} 1)x^2dx \equal{} \frac 23 (B \plus{} 1)\Longleftrightarrow 3A \minus{} 2B \equal{} 2\ \cdots [1]$.

$ B \equal{} \int_{ \minus{} 1}^1 tg(t)\ dt \equal{} \int_{ \minus{} 1}^1 \{At^4 \plus{} (B \plus{} 1)t^3\}\ dt \equal{} 2\int_0^1 At^4\ dt$

$ \equal{} \frac 25A\Longleftrightarrow 2A \equal{} 5B\ \cdots [2]$

Solving the system of equation of $ [1]$ and $ [2]$ gives $ A \equal{}\boxed{ \frac {10}{11}},\ B \equal{} \frac {4}{11}$, yielding $ g(x) \equal{} \frac {10}{11}x^3 \plus{} \frac {5}{11}x^2$

please illustrate(explain) how you get eq.1 ,eq.2


geometry
incenter
circumcircle
geometric transformation
reflection
conics
analytic geometry

Let ABC be a scalene triangle. A point P is called [i]nice[/i] if AD,BE,CF are
concurrent, where D,E, F are the projections of P onto BC, CA, AB, respectively.
Find the number of [i]nice [/i] points  that lie on the line OI

I think there are three nice points. The incenter, the circumcenter and the reflection of the incenter in the circumcenter.

[b]Lemma:[/b] Let P be a point on an arbitrary line d cutting a triangle $\triangle ABC$ and let N, M be the feet of normals from P to AB, AC. Let BM meet CN at Q. The locus of Q is a conic passing through B, C.

Let d meet AB at O. Set up the coordinate system with the origin O and x-axis identical with AB. Let A = (a, 0), B = (b, 0), C = (c, c') be the coordinates of A, B, C, let k be the slope of d, and let P = (t, kt) be the coordinates of P. The coordinates of N are then N = (t, 0). Equation of AC is

$AC: \ \ y = \frac{c'}{c-a}(x-a)$

Normal to AC through P has the slope $\frac{a-c}{c'}.$ Since it passes through P, its equation is

$PM: \ \ y = \frac{a-c}{c'}x+kt-\frac{a-c}{c'}t$

Solving the equations of AC, PN, we get the coordinates of M as M = (m(t), m'(t)), where m(t), m'(t) are some linear functions of t. Equations of BN, CM are then

$BM: \ \ y = \frac{m'(t)}{m(t)-b}(x-b)$

$CN: \ \ y = \frac{c'}{c-t}(x-t)$

Solving these 2 equations, we could get get the coordinates of Q in terms of t. But to get the locus of Q, we eliminate t instead. Calculating t from the equations of BM, CN leads to

$BM: \ \ t = \frac{p(x, y)}{q(x, y)}$

$CN: \ \ t = \frac{u(x, y)}{v(x, y)}$

where p(x, y), q(x, y), u(x, y), v(x, y) are some linear functions of x, y. Eliminating t, we get

$\frac{p(x, y)}{q(x, y)}= \frac{u(x, y)}{v(x, y)}$

This clearly is general equation of a conic. This conic passes through B, C. To see this, it is sufficient to move $P \in d,$ so that $\angle PBA$ or $\angle PCA$ are right, when $Q \equiv N \equiv B$ or $Q \equiv M \equiv C.$

Let $d \equiv OI$ be the diacentral line of $\triangle ABC.$ Since the triangle is scalene, the diacentral line does not pass through any vertex. Let M, N, K be the feet of normals from P to AB, AC, BC. Let $Q \equiv BM \cap CN,$ $R \equiv AK \cap BM.$ The locus of Q is a conic $q_{BC}$ passing through B, C, while the locus of R is a conic $r_{AB}$ passing through A, B. These 2 conics intersect at B and they can have at most 3 other intersections, where Q, R coincide. Thus there are at most 3 nice points on the line $d \equiv OI.$ But there are at least 3 nice points on the diacentral line: the incenter I, the circumcenter O, and the reflection I' of I in O. The feet of normals from the incenter are the tangency points of the triangle sides with the incircle (I) and the corresponding cevians meet at the Gergonne point. The feet of normals from the circumcenter are the midpoints of the triangle sides and the corresponding cevians (medians) meet at the centroid. The reflection I' of I in O is the circumcenter of the excentral triangle, the feet of normals from I' are the tangency points of the 3 excircles with the triangle sides, and the corresponding cevians (perimeter splitters) meet at the Nagel point.

What's the point of your post, arline (besides showing that you know Ceva's theorem)? It's absolutely useless.

Also [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=110743]this lemma[/url] can be useful.  :)


calculus
integration
function
limit
complex analysis
complex analysis unsolved

Show that for any real numbers $ a, b, c$ with $ a < b < c$ we have

$ PV \int^{\infty}_{\minus{}\infty} \frac{1}{(x\minus{}a)(x\minus{}b)(x\minus{}c)}dx \equal{}0.$

I know, in general that the Cauchy principal value of a doubly infinite integral of a function $ f$ is

$ PV \int_{\minus{}\infty}^{\infty}f(x)dx\equal{}\lim_{R \rightarrow \infty}\int_{\minus{}R}^Rf(x)dx$.

Also, here we have three consecutive real numbers. However, I do not see how we get the Cauchy principal value from all this. I need a bit of help on how this one.

Think a bit here: $ \int_{ \minus{} R}^Rf(x)dx$ fails to exist for large $ R$. You'll need to rethink your definition of the Cauchy principal value here.

You want the version of the principal value for singularities in the finite plane; if there's a simple pole at $ p\in (u,v)$ (on the real line), the principal value of $ \int_u^v f(x)\,dx$ is $ \lim_{t\to 0^\plus{}}\int_u^{p\minus{}t} f(x)\,dx\plus{}\lim_{t\to 0^\plus{}}\int_{p\plus{}t}^{v} f(x)\,dx$.
There's a similar version for straight lines in $ \mathbb{C}$ passing through poles.


$ The \ vectors
\\
\\
x_{1} \ = \ \left(\begin{array}{}1&2&2\end{array}\right) \ x_{2} \ = \ \left(\begin{array}{}2&5&4\end{array}\right) \ x_{3} \ = \ \left(\begin{array}{}1&3&2\end{array}\right) \ x_{4} \ = \ \left(\begin{array}{}2&7&4\end{array}\right) \ x_{5} \ = \ \left(\begin{array}{}1&1&0\end{array}\right) \\
\\
span \ R^3. \ Pare \ down \ the \ set \ \{x_{1}, \ x_{2}, \ x_{3}, \ x_{4}, \ x_{5}\} \ to \ form \ a \ basis \ for \ R^3.$

Does this mean I should just take out the vectors that contain free variables?
Here's another:

$ Let \ A \ be \ a \ 4 \ \times \ 5 \ matrix. \ \ If \ a_{1}, \ a_{2}, \ a_{4} \ are \ linearly \ independent \ and  \\ \\ \ \ \ \ a_{3} \ = \ a_{1} \ + \ 2a_{2} \ \ \ \ \ a_{5} \ = \ 2a_{1} \ a_{2} \ + \ 3A_{4} \\ \\ determine \ the \ reduced \ row \ echelon \ form \ of \ A.$

Thank You

"Pare down" just means "reduce the size of."  Remove two vectors so that the three remaining vectors span $ \mathbb{R}^3$.

You have a $ 3\times 5$ matrix which allegedly has rank $ 3.$ We would like to find a basis for its column space. (If it does have rank $ 3,$ then its column space is $ \mathbb{R}^n.$)

There is a standard procedure that can be followed.

Start with the matrix whose columns are the vectors in questions. (In this case, that will be the $ 5\times 3$ matrix referred to above.)

Perform row elimination to arrive at a row echelon form.

Find the pivots in that row echelon form. Then go back to the original matrix, before any row operations. The columns of the original matrix corresponding to those pivot elements will be a basis for the column space.

(Note that we will know the dimension of the column space, hence know whether the claim that this spans $ \mathbb{R}^3$ is correct.)

Note also there there is no sense in which our answer is anywhere close to unique. Following this algorithm will lead in a particular direction, but that's still an accident depending on our choice of which order to list the vectors in.


i got this question in a maths competition i was in, i know the answer but i want to know how to solve it. the question is...

If x is a positive real number, simplify:
[cos(arctan(sin(arccot(x))))]^2


10.7 Find all pairs $(a,b)$ of natural numbers s.t. $a^n+b^n$ is a perfect $n+1$th power for all $n\in\mathbb{N}$.
([i]V. Senderov[/i])

suppose $a^n+b^n$ has an odd prime factor $p$ that doesnt divide $a$ neither $b$. then $p^{k(n+1)}||a^n+b^n$ now we must have that $a^{npq}+b^{npq}$ is a perfect $npq+1$ power for $q$ an odd prime different from $p$, the order of $p$ here is $kn+k+1$ so for big enough $q$ we have $npq>kn+k$, and thus the number is not a perfect $npq+1$ power...

So let $p_1,p_2...,p_m$ be the common prime factors of $a$ and $b$, we must then have that: $a^n+b^n$ can only have the prime factors $p_1,p_2...,p_m$ and $2$, now if $n$ is odd we get all primes factors of $a+b$ are a subset of those from $a^n+b^n$ from here, let $d=(a.b)$, then $a+b=2^kd$, but then $a^q+b^q$ for q different from $2$ and $p_i$, has a prime factor different from those of $a+b$, and this new prime doesnt divide $a$ neither $b$...so we get the initial contradiction, unless $a=b=1$ but this case doesnt work....

Please, can someone chek it???? thanks!

there are gaps in the above post....

Until now we just have to say, that $a^n+b^n$ has all prime factos divisors of $2ab$....some cases may be dealed here, but i dont have time right now...

If $a=b$ it is easy...
Let $ a\geq b+1 $ and $a^n + b^n = (k_n)^{n+1}$ for $ k_n \in N $
Since $a \geq 2$ we get $ a^{n+1}\geq 2a^n \geq (k_n)^{n+1} + 1$ so $a\geq k_n +1$
As a result there is a $k\in N$ such that $a^n + b^n = (k)^{n+1}$ for infinetely many n.
But then $ak^n\geq k^{n+1}\geq a^n$ so $ a \geq (\frac{a}{k})^n \rightarrow \infty$ since $a\geq k+1$ which is a contradiction. The rest follows.

[quote="Anto"]As a result there is a $k\in N$ such that $a^n + b^n = (k)^{n+1}$ for infinetely many n.
[/quote]

Why?  :?

Because $a$ is fixed and $k_n \leq a -1$ is bounded for all $n$...


A flag is made of three horizontal strips of fabric, each of a solid color, either red, white, blue or green. If no two adjacent strips can be the same color, how many distinct flags are possible? These two flags are different.

[asy]draw((0,0)--(0,6),linewidth(1));
filldraw((0,3)--(4,3)--(4,4)--(0,4)--(0,3)--cycle,white);
filldraw((0,4)--(4,4)--(4,5)--(0,5)--(0,4)--cycle,red);
filldraw((0,5)--(4,5)--(4,6)--(0,6)--(0,5)--cycle,green);
filldraw(circle((0,6),1/6),white);
label("Green",(2,5.5));
label("Red",(2,4.5));
label("White",(2,3.5));

draw((5,0)--(5,6),linewidth(1));
filldraw((5,3)--(9,3)--(9,4)--(5,4)--(5,3)--cycle,green);
filldraw((5,4)--(9,4)--(9,5)--(5,5)--(5,4)--cycle,red);
filldraw((5,5)--(9,5)--(9,6)--(5,6)--(5,5)--cycle,white);
filldraw(circle((5,6),1/6),white);
label("White",(7,5.5));
label("Red",(7,4.5));
label("Green",(7,3.5));[/asy]

There are 3 choices for the first strip, 2 for the second strip, and 2 for the third. Thus
\[ 3\cdot2\cdot2\equal{}\boxed{12}
\]

i believe there's 4 colors  :)

4*3*3=[b]36[/b]


my answers:

1. 592
2. i got this wrong, but the answer is 441
3. 384
4. i got 948 or something, but maybe 927
5. 766
6. 408?
7. 293
8.  54, but stupid me put down 44 cuz i cant ADD
9. 973
10. 133 i think
11. 625
12. 788 pretty sure, but m = 1, which was pretty sketchy
13. 16 (the fraction was 1/15), but it was a wild guess
14. 460, but unsure, combinations could possibly overlap, so 460 could be an upper bound.
15. 681, total guess...

Hmm.. can someone post up the problems as well?
I don't remember the problems so i can't rework on the problems.

My 4,5, and 14 match yours ************...

For #5:

[hide]To people that got 500 as an answer, the thing I believe we did differently is that you did not consider how much time was being lost in the quarters in which they were lagging behind schedule.

If for each quarter they spent the same amount of time, and just let the unfinished work remain unfinished as they began their next quarter AND THEN finished the unfinished work all in the fourth quarter, the answer would be 500 (200 to make up for the workers below 1000 for a full quarter and 300 additional for the work left over from quarters 2 and 3).

However, here we must consider that b/c of the 2nd quarter taking 10/9 of it's scheduled time, and 3rd quarter taking 5/4 (10/8) or it's scheduled time, the schedule is lagging by 13/36 of a quarter when the fourth quarter begins. This means that we must complete the work of the fourth quarter is only 23/36 of the time. So, we will need 36/23 of the 1000 workers, putting us at needing a little more than 1565 workers. Because we can't hire a fraction of a worker, we will use 1566 workers - 800 that we already have = 766 workers hired.[/hide]

HMM...
My head became blank at the last hour....
So I only answered only 8. (just an excuse)

1. 592
2. 441
3. 384
4. 927
5. 766
6. 408
7. 207
8. 22

Just to note: we have several different answers from each of the complete lists posted.  We will be going over the problems in a Math Jam in an hour.


Gauss
Euler
geometry
inequalities
summer program
Mathcamp
AwesomeMath

what do u guys think requires more intelligence and praise, learning a lot of advanced math concepts early or winning contests? I know contests require something extra but time constraints and silly errors can limit one from doing optimally, and learning new things on your own requires an ability to grasp and apply concepts easily, both require initiation(preparing for contests, doing the independant study), because it is rare for somebody to just be able to win contests out of nowhere, and it is impossible(unless ur ramanujan) to just know subjects.  I've been thinking about this lately and contest wise, I'm nothing compared to so many people, but I feel I am more advanced in my subject areas. What do you guys think?

I know freshmen in calculus who didn't even qualify for AIME through the AMC12 when I got a 6 on AIME as an 8th grader in geometry - anyone can master a concept eventually, it takes certain skills to use those concepts and some creativity to solve contest problems.

Thinkers>Learners

[quote="Dvagner"]i agree, however contests require preparation, which a lot of kids were never exposed to(just like new subjects), and contests can judge you better in some cases, however when u get beyond calc, and get more theoretical it gets harder to understand especially without a teacher. contest problems however do require more creativity however there are certain "tools" which so many people just don't know wh ocould master them[/quote]
No matter how advanced stuff gets, I'm pretty sure most AoPSers can learn school/college math on their own. I think that it is more impressive to qualify for USAMO than to learn calculus in middle school. So,
problem solving > 5*school math

Gaining good performance in contests is generally a better idea than learning a great deal of advanced math that is still not of great difficulty.  See [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php]this article[/url].  It is quite possible to complete a course without having actually becoming significantly more intelligent, whereas improvement in contests ideally requires nothing but an increase of intelligence.  Plus, contests never put you under the delusion that you have "finished" something. :wink:

although i've definitely somewhat fallen into this trap(calc 1, 2, 3, DE, some math analysis, linear algebra, some complex variables by begining of jr yr in highschool), i don't just learn subjects; eg. i proved most of the calculus equations and theories and figured out myself how to use them in more advanced topics, because i love playing around with the topics and enhancing my ability to use and apply them.  I know a kid who is a senior and knows topology(and is doing algebraic topology now) and he got like a 4 on the AIME, so i guess hes not that smart.

subject math seems to pwn contest math pretty badly

like both are hard (for me at least probably for some people like a certain GAUSS or EULER this might not be true) but subject math is like all the coolness that ever exists

Well in my opinion getting good at contests definitely will help you to understand more advanced math topics.

Although, he's right in that contest math does not measure creativity directly anymore. At this point, it's also largely dependent on preparation. Anything that begins as purely innovative eventually gets sucked into this "industrial" cycle of former winners writing books, etc. and the new generation desperately buying all sorts of preparation materials. This very site is a prime example. It's simply impossible for a very intelligent person to walk into a contest like the IMO and compete with a perhaps less intelligent, but far more prepared student who has read books like AoPS, Engel, Zeitz, been to MathCamp or Awesomemath, taken WOOT and Olympiad Problem Solving, and done practice problems and previous tests.

Of course, I still think that mastering concepts through high-level problem solving is far more difficult than just having a load of material pounded into your head and knowing how to solve a very limited range of very basic problems. I don't see why middle schoolers wouldn't be able to learn how to differentiate and integrate through some boring, uninnovative program. I can imagine it- one chapter devoted to the Power Ruler, $x^{n}\rightarrow nx^{n-1}$ with repetitive examples like "Find the derivative of $2x^{3}, x^{7}, 14x^{2}$, etc." and walah, your 7th grader knows how to take the derivative of any polynomial.

I believe preparation is the key to success.

As well as a set time to practice each and every single day.

I agree with Phelpedo that contest math does not measure creativity.

Some contests are strictly measurements of how much you know about a certain subject. That is something that anyone can study and learn.

that's the point I was trying to prove, that these  days with things like aops winning contests isn't that much more incredible than learning new math.  contests measure how well prepared u r for a certain set of problems on a certain day and limit your time; if i were to figure out a bunch of aime problems after i already handed it in, it would not make me any dumber than the people who got them there. Also, although simple differentiation techniques may be learnable to almost all, math gets more difficult.  Certain integration problems, solving DE's, and so on math gets more difficult and theoretical and requires a high intelligence to absorb and use all the material.

At a certain point there is really no difference between "contest" math and "real" math.

It is just that in high school math is generally not taught to be highly challenging. There is much more to calculus than differentiating polynomials. Characterizing the conditions for saddle points, minima, and maxima is harder than at the very least all contest math that is below the olympiad level, as is proving that Jacobians do what the textbook says they do.

I do, however, believe that contest math, at least at the Olympiad level, is very enriching as well, and in the process you learn a lot of the basic undergraduate material, i.e. inequalities, number theory, combinatorics, non-euclidean geometry, as well as many of the connections between subjects, such as geometric inequalities, combinatorial number theory, generating functions (algebra + combo), and combinatorial geometry. And to get to the Olympiad level you generally need at least somewhat of a grounding in lower-level contest math.

Personally, I would try to do both.

Dude yeah. Real math does not involve differentiating. Nobody cares about the answers. Well some people do but they are not mathematicians they are like physicists or something. Real math means proving all the stuff. And it is hard. Like seriously pretty much all the IMOers who have now gone to college would tell you real math pwns if you asked them now.

It's not hard at all to get access to problem books and stuff so long as you have access to a computer (and if someone cannot afford any sort of method of using a computer at all they're probably too poor to go to tournaments in the first place so trivially we can discount that case). I know of several places where you download all of zeitz, PSS, Problem solving through problems, and several other books...and that's not even counting the free stuff, such as ALL those geometry and inequalities PDFs as an example as well as all the contest math that's just sitting on the hosting school's website.

You know places where you can download those problem-solving books?  :roll:

I'm pretty sure that most of the copies of people's books that are available online are illegal unless they are published on the author's own sites or with their permission.

But for free stuff that is definitely legal, there is the things in the AoPS resources, and Mildorf's lectures, and various other lectures from TJ people, as well as Naoki Sato's stuff, and some online collection of Number Theory problems that I can't quite remember the link to (I'll post it if I figure it out).

Those are Hojoo Lee's problems: http://myhome.naver.com/hojoolee/orange.html

[quote="now a ranger"]i definitely agree with elemmnop

the person who gives more effort + time deserves to be rewarded more.[/quote]
People do not "deserve" anything. What determines if you get a high score, or what should (IMO), is whether you solve the problems or not. The problems should not be rigged so that the ones that know all the mathy shortcuts or have gone through the standard AoPS 2 problems automatically do better (AIME, I look at you). Obviously, you should be expected to learn the material. And obviously, life favors the better prepared. But when some contests (cough cough) but some trivial variations on standard training problems and put them on the contest, it gives a severe advantage to those who have studied that particular problem.

Two examples I can pull out of the first chapter of AoPS: the ones where knowing the chain rule or log reciprocality rule makes a #20 AHSME trivial, but make the unprepared student rewrite one as log_x x (significantly harder to see), and the two almost identical problems coming from MAO and the AHSME.

So, um, yeah. I want more unique problems, and less recycled telescoping junk.

@Phelpedo: Thanks for the link, those actually weren't what I had in mind but the source appears to be excellent!

@PenguinIntegral: Try Olympiad problems then. Or write your own problems.

Hm, so you weren't referring to his number theory [url=problems]http://myhome.naver.net/hojoolee/pen2005.pdf[/url]?

@penguinintegral, just because a problem can be solved easily by more advanced techniques doesn't mean it is recycled.

[quote="JSteinhardt"]
@PenguinIntegral: Try Olympiad problems then. Or write your own problems.[/quote]
Sorry, I ment for the AIME :roll:

[quote]@penguinintegral, just because a problem can be solved easily by more advanced techniques doesn't mean it is recycled.[/quote]
Elaborate please?

[quote="Elemennop"][quote="JSteinhardt"]I'm pretty sure that most of the copies of people's books that are available online are illegal unless they are published on the author's own sites or with their permission.[/quote]

Oh, of course they are, but that doesn't stop alot of people from downloading it, and if one really wanted to become better at contests and did not have the money to buy those books I don't think they should have any reason NOT to download those.[/quote]

Well they must be fairly difficult to find (I've never seen one).

[quote="PenguinIntegral"][quote="JSteinhardt"]
[quote]@penguinintegral, just because a problem can be solved easily by more advanced techniques doesn't mean it is recycled.[/quote]
Elaborate please?[/quote][/quote]
Well you were complaining about how the certain problem could be solved by the chain rule...

[quote="Phelpedo"]
Well they must be fairly difficult to find (I've never seen one).

[/quote]

Well, you certainly haven't looked hard enough, or at least not in the right circles.  :wink:

[quote="Elemennop"]Oh, of course they are, but that doesn't stop alot of people from downloading it, and if one really wanted to become better at contests and did not have the money to buy those books I don't think they should have any reason NOT to download those.[/quote]

Well except for, you know, the ethical/moral implications.

@Phelpedo: Yeah I was referring to the NT problems when I praised the source, though his "oranges" are also pretty interesting. Mildorf has given me enough Inequalities to last me for a long time.

I think Joshua Nichols-Barrer gave us an adaptation of the NT problems I was referring to at MOP, I've e-mailed him, hopefully he remembers where he got them from.

[quote="Phelpedo"]
[/quote][quote="PenguinIntegral"][quote="JSteinhardt"]
[quote]@penguinintegral, just because a problem can be solved easily by more advanced techniques doesn't mean it is recycled.[/quote]
Elaborate please?[/quote]
Well you were complaining about how the certain problem could be solved by the chain rule...[/quote][quote="Phelpedo"][/quote]
For logs.

[quote="Phelpedo"][quote="Elemennop"]I don't see what's so bad about the more prepared student winning contests. Being a genius certainly helps, but the person who's willing and able to work to improve himself will perform better and should deserve to win.[/quote]

But this means that talented students that are less privileged don't really have a chance, just like talented students in schools that simply don't have accelerated math programs don't hace the opportunity to take higher math courses. A brilliant student who works hard but doesn't have access to past problems and training books and online courses is too disadvantaged.[/quote]

Dude, that's what AoPS is for.  :D

[quote="ThAzN1"]
Dude, that's what AoPS is for.  :D[/quote]
True dat. I took calc in 8th grade and was really lazy as a middle schooler, anyone can get A in calc by middle school, easily if you're a mathematical person. However, last year, as a 10th grader, I didn't even qualify for AIME :rotfl: Partly bcuz I'm not that good of a test-taker, but mainly because (with the exception of some mathcounts) I hadn't been exposed to much math outside of textbooks. But after messing around on AOPS for a couple months I could easily get in 130's on amc12 and haven't had the time to take a past AIME, but from looking at sum past AIME's I can safely say that I could consistenly get 5+. And even though these contests are composed of material I covered in 7th grade, I love it! Of course, learning new higher level concepts is fun too. I personally try to do both.


Let $ L$ be the lattice generated by $ {(1,3,\minus{}2),(2,1,0),(\minus{}1,2,5)}$. Sketch a fundamental domain for $ L$ and find its volume.

is the volume just the $ |\det(L)|$? thank you.

Yes, the volume of a fundamental region is the determinant of a basis.

this topic is to sort
we can post some excercise


rotation
symmetry
ratio
geometry
rectangle
angle bisector

Let $ABCD$ be a square, and $C$ the circle whose diameter is $AB.$ Let $Q$ be an arbitrary point on the segment $CD.$ We know that $QA$ meets $C$ on $E$ and $QB$ meets it on $F.$ Also $CF$ and $DE$ intersect in $M.$ show that $M$ belongs to $C.$

Very strange, that it's from Moldova since I don't remember it.
Anyway the first thing that comes in my mind is an analitic solution, it's really not hard to compute the equation of the two intersecting lines(CF and DE), and you just check that it belong to C(here you may choose the origin the midpoint of $AB$)!

Let BE and AD intersect at S, and let AF and BC intersect at T, and define $\alpha=\angle BAQ$ and $\beta=\angle QBA$. Also let Y be the center of the square. Since $\angle SBA = 90-\angle BAE = \angle QAD$, and since AB = DA, right triangles QDA and SAB are congruent. This means that S is the image of Q when rotated 90 degrees around Y. By similar arguments, T is the image of Q when rotated around Y 90 degrees the other way. So TQS is an isosceles right triangle. Let $\angle QSD = \delta$, so that $\angle CTQ = 90-\delta$.

Let's calculate angle CMD. First, since QDSE is cyclic, $\angle MDC = \angle ESQ = 180-\angle ASB-\angle QSD = 180-\alpha-\delta$. (We have used the fact that $\triangle ASB\sim\triangle EAB$ which implies $\angle ASB = \angle EAB = \alpha$.) Likewise, $\angle DCM = \angle QTF = 180-\beta-(90-\delta)$. We then have $\angle CMD = 180-\angle MDC - \angle DCM$, and after some calculation this comes out to $\alpha+\beta-90$.

Now lets calculate angle FBE. We have $\angle FBA = \beta$ and $\angle EBA = 90-\alpha$, so $\angle FBE = \beta-(90-\alpha)=\alpha+\beta-90$.

Since we have $\angle FBE = \alpha+\beta-90 = \angle FME$, EFBM is cyclic, so M is on the circle as desired.

[quote="wisemaniac"]believe me, an analityc solution can't be done by a human,i ve already tried , i am not sure of the source(maybe bosnia) but it remains a very hard problem, much more than it looks . can someone have a try?[/quote]

Well it took ''only'' 2 pages for me to write the solution, so there is a human analitic approach!

a really beautiful problem as it is a special case of a special case you get from a special case ;).

so, the general problem is the following:

suppose we have a cubic $E$ in the plane and points $A,B,C,D$ on it s.t. the line $AC$ and $BD$ meet on $E$. let $X$ be an arbritary point of $E$. let $P$ be the third point of intersection of $XA$ with $E$ and let Q be the third points of intersection of $XB$ with $E$. then $PD$ and $QC$ concur on $E$.

proof: i will do the proof in the language of divisors and linear equivalence; it's quite the same as the language of addition of points on cubics. so we know that $A+C\sim B+D$ by the first condition. furthermore we know that $P+X+A\sim P+R_1+D$ where $R_1$ is the third point of intersection of $PD$ with $E$; this is clear since both sides are the zeroes of some linear functions(the intersections with lines). Thus we have $R_1\sim A-D+X$. defining $R_2$ analogously, we get $R_2\sim B-C+X$. Thus we have, since $A-D\sim B-C$ that $R_1\sim R_2$ and this means that $R_1=R_2$ (this is the crucial step; this implication holds for all curves of genus at least 1).

now for our problem we take $E$ as the union of the $C$ and the line $CD$. it is clear that $AC$ and $BD$ meet on $E$. therefore we immediately get the result.

The problem is a one-liner using Pascal, if one lets the center O of the square ABCD join the picture. Here is an extended version of the problem:

[color=blue][b]Extended problem.[/b] Let ABCD be a square, and k the circle with diameter AB. Let Q be an arbitrary point on the segment CD. Let the line QA meet the circle k at a point E (apart from A), and let the line QB meet the circle k at a point F (apart from B). Furthermore, let O be the center of the square ABCD, and let R be the orthogonal projection of the point Q on the line AB. Then, prove that the lines CF, DE and OR intersect at one point which lies on the circle k.[/color]

[i]Solution of the extended problem.[/i] The center O of the square ABCD clearly must lie on both diagonals AC and BD of the square and on the circle k with diameter AB (since < AOB = 90°). Now, let the line OR meet the circle k at a point M (apart from O). We will show that this point M lies on the lines CF and DE; then, of course, the problem will be solved.

Since the points Q and R lie on the sides CD and AB of the square ABCD, respectively, and satisfy $QR\perp AB$, they are symmetric to each other with respect to the parallel to the side AB of the square through its center O. Of course, the diagoanls AC and BD are also symmetric to each other with respect to this parallel. Since the point O lies on this parallel, it thus follows from symmetry that < QOC = < ROB.

Let the line OQ meet the circle k at a point N (apart from O). Then, < QOC = < ROB becomes < AON = < MOB. But since the points A, O, N, B lie on one circle (namely, on the circle k), we have < AON = < ABN, and since the points M, O, B, N lie on one circle (namely, on the circle k), we have < MOB = < MNB. Thus, < ABN = < MNB, and hence NM || AB. But since ABCD is a square, we have AB || CD. Thus, the lines NM, AB and CD are all parallel to each other; hence, they concur at one point (namely, at an infinite point). In other words, the point of intersection $AB\cap NM$ lies on the line CD.

The point of intersection $EA\cap ON$ is the point Q and also lies on the line CD.

Now, applying the Pascal theorem to the cyclic hexagon EABONM, which is inscribed in the circle k, we see that the points of intersection $EA\cap ON$, $AB\cap NM$ and $BO\cap ME$ are collinear. Since the points $EA\cap ON$ and $AB\cap NM$ both lie on the line CD, it follows that the third point $BO\cap ME$ must also lie on the line CD. In other words, the lines BO, ME and CD concur. But we know that the lines BO and CD intersect each other at the point D; thus, the line ME passes through the point D. In other words, the point M lies on the line DE. Similarly, the point M lies on the line CF. Thus, the problem is solved.

[b]EDIT:[/b] We can show a bit more: [color=blue]We have $\frac{AM}{BM}=\frac{DQ}{CQ}$, where M is the point of intersection of the lines CF, DE and OR.[/color]

[i]Proof.[/i] Since the points B, O, A and M all lie on the circle k, we have < BMO = < BAO. But since ABCD is a square and O is its center, < BAO = 45°. Thus, < BMO = 45°. Similarly, < AMO = 45°. Hence, < AMO = < BMO, so that the line MO is the angle bisector of the angle AMB. Now, in a triangle, the angle bisector of an angle divides the opposite side in the ratio of the adjacent sides. Hence, in the triangle AMB with the angle bisector MO of its angle AMB, we have $\frac{AR}{BR}=\frac{AM}{BM}$. But since the quadrilateral BRQC is a rectangle (in fact, it has three right angles: < RBC = < ABC = 90°, < BCQ = < BCD = 90° and < QRB = 90°), we have BR = CQ, and similarly AR = DQ, and thus this equation becomes $\frac{DQ}{CQ}=\frac{AM}{BM}$. This completes our proof.

  Darij

Dear Mathlinkers,
an article concerning "Geometric miniatures on a square" with original proofs has been put on my website
http://perso.orange.fr/jl.ayme    vol. 7 Miniatures géométriques sur un carré  p. 10
Sincerely
Jean-Louis


[quote="rustam"]Find the number of all non-negative ordered integer triples (x,y,z) such that x+y+z=2006.[/quote]

[hide]$2004^2 ???$(Considering that the same numbers in different letters is a different solution)[/hide]

How about using generating function?
Consider $(1+x+x^2+....)^3$.
The required number should be the coefficient of $x^{2006}$ in this generating function, which is equal to
$-3 \choose 2006$ = $2008 \choose 2006$ = $2008 \choose 2$
Answer: 2015028.

Answer:
$\binom{2008}{2}$

Link to similar problem [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=54188[/url] There is my explanation to the problem

You can also consider it as a partition. 2006 *'s with 2 |'s placed somewhere in between them. 2 ways to choose the |'s and 2008 total places. $\binom{2008}{2}$. This isn't pre-olympiad level. :)

[quote="rustam"][quote]
(Considering that the same numbers in different letters is a different solution)[/quote]
no, i meant that if (x,y,z) is a solution then u mustn't count as a solution the  (y,x,z),....[/quote]

Ohh sorry


I NEED ALL OF YOU THAT ARE INTERESTED TO PLEASE CONTACT YOUR SPONSORS AND REGISTER!!! TOURNAMENT IN 1 WEEK!

Walton is coming. I'll get the paperwork together over the weekend.

Norcross is bringing one team.  The paperwork will come later if it hasn't been done yet.

I think Parkview will also bring a team...does anyone know where I might be able to find a registration form and info about the competition?

Anyone that needs tournament information or registration form...email me : Heng Tong at heng825@yahoo.com

Hi, I have two questions:

1. Is on-site registration acceptable?

2. Can we compete as individuals? If so, how much does it cost, and how should we indicate that on the registration form? (in case that we cannot get a full team)

Tianqi (Tim), this is Heng from Northview.

Yes, on-site registration is available, it will be from 8:15 to 8:45.

On the registration form, there is a place for you to check off if you are competing as individual or not, it is 10 dollars as an individual.

I believe on-site registration will cost a little more though.

Argh, I totally failed this competition. I felt I wasn't even awake at the beginning of the written, and I just can't stop thinking about the failure of written when doing the ciphering, which brought further failure...

Two questions for the ciphering part:
#4, I thought "bouncing" distance only meant when it was bounced upwards but the solution seemed including the distance falling downwards too.
#5, was the answer corrected? It seemed not by judging from the score sheet.

Anyways, overall, the written questions were pretty challenging, though I would personally prefer to have them in a relatively predictable order of difficulty.

Yea, being awake that early is not something I'm used to (school doesn't affect this).

Bouncing would include when it goes up then down, wouldn't it?
Which was #5? Besides, a lot of them needed to be corrected...

Even vaguely in order would have been nice... But besides that, that was a very good competition. The written was incredible, which was a good change. It would have been even nicer for the answers to be on the choices, though  :P . Walton totally failed relay though...

One thing I didn't like about the ciphering was the brute force needed.  For example, I memorized 9!, but I don't think people will automatically know it to be 362880.  The same with number 10.  The probability is easy to set up, but lots of multiplication is needed.  And naturally, I didn't do it correctly.

My compliments on the writing.  It reminds me of the GACS tourney last year when Norcross got 0 on the ciphering.

The relay was well designed.

Does anyone have the results?

Bahhh! You guys are wimps! Questions in order of difficulty! Read the test and decide which ones to do! Perhaps that's [b]PART [/b]of the test!  :lol:

Thanks again to the Northview folks. I was a good tournament and should only get better.

Individual:
1. Santhosh - 185
2. Eric - 167
3. Robert - 163
4. Miles - 159
5. Claus - 157
6. Abubakar - 144
7. Jerry - 142
8. Lillian - 142
9. Allen - 138
10. Maja - 129

Team was:
1. Chamblee - 630 (They beat us by 90 on the relays!)
2. Walton - 579
3. Alpharetta - 552
4. Chattahoochee - 512
5. Woodward

Take care,
Tom

lol, it was funny when Heng couldn't pronounce Maja's last name.  :)  It was also funny when Robert was the only individual to walk onto the stage through the door on the side of the stage.  :)  It was not funny when Karthik came without his signature mustache. :(

Good times.

[quote="gtf"]Bahhh! You guys are wimps! Questions in order of difficulty! Read the test and decide which ones to do! Perhaps that's [b]PART [/b]of the test![/quote]

Fact.  :lol:

[hide="a couple errors in written solutions, methinks"]
#1: it should be $ 2009!$ instead of $ 2008!$, which gives $ 9$ as an answer (still giving E as the anwer, though)
#4: it should be $ f(7) \equal{} 67$ instead of $ f( \minus{} 1/11)$ (amazingly neither is on the answer choices)
#12: it looks that $ \angle DAB$ in the problem statement should be $ \angle ADB$, otherwise $ D$ will be inside the triangle (and gives ugly answer not on the answer choice)
[/hide]
[quote="gtf"]Bahhh! You guys are wimps! Questions in order of difficulty! Read the test and decide which ones to do! Perhaps that's [b]PART [/b]of the test! [/quote]
Our proctor actually suggested we should look at the later questions first, but... Oh well.

BTW, for the polyhedron problem, why is the number of vertices equal to $ (5n \plus{} 60)/4$ (because each vertex is connected to 4 faces? But again why this)?

That was actually a typo.  The problem was going to say each vertex as the intersection of [b]four[/b] faces, not two like in the test paper.  But that question was thrown out, so that failure of ours didn't really matter.

I have no idea what happened on number 4.

The rest were not my fault. :)

The number of errors in the written test was reminiscent of the written test for North Springs, but oh well.  We'll do better next year :!:

Mr. Fulton, I lied. Chamblee only got 90 on the relays whereas you guys got 30, so beat you guys by 60 only. Sorry.

Still, the difference in he tournament. Chamblee did a great job and deserved the win. Congratulations to them and Santhosh!


[hide="Answer"]
EDIT: I thought numbers the 4 numbers had to be next to each other.

Answer-  $444$
[/hide]

uhh are all you mistaken or am i giong crazy?!?! isn't it 444 hahaha... 62222 + 444 = 62666...  :?:

[hide]i think its 444 also
 :?:
 :roll: [/hide]

I have fixed my error.

That makes me suspicious about math92 (same wrong answer??) :rotfl:

[quote="colts18"]The odometer in Mrs. Tullo's car shows she has traveled 62,222 miles. What is the fewest number of miles that the car must travel in order for the odometer again to display four of the five digits the same as each other?[/quote]

[hide="use this thing"]
 62666
-62222
--------
     444 miles
[/hide]

are you ever going to learn :dry:
[hide]use this![/hide]

[quote="math92"][quote="math92"][hide]333[/hide][/quote]
whoa, what did i do?[/quote]

I think you copied me, since that's what i put too.  :D


I'm doing a site to gather IMO photos here: http://www.imophotos.fr.st
If you have a link of other photos please send it to me !

I don't see anything when I click your link.  The official USA photos are [url=http://www.unl.edu/amc/e-exams/e9-imo/e9-1-imoarchive/2004-ia/2004imopictures.html]here[/url].  Maybe I'll put my pics online at some point.

For me it works...
Try here: http://membres.lycos.fr/imo04


Prove that for $n \in \mathbb{N}$,
\[ \varphi ( n ) = \pi ( n ) \Longleftrightarrow n \in \{ 2, 3, 4, 8, 14, 20, 90 \}  \]
Can you use Bertrand Postulate ?


rotation
analytic geometry
AMC
AIME
search

A bug sits on the point (5,0) on the plane. Each second, it rotates by 45 degrees clockwise and move right 5 units. Find its coordinates after 100 seconds.

What do you mean by "rotate 45 degrees clockwise and move 5 units to the right"? Is this to the right of its new orientation or just in the rightwards (x+) direction? Also, what is its initial orientation?

[b]@ aleph0:[/b] The problem is really trivial if you consider the bug actually turning around its own axis 45 degrees after each translation. Most likely, OP means the bug is rotated 45 degrees around the origin.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1088469]See this AIME problem[/url]

See also the discussion [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=243759842&t=232565]here[/url].


Solve the equation: $ \sqrt{3x\plus{}1}\equal{}\minus{}4x^2\plus{}13x\minus{}5$

here is my solution
with the condition $ 4x^2\minus{}13x\plus{}5\le0$
squaring 2 sides of the equation we get
$ 3x\plus{}1\equal{}(4x^2\minus{}13x\plus{}5)^2$
$ \Leftrightarrow (4x^2\minus{}13x\plus{}\frac{11}{2})^2\equal{}(2x\minus{}\frac{5}{2})^2$
this equation can be solved easily
what about your solution tdl


topology
real analysis
real analysis unsolved

Let $ A$ be the subset of real line with non-zero Hausdorff dimension. Does it follow that $ \text{card}(A) \equal{} \mathfrak{c}$?

Let $ S \equal{} \lbrace s_k\rbrace_{k\in \mathbb{N}}$ be a countable subset of $ \mathbb{R}$, then for each $ \alpha > 0$, $ S\subset \bigcup _k B\left(s_k,2^{ \minus{} \alpha k}\right)$.  Then for any $ \alpha > 0$ and $ d > 0$ the $ d$-dimensional Hausdorff content of $ S$ is less than $ \sum_k 2^{ \minus{} \alpha d k} \equal{} \mathcal{O}\left(2^{ \minus{} d\alpha}\right)$.  Taking $ \alpha\to \infty$ implies the $ d$-dimensional Hausdorff content of $ S$ is $ 0$.  Since $ d$ was arbitrary, the Hausdorff dimension of $ S$ is $ 0$.

Since we have shown that all countable subsets of $ \mathbb{R}$ have $ 0$ Hausdorff dimension, $ |A| > \aleph _0$.  Since $ A\subset \mathbb{R}$, $ |A|\leq \aleph _1$, and hence by the continuum hypothesis $ |A| \equal{} \aleph _1$.

I'm aware of fact that every set with non-zero Hausdorff dimension have cardinality greater or equal $ \aleph_{1}.$
What I'm interested in is: what can be shown about cardinality of set with non-zero Hausdorff dimension [b]without assuming CH[/b]?

The above post shows that my choose for name of this topic was somewhat misleading. I'm truly sorry.


function
limit
topology
complex analysis
complex analysis unsolved

Let $ (f_n)_{n \equal{} 0}^\infty$ be a sequence of analytic functions defined in the unit disk and
let
\[ f_n(z) \equal{} \sum_{k \equal{} 0}^\infty a_{n,k} \ z^k, \qquad |z| < 1, \qquad n\geq 0,
\]
be their Taylor series.
Assume that there exists a constant $ C > 0$ such that
\[ |a_{n,k}| \leq C \qquad \text{for all} \qquad k,n \geq 0,
\]
and also assume that
\[ a_k \equal{} \lim_{n \to \infty} a_{n,k}
\]
exists for all $ n \geq 0.$

Prove that the sequence $ (f_n)$ converges uniformly on every compact subset of the unit disk.

Proof:  Let K be a compact subset of the unit disc in the complex plane.

Let $ \delta \equal{} dist(K, boundaryD(0,1))$ then $ \delta > 0$ since the unit circle is closed set disjoint from the compact set K.

Now, it suffices to show that the $ f_n$ are uniformly cauchy on K.

So let $ \epsilon > 0$.

Then for $ z \in K$ $ \mid f_n(z) \minus{} f_m(z) \mid \leq \sum_{k\equal{}0}^{\infty} \mid a_{n,k} \minus{} a_{m,k} \mid |z|^k \leq \sum_{k\equal{}0}^{\infty} \mid a_{n,k} \minus{} a_{m,k} \mid {\delta}^k$

Now since $ lim_{n \to \infty} a_{n,k} \equal{} a_k$ and $ l im_{m \to \infty} a_{m,k} \equal{} a_k$ there exists and $ N \in \mathbb{N}$ such that for all $ n,m  \geq N$ we

have that $ \mid a_{n,k} \minus{} a_{m,k} \mid \leq (1\minus{}\delta)(\epsilon)$

So that $ \sum_{k\equal{}0}^{\infty} \mid a_{n,k} \minus{} a_{m,k} \mid {\delta}^k \leq (1\minus{}\delta)(\epsilon)\sum_{k\equal{}0}^{\infty} {\delta}^k \leq \epsilon$ whenever we
have that $ m,n \geq N$ for any $ z \in K$


function
parameterization
LaTeX
real analysis
real analysis unsolved

Hello,
Some paper I am reading analizes the asymptotic behaviour of some algorithm.
I shows that some quantitiy is smaller than some function (of m,n,d) which is O(n*m^((d-2)/3))
[url]http://img516.imageshack.us/my.php?image=bigo1cb0.gif[/url]

(n,m,d are some parameters, what they represent is irrelevant to my question, as you will soon see)

Then they claim that since m<= C(d,3)*n*(2^3)
[url]http://img185.imageshack.us/my.php?image=bigo2yh3.gif[/url]

(the parentheses thing is the binomial coefficient)

It follows that the bound can be expressed in terms of d & n as:
[url]http://img171.imageshack.us/my.php?image=bigo3wq7.gif[/url]

They claim that without proving it.

*) My Question Is: - How Do I Prove it? I am trying for some time with no success.

**) Reminder, big O notation.
f(n) = O(g(n))
iff  exists C such that  f(n) <= C g(n) for sufficiently large n.
similarly in 3 variables it's something on the lines of:
f(n,m,d) = O(g(n,m,d)) iff  exists C such that  f(n,m,d) <= C g(n,m,d) for sufficiently large n m & d.

Thank you very much!

Apparently, either they made a mistake in their computation, or you copied some formula incorrectly. The power of $ n$ is fine but the rest should be $ d^{d - 2}\left(\frac43\right)^{\frac {d - 2}{3}}$, which is much larger than the declared bound for large $ d$.

By the way, we all will appreciate it greatly if you spend half an hour learning basic $ \text{\LaTeX}$ (all needed instructions are right on this site). It'll spare both your and our time :).

Thank you, I have double checked, I copied just as written in the paper, "Binary Space Partitons of Orthogonal Subdivisons" by Hershberger, Suri & Toth.

I got the same result as you got, that's why I was confused and posted here

I will have this thing checked out further.

Thanks again


geometry rectangle ratio MATHCOUNTS LaTeX

Muppy wanted to build a corral for his horses, cows, and ducks. He has 600m of fencing material and he wants to make a corral sort of in this shape:
_______________________
|******|******|******|
|******|******|******|
|******|******|******|
|******|******|******|
|_______|_______|_______|

He wants the three sections to be congruent rectangles, and he wants the area to be fenced to be the greatest possible. What is this area, in square meters?

[hide]I believe the big rectangle has to be in a 2:1 ratio so 150*75 = [b]11250[/b]?[/hide]

that is so stolen from a mathcounts target...

[quote="biffanddoc"][quote="lotrgreengrapes7926"][hide] Call the 4 vertical fences length $x$. There is $600-4x$ left, so each horizontal fence is $300-2x$. The area is $x(300-2x)=-2x^2+300x$. The vertex is at $-\frac{b}{2a}=-\frac{300}{-4}=75$. $-2(75)^2+300(75)=\boxed{11250}$. :D [/hide]
Please hide your answers.[/quote]
woah thats weird $\LaTeX$[/quote]
Not my fault. I don't know why it's like that. But simple expressions like $3x$ are still the same.