Just thought some people might be interested in the problems we do over here.
http://www.cs.otago.ac.nz/staffpriv/malbert/IMO/index.html
It has the 4 assignments and 6 mock tests we have been doing, plus the questions that decided the 24 people who should go to the selection camp.

How many people compete in the selection process in NZ? what is the training camp like over there? btw i know your name  :lol:

could it be Stephen. just maybe. i dunno. maybe if you look at his signature. or. i could be totally off

Great work, lol.
Its nothing like what I can make of America. The 'september problems' are sent out to schools, nothing hugely difficult for people who know that sort of stuff but if you have only ever learnt things at school you'd probably find it pretty tough. You get a whole 10 days to work on them. Top 24 are selected, sent to Christchurch for the week camp. Here we just do lectures (in two groups) in the morning, go over some problems and then after lunch work on some problems and then go over the answers. Nights are the fun stuff (not saying maths isn't fun! :D) tenpin bowling, stuff like that. Finish the week with a 4.5 hour test with a shorter one in the middle of the week too actually, and then they pick 6 from that. Seems a lot shorter than all these AIMEs and all these other things I hear about.

In the US, the first round is the American Math Competition, which consists of 25 multiple choice questions to be answered in 75 minutes. The next round is the American Invitational Math Exam, which consists of 15 questions with integer answers from 0 to 999, and you are given 3 hours for this one. The next round is the USA Math Olympiad, which is in IMO format. Then the top 12 on the USAMO take the Team Selection Test, and I believe the top 6 from the TST go to the IMO, but it might be the average of USAMO and TST. I never did well enough on the USAMO to know how IMO selection works among the top 12.

[quote="ComplexZeta"] I never did well enough on the USAMO to know how IMO selection works among the top 12.[/quote]

That scares me. And yeah tripleM, theres wayyyyy too many tests over here.

Why does it scare you that there are more than twelve people in the US who are better than I am at olympiads? In fact, there are quite a lot more than twelve (unfortunately). As for the number of competitions we have, I think it is good. More competitions tend to sharpen a person's skills (I imagine; it's certainly true for me), and I don't really do very many other things that are more interesting for me.

More competitions also weeds out [b]all[/b] who are unfit for the IMO.

Sort of, but I don't think it is so clear that the top 6 in the nation will be in the top 12 in the USAMO.

Yeah someone could mess up on computation or something and not make it but still be better than someone on the squad.

Yeah. Speaking of someone (nearly) messing up..

we had a brilliant guy in our team for the last couple of years, he got a silver his first year and then 2002 was his last one. Our first ever hope for a gold medal - question one he completely misunderstood and got 0. The majority of people got that right. OK well maybe not majority but it definitely was an 'easy' question in IMO terms. Well somehow he managed to get a gold medal anyway.. just! I think if he had got question one he would have come 6th in the world or something? Yeah anyway, theres a little story..

Yeah, I seem to remember that #1 on last year's IMO wasn't very hard (although it was more difficult than #4). How did you do on the IMO (I'm assuming you've taken it at least once)? I wish I could have taken it, but it's just too hard in the US.

Not brilliant, but I'm hoping for something this year.
2000 - 4  :D
2001 - 13, bronze
2002 - 11, 2 marks off a bronze (same mark as two other members in my team).

At least you were on your country's squad for three years!  And that story was....quite touching...it should be made into a TV movie  :cry:

Great job, thanks for the site.  I'm always looking for olympiad-type problems, as many of my other posts may or may not have mentioned.

Need a problem?  Or a bunch of problems?  See the "Problems on this Site in a large downloadable file" thread for two good sites.  Happy solving. 8-)

[quote="Syntax Error"]we could get richard nixon to play the lead role of the guy. woooooooooooooooooooooooooot.[/quote]

yes syntax 3M's friend can be portrayed best by a 9 yr old corpse  :P
lotta problems to try.


let $ABCD$ be a convex quadrilateral and $E$ the common point of it`s diagonals. it is known that $AB=BC=CD$ and $AC\neq BD$.
show that $AE=DE$ if and only if $\angle BAD+\angle CDA=120^{\circ}$.


limit
calculus
integration
inequalities
function
real analysis
calculus computations

calculate:

lim (n->infinity) n*sigma(1/k^2) - ?      ;   k=n,n+1,n+2,n+3,....

Moderator edit: $\lim_{n\to\infty}n\sum_{k=n}^{\infty}\frac1{k^2}$

The integral test does what we need here.

$\int_n^{\infty}\frac1{x^2}\,dx\le\sum_{k=n}^{\infty}\frac1{k^2} \le\frac1{n^2}+\int_n^{\infty} \frac1{x^2}\,dx.$

Multiply by $n$ to get

$1\le n\sum_{k=n}^{\infty}\frac1{k^2}\le 1+\frac1n.$

So the limit is 1.

Alternative idea: write this as

$\lim_{n\to\infty}\sum_{k=n}^{\infty}\frac1{\left(\frac kn\right)^2}\cdot\frac1n$

and then see it as an (improper) Riemann sum for $\int_1^{\infty}\frac1{x^2}\,dx.$

I understand the second explaination,

but I don't understand what leads you to getting:

$\int_n^{\infty}\frac1{x^2}\,dx\le\sum_{k=n}^{\infty}\frac1{k^2} \le\frac1{n^2}+\int_n^{\infty} \frac1{x^2}\,dx.$

?

thanks!!

If you draw the 'boxes' around the curve $y=1/x^2$, the reasons for those inequalities should become clear (one is like a left endpoint sum, the other a right endpoint sum...)

in other words:

for every x in [m,m+1] (m>=n, integer)

1/(x-1)^2  <=  f(m) <= 1/x^2

and therefore:

INT (n->n+1) 1/(x-1)^2 <=   INT (n->n+1) f(m)  = f(m) <= INT (n->n+1) 1/x^2

and therefore the sum of f(m) from n to infinity is between both  integrals mentioned in the question.

Right. The $\frac1{n^2}$ on the right is the extra box that slipped out from under when we slid the stack of boxes to the left to be under the curve.

This is what I refer to as the "careful" version of the integral test:

If $f(x)$ is a positive, decreasing function such that $\lim_{x\to\infty}f(x)=0,$ then

$\int_n^{\infty}f(x)\,dx-f(n)<\sum_{k=n+1}^{\infty}f(n)<\int_n^{\infty}f(x)\,dx.$

This then implies the standard conclusion to the integral test:

$\sum_{n=a}^{\infty}f(n)$ converges iff $\int_b^{\infty}f(x)\,dx$ converges for some $a$ and $b.$


Can anyone help me the following? Thx in advanced!! ^^

Suppose $A= (a_{ij})_{n \times n} \in R^{n \times n} (n \geq 3)$ is a tridiagonal matrix. i.e $a_{ij}=0$ for $\vert i-j \vert > 1$, $Q \in R^{n \times n}$ is an orthogonal matrix and $R \in R^{n \times n}$ is an upper-triangular matrix such that $A=QR$ .

Must $RQ$ be a tridiagonal matrix?
If yes, give a proof; otherwise, construct a counterexample.

$\left(\begin{array}{cccc} 1 & 3 & 0 & 0 \\ 2 & 1 & 3 & 0 \\ 0 & 2 & 1 & 3 \\ 0 & 0 & 2 & 1 \end{array}\right)$
Is tridiagonal but when QR Factored, the product RQ is not a tridiagonal matrix anymore.
Now, QR decomposition is not unique unless you want the diagonal elements of R to be all positive, than there is some possibility that there is a certain QR decomposition such that RQ is still tridiagonal.
Anyway requiring RQ to be tridiagonal is the same as requiring $Q^HAQ$ to be tridiagonal, now I don't remember if such structural property as tridiagonality is preserved under Unitary Similarity transformation.


Let $ A$ be the numbers of 5-digit positive numbers satisfying following condition:
The first digit is odd. Remaining $ 0$, or $ 2$ or $ 4$ digit/digits are even.

Let $ B$ be the numbers of 5-digit positive numbers satisfying following condition:
The first digit is even. Remaining $ 0$, or $ 2$ or $ 4$ digit/digits are even.

$ A \minus{} B \equal{} ?$

$\textbf{(A)}\ 5000  \qquad\textbf{(B)}\ 4640  \qquad\textbf{(C)}\ 3200  \qquad\textbf{(D)}\ 0  \qquad\textbf{(E)}\ \text{None}$

[hide="Solution"]$ X\square\square\square\square$
There are $ 10^4$ different numbers starting with $ X$.
If only one of them is odd, there are $ {{5}\choose{1}}\cdot 4 \cdot 5^3$ numbers.
If only three of them is odd, it is equal to only one of them is even, so again $ {{5}\choose{1}}\cdot 4 \cdot 5^3$ numbers.
If $ 0$, $ 2$, or $ 4$ digits are even, it is $ 10^4 \minus{} 2\cdot {{5}\choose{1}}\cdot 4 \cdot 5^3 \equal{} 5000$.
For the first situation $ X$ can be one of the $ 5$ odd numbers.
For the second situation $ X$ can be one of the $ 4$ even numbers.
So the result is $ 1 \cdot 5000 \equal{} 5000$ which is $ \boxed A$[/hide]


Is this problem solved:
Find all a,b and c that we have $a,b,c,k \in N$(k is odd):
$2a^k=b^k+c^k$

This is the problem of Denes, and was solved by Darmon and Merel: the are only the obvious solutions and those with k=1 or 2.  See the exposition [i]The arithmetic of elliptic curves and diophantine equations[/i] on [url=http://www.math.jussieu.fr/~merel/Recherche.html]Loic Merel's research page[/url].


AoPSwiki geometry ARML Columbia Putnam Support HMMT

Go check the AoPSwiki - I already entered it there.

Anyway ...

Division A:

1. NYC A - 170
2. Phillips Exeter Red - 168
3. North Carolina A - 162
4. SFBA A - 161
5. Southern California A - 160

Divison B:

1. Vermont All Stars - 122
2. AAST Mu B - 115
3. SFBA B - 114
4. Chesapeake - 109
5. Oregon - 108

Site winners for Las Vegas were San Diego Surf (145 points) in Division A and Desert Southwest N (104 points) in Division B. I haven't taken the trouble yet to figure out who the site winners would have been at the other three sites.

I believe Montgomery A was site winner at Penn State with 151.

Montgomery A was 9th overall. What does "national honors" in Division A mean? Perhaps it's the top 8, in which case:

6. Massachusetts A - 157
7T. Chicago A - 154
7T. Georgia A - 154

If I'm right about that top 8, then Division A site honors for Iowa should have been Texas Gold (149, 10th place) and for the Georgia site, Florida A (136, tied for 17th place). San Diego's 145 was good for 12th place.

Site honors for Division B:

(Las Vegas) Desert Southwest N - 104
(Iowa) Minnesota Maroon - 100
(Georgia) Kentucky Gold - 92
(Penn State) NYC Y - 107

(Probably. Correct me if I'm wrong.)

Six teams in division B were given "national honors", and NYC Y placed 6th. I think TJ B and one other team tied for the site award at Penn State.

OK, then TJ B and Connecticut A at 106 for Penn State Div. B site honors.

Is that surprising as in very low or very high? As far as I'm aware many top division B teams had 0; the report from one of our team's parents was that our team (Chesapeake) was 7th before relay scores were counted, when we jumped to 4th by getting a 6.

I was under the impression that that was low, but it appears I may have been wrong.

Agh, if I hadn't screwed up the snake problem we'd have had a 6 as well.

I know that the national awards for the A division are top 7 and that there is no tiebreaker for teams. That's why Ga. A and Chicago A were both listed. Everything else is right for the Georgia site.

Tom

National awards for division A were the top 8 teams (although gtf may be correct that this was only due to a tie) and for division B were the top 6 teams (although why NYC Y was in B division is unclear: the NYC coach intended to register them as A division and may not have been entitled to have as many B division teams as they ended up giving him); the 6 on relays was tied for first place in division B with about 4 or 5 teams.  (A single team got 0/6; the others went 2/4.)

And a celebratory shout-out to NYC A!

It says [url=http://arml.com/arml_registration/index.php]here[/url] that awards were increased to 8 in A this year.

Here's a category I like because we did well in it this year: the best "B team" meaning best second team of an organization rather than Division B:

New York City N: 126
Southern California B: 126
Massachusetts E: 122
Montgomery B: 117
Georgia B: 116
Chicago B: 115
AAST Mu B: 115
San Francisco Bay Area B: 114
Texas Silver: 108

Referring back to JBL's post above, it looks like NYC Y at 107 is the best third team of an organization.

I wonder how many of the individual 7's and 8's were on second teams (or third teams)? We had one 7 on SoCal B.

Harriet Li from NYC N got a 7

It's weird.  I registered 4 team for the A division (NYC A,N,Y,C) and two for the B division (H, J).  The day of ARML, one of my coaches reiterated this request in the grading room.  I was as shocked as anyone when they read our team as placing in the B division.  (Both Y and C were put in the B division along with H and J)  I first I thought that they only let 2 teams per region into the A division, but then Chicago A B and C are A division teams.  I emailed people at ARML about this but haven't heard anything.

I just have to give some kudos to Prof Merryfield for the way he handles the So Cal team. He tries to accomodate as many students as he can, even adding a 4th team this year, which required a fair amount of coordination since not everyone could fit on the bus. While he uses scores at math competition to decide who is on which team, he also gives a fair amount of credit for kids who show up at the practice regularly.

 It is too bad that kids who want to go cannot. Everyone who goes seems to enjoy it so much. Few math competitions have the social component that ARML has.

[quote="Kent Merryfield"]
States that haven't had ARML teams recently (if ever): MT, ND, SD, OK, AR, MS, AK, HI, DE[/quote]

This past year in Hawaii chi^2 tried to work on a team, but the cost to get to ARML was a bit prohibitive (besides being difficult to find enough people, and to practice together). If anyone wants to help out over there / has ideas, I imagine she'd appreciate the help (especially if you're a teacher!).

As a former GA B-teamer, we liked being in A division because it meant we directly competed with GA A. Actually, iirc, GA B beat GA A one year. Thus being in A division with our organizations' A team was a matter of pride for us. I believe it would be better to allow organizations to win multiple trophies because it makes the B-teamers actually _want_ to be in the A division, instead of the freshmen wanting to move down so their senior year they can win B.

No B teams (in the sense solafidefarms was just talking about, second teams of multi-team organizations) beat their own A teams this year, at least not overall. You could find specific rounds where that happened. For a ranking among such teams, see post #13 on the first page of this topic.

The team listed as "Taiwan B" outperformed the team listed as "Taiwan A". On the other hand, the Taiwan teams are not really a single stratified organization in the US sense, and their "B" team doing better than their "A" team is quite common.
Also, Desert Southwest N was better than Desert Southwest A. In this case, "A" stands for Arizona and "N" stands for New Mexico/Nevada, so the difference is geographic rather than a quality selection.

Georgia first brought two teams in 1991. Every year since then, the A team has outperformed the B team, by margins ranging from 18 to 80 points.

Actually, one thing that's cool about our teams here in SFBA is that our teams are so young. We have an astounding number of freshmen and sophomores on our A and B teams. We have a lot of potential to get back up really high again for the next few years... :D

@lfm: SFBA also had a similar problem; a fourth team was added for us this year because we had so many people try out. We had a difficult time fitting on the bus. ARML is awesome, and I think our coaches also deserve some applause for their efforts in accommodating everyone this year.

Btw, on the subject, I know that my school does not consider ARML a worthwhile activity <_<  I would think the majority of our region operates the same way, but I don't know.

Catherine, congratulations on starting a successful team essentially from scratch in just a few years!

Based on my previous experiences with ARML, it seems completely absurd to apply Maryland's policies to ARML teams.  For both of the teams I've been with (Minnesota and SFBA), prohibiting homeschoolers from joining the team would be like saying that homeschoolers and public school students were not allowed to compete together on club sports teams.  Both run as non-school activities.  In the Bay Area, the ARML team is a fully independent organization, with loose ties to the Stanford math club (mainly in that the club provides volunteer support and practice space).  In Minnesota, the team is run by the state math league, but the league is an independent organization not connected to the league that runs school sports.  But then again, it sounds like the teams in Maryland are sponsored directly by the appropriate counties, so it's more reasonable if they restrict it to public-school students.  I'm glad ARML is being flexible about the geography rules so that students that are barred from their region's teams for non-mathematical reasons can still compete.

[quote="xscapezaer"]Btw, on the subject, I know that my school does not consider ARML a worthwhile activity <_<  I would think the majority of our region operates the same way, but I don't know.[/quote]
That's sad.  I hope they don't give you trouble about skipping a day of school to go.  Mine never did--in fact, in at least one year, they found some money in the gifted-education fund to cover part of the cost for me and one other student.

By the way, Kent, it looks like Chicago sent four teams this year, as they always have.  A, B, and C placed 7th, 31st, and 40th in divA, while D placed 28th in divB.

Thanks, Si. It turns out that it was Chicago B that was missing from my spreadsheet. I've edited the post you were responding to - so, 127 teams, 120 of them from organizations. And if you go back to post #45 on this topic, about 6-year ranks, Chicago B (or whatever it calls itself - the exact name has varied) would have a 6 year rank sum of 169.5, putting it ahead of the A teams from Wisconsin and Missouri and as the highest-ranking B team over those 6 years.

We had to fit 43 students and 6 chaperones on a 47-person bus... that was interesting.

Catherine,

Thank you for a very thorough explanation of how your team came about.  Let me first assure you that there is no law in Maryland prohibiting any of the county teams from accepting your students onto their ARML teams.  i am confident that what is happening instead is that the teams are being funded by the state and the state prohibits the [i]funding [/i]from being spent on homeschoolers.  Of course, for most of us coaches in the country this is totally irrelevant because we receive no funding anyway.  But i can certainly understand the position of a coach who didn't want to jeopardize their funding by spending it on homeschoolers.  If any of the Maryland coaches are interested, i will be more than happy to work with all of you to come up with creative solutions to allow Maryland to field its most competitive teams while still being able to take advantage of any funding that is available.  If--as sounds likely from the tone of your email--none of them are interested in working with you even if it is to their advantage, i would be glad to help you build a geographically coherent team and work through some of the related boundary and recruitment issues.  As someone who has organized ARML teams in six states and who authored ARML's current boundary rule, i am confident i can be of some help to you in moving forward. Feel free to email me at tim@mathleague.org to discuss this further..

BTW my offer above stands for people in other parts of the country as well:  anyone who is interested in starting a team where there is none currently, or in finding a way around quirky local rules that keep them off a local ARML team for bureaucratic reasons, is invited to contact me so i can help you come up with solutions..

As a former member of the Montgomery County teams, I believe the entire trip is funded by the county board of education.  This gives the coaches a lot more flexibility in bringing students (since funding isn't an issue, and the coaches can bring pretty much all of the interested students.)

Yeah, I don't think the county teams are state-funded.  I know that the Baltimore County team was funded by a grant their coach got from NSF.

Also, I'm pretty sure there is actually a law prohibiting homeschoolers from participating in school activities.

My understanding is that the prohibition against Maryland homeschoolers participating in extracurricular activities, sports, or other programs with the schools is a policy of the school board rather than a law.

When it comes to the ARML, Montgomery County has always been more flexible and supportive of students who don't fall into the usual categories.  I think the results speak for themselves.

As to why Maryland education officials make it so difficult for some of the state's best math students to attend the ARML, I really don't know.  When the homeschool students participate, they do well, which makes the state and counties look good.  What's not to like?  Just that they're homeschooled.  It's a hot button issue in Maryland for some reason.

Not only Maryland.  We were turned down, after a long application/appeal process, from joining Mu Alpha Theta.  Initially, our application was approved, but the approval was rescinded within a few days.  Apparently objections arose over our being homeschooled.  I was given a list of questions and concerns, which I answered at length.  It went to appeal. The appeal was rejected on grounds different from what was in the list of concerns.  Basically, they said the by-laws for the honor society required me as faculty advisor to be employed by a school.  I was told that if my students ever went back to a regular schooling situation, they would be welcome to join Mu Alpha Theta.

[quote="Sly Si"]Sounds like it may be less of an issue soon if a bill passes the MD legislature:
http://www.hslda.org/hs/state/md/200803140.asp

That bill adds section 7-119 to the state code on education.  I looked through the rest of section 7 (at [url]http://www.dsd.state.md.us/comar/Annot_Code_Idx/EducIndex.htm[/url]) and found no references to extracurricular activities.  If there is such a law regarding homeschoolers, it's either in another section, or it's prohibited by the schools or counties because there's no law authorizing it.  If passed, the bill above will take effect in October, in plenty of time for ARML 2009, assuming it applies to the ARML teams.[/quote]

[quote="dnkywin"][quote="n^4+4"]Haha why six years?  I think someone has a bias against North Carolina.  Well in 3 years we will be the undisputed #1 average ^^

The A team was fairly young for us this year.[/quote]
Well, the A Team was not nearly as young as the B Team, which had 5 middle schoolers. I almost ended up on the B Team too, which would have made it 6, but I got on the A Team :D[/quote]

And the NC Arml team will have a lot of new members next year, too.


vote for president and i suppose political discusiion would be allowed in this part of the forum

acutally i would love people to discuss cause i might be able to use it for my civics class :D.

Obama, because he is simply better than mccain.

agreed there aime obama wants to support math and science in school :D ( he never mentioned english and history :D:D)

Please, if you want a topic to be taken seriously, use correct punctuation and grammer, and try to support your thoughts with real reasons.

1) Barack Obama has a better economic plan, which is important given the current economic plan.

On the other hand, as national security is also a issue, McCain seems much more experienced on foreign policy.

Gramm[b]a[/b]r. :wink:

I just don't like how McCain never talks about his own campaign and just tries to demote Obama's. (Along with other things.)

[quote="hunter34"]I don't think Marxism is going to help a Capitalist economy[/quote]

I agree socialism sucks, which is one reason why I hate McCain.

Bob Barr '08! And Ron Paul '12!

Sorry about my grammar I dont like to use it outside of school but I see your point. Obama does have a better economic plan McCain just gives tax cuts to the rich mostly 2.12% if I recall correctly while obama give roughly 4% if I recall to the middle class. Also Obama actually cares about education I haven't heard McCain say a word about education. Not to mention that when you look at their vice presidential candidate selection you cannot deny that since Joseph Biden has been a senator for about 35 years while Sarah Palin hasn't been in politics nearly as long. Also Biden is the chairman of the senate of international relations which definetly gives Obama an edge over McCain in international affairs.

Just consider the following facts about Sarah Palin.
1)She didn't know the job of vice president, and she thought that vice president is the leader of congress.
2)She spent $150,000 of campaign money for clothes.
3)She has a bachelor in journalism (please tell me how this will help in government).

I don't mean to be pruposly digging up dirt on Palin, but i just hear this in the news I don't look for it whereas for the other candidates i don't hear as much of a downside about them. Although I did hear one thing about Biden which was that he talks too much which happens to get him in trouble.

oops I forgot to mention this in my last post. I don't believe that I am mistaken when I say Obama actually does support nuclear power.

This in my opinion is one of the worst elections in U.S. history in terms of candidates. What one has, the other lacks. But for me, (if I were forced to vote) I would vote McCain. I would feel safer with the country in his hands.

The Marxism arguments are ridiculous.
 [i][color=blue]Obama[/color] = big government
 [color=red]McCain [/color]= big government[/i]

That's not a problem for me because I don't mind big government, so long as it works.

What [i]would be[/i] a problem for me is if John McCain gets elected and makes comparably brainless nominations for other government positions, such as justices to the supreme court, his cabinet, his AG, etc...

by the way, "educated people" generally refers to people who are extensively educated, possessing a bachelor's and perhaps a graduate degree. this is probably not the audience from whom you are getting responses, for the most part.

McCain just comes off as someone who is NOT likable.
Also, what if the old guy dies?  We have Sarah Palin as president...and we all know that's not good...

[quote="hunter34"]I don't think Marxism is going to help a Capitalist economy.

Mccain++[/quote]I don't either. How is this relevant? I fail to see the connection. What relevance does Marxism have? Yes, it's a good reason to vote against candidates for the Communist party, but I've heard they really don't hold much sway over here.

I really haven't heard either candidate talking about the proletariat taking control of the means of production from the bourgeoisie. Neither candidate has talked about class warfare. Neither lives in 19th-century Manchester. I'm really having trouble with this. Enlighten me.

It doesn't matter who we vote for. He's gonna die(Ralph Nader doesn't stand a chance). Vote on the vice-president.

Joe Biden FTW

There's this comic strip that goes like this:

FRAME 1: Person says to Obama: "I have good news and bad news"

Obama says, "Tell me the good news first"

FRAME 2: Person: "Okay, the good news is that you've been elected president. Congratulations"

Obama: "Great! But, what's the bad news?"

FRAME 3: Person: "The bad news is that you've been elected president."

-RIMSHOT-

lol

people plz come to my poll not someone elses and i just posted this so thered be a yellow sticky note next to my thread

This is flawed. Bob Barr is not on here, so the proper results won't be given.

There are at least three other candidates who are also not on here.

just another point against McCain:

[youtube]k4aHL12vtEM[/youtube]

edit: stupid me i meant Palin

on pg 1 that joke: Frame 3 makes the most sense

Edit: M... the guy with bread under his armpits? :rotfl:
She fell for that call so badly...


trigonometry
inequalities proposed
inequalities

1.
Let a,b,c: $ a^2\plus{}b^2\plus{}c^2\equal{}2$. Prove that:
$ a \plus{} b \plus{} c \le 2 \plus{} abc$

2.
Let x,y such that: $ {x^2} \minus{} xy \plus{} {y^2} \le 3$. Prove that:
$ \minus{} 1 \minus{} 2\sqrt 7  \le {x^2} \plus{} xy \minus{} 2{y^2} \le  \minus{} 1 \plus{} 2\sqrt 7$

[quote="Gnaht"]
2.
Let x,y such that: $ {x^2} \minus{} xy \plus{} {y^2} \le 3$. Prove that:
$ \minus{} 1 \minus{} 2\sqrt 7 \le {x^2} \plus{} xy \minus{} 2{y^2} \le \minus{} 1 \plus{} 2\sqrt 7$[/quote]

let $ x \equal{} a \plus{} b,y \equal{} a \minus{} b$.

$ {x^2} \minus{} xy \plus{} {y^2} \equal{} a^2 \plus{} 3b^2\le3$.

$ S \equal{} {x^2} \plus{} xy \minus{} 2{y^2} \equal{} 6ab \minus{} 2b^2$.

let $ a \equal{} \sqrt {3}r \cos{t} , b \equal{} r\sin{t}, 0\le r\le 1$.

$ S \equal{} 6ab \minus{} 2b^2 \equal{} 6\sqrt {3}r^2\sin{t}\cos{t} \minus{} 2r^2\sin^2{t} \equal{} r^2(2\sqrt {7}\sin{(2t \plus{} w)} \minus{} 1)$.

so...

[quote="Gnaht"]1.
Let a,b,c: $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2$. Prove that:
$ a \plus{} b \plus{} c \le 2 \plus{} abc$
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32399


A pair of 8-sided dice have sides numbered 1 through 8. Each side
has the same probability (chance) of landing face up. What is the
probability that the product of the two numbers on the sides that
land face-up exceeds 36?

First we need to find the number of ways to roll a product greater than 36.

6,7
6,8
7,6
7,7
7,8
8,5
8,6
8,7
8,8

There are $ 9$ ways to get such a product, and $ 64$ ways to get a product. Our answer is $ \frac{9}{64}$.

The only way I know how to do this is casework:

If the largest number is 8: (7 ways)

5,8
6,8
7,8
8,8
8,7
8,6
8,5

If the largest number is 7: (3 ways)

6,7
7,7
7,6

If the largest number is 6: (No ways)

Since there is no way for the largest number to be 6, we can forget about counting the numbers less than that. Thus, there are 7+3=10 ways for the product of the number to be greater than 36. There are a total of 64 possibilities, so the chance of that happening is $ \frac {5}{32}$.

@izzy: You missed out 5,8


$P(x)$ is a hundred-degree polinomal, $P(x)=x^{100}+a_{99}x^{99}+....+a_1x+a_0$ satifying
$P(1)=1, P(2)=2,...,P(100)=100$.
Compute for $P(101)$
 :rotfl:
(Don't say it is101! I won't believe you!) ;)  :lol:

[quote="towersfreak2006"]Uh...is P(101) uniquely determined? You gave 100 points for a polynomial of degree 100...[/quote]

I think maybe forcing the first coefficient to be a $1$ might be it? Not sure, though.

Oh yeah, you're right. Heh, I didn't notice that.  :blush:

[quote="towersfreak2006"]Uh...is P(101) uniquely determined? You gave 100 points for a polynomial of degree 100.  :?[/quote]
yea, its like finding a straight line with just one point....  :icecream:  :trampoline:

Easy.

[hide]
P(x)-x is also a 100 degree polynomial.  It factors as

P(x) - x = (x-1)(x-2)...(x-100)
P(x) = (x-1)(x-2)..(x-100) + x
P(101) = 100! + 101[/hide]

By the way,

[hide]If the leading coefficient had been $a$ instead of 1, then P(x)-x would factor into a(x-1)(x-2)...(x-100), so the answer would be 100! * a  + 101.  The fact that the leading coefficient is given to be 1 makes the polynomial unique.[/hide]

The places in between two integers will alternate between positive and negative.  So you can place the points $(1,1) (2,2) (3,3) ... (100,100)$ there
From before 0, it comes from infinity down to (1,1).  It goes through (1,1) and goes down to some very low point, then curves back up, passes through (2,2) and goes to some very high point.  Repeats until it passes through (100,100), and it then goes up to infinity.


How do you create a centered title in the document? In general, how do you center/align right/align left?

It depends what you mean by title. Do you mean a chapter or section heading? The [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=secthead]TeX FAQ[/url] lists a number of packages which help to change the style of chapter or section headings.

i use:

[code]
\begin{center}
...your stuff here...
\end{center}[/code]

It works fine. Inside the center, you can use bold and size like:

[code]
\begin{center}
\bfseries{\LARGE{Title Here}}
\end{center}[/code]


[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107262]IMO 2007 HKTST 1[/url]
Problem 6

Determine all pairs $(x,y)$ of positive integers such that $\frac{x^{2}y+x+y}{xy^{2}+y+11}$ is an integer.


geometry
circumcircle
trigonometry
geometry unsolved

Let the sides of a scalene triangle $ \triangle ABC$ be $ AB \equal{} c,$ $ BC \equal{} a,$ $ CA \equal{}b,$ and $ D, E , F$ be points on $ BC, CA, AB$ such that $ AD, BE, CF$ are angle bisectors of the triangle, respectively. Assume that $ DE \equal{} DF.$ Prove that

(1) $ \frac{a}{b\plus{}c} \equal{} \frac{b}{c\plus{}a} \plus{} \frac{c}{a\plus{}b}$

(2) $ \angle BAC > 90^{\circ}.$

Circumcircle $ \mathcal P$ of $ \triangle DEF$ cuts line $ AD$ again at a point $ A'.$ Since $ DE \equal{} DF,$ line $ AD$ bisects both the angles $ \measuredangle EAF \equal{} \measuredangle CAB$ and $ \measuredangle EA'F.$ This is possible only when $ AD \perp EF,$ which means that $ \triangle AFE, \triangle ABC$ are A-isosceles, or when points $ A' \equiv A$ are identical, which means that quadrilateral $ AFDE$ is cyclic. The first possibility is eliminated by the condition that $ \triangle ABC$ is scalene.
_____________________________________________________________________________________________

Circumcircle $ \mathcal P$ of $AFDE$ cuts $ BC$ at $ D$ and again at a point $ D'.$ Denote $ d \equal{} \overline{DD'}.$ On one hand, powers of $ B, C$ to $ \mathcal P$ are

$ \text{pow} (B \to \mathcal P) \equal{} \overline{BA} \cdot \overline {BF} \equal{} c \cdot \frac {ca}{a \plus{} b},$ $\quad \text{pow} (C \to \mathcal P) \equal{} \overline{CA} \cdot \overline {CE} \equal{} b \cdot \frac {ab}{c \plus{} a}.$

On the other hand,

$ \text{pow} (B \to \mathcal P) \equal{} \overline{BD} \cdot \overline {BD'} \equal{} \frac {ca}{b \plus{} c} \cdot \left(\frac {ca}{b \plus{} c} \pm d\right),$ $\quad \text{pow} (C \to \mathcal P) \equal{} \overline{CD} \cdot \overline {CD'} \equal{} \frac {ab}{b \plus{} c} \cdot \left(\frac {ab}{b \plus{} c} \mp d\right),$

where upper signs apply for $ b < c$ and lower signs for $ b > c$.
_____________________________________________________________________________________________

Comparing the two expressions for $ \text{pow} (B \to \mathcal P)$ and the two expressions for $ \text{pow} (C \to \mathcal P)$ $\implies$

$ \frac{ca}{b \plus{} c} \cdot \left(\frac {ca}{b \plus{} c} \pm d\right) \equal{} c \cdot \frac {ca}{a \plus{}  b},$ $\quad \frac {ab}{b \plus{} c} \cdot \left(\frac {ab}{b \plus{} c} \mp d\right) \equal{} b \cdot \frac {ab}{c \plus{} a}.$

Reducing these equations by factors [color=white].[/color]$ ca,\ ab,$ [color=white].[/color]resp. $\implies$

$ \frac{1}{b \plus{} c} \cdot \left( \frac {ca}{b \plus{} c} \pm d \right) \equal{} \frac {c}{a \plus{} b},$ $\quad \frac{1}{b \plus{} c} \cdot \left( \frac {ab}{b \plus{} c} \mp d \right) \equal{}  \frac {b}{c \plus{} a}.$

Adding the two equations to eliminate segment $ d$ $\implies$

$ \frac{1}{b \plus{} c} \cdot \left( (c \plus{} b) \cdot \frac {a}{b \plus{} c} \right) \equal{} \frac {c}{a \plus{} b} \plus{} \frac {b}{c \plus{} a}.$

Reducing left side of the last equation by factor $ (b \plus{} c) = (c \plus{} b)$ $\implies$

$ \boxed{ \frac {a}{b \plus{} c} \equal{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}} \ .$
_____________________________________________________________________________________________

Multiplying this through $\implies$ [hide=(details)]$a \cdot (c \plus{} a) \cdot (a \plus{} b) \equal{} b \cdot (b \plus{} c) \cdot (a \plus{} b) \plus{} c \cdot (b \plus{} c) \cdot (c \plus{} a),$

$ (ac \plus{} a^2) \cdot (a \plus{} b) \equal{} (b^2 \plus{} bc) \cdot (a \plus{} b) \plus{} (cb \plus{} c^2) \cdot (c \plus{} a),$[/hide]

$ a^2c \plus{} a^3 \plus{} abc \plus{} a^2b \equal{} b^2a + b^3 \plus{} abc \plus{} b^2c \plus{} c^2b \plus{} c^3 \plus{} abc \plus{} c^2a.$

Collecting terms with factors [color=white].[/color]$ a^2,\ b^2,\ c^2$ $\implies$

$ a^2(c \plus{} a \plus{} b) \plus{} abc \equal{} b^2(a \plus{} b \plus{} c) \plus{} abc \plus{} c^2(b \plus{} c \plus{} a) \plus{} abc,$

$ a^2 \equal{} b^2 \plus{} c^2 \plus{} \frac {abc}{a \plus{} b \plus{} c} > b^2 \plus{} c^2.$

By reverse Pythagorean theorem, angle $\boxed {\measuredangle CAB > 90^\circ}$ is obtuse.
_____________________________________________________________________________________________

it can be done using cosines lemma in $\triangle CED$ and $\triangle BFD$ on $DE,DF$,respectively.
and of course doing a lot of calculations.
(the solution isn't mine,but i have a sketch of it's proof in this way)

See the proposed problem P5 from [url=https://artofproblemsolving.com/community/c1090h1519538_459working_page_v][color=darkred][b]here[/b][/color][/url].


modular arithmetic
number theory unsolved
number theory

Find all natural solutions
$ {n}^{4}(n \plus{} 1) \equal{} {7}^{m} \minus{} 1$

$ {n}^{4}(n \plus{} 1) \equal{} {7}^{m} \minus{} 1$
$ \Longleftrightarrow (n^2\plus{}n\plus{}1)(n^3\minus{}n\plus{}1)\equal{}7^m$,hence we have

$ n^2\plus{}n\plus{}1\equal{}7^a,n^3\minus{}n\plus{}1\equal{}7^b$ with $ a,b\in \mathbb{N}_0$ and $ m\equal{}a\plus{}b$.

Suppose $ a\ge 2$ and we lead a contradiction as following.
$ n\equal{}1$ is obviously ruled out,so we have $ b\ge 2$ since $ 7^b\equal{}n^3\minus{}n\plus{}1\ge n^2\plus{}n\plus{}1\equal{}7^a$.
$ n^2\plus{}n\plus{}1\equiv 0$ and $ n^3\minus{}n\plus{}1\equiv 0\pmod{7}$ hence we have $ n\equiv 2\pmod{7}$,so let $ n\equal{}7u\plus{}2$,
then $ n^2\plus{}n\plus{}1\equal{}49u^2\plus{}35u\plus{}7\equal{}49u^2\plus{}7(5u\plus{}1)\equiv 0\pmod{7^2}$ since $ a\ge 2$ so we have $ 5u\plus{}1\equiv 0\pmod{7}$ thus we have
$ u\equiv 4\pmod{7}$,so let $ u\equal{}7v\plus{}4$ hence $ n\equal{}49v\plus{}30$ then we have
$ n^3\minus{}n\plus{}1\equiv 30^3\minus{}30\plus{}1\equiv 21\equiv 0\pmod{49}$ this is contradiction.

so we have $ a\equal{}0,1$ and $ n\equal{}2$,hence $ (m,n)\equal{}(2,2)$ is the only solution.


Can someone ([i]again[/i]) kindly help me solve these problems? I'm still in the learning process so any form of help will be much appreciated.

1. Find all the possible positive integers $ (x,y,z)$ such that $ ( {1 + \frac {1}{x}} )( {1 + \frac {1}{y}} )( {1 + \frac {1}{z}} ) = 2$

I'm lost in here..
[hide]
So far, this is my work and I am stuck..

$ ( {1 + \frac {1}{x}} )( {1 + \frac {1}{y}} )( {1 + \frac {1}{z}} ) = 2$

$  {\frac {{(x + 1)(y + 1)(z + 1)}}{{xyz}}}  = 2$

What should I do next?

I came up with the idea that since $ x,y,z > 1$

Then $ x = y + 1$, $ y = z + 1$ and $ 2z = x + 1$ is possible.

After some algebra, I got $ (3,4,5)$ and its permutations (like $ (4,3,5), (5,4,3)$) as solutions.

But $ (9,5,2)$ and its permutations are also solutions. How can I arrive at these solutions. Maybe my steps are wrong. Can someone teach me the proper way of solving this?
[/hide]

2. Determine all sets of non-negative integers $ x,y,z$ which satisfy the equation ; $ 2^x + 3^y = z^2$

Here, I'm so $ awfully$ lost
[hide]
In here, I'm totally lost. I've found some solutions, (by trial and error)

$ (0,1,2),(3,0,3), (4,2,5)$

but I'm not sure if these are the only solutions.

Can someone teach me how to work around this problem? Thanks.
[/hide]

I think the idea to first problem is ok , so you can use $ x\plus{}1\equal{}yz$ and $ (y\plus{}1)(z\plus{}1)\equal{}2x$ and you win the solutions $ (9,5,2)$
 By the way I think these problems are not very Basic

I haven't finished solving number one, but enndbox has the right idea: since you have
\[ 2xyz \equal{} (x \plus{} 1)(y \plus{} 1)(z \plus{} 1)
\]
we can equate factors.

For example, when we split the factors to $ x \plus{} 1 \equal{} yz$ and $ (y \plus{} 1)(z \plus{} 1) \equal{} 2x$, we have
\[ x \plus{} 1 \equal{} yz \Longleftrightarrow 2x \equal{} 2yz \minus{} 2 \Longleftrightarrow 2x \equal{} 2yz \minus{} 2 \equal{} (y \plus{} 1)(z \plus{} 1) \Longleftrightarrow yz \minus{} y \minus{} z \minus{} 3 \equal{} 0.
\]
Here we do something clever: we use the fact that $ y$ and $ z$ are integers, so we know that that greatly limits the possible solutions for $ y$ and $ z$. Also, the left hand side looks like its beggin to be factored, but it doesn't quite work. However, we can make our wishes come true in math, and we try to factor anyways, by adding 4 to both sides:
\[ yz \minus{} y \minus{} z \plus{} 1 \equal{} 4 \Longleftrightarrow (y \minus{} 1)(z \minus{} 1) \equal{} 4.
\]
Since $ y$ and $ z$ are positive integers, we know that the factors $ y \minus{} 1$ and $ z \minus{} 1$ are positive factors of $ 4$, namely, $ y \minus{} 1 \equal{} 1$, $ z \minus{} 1 \equal{} 4$, which gives $ (9,5,2)$, OR $ y \minus{} 1 \equal{} 2$, $ z \minus{} 1 \equal{} 2$, giving $ (3,3,8)$.

We can go on matching factors, like maybe $ x \plus{} 1 \equal{} 2yz$, $ (y \plus{} 1)(z \plus{} 1) \equal{} x$, which gives the solution $ (7,4,2)$.

And it's getting late for me tonight, so I'll let you work from here.

For number 2, note that when $ x \ge 2$, $ y$ must be even, or the left hand side becomes $ 3 \pmod 4$. Similarly, when $ y \ge 2$, $ x$ must be even.

I don't think you can find more solutions (other than what you already have) with $ x < 2$ or $ y < 2$, but it's late, and I'm not sure how to best explain this. Assuming this, we will proceed with assuming $ x$ and $ y$ to be even and let $ x \equal{} 2a$, so that we have
\[ 2^{2a} \plus{} 3^y \equal{} z^2 \Longleftrightarrow z^2 \minus{} 2^{2a} \equal{} 3^y \Longleftrightarrow (z \minus{} 2^a)(z \plus{} 2^a) \equal{} 3^y.
\]
From here, we see that the factors $ z \minus{} 2^a$ and $ z \plus{} 2^a$ are both powers of $ 3$, so the difference of these 2 powers of 3 is $ 2(2^a) \equal{} 2^{a \plus{} 1}$. Let's let the 2 powers be $ 3^m$ and $ 3^n$ with $ m > n$, so that we have $ 3^m \minus{} 3^n \equal{} 2^{a \plus{} 1} \Longleftrightarrow 3^n(3^{m \minus{} n} \minus{} 1) \equal{} 2^{a \plus{} 1}$.

Again, both factors on the left hand side are powers of 2, so that $ n \equal{} 0$, and we have $ 3^m \minus{} 1 \equal{} 2^{a \plus{} 1}$. Like before, $ m$ is even for large enough $ a$, so that we have $ m \equal{} 2b$ and $ 3^{2b} \minus{} 1 \equal{} 2^{a \plus{} 1} \equal{} (3^b \plus{} 1)(3^b \minus{} 1)$. Again, the 2 factors $ 3^b \plus{} 1$ and $ 3^b \minus{} 1$ are both powers of 2. It's now clear that $ b \equal{} 1$, and I think you can finish from here.

Don't worry too much about this being a little too hard for this forum. I usually have no clue which forum to post a problem I can't solve either. :)

The key to the first problem is to see that you CAN assume $ x\leqslant y\leqslant z$, because the expression is symmetric.

Now $ x\leqslant y\leqslant z\implies\left(1\plus{}{1\over x}\right)\left(1\plus{}{1\over y}\right)\left(1\plus{}{1\over z}\right)\leqslant\left(1\plus{}{1\over x}\right)^3$, which yields $ 1\plus{}{1\over x}\geqslant\sqrt[3]{2}\iff x\leqslant{1\over{\sqrt[3]{2}\minus{}1}}$, therefore $ x\leqslant 3$.

Now that you know that you have but two possibilites for $ x$, it shouldn't be difficult to continue this reasoning to $ \left(1\plus{}{1\over y}\right)\left(1\plus{}{1\over z}\right)\equal{}{4\over 3}$ and $ \left(1\plus{}{1\over y}\right)\left(1\plus{}{1\over z}\right)\equal{}{3\over 2}$, and check all the cases, to find the solutions: $ (2,4,15),(2,5,9),(2,6,7),(3,3,8),(3,4,5)$ (and all the permutations).


[quote="catalan"]Hello,

The question asks me to simplify radicals where possible and factor for:

10 - 2√8

so,

   10 - 2√8

= 10 - 2√4 x 2

= 10 - 4√2

= 2(5 - 2√2)

However, the answer in the textbook is 2(5 - 12√2).

Thanks in advance![/quote]
well...we know
that(srry no latex i am using)
10-4 sqrt 2
2(5-2 sqrt2)
and how the hackl would you get 12? the original would've been
10-24sqrt2
10-sqrt1152
so i think there is a problem with the solution given to you

Thanks for your confirmations!

This is one of the times a textbook is wrong.

Oh, I am just reviewing radicals and roots in my free time.

[quote="xpmath"]It's definitely a typo...
Is this in your own free time? Or in school?[/quote]
Not necessarily... they could have put the answer for another question, maybe even from another book in. But yeah, it's definitely wrong.

$2\left(5-12\sqrt{2}\right) = 10-2\sqrt{288}$

in case anybody really cared.


$ A_1, A_2,\ldots A_{n^2}$ are points taken inside a circle with a radius of $ 10n$. Prove that for some $ i>j$, the distance between the points $ A_i$ and $ A_j$ is not more than $ \sqrt{i\minus{}j}$.

It is wrong, for example for $n=3$ we can build inscribed in circle right 9-gon.


calculus
integration
real analysis
real analysis unsolved

Given $\epsilon>0$ do you know an explicit $\delta=\delta(k)$ (as small as possible) such that
$\frac{1}{k!}\int_{k-\delta}^{k+\delta}e^{-t}t^kdt>1-\epsilon$ for $k$ large enough?
any hint is welcome.

Seems like a homework/warm-up problem, hints only. "Near" $t=k$, $e^{-t}t^k\sim\left(\frac{k}{e}\right)^ke^{-\frac{(t-k)^2}{2k}}$. and $k!\sim\sqrt{2\pi k}\left(\frac{k}{e}\right)^k$

thanks for the hint. actually you need $\delta(k)/\sqrt k\to+\infty$ but i'm not sure if this is the best.


MATHCOUNTS
geometry
FTW
blogs
analytic geometry
graphing lines
slope

How do you prepare for MATHCOUNTS? Should you take a bunch of tests from previous years, or read a math book...?

I've got about 26 tests from previous years to take, and 4 handbooks.

Searching the forums gives very useful results.

My advice:

Do lots of problems from old tests. Make sure you understand which ones you got wrong and why. After you read a solution, go back to the problem a day or two later and see if you are able to solve it. If you notice that you get a certain kind of problem wrong very frequently(e.g. geometry, number theory, etc.), it's a good idea to use the AoPS books or other good source and read up on it.

I would take a lot of old tests. The ones that you get wrong are a lot more important than the ones you get right (at practice, at least =P). Go back and figure out your error, and if you still can't identify your error or have no idea how to do a problem, look at the solution. To be absolutely sure you understand it, do the problem again in a few days. I've found that the AoPS classes helped me a lot; maybe that could be something to look into?

[quote="akhil0422"]As they say, Practice make the man/woman perfect![/quote]

.... never heard it like that before and PERFECT practice makes perfect.

[quote="jonathanchou711"]My advice is that you work on the old tests... Once you feel comfortable with that and you are sure to get in the top 10, work on FTW!!!! It's my favorite!!! FTW helps you a lot!! You won't get nervous! I had my first mock mc competition, I got RLLY nervous. I like died at countdown even thought my performance could have been better. So that happened for the next few mock mc competitions. But then I found FTW and then I practiced and practiced and I felt confident!!! But my mom constantly tells me that I need to work harder on written tests... :(  But if you are like a guy like me, who places in average in written, (3rd place :D) then just work on your countdown. My brother says countdown is no use, but if you are hopeless at written (which is a guy like me who never gets first) then just put your efforts into countdown cause you can advance from 10th to first!! w00t!!![/quote]

This is some of the worst advice I've seen.  If you are "hopeless" at written, then practice for written.  You won't even qualify for countdown if you are "hopeless" at written.  Also, placing high isn't the only important thing in MATHCOUNTS.  It's important to learn.  Anyway, in many chapters/states, the countdown round isn't even official.

I'm not doing mathcounts this year (9th grader) but I got 1st countdown and 2nd written in my state, so here's my advice.

If you haven't taken chapter yet, skim over all of the school problems and make sure you can easily do all of them. (this should take about a hour or two.) Then move on to the chapter ones. Do the first 10 problems in 8 minutes, the next 10 in 12, and the last 10 in 16. Then use 4 minutes to check. When done, review all of your answers and make [b]sure[/b] you learn how to do the ones you got wrong. Onto target, just skim through the two problems and do the easy one first. If you have time, check the easy one first. If not, go straight to the 2nd. Once again make sure you know how to do all of these. Onto the team. Give yourself 60 minutes to do all the problems. It's ok if you don't do that well. If you have time at the end, check over them. It's not as important to know how to do all of these questions, just make sure at least 2 people on your team do. For countdown, you can either just quickly do them or play with your friends.

If on states, do the same thing except skim over the chapter problems and then move to past state tests.

I usually take a bunch of tests from previous years. This really helps me prepare.  Also, if you live in the Bay Area, CA, there are some contests like the Miller Math Marathon, and more recently, Mathleague that you can take in your teams.  These are already over, but there might be another one before State.   If you are one of the top mathletes, consider For The Win, to help your countdown.  And if you aren't, it's fun anyway!

Geez AIME15. Do you need me to create a fan site for you too? Man.. I'm not hopeless, I was just being modest. DUH     :bruce:

[quote="jonathanchou711"](3rd place :D)[/quote]

Quite modest, indeed.

Um, I might also add that [b]careless mistakes kill[/b]. If I don't check at all and just speed through for practice, I can make like 5 of them. The Target Round is where you really don't want to make careless mistakes, because they matter a lot. It will make a big difference if somebody is about the same level as you, you make a careless mistake, and they don't...

[quote="jonathanchou711"]My advice is that you work on the old tests... Once you feel comfortable with that and you are sure to get in the top 10, work on FTW!!!! It's my favorite!!! FTW helps you a lot!! You won't get nervous! I had my first mock mc competition, I got RLLY nervous. I like died at countdown even thought my performance could have been better. So that happened for the next few mock mc competitions. But then I found FTW and then I practiced and practiced and I felt confident!!! But my mom constantly tells me that I need to work harder on written tests... :(  But if you are like a guy like me, who places in average in written, (3rd place :D) then just work on your countdown. My brother says countdown is no use, but if you are hopeless at written (which is a guy like me who never gets first) then just put your efforts into countdown cause you can advance from 10th to first!! w00t!!![/quote]

*cough*...from oklahoma...must be easy to qualify for nationals there :roll:

but yeah practice on old tests and find out [b]WHY[/b] you missed the ones you did

[b]Do problems.
And more problems.
And more problems.
[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=100]This[/url] has lots of problems.
As well as [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=202]this.[/url]
And [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=204]this.[/url]
And [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=206]this.[/url]
And [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=200]this.[/url]
And [url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=37]this.[/url]
And [url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=5]this.[/url]
And [url=http://www.artofproblemsolving.com/Forum/index.php?f=298]this.[/url][/b]

Comment: [quote="AIME15"]Anyway, in many chapters/states, the countdown round isn't even official.[/quote]
Same here. HIGH FIVE! :D

All those problems cost 50 dollars each, so like $ \$$500, which is a lot if you're gonna study and not make nats (like me). :(

I heard no one scored 30 or more in the Oklahoma state competition last year.

That's actually true.. It's not my fault I live in OK. I would move to CA but my parents won't. So I guess I have it lucky. 28 was actually the highest score.. Kind of sad.. What.. was 46/46 the highest score for Texas or something? You guys probably have a better advantage cause you have a coach that is probably more experienced. Luckily, I have a coach that has only been teaching for 5 years, and he seems like he has been teaching for 238429384729384^238420394802394 yrs...!   :huh:

Seriously guys, nobody cares about what the highest scores in any states were (well the people who usually do pretty well don't).

I think [b]1=2[/b] has probably posted the best advice, or at least the main idea of it.

The best thing you can do to prepare for Mathcounts is [b]solve as many problems as you can[/b]. Actually, the best thing you can do for math competitions in general is to solve as many problems as you can. There are a lot in the contest section or on the Wiki, along with on the internet. They don't have to just be Mathcounts. For the ones that you can't solve:

1) [b]Understand the solution[/b] and how you would have thought of stuff. In other words, ask yourself what the motivation for a particular step was. Make sure you understand the whole solution. The best way is to probably think how you would explain the solution to someone else. If you can explain it well, then you understand it. Teaching can be extremely helpful in making sure you understand topics. (To go off on a bit of a tangent, I remember early last year when Mr. Rusczyk played a game on For the Win with several people. He won quite easily. Do you think that he spends much time practicing for Mathcounts?)

2a) If you don't understand something in a solution, [b]look it up[/b]. The Wiki isn't just for finding old problems. Also, you can definitely ask on the forum.
2b) This is also where the AoPS books come in. After solving a lot of problems, if you find that many of the problem you're getting wrong are from a particular topic, buy the corresponding intro book. It's not really worth it to buy every book and work on it. I'd suggest Volume 1 first, along with lots of problems. That may be all you need.

3) [b]Hard work is the most important thing.[/b]You won't get anywhere without working hard. The person who works the hardest has the best chance of winning. How do you suppose Coach Boyd keeps winning? Read some of his blog posts to see the amount of time his students put in.

If you don't think you have a chance of competiting against several people, there are a lot of great counterexamples. Allen Yuan (if I recall correctly) didn't even make state Mathcounts in sixth grade. He ended up making Nationals the next year in 7th grade. In 8th grade, even though he didn't place in the top 57 in 7th grade, he came 2nd written, which is qutie difficult to do  :P . There have been several people who've made National Countdown one year and didn't even make Nationals the year before. So don't think that you have no chance against people who've been competing for years. Just work harder than them.

By the way, I'm being a huge hypocrite. The reason I never win anything is because I'm a lazy bum.

Also, although AoPS is a [b]great[/b] resource, don't spend all your time on it. If you really want to succeed, spend more of your time solving problems or learning. When you are on AoPS, try to solve the problems posted in the forums or ask something you don't know.

(I'm being a hypocrite again).

Also, I know it's what everybody says, but it doesn't matter if you don't win. To be honest, math competitions aren't really the point. They're mostly for getting people introduced and interested to math, but real mathematics isn't like this at all. Most competition problems involve tricks and are meant to be solved relatively quickly. In mathematcis, your goal is probably to solve research problems, where tricks won't help you at all. Also, time doesn't matter much. It's not very likely that you'll be able to solve a research problem soon after you see it.
By the way guys, 5849206328x just mentioned a pretty interesting observation to me. There are only a few weeks until state, but you all are still talking about how you did at chapter :wink: . Although I guess that many of you haven't taken chapter yet so... go prepare.

well said xpmath...anyway, last year i didn't even make top 20 in my chapter, and this year i'm going to states, hopefully to nats :roll:

How i review is that i take the old test like i'm suppose to in a real MC competition, then i put an x, or a check mark to the ones i missed or got correct. After that, i go over and try to work out the problems i've missed. Then, i'll see if i still got the wrong answers wrong, and if so, i will see learn how to actually solve it.

1) Understand the solution; if you get it wrong, make sure you won't get that type of problem wrong again. Understand the solution and understand what made you get it wrong then resolve it without [b]any[/b] help. Make of note of [i]why[/i] you got it wrong and devise strategies to overcome that. For example, if you're careless, underline the question, read the problem more carefully, etc. However, don't spend too much time on a problem. You can always resort to the good old AoPS forum  :wink:! Also, if the reason for your mistake was lack of knowledge, then research the whole topic, not just the problem. For example, if you got a slope problem wrong, then go through [b]all[/b] of coordinate geometry, not only slopes.

2) Make a study schedule and [b]stick to it[/b]. It will make studying less tedious and you will do better. Don't set too much time for studying or too less (this depends on how much you think you can handle without going crazy; believe or not, I actually did become mentally retarded last year (a few months, not anymore) because I tudied too much. And also, take a lot of breaks (not too many  :)).

3) Do many, many, many, many, many problems!

4) Do your best and have a laid-back attitude; it's not the end of the world if you fail!

5) Keep an open mind.

6) [b]Be confident[/b]. And also maintain personal hygiene and good grades and blah and blah and blah.

7) Follow this!

[quote="1=2"][b]Do problems.
And more problems.
And more problems.
[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=100]This[/url] has lots of problems.
As well as [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=202]this.[/url]
And [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=204]this.[/url]
And [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=206]this.[/url]
And [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=200]this.[/url]
And [url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=37]this.[/url]
And [url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=5]this.[/url]
And [url=http://www.artofproblemsolving.com/Forum/index.php?f=298]this.[/url][/b][/quote]
I find it hilarious that AoPS gets free advertising out of its users.

[quote="akhil0422"]

2) Make a study schedule and [b]stick to it[/b]. It will make studying less tedious and you will do better. Don't set too much time for studying or too less (this depends on how much you think you can handle without going crazy; believe or not, I actually did become mentally retarded last year (a few months, not anymore) because I tudied too much. And also, take a lot of breaks (not too many  :)).

[/quote]

This depends on what kind of person you are. If study schedules work for you, great. Use them. If you're like me and just do math when you feel like it, don't use a schedule. I've tried schedules countless times before, but they just don't work for me.

Also, try to work your math time around the rest of your life, not the other way around. As much of an impact math competitions appear to have in your life, they really don't matter as much as you think.

As Temperal just mentioned to me, "much more important than the formulas and tricks you learn is the food you get." Perhaps it should also be added that some of the best things you can learn from math competitions are learning how to work hard, getting better at justifying and explaining yourself, learning better logic, and meeting new people. Perhaps one of the best results of math competitions such as ARML and Mathcounts is the social part: you get to meet new people who share a common interest.

Reworded.

What I do is reveiw the basics and old tests, and get on For the win, but i failed yesterday so im not even going to state, and advice?

If someone said this already, than this is just emphasis-
Don't memorize formulas-
understand them.

[quote="diophantient"]After I tried doing USAMO problems, mathcounts ones became easy and obvious.
Not that I recommend this for everyone, but for some people it really works (I made nationals twice).[/quote]

That is pretty genius. Although, you should probably do some more speed oriented practices as well, since a great deal of Mathcounts needs speed.

To train speed on the sprint round:

-Do it standing up with a clipboard

-Do it without eraser and with really blunt pencil

-Do w/o scratch paper

The general tip is to work in as bad conditions as possible while not being completely inhibited. For example, try sitting somewhere really uncomfortable, with little writing space. If you have siblings, let them annoy you and make lots of noise, too. If you can learn to work in these conditions, then it should be much easier at the actual competition with the addition of pressure.

My Chapter Competition is in the morning.  I have discovered that I must get a good nights sleep before morning math contests to do well.  This is also important on the high pressure days.  If at all possible, go to bed earlier before any math contest.  Otherwise, my brain is turned off and I don't do very well.:sleeping:

i would just like to add something...TAKE CARE OF YOURSELF. i took the AMC 12B, which i got a decent enough score on, but probably would have done much better if i didn't have an annoying canker sore in my mouth from lack of sleep, as well as a really stuffy nose.i kinda learned it the hard way... :(

I might also add, since the state competitions are coming up, that cramming is usually detrimental (i.e. other than pseudo-studying like doing stuff with your calculator for the calculator rounds--that can even help you relax).

i just go do some state tests when i feel like it...
something you should NOT do is trying to cram in 7 tests in one weekend (like im doing right now :D)

I'm cramming 20 years of state sprints, targets, and teams. :P

[quote="Temperal"][quote="xpmath"]As Temperal just mentioned to me, "much more important than the formulas and tricks you learn is the food you get." [/quote]

as everyone knows, the most important part of mathcounts is devouring as much food as you can before they force you to take a math test in a ridiculously short amount of time.[/quote]

 :rotfl:  spoken like a true asian (i would know :D )

I failed chapter because of too many careless errors. Does anyone have any advice as to how to not make so many. I double checked most of my answers and still made a ton.

Read [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php]this article[/url]? I don't know; I can make lots of careless mistakes too. I guess what helps the most is to work at a pace that allows you to do the problems quickly enough, but still have a very good chance of getting it right the first time.

Go over the topic, and prepare.  In the chapter competition, despite my terrible math skills, I got 1st place (Also:  At my school I skipped 10 problems on the qualifying School Sprint Round and still made it. Amazing, isn't it?  :wink:  ).  Unfortunately, I got 63rd(GASP!) in the state competition.  My big mistake was not checking over the problems I already did.  Instead, I tried to tackle those problems I basically knew I couldn't solve.  I'm probably never going to do very well in Mathcounts.  My goal is to get to the Countdown Round in the State competition.  If I'm really, really lucky, that may happen in either 7th or 8ith grade, and maybe these strategies may help:
              1. Practice sprint rounds, and then do them again and again until you can answer all 30 in 40 minutes, as well as be able to explain the answers to all of them and how you got them(easier said than done)
               2. Make sure you know Algebra I, Geometry, Discrete Math, Combinatrics, and Complex Probability.
                3. Get a good night's sleep and hope you don't get stuck with a crummy bed in the Memorial Union at the University(This may have affected my performance.
                 4. Skip all of the questions that you know will take too long and put them off until the end.
                  5. Take a deep breath and acknowledge that it's not that big of a deal if you do badly on the test.

And, on a more humorous note:  If none of that works out, you can try some of these things(I'm joking!)  They probably won't work, and some will get you disqualified, but it's fun for a little humor, eh?
                   1. Try to smuggle in a caclulator wristwatch.
                   2. Try to learn the quadratic formula right before the test starts.
                   3. Try to take a look at the answer key before you hand in the test.  Then, if the proctor says you're disqualified, say "I know you are, but what am I?"
                    4. On the countdown round, say the answer is "square root of 2 over 2" for every question
                    5. Show up at the national competition and use a false name to convice the officials you're actually a competitior.
                    6. Bribe the moderator into giving you the answers.


How do you divide a number by a fraction with out a calculator. example: 12/10.9 percent

also how do you divide decimals. example: 8/4.2

Well....to divide a number by a fraction is simple.  To do, you must convert both the numbers you are dividing into fractions, like if i was doing $ 12\div\frac12$, then I would make $ 12\rightarrow\frac{12}{1}$.  Then we divide.

$ \frac{12}{1}\div\frac{1}{2}\equal{}$

Because we are dividing, we find the reciprocal of the number that divides into 12, which is 1/2.  So our problem becomes:

 $ \frac{12}{1}\div\frac{1}{2}\equal{}\frac{12}{1}\cdot\frac{2}{1}\equal{}\frac{24}{1}\equal{}24$

Thats how we divide a number by a fraction.

To divide by a decimal, you could convert the decimal to an fraction, or you could simple divide it at it is.

Like to do $ 8\div4.2$, we set it up in the long division method,
    ________
4.2|8

then we get rid of the decimal by multiplying by 10:
     ________
42| 80

then you divide as you would to get .525.

I hope this helps.  If you have any questions, you can post them here.

oh sorry i know how to divide fractions. if you see my example i meant percents.

Make the percent into a decimal or a fraction over 100. Thus, $ 12 \div 10.9$ percent is $ 12 \div 0.109$, which is (if you multiply both sides by  1000) $ 12,000 \div 109$. Now you can do the math, I'm sure.  :)

To convert a percent into a decimal, drop the percent sign and multiply by 100. To convert a percent into a fraction, drop the percent sign and put the number over $ 100$.

To divide by a decimal, multiplying both sides by a number until you have whole numbers or numbers that are easy to work with. Remember that $ 0.a$ is the same as $ \frac{a}{10}$


[hide="Dunno if this is the only n"]

You know that if $ n \equal{} \prod p_i^{n_i}$ ,       $ p_i$ s being primes

Sum of divisors of n = $ \prod \frac{p_i^{n_1 \plus{} 1} \minus{} 1}{p_i \minus{} 1}$

here [b] Sum = 1240 = [/b] $ 31 \times 40 \equal{} \frac{2^5\minus{}1}{2\minus{}1} \times \frac{3^4\minus{}1}{3\minus{}1}$

So, n can be  $ 2^4 . 3^3 \equal{} 432$

[/hide]

[quote="manjil"][quote="srinath.r"]Find all natural numbers $ n$ such that the sum of divisors of $ n$ is $ 1240$[/quote]

maybe you meant the smallest $ n$?[/quote]
Find ALL natural numbers n.

there may be  other $ n's$ apart from what Nora91 says   ,I didn't mean the smallest $ n$ anyway

The solution is kind of bruteforce; and I can't imagine solving it shortly. I made it after loads of calculations, so if you find little errors, just tell it.  :)

[hide="Solution"]We shall use the formula $ S(n) \equal{} (1 \plus{} p_1 \plus{} p_1^2 \plus{} \ldots \plus{} p_1^{\alpha _1})(1 \plus{}  p_2 \plus{} p_2^2 \plus{} \ldots \plus{} p_2^{\alpha _2})\cdot \ldots \cdot (1 \plus{} p_n \plus{} p_n^2 \plus{} \ldots \plus{} p_n^{\alpha _n})$, where $ S(n)$ is the sum of the $ p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdot \ldots \cdot p_n^{\alpha_n}$ divisors. All the divisors of $ 1240$ are \[ 2, 4, 5, 8, 10, 20, 31, 40, 62, 124, 155, 248, 310 \equal{}A\] Thus each of the above formula's sums is one of these numbers.
Now we shall define the function $ h(x)$. If $ x\equal{}p^\alpha$ then $ h(x)\equal{}\sum_{i\equal{}0}^{\alpha}p^i$. Now for each of $ A$ elements we will find $ h^{\minus{}1}(x)$ (all the values for each $ x$):
$ h^{\minus{}1}(2)$ is not defined;
$ h^{\minus{}1}(4) \equal{} 3^1$;
$ h^{\minus{}1}(5) \equal{} 4^1$;
$ h^{\minus{}1}(8) \equal{} 7^1$;
$ h^{\minus{}1}(10) \equal{} 9^1$;
$ h^{\minus{}1}(20) \equal{} 19^1$;
$ h^{\minus{}1}(31) \equal{} 2^4; 5^2; 30^1$;
$ h^{\minus{}1}(40) \equal{} 3^3; 39^1$;
$ h^{\minus{}1}(62) \equal{} 61^1$;
$ h^{\minus{}1}(124) \equal{} 123^1$;
$ h^{\minus{}1}(155) \equal{} 154^1$;
$ h^{\minus{}1}(288) \equal{} 287^1$;
$ h^{\minus{}1}(310) \equal{} 309^1$;
$ h^{\minus{}1}(620)$ is not necessary for us, because $ h^{\minus{}1}(2)$ is not defined;
$ h^{\minus{}1}(1240) \equal{} 1239^1$.

Now we can divide $ 1240$ in 'good' divisors in $ 9$ different ways:
$ 4 \cdot 310$;
$ 5 \cdot 248$;
$ 8 \cdot 155$;
$ 10 \cdot 124$;
$ 20 \cdot 62$;
$ 31 \cdot 40$;
$ 4 \cdot 5 \cdot 62$;
$ 4 \cdot 10 \cdot 31$;
$ 5 \cdot 8 \cdot 31$;
$ 1 \cdot 1240$.
Inserting the corresponding $ h^{\minus{}1}$ value in each of these divisors, we obtain all of such numbers (I won't do it here  :) ).[/hide]

Is $ p$ prime in $ h(x)\equal{}\sum_{i\equal{}0}^{\alpha}p^{i}$?
If so, I think there are some errors where you treated $ p$ as composite (4, for example).

Yes, you are right! Then we must exclude all these composite $ p$'s from $ h^{\minus{}1}$ values.
So the only numbers will be
$ 2^4 \cdot 3^3 \equal{} 432$
$ 5^2 \cdot 3^3 \equal{} 675$
$ 19 \cdot 61 \equal{} 1159$
Thanks!  :)


Let  $[x]$ denote great integer that is equal or less from $x$. Is $\lim_{x\rightarrow 0}[0,99999...]$ equal to $1$ or $0$.

There is no $x$ occuring in the term ;)
$1=0.999...$, thus $1$.

I am sorry.  :blush: Where from did it come? There is no limit in this question. True version of this question is: Is $[0,99999...]$ equal to $1$ or $0$?.

Thanks Zetax. But,  $0,999...$  not equal to $1$. It aproximately equal. I think $\lim(0,999...)=1$. So that.  $[0,999..]=0$. If it is not so that what is the explanation of this statement.

Oh not again  :roll:
$0,9999...=1$, nothing else.
There is no "approximately equal", and limites without a "variable" (that thing that goes to $\infty$ or $0$ or whatever) make no sense.

Things change if you consider $\lim_{n \to \infty}[0,\underbrace{999...99}_{n \text{ times}}]$, because this really is $0$.

But $\lim_{n \to \infty}[0,\underbrace{999...99}_{n \text{ times}}] = \lim_{n \to \infty}0 = 0 \neq 1= [1] = [ \lim_{n \to \infty}0,\underbrace{999...99}_{n \text{ times}}]$, or in other words: $[\cdot ]$ is not continuous at $1$.

ZetaX has already beat this dead, but if [$0.999\dots$]$\neq 1$, then there is some real number between the two.  It's easy to show this isn't possible.


In triangle $ ABC$, points $ M,N$ lie on line $ AC$ such that $ MA\equal{}AB$ and $ NB\equal{}NC$. Also $ K,L$ lie on line $ BC$ such that $ KA\equal{}KB$ and $ LA\equal{}LC$. It is know that $ KL\equal{}\frac12{BC}$ and $ MN\equal{}AC$. Find angles of triangle $ ABC$.

You sure MN = AC? Or MN = AC/2 just like that of KL = BC/2?

Dien


geometry
3D geometry
algebra unsolved
algebra

Determine all triplets of integers (x,y,z) satisfying the equation x3+y3+z3 = (x+y+z)3.

Well, $(x+y+z)^3-(x^3+y^3+z^3)=3(x+y)(y+z)(z+x)$, so two of the necessary and sufficient condition is for two of them to have sum $0$.


[b]a.)[/b] Let $a,b$ be real numbers. Define sequence $x_k$ and $y_k$ such that

\[x_0 = 1, y_0 = 0, x_{k+1} = a \cdot x_k - b \cdot y_l,  \quad y_{k+1} = x_k - a \cdot y_k \text{ for } k = 0,1,2, \ldots \]

Prove that

\[x_k = \sum^{[k/2]}_{l=0} (-1)^l \cdot a^{k - 2 \cdot l} \cdot \left(a^2 + b \right)^l \cdot \lambda_{k,l}\]

where $\lambda_{k,l} = \sum^{[k/2]}_{m=l} \binom{k}{2 \cdot m} \cdot \binom{m}{l}$

[b]b.)[/b] Let $u_k = \sum^{[k/2]}_{l=0} \lambda_{k,l} $. For positive integer $m,$ denote the remainder of $u_k$ divided by $2^m$ as $z_{m,k}$. Prove that $z_{m,k},$ $k = 0,1,2, \ldots$ is a periodic function, and find the smallest period.

Anyway,the recursion of Xn should be changed to Xk+1=aXk-bYk.

[hide = Ignore]
Incomplete solution for a). Can someone help me finish? I think I've made an error somewhere.

We will use generating functions. We will use the fact that
\[ \frac{X^m}{(1-X)^{m+1}} = \sum_{k \geq m} \binom{k}{m} X^k \tag{1} \]
liberally. This fact is well known, so we shall not prove it.

Let $F(X) = \sum_{k \geq 0} x_k X^k$ and $G(X) = \sum_{k \geq 0} y_k X^k$. Then
\[ F(X) = 1 + \sum_{k \geq 1} (ax_{k-1} - by_{k-1}) X^k = 1 + aXF(X) - bxG(X). \]
Likewise
\[ G(X) = XF(X) - bXG(X) . \]
We follow the standard elimination procedure to solve for $F(X)$:
\[ F(X) = \frac{aX + 1}{1 + (a^2+b)X^2} . \]
This implies that
\[ F(X) = \sum_{\ell \geq 0} (-1)^{\ell} (a^2+b)^{\ell} (1+aX) X^{2\ell}. \]
We want to show that
\[ F(X) = \sum_{k \geq 0} \sum_{\ell = 0}^{\lfloor k/2 \rfloor} (-1)^{\ell} a^{k - 2\ell} (a^2 + b)^{\ell} \sum_{m=\ell}^{\lfloor{k/2} \rfloor} \binom{k}{2m} \binom{m}{ \ell} X^k. \]
Swapping sums, we obtain that that sum is equal to
\[ \sum_{\ell \geq 0} (-1)^{\ell} (a^2 + b)^{\ell} a^{-2\ell} \sum_{m \geq \ell} \binom{m}{\ell} \sum_{k \geq 2m} \binom{k}{2m} (aX)^k. \]
Thus, it suffices to show that
\[ a^{-2\ell} \sum_{m \geq \ell} \binom{m}{\ell} \sum_{k \geq 2m} \binom{k}{2m} (aX)^k = (1+aX) X^{2\ell}\]
Call the left hand side $H(X)$. From (1), we have that
\[ \sum_{k \geq 2m} \binom{k}{2m} (aX)^k  = \frac{(aX)^{2m}}{(1 - aX)^{2m+1}} . \]
Let $Y = \frac{(aX)^2}{(1-aX)^2}$. Then
\[ H(X) = a^{-2\ell} \sum_{m \geq \ell} \binom{m}{\ell} \frac{(aX)^{2m}}{(1-aX)^{2m+1}} = (1-aX)^{-1} a^{-2\ell} \sum_{m \geq \ell} \binom{m}{\ell} Y^m. \]
Applying (1) again gives
\begin{align*}
H(X) &= (1-aX)^{-1} a^{-2\ell} \frac{Y^\ell}{(1-Y)^{\ell+1}} \\
&= (1-aX)^{-1} a^{-2\ell} \frac{\frac{(aX)^{2\ell}}{(1-aX)^{2\ell}}}{(1 - \frac{(aX)^2}{(1-aX)^2})^{\ell+1}} \\
&= (1 - aX) \frac{X^{2\ell}}{(1 - 2aX)^{\ell + 1}}
\end{align*}
We want this to be equal to $(1 + aX) X^{2\ell}$, which it is not!  :mad:
[/hide]

Alas, I think the problem statement is incorrect. If $a = 2$ and $b = 1$, then $x_3 = 6$, but
\[ \sum_{\ell = 0}^{[3/2]} (-1)^{\ell} 2^{3 - 2 \ell} 5^{\ell} \lambda_{k,\ell} = 2 . \]
If someone could fix the problem statement, that would be much appreciated.

I have done some sleuthing on the internet and found the correct problem statement. It should be as follows
[quote=orl][b]a.)[/b] Let $a,b$ be positive real numbers. Define sequence $x_k$ and $y_k$ such that

\[x_0 = 1, y_0 = 0, x_{k+1} = a x_k - b y_k,  \quad y_{k+1} = x_k + ay_k \text{ for } k = 0,1,2, \ldots \]

Prove that

\[x_k = \sum^{[k/2]}_{l=0} (-1)^l a^{k - 2 l} \left(a^2 + b \right)^l \lambda_{k,l}\]

where $\lambda_{k,l} = \sum^{[k/2]}_{m=l} \binom{k}{2m} \binom{m}{l}$
[/quote]
In short, the recurrence relation for $y_k$ should be $y_{k+1} = x_k + ay_k$. (I also fixed the typos in the original problem statement.)

We will use generating functions. We will use the fact that
\[ \frac{X^m}{(1-X)^{m+1}} = \sum_{k \geq m} \binom{k}{m} X^k \tag{1} \]liberally. This fact is well known, so we shall not prove it.

Let $F(X) = \sum_{k \geq 0} x_k X^k$ and $G(X) = \sum_{k \geq 0} y_k X^k$. Then
\[ F(X) = 1 + \sum_{k \geq 1} (ax_{k-1} - by_{k-1}) X^k = 1 + aXF(X) - bxG(X). \]Likewise
\[ G(X) = XF(X) + aXG(X) . \]We follow the standard elimination procedure to solve for $F(X)$:
\[ F(X) = \frac{1-aX}{1 -2aX + (a^2+b)X^2} . \]
Mow we just want to show that $F(X) = F_1(X)$, where
\[ F_1(X) == \sum_{k \geq 0} \sum_{\ell = 0}^{\lfloor k/2 \rfloor} (-1)^{\ell} a^{k - 2\ell} (a^2 + b)^{\ell} \sum_{m=\ell}^{\lfloor{k/2} \rfloor} \binom{k}{2m} \binom{m}{ \ell} X^k. \]
Swapping sums, we obtain that that
\[ F_1(X) = \sum_{\ell \geq 0} (-1)^{\ell} (a^2 + b)^{\ell} a^{-2\ell} \sum_{m \geq \ell} \binom{m}{\ell} \sum_{k \geq 2m} \binom{k}{2m} (aX)^k. \]
From (1), we have that
\[ \sum_{k \geq 2m} \binom{k}{2m} (aX)^k  = \frac{(aX)^{2m}}{(1 - aX)^{2m+1}} . \]
Let $Y = \frac{(aX)^2}{(1-aX)^2}$. Then
\[ F_1(X) = \sum_{\ell \geq 0} (-1)^{\ell} (a^2 + b)^{\ell} a^{-2\ell} \sum_{m \geq \ell} \binom{m}{\ell} \frac{(aX)^{2m}}{(1-aX)^{2m+1}} = (1-aX)^{-1} a^{-2\ell} \sum_{m \geq \ell} \binom{m}{\ell} Y^m. \]
Applying (1) again gives
\begin{align*}
F_1(X)  &= (1-aX)^{-1} \sum_{\ell \geq 0} (-1)^{\ell}(a^2+b)^{\ell} a^{-2\ell} \frac{Y^\ell}{(1-Y)^{\ell+1}} \\
&= (1-aX)^{-1} \sum _{\ell \geq 0} (-1)^{\ell} (a^2+b)^{\ell} a^{-2\ell} \frac{\frac{(aX)^{2\ell}}{(1-aX)^{2\ell}}}{(1 - \frac{(aX)^2}{(1-aX)^2})^{\ell+1}} \\
&= (1 - aX)  \sum_{\ell \geq 0} (-1)^{\ell} (a^2+b)^{\ell} \frac{X^{2\ell}}{(1 - 2aX)^{\ell + 1}} \\
&= \frac{1-aX}{1 - 2aX} \sum_{\ell \geq 0} (-1)^{\ell} (a^2+b)^{\ell} \left( \frac{X^2}{1 - 2aX}\right)^{\ell}  \\
&= \frac{1-aX}{1-2aX} \cdot \frac{1}{1 + (a^2+b) \frac{X^2}{1-2aX} } \\
&= \frac{1- aX}{1 - 2aX + (a^2+b) X^2} = F(X).
\end{align*}
Done.

I have translated the official solutions if anyone is interested. Hopefully, it is clear.

Solution 1: Trigonometry
Because
\begin{align*}
x_{k+1} + i \sqrt{b} y_{k+1} &= (a + i \sqrt b) x_k  + (i a \sqrt{b} - b)y_k \\
&=(a + i \sqrt{b})x_k + i \sqrt{b}  ( a + i \sqrt{b}) y_k \\
&= (x_k + i \sqrt b y_k) (a + i \sqrt b)
\end{align*}
and $x_0 = 1, y_0 = 0$, then
\[ x_k + i \sqrt{b} y_k = (a + i \sqrt b)^k  . \]
Similarly, $x_k - i \sqrt{b} y_k = (a - i \sqrt{b})^k$. Therefore
\[ x_k  = \frac{1}{2} (( a + i \sqrt{b})^k + (a - i \sqrt{b})^k.) \]
Take $\theta \in [0, \pi]$ such that
\[ \cos \theta = \frac{a}{\sqrt{a^2+b}}, \cos \theta = \frac{\sqrt{b}}{\sqrt{a^2 +b}} . \]
Then $x_k = (\sqrt{a^2+b})^k cos k \theta. $ Because $\cos k\theta + i \sin k \theta = (\cos \theta + i \sin \theta)^k$,
\begin{align*}
\cos k \theta  &= \sum_{m=0}^{[k/2]} \binom{k}{2m} (i \sin \theta)^{2m} (\cos \theta)^{k - 2m} \\
&= \sum_{m=0}^{[k/2]} \binom{k}{2m}(\cos \theta)^{k-2m} (\cos^2 \theta - 1)^m \\
&= \sum_{m=0}^{[k/2]} \binom{k}{2m} (\cos \theta)^{k - 2m} \sum_{\ell =0}^m \binom{m}{\ell} (-1)^{\ell} (\cos \theta)^{m - \ell} \\
&= \sum_{\ell =0}^{[k/2]} (-1)^{\ell} (\cos\theta)^{k - 2\ell} \sum_{m = \ell}^{[k/2]} \binom{k}{2m} \binom{m}{\ell} \\
&= \sum_{\ell = 0}^{[k/2]} (-1)^{\ell} (\cos \theta)^{k - 2\ell} \lambda_{k,\ell} .
\end{align*}
Therefore
\begin{align*}
x_k &= (\sqrt{a^2+b})^k  \sum_{\ell = 0}^{[k/2]} (-1)^{\ell} \left( \frac{a}{\sqrt{a^2+b}} \right)^{k - 2\ell} \lambda_{k,\ell} \\
&= \sum_{\ell = 0}^{[k/2]} (-1)^{\ell} a^{k-2\ell}(a^2 + b)^{\ell} \lambda_{k,\ell} .
\end{align*}

Solution 2: Characteristic Equations
From the first recursive relation, we rewrite $y_k = \frac{1}{b} (ax_k - x_{k+1})$. Substituting this back into the second recursive relation gives
\[ \frac{1}{b} (x_{k+1} - x_{k+2})  = x_k + \frac{a}{b} (ax_k - x_{k+1}) . \]
Simplifying,
\[ x_{k+2} = 2ax_{k+1}  - (a^2 +b) x_k. \]
This recursive relation has the characteristic polynomial
\[ t^2 - 2at + a^2 +b . \]
It has two distinct roots $a \pm i \sqrt{b}$. Using the initial values $x_0 =1$ and $x_1 = ax_0 - by_0 = a$ gives
\begin{align*}
x_k  &= \frac{1}{2} (( a + i \sqrt{b})^k + (a - i \sqrt{b})^k). \\
&= \sum_{s=0}^{[k/2]} (-1)^s \binom{k}{2s} a ^{k - 2s} b^s .
\end{align*}
To finish the proof, we just need to compute the coefficient of $a^{k-2s} b^s$ in the second expression. Because $(a^2+b)^{\ell} = \sum_{r = 0}^{\ell} \binom{\ell}{r} a^{2(\ell - r)} b^r$, each term contributes to the $a^{k-2s} b^s$ only when $r = s$. From this, we determine the coefficient of $a^{k-2s} b^s$ is
\begin{align*}
&\sum_{\ell =0}^{[k/2]} (-1)^{\ell} \binom{\ell}{s} \lambda_{k,\ell} \\
&= \sum_{\ell = s}^{[k/2]} (-1)^{\ell} \binom{l}{s} \sum_{m=1}^{[k/2]} \binom{k}{2m} \binom{m}{\ell} \\
&= \sum_{\ell = s}^{[k/2]} \sum_{m=1}^{[k/2]} (-1)^{\ell} \binom{\ell}{s} \binom{k}{2m} \binom{m}{\ell} \\
&= \sum_{m=1}^{[k/2]} \binom{k}{2m} \sum_{\ell =0}^m (-1)^{\ell} \binom{\ell}{s} \binom{m}{\ell}.
\end{align*}
It is easy to see that $\binom{\ell}{s} \binom{m}{\ell} = (-1)^{\ell} \binom{m}{s} \binom{m-s}{\ell - s}$. So when $m > s$,
\begin{align*}
&\sum_{\ell = s}^m (-1)^{\ell} \binom{\ell}{s} \binom{m}{\ell} \\
&= (-1)^s \binom{m}{s} \sum_{\ell = s}^m (-1)^{\ell - s} \binom{m - s}{\ell - s} \\
&= (-1)^s \binom{m}{s}(1-1)^{m-s}  = 0.
\end{align*}
Thus our coefficient is the desired $(-1)^s \binom{k}{2s}$.

And now the solution for b)

Let $T_n$ be the length of the period of $z_{n,k}$. The answer is
\[ T_n = \begin{cases} 2^{n-1} &\text{if $n \neq 2$,} \\
4 &\text{if $n = 2$.} \end{cases} \]

For $n \leq 3$, we check this by hand. Henceforth, assume that $n > 3$. First, swap sums to obtain
\[ u_k = \sum_{m = 0}^{\lfloor k/2 \rfloor} \binom{k}{2m}  \sum_{\ell = 0}^m \binom{m }{\ell} = \sum_{m=0}^{\lfloor k/2 \rfloor} \binom{k}{2m} 2^m . \]
Immediately, we have $u_0 = u_1 = 1$. From the binomial theorem, $u_k = \frac{1}{2} ( (1 + \sqrt{2})^k + (1 - \sqrt{2})^k).$ Then $u_k$ satisfies the recurrence relation
\[ u_{k+2} = 2u_{k+1} + u_k . \]
This straight away implies periodicity of $z_{n,k}$; what is left is to find the length of the period. We are going to show that $z_{n,2^{n-1}} = z_{n,2^{n-1}+1} = 1$. This will imply that the period divides $2^{n-1}$ because $z_{n,k}$ is decided entirely by the two preceding terms of the sequence. Let $f(a,b) = \binom{a}{2b} 2^b$. We claim that $\nu_2(f(a,m)) \geq n$, where $a \in \{ 2^{n-1}, 2^{n-1} + 1\}$ with equality if and only if $m = 4$. First, we deal with $a = 2^{n-1}$. Then
\[ \nu_2(f(2^{n-1},m)) = \nu_2 \left( \binom{2^{n-1}}{2m} 2^m \right) = m + n - 1 - \nu_2(2m) + \sum_{\ell =1}^{2m-1} \nu_2(2^{n-1} - \ell) - \nu_2(\ell) . \]
Since $\nu_2(\ell) < n-1$, it follows that $\nu_2(2^{n-1} - \ell) = \nu_2(\ell)$, so this reduces to
\[ \nu_2(f(2^{n-1},m)) = m + n - 2 - \nu_2(m) .\]
It is easy to verify that $m \geq 2 + \nu_2(m)$ for $m \geq 3$ with equality if and only if $m = 4$. We repeat an identical line of logic for $a = 2^{n-1}+1$. This establishes the claim.

Next, we show that $2^{n}$ divides $f(2^{n-1}, 1) + f(2^{n-1}, 2)$. Just directly compute that
\[ f(2^{n-1}, 1) + f(2^{n-1}, 2) = 2 \binom{2^{n-1}}{2} + 4 \binom{2^{n-1}}{4} = 2^{n-1} \cdot \frac{3 \cdot( 2^{n-1}-1) + (2^{n-1}-1)(2^{n-2}-1)(2^{n-1}-3) }{3}. \]
The coefficient of $2^{n-1}$ here is clearly even, so we get that $2^n$ divides the whole thing.

These facts taken together show that $2^n$ divides
\[ u_{2^{n-1}} - 1 = \sum_{k=1} f(2^{n-1}, m) . \]
This implies that $z_{n,2^{n-1}} = 1$, as desired.

Our argument about $z_{n, 2^{n-1}+1}$ will be little more subtle. We shall show that $\nu_2(u_{2^{n-1}+1-1})$ is exactly $n$. This will prove that the period is precisely $2^{n-1}$ since $2^n$ does not divide $u_{2^{n-2} + 1} - 1$. To do this, we first show that $2^{n+1}$ divides $f(2^{n-1}+1,1) + f(2^{n-1}+1,2)$.Like before, this is a direct computation.
\[ f(2^{n-1}+1,1) + f(2^{n-1}+1,2) = 2 \binom{2^{n-1}+1}{2} + 4 \binom{2^{n-1}+1}{4} = 2^{n-1} \cdot \frac{ 3 \cdot (2^{n-1} + 1) + (2^{n-1} + 1)(2^{n-1} - 1)(2^{n-2} + 1)}{3} . \]
Since
\[ 3 \cdot (2^{n-1} + 1) + (2^{n-1} + 1)(2^{n-1} - 1)(2^{n-2} + 1) \equiv 3 + 1 \equiv 0 \pmod{4}, \]
the conclusion follows. Then $f(2^{n-1} + 1, 4)$ is the only term in the sum
\[ ( f(2^{n-1} + 1, 1) + f(2^{n-1} + 1, 2) ) + f(2^{n-1} + 1, 3) + f(2^{n-1} + 1, 4) + \cdots + f(2^{n-1} + 1, 2^{n-2})  = u_{2^{n-1} + 1} - 1 \]
whose exponent of 2 is equal to $n$. It follows then that $\nu_2(u_{2^{n-1}+1} - 1) = 2^n$, so we are done.


I put myself at probably <10.

Aargh, I put an option for 'all of my life' but it does not show :(

[i][Moderator Note: Poll edited to include 'all of my life' and to add MathLinks to the title/poll][/i]

damn, I should have included MathLinks in the poll as well, oh well...Valentin the pollmaster will change accordingly :D

i'm on AoPS for <10 hrs. a week cause my school is full of incompetent people!!! :mad:

The vast majority of people at my school are even more mathematically challenged than I am.
AoPS is my refuge... :)

i'd have to say <20. i spend between 1-2 hours a day on AoPS on weekdays and 2-3 on weekends depending on what's happening.

Is there always a admin on?  How about a moderator?

The admins are not always on.  There are a bunch of moderators from a bunch of locations so the chances of there being either an admin or moderator on is pretty high at any given time.  I'm on about 3 hours weekdays and maybe 10 on the weekends.  I usually do everything at my computer.  So there is a lot of time where I will just go and check to see about new posts etc. and then go back to my other work.  So it's kinda hard to give a good approximante.

I'm on a whole lot, usually a bunch of my free time because there really isn't anything else I do on the internet.  And even if I'm doing something else, this is my homepage, so I kind of have to go here whenever I go on the internet.

[quote="kueh"]I put myself at <20. But its hard to tell sometimes.[/quote]

Yeah, it is really hard to say!!
Hey, does the server have some kind of stadistics about this??

Best,

[quote="Valentin Vornicu"][quote="mcalderbank"]how many hours are you on, then? :D[/quote]Secret :)[/quote]

But in any case you are not online all the time... it says offline !

Maybe we should get a group to help determine when Valentin is off? Then subtract that from 168?

My content isn't removed because it's too 1337 to be offensive ;) .

im talking something like a post with all swear words with a dash inserted between them so the filter doesnt recognize it

if you hide yourself, will it say online or offline

[quote="Scrambled"][quote="Valentin Vornicu"][quote="joml88"]The admins are not always on.[/quote]Are you absolutely sure about that? :D :D[/quote]
im quite certain: ive seen offensive content not removed in 15 minutes (which would be the time i assume the admin should get to that forum)[/quote]

C'mon, according to the main administrator page, there are four of them (even when I think Valentin recently mention five) therefore if you need an administrator premanently online, you would need to have each of them in different hours at least six hours a day, seven days a week, and all that time just patrol the place to find the offensive content to remove it! (Well, I don't know if the moderators can do that too, to remove some messages). I suppose they also want to take their time and try one of the problems once in a while, or read others opinions carefully.

Best,


MATHCOUNTS
SFFT
trigonometry
geometry
3D geometry
algorithm
inradius

How did you guys do?
The answers were
1. 32
2. SOMETHING i think it was A=0,B=1,C=7,L=6,N=2,O=4,R=4,S=8,T=3
3. a)-1/2 b)5/2
4. A=2,B=3
5. a)$24^{2}$ b)$n^{2}$
6. 2 and 3
7. 15/2 or 7.5

I think I got them all right  :D

Anyone know when the results are supposed to be up?
It's been a month already....

Dude, I got 2nd place in the frosh/soph division or whatever that is.
How were the tiebreakers graded? was it like "oh, I like this solution, I'll give it grand prize!"? I can't see how else it was graded.....oh and they better give me a prize or a trophy

What competition is this? I've never heard of it.

[quote="junggi"]It's some retarded, unknown, incredibly easy competition where they give you 3hours for 7 mathcounts chapter(state AT MOST) level questions and has tiebreaker according to the "elegantness" of your solution to your #7.[/quote]

What is the competition called?? Can middle school kids do it too?

Never mind MathCounts level, I think some of them [i]are[/i] old state/national sprint problems. Though #5 is slightly interesting if you didn't know the trick already.

It's called the SCUHSMC, or something. There might be an I in it somewhere. I'm sure you can take it if you talk to your local high school.

[quote="bookaholic"]Never mind MathCounts level, I think some of them [i]are[/i] old state/national sprint problems.
[/quote]
LOL niceeee
[quote]
Though #5 is slightly interesting if you didn't know the trick already.
[/quote]
Yea I guess it is. But I did know the trick beforehand. From w00t.  :wink:
[quote]
It's called the SCUHSMC, or something. There might be an I in it somewhere. I'm sure you can take it if you talk to your local high school.[/quote]
Well but it's over already  :|
So you'll be in high school anyway when you can take it. (you are an 8th-grader now right?)

Does anyone know what date it is this year? Nov 10, perhaps?

It's on saturday, november 10, 2007
at santa clara university, somewhere on campus
50th annual! yay!

How did you guys do? are we allowed to discuss teh answers yet?

anyways..
i got:
1 a) 108 b) 1029
2 1815
3 140
4 (33,-36)
5 13
6 20/3
7 (34,85)

What's your solution for 7? Mine:
[hide]
$ mn\equal{}10(m\plus{}3n)$
$ n|mn \Rightarrow n|10(m\plus{}3n)$
Let $ n\equal{}rs$ such that $ r|10$ and $ s|m$.
Let $ m\equal{}ks \Rightarrow \dfrac{n}{m}\equal{}\dfrac{r}{k}>1$ because $ n>m$.
Casework: if $ r\equal{}1$, $ \dfrac{r}{k}$ cannot $ >1$.
if $ r\equal{}10$, $ n\equal{}10s$.
if $ r\equal{}2$, $ k\equal{}1$ causing $ 10|n$ (work it out)
if $ r\equal{}5$, $ k\equal{}1,2,3,4$
$ k\equal{}1,3 \Rightarrow 10|n$
$ k\equal{}4 \Rightarrow n$ is not an integer.
so that leaves $ k\equal{}3 \Rightarrow m\equal{}34,n\equal{}85$
you get the idea.
[/hide]
everyone i asked got a different solution...

Interesting... there's a [b]much[/b] more famous [url=http://math.scu.edu/putnam/]competition[/url] connected to Santa Clara University.

[quote="bowei"]How did you guys do? are we allowed to discuss teh answers yet?

anyways..
i got:
1 a) 108 b) 1029
2 1815
3 140
4 (33,-36)
5 13
6 20/3
7 (34,85)

What's your solution for 7? Mine:
[hide]
$ mn \equal{} 10(m \plus{} 3n)$
$ n|mn \Rightarrow n|10(m \plus{} 3n)$
Let $ n \equal{} rs$ such that $ r|10$ and $ s|m$.
Let $ m \equal{} ks \Rightarrow \dfrac{n}{m} \equal{} \dfrac{r}{k} > 1$ because $ n > m$.
Casework: if $ r \equal{} 1$, $ \dfrac{r}{k}$ cannot $ > 1$.
if $ r \equal{} 10$, $ n \equal{} 10s$.
if $ r \equal{} 2$, $ k \equal{} 1$ causing $ 10|n$ (work it out)
if $ r \equal{} 5$, $ k \equal{} 1,2,3,4$
$ k \equal{} 1,3 \Rightarrow 10|n$
$ k \equal{} 4 \Rightarrow n$ is not an integer.
so that leaves $ k \equal{} 3 \Rightarrow m \equal{} 34,n \equal{} 85$
you get the idea.
[/hide]
everyone i asked got a different solution...[/quote]
um use simon's favorite factoring trick for 7
the answers look right

[quote="junggi"][quote="archimedes1"]Ugh, I got number 1b wrong  :roll: Perhaps I'll get "partial credit"  :rotfl:
I used SFFt for 7 as well.

Is it just me or can number 5 be killed by Ptolemy's in like 2 seconds?[/quote]
it was a "problem" testing whether you knew the theorem or not
if you didnt then it was actually a problem where you had to think(gasp)[/quote]
Or use Law of Cosines, which wouldn't take much longer.
I had to think for problem 1.

Is there a special way to do number 2?

[quote="archimedes1"][quote="junggi"][quote="archimedes1"]Ugh, I got number 1b wrong  :roll: Perhaps I'll get "partial credit"  :rotfl:
I used SFFt for 7 as well.

Is it just me or can number 5 be killed by Ptolemy's in like 2 seconds?[/quote]
it was a "problem" testing whether you knew the theorem or not
if you didnt then it was actually a problem where you had to think(gasp)[/quote]
Or use Law of Cosines, which wouldn't take much longer.
I had to think for problem 1.[/quote]
hrm did i see you there?

i saw i like pie xscapezer and vishalarul
i think

[quote="junggi"][/quote][quote="archimedes1"][quote="junggi"][quote="archimedes1"]Ugh, I got number 1b wrong  :roll: Perhaps I'll get "partial credit"  :rotfl:
I used SFFt for 7 as well.

Is it just me or can number 5 be killed by Ptolemy's in like 2 seconds?[/quote]
it was a "problem" testing whether you knew the theorem or not
if you didnt then it was actually a problem where you had to think(gasp)[/quote]
Or use Law of Cosines, which wouldn't take much longer.
I had to think for problem 1.[/quote][quote="junggi"]
hrm did i see you there?
[/quote]
You don't know me.

[quote="archimedes1"][/quote][quote="junggi"][/quote][quote="archimedes1"][quote="junggi"][quote="archimedes1"]Ugh, I got number 1b wrong  :roll: Perhaps I'll get "partial credit"  :rotfl:
I used SFFt for 7 as well.

Is it just me or can number 5 be killed by Ptolemy's in like 2 seconds?[/quote]
it was a "problem" testing whether you knew the theorem or not
if you didnt then it was actually a problem where you had to think(gasp)[/quote]
Or use Law of Cosines, which wouldn't take much longer.
I had to think for problem 1.[/quote][quote="junggi"]
hrm did i see you there?
[/quote][quote="archimedes1"]
You don't know me.[/quote]
yes that is why im asking did i see you there

[quote="junggi"][/quote][quote="archimedes1"][/quote][quote="junggi"][/quote][quote="archimedes1"][quote="junggi"][quote="archimedes1"]Ugh, I got number 1b wrong  :roll: Perhaps I'll get "partial credit"  :rotfl:
I used SFFt for 7 as well.

Is it just me or can number 5 be killed by Ptolemy's in like 2 seconds?[/quote]
it was a "problem" testing whether you knew the theorem or not
if you didnt then it was actually a problem where you had to think(gasp)[/quote]
Or use Law of Cosines, which wouldn't take much longer.
I had to think for problem 1.[/quote][quote="junggi"]
hrm did i see you there?
[/quote][quote="archimedes1"]
You don't know me.[/quote][quote="junggi"]
yes that is why im asking did i see you there[/quote]
I don't know you.

[quote="archimedes1"][/quote][quote="junggi"][/quote][quote="archimedes1"][/quote][quote="junggi"][/quote][quote="archimedes1"][quote="junggi"][quote="archimedes1"]Ugh, I got number 1b wrong  :roll: Perhaps I'll get "partial credit"  :rotfl:
I used SFFt for 7 as well.

Is it just me or can number 5 be killed by Ptolemy's in like 2 seconds?[/quote]
it was a "problem" testing whether you knew the theorem or not
if you didnt then it was actually a problem where you had to think(gasp)[/quote]
Or use Law of Cosines, which wouldn't take much longer.
I had to think for problem 1.[/quote][quote="junggi"]
hrm did i see you there?
[/quote][quote="archimedes1"]
You don't know me.[/quote][quote="junggi"]
yes that is why im asking did i see you there[/quote][quote="archimedes1"]
I don't know you.[/quote]
ah i see

1) I dunno, guess it shoulda been easy, but I got it wrong. Both parts. I got part (b) wrong because I counted 0000 as a valid 4-digit #. :oops: Not quite sure about part (a). I put 117.

2) Easy, almost trivial

3) Pretty darn easy

4) Okay, fine, I used long division. But still, I got it right...

5) Easy (especially awesome w/Ptolemy's Thm)

6) I am the stupidest... well, whatever. I thought the Pythagorean triple was 10, 24, 25. Don't ask. I put 240/35 = 48/7 instead of 240/36 = 20/3

7) Erm, well... I got it right, at least...

8) Eh, I'm not quite sure what I put there anyway.

So basically, I psuedo-failed, as usual.

1 a) 120... I thought the problem was easier than 1 thought. Anyway, I overcounted things like 353 and 535 and 335, etc.

b) 1031... forgot 2003 and 2005

2 Yeah, pretty easy $ 5\cdot 19^{2}$

3 Just cube and add 15, so $ 140$

4 Division algorithm kills problem

5 Ptolemy kills problem

6 Just consider the cross-section, inradius

7 SFFT

8 Did anyone consider negative factors of 300? Because $ m\minus{}30$ and $ n\minus{}10$ could be less than $ 0$...

i got second place again XD (among frosh soph)

Could you post a link to the winners page, if there is one?

[quote="bowei"]grr honorable mention
do you think the graders grade on whose handwriting looks neatest?
i heard you could get perfect score and still not make honorable
it all depends on the tiebreaker, eh?[/quote]
everything depends on the tiebreaker...i doubt that only 15-20 got perfects
the handwriting theory doesnt seem plausible; if that was the case, id never be even close to getting an award

@arch: the winners got sent a certificate and a list of winners

Well, could you post the winners? I want to know what happened. How much do they show?

EDIT: What kind of scores did you get?

edited for spelling

I think the basic point is that the prize-winners might as well be determined by lottery from the list of perfect scorers, and there wouldn't be much difference. Though I'm still happy that for once in my life, somebody actually liked my proof-writing style.

[quote="junggi"]i gots my prize
a book (visions of symmetry; should be an annoying book...) and 100 bucks

yo lynnelle how much money did you get[/quote]
HOLY CRAP THAT MUCH MONEY? I hella gotta not read 6 questions wrong next year.