Here's an problem I made. Very easy, but hope you appreciate it:

If $ {a, b, c, d, e, f, g, h}$ are real numbers such that
$ a\plus{}b\equal{}c\plus{}d\equal{}e\plus{}f\equal{}g\plus{}h\equal{}2$,
$ a^2*b\plus{}b^2*a\plus{}c^2*d\plus{}d^2*c\plus{}e^2*f\plus{}f^2*e\plus{}g^2*h\plus{}h^2*g\equal{}3$,
$ a^3\plus{}b^3\plus{}c^3\plus{}d^3\plus{}e^3\plus{}f^3\plus{}g^3\plus{}h^3\equal{}5$,
then find $ (a\minus{}b)^2\plus{}(c\minus{}d)^2\plus{}(e\minus{}f)^2\plus{}(g\minus{}h)^2$.

I might have made a mistake in devising the problem, so this is just a test drive. If you find flaws in the problem, please let me know!

Sorry, I overlooked something very simple. What I was trying to should have said was [u]prove this system can not exist.[/u] I found a contradiction. Anyway, if you want to do that, it should be a breeze.

How come nobody did it yet, or maybe you didn't try to do it. Anyway, here's the solution if you want it. I'll hide it if anyone else wants to do it:

[hide]There are many different ways of doing it. One of the flowery ones is
1. $ a^2*b\plus{}b^2*a\plus{}c^2*d\plus{}d^2*c\plus{}e^2*f\plus{}f^2*e\plus{}g^2*h\plus{}h^2*g\equal{}ab(a\plus{}b)\plus{}cd(c\plus{}d)\plus{}ef(e\plus{}f)\plus{}gh(g\plus{}h)\equal{}3$
Also, $ a^3\plus{}b^3\plus{}c^3\plus{}d^3\plus{}e^3\plus{}f^3\plus{}g^3\plus{}h^3\equal{}(a\plus{}b)(a^2\minus{}ab\plus{}b^2)\plus{}(c\plus{}d)(c^2\minus{}cd\plus{}d^2)\plus{}....\equal{}5$
Subtracting the 1st equation from the second gives $ (a\plus{}b)(a\minus{}b)^2\plus{}(c\plus{}d)(c\minus{}d)^2\plus{}(e\plus{}f)(e\minus{}f)^2\plus{}(g\plus{}h)(g\minus{}h)^2\equal{}2$. Since it has been established that $ a\plus{}b\equal{}c\plus{}d\equal{}e\plus{}f\equal{}g\plus{}h\equal{}2$, $ (a\minus{}b)^2\plus{}(c\minus{}d)^2\plus{}(e\minus{}f)^2\plus{}(g\minus{}h)^2\equal{}1$.
However this contradicts the value given forth by $ ab(a\plus{}b)\plus{}cd(c\plus{}d)\plus{}ef(e\plus{}f)\plus{}gh(g\plus{}h)\equal{}2(ab\plus{}cd\plus{}ef\plus{}gh)\equal{}3$, and since $ (a\plus{}b)^2\plus{}(c\plus{}d)^2\plus{}(e\plus{}f)^2\plus{}(g\plus{}h)^2\minus{}4(ab\plus{}cd\plus{}ef\plus{}gh)\equal{}(a\minus{}b)^2\plus{}(c\minus{}d)^2\plus{}(e\minus{}f)^2\plus{}(g\minus{}h)^2\equal{}10$,  the system is not possible.
2. A less flowery and more practical approach would be that $ (a\plus{}b)^3\plus{}(c\plus{}d)^3\plus{}(e\plus{}f)^3\plus{}(g\plus{}h)^3\equal{}32$, so
$ (a\plus{}b)^3\plus{}(c\plus{}d)^3\plus{}(e\plus{}f)^3\plus{}(g\plus{}h)^3\minus{}3(ab(a\plus{}b)\plus{}cd(c\plus{}d)\plus{}ef(e\plus{}f)\plus{}gh(g\plus{}h))\equal{}32\minus{}9\equal{}23$; this is equal to $ a^3\plus{}b^3\plus{}c^3\plus{}d^3\plus{}e^3\plus{}f^3\plus{}g^3\plus{}h^3$, and that contradicts the fact that $ a^3\plus{}b^3\plus{}c^3\plus{}d^3\plus{}e^3\plus{}f^3\plus{}g^3\plus{}h^3\equal{}5$.[/hide]

Let the second equation*3 and then add to the third equation you'll find(a+b)^3+(c+d)^3+(e+f)^3+(g+h)^3=3*3+5=14

but it should be 32 because of the first equation


[quote="bos1234"]If $ 1,w,w^{2}$ are the cube roots of unity, state the value of $ w^{3}$

I know that $ w^{3}\equal{} 1$, but how do I prove this??[/quote]
Didn't you define $ \omega$ to be a cube root of unity?  So $ \omega^{3}\equal{}1$.

Uh, by definition, a $ n^$th root of unity is a solution to the equation $ x^{n}-1= 0$. So if we're given that $ \omega$ is a solution to the equation $ \omega^{3}-1 = 0$, isn't it obvious that $ \omega^{3}= 1$?


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

Let $R$ be a commutative ring.  Prove that if for every $x \in R$ there is a $n > 1$ such that $x^n = x$, then every prime ideal is maximal.

Let $\underline p$ be a prime ideal. Then $D=R/{\underline p}$ is an integral domain s.t. for each $\bar x\in D$ there is some $n>1$ for which $\bar x^n=\bar x$. Since $D$ is an integral domain, if $\bar x\ne 0$, this implies that $\bar x^{n-1}=\bar 1$, so every $\bar x\ne 0$ from $D$ is actually invertible, meaning that $D$ is actually a field and $\underline p$ is maximal.


Hey guys;
           Since I am planning to apply to the schools in the SAtes, this contest is pretty iimportant to me. But my school hasn't registerd in this contest and won't either. Can I sign up on my own? Or if not, how can I still take this contest?

Pretty urgent. anyone pleasseeeeee help.!
thx in advance.!:)

[quote="lowest_reynolds"]DUDE MAN WHAT CONTEST MAN.[/quote]

oh, yeah/sorry. I mean ah... AMC.

[quote="lowest_reynolds"]OH YEAH DUDE, YOU CAN PAY FOR A CONTEST PACKAGE AND JUST GET A TEACHER TO PROCTOR YOU.

OR YOU CAN FIND ANOTHER SCHOOL THAT'S ADMINISTERING IT AND JUST GO THERE DUDE.

JUST CALL THEM EARLY AND BE LIKE, "HEY DUDE I WANNA TAKE THE AMCS MAN."[/quote]

ok, thank you so much

ya i was thinking of writing it too but my school's not offering it, so i guess i can try the above options

ya, i didnt know i could order for a contest package myself, thx lowest_reynolds

yeah, I will try to find a school ............lol

the deadline is jan 4th, so we need to hurry up.

[quote="rem"][quote="lowest_reynolds"]OH YEAH DUDE, YOU CAN PAY FOR A CONTEST PACKAGE AND JUST GET A TEACHER TO PROCTOR YOU.

OR YOU CAN FIND ANOTHER SCHOOL THAT'S ADMINISTERING IT AND JUST GO THERE DUDE.

JUST CALL THEM EARLY AND BE LIKE, "HEY DUDE I WANNA TAKE THE AMCS MAN."[/quote]
stop typing like Mysticterminator! It's annoying[/quote]
eh?
do all americans say dude?


modular arithmetic
number theory
relatively prime

Find the remainder when 123456789^123456789+987654321 is divided by 61.

I don't understand modular arithmetic so I have no clue how to do this...

You can use [url=http://www.artofproblemsolving.com/Wiki/index.php/Fermat%27s_Little_Theorem]Fermat's Little Theorem[/url], which gives that $ a^{60}\equiv1\pmod{61}$ for all $ a$ that are relatively prime to $ 61$, so $ 123456789^{123456789}\plus{}987654321\equiv48^9\plus{}27\equiv(\minus{}13)^9\plus{}27\equiv\minus{}13((13^2)^2)^2\plus{}27$
$ \equiv\minus{}13(47^2)^2\plus{}27\equiv\minus{}13(14^2)^2\plus{}27\equiv\minus{}13(13)^2\plus{}27\equiv\minus{}13(\minus{}14)\plus{}27$
$ \equiv\boxed{26}\pmod{61}$, where we have used the properties of [url=http://www.artofproblemsolving.com/Wiki/index.php/Modular_arithmetic]modular arithmetic[/url] many times.


One day ... i thought about 2 integer a and b (which is less than 30 )
i gave the sum of them to professor A and the product of them to professor B
^_^ I wanted to challenge them .
Then ,
the professor A asked professor B : "do u know a and b ? "
professor B :" i don't know ... do you know it ? "
professor A :" i don't know ,too "
professor B :" :lol:  ok , i know them "
professor A :" :D  i know them ,too "

And here is a question for u : " what is a and b ?"

[quote="grobber"]They don't know which is the sum and which is the product. They just know two numbers and the fact that one of them is the sum and one of them is the product.[/quote]

No. Actually, they know which one is the sum and which one is the product, but A only knows the sum and B only knows the product, and A and B don't exchange this information with each other.

  Darij

There was whole article in Kvant Magazine, but I can't find it :(

Yes, this is a really well-known problem.
This can been solved for $a,b$ bounded by much more than $30$. But, as far as I know, the unbounded case is an open problem.

Pierre.

[quote="Nuln"]Hello,

if I know the sum S and the product P of your integers, I can find them by solving x²-Sx+P=0. But that's too easy, so I think I missed something.  :)[/quote]
:) they only said that , they don't  exchange their number ...

[quote]you say a, b are integers,
but are a, b supposed to be greater than 1?[/quote]
sorry , a,b  >= 1 ... i thought that an integer is greater than 0 ...

[quote]Yes, this is a really well-known problem.
This can been solved for  bounded by much more than . But, as far as I know, the unbounded case is an open problem.

Pierre.[/quote]
... i can find only one root for (a,b)
 i am trying to prove that .. there is only one root for this problem !

--
can somebody find it ? and may be .. prove that ..


function
real analysis
real analysis unsolved

Let $ f_n : [0,1] \rightarrow \Re$ be a sequence of functions defined for $ n >1$ by

$ f_n(x) \equal{} n^2 x$     ;  $ 0 \le x \le \frac {1}{n}$

             $ \minus{}n^2(x\minus{} \frac {2}{n})$ ; $ \frac {1}{n} \le x \le \frac {2} {n}$

$ 0$ ; $ \frac {2}{n} \le x \le 1$

Prove that the sequence of functions convergence pointwise to $ f(x)\equal{}0$ for all $ x \in [0,1]$

My work:
When $ n \rightarrow \infty$ , $ \frac {2}{n} \rightarrow 0$,
so $ f_n \rightarrow 0$ for $ x \in (0,1]$
at $ x\equal{}0$, $ f_n(0)\equal{}0$ for all $ n$,so $ f_n \rightarrow 0$
so, $ f_n \rightarrow 0$ for all $ x \in [0,1]$

Is this the way to prove?

It's correct, but since the proof is so simple you may be demanded to use the definition of limit.

Let $ \epsilon > 0$.

1) Suppose $ x\in(0,1]$. Choose $ N\in\mathbb{N}$ so that $ \frac {2}{n} < x$. Then we have $ n > N\Rightarrow \frac {2}{n} < \frac {2}{N} < x\Rightarrow |f_n(x)| \equal{} |0| < \epsilon.$
2) Suppose $ x \equal{} 0$, then, for every $ n$, $ |f_n(x)| \equal{} |n^2\cdot 0| \equal{} |0| < \epsilon.$

So $ (\forall x\in[0,1])(\forall\epsilon>0)(\exists N\in\mathbb{N})(n>N\Rightarrow |f_n(x)|<\epsilon)$.


How many ways can $ \$16$ be changed into quarters and dimes, provided that at least one of each coin is used?

We can have from $ 2$ to $ 62$ quarters, but we need an even number.  Thus, the answer is $ \frac{62}{2}\equal{}\boxed{31}$.


Let $ x$ be a real number such that $ x\plus{}x^{\minus{}1}$ is an integer. Prove that $ x^n\plus{}x^{\minus{}n}$ is an integer, for all positive integers n.

Let $ x \plus{} x^{\minus{}1} \equal{} a \in \mathbb{N}$.
So, $ (x \plus{} x^{\minus{}1})^n \equal{} \sum_{r\equal{}0}^n \binom{n}{r}x^{n\minus{}r}\left(x^{\minus{}1}\right)^r \equal{} \sum_{r\equal{}0}^n \binom{n}{r}x^{n\minus{}2r} \equal{} a^n$.

If $ n\equal{}2$,
$ (x \plus{} x^{\minus{}1})^2 \equal{} x^2 \plus{} x^{\minus{}2} \plus{} 2 \equal{} a^2 \iff x^2 \plus{} x^{\minus{}2} \equal{} a^2 \minus{} 2 \in \mathbb{N}$

If $ n\equal{}3$,
$ (x \plus{} x^{\minus{}1})^2 \equal{} x^3 \plus{} x^{\minus{}3} \plus{} x \plus{} x^{\minus{}1} \equal{} a^3 \iff x^3 \plus{} x^{\minus{}3} \equal{} a^3 \minus{} a \in \mathbb{N}$

If $ n\equal{}4$,
$ (x \plus{} x^{\minus{}1})^2 \equal{} x^4 \plus{} x^{\minus{}4} \plus{} x^2 \plus{} x^{\minus{}2} \plus{} 6 \equal{} a^4 \iff x^4 \plus{} x^{\minus{}4} \equal{} a^4 \minus{} (a^2 \minus{} 2) \minus{} 6 \in \mathbb{N}$

If $ n\equal{}5$,
$ (x \plus{} x^{\minus{}1})^2 \equal{} x^5 \plus{} x^{\minus{}5} \plus{} x^3 \plus{} x^{\minus{}3} \plus{} x \plus{} x^{\minus{}1} \plus{} \equal{} a^5$
$ \iff x^5 \plus{} x^{\minus{}5} \equal{} a^5 \minus{} (a^3 \minus{} a) \minus{} a \in \mathbb{N}$

$ \vdots$

By PMI.

$ \left(\because \forall \text{predicates}P, (P(1), P(2), \cdots, P(k) \land \forall k \in \mathbb{N}\,[P(1), P(2), \cdots P(k) \Rightarrow P(k\plus{}1)]) \Rightarrow \forall n \in \mathbb{N}\, P(n) \right)$


nat mc is right, but you guys should show work as this is the classroom math level forum :)


function
trigonometry
limit
calculus
calculus computations

[quote="streeler"]Find the limits of the following functions as $ x\rightarrow 0$
a) $ x\sin(1/x)$ b) $ (\tan (ax))/x$[/quote]

a)
$ \minus{}1\le\sin\left(\scriptstyle\frac{1}{x}\right)\le 1$
$ \minus{}x\le x\sin\left(\scriptstyle\frac{1}{x}\right)\le x$
$ \lim_{x\to 0}(\minus{}x)\le \lim_{x\to 0}x\sin\left(\scriptstyle\frac{1}{x}\right)\le \lim_{x\to 0}x$
$ 0\le \lim_{x\to 0}x\sin\left(\scriptstyle\frac{1}{x}\right)\le 0$

Therefore $ \lim_{x\to 0}x\sin\left(\scriptstyle\frac{1}{x}\right)\equal{}0$ by Squeeze theorem.

b) $ \lim_{x\to 0} \frac{\tan ax}{x} \equal{}^{H}  \lim_{x\to 0} \frac{a\sec^2 ax}{1} \equal{} a$ where $ \equal{}^{H}$ denotes the use of l'Hospital's rule.

$ x\sin{\frac{1}{x}}\equal{}\frac{\sin{\frac{1}{x}}}{\frac{1}{x}}$

As $ x$ closes in on zero, $ \alpha\equal{}\frac{1}{x}\to\infty$.
Thus, $ \lim_{\alpha\to\pm\infty} \frac{\sin{\alpha}}{\alpha}\equal{}0$

Thanks, try to find b) without using  l'Hospital's rule.

Easy enough,

$ \displaystyle \lim_{x\to 0} \left [\frac{\tan{kx}}{x}\right ]\equal{}\lim_{x\to 0} \left [\frac{\sin{kx}}{kx}\cdot \frac{k}{\cos{kx}}\right]\equal{}k$


geometry
3D geometry
Putnam
ARML
HMMT
MATHCOUNTS
AMC

NYSML 2007 is being held from April 20th to the 22nd (I don't know details), at Stuyvesant High School.

So...  Who's going?

I'll be there with the Monroe County team.

I'm going with Albany Area Math Circle.

sidenote:
Do you know where they sell rubix cubes with stickers that CAN'T peel off??? :P

I'll be there with Westchester(-Putnam, i guess, though I'm pretty sure Westchester is a huge majority of the team).  We should all meet up at some point.

1=2: Cubesmith.com.  The tiles stay on really well but have a different feel; the stickers look ratty after a while but don't turn white like the stickers the cubes come with and don't fall off.  My (cubesmith) stickers have lasted almost 18 months, and I don't plan on changing them any time soon.  They cost like $1.50 per set of stickers plus$1.00 shipping for orders <$10.00. Thus, if your order is at least$9.25, you can save 25 cents by ordering the 75 cent Scrape-rite plastic razor blade (which is quite useful).

[quote="TimReyn"]We should all meet up at some point.[/quote]

Agreed.  Especially since I suspect that most of the UNYers who read this are going to go to ARML.

Tim- is the Westchester-Putnam team staying at the Embassy Suites, or leaving Saturday morning?  I know that MCML and DUSO are staying, but Rockland is leaving insanely early Saturday morning.

We should probably have two reunions either way: a short one during the lunch break at NYSML, and a longer one Friday or Saturday evening.

[quote="perfectnumber628"]I'll be there with Monroe County.  I heard a rumor we're staying in a hotel and it's going to be amazing.

Dude, why would you not peel off the stickers?  That's how you're supposed to solve it.  Just kidding.   :maybe:[/quote]

You are correct.  The NYC NYSML committee has spent a long time organizing a good time for you guys!

I just updated http://www.nysml.org with the planned event schedule.  Hope to see you guys there!

Cheers,
JackPo

[quote="B-tone_Party"]Especially since I suspect that most of the UNYers who read this are going to go to ARML.[/quote]

<bump>

I'm not going to ARML no matter what. :(

[quote="JackPo"]You are correct.  The NYC NYSML committee has spent a long time organizing a good time for you guys!

I just updated http://www.nysml.org with the planned event schedule.  Hope to see you guys there!

Cheers,
JackPo[/quote]

And I am seriously impressed.  This almost makes up for the fact that you consistently beat us at NYSML and ARML.  Thanks!

I am a bit surprised, though, that the meet is in NYC this year, after being in Rockland last year.

Woah back up. You have met Beth? From my team?

P.S. Where are we meeting???

[quote="B-tone_Party"]So, um, AoPS meeting at lunch Saturday?  I think 1=2 is the only person coming whom I haven't met, so we'll

Also, Beth and I are in the planning stages of an ARML reunion thing Friday night during a freeish time, though I think that only Monroe, some of AAMC, and a few people from other places (DUSO and whichever team is north of Syracuse) will be able to come.[/quote]

Chris, you have my cell-phone number, right?  If we don't find each other at lunch, call me.

Why can't I go to the ARML reunion?  And we should let anyone from PUMaC and HMMT (i.e. including Emil) come as well.

Tim: Is Westchester coming the night before?  Andre informed me that Rockland was leaving mega early Saturday morning and returning Saturday night, which would put a bit of a restriction on reunion times, so I assumed Westchester was doing the same.  Also, concerning who can come, the boundary between AoPSers, ARMLers, and hardcore mathletes in general can be kinda blurred.  Torin went to the ARML AoPS meeting last year, and I'm still not sure whether or not he's an AoPSer.

Also, I do have your cell phone number.

1=2, I know Beth from ARML, as well as quite a few other AAMCers.  If lunch is consolidated in one place, just ask Beth or Drew (or pretty much any AAMCer who went to ARML) to point me (Chris Kauffman) out to you.  Barring that, just look for a small group of people from ~4 different teams.

That was my assumption as well, but for some reason we are going in Friday night.  Not sure why...

There is a special event prepared for NYSML participants the night of NYSML.  We have reserved part of the American Museum of Natural History for you guys.  So you guys can always meet there in the "Hall of the Universe".

When is the ARML/HMMT/PUMaC reunion? during lunch?

I got a 7, missing number 6 (like many).

My team (southern tier) was completely unmentioned.

Did any of you have to wake up really early to go to the meet? I had to wake up at 3:45am  :(

Sorry, B-tone, I had to wake up at 3:00(AM, of course),(I live an hour north of Albany, maing it a four hour drive(!))so I forgot to ask Beth to point you out. :(

I was Matthew "Bobbitt". My real last name is Babbitt.

I also got a 7, missing #6. I tied with 6 other people on my team, and they all missed #6.

I was expecting something like a.................... say...........  5.

Sorry I couldn't attend any of this years meetings, I'm so busy this year. :cleaning:

So, did anyone hear Reuter say that he had a good chance of getting 65 applications?  'Cuz that would be amazing.

Also, problem 6 was not at all fun.  Though it might have been necessary to make sure we didn't have eleven billion 8s.  I'd have rather had a few problems that were hard enough to weed people out, though, instead of one freakishly annoyingly hard one.

Yeah, I'm pretty sure our team will do awesomely at ARML.  Though I don't see how we can reasonable fit 65 people on the bus.  We may need to bring a few limos too.

Actually I'd like to find out how many people got 7s.  I thought all the problems were easy except #6, but it turned out only 3ish people on our team got a 7.  Like we were all sitting there boredly after solving 1 and 2, thinking, "maybe I should check my work again.  No, instead I'll draw a duck."

I was really scared after answer the questions with answers $3, 1$ and $\sqrt{17}$. I checked my work at least 50 bajillion times in those 600 seconds. I have never assigned a specific number of zeros to my number "1 bajillion," so the average I have spent on checking over the problem once ranges from $12$ seconds to $1.2 \cdot 10^{-\infty}$ seconds. Cool!

http://www.kodakgallery.com/I.jsp?c=efh7xhl.bilgcc0t&x=0&y=tjxxsg

There's a bunch of awful pictures in there, but I'm too lazy to delete them.

Nice pictures!!!

How long did it take you to get to NYSML from your home/school?

For me it took about 4 or 5 hours. :(

Well, I live in Westchester, so it *should* have taken about 1.5 hours.  Given that we got horribly lost in NYC, it took about 3 hours.

That might be one of the best pictures of Reuter that I have ever seen.

Also, one of the worst pictures of me that I have ever seen (The one at the ARML reunion table.)

Also, it took us 6-7 hours, I think.

Hmm, Steph doesn't look great either.  Really, only Jarrett looks okay.  Maybe I should have announced that I was taking a picture so that people would close their mouths and open their eyes.  Well, I was standing up with my camera...

I may make the Reuter picture my avatar...cropped so that his face is the entire picture.  Only the Upstate New York people would know who it is...but we're the only people who matter anyways.  :lol:

[quote="TimReyn"]I may make the Reuter picture my avatar[/quote]
You definitely need to do that.   :rotfl:  :rotfl:

[quote="TimReyn"]I may make the Reuter picture my avatar...cropped so that his face is the entire picture.  Only the Upstate New York people would know who it is...but we're the only people who matter anyways.  :lol:[/quote]

Not [i]only[/i] upstaters, just [i]mostly only[/i].  It looks good!

Ah.  Remove the word "upstate" and we're good.

I've updated nysml.org into a wiki, and now you guys can edit parts of the site and also upload your photos onto NYSML's gallery (like George's photo!).

BTW, George is my hero.  (Joel is my lover).

[quote="JackPo"]BTW, George is my hero.[/quote]

Actually, Reuter said that, now that Frank Fan is in college, you're his new hero, which makes things king of recursively interesting.


geometry
perimeter
ratio
geometric transformation
reflection
rectangle
geometry proposed

Let S be a square, Q be the perimeter of the square, and P be the perimeter of a quadrilateral T inscribed within S such that each of its vertices lies on a different edge of S.
What is the smallest possible ratio of P to Q?

[quote="grobber"]Is it $\frac 1{\sqrt 2}$? That's what I keep getting, and I think I can prove it, but I'm not sure it's Ok. I'd like to know if what I'm trying to show is correct first :).[/quote]

Yes! Your claim is correct, or to be more precise is the same result I found:-) : I don't know the "ufficial" solution.

I faced this nice problem few years ago: it was proposed as one of selection tests for the admission to the "Scuola Normale di Pisa".

Just this days I seen it again on the web from the source I referenced. But I don't know who is the original proposer of the problem: russian or italian or others.

I found a very short, purely geometric and extremely elementar solution of this problem.

I had a rather ugly (although not too ugly :)). I think I also have a geometric solution now: use three reflections in the sides of the square: reflect $ABCD$ in $CD$ to get $A'B'CD$, then $A'B'CD$ in $B'C$ to get $A''B'CD''$, and then this last square in $A''B'$ to get $A''B'C'''D'''$. The perimeter we're of the quadrilateral is the length of a broken line $MN$ with $M\in AD,\ N\in A''D'''$ s.t. $MD=ND'''$, and this line has the smallest length when it's straight, and it's easy to see that its length is, in this case, the length of a diagonal of a square with the side $2AB$, which is $2\sqrt 2 AB$, and since the perimeter of the square is $4AB$, that's pretty much it.

The equality is reached when the inscribed quadrilateral is a rectangle having its sides parallel to the diagonals of the square.

very beautiful proof! Fagnano style?

I have a different one.

If ABCD is the square and EFGH the quadrilateral inscribed with E on AB, F on BC, G on CD and H on DA, let E',F', G', H' be the orthogonal projection of EFGH on diagonal AC.

It holds that:

HE > H'H + EE' = AH' + AE'

EF > E'F'

FG > F'F + GG' = CF' + CG'

GH > G'H'.

Given that CG' + G'H' + H'A = CA and AE' + E'F' + F'C = AC

it results that P >= 2AC.


[quote="minhphuc.v"]Let $ a,b,c > 0,\;\;and \;\;a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$
show that:
\[ \frac {a}{b^2 \plus{} c^2} \plus{} \frac {b}{a^2 \plus{} c^2} \plus{} \frac {c}{a^2 \plus{} b^2}\geq\frac {3}{2}\sqrt {3}\]
[/quote]
It is not hard
We rewrite it into
$ \sum {\frac{a}{{1 \minus{} a^2 }}}  \ge \frac{{3\sqrt 3 }}{2}$
Then
We easily can prove that$ \sum {\frac{a}{{1 \minus{} a^2 }}}  \ge \frac{{3\sqrt 3 }}{2}a^2$
which equivalent to$ (a \minus{} \frac{{\sqrt 3 }}{3})^2 (a \plus{} \frac{2}{{\sqrt 3 }})a \ge 0$ :D

thanks you. you are good at that fied. i am learning maths in taynguyen university . i hope you can hepl me in next times.

TRAN THAI HUNG, I do not understand how you have proved the inequality.
Could you explain it to me again.

varunrocks: he was able to reduce a term in the inequality into a single variable term using $ a^2\plus{}b^2\plus{}c^2\equal{}1$. Note that $ a,b,c\in(0,1)$.

So this reduces to his claim
$ \sum_{cyc}\frac{a}{b^2\plus{}c^2}\equal{}\sum_{cyc}\frac{a}{1\minus{}a^2}\ge\sum_{cyc}\frac{3\sqrt3}{2}a^2$.

Strip the $ \sum_{cyc}$ and the rest is straightforward.


modular arithmetic
inequalities
quadratics
number theory unsolved
number theory

Let $p$ be a prime that $p=1(mod 4)$. Prove that for all integers $a$ that $p$ doesn't divide $a$ there are integers $x,y,z$ that $p$ divides none of them and $a = x^2+y^2+z^2(mod p)$

Apparently, $p$ does not need to be $1\pmod 4$. We just need to disregard some special cases (like $p=3$, for example). Use the Cauchy-Davenport inequality: for $A,B\in\mathbb Z_p$, we have $|A+B|\ge \min(p,|A|+|B|-1)$. We use this twice. First, for $A,B$ choose the set $Q$ of all quadratic residues (non-zero). We get $|Q+Q|\ge p-2$. Then, choose $A=Q+Q,B=Q$. We get $|Q+Q+Q|\ge\min(p,p+\frac{p-1}2-3)$. That minimum is $\ge p$ when $p\ge 7$, so that $\frac{p-1}2\ge 3$.

This means that the assertion is true for all $p\ge 7$ (because $|Q+Q+Q|\ge p$, and it's also $\le p$, since $Q+Q+Q$ is a subset of $\mathbb Z_p$, which has $p$ elements), and we check it by hand for $p=5$ (it's false for $p=3$).


geometry
circumcircle
incenter
geometry proposed

A circle having common centre with the circumcircle of triangle $ABC$ meets the sides of the triangle at six points forming convex hexagon $A_{1}A_{2}B_{1}B_{2}C_{1}C_{2}$ ($A_{1}$ and $A_{2}$ lie on $BC$, $B_{1}$ and $B_{2}$ lie on $AC$, $C_{1}$ and $C_{2}$ lie on $AB$).
If $A_{1}B_{1}$ is parallel to the bisector of angle $B$, prove that $A_{2}C_{2}$ is parallel to the bisector of angle  $C$.

[i]Proposed by S. Berlov[/i]

I think that, something is not correct in the notations as the problem states.

Kostas Vittas.

If circumcenter is replaced by incenter $I$ then the result is true. In this case by the simmetry we have $BI\perp A_1C_2$ which implies, by the condition, that $A_1C_2\perp A_1B_1$, so $B_1C_2$ is a diameter and then $A_2C_2\perp A_2B_1\perp CI$.


geometry
circumcircle
inequalities
trigonometry
absolute value
geometry solved

Let be given triangle $ABC$ with its circumcircle $(O,R)$ and incircle $(I,r)$.$M$ is a point inside $ABC$.Prove that:
$2r \leq \frac{MA.MB.MC}{R^{2}-MO^{2}}$

[quote="James Potter"]Let be given triangle $ABC$ with its circumcircle $(O,R)$ and incircle $(I,r)$.$M$ is a point inside $ABC$.Prove that:
$2r \leq \frac{MA.MB.MC}{R^2-MO^2}$[/quote]

Since O is the center and R is the radius of the circumcircle of triangle ABC, the expression $R^2-MO^2$ is the absolute value of the power of the point M with respect to the circumcircle of triangle ABC. Denoting this absolute value by p, our inequality becomes $2r\leq\frac{MA\cdot MB\cdot MC}{p}$, or, equivalently, $\frac{MB\cdot MC\cdot MA}{p}\geq 2r$. But this was proven in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=19189#p19189]http://www.mathlinks.ro/Forum/viewtopic.php?t=5658 post #2[/url].

  darij

Dear Darij.

I have read your solution and I think it too long. I have another solution. It quite short and nice:

Denote that
+) the line MA,MB,MC intersect (O) at A',B',C' respectively
+) X,Y,Z are the orthogonal projections of M on BC,CA,AB respectively
+) k is the circumradius of triangle A'B'C'
We have:
1) $\frac{MA.MB.MC}{R^{2}-OM^{2}}=\frac{MA.MB.MC}{MA.MA'}=\frac{MB.MC}{MA'}=\frac{MB.AC}{A'C'}=\frac{MB.\sin\widehat{ABC}}{\sin\widehat{A'B'C'}}$
2) It's easy to show that $\widehat{A'B'C'}=\widehat{XYZ}$
3) $MB\sin\widehat{ABC}=XZ$
From 1,2 and 3 we have:
$\frac{MA.MB.MC}{R^{2}-OM^{2}}=\frac{XZ}{\sin\widehat{XYZ}}=2k$
You and me know $k \geq r$ thus $\frac{MA.MB.MC}{R^{2}-OM^{2}}\geq 2r$
This solution is a bit similar yours but it's shorter and more beautiful.Do you agree?

                                                                              James


inequalities
calculus
derivative
trigonometry
vector
geometry unsolved
geometry

[color=darkblue]Let $ ABC$ is a triangle, known that three angle $ \angle A, \angle B, \angle C$ are acute angle and $ tan^nA \plus{} tan^nB \equal{} 2tan^n(\frac {A \plus{} B}{2})$. Prove that $ ABC$ is isosceles triangle.[/color]

I think this is wrong and may be this:
\[ tan^nA\plus{}tan^nB\equal{}2tan^n(\frac{A\plus{}B}{2})\]

[quote="m bahador"]I think this is wrong and may be this:
\[ tan^nA \plus{} tan^nB \equal{} 2tan^n(\frac {A \plus{} B}{2})
\]
[/quote]
yes. I think so. But I'm not sure.

[quote="HTA"][quote="m bahador"]I think this is wrong and may be this:
\[ tan^nA \plus{} tan^nB \equal{} 2tan^n(\frac {A \plus{} B}{2})
\]
[/quote]
definitely , i think , 'cause if this condition is true then the isosceles triangle must have A=B , since then it should be $ tan^nA \plus{} tan^nB \equal{} 2tan^n(\frac {A \plus{} B}{2})$  :)

if so , Jensen's inequality works  :)[/quote]

[color=darkblue]Hi HTA. How you use Jensen's inequality to done this problem?[/color]

[quote="Nemion"]Making the computations, the second derivative of $ \tan^{n}{x}$ is:

$ 2n\frac {\sin^{n \minus{} 2}{(x)}(\sin^{2}{(x)} \plus{} \frac {n \minus{} 1}{2})}{\cos^{n \plus{} 2}{(x)}} > 0$ since $ x\in (0,\pi)$

Using Jensen, equality can hold only when $ A \equal{} B$[/quote]

[color=darkblue]Thank Nemion. Can you help me understand: [hide]$ \overrightarrow{IA} \equal{} \overrightarrow{A} \minus{} \overrightarrow{I}$[size=150]?[/size][/hide][/color]

[quote="Nemion"]In vectors notation, the vector that goes from $ I$ to $ A$ is $ \overrightarrow{IA}$ and it is equivalent to the subtraction: $ \overrightarrow{A} \minus{} \overrightarrow{I}$.
Look here, in the [b]Addition and scalar multiplication[/b] section the author explains what you want to know.
[u][color=blue]http://en.wikipedia.org/wiki/Vector_(spatial)[/color][/u][/quote]
Thank you very much. I see.


function
integration
inequalities
calculus
real analysis
real analysis unsolved

If $f, g$ are two integrable functions on $[a, b]$, $f$ is non-increasing and $0\leq g\leq 1$. Prove that \[\int_{b-\lambda}^{b}f\leq\int_{a}^{b}fg\leq\int_{a}^{a+\lambda}f,\] where $\lambda=\int_{a}^{b}g.$

[quote="Cezar Lupu"]If $f, g$ are two integrable functions on $[a, b]$, $f$ is non-increasing and $0\leq g\leq 1$. Prove that

\[ \int_{b-\lambda}^{b} f\leq\int_{a}^{b} fg\leq\int_{a}^{a+\lambda}f,  \]

 where $\lambda=\int_{a}^{b}g.$[/quote]

$\int_{a}^{b} fg = \int_{a}^{b-\lambda} fg + \int_{b- \lambda}^{b} fg = \int_{b- \lambda}^{b} f + \int_{a}^{b-\lambda} fg - \int_{b- \lambda}^{b} f(1-g) \geq \int_{b- \lambda}^{b} f + f(b-\lambda)\int_{a}^{b-\lambda} g - f(b-\lambda)\int_{b- \lambda}^{b}(1-g) =\int_{b- \lambda}^{b} f +f(b-\lambda)(\int_{a}^{b-\lambda} g +\int_{b- \lambda}^{b}g - \lambda) = \int_{b- \lambda}^{b} f$

So the left-side inequality is proved. In the same way splitting the integral in two parts we can prove the right side inequality.


What is the value of the product $(\sin{\frac{\pi}{5}})(\sin{\frac{2\pi}{5}})(\sin{\frac{3\pi}{5}})(\sin{\frac{4\pi}{5}})$?

[hide="Partial answer"]$\sin \frac{\pi}{5}=-\sin \frac{4\pi}{5}$, $\sin \frac{2\pi}{5}=-\sin \frac{3\pi}{5}$, so $(\sin \frac{\pi}{5})(\sin \frac{2\pi}{5})(\sin \frac{3\pi}{5})(\sin \frac{4\pi}{5})=\sin ^2\frac{\pi}{5}\cdot \sin ^2\frac{2\pi}{5}$.[/hide]

I don't know $\sin \frac{\pi}{5}$ offhand, though...

[hide]It is  (5/16)[/hide]For any of this kind of question look here: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53316&postorder=asc&highlight=sin+1%5Eo&start=20]URL[/url]

Another fun way ...

[hide]We know that (use Demoivre)) $\sin (5\theta) = 5 \cos^4 \theta \cdot \sin \theta -10 cos^2 \theta \cdot \sin^3 \theta + \sin^5 \theta$

Let $x=sin \theta$  we get:
RHS = $x(5(1-x^2)^2 -10(1-x^2)x^2 + x^4) = x(16x^4+ \dots x^2 +5)$

When $x=sin (\pi/5 )$ or  $sin (2 \pi/5)$ or $sin (3 \pi /5)$  or $sin (4 \pi/5)$  LHS is obviously zero, Hence the product of the roots

 = $5/16$ :)  [/hide]

Another way given in the previous thread
[hide]
   Use   $\sin (5\alpha) = 2^4 \ \sin (\alpha) \ \sin (\alpha + \pi/5) \ \sin (\alpha+2\pi/5) \ \sin(\alpha+3\pi/5) \ \sin(\alpha+4\pi/5))$
and take the limit when $\alpha$ tends to zero. [/hide]

One can us similar logic if this were not pi/5 but, say pi/7  (and there were 6 terms ---the answer would be (7/2^6)  etc...)

(BTW [hide]$sin(\pi/10) = (\sqrt 5 - 1)/4$  and that  is not so hard too but not that much fun ) [/hide]

I was wondering where this thread went.   :roll:
What is DeMoivre's formula, and what is it used for? (anything about complex numbers?)

Yes, something about complex numbers ;)

http://mathworld.wolfram.com/deMoivresIdentity.html
http://en.wikipedia.org/wiki/De_Moivre's_formula
http://planetmath.org/encyclopedia/DeMoivresFormula.html

[quote="Gyan"]Another fun way ...

[hide]We know that (use Demoivre)) $\sin (5\theta) = 5 \cos^4 \theta \cdot \sin \theta -10 cos^2 \theta \cdot \sin^3 \theta + \sin^5 \theta$

Let $x=sin \theta$  we get:
RHS = $x(5(1-x^2)^2 -10(1-x^2)x^2 + x^4) = x(16x^4+ \dots x^2 +5)$

When $x=sin (\pi/5 )$ or  $sin (2 \pi/5)$ or $sin (3 \pi /5)$  or $sin (4 \pi/5)$  LHS is obviously zero, Hence the product of the roots

 = $5/16$ :)  [/hide]
[/quote]
If you don't know De Moivre's theorem, here's another (simple but long) way to find the formula for $\sin{5\theta}$.
[hide]
$\sin{5\theta}$
$=\sin{4\theta}\cos{\theta} +\cos{4\theta}\sin{\theta}$
$=2\sin{2\theta}\cos{2\theta}\cos{\theta} +(1-2\sin^2{2\theta})\sin{\theta}$
$=4\sin{\theta}\cos^2{\theta}(1-2\sin^2{\theta}) +\left(1-2(2\sin{\theta}\cos{\theta})^2\right)\sin{\theta}$
$=4\sin{\theta}(1-\sin^2{\theta})(1-2\sin^2{\theta}) +\sin{\theta}\left(1-8\sin^2{\theta}(1-\sin^2{\theta})\right)$
$=16\sin^5{\theta}-20\sin^3{\theta}+5\sin{\theta}$

[/hide]


what is the smallest multiple of 99 whose sum of digits is 99 that begins and ends with 97

Is the answer :
9799999890597

I just used the divisibility properties of 11 and 9. Is the answer right?

He asked for the [i]smallest[/i] number with these properties. Therefore, if $ 9799999890597$ works, then so does $ 9750989999997$, which is smaller and is also divisible by $ 9$ and $ 11$. However, to prove it, we must show that there is no 12-digit number that satisfies these conditions. Consider $ 97x999999997$ and $ 979y99999997$, where $ x$ and $ y$ are digits that would make each number divisible by $ 99$. By adding up the digits of each number, $ x,y\equiv 4 (mod 9)$. By taking the alternating sums of the digits of each number, $ x\equiv 2 (mod 11)$ and $ y\equiv 5 (mod 11)$. There are no solutions for $ x,y\in\{0,1,2,3,4,5,6,7,8,9\}$, so there are no 12-digit number that satisfies these conditions.

Oops, i messed up :oops: . But my solution was similar to yours (taking alternating sums).

similarly:,[hide]what is the smallest multiple of 999 that begins and ends with 997[list][/hide][/list]


calculus
integration
function
logarithms
absolute value
calculus computations

Hello, I was hoping someone could help me with this simple question:

Set up the complete integral from 1 to 3 of | ln x/2 | dx

Usually when I have an absolute value problem I determine where the function is negative and positive and break up the integral into these corresponding parts. However, the ln throws me for a curve since a negative x value (without absolute value of course) would yield an error. Can someone help me out please?

Thanks

Well, first we have to dertermine when $\ln\left(\frac{x}{2}\right)=0$, we have
$\frac{x}{2}=1\Rightarrow x=2$, so the function is negative for $x<2$, thus the integral can be written as
$\int_2^3 \ln\left(\frac{x}{2}\right) \ dx - \int_1^2 \ln\left(\frac{x}{2}\right) \ dx$


AMC
AIME
geometry
USA(J)MO
USAMO
Support
AMC 10

Dear Mr. Dunbar, is AMC B harder that AMC A this year?

How would he know?  Presumably the AMC hasn't even received most of the B tests yet, much less graded them.

Thanks for the update. I have some friends who will be excited to hear that the qualifying score is below $ 100$. Also makes me glad I advised them not to game the scoring system and answer exactly $ 14$ questions or whatever.

[quote="AMCDirector"]We do not yet have complete statistics on the 2008 AMC 10 [b]A[/b] and 2008 AMC 12 [b]A[/b].  Those will be completed after all the school reports are generated. [/quote]
Typo? EDIT: Ah I see, I thought you meant stuff like cutoffs by "statistics".

No typo, I truly mean AMC 10 A and AMC 12 A.  We do not yet have the statistics such as mean, item analysis, averages by grade and gender, etc that we include in the Summary of Results and Awards.  Those are the kinds of statistics that indicate the relative difficulties between tests and across years.   We will run those statistics after we get the school reports out, the school reports are far more important right now.

At this time (Thursday, Feb 28), I would estimate that we have received back 10 B and 12 B answer from fewer than 5% of all the participating schools.  We will not be able to announce or even estimate AIME invitation scores afor about 2 weeks.

Steve Dunbar
MAA Director, American Mathematics Competitions

This may sound like a stupid question, but does the AMC mail the official score reports to schools in any order(for example would AMC mail a school like Apple Middle school before Zebra High School)?

Generally speaking, there is no order in which the reports are emailed or mailed.  States with large enrollments (California, Texas, Florida, New York, etc. ) tend to get done first, but that is not a hard and fast rule.   It really amounts to the stack of tests that is first at hand gets done first.

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote="maomao13"]Hello, e^(pi*i)+1=0. We are from same state. May I ask which school are you going to?[/quote]

I'm a member of the Howard Area Homeschoolers.

oh I heard that before. But is that a school or you are a homeschooler?
Your Name initial is MP?

I'm a homeschooler.

EDIT: Um, no, those aren't my initials, and I don't know whose they would be.

I just got my AMC 10A score. It gives me two results for the same contest and I don't know why. The e-mail my teacher received says my scores are 136.5 and 127.5. I only filled out one answer form and I left none blank so I don't know how I got a .5. Also, when I looked at the official solutions after I had submited my answers to to the teacher, I saw that I got everything right.
Is there any way to get my paper remarked?
I don't want my USAMO index to be made lower.

[quote="mathgirl_fractals"]Is there any way to get my paper remarked?
I don't want my USAMO index to be made lower.[/quote]There are procedures for having your AMC rescored.  See page 12 of the Teachers' Manual (your teacher should have a copy, or it is available on the [url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/1012TM/2008AMC1012A-TM.pdf]AMC web site[/url]).  But I would probably call the AMC first and determine whether what you are viewing is a reporting error or a scoring error.  (The contact info for the AMC is also in the Teachers' Manual.)

I know someone who will be happy to hear the cut-off scores. It also shows that I got a bit lucky, because if I hadn't guessed any quesitons right (I guessed 2 right), I would've fallen short of the cut-off. Glad I still made the AIME, though. Surprisingly it shows that the AMC 12 didn't make the cut-off lower than last year, despite that many people thought it was harder.

[quote]I just got my AMC 10A score. It gives me two results for the same contest and I don't know why. The e-mail my teacher received says my scores are 136.5 and 127.5. I only filled out one answer form and I left none blank so I don't know how I got a .5. Also, when I looked at the official solutions after I had submited my answers to to the teacher, I saw that I got everything right. [/quote]

The AMC office also got a call from the teacher.   Having the teacher or Contest Manager handle the problem through our office is the best way to address a problem of this sort.

The explanation is that this school's answer forms got wet somehow, and arrived here wrinkled.  The wrinkled answer sheets were put through the scanner, and not surprisingly were not read correctly.   In fact, this particular sheet must have gotten stuck in the scanner and was put through again, and that accounts for the double listing, and the wrinkled sheets being misread accounts for the score errors.  The answer forms have now been re-read by hand and scores entered correctly.

This is the first time that I can re-call that particular set of circumstances occurring.  In that time, we have successfully scanned several million answer forms.  The error is entirely due to the answer forms getting wet and not passing through the scanners.  The scanners are very particular, since they pick up a single sheet approximately once each second and read it completely in one pass.  That's why we emphasize the need to darken the circles completely, and be careful with erasures.

If you believe something is wrong with our scores, the Teachers'  Manual has a Rescoring Request Form and submitting that to our office will address the problem.

Steve Dunbar
MAA Director, American Mathematics Competitions

Dear Mr. Dunbar,

I am a 7th grader. My school is not releasing my AMC 10A score but they are telling me I am moving on to the AIME. My school also said that I had a score above 90, but they can't tell me my score. The only score they are releasing is the highest score in the school (117.0, which is not my score). If mine is not the highest score and the cutoff is 117, how is it possible for me to move on to the AIME? Is there a phone number that I can call to find out my score?

[quote]I am a 7th grader. My school is not releasing my AMC 10A [/quote]

With only this information, this sounds to me like a school policy being applied.  The best way to resolve this would be to have an adult, probably your parents, contact the school where the contest was administered.  I would suggest first contacting the School Contest Manager who adminstered the contest and therefore received the school report with your score information, then the Math Chair, then the principal, until you receive either your scores or an understanding of the policy being applied.

As a general rule, the AMC office does not give out scores for individuals.   We especially do not give out individual score reports in response to emailed or phoned requests.  The reasons are several:
(1) As a practical matter, with over 250,000 students taking the AMC 10 and AMC 12, even a 1% request rate would quickly overwhelm our office.  The best way to handle this is distributed information through the schools who administer the contests.
(2) The Federal Educational Right to Privacy Act (FERPA) is a federal law that limits who may receive information about a student.  Basically, test score and grade information may only be given to parents or legal guardians for minors and only to the individual for non-minors.  While the law probably doesn't directly apply to the AMC, it provides a consistent framework for our policies.  For a score request, we usually would need a formal request letter from a parent, with a documented reason for needing that knowledge.
(3) It is our intention and process design to work primarily with teachers and schools in support  of math education.  Our systems are designed around schools and teachers as the base level of registration, administration, and reporting.

Steve Dunbar
MAA Director, American Mathematics Competitions


I'm very interested in Astronomy but we weren't taught it at school.

I have tried learning it by looking for information on the internet but there's too much information that I really don't know where to begin.

Can anyone help me? I think I need a specific material in pdf-file or in webpage.

Thanks for any help.

http://space.about.com/od/basics/a/whatisastro.htm

That is for what Astronomy is.

:) Astronomy is basically the study of the solar system and the universe, how planets, stars, and galaxies form and interact.
Try google and wikipedia.
There are some other sites in other topics posted in this forum also.

:maybe: I have try google and wikipedia but there's too much information with no order and I don't know what to read first because there are some words that I don't understand and when I try to find the meaning of these words, I have to read the whole definition which may contain other difficult words.

I think I need some materials with order (like the textbooks) :roll:

Okay. Then, it depends on what you want to read about.
Do not know of any E-books for Astronomy though.

I'm a person who like astronomy as you!Nice to meet you! :)
Nobody has any textbook about astronomy?(I ask because i wasn't taught it in my school :( )

Good places to get astronomy books are at local colleges. Sometimes professors are willing to lend out books they have that they aren't using anymore.

Another good place to start learning about astronomy is to watch television series--such as "The Universe" on the History Channel in the U.S.--because they give an overview on many topics, and that makes it easier to choose a topic that interests you.

Many elementary Physics books like Halliday & Resnick contain a bit of Astronomy (In chapters as well as through problems). In any case you will need at least that much Physics to do real Astronomy. After this, Introduction to Astrophysics by Zeilik & Smith/Gregory is a good text, but a little above High School level.

I know a book about Introduction to astrophysics but by Nguyen Quang Rieu.It's good book too.
By the way,I want to ask you about branch of astronomy in US or France,universitys which educate it well?because In my country,university have not it! :(
Thank.

Not that many universities offer astronomy as a major here (the US), or atleast most of the ones I have looked at.

Thanks you so much.
I live in Vietnam.When I was child,i uesd to look at the sky, looked at far stars in universe.
I believe in having civilization in universe.maybe, your love for astronomy started from this,too.All right? :lol:
Send to you the best wish!

Many high rate American universities do not offer an Astronomy major, but instead offer a Physics major with astronomy concentration. Even those which do have an astro major follow a similar pattern. (I ought to know, cause I am applying myself :) ). For more information on college education for AOPSers see [url=http://www.artofproblemsolving.com/Forum/index.php?f=143]here[/url]

But when we apply for the university, will they require any astronomy or physics knowledge?

As far as I know, people who are looking to major in astrophysics or astronomy are usually expected to have taken advanced mathematics, physics and other sciences, and computer courses before college. Since astronomy isn't offered as a class in many U.S. high schools, I don't think it would be required as a prerequisite in colleges there. I'm not sure about other countries, though.


topology
function
algebra
domain
Functional Analysis
real analysis
real analysis unsolved

mabe it seems easy or classical but it is hard for me :
prove that being path-connected and arc-connected is equivalent in hausdorf space specially for metric space.

Sorry for my ignorance, but what is arc-connected?

For any two distinct points $x,y$, there is an embedding $f:[0,1]\to X$ s.t. $f(0)=x,f(1)=y$.

Replace "embedding" with "continuous function".

Here's a link:

[url]http://encyclopedia.laborlawtalk.com/Connectedness[/url]

Of course, I have a stupid question: what meaning for "embedding" do we use here?

I meant that $f$ is a homeomorphism from $I=[0,1]$ to $f(I)$.

I ahve asked this question some time back in the forum but I didnt get any answer. :(

I'll try something, but it has high chances of being flawed, I'm afraid.. :(

If $X$ is our space, and $f(0)=x_0,f(1)=x_1\in X$, we may as well assume, for the sake of simplicity, that $f$ is surjective, so all of $X$ is a continuous image of $I=[0,1]$. The Hahn-Mazurkievicz Theorem states that for Hausdorff spaces this is equivalent to $X$ being compact, connected, locally connected, and metrizable. We may thus consider $X$ to be a metric space with some metric $d$.

We can cover $X$ with finitely many connected open balls of small enough radius because of the local connectedness and compactness. The graph having these balls as vertices and in which two vertices are connected iff the corresponding balls intersect must be connected, for otherwise $X$ would be disconnected. Choose a path in this graph from a ball covering $x_0$ to a ball covering $x_1$. Let these balls be $x_0\in B_0,\ldots,B_n,x_1\in B_{n+1}$, and take $a_k\in B_{k-1}\cap B_k$. If the radius of the balls is small enough, there there will necessarily be many of them, and we can take many points in $I$, $\tilde a_k$, s.t. $g(\tilde a_k)=a_k$, $\tilde a_k$ is an increasing sequence, and the distance between $\tilde a_k,\tilde a_{k+1}$ is small. Now, what we did for $x_0,x_1$, we do for each $a_k,a_{k+1}$, working in the ball $B_{k+1}$, which contains both $a_k,a_{k+1}$ (i.e. we take a sequence of small balls from $a_k$ to $a_{k+1}$, and so on). We get more points in $I$ between $\tilde a_k,\tilde a_{k+1}$, and repeat the process.

We'll get a dense countable set of points in $I$, and a function $g$ from the set formed by them to a countable set in $X$, and it looks right now to me like this function can be extended to the entire $I$ so that the resulting function $g:I\to X$ is both continuous and injective, and $f(0)=x_0,f(1)=x_1$, if the radii of the balls we use at each step decrease to $0$.
Any opinions? Could something like this work? :?

I think the result holds for non-metrizable Hausdorff spaces also.

But if the "solution" above is correct, then it proves the general case: I only looked at $f(I)$ (or, in other words, I assumed WLOG that $f$ was surjective), which [b]must[/b] be metrizable due to the theorem I cited.

Hmm all right but the proof is pretty compicated.Doesnt it have any elementary proof.By the way Grobber, I dont know the theorem you stated.I think the theorem requires a proof.I would be more than happy if you kindly give me some hints regarding the proof of Hahn-Mazurkievicz theorem.

If we're talking about metric spaces and such then we're out of the elementary domain, so I don't know what "elementary solution" might mean :). As for the theorem, itself (or, more specifically, the direction we use here, i.e. $X$ is a continuous image of $I\Rightarrow$ $X$ is compact, connected, locally connected and metrizable), the compactness and connectedness are immediate, but I would have to work the rest out. I do not have the time to do this now. Unfortunately, there do not seem to be any easy-to-find refrences reachable through Google :(.

I had posted a totally bogus proof here. Luckily, nobody seems to have bothered to read it :).

Ok, so we wanted to prove the half of the mentioned theorem that was used above, namely: if the Hausdorff topological space $X$ is a continuous image (through $f$) of $I=[0,1]$, then it’s compact, connected, locally connected, and metrizable.

Compactness and connectedness follow immediately from the fact that a continuous image of a compact (connected) topological space is compact (connected, respectively).

Local connectedness is equivalent to the fact that the onnected components of open sets are open. Let $U\subset X$ be an open subset of $X$, and let $C$ be a connected component of $U$. We want to prove that $C$ is open:

$f^{-1}(U)$ is open in $I$, so it’s a countable union of disjoint open intervals $(I_n)_n$. If $x\in C$ and $y\in U$ s.t. $f^{-1}(x),f^{-1}(y)$ both have non-empty intersection with some interval $I_n$, then there is a path in $U$ from $x$ to $y$, so $y\in C$.This shows that $f^{-1}(C)$ consists precisely of those open intervals $I_{n_i}$ which have non-empty intersection with $f^{-1}(C)$. We will then have $f(I\setminus\bigcup I_{n_i})=X\setminus C$, so $X\setminus C$ is an image of a closed and thus compact subset of $I$, meaning that it’s compact, and thus closed in $X$, or, in other words, $C$ is open in $X$, as desired.

  Now for the metrizability:

Our space $X$ is compact Hausdorff, so it’s normal (and thus regular). By Urysohn’s Metrization Theorem (a regular space with a countable basis is metrizable), it suffices to prove that $X$ has a countable basis. Notice that $f: I\to X$ is what is called a quotient map: it’s surjective, and a subset $U\subseteq X$ is open iff $f^{-1}(U)$ is open. The map $U\mapsto f^{-1}(U)$ is a bijection between the open sets of $X$ and the elements of a Hausdorff topology on $I$ weaker than the usual one. If we prove that this topology is actually the usual one, we’re done, since then, given a countable basis $(V_n)_n$ of $I$, there will be open sets $(U_n)_n$ in $X$ s.t. $f^{-1}(U_n)=V_n,\ \forall n$, and these will form a countable basis for the topology of $X$. It thus suffices to prove that

There is no Hausdorff topology $\tau_1$ on $I$ strictly weaker than the usual one (denoted by $\tau$).

Take any $V\in\tau$. It's complement $I\setminus V$ is closed and thus compact in $\tau$. Given any covering of $I\setminus V$ with elements of $\tau_1$, we can find a finite subcovering, because all the open sets in $\tau_1$ that we're working with are also elements of $\tau$. This means that $I\setminus V$ is compact in $\tau_1$, and since $\tau_1$ is a Hausdorff topology, $I\setminus V$ must also be closed in this topology, meaning that $V\in\tau_1$, as desired.

I hope it’s Ok this time :).

[quote="grobber"]
Now for the metrizability:

Our space $X$ is compact Hausdorff, so it’s normal (and thus regular). By Urysohn’s Metrization Theorem (a regular space with a countable basis is metrizable), it suffices to prove that $X$ has a countable basis. Notice that $f: I\to X$ is what is called a quotient map: it’s surjective, and a subset $U\subseteq X$ is open iff $f^{-1}(U)$ is open. The map $U\mapsto f^{-1}(U)$ is a bijection between the open sets of $X$ and the elements of a Hausdorff topology on $I$ weaker than the usual one. If we prove that this topology is actually the usual one, we’re done, since then, given a countable basis $(V_{n})_{n}$ of $I$, there will be open sets $(U_{n})_{n}$ in $X$ s.t. $f^{-1}(U_{n})=V_{n},\ \forall n$, and these will form a countable basis for the topology of $X$. It thus suffices to prove that

There is no Hausdorff topology $\tau_{1}$ on $I$ strictly weaker than the usual one (denoted by $\tau$).

Take any $V\in\tau$. It's complement $I\setminus V$ is closed and thus compact in $\tau$. Given any covering of $I\setminus V$ with elements of $\tau_{1}$, we can find a finite subcovering, because all the open sets in $\tau_{1}$ that we're working with are also elements of $\tau$. This means that $I\setminus V$ is compact in $\tau_{1}$, and since $\tau_{1}$ is a Hausdorff topology, $I\setminus V$ must also be closed in this topology, meaning that $V\in\tau_{1}$, as desired.
[/quote]

This, unfortunately, is completely wrong. The topology $\{f^{-1}(U)\}$ as $U$ ranges through the open subsets of $X$ is not, in general, Hausdorff, simply because $f$ is not necessarily one-to-one.

I'm posting another proof. First let's restate the problem we're concerned with:

Let $f: I=[0,1]\to X$ be a continuous surjection, $X$ being Hausdorff. Prove that $X$ is metrizable.

A compact Hausdorff space $X$ is metrizable iff the Banach space $\mathcal C(X;\mathbb R)$ consisting of real-valued continuous functions on $X$ normed with the supremum norm is separable (this has been discussed on the forum). In our case, $\mathcal C(X;\mathbb R)$ is embedded isometrically in $\mathcal C(I;\mathbb R)$ through $\varphi\mapsto\varphi\circ f,\ \forall\varphi\in\mathcal C(X;\mathbb R)$. Since $\mathcal C(I;\mathbb R)$ is separable, $\mathcal C(X;\mathbb R)$ must be separable as well (being a subspace of a separable Banach space), so we're done.


This probably *should* be an easy problem...

Three circles having equations $ x^{2}\plus{}y^{2}\equal{}1$, $ x^{2}\plus{}y^{2}\minus{}6x\plus{}5\equal{}0$, and $ x^{2}\plus{}y^{2}\minus{}2ky\plus{}2k\minus{}1\equal{}0$ are pairwise tangent (that is, each pair of circles is tangent). Determine the value of $ k$.

Answer

Let's call the circles $ X$, $ Y$ and $ Z$ respectively.

Using the appropriate circle forumlae:

$ r_{X}\equal{} 1$ and $ C_{X}\equal{} (0,\,0)$

$ r_{Y}\equal{}\sqrt{3^{2}\minus{}5}\equal{} 2$ and $ C_{Y}\equal{} (3,\,0)$

$ r_{Z}\equal{}\sqrt{k^{2}\minus{}(2k\minus{}1)}\equal{}\sqrt{(k\minus{}1)^{2}}\equal{} |k\minus{}1|$ and $ C_{Z}\equal{} (0,\,k)$

Lets assume that $ Y$ and $ Z$ externally tangent.

$ \Rightarrow C_{Y}C_{Z}\equal{} r_{y}\plus{}r_{z}$

$ \Rightarrow\sqrt{3^{2}\plus{}k^{2}}\equal{} 2\plus{}|k\minus{}1|$

Lets consider $ k\geq 1$

$ \Rightarrow\sqrt{9\plus{}k^{2}}\equal{} 2\plus{}k\minus{}1 \equal{} k\plus{}1$

$ \Rightarrow k^{2}\plus{}9 \equal{} k^{2}\plus{}2k\plus{}1$

$ \Rightarrow k \equal{} 4$

$ k \equal{} 4$ gives $ C_{X}C_{Z}\equal{} r_{X}\plus{}r_{Z}$ such that $ X$ and $ Z$ tangent. Hence $ k \equal{} 4$ is a soution.

Lets consider $ k < 1$

$ \Rightarrow\sqrt{9\plus{}k^{2}}\equal{} 2\plus{}1\minus{}k \equal{} 3\minus{}k$

$ \Rightarrow k^{2}\plus{}9 \equal{} k^{2}\plus{}6k\plus{}9$

$ \Rightarrow k \equal{} 0$

This is not possible as $ k \equal{} 0$ would mean that $ X$ and $ Z$ are coincident.

Now lets assume that $ Y$ and $ Z$ internally tangent.

$ \Rightarrow C_{Y}C_{Z}\equal{} |r_{Y}\minus{}r_{Z}|$

$ \Rightarrow\sqrt{3^{2}\plus{}k^{2}}\equal{} |2\minus{}|k\minus{}1|\,|$

$ \Rightarrow k^{2}\plus{}9 \equal{} 4\minus{}4|k\minus{}1|\plus{}(k\minus{}1)^{2}\equal{} 4\minus{}4|k\minus{}1|\plus{}k^{2}\minus{}2k\plus{}1$

$ \Rightarrow 4 \equal{}\minus{}2k\minus{}4|k\minus{}1|$

$ \Rightarrow\minus{}k\minus{}2 \equal{} 4|k\minus{}1|$

Hence $ \minus{}k\minus{}2\geq 0\Rightarrow k\leq\minus{}2\Rightarrow |k\minus{}1| \equal{} 1\minus{}k$

$ \Rightarrow\minus{}k\minus{}2 \equal{} 4(1\minus{}k)$

$ \Rightarrow k \equal{}\minus{}2$

This contradicts the fact that $ k\leq 2$. Hence the assumption that $ Y$ and $ Z$ internally tangent is false. Thus $ k \equal{} 4$ is the only solution.

[hide]
getting those three into center-radius form

$ x^{2}\plus{}y^{2}\equal{} 1$
$ (0,0)$
$ r \equal{} 1$

$ (x\minus{}3)^{2}\plus{}y^{2}\equal{} 4$
$ (3,0)$
$ r \equal{} 2$

$ x^{2}\plus{}(y\minus{}k)^{2}\equal{} (k\minus{}1)^{2}$
$ (0,k)$
$ r \equal{} k\minus{}1$

From these first two, you can see that the two circles are externally tangent. From here, it is easy.
Then, applying the distance formula between centers and using the sum of the radii as the distance, we have

$ (k\plus{}1)^{2}\equal{} k^{2}\plus{}3^{2}$

$ k^{2}\plus{}2k\plus{}1 \equal{} k^{2}\plus{}9$

$ 2k \equal{} 8$

$ \boxed{k\equal{}4}$
[/hide]

The fact that the first two externally tangent is not enough to say that they all externally tangent. Plus you have ignored the case where k is less than or equal to one. Remember, r = |k - 1|, so the case where r = 1 - k has to be eliminated.


[quote="jtw"]When will I get my scores?  :sleeping: How long does it normally take?[/quote]

Have some patience. Believe it or not, people running the USAMTS have other things going on in their lives other than grading your solutions.

The scores and feedback should be out later this week.

1. It was said that the USAMTS staff hoped to get the scores up by November 3rd to November 7th.

2. The other topic says (something like):

"We are currently scanning the scores. As it is late in the day, we will not complete it today, hopefully it will be done tomorrow, and if not, most certainly Monday"

[quote="mathemonster"]why isn't it up yet?[/quote]It's still "this week", is it not?

"nov 3 - nov 7 " so theyll be up 2nite? but take as much as u need im just wondering cause ive checked like ten times  :D

[quote="BOGTRO"]Uh... for #3 my grading feedback tells me to provide an example, which I did...

Also for #5 I got "You didn't isolate x at all. What happened to sqrt(2x[x+1])?" when I specifically stated that I was equating the rational parts, and that I could have used that, but it would be easier to use the rational parts.[/quote]

Therefore, send a protest. :)


How did everybody do?  This was a really hard contest as far as the gold division goes.  For "secret", I implemented a modified Kruskal's minimal spanning tree algorithm, a flood-fill algorithm, and a breadth-first search and still only got 3 test cases right on that problem!  :what?:

I had very simple mistakes, I changed them and in jpol it runs 7 test and in secret 8 tests. Here is my algorithms:

********
jpol: firs we must sort this 3k numbers in non decreasing order and first k city must be in first group. Then with DP we must find such K city, such that the sum of Jersey Cows must be more then 500k and must be as near as it is possible. This K city must be in second group and the last K city must be in third group. I think this algorithm is right, but it runs only 7 tests.

********
secret: first we mast sort this trails with their lengths. It’s true, that if with first K trail we can get from first landmark to N-th landmark T times, the solution is less then K-th trail’s length. We must search K with “binary search? from 1 to N. We can say that form first landmark to N-th landmark we can get T times if maximum flow from 1 to N is not less than T, and what maximum flow we can get with “Lift To Front? algorithm. This algorithm runs only 8 tests.

If anyone knows more good algorithm please post here.

I did the silver division, and got

acquire     *  *  x  *  t  t  t  t  t  *
fcount      *  *  *  *  *  *  *  *

I tried to do the other problem with flood-fill, but it was really hard becuase a lot of time, the program would do some sections more than once.  It was kind of weird though how I did better than people who usually do good (like one person got a lower score got a 812 one time), but people who usually do worse than me did better than me.

I attempted Aggressive Cows using heaps and a somewhat greedy algorithm during the contest and ended up failing most of the cases.

The following solution however passes all the given test cases. First, do a binary search for the minimum distance between the cows. Let this value be X. To test to see if C cows can be placed in the stalls with X minimum distance, take the left stall as one of the chosen ones. Ignore any stall that is closer than X to the left stall and choose the leftmost stall that is more than X distance away from the first one. Repeat the algorithm for this stall and if, at the end, mroe than C cows have been successfully placed into stalls, the value of X is an acceptable distance between cows.

Throught the binary search, you can find the maximum X that results in C or more cows being placed inside the given stalls.

It'd be great if perfect solutions for Jersey Politics and Secret are also posted here.

I did silver too. I got full score on aquire and fcount (really easy?), but couldn't do the last one. I would appreciate if someone could post a solution to that too (jersey politics)

I thought fcount was really easy also. I also got all test cases for acquire, but had some trouble with cowrig. I used a variation of the floodfill algorithm, but it took way too long on most of the cases and I only got a few right. I'm still not sure how to do that one either.

i got all cases in acquire and fcount, and got only 4 in the cowrig. i used a full search algorithm..
the following algorithm works however:
what we need to do actually, is select 7 squares from the map, check if the jerseys outnumber holsteins, and that these 7 squares are connected. If these conditions are ok, then we can count a solution.
for this, we'll identify each square of the map by a number. for example:
[code]
1  2  3  4  5
6  7  8  9  10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25
[/code]
then what we need to do, is to generate a sorted (so that we don't count the same sequence more than once) sequence of 7 integers between 1 and 25, and so, we 'll choose 7 squares among the 25. The number of possibilities are not very big, so a recursive solution will work. then, we have to make sure that the 7 selected squares are connected. we use a small floodfill routine.

by the way, if you solved the jpol and secret (gold division), please post here your solution..
thanks,


function
limit
integration
real analysis
real analysis unsolved

Let $g: \mathbb{R}_+\to \mathbb{R}$ be a continuous function such that $\lim_{x\to \infty}g(x)+k\int_0^xg(t)dt=a\in\mathbb{R}$, where $k>0$. Prove it's existence and find $\lim_{x\to \infty} \int_0^xg(t)dt$.

[hide="my approach"] for any $\epsilon$ we have $g(x)+k\int_0^xg(t)dt>a-\epsilon$ for all $x\ge x_1(\epsilon)$.
Multiplying by $e^{kx}$ and get that $(e^{kx}\int_0^xg(t)dt)'>e^{kx}(a-\epsilon)$. Integrating and dividing by $e^{kx}$ we get that $\int_0^x g(t)dt \ge \frac{a-\epsilon}{k}$ for all $x\ge x_1(\epsilon)$. Similarly for $\int_0^x g(t)dt \le \frac{a+\epsilon}{k}$ for all $x\ge x_2(\epsilon)$. Hence the limit is $\frac a k$.

Am I wrong somewhere? :? [/hide]

I don't think you are wrong.You are right :lol:
Here is my solution:
Solution:Use L'Hospital's:
\[ \lim_{x\to \infty} \int_0^xg(t)dt=\lim_{x\to \infty} \frac{e^{kx}\int_0^xg(t)dt}{e^kx}=\lim_{x\to \infty} \frac{e^{kx}\left(g(x)+k\int_0^xg(t)dt\right)}{ke^{kx}}=\frac a k  \]

very elegant solution :lol:  thank you very much!


Let $ a$ and $ b$ be positive real numbers with $ a>b$. Prove that

$ \lim\sqrt[n]{a^n\plus{}b^n}\equal{}a$

Hint: $ \left(a^n \plus{} b^n\right)^{\frac{1}{n}} \equal{} a\left(1 \plus{} \left(\frac{b}{a}\right)^{n}\right)^{\frac{1}{n}}$.
Now the condition that $ a > b \implies \frac{b}{a} < 1 \implies \left(\frac{b}{a}\right)^n \to 0$ as $ n \to \infty$.

Another variation: $ a^n<a^n\plus{}b^n<2a^n,$ so $ a<\left(a^n\plus{}b^n\right)^{1/n}<\sqrt[n]{2}\,a.$

Now all you need to know is that $ \sqrt[n]{2}\to 1$ and we have the squeeze theorem.

$ \boxed{\lim_{n\to\infty} \sqrt[n]{pa^n\plus{}qb^n}\equal{}max\{a,\ b\}}$ for $ a,\ b,\ p,\ q>0$ and $ n\in\mathbb{N}$.

I just managed to solve this today's morning. My solution used precisely akech's method. Kent Merryfield's solution is really nice! I think it can be easily extended:

$ pa^n<pa^n\plus{}qb^n<(p\plus{}q)a^n$

$ a\sqrt[n]{p}<\sqrt[n]{pa^n\plus{}qb^n}<a\sqrt[n]{p\plus{}q}$

Both $ \sqrt[n]{p}$ and $ \sqrt[n]{p\plus{}q}$ converge to $ 1$, so use the squeeze theorem (I used to know it as the "sandwich theorem").


A friend of mine who went to a Russian polytechnic university to study physics for a few weeks showed me this problem today.  It was intended to show the professor who was good at math beyond the placement test they took (of course, that's assuming my friend rememered the problem right), but it's not too hard.  My solution is a little funny, but it gets the job done.

Find the limit as n approaches infinity of ${(|x|^n+|y|^n)^{\frac{1}{n}}}$.

It's $\max(|x|,|y|)$, right? :) Suppose $|x|\ge|y|$, and write it as $|x|\cdot (1+\frac{|y|^n}{|x|^n})^{\frac 1n}$. The stuff multiplied with $|x|$ goes to $1$.

He is completely right. Note that $1<1+\frac{|y|^n}{|x|^n}\leq2$ with the grobber's hypothesis, hence $1<\left(1+\frac{|y|^n}{|x|^n}\right)^{\frac1n}\leq2^{\frac1n}\rightarrow1$.  :D


I think this problem must to post on Number Theory box . :)

No, I think post in this box is exactly because it isn't a number theory problem! Try to sovle its!


[u]Question:[/u] Write a C language program to print a truth table for the Boolean expression i&&(j||k), where i,j and k range from FALSE to TRUE.

[u]My work:[/u]
[code]#include<stdio.h>
#define TRUE 1
#define FALSE 0
#define T "true"
#define F "false"
main ()
       {
         int i=FALSE,j=TRUE,k=FALSE;
         while (i<TRUE)
                {
                  while (j<TRUE)
                        {
                          while (k<TRUE)
                                {
                                  printf("%s",i&&(j||k));
                                  k++;
                                 }
                          j++;
                        }
                 i++;
                 j=FALSE;
                }

         }
[/code]

what you say guys ,,, is it appropriate to that question ,,, ??

Umm no it isn't because it doesn't print anything at all!

Try this instead:

[code]
#include<stdio.h>

int main()
{
    printf("i j k i&&(j||k)\n");
    printf("---------------\n");
    for (int i = 0; i <= 1; i ++)
    {
        for (int j = 0; j <= 1; j ++)
        {
            for (int k = 0; k <= 1; k ++)
            {
                printf("%d %d %d     %d\n", i, j, k, i && (j || k));
            }
        }
    }
    getchar();
    return 0;
}
[/code]

Hey ,,, I realized it's due to a small mistake that it's written as "<" instead of "<=" ,,, !!
Thanks anyway ,,, !


Let (O',R') be the circle that intouches the circle (O,R) at C.
  AB is a chord of (O) and touches (O') at M.
  CA and CB intersect (O') respectively at D and E.
  Denote K the intersection of the tangent at D and E of (O').
  Call I, H be the intersections of AK and (O), BK and (O).
  Let U be the intersection of AH and BI.
  Prove that C, U, K are collinear.

I have a proof of the fact that they're collinear even if $AB$ isn't tangent to $(O')$, but I'll post it some other time.

Yes ! It is also right if AB doesn't touch (O'). I guess you must use polar. :D


Graph the function below.

R(x) = (x - 1)(x + 2)(x - 3)/(x)(x - 4)^2

How would I graph similar rational functions given in my textbook?

I would find and plot it's x-intercept(s) (zeros) and its y-intercept(s).  Then find it's asymptotes and account for those on your graph.  You could also find the sign of the function at x-values between its zeros to determine whether the graph is above or below the x-axis on those intervals.

Those should serve as a guide to drawing the graph.

Calculus could also be applied to find its extrema and concavity.

You said:

"Then find it's asymptotes and account for those on your graph. You could also find the sign of the function at x-values between its zeros to determine whether the graph is above or below the x-axis on those intervals."

The above is not too clear for me.

Can you explain it in easy terms?

sometimes there are values of x on a graphy where that x can't go into the function (it's not in the domain). when this happens there's either a hole in the line or an asymptote: a vertical line that the graph approaches, but never touches, and bends away either down or up.

the second part: test between the zeros you found for the function. If R(x)>0, then the graph is above the y-axis, if R(x)<0 then the graph is somewhere below the y-axis.


There are too many errors in the above code to try to compile it. Could you post the code that [i]does[/i] compile (so no errors when you compile) even if it doesn't give the effect you want? If you put the document here between [code ] and [/code ] then it won't mess up dollar signs or other things.

If you are having trouble getting your document to compile without errors and need help post a shorter document with the errors and say what the problem is.

There isn't a package called ams - do you mean amsmath?

Finally, eqnarray will not produce good results (see [url=http://tug.org/pracjourn/2006-4/madsen]Avoid eqnarray![/url] for the reasons) so start with align and see how far you get.


A number of elements from a set $S$ is colored red, all of the rest blue. If $S_{1},\dots,S_{m}$ is a family of subsets of $S$, prove that it has SDR (system of distinct representatives) with $\leqslant t$ red elements iff the next conditions hold:
a) $S_{1},\dots,S_{m}$ has SDR;
b) eack $k$-union has $\geqslant k-t$ blue elements.

Is $k$ fixed in part b) or does part b) hold for arbitrary $k\in [1,m]$?


The Problem. Not just for the simple or the easy or the intelligent or the dumb but for all.

The Problem.

Numbers.

3 sets of Numbers of 4 as shown.

The 1st Set Given:

418
4241
6757
262
1132
1742
4334
5082
401
1264
7120
8138
9710
521
1664
4053
4809
5911
829
2524
7380
9548
9995

Followed by the 2nd:

212
9696
7310
1348
2569
4979
8220
9546
1047
2271
4679
8289
9455
1179
2082
4362
5713
8346
523
2248
4151
7573
8981

Lastly the 3rd:

3948
1582
7132
501
5426
7048
7999
8642
196
2246
4422
5738
9269
3028
6519
7245
8094
9125
320
3843
5606
7526
9541

The Question.

To Provide.

To Provide an equation or perhaps a set of equations, to an arrival of what the 4 set of Numbers should be.

I have been working at it for years, months, days, hours and minutes and yet have arrived at a most depressing outcome. Nothing. Now lost and perhaps at my wits very end I ask, is someone here able to help?

With all my heart. My Thanks.

What is the source of the numbers?  This question is meaningless in isolation.

The source of the numbers given to me were not explained by the person that gave the question and therefore are deemed unknown. The task however is finding a correlation , a pattern , an equation from infinite possibilities which elevates it from one set of numbers to the next and so on.

I know it may sound impossible to solve and its like trying to find a pin in a haystack but I thought it was worth giving a shot here to new possibilities that I haven't been able to come up with.

Thank you.

What pattern do you want?

We can interpolate a polynomial that will go through the points (with the x-value being the order in the list).  Then with that, we can predict the 4th set of numbers.

The above post is correct though.  Without context the problem is rather meaningless.

A simple glance reveals the 4th, 9th, 14th, and 19th numbers all spike downward.

There's a million "possibilities," but the real question is if there is even a unique answer?  One can find patterns in almost anything.  It just depends on what you're looking for.  If you want to believe those numbers are a code, you can try ciphering the numbers into letters.

If you want to believe they are simply data to be interpolated, you can create a polynomial to make a prediction.

Is the 4th set of numbers even supposed to be a unique set?  Or perhaps it is simply the behavior that is important.  Graphing the three sets reveals similar increases and decreases.  So any 4th set with similar behavior will suffice if that is what you're looking for.

If you want to believe they are a sequence, you can predict the behavior of the sequence.

It's up to you what you want to do with it.  That's the beauty of mathematics.  To be honest, it feels like someone just has too much time on their hands and is trying to play a joke.


how do you set up a "click to reveal hidden content" link?

You need to surround the text you want to hide with hide tags.  For example

[code]
[hide] Put hidden text here[/hide][/code]

You can either just type the code tags or you can select the text you want to hide and click the "Hide" button.

If you want to change the "Click to reveal hidden content" to say something else you need to do this:

[code][hide=put what you want here]Put hidden text here[/hide][/code]


calculus
integration
limit
function
trigonometry
real analysis
real analysis unsolved

Okay, so a question from Rudin's Principles begs for this:

In what conditions do we have that $ \lim_{x \to 0^\plus{}} x \sum_{n \geq 1} f(nx) \to \int_0^\infty f(t) \, dt$?
(Call the series $ S_x$ and the integral $ I$.)

One thing that can be done is to define $ f_x : \mathbb R \to \mathbb R$ by $ f_x (t) \equal{} f(nx)$, for all $ t \in ( (n\minus{}1)x, nx]$. Making some pretty strong assumptions, we can say that $ f_x \to f$ pointwise and that $ |f_x| \leq g$ for a nice $ g$, so the claim is true by the Dominated Convergence Theorem. (I hope this works, I struggled quite a lot to work it out.)

What happens in other cases when $ f$ is not so nice? One can easily construct an $ f$ such that the formula fails for a certain subsequence $ x_n \to 0$, but can it fail completely? That is, can we always find $ x_n \to 0$ such that the limit equals the integral? If that's true (at least for a large class of functions), then can we say that the $ x$'s from $ (0,\delta)$ that verify $ |S_x\minus{}I| < \epsilon$ are many (uncountably many / of positive measure / anything like that)?

Clearly for your DCT argument, you are assuming that $ f$ is at the very least absolutely integrable. (You really need $ g,$ whose construction you didn't specify, to be absolutely integrable, and $ g$ is at least as big as $ f.$)

So to destroy things, we take an improper integral which does not converge absolutely. Here's my example:

$ f(x)\equal{}\sin(x^2)$ on $ [0,\infty).$ We know that $ \int_0^{\infty}\sin(x^2)\,dx$ converges (conditionally) as an improper integral.

What about $ x\sum_{n\equal{}1}^{\infty}\sin(n^2x^2)\ ?$

I think that (at least for almost every $ x$), this sum diverges as its terms do not even tend to zero.

Here's an absolutely integrable $ f$ which misbehaves:

Define the "tent function" $ t(x)\equal{}\max(1\minus{}|x|,0).$

Let $ f(x)\equal{}\sum_{k\equal{}1}^{\infty}t(k^2(x\minus{}k)).$

We have a spike centered at each integer, each spike the same height but getting narrower. The whole integral of $ f$ is $ \sum_{k\equal{}1}^{\infty}\frac1{k^2},$ which is finite.

But $ S_x\equal{}x\sum_{n\equal{}1}^{\infty}f(nx)$ is infinite for every $ x\in\mathbb{Q},$ and hence cannot have a finite limit as $ x\to 0.$

This example also sheds some light on perfect_radio's dominating function $ g.$ We can't necessarily take $ g$ to be something simple like a multiple of $ f,$ and in this particular case, our $ g$ must not be integrable.

Based on my last two examples, it seems that $ \lim_{x\to\infty}f(x)\equal{}0$ would be a useful condition. If you combine that condition with $ f$ being continuous, you would also have that $ f$ is uniformly continuous. (Both of my examples were continuous but not uniformly continuous.) That might also be useful.

1. Yes, for that DCT I need uniform continuity and a good dominating function. In the case $ f(t) \equal{} \frac {\sin^2 t}{t^2}$ that appears in Rudin, we have $ f(t) \leq h(t) \equal{} \frac1{t^2}$, where $ h$ is a decreasing integrable function, so it's easy to find a dominating $ g$ for $ f_x$. However, I don't think (but I may be wrong) that in the general case (when $ f(x) \to 0^\plus{}$ as $ x \to \infty$) we can always find a decreasing $ h$ to bound our $ f$.

2. Okay, so it may not work for that function.... But what happens in a case like this: $ \delta \sum_{n \equal{} 1}^\infty \frac {\sin \left( n \delta \right)}{n \delta} \equal{} \frac {\pi \minus{} \delta}2$? $ f(t) \equal{} \frac {\sin t}{t}$ isn't absolutely integrable, but fortunately it's uniformly continuous, unlike $ f(t) \equal{} \sin \left( t^2 \right)$.

3. I also thought about basically the same function with spikes at the integers (I guess it's pretty natural) when I asked the last two questions from my original post. However, I cannot really work out what happens when $ x \in \mathbb R \setminus \mathbb Q$, hence I thought there was still some hope.


How'd Westpac 2007 go?

Pls refrain from posting problems till July 27.

Well, I think it was easy. I took the junior paper.

I took the Senior paper..

I think this year was about as hard as last year one.

How did others go?

Intermediate was harder than last year's, I think. Although, the second last question (9 marks) was a giveaway. Some of the questions had really tricky wording though *cringe*, a good example being the elevator question.

I took the seniors and now i think.. oh shivers... there goes my distinction streak. damn it was hard.

i am in year 12 (made a mistake earlier in one of the forums posting for a friend in year 9, so it appears that i am in year 9 :blush:), so i did the senior paper, and i agree with teatea it was quite hard, but i think ill get a Distinction.


[quote="anuj kumar"]let $a,\ b,\ c$ be positive reals then prove that
\[\sqrt[3]{(b^{2}+c^{2}-a^{2})(c+a-b)}+\sqrt[3]{(c^{2}+a^{2}-b^{2})(a+b-c)}+\sqrt[3]{(a^{2}+b^{2}-c^{2})(b+c-a)}\le (a+b+c) \]
. when does equality holds?.[/quote]
Fall ! Try $a=1$ and $b=4$ and $c=7$
or $a=b=1$ and $c=0.8$  :blush:

sorry i forgot to mention that $a,b,c$ are sides of a triangle .
thanks

so for the last case it must holds too.for $a=b=1$and $c=0.8$ it holds
please check it again.
thanks

also one more condition is that
\[a^{2}+b^{2}-c^{2}\ge 0\] and cyclically
thanks

[quote="anuj kumar"]let $a,\ b,\ c$ be positive reals then prove that
\[\sqrt[3]{(b^{2}+c^{2}-a^{2})(c+a-b)}+\sqrt[3]{(c^{2}+a^{2}-b^{2})(a+b-c)}+\sqrt[3]{(a^{2}+b^{2}-c^{2})(b+c-a)}\le (a+b+c) \]
. when does equality holds?.[/quote]
$\sum_{cyc}\sqrt[3]{(b^{2}+c^{2}-a^{2})(a+c-b)}\leq\sqrt[3]{9\cdot\sum_{cyc}(b^{2}+c^{2}-a^{2})(a+c-b)}.$
Let $a=y+z,$ $b=x+z$ and $c=x+y.$ Then remain to prove that
$2(x+y+z)^{3}\geq9\cdot\sum_{cyc}(x^{2}+xy+xz-yz)y,$ which true.

your last is not so trivial and involves sharp AM GM s i suppose.


If a and b are positive constants, where is ax^2+by^3=1 increasing?

[hide]
Deriving, we get
$ ax \plus{} 3by^2\frac{dy}{dx} \equal{} 0$
$ \frac{dy}{dx} \equal{} \frac{\minus{}ax}{3by^2}$

We are looking for the part of f where $ \frac{dy}{dx} > 0$
Since $ y^2 > 0$ always, and the constants are positive, wherever $ x \leq 0$ is where $ \frac{dy}{dx} > 0$

So, the function is increasing on $ (\minus{}\infty, 0)$

[/hide]


MATHCOUNTS
quadratics
trigonometry
number theory
modular arithmetic

Hey guys, I had a quick question for yall.  I've searched the forums for litterally two days, and have found no answer to my question lol.  My question is that I recently purchased AoPS Vol 1, 2, and the solution manuals.  I am able to solve the problems in the Chapter exercises, but when it comes to the end of chapter questions I have a little trouble (especially with Chapter 5 Number Theory :) ).  Is it normal when first starting out in competitive math (I've only done MathCounts) to only be able to solve a few of these.  How did you guys do when you first went through the books?

I had trouble the first time too. You slowly develop the problem solving skills and then it becomes easier. :)

Generally, I liked saving a few problems (instead of doing them all at once). That way, you can come back and refresh your brain with a few more problems of a previous subject.

But I didn't have that much trouble doing many of the end-of-the-chapter problems (in AoPS 1) due to the fact that I started when I was older than you are now (you're lucky for starting early) and I already covered a lot of the material.

[quote="Jashper"]Should I just read through the chapters, try to do some of the end of chapter problems, and then go back once I finish the book and do all the problems?[/quote]

This is a good strategy - try some of the problems once, then come back to them after having more exposure to problem solving in general.

Excellent, thanks for the advice everyone.  Time to go work on some quadratics lol :)

Just try to understand each problem.

Go through the book several times to fully understand them.

[quote="Jashper"]One more quick question for yall, should I just skip the number theory end of chapter problems (Vol 1 Chapter 5) for now, and then go back to them at the end, cause I can barely do any of them lol.[/quote]
I know what you mean, I bought a number theory book so I could try to teach myself, and it is extremely difficult (mostly because all of the excersises are proofs and I have never really proved anything before now). Hopefully it gets easier for both of us.  :P
I'm trying not to skip any of the problems, but some of them just seem insoluble so I get stuck on them for days (and I'm too stubborn to go on to the next one).

the number theory ch was a bit hard in vol.1. But once you try the number theory ch. in vol.2, it becomes a lot clearer. It good that you're starting young, you got lots of time to work on your skills. You might even make it to IMO. :lol:
If you have any other questions, feel free to ask.

[quote="Jashper"]I'm only going into 10th grade, I'm not that young ;)[/quote]
I see your point. :)
I'm going into 11

[quote="Jashper"]One more quick question for yall, should I just skip the number theory end of chapter problems (Vol 1 Chapter 5) for now, and then go back to them at the end, cause I can barely do any of them lol.[/quote]

You may want to look into our Intro Number Theory text, which goes through these topics in much more detail and depth.

[quote="Totally Zealous"]Wow you guys are old.[/quote]
lol, what grade are you tz? :D

Exellent, more for me to look forward to :)  I wish I had known about this website when I was in MathCounts :(

[quote="rrusczyk"][quote="Jashper"]Yeah, for Christmas I was gonna get those 3 Intro books.[/quote]

If I keep going at the pace I'm going, there will be 4 Intro books by then ;)[/quote]

Does that mean that the Algebra book is coming out soon?

The good news is that I'm getting better at the problems and am pretty much able to the majority of them at the end of each chapter :)

Will there be other Intermediate books coming out? If so, on what topics?

[quote="rrusczyk"][quote="mtms5467"]Does the Intro to Algebra book cover material in both Alg I and II?[/quote]

Mostly Algebra 1, though there are some Algebra II topics as well, and there are many much more challenging problems than you see in the typical curriculum in either class.  The Intermediate Algebra book will cover what's left of Algebra II and go much beyond a standard Algebra II curriculum.

I hope to have the Intro Algebra book out by the end of the year.  The Intermediate Algebra book will hopefully be out by the end of next spring.[/quote]
When will the algebra online classes be offered?

[quote="daermon"]do the AoPS volumes 1 and 2 cover much algebra II? :maybe:[/quote]

I would say Vol 2 does

[quote="ragnarok23"]Is there going to be an intermediate trig book?[/quote]

there isn't even an intro trig book yet

i would love a book on trig too

I personally liked the number theory and quadratics chapter the best, because I was able to understand most of it. The rest of the chapters were ok, but Linear Equations was tough  :blush: that and propotions... ouch, could barely solve the problems at the end. I'm doing much better with quadratics though  :lol:

I'll be going into 9th grade in a month. I got the book a week ago, I'm about to start chapter seven. Hopefully I'll get better  :D

you're going at a relatively fast pace chaostheory. :)
btw, i like your screename.

Yeah the word problems are my weakness as well lol, but I can solve most of the other end-of-chapter problems.

That's funny, I got them a week ago too lol :)

Thanks, hopefully I'm not going too fast so I'll forget the material
Thanks, I like my sn too   :D
Such a fascinating concept...

What other things do you recommend I study? I'm thinking about pulling out my College physics book.  :lol:

Wouldn't that be too hard? :maybe:

I was thinking about getting the number theory book.

Yes, it would be difficult if you're on AoPS Volume 1.

You should consider getting Intro to Number Theory. :D

[quote="I said"]Try ACops, art and craft of problem solving. Its a little harder than vol.2, i think. :)[/quote]
I didn't do Acops, so i was just guessing. I heard from other people that its hard though. :P

I'll try to get number theory; modular arithmetic is one of the more interesting concepts to me. Can't wait to learn infinite series and stuff though...

 :lol:


calculus
integration
Miscellaneous Problems

Alice and Bob play the following number-guessing game. Alice writes down a list of positive integers $x_{1}$, $\cdots$, $x_{n}$, but does not reveal them to Bob, who will try to determine the numbers by asking Alice questions. Bob chooses a list of positive integers $a_{1}$, $\cdots$, $a_{n}$ and asks Alice to tell him the value of $a_{1}x_{1}+\cdots+a_{n}x_{n}$. Then Bob chooses another list of positive integers $b_{1}$, $\cdots$, $b_{n}$ and asks Alice for $b_{1}x_{1}+\cdots+b_{n}x_{n}$. Play continues in this way until Bob is able to determine Alice's numbers. How many rounds will Bob need in order to determine Alice's numbers?

We assume $n>1$, $n=1$ is trivial.
Two rounds suffice:

In the first round, Bob gives the list $1,1,1,...,1$ to get a bound $x=x_1+x_2+...+x_n$ on the size of the numbers $x_i$. Then he gives the list $1,x,x^2,x^3,...,x^{n-1}$, the digits of the new result written in base $x$ are exactly the numbers of Alice (this is gurantated because of $x>x_i$).

Edit:
And well, one round doesn't suffice, Alice could have choosen such that the result can be represented in more than one way as integral linear combination of the given $a_k$ (e.g. $2 a_2 \cdot a_1 +  a_1 \cdot a_2 = a_2 \cdot a_1 +  2 a_1 \cdot a_2$, the rest of the sum neglected and both times the same).


i cant get what do want to ask. please clearfy it.

[hide]$(-2)(6)(a^6)(a^2)=-12a^{6+2}=-12a^8.$

[quote="ashwinrk_jain"]What to do in this[/quote]
Here are some links that may help:
http://purplemath.com/modules/exponent.htm
http://purplemath.com/modules/polymult.htm[/hide]

[quote="math92"]Simplify:
$(-2a^6)(6a^2)$[/quote]

[hide]We can split this up to get $-2*6*a^6*a^2$.  Then by computation and properties of exponenents, we get $-12a^8$

$-12a^8$[/hide]

Oh you want to multiply them.
I got you.

[quote="José"][quote="kunny"][quote="ashwinrk_jain"]no $\frac{-a^{4}}{3}$ is a wrong answer[/quote]

No. you are wrong. bpms is right.[/quote]

Yes. That's the correct answer. You have to do $6-2$. Maybe you think you have to do $\frac{6}{2}$, but you have to rest[/quote]
Exactly :ninja:

[quote="José"][quote="math92"]Simplify:
$(-2a^{6})(6a^{2})$[/quote]

Let me change a bit the question: what about this?: $\frac{-2a^{6}}{6a^{2}}$[/quote]

[hide="answer"]$-\frac{a^{4}}{3}$[/hide]

[quote="José"][quote="kunny"][quote="ashwinrk_jain"]no $\frac{-a^{4}}{3}$ is a wrong answer[/quote]

No. you are wrong. bpms is right.[/quote]

Yes. That's the correct answer. You have to do $6-2$. Maybe you think you have to do $\frac{6}{2}$, but you have to rest[/quote]
One of the basic exponential properties-- $\frac{x^{m}}{x^{n}}=x^{m-n}$


I was wondering if someone could make another amc 8 or even a 10, because the last one done was so helpful for me to find where I am this year in math and it has already helped me to improve, so a second one couldn't hurt  :lol:

                                            -Thanks

I am almost finished with a Mock AMC 12 thing.  I have been taking the problems from various places meaning that none of them are original but it is still practice nonetheless.  I will need/want someone to check it for me.  Are there any takers who have extra time and don't really want to take the mock amc?  If I were to have this it wouldn't be for a good month or so.

I would be glad to do the checking. I'd have to hide it from my son, but I never do these online contests, so I might as well help you with the editing of the contests you prepare.

[quote="Rep123max"]And another Mock AMC 12 would be awesome too. I would make one, but it would take about a year for me to finish it. I've gone from being on AoPS over an hour each day during the summer, to being on about 5-10 minutes each day.[/quote]

That's true. Creating a Mock AMC or AIME or any kind of test takes a really long time. Now that school's started, most of us are really tight with schoolwork and activities, which would make it really tough to make a test. I've also seen my time on AoPS drop dramatically. Making a Mock AMC would probably be much harder than making the Mock MC I did, and that alone took about a week, working an hour or so a day. I really respect all the people who are making these tests, for their hard work.

I'd be glad to try to help with some problems for some of the tests, preferably easy ones though. I just can't work in most of October. I've got to watch baseball's playoffs and root for my team...

For your information,

Myself and singthesorrow have actually created 25 questions, roughly a Mock AMC 11 level... we are about 80% done the test but due to school, progress on it has been delayed for a while... one day, if (and hopefully when) we finish, we can administer the test.


Find all of the values of the digits so that
[code]
 WRONG
 WRONG
+-----
 RIGHT
[/code]

[quote="sponge008"]I saw the problem and couldn't find a good way to solve it...apparently it's just guess and check -_-[/quote]

Not necessarily; After brute forcing a few initial values, logic can be used to find the rest of the values.


Let $ d(n)$ denote the largest odd divisor of a positive integer $ n$. The function $ f: \mathbb{N} \rightarrow \mathbb{N}$ is defined by $ f(2n\minus{}1)\equal{}2^n$ and $ f(2n)\equal{}n\plus{}\frac{2n}{d(n)}$ for all $ n \in \mathbb{N}$. Find all natural numbers $ k$ such that: $ f(f(...f(1)...))\equal{}1997.$ (where the paranthesis appear $ k$ times)

$ f(m*2^k)\equal{}(m\plus{}2)*2^{k\minus{}1}$ if $ m$ odd and $ k>0$.
Let $ t(m*2^k)\equal{}m\plus{}2k$ for odd m $ k>0$, then $ t(f(n))\equal{}t(n)$ for any even $ n$, if n odd, then $ f(n)\equal{}2^{(n\plus{}1)/2}$ and $ t(f(n)\equal{}1\plus{}(n\plus{}1)\equal{}n\plus{}2\equal{}2\plus{}t(n)$.
Therefore before 1997 all numbers with $ t(n)\le 1997$. Include 1997 ($ k\equal{}0$ ) number of them is $ 999*500$


$ \begin{gathered} {\text{Find non \minus{} negative integers }}a{\text{ and }}b{\text{ such that }} \hfill \\
2(a \plus{} b) \equal{} 3^a \minus{} 3^b \hfill \\
\end{gathered}$

[b]An extension[/b]:
$ \begin{gathered} {\text{Find non \minus{} negative integers }}a{\text{ and }}b{\text{ such that }} \hfill \\
2(a \plus{} b) \equal{} p^a \minus{} p^b {\text{ , where }}p{\text{ is a prime}} \hfill \\
\end{gathered}$

One only needs $ p>2$ (not necessarily prime).

From $ a\equal{}b$ follows $ a\plus{}b\equal{}0$, so $ a\equal{}b\equal{}0$. From $ b\equal{}0$ follows $ 2a \equal{} p^a \minus{}1 \geq 1 \plus{}(p\minus{}1)a \minus{}1 \geq 2a$, with equality only for $ p\equal{}3$ and $ a\equal{}1$.

For $ a > b \geq 1$ we have $ 2(a\plus{}b) \equal{} p^a \minus{} p^b \equal{} p^b(p^{a\minus{}b} \minus{} 1) \geq$ $ (1\plus{}(p\minus{}1)b)(1 \plus{} (p\minus{}1)(a\minus{}b) \minus{} 1) \geq 2(a\minus{}b) \plus{} 4b(a\minus{}b) \equal{}$ $ 2(a\plus{}b) \plus{} 4b(a\minus{}b\minus{}1)$, with equality only for $ p \equal{} 3$, $ b\equal{}a\minus{}b\equal{}1$, so $ a\equal{}2, b\equal{}1$.

Anyway, this is actually a rephrased version of the following problem:
$ {\text{Find all natural numbers }}n \geqslant 1{\text{ for which }}n^2 \plus{} 3^n {\text{is a square}}{\text{.}}$ :wink:


Why are discrete topologies said to be "discrete"? From the definition of a discrete topology I do not have any idea about it.

Because they are induced by a discrete metric, maybe?

The discrete topology gives you no extra information about "closeness."  The points are discrete: the topology gives you no way to decide which points are close to which points.  It should be contrasted with the [i]indiscrete topology[/i], where every element of the set is lumped together and points are not topologically distinguishable.

Every function from a discrete topology is automatically continuous.  Every group is a topological group with the discrete topology.  So in that sense, the discrete topology gives you no extra "control."

Also the integers as a subspace of the reals have the discrete topology, FWIW.


Sorry if this issue has been posted.

In Classroom Math, a new user who joined yesterday has a rating of 4 as his/her overall user rating. However, this particular user only has one post and that post was not rated at all.

Is the 4 rating meant to be a default rating for new users? I thought that the default simply said "user not rated yet"

The URL to the thread is found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=168726]here[/url]

Bug fixed, ratings resynced and corrected.


A regular hexagon and an equilateral triangle have equal areas. What is the ratio of the length of a side of the triangle to the length of a side of the hexagon?

$ \textbf{(A)}\ \sqrt 3\qquad
\textbf{(B)}\ 2\qquad
\textbf{(C)}\ \sqrt 6\qquad
\textbf{(D)}\ 3\qquad
\textbf{(E)}\ 6$

Let the side of the hexagon be x.  Since a regular hexagon is made from 6 equilateral triangles with the same side length as the hexagon, the area will be 6(x^2)(sqrt3)/2.  The triangle has the same area, so we work backwards through the formula to get the side length of the triangle.  (s^2)(sqrt3)/2=6(x^2)(sqrt3)/2.  This simplifies to a ratio of sqrt6:1, or sqrt6, C.

[hide]
Call the length of a side of the hexagon S(1) and the length of a side of the triangle S(2).
The area of the triangle is:
(S(2)^2*sqrt3)/4

The area of the hexagon is:
6(S(1)^2*sqrt3)/4

Setting those two equal, we get:
(S(2)^2*sqrt3)/4=6(S(1)^2*sqrt3)/4

Simplifying, we get:
S(2)^2/S(1)^2=6

So S(2)/S(1)=sqrt6

Therefore the answer is C, sqrt6.

Q.E.D.

[/hide]

[hide]Call the length of a side of the equilateral triangle m, and the length of the side of the hexagon n. The area of the triangle is now (m^2)(sqrt3)/4, and the area of the hexagon is (6)(n^2)(sqrt3)/4, since the hexagon is made up of six equilateral triangles that have the same side length as the hexagon. As both areas are equal,           (m^2)(sqrt3)/4=(6)(n^2)(sqrt3)/4. Multiplying both sides by 4/sqrt3 gets m^2=(6)(n^2), or m=nsqrt6. The answer is [b]C[/b].[/hide]

[hide="Answer"]A regular hexagon is composed of six equilateral triangles. The area of an equilateral triangle has a square of $s$ in it, so to have the same area as the hexagon, the side length must be $\sqrt{6}$ times that of the hexagon. $\boxed{C}$[/hide]


Find the coordinates for the vertices of a lattice rectangle (with vertices on lattice points) whose long-side length is triple its short-side length, and none of whose sides is parallel to the horizontal axis.

Thanks

Erm... if I correctly understood what you're asking for, then $ A\equal{}(0,1); B\equal{}(4,5); C\equal{}(5,4); D\equal{}(1,0)$ works (side lenghts $ \sqrt{2}$ and $ 3\sqrt{2}$).


induction
modular arithmetic
quadratics
number theory unsolved
number theory

Find the number of squares in the sequence given by $ a_0\equal{}91$ and $ a_{n\plus{}1}\equal{}10a_n\plus{}(\minus{}1)^n$ for $ n \ge 0.$

[quote="moldovan"]Find the number of squares in the sequence given by $ a_0 \equal{} 91$ and $ a_{n \plus{} 1} \equal{} 10a_n \plus{} ( \minus{} 1)^n$ for $ n \ge 0.$[/quote]

We can prove by induction that for $ k \ge 2$,

$ a_{2k} \equiv 909 \pmod{1000} \rightarrow a_{2k} \equiv 5 \pmod{8}$ , which is not a quadratic residue mod $ 8$.

$ a_{2k\minus{}1} \equiv 091 \pmod{1000} \rightarrow a_{2k\minus{}1} \equiv 3 \pmod{8}$, which is not a quadratic residue mod $ 8$.

So $ a_n$ is not a square for $ n \ge 3$, with also $ a_0,a_1,a_2$ are not squares we have the number of squares in the sequence is $ 0$.


modular arithmetic
trigonometry
number theory proposed
number theory

Prove that for any n>=1 natural
$\frac{2^{4n+2}+1}{5}$
integer and composite number.

can somebody help me with this problem??
I  can prove first part of this problem (integer)but I can't prove that it's composite number :
my solution :
\[ 2{}^4{}^n^+^2+1=4{}^2{}^n^+^1+1  \] so I used this identity and write :\[ (4+1)(4{}^2{}^n-4{}^2{}^n{}^-{}^1+.......\pm1)  \]
and we can under stand that :\[ (5)(4{}^2{}^n-4{}^2{}^n{}^-{}^1+.......\pm1)/5=integer  \]
but I can't prove that it is composite number!!

[quote="_el_doopa"]prove is above, isn't it?[/quote]
I know but I want to know how I can prove it with my solution for composite number, because my solution isn't complete!!and I want to complete it!!

[quote="Davron"]parinaz.sa your solution is true sine 2n+1 is an odd integer.

Davron[/quote]
thanks DAVRON  :)