AMC USA(J)MO USAMO AIME USAMTS geometry HMMT

I'm gonna be a senior in high school next year and I got into all this competitive problem solving stuff kinda late, but I really like it.  Its seems kinda pointless to go after USAMO and stuff because I would guess all of the people that get that far have probably been preparing for years.  So why should I keep learning and studying math if I'll always be a step behind a lot of other people?  I would like to think that because "I enjoy doing it" is enough but its not.  I think I just need some kind of goal, something to work toward, but I don't know what that is.  Any thoughts?

It's never too late to learn math.  While it definitely seems more rewarding starting young, you can always benefit.  Unless you want to major in English or something in college, math will certainly help, and it's the problem solving ability that really matters.  The guy who learns Linear Algebra, Differential Equations, and all that other stuff by 8th grade yet still has no clue on AIME problems is not going to be able to do much as far as research goes, which is what really matters (well you'd think, anyway).  Plus, there are many college competitions that you can participate in.

All this can help with your schoolwork (if you take an advanced class), and even if you're not an IMO gold medalist, you can at least actually use your knowledge to solve problems with our troubled world.

So you might've come a little late, but think of how much farther you'll be than most people; it's better late than never, and there's nothing to lose.  I mean, most people wouldn't refuse to take $100 just because they knew that other people were getting $1000, right?

What's even better for you is that you're not too late yet.  There's a whole year until the next USAMO!  If you work really hard, I think you have a good chance of at least getting in.  I know I spent countless hours on AIMEs, although still probably not as many as the more diligent of us would.  I made the USAMO and this is only my first serious year doing the AMC.  Even if you don't do so well, colleges will be more likely to accept you.

Something else you could try is to make your own little competition or try to help someone else learn about problem solving.  That's always very rewarding.  Maybe AoPS could hold some competition where [b]everyone[/b] can enter (maybe even for prizes!).

I'm also pretty sure you'll get addicted once you try.  Just read my signature!  Then you'll have to take math-quitting counseling :-D.

I began to like math in the 4th grade when my teacher recognised that I was good at math and put me on the school's math team.  That year I competed in the MCTM(I think that is what it was called) and had so much fun that I joined the math club my 5th grade year.  Sadly, I moved to California for my 6th grade year and found myself in a class that was less than stimulating and despite my persistant asking I was not able to locate any competitions.  I spent the next few years fighting the school district (they tried to tell me that 7th graders could not take Algebra...sure...) and did not really find any competitions until my Freshman year when I fought my way onto the Academic Decathlon team.  AcaDec was awesome and I got the highest math score on my team, despite the fact that I was the youngest.  I was the only one who took the time to learn the basic statistics that it was over.  Then the summer after my Freshman year my family moved to Colorado and the school here does not have AcaDec.  During the summer after my sophomore year I discovered the USAMTS competition.  I competed in that last year and did relatively well, despite the fact that I was new to that type of competition.  I only really learned anything about the other competitions that many other people on this board have been competing in for years in this last month.  I too am getting ready for my senior year of high school next year and am new to these competitions.  The way I see it, anything that will help me to learn more math is a good thing.  For the time being at least, I intend to major in mathematics when I get to college.  I haven't even had much chance to compete but, like Probability1.01, I am addicted to math.  I have been since 4th grade, and not even troublesome school districts could break me.  If you give it a try I think you will find that the fact that you enjoy it is enough, I know it will be for me even if I do not get to do well in the competition.  As for always being a step behind...my theory is that I might have to work a little harder than some of the others, but that I should be able to catch up before too terribly long.  If you really do enjoy math I see no reason why it should not hold for you too!

I too believe that math will help you get anywhere.  I only started to really like math though when I got into public schools after struggling through really fast moving and avdanced math programs at a private school.  After I got down the basics I began to really enjoy math and recentyl joined this site to be able to do more since the only math competition that my school is/was involved in was tmsca which is a really easy contest for my state.  I think that the high school here does the AMC tests but they wouldn't allow any of the jr high kids to take the tests.  They wouldn't allow me to take algebra this year either.  They only let me take the 8th grade math along with other advanced students.  The advanced 8th ga=raders were allowed to taek algebra 1 which here is a freshman class.  I think they realized their mistake this year and I wouldn't be surprised to see a few 7th graders in my algebra class next year.  The point that I am trying to make is that it is never too late to start loving and competing in mathmatics.  there is always something new to learn and even if you major in like english in college the advanced mathmatics that you learn now will help you when you are an adult.  You will still have to be able to calculate taxs and things like that.  Even if your math education is a little above what you would need to do that what you have learned will help you be able t do it faster and easier.

Only around 100 students in the entire nation graduate high school having taken the USAMO.

A few observations I've made:

-- A small proportion of the professional mathematicians I have met were particularly successful with competition mathematics.

-- The most useful math (in terms of opportunity) is the math you would study preparing for the AMC or the AIME.  Learning to diagnose and model algebra problems is useful in mathematics, science, business, government, and anywhere you might find a quantitative problem.

-- You can always learn more math.  While you might feel behind the very top students, it is simply not true that you will necessarily remain behind within the realm of academic specialization or career you might eventually choose.  The work you do now will reward you tomorrow.  Besides, if you choose an area of specialization, you'll still be ahead of nearly all of your peers in that specialization if you have studied mathematics (and surely most USAMO participants will not have chosen your specialization).

Don't think of the goal so much as learning mathematics.  Think of the goal as learning how to solve problems.  This is what gets you somewhere.  I do believe that studying mathematical problem solving first is the way to go because of the extremely large number of problems in our world that can be modeled mathematically (even problems that most people see as being outside the realm of mathematics entirely -- such as politics, but that's often where we apply game theory and other ideas).

I encourage you to explore your new found interest in mathematics.  Set yourself some reasonable goals and then see what you can do with the knowledge and experience you accumulate.

[quote="MCrawford"]Only around 100 students in the entire nation graduate high school having taken the USAMO.[/quote]
Are you sure about this? I thought about 250 take it each year.

As to the original question, I agree with what everyone else has said so far. So what if there are people who are better than you or who studied more than you? There will always be. The point is to get as far ahead as you can. Whatever work you put in, you will always get something out of it. Whatever you learn, even if it doesn't get you into the USAMO, will help you later on. Although high school competitions can affect the college you get into, in the long run they won't really matter except for their own sake. A lot of people go into graduate school without having participated in any math competitions.

[quote="Yasha"][quote="MCrawford"]Are you sure about this? I thought about 250 take it each year.[/quote][/quote]

250 do take it but not all of them graduate.  So there are probably only about 1/3 of the students in the USAMO who are seniors and maybe a few juniors who want to graduate early.  I don't know the official stats for year to year but 100 would seem about right.  Also you can go to the amc website and look at the USAMO participants to see what the grade levels look like   [url]http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2004-ua/04usamoqualstate.html[/url]

[quote="Yasha"][quote="MCrawford"]Only around 100 students in the entire nation graduate high school having taken the USAMO.[/quote]
Are you sure about this? I thought about 250 take it each year.[/quote]

Yeah, as joml88 said, there are many repeats.  That decreases the number graduating rather significantly.

I am in the same situation as you are Robert0201, except I may be a little bit better off since I found out about this at the end of sophomore year when I got into PROMYS. So I have only had one year to do this, but I have taken every opportunity I have gotten: USAMTS, HMMT, AMC12, AIME, started a math team at my school, and joined the state ARML team. I absolutely agree with you; if you are starting late, then it is kind of pointless. You get to participate in all this stuff, but every time you go to a competition you perform poorly-- at least I do-- and lose. In the end, it feels like you are going to all these competitions to be stepped on and so that there will be a #1, who will not be you. Competition is a HUGE part of my motivation for studying math, and, when you know that you are never going to win or even do well, it hurts. When you think about all those other people who have been doing it for years, go to posh private schools or some sort of "academy for math and science," and learned calculus in 7th grade b/c they had the resources to do it AND then you compare it to yourself and you realize that you have nothing like that it is really frustrating. I feel exactly as you do. I have this summer and next year to start preparing again and I will try my best, but, realistically, I know I am going to lose and I will be depressed and I will look back on all this time spent as time wasted. I know some or most people here will feel that I have the wrong attitude or am too pessimistic, but the truth is you might feel the same if everytime you go to a competition, you are going, not to win, but to lose. I absolutely enjoy math and problem solving, but it feels like a slap in the face every time you have to do poorly at a competition because I want to compete too and I want to do well too and I want to get up on the stage and hear my name too, but I know that that will never happen. I am good at math, but, in terms of being one of the best, I can only say that that would apply to the limited scope of the population of my small public school.

This was a very unorganized and unedited response, but I really do get upset when I think about this. I try not to think about it that much and keep a positive attitude, but your comment really unleashed a lot of anger and frustration I have in me.

My best advice to you is just try not to think about it and do what you can do. You might do well in college math competitions like the Putnam if you start now. Good luck!

I think that you both have a right to think like you do.  I got into math competitions this year and I'm several years yonger than yall but for the state contest I do all the others have been doing it for 4 years and I feel bad when I go to meets and freeze because I am not used to them.

The best thing that people can do though is keep trying and do it fro the love of the "sport".

When I went to ARML last weekend, I wasn't expecting neither myself nor my team to win anything. I was playing hearts when the awards were announced. But the fact that I was not going there to win doesn't mean I was going there to lose. The main reason I wanted to go there is to see a lot of people I knew and to have fun, and I succeeded in that: I played frisbee and hearts and went to Hershey Park with my team.

Also, if it's a good math competition, the time studying for it won't be wasted if you lose. Everything you learned studying is useful information by itself.

One year or two is great for math if youre dedicated. Theres this guy i know who participated in the MMPC, our local state contest. When he was in 8th grade, he ranked ~4000th. When he was in 9th, he was 125th. When he was in 10th he was 13th. So theres definately lots of improvement possiblilities in a year or two.

[quote="tetrahedr0n"]... our local state contest. [/quote]
lol
[quote]When he was in 8th grade, he ranked ~4000th. When he was in 9th, he was 125th. When he was in 10th he was 13th. So theres definately lots of improvement possiblilities in a year or two.[/quote]
Yeah.  And that knowledge always helps when you compete at the next level.  If you think you really can't get first (you shouldn't think that way, but anyways) you can always go for the top 10 or something.  It'll at least help you get into a good college.

Just imagine how far behind the 99% of the world that never learns serious math is.

If I had one MAJOR COMPLAINT about AoPS it is that you guys didn't start up earlier! I found out at about the time the AoPS website got up, but, by then, I was at the END of my SOPHOMORE year. If AoPS had started when I was a freshman, my life would be so much better now. When I learned about AMC and AIME in 9th grade I searched all over the Internet for "math contest preparation," but I couldn't find anything. I also didn't know how math contests worked until I got to AoPS. For example, I didn't know that I could still take the AMC12 even if my school didn't do it. Thus, I gave up on it for a few years until I read about PROMYS in Imagine and so I once again searched for math contest preparation and stuff. This time I found AoPS, and I have been on AoPS website nearly everyday since then. I also learned a lot from other PROMYSers. BUT, if I had known all this stuff earlier I would be so better off. Arghhh............... so upset.... I guess that it is better late than never... but sometimes I think about what my life could have been if AoPS had started up just two years earlier and I get very depressed.  :(

I tried to find something like this when I was in 9th grade also. Of course, nothing existed, so I wasn't successful. AoPS didn't start until the end of my senior year, so you still have an edge over me...

You have a point-- I should probably quit whining and start working.


geometry
circumcircle
trigonometry
trig identities
Law of Sines

(1) In $ \triangle{ABC}$ we have $ \angle{A}\equal{}60^\circ$, $ \angle{b}\equal{}40^\circ$, and $ \angle{C}\equal{}80^\circ$.  If $ O$ is the center of the circumcircle of $ \triangle{ABC}$ and the radius of the circle $ O$ is $ 1$, find the radius of the circumcircle of $ \triangle{BOC}$.

(2) In triangle $ ABC$, $ CD$ is the bisector of angle $ C$, and $ D$ on $ AB$.  If $ \cos{(C/2)}\equal{}1/3$ and $ CD\equal{}6$, compute $ \frac{1}{a}\plus{}\frac{1}{b}$.

Thanks!

[hide="1"]Notice that $ OA,OB,OC$ cuts up the triangle into iscoceles triangles. Let the base angles of these iscoceles triangles be $ x,y,z$ so that $ A\equal{}x\plus{}y,B\equal{}y\plus{}z,C\equal{}z\plus{}x$.  Thus, we have from adding up the equations $ x\plus{}y\plus{}z\equal{}90$.  Since $ A\equal{}x\plus{}y\equal{}60$, we have $ z\equal{}30$.

From the law of sines, we have $ \frac{a}{\sin A}\equal{}2\Rightarrow a\equal{}\sqrt{3}$.  Since the iscoceles triangle has base angles of $ 30$ degrees, we have $ \angle BOC\equal{}120$ so by the law of sines, $ \frac{a}{\sin 120}\equal{}2R\Rightarrow R\equal{}1$.

[/hide]

[hide="2"]This should help:  http://www.artofproblemsolving.com/Forum/viewtopic.php?p=511031#511031[/hide]

[i]A Solution for 1.[/i] Notice that $ 4R\equal{}\frac{abc}{[ABC]}$ for all $ \triangle ABC$.  It follows (by the extended sine law) that $ [ABC]\equal{}2R^2\prod_{cyc}\sin A$.  Since $ [BOC]\equal{}\frac{\sin 120^{\circ}}{2}\equal{}\frac{\sqrt{3}}{4}$ and its angles are $ 120^{\circ}, 30^{\circ},30^{\circ}$, $ R_{\triangle BOC}\equal{}\sqrt{\frac{\frac{\sqrt{3}}{4}}{2\sin 120^{\circ}\sin^230^{\circ}}}\equal{}1$


modular arithmetic
quadratics
function
absolute value
number theory proposed
number theory

If $ c$ is a positive integer and $ p$ is an odd prime, what is the smallest residue (in absolute value) of \[  \sum_{n=0}^{\frac{p-1}{2}} \binom{2n}{n}c^n \;(\textrm{mod}\;p\ ) \ ?\]

J. Suranyi

I'm not sure what the question is asking but the number is clearly $ \equiv (1 \minus{} 4c)^{\frac {p \minus{} 1}{2}}\pmod{p}$

you have a proof ? this problem seems realy exciting

Of course! It's not that hard. Just expand and work separately with coefficients of $ c^n$

So I guess the answer is $ 0$ when $ c\equiv 4^{\minus{}1}\pmod{p}$
$ 1$ if $ 1\minus{}4c$ is a quadratic residue $ \pmod{p}$ and $ p\minus{}1$ if it's not.

All you really need here is the generating function $ F(x) \equal{} \sum_{n\equal{}0}^{\infty} {2n \choose n} x^n$.  (The terms after $ n \equal{} \frac{p\minus{}1}{2}$ drop out $ \bmod p$.)  This generating function is, in fact, $ \sqrt{1 \minus{} 4x}$, which gives the result immediately.  See the related generating function for the [url=http://en.wikipedia.org/wiki/Catalan_number]Catalan numbers[/url].

Yes I also thought of the generating function and that makes you think directly of $ 1\minus{}4c$. but notice that $ (1\minus{}4c)^{\frac12}\not \equiv (1\minus{}4c)^{\frac{p\minus{}1}2}\pmod{p}$
Indeed how can you say that terms after $ \frac{p\minus{}1}2$ drop out?  :wink:


In a convex quadrilateral ABCD $\angle{ABC}=\angle{ADC}=135^\cdot$. Let X and Y be points on AB respectively AD such that $\angle{XCD}=\angle{YCB}=90^\cdot$. If the circumcircles of AXY and ABD intersect at $Z$ then prove that ${AZ}\perp{ZC}$.

I'm sure this one was proposed a couple of times on the forum but i can't find it :blush:

Posted by... you!  :)  See http://www.mathlinks.ro/Forum/viewtopic.php?t=16899 or http://www.mathlinks.ro/Forum/viewtopic.php?p=21792 .

  darij


$ (\forall)n\in \mathbb{N}^*$ , show that :

$ \left[\frac{n}{\sum_{k\equal{}1}^n\{\sqrt{4k^2\minus{}2k}\}}\right]\equal{}2$

Sorry because is not so clear but i think you will understand :clap2:

No one wants to "attack" this problem ?  :lol:

Since for $ k\geq 1$ we have

$ 4k^2 \minus{} 4k \plus{} 1 < 4k^2 \minus{} 2k < 4k^2\implies 2k \minus{} 1 < \sqrt {4k^2 \minus{} 2k} < 2k$, we get

$ \{\sqrt {4k^2 \minus{} 2k}\} \equal{} \sqrt {4k^2 \minus{} 2k} \minus{} 2k \plus{} 1$

Now it's easy to show that

$ {1\over 3} < \sqrt {4k^2 \minus{} 2k} \minus{} 2k \plus{} 1 < {1\over 2}$

The lefthand part boils down to $ \sqrt {4k^2 \minus{} 2k} > 2k \minus{} {2\over 3}\iff\dots\iff k > {2\over 3}$, and the righthand part gives $ \sqrt {4k^2 \minus{} 2k} < 2k \minus{} {1\over 2}\iff\dots\iff 0 < {1\over 4}$

Therefore

$ {n\over 3} < \sum_{k \equal{} 1}^n\sqrt {4k^2 \minus{} 2k} \minus{} 2k \plus{} 1 < {n\over 2}$, which yields

$ 2 < \frac {n}{\sum_{k \equal{} 1}^n\sqrt {4k^2 \minus{} 2k} \minus{} 2k \plus{} 1} < 3$

and the desired result follows.


inequalities function symmetry trigonometry

Given a,b,c such that :$ \left\{ \begin{array}{l}
 a,b,c \ge 0 \\
 a \plus{} b \plus{} c \equal{} 2 \\
 \end{array} \right.$
Prove that :
$ (a^2  \plus{} ab \plus{} b^2 )(b^2  \plus{} bc \plus{} c^2 )(c^2  \plus{} ca \plus{} a^2 ) \le 3$

[hide="just a thought"]
this is symmetric, but when a=b==c2/3, the RHS is equal to 64/27, not 3...
[/hide]

I think the equality occurs when one of 3 numbers is equal to 0 and the two others are equal to 1 :roll:

The function $ f(a,b,c) \equal{} (a^2\plus{}ab\plus{}b^2)(b^2\plus{}bc\plus{}c^2)(c^2\plus{}ca\plus{}a^2)$ in convex in $ a,b,c$. Fix $ c$, and WLOG $ a\ge b$ (due to symmetry); then the function is maximized when $ a,b$ are as far apart as possible [by Jensen/convexity], i.e. when $ b\equal{}0$, $ a\equal{}2\minus{}c$. So, we just need to maximize
$ f(c) \equal{} (2\minus{}c)^2c^2(c^2\plus{}c(2\minus{}c)\plus{}(2\minus{}c)^2) \equal{} (c(2\minus{}c))^2((c\plus{}(2\minus{}c))^2 \minus{} c(2\minus{}c)) \equal{} (c(2\minus{}c))^2(4\minus{}c(2\minus{}c))$ for $ 2\ge c \ge 0$. Let $ c(2\minus{}c)\equal{}x$. By AM-GM, $ x\equal{}c(2\minus{}c) \le \frac{c\plus{}(2\minus{}c)}{2} \equal{} 1$. Since $ 2\ge c\ge 0$, we also have $ x\equal{}c(2\minus{}c) \ge 0$. But $ f(c) \equal{} x^2(4\minus{}x)$ is increasing in $ x$ on $ [0,1]$, so it is maximized when $ x\equal{}1$, which is when $ c(2\minus{}c)\equal{}1 \implies c\equal{}1, a\equal{}1, b\equal{}0$. But at this point, $ f(c) \equal{} 3$, so $ f(a,b,c) \le 3$ as desired.

darn I should actually remember how to use classical inequalities?

not_trig, can you explain why convexity implies that the function is maximized when $ a$ and $ b$ are as far away as possible?


What is the least positive integer divisible by each of the first five composite numbers?


Prove that the number $ 10^{10}$ can't be written as the product of two natural numbers which do not contain the digit "$ 0$" in their decimal representation.

[hide="Solution"]Let $ ab\equal{}10^{10}$ and suppose $ a$ and $ b$ have no zeroes in their decimal representation.  One of $ a,b$ must be divisible by 2 and one must be divisible by 5, and neither can be divisible by both.  Therefore $ a\equal{}2^{10}$ and $ b\equal{}5^{10}$.  But then $ a\equal{}1024$, which has a zero, contradiction.[/hide]


trigonometry
geometric series
algebra unsolved
algebra

Compute

$\cos^2x+\cos^22x+\cos^23x+\cdots +\cos^2nx$  for  $n\in\mathbb{N}$ , $x\in\mathbb{R}$

That's just \[\sum_{i = 1}^n \frac{\cos{2ix} + 1}{2}\] and $\sum_{i = 1}^n \cos{2ix}$ can be easily computed using De Moivre and geometric series (or you can use some well-known formula to do this).

Let $D= \sum_{k=1}^n cos^2 kx=\sum_{k=1}^n \frac{cos 2kx+1}{2}=\frac{n}{2}+\frac{1}{2}\sum_{k=1}^n cos 2kx$

Now, let $C=\sum_{k=1}^n cos 2kx$ and $S=\sum_{k=1}^n sin 2kx$.

So, $\sum_{k=1}^n e^{2kix}=C+iS$

But, $\sum_{k=1}^n e^{2kix}= e^{2ix} \frac{e^{2inx}-1}{e^{2ix}-1}= e^{2ix} \frac{e^{inx}}{e^{ix}} \frac{sin 2nx}{sin x} = e^{i(n+1)x} \frac{sin 2nx}{sin x}$

So, $\sum_{k=1}^n e^{2kix} = (cos ((n+1)x) +isin((n+1)x)) \frac{sin 2nx}{sin x}$

So, $C= \frac{sin 2nx}{sin x}cos ((n+1)x)$

And finally,

$D=\frac{\frac{sin 2nx}{sin x}cos ((n+1)x)}{2} + \frac{n}{2}$


superior algebra
superior algebra solved

Let A be a ring s.t. x 2 =1 or x 2 =x for all x in A. Prove that if A has at least 2 invertible elements then A is isomorphic to Z_3.

[i]Romanian olympiad 1998/2'nd round/12'th grade[/i]

Let x be an invertible element of A, which is not 1. If x^2=x, then x(x-1)=0 so x=1, a contradiction. Thus x^2=1.

Let us prove that 1+1+1=0. If (1+1)^2 = 1+1 => 1+1=0. Since there are at least two invertible elements, there is one of them with y^2=1, and y is not 1. Then let z=y-1, so z^2+2z+1=1. Since z(1+1)=0, z^2=0, and since we are not in the trivial case 0=1, z^2=z=0 => y=1, contradiction. Thus 1+1+1=0.

Let x not in {0,1} be an element such that x^2=x. If (-x)^2 = 1 = x^2, then x=1, a contradiction. Therefore (-x)^2 = -x = x, so x+x=0. But x+x+x=0, so x=0. This is a contradiction.

Therefore, the equation x^2=x has only 0 and 1 as roots in A. So, for any x in A\{0}, x^2=1.

Let y in A be an element not in {-1,0,1} => y^2=1
(y+1)^2 = y^2 + 2y + 1 = 2y+2
If (y+1)^2 = y+1 => y=-1, contradiction.
If (y+1)^2 = 1 => 2y = -1, and since 2^2=1 we get y=-2=1, contradiction.

So it has at most 3 elements, but it must have at least 3, since 0 and 2 invertible elements are in it. Thus A = Z_3.

Hope it's correct :)


Bobby goes on vacation from New York City to an island called Kant.  When he arrives, it is 4 hours later than when he started, but he goes to the west across 5 zones.  His plane flies at 550 mph.  He then returns after spending 1 week on Kant.  How far does the plane he rides fly (to Kant and back) and how many total hours does he spend on vacation? Include the time zone differences.

[hide]Umm.... 188 hours and 11000 miles?[/hide]

It seems you have gotten how long the plane ride is incorrect.

[hide="hint"]When you go across a time zone going west, you subtract an hour from your total time you spent going.  i.e if it is 2 hours later when you arrive at the new time zone going west, you really spend 3 hours travelling.[/hide]

[hide][b]My answer was: 9900 hours round trip and 186 hours on vacation.[/b]
Here's my process.
So you know that he finds it 4 hours flight. But you also know he has a 5 hour time difference. 4+5=9 hour real flight ride. So we know that we have a 18 hour round trip flight ride.

We know:
[b]18 hours round trip[/b]

We also know that the plane travels 550 MPH. So we do $550*9=4950$ miles per hour on for one trip, and $4950*2$ to find round trip is 9900 miles round trip.

We know:
[b]18 hours round trip[/b]
[b]9900 miles round trip[/b]

Okay, let's go on to how many hours he spends on Kant. He spends $24*7=168$. Adding it to how many hours he spends on the round trip: $168+18=186$ hours.

We now know:
[b]186 hours for his whole vacation[/b]
[b]9900 hours round trip[/b]

So there's your answer.

[/hide]

Hope this helped!  :)

[hide]If it is 4 hours later when Bobby arrives in Kant and he flew over 5 time zones, then his flight took 9 hours. The whole round trip distance between NY and Kant is then 2*9*550=9900 miles. A week is 24*7=168 hours, plus the time he spent on the plane (18 hours) equals [b]186[/b] total hours Bobby spent on his vacation.[/hide]


Suppose $a\ge b\ge c$.  If $c=0$, then obviously $a=b=c=0$, which is one solution - otherwise, assume $c\ge1$.
If $b>2$, $abc\ge3a\ge a+b+c$, with equality iff $a=b=c$ and $c=1$ - a contradiction.  If $b=1$, $a=a+2$, also a contradiction, so $b=2$, and $c\in\{1,2\}$.  $c=2\Rightarrow 4a=a+4$, or $3a=4$, which is impossible, so $c=1$, $2a=a+3$, and $a=3$.

So the only solutions are $(a,b,c)=(0,0,0)$ and $(a,b,c)=(1,2,3)$.

ok, now solve \[a_1+a_2+a_3+a_4+a_5+a_6+a_7+a_8=a_1a_2a_3a_4a_5a_6a_7a_8\] in the set of non-negative integers.

Wlog, we may assume that $a_1 \leq \cdots \leq a_8.$
If $a_8 = 0$ then $a_i = 0$ for all $i$.
Otherwise, we must have $a_1a_2 \cdots a_7 \leq 8$ because the LHS is not greater than $8a_8.$
This leads to a finite and not too big direct (but boring) checking by hands...

Pierre.

A more interesting problem (in terms of cutting out the tedium) is to solve it for distinct integers - it has a much more concise solution.

[quote=mathematica]Find all the solutions of the equation \[a+b+c=abc\] in the set of non-negative integers.[/quote]

as one of the number equals to 0 others will automatically be equal to 0. so let's assume that $a, b, c$ are positive integers. we can write the equation as $$ \frac{1}{ab}+ \frac{1}{bc} + \frac{1}{ac} =1$$. if $ a, b,c \geq 2$ then $\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac} \leq \frac{3}{8}$. so one of them must equals to 1. let's assume that the number is $c$. the $1+a+b=ab \rightarrow 2=(a-1)(b-1)$ rest is easy


Let $ a,b,c$ be the side lengths of a triangle. Prove that
\[ \frac{a(a\plus{}b)}{a^2\plus{}b^2}\plus{}\frac{b(b\plus{}c)}{b^2\plus{}c^2}\plus{}\frac{c(c\plus{}a)}{c^2\plus{}a^2} \le 2\plus{}\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}\]
:)


function
complex analysis
complex analysis unsolved

Let f(z)=sum(a_n*z^n)  (n goes to zero to infinity) be an analytic function for |z|<1 and satisfies |f(z)|<M.

Show that sum(|a_n|^2) < M^2

Hint:
First consider the inegral of $ |f(z)|^2$ over circle with radius $ r<1$ and compute them in terms of $ a_i$ :wink:


Anyone know if AIME ever qualified anyone who had slightly less than 120 on AMC 10? If so, what score was needed and what year did this occur??

AMC 10A , 2004 AIME qualifying score: 110.00

All this data is in the [url=http://www.unl.edu/amc/e-exams/e5-amc10/archive10.shtml]AMC archives[/url].  Click on the entries in the line that says "General Statistics".

The last time it happened was 2004, when the 10A qualifying score was lowered to 110.

thanks - does anyone think changing the blanks from 2.5 to 1.5 will significantly change the qualifying numbers?

Personally, I think it will have no effect for the AMC12.  There's an excess of AIME qualifiers every year--the minimum is 5%, but it's usually closer to 10% that make it if I remember right.  However, the new scoring makes it much more difficult to get above 120.  You need a minimum of 19 correct answers (with 4 blanks).  So I predict that there's a good chance the AMC10 cutoff will be lowered.

No as almost everyone I talked to thought this 10 was the easiest they have ever took.

[quote="tjhance"]Is it possible for the qualifying value to be raised?[/quote]

Well it would go against everything the AMC has printed in terms of instructions so 99.99999% no.

For AMC 10, I think it is either top 1% or 120.  Whatever is lower.

For AMC12 , I think it is either top 5% or 100.  Whatever is lower.

Even if I got those percents wrong, the answer is still no.

[quote="h_s_potter2002"][quote="tjhance"]Is it possible for the qualifying value to be raised?[/quote]

No, but is has been done in the past, I think in like 2003 maybe?[/quote]

It was done before when it was only like 1%

2003, for the 10A it was 119, for 10B it was 121.

The AMC10 has not always had a 120 cutoff.  In the beginning, it was just the top 1%, and if that made a cutoff over 120, that was still the cutoff.

[quote="Piggypi314"][quote="h_s_potter2002"][quote="tjhance"]Is it possible for the qualifying value to be raised?[/quote]

No, but is has been done in the past, I think in like 2003 maybe?[/quote]

Nah dude, I remember because in 2003 I made it with a 119.  I'm pretty sure it was 2003.[/quote]

It was 2003 B where you needed a 121.


Prove that for x,y,z >0 it holds that

$ 2(x^2\plus{}y^2\plus{}z^2)^3\geq (x^4\plus{}y^4\plus{}z^4\plus{}x^2y^2\plus{}y^2z^2\plus{}z^2x^2)(x\plus{}y\plus{}z)^2$

Thank you silouan for your gift, but your inequality is wrong. :(
Try $ x\equal{}3$ and $ y\equal{}z\equal{}2.$ :wink:


limit
calculus
integration
logarithms
calculus computations

[hide="First hint"]Factor $\frac1n$ out of the sum.[/hide]
[hide="Second hint"]Make it a Riemann sum for an integral[/hide]

[hide="Another idea"]Let $x_{n}= 1+\frac12+\ldots+\frac1{n}-\ln n$. Then we can write \[\left. \frac1{n}+\frac1{n+1}+\ldots+\frac1{kn}= x_{kn}+\ln kn-x_{n-1}-\ln (n-1) = x_{kn}-x_{n-1}+\ln k+\ln \left( \frac{n}{n-1}\right) . \right.\] But $\lim_{n \to \infty}x_{n}= \gamma$, so the sought limit is $\ln k$.[/hide]


geometry
3D geometry
number theory
prime factorization

Below is a multiplicative magic square, where the product of the numbers in each row, column and diagonal is the same and equal to four-digit number ABCD, where each letter represents a digit and each box contains a number integer. If AC is the two-digit number in the center of the square, the sum A + B + C + D equals:

[img]http://img27.imageshack.us/img27/6638/calsa.jpg[/img]

A) 17
B) 18
C) 19
D) 20
E) 21

The product of the middle row tells us that AC|$ 96$., so that AC can equal 96, 48, 24, 16, or 12.  We don't expect AC too be too big, because that would make the product of each individual row greater than a four-digit number.  Moreover, notice that AC must have a greater power of $ 2$ in its prime factorization compared to C, as well as a power of 3.  If AC=12, which satisfies all of these properties, the product would have to be $ (6)(12)(24) \equal{} 1728$, and the sum of the digits is $ 18$.

Ah ok ;)
Nice logic. Thanks for your help.


I updated Asymptote on Vista recently. Now all the file type associations are gone and I can't reset them. That is, when I say "Open With..." the Asymptote option is gone. When I try to set it manually to ASY nothing happens. I re-created the config.asy file but this doesn't change anything. I would like to get back to simply double-clicking on files to compile them. Any help would be appreciated, thanks!

Technically speaking, it is not an Asymptote problem, but a Vista problem. I prefer to stick to XP, so I don't really know much about Vista but see if [url=http://articles.techrepublic.com.com/5100-10878_11-6172036.html]this[/url] can help :).

I have had very few issues with Vista since getting this laptop. I updated to 1.63 today and the problem persists. It was not this way for the 1.4-ish version of Asymptote I had months ago. I know how to change file types, but when I double-click on Asymptote the computer doesn't respond. This leads me to believe that something about the newest versions of ASY causes it to get stuck behind Vista's newfangled security measures.

Has anybody else out there had issues (or success) getting the latest versions of ASY to run in Vista?

Ah, yes, one more important thing: with Vista and the newest versions of Asymptote you cannot call asy.exe directly; instead you have to call asy.bat that takes care of some settings first and then calls Asymptote. Were you aware of that when (re)setting your extension associations?


Hi everyone,

Could you please tell me if this is correct? I have to negate the following sentences and prove them.

1. I will do my homework and I will pass this class.-I will not do my homework and I will not pass this class.
2. If T is continuous, then T is bounded-T is continuous and T is not bounded.
3. If x is odd, then x is prime-X is odd and X is not prime
4. If she is not home, then Sam will answer the phone and he will tell you how to reach her-She is not home, Sam will not answer the phone and he will not tell you how to reach her.

Thank you in advance

That definitely does not belong to advanced fields but I'm at loss where to move it :?

OK, here are a few rules that you should look at in each case:

1) NOT (A AND B) = (NOT A) OR (NOT B)

2,3) NOT (IF A(x) THEN B(x))=THERE EXISTS x SUCH THAT A(x) AND NOT B(x)

3)  NOT(IF A THEN (B AND C))=IT MAY HAPPEN THAT A AND NOT(B AND C) (now look at (1) to negate (B AND C))

Once you are through that, try to negate the following sentence

If a cat wins a fight with a dog, then either the cat is big and agressive or the dog is small and timid.

[quote="chocolatelover"]I will do my homework and I will pass this class.[/quote]
I thought the negation of this was, "I will post my homework on the internet in the hopes that other people will do it for me, but because I am an obnoxious person I will put it in an obviously wrong forum, despite the fact that there are a wide variety of clues available that what I am doing is not correct."

[quote="fedja"]I'm at loss where to move it :Confused:[/quote]
Probably Intermediate forum is best, or maybe HSB.  This is pretty standard high school material.

[quote="JBL"][quote="chocolatelover"]I will do my homework and I will pass this class.[/quote]
I thought the negation of this was, "I will post my homework on the internet in the hopes that other people will do it for me, but because I am an obnoxious person I will put it in an obviously wrong forum, despite the fact that there are a wide variety of clues available that what I am doing is not correct."[/quote] :rotfl:

[quote="JBL"][quote="fedja"]I'm at loss where to move it :Confused:[/quote]
Probably Intermediate forum is best, [b]or maybe HSB[/b].  This is pretty standard high school material.[/quote]HSB is sensitive for the material may be too new for some students. I remember administrators have discussed this before and that if the material is about standard high school level and unsure of the difficulty, Intermediate Topics forum is the place for it.


Harvard college Princeton search Stanford MIT Duke

Just wanted general opinions on the pros and cons of ivy league or top-ranked colleges (such as Harvard, Princeton, etc.) verses public or state schools (such as U of Michigan- Ann Arbor, U of Florida, etc.).

umm one thing i would say that ivy leauge is better at is reputation.

In general, yes. but you can find some public schools with great reps for example....Berkeley

The biggest factor has to be the money...
But yes, prestige is a large factor as well. Saying you went to Harvard sounds better to people than saying you went to UMich Ann Arbor....even though they are both spectacular schools....

If prestige is worth $\$100000 to you, then go to an Ivy League school. Otherwise don't. (There are a few other reasons in my mind for someone to go to Ivy League schools as an undergraduate, but the point is that they are extremely overrated.)

This conversation has happened here before.  It's probably worth a search to go see what people said last time.

My view: There are lots of really good schools in this country, Ivy League, public, or private-but-not-in-the-Ivy-league-and-not-Stanford-or-MIT-either.  Each of those three groups is very broad.  Each group has colleges with vastly different strengths, weaknesses and learning experiences.  Brown is a very different type of school than Columbia.  Cornell is both an Ivy League school and (in part) a public university.  What you've asked, as a stand-alone question, isn't really a "good" question if you want to learn about the sort of school that you should seek for your (or your children's, I suppose) education.

most people wanna go to ivy rather than public(most don't make it to ivy, so they settle for public)

I know it's not a great question and it's kinda broad.  The reason I asked was because, right now, I really would like to go to MIT and I'm mainly looking at other Ivy schools for college, but I'm not sure if that's a good thing, to get locked into the "Ivy track" just because everyone says it's the best.  I guess the better question to ask would be for people to give the college they attended or are attending and the reason they choose that college over all the others.

There are a lot of good schools out there, each with their own particular pros and cons. You really have to look at each school on an individual basis and weight its pros and cons against your personality and your long term goals.

If your goal is to go to graduate or professional school after college, you'll find that a lot of people who successfully took that path come from a wide variety of undergraduate institutions. In that case it's more about what you can do with the school, than what the school is able to do for you.

If you want to go into business or the work force right after college, then the name and reputation of the school could be more important to you. But that doesn't necessarily mean that its essential. There can be benefits to paying less for college and saving that money for future ventures.

Either way, you shouldn't get locked into the mentality of "its either this school or bust for me." Ranking and people's opinions can't tell you what schools best for [i]you[/i]. You have to make that decision for yourself - and for almost everyone there's more than one right choice.

I remember when I applied to college, I had a list of about 6 or 7 that I thought I would definitely be happy going to. They ranged from Private to State. The important thing is to be able to think of [i]specific[/i] things about each school that are genuinely good for you. Not for people who do ranking or for people as a whole, but for you specifically.

I would also think about what makes a good college for YOU. Many of the "top" colleges in the country are so because they have a lot of researchers, etc. However, the problem with research professors is that, unless you are in one of their upper-level courses, your quality of education is probably going to be compromised to an extent. however, there are some (lesser-known) universities that are "teaching schools"- basically, their main purpose for existence is for the continuing education of young people, not for research. while perhaps for grad schools these may not be the place for you, if you want an great extensive education, you might want to check out schools that boast this factor. its also probably going to be a lot cheaper, as they don't have to fund as much research.

Ivy league schools are generally better, but if you want to save money and go to a public school out of state than you might as well go to a public school because the cost is almost the same. I live in a state where my state system is ok but not great so my plan is to go to an ivy league or private school out of state.(princeton

What are the chances to get admitted into an Ivy league graduate school if you go to a public school?
If you are in the Ivy's leagues undergraduate school are you automatically accepted into graduate if you are doing well in your classes?

Just asking in general, but if there are huge variations in each of the school's policies use Princeton to make it specific.

If you go to an Ivy League school for undergrad, generally speaking, you have a leg up on getting into other Ivy League schools for grad school.  A lot of universities don't generally allow people to do undergrad and grad studies at the same place.

Princeton's math grad students come from a lot of different schools. Quite a few are from foreign Universities. The most from an American University seemed to come from Harvard, but I also remember there were ones from Duke, Stanford, MIT, Cal Tech, Georgia Tech, U of Chicago, Washington University, U of Michigan, Harvey Mudd, Williams College, Rice, and Berkeley. And I'm sure there were also ones from other public schools that I can't remember.

On the flip side there were some Princeton undergrads who didn't get into the very best grad schools (many did, but not all). So the school you go to isn't necessarily as important as what you do when you are there, and how much you impress the professors who write your recommendations.

Typically if you go to an Ivy League school for undergrad you won't be allowed to stay for graduate school. They really discourage that. As an example, the famous physicist Richard Feynman went to MIT for his undergrad degree and wanted to stay for graduate school, but they wouldn't let him. So he went to Princeton for grad school instead.

It does happen on some occations though, if you have a really compelling reason to stay or if you have a strong connection with a particular professor. I've known of three of four students who stayed at Princeton (either in math or physics) in the last couple of years.

If you go to Harvard and become a Putnam fellow, they have to let you back.

I'm sure that applies to many of us here.

[quote="gauss202"]Typically if you go to an Ivy League school for undergrad you won't be allowed to stay for graduate school. They really discourage that.[/quote]

Say what?!  So, if I got into MIT and decided to go there for undergrad, I couldn't stay there for grad??  That's bogus!

Its not impossible but its discouraged. The reason being that you will get a better education by being exposed to different faculty in a different place.

I wouldn't worry too much about it though. If you get into MIT and that's where you really wanted to go for undergrad, then you should go. The grad school thing will work itself out when the time comes. And like I said, there are always a couple of people who stay in the same place each year, but i don't know how or why.

I will be at MIT for four (..maybe more) years, but as of now, MIT grad is not one of my top choices. I believe that it would be much nicer if you see more people from various places, and learn more from them. (Mr. Richard Feynman once said something similar to that I think - He went princeton after MIT)

I'll agree with the other people here who have reported that generally top math schools discourage their own undergraduates from studying at the same school for graduate school. Harvard is very explicit about this, and has made very few exceptions over the years. But that's not a problem, because as several participants on this forum repeatedly say, there is more than one good math school. Apply to strong undergraduate programs (PREPARE for them while you are still in high school) and don't worry too much about which program you get into. Then, if you decide to go on to graduate school after your undergraduate degree, apply for the better independent fellowships (your undergraduate school should be able to advise you about that) and apply to OTHER strong graduate programs.

A funny example I read online is a [url=http://www.jhu.edu/gifted/set/members.html]member of the Study of Exceptional Talent[/url] who once had on his home page a statement that the best way to study math is to do what he did: get an undergraduate degree from Harvard and do graduate school at MIT. He has removed that statement from his Web page since becoming a lecturer at Stanford.  :oops:

It is a GOOD thing to study at more than one school, just as it is a good thing to read about, say, calculus from more than one book. Don't limit yourself in what learning opportunities you pursue.

[quote="ComplexZeta"]But if there are $n$ good grad schools and you go to Harvard, then you only have $n-1$ good grad schools open to you. So you'll still be worse off, especially if you would really like to work with a specific professor there. (And how would you know when you are only in high school?)[/quote]

But *you* only need [i]one[/i] good grad school. I presume that if you could get into Harvard grad school, you could probably get into most of the other n.

And if you go to Harvard, and there is a professor that you [i]really, really[/i] want to work with, it is possible to stay on for grad school. It does happen every now and then.

Yeah, Harvard's strong, consistent discouragement (but not absolute ban) of its own undergrads going to Harvard grad school is hardly a reason to turn down an acceptance to Harvard, if you obtain one. After all, a strong undergraduate program (and Harvard is undoubtedly one of those) both increases the odds for acceptance into a strong graduate program (not to mention increasing the odds of getting a top fellowship) but also PREPARES the learner for making the most of whatever graduate program the learner pursues. The same, of course, could be said about other strong undergraduate programs in math at schools that also have graduate programs.

In a lot of cases, going to an Ivy League school does not mean you will be spending more money.  In California, Schwarzenegger and the Republicans have tightened the belt of the state schools, and offered little financial incentive for me to go there.  For example, a professor at UCSB could only guarantee me about $10,000 a year in help.  Tuition for out of state public schools are significantly greater, and they focus on taking care of their own residents.  I was essentially forced to go to the top private schools, which didn't really bother me, because they have an excellent commitment to the financial aid of middle/lower income students.  Last year, I paid $8,000 to go to an Ivy and it would have been at the same level regardless of how many outside scholarships I brought in.  I believe for families with incomes under $40,000 it is very close to free.  So don't let people throwing six figure numbers at you scare you into staying in your state.  If you have to pay the full $40,000/year, most likely you can afford it anyway.

While financial aid certainly helps, also be aware that it might not be enough.  You have to have the right amount of assets and the right type, as well as the right proportion of water on your left thumb (just kidding ;) ).

[quote="gauss202"]One very nosey thing you can do is to look at the current [url=http://www.math.princeton.edu/graduate/gradstudents.04.05.html]Princeton graduate students[/url] and google them to see where they went for undergrad. The first one I tried (Paul Pollack) went to the University of Georgia.[/quote]

I know of that guy, he used to live in Florida (did really well in Mu Alpha Theta contests and such).  Somewhat of a legend I guess.


advanced fields
advanced fields unsolved

I'm reading James Ax's paper "The elementary theory of finite fields". On page 246 there is a following proposition:
Let $ E$ be a quasi-finite and let $ F,F^\prime$ be extensions of $ E$ such that
(a) $ E$ is relatively algebraically closed if $ F$ and $ F^\prime$;
(b) $ F$ and $ F^\prime$ are hyper-finite;
(c) $ |F|\equal{}|F|^\prime>|E|$.
Then $ F\cong_E F^\prime$.

On the proof it is said that:

If $ \rho$ is an odd positive integer plus a limit ordinal $ F_\rho\equal{}$ relative algebraic closure of $ F_{\rho\minus{}1}(t_\sigma )$ where  $ \sigma$ is the smallest such that $ t_\sigma\not\in F_{\sigma\minus{}1}$

What does the sentence "an odd positive integer plus a limit ordinal" mean?

you know how ordinal addition defined? what's unclear?

I'm not familiar with ordinal arithmetic.

there is "some" ;) literature about it.
you may ask more concrete questions ;).

Can I ask a more concrete question (about the question that was asked)?

I thought that to be a limit ordinal you don't have a predecessor, and thus are necessarily infinite. So if you take an odd integer $ \rho$ and a limit ordinal $ F_\rho$ shouldn't the adding $ \rho \plus{} F_\rho\equal{}F_\rho$?


First, I am using windows xp home, with TeXnicCenter (Version 1 Beta 7.01), and using the free Adobe Acrobat version 8.

Whenever I close Acrobat, after viewing a perfectly fine built tex file, the program (Acrobat) crashes.  I am not sure what is causing the crashes.   Before I updated, I used Acrobat 7 and everything was fine.

Any ideas/tips?

Thanks.

edit:  I think it definitely has something to do with TeXnicCenter, as if I open the same file (after it has been converted to pdf) Acrobat does not crash on closing.

edit2:  I suppose I will just preview my LaTeX with Yap (dvi), then convert it to pdf when necessary.  So as to avoid the problem.  However, if anyone knows of a solution I would love to hear it.

It appears to be an Adobe Reader bug not confined to TexnicCenter, as others have reported the same problem; see [url=http://groups.google.co.uk/group/adobe.reader/browse_thread/thread/668e0ee823a0aa47/6dbf73306c34af69?lnk=st&q=texniccenter+adobe+reader+8+crash&rnum=2&hl=en#6dbf73306c34af69]here[/url]. There is a workaround posted there of compiling another tex file and leaving it open.

maybe cuz if you compile a tex file with the same file name,


geometry
combinatorics unsolved
combinatorics

For a point set P in d, let fd(P) be the number of unit-distance point pairs:

fd(P) = {(u, v) | u, v  P, | u - v| = 1}  ;

and let fd(n) be the maximum over all sets of n points:

fd(n) = fd(P)  .

Further, let gd(P) denote the number of distinct distances induced by a set of points P:

gd(P) = {| u - v| | u, v  P}  ;

and let gd(n) be the minimum over all sets of n points:

gd(n) = gd(P)  .

Give upper and lower bounds on fd(n) and gd(n), particularly for d = 2 and d = 3.

solution-
f2(n) = O(n^4/3) , and f2(n) = (n^1+c/loglog n).f3(n) = O(n^5/3) and f3(n) =(n^4/3loglog n). For g2(n), the best result is that g2(n) = (n6/7)
the correct answer here is n/squared root Log N. this bound is achieved on the grid.

 :D  :D

So, what are you asking for?

Moreover, this is combinatorial geometry thus I move it into the combinatorics section. Thre you will find other threads about these problems.

Pierre.

Start with a look at :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=distances&t=22401
and
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=distances&t=4889

But there are other threads...

Pierre.

[quote="Azat123"]For a point set P in d, let fd(P) be the number of unit-distance point pairs:

fd(P) = {(u, v) | u, v  P, | u - v| = 1}  ;

and let fd(n) be the maximum over all sets of n points:

fd(n) = fd(P)  .

Further, let gd(P) denote the number of distinct distances induced by a set of points P:

gd(P) = {| u - v| | u, v  P}  ;

and let gd(n) be the minimum over all sets of n points:

gd(n) = gd(P)  .

Give upper and lower bounds on fd(n) and gd(n), particularly for d = 2 and d = 3.

solution-
f2(n) = O(n^4/3) , and f2(n) = (n^1+c/loglog n).f3(n) = O(n^5/3) and f3(n) =(n^4/3loglog n). For g2(n), the best result is that g2(n) = (n6/7)
the correct answer here is n/squared root Log N. this bound is achieved on the grid.

If indeed, as Hilbert asserted, mathematics is a meaningless game played with meaningless marks on paper, the only mathematical experience to which we can refer is the making of marks on paper.
E. T. Bell


parameterization
number theory
combinatorics proposed
combinatorics

A graph has 12k points. Each point has 3k+6 edges. For any two points the number of points joined to both is the same. Find k.

could anyone please explain what does it mean by each point has 2k+6 edges??

every point is joined to $ \binom{3k\plus{}6}2$ pairs of points. so the total number of edge joining a point and two other points is $ 12k\cdot\binom{3k\plus{}6}2$. this number must be divisible by $ \binom{12k}2$ since the number of points joined to any pair is the same. thus $ \frac{(3k\plus{}6)(3k\plus{}5)(12k)}{(12k)(12k\minus{}1)}$ is an integer. some simple number theory analysis gives $ k\equal{}3$ as the answer.

is the above solution correct? :maybe:

The problem describes a graph $ G \equal{} (V,E)$ with $ |V| \equal{} n \equal{} 12k$ vertices and $ d \equal{} 3k \plus{} 6$, the degree of each vertex (i.e. the number of edges incident in that vertex). Such a graph is called $ d$-regular. Moreover, it is given that for any two vertices, the number of other vertices joined to both those two is the same, be that constant $ \lambda$. Usually we denote the degree of a vertex $ x$ by $ \deg x$, and the set of its neighbors (joined to him by edges) by $ N(x)$, so we are given that $ \deg x \equal{} d$ for all $ x \in V$, and $ | N(x) \cap N(y) | \equal{} \lambda$ for all distinct $ x,y \in V$.

Solution.
[hide]In fact what it is described here is a symmetric balanced incomplete blocks design (BIBD), with parameters $ (n, d, \lambda)$. But it is immediate, by a little double-counting, that the relation $ \lambda(n \minus{} 1) \equal{} d(d \minus{} 1)$ is necessary; in our case $ \lambda(12k \minus{} 1) \equal{} (3k \plus{} 6)(3k \plus{} 5)$. It is relatively easy to prove that this may happen only for $ k \equal{} 3$ (therefore $ \lambda \equal{} 6$).
The question now is if a $ (36, 15,6)$ symmetric BIB actually exists (the fact that the numbers check the relation above is not always sufficient). Probably some research in tables of known designs is the easiest way; or to find a convenient model, but that may be arduous. The good news is that these numbers verify the condition of the Bruck-Ryser-Chowla theorem ($ 36$ is even, and $ 15 \minus{} 6 \equal{} 9$ is a perfect square), but that is also just necessary, not sufficient. Maybe someone may find the model.[/hide]

YES, Johan Gunardi, check my post (written as you posted yours)  :)

Re-EDIT. I found  the reference for what I hinted in my solution.
[hide]Up to isomorphism there are $ 4$ symmetric $ (36, 15, 6)$ designs with automorphisms of order $ 7$, and $ 38$ symmetric $ (36, 15, 6)$ designs with automorphisms of order $ 5$. For those designs full automorphism groups are determined. Also, all symmetric $ (36, 15, 6)$ designs having automorphisms of order $ 3$ acting with $ 9$ and $ 6$ fixed points, or cyclic automorphism groups of order $ 4$ acting standardly are constructed and orders of their full automorphism groups are determined.
[url]http://crosbi.znanstvenici.hr/prikazi-rad?chset=ASCII&lang=EN&rad=56134[/url]
Also, best [url]http://www.uow.edu.au/~jennie/WEB/WEB69-93/max/005_1969.pdf[/url]
[/hide]


function
integration
Support
limit
topology
real analysis
real analysis theorems

Is is true that $f \in L_{loc}^{1}(\Omega) \mapsto T_{f}\in D'(\Omega)$, the mapping defined on the locally integrable functions on an open set $\Omega \subseteq \mathbb{R}^{n}$ which associates to $f$ the distribution $T_{f}(\phi)=\int_{\Omega}{f\phi}$ is injective ?

I wanted to use the density of $C_{c}^{\infty}(\Omega)$ in $L_{loc}^{2}(\Omega)$, but $f$ is only $L_{loc}^{1}$. So, how do I prove that if $\int_{\Omega}f\phi=0$ for all $\phi$ whose support is compact and included in $\Omega$, then $f=0$ ?

The definition you're trying doesn't work globally. If our distributions are merely locally $L^{1}$, we need the test functions to be at least $L^\infty$ and compactly supported, which is much stronger than being locally $L^{1}$.
The $\phi$ should probably be at least continuous and compactly supported. ($C^\infty$ compactly supported is standard in some contexts)

Assuming that, here's what I'd do. For any pair $K,U$ of sets in $\mathbb{R}^{n}$ with $K$ compact, $U$ open, and $K\subset U$, we can find a "bump" $\phi$ with $0\le\phi(x)\le 1$, $\phi(x)=1$ if $x\in K$, and $\phi(x)=0$ if $x\not\in U$.
Suppose $f$ is locally $L^{1}$, and nonzero on a set of positive measure. Since $\lim_{r\to 0}\frac1{\mu(B_{r}(x))}\int_{B_{r}(x)}f=f(x)$ a.e., there is some ball $B$ with $\int_{B}f\neq0$; this ball can be made arbitrarily small, and we shrink it if necessary so that it has compact closure. Take sequences of compact sets $K_{n}$ and open sets $U_{n}$ with compact closure such that $K_{n}\subset K_{n+1}\subset B$ and $B\subset U_{n+1}\subset U_{n}$, with $\cup K_{n}=\cap U_{n}=B$. Let $\phi_{n}$ be a bump associated with $K_{n}$ and $U_{n}$. The $\phi_{n}$ converge to the characteristic function of $B$ pointwise. Since $|f\cdot \phi_{n}|\le f\cdot \chi_{U_{1}}$ and the latter is integrable, the $\int f\cdot \phi_{n}$ converge to $\int f\cdot\chi_{B}=\int_{B}f\neq 0$.
In particular, some $T_{f}(\phi_{n})$ is nonzero, and $T_{f}$ is nonzero.


modular arithmetic
function
number theory
totient function
relatively prime
number theory unsolved

[quote="Ibrogimov"]Show that for any $ m\in N$ there exist $ n\in N$ such that $ m$ divides $ n$ and every digit of $ n$ is 0 or 1.[/quote]

Let $ A$ the set of positive integers whose all digit in decimal representation are 0 or 1.

The problem is solved if I can show that for any $ a\in A$ and any prime $ p$, it exists $ b\in A$ such that $ ab\in A$ and $ b\equal{}0\pmod{p}$ :
I just have to take $ a\equal{}1$ and apply this property for each prime divisor of $ m$ in order to build $ n$.

In order to show the property :
$ 1)$ if $ p\equal{}2$ or $ p\equal{}5$, take $ b\equal{}10$
$ 2)$ if $ p\neq 2$ and $ p\neq 5$, we have $ 10^{k(p\minus{}1)}\equal{}1\pmod{p}$ $ \forall k\geq 0$
Let then $ d$ the number of digits (0 or 1) of number $ a$
Let then $ c$ the fewest positive integer such that $ c(p\minus{}1)\geq d$
Let then $ b\equal{}\sum_{i\equal{}0}^{p\minus{}1}10^{ic(p\minus{}1)}$
We have :
$ b\in A$
$ b\equal{}0\pmod{p}$ since $ b$ is sum of $ p$ numbers, each $ \equal{}1\pmod{p}$
$ ab\in A$ since $ c(p\minus{}1)\geq d$

So the property is true,

Q.E.D.

We need to prove that for every $ m\in\mathbb N$ there exists a multiple of $ m$ with only the digits $ 0,1$.

Consider the numbers $ 1,1 \plus{} 10,1 \plus{} 10 \plus{} 10^2,\dots,1 \plus{} 10 \plus{} \dots \plus{} 10^m$. There are $ m \plus{} 1$ numbers. Hence two of them must yield the same remainder upon division by $ m$. Then we have
\[ \underbrace{111\dots 11}_{a}\equiv \underbrace{111\dots 11}_{b}\pmod m\implies \underbrace{111\dots 11}_{a \minus{} b}\underbrace{000\dots 00}_{b}\equiv 0\pmod m
\]
Which was what we wanted.

Let T(A): Totient function of A.
and let 10 is relatively prime to m, then 10^T(m) = 1 mod(m), if m>2 T(m) is even
=> 10^(T(m)/2)=-1 mod m
=> n = 10^T(m)+10^(T(m)/2) if m is relatively prime to 10.
Otherwise it is direct.


Apparently this article appeared on the front page of a recent Wall Street Journal

http://online.wsj.com/article_email/SB113236377590902105-lMyQjAxMDE1MzEyOTMxNjkzWj.html

Your thoughts on it?

Asian people and a LITTLE competition don't mix.

To them, competition is in grand quantities.

They could have some fun once in a while...

[quote="WarpedKlown1335"]
They could have some fun once in a while...[/quote]

Err... except they do?

Whether they're stereotypes or not, it seems to work. I go to Vincent Massey Secondary School. A lot of you have probably heard of it -- we're one of the top schools in Canada, and we've been in the top 3 of the AMC worldwide the last couple of years. The truth is, we're about 60% Asian, and among the math teams, we're about 95% Asian.

Our teacher is Dr Bruce White (you might have heard of him too), and he holds weekly "Night Math" sessions where we do problem solving, contest preparation, that sort of thing. There are about 100 people for each of the three classes every week. I am one of only 3 non-asians. Three out of around three hundred.

AAAH! this is so disturbing. ... really really disturbing.

a. my school is becoming more asian every year. now there's nothing wrong with that, it's just that the asian:non-asian ratio at my (math/science based) school is MUCH higher than in the local community.
b. there was recently an article in reader's digest about how far behind the U.S. is falling in the great rat race- especially in math and science, and especially in middle and high school. What the heck? hello!
c. competition is WHAT MAKES US STRONGER.
d. i love the crap about the "whole child" business. i.e. SPORTS. because YOU KNOW WHAT? FOOTBALL IS GOING TO BE *SO* IMPORTANT TO YOUR CHILD WHEN HE'S TWENTY AND A MORON.
e. *bashes head on desk* why. are these people. insane?!

now, there is some "point" to this, however. stereotypes dont hold true for every person on this planet (luckily for me), but they do have a basis in fact. i've seen incredibly competitive asian parents, and i've seen incredibly unhappy asian children that aren't up to their parents' "standard," even though they are smart, beautiful, and incredibly (INCREDIBLY) talented in non-math areas. so yes, sometimes, it happens.

but still. its depressing to think that these parents are paying to take their children out of what is probably one of the best public educations in the country. that's insane.

No it's not, sometimes the best public education systems drain a child of his/her livelihood, and personally if my kid only talked about science bowl, AMC competitions, and was at a school which didn't promote stuff outside academics, I'd scream.

I'm not saying that's what Lynbrook or some other high schools are, but they're close.

[quote="WarpedKlown1335"]and was at a school which didn't promote stuff outside athletics[/quote]

Are you sure you meant athletics?

Hey Theone... it says that article is only for subscribers.... could you quote the important sections?

[quote][In Silicon Valley, two high schools with outstanding academic reputations are losing white students as Asian students move in. Why?]

Monta Vista High School's parent-teacher association, recently dissuaded a family with a young child from moving to Cupertino because there are so few young white kids left in the public schools. "This may not sound good," she confides, "but their child may be the only Caucasian kid in the class." All of Ms. Gatley's four children have attended or are currently attending Monta Vista. One son, Andrew, 17 years old, took the high-school exit exam last summer and left the school to avoid the academic pressure. He is currently working in a pet-supply store. Ms. Gatley, who is white, says she probably wouldn't have moved to Cupertino if she had anticipated how much it would change.

In the 1960s, the term "white flight" emerged to describe the rapid exodus of whites from big cities into the suburbs, a process that often resulted in the economic degradation of the remaining community. Back then, the phenomenon was mostly believed to be sparked by the growth in the population of African-Americans, and to a lesser degree Hispanics, in some major cities.

But this modern incarnation is different. Across the country, Asian- Americans have by and large been successful and accepted into middle- and upper-class communities. Silicon Valley has kept Cupertino's economy stable, and the town is almost indistinguishable from many of the suburbs around it. The shrinking number of white students hasn't hurt the academic standards of Cupertino's schools -- in fact the opposite is true.

This time the effect is more subtle: Some Asians believe that the resulting lack of diversity creates an atmosphere that is too sheltering for their children, leaving then unprepared for life in a country that is only 4% Asian overall. Moreover, many Asians share some of their white counterpart's concerns. Both groups finger newer Asian immigrants for the schools' intense competitiveness.

Some whites fear that by avoiding schools with large Asian populations parents are short-changing their own children, giving them the idea that they can't compete with Asian kids. "My parents never let me think that because I'm Caucasian, I'm not going to succeed," says Jessie Hogin, a white Monta Vista graduate.

The white exodus clearly involves race-based presumptions, not all of which are positive. One example: Asian parents are too competitive. That sounds like racism to many of Cupertino's Asian residents, who resent the fact that their growing numbers and success are causing many white families to boycott the town altogether.

"It's a stereotype of Asian parents," says Pei-Pei Yow, a Hewlett- Packard Co. manager and Chinese-American community leader who sent two kids to Monta Vista. It's like other familiar biases, she says: "You can't say everybody from the South is a redneck."

Jane Doherty, a retirement-community administrator, chose to send her two boys elsewhere. When her family moved to Cupertino from Indiana over a decade ago, Ms. Doherty says her top priority was moving into a good public-school district. She paid no heed to a real- estate agent who told her of the town's burgeoning Asian population.

She says she began to reconsider after her elder son, Matthew, entered Kennedy, the middle school that feeds Monta Vista. As he played soccer, Ms. Doherty watched a line of cars across the street deposit Asian kids for after-school study. She also attended a Monta Vista parents' night and came away worrying about the school's focus on test scores and the big-name colleges its graduates attend.

"My sense is that at Monta Vista you're competing against the child beside you," she says. Ms. Doherty says she believes the issue stems more from recent immigrants than Asians as a whole. "Obviously, the concentration of Asian students is really high, and it does flavor the school," she says.

When Matthew, now a student at Notre Dame, finished middle school eight years ago, Ms. Doherty decided to send him to Bellarmine College Preparatory, a Jesuit school that she says has a culture that "values the whole child." It's also 55% white and 24% Asian. Her younger son, Kevin, followed suit.

Kevin Doherty, 17, says he's happy his mother made the switch. Many of his old friends at Kennedy aren't happy at Monta Vista, he says. "Kids at Bellarmine have a lot of pressure to do well, too, but they want to learn and do something they want to do."

While California has seen the most pronounced cases of suburban segregation, some of the developments in Cupertino are also starting to surface in other parts of the U.S. At Thomas S. Wootton High School in Rockville, Md., known flippantly to some locals as "Won Ton," roughly 35% of students are of Asian descent. People who don't know the school tend to make assumptions about its academics, says Principal Michael Doran. "Certain stereotypes come to mind -- 'those people are good at math,' " he says.

In Tenafly, N.J., a well-to-do bedroom community near New York, the local high school says it expects Asian students to make up about 36% of its total in the next five years, compared with 27% today. The district still attracts families of all backgrounds, but Asians are particularly intent that their kids work hard and excel, says Anat Eisenberg, a local Coldwell Banker real-estate agent. "Everybody is caught into this process of driving their kids." Lawrence Mayer, Tenafly High's vice principal, says he's never heard such concerns.

Perched on the western end of the Santa Clara valley, Cupertino was for many years a primarily rural area known for its many fruit orchards. The beginnings of the tech industry brought suburbanization, and Cupertino then became a very white, quintessentially middle-class town of mostly modest ranch homes, populated by engineers and their families. Apple Computer Inc. planted its headquarters there.

As the high-tech industry prospered, so did Cupertino. Today, the orchards are a memory, replaced by numerous shopping malls and subdivisions that are home to Silicon Valley's prosperous upper-middle class. While the architecture in Cupertino is largely the same as in neighboring communities, the town of about 50,000 people now boasts Indian restaurants, tutoring centers and Asian grocers. Parents say Cupertino's top schools have become more academically intense over the past 10 years.

Asian immigrants have surged into the town, granting it a reputation -- particularly among recent Chinese and South Asian immigrants -- as a Bay Area locale of choice. Cupertino is now 41% Asian, up from 24% in 1998.

Some students struggle in Cupertino's high schools who might not elsewhere. Monta Vista's Academic Performance Index, which compares the academic performance of California's schools, reached an all-time high of 924 out of 1,000 this year, making it one of the highest- scoring high schools in Northern California. Grades are so high that a 'B' average puts a student in the bottom third of a class.

"We have great students, which has a lot of upsides," says April Scott, Monta Vista's principal. "The downside is what the kids with a 3.0 GPA think of themselves."

Ms. Scott and her counterpart at Lynbrook know what's said about their schools being too competitive and dominated by Asians. "It's easy to buy into those kinds of comments because they're loaded and powerful," says Ms. Scott, who adds that they paint an inaccurate picture of Monta Vista. Ms. Scott says many athletic programs are thriving and points to the school's many extracurricular activities. She also points out that white students represented 20% of the school's 29 National Merit Semifinalists this year.

Judy Hogin, Jessie's mother and a Cupertino real-estate agent, believes the school was good for her daughter, who is now a freshman at the University of California at San Diego. "I know it's frustrating to some people who have moved away," says Ms. Hogin, who is white. Jessie, she says, "rose to the challenge."

On a recent autumn day at Lynbrook, crowds of students spilled out of classrooms for midmorning break. Against a sea of Asian faces, the few white students were easy to pick out. One boy sat on a wall, his lighter hair and skin making him stand out from dozens of others around him. In another corner, four white male students lounged at a picnic table.

At Cupertino's top schools, administrators, parents and students say white students end up in the stereotyped role often applied to other minority groups: the underachievers. In one 9th-grade algebra class, Lynbrook's lowest-level math class, the students are an eclectic mix of whites, Asians and other racial and ethnic groups.

"Take a good look," whispered Steve Rowley, superintendent of the Fremont Union High School District, which covers the city of Cupertino as well as portions of other neighboring cities. "This doesn't look like the other classes we're going to."

On the second floor, in advanced-placement chemistry, only a couple of the 32 students are white and the rest are Asian. Some white parents, and even some students, say they suspect teachers don't take white kids as seriously as Asians.

"Many of my Asian friends were convinced that if you were Asian, you had to confirm you were smart. If you were white, you had to prove it," says Arar Han, a Monta Vista graduate who recently co-edited "Asian American X," a book of coming-of-age essays by young Asian- Americans.

Ms. Gatley, the Monta Vista PTA president, is more blunt: "White kids are thought of as the dumb kids," she says.

Cupertino's administrators and faculty, the majority of whom are white, adamantly say there's no discrimination against whites. The administrators say students of all races get along well. In fact, there's little evidence of any overt racial tension between students or between their parents.

Mr. Rowley, the school superintendent, however, concedes that a perception exists that's sometimes called "the white-boy syndrome." He describes it as: "Kids who are white feel themselves a distinct minority against a majority culture."

Mr. Rowley, who is white, enrolled his only son, Eddie, at Lynbrook. When Eddie started freshman geometry, the boy was frustrated to learn that many of the Asian students in his class had already taken the course in summer school, Mr. Rowley recalls. That gave them a big leg up.

To many of Cupertino's Asians, some of the assumptions made by white parents -- that Asians are excessively competitive and single-minded -- play into stereotypes. Top schools in nearby, whiter Palo Alto, which also have very high test scores, also feature heavy course loads, long hours of homework and overly stressed students, says Denise Pope, director of Stressed Out Students, a Stanford University program that has worked with schools in both Palo Alto and Cupertino. But whites don't seem to be avoiding those institutions, or making the same negative generalizations, Asian families note, suggesting that it's not academic competition that makes white parents uncomfortable but academic competition with Asian-Americans.

Some of Cupertino's Asian residents say they don't blame white families for leaving. After all, many of the town's Asians are fretting about the same issues. While acknowledging that the term Asian embraces a wide diversity of countries, cultures and languages, they say there's some truth to the criticisms levied against new immigrant parents, particularly those from countries such as China and India, who often put a lot of academic pressure on their children.

Some parents and students say these various forces are creating an unhealthy cultural isolation in the schools. Monta Vista graduate Mark Seto says he wouldn't send his kids to his alma mater. "It was a sheltered little world that didn't bear a whole lot of resemblance to what the rest of the country is like," says Mr. Seto, a Chinese- American who recently graduated from Yale University. As a result, he says, "college wasn't an academic adjustment. It was a cultural adjustment."

Hung Wei, a Chinese-American living in Cupertino, has become an active campaigner in the community, encouraging Asian parents to be more aware of their children's emotional development. Ms. Wei, who is co-president of Monta Vista's PTA with Ms. Gatley, says her activism stems from the suicide of her daughter, Diana. Ms. Wei says life in Cupertino and at Monta Vista didn't prepare the young woman for life at New York University. Diana moved there in 2004 and jumped to her death from a Manhattan building two months later.

"We emphasize academics so much and protect our kids, I feel there's something lacking in our education," Ms. Wei says.

Cupertino schools are trying to address some of these issues. Monta Vista recently completed a series of seminars focused on such issues as helping parents communicate better with their kids, and Lynbrook last year revised its homework guidelines with the goal of eliminating excessive and unproductive assignments.

The moves haven't stemmed the flow of whites out of the schools. Four years ago, Lynn Rosener, a software consultant, transferred her elder son from Monta Vista to Homestead High, a Cupertino school with slightly lower test scores. At the new school, the white student body is declining at a slower rate than at Monta Vista and currently stands at 52% of the total. Friday-night football is a tradition, with big half-time shows and usually 1,000 people packing the stands. The school offers boys' volleyball, a sport at which Ms. Rosener's son was particularly talented. Monta Vista doesn't.

"It does help to have a lower Asian population," says Homestead PTA President Mary Anne Norling. "I don't think our parents are as uptight as if my kids went to Monta Vista."[/quote]
-found on proquest newstand

oh my GOD!  [b]I[/b] go to the Thomas Wootton High School mentioned in the article!  what the heck?  there's nearly no pressure here at all.

But our extremely awesome principal got quoted, which I do love.

[quote="joml88"][quote="agolsme"]Asian parents are uptight in general - that's true. Having gone to middle school in silicon valley(Almaden Valley in San Jose specifically - about 30 minutes away from those two schools) much of what the article says about asians being concerned about test scores rings true. However, I find it unlikely that asians would be seen as de facto smarter than white students. From my own personal experiences, smart kids were given credence regardless of their race.

In fact, I think the article is sort of offensive in the sense that it possesses the underlying tone of assimilation - "crap, those asians are too competitive; they're isolating themselves from the beer, parties, and stupor of regular high school!"

Anyways, what's wrong with a little competition? Just because the school is competitive doesn't mean you need to buy into the general culture.[/quote]

There's a difference between what the article asserts to be fact and what the article asserts as the reasoning behind the white flight.  Even thought a lot of stereotypes are discussed, the author doesn't claim any of them to be true.

Also note that the author has an asian name.  I highly doubt that the author holds the Asian stereotypes to be true.[/quote]

While you're right that the author doesn't claim any particular stereotypes to be true, she/he disproportionately mentions one perspective.

There are 3 positive/neutral views of "the white flight" - 1. Pei Pei Yow 2. The principals of the schools (April Scott and something else) 3. Judy Hogin

The rest of the article, which quotes quite a few more people, offers discouraging views of asian "competitiveness" and culture.

I'm moving this discussion into the Round Table since it's a tad more involved than a normal G&F thread would be.

Also, the link provided in the first thread requires subscription to access. Where might people who do not subscribe to the WSJ read the article?

(Or is that what was posted earlier?)

It looks like agolsme posted the article.  WSJ must have a policy that after so many days they archive the articles so that only registered users can view it because the article was viewable to non-subscribers when this thread originated.

[quote="agolsme"]While you're right that the author doesn't claim any particular stereotypes to be true, she/he disproportionately mentions one perspective. [/quote]The whole article is devoted to a "flight" phenomenon which itself disproportionately involves just one perspective: that it is better for certain children to avoid being placed in an Asian pressure cooker.   Whatever the realities of the underlying educational situation, it is the existence of that negative perception and the resulting flight which is recent and newsworthy.  It is a "second-order effect" of the endlessly documented Asian scholastic achievement, which has for years been the subject of glowing media reports.  I don't think there is a major issue of balance here, if you also consider those hundreds of other (overwhelmingly positive) reports and that the WSJ author is dealing with the new secondary phenomenon.

I don't have much sympathy for the American parents in the story, but for the Asian parents there are other politically incorrect complications that the story does not mention, such as the huge social benefit of a majority Asian school (less assimilation to the awful American average, child not raised with "minority" or immigrant neuroses, girls dealing less with Asian fetishes, etc), against the comparably awful possibility of placing the child in an harmful educational arms race.

I can't veiw the article. It says I have to be a subscriber. :( EDIT: Nvm I just read what you said later above.

[quote="bleumoose"]AAAH! this is so disturbing. ... really really disturbing.
b. there was recently an article in reader's digest about how far behind the U.S. is falling in the great rat race- especially in math and science, and especially in middle and high school. What the heck? hello!
[/quote] Yeah I read that article too.

[quote="daveed"]You know, I talked to my principal a few days ago about his quote in the article.  Apparently, the author didn't use any of the things he said(that, though wootton's a diverse school with many asians, it doesn't turn out as a problem).  The only reason they interviewed him was because of the "wonton" thing.

that makes me mad.  stupid journalists[/quote]The journal(ist) is not obligated to provide a forum for the principal or any other interviewed person, and the article was not devoted specifically to his school.   Wootton's principal was quoted as contesting certain perceptions of the school as being stereotypes, which along with the remarks you mention is just the kind of bland PC mush that one would have expected from a principal and also appears several times from other interviewees.  Thus, he appears to have been quoted [i]ex officio[/i], and none of his other remarks sound worthy of space in the article.  Had he said something more informative about the subject of the article, he probably would have been quoted at greater length.

Lmao, I go to Monta Vista. 81% Asian guaranteed.

I go to Highland Park Elementary. $\frac 12$ minority groups, $\frac 12$ white. It is very diverse.


limit
real analysis
real analysis unsolved

Please take a look and tell me if it's something wrong... If somebody knows another way please tell me! Thx!
I must prove that $\lim_{x\to\infty}\left( 1+\frac{k}{x}\right)^{x}=e^{k}$

Let $\frac{k}{x}=\frac{1}{u}$ and $x=ku$
Then $\lim_{x\to\infty}\left( 1+\frac{k}{x}\right)^{x}=\lim_{u\to\infty}\left( 1+\frac{1}{u}\right)^{ku}=\lim_{u\to\infty}\left[ \left( 1+\frac{1}{u}\right)^{u}\right]^{k}=e^{k}$

It might be a stupid question but: is there a problem doin $ku\to \infty\Rightarrow u\to \infty$ ? Can I just change?
 :blush:

I think that if K is fix then it's ok :)


geometry
3D geometry
calculus
derivative
function
inequalities

[hide="1"]
Call the numbers a and b.

A. $ ab$ is a maximum when the two numbers are as close as possible.  Hence $ 10\cdot10\equal{}100$

B. $ a^2\plus{}b^2$ is a minimum when $ ab$ is a maximum because $ (a\plus{}b)^2\equal{}a^2\plus{}b^2\plus{}2ab\equal{}400$ is constant.  Hence $ 400\minus{}200\equal{}200$

C. $ a^5\minus{}40a^4\plus{}400a^3$, graph it.  In other words, I don't know.
[/hide]
[hide="2"]
$ f(x)\equal{}ax^2\plus{}bx\plus{}c$ reaches its maximum at $ (\frac{\minus{}b}{2a},f(\frac{\minus{}b}{2a}))$
$ \equal{}15$ in this case.  This is where the function begins to go down, i.e. CK's car is rolling down.
No, it's not a trick, there's always an easy problem hidden in the lot.
[/hide]

To find the points you could always use basic calculus

First derivative at 0 (or undefined) indicates a critical point (either maximum, minimum, or neither)
Second derivative indicates whether the critical point is a maximum or minimum

for example for 1c

With $ a^5\minus{}40a^4\plus{}400a^3$ simply take the derivative and set to 0:
$ 0\equal{}5a^4\minus{}160a^3\plus{}1200a^2$
$ 0\equal{}a^4\minus{}32a^3\plus{}240a^2$
solving you get
$ a\equal{}0$, $ a\equal{}12$, $ a\equal{}20$

then you take the second derivative
$ 4a^3\minus{}96a^2\plus{}480a$
$ a^2\minus{}24a\plus{}120$
(You can factor out $ a \geq 0$ and 4)

Which checking you find the solution is $ a\equal{}12$ for maximizing it.

Kevin

There is a calculus forum.  Please put your calculus problems there in the future.

[quote="stack3d"]Ugh, I remember doing this before but can't find my notes on it and can't think of anything right now.

The sum of two positive integers is twenty.Find the numbers if
a. their product is a maximum
b. the sum of their squares is a minimum
c. the product of the square of one and the cube of the other is a maximum

Also I have this problem.

CK's car runs out of gas as it is going up a hill. The car rolls to a stop then starts rolling backwards. As it rolls it's displacement d(t) in meters from the bottom of the hill at t seconds since CK's car ran out of gas is given by d(t) = 99+30t - t^2. At what time did CK's car stop rolling up and start rolling down
the hill? How far was he from the bottom of the hill at this time?

I just found the vertex for this one, but it seems too easy for the last problem compared to the rest of the assignment which are mostly about derivatives and critical points. Am I missing something?[/quote]

[hide="1"]AM-GM: $ 10\equal{}\frac{a\plus{}b}{2}\ge\sqrt{ab} \implies ab\le 100$.[/hide]
[hide="2"]RMS-AM: $ \sqrt{\frac{a^2\plus{}b^2}{2}}\ge\frac{a\plus{}b}{2} \equal{} 10\implies a^2\plus{}b^2 \ge 200$.[/hide]

3. $ a^3(20\minus{}a)^2 \equal{} a^5\minus{}40a^4\plus{}400a^3$... yes, calculus works... there is probably a better way with weighted power mean though...


Maybe there should be an astronomy board.  :)

Yes that would be nice, however I guess one may always make some attempt to say we may not need one.  :wink:

:lol: Thanks funcia. Though some will probably come up with reason against it.

You need to gather supporters for that :)

Oh yeah, I better [i]write[/i] I support...
Even though I am not that interested in astronomy a few people I know are obsessed with it.
Maybe they'd want to AoPS!

Astronomy can be used to tell if the sun will explode or not... :P

I definitely support.(I have a Mac, and I just got a star-gazing widget.)

If this matters to anyone, there a meteor shower peaking on May 6th.
They are a remenant from Halley's Comet and called the eta Aquarids.

Stuff like that could be discussed there  :) among other things.

An astronomy board's a great idea. People can talk about astronomy news (like AstroPhys was just doing) and cosmology topics that don't really fit in anywhere else.

For example, I wanted to post a random thought I had about dark matter (see "Coffee Cup Analogy"), but I didn't know where to put it. After thinking for a while, I finally decided to put it in Introductory Physics because the post guidelines there mentioned "fun" topics. As I didn't have any equations or anything--it was just an idea--I thought maybe the topic could be considered to be at an introductory level. Now it's sitting among a bunch of friction, light refraction, and harmonics problems; it doesn't really seem like it fits there.

I think an astronomy board would be an excellent place for topics like my dark matter one, as well as for news and for astronomical events. :thumbup:

[quote="AstroPhys"]News
Scientists have observed the biggest supernova than any that had ever been previously.[/quote]

Where? When? What star?

It happened in NGC 1260, 240 million years ago, and they are calling it SN 2006gy.
 :)

Another thing. I just noticed that AoPS has an engineering board, in which there are only two threads and eight posts, and the most recent post in it is from the tenth of April. I know an astronomy board would have more activity than that, so why does AoPS have an engineering board and not an astronomy one? :roll:

Okay, I've made it so that forum will also be about Astronomy: http://www.mathlinks.ro/Forum/index.php?f=449


Let $ \omega$ be a circle.
$ \omega$ is tangent on the sides $ AC$ and $ AB$ of the triangle $ ABC$ in the points $ B'$ and $ C'$.
$ \omega$ intersects  the side $ BC$ in the points $ A'$ and $ A"$, in the order $ B,A',A",C$.Let $ P_1,P_2,P_3$ the intersections of the lines $ BC$ and $ B'C'$ ,
$ AC$ and $ A'C'$ ,  $ AB$ and $ A"B'$. Prove that $ P_1,P_2,P_3$ are collinear points.


In the square shown, each curve is an arc of a circle with radius 4, having centers at the vertices of the square. Find the number of square units in the area of the football-shaped portion of the figure. Express your answer in terms of $ \pi$.
[asy]draw((0,0)--(4,0)--(4,4)--(0,4)--cycle);
draw((0,0)..(.8,2.4)..(1.6,3.2)..(4,4));
draw((0,0)..(2.4,.8)..(3.2,1.6)..(4,4));
draw((0,4)--(2.828,1.172));
label("4",(1.414,2.586),SW);[/asy]

The shape at the corner bordering the football is equal to the squares area - the quarter circle, which is
$ 4^2 \minus{} (\frac 14\times 4^2 \pi )\equal{}16 \minus{} 4\pi$.
Since there are 2 of those curvy shapes, the total area of both of them is $ 32\minus{}8\pi$. The football area is the area of the square - the two curvy places, so the area of the football shape is $ 16\minus{}(32\minus{}8\pi)\equal{}8\pi\minus{}16$


How to derive $x(t)=A\cos (\omega t+\phi)$ from $F=-kx$ mathematically, i.e. how to solve $m\ddot x=-kx$?

You can either plug the solution into the differential equation and prove that the identity holds (this is a perfectly legitimate method!), or you can separate variables:
\[m\ddot x =-kx \Leftrightarrow \]

\[m \frac{\text dv}{\text dt}=-kx\Leftrightarrow \]

\[mv \frac{\text dv}{\text dx}=-kx\Leftrightarrow \]

\[mv\text dv =-kx \text dx. \]
Integrating this gives
\[\frac{1}{2}mv^{2}+\frac{1}{2}kx^{2}= E = \text{const.}\]
which is in fact conservation of energy. But,
\[\text dt = \frac{\text dx}{v}= \frac{\text dx}{\sqrt{\frac{2E}{m}-\frac{k}{m}x^{2}}}. \]
Integrating this, by using
\[\int\frac{\text dx}{\sqrt{1-x^{2}}}= \arcsin x+\text{const.}\]
we get the dependence $t(x)$ which has the form of an arcsin function; inverting this we get the sine (or cosine) dependence $x(t)$.

or you could solve the ODE by substituting $y = e^{(}rt)$ and changing complex exponents to trig functions...

[quote="Djole"]... which is in fact conservation of energy. But,
\[\text dt = \frac{\text dx}{v}= \frac{\text dx}{\sqrt{\frac{2E}{m}-kx^{2}}}. \]
Integrating this ...
[/quote]
I think it should be
\[dt=\frac{dx}v=\sqrt{\frac m{2E-kx^{2}}}\, dx\]
And, if you set $E=\frac12 kA^{2}$, then
\[dt=\frac{dx}{\sqrt{\frac{k}{m}(A^{2}-x^{2})}}\]
Hence
\[t=\sqrt{\frac{m}{k}}\arcsin\frac{x}{A}+\text C \]
i.e.
\[x=A\sin\sqrt\frac{k}{m}(t-\text C) \]

Yes, I forgot to add another $m$ in the equation. I'll fix that now.


Let $ a,b,c \in [\minus{}1,1]$ such that $ a\plus{}b\plus{}c\equal{}0.$ Prove that
$ \sqrt{1\plus{}a\plus{}\frac{b^2}{6}}\plus{}\sqrt{1\plus{}b\plus{}\frac{c^2}{6}} \plus{}\sqrt{1\plus{}c\plus{}\frac{a^2}{6}} \le 3.$
My solution used Cauchy Schwarz but it is not nice, so I am looking for a nicer solution. :)

[quote="phymax"]we know that,

$ \frac {(1 \plus{} a \plus{} \frac {a^2}{6}) \plus{} (1 \plus{} b \plus{} \frac {b^2}{6}) \plus{} (1 \plus{} c \plus{} \frac {c^2}{6})}{3} \geq\ (\frac {\sqrt {1 \plus{} a \plus{} \frac {b^{2}}{6}} \plus{} \sqrt {1 \plus{} b \plus{} \frac {c^{2}}{6}} \plus{} \sqrt {1 \plus{} c \plus{} \frac {a^{2}}{6}}}{3})^2$

$ \rightarrow\ \frac {3 \plus{} 0 \plus{} \frac {1}{6}(a^2 \plus{} b^2 \plus{} c^2)}{3} \geq\ (\frac {\sqrt {1 \plus{} a \plus{} \frac {b^{2}}{6}} \plus{} \sqrt {1 \plus{} b \plus{} \frac {c^{2}}{6}} \plus{} \sqrt {1 \plus{} c \plus{} \frac {a^{2}}{6}}}{3})^2$

the min. of $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 0$ (as per your set)

so the ineq. becomes
$ 1 \geq\ (\frac {\sqrt {1 \plus{} a \plus{} \frac {b^{2}}{6}} \plus{} \sqrt {1 \plus{} b \plus{} \frac {c^{2}}{6}} \plus{} \sqrt {1 \plus{} c \plus{} \frac {a^{2}}{6}}}{3})^2$

$ 1 \geq\ \frac {\sqrt {1 \plus{} a \plus{} \frac {b^{2}}{6}} \plus{} \sqrt {1 \plus{} b \plus{} \frac {c^{2}}{6}} \plus{} \sqrt {1 \plus{} c \plus{} \frac {a^{2}}{6}}}{3}.$

and hence the result
(hope it is satisfactory  :) )[/quote]

This is not correct since you have to find upper bound for $ a^2 \plus{} b^2 \plus{} c^2$ not lower

[quote="Potla"]Well, I am doubtful if my solution is right,.....
[hide]
From CS,
$ \left(\sum \sqrt {(1 \plus{} a \plus{} \frac {a^2}{6})}\right)^2\leq \left(1 \plus{} 1 \plus{} 1 \plus{} a \plus{} b \plus{} c \plus{} \sum\frac {a^2}{6}\right)(1 \plus{} 1 \plus{} 1)$.
Now $ a^2 \plus{} b^2 \plus{} c^2\ge 0\implies :$
$ (\sum \sqrt {(1 \plus{} a \plus{} \frac {a^2}{6})})^2\le 3\cdot3\implies \sum \sqrt {1 \plus{} a \plus{} \frac {a^2}{6}}\le \sqrt {9} \equal{} 3$[/hide]
The solution of Phymax & mine are similar; but I am doubtful of my solution because you rarely present easy problems, can_hang! :o  :(
[hide="@can_hang"]I hope you will talk to me through pm again[/hide][/quote]Same comment as cefer. :wink:

[quote="phymax"]if checked carefully,
by applying upper bound of $ a^2 \plus{} b^2 \plus{} c^2$ the $ LHS$ shifts towards more positive than the $ RHS$ : the reason bieng that in $ LHS$ the quantities are open while in $ RHS$ the are confined in a root..
so it practically is of not much use..(thats what i feel :) )[/quote]

If you check carefully you can not take $ a^2\plus{}b^2\plus{}c^2\equal{}0$
You have two things $ A \geq B$ and $ A \geq C$. From here you can not say that $ B \geq C$ or vice versa.

check your solution one more time:)

@phymax: you can't assume $ a^2\plus{}b^2\plus{}c^2\equal{}0$. If you do that, then $ a\equal{}b\equal{}c\equal{}0$ and nothing is proved except this case.

[quote="phymax"]you are right my dear friend. but in this case we dont observe anyhing like this explcitly.
(or if not so then please let me know where i am mistaken :oops: )

i still feel that my proof is very much complete..and that we should consider the minimum value of $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 0$
rather than the upper bound which is nothing but the maximum possible value of $ 2(c^2 \minus{} ab)$ with $ a,b,c \in [ \minus{} 1,1]$ & shifting the $ LHS$ to more positive and making it weaker.~[/quote]

But I think your solution is not complete and I am sure!
Just if you had an additional information $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 0.5$
can you still prove this inequality with your method? No.
So your solution works only when $ a\equal{}b\equal{}c\equal{}0$.
Just think few minutes carefully and you will see where is the problem:)

So, phymax, please tell me why the rest gives the inequality.

[quote="phymax"]yes obviously because all the rest give greater than $ 0$.
and this case proves the $ equality$. rest all give the $ inequality$!.
please check if i am wrong in saying this :?:

or please send a complete solution so that my understanding becomes broader~[/quote]
When you solve an inequality $ A \geq B$ you can prove that  $ A\geq C$ & $ C \geq B$ for some $ C$

Here you have to prove:$ A \equal{} 3 \geq \sqrt {1 \plus{} a \plus{} \frac {b^{2}}{6}} \plus{} \sqrt {1 \plus{} b \plus{} \frac {c^{2}}{6}} \plus{} \sqrt {1 \plus{} c \plus{} \frac {a^{2}}{6}} \equal{} B$

You just proved that $ C  \geq B$
Now to complete you have to prove that $ A \equal{} 3 \geq  C$
which is not always correct.

[quote="can_hang2007"]Let $ a,b,c \in [ \minus{} 1,1]$ such that $ a \plus{} b \plus{} c \equal{} 0.$ Prove that
$ \sqrt {1 \plus{} a \plus{} \frac {b^2}{6}} \plus{} \sqrt {1 \plus{} b \plus{} \frac {c^2}{6}} \plus{} \sqrt {1 \plus{} c \plus{} \frac {a^2}{6}} \le 3.$
My solution used Cauchy Schwarz but it is not nice, so I am looking for a nicer solution. :)[/quote]
Thank you very much everyone for spending time on this problem. Here is the ugly proof which I mentioned


what is the formal definition of a limit ?

[quote="MathWorld"]A function f(z) is said to have a limit [tex]\displaystyle \lim_{z\to a}f(z) = c[/tex] if, for all [tex]\epsilon > 0[/tex], there exists a [tex]\delta > 0[/tex] such that [tex]|f(z) - c| < \epsilon[/tex] whenever [tex]0 < |z - a| < \delta[/tex].[/quote]

This version is a bit hard to understand.  I'll go find another definition unless someone else does first.


The sequence 2, 3, 5, 6, 7, 10, 11, $ \ldots$ contains all the positive integers from least to greatest that are neither squares nor cubes. What is the $ 400^{\mathrm{th}}$ term of the sequence?

Obviously, $ n\geq400$.

First, assume that $ 400\leq n < 441$.

For all $ n$ in this range, there are $ 20$ squares, $ 7$ cubes, and $ 2$ perfect sixth powers, and $ 20\plus{}7\minus{}2\equal{}25$ numbers not in the sequence.

Then we must have $ n \minus{} 25 \equal{} 400\implies n \equal{} \boxed{425}$, which indeed satisfies the aforementioned range.

There are no other possible values of $ n$, since we have already passed the $ 400$th term! :P

Official answer:

Let's try counting the number of perfect squares and cubes less than 441=21^2. There are twenty perfect squares less than 441: 1^2, 2^2, \ldots, 20^2. There are also seven perfect cubes less than 441: 1^3, 2^3, \ldots, 7^3. So there would seem to be 20+7=27 numbers less than 441 which are either perfect squares and perfect cubes.

But wait! 1=1^2=1^3 is both a perfect square and a perfect cube, so we've accidentally counted it twice. Similarly, we've counted any sixth power less than 441 twice because any sixth power is both a square and a cube at the same time. Fortunately, the only other such one is 2^6=64. Thus, there are 27-2=25 numbers less than 441 that are perfect squares or perfect cubes. Also, since 20^2=400 and 7^3=343, then all 25 of these numbers are no more than 400. To compensate for these twenty-five numbers missing from the list, we need to add the next twenty-five numbers: 401, 402, \ldots, 424, 425, none of which are perfect square or perfect cubes. Thus, the 400^{\text{th}} term is \boxed{425}.

Why do we pick the number 441? What is the reason for this?

It's $21^2$, a good starting point. It's unlikely from the start that it will be more than that.

I guess you have to find the perfect squares and perfect cubes first.

[quote=xiaomage]Please hide your solutions.[/quote]

omg. these posts are from 2010 by people who are probably inactive. this is a useless time to mention it.

[hide=Solution]
We know that the 400th term of the sequence 2, 3, 4, 5, 6, ... is 401. So if we remove all the squares and cubes from this sequence, the number we subtract is the amount that the 400th term goes up by.

The squares of the numbers from 2-20 are all under 401, as well as the cubes of the numbers from 2-7. (Note that we do not include 1 in these ranges because 1 is not included in the given sequence.) So it would seem that we need to remove 19+6=25 squares or cubes from the sequence. But, since the cube of 4 is equal to the square of 8, we count one number twice, so we need to remove 24 numbers instead. Therefore, the 400th term of the given sequence is 401+24=[b]425[/b].
[/hide]