ARML MATHCOUNTS calculus AMC USA(J)MO USAMO AIME

Hi, my name is Jeffrey and I'm the moderator of this forum.

I don't see too many others from Washington post on the forums, but who knows? If you see this post, please respond so I can get a feel of who comes on.

A little about myself. I go to Lewis and Clark High School in Spokane. We're a little out of the way of the main competitions on the west side and we don't have a math team anymore. However, in the past we were pretty active (we started Math is Cool and wrote all the nonhigh school  problems until this year).

If you see this, post and tell us a little about yourself and we'll go from there.

Thanks.
Jeffrey

Hello, I'm Adam Hesterberg. I check the for[b]a[/b] about once every 4 months, but I saw this.  I go to Garfield High School. I won National MathCounts in 2003 and went to MOP as a ninth grader (the real MOP, not red) last year.

-Adam

Hi, I'm Nathan Pflueger. I am the President of the Garfield High School math team, and generally place pretty well at state-wide contests. I also attended MOP in 2003. Other than that I haven't really done anything notable, like Kamior has with Mathcounts.

By the way, Kamior, as much as you may like Germany, I think the flag is supposed to be your actual country...

[quote="Kamior"]The entry in profile just said "country." If everyone in aops did that, most of the flags would be from the US, so I chose the country I'd want to move to instead.

-Adam[/quote]Not quite. You're forgetting that AoPS merged with MathLinks about 6 months ago, so the number of US - non-US users has been equilibrated. Country field should thus refer to your current residence, and not your natal country ;). We will soon add state flags for you to choose if you select US as your current residence.

PS Sorry for barging in like this Jeffrey :P If this is out of order, feel free to forget that I'm an admin, and delete this post.

Well, while evangelical Christians control that flag, I won't use it, so I've just reset the country field.

Well, its nice to see the two top math students in Washington here. But it doesn't look like there's too many others.

By the way, I think Washington has an ARML team, but how do you get selected for it?

My name is Nathan and I am homeschooled. I have won the shoreline math olympiad in 2002, 03 and 04 and our team got 3rd, 1st and 4th in those years. I am in 7th grade.

Edit: 05 as well and team got i think 3 or 4

My name is Meir Lakhovsky. I go to Redmond Junior High. I do some statewide contests.

hi i am young jeong. i do some statewide math contest also, and i just found out about this awshington site.

LOL Adam - exactly the same reason why I'm not using the flag of the US.

BTW - what exactly is the math curriculum at Garfield High? Mr. Pounder told me that it went from Calc AB to Calc BC to AP Stats but that's like 2 years of wasted time considering that all three of these topics can be covered in the same year. Do you take math at the UW or something? Because I'm just about to exhaust my math at Inglemoor and would like to take a math course that colleges give credit for (A lot of colleges aren't fans of Stanford EPGY for some reason).

-----

Alex C here - I'm still not exceptional but am making rapid progress (even though I still waste so much time on collegeconfidential and other forums). I'm one of those sophomore AIME qualifiers taking calculus - while I'm still well-above average, I'm nothing like you AOPS folks. I go to the same school as William Johnson, a USAMO qualifier from Inglemoor. I want to be second place in the school although it'll still take some more improvement from my erstwhile hardcore computer gaming habits. Unfortunately I still have to rely on external motivation but hopefully it'll change for next summer.

I really want to see Inglemoor do better in math competitions next year (when WIll Johnson's the only exceptional one from the school, it doesn't help much)  but sadly, the team's not very much into math, meeting only during its occasional competitions. Neither is the school. I dn't exactly know how I could promote interest in math considering that the people there are into "well-roundedness" and sports.

Came from hardcore computer gamer with 540 on Math SAT I and only taking Integrated I (I didn't even make the MATHCOUNTs team and my school was still owned)  in 8th grade to 640 Math SAT I and less than 80 on AMC 10 in 9th grade to 104.5 AMC 12A taking Calculus 76 PSAT Math and 800 Math IIC in 10th. Yeah I still suck compared to most of you but if I can stop wasting so much time I might have a shot at USAMO next year (that student whose AIME went from 0 to 7 in one year inspires me).

hmm.... from Inglemoor... i sure remember Will Johnson but not you. ANuyways, welcome to AOPS.

I barely missed top 10 on the Northwest Championships(and it hurts that I got #1 wrong too). Should have actually prepared. But not a big deal 'cuz the competition really ins't very useful.

BTW reddevil - is Thomas Jefferson really into math? There were 13 teams - and a lot of them placed quite poorly -  seems like the school made a pretty good effort in getting people into the competition.

we are into math, except not that many people are very prepared at hard competitions like NMC or Math Is Cool or AMC. We are mainly focused on Mu Alpha Theta competitions.... I was lucky enough to find this site early and study on my own. I think I am the best among my school in AMC-type test, but i do suffer a little bit on Mu Alpha Theta competition tests, because they do not test on fundamental skills, but how much you learned at school (known as the "class" tests). Again, I was lucky enough to find out that MAT is not the best competition and reacted fast b4 i ran out of time (buying useful books, studying AMC alot, among with AIME).... I am bound to qualify for AIME this year. the only one with even a chance at our school.

Ahh, I see. Yeah - it sucks that sometimes coaches trust these classroom based competitions more than the real problem-solving competitions. By talking about qualifying for the AIME - do you mean USAMO? Since we all have taken the AIME by now.

Ah, didn't think so. Yes, Eric plays tennis. And pretty much everything else. My teams always have people from completely different schools.

Hey, I'm Qui Nguyen. I'm a crazy middle schooler at Washington (go Pounder), but I'm going to Garfield in the fall.

Hi Qui! It's Jane. I told you to join like last year at Gresham (hope I'm not mistaking you for your sister).

Pounder says he'll try to get me into Garfield.

Hey, you are mistaking me for my sister!  :mad: Don't worry, I'm not really mad. But Qui = alto sax and Khoi = viola, and Khoi hasn't joined yet.

You should get into Garfield next year! Then you can be on our 9/10 team, since everyone else I know in 8th grade is trying to go to TS (Andy M, Jack C, Kenneth L, and Sam B, and Mark B is already there).  :(

Oh, I'm sorry about mixing you two up!  :blush:
Khoi=viola, Qui=alto sax (have to remember that!)

About the whole TS thing, I don't think you should be that worried because the chances all or most of them will get in are very very very slim.

[quote="janieluvsmusic"]Kaiying said Interlake wasn't very good...[/quote]
wow.. sorry for not replying to this. Yeah, our #1 team has strong people, but we started sucking lol. And the rest of our teams aren't supergood, most people don't even show up \=
As for recruiting, none of us on team 1 can transfer schools cuz we don't even live in the BSD area and are only at IHS because of the whole GATE thing.  :P

[quote="PigsOnTheWing"]As for recruiting, none of us on team 1 can transfer schools cuz we don't even live in the BSD area and are only at IHS because of the whole GATE thing.  :P[/quote]

Moving is a necessary sacrifice.   :P

Psh. Joy should SO go to Interlake. All because I'm there ;P
Joy! I never see you in the morning anymore! (Though I've only seen you like... 10 times in the hall)

*Oink* *Oink* *Snort* *Snort*
PigsOnTheWing, do I even know who you are?
(I'm hopeless at keeping track on anything, much less people's usernames)

Hello. I'm Andrew Morgan, from Washington Middle School (Seattle), but only for one more month or so...

*visits*

hm, this forum isn't much more active than it was back in 2005.


calculus
integration
calculus computations

Im having trouble with this problem:

the integral of (x^2)(sin[2x])dx

what I wind up with is:

(x^2)(1/2)cos[2x] - sin[2x]

is that right? something seems off when I evauluate that.

[quote="amirhtlusa"]you have to use integration by parts twice, for the first one let
$u=x^2,dv=sin2xdx$. then you will end up with another integral that you  have to use by parts again. you let $u=x, dv=\frac {cos(2x)}{2}$.
the final answer would be
$\frac{-x^2cos(2x)}{2}+\frac {xsin2x}{2}+\frac  {cos2x}{4}$
you can simplfy it...[/quote]

edited by kunny


$\textrm{A farmer has }$ $60$ $\textrm{ meters of fence with which to build a rectangular animal run for her cows, horses, and pigs.}$ $\textrm{She wants each type of animal to have the same area.}$ $\textrm{What is the largest number of square meters that can be enclosed?}$ $\textrm{Express your answer as a decimal.}$


hey just for your info im really bad at searching the web and i cant find any sites with sprint rounds before 2000, and ive already exhauseted yashar.biz/mathcounts so are there any that you all would recomend???

2000 sprint was the earliest I could find, sorry.

Well if you're desperate, I remember EducAide Softwares had the complete MathCounts library for sale for about $ \$$120 (thats a lot of tests, and a lot of money).

Like how many tests and where can i find that.  Do they come on CD or like actual tests.


Let n be a positive integer and $M$ be a set containing $n^2+1$ positive integers and having the following property:in each $n+1$ numbers from the set $M$ there are two numbers for which one is divisible by the other. Prove that the set $M$ contains different numbers $a_{1},a_{2},...,a_{n+1}$ for which $a_{i+1}$ divisible $a_{i}$ for $i=1,...,n$
if it has appeared in the forum ,can you tell my the number of the forum.Thanks  :(  :(

It's a particular case of Dilworth's Lemma:

In a set with $mk+1$ elements of a partially ordered set, we can find either a chain with $k+1$ elements of an antichain with $m+1$ elements.

Here, the partially ordered set is the set of naturals, where we put $a\le b$ iff $a|b$. We also have $m=k=n$, and we are told that in every subset with $n+1$ elements, two of them are comparable, so there are no antichains with $n+1$ elements. By the lemma, there must be a chain with $n+1$ elements, and this is precisely what we have to show.

For each member $ x \in M $ look on all sequences $ x_{1} := x, x_{2} , \ldots , x_{m} $,
such that $ x_{k+1} | x_{k} $ for $ k=1,2, \ldots , m-1 $. Define a degree of such an
element $ x \in M $ the maximal length of such sequences. If degree of some element
is at least $ n+1 $, we are done. Otherwise all degrees are among $ 1,2, \ldots ,n $.
Because we have $ n^{2} + 1 $ elements, there is some $ k $ and distinct elements
$ y_{1}, y_{2}, \ldots , y_{n+1} $ which have degree exactly $ k $.
By conditions of the problem, there are such $ i , j $ such that $ y_{i} | y_{j} $, but
then we have that degree of $ y_{j} $ is at least $ k+1 $ and we come to contradiction.


Given $ a,b,c\geq0$ such that $ a\plus{}b\plus{}c\equal{}3$.Prove that
$ \frac a{b^3 \plus{} c^2} \plus{} \frac b{c^3 \plus{} a^2} \plus{} \frac c{a^3 \plus{} b^2}\geq\frac3{2}$

I believe that a condition is required here otherwise it's wrong .See $ a\equal{}b\equal{}c\equal{}3$

[quote="Svejk"]I believe that a condition is required here otherwise it's wrong .See $ a \equal{} b \equal{} c \equal{} 3$[/quote]
Sorry,$ a \plus{} b \plus{} c \equal{} 3$. I edited my post :D Thank you Svejk!

[quote="dduclam"]Given $ a,b,c\geq0$ such that $ a \plus{} b \plus{} c \equal{} 3$.Prove that
$ \frac a{b^3 \plus{} c^2} \plus{} \frac b{c^3 \plus{} a^2} \plus{} \frac c{a^3 \plus{} b^2}\geq\frac3{2}$[/quote]
Try $ a\equal{}b\equal{}1.5$ and $ c\equal{}0.$ :wink:

[quote="arqady"][quote="dduclam"]Given $ a,b,c\geq0$ such that $ a \plus{} b \plus{} c \equal{} 3$.Prove that
$ \frac a{b^3 \plus{} c^2} \plus{} \frac b{c^3 \plus{} a^2} \plus{} \frac c{a^3 \plus{} b^2}\geq\frac3{2}$[/quote]
Try $ a \equal{} b \equal{} 1.5$ and $ c \equal{} 0.$ :wink:[/quote]
Thank you arqady,I make mistake   :(
And I think the following inequality is true
$ \frac {a^2}{b^3 \plus{} c} \plus{} \frac {b^2}{c^3 \plus{} a} \plus{} \frac {c^2}{a^3 \plus{} b}\geq\frac3{2}$

[quote="arqady"][quote="dduclam"]
And I think the following inequality is true
$ \frac {a^2}{b^3 \plus{} c} \plus{} \frac {b^2}{c^3 \plus{} a} \plus{} \frac {c^2}{a^3 \plus{} b}\geq\frac3{2}$[/quote]
You have the proof or it's your assumption?[/quote]
Yes,but very long and not nice :( Hope a nice solusion from all you :D


This year is going to be the first time I'm taking the AMC (my school-despite all it's blabbering about being so good-does not offer it to anyone younger than 9th grade). I was wondering if anyone could give me some advice on how to spend my time while taking it, the material I should focus on most (especially so close to date), the best ways to prepare, etc.

Also, I'm not sure if I'm taking the AMC10 or 12 yet (I might end up taking both if it's allowed).

[quote="horselvr101"]how to spend my time while taking it[/quote]

Don't waste your time on questions you can't do.  Skip around to questions you know methods for, then come back to the others (this is just an overall good test-taking strategy).

[quote="horselvr101"]the material I should focus on most (especially so close to date)[/quote]

Past AMCs are still tried-and-true unless you can't understand the solutions (the ones posted in the Contest section of this site).  In that case, ask for help from a more experienced problem-solver (either in real life or in a topic on the forum) or consult the Art of Problem Solving book.

[quote="horselvr101"]the best ways to prepare[/quote]

Taking past AMCs will let you get a feel for the types of questions they ask, so I still think for most people this is the best preparation.


How many different routes are possible from A back to A via B?
[asy]import graph;
size(200);
pair A = (0,0), B = (10,0);
draw(ellipse((5,0),5,1));
draw(A--B);
label("$A$",A,W);
label("$B$",B,E);[/asy]

There are $ 3$ ways to go there and $ 3$ to come back.  Our answer is therefore $ 3 \times 3\equal{}\boxed{9}$.


(c)(i) Show that $ 1+cos2\theta+isin2\theta = 2cos\theta(cos\theta+isin\theta)$

(ii) If A,B,C are the angles of a triangle, show that the expression
$ \frac{(1+cos2A+isin2A)(1+cos2B+isinB)}{1+cos2C-isinC)}$ is real

For the first one:

we have that $ \cos 2\theta+\sin 2\theta = (\cos \theta+i\sin \theta)^{2}$ by de moivre's theorem.

So the left side of the equation can be expanded into $ 1+\cos^{2}\theta-\sin^{2}{\theta}+i2\sin \theta \cos\theta$.

Because $ 1 = \cos^{2}\theta+i\sin^{2}\theta$ the above equation can be further modified to $ \cos^{2}\theta+\sin^{2}\theta+\cos^{2}\theta-\sin^{2}\theta+i2\sin\theta\cos\theta$.

Cancel and collect like terms and we have arrived at  $ 2cos\theta(cos\theta+i\sin\theta)$

[quote="bos1234"](ii) If A,B,C are the angles of a triangle, show that the expression
$ \frac{(1+\cos 2A+i\sin 2A)(1+\cos 2B+i\sin 2B)}{1+\cos 2C-i\sin C}$ is real[/quote]
[hide="Solution ii)"]
$ \frac{(1+\cos 2A+i\sin 2A)(1+\cos 2B+i\sin 2B)}{1+\cos 2C-i\sin 2C}=$
$  =\frac{(2\cos^{2}A+2i\sin A\cos A)(2\cos^{2}B+2i\sin B\cos B)}{2\cos^{2}C-2i\sin C\cos C}=$
$  =\frac{4\cos A\cos B(cos A+i\sin A)(\cos B+i\sin B)}{2\cos C(\cos(-C)+i\sin(-C))=}$
$  =\frac{2\cos A\cos B}{\cos C}(\cos(A+B+C)+i\sin(A+B+C))=-\frac{2\cos A\cos B}{\cos C}$
[/hide]


calculus
integration
logarithms
trigonometry
calculus computations

Find

 :huh:

$ \int{\frac{2}{sin(2x) ln(tanx)}dx}$

$ \int{\frac{2}{\sin(2x)\ln(\tan x)}dx}$

$ u =\ln(\tan x)$

$ du =\frac{\sec^{2}x}{\tan x}dx =\frac{1}{(\cos^{2}x)(\frac{\sin x}{\cos x})}dx =\frac{2}{2\sin x\cos x}dx =\frac{2}{\sin(2x)}dx$

$ \int\frac{1}{u}du$

$ \ln|u|+c$

$ \ln|\ln(\tan x)|+c$


Prove that for all $ a,b,c\in(0;1)$ we have:
$ \frac{c}{1\plus{}ab}\plus{}\frac{a}{1\plus{}cb}\plus{}\frac{b}{1\plus{}ca} \leq 2$

[quote="nkht-tk14"]Prove that for all $ a,b,c\in(0;1)$ we have:
$ \frac {c}{1 \plus{} ab} \plus{} \frac {a}{1 \plus{} cb} \plus{} \frac {b}{1 \plus{} ca} \leq 2$[/quote]
I think, "=" don't happen ! :maybe:
we have :
$ \frac{c}{1\plus{}ab}\prec  \frac{2c}{a\plus{}b\plus{}c}$
done !


MATHCOUNTS
articles
videos
geometry
rectangle
3D geometry
geometric transformation

hmm i think the countdown round is over by now. I was just wondering about the level of the test. post any comments you have about it. and yes, this can be a place to vent your anger. :)

and please be honest with the poll

When are scores going to be on the mathcounts website?
Eh, I'll just ask Mr Fischer when they get back.

I didnt go to nats.
Or states
Or regionals
Or made the team
 :(

Can someone post the top 4 individuals and teams? Thnks.

I'm mad I didn't come close to nats.

I would have gotten to nats if I solved a few more problems (too lazy) and did 50-36 correctly.

well according to wikipedia Texas got first(expected).

[url=http://news.yahoo.com/s/usnw/20070512/pl_usnw/texas_eighth_grader_and_texas_team_awarded_mathematics_champions_at_lockheed_martin_mathcounts_national_competition_2007]Click link for story.[/url]
(This was a valid link at 21:10 EST, 11 May 2007.  Often news links go stale in a few hours or days.)

[url=http://sev.prnewswire.com/education/20070511/DCF30312052007-1.html]Same story[/url], formatted a little more nicely.
(This was a valid link at 19:45 EST, 12 May 2007.)

[quote="indianamath"]someone edited it. it said kevin chen before.[/quote]
No kevin won, wenyu(sp?) got second,

hey indidanamath, do you know the kid that got first on written he was from indiana.

It appears to me, in the yahoo article, that wenyu and allan got 3rd/4th, and ben kraft of Pennsylvania got 2nd.

Wikipedia can be edited and/or defaced by anyone.  The article has been defaced several times today.  I would suggest you trust the bona-fide news media and the information in the link posted above.

A list of winners will eventually be posted to the official MATHCOUNTS Web Site, but since MATHCOUNTS runs on a staff of about 6 people, they are probably a little busy right now.

Hello, all!

I'm not official media, but I am in Ft. Worth, and attended the Countdown Round, so I can report the following (I just watched it on my video of the event to confirm):

Individuals:
1. Justin Amman (Indiana)
2. Allen Yuan (Michigan)
3. Kevin Chen (Texas)
4. Abubakhar Abbis (Minnesota?)
5. Ben Kraft (Pennsylvania)
6. Mario Choi (Maryland)
7. Elizabeth Synge (Massachussetts)

I had to include Elizabeth in my list because I was SO happy to see a girl up there!  The top three teams were Texas, Indiana, and New Jersey.  The Pennsylvania coach was telling us later that since Pennsylvania and NJ were tied in their total score, they had a tie-breaker.

Lots of fun!  Masters Round will be in the morning.

--Nancy Hansen
(D.C. Team mom)

p.s.-Almost forgot to mention--the above are written results.  In countdown, it came down to Ben and Kevin in the final and Kevin won.  Allen faced Kevin in the semi-finals, but I don't remember who faced Ben.

The initial list of winners is up on the Mathcounts site, more to come soon.

[quote]Individuals:
National Champion - Kevin Chen of Texas
Runner-Up - Ben Kraft of Pennsylvania
Semifinalist - Wenyu Cao of New Jersey
Semifinalist - Allen Yuan of Michigan
Quarter-finalist - Justin Ahmann of Indiana
Quarter-finalist - Abubakar Abid of Arkansas
Quarter-finalist - Mario Choi of Maryland
Quarter-finalist - Gaurav Singh of Minnesota
Countdown Round Participant - Elizabeth Synge of Massachusetts
Countdown Round Participant - Nathan Salo of Minnesota
Countdown Round Participant - Paul Davey of Nebraska
Countdown Round Participant - Michael Zhao of Missouri

Teams:
1 Texas
2 Indiana
3 New Jersey
4 Pennsylvania
5 Michigan
6 Massachusetts
7 Washington
8 North Carolina
9 Kentucky
10 Maryland[/quote]

Two from Minnesota!  Yes!

Anyone have a full Countdown bracket, by the way?

[quote="ZzZzZzZzZzZz"][quote="13375P34K43V312"]I may have said this already, but this basically how the test went down...

Sprint: Everything was easy except 27, 28, and 30
Target: Everything was easy except 8
Team: Everything was way easy but 10 was alot easier if you had a TI-89
Masters: Pretty boring problem, the 2*n case was stupid, because either you got it wrong or you got really lucky and saw the random subtlety. The 3*5 case was even worse, because it was just random cases.[/quote]

If they were so easy, why didn't you get them all right?[/quote]

Sprint #6 (perfect squares/cubes): read wrong
Sprint #10 (arithmetic sequence): didn't bother checking, apparently forgot to divide by 2
Sprint #16 (cube formations): screwed up with my rotations, didn't think to model it with my erasers
Sprint #23 (direct variation): misinterpreted
Target #4 (1/b^2-1/a^2): read wrong
Team #7 (rectangle dissection): read wrong

answer your question?

[quote="kyyuanmathcount"][quote="uldivad9"][quote="kyyuanmathcount"]It really isn't that difficult.  Perhaps he did have the test - it's handed out to all the coaches.

#6 was the one about the clocks, where the guy left, then came back in the same hour and noticed both times that the degree measure between the minute and hour hands was 110 degrees.[/quote]

Did anyone think that was hard? [size=50](I did)[/size][/quote]

Like last years national, it feels hard when you do the real thing, but if you come back the year after and do it again, it usually feels about the same as the other practices.

Ex:  Last year, I scrambled and barely finished the sprint (skipped a few), which had never happened on any of my practices.  However, when I practiced it this year, it was the same as all the previous ones.[/quote]

Couldn't that just be true because you've already seen the problems? Even if I don't remember the answers, I always find that the problems are easier when you've seen and/or practiced them before.

I was rather surprised (and not in a good way) by how I did. I was, on a slightly lighter note, the only person to have a score equivalent to my rank (33 = 21 + 12). I would agree that the tests were very easy, with the exception of a few sprint problems and target #8, and I was really disappointed by how easy the team was this time around (too many teams got a perfect score), not to mention the sheer number of times that I made mistakes such as adding instead of multiplying (Sprint #29) or thinking that 144 divided by 8 equaled 16 (Sprint #21), among far too many others.

No, that's why I compared it to other practice tests rather than other real tests.  I had seen each set of problems before.

My coach was proctoring the CA team and he said that they filed like 4 protest forms. Then he said that he was working out the problem on scratch paper and a floor team member b^*#%ed at him and then he's like "well you said we couldn't write on test..." Funny story. I thought that would help CA win. lol. but not really.

[quote="CircleSquared"]Actually, we didn't file a single protest :D

Did someone really !@#^% at the proctor? Funny, I agree...

Your coach actually did probably improve our performance, I almost imagined a huge jailer towering over our team.  :rotfl:[/quote]

[quote="CircleSquared"]

I did horribly, but at least there's next year, if I make it there.
[/quote]

You are the only one in CA team who made top 50 in the National, you
should be so proud of yourself!

good job kevin chen for pwning everyone in general
(expected, but still friggin awesome)
i hope we can repeat next year =)

[quote="vector2101"]Oh. Okay. But isn't it still harder the first time you take the test (in the competition) because you haven't seen the problems before and might not quite know how to solve them (if they're hard enough, that is...)? I would hazard a guess that that's the reason that I always do better at practices than at competitions, especially on the last problems in the sprint round.[/quote]

Oh definitely!  I averaged....a 29ish on national sprints, and 16 on targets, so usually around 45 on national rounds.  yeah, so 26/14 = 40 on the real thing.  The test writers are getting trickier every year... and the tests get harder overall (generally, I'm not saying that 2007 was harder than 2005, but I am saying that the 2004-2007 tests were much harder than the 1997-2003 tests.)

Bobby shen and Kevin Li i think? one of them got 13 and 17or 18, jeff cheng got around 40.

yeah go tx :starwars:

who's gonna win the team next year? any guess?

i have no idea, i just put it up there cause i saw a question about it


integration
logarithms
real analysis
real analysis unsolved

$ \textbf{(i)}\;\;\;1\minus{}\frac{1}{3}\plus{}\frac{1}{4}\minus{}\frac{1}{6}\plus{}\frac{1}{7}\minus{}\frac{1}{9}\plus{}\frac{1}{10}\minus{}\frac{1}{12}\plus{}\;\;\;\ldots\;\;\;\ldots\;\;\;\plus{}\;\minus{}\;\;\;\ldots$
Your type is in the attached document.

[hide="Answer for (i)"]We have $ S \equal{}\sum_{n\equal{}0}^{\infty}\left(\frac{1}{3n\plus{}1}\minus{}\frac{1}{3n\plus{}3}\right)$. As usual, now define

$ f(x) \equal{}\sum_{n\equal{}0}^{\infty}\left(\frac{x^{3n\plus{}1}}{3n\plus{}1}\minus{}\frac{x^{3n\plus{}3}}{3n\plus{}3}\right)$.

Then f(1) = S, f(0) = 0, $ f'(x) \equal{}\sum_{n\equal{}0}^{\infty}\left(x^{3n}\minus{}x^{3n\plus{}2}\right) \equal{}\frac{x\plus{}1}{x^{2}\plus{}x\plus{}1}$, and

$ S \equal{}\int^{1}_{0}\frac{x\plus{}1}{x^{2}\plus{}x\plus{}1}\,dx \equal{}\frac{\ln 3}{2}\plus{}\frac{\pi}{6\sqrt{3}}$.[/hide]


logarithms
complex analysis
complex analysis unsolved

[size=150]$i^i$[/size]

that's it. hint :arrow: use logarithms.

Because the logarithm has infinitely many branches, this actually has infinitely many possible answers.

But they're all real.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Classify the finite groups G such that G has a proper subgroup P, which contains all other proper subgroups of G.

[hide="Solution"]$ \mathbb Z/p^n\mathbb Z$, $ p$ prime[/hide]

[hide="proof"]take $ g \in G \backslash P$. the maximal property of $ P$ shows $ G \equal{} <g>$ is cyclic, say of order $ n$. the property of $ P$ says that $ n$ has a nontrivial divisor $ m$ such that every nontrivial divisor of $ n$ divides $ m$. thus $ n$ is a prime power.[/hide]


Given a $15\times 15$ chessboard. We draw a closed broken line without self-intersections such that every edge of the broken line is a segment joining the centers of two adjacent cells of the chessboard. If this broken line is symmetric with respect to a diagonal of the chessboard, then show that the length of the broken line is $\leq 200$.

color the board black and white like a chess board.

notice that the diagonal is all with same color(say black)

obviously, the segment can only pass through 2 of the 15 blocks on diagonal(otherwise it is self-intersected). Thus, the segment can pass through at most 113-13=100 black blocks.

Notice the segment pass through black block and white block alternately, it must pass through same number of black blocks and white blocks, gives a total of at most 200 blocks.

I dont know another solution,  it's official solution on ARO

Yes, this is the official solution.

By the way, this problem is too hard for the first problem for ninth grader, is it?

Just observe that every center can have at most two edges(color black-white and if the main diagonal is white count edges from the white ones),and that there are at most two centers that have any edges from the main diagonal,so from these we easy obtain the answer.

Can anyone give picture for "closed broken line"?
İ think İ understood what problem means but I am afraid I will try for wrong problem for hours...
Please help me...
Thanks!

nice problem
[b]S[/b]olved with [b]mxlcv[/b]
[hide = solution]
Let $\ell$ be the main diagonal the path is symmetric about. Obviously the broken line meets $\ell$ in some cell $C$. Moreover, there is at most one such cell $C$ since the path is not self intersecting. Now observe that the broken line can be broken into two parts $P_1, P_2$, both containing $C$ which are reflections of each other about $C$. Therefore it suffices to show that the length of $P_1$ is at most $100$. Color the $15 \times 15$ like a chessboard with $\ell$ colored black and observe that the length of the broken path is at most $1$ plus the number of cells visited by the path. This number is at most $2 \cdot W+1$, where $W$ is the number of white cells in the region below $\ell$. There are $1+3+5+\dots+13 = 49$ such cells, so the length of the broken line is at most $2 \cdot 49 + 1 + 1 = 100$, as desired.


[quote="shobber"]In quadrilateral $ABCD$, $\angle{D}=90^o$, $\angle{BCA}=78^o$, $\angle{CAB}=\angle{CBA}$, $AB=2AD$.

Find $\angle{CAD}$.[/quote]
Shouldn't be $51^o$ ?  :?:

The triangle $CAB$ is isosceles $(CA=CB)$. If $CM$ is its median, then $\triangle CMA$  is right and congruent with $\triangle CDA \ (CA=CA,AD=AM).$ Hence $\angle CDA = \angle CMA = 51^o$.


algebra polynomial function calculus derivative

[hide]I'm just gonna go for a rough proof.  Nothing too comprehensive, mostly intuitive.

Lim f(x)=A

x-> :inf:

A is some value.

Obviously

Lim f'(x)=-A

x->  :inf:

A can only exist if -A is 0, otherwise the initial limit does not exist.  QED.[/hide]

Magnara, I don't get your proof. I think you basically assumed the limit was 0 and then said QED.

Edit: I shouldn't say I proved it for polynomials f(x), since that's not even true. I proved it using polynomials in a contradiction or something weird like that :).

[quote="rlee"]Magnara, I don't get your proof. I think you basically assumed the limit was 0 and then said QED.

I've proved it for [b]polynomials[/b] f(x), but I'm still working on the general proof.[/quote]

No...I said first the limits are opposites.  Then I said that if the lim f'(x) was not equal to 0, then the limit of f(x) would not exist.  For an infinite limit to exist, the functions derivitive must have an infinite limit that goes to 0.  Therefore, f'(x) is 0, and f(x) is -0=0.

If you don't believe my limit of f'(x) must be 0 thing, find me a function that has a finite limit as it goes to infinity whose derivitive has a non-zero limit as it goes to infinity.

You can't say in a proof ``if you don't believe what I'm saying, then try to find me a counterexample.'' I'm trying to find one right now because I'm convinced that one exists.

Well, intuitively it's Ok, but here's a more rigorous approach:

Let h(x)=ex*f(x) and g(x)=ex. We can see that h'(x)=ex(f(x)+f'(x)), so from the hypothesis we have lim{x->00}h'(x)/g'(x)=0, so, by L'Hospital's rule, lim{x->00}h(x)/g(x) exists and it's equal to lim{x->00}h'(x)/g'(x)=0, so f(x)->0 as x->00, so f'(x)->0 as x->00.

I don't think we need to know that f has a limit when x->00 (I mean to say that it's true, but we don't need to know it in advance in order to prove that f'(x)->0 as x->00).

Oh, I thought if lim x-> :inf:  f(x) =  :inf: , then the limit exists and is just :inf: .  :inf:  implies that the limit DNE?

Taking that into consideration, it makes conceptual sense. Still, it's not  extremely rigorous.

Lemma 1:

For continuous, differentiable f

Lim x- :inf:  f(x) = A implies that f'(x)=0

Proof by contradiction:

Apply the mean value theorem to two points a,b 'close' to infinity.  Then because Lim x- :inf:  f(x) = A:

f(b)-f(a)/(b-a) = (A-A)/(b-a) = 0 = f'(c) for c E (a,b)

But since we can choose (a,b) arbitrarily close to each other, f'(c)=0 for all c sufficiently close to infinity.

Now, since Lim x- :inf:  f'(x) = 0, it follows from the additive law of limits that Lim x- :inf:  f(x) = 0

Edit:

There is also another case, namely, that f(x)=-f'(x) everywhere so that (f(x)+f'(x)) is uniquely zero.

But the only f which satisfies this is f(x)=a*exp(-x) (from the diff eq. y=-y'), which has f, f' going to zero anyways as is stated in the problem.

[quote="Magnara"]For an infinite limit to exist, the functions derivitive must have an infinite limit that goes to 0...

If you don't believe my limit of f'(x) must be 0 thing, find me a function that has a finite limit as it goes to infinity whose derivitive has a non-zero limit as it goes to infinity.[/quote]
What kind of attitude is that?  It is the writer's obligation to prove his or her claims, not the reader's obligation to disprove them.

But if I understand your claim, here is a counterexample to it (and blahblahblah's lemma).  Let f(x) = sin(x2) / x.  To get a result like blahblahblah's lemma, you need an extra smoothness condition.  For example, assuming that f'' is bounded is enough.

[quote="churchilljrhigh"]Another problem from the Calc Challenge 4![/quote]

4a - Use AM-GM.

Yup. Or use the solution that was given underneath the question :D

Yep 4a was really easy. But I included it because it helps do 4b. Good Luck!

4b:

[hide]apply 4a repeatedly to find that a_n <= 2/2+3/4+4/8+5/16+6/32+... which converges.[/hide]

xn converges to be less than or equal to 3.


vector
geometry
linear algebra
matrix
LaTeX
analytic geometry

If we take $\overrightarrow{v}=(\begin{array}{ccc} 0 & 1 & 0 \\ \end{array} )$ and $\overrightarrow{w}=(\begin{array}{ccc} 1 & 0 & 0 \\ \end{array} )$, then find $\overrightarrow{v}\times\overrightarrow{w}$ (the cross product).

Move this if necessary....

As a suggestion (to everyone, not just to 236!), any time you think something might need to be in a more advanced forum than the one you are posting in, you should be posting in a more advanced forum.  (This was originally posted in Getting Started; I have moved it).

By the well-known formula from analytic geometry, if $\vec a=\langle a_x,a_y,a_z\rangle$ and $\vec b=\langle b_x,b_y,b_z\rangle$, then

$\vec a\times\vec b=\left|\begin{matrix}\vec i & \vec j & \vec k\\ a_x & a_y & a_z\\ b_x & b_y & b_z\end{matrix}\right|=(a_yb_z-a_zb_y)\vec i+(a_zb_x-a_xb_z)\vec j+(a_xb_y-a_yb_x)\vec k$

In the given example we get $\vec v\times \vec w=\langle 0, 0, -1\rangle$

or simply use the "right hand rule" for cross products

make a fist with your thumb pointing straight out, and curl your fingers from the first vector to the second vector. your thumb should point in the direction of the cross product.

so let's say positive x is coming out of the computer screen, towards you
and positive y is right
and positive z is up

so you would curl your fingers from the positve y direction to the positive x direction, and your thumb points down, which is in the negative x direction.

another way to do it is, outstretch your fingers, and your thumb pointing perpendicular to your outstretched fingers, in the same position it was before. it kinda makes an 'L' shape now.

the first vector goes in the direction of your thumb, the second vector goes in the direction of your outstretched fingers, and the cross product goes in the direction your palm faces.

the curling fingers seems better to me.

$i \times j=k$ $j \times i=-k$
$j \times k=i$ $k \times j=-i$
$k \times i=j$ $i \times k=-j$
for the right hand rule! Because the cross product is anti-commutative

In this case, the easiest way to do the problem is to realize that

$\overrightarrow{v} = \mathbf{j}$

and that

$\overrightarrow{w} = \mathbf{i}$

Then, $\overrightarrow{v} \times \overrightarrow{w} = \mathbf{j} \times \mathbf{i} = -\mathbf{i} \times \mathbf{j} = -\mathbf{k}$

The cross product of any two unit vectors will be a new unit vector orthagonal to the original two, and as stated above by The Master, the direction will be given by the right hand rule.

EDIT: LaTeX issues, sorry


How can I prove that there are just three subgroups of order 8 in $ S_4$? I know by Sylow Theory that the number of subgroups of order 8 must be either 1 or 3, but I do not know how to conclude that there are 3 and not 1. Thanks in advance

[quote="ZetaX"]Find one of them. Then show that it isn't normal.[/quote]

I could also find two and conclude that there must be three, but this is not very 'elegant', so I would like to know another manner to solve this.

You may also count elements of order a power of $ 2$ (showing that there are more than $ 8$ is enough).

Let P be a Sylow 2-subgroup of S_4. If P is normal in S_4, we consider the fact that S_4 has no element of order 6, then S_4 is a Frobnius, which kernel  is P, then 3| |P|-1=7, a contradiction.


group theory
abstract algebra
Galois Theory
superior algebra
superior algebra unsolved

Let $H_{1},H_{2}\leq G$ be solvable. Is $H_{1}H_{2}$ again a subgroup of $G$? If so, is it solvable?

The product $H_{1}H_{2}$ needs not be a subgroup. Note that the non-solvable group $G=S_{5}$ is generated by two cyclic subgroups $(12)$ and $(12345)$.

But if we [b]assume[/b] that $H_{1}H_{2}$ is a subgroup and ask if it's solvable... I suppose you can start by multiplying an abelian tower for $H_{1}$ by $H_{2}$, but I don't know if this works.

[quote="mlok"]The product $H_{1}H_{2}$ needs not be a subgroup. Note that the non-solvable group $G=S_{5}$ is generated by two cyclic subgroups $(12)$ and $(12345)$. [/quote]But being generated by two subgroups and being their product is something different in general.

[quote="sissa"]But being generated by two subgroups and being their product is something different in general.[/quote]
My point exactly.  :)  If $H_{1}H_{2}$ is a subgroup, then it is the smallest subgroup containing both $H_{1}$ and $H_{2}$, i.e. it's the subgroup generated by $H_{1}$ and $H_{2}$.

Here is a better example: $H_{1}$ is $S_{4}$ considered as a subgroup of $G=S_{5}$, and $H_{2}$ is generated by $(12345)$. Both $H_{1}$ and $H_{2}$ are solvable, and $H_{1}H_{2}$ is a subgroup of $G$ (namely, $G$ itself), but $H_{1}H_{2}$ is not solvable.

[quote="mlok"][quote="sissa"]But being generated by two subgroups and being their product is something different in general.[/quote]
My point exactly.  :)  If $H_{1}H_{2}$ is a subgroup, then it is the smallest subgroup containing both $H_{1}$ and $H_{2}$, i.e. it's the subgroup generated by $H_{1}$ and $H_{2}$.[/quote]Ok, I understand now :)


calculus
integration
trigonometry
calculus computations

[hide="not quickly"] \[\int_{0}^{\pi}\frac{dx}{\cos x-a}=\int_{\pi}^{0}\frac{d(\pi-x)}{\cos(\pi-x)-a}=1/2\int_{0}^{\pi}\left(\frac1{\cos x-a}-\frac1{\cos x+a}\right)dx=\int_{0}^{\pi}\frac{a\,dx}{\cos^{2}x-a^{2}}=2\int_{0}^{\pi/2}\frac{a\,dx}{\cos^{2}x-a^{2}}=2\int_{0}^{\pi/2}\frac{a\,d(\tan(x))}{\frac{a^{2}}{\cos^{2}x}-1}.\] The rest is obvious. Not very quick though :blush: [/hide]


Four people are sitting around a table.  A name tag is made for each person.  In how many ways can these name tags be passed out so that every person gets the wrong name tag?

Person 1 can recieve $ 3$ nametags WLOG, say he gets person 2s nametag.
Person 2 can receive $ 3$ (1,3,4)

WE now have 3 scenarios:

He gets 1: 3 muts get 4 and 4 must get 3
He gets 3: 3 must get 4 and 4 must get 1
He gets 4: 3 must get 1 and 4 must get 3

As you cansee, regardless of what 2 gets, 3 and 4 have only 1 option.

So the answer is $ 3\cdot{3}\equal{}9$

There are 24 ways to give out the name tags, and for the first person, 3 ways to get the wrong person.

2 Cases:

[hide="If the pile doesn`t have 2nd person`s name tag:"]
There would be 3 ways

2 cases:

[hide="If the pile doesn`t have the 3rd person`s name tag:"]There would be 2 ways[/hide]
[hide="Else if the pile does have the 3rd person`s name tag:"]There would be 1 way[/hide]

A total of 6+3=9 ways
[/hide]
[hide="If the pile does have 2nd person`s name tag:"]
There would be 2 ways

2 cases:

[hide="If the pile doesn`t have the 3rd person`s name tag:"]There would be 2 ways[/hide]
[hide="Else if the pile does have the 3rd person`s name tag:"]There would be 1 way[/hide]

A total of 4+2=6 ways
[/hide]

If you looked through the hide tags (for organization, sorry if its confusing), there are a total of 15 ways to get everyone's name tags wrong. But I have this feeling that the answer is 9, not 15.

[quote="RightAnnihilator"]In short:
Person A has three choices.  Let's call the person he picks person B.

Person B now has three choices.  Let's call the person he picked person C.

There are two people left, persons C and D.  The only two tags left are A and D.  Only one possibility here:  D is forced to chose tag A and person C has no option left but tag A.
So, 3*3*1=9.[/quote]

Oh. That is much clearer than mewto's explanation. Thanks.

But what did I do wrong?

[quote="dragon96"]There are 24 ways to give out the name tags, and for the first person, 3 ways to get the wrong person.

2 Cases:

[hide="If the pile doesn`t have 2nd person`s name tag:"]
There would be 3 ways

2 cases:

[hide="If the pile doesn`t have the 3rd person`s name tag:"]There would be 2 ways[/hide]
[hide="Else if the pile does have the 3rd person`s name tag:"]There would be 1 way[/hide]

A total of 6+3=9 ways
[/hide]
[/quote]
You just needed the first part (I put it above)

Hi Dragon69

I tried to track down your error, but I can't really figure out what your labels mean on your hide tags.  If you really want to pursue it, please PM me a rewrite of your reasoning.


An amusement park ride consits of a rotating vertical cylinder with rough canvas walls. After the rider has entered and the cylinder is rotating sufficiently fast, the floor is dropped down, yet the rider does not slide down, The rider has a mass of 50 kilograms, the radius R of the cylinder is 5 meters, the angular velocity of the cylinder when rotating is 2 radians per second, and the coefficient of static friction between the rider and the wall of the cylinder is 0.6.

(b) Calculate the centripetal force on the rider when the cylinder and state what provides that force.

(c) Calculate the upward force that keeps the rider from falling when the floor is dropped down and state what provides that force.

(d) At the same rotational speed would a rider of twice the mass slide down the wall? Explain you answer.

I need all the posible help and explanation, Thank You!

(b) centripetal force is mr*w^2 = 1000N, mass m, radius r, angular velocity w, it is provided by the normal force [b]N[/b] of the cylinder on the man
(c) upward force = mg = 490N, provided by friction (whose max. value is [b]Nu[/b] = 600N
(d) Weight of second man = 980N, Normal force [b]N[/b]= 2000N; max. friction force is 1200N, hence, he won't fall down

This is actually a solved problem in Halliday & Resnick's book (6th edn).


Prove whether or not you can trisect a 60 degree angle using only constructions with a unmarked straightedge and a compass.

A detailed explanation of this would take a while, but in the end, it is possible to trisect a 60 degree angle. The construction consists of drawing a bunch of circles, finding intersections, and connecting those intersections. Though the attached image is pretty cluttered, it may be possible to understand what I did. Those red line segments trisect the angle formed by those two rays.

that solution does not work... the angle between the red lines is

$2\sin^{-1}\frac{\sqrt{7}}{14}$

which is about

$21.79^\circ$

--

generally, this is not possible since $20^\circ$ is a solution to a cubic, and those lengths have been proven to be impossible to construct

This is a famous problem, and the proof that it is impossible to trisect a general angle rests on showing that this implies one could trisect a 60 degree angle and construct a 20 degree angle. Then, we see that a 20 degree angle satisfies an unfactorable cubic which implies we can't construct a 20 degree angle. It's something like that, it either uses cos20 or sin20

You can however construct  $20$ grad  (You know the kind of unit they use in france -  where a right angle = 100 grads )   angle. :) .

$20 \text{grad} = \frac{\pi}{10}$

And $sin \frac{\pi}{10} = \frac{\sqrt 5 - 1}{4}$ which is perfectly easy to construct. But I guess you all knew that. :lol:

Yes, and   one can also construct   ($\frac{pi} {17}$)  and  ( $\frac{\pi} {257}$ )  not to  mention   ($\frac{\pi} {65537}$ )
And for those who are interested in trivia .
Rumor has it that the method for  the last one  (some one's thesis)  occupies many boxes of paper and took years to write it down..:)

yes, wasn't it gauss that found out how to construct for some numbers of the form $2^n + 1$... in fact they might be mersenne primes...
more specifically, those angles mean you can construct 17 gons and 257 gons etc.. for trivia, gauss wanted a 17gon inscribed on his tomb stone (but it wasn't)


function
integration
algebra
domain
real analysis
real analysis unsolved

Please help me in this problem   :D
Let $U \subseteq \mathbb{R}^2$ a open convex set and $f: U \mapsto \mathbb{R}$ twice differentiable on $U$. Prove that  $\frac{\partial^2f}{\partial x\partial y}$  is identically zero on $U$ iff exists functions $\phi : I \mapsto \mathbb{R}$,  $\psi :J \mapsto \mathbb{R}$  twice differentiable on the intervals I, J,respectively,  such that $f(x,y)= \phi(x)+\psi(y)$,  $\forall (x,y) \in U$.

[quote="liyi"]$\frac{\partial^2f}{\partial x\partial y} = 0$
Integrate it with respect to $y$, we get
$\frac{\partial f}{\partial x} = g(x)$
Integrate it again, we get
$f = \int g(x) dx + \psi(y)$.[/quote]
thanks  :D ,.... but is it necessary the convexity?  :huh:

[quote="tipe"]thanks  :D ,.... but is it necessary the convexity?  :huh:[/quote]
Er... I don't know how to explain it...

The derivative's being $0$ implies that the the function is constant only if the domain consists of one interval (if it consists of several disjoint intervals, the constants may be different on different intervals). So, for [b]liyi[/b]'s reasoning to work, we need all horizontal and all vertical sections of $U$ to be intervals, which is automatically true if $U$ is convex but may be false otherwise (though there are some non-convex regions that satisfy this condition)


1).  Solve in natural numbers the equation:

7x+x4+47=y2.

2).  Let p>3 a prime number. Show that 43 divides k=7p-6p-1

Second one :

[hide]

In fact it's true for every integer of the form 6n+1 or 6n+5 (and

every prime number > 3 is of one of these forms). Perhaps the hypothesis 'p prime' is here just to (try to) confuse us (and it works perfectly with me !), suggesting some use of Fermat little theorem ???

n : 7^n and 6^n mod 43 :

1 : 7 6

2 : 6 36

3 : 42 1

4 : 36 6

5 : 37 36

6 : 1 1

repeat .... solved.[/hide]

good job.  The fact that it is prime is not meant to confuse you, but it could lead you in the wrong direction and (spoiler) [hide] if i said = 1 or 5 mod 6, it would have made the problem too trivial =P[/hide]

the second one is still up for grabs

Did you invent these nice problems ?

[hide]

I worked hard on the first one, but failed miserably !

For now I have only proved that 2 divides x, 4 divides y. (4,4*13) is the smallest solution.

[/hide]

I didn't invent these problems, I remember i saw them somewhere and I liked them so i decided to post them.

zscool wrote:1).  Solve in natural numbers the equation:

7x+x4+47=y2.

2).

Here goes

[hide]

Case 1: x odd.  Then 7^x is 3 mod 4, x^4 is 1 mod 4, and 47 is 3 mod 4.  This means that y^2 is 3 mod 4, which can't happen.

Case 2: x even = 2k. Then for k large enough, (2k)^4 + 47 < 2*7^k+ 1.  That is, (7 ^ k)^2 = 7^2k < 7^2k + (2k)^2 + 47 = y^2 < 7^2k + 2*7^k + 1 = (7^k + 1)^2.  This sandwiches y^2 between two consecutive squares, making for another contradiction.

Now all we're left with is the case of x small even.  This is finite and I'm feeling lazy, so I'll leave it as an exercise for the reader ;-)

[/hide]

Bother, I'm 15 minutes too slow to catch Alison.  And I was feeling so happy for getting it.

[hide]Yes, you just have to use the gaps between the squares.  We dispatch with the odd case by a mod 4 analysis, and we then know that eventually

(2k)^4 + 47 < 2*7^k+ 1

8k^4 + 23 < 7^k

When k = 4, 2^11 + 23 = 2071, while 7^4 = 49^2 = 2401.  So we just have to check x = 2, 4, 6 and 8 by hand, giving us LHS's equal to 112, 2704, 118992, and 5768944, giving us the unique solution x = 4, y = 52.[/hide]


I have a problem forming the index.

I have some words using different letters such as \"{o},  \c{C}, $\Omega$. These letters\symbols are given priority in the index.

How can I get rid of this priority and form a more logical order? I don't want S\"{o}nmez to come before Sasaki in the index.

Thank you very much for your help in advance.

Best,
C.K.

If there aren't too many awkard index entries then use the \index{sortkey@printkey} method which prints printkey but sorts as sortkey. So \index{Sonmez@S\"{o}nmez} will print S\"{o}nmez but sorts it as if it were Sonmez and it therefore comes after Sasaki.

You have complete freedom to sort names in any order but it would be tedious to do hundreds of them. In that case you need a system where you can tell it what order you want the alphabet sorted, which means using a program like [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=xindy]xindy[/url].

Thank you so much stevem.

Best,

gametheorist


invariant
search
combinatorics proposed
combinatorics

This is one of the most beautiful problems of this type I have ever seen:

We have $\frac{n(n+1)}{2}$ stones and they are divided into a few groups. In each move
we take one stone from each group and form a new group with taken stones. Prove that after
a few moves we will get groups with $1,2,3,\dots,n$ stones (and after that position doesn't change).

(I'd like to get any information about the source of this problem.)

Enjoy solving!

[quote="Myth"][quote="fleeting_guest"]You can find more information by searching for "Bulgarian Solitaire".[/quote]
If you know the solution you can post it here.[/quote]

The shortest paper about this subject is longer than what I can post in one article.

There are quite a few papers about this problem since the 1980's when it was publicized in a Martin Gardner article (reprinted in [i]The Last Recreations[/i]). Some articles are available online (again I recommend a WWW search), and there are some solved and unsolved conjectures by Knuth and others.

[quote="fleeting_guest"][quote="sbos"]I now that there's one not so long solution in 'olympiad' style, I didn't know that it is a serious problem.[/quote]

I don't know how hard the problem is, but the solutions I've seen developed a long series of lemmas.  If there's a short solution it would be interesting to read it.[/quote]

well, i saw this paper in  ADVANCES IN APPLIED MATHEMATICS 21, page:205-227 (1998) titled 'The Cycling of Partitions and Compositions under Repeated Shifts';the proof in this is pretty short-less than half a page.the entire paper deals with some more variations and some bounds on the periodicity of cycles in Bulgarian Solitaire in the general case as well(non-triangular numbers).i got the paper through Google anyway...


calculus
integration
factorial
function
LaTeX
complex analysis
complex analysis unsolved

$ \int_{|z|=1}\frac{e^{kz}}{z^{n+1}}dz=\frac{2\pi ik^{n}}{n!}$
$ \int_{|z|=2}\frac{z^{3}}{z^{2}-2z+2}dz$ where $ \frac{z^{3}}{z^{2}-2z+2}=\frac{i}{2}*(\frac{1}{z-[1-i]}-\frac{1}{z-[1+i]})$ to obtain $ -\pi [(1-i)^{3}-(1+i)^{3}]=4\pi i$
$ \int_{|z|=2}\frac{e^{z}}{\pi\imath-2z}dz=-\pi ie^{(\frac{\pi i}{2})}=\pi$
Can anyone show solutions to these? Thanks in advance. PS.. I finally took the time to understand Latex. I thought I needed a seperate program to post in mathematical script! I'm going to really impress my profs next semester!

There is only one pole in the first one, so if $ f(z)=\frac{e^{kz}}{z^{n+1}}$, then

$ \int_{|z|=1}f(z)dz=2\pi i\text{Res}(f ; 0)$. To get the residue I used the Laurent series, which you should double check (I did it in my head) as $ f(z)=\frac{1}{z^{n+1}}+\frac{k}{z^{n}}+\frac{k^{2}}{2! z^{n-1}}+\cdots$, so the coefficient of $ \frac{1}{z}$ is $ \frac{k^{n}}{n!}$ and thus $ \int_{|z|=1}f(z)dz=\frac{2\pi ik^{n}}{n!}$

(edited: forgot the factorials and messed up the exponents)

All three of these are residue problems; the functions have one or more poles inside the circle, and we are to evaluate the integral by the residue theorem or an equivalent.
For the second, there are simple poles at $ 1\pm i$. For the third, there is a simple pole at $ \frac{\pi}{2}i$.
For the first, we have a pole of order $ n+1$ at the origin; to handle this, I'd write $ e^{kz}$ as its power series. The integrand becomes
$ \sum_{j=0}^{\infty}\frac{(kz)^{j}}{j!}z^{-n-1}=\sum_{j=0}^{\infty}\frac{k^{j}}{j!}z^{j-n-1}$
The integral of $ z^{m}$ around the circle is zero unless $ m=-1$, so only one term contributes to the integral.

Oh, and if you want to use $ \text{\LaTeX}$ outside this forum, you will need to download an implementation. I've done my homework in $ \text{\LaTeX}$ for the last two years.

can anyone show how to evaluate the third integral?
-Lauren

Hint: It may be easier to see in the form $ f(z)=\frac{e^{z}}{\pi i-2z}=\frac{e^{z}/2}{i\pi/2-z}$. Now you don't even need Residue Theorem. You can just use Cauchy Integral Formula.

I guess I got confused because the residue theorem isn't presented until later in the chapter, so only Cauchy was expected to be used, I believe. I've posted the solutions in the original post, but I'm still having difficulty. I think it stems from not quite understanding how to setup the Cauchy integral which is why I was hoping someone would be able to show the steps.
The complex analysis class that I will take next semester uses the Brown Complex Analysis with applications book, but the Howie book seems to go into more depth on the subjects, hence I've been studying mainly from that text. However, the Cauchy integration has been giving me trouble due to the lack of examples that I have been able to come across in books from the university library.
I'll look through internet findings, and hope that some examples will help give me a better understanding.

Let  $ f(z)=\frac{e^{z}}{\pi i-2z}=\frac{-e^{z}/2}{z-i\pi/2}$, so we somehow want a formula for $ \int_{|z|=2}f(z)dz$. Now Cauchy's Integral Formula says: $ g(z)=\frac{1}{2\pi i}\int_\gamma \frac{g(\zeta)}{\zeta-z}d\zeta$, with all the usual conditions on $ \gamma$ etc.

Plug in what you know in the formula. Let $ g(z)=\frac{-e^{z}}{2}$

$ g(i\pi/2)=\frac{1}{2\pi i}\int_{|z|=2}\frac{g(\zeta)}{\zeta-i\pi/2}d\zeta=\frac{1}{2\pi i}\int_{|z|=2}f(z)dz$

So if we multiply both sides by $ 2\pi i$ we get $ 2\pi ig(i\pi/2)=\int_{|z|=2}f(z)dz$ a formula we can evaluate: $ \int_{|z|=2}f(z)dz=2\pi i \left(-e^{(i\pi/2)}/2\right)=\pi$.

Hope that gives good step-by-step on how to apply the formula. It probably has too much stuff to be really clear though.

[quote="laurenviv"]THANKS! It was actually very helpful in understanding what I was missing! :idea:[/quote]

What I dont understand is that you have Howie's Complex Analysis book, I read it already :). It he does a great job explaining [b]all[/b] the excercises in the back of the book. So why didnt you just look there?

I agree that Howie does a much better job of explaining the material than the Brown book used in the class, as brown only gives final answers to a small amount of material.
I didn't quite grasp the answers in the back of Howie. The answers were not as thorough as I have been shown here. I think part of my trouble comes from the way I understand things. I like a large dose of examples, and an equally large dose of practice problems. I've found that often times the more advanced classes offer a minute amount of examples and practice problems. So, sometimes a thorough run-through of the problems clears the connections in my brain and it's like a light switch going off. I just have to keep trying and looking at examples until that light switch finally trickles on, but... that's just the way I understand things I guess is the short answer.


geometry rhombus calculus integration inequalities

For the rhombus problem:

[hide]We can divide the rhombus into four congruent 30-60-90 triangles with hypotenuse $ 4$, so the legs are $ 2$ and $ 2\sqrt{3}$. We want the distance between the focal points. If the semimajor and semiminor axes are of length $ a$ and $ b$, respectively, then $ c^2 \equal{} a^2\minus{}b^2$, where $ c$ is the distance between the foci. $ \sqrt{12\minus{}4} \equal{} 2\sqrt{2}$[/hide]

For the first:
[hide="Hint"]
AM-GM.
[hide="Further Hint"]
Look at the equality values closely . . .
[/hide]
[/hide]

I think it's not AM-GM, x,y is real number, we're not sure it's positive. I get $ (x,y)\equal{}(\minus{}3,5)$

AM-GM can be extended to negative numbers by $ \frac{|x| \plus{} |y|}{2} \ge \sqrt{|xy|}$. Unfortunately it doesn't help here because the given equation cannot be bounded; there are easily an infinite number of solutions unless there's a further restriction (integers?).

What is the source? For example, if I randomly let $ x \equal{} 9$, then we get $ 10 \plus{} y \plus{} \frac{25}{y} \equal{} 4 \Longrightarrow y^2 \plus{} 6y \plus{} 25 \equal{} 0 \Longrightarrow y \equal{} \frac{\minus{}6 \pm \sqrt{11}}{2}$, which already gives us two distinct values for $ 4a \plus{} 7b$.

[hide="#1."]
$ x\plus{}\frac{9}{x}\plus{}y\plus{}\frac{25}{y}\equal{}4$

$ x^2y\plus{}9y\plus{}xy^2\plus{}25x\equal{}4xy$

$ y(x\plus{}3)^2 \plus{} 25x \minus{} 10xy \plus{} xy^2 \equal{} 0$

$ y(x\plus{}3)^2 \plus{} x(y\minus{}5)^2 \equal{} 0$

One integral solution is $ (\minus{}3, 5)$.
[/hide]

[hide="if integral"] From AM-GM, we know both x and y can't be positive. If x is negative, than 9 divided by x is also negative. Similarly, if y is negative, 25/y is also negative. We can now factor out negatives when using AM-GM, so we have ${ |x+\frac{9}x}|\ge{6}$ and $ |y+\frac{25}{y}|\ge{10}$. Since -6+10=4, we must have equality achieved and have x be negative and y be positive, which gives the only integral solution (-3,5). Just plug in now.[/hide]

Or I think we could find $ y \equal{} .....x$ and then calculate $ 4a \plus{} 7b$ by $ x$.

[quote="xpmath"][hide="if integral"] From AM-GM, we know both x and y can't be positive. If x is negative, than 9 divided by x is also negative. Similarly, if y is negative, 25/y is also negative. We can now factor out negatives when using AM-GM, so we have ${ |x + \frac {9}x}|\ge{6}$ and $ |y + \frac {25}{y}|\ge{10}$. Since -6+10=4, we must have equality achieved and have x be negative and y be positive, which gives the only integral solution (-3,5). Just plug in now.[/hide][/quote]

Your inequality is backwards as I noted above.  :wink:

Okay, what I meant by AM-GM was to notice that if you use it on $ x \plus{} \frac {9}{x}$ you get a minimum of $ 6$, and if you use it on $ y \plus{} \frac {25}{y}$ you get a minimum of $ 10$.  $ 10 \minus{} 6 \equal{} 4$, so take the equality value for $ y$ and the negative of the equality value for $ x$ and you have a solution.

EDIT: Whoops, didn't notice that xpmath had already explained this method.


What is x if the reciprocal of x+1 is x-1?


vector
abstract algebra
linear algebra
linear algebra unsolved

let T be a linear operator defined on a finite dimensional vector space. show that there exists 'n' such that ker( T^n) and im(T^n) contain exaclty the 0 vector in commmon. ie intersection contains just 0.

[quote="julien_santini"]It is clear that $Ker(T^{n})$ and $Im(T^{n})$ are ultimately constant. If $N$ is such that  $Ker(T^{N})=Ker(T^{N+1})$ and $Im(T^{N})=Im(T^{N+1})$, then $Ker(T^{n}) \cap Im(T^{n}) = \{0\}$, otherwise we would have $Im(T^{N+1}) \subsetneq Im(T^{n})$.[/quote]

may be i'm stupid, but i dont understand why, ker (T^n) and im (T^n) should be constant ultimately. i'm really sorry, could you give a little more detailed solution? thanks in advance

it's obvious that it can't be strictly increasing as your dimensions are finite

so there has to be a n for which $Im(T^{n+1})=Im(T^n)$
now the point : it doesn't ever strictly increase after that :

if v is in $Im(T^{n+2})$, then $v=T(w)$ for some $w\in Im(T^{n+1})=Im(T^n)$
this means $v\in Im(T^{n+1})=Im(T^n)$

it goes almost the same for the kernels


let a be a positive real number and sequence with $ u_{n}\equal{} 1\plus{} \frac{1}{2^a} \plus{} \frac{1}{3^a} \plus{}...\plus{}  \frac{1}{n^a}$
Prove, exist $ x$ diffrent from $ \infty$ such that  $ limu_{n}\equal{}x$

This is true if and only if $ a > 1$; in particular, it fails for $ a \equal{} 1$.  What do you know about series that you can use here (since you are being asked to prove the power test)?  One great idea is [url=http://en.wikipedia.org/wiki/Cauchy_condensation_test]Cauchy condensation[/url].


vector
function
trigonometry
analytic geometry
geometry
3D geometry

In the classical version of the Hahn-Banach extension theorem, as well as in other versions dealing with vector-valued or set-valued maps that I've seen, an assumption on the majorant is that it is subadditive. Are there versions where the assumption is weaker, such as $ p(x\plus{}y) \leq K(p(x)\plus{}p(y))$ for some $ K \geq 1$?

The assumption that the majorant is sublinear is motivated by the fact that the supremum of any family of linear functions is sublinear. It looks quite reasonable to me.

You'll find counterexamples in $ \mathbb R^2$ by defining $ p$ so that it agrees with $ r|\cos\theta|$ at the coordinate axes but is smaller otherwise.

The reason I'm asking is because recently I read an article where the norm $ \parallel{}\cdot\parallel{}: X \rightarrow \mathbb{R}$ was replaced by a cone norm $ \parallel{}\cdot\parallel{}_{P}: X \rightarrow E$ where $ E$ is a real Banach space with a cone $ P$. The cone is assumed to be normal, that is, $ 0 \leq x \leq y$ implies $ \parallel{}x\parallel{} \leq M \parallel{}y\parallel{}$ where $ M \geq 1$ is the normality constant. I was trying to see if a version of the theorem would be valid in this setting.


Putnam
logarithms
inequalities
college contests

Prove that there exists a constant $ c>0$ such that in every nontrivial finite group $ G$ there exists a sequence of length at most $ c\ln |G|$ with the property that each element of $ G$ equals the product of some subsequence. (The elements of $ G$ in the sequence are not required to be distinct. A [i]subsequence[/i] of a sequence is obtained by selecting some of the terms, not necessarily consecutive, without reordering them; for example, $ 4,4,2$ is a subesequence of $ 2,4,6,4,2,$ but $ 2,2,4$ is not.)

Solution due to JSteinhardt:

Take some sequence and look at the set $ S$ generated (in the subsequence sense) by it.  If the set $ S^{ \minus{} 1} S$ is all of $ G$, we can take the sequence that begins with the elements of $ S^{ \minus{} 1}$ and continues with the elements of $ S$, and that doubles the length of our sequence.  Otherwise, there is some $ x \not \in S^{ \minus{} 1} S$, which implies that $ S$ and $ Sx$ are disjoint.  Then $ S$ with $ x$ appended to it generates twice as many elements as $ S$ did.  Repeating the second move some number of times and then ending with the first move gives the bound $ c \equal{} \frac {2}{\ln 2}$.

I went one element at a time. Start by defining the empty product to be $ e$; if you don't like that, start with the sequence $ e$. Let $ A_k$ be the set of all products of our $ k$-element sequence. The cosets $ A_k x$ cover $ G$ evenly, so there is some $ x$ with $ |A_k\cap A_k x|\le \frac {|A_k|^2}{|G|}$, and thus $ |A_k\cup A_k x|\ge 2|A_k| \minus{} \frac {|A_k|^2}{|G|}$. We append this $ x$ as the $ k \plus{} 1$st element of the sequence, and get $ A_{k \plus{} 1} \equal{} A_k\cup A_k x$.
If $ \frac {|A_k|}{|G|} < \frac12$, we have $ |A_{k \plus{} 1}| > \frac32|A_k|$. If $ \frac {|A_k|}{|G|}\le\frac12$, we have
$ |G \minus{} A_{k \plus{} 1}|\le|G| \minus{} 2|A_k| \plus{} \frac {|A_k|^2}{|G|} \equal{} \left(\frac {|G \minus{} A_k|^2}{|G|}\right)$
$ \equal{} \frac {|G \minus{} A_k|}{|G|}\cdot |G \minus{} A_k|\le\frac12|G \minus{} A_k|$

It takes less than $ \frac {\log \frac n2}{\log \frac 32}$ steps for $ A$ to go from size 1 to size $ \frac n2$, and at most $ \frac {\log\frac n2}{\log 2}$ steps for $ |G \minus{} A|$ to go from size at most $ \frac n2$ to size at most $ 1$. That gives $ c \equal{} \frac1{\ln 2} \plus{} \frac1{\ln\frac32}$; we may add one to go from one missing element to zero and/or the identity element to prime things, but taking those $ \ln\frac {n}{2}$ to $ \ln n$ counters those.

This is also an overestimate of how long the method takes, especially when $ |G \minus{} A_k|$ is small.

I did the first part (the averaging method) as you did, jmerry.  The only extra thing I noticed is that the inequality
\[ |A_{k \plus{} 1}| \ge 2|A_k| \minus{} \frac {|A_k|^2}{|G|}
\]
is equivalent to the inequality
\[ \frac {|\overline{A_{k \plus{} 1}}|}{|G|} \le \left( \frac {|\overline{A_k}|}{|G|} \right)^2.
\]
In other words, the missing fraction will square (at least) at each step.  After $ m$ steps, we have
\[ \frac {|\overline{A_m}|}{|G|} \le \left( \frac {|\overline{A_0}|}{|G|} \right)^{2^m} \equal{} \left( 1 \minus{} \frac {1}{|G|} \right)^{2^m} < e^{ \minus{} 2^m / |G|}.
\]
So as long as $ 2^m \ge |G| \ln |G|$, the set $ \overline{A_m}$ will be empty, which means $ A_m$ is all of $ G$.  Thus we can choose the length $ m$ to be asymptotic to $ \log_2 |G|$, which is optimal.

Let $a_1, a_2, \dots$ denote the sequence, and $P_k$ denote the set of products of the subsequences of $a_1 \cdots a_k$ (set $P_0 = \{1_G\}$). Apply the following [i]key[/i] as many times as possible:
[list]
[*] If there exists a $g \in G$ with $P_k \cap P_k g = \emptyset$, set $a_{k + 1} = g$.
[/list]
Initially, $|P_k|$ is one and doubles with each application of the key, so the key is applied at most $\log_2 |G|$ times.

Consider the final set $P_k$ after the key has been applied. Since the key cannot be applied again, for each $g \in G$, there exists $p \in P_k$ with $pg^{-1} \in P_k$ too.

But now we claim $P_k^{-1} P_k = G$. Indeed, by the above, we have $(pg)^{-1} \in P_k^{-1}$ and $p \in P_k$, so $(pg^{-1})^{-1} p = g$ is in $P_k^{-1} P_k$.

Thus, if the key generates the sequence $a_1, \dots, a_k$, then the sequence
\[a_k^{-1}, \dots, a_1^{-1}, a_1, \dots, a_k\]
is valid and has length $2k \le 2 \log_2 |G| = \tfrac{2}{\ln 2} \ln |G|$. Thus, $c = \tfrac{2}{\ln 2}$ works.

The key is to consider cosets. They behave like modulo.

We consider this process:

Let $T$ be a subset of elements that can be generated by $S$. Initially, $T=\{e\}$. As $T$ changes, call two elements $a,b$ equivalent if there exists $t\in T$ such that $at=b$. (In short, $aT=bT$; think of right cosets)

If $T \ne G$, pick an element in $G\backslash T$, call $x$. Consider $H_x=\{e,x,\cdots,x^{n-1}\}$, where $n$ is the smallest integer such that $x^nT=T$. Put $x, x^2, \cdots, x^{2^k}$ in $S$ where $k = \lfloor \log_2(n) \rfloor$. Replace $T$ with $T\cup xT\cup \cdots \cup x^{n-1}T$. The size of $T$ gets multiplied by $n$.

Suppose I do this $k$ times, each time $|T|$ increases by factors of $n_1,\cdots,n_k$. Then $|G| = \prod\limits_{j=1}^k n_j$ and $|S| = \sum\limits_{j=1}^k  \lfloor \log_2(n_j) \rfloor \le \sum\limits_{j=1}^k \log_2(n_j) = \log_2(|G|)$.

As needed.


Hey. This is my first year going to nats, and I would appreciate lots of help about what to expect. Thanks in advance!!!!!!! :lol:

1. What are the hotels like?

2. Do you always room with your requested person?

3. Do we test in a big room together or separate smaller rooms?

4. What is the yearbook?

5. What places do they award up to at the awards ceremony?

6. How fast does the countdown reader read (I don't think I'll make nats countdown. I really doubted even making state countdown, so making nats was a surprise.)?

7. When do they give you calculators? In time to use for the target and team rounds?

8. What time intervals do they give you on sprint/team/target?

Again, thanks!!!!!!  :lol:

1. The hotels are decent, I mean, it doesn't really matter does it?

2. I believe if you put down each other, you should be roomed together, I'm not sure about if only one of you does...

3. You test in a big room.  To elaborate, you sit together as a team, with another team coach at your table.  For the indiv round, ppl r separated by dividers.

4. The yearbook is a yearbook.  It has photos of all the participants, and some basic info about them.

5. I believe all the people who make the cd round are recognized in the awards ceremony, along with top 2 written, masters round, teams, spirit award, 3 time participants, etc...

6. I really don't remember anything about this.  The level of the participants is so high, i don't think it matters.

7. For the last 2 yrs, calcs were distributed along with all the other stuff when we arrived, given to the coaches, and then handed out.  You will have it in time to use for the comp.

8. Time intervals are pretty much the same.  Maybe a little longer, b/c after every round you are allowed to see the answers, and may file a protest.

Hope this Helps!

P.S. I was a participant in '06, '07, and will be going to this yrs nats, so the info should be right, if they don't make any drastic changes this yr

[quote="Quickster94"]1. The hotels are decent, I mean, it doesn't really matter does it?[/quote]

I think the hotels are really good.  They're not like uber-fancy, but that's not to say they weren't nice.  And the vastness of the hotels makes for a great time of lapel pin-trading! :lol:

[quote="dynamo729"]4. Who gets a yearbook?

5. Does everyone do masters round?

8.  I mean what intervals do they call when you take the test like "5 minutes left" or something.[/quote]

4. Everyone who participates in nats.
5. No, only the top two written and top two cd, I think.  I don't know, cuz I had to fly from Fort Worth back to GA on the 2nd day for a state piano competition.  I'll have to do that this year, too. :|
8. They have a giant stopwatch thing on a screen in front of everyone, so it doesn't matter, really.

The hotel is normally extremely nice, considering it usually costs around 250 dollars a night.  Youre roomed with your requested person unless they put someone else.  They award first through twelfth.  Top two in written and top two countdown to masters.  And if you make countdown, you should know that the speed at which the announcer reads is irrelevant.  Everyone answers long before the the announcer finishes.

I'm not sure what nats is like (for obvious reasons!)  :P

Anyway, good luck to you people! Rawr!

I have some questions also. (Considering this is my first and only year at nats.)

Do all top 12 from countdown get trophies?
If not, what do they get?
In countdown, after the first round (5v12,6v11....) who plays who in quarters and who plays who in semis?
How many of the top teams get awards/trophies?
Any other awards other than top written guys, top countdown guys and top teams?

What do you do with the masters round?(hand it in?)
How do they decide 3,4,5,6,7,8,9,10,11, and 12th place?
Lastly: Around how long is the average answering time for a countdown question at nats?

(Response)

Do all top 12 from countdown get trophies?
Yeah.

In countdown, after the first round (5v12,6v11....) who plays who in quarters and who plays who in semis?
Randomized, I believe.

Any other awards other than top written guys, top countdown guys and top teams?
Top masters round, and the Spirit award.

What do you do with the masters round?(hand it in?)
It's a proof, and you present the proof in front of a panel of judges and most of the Mathcounts contestants.

How do they decide 3,4,5,6,7,8,9,10,11, and 12th place?
Well, score, obviously, and tiebreakers. Basically, if P1 gets #30 wrong and P2 gets #29 wrong, P2 would be in front of P1.

Lastly: Around how long is the average answering time for a countdown question at nats?
Very, very short. If you expect to do well in countdown, you have to be able to answer almost every question in less than, say, 10 seconds. Most questions are done in 3 seconds though. I think.


geometry
parallelogram
geometry proposed

[i]This is, to the best of my knowledge, not an original problem - it was probably used some time in the past in a competition or in a book of some kind. This was Problem 3 (Geometry) for 10th grade on 1st Balkan Student Mathematical Competition 2008 - a competition 11 schools from Balkan area took part of during November 2008. All information about the competition can be found on http://www.mathlinks.ro/viewtopic.php?t=243614.[/i]

Let $ M_n$ and $ N_n$ be points on sides $ \overline{CA}$ and $ \overline{CB}$ of triangle $ ABC$, respectively, such that $ |CM_n| \equal{} \frac {1}{n}|CA|$, $ |CN_n| \equal{} \frac {1}{n \plus{} 1}|CB|$,  $ \forall n \in \mathbb{N}$.
Find the locus of points $ M_iN_i \cap M_jN_j$, where $ i$ and $ j$ are different natural numbers.

Note. $ M_iN_i \cap M_jN_j$ denotes the intersection of lines $ M_iN_i$ and $ M_jN_j$.

I'll write the solution when I have time, and yes this problem can be found in books. [hide="The answer"] Point D such that ABDC is paralellogram [/hide]

HOHOHO you don't have any time but I have some to borow :D :D

When we concider that $ B\in s\parallel CA$, $ C\in r\parallel AB$ and $ F=s\cap r$ therefor $ [ABFC]$ it's a paralelogram and we have some stuff like the colinearity of $ A, M_2$ and $ F$ or $ \frac{M_2F}{FA}=-\frac{1}{2}$
We assume that $ M{}_iN{}_i%Error. "capM" is a bad command.
_{j}N_{j}=F$

Proof: notice that $ AM_i=AC-M_iC=AC\left( 1-\frac{1}{i}\right) =AC\left( \frac{i-1}{i}\right)$ and that $ M_iC=AC\frac{1}{i}$ and $ CN_1=CB\frac{1}{i+1}$ and finaly that $ N{}_iN{}_2=CB-CN_i-N_2B=CB\left( 1-\frac{1}{i+1}-\frac{1}{2}\right) =CB\frac{i-1}{2(i+1)}$
As well we have that $ \frac{AM_i}{M_iC}\times \frac{CN_i}{N{}_iN{}_2}\times \frac{N_2F}{FA}=\frac{AC\left( \frac{i-1}{i}\right)}{AC\frac{1}{i}}\times\frac{CB\frac{1}{i+1}}{CB\frac{i-1}{2(i+1)}}\times\frac{-1}{2}=(i-1)\times\frac{2}{i-1}\times \frac{-1}{2}=-1$ and by the menelaus teorem the colinearity of $ F, M_i$ and $ N_i$ is proven for any $ i$ and that means $ F\in M{}_iN{}_i$ for any $ i$ as we want to proof :D


analytic geometry
calculus
derivative
real analysis
real analysis unsolved

Show that all non-trivial solutions to the system
$ x_1' \equal{} 2x_2 \plus{} x_1(x_1^2 \plus{} 2x_2^2)^2$
$ x_2' \equal{} \minus{} x_1 \plus{} x_2(x_1^2 \plus{} 2x_2^2)^2$
go to infinity in finite time

~~>
\[ \begin{gathered} x'_1 \equal{} 2\frac {{x_2 }} {{\sqrt 2 }} \plus{} x_1 \left( {x_1^2 \plus{} 2\left( {\frac {{x_2 }} {{\sqrt 2 }}} \right)^2 } \right)^2 \hfill \\
\frac {{x'_2 }} {{\sqrt 2 }} \equal{} \minus{} x_1 \plus{} \frac {{x_2 }} {{\sqrt 2 }}\left( {x_1^2 \plus{} 2\left( {\frac {{x_2 }} {{\sqrt 2 }}} \right)^2 } \right)^2 \hfill \\
\end{gathered}
\]
~~>
\[ \begin{gathered}
  x'_1  \equal{} \sqrt 2 x_2  \plus{} x_1 \left( {x_1^2  \plus{} x_2^2 } \right)^2  \hfill \\
  x'_2  \equal{}  \minus{} \sqrt 2 x_1  \plus{} x_2 \left( {x_1^2  \plus{} x_2^2 } \right)^2  \hfill \\
\end{gathered}
\]

Converting to polar coordinate: $ r' \equal{} r^5$ which yields general solutions look like
\[ r\left( t \right) \equal{} \frac{{ \pm 1}}{{\sqrt[4]{{4C \minus{} 4t}}}}
\]
which completes the proof.

[quote="Kalle"]I don't understand how you got this transformation. I kind of doubt that this can be solved explicitly, but very nice job if you actually did. A less explicit solution can be done by studying the derivative of the square expression.[/quote]

Actually, $ (x_1, x_2)$ ~~> $ (x_1, x_3)$. The key here is to simplify $ x_1^2\plus{}2x_2^2$ to $ x_1^2\plus{}x_3^2$ in order to use polar coordinate.


What is the sum of all values of $ y$ for which the expression $ \frac{y\plus{}6}{y^2\minus{}5y\plus{}4}$ is undefined?

$ \frac{y\plus{}6}{y^2\minus{}5y\plus{}4}\equal{}\frac{y\plus{}6}{(y\minus{}1)(y\minus{}4)}$,
Thus, $ 1\plus{}4\equal{}\boxed{5}$.

This is when $ x^2 \minus{} 5x \plus{} 4 \equal{} 0$ We apply Vieta's and get $ \frac { \minus{} (\minus{} 5)}{1} \equal{} 5$


logarithms
advanced fields
advanced fields unsolved

Prove that:
\[ \sigma(n)=O(n \log n)  \]

P.S. sorry if this is trivial, it's a little bit late and nothing comes to my mind

$\sigma (n) \leq n (\frac 11 + \frac 12 + \frac 13 + ... \frac 1n) \approx n \log (n)$

Perhaps
\[ \log \log n - 1 \leqslant \sum_{2 \leq p \leq n} \frac{1}{p} \leqslant \log \log n + c  \]
(where $p$ runs over primes and $c$ is a constant.) Is a bit more challenging ? ;)


Here's a nice question to ponder. It may be trivial (I haven't yet spent any time on it), but I don't have a solution, so I'll put it in the 'Unsolved' section.

Let $a_1,\ldots,a_k \in \mathbb{R}^+$ be pairwise distinct positive reals, and let $c_1,\ldots,c_k \in \mathbb{C}$ be constants such that
\[
c_1a_1^{1/n} + c_2a_2^{1/n} + \cdots   + c_ka_k^{1/n} = 0
\]
for infinitely many $n \in \mathbb{N}$. Prove that $c_1 = \cdots = c_k = 0$.

--Vesselin

I am quite tired now, but I am sure the following method works: write $ a_i=1+b_i$ and develop each root with Taylor formula. At each stage pass to limit as n is large and delete all factors equal to 0 that you have already obtained (if you want, it is the reasoning to prove the unicity of the writing as an entire series adapted). You will find that the sums of the numbers $ c_i b_i^k$ are 0 for all k, and from here the conclusion is not far.  ;)


geometry
MATHCOUNTS
analytic geometry
AMC
USA(J)MO
USAMO
circumcircle

Here's the test results. In all lists, 2nd through 5th were pretty close while 1st was ahead by at least a substantial margin.

(AOPS names used)

[hide="Red MOP test winners"]
5th: Ubemaya
4th: Sean Gu
3rd: kyyuanmathcount
2nd: bookaholic
1st: CatalystOfNostalgia[/hide]

[hide="Blue MOP test winners"]
5th: gighiuhui
4th: jb05
3rd: TheBernstinator
2nd: ductoan0503
1st: SplashD[/hide]

[hide="ELMO Winners"]
5th: J. Iglesias (sorry, don't know AOPS name)
4th: minsoens
3rd: SplashD
2nd: Ubemaya
1st: ductoan0503[/hide]

[hide="Black MOP test winners"]
5th: K81o7
4th: SamE
3rd: 101101101
2nd: HariSeldon(insert numbers here)
All with less than 60 MOP points. With about 90...
1st: probability1.01[/hide]

Quotes, hall of fame, and other results like USEMO and style to follow.

Good luck to this year's IMO team, and now everyone get cracking so you can make MOP next year.

[quote="Sean Gu"][quote="junggi"][quote="Sean Gu"]usually how many red MOPpers make blue or black next year... :?:[/quote]
this year....like uh 3 or 4? lol

oh man mop was pretty intense
i kinda failed the tests hmmm i guess i should study a lot this year[/quote]

I think this year was pretty special since there were no 10th graders in black....[/quote]

No actually it's pretty common to only have like, 4-5 red MOPpers make it back the next year.

In '07, 5 of the '06 red MOPpers made it back: Sergei, Eric, Delong, David B. and Christian (and Eric and Delong really should have been in Blue instead of red). (At the same time, only 3 '06 [i]blue[/i] MOPpers made it back...)

In '06, 5 of the '05 red MOPpers made it back: Joseph, Alex, Shui, Shaunak and Yi Han.

(sorry if I missed anyone...)

Basically, it's [i]much[/i] harder to to get into blue than it is to get into red. Red MOP alone is not even close to enough to enough to get you into blue. To get in, you need to improve [i]a lot[/i], and you need to work pretty hard. And even then, you don't have much of an advantage over the non-MOPpers who are also working hard.

I want to see these usemo results.

[quote]
Just wondering...has anyone ever made black MOP as a senior without ever having made MOP before?
[/quote]

Yes, Keenan did it in 05 and Oleg did it in 04 (though this doesn't really count).

[quote]
Basically, it's much harder to to get into blue than it is to get into red. Red MOP alone is not even close to enough to enough to get you into blue. To get in, you need to improve a lot, and you need to work pretty hard. And even then, you don't have much of an advantage over the non-MOPpers who are also working hard.
[/quote]

This is definitely true -- you should aim for problems at least a tier harder that what you can do now.  There are plenty of these at the end of the handouts you guys have.  And also patch up your weak areas; it's hard to make it back to MOP if you can't solve any problems from a certain area.

[quote="Klebian"][quote="MellowMelon"]

[hide="Red MOP test winners"]
5th: Ubemaya
4th: Sean Gu
3rd: kyyuanmathcount
2nd: bookaholic
1st: CatalystOfNostalgia[/hide]

[/quote]
bookaholic is in 10th grade[/quote]

she was still in red mop

[quote="junggi"]oh man mop was pretty intense
i kinda failed the tests hmmm i guess i should study a lot this year[/quote]

Or just play less Starcraft.  :lol:

I got home at 2 PM yesterday and slept from 2 PM to 4 AM.  I guess I was really tired.

MOP was amazing by the way.  Except I can't believe I lost to David Bay Rush for 5th in blue...

I don't have full USEMO results, just the top 5 and about what score they said. Ask Sam or someone for more.

(I can't find the contact sheet and I know very few AOPS names of people in these lists, so using actual names)

[hide="USEMO Top 5"]5. Palmer Mebane ~68 (woo!)
4. Lynnelle Ye ~81
3. Po-Ru Loh ~82
2. Gabriel Carroll ~84
1. Alex Zhai ~111 (ouch)[/hide]

[hide="USEMO Special Awards"]Unfortunately I missed copying the special round awards, so I'll list as many as I can remember.

Bryce Taylor got >300 solutions on the "secret" round on Diophantine Equations (writing $ (0,n)$ solutions on 4). This was the most by far.

Qinxuan Pan, however, proved the completeness of his solutions for 2 problems; that was a lot of points.

This wasn't actually mentioned, but I was told this personally and my ego is big. I got 20 on the 5-minute Mathcounts sprint round. Matt Superdock was second with a 17. He also did them in order, so he was probably faster on average than me.

Po-Ru Loh was on average a couple millimeters off on Find Your Center. I think it is no coincidence that he taught the drawing diagrams class (surprisingly informative).

Two people got three parts of Mathematical Astrology, but I remember neither.[/hide]

Let's throw some more stuff out.

[hide="Best and Worst Style"]- Best -

Red: Carolyn Kim 0.797
(RU: Colleen Lee 0.795)

Blue: Minseon Shin 0.8??, missed this
(RU: Sam Keller 0.798)

Black: Sam Elder 0.81?
(RU: Alex Zhai 0.79?)
I think those are wrong.

- Worst -

Red: Gray Symon 0.662 :o
(RU: Amol Aggarwal 0.668)

Blue: Taylor Han 0.707
(RU: Justin Brereton ??)

Black: Colin Sandon ??
(second not mentioned)[/hide]

[hide="Best Team Contest Team"]David Bay Rush
Minseon Shin
Johnny Li
Joy Zheng
Lynnelle Ye

This is probably not valuable because I don't have the list of other teams.[/hide]

Will eventually type up Hall of Fame and quotes... plenty of good quotes...

About this whole getting to Blue from Red thing, this year the Red cutoff was 10 (and that's high) and the Blue cutoff was 20. The difference between these scores is much more than just a factor of two, because getting a 20 means solving a problem that isn't 1 or 4. You red MOPpers should listen to Yi Sun for that reason.

[quote="MellowMelon"]
Bryce Taylor got >300 solutions on the "secret" round on Diophantine Equations (writing $ (0,n)$ solutions on 4). This was the most by far.
[/quote]

So it seems that >300 solution did not bring him >300 points?

[quote="1=2"][quote="MellowMelon"]5-minute Mathcounts sprint round[/quote]

 :o  :o  :o

HOW CAN ANYONE DO THAT!?!?!?!?[/quote]

If they can make MOP black...I think they can do that.

[quote="1=2"][quote="MellowMelon"]5-minute Mathcounts sprint round[/quote]

 :o  :o  :o

HOW CAN ANYONE DO THAT!?!?!?!?[/quote]

It's not supposed to be easy...

Argh...if just I hadn't bombed those three tests in the middle of MOP...

Anyways...some more info...

The two people who got three of mathematical astrology: I remember one was Ian, don't remember the other.

Also, David Rolnick won Miscellaneous (I know because I graded it all), but I don't know whether we announced that.

[quote="matt276eagles"][quote="not_trig"]in other news, Delong was definitely later to class than I was?[/quote]

Um, Delong didn't even come to the morning classes that day, so he wasn't late, he was just... not there.[/quote]

what? how is that not "late"  :maybe:

missing a whole class and arriving at lunch is less late than arriving at 10? I am pretty confused o.O

:P

He never arrived at the class, so you can't say he was late.  This is a really pointless argument anyway.  :lol:

Also, I find it hilarious that David still remembers his MOP score from last year  :P

@M@

oh man that was awesome--i just got to go like (at)M(att)

but anyway M@, I saved my MOP tests and added.  Or maybe I just made the score up ;)  My memory is pretty impeccable, don't get me wrong, but... well not that impeccable ;)

So you have to be in 9th grade when you take USAMO? To be eligible for Red MOP?

Only $ 9$th graders can make red MOP. A high scoring $ 8$th or $ 10$th grader or any other grade won't be eligible. (excluding CGMO considerations, which may not be the same next year anyway)

Can you go to Blue or Black MOP in 8th grade?

MOP eligibility (this has been mentioned in the thread by the way)

[color=red]Red[/color]: 9th grade only, cutoff usually near 9
[color=blue]Blue[/color]: Any grade below 12th (including 8th, 7th, whatever), cutoff usually near 20
Black: Any grade, cutoff usually near 28

blargh I don't think so

of course a lot may change in a year

i think the red MOP rules are pretty pointless...

They have their reasons, but this has been discussed rather heatedly already in other threads.

He's referring to the fact that only 9th graders are invited to attend red MOP (with the exception of the CGMO team this year).

just out of curiosity, if somebody has a legal sex change to become female, are they eligible for CGMO? hmm.  :idea:


Find a formula for a quadratic that passes through the points (8,0) and (-2, 0). Explain your reasoning behind your formula choice.

What I did:

(8,0)
y=ax^2 + bx + c
so
0=64a + 8b + c

(-2,0)
0=4a-2b+c

64a + 8b + c=0
-
4a-2b+c=0
_____________
60a + 10 b = 0

Then I stuck.

Help please!

[hide]First graph the points and note that the y-coordinates are the same; therefore the line must be in the form y=a. Now plug in our x and y values to learn 0=a, so the desired formula is y=0. [/hide]

Billy

there are infinite quadratics that pass through those 2 points...3 points determine a quadratic, perhaps one of the given points was the vertex, but what you can have is a family of solutions using the formulas that you got

[hide]Since $x=-2$ and $x=8$ are the roots of the quadratic, we can write the quadratic function as $y=a(x+2)(x-8)$.

As Altheman mentioned, there are infinitely many possible quadratic functions that satisfy this condition because $a$ could be any real number.

Since we only need to find one of such quadratic function, just pick any number you want for $a$ :P
[/hide]


Consider a quadrilateral $ABCD$ such that $\angle BCD = \angle ADC$. If $E$ lies on segment $CD$ such that $BE$ bisects $\angle ABC$ , prove that $\angle AEB$ = $\pi/2$ iff $AD+BC = AB$

$S(BE)$ then $A \to A'$. Thus $AB=BC+AD \Longleftrightarrow A'B=BC+AD \Longleftrightarrow A'C=AD$. Since we have <BCE=<ADE, so <ECA' + <AED = $180^{o}$. Hence $\sin{\angle{ECA'}}=\sin{\angle{ADE}}$, thus: \[\frac{A'C}{\sin{\angle{CEA'}}}=\frac{A'E}{\sin{\angle{A'CE}}}=\frac{AE}{\sin{\angle{ADE}}}=\frac{AD}{\sin{\angle{AED}}}\] Hence $A'D=AD \Longleftrightarrow \sin{\angle{CEA'}}=\sin{\angle{AED}}\Longleftrightarrow \angle{AED}=\angle{A'EC}$
$\Longleftrightarrow A', E, A$ collinear $\Longleftrightarrow AE \perp AB$.


Someone posted this a while ago and I don't believe it was answered.

On the plane with lattice points,there is a frog.
First it is on the point(0,0).At each second if it is on point (x,y) it goes to point (x+1,y) or (x,y+1).
Prove that for each n there are n collinear points on the frog's path.

Thank you!

It was solved by pbornsztein, so I suggest thinking of a good keyword ("frog" should work :)), and searching pbornsztein's posts which contain that word (there certainly can't be that many, right? :)).

The problem statement does not make sense.
What does it mean that "for each $n$ there are $n$ collinear points on the frog path"?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5294]Here's[/url] the original thread.


function
algebra
polynomial
geometry
3D geometry
tetrahedron

can anyone simplify this if n is odd?

If it's even it can be simplified since n can be 2x, which then becomes
(a^x+b^x)(a^x-b^x).

But, then, x could be odd unless n is an exponential function of 2.
Since that is one peek, what's the rest?

 :cap:  :D  :cup:

When a = b, $a^n - b^n = 0$, so there should be a factor of $(a-b)$ in there somewhere.  Using synthetic division, we divide $\frac{a^n - b^n}{a-b}$ by considering $a^n - b^n$ a polynomial in a:

[code]
b | 1  0    0    0 ......  0     -b^n
---    b   b^2  b^3     b^(n-1)   b^n
----------------------------------------
   1   b   b^2  b^3 ... b^(n-1)   0
[/code]

So $\frac{a^n - b^n}{a-b} = a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \ldots + b^{n-1}$, or
$a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \ldots + b^{n-1})$

This works for any $n$, odd or even.

[quote="Scrambled"]geometric sequence:
$(a+b)(a^n+a^(n-1)b+a^(n-2)b^2...+b^n)$[/quote]

Scrambled, you need to use {}s instead of ()s, the thing in the second parenthesis should have highest term $a^{n-1}$ rather than $a^n$, and I think $(a-b)$ instead of $(a+b)$.

[quote]can you simplify what's on the right?[/quote]

In the general case, we cannot factor

$a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1})$

any further than it already has been. In specific cases, it is possible to be further factored. Consider b = 1. Then we have that

$a^n - 1 = (a-1)(a^{n-1} + a^{n-2} + \cdots + a + 1)$

which can be further factored as

$(a-1)(a- \omega )(a- \omega ^2)\cdots(a- \omega ^{n-1})$

where $1, \omega, \omega ^2 ,\ldots, \omega ^{n-1}$ are the nth roots of unity.

In a certain problem, I took out the (a-b) as a factor.

Can the rest still be factorized?

Or proven to be divisible by something?

Cheat...

the only thing i can think of is that its divisible by gcd(a,b)...
other than that, i believe it is what tetrahedron said.

I still need to know if it's divisible by something.

a's and b's are defined by certain values, too.

two pairs, but I will not say them right now.


Find all positive integers such that the sum of their digits in base ten equals their square root, e.g. $1=\sqrt{1}$, but $2+0$ does not equal $\sqrt{20}$.

Another example: $81$.  $8+1=\sqrt{81}$.  Are there any positive integers other than $1$ and $81$ that satisfy this property, or are those two the only ones?

[quote="chess64"][hide]Well modulo 9, $N\equiv \text {(sum of N's digits)} \pmod {9}$, so $\sqrt{N} \equiv N \pmod {9}$. Squaring gives \[ N^2\equiv N \pmod {9} \qquad (*).  \] I think it's straightforward from here.[/hide][/quote]
Could you lay out your conclusion please?  You say it's "straightforward," but it's not so straightforward for me.  You have that if $N$ meets the criteria, then $N^2\equiv N\pmod {9}$.  I get that part, but you haven't established that $N$ meets the criteria if and only if $N^2\equiv N\pmod {9}$ (in fact, that's obviously false, e.g., $18^2\equiv 18\pmod {9}$, but $18$ clearly does not meet the criteria).  So which $N$ that satisfy $N^2\equiv N\pmod {9}$ satisfy the criteria?

Well 18 is not a perfect square... ;) I'll finish for you though.

Let $M=\sqrt{N}$.

$M^2\equiv M \pmod {9}$

$0^2,1^2,2^2,3^2,4^2,5^2,6^2,7^2,8^2\equiv 0,1,4,0,7,7,0,4,1\pmod {9}$

So if $M^2\equiv M\pmod {9}$, then $M\equiv 0,1 \pmod {9}$.

OK so now you have to consider cases. However, not all solutions will work because even if $a\equiv b$, $a=b$ is not necessarily true. There is probably a better solution though.

Sorry!
-chess64

OK, that part's clear, but I'm still wondering about your conclusion.  Which numbers work?  Are there are an infinite number, or a finite number?  If a finite number, which ones?

no its not infinite cuz say if theres 4 integer places the largest it can be is 999 and $sqrt{1999}$ is greater than  1+9 + 9 + 9

which says that the sqrt of a number will over take the suum of the digits in it

let $a$ be the one's place, $b$ be the tenth's place, and $c$ be the hundredth place

if theres 1 digit place then

$a = sqrt(a)$

if theres 2 digit place then

$a + 0.1b = sqrt(a+b)$

and find integer solutions or something lol

well, it cannot be more than a 4 digit number, because, the highest sum of digits is 36, yet the lowest square root is 10\sqrt{10}, so we need to look at 3, 2, 1 digits

but we can really just do casework with all the perfect squares that are less than 1000, using the difference between (n+1)^2 and n^2 =2n+1, gives us an easy to write out a table real fast


if $a,b,c,d$are natural numbers and we have  $ab=cd$
 can $S=a+b+c+d$be prime number or not???

[hide]Looking at the definition of a natural number, I will assume 0 is not included: 0+1+0+2=3

ab=cd
ab/c=d

a+b+c+ab/c=S
ac+bc+c^2+ab=Sc
(a+c)(b+c)/c=S

Since S is a whole number b/c or a/c must be whole as well.  Without loss of generality, let a be divisible by c

(a/c+1)(b+c)=S

Both of the factored parts are greater than 1 so no (I hope)[/hide]

[quote="ehehheehee"]Since S is a whole number b/c or a/c must be whole as well.  Without loss of generality, let a be divisible by c[/quote]

This is false.  For instance, take $(a, b, c, d) = (2, 2, 4, 1)$.  However, this flaw is not too hard to fix: let $c = e\cdot f$ where $e | a$ and $f | b$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=33417 (see the posts below, not only the first one)

[quote="ZetaX"]http://www.mathlinks.ro/Forum/viewtopic.php?t=33417[/quote]
uhh...this is different, because in your link there's squares on the variables while in this one there's none.

[hide]
we can still use the same idea, $ab = cd \implies \frac{a}{c}= \frac{d}{b}= \frac{x}{y}, (x, y) = 1$.  So $a = ux, c = uy, d = vx, b = vy$, and we have:
$S = ux+uy+vx+vy$
$S = (u+v)(x+y)$
since $u, v, x, y \ge 1$, this will never be a prime number
[/hide]

Read the link again, especially all following posts.


There are n > 2 lines in the plane, no two parallel. The lines are not all concurrent. Show that there is a point on just two lines.

Suppose, for a contradiction, that any common point of two lines lies on at least 3 lines.
Let A be a common point of the lines D_1 and D_2, and then A is also a point of a line D_3. Since the lines are not all concurrent we may consider a line D which does not pass through A.
Since there is a finite number of lines, there is a finite number of common points between two of the lines. So, there is a finite and non-zero number of points which are common to at least two (and then 3) lines but which are not on D. Wlog, we may assume that among these points A is such that the distance from A to D is minimal.
Let A_i be the common point of D_i and D (since they are not parallel), for i = 1,2,3.
Wlog, we may assume that A_1, A_2, A_3 are in that order on D. Note that these 3 points are pairwise distincts. From our hypothesis, there is another line than D and D_2, through A_2. This line, say D', cannot pass through A, then it meets the open segment ]AA_1[ or ]AA_3[, say ]AA_1[, in a point B.
This point B is then a common point of two of the lines, but it is not on D. Moreover d(B,D) < d(A,D), which contradicts the minimality of A.

Pierre.

This is a problem which has actually been unsolved for a long time. I remember reading it in a book about FLT.

The dual is also nice (they're equivalent, of course):

We have n>2 points in the plane, not all of them collinear. Show that one of the lines formed by pairs of points has at most 2 points on it.

To answer the question in the title of this topic, I think it is geo :)

dunno, I'd rather classify it as combinatorics. It may have a geometric look, but the essence is still combinatorics.

Well heck let's just call this "geometric combinatorics" :D

The expression 'Combinatorial Geometry' already exists....

Pierre.