AMC
USA(J)MO
USAMO
AIME
AMC 12
USAMTS
MIT

In response to a recent proposal and increased sponsorship by the Art of Problem Solving, the AMC will invite approximately 50% more students to take the USAMO this year, for a total of around 375 students.

Time to start working on those olympiad problems!

Wow! This is awesome!

$\textsc {Thanks AoPS!!!}$

:coolspeak:

How are the extra people picked exactly?

Excellent!

Now, how are the extra people picked?

Thanks so much for the extra sponsorship!

Darn it would be better if it's MOP expansion oh well

[quote="beta"]Whoa huh??

But then that's nice too!! Now I can concentrate on olympiad problems and not have to worry about not qualifying for USAMO!![/quote]

That was a large part of the motivation for sponsoring the increase in students invited to sit for the USAMO.

[quote="h_s_potter2002"]Wow, this is awesome.

Does anyone have any idea how this will affect floor scores, and the AIME qualifying scores?[/quote]

Well, I think its going to work the same way... get all juniors and above in top whatever, and then out of the lowest AIME score of them take all students younger than juniors who got that score to get a total of about 375.

IMO, this won't make the floor for AIME THAT much lower.. probably 1 point. So this basically affects all the people who are borderline making it.

YAY!! I finally have a chance... Time to hit those AIME problems

Props to AoPS for making USAMO a more reachable target in terms of winnowing down the AIME-taking population. (Of course, it's all AoPS's fault that more and more younger people are qualifying for AIME, so this is poetic justice.  :P  )

[quote="probability1.01"]I think this kind of thing is generally a good idea, but why does it take any sponsoring? It's not like the AMC can't even afford to mail those extra 125 forms, contrary to how it might appear. :D[/quote]

I think you are underestimating the cost of the process -- the USAMO is more carefully administered than other contest exams and the grading process alone is time consuming.  It requires the minds of many talented (not to mention patient) mathematicians and problem solvers.

YAY! AIME should be easy to qualify for, but qualifying for USAMO is going to be a bit harder. Now I have a reasonably better chance  :)
If I manage to make USAMO maybe I can start to think about problems at that level. I haven't even attempted one of those let alone  solve one, so maybe I can eventually start learning them and well, solve much harder problems than I've solved before.

[quote="MithsApprentice"]Does anyone know if the ratio of first selection (USAMO indices) and second selection (AIME floor) will be the same?

That should make a pretty significant mark on how much the index is expected to drop below "normal". Assuming something like 250 index qualifiers - not a ridiculous number - I would agree with jackb. 6 - 10 seems shy of where the index will actually land. 12 - 15, on the other hand, is a much more realistic guess. Based on what I know about Texas qualifiers and almost-qualifiers, this would be a good estimate.

The effect on the AIME floor will be proportionally much greater than the effect on the USAMO index, in my opinion. An index dropping 12 - 15 points could easily lower the floor by 2 problems. And wow...if you took all the sophomores and below who scored 7 and up last year, we're talking a much larger field of young qualifiers.[/quote]
*sniffle*
ah the wonders of qualifying in 8th grade but not the year after...
well its a good thing i can focus on olymp problems this year :D (or rather the next 3 months)

[quote="MCrawford"][quote="smbelcas"]Referring to USAMO qualification, MCrawford says:

[quote="MCrawford"] It's a very good marker in a world where a typical math major doesn't need to know how to write a decent proof.[/quote]

While it [i]may[/i] be true that a typical math major doesn't need to know how to write a decent proof (what's a typical math major, this depends on the institution, etc.) I haven't seen a strong correlation between people who know how to write decent proofs and people who qualify for the USAMO.  I've had plenty of high-school students who are excellent at proofwriting and who weren't able to qualify for the USAMO, and those students I've had who qualified for the USAMO have not always been the best proofwriters.  There are stories which circulate about students who scored well in contests and couldn't survive as mathematics majors in college.
[/quote]

I must have mistyped.  I means to see that a typical math major doesn't know how to write a decent proof.  I might have used the word "need" as in "can get away with never learning it."

I'm not sure what you're trying to convey in your response however.[/quote]

I'll assume that in your last post you meant to say "I meant to say that a typical math major doesn't know how to write a decent proof." *grin*

As you weren't sure what I was trying to say... I'll be blunt.

(1) I'm rather surprised to hear you say that a typical math major graduates from college without learning to write a decent proof.  That's a pretty strong allegation to make.  Do you have any experience on which to base this claim?  I wonder whether your ideas of "typical" and "decent proof" are different from mine or whether we really perceive students as performing differently.

(2) I don't think that USAMO qualification says anything about proofwriting skills.  It may be "a very good marker" of something, but that something is not proofwriting skills.

I agree that a [i]fourth[/i] stage is just going too far. This would create too much work and hassle, and it would throw off timing. However, I do agree that the difference in styles of the USAMO and AIME (which determines qualification for USAMO) is a problem.

I would suggest giving a special easier USAMO qualification opportunity for first-time AMC/AIME takers who are in at least 9th grade (to avoid say, someone unknowingly taking the AIME in 7th grade and blowing their opportunity). Out of those people, whoever makes it to MOP (or potentially whoever scores above a certain amount) will automatically qualify for USAMO in future years.
This way, gifted proof-writers will have a much better chance of being recognized, by being given a chance to take the USAMO, and will continue to be allowed to take it (and then move on to higher contests such as MOP/TST and IMO) if they prove to be talented at it.
As this opportunity will only be available to first-time participants (mostly 9th graders), it will not cause a major financial flaw among the AMC organization. On the other hand, it will give people at least one chance to show their proof-writing (well, and proof-solving) abilities without having to do extraordinarily well on a contest that does not test the same skills.

Well, in all honesty, if you can solve USAMO problems and structure your thoughts into a well-organized proof, you can pretty much solve enough AIME problems. However, I do think there is the problem that some people who have potential at USAMO don't make it past the AIME due to stupid mistakes. They should figure out a way to reduce that, since just two or three stupid mistakes make a huge difference in AIME. On the other hand, not making stupid mistakes is something that people can (pretty easily) work on themselves.

[quote="MCrawford"][quote="smbelcas"]
I'll assume that in your last post you meant to say "I meant to say that a typical math major doesn't know how to write a decent proof." *grin*

As you weren't sure what I was trying to say... I'll be blunt.

(1) I'm rather surprised to hear you say that a typical math major graduates from college without learning to write a decent proof.  That's a pretty strong allegation to make.  Do you have any experience on which to base this claim?  I wonder whether your ideas of "typical" and "decent proof" are different from mine or whether we really perceive students as performing differently.

(2) I don't think that USAMO qualification says anything about proofwriting skills.  It may be "a very good marker" of something, but that something is not proofwriting skills.[/quote]

I base my opinion on my experiences as an undergraduate student at a school that was often in the top 5 on the Putnam and nearly always in the top 10.  I doubt that half the students who graduated with a math major could conjure a proof of more than a curricular problem.  Even then, many certainly cannot.

A great majority of USAMO qualifiers got a full problem or more last year.  It's not qualification, but performance from which I draw my conclusions.[/quote]

Based upon my experience of grading/looking at some college math tests, most students have little proof writing ability and the ones that do are often previous USAMOers or other olympiad qualifiers.

[quote="MCrawford"]I doubt that half the students who graduated with a math major could conjure a proof of more than a curricular problem. [/quote]

Of course, fewer than half the students who graduate with a math major are graduate-school bound, and thus couldn't be expected to produce original mathematics research (which is what I suppose you mean by "more than a curricular problem").  The bulk of my math-major students are able, by the time they graduate, to produce proofs for many curricular problems.

[quote="DPopov"]Based upon my experience of grading/looking at some college math tests, most students have little proof writing ability and the ones that do are often previous USAMOers or other olympiad qualifiers.[/quote]

That's interesting---I haven't worked with college students at institutions which draw students with contest experience.  (My work with contest-experienced students has been while they're in high school.)

[quote="smbelcas"][quote="MCrawford"]I doubt that half the students who graduated with a math major could conjure a proof of more than a curricular problem. [/quote]

Of course, fewer than half the students who graduate with a math major are graduate-school bound, and thus couldn't be expected to produce original mathematics research (which is what I suppose you mean by "more than a curricular problem").  The bulk of my math-major students are able, by the time they graduate, to produce proofs for many curricular problems.
[/quote]

What I really meant was frame and solve a proof that wasn't taught to them -- not just original research, but original approaches to any of the kinds of problems they may be called upon to solve mathematically in any line of work (that isn't more than just using the curriculum).

[quote="probability1.01"]Well, in all honesty, if you can solve USAMO problems and structure your thoughts into a well-organized proof, you can pretty much solve enough AIME problems. [/quote]
Errr... I know a kid who first took AMC-12 in 6th grade and qualified for AIME;  got USAMTS Bronze prize that same year; continued competing through the high school -- but  he never made it past AIME. He definitely was good at proof-writing -- he took upper-level Math courses in one of the top colleges while in high school, and got into top 100 in Putnam as a college student. He also did not make THAT many careless mistakes (region winner in AMC).

Looks like... AIME has a pretty specific format (and tests pretty specific kind of Math), and some kids just are not so good in that kind of Math, or they are not interested in it. And I am pretty sure that there are other kids who might do pretty well on USAMO had they make it through this AIME barrier... and many of them are the kids who started doing college-level Math by 9th grade and just switched their minds off the AIME-type stuff. Counting prime numbers is not among their favourite things to do. :roll:

I guess what I'm trying to tell is: the low correlation between AIME and USAMO scores may follow from the fact that  AIME and USAMO are based on different kinds of Math, and the careless mistakes may not play the role you think they do.

I hope that USAMO and MOP expansions made lately will help to attract these kids to the competitions... and maybe they'll forget for a week about their Analysis homework and try to concentrate on the AIME-type problems.  Or maybe they will not. So far most of them just say "I am not interested in competitions"...

Yeah...the careless mistakes have always killed me in competitions...that's why I really want to make USAMO...
I just feel that after I know HOW to solve a problem, actually chugging it through gives little or no satisfaction...and I don't know a lot of contest-math kids who are like that...
But I know a lot of non contest, but still very smart kids who are like that.

[quote="marma103"]
Errr... I know a kid who first took AMC-12 in 6th grade and qualified for AIME;  got USAMTS Bronze prize that same year; continued competing through the high school -- but  he never made it past AIME. He definitely was good at proof-writing -- he took upper-level Math courses in one of the top colleges while in high school, and got into top 100 in Putnam as a college student. He also did not make THAT many careless mistakes (region winner in AMC).
[/quote]

Then apparently he wasn't very good with the time limit...

[quote="K81o7"]
I just feel that after I know HOW to solve a problem, actually chugging it through gives little or no satisfaction...and I don't know a lot of contest-math kids who are like that...
But I know a lot of non contest, but still very smart kids who are like that.[/quote]

How much satisfaction does writing your solution give?

Actually figuring out a solution is fun.  Chugging through calculations isn't...
Writing out a solution isn't all that fun either...but it is somewhat fun to look through it and simplify it...
It's always fun to find a solution which is easy to understand, even intuitive, over a complicated one.

[quote="beta"][quote="marma103"]
Errr... I know a kid who first took AMC-12 in 6th grade and qualified for AIME;  got USAMTS Bronze prize that same year; continued competing through the high school -- but  he never made it past AIME. He definitely was good at proof-writing -- he took upper-level Math courses in one of the top colleges while in high school, and got into top 100 in Putnam as a college student. He also did not make THAT many careless mistakes (region winner in AMC).
[/quote]Then apparently he wasn't very good with the time limit...
[/quote]
A top 100 Putnam finisher has to be good with time limit!

The fact remains that AIME along with the overall AMC scoring system is not the ideal way to select USAMO qualifiers, but it is not so bad either. If there ever is a consensus at AMC to want to improve, it can definitely be done without much investment.

I doubt any major revision will happen. So it is purely up to each individual to formulate one's strategy with the given scoring system in mind. If AIME is the stumbling block against one’s strong desire to qualify for USAMO, then one should practice AIME more. While that presents no guarantee against AIME's brutal nature, it should at least increase one's chances.

I know several who did have that strong desire and did invest a bit more time to prepare for this year’s AIME had come away with very amazing results. For those without such strong desire, why waste time and emotion? I wouldn’t call (potential) top performance in Putnam a bad consolation!

[quote="marma103"]... and many of them are the kids who started doing college-level Math by 9th grade and just switched their minds off the AIME-type stuff. Counting prime numbers is not among their favourite things to do. :roll: [/quote]Here here.

This is an object lesson of what is spam.  The previous post is spam.

Resurrecting a 6-month-old dormant thread just to say "here here" to someone's point is pointless and unnecessary.

Some guidelines:

- every post should have content.  Don't post just to say "I agree" or "You're right" or something without any content.

- don't resurrect long-dormant threads unless you have a very good reason for doing so.


quadratics
algebra
quadratic formula
polynomial
sum of roots
product of roots
Rational Root Theorem

[hide]The roots of the first equation by factoring are -5 and 1.

Applying synthetic division, since the roots are common,

Umm... someone can show the division, but we get (2x+1), and the third root is -1/2.
     [/hide]

[hide]the roots of x^2+4x-5 are -5 and 1. therefore, we know that -5 and 1 are also roots of the second equation. if we use synthetic division on the second equation and -5, we get 2x^2-x-1. applying the quadratic formula on this new equation, we find that the roots are 1 and -1/2. since 1 and -5 were the roots of the first equation, -1/2 must be the third root of the second equation.[/hide]

Great.

But here's another thought. Did we need to find the roots of the first equation?

What if the roots were something like $4+3i$ and $4-3i$? That wouldn't be a neat numbers to use in a synthetic division. Is there anyway we can do this problem WITHOUT using synthetic division?

We could simply do a long division ;)

$\frac{2x^3+9x^2-6x-5}{x^2+4x-5}=2x+1$

:) Yes.

Although it doesn't hurt you to find the roots, frt's method is quicker and what the CMO solution said (and how I did.. I actually did frt's method and used synthetic division to check my answer).

i think that's similar to what 236factorial did.

[hide="strange way"]The first equation is $x^2+4x-5=0$, The second $2x^3+9x^2-6x-5=0$. The roots of the first equation are in the form (x+a) and (x+b). Somehow, the third root (cx+b) must bring the leading coefficient from 1 to 2. Only (2x+d) will satisfy this.

Now look at the final term (the d) and we realize that d must be a number so that -5*d=-5. Only 1 works. d=1.

So the third root is the result from (2x+1). The root is -1/2

Yeah, this is strange. [/hide]

That's not a strange way.

It is pretty clever actually. Noticing the first equation is MONIC (leading coefficent is 1) and the second equation has coefficent of 2 helps you to know that it's in form of $(2x+d)$. From here, like you did, it's not that hard to figure it out.

Good job! :)

Yeah, I think that's actually faster than doing a long division or synthetic division.

[quote="ajai"]for synthetic division I thought that you can only divide by (x-a), how do you divide by
$x^2+4x+5$?[/quote]

Do you mean $x^2+4x-5$?
Factoring, that makes (x+5) and (x-1). Now you can use synthetic division.

[quote="Silverfalcon"]The roots of the equation $x^2+4x-5 = 0$ are also the roots of the equation $2x^3+9x^2-6x-5 = 0$. What is the third root of the second equation?

Please post full solution to the problem. If you consider this problem easy, leave to other people who can benefit from trying this.[/quote]

$x^2+4x-5 \Longrightarrow (x-1)(x+5) \Longrightarrow x=1, -5$, now use one of the roots for the second equation with synthetic division, I chose $1$ and then I factored the second equation to: $(x-1)(2x^2+11x+5) \Longrightarrow 2x^2+11x+5 \Longrightarrow (2x+1)(x+5) \Longrightarrow x=-5,\boxed{-\frac{1}{2}}$.

Let me add that this question can be solved by another method too. Solution by Viete formulas

1) Sum of roots of quadratic is -4. Sum of roots of cubic is $-4+r=-\frac92$. Third root is $r=-\frac12$

2) Product of roots of quadratic is -5. Product of roots of cubic is $-5r=\frac52$. Third root is $r=-\frac12$

[quote="Iversonfan2005"][quote="AntonioMainenti"][quote="Iversonfan2005"]$x^2+4x-5 \Longrightarrow (x-1)(x+5) \Longrightarrow x=1, -5$, now use one of the roots for the second equation with synthetic division, I chose $1$ and then I factored the second equation to: $(x-1)(2x^2+11x+5) \Longrightarrow 2x^2+11x+5 \Longrightarrow (2x+1)(x+5) \Longrightarrow x=-5,\boxed{-\frac{1}{2}}$.[/quote]I hate to say this over and over, but this is a misuse of the $\Longrightarrow$ symbol. Statements imply other statements. An expression on its own cannot imply anything.[/quote]

Does it really matter? ;)  :|[/quote]
It doesn't matter a lot if used in forums, but I think it matters elsewhere :D
OK here is one suggestion to use it better:

$x^2+4x-5=(x-1)(x+5)=0\Longrightarrow x=1, -5$

$(x-1)(2x^2+11x+5)=0\Longrightarrow x-1=0\text{ or }2x^2+11x+5=0\Longrightarrow x=1\text{ or }(2x+1)(x+5)=0\Longrightarrow x=1,-5,\boxed{-\frac{1}{2}}$

[quote="Iversonfan2005"]Does it really matter? ;)  :|[/quote]Well, you're using an implication symbol where you really mean $=$. What if someone used $+$ where they meant $-$, or $\in$ where they meant $\le$. It's not any different to me at least. Anyway, I'm sure your math professors won't like it either, as 10000th User hinted at.

[quote="anirudh"]what is a third root.  :?[/quote]Cubic equations have $3$ roots.  In the question it gives you the first two roots (same as the quadratic equation's) and asks for the remaining (3rd) root.

[quote="anirudh"]What is synthetic divisioin?[/quote]

It's a way of factoring by using trial and error using the last term over the first term to find the roots.

[quote="Iversonfan2005"][quote="anirudh"]but i don't like trial and error.[/quote]

You must like spamming though.. ;)

Go read about synthetic division, it's very useful.[/quote]

Synthetic division isn't all about guess and checking though (especially when you have a range of numbers using another theroem I can't remember, rational root theorem?).

In this case, you knew the roots, so synthetic division is perfect. Just remember that the divisor has to be in the form x-a.

the equation is  x^2 +4x-5     roots r 1 and -5             concerning 2nd equation

2x^3+9x^2-6x-5=0          the sum of the roots is      1-5+p=4.5           p-4=4.5
  therefore     p=9.5       ( i may be mistaken though   if  p-4 =-4.5    then  p=-0.5) :D  :D

[hide]sum of roots of first equation is -4.

sum of roots of second equation is -9/4.

third root equals the difference. -1/2[/hide]

[hide=alternate solution]
We see that the roots of this equation:
$$x^2+4x-5 = 0$$
are $1$ and $5$.

We know that this equation:
$$2x^3+9x^2-6x-5 = 0$$
has 3 roots because it is a cubic.

By Vieta's formula, the product of the roots of the second equation is $\frac{5}{2}$.

We know $2$ of the roots of the 2nd equation ($1$ and $-5$) so our answer is just $\frac{\frac{5}{2}}{1\cdot-5} = \boxed{\frac{-1}{2}}$

if you have a new solution to a problem, it is encouraged to share it.

The simplest solution:
Product of roots in cubic =$\frac{-5}{2}$
Product of roots in quadratic = $5$
Remaining root = $\frac{\frac{-5}{2}}{5}= \frac{-1}{2}$


Let $ H$ be the orthocenter of acute triangle $ ABC$. The tangents from $ A$ to the circle with diameter $ BC$ touch the circle at $ P$ and $ Q$. Prove that $ P,Q,H$ are collinear.


Find $f : Z \to Z$ such that :
 $f(x+y+f(y))=f(x)+2y$

I don't think there are any. Let $g(y)=y+f(y)$, and in the equation given, let $x=g(y)$ and consider $y\neq0$. Then we get $f(2g(y))=f(2g(y))+2y\Rightarrow y=0$. Contradiction.

Sorry for my stupid typo !
 The correct version :
 $f(x+y+f(y)) = f(x)+2y$


Harvard
college
Princeton
Yale
MIT
University of Chicago
Stanford

Okay, this is getting a bit silly .....

And also, lowest_reynolds, you don't even live in Canada .....

[quote="lowest_reynolds"]
I got into:

Harvard
Princeton
Yale
MIT
...
[/quote]

Yay! Now all you've got to do is create some clones of yourself, and send them to all those places. And see which of them gets better grades.

[quote="lowest_reynolds"]Oh I live in Canada!  O Caaaaanadaaa...

Hoorah first post!  I'll start:

I got into:

Harvard
Princeton
Yale
MIT
Cornell
UPenn
Brown
UChicago
Stanford
Duke
Caltech
Oxford
Oxbridge
Ox-something I forgot
Cambridge
U. College London
Columbia
Dartmouth
Southern Polytechnic

Buuuuuuttt, I've ultimately decided to attend Southern Polytechnic.  Ahhh...we all know colleges bow down to my godlike presence anyways! :lol:

Lalalala...

Anybody else going to Southern Polytechnic?  OMG I'LL SEE YOU GUYS THERE.[/quote]

LMFAO!  That's the funniest post I've read in a while.

In Canada, it costs a lot of money per application, so most people only apply to about 3 schools.  I only applied to Waterloo.  ;)

[quote="lowest_reynolds"]Oh I live in Canada!  O Caaaaanadaaa...

Hoorah first post!  I'll start:

I got into:

Harvard
Princeton
Yale
MIT
Cornell
UPenn
Brown
UChicago
Stanford
Duke
Caltech
Oxford
Oxbridge
Ox-something I forgot
Cambridge
U. College London
Columbia
Dartmouth
Southern Polytechnic

Buuuuuuttt, I've ultimately decided to attend Southern Polytechnic.  Ahhh...we all know colleges bow down to my godlike presence anyways! :lol:

Lalalala...

Anybody else going to Southern Polytechnic?  OMG I'LL SEE YOU GUYS THERE.[/quote]

*falls over and dies*

[quote="Elyot"] In Canada, it costs a lot of money per application, so most people only apply to about 3 schools.  I only applied to Waterloo.  ;)[/quote]

I wouldn't make that generalization. Most of my friends applied to a lot more than just the three choices provided by the OUAC.

As well, as you probably never considered the U.S. schools, it costs much more to apply there (~$\$$65 for U.S. vs ~$\$$33 for Canadian).

[quote="lowest_reynolds"]Oh I live in Canada!  O Caaaaanadaaa...

Hoorah first post!  I'll start:

I got into:

Harvard
Princeton
Yale
MIT
Cornell
UPenn
Brown
UChicago
Stanford
Duke
Caltech
Oxford
Oxbridge
Ox-something I forgot
Cambridge
U. College London
Columbia
Dartmouth
Southern Polytechnic

Buuuuuuttt, I've ultimately decided to attend Southern Polytechnic.  Ahhh...we all know colleges bow down to my godlike presence anyways! :lol:

Lalalala...

Anybody else going to Southern Polytechnic?  OMG I'LL SEE YOU GUYS THERE.[/quote]

so you sacrifised something like $\$$2000 to apply to all those universities?....Not to mention the amount of hours you wasted on writing all those 700-word essays

when i was gr 12 (long time ago), i applied for 5 programs, i was forced to apply for 8. lol... i dunno what my parents were worried about... lol.... i guess it would be interesting to hear ppl's choices... i'll start!
uw mech eng
uw cs
uw physics
ut elec eng
queens eng
queens cs
guelph system eng
guelph cs

btw, it turned out that you only need to apply to one of guelph's program. i wasted 25 dollars!!! why can't parents listen for once? "i just want to apply for 5!!" yeah last three choices were just what i picked randomly looking at info. definitly got a lecture on why would i apply for guelph engineering when guelph is good for bio.  :|  :|
i was thinking of uw cs, but then i remembered i hate programming, i just want to go for the name... then i pursued my dream since gr 6... uw mech eng...  :D and i love it! i never considered US schools cause i can't afford them.

[quote="Elyot"]Heheh... almost all my friends only applied to 3.  The OUAC actually used to allow a minimum of 2 applications, but they switched to 3 for my year.  If you know where you want to go and know you'll be accepted, there's no reason to apply to any more.  I could understand if you still need a few more months to pick which program you want, or you're not sure about getting in... but I'd think people on here would have their minds made up by January about that :P

I'd like to see how many of your friends applied to >3 universities because THEY wanted to, or because THEIR PARENTS wanted them to.  Your high school has a very different demographic than mine!!! ;)[/quote]

hahaha... everyone i know in my year applied for at least 4. i think it was cause i'm the year after double cohort, so everyone is afraid that the ppl who did victory lap from the year before would take our spots. but it turned out that it's not too bad.

i think my bro got it bad. i think he applied for 5 or 6. he was in the double cohort year. that was pretty intense, cause we're all afraid that he wouldn't get into anything. but it turned ok, he's in comp eng. he rejected 17k from cs. his teacher was furious about it lol...
i think he applied to
uw comp eng
uw cs
ut eng sci
queens eng
?
?
again, my parents forced him to apply as well...

i know someone who applied to 12. lol... do u really want to go to ur 12th choice? i don't even want to go to my 3rd choice!

what about u Elyot?

[quote="J-I-M-B-O"]
what about u Elyot?[/quote]

Having 3 applications, I applied to Waterloo Math, Waterloo CS, and Waterloo Software Eng.  Of course, I had no intent of actually doing software, as it's not one of Waterloo's stronger engineerings anyway, so I didn't even do any of the forms for engineering because I really had no desire to go there.  I ended up doing CS because there were 2 25k scholarships for CS and only 1 for math, so I figured my chances were better.  I wanted a double major in Math/CS anyway.

[quote="J-I-M-B-O"]nice... hows the workload for double degree?[/quote]

Truthfully, I don't like electives anyway.  Currently, I'm doing 2 maths, 2 cs, and physics (need for applied math & it counts as an elective), and it's not too bad.

CS alone is pretty easy, in fact.

I'm still not sure whether I'm gonna add pure or applied math.  I don't think I want to add C&O or something trivial like that.

[quote="Elyot"]

I'd like to see how many of your friends applied to >3 universities because THEY wanted to, or because THEIR PARENTS wanted them to.  Your high school has a very different demographic than mine!!! ;)[/quote]

I just wanted uwaterloo math, kind of wanted too see what else I wanted to get into, but didn't care to much.
when I had filled out four ( uw math, uw cs, uw eng, queens math ), my dad told me to keep going. I told him it costs money, and that I wasn't too much more interested. He made me sighn up for 10!!!

So I ended up getting alot of early acceptances from places I didn't care about. felt wierd

[quote="rzsolt"]
my dad told me to keep going. I told him it costs money, and that I wasn't too much more interested. He made me sighn up for 10!!!

So I ended up getting alot of early acceptances from places I didn't care about. felt wierd[/quote]

Lol that is so lame, why pay an extra $\$$ 200 for nothing

Pffft...200 bucks?  That's nothing.

He could be like...YEAH I GOT ACCEPTED INTO TEN DIFFERENT PROGRAMS.  HOW MANY DID YOU GET INTO?  WHAT'S THAT?  ONE, TWO?  PWN-LOBSTER!!!

It's the ego that counts.

[quote]
[b] PWN-LOBSTER!!!!!!!!!!!! [/b]
[/quote]

Thanks.  I'll start using that.

PS.  you are up at 214am?


So say you had to use USAMTS as a fallback to get to the AIME. What score would you need to get to the USAMO?

If you make AIME through USAMTS, you get a score of 100. Hence, you'd probably need ~10-11 on aime, or so, depending on the year.


all the forms are due tomorrow!!! ahhh. need a topic
anyone here knows a good mathematic or physics topics i could do.
it needs to have a independent and dependent variables.
any suggestions or ideas .. please help thanks

Elliptic curves? Though from the very little that I know the dependent and independent variables are a bit fuzzy.

catapult. I guess it could relate with projectile motion in physics.  :?


inequalities
geometry
incenter
geometry proposed

[quote="quykhtn-qa1"][quote="leedt26"][quote="quykhtn-qa1"]Let $ G$ and $ I$ be the centroid and incenter of a triangle $ ABC$ .Prove that
$ GA \plus{} GB \plus{} GC \ge \sqrt {\frac {2(a^2 \plus{} b^2 \plus{} c^2 \plus{} 4\sqrt {3}S}{3}} \ge IA \plus{} IB \plus{} IC$
( $ a,b,c$ be the sides triangle and $ S$ is area)[/quote]
Dear quykhtn-qa1, Can you have a simple solutions for this inequality?[/quote].
This inequality stronger than inequality of my teacher: Nguyen Minh Ha , I created it and it 's true and difficult[/quote]
Why do you say it's true?

[quote="litongyang"][/quote][quote="quykhtn-qa1"][quote="leedt26"][quote="quykhtn-qa1"]Let $ G$ and $ I$ be the centroid and incenter of a triangle $ ABC$ .Prove that
$ GA + GB + GC \ge \sqrt {\frac {2(a^2 + b^2 + c^2) + 4\sqrt {3}S}{3}} \ge IA + IB + IC$
( $ a,b,c$ be the sides triangle and $ S$ is area)[/quote]
Dear quykhtn-qa1, Can you have a simple solutions for this inequality?[/quote].
This inequality stronger than inequality of my teacher: Nguyen Minh Ha , I created it and it 's true and difficult[/quote][quote="litongyang"]
Why do you say it's true?[/quote]
Can you solve it? . :wink:
It 's true and true :lol:

[quote=quykhtn-qa1]Let $ G$ and $ I$ be the centroid and incenter of a triangle $ ABC$ .Prove that
$ GA \plus{} GB \plus{} GC \ge \sqrt {\frac {2(a^2 \plus{} b^2 \plus{} c^2) \plus{} 4\sqrt {3}S}{3}} \ge IA \plus{} IB \plus{} IC$
( $ a,b,c$ be the sides triangle and $ S$ is area)[/quote]

My beautiful inequality from [url=http://artofproblemsolving.com/community/c6h608080]here[/url] gives a stronger upper bound for $\textstyle\sum AI$ :-D. In other words, the following chain of inequalities holds in any triangle $ABC$ with incenter $I$, centroid $G$ and area $S$:

\[\boxed{IA + IB + IC \stackrel{(1)}{\le}s + r\cdot \frac {s + 2R - r}{s + 2R + r} \stackrel{(2)}{\le} \sqrt{\frac {2\left(a^2 + b^2 + c^2\right) + 4S\sqrt{3}}3} \stackrel{(3)}{\le} GA + GB + GC}.\]

Since inequality $(1)$ is already established, it remains to prove $(2)$ as well as $(3)$. Note that inequality $(3)$ is just the dual of Hadwiger-Finsler inequality applied to a triangle with sides $m_a$, $m_b$ and $m_c$ (see [url=http://artofproblemsolving.com/community/c1756h1021231_lower_and_upper_bound_for_the_sum_gagbgc_in_any_triangle]here[/url] for example, where you can also find an upper bound for $\textstyle \sum GA$). Finally, we are left to show inequality $(2)$. This is a bit messy, but definitely doable. After squaring both sides and rearranging we end up with:

\[f(s)\equiv 4\left(s + 2R + r\right)^2\left(s^2 + rs\sqrt{3} - 4Rr - r^2\right) - 3\Big[\left(s + 2R\right)\left(s + r\right) + r\left(s - r\right)\Big]^2 \ge 0.\]

It is possible to show that $f^{\prime}(s)\ge 0$, which means that $f$ is an increasing function. Now all we need is Gerretsen's inequality, i.e. $s^2\ge 16Rr - 5r^2$ which allows us to write:

\[f(s)\ge f\left(\sqrt{16 Rr - 5r^2}\right),\]

so it is enough to prove that:

\[f\left(\sqrt{16 Rr - 5r^2}\right)\ge 0.\]

Now, upon the substitution $t=\tfrac Rr \ge 2$ the latter is transformed into a one-variable inequality, which is true according to [url=http://www.wolframalpha.com/input/?i=4+(x+%2B+2+t+%2B+1)%5E2+(x%5E2+%2B+x+sqrt(3)+-+4+t+-+1)+-+3+(x+(x+%2B+2+t+%2B+1)+%2B+(x+%2B+2+t+-+1))%5E2+where+x+%3D+sqrt(16+t+-+5)]Wolfram|Alpha[/url] ;).

[quote=quykhtn-qa1]Let $ G$ and $ I$ be the centroid and incenter of a triangle $ ABC$ .Prove that
$ GA \plus{} GB \plus{} GC \ge \sqrt {\frac {2(a^2 \plus{} b^2 \plus{} c^2) \plus{} 4\sqrt {3}S}{3}} \ge IA \plus{} IB \plus{} IC$
( $ a,b,c$ be the sides triangle and $ S$ is area)[/quote]

one  of is:

\[m_a+m_b+m_c\geq \frac{1}{2}\sqrt{6b^2+6c^2+6a^2+12\sqrt{3}S}\]

No say something,
must be nice!

[quote=xzlbq]\[f(s)\equiv 4\left(s + 2R + r\right)^2\left(s^2 + rs\sqrt{3} - 4Rr - r^2\right) - 3\Big[\left(s + 2R\right)\left(s + r\right) + r\left(s - r\right)\Big] \ge 0.\]

Dear Mateescu Constantin:

But if $s=3\sqrt{3},R=2,r=1$

$f(s)$ is not $0$?[/quote]

Thanks for pointing that out. I missed a power of $2$ in the very last term, which I fixed now ;).

[quote=xzlbq]
one  of is:

\[m_a+m_b+m_c\geq \frac{1}{2}\sqrt{6b^2+6c^2+6a^2+12\sqrt{3}S}---(1)\]

No say something,
must be nice![/quote]

(1) with Klmkin,get this:
\[\sum{r_a}\geq \sqrt{3}s\]

[quote=quykhtn-qa1]Let $ G$ and $ I$ be the centroid and incenter of a triangle $ ABC$ .Prove that
$ GA \plus{} GB \plus{} GC \ge \sqrt {\frac {2(a^2 \plus{} b^2 \plus{} c^2) \plus{} 4\sqrt {3}S}{3}} \ge IA \plus{} IB \plus{} IC$
( $ a,b,c$ be the sides triangle and $ S$ is area)[/quote]
The left inequality it's just $2(ab+ac+bc)-a^2-b^2-c^2\geq4\sqrt3S$, which is obvious.

[color=darkblue]As a final remark, note that for a [b]non-obtuse angled[/b] triangle $ABC$ the following inequality holds:

\[\boxed{ IA + IB + IC \le \sqrt{\frac {3 \left(a^2 + b^2 + c^2\right) + 8S\sqrt{3}}{5}} }\ .\]

Also compare against [url=https://artofproblemsolving.com/community/u61896h1334757p7214163][b]this even sharper inequality[/b][/url] for which I am still waiting for a solution :D. So for all [b]acute[/b] or [b]right-angled[/b] triangles $ABC$ we have the following [u]strong[/u] chain of inequalities:

\[\boxed{ IA + IB + IC \le s + r\cdot\left(4 - 2\sqrt{3} + \frac 13\cdot\frac {s + 2R - r}{s + 2R + r}\right) \le  \sqrt{\frac {3 \left(a^2 + b^2 + c^2\right) + 8S\sqrt{3}}{5}} }\ .\]
[/color]


complex analysis
complex analysis unsolved

[quote="Enea"]Solve for $ t>0$

$ y^{''}+y^'+y=e^t*chi_{[0,1]}(t)$
where the initial conditions are:
$ y^'(0)=y(0)=0$,and $ chi_{[0,1]}$ is a function which value $ 1$ on $ [0,1]$,$ 0$ otherwise[/quote]


probability
ratio
combinatorics unsolved
combinatorics

part1:
A is visiting 6 cities, each one twice and no same city consecutively, find the number of ways A can do so.
part2:
B is chasing to kill A. B visits each of the 6 cities once but stays at each of the 6 cities for twice the duration as A stays in any city. Suppose A stays at each of the 12 locations for the same length of time. If you ignore traveling time between cities, A and B do not go anywhere else, and A is dead whenever A and B are in the same city at the same time, find the probability that A will survive. A travels as it did in part 1.

e.g.
A and B start off in a city but B first moves at the same time as A going to his third location

Let $ 6 \equal{} n$.

Part 1 is [url=http://www.research.att.com/~njas/sequences/A114938]A114938[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].  A straightforward inclusion-exclusion argument gives the formula listed there.  The first few terms are 0, 2, 30, 864, 39480, 2631600, ....

For part 2, we may rephrase it as the enumerative problem asking for the number of permutations of the multiset $ \{1, 1, 2, 2, \ldots, n, n\}$ such that for all $ i$, the two copies of $ i$ are nonconsecutive and neither copy of $ i$ lies in position $ 2i \minus{} 1$ or $ 2i$.  With only the second restriction, this is some multiset analogue of derangements.  The first few values are 0, 0, 5, 86, 4181, 293178, ....  Neither this sequence nor its complementary sequence appear in the OEIS.

If we do part 2 without the restriction that A must change towns, the total number of paths for A is $ \frac {(2n)!}{2^n}$ and the total number that escape is [url=http://www.research.att.com/~njas/sequences/A000459]A000459[/url] or equivalently [url=http://www.research.att.com/~njas/sequences/A059072]A059072[/url], which apparently satisfies the recursion [hide="might be fun to try to find on your own."]$ a(n) \equal{} n(2n \minus{} 1) a(n \minus{} 1) \plus{} 2n(n \minus{} 1)a(n \minus{} 2) \minus{} (2n \minus{} 1)$.[/hide]  The convergence of the ratio $ \frac {a(n)}{\frac {(2n)!}{2^n}}$ seems to be awfully slow, but I'd be willing to bet (based on some numerical monkeying around) that the limit value is no larger than $ 0.13533528\ldots$.


Is there another way to do it other than substituting a variable in for a number?

Yes, there is.

[hide="solution"]We see that it is equal to

$\frac{(1+2005^3)}{2006^3}+\frac{3*2005}{2006^2}$

We know that the sum of the cubes equals the sum of the squares minus the product of the two numbers all times the sum of the numbers, so the equation is now

$\frac{2006(1+2005*2004)}{2006^3}+\frac{3*2005}{2006^3}=\frac{1+2005*2007}{2006^2}$

If you know that $(n+1)(n-1)=n^2-1$, you get the equation

$\frac{2006^2-1+1}{2006^2}=\frac{2006^2}{2006^2}=$ [b][u]1[/u][/b][/hide]

[quote="236factorial"][hide]Let $a=\frac{1}{2006}$ and $b=\frac{2005}{2006}$.

Then $a^3+b^3+3ab=(a+b)(a^2-ab+b^2)+3ab$. Since $a+b=1$, the expression becomes

$a^2-ab+b^2+3ab=a^2+2ab+b^2=(a+b)^2=1$. [/hide][/quote]

[hide]I did it the same way except I used x and y instead of a and b, respectively.  So that is why my answer is 1.

$1$[/hide]


let $ x,y,z >0. x\plus{}y\plus{}z\equal{}12$. find min :
$ \frac{x}{\sqrt[]{y}}\plus{} \frac{y}{\sqrt{z}}\plus{}\frac{z}{\sqrt{x}}$

By am-gm,
$ \sum_{cyclic} 2\frac{x}{\sqrt{y}} \plus{} xy \ge 3x$
to minimize - $ \sum_{cyclic} \frac{x}{\sqrt{y}}$ we will maximise - $ xy\plus{}yz\plus{}zx$
by AM-GM,$ 48 \ge xy\plus{}yz\plus{}zx$
therefore the minimum of the given expression is - $ \frac{36 \minus{} 48}{2}$,
but since they are positive the answer is - $ 0$ ??

[quote="The Enigma"]By am-gm,
$ \sum_{cyclic} 2\frac {x}{\sqrt {y}} \plus{} xy \ge 3x$
to minimize - $ \sum_{cyclic} \frac {x}{\sqrt {y}}$ we will maximise - $ xy \plus{} yz \plus{} zx$
by AM-GM,$ 48 \ge xy \plus{} yz \plus{} zx$
therefore the minimum of the given expression is - $ \frac {36 \minus{} 48}{2}$,
but since they are positive the answer is - $ 0$ ??[/quote]

Thanks you very much !  :D

[b]@The Enigma[/b]: In your first inequality, equality holds when $ xy\equal{}\frac{x}{\sqrt{y}}$ ... ,that is $ x\equal{}y\equal{}z\equal{}1$, which is not possible according to the given condition. So the value you've arrived at is clearly wrong.

[quote="bigbang195"]let $ x,y,z > 0. x + y + z = 12$. find min :
$ \frac {x}{\sqrt []{y}} + \frac {y}{\sqrt {z}} + \frac {z}{\sqrt {x}}$[/quote]
$ A = \sum_{cyclic}\frac {x}{\sqrt {y}}\ \ B = \sum_{cyclic}xy$

$ \right A^2B\ge (x + y + z)^3$

$ \rightarrow A^2\ge \frac {(x + y + z)^3}{ab + bc + ca}\ge \frac {12(a + b + c)^2}{ab + bc + ca}\ge 36$

$ \rightarrow A\ge 6$

[b]We are done![/b]

[quote="keyree10"][b]@The Enigma[/b]: In your first inequality, equality holds when $ xy \equal{} \frac {x}{\sqrt {y}}$ ... ,that is $ x \equal{} y \equal{} z \equal{} 1$, which is not possible according to the given condition. So the value you've arrived at is clearly wrong.[/quote]
i am sorry for my wrong solution.. you are right.. i will remeber to not do that mistake next time... thanks


Squares $S_1$ and $S_2$ are inscribed in right triangle $ABC$, as shown in the figures below.  Find $AC + CB$ if area$(S_1) = 441$ and area$(S_2) = 440$.

[asy]
size(250);
real a=15, b=5;
real x=a*b/(a+b), y=a/((a^2+b^2)/(a*b)+1);
pair A=(0,b), B=(a,0), C=origin, X=(y,0), Y=(0, y*b/a), Z=foot(Y, A, B), W=foot(X, A, B);
draw(A--B--C--cycle);
draw(W--X--Y--Z);
draw(shift(-(a+b), 0)*(A--B--C--cycle^^(x,0)--(x,x)--(0,x)));
pair point=incenter(A,B,C);
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$A$", (A.x-a-b,A.y), dir(point--A));
label("$B$", (B.x-a-b,B.y), dir(point--B));
label("$C$", (C.x-a-b,C.y), dir(point--C));
label("$S_1$", (x/2-a-b, x/2));
label("$S_2$", intersectionpoint(W--Y, X--Z));
dot(A^^B^^C^^(-a-b,0)^^(-b,0)^^(-a-b,b));[/asy]

[hide="Solution"][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=786&sid=381a70b94ead6fb11f3b600d764eba31[/img]

Because all the triangles in the figure are similar to triangle $ABC$, it's a good idea to use area ratios. In the diagram above, $\frac{T_1}{T_3}=\frac{T_2}{T_4}=\frac{441}{440}.$ Hence, $T_3=\frac{440}{441}T_1$ and $T_4=\frac{440}{441}T_2$. Additionally, the area of triangle $ABC$ is equal to both $T_1+T_2+441$ and $T_3+T_4+T_5+440.$

Setting the equations equal and solving for $T_5$, $T_5=1+T_1-T_3+T_2-T_4=1+\frac{T_1}{441}+\frac{T_2}{441}$. Therefore, $441T_5=441+T_1+T_2$. However, $441+T_1+T_2$ is equal to the area of triangle $ABC$! This means that the ratio between the areas $T_5$ and $ABC$ is $441$, and the ratio between the sides is $\sqrt{441}=21$. As a result, $AB=21\sqrt{440}=\sqrt{AC^2+BC^2}$. We now need $(AC)(BC)$ to find the value of $AC+BC$, because $AB^2+2(AC)(BC)=(AC+BC)^2$. (Does $(AC)(BC)$ suggest what the next step is? :P )

Let $h$ denote the height to the hypotenuse of triangle $ABC$. Notice that $h-\frac{1}{21}h=\sqrt{440}$. (The height of $ABC$ decreased by the corresponding height of $T_5$) Thus, $(AB)(h)=(AC)(BC)=22\cdot 21^2$. Because $AB^2+2(AB)(BC)=(AC+BC)^2=21^2\cdot22^2$, $AC+BC=(21)(22)=462$.[/hide]

[hide="screw your courage to the sticking place"]
Because we embark on a journey that will require immense calculational fortitude.

Label the legs $ a$ and $ b$. Using some similar triangle stuff that I will omit, we get that the square in the first config has side $ \frac {ab}{a \plus{} b} \equal{} 21$. For the second one, we get $ \frac {abc}{a^2 \plus{} ab \plus{} b^2} \equal{} \frac {ab\sqrt {a^2 \plus{} b^2}}{a^2 \plus{} ab \plus{} b^2} \equal{} \sqrt {440}$. That was only the easy part, I'm afraid, for we must now pursue the values $ a$ and $ b$ through whatever dangerous obstacles may confront us.

Anyway, squaring the second equation, we get
\[ a^2b^2(a^2 \plus{} b^2) \equal{} 440(a^4 \plus{} b^4 \plus{} a^2b^2 \plus{} 2a^2b^2 \plus{} 2ab^3 \plus{} 2a^3b) \]
\[ \equal{} 440 (a^4 \plus{} b^4 \plus{} 3a^2b^2 \plus{} 2ab^3 \plus{} 2a^3b) \]
Now subbing $ ab \equal{} 21(a \plus{} b)$ into the left side,
\[ 441(a^2 \plus{} 2ab \plus{} b^2)(a^2 \plus{} b^2) \equal{} 440 (a^4 \plus{} b^4 \plus{} 3a^2b^2 \plus{} 2ab^3 \plus{} 2a^3b) \]
\[ 441(a^4 \plus{} b^4 \plus{} 2a^3b \plus{} 2ab^3 \plus{} 2a^2b^2) \equal{} 440 (a^4 \plus{} b^4 \plus{} 3a^2b^2 \plus{} 2ab^3 \plus{} 2a^3b) \]
Now we get some decent cancellation, resulting in
\[ a^4 \plus{} b^4 \plus{} 2a^3b \plus{} 2b^3a \minus{} 438a^2b^2 \equal{} 0
\]
We might prefer to write this as
\[ (a^4 \plus{} 2a^2b^2 \plus{} b^4) \plus{} 2ab(a^2 \plus{} b^2) \minus{} 440a^2b^2 \equal{} 0
\]
Whereupon the form
\[ (a^2 \plus{} b^2)^2 \plus{} 2ab(a^2 \plus{} b^2) \minus{} 440(ab)^2 \equal{} 0
\]
draws our attention. Now dividing by $ (ab)^2$ and making the substitution $ t \equal{} \frac {a^2 \plus{} b^2}{ab}$,
\[ t^2 \plus{} 2t \minus{} 440 \equal{} 0
\]
Aha! This factors as $ (t \plus{} 22)(t \minus{} 20) \equal{} 0$, so $ t \equal{} 20$, so $ \frac {a^2 \plus{} b^2}{ab} \equal{} 20$. In addition, recall that $ \frac {ab}{a \plus{} b} \equal{} 21$, so we get
\[ a^2 \plus{} b^2 \equal{} 20ab
\]
and
\[ ab \equal{} 21(a \plus{} b)
\]
Substitute $ p \equal{} a \plus{} b, q \equal{} ab$ just to make things prettier. Then $ p^2 \minus{} 2q \equal{} 20q$ and $ q \equal{} 21p$, so
\[ p^2 \equal{} 22q \equal{} 22(21)p
\]
implying that $ p \equal{} 22(21) \equal{} 462$.

[/hide]

[hide="a coward approach"]
$ \frac {ab}{a\plus{}b}\equal{}21$
$ \frac {ab\sqrt {a^2\plus{}b^2}}{a^2\plus{}ab\plus{}b^2}\equal{}\sqrt{440}$

Substitute $ x\plus{}y\equal{}u, \; xy\equal{}v, \; 21\equal{}m;$ to obtain
$ v\equal{}mu$
$ \frac {v\sqrt{u^2\minus{}2v}}{u^2\minus{}v}\equal{}\sqrt{m^2\minus{}1}$
Replace v to get
$ \frac {mu\sqrt{u^2\minus{}2mu}}{u^2\minus{}mu}\equal{}\frac {m\sqrt{u^2\minus{}2mu}}{u\minus{}m}\equal{}\sqrt{m^2\minus{}1}$ and square both sides
$ m^2(u^2\minus{}2mu)\equal{}(m^2\minus{}1)(u^2\minus{}2mu\plus{}m^2)$ getting
$ u^2\minus{}2mu\minus{}m^2(m^2\minus{}1)\equal{}0$ with positive root
$ u\equal{}m\plus{}m^2\equal{}m(m\plus{}1)$.[/hide]

Parallel-line ratios and monstrous algebra...
[hide]Let $ AC, AB$ be $ x, y$ respectively.
From diagram 1, by parallel-line ratios, we easily observe
$ \frac {21}{x} \equal{} \frac {y \minus{} 21}{y} \Rightarrow \frac {x \plus{} y}{xy} \equal{} \frac {1}{21}$ (1)
In diagram 2, drop an altitude on hypotenuse. The hypotenuse has length $ \sqrt {x^2 \plus{} y^2}$, and the altitude $ \dfrac{xy}{\sqrt {x^2 \plus{} y^2}}$.
With parallel-line ratios,
$ \dfrac{\sqrt {440}}{\dfrac{xy}{\sqrt {x^2 \plus{} y^2}}} \equal{} 1 \minus{} \dfrac{\sqrt {440}}{\sqrt {x^2 \plus{} y^2}} \Rightarrow \frac {\sqrt {x^2 \plus{} y^2}}{xy} \plus{} \frac {1}{\sqrt {x^2 \plus{} y^2}} \equal{} \frac {1}{\sqrt {440}}$ (2)
which suggests the following step: squaring!
$ \frac {1}{x^2} \plus{} \frac {2}{xy} \plus{} \frac {1}{y^2} \equal{} \frac {1}{441}$
$ \frac {1}{x^2} \plus{} \frac {1}{y^2} \plus{} \frac {2}{xy} \plus{} \frac {1}{x^2 \plus{} y^2} \equal{} \frac {1}{440}$
Clear all this mess with a substitution
$ \frac {1}{x^2 \plus{} y^2} \equal{} \frac {1}{440} \minus{} \frac {1}{441} \Rightarrow x^2 \plus{} y^2 \equal{} 441\cdot 440$.
Substitute it back into (2)
$ \frac {21\sqrt {440}}{xy} \equal{} \frac {1}{\sqrt {440}} \minus{} \frac {1}{21\sqrt {440}} \Rightarrow xy \equal{} 441\cdot 22$.
Finally, plugging this back into (1)
$ x \plus{} y \equal{} (441\cdot 22) / 21 \equal{} 22\cdot 21 \equal{} 462$.
[/hide]


geometry
geometric transformation
rotation
topology
real analysis
real analysis unsolved

Prove that for any $n$ we can find a partition of the plane using $n$ congruent connected sets.

A partition of a set $S$ is a collection of (nonempty) sets $S_1,\ldots,S_n$ so that:

$S_i\subseteq S\ \ \forall \ i$
$S_i\cap S_j = \emptyset \ \ \forall \ i,j, \ i\neq j$

The $n$ sets which are supposed to partition the plane will be denoted by $S_0,\ldots,S_{n-1}$, and they will be translates of one another. In fact, I’ll describe $S=S_1$, whose intersection with every horizontal line will be a disjoint union of segments closed to the left, of length $1$, and for $j\in\overline{1,n-1},\ S_j$ will be equal to $S_1+(j,0)$. This means that the distance between two consecutive segments in $S$ lying on the same horizontal line is $n-1$.

We construct $S$ by transfinite induction. Let $\alpha$ be the first ordinal with cardinality $c$ (continuum cardinality), and let $(A_\beta)_{\beta<\alpha}$ be a well-ordering of the closed plane sets which arise as borders of open sets whose complements in the plane have non-empty interior.

Let $\beta<\alpha$ be an ordinal, and suppose you’ve performed the steps with index $<\beta$ in the induction. Now look at $A_\beta$. It’s easy to prove that either it intersects $c$ horizontal lines, in which case you can choose the segments of the intersection between $S$ and one of the lines which we haven’t used in the previous inductive steps so that $S\cap A_\beta\ne\emptyset$ (on that horizontal line), or it contains an entire horizontal line, in which case no matter what we do, $S$ will intersect $A_\beta$. This construction ensures that $S$ intersects every closed set of the type mentioned in the previous paragraph, and the connectedness of $S$ follows from this property.

Excuse me if this is moronic, but what's wrong with dividing the plane by $n-1$ rays, each starting from the origin and evenly spaced (so the angle in between is $\frac{2\pi}{n}$)? Then we just let $S_k$ be the region between ray $k$ and ray $k+1$ (where we identify ray $n$ to be ray $1$) plus ray $k$. Then each region is obviously simply connected and the whole situation is symmetric under rotation by $\frac{2\pi}{n}$ radians.

Perhaps I misunderstand "congruent" and the second part is what makes them non-congruent?

I'm sure there are correct constructions a good deal simpler than what I did, but this is not one of them :). Where does the origin go in this example? This division of the plane has to be a partition.

Oh yeah, good point.

(Sorry, that's a terrible pun)


[quote="boukharfane"]If $ x_1$, $ x_2$, ..., $ x_n$ are positive reals, show that

$ \frac{1}{x_1}\plus{}\frac{2}{x_1\plus{}x_2}\plus{}...\plus{}\frac{n}{x_1\plus{}x_2\plus{}...\plus{}x_n}<\frac{4}{x_1}\plus{}\frac{4}{x_2}\plus{}...\plus{}\frac{4}{x_n}$.[/quote]
Rewrite into
$ \sum_{i \equal{} 1}^{n}\frac {1}{\sum_{j \equal{} 1}^{i}x_{j}} < 4\sum_{i \equal{} 1}^{n}x_{i}$
And with Cauchy we have
$ \sum_{i \equal{} 1}^{n}\frac {1}{\sum_{j \equal{} 1}^{i}x_{j}} \\
\leqslant\sum_{i \equal{} 1}^{n}\frac {1}{i^{2}}\sum_{j \equal{} 1}^{i}\frac {1}{x_{j}} \\
\equal{} \sum_{j \equal{} 1}^{n}\frac {1}{x_{j}}\sum_{i \equal{} j}^{n}\frac {1}{i^{2}}$
So we just need to prove
$ \sum_{i \equal{} j}^{n}\frac {1}{i^{2}} < 4\ (j \equal{} 1,2,\ldots,n)$
Actually
$ \sum_{i \equal{} j}^{n}\frac {1}{i^{2}} < \sum_{i \equal{} 1}^\infty\frac {1}{i^{2}} < 1 \plus{} \sum_{i \equal{} 2}^\infty\frac {1}{(i \minus{} 1)i} \equal{} 2 < 4$

[quote="shalex"]If $ x_1$, $ x_2$, ..., $ x_n$ are positive reals, show that

$ \frac {1}{x_1} \plus{} \frac {2}{x_1 \plus{} x_2} \plus{} ... \plus{} \frac {n}{x_1 \plus{} x_2 \plus{} ... \plus{} x_n} < \frac {4}{x_1} \plus{} \frac {4}{x_2} \plus{} ... \plus{} \frac {4}{x_n}$.[/quote]
Rewrite into
$ \sum_{i \equal{} 1}^{n}\frac {1}{\sum_{j \equal{} 1}^{i}x_{j}} < 4\sum_{i \equal{} 1}^{n}x_{i}$
And with Cauchy we have
$ \sum_{i \equal{} 1}^{n}\frac {1}{\sum_{j \equal{} 1}^{i}x_{j}} \\
\leqslant\sum_{i \equal{} 1}^{n}\frac {1}{i^{2}}\sum_{j \equal{} 1}^{i}\frac {1}{x_{j}} \\
\equal{} \sum_{j \equal{} 1}^{n}\frac {1}{x_{j}}\sum_{i \equal{} j}^{n}\frac {1}{i^{2}}$
So we just need to prove
$ \sum_{i \equal{} j}^{n}\frac {1}{i^{2}} < 4\ (j \equal{} 1,2,\ldots,n)$
Actually
$ \sum_{i \equal{} j}^{n}\frac {1}{i^{2}} < \sum_{i \equal{} 1}^\infty\frac {1}{i^{2}} < 1 \plus{} \sum_{i \equal{} 2}^\infty\frac {1}{(i \minus{} 1)i} \equal{} 2 < 4$[/quote]
I'm sorry to make a stupid mistake in this solution while now I'll post the right one.
Rewrite into
$ \sum_{i \equal{} 1}^{n}\frac {i}{\sum_{j \equal{} 1}^{i}x_{j}} < 4\sum_{i \equal{} 1}^{n}x_{i}$
And with Cauchy we have
$ \sum_{j \equal{} 1}^{n}a_{j}\sum_{j \equal{} 1}^{i}\frac {j^{2}}{a_{j}}\geqslant\left(\sum_{j \equal{} 1}^{i}j\right)^{2} \equal{} \frac {i^{2}(i \plus{} 1)^{2}}{4} \\
\Longrightarrow\sum_{i \equal{} 1}^{n}\frac {i}{\sum_{j \equal{} 1}^{i}a_{j}} \\
\leqslant\sum_{i \equal{} 1}^{n}\frac {4}{i(i \plus{} 1)^{2}}\sum_{j \equal{} 1}^{i}\frac {j^{2}}{a_{j}}$
And use the similar method in the preceding given solution (wrong post by me :oops: ) and get the result and the constant $ 4$ seems can be replaced by $ 2$ and $ 2$ seems to be a very strong one.


integration
function
real analysis
real analysis unsolved

Assume $f$ is continuous on $I =[a,b]$ and that $\int_{a}^{b}f \cdot g = 0$ for every continuous function $g$ on $I$.  Prove $f(x) = 0$ $\forall x \in I$.

If $g=f$ you have that

$\int_{a}^{b}f^{2}dx=0$ from which if follows that $f=0$. More precisely if $f$ is a nonnegative continuous function on $[a,b]$ such that

$\int_{a}^{b}f(x)dx=0$ then $f=0$.

The proof follows by contradiction. Let assume that $f\neq 0$ i.e. there is a point $x_{0}\in[a,b]$ such that $f(x_{0})>0$. Now because of the continuity of $f$ there is an interval $I$ containing $x_{0}$ such that $f>0$ on $I$.

Now

$0=\int_{a}^{b}f(x)dx\geq \int_{I}f(x)dx>0$ and hence the contradiction.

For the uniform convergence,  if $G$ is dense in $C_{[a,b]}$ (continuous real  function on $[a,b]$) and $f\in C_{[a,b]}$ ,  then  :  $(\forall_{g \in G}, \int_{a}^{b}f(t) g(t)dt=0) \Longrightarrow (f=0)$
:cool:

I first saw the problem in the form "If $\forall n\in \mathbb{N}, \int_{a}^{b}f(t)t^{n}dt = 0$, then $f=0$" and I think it´s a bit more difficult in this way, although the proof is exactly the one given above.


inequalities
real analysis
real analysis unsolved

Prove that for all $ 1\leq k\leq m$ where $ m\geq 1$ is an integer, we have
\[ \binom{2m}{m\plus{}k}\leq 4^{m}\exp\left( \minus{}\frac{k^2}{8m}\right).\]

I found a very technical solution that can apply to generalize this, but I suspect there exists a very elementary solution. I would be very thankful to anyone giving me a hand here.

Comparing $ {2m\choose m \plus{} k}$ to $ {2m\choose m}\le 4^m$, we get the factor $ \prod_{j \equal{} 0}^{k \minus{} 1} \frac {m \minus{} j}{m \plus{} j \plus{} 1}\le \prod_{j \equal{} 0}^{k \minus{} 1}e^{ \minus{} j/m} \equal{} e^{ \minus{} (k \minus{} 1)k/2m}\le e^{ \minus{} k^2/4m}$ if $ k\ge 2$. For $ k \equal{} 1$ just use the inequality $ \frac m{m \plus{} 1}\le e^{ \minus{} 1/8m}$.


x1,x2,....,xn are distinct points in [-1,1]. Pk is product of the distance between xk and xi (for each i=/=k)prove :
 \sum {k=1 to n}1/Pk \geq 2^(n-2)

A magnificient problem. An indirect approach. Of course, we must think at the Lagrange interpolation. So, let us take an arbitrary polynomial f of degree n-1. Then f(x)= \sum (k=1,n) f(x_k)*(x-x_1)...(x-x_(k-1)(x-x_(k+1))...(x-x_n)/(x_k-x_1)...(x_k-x_(k-1)...(x-x_n). THus, |f(x)|<=  \sum (k=1,n) |f(x_k)|x-x_1|...|x-x_(k-1)||x-x_(k+1)|...|x-x_n|/P_k. Take the limit of this when x goes to infinity. It follows that | leading coeficcient of f|<=  \sum |f(x_k)|/P_k. Thus, we need a polynomial f with a huge leading coeficcient and small values in the unit circle. What is this? Of course, Cebasev polynomial. So, we take this as the n-1-th Cebasev polynomial and we find that 2^(n-2)<= sum |f(x_k)|/P_k<=sum 1/P_k.


prove that   the equation : $x^2 -2^y =2005$ has no integer solution.

We obtain that y is odd. (from the divisibility by 3).
2^y can be M5+2, M5+3.
x^2-2^y=2005 so the difference must be divisible by 5.
But x^2 cannot be M5+3 or M5+2.
So we have no solutions.


inequalities
function
number theory proposed
number theory

suppose that $ k \neq 1$.Prove that if $ m$ is the smallest number such that $ m^m \ge k$ then the equation $ x^y \minus{} y^x \equal{} k$ has at most $ m$ solutions in $ \mathbb{N}^2$ where $ \mathbb{N}\equal{}\{1,2,...\}$.
note that for the case $ k \equal{} 1$ we have only two solutions.

First note that from $ x^y>y^x$ we deduce that $ y>x$ (unless $ x\equal{}2$) and therefore for all $ x\geq3$,
$ m^m\geq k\equal{}x^y\minus{}y^x\equal{}x^x\Big(x^{y\minus{}x}\minus{}\left(1\plus{}\frac{y\minus{}x}x\Big)^x\right)$
$ > x^x\Big(x^{y\minus{}x}\minus{}e^{y\minus{}x}\Big)>x^x$
the last inequality is true except for $ x\equal{}3$ and $ y\equal{}4$, thus $ x\leq m\minus{}1$.
second, for every $ x,k$ there is at most one $ y$ such that $ x^y\minus{}y^x\equal{}k$. to prove of this statement, it is enough to show that for each $ y>x$, $ x^{y\plus{}1}\minus{}(y\plus{}1)^{x}>x^y\minus{}y^x$.
we have
$ x^{y\plus{}1}\minus{}{(y\plus{}1)}^{x}\equal{}x^x\left(x^{y\plus{}1\minus{}x}\minus{}\Big(1\plus{}\frac{y\plus{}1\minus{}x}x\Big)^x\right)>x^x\Big(x^{y\plus{}1\minus{}x}\minus{}e^{y\plus{}1\minus{}x}\Big)$
and
$ x^y\minus{}y^x\equal{}x^x\left(x^{y\minus{}x}\minus{}\Big(1\plus{}\frac{y\minus{}x}x\Big)^x\right)<x^x\Big(x^{y\minus{}x}\minus{}1\Big)$
and also for all $ y>x$, $ x^{y\plus{}1\minus{}x}\minus{}e^{y\plus{}1\minus{}x}>x^{y\minus{}x}\minus{}1$ (except for small values of $ x,y$ that can be checked easily).
Therefore for each $ k$, we have $ x\leq m$ and there is at most $ m$ solutions for equation.

clear ... :D
same as mine ...
the whole idea was using the fact that the function $ f(n) : N \rightarrow N ,f(n) \equal{} x^n \minus{} n^x$ is strictly increasing for $ n \in [x \plus{} 1, \infty]$
but i think that the bound can be stronger .... :maybe:


This was the only problem i wasn't able to do on the MAO individual test today, so i was wondering how to do it...

If $\displaystyle f_{0}(x)=\left(\frac{x}{x+1}\right)$ and $\displaystyle f_{(n+1)}=f_{0} \circ f_{n}$ for $n=\{0,1,2...\}$. Solve for $f_{49}(12)$. Put your answer in decimal form, do not round, and give the integer value of the $5^{th}$ decimal place.

$A.$ $3$

$B.$ $6$

$C.$ $7$

$D.$ $8$

$E.$ $\text{NOTA}$

[hide="the answer if it will help you explain it"] B [/hide]


real analysis
topology
integration
real analysis unsolved

Let $ \mu$ be a locally finite Borel doubling measure in $ \mathbb{R}^n$ (doubling: $ \mu(B(x,2r))\leq C\mu(B(x,r))$).

Prove that given an open set $ U$ and a $ \delta>0$, we can find a countable collection of closed balls, $ \mathcal{G}$, such that for any $ B\in \mathcal{G}$ we have $ diam (B)\leq \delta$ and

$ \displaystyle \mu\left(U\setminus\bigcup_{B\in \mathcal{G}} B\right)\equal{}0$.

I realize that this is almost a precise restatement of Vitali's Covering Theorem, but I don't see how to use the doubling property to make the Vitali covering of $ U$. Also, I think the Theorem only applies to Lebesgue measure, so there needs to be a way to relate to that.

Of course you can't use the Vitali covering theorem for Lebesgue measure to prove this. However, the doubling condition is precisely what is needed to effectively use the standard covering theorem: If $ \mathcal{F}$ is a collection of balls for which $ \sup\{d(B): B \in \mathcal{F}\} < \infty$, then one can choose disjoint balls $ B_1, B_2, \ldots \in \mathcal{F}$ so that $ \bigcup_{B \in \mathcal{F}} B \subset \bigcup_i 5B_i$. You need the doubling condition to control the measure of the balls $ 5B_i$. Now you can essentially repeat the proof of Vitali covering theorem for Lebesgue measure (note also that locally finite implies that $ \mu$ is a Radon measure (so you can approximate by open sets which is needed in the standard proof of the Vitali covering theorem for Lebesgue measure)). Here if $ U$ is open and $ \delta > 0$ is given you can then just apply the Vitali covering theorem to the collection of balls $ \bar B(x,r) \subset U$ where we also demand that $ r/2 \le \delta$. That is clearly a Vitali cover for every given $ \delta > 0$.

I note that you can actually drop the assumption of the doubling property and use the Besicovitch covering theorem instead of the  basic covering theorem to prove Vitali covering theorem for all locally finite Borel measures (Radon measures) in $ \mathbb{R}^n$ (the following is true: If $ \mu$ is a Radon measure on $ \mathbb{R}^n$, $ A \subset \mathbb{R}^n$ and $ \mathcal{B}$ is a family of closed balls such that each point of $ A$ is the centre of arbitrarily small balls of $ \mathcal{B}$, then there are disjoint balls $ B_i \in \mathcal{B}$ such that $ \mu\left(A \setminus \bigcup_i B_i\right) \equal{} 0$. You need closer control of the centers of the balls because you need to use Besicovitch instead of the basic covering theorem (which is not effective without the doubling condition)).

The definition I have of Radon measure is that it is a measure on the Borel sets that is locally finite and inner regular. You seem to indicate that the "inner regular" part is redundant. Is there a way to prove inner regular from locally finite, or is it the doubling condition that gives inner regular?

If $ X$ is a locally compact Hausdorff space in which every open set is $ \sigma$-compact (such as $ \mathbb{R}^n$), then every Borel measure that is locally finite (that is $ \mu(K) < \infty$ if $ K \subset X$ is compact) is actually both inner and outer regular (for every Borel set). This is because the Riesz representation theorem (which works in locally compact Hausdorff spaces) gives locally finite measures that have the outer regularity for every Borel set and the inner regularity for every open set and for every measurable set of finite measure. If we demand that $ X$ is $ \sigma$-compact such measures given by the Riesz representation theorem actually satisfy the inner regularity for all Borel sets. So to prove the claim made in the first sentence, one can define a positive functional $ f \mapsto \int_X f\,d\mu$ ($ f$ continuous and compactly supported) and note that the Riesz representation theorem gives a regular measure $ \lambda$ (see the comment above) for which $ \int_X f\,d\mu \equal{} \int_X f\,d\lambda$ for all continuous and compactly supported $ f$. With the additional hypothesis that not only $ X$ but every open set is $ \sigma$-compact it is not hard to show that $ \mu \equal{} \lambda$ and so $ \mu$ is actually very regular. So the point here is that quite strong regularity properties follow for all the locally finite Borel measures in reasonable spaces.


combinatorial geometry
linear algebra
linear algebra unsolved

Let $ A \in\mathcal{M}_{m,n}(\mathbb{Z})$.
Show that there exists $ X \in \mathbb{R}_\plus{} ^n \setminus \{0\}$ with $ AX\equal{}0$ iff the convex hull of the columns of $ A$ contains $ 0$.

That's pretty straightforward, isn't it? Given such an $ X,$ we can always normalize it to $ (\sum x_i)^{-1}X,$ which would then be the coefficients of a convex combination. And the coefficients of a convex combination are already such an $ X.$

The only issue I see is one of definition: this only makes sense if by $ \mathbb{R}_+^n$ you mean $ [0,\infty)\times[0\infty)\times\cdots\times[0,\infty).$ If you try to exclude the zero coefficients, things stop making sense.

[quote="Kent Merryfield"]... by $ \mathbb{R}_ \plus{} ^n$ you mean $ [0,\infty)\times[0\infty)\times\cdots\times[0,\infty).$ [/quote]

Yes, that's correct.


The morning after the the supposed "Mars Approach-Hoax" the moon will have a total eclipse.   :)
Times for viewing depend on where you live
[quote="NASA.gov"]
[img]http://science.nasa.gov/headlines/y2007/images/dreamyeclipse/table.gif[/img]
[/quote].
The times are quite early in the morning but if you have never seen the moon look like
[img]http://science.nasa.gov/headlines/y2007/images/dreamyeclipse/Finazzi1_med.jpg[/img]
then you should not miss it.

Drawing of this Year's Eclipse
[quote="NASA"]
[img]http://sunearth.gsfc.nasa.gov/eclipse/LEmono/TLE2007Aug28/image/TLE2007Aug28-PDTs.GIF[/img]
[/quote].

The two next lunar eclipse occur on February 21 2008 and December 21 2010 , if you happen to miss this one.

[b]Information for Lunar Eclipses in General[/b]
A lunar eclipse occurs when the Earth goes in between the moon and the sun blocking the light from what would be a full moon.
[img]http://upload.wikimedia.org/wikipedia/en/thumb/0/0d/Umbra01.svg/200px-Umbra01.svg.png[/img]
At totality, or when moon is completely under the Earth's shadow, during a total lunar eclipse the moon does appear to turn red.  The reason for this is the same as the reason for the red sun rises or sets on Earth, the light from the sun passes through the Earth's atmosphere and it gets scattered, so only the red light with longer, more resilient wavelengths shows.

[b]Links[/b]
[url=http://science.nasa.gov/headlines/y2007/03aug_dreamyeclipse.htm]NASA's Article[/url].
[url=http://sunearth.gsfc.nasa.gov/eclipse/eclipse.html]NASA's Eclipse Homepage[/url].
[url=http://en.wikipedia.org/wiki/Lunar_eclipse#Appearance]Wikipedia's Article[/url].
[url=http://www.hermit.org/Eclipse/when_lunars.html]Hermit.org - List of Lunar Eclipses Up Until 2020[/url].

Nice post and usefull information. Just a comment:

[quote="Astrophys"]The reason for this is the same as the reason for the red sun rises or sets on Earth, the light from the sun passes through the Earth's atmosphere and it gets scattered, so only the red light with longer, more resilient wavelengths shows.[/quote]

And also because the light from the sun have to cross a long layer of the Earth's atmosphere. Due to the size of the particles in the atmosphere, shorter wavelenghts are more readily scattered: that's why during daylight the sky appears blue. But at the sunrise, sunset, and lunar eclipse, sunlight must cross a longer portion of the atmosphere and so mainly the longer wavelenghts will "survive".

Meh,

I have seen the moon like that. My mom calls it hunting moon.

From SpaceWeather.com, a nice animation of the eclipse.
[img]http://www.spaceweather.com/swpod2007/17aug07/koehn.gif[/img].

Just a post to remind everyone that the eclipse is tomorrow morning.  :wink:
Here is another picture from SpaceWeather.com, since it is big I will hide it.
[hide]
[img]http://science.nasa.gov/headlines/y2007/images/dreamyeclipse/LE2007Aug28-Fig3.GIF[/img].
[/hide]

Today, I saw the Moon really close to the Sun around 7 am PST (15 GMT), but I don't think I'll see the actual thing.

How does Africa have "no eclipse visible" when the times listed are around morning in England?

The eclipse is happening right now, how exciting!

I have images, but it seems AoPS hates me and won't let me upload them.

BanishedTraitor and hunter34 that is cool that you saw it  :) .  Try to see if you can upload those pictures again hunter34.

i_like_pie , I do not really understand what you are trying to say about the moon being close to the Sun.  About the times, in England the Moon would have already set by that time.

Forgot to wake up so I missed the eclipse :( . Though at most what I would have seen would have been a partial shadow on the Moon since totality was reached when it was on the horizon here.  The houses on the horizon would not have help either :lol: .

How many times how I quoted Spaceweather.com or posted links from there?
Here are some more of pictures of the eclipse
http://spaceweather.com/submissions/map.php
http://www.spaceweather.com/eclipses/gallery_28aug07.htm .

I got the Images!,

For the last 2 photos, I circled the position of the moon with red ink.

http://www.artofproblemsolving.com/Forum/album.php?mode=attach&id=10378

http://www.artofproblemsolving.com/Forum/album.php?mode=attach&id=10379

http://www.artofproblemsolving.com/Forum/album.php?mode=attach&id=10380

http://www.artofproblemsolving.com/Forum/album.php?mode=attach&id=10381

http://www.artofproblemsolving.com/Forum/album.php?mode=attach&id=10382

It was pretty cool this morning!
Eventually I couldn't see it because of the neighbors trees but it was a nice sight!
:D

Thank you folks ! for the images !

In France we must wait for February 21, 2008....


Proved that there are infinite possitive numbers in the form
(n^p-1)/(n-1)
with p be prime number and p>2
n is  any posstive number
SOrry for my poor english

are you sure this is correct :huh:
$n^{p-1}$ is not divisible by $n-1$.

I think it is $\frac{n^{p}-1}{n-1}$.

-though this is of course trivial as $n^{p}-1$ is always divisible by $n-1$.


Given that $ a$, $ b$ and $ c$ satisfy the triangle inequality, show that $ \frac{a}{a\plus{}1}\plus{}\frac{b}{b\plus{}1} \ge \frac{c}{c\plus{}1}$. I'm almost sure that this is actually true. If you know how to solve this or know that it isn't true, I'd be grateful for at least 15 minutes if you'd post soon.

Do you mean a+b>c ? If that is the case :
2abc+2ab+ac+a+bc+b>a+b>c .
Dividing by (c+1) you will get to intended inequality .

Let
$ x\equal{}a\plus{}1 , y\equal{}b\plus{}1 ,z\equal{}c\plus{}1$

$ \frac{a}{a\plus{}1} \plus{}\frac{b}{b\plus{}1} \ge \frac{c}{c\plus{}1} \iff \frac{x\minus{}1}{x} \plus{}\frac{y\minus{}1}{y} \ge \frac{z\minus{}1}{z}$

$ \iff \frac{x\minus{}1}{x} \plus{}\frac{y\minus{}1}{y} \plus{}\frac{1\minus{}z}{z}\ge 0$

Consider

$ (x\plus{}y\plus{}z)(\frac{x\minus{}1}{x} \plus{}\frac{y\minus{}1}{y} \plus{}\frac{1\minus{}z}{z})\ge (x\minus{}1)\plus{}(y\minus{}1)\plus{}(1\minus{}z)  \equal{} a\plus{}b\minus{}c >0$

$ \implies \frac{x\minus{}1}{x} \plus{}\frac{y\minus{}1}{y} \plus{}\frac{1\minus{}z}{z}\ge 0$

the equailty holds when  $ x\equal{}y\equal{}z\equal{}1 \iff a\equal{}b\equal{}c\equal{}0$

[quote="vinskman"]$ (x \plus{} y \plus{} z)(\frac {x \minus{} 1}{x} \plus{} \frac {y \minus{} 1}{y} \plus{} \frac {1 \minus{} z}{z})\ge (x \minus{} 1) \plus{} (y \minus{} 1) \plus{} (1 \minus{} z) \equal{} a \plus{} b \minus{} c > 0$
[/quote]

Why $ (x\plus{}y\plus{}z)(\frac{x\minus{}1}{x}\plus{}\frac{y\minus{}1}{y}\plus{}\frac{z\minus{}1}{z})\ge(x\minus{}1)\plus{}(y\minus{}1)\plus{}(z\minus{}1)$ ?

Another inequalities theorem?

I’m sorry to make a mistake because of hurry .If you use a+b>c:
a(b+1)(c+1) + b(a+1)(c+1)= 2abc + 2ab + ac + bc+ a + b> abc + ac + bc + a + b > abc + ac + bc + c = c(a+1)(b+1)   .
   Dividing two sides  by (a+1)(b+1)(c+1) ,you will get to the intended inequality .Of course two similar inequalities would be yielded if you used a+c>b or b+c>a   .

[quote="mrosett"]Given that $ a$, $ b$ and $ c$ satisfy the triangle inequality, show that $ \frac {a}{a \plus{} 1} \plus{} \frac {b}{b \plus{} 1} \ge \frac {c}{c \plus{} 1}$. I'm almost sure that this is actually true. If you know how to solve this or know that it isn't true, I'd be grateful for at least 15 minutes if you'd post soon.[/quote]
[hide]I think I get an interesting solution that I want to share it now!
We will have 3 cases(and one line for each case) :lol:
1st case: c=max(a,b,c). Then we have: a/(a+1)+b/(b+1)>=a+b/(c+1)>=c/(c+1)
2nd case: c=min(a,b,c). Then we have: a/(a+1)-c/(c+1)+b/(b+1)=(a-c)/(a+1)(c+1)+b/(b+1)>0
3rd case: c is between a and b. Then we easily see that it's a proved exactly like the 2nd case was.
From 3 cases above, we can conclude that the original inequality is proved(Q.E.D)[/hide]

The problem is equivalent to $ 1 \minus{} \frac{1}{a\plus{}1} \plus{} 1 \minus{} \frac{1}{b\plus{}1} \geq 1 \minus{} \frac{1}{c\plus{}1}$, or $ 1 \plus{} \frac{1}{c\plus{}1} \geq \frac{1}{a\plus{}1} \plus{} \frac{1}{b\plus{}1}$. Increasing $ c$ makes it sharper, so WLOG $ c \equal{} a \plus{} b$. Then $ \frac{a\plus{}b\plus{}2}{a\plus{}b\plus{}1} \geq \frac{a\plus{}b\plus{}2}{(a\plus{}1)(b\plus{}1)}$, which reduces to $ ab \geq 0$.

[quote="MellowMelon"]Increasing $ c$ makes it sharper[/quote]

Could you word that in another way, I'm not quite sure I understand what you're saying

As $ c$ increases the left hand side decreases, and since we have $ \geq$ that makes the two sides closer together.

Another way I could have done the step:

$ 1 \plus{} \frac{1}{c\plus{}1} \geq 1 \plus{} \frac{1}{a\plus{}b\plus{}1} \geq \ldots$.


Suppose Alex and Alexia are equally strong Ping-pong players. Is it more probable that Alex will beat Alexia in 3 games out of 4, or in 5 games out of 8.

exactly 3 games out of 4 or at least 3 games out of 4?

Im thinking its exactly 3 out of 4.

[hide]
Shouldn't the correct answer be 5 out of 8? Since they are equally good, each should win half of the total matches played (or close to half anyway).
[/hide]

$ \frac{\binom34}{2^{4}}> \frac{\binom58}{2^{8}}$

Exactly 3 of 4 is more likely than exactly 5 of 8.

right, the more probable is 3 games out of 4.
for 5 games out of 8=7/32
for 3 games out of 4=8/32

Wait but if you said that they were equally strong, then they should tie or at least have very close results....?

He means that they have an equal chance of winning against each other.


$ {\rm{Let a,b,c  > 0 , }}a^2  \plus{} b^2  \equal{} 1.\Pr {\rm{ove that}} \\
 \frac{1}{a} \plus{} \frac{1}{b} \minus{} \left( {\sqrt {\frac{a}{b}}  \minus{} \sqrt {\frac{b}{a}} } \right) \ge 2\sqrt 2  $

 :lol:

[quote="SHIRO_DHKT"]$ {\rm{Let a,b,c > 0 , }}a^2 \plus{} b^2 \equal{} 1.\Pr {\rm{ove that}} \\
\frac {1}{a} \plus{} \frac {1}{b} \minus{} \left( {\sqrt {\frac {a}{b}} \minus{} \sqrt {\frac {b}{a}} } \right) \ge 2\sqrt 2$

 :lol:[/quote]
Where is c? :(

If $ a \equal{} \frac {1}{2}, b \equal{} \frac {\sqrt {3}}{2} \longrightarrow f(a,b) \equal{} 2.598... < 2\sqrt {2}$

So should we assume $ a\le b$, if so...
[hide="solution"]
From AM-GM
$ \frac {1}{a} \plus{} \frac {1}{b} \minus{} \left({\sqrt {\frac {a}{b}} \minus{} \sqrt {\frac {b}{a}} } \right) \ge \frac {2}{\sqrt {ab}} \minus{} \frac {a \minus{} b}{\sqrt {ab}}$

Therefore we need to proove

$ (2 \minus{} a \plus{} b) \ge (2\sqrt {2ab})$

$ 4 \plus{} 4(b \minus{} a) \plus{} a^2 \minus{} 2ab \plus{} b^2 \ge 8ab$

$ 5 \plus{} 4(b \minus{} a) \ge 10ab$

This is true because $ b\minus{}a \ge 0$ and $ a^2 \plus{} b^2 \ge 2ab \Leftrightarrow ab \le \frac {1}{2}$ [/hide]

[quote="ocha"]If $ a \equal{} \frac {1}{2}, b \equal{} \frac {\sqrt {3}}{2} \longrightarrow f(a,b) \equal{} 2.598... < 2\sqrt {2}$

So should we assume $ a\le b$, if so...
[hide="solution"]
From AM-GM
$ \frac {1}{a} \plus{} \frac {1}{b} \minus{} \left({\sqrt {\frac {a}{b}} \minus{} \sqrt {\frac {b}{a}} } \right) \ge \frac {2}{\sqrt {ab}} \minus{} \frac {a \minus{} b}{\sqrt {ab}}$

Therefore we need to proove

$ (2 \minus{} a \plus{} b) \ge (2\sqrt {2ab})$

$ 4 \plus{} 4(b \minus{} a) \plus{} a^2 \minus{} 2ab \plus{} b^2 \ge 8ab$

$ 5 \plus{} 4(b \minus{} a) \ge 10ab$

This is true because $ b \minus{} a \ge 0$ and $ a^2 \plus{} b^2 \ge 2ab \Leftrightarrow ab \le \frac {1}{2}$ [/hide][/quote]
I can't understand what you said. When $ a \equal{} \frac {1}{2}, b \equal{} \frac {\sqrt {3}}{2}$, LHS =3,71... $ >2\sqrt {2}$. So when a>b $ ({\sqrt {\frac {a}{b}} \minus{} \sqrt {\frac {b}{a}} })$ >0 :o

[quote="ocha"][quote="SHIRO_DHKT"]
I can't understand what you said. When $ a \equal{} \frac {1}{2}, b \equal{} \frac {\sqrt {3}}{2}$, LHS =3,71... $ > 2\sqrt {2}$. So when a>b $ ({\sqrt {\frac {a}{b}} \minus{} \sqrt {\frac {b}{a}} })$ >0 :o[/quote]

Sorry i meant $ b \equal{} \frac {1}{2}, a \equal{} \frac {\sqrt {3}}{2}$[/quote]
Sorry. Long time no see. I think it is
$ {\rm{Let a,b,c > 0 , }}a^2 \plus{} b^2 \equal{} 1.\Pr {\rm{ove that}} \\
\frac{1}{a} \plus{} \frac{1}{b} \minus{} \left( {\sqrt {\frac{a}{b}} \minus{} \sqrt {\frac{b}{a}}}\right)^2 \ge 2\sqrt 2$
 :blush:  :blush:

$ \begin{array}{l}
  \Leftrightarrow \frac{1}{a} \plus{} \frac{1}{b} \minus{} \frac{a}{b} \minus{} \frac{b}{a} \plus{} 2 \ge 2\sqrt 2  \\
  \Leftrightarrow (\frac{{2(a \plus{} b)}}{{\sqrt {2(a^2  \plus{} b^2 )} }} \minus{} 1)\frac{{(a \minus{} b)^2 }}{{ab}} \ge 0(true) \\
 \end{array}$
It's easy :) .

[quote="TRAN THAI HUNG"]$ \begin{array}{l} \Leftrightarrow \frac {1}{a} \plus{} \frac {1}{b} \minus{} \frac {a}{b} \minus{} \frac {b}{a} \plus{} 2 \ge 2\sqrt 2 \\
\Leftrightarrow (\frac {{2(a \plus{} b)}}{{\sqrt {2(a^2 \plus{} b^2 )} }} \minus{} 1)\frac {{(a \minus{} b)^2 }}{{ab}} \ge 0(true) \\
\end{array}$
It's easy :) .[/quote]

$ \boxed{\Leftrightarrow (\frac {{2(a \plus{} b)}}{{\sqrt {2(a^2 \plus{} b^2 )} }} \minus{} 1)\frac {{(a \minus{} b)^2 }}{{ab}} \ge 0}$

Sorry , can't get this equivalence,please give more details. :oops:


[color=blue][b]Express the following quadratic expression in factored from by using the quadratic formula:[/b][/color]
$x^{2}-2x+3$

[hide="Pah"] We will instead complete the square.

$x^{2}-2x+3 = (x-1)^{2}+2 = (x-1+i \sqrt{2})(x-1-i \sqrt{2})$

Here I use the identity

$a^{2}+b^{2}= (a+bi)(a-bi)$ [/hide]

Well...using the quadratic formula:

[hide] $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}= \frac{2\pm\sqrt{4-4\cdot 1\cdot 3}}{2\cdot 1}= \frac{2\pm\sqrt{-8}}{2}= 1\pm\sqrt{2}\,i$.

Thus, in factored form, $x^{2}-2x+3$ is

${(x-(1+\sqrt{2}\,i)(x-(1-\sqrt{2}\,i)}=(x-1-\sqrt{2}\,i)(x-1+\sqrt{2}\,i)$
[/hide]


For a positive integer $n$, let $r(n)$ denote the sum of the remainders when $n$ is divided by $1,2,...,n$ respectively. Prove that $r(k)=r(k-1)$ for infinitely many integers.

[hide="Solution sketch"]For any $p \leq 2^{n}-1$ which is not a power of 2, the remainder when $p$ divides $2^{n}$ is exactly one more than the remainder when $p$ divides $2^{n}-1$.

For $p = 2^{m}\leq 2^{n}-1$, the remainder when $p$ divides $2^{n}$ is 0 and the remainder when $p$ divides $2^{n}-1$ is $2^{m}-1$.

Thus $r(2^{n})-r(2^{n}-1) = 1\cdot (2^{n}-n-1)-\sum_{m = 0}^{n-1}2^{m}-1 = 0$, so we're done.

At least up to $n = 4096$ the powers of 2 are the only numbers with this property.[/hide]

The values of this function are [url=http://www.research.att.com/~njas/sequences/A004125]A004125[/url] at the OEIS.


geometry
geometric transformation
reflection
analytic geometry

Point $ (a, b)$ is reflected over the x-axis and then is reflected over the y-axis. After these reflections, the new coordinates of the point are $ (c, d)$. What is the value of $ ab \minus{} cd$?

After double reflection point (a,b) becomes (-a,-b). So c=-a and d=-b.
Hence ab - cd = ab - ab = 0

Note that this reflection is the same as flipping over the line y=-x or quadrants I, III, and y=x for II,IV.


Let the function $ S(n)$ be defined as the sum of the digits of a positive integer $ n$. For example, $ S(42) \equal{} 6$ and $ S(1337) \equal{} 14$.

1. For how many positive integers $ k < 100$ is $ S(k)$ a single-digit number?

2. For how many positive integers $ k < 100$ is $ S(S(k))$ a single-digit number?

[hide="1"]The amount of possibilities for $ S(k) \equal{} 9$ is 10
The amount of possibilities for $ S(k) \equal{} 8$ is 9
...

$ \frac {10\cdot11}{2} \minus{} 1 \equal{} 54$[/hide]

[hide="2"]The largest possibility for $ S(k)$ is 18...which is if $ k \equal{} 99$.

Since in this case, $ S(S(k)) \equal{} 9$...it's $ \le9$ for all values of $ k<100$.

So the answer is 99.[/hide]


algebra
polynomial
function
set theory
algebra theorems

Wel, this is not a formula or theorem, but, we know that in analysis, $ 0^0$ is an indeterminate form or excepted case, or we say the operation has no sense. But there $ 0^0$ means that the base and the exponents tend to 0, they are not actually equal to 0.On the other part, in algebra, we still say that $ 0^0$ has no sense??
Cannot we simply put $ 0^0\equal{}1$?? I don;t see any problems.
For example, if we have a polynomial function $ p(x)\equal{}ax^2\plus{}bx\plus{}c$
if we want to say that $ p(x)\equal{}ax^2\plus{}bx^1\plus{}cx^0, \forall x \in R$ it appears a problem with $ p(0)$ if we do not define $ 0^0$!

So, it is $ x^0\equal{}1, \forall x \in R$??
Personal i was taught in the 5-th grade, at natural powers, that $ 0^0$ has no sense. Is there a real reason for that??

There are some conventions in arithmetic and set theory that are convenient for unifying formulae.

$ 0^0\equal{}1$, $ \sum_{i\in\emptyset} a_i \equal{} 0$, $ \prod_{i\in\emptyset} a_i \equal{} 1$.

$ \bigcup_{i\in\emptyset} A_i \equal{} \emptyset$, $ \bigcap_{i\in\emptyset} A_i \equal{} A$ if all sets $ A_i$ are subsets of some universal set $ A$.

The king of formalization, Bourbaki, uses them (with some justification).

Well it is true that most often, the limit $ 0^0$ tends to 1. But take a function like $ f(x)\equal{}0^x$. There are also non trivial functions that are continuous for $ x\to 0$ but tending to something different. If you like, try

\[ f(x)\equal{}(e^{\minus{}\frac1{x^2}})^{(e^x\minus{}1)^2}\]

;)


geometry
geometric transformation
reflection
Euler
circumcircle
geometry unsolved

Show that two reflections: first in Euler line, then in Lemoine line will
    take Tixier point to far-out point.

    Maj. Pestich

Just one remark: the reflection of the Tixier point in the euler line is $X(110)$, the point where the reflections of the Euler line in the sides of the triangle concur. Actually, I know this as the definition of the Tixier point :). We should thus show that $X(110)$ reflected in the Lemoine line (the trilinear polar of the symmedian $K$ and also its polar wrt the circumcircle) is the far-out point (the inverse of the centroid wrt the circumcircle).


function
floor function
special factorizations
algebra unsolved
algebra

Find all real numbers $ x$ such that:
$ [x^2\plus{}2x]\equal{}{[x]}^2\plus{}2[x]$

(Here $ [x]$ denotes the largest integer not exceeding $ x$.)

Let $ x \plus{} 1 \equal{} n \plus{} \alpha$, where $ \alpha \equal{} \{x \plus{} 1\} \equal{} \{x\}$.

$ \lfloor x^2 \plus{} 2x \rfloor \equal{} \lfloor x\rfloor^2 \plus{} 2\lfloor x\rfloor \plus{} \lfloor\alpha^2 \plus{} 2\alpha \lfloor x \rfloor \plus{} 2\alpha\rfloor \equal{} \lfloor x\rfloor^2 \plus{} 2\lfloor x\rfloor.$
$ \Rightarrow 0\leq \alpha^2 \plus{} 2\alpha n < 1$.

$ \Rightarrow \alpha\in \left[0; \minus{} n \plus{} \sqrt {n^2 \plus{} 1} \right)$ for every $ n\in \mathbb{N}$.
$ \Rightarrow x\in \left[ n \minus{} 1 ; \sqrt {n^2 \plus{} 1} \minus{} 1 \right)$ for every $ n\in \mathbb{N}$.

For negative numbers it's similar.

[b]REMARK:[/b]
$ \left[ x^2 \plus{} 2x\right] \equal{} [x]^2 \plus{} 2.[x]\Leftrightarrow \left[ x^2 \plus{} 2x\right] \plus{} 1 \equal{} [x]^2 \plus{} 2.[x] \plus{} 1\Leftrightarrow \\
\Leftrightarrow \left[ x^2 \plus{} 2x \plus{} 1\right] \equal{} \left( [x] \plus{} 1\right)^2\Leftrightarrow \left[ (x \plus{} 1)^2\right] \equal{} \left( [x] \plus{} 1\right)^2\Leftrightarrow \\
\Leftrightarrow \left( [x] \plus{} 1\right)^2\le (x \plus{} 1)^2 < \left( [x] \plus{} 1\right)^2 \plus{} 1; \ (\forall)x\in\mathbb{R}.\ (1)$
[b]1).[/b] $ \mbox{If } x\ge \minus{} 1, \mbox{ then: }$
$ (1)\Leftrightarrow [x] \plus{} 1\le x \plus{} 1 < \sqrt {\left( [x] \plus{} 1\right)^2 \plus{} 1}\Leftrightarrow [x]\le x < \sqrt {\left( [x] \plus{} 1\right)^2 \plus{} 1} \minus{} 1\Leftrightarrow \\
\Leftrightarrow x\in \left[ k; \sqrt {k^2 \plus{} 2k \plus{} 2} \minus{} 1\right); (\forall)k\in \mathbb{N}\cup\{ \minus{} 1\}.$
[b]2).[/b] $ \mbox{If } x<\minus{}1, \mbox{ then: } (1)\Rightarrow  [x] \plus{} 1\ge x \plus{} 1\Leftrightarrow [x]\ge x \Leftrightarrow x\in \mathbb{Z}.$

the proof of my method and the solution is 3 pages long and thats why i only give the result and critical ideas.
>this equation is an identity for all integers and hence the set of integers is a solution to the equation.
>Use the idea that [x]= x - {x} where {} means fractional part function. convert this into this form and then try to simplify it. u will find a range and this will come out to be
|x{x+n}|<0.2, where n is
i have checked all the soln, and is yet to find a flw in it.
an other idea to be used is {x}={x+n} for all natural n. and thats all. :wink:

both sides add +1
let x+1=a=>
[a^2]=[a]^2
let a=b+c where b=[a]
<=>[(b+c)^2]=b^2<=>[2bc+c^2]=0 =>0=<2bc+c^2<1
b is any integer and c is satisfying 0=<2bc+c^2<1 :)

[hide="Try Completing the square by puting constant integers inside GInt function"]
$ \& \left[ {{x^2} + 2x} \right] = {\left[ x \right]^2} + 2\left[ x \right] \\
 \\
\& \implies \left[ {{x^2} + 2x + 1} \right] = {\left[ x \right]^2} + 2\left[ x \right] + 1 \\
 \\
\& \implies \left[ {{{\left( {x + 1} \right)}^2}} \right] = {\left( {\left[ x \right] + 1} \right)^2} \\
 \\
\& \implies \left[ {{{\left( {x + 1} \right)}^2}} \right] = {\left[ {x + 1} \right]^2}$

let $ x + 1 = y$.

$ \& \implies \left[ {{y^2}} \right] = {\left[ y \right]^2} = {n^2}{\text{ Where }}n \in {\Bbb N} \\
 \\
\& \implies {n^2} \leq {y^2} < {n^2} + 1$

$ \implies n\le y \le \sqrt {n^2 + 1} \\
 \\
\implies n\le {x + 1} \le \sqrt {n^2 + 1} \text{ OR } - n\ge {x + 1} > - \sqrt {n^2 + 1} \\
 \\
\implies \boxed{n - 1\le {x} \le \sqrt {n^2 + 1} - 1 \text{ OR } - n - 1 > x > - \sqrt {n^2 + 1} - 1} \text{ } \forall \text{ } n\in {\Bbb N} \cup \{0\}$[/hide]

my solution is almost same as that of mihai miculita


Well, somebody in our school aced the ACT, so while it's on my mind, I might as well ask...what's your highest result?

mines suck its 1300, and my ACT score is 27 but that was in 8th grade (not too bad). i'm going for a 1400 and 30 for 9th grade and a >1500 and >33 for the rest of high school untill they add the writing section.

my cousin aced the SAT in 10th grade and he came to the U.S. when he was 10!!!!!!!!! that is very impressive. he did way too many SAT practice tests and memorized every vocab word.

Lol, I knew you would be one of them. It's just that they're rare and I didn't think you would respond that fast...oh well, ignore my previous comment.

No, wasn't me actually.  I never took the ACT, so although I did get every ACT question I ever answered correct, I still didn't get a perfect score in the exam.

Ah, there's another one.  I think the probability they're both genuine is pretty low.

This year there were three girls in one high school in Minnesota who all aced both the SAT I and the ACT. They got on the TV news, and it was reported that none had (as of then) decided where to go to college next year.

But I don't suppose any of those high school seniors are posting here, so I don't know who's answering the poll.

It is definitely possible. One of my friends did it this year (36 ACT, 1600 SAT I). He also has 3 800s on SATII tests already (not fake ones like Mandarin either-IIC, Chemistry, and Biology :)).

It's definitely possible to score very well on standardized tests even if you're not that good right now. It just takes tons of practice. Even if you can squeak out answers to a few of the intermediate problems here, I think the math section should be an easy 800 for you [b]as long as you don't make careless mistakes[/b]. The SATI math is all about not making stupid mistakes.

The verbal is an entirely different story, especially if you weren't born in America. Basically, the verbal section is about mass-memorization of words and reading.

For mass-memorization, just buy a few vocab books (I recommend Word Smarts I and II from the Princeton Review) and read those a few times through over the course of a year or so. For reading, the best thing that worked for me was AP Junior English (Language and Composition). The AP system (maybe IB too) stresses so much on reading for tone and author purpose that scanning for tone and stuff in passages becomes instinctive. It made those critical reading sections so much easier for me.

For sophomores and under, the SAT is changing though. For the new writing section, I would recommend buying a book for SAT II Writing like Barron's, which is great. The thing about writing is that it is terribly named. What it should be called is [b]Grammar[/b]. The entire test is basically focused on your ability to spot grammatical mistakes. The essay is also generally an easy question that doesn't require any thought-provoking analysis. Having read the books, my scores on the writing section increased by a large margin. If you can't catch the grammatical error in the previous sentence (Having..dramatically), you need to buy it :).  There are so many obscure grammatical errors out there that most people have never heard of that you absolutely cannot take the test without studying.

The math shift towards Algebra II should be helpful for people on this board generally. Basically, it's becoming more like the IIC exam where the questions are markedly harder but you can afford to make a few careless mistakes. The disappearance of the analogies section also puts less emphasis on memorization of vocabulary, so that could be another good thing.

Wow, that post was long :).

my cousin got a perfect score when he was 16.

[quote="rlee"]The verbal is an entirely different story, especially if you weren't born in America. Basically, the verbal section is about mass-memorization of words and reading.[/quote]

I wouldn't characterize what you have to do as "mass-memorization." I think you are correct that READING is the key strategy. Some of the short answer questions are consciously designed, as are some of the long reading passage questions, to trip up people who memorize a particular definition for a word without thinking about what a word means in context. Nothing but reading, reading, reading, and reading will advance your ability to see word meanings in context.

I would rather characterize what you need to do to rapidly boost vocabulary as learning how analyze words and memorizing a selected group of word roots from Latin and Greek, something which is offered by many vocabulary courses. My wife is a native speaker of Chinese from Taiwan who did her undergraduate degree here in the States, and she found the book Webster's New World Power Vocabulary helpful for preparing for her studies. My son is taking the Center for Talent Development course in English Vocabulary, which is all based on etymology, mostly Latin and Greek roots, and that seems to be helping him immensely. (He, of course, has one parent who is a native speaker of English, but one who is not, and he lived in Taiwan from age six to age eight-going-on-nine.) Rather than memorizing huge word lists, learn groups of words of related origin, being aware of how word roots can change meaning, and then spend the rest of your time reading.

[quote="rlee"]For reading, the best thing that worked for me was AP Junior English (Language and Composition). The AP system (maybe IB too) stresses so much on reading for tone and author purpose that scanning for tone and stuff in passages becomes instinctive. It made those critical reading sections so much easier for me.[/quote]

Thanks for sharing that specific tip. That makes a lot of sense. Again, a native speaker of English could, and often does, pick up that kind of reading skill just from recreational reading, but your advice would be a good plan for getting faster results in gaining critical reading skill.

[quote="rlee"]The thing about writing [on the forthcoming new version of the SAT I] is that it is terribly named. What it should be called is [b]Grammar[/b]. The entire test is basically focused on your ability to spot grammatical mistakes. . . . There are so many obscure grammatical errors out there that most people have never heard of that you absolutely cannot take the test without studying.[/quote]

Again, native speakers of English can often rely on their "gut reactions" IF AND ONLY IF they are well-read in professionally edited books. But for everyone else, it is an excellent idea to specifically study grammar and usage of English. Besides taking the vocabulary course he's taking this year, my son also took the EPGY Grammar of the English Sentence course, and that boosted his verbal SAT I score above what I expected him to get at age eleven. He enjoys recreational reading in harder books now too, so I consider those courses a good investment. There are lots of choices for studying grammar and vocabulary, but studying those subjects intensively for a while is definitely a good idea for getting ready for the SAT I, all the more so for young people whose parents didn't grow up speaking English.

Thanks for sharing the tips.

[quote="kel227"]I would like to skip to college......I wonder how I can take the sat?[/quote]

You can register for the SAT I online, although to do that if you are currently enrolled in school you should try to find out the [url]school code of your high school[/url] before you start the rest of the process. Anyone can register for the SAT I at your age. All it takes is money  :)  which I suppose you would ask your parents for in this case.

But your more important question, about early college entrance, is more complicated. You can get into college without ever taking an SAT I or ACT--many people do, and some colleges have policies of open admission. Is anyone else interested in discussing this issue of early college admission in more detail? If yes, I will open a new thread on this issue.

Going to college early is a good idea if your high school isn't keeping you challenged and learning to your full potential. It might be a bad idea if you don't have a plan for how the college studies will lead on to you being a well-educated person and to achieving your future goals. What makes you desire trying to get into college early?

[quote="applepuppy"]I know someone who got a 1600 SAT... just recently someone at a local school got a 36 ACT...

I got 1490/33 and I'm a sophomore...

SAT (current one) is much easier to ace than the ACT, right?[/quote]

It all depends on the person... they definitely test different aspects of a person. That is why it's a rare occurance to find someone who aces both... It of course happens though, just not often.

Last year, in my school, there was only one 1600 SAT and three 36 ACTs. The person with the perfect SAT score also got a perfect ACT.

[quote="bleumoose"]you guys have [i]got[/i] to remember, though, that a good, or even perfect, SAT score isn't your key to college. if all you are is a good test taker, thats not good enough. in fact, many students with perfect scores get turned away from universities, and students with lower scores get in. just dont make the SAT or the ACT your life.[/quote]
very good point...though these standardized tests are still just about the only way colleges can compare u with everyone else...so it's still fairly important...

Which is rather sad, since SAT/ACT only tests shallow material requiring relatively little skill.  I had an estimated score on SAT math in the 500s in 5th grade, not having even finished algebra.  That goes to show how little problem solving ability the SAT math requires.

[quote="MithsApprentice"]
very good point...though these standardized tests are still just about the only way colleges can compare u with everyone else...so it's still fairly important...[/quote]

That's not true. There is a lot more to distinguish candidates than just SAT Scores. Colleges that don't get overwhelmed with applications, as some state schools do, have well staffed admissions departments that try to get to know each applicant to some degree.

I agree with Bluemoose that SAT I scores are not the absolute most important thing to worry about (though for some scholarships they may be). As long as you get in a good "range" you'll be in good shape.

by the way, its bleu, french, like the cheese. (my friends and i had a cheese fascination when i made this username for something else, and now everything is bleumoose).

I just finished 8th grade, took the May 1st SAT and made and 800 on math, and 690 on Verbal. I don't know how I should contact someone about the ACT--I don't know anyone who has taken it, either.

My uncle gave me the math part of an ACT test and I got a 34/36.  I would have gotten a 36 if I had another minute (I guessed on the last two when he announced that there were 10 seconds left).

My daughter just finished 4th grade.  I was curious how she would do on the SAT I Math, so I gave her an old test.  She got a 550.  It's hard to believe sometimes that this is a test for college.

Have all of you who received high SAT scores at an early age heard about the Study of Exceptional Talent ([url=http://www.jhu.edu/gifted/set/index.html]SET[/url])? Young people who take the SAT through the Johns Hopkins CTY regional talent search on the east and west coasts seem to hear about SET automatically, but if you took the SAT through other channels, and got a qualifying score, you should self-report your score to the SET office in Baltimore.

[quote="akatookey"]The SAT is the more challenging of the 2...I have taken both and have a 1550/34...

But if you take the time to study for them...yes...they require studying...you will do well on both[/quote]
You claim that the SAT is the harder, but yet, based on your percentage of total score possible, you did better on the SAT than the ACT. That's somewhat meaningless without looking at standard deviations and means, but I thought I'd point it out anyway.

the sat is harder cuz theres only 2 parts, verbal and math... and everyone sucks at verbal.. but for act, theres many sections and everyone still sucks at verbal.. but the other sections balance it out more..

uh.. tokenadult, i would like to know more about skiping high school.. not tat im gonna do it or anything, but its interesting to know..

and i agree with the fact tat the sat is all crap.. no skill required.. only hard work.. and most of AP tests are crap too.. i have like no respect for these standardized tests..
anyknow know like other national subject tests or competitions?
especially physics.. im good at tat..

USAPhO is the physics equivalent of USAMO.  Google it.

One thing I noticed on the poll...no PSAT (even though it's the same as the SAT I and the Writing SAT II).  How many 240s are out there?

eum... how and where do I take this test?

Here is a link [url]http://www.collegeboard.com/student/testing/sat/about.html[/url]

I don't know much about taking it in other countries.  Most of the ppl who take it live in America but there are a bunch of people who take it in different countries.  Maybe some other people have addititonal info. from first hand experience or something like that.