USAMTS
function
summer program
Mathcamp
conics
ellipse
LaTeX

come and showoff/complain/give suggestion for USAMTS.

round 4 mess up on number 3...Oh well (Erik Madisen I still have the same score as you!!!)

round 4: [b]5[/b] [b]5[/b] 2 [b]5[/b] [b]5[/b]

SUPER happy because THIS IS THE FIRST TIME I GOT 4 COMMENDED SOLUTIONS!!!! :first:

boost my number of commended solutions to 6...GREAT!!! (before I got no commended for 1st round, 1 commended for each of 2nd and 3rd round)

I suppose we were so fated...  ;) I got 5 5 [b]5[/b] 5 2.  Darn I made such a stupid assumption on problem 5 - never again will I make an assumption because it's easy (and when even my diagram indicated that it wasn't true!).  One point worse than last year, but comparatively I'm higher in the rankings than I was last year.  I feel like I may have missed out on a commended for problem 1 because the grader commented that I didn't have a diagram even though I spent half an hour making a really nice one for it  :mad: Ah well, 8 total commendeds still makes me happy.

This round:
[b]5[/b] 5 5 3 5

Total:  93

3 commended solutions in the end (one on each round 2, 3, and 4), and a 25 on round 1.  Overall, each round brought pleasant surprises, and I definitely didn't expect to do this well in my first year.  Hopefully neither Erik nor Eric is a senior, so we can have a nice rematch for next year ;).  At the end of the year, I think my favorite problem is definitely round 3 #5, I love problems that don't look like math at all but turn out to be highly mathematical (If anyone didn't do so already, all should read the solution that someone submitted using functions to describe the moves to 3/5 'twas sicknasty.)

Congrats to everyone who saw the contest through to the end.  I say congrats to so many people because this contest is almost entirely self-driven.  I mean, yeah, there are some parents that will try to force it, but ultimately, the work is self-motivated.  And, of course, thanks to AoPS for running the show (large-scale contests can be very headache inducing.), and much gratitude to the sponsors, NSA (massive thanks to the graders), C/USA Mathcamp, and Key Curriculum Press.  Thankee for improving my (and presumably many others') proof-writing skills (and toolboxes against expected value problems) over the year.

[quote="pkerichang"]yeah I guess according to the rubic you should get 98, huh... (you're still doing this next year, right??? Don't tell me you get lazy your senior year).

First time doing USAMTS and I enjoyed a lot (although the internet policy should change a little, as I pm Mr. Ruszck about).

I guess this applies to USAMTS too: wait until next year for perfect score

side note: It says we choose prize on April 17, how do we go about doing so???[/quote]
I'll definitely be doing this again next year - I was shooting for 100 this year, but I guess that and 15 commendations (to beat Zach Abel's 14 commendations last year) will have to wait until next year  ;) .  I don't know what was up with my score - I definitely should have gotten a 98, given nsato's interpretation of the rubric, and my solution was well-presented (albeit incorrect), so I don't see any reason for a discretionary point to be taken off.  I suppose I see the rationale for the graders having some wiggle room in their grading, but I still think that the extra point taken off is unwarranted.  Also, once prizes are released, you'll get a list in your USAMTS account with check boxes - you check which prizes you want and then submit your choices.

On an unrelated note, did anybody notice that USAMTS participation dropped by almost half from last year?   428 people particpated this year, and 718 participated last year.  That's a pretty concerning drop in participation.

Why are there so few participants??!?!  I didn't compete in it this year, but then again this year was the first time I was ever asked to write a proof for a math competition.  Free proof practice with detailed comments on your solutions? I can't turn that down next year.  Only 400 out of 50 states, that is surprisingly little.
Btw, what are the prize choices like anyway?

Round 4: [b]5[/b] 5 5 3 4

The grader must have liked my diagram and extremely short solution for #1  :D .  I made an incorrect assumption on #4, and I made a mistake in finding the minor axis of the ellipse on #5.

My overall score is 91.  It was interesting watching the participation drop round by round.  I think there are very few people who actually stayed with it until the end, judging by the leaderboard.  Congrats to those who did!

I have to say that USAMTS has been a learning experience for me.  It is the only proof-based competition I have ever participated in.  I learned from the problems that I struggled with and eventually solved, and even more from the problems I couldn't solve.  My proof-writing and non-procrastination skills have definitely increased.

Hmmmm... only 400 people?  We need to advertise this contest more prominently in our schools, like... start getting word out about the USAMTS before summer break begins (and give somewhat frequent reminders to participate).  The last thing we want is this contest going under due to low participation, and therefore loss of funding.  I think it's up to all of us to make sure we bring participation up to (if not past) 700.  I'm sure there are plenty of people who'd love to do this thing (I sure would have enjoyed it last year) that just haven't heard about it.  Get your math team interested in this!  Showing a full, TeX'd up solution set is sure to impress plenty of people (I know the people who've seen my solutions are always impressed, because of the typesetting if for no other reason.  A couple even asked how I got ahold of the answers before they were posted, because of how professional LaTeX looks).  Bottom line is, we can't let such an awesome oppertunity vanish (a free, proof based contest?  Thanks a bunch, graders!).

Not that I'm saying anything new, of course... but just re-iterating the obvious =p.

To Matt:  Wow... about 75 people dropped out each round... I didn't even notice that while it was in progress...  Percentage wise, dropout looks a bit worse than in year 16.  Fortunately, our dropout percentage change is no where near the massive change in number of initial perticipants.

Complaints, i suppose...

These graders often seem to take as much as possible (this time, some were a little nicer and didn't penalize me for not reducing my answers completely).  On number 1, for example, I immediately saw that $RQ$ was the radius of $\mathcal{C}_4$.  I just never thought to prove it, because it's that kind of thing you see so much and never really have to prove...so much for being able to recognize things.  Anyways, my grader gave me a 3 on that one, so I don't even get to protest for a 4.  Other than that one flaw, I made the proof as simple and clear as possible...

On other problems, they took off points simply because the proof was "not clear", even though it was completely clear to me and I couldn't see a way to clear it up.  And they took off 1 point instead of 2, so i never could protest.
Even with nice graders, I would have gotten maybe 95 max though...so it's okay...

So now, my final score is 88.  For those who want to know, it's
5/5/2/[b]5[/b]/5
4/5/5/4/3
5/5/5/5/2
3/5/5/[b]5[/b]/5
My first year on it...

I'll go for a gold next year...and aim for...8 commendations.

To K81o7:  I've found the graders taking fewer points from me than I expected (I actually got a 5 with a wrong answer once because the mistake was simply an addition error).  Keep in mind, a big part of this contest is learning how to communicate your proofs both clearly an concisely.  That's why the "discretionary point" exists, so that graders can show you what problems were unclear, so that you can learn to fix it later.  Remember, the main goal of this contest (and all math contests) isn't to win medals/score well/etc, but rather to learn from it as much as possible.  If you want, post up a few of the solutions that lost points for being unclear, and I'll see if I can find out where they were unclear or could be improved.

round 4 i got 5 5 5 [b]5 5[/b] and i was so excited about that!!

so my total is 86.  i heard we get t-shirts.  that's awesome!!!  anybody know what sort of prizes the prizes are??

A few people have asked about prizes, so I'll recount what I remember about last year's prizes.  The prizes are a selection of math-related books and software.  The books include both specialized books (like books on Diophantine equations, inequalities, etc.), general problem books (like USAMO-style problem books, compilations of problems from various math journals and contests, things like that), and general interest math books (if I remember, In Code was offered last year - a really good, accessible intro to coding theory narrated by a girl who developed her own coding method).  The software last year was Geometer's Sketchpad.  Of course the prizes are subject to change every year, but probably the mix will be similar.  I really liked what they had available - they're good prizes for budding math students  :)

I got my highest score ever, plus I discovered how much I enjoyed geometry. (Namely a lot).

So I got a 21, which is the highest score of anyone in Ga on that the last round.

I'm super new at this. I've never done the usamts before but I got something in the mail for it and I tried it. I really enjoyed it, just because I'm so used to math being... less than challenging. Mostly, I just don't know what constitutes a 'good' score. I mean, looking at these forums and everything, most people have really [i]really [/i]high scores and multiple commended scores and such. Obviously there are prize levels and things, but it seems to me that everyone in the top half gets a prize, and the other people probably didn't even submit for all four rounds, so... I guess I have two main questions. First, what kind of people are involved in this to begin with and second, what [i]is [/i]considered a good score? I don't know. I got a 60 but I didn't get commended scores or anything... Insight?[/i]

[quote="pkerichang"]hmm...are the problems this year harder (except round 4)??? Because I try out some past problems and they seem to be easier.. (but that may also be that I improved)

Hope I get some cool math stuff for the prizes[/quote]
My memory of last year is a little fuzzy, but I'd say in general the problems this year were similar in difficulty to those last year, except that the last problem (#5) of each round this year was a bit harder than last year - but in a good way; they were well-varied and made you think (I was especially impressed by the table-rotating problem, which was unusual and challenging).  Also, some problems this year (namely the expected value problems) could have been tough if you didn't really grasp the meaning of it, but I think that the "intro to expected value" posted beforehand did a wonderful job bringing everybody (including myself) up to speed.  I'm looking forward to seeing next year's theme.

[quote="rrusczyk"]For those asking about fewer participants and talking about getting out the word, [b]please do so![/b] ...
[/quote]Can having divisions A and B help draw less experienced audience? Perhaps with 2 or 3 common problems between the A and B divisions each round (like Mandelbrot)? Is it feasible in terms of the additional work load? I don't think that should actually add too much work -- still grading 5 problems per paper.

I suggest this because the posted scores may be intimidating to potential newcomers. Unlike 2 years ago and prior, when scores were unknown, or much lesser known, to the public.

Well, I got 16 like I predicted (I always get exactly the score I predict, yay) but not in the way I expected. I was thinking 5 5 4 1 1 and got 4 5 3 3 1. I am completely and utterly furious about #1--out of laziness, I randomly assumed a diameter of 2 for circle C_1 instead of using, say, "d"...in hindsight, I really have no clue what I was thinking. Jeez. Bronze it is, not that I was ever in the running for silver...but I am so getting it next year.

EDIT: if it's possible, having two divisions sounds like a great idea. I know some people who probably can't solve most of the current USAMTS problems but who would probably jump to participate in an easier version. So yeah.

Seriously, PLEASE LET THE NEXT THEME BE ASSOCIATED WITH COUNTING!!! (one of my WORST field in mathmatics).

To be honest, ust by doing USAMTS makes me better at math (seriously, for the first 2 rounds, I'm constantly thinking about USAMTS problems when I'm awake for the WHOLE month.  Ends up round 3 and round 4 getting easier for me.)

I must again say thanks to the graders, sponsors, and the organizers for this wonderful competition (especially the one grader who gave me a five on my COMPLETELY brute force solution for round 3 problem 5, I work out ALL cases).

The two divisions idea sounds good to attract people, but if there are more honor/recognition to this contest, I believe more people will participate (like scholarship and stuff, I think most of the 400 participants just do this out of self interest).  Also, if we have two divisions, wouldn't grading be more stressful for the already busy graders??? (because there's more problems).

At last I know most of the people who didn't participate because they give up after attempting for 1 hour and just say "oh, I'm completely hopeless."  For these people, I want to say that it's not true.  For the first two round I solved the last one or two problems like 3 days before the due date (and as mention above, I kept on trying for the whole month).  Having "mental toughness" (quote by Paul Zeitz) is definitely one of the important key to USAMTS problems.  Hey, they're designed so that you have a whole month for them.

I disagree with creating a "tier" system for the USAMTS - I think it would betray the principles of the USAMTS.  This contest is meant to be accessible to students of all skill levels,  yes; however, I don't think that splitting it into two contests would help to accomplish that.  Instead, it would only accentuate the differences between the "lower level" less experienced problem solvers and the "upper level" more experienced ones, without really giving the less experienced solvers a taste of more difficult problems.  Furthermore, the way the USAMTS is currently designed, every contest has easier problems (problems 1 and 2 are almost always accessible even to those with very little prior experience) and harder ones, so that those with less experience and skill can still feel that they've accomplished something, while at the same time allowing them to play with some more advanced problems.  So, this setup actually helped to improve the skill levels of all involved - the newcomers can get exposure to intermediate-level problems, and the old hands can cut their teeth on the advanced problems, and on crafting elegant solutions to them.

Another thing to remember is that the contest is designed to make you think deeply about problems and express yourself clearly when writing solutions to them.  Creating a lower-level contest would also work against this by encouraging newer students not to think as deeply about hard problems by giving them an "easy way out," so to speak, and would put them at a disadvantage in the grading system; as it is, all solution-writers are compared equally.  This gives newcomers a tremendous opportunity to learn from more experienced solution-writers and polish their skills.  If the contests were segregated, differing grading standards would undoubtedly apply, lessening the potential experience gained, and the newer solvers would not interact as much with those taking the more advanced contest.

On balance, I think that segregating the contest would not be beneficial - one of the great strengths of the contest is that it brings together those of us who have been doing contest math for years with students just looking into advanced problem-solving for the first time; by working together on the same problem sets and sharing thoughts and solutions, it helps all parties involved to grow personally and gain knowledge and experience more effectively.

[quote="zanttrang"]... the newcomers can get exposure to intermediate-level problems ...[/quote]If there are enough newcomers.

I agree with your philosophy in general. My suggestion was strictly a reaction to what seemed to be a rally for higher future turnout.

It is ironic that I suggested in another thread that AMC10 should be abolished in favor of a single, recombined AMC10-12 test (like the old AHSME), versus my suggesting a division here!

For AMC, since the selection to higher acclaims (AIME, USAMO, MOP, IMO) are at stakes, I feel that the advancing mechanism should be, and can be, more equitable. AMC can still rightfully adjust the advancing criteria to afford the 10th grades and below better chances. Also, with 25 problems in the test, more can be done to accommodate a much wider range of audience.

For USAMTS however, the idea is get students involved in problem solving in a well thought out manner without acute time pressure. The current format had certainly proven to be right for many upperclassmen and a selected few underclassmen. It had also been alienating to prospective underclassmen who were not quite ready, and alienating to even prospective upperclassmen who were late to discover USAMTS. Meanwhile, it is indicated that participation has dropped.

If USAMTS attracts a lot of students (I don’t know what a lot is), then whether they attempt all or some of the problems, the main USAMTS objective will be well served. I am guessing though, that once a student finds that 2 or 3 problems in one round (especially round one) are beyond their willingness, they are likely to give up on USAMTS for the remainder of the year -- especially in light of the high scores that are visibly posted. If that is the case, then USAMTS loses out on a fair number of participants – a number that might be the difference between adequate and inadequate turnout.

I got 5/5/5/3/1 on this round, but I'm challenging #4.  So 82 overall (hopefully 84).  It is my first time, so I am happy with a silver.  Would have liked more than 2 commendations though . . .

[quote="rrusczyk"]For those asking about fewer participants and talking about getting out the word, [b]please do so![/b]

I think one reason we had so many fewer participants this year is that we got the first round problems out much later than in past years.  We'll be much better about that next time around.  We expect to have the first round up by the end of June.[/quote]

The end of June! :o  What about those of us going to MathCamp?

EDIT: Whew, #4 was resolved in my favor.  So 84 overall.

[quote="E^(pi*i)=-1"]I got 5/5/5/3/1 on this round, but I'm challenging #4.  So 82 overall (hopefully 84).  It is my first time, so I am happy with a silver.  Would have liked more than 2 commendations though . . .

[quote="rrusczyk"]For those asking about fewer participants and talking about getting out the word, [b]please do so![/b]

I think one reason we had so many fewer participants this year is that we got the first round problems out much later than in past years.  We'll be much better about that next time around.  We expect to have the first round up by the end of June.[/quote]

The end of June! :o  What about those of us going to MathCamp?[/quote]

Yeah, I'm sure that many people would be away for the summer (like me), either for camps or family vacations.  So that might cause even less people to participate (at least in Round 1).  A good time would probably be around August-September, after school as started.  It would still be earlier than the time this year, which was October, I believe?

The first round will most likely not be due during the summer, even if they release it early.  It will give people who know about the contest to begin wtih a lot of extra time, yes, but it will also give more opportunities for newcomers to discover the contest over the summer.  For reference, the first round of year 16 was still due in October, and it was released earlier than this year's round (during the summer sometime; I don't remember when).

78 - 5 5 5 3 4

I found this round somewhat strange - on problem 4, I made a very very minor, trivial mistake (left out $\pm \sqrt{6}$), a mistake where the answer should be screaming out to anyone who reads my solution and hasn't already solved the problem, and...I got a 3.
On the other hand, on problem 5, I made a more drastic error of method (messed up minor axis), where my solution gives no obvious way to fix the error, yet I got a 4. Weird huh?  :?

YEAH USAMTS!!!! A great way to practice proof-writing and problem-solving. And that is, problem-solving of a kind where the solution isn't so obvious and can take a long time to find (unlike AMC/AIME).

In particular, I would say the USAMTS is a good way to transition to the olympiad level. Although USAMTS problems tend to be easier than USAMO problems, they're of the same kind; [i]proof-style problems that take a very long time and a lot of mental toughness to solve[/i]. Once you get good enough at problems this hard, even if it does take you hours to find the solutions (but you do eventually find them on your own), you're ready to begin to tackle olympiad level problems. (AoPS olympiad classes or certain other resources are probably the other step important in beginning to tackle olympiads.)

5/5/5/2/3....95...ouch

The prizes have been discussed recently in this forum.  I believe they are books and computer software (geometer's sketchpad).  The difference between gold, silver, and bronze getting prizes is that gold gets the first pick.

[quote="tongchen1226"]5, 5, 2, [b]5[/b], 5
82
im actually surprised that i didnt get commended for 1, 2, and 5. in my opinion, my #4 was rushed and not as great as my other ones. but getting commended is always good.
silver, what are the prizes?[/quote]

Hey I always find the same thing happens to me, I get the right score and the right number of commendations, but they're on the wrong problems!

[quote="K81o7"][quote="joml88"]5/5/5/2/3....95...ouch

[/quote]

Ow, that hurts...what were the mistakes?[/quote]

I only found 2 solutions for #4 and for #5 I made a wrong geometrical assumption I believe.

Sucks to lose it right there at the end...

But I agree with calc rulz, I often get commendations on problems, my solutions to which were very messy in my opinion.

5/5/5/[b]5[/b]/5

100 total, 1 total commended.

Looks like I almost got 97, but the graders changed two of my scores on re-grade.

Anyway, my family is complaining, so I can't weigh in on some of the other issues discussed in this thread.

JB[/b]

jackb1115 is the bomb

(oh, I have a 47; I didn't do the last two rounds.)

=)

Thanks Sonar!

Anyway, I first wanted to extend my thanks and gratitutde to the Art of Problem Solving Foundation, the NSA, the graders and problem writers, and all the other sponsors of the USAMTS.  I really enjoyed this contest, and hopefully speak for everyone when I say "Thank you for an awesome group of problems."

As for the declining "membership," I think that might partially due to the fact that this year's round one problems were, on the whole, more difficult than last year's round one problems.  I think some people might have looked at round one this year and had a feeling of "I can't do this."  Perhaps if the round one problems were kept at the difficulty levels they were at in previous years, it would help get more people turn in round one solutions and stick with the contest.  I understand, however, that controlling the difficulty of a group of problems is more difficult than it sounds.

Another reason USAMTS membership is not high is the amount of work involved.  Unlike the AMC, which takes up 75 minutes (and you get to miss school to do it), USAMTS requires solving and writing up 5 problems over the course of a month.  The extra workload makes the contest seem more difficult and less fun to a lot of students, especially upperclassmen who have enough on their plates as it is.

Though it is impossible to reduce the amount of work that participants need to go through to participate without compromising the integrity and "Feel" of the contest, I beleive that good advertising could counteract this problem.  I only knew about USAMTS because of AoPS and because I attended HCSSiM.  Not many students, even those who are mathematically inclined, get the word about USAMTS.  If flyers were sent to high schools to be given to their strong math students, more students would know about the contest.

I hope my suggestions help, and thank you again for a wonderful contest.

JB

[quote="jackb1115"]
Though it is impossible to reduce the amount of work that participants need to go through to participate without compromising the integrity and "Feel" of the contest, I beleive that good advertising could counteract this problem.  I only knew about USAMTS because of AoPS and because I attended HCSSiM.  Not many students, even those who are mathematically inclined, get the word about USAMTS.  If flyers were sent to high schools to be given to their strong math students, more students would know about the contest.
[/quote]

I thought that USAMTS had some very good advertising.  I remember getting a little flyer about it at least two (maybe some more) competitions.  Duke was one of them and I think C of C was another one.  This seems to be the correct route to go in my opinion.  Sending flyers to high schools might work...but for the most part I don't see this as being very cost efficient.  Most high schools would just toss them away, not hand them out, etc.  Perhaps sending flyers to certain high schools which you already know have a strong math foundation would be a good idea (e.g. TJ, Exeter, Vestavia, etc).  But then again, a lot of the coaches and more experienced students at these schools might already be spreading the word regardless.

Ok, so here are my suggestions then:

$\bullet$ Send flyers to all the USAMO qualifiers from the previous year.
$\bullet$ Send flyers to all math summer programs, i.e. promys, math camp, awesomemath, etc.
$\bullet$ Continue to send flyers to the larger competitions.
$\bullet$ Make the first round easier next year (not trivial...but, you know, a tad bit easier than this years first round)


geometry
vector
calculus
analytic geometry

Does anyone know any good books dealing with 3-dimensional geometry? I can't seem to find any.

I don't think this subject really merits a book of its own, since 3-D Geometry can really be solved using the techniques of 2-D Geometry. However, learning some vector calculus could help in a few problems.

[quote="1234567890"]I don't think this subject really merits a book of its own, since 3-D Geometry can really be solved using the techniques of 2-D Geometry.[/quote]

What types of 3-d geometry problems are reducible to 2-d?  There are many, many situations where 3-d is qualitatively harder and not just an extra coordinate in the calculations.

For a ton of problems (with solutions) you might want to try this:

http://www.math.su.se/~mleites/Prasolov/Solid/


show that if$ x,y,z$are positive integers, then$ (xy\plus{}1)(yz\plus{}1)(zx\plus{}1)$is a perfect square if and only if$ xy\plus{}1,yz\plus{}1,zx\plus{}1$are all perfect squares.

Hi yildobekir. Thanks for your message.

All I can offer at the moment is the following.

Clearly, if $ (xy \plus{} 1) \equal{} a^{2}, (yz \plus{} 1) \equal{} b^{2}$ and $ (zx \plus{} 1) \equal{} c^{2}$ then

$ (xy \plus{} 1)(yz \plus{} 1)(zx \plus{} 1) \equal{} (abc)^{2}$.

For the converse, I'm thinking of looking at none, one or two of the factors being squares and obtaining contradictions.

If two of the factors are squares then the third one must be.

If only one of the factors is a square , then we have, say,

$ xy \plus{} 1 \equal{} a^{2}$
$ yz \plus{} 1 \equal{} b$
$ zx \plus{} 1 \equal{} b$,

so $ x \equal{} y$ and $ x^{2} \plus{} 1 \equal{} a^{2}$

So $ x \equal{} 0$ and $ a \equal{} 1$, but $ x$ has to be a positive integer.

So the only possibility left is that none of the factors is a square and I suspect that this can be dealt with in the same way.

Hope this is helpful. I'm hoping that someone else will point out an error in my thoughts and/or provide a full solution.

See [url]http://www.mathlinks.ro/viewtopic.php?p=849187#849187[/url]. It is a hard problem.


Something like this has been posted before, I believe, but it's a fun problem.

You are given 19 chips, and the information that 18 of the chips are identical and one is heavier than the rest. Using only a balancing scale, show that it is possible to discover which chip is the heaviest in three or fewer weighings.

And while you're at it, can 19 be improved? (i.e. can you still do it with 20 chips and 3 weighings? If so, what's the maximum?)

First, you pick twelve random ones.  Then you put six on each side and weigh it.
            IF: one side is heavier than the other side, take all 6 and split them and put them on the scales again.  Whichever one is heavier, take that group of 3 and pick two random ones.  if one is heavier than the other, then that is the heavy one.  If they are both the same, the one you excluded from the last weighing is the one.
           IF: both of them are the same, then take the 7 you excluded and pick 6 out of that group and put 3 on each side.  If the weighing is the same, the one you excluded is the heavy one.  If not, take the heavier group of 3 and pick 1 for each side.  If one is heavier, that is the chip.  If not, the one you excluded is the chip.

No. I couldn't find a way to do it in 3 steps.
Split 20 into two groups of 10 chips.
Take the heavier group and split that into 5 chips and put 5 on each side of the balance.
There will be a heavier side.
Take the five chips on that side and split it into 2,2, and 1 chips. \
Place two on each side of the balance.

IF:One side is heavier, take that side and split the chips so that there is 1 on each side, and you will find the heavier one. (this is the fourth weighing)
IF NOT: The excluded chip is the heavy one.

You're not thinking hard enough. Try reading the earlier solution.

By the way, yes nineteen can be improved.  I believe it can be done up to 22.

27? If you split each round into 3 piles, you can definitely do up to 27. Not sure if you can go any higher.

As a side note (question), what would happen if we didn't know whether the odd coin was heavier or lighter than the others?

Try proving that you can't go higher. I haven't actually tried it, so I don't know how difficult it is.

Once you understand the most mechanism, it's fairly obvious that you can't go higher than [hide]27. With a single weighing, you can determine which of three sets contains the heaviest coin. On the third set of weighings, this means you can have at most three coins left. On the second set, you can have three sets of three, which is nine coins, left. Therefore, on the first weighing, you can have three sets of nine, or 27 coins, and no more.[/hide]

As for the side note, [hide]I haven't thought about it carefully, but it seems like you would require at least one extra weighing. Perhaps only one extra.[/hide]

I think it will always take you two weighings to figure out if it's heavier or lighter: whether or not you can gain additional information from that, I'm not sure.

Well, actually, I think it depends what you want:

With two weighings, can get 4 coins, I think, but you won't know at the end necessarily whether the counterfeit is heavier or lighter: consider coins 1, 2, 3, and 4.
Weigh 1 vs. 2.  If equal, weigh 2 vs. 3.  If equal, 4 is counterfeit.  If unequal, 3 is counterfeit.
Weigh 1 vs. 2.  If unequal, weigh 1 and 2 vs. 3 and 4.  If 1 and 2 are lighter, then the lighter of 1 and 2 is counterfeit.  If 1 and 2 are heavier, than the heavier of 1 and 2 is counterfeit.

However, if you want to know at the end for sure whether the counterfeit is heavier or lighter, you can only get 3 in 2 weighings.

Confuted, i think it will work past 27.  28 will work.

Um, Scrambled:  2 of 8 and 3 3's is (first of all) five, not four groups and (second of all) only 25 coins.

This is what mathworld has to say:
http://mathworld.wolfram.com/Weighing.html

"n weighings are sufficient to find a bad coin among (3n - 1)/2 coins.  vos Savant gives an algorithm for finding a bad ball among 12 balls in three weighings (which, in addition, determines if the bad ball is heavier or lighter than the other 11), and Steinhaus gives an algorithm for 13 balls."

The 12 coin problem is a classic (and pretty hard).  See if you can do it on your own...


Let $a,b,c$ be positive real numbers with $abc=1$.Prove that the following inequality occurs: \[a^{3}+b^{3}+c^{3}+a^{2}+b^{2}+c^{2}+a+b+c\geq 3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\] I have two nice solutions for this ineq,and I think there are more :)

[quote="andyciup"]Let $a,b,c$ be real numbers with $abc=1$.Prove that the following inequality occurs: \[a^{3}+b^{3}+c^{3}+a^{2}+b^{2}+c^{2}+a+b+c\geq 3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\] I have two nice solutions for this ineq,and I think there are more :)[/quote]

[hide]$a^{3}+a^{2}+a\ge 3a^{2}$ by AM-GM

add similar inequalities to get

$a^{3}+b^{3}+c^{3}+a^{2}+b^{2}+c^{2}+a+b+c\ge 3(a^{2}+b^{2}+c^{2})\ge 3(bc+ca+ab)=3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$[/hide]

I posted the same solution, but I deleted it soon afterwards, because andyciup didn't say that $a,b,c$ are positive. After that, I was too lazy ( :( ) to check if it's true when two of the numbers are $< 0$. Maybe finding a counterexample is very easy. Who knows?

[quote="D.P.L"][hide="click not to reveal the hidden document"]Expanding and simplifying.

$a^{3}+b^{3}+c^{3}+a+b+c \ge 2(a^{2}+b^{2}+c^{2})$

NOte that $(a-1)^{2}+(b-1)^{2}+(c-1)^{2}\ge 0$[/hide][/quote]

actually, you would get $a(a-1)^{2}+b(b-1)^{2}+c(c-1)^{2}\ge 0$ which isn't necessarily true when a,b,c are real numbers. It has the same flaw as my solution.

Ok,sorry `bout that,I forgot the "positive" condition..My solutions also use the fact the $a,b,c$ are positive.Here is my first solution,since the second one was allready posted:
By AM-GM we have \[a^{3}+b^{2}+b\geq3\sqrt[3]{a^{3}b^{3}}\] ,so \[a^{3}+b^{2}+b\geq3\frac{1}{c}\] .By the same manner we have \[b^{3}+c^{2}+c\geq3\frac{1}{a}\] and \[c^{3}+a^{2}+a\geq3\frac{1}{b}\] Summing these up gives us the requested inequality :lol:

[quote="perfect_radio"]It ain't a flaw after all...
For $a =-1, \, b =-2, \, c = \frac12$, Mathematica tells me that $LHS-RHS =-\frac{61}8 < 0$, so it's not true if two of $a,b,c$ are $< 0$.[/quote]

sorry if this is off-topic but did you use Mathematica to find the counterexample? If so, how?

Actually, I first used a C++ program
[code]float a,b,c,s,min=100;
for(a=-.001;a>=-30;a-=.1)
 for(b=.001;b<=30;b+=.1)
  {
   c=1./(a*b);
   s=pow(a,3)+pow(b,3)+pow(c,3)+a*a+b*b+c*c+a+b+c-3*(1./a+1./b+1./c);
   if(s<0)
	cout<<a<<' '<<b<<' '<<c<<endl;
  }
cout<<min<<endl;[/code]
to find a reasonable counterexample.
Then, I checked with Mathematica.

I think that you can do the same thing with Mathematica, but I'm not too sure...

[quote="D.P.L"][hide="click not to reveal the hidden document"]Expanding and simplifying.

$a^{3}+b^{3}+c^{3}+a+b+c \ge 2(a^{2}+b^{2}+c^{2})$

NOte that $(a-1)^{2}+(b-1)^{2}+(c-1)^{2}\ge 0$[/hide][/quote]
[hide]This is stronger than the original problem: \[a^{3}+b^{3}+c^{3}+a^{2}+b^{2}+c^{2}+a+b+c\geq 3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \\ \Leftrightarrow a^{3}+b^{3}+c^{3}+a^{2}+b^{2}+c^{2}+a+b+c\geq 3\left(ab+bc+ca\right)\] But in fact he proved the following: \[a^{3}+b^{3}+c^{3}+a^{2}+b^{2}+c^{2}+a+b+c\geq 3\left(a^{2}+b^{2}+c^{2}\right) \geq 3\left(ab+bc+ca\right)\] So in fact we can change the RHS of the original inequality to the stronger: \[a^{3}+b^{3}+c^{3}+a^{2}+b^{2}+c^{2}+a+b+c\geq 3\left( \frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)\] Easy problem...[/hide]

[quote="perfect_radio"]

Then, I checked with Mathematica.

[/quote]

How can you check inequalities with Mathematica ?
 :maybe:

For me@home:I know the original inequality can be improved,but I  sticked to the easier but nicer form(I liked the decreasing powers :) )

[quote="silouan"]How can you check inequalities with Mathematica ?
 :maybe:[/quote]

If you write $a>b$ in Mathematica, where $a,b$ are some numerical expressions, then you'll either get True of False.
You can also do the same thing with $a > = b$.

[quote="perfect_radio"]If you write $a>b$ in Mathematica, where $a,b$ are some numerical expressions, then you'll either get True of False.
You can also do the same thing with $a > = b$.[/quote]
How good is it? You can do this in Maple also, but it's really pretty stupid. Most of the time it can't figure it out. As far as I can tell, it doesn't know AM-GM. It probably only uses calculus methods to check.


geometry
perimeter
ratio
geometry unsolved

Can someone help me to solve these problems please

1. Two tangents are drawn to a circle from an exterior point A, touching the circle at points B and C. A third tangent to the circle intersects segment AB at P and segment AC at R and touches the circle at Q. If AB=20, then the perimeter of triangle APR is?

2. A circle is inscribed in an equilateral triangle, and a square is inscribed in the circle. The ratio of the area of the triangle to the area of teh square is?

3. A circle inscribed in a triangle with sides of lengths 8, 13 and 17. Let the segments of the side of length 8, made by the point of tangency, be r and s with r<s. Then the ratio r to s is?

Here's a very useful formula that can help you for 2: if there's an incircle in a triangle, then you can work out it's radius by using a=rs, where a is the area of the triangle, s it's semi-perimeter and r the radius of the circle. The area of the equilateral triangle of side x is (root3x)/4 and the s is 3x/2. Plug in the values to find the radius. The rest is easy.[/u]


OK, any of you peeps with even an ounce of creativity or experience within you, I need some help...

Basically I wanna do a project in mathematics for Westinghouse/Intel. The problem is, I have no idea what genre of mathematics the winning projects usually come from so I need you guys to tell me.

Also, if you have any kewl ideas as to what would be a good project.. I'm all ears..

Here's a site where you can see descriptions of past projects that have done well - http://www.sciserv.org/sts/results/.

Also here is some other information about past projects - http://www.infoplease.com/ipa/A0777576.html

There is a book that was put out, I believe by the MAA (Mathematical Association of America), about topics of investigation for high school students. It must have been made a long time ago because I remember seeing it when I was in high school (back in the early 90s). But I do remember that it had some good ideas that could be extended into full scale research projects. Unfortunately I cannot remember the exact title, but it was definitely by one of the mathematical associations, like the MAA or maybe the AMS. You might want to check their websites for a list of books published by them.

I've also always thought there might be some room for investigation on Mahler Classification of numbers (See A Course in Number Theory, p. 148)that may be possible by high school students. There are also many problems in Partition Theory which are still open and amenable to computer investigation.

The best thing you could do though would be to find a mentor who would lead you in your investigation. I imagine that almost all of the projects that do well in the ISTS are mentored by someone.

Hmm, does anyone know where I can find some past winning intel/siemens project writeups? I want to see how involved and advanced the mathematics ones get, but I can't seem to find any samples to download off the main sites.

Fierytycoon

Come on... there must be more than 2 creative people...

I really need some ideas here guys!

Thanks,
The Idiot without a village

Hmm this may be stupid, but this year for my middle school science fair, I did my project on how pascal's triangle comes into play during pachinko and it got me all the way to 3rd in state.  That perhaps?


An altitude $h$ of a triangle is increased by a length $m$.  How much must be taken from the corresponding base $b$ so that the area of the new triangle is one-half that of the original triangle?
(A) $\frac{bm}{h+m}$
(B) $\frac{bh}{2(h+m)}$
(C) $\frac{b(2m+h)}{m+h}$
(D) $\frac{b(m+h)}{2m+h}$
(E) $\frac{b(2m+h)}{2(h+m)}$

[hide]Set up some equations, let x be the amount that the base is decreased by.

$2(b-x)(h+m)=bh$
$2(bh+bm-xh-xm)=bh$
$2bh+2bm-2xh-2xm=bh$
$bh+2bm=2xh+2xm$
$b(h+2m)=x(2h+2m)$
$x=\frac{b(h+2m)}{2(h+m)}$

So the answer is E[/hide]

i agree but am to lazy to do it myself right now.  Nice formula.  I liked how you put x in there.


Is the dikaryon stage the same as the heterokaryotic stage in basidiomycetes?

nearly...

but i think the 2 nuclei in dikaryon r similar.....


I think we need to find the maximum of P (not the minimum ) ??!!!

No. It's minimum. Maximum is +infinitive

For $b=c=1$ and $a\rightarrow\infty$, from the inequality
$\frac{1}{\sqrt{a^2+bc}}+\frac{1}{\sqrt{b^2+ac}}+\frac{1}{\sqrt{c^2+ab}} \ge \frac{C}{\sqrt{ab+bc+ca}}$
we get $C \le 2\sqrt{2}$.
I think the inequality is true for $C=2\sqrt{2}$, but I have not a solution.:mad:
Hungkhtn, do you have a solution?

[quote="Diarmuid"][quote="Soarer"]Replace $ab$ by $z$ and so on and then you get $x+y+z=1$ and you have a new inequality to prove. I htink the equality case of the new inequality is $(x,y,z)=(0,0.5,0.5)$ and maybe my computations are wrong.[/quote]

I think there's a mistake here - if you try to convert back from $x,y,z$ to $a,b,c$, you'll find $x=0\Rightarrow a=0$ or $b=0\Rightarrow y=0$ or $z=0$, which contradicts $y=z=\frac{1}{2}$.  In other words, you've got zero divisors.[/quote]

Not really. $(x,x,\frac{1-x^2}{2x})$ gives a good solution, and in the limit to zero we have $2\sqrt2$? He just takes the limit right away I think.

[quote="Vasc"]I think your solution is wrong, Gabriel. ;)
Easier is the following nice inequality:
$\frac{1}{a^2+bc}+\frac{1}{b^2+ca}+\frac{1}{c^2+ab} \ge \frac{3}{ab+bc+ca}$.
I wait a nice solution.[/quote]

Nice,too Vasc.
But I don't know if this problem can be solved without computing?

I found a nice solution, without computing. ;)

Much more difficult is the follwong:
If $a,b,c$ a non-negative numbers, then
$\frac{1}{a^2+2bc}+\frac{1}{b^2+2ca}+\frac{1}{c^2+2ab} \ge \frac{2}{ab+bc+ca}$.

I will post it in a new topic.

[quote="hungkhtn"]OK. Lastnight I checked this one and see that easy. (Without computing)

Otherhand, I think this one is true:

$\frac{1}{ka^2+bc}+\frac{1}{kb^2+ac}+\frac{1}{kc^2+ab}\ge \frac{M}{ab+bc+ca}$
With $M=\frac{1}{k}+2$[/quote]

  Can you post your solution for Vasc's ineq , Hungkhtn ?

[quote="hungkhtn"]Otherhand, I think this one is true:

$\frac{1}{ka^2+bc}+\frac{1}{kb^2+ac}+\frac{1}{kc^2+ab}\ge \frac{M}{ab+bc+ca}$
With $M=\frac{1}{k}+2$[/quote]
I think $M=1+\frac{2}{k}$, but only for $\frac{2}{3} \le k \le 3+\sqrt{7}$.
For $k=\frac{2}{3}$ we get a very difficult and interesting inequality.

OK. I'll check it VASC.

It's interesting me.

But maybe the way isn't vaild for my first Problem. Are you have a solution for it?

Sorry, but how did you prove this one ?

$64abc(a^3+b^3+c^3) \le 6(a+b)^5c$

I tried all last night, but I failed.

Thank you very much to anyone can help me

Dear Manlio. I will think about this question. Wait me and I will post as soon.

At first I was wrong when thinking that it's an easy problem. But now I was correct the solution. It's use expanding+ $(a-b)^2(b-c)^2(c-a)^2 \ge 0$ + S.O.S method.

[quote="hungkhtn"]Suppose that $ a,b,c$ positive and $ ab \plus{} bc \plus{} ca \equal{} 1$.

Find the minimum of:

$ P \equal{} \frac {1}{\sqrt {a^2 \plus{} bc}} \plus{} \frac {1}{\sqrt {b^2 \plus{} ac}} \plus{} \frac {1}{\sqrt {c^2 \plus{} ab}}$[/quote]

Have you thought about a simple solution by Cauchy-Schwarz for this monster?

See [b]Problem 2.111[/b], page 165 of new book:

Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, [url=http://canhang2007.wordpress.com/2009/12/08/new-book-inequalities-with-beautiful-solutions-v-cirtoaje-v-q-b-can-t-q-anh/][b][i]Inequalities with Beautiful Solutions[/i][/b][/url], GIL publishing house 2009.

[quote=arqady][quote="hungkhtn"]Suppose that $a,b,c$ positive and $ab+bc+ca=1$.

Find the minimum of:

$P=\frac{1}{\sqrt{a^2+bc}}+\frac{1}{\sqrt{b^2+ac}}+\frac{1}{\sqrt{c^2+ab}}$[/quote]
We'll prove that $\frac{1}{\sqrt{a^2+bc}}+\frac{1}{\sqrt{b^2+ac}}+\frac{1}{\sqrt{c^2+ab}}\geq\sqrt{\frac{8}{ab+ac+bc}}$.
By Holder $\left(\sum_{cyc}\frac{1}{\sqrt{a^2+bc}}\right)^2\sum_{cyc}(a^2+bc)(b+c)^3(3a+b+c)^3\geq\left(\sum_{cyc}(b+c)(3a+b+c)\right)^3=$
$=8\left(\sum_{cyc}(a^2+4ab)\right)^3$.
Hence, it remains to prove that
$(ab+ac+bc)\left(\sum_{cyc}(a^2+4ab)\right)^3\geq\sum_{cyc}(a^2+bc)(b+c)^3(3a+b+c)^3$,
which is obvious after full expanding:
$\sum_{sym}(5a^6b^2+3.5a^6bc+12a^5b^2c+7a^4b^4+181a^4b^3c+162a^4b^2c^2+317a^3b^3c^2)\geq0$.[/quote]

[img]http://s16.sinaimg.cn/middle/006ptkjAzy77gdY7UCj8f&690[/img]

[quote=jokehim]The following is also true: $$\frac{1}{\sqrt{4a^2+bc}}+\frac{1}{\sqrt{4b^2+ca}}+\frac{1}{\sqrt{4c^2+ab}}\ge\frac{2}{\sqrt{ab+bc+ca}}$$
holds for $a,b,c\ge0: ab+bc+ca>0$
Is there any nice proof( eg: C-S, AM-GM,..) for those nice inequalities?[/quote]

Ugly proof:

Let $a\ge b\ge c$.

$$\frac{2}{\sqrt{(4a^2+bc)4(ab+bc+ca)}}+\frac{2}{\sqrt{(4b^2+ca)4(ab+bc+ca)}}+\frac{1}{\sqrt{(4c^2+ab)(ab+bc+ca)}}\ge\frac{2}{ab+bc+ca}$$

After using AM-GM, we need to prove

$$ \frac{4}{(4a^2+bc)+4(ab+bc+ca)}+\frac{4}{(4b^2+ca)+4(ab+bc+ca)}+\frac{2}{(4c^2+ab)+(ab+bc+ca)}\ge\frac{2}{ab+bc+ca}$$

$\iff 2 c (4 a^4 b + 8 a^4 c + 56 a^3 b^2 + 34 a^3 b c - 30 a^3 c^2 + 56 a^2 b^3 + 51 a^2 b^2 c + 14 a^2 b c^2 - 8 a^2 c^3 + 4 a b^4 + 34 a b^3 c + 14 a b^2 c^2 - 20 a b c^3 + 8 b^4 c - 30 b^3 c^2 - 8 b^2 c^3)\ge 0$

which is true.

[quote=jokehim]The following is also true: $$\frac{1}{\sqrt{4a^2+bc}}+\frac{1}{\sqrt{4b^2+ca}}+\frac{1}{\sqrt{4c^2+ab}}\ge\frac{2}{\sqrt{ab+bc+ca}}$$
holds for $a,b,c\ge0: ab+bc+ca>0$
Is there any nice proof( eg: C-S, AM-GM,..) for those nice inequalities?[/quote]

It's Can's inequality.
By Holder:
$$\left(\sum\limits_{cyc}\dfrac{1}{\sqrt{4a^2+bc}}\right)^2\left[\sum\limits_{cyc}(b+c)^3(4a^2+bc)\right] \ge \left[\sum\limits_{cyc}(b+c)\right]^3=8\left(a+b+c\right)^3$$
It's enough to prove that:
$$2(a+b+c)^3(ab+bc+ca) \ge \sum\limits_{cyc}(b+c)^3(4a^2+bc)$$
or
$$\sum\limits_{cyc}ab(a-b)^2(a+b)+14abc(a^2+b^2+c^2) \ge 0$$
which is true.
Inequality holds iff $a=b, c=0$ and per.

[quote=jokehim]The following is also true: $$\frac{1}{\sqrt{4a^2+bc}}+\frac{1}{\sqrt{4b^2+ca}}+\frac{1}{\sqrt{4c^2+ab}}\ge\frac{2}{\sqrt{ab+bc+ca}}$$
holds for $a,b,c\ge0: ab+bc+ca>0$
Is there any nice proof( eg: C-S, AM-GM,..) for those nice inequalities?[/quote]
For all $a,b,c\ge 0: ab+bc+ca>0,$ by AM-GM inequality
$$\frac{(a+b)(a+c)}{4a^2+bc}\ge \frac{ab+bc+ca}{\dfrac{(4a^2+bc)(ab+bc+ca)+a^2bc}{(a+b)(a+c)}}=\frac{ab+bc+ca}{\dfrac{4a^2(ab+bc+ca)}{(a+b)(a+c)}+bc}  $$
$$\ge\frac{ab+bc+ca}{2a\sqrt{ab+bc+ca}+bc}\ge \frac{ab+bc+ca}{(ab+bc+ca)\left(\dfrac{a}{b+c}+\dfrac{a(b+c)}{ab+bc+ca}\right)+bc}=\frac{b+c}{a+b+c}$$
$$\implies \frac{1}{4a^2+bc}\ge \frac{b+c}{(a+b)(a+c)(a+b+c)}.$$
Or$$\frac{1}{\sqrt{4a^2+bc}}\ge \frac{b+c}{a+b+c}\sqrt{\frac{a+b+c}{(a+b)(b+c)(c+a)}}\ge \frac{b+c}{a+b+c}\sqrt{\frac{1}{ab+bc+ca}}.$$

Does equality hold for 2×sqrt2? I think that equality (and actually minimum ) is when a=b=c=1/sqrt3.
Then minimum of P will (3×sqrt3)÷2.
Where is the wrong in my solution?


im sorry if this has already been posted.

when is the due date to fax my entry form?

There's no specific due date.  Fax it in at the time at which you submit your Round 1 solutions.


[quote="xzlbq"][size=150]For $ x,y,z,a,b,c>0$,prove that:

$ \frac{x^2\plus{}a}{yz\plus{}b}\plus{}\frac{y^2\plus{}b}{zx\plus{}c}\plus{}\frac{z^2\plus{}c}{xy\plus{}a} \geq 3$[/size][/quote]
nice,
rise....


analytic geometry
graphing lines
slope
ratio

Find the distance between the lines whose equations are: 3x-4y=7; 6x-8y=-7

[hide="answer"]
first of all rewrite each equation in terms of y

$3x-4y = 7 \Rightarrow y = \frac{3x-7}{4}$
and
$6x-8y =-7 \Rightarrow y = \frac{6x+7}{8}$

notice that both lines have a slope of $\frac{3}{4}$, thus their distance apart can be found by finding the difference of their y-intercepts

$y_{2}-y_{1}= \frac{7}{8}-\left(-\frac{7}{4}\right) = \frac{21}{8}$

So their distance apart is $\framebox{2.625}$[/hide]

[quote="Ignite168"]Incorrect.[/quote]
[hide]doubling the the first, i'm going to go with 21?[/hide]

[hide]Because the lines are parallel, the distance between them will be the right triangle that has the y-intercepts as vertices's on the hypotenuse.

The slope of each line is $\frac34$, so the ratio of the hypotenuse to the distance between the lines is $\frac54$, so the distance is $\cfrac{\frac{21}8}{\frac54}=\frac{21}{10}$.[/hide]

[hide="who's right?"] Subtracting we get 6x-8y=21 6^2+8^2=36+64=100 so we write it as .6x-.8y=2.1 so 2.1 is the correct answer[/hide]

I'm going with b-flat on this one.

I wouldn't recommend using ' in hide tags  :wink:

[hide="who is right?"]insert text here[/hide]

b-flat is definatly right.

How do you know that since the slope is 3/4 the ratio of the hypotenuse to the distance between the lines is 5/4??

Draw the line that is perpendicular to the two lines & look at how it is nicely similar to a 3-4-5 right triangle :D

and you can just tell this by looking at a rough sketch?

unless you graphed it with a calculator..

No, I simply found the lengths of two of the sides & used the Pythagorean Theorem.

Let 1 be the distance between the y-intercepts (the hypotenuse).

Then you should be able to find the other side because the lines are perpendicular.

ahh okay, I see now, I thought maybe you knew some trick or theorem to help you get that other than pythagorans.

thanks :blush:


blogs
search
geometry
geometric transformation
reflection

I'm planning to host a graphic design contest.  I saw one down this forum, but I don't think that was ever completed  :?:  :huh:  So before I list all the rules and ect. I have a couple of questions for possible participants.

1>  Should there be a specific "theme" or "topic" for the graphics? Ex: Nature, anime, etc.

2>  Should there be judges or should this be a poll based contest or a combination of both?

Sure ^_^ I'll try my hand at some graphics.

:D  It might take a while to get participants since this forum is mostly deserted...  :(

It would help if we both put the link in our signatures. Maybe stir up a little excitement in our blogs and other peoples' shout box.

That's a good idea :D  I'll put a post on my blog later.  Edit:  I put a link in my signature but it's not showing up?

Go to Edit Post --> Check the Box "Attack Signature" --> Voila! I think you're done, but that way you may have to check the box every time. I think you don't have to do that if you changed preferences in your Profile.

Edit:

There should be a "qualifying" round, first. Anyone can put in graphics, and you don't have to be bad or good but you can't just text-slap.

So you can't get random images and just add a text and a small border to get through to the competition. Feel free to add this in if you want.

That's true.  But I have an idea for each of the rounds you will be given a picture (chosen by me) and you must work with that picture.  I think this will definitely eliminate 'text-slapping'.  Thanks for telling me.  I probably would have never thought of this idea otherwise.  Also this idea challenges the artists creativity.  The only thing now is to get participants which is probably going to be the hardest part in running this contest.  :wink:

Also if anybody would like to be a judge please PM me.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=214910][img]http://img390.imageshack.us/img390/13/aopspandakc6.jpg[/img][/url]
[u]
AOPS Graphic Design Contest[/u]

Hello everyone I'm hosting a graphic design contest.  The rules are as follows.

-  To join you must PM me or post on the topic (click the above image to go to the topic)
-  Beginners [i]are[/i] encouraged to participate.
-  There will be no saying other peoples graphics "that sucks" or "....wow...".  You will be penalized.
-  25% of your score will be chosen by the public.  (e.g. a poll)
-  75% of your score will be chosen by the judges.
-  The judges will grade you based on creativity,  use of brushes (e.g. not just splattering brushes on an image),  work put into the image,  the background, extra touches, and overall appeal.
-  You do not have to use brushes, gradients, or anything like that.  Do what you feel will create the best graphic
-  All participants will be given the same image to work with.
-  There will be a number of rounds that is based upon how many participants there are.  The amount of participants eliminated in each round also depends on how many participants there are.

[u]Resources[/u]
Like I said before beginners [i]are[/i] encouraged to participate.  If you've never heard of graphic design or have little experience in it.  Here are some good things to download.

[url=http://www.gimp.org]GIMP[/url]  :  Gimp is a [i]free[/i]  graphic design program.  You will definitely need some sort of graphic design program if you want to enter this contest.  Currently I am using GIMP.  I made the above with GIMP and if you scroll down in my blog.  (maybe it's on page 2 now)  you will see another example of what I made in GIMP.  GIMP probably isn't as good as say photoshop (a program you have to buy)  but it's still GREAT.  Also it's really fun! =D

[url=http://browse.deviantart.com/resources/applications/gimpbrushes]GIMP brushes at Deviant Art[/url]  : GIMP comes packed with brushes but they aren't exactly..good for graphic design.  You'll see what I mean by 'brushes' when you download GIMP.  There are [i]loads[/i] of brushes for graphic design that you can download for GIMP at Deviant Art.

[url=http://gimp-tutorials.net]Gimp Tutorials[/url]  :  Check this site out.  It's great for learning some cool tricks in GIMP.

Remember there's not really a way to teach graphic design.  You have to play around by yourself.  But I can post tips and some cool tricks which I will be doing later.  Also if anybody would like to be a judge please PM me. :D

I'm interested. =)

I have just one question. Is this the topic we're supposed to post on? Because the link you have from the (adorable) panda picture is this same topic.

Yes this is the right topic. :D  (I posted the same thing on my blog and the link was for people there I'll edit it.  Sorry for the confusion)  So far..

D3m0n Shad0w
KrazyNerd

This looks interesting. Any idea when it'll start?

Hmm, I've had literally no experience with graphic design, but I might consider once I find out what the first theme is.

[quote="D3m0n Shad0w"]The first one should be free design, to let the judges see how well you do for the first time.
I want to be a judge. I know I'm entering, but I always compare my work to others before thinking how it is  :D[/quote]
Will we be marked down if we use jokes that nobody understands?  :P . Since it doesn't seem like many people watch YGOTAS on AoPS (yep, I bet you have no idea what I'm talking about).

Thanks for the tip, Dem0n Shad0w. I'm still kind of new at this. :)

Here's my second try. I just colorized the render's layer because it was fastest, but I might go back later to tweak it a bit when I have more time.

Very, very nice! A good choice of color too. Great job  :D

Do you two want to try merging the two layers together?

Eh he he...*long, awkward silence*...Sorry that took so long. Here's my first try:

Nice! I love it! The only advice I would give is that if you look real closely, it looks like there's an edge where the shade of the color suddenly changes (it's a diagonal starting a little bit to the right of the word "of" and ending around the beginning of the word "may"). Maybe I'm just seeing things though. Good job taking the time to actually do it!

I see what you're talking about. I notice it in the original now, too. >_< For some reason it's a lot more apparent in the blend. I'll try to fix it.

Great job  :lol:  :lol: That's really nice! You know, you could submit it to a real online graphics competition for judging. I know a good forum where the players are critical and offer patience with improving graphics, as well as links to online help guides. If you were interested, PM me.

Also, go to fonts.com, as they have better fonts that might match. You can play around with the search option (my favorite font is Av Qest, but I don't think it would suit well in this case) and Lulze and KrazyNerd could find a suitable font.

Okay, thanks. I'm not really interested in a "serious" graphics forum right now, but maybe I will be once I have some more time on my hands. :)

Now, was suma_milli going to make another thread someplace, or did that idea get thrown out?

Maybe you should ask him. I think we can continue the graphics competition now, because I'm finally home.  :)

Maybe d3m0n shad0w can make the next contest?  Sorry guys,  I'm too busy now that school's started.

I'm kind of busy, but I have about an hour to schedule the next contest for most of the week.

I'm going to quickly make one up, and I'll send PM reminders to people who participated in the last contest. Here are the guidelines:

1) The work must be purely original.
2) [b]This contest's theme is a collage that reflects you (more of a graphic collage)[/b] There can be items that you like, such as a favorite teddy bear, and sports you play that represent things.
3) It must have a background
4) It must be creative
5) Try your best!
6) Have fun

Deadline is by Halloween.

If you have any questions, post here!

Excellent idea! I'll definitely get working on it! :)

Is there any way we can get the word out?

If you didn't mind, I could quickly whip up a link for your signature if you had space.

Here:

[code][color=green][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1246323#1246323]The Graphics Design Contest[/url] is now up and running! Join Today![/color][/code]

We'll try that for now.  :)

[b]Also, I forgot to mention:

The points on your graphic will be determined by the public 100%, and the voting topic will be posted in G&FF. Please do not post your graphics here; instead PM them to me.[/b]

Well its not due for like 11 mor days...

ooh ooh ooh! What is this? It looks fun! Can I join in? XD


determine all  real solution of the equation
$\frac{4^{x}}{9^{\frac{1}{x}}}+\frac{9^{x}}{4^{\frac{1}{x}}}=\frac{275}{6}$

just rewrite the problem and my solution is ok

the original problem dont have a explicit formula, its impossible

i only forgotten to rewrite this

sorry

[quote="Thales418"]
${\left(\frac{3^{x}}{2^{1/x}}\right)^{2}+\left(\frac{2^{x}}{3^{1/x}}\right)^{2}=\frac{275}{6}}$

${\frac{2^{x}}{3^{1/x}}=y}$

${\left(\frac{1}{y}\right)^{2}+y^{2}=\frac{275}{6}}$
[/quote]

This is wrong. why do you think $\frac{3^{x}}{2^{1/x}}=\frac{3^{1/x}}{2^{x}}$?

anyhow, the answer is $2,-\frac{1}{2}$. in the case of $x>0$, the LHS is increasing function, and therefore the only answer is 2. in the case of $x<0$, the LHS is increasing function again, and therefore the only answer is $-\frac{1}{2}$.


let $a,b,c$ be potisive real  such that $abc \geq 1$ .
 prove that :
                $(\sqrt{a}+\sqrt{b}+\sqrt{c})^{2}\geq 3\sqrt{3(a+b+c)}$
 [b] Is that right ? [/b]


limit
function
trigonometry
calculus
calculus computations

How is the statement

$ o(o(g(x))) = o(g(x))$ as $ x \rightarrow a$

to be interpreted?

I think it's not correct because:
let $ f(x)=o(g(x))$
then we have
$ f(x)=o(f(x))$
what does it mean? it' mean that :
$ \lim_{x\to a}\frac{f(x)}{f(x)}$ must be zero, but it's not true for all $ f(x)$, I mean there is no exist such $ f(x)$ or such $ g(x)$ for which your expression will be true.

exactly... that's why I don't think i'm interpreting it right.  But it's stated in my book as something that's trivially true. :huh:

Here's the interpretation: Consider two functions $ A(x)$ and $ B(x)$. If $ A(x)$ is $ o(f(x))$ and $ B(x)$ is $ o(A(x))$, then $ B(x)$ is $ o(f(x))$. It's a one-way implication, and it is indeed trivial; $ \lim \frac{B(x)}{f(x)}=\lim\frac{B(x)}{A(x)}\cdot\lim\frac{A(x)}{f(x)}=0\cdot 0=0$.

Right, now that first thing makes sense, thanks jmerry...

But i think I'm going crazy.  I don't follow this:
"
$ \frac{1}{\cos x}= \frac{1}{1-\frac{1}{2}x^{2}+o(x^{3})}= 1+\frac{1}{2}x^{2}+o(x^{2})$ as $ x \rightarrow 0$

by taking $ g(x) =-\frac{1}{2}x^{2}+o(x^{3})$ in

$ \frac{1}{1+g(x)}= 1-g(x)+o(g(x))$ if $ g(x) \rightarrow 0$ as $ x \rightarrow a$. "

In particular, how does $ o(x^{3})+o(-\frac{1}{2}x^{2}+o(x^{3}) )= o(x^{3})$ ?
 :huh:

You need a more accurate expansion of $ \frac1{1+u}$; it's $ 1-u+O(u^{2})$, and that $ O(u^{2})$ term becomes $ O(x^{4})$, which is $ o(x^{3})$.

good, then it was just a mistake in the book; i'm still sane.


This problem probably has been already posted here, but it's woth mentioning for it's beauty and difficulty

Given is a set of $n$ elements and a set $S$ of it's subsets such that for any $A,B,C,D \in S$ we have
$|A\bigcup B \bigcup C \bigcup D|\leq n-2$.

Prove $|S|\leq 2^{n-2}$

see this post :
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=4400


absolute value
real analysis
real analysis unsolved

Here are two problems:

For which real $x$ does $\sum_{n=1}^{\infty}\frac{x^{n}}{1+x^{2n}}$ converge?  For which intervals $I$ is the series uniformly convergent on $I$?

Show that $\sum_{n=1}^{\infty}\frac{x^{2}+n}{n^{2}}$ is uniformly convergent on every bounded interval in $\mathbb R$, but not absolutely convergent for any $x$.

Is the second series supposed to be $\sum_{n=1}^{\infty}\frac{x^{2}+(-1)^{n}n}{n^{2}}\ ?$ Or something like that?

oops, you're right... typo.  It should be
$\sum_{n=1}^{\infty}(-1)^{n}\frac{x^{2}+n}{n^{2}}$
sorry about that

First series: $\left|\frac{x^{n}}{1+x^{2n}}\right|<|x|^{n}$ for $|x|<1$ and $\left|\frac{x^{n}}{1+x^{2n}}\right|<|x|^{-n}$ for $|x|>1$.

Second series: That's an alternating series of terms decreasing in absolute value- if you use the error estimate for alternating series (the error is no more than the next term), you get uniform convergence on every interval $[-A,A]$.

The first series should be uniformly convergent on all intervals not containing or having endpoints at $-1$ or $+1$ then, right?


Sam is younger Sammy, older than Tom, but younger than Tommy.  Sammy is older than Tommy.  Who is the oldest?

[hide="my answer"]sammy is oldest[/hide]

yea i just made this up five minutes ago.

Since Sam is younger than Sammy, Sammy is older than sam. Sammy is also older than Tommy. Since Sam is older than Tom, Sammy also has to be older than Tom. So Sammy is older than Tom, Tommy, and Sam. So $\boxed{Sammy}$ is the eldest

Wait WHAT? Nooo not another 2005 revival

Wow.  Please guys, stop trolling and spamming the forum.  It is very detrimental.

[quote="tmathman"]In case you wanted to make your answer look like $\boxed{\text{Sammy}}$,

[code]$\boxed{\text{Sammy}}$[/code][/quote]

As this is starting to come up often, I'd like to mention this. Please don't revive topics older than 2 weeks. If it has been dormant for that long, the question has been answered or the problem solved. It is completely unnecessary to make another post to that topic. It clogs up the forum and makes life for the mods that much harder. Please refrain from reviving topics. -tmathman
Why post then?


algebra polynomial trigonometry calculus integration search

The following fact is fairly well-known, but certainly interesting.

Theorem. Let $f \in \mathbb{Z}[x]$ be a monic  polynomial with integer coefficients such that all roots of $f$ are real and lie in the interval $[-2,2]$. Then, these roots have the form $2 \cos{\frac{2\pi m}{n}}$, where $m,n \in \mathbb{N}$.

It bears a connection with the well-known fact that every $z\in \mathbb C$ with unit modulus which is the root of a monic polynomial in $\mathbb Z[x]$ satisfies an equation of the type $x^n-1=0$.

All the roots are of the form $2\cos x_i$. In this case, all the Viete sums involving the cosines are integral. We can construct another polynomial, $\prod (x^2-2\cos x_i\cdot x+1)$ which has as roots exactly the numbers $z_i=\cos x_i+i\sin x_i,\ \overline{z_i}=\cos x_i-i\sin x_i$ (here the $x_i$ might be repeated, i.e. we could have multiplicities). This means that the coefficients of this new polynomial are also integral, so $z_i$ is a complex number of modulus one satisfying a monic integral polynomial equation. If we use the stated result, we find that $x_i$ must be rational multiples of $\pi$.

However, I don't know the proof of the "well-known fact": :D. This is just an observation. I wasn't even sure it was true, but I found it as problem $55$ on [url=http://www.math.columbia.edu/~zare/prob_all.html]this[/url] page. There are quite a few beautiful problems on that page.

I think I saw another problem which could solve this one:
    If all roots of a monic polynomial have modulus 1 and the coefficients are integers, then all roots are roots of unity. From what grobber wrote there, it seems that we could deduce the conclusion from this problem, which can be solved in the following way:
   Suppose f is the polynomial and $ z_1,...,z_n$ are its roots. Then consider the polynomials $ f_k=(X-z_1^k)...(X-z_n^k)$. One can prove that all these polynomials have integer coefficients and their coefficients are uniformly bounded (that is there exists an absolute constant which majorizes them!) and thus two such polynomials are identical. From now on it's not hard.

This is really cool, but I wanted to show that stronger thing: if $z$ has modulus one and is the root of a monic integer polynomial then it's the root of unity. I wasn't sure about it, but now I am because I saw it as a proposed problem on that web page (I popsted the link in my previous message). Any idea about that one? (I think the proof should be algebraic, rather than analytic, such as the beautiful one you've just presented). It's also obvious that this result implies the one you've mentioned.

I think I can remember someone asking about this problem on the forum, but I have no idea which key words to use in order to search for the topic.

I also tried to solve some time ago this problem you speak about, but I failed. I really don't know how to start. Maybe someone will help us. By the way, Grobber, I really don't think that a list with your mistakes should be written, but a list with lots of your remarkable results! Keep going!

Yes, harazi, your solution is what I expected to be given; this, in essense, is Kronnecker's theorem.

About the more general assertion. If I understand correctly, you claim that $e^{i\pi x}$ is transcendental (!) for all $x \notin \mathbb{Q}$, right? (Or, do you just claim that $e^{i \pi x}$ cannot be an algebraic integer?) I tried to prove this, but got stuck.

As we all know, the Hermite-Lindemann theorem asserts that $e^a$ is transcendental whenever $a \neq 0$ is algebraic. Because, by the transcendence of $\pi$, $\frac{a}{\pi}$ can certainly not be rational for algebraic $a \neq 0$, the Hermit-Lindemann theorem would be a very special case to our claim. It is therefore somewhat weird to have this problem in a set (see grobber's reference) of other not *that* difficult problems. Can anyone suggest another reference?

Consider equation $z^2+1.5z+1=0$. It has two solutions z1 and z2, |zi|=1, and $z_i^n-1\neq 0$ for all $n\in \mathbb{N}$. Thus assumption about unitary of polynomial has great importance.

Oh, yes, surely there exist algebraic numbers on the unit circle that are not roots of unity; apologies for misunderstanding grobber's claim.  So, what we're trying to prove is this:

If $x \in \mathbb{R} \setminus \mathbb{Q}$, then $e^{i\pi x}$ is not an algebraic integer.

Equivalently, $e^x$ is not an algebraic integer whenever $x$ is an irrational multiple of $\pi$. This implies at once that $e$ is not an algebraic integer (though not the transcendence of $e$.)

you can get some idea from "simple proof for a theorem of kronecker " ,AMM,1978,P.756

Is that by any chance available online? (I mean for free :D). I could find nothing by searching (I found the tables of contents of old issues of AMM, but I couldn't read the article, of course), but maybe I didn't know what to look for..

[quote="harazi"]consider the polynomials $ f_k=(X-z_1^k)...(X-z_n^k)$. One can prove that all these polynomials have integer coefficients .[/quote]

hi harazi, i have a funny proof of this fact with linear algebra .. what's your proof ?

To alekk.
I think harazi's proof is following. All coefficients of $(x-z_1^k)\dots(x-z_n^k)$ are symmetric polynomials of $z_1,\ z_2,\ \dots,\ z_n$, so they can be expressed as polynomials with integer coefficients of basic symmetric polynomials $z_1+\dots+z_n,\ z_1z_2+\dots+z_{n-1}z_n,\ \dots, z_1z_2\dots z_n$, which are integer numbers due to Viete's theorem.

now i write the outline ,assume that $\omega$ is the root of $f$ :
$f=x^n+\sum_{i=0}^{n-1}a_ix^i     (a_i \in Z)$
we know that f is irreducible so f and f' are coprime ,now assume that $M_n(Z)$ is the ring of n*n matric with integer array and $M_n(C)$ is the ring of n*n matric with complex array and assume that $|M|$ is the matric which its array is absolute value of $M$ u can conclude that the eigenvalue of $A$ is exactly n roots of f :
$A=\left(\begin{array}{cccc}0&0&  ....  &-a_0\\\ 1&0&  ....  &-a_1\\\ 0& 1 &  ....  &-a_2\\\ .  &.  &  ...  &.\\\ .  &  .  &   ...  &.\\\ .  &  .   &  ... &.\\\ 0&0&  ....  &-a_{n-1}\end{array}\right) \in M_n(Z)$ so $A$ is similar to diagonal  matric $D \in M_n(C)$ s.t  $ D_{1,1}=\omega$ and$  D_{2,2},...,D_{n,n}$ are conjugates of$ \omega$.now assume that $V\in M_n(C)$ is invertible matric s.y $A=VDV^{-1}$ so $|VD^n|=|V|$ and note that {$A^n=VD^nV^{-1} | n \in N$} is bounded (means that its aray are bounded) so it is finite:  mean there exist t ans s  s.t $A^t=A^{t+s}\Longrightarrow D^t=D^{t+s}\Longrightarrow \omega ^s=1$

I thought that sam-n solved grobber's advanced problem :(
Though his approach is quite unusual.

Wait, I think there's something wrong here. Actually, I think I can disprove the statement that all algebraic integers on the unit circle are roots of unity. For each $n$, we can construct a monic irreducible polynomial $f(x)$ of degree $2n$ with precisely $2n-2$ roots on the unit circle as follows. Take an arbitrary integer polynomial $g(x) \in \mathbb{Z}$ of degree $n-1$ with $n-1$ real roots in the interval $(-2,2)$, and put $h(x) = x^n + Ag(x)$, where $A$ is a sufficiently large integer (to be chosen). We claim that for $A$ large enough the monic integer polynomial $f(x) = x^nh\big(x + \frac{1}{x} \big)$ of degree $2n$ has precisely $2n-2$ roots on the unit circle (this is obvious because for large enough $A$, $h(x)$ has precisely $n-1$ real roots in $(-2,2)$, each of which accounts for a pair of conjugate roots of $f$ of absolute value $1$), and is irreducible (again, for large enough $A$). Indeed, assume that $f(x)$ is reducible. As it has just two roots not on the unit circle (which must be algebraic conjugates), one of the non-constant factors of $f(x)$ must have all its roots on the unit circle; by the Kronecker theorem, $f(x)$ is divisible by a cyclotomic polynomial of order $m<2n$, implying that $f(e^{2\pi i/m}) = 0$, and hence that $Ag(2\cos{2\pi/m}) = -2^n\cos^n{2\pi/m}$. Hence, for each $m<2n$ there is at most one $A$ for which $f(x)$ is divisible by $\Phi_m(x)$, so this can happen for only finitely many $A$. Hence, if $A$ is large enough $f(x)$ is certainly irreducible, and moreover it is monic and has precisely $2n-2$ roots on the unit circle (by the way, the other two roots are real and not on the unit circle).

What's wrong with this construction? I don't think that sam-n's proof is correct ... .

I think you'd better write everything out in more detail. For instance, why does $h$ have, for large $A$, $n-1$ real roots in $(-2,2)$?

Because it changes sign $n-1$ times in $(-2,2)$. Indeed, if $-2 < w_1 < \cdots < w_{n-1} < 2$ are the roots of $g(x)$, and if we choose arbitrary $b_i \in (w_i,w_{i+1})$, then $g(b_i)g(b_{i+1}) < 0$ for $0 \leq i \leq n-1$ (where $w_0=-2$ and $w_n=2$); for $A$ sufficiently large, we have $\text{sgn} \, {h(b_i)} = \text{sgn} \, {g(b_i)}$. It follows from the intermediate value theorem that $h(x)$ has $n-1$ roots in $(-2,2)$.

I see. Sorry about that, I'm a bit confused right now :). It seems Ok right now, but I'll have to look again and again. I was so sure the problem was correct.. :(

:)  Have you seen it formulated elsewhere (save for the list of problems given above)? I think it seems very strange to have such obvious strengthening of the Kronecker result; that's what shocked me initially ...  :)

I remember having seen it before. The statement was not new to me when I saw it on that list. I think... :)

Well, perhaps I'm too tired now, but I find no gaps in the construction I gave; I'll revise it later, of course  :D . Perhaps others will help to explain this mystery  :lol:  , but it seems to me very clear that the assertion is wrong ... unless I miss something stupid. Perhaps we should organize a poll to decide by voting whether the construction is correct or not  :D .

And it looks like sam-n went wrong when he asserted that the set $\{A^n\}$ was bounded. I'm also tired.. :)

For a simple numerical counterexample, consider the polynomial
\[
f(x) = x^4 + 2x^3  + 2x + 1.
\]
This polynomial has two conjugate roots on the unit circle, namely $e^{\pm i t}$, where $t = \arccos{\frac{\sqrt{3}-1}{2}}$. Indeed, $e^{it} + e^{-it} = 2\cos{t} = \sqrt{3}-1$ is a root of $z^2 + 2z - 2$; after a change of variables $z \mapsto x + \frac{1}{x}$, we see that $e^{it}$ fulfills the equation $f(x) = 0$. Needless to say, $t \notin \mathbb{Q}$, and $e^{i t}$ is an example of an algebraic integer on the unit circle which is not a root of unity.  :)

Yes, I also think that sam-n's mistake is in the assumption that $\{A^n\}$ is bounded ...

--Vesselin


1/Let a,m,n are positive integers and a>1.Prove that
$ (a^m \minus{} 1,a^n \minus{} 1) \equal{} a^{(m,n)} \minus{} 1$
2/Find the positive integer n satisfies that 150<n<1997 and $ \frac{2^n \plus{} 2}{n}$ is an integer
3/Let p is the prime integer and p>5.
Find the max value of k for
$ (p \minus{} 2)^{2p \minus{} 2} \minus{} (p \minus{} 4)^{p \minus{} 1} \setminus p^k$


im a sophmoretaking differential equations. This is my group project and we have been on this for two days and couldnt get even our graduate student neighbor to solve it.we are currently using integrating factors to solve stuff.

it starts snowing at a constant rate(r) at 1200. at one oclock a snowplow
(x) leaves the house with a speed inversely proportional to the amount of snow its pushing. at 2 a second snowplow(y)(also speed inversely proportional to snow its clearing) leaves the house. when do they collide? when if the second didnt leave till 3?

heres what we got so far:  dx/dt=1/(rt)
  x(t)=(1/r)ln(t)
   t(x)= e^(xr)
 dt/dy=(t-e^(yr))r

then we rearange: dt/dy-rt=-re^yr  which is the formdx/dy+P(x)y+Q(x)
and then it goes downhill from there. either all the t's cancel out and we got 0=ln(2) which is wrong or something that makes no sense.

For this problem, we want two time variables, not two position variables. Let $x(z)$ be the time at which plow $x$ reaches $z$ and $y(z)$ be the time at which plow $y$ reaches position $z$. We have $x'(z)=rx(z)$ and $y'(z)=ry(z)-rx(z))$ with initial conditions $x(0)=1$, $y(0)=2$ or $y=3$ in the second case. Solving this system and setting $x=y$ will naturally tell you where they collide; evaluating the functions will tell you when.

While your approach was correct as far as you showed it, it was hard to understand and easy to get lost in the inverse functions required. The problem is much easier to handle with two time variables.

Ok solved it:
HEres how my group did it.. it was so easy afterall .. it was just elusive

x(t) = (1/r)*ln(t)
t=e^(rx)
dt/dy=r(t-e^(yr))
dt/dy-rt=-r*e^(yr)
==t(y)=(1/e^(-ry))*(  int(e^(-ry)*-re^(ry)dy   +C)
for x=y=(1/r)ln(t)

t(y)=e^(ry)(C-ry)

after inputing multiple conditions
t(collision)=e^((time truck2 leaves)-1)
collision if truck 2 leaves at 2pm is 2.718pm
collision if truck 2 leaves at 3pm is 7.389pm


Q 1) there is a triangle ABC.  one line l passing through pt C which is || to AB. angle bisector of A cut side BC at pt D & cut line l at pt X. similarely angle bisector of B cut side AC at pt E & cut line l at pt Y such that DX = EY .prove that AC=BC[/code][/quote]

It's [color=red]QUITE [/color]easy.
Let $AB=c,AC=b,BC=a,EY=DX=x,AD=ta,BE=tb$
As $\angle CAX = \angle BAX = \angle CXA$
$AC=CX$
similarely
$BC=CY$

$\frac{BE}{EY}=\frac{AB}{CY}$
$tb=BE=\frac{AB \cdot EY}{CY}=\frac{cx}{a}$
similarely
$ta=\frac{cx}{b}$
If $a>b$ then
$\frac{cx}{b}>\frac{cx}{a}$
$ta>tb$
It's impossible
then $a\leq b$
If $a<b$ then
$ta<tb$
It's also impossible

So $a=b$

see http://www.mathlinks.ro/Forum/viewtopic.php?t=59227 also.

  darij


trigonometry
analytic geometry
ratio
function

Let $ r$ be the distance from the origion to a point $ P$ with coordinates $ x$ and $ y$. Designate the ratio $ \frac{y}{r}$ by $ s$ and the ratio $ \frac{x}{r}$  by $ c$. Then the values of $ s^2 \minus{} c^2$ are limited to the numbers:

$ \textbf{(A)}\ \text{less than }{\minus{}1}\text{ are greater than }{\plus{}1}\text{, both excluded}\qquad\\
\textbf{(B)}\ \text{less than }{\minus{}1}\text{ are greater than }{\plus{}1}\text{, both included}\qquad \\
\textbf{(C)}\ \text{between }{\minus{}1}\text{ and }{\plus{}1}\text{, both excluded}\qquad \\
\textbf{(D)}\ \text{between }{\minus{}1}\text{ and }{\plus{}1}\text{, both included}\qquad \\
\textbf{(E)}\ {\minus{}1}\text{ and }{\plus{}1}\text{ only}$

By trig ratios on the unit circle,

$ \frac {y}{r} \equal{} s \equal{} \sin{\theta}$
$ \frac {x}{r} \equal{} c \equal{} \cos{\theta}$

$ s^2 \minus{} c^2 \equal{} \sin^2{\theta} \minus{} \cos^2{\theta} \equal{} \minus{} \cos{2\theta}$ (double angle formula for cosine)

Since the result is just a standard cosine function, its range is simply between -1 and 1, inclusive.

Answer: D


Consider the region $A$ in the complex plane that consists of all points $z$ such that both $\frac{z}{40}$ and $\frac{40}{\overline{z}}$ have real and imaginary parts between $0$ and $1$, inclusive. What is the integer that is nearest the area of $A$?

[hide]Let's approach this problem algebraicly. Notice that $\frac{40}{\overline{z}}=\frac{40}{a+bi}=\frac{40a+40bi}{a^2+b^2}$. Examining this form of $\overline{z}$, we see that $a>0$, $b>0$, and $a^2+b^2>40a\implies (a-b)^2>0$.

This gives us a nice semi-circle: $(a-20)^2+b^2>20^2$. Similarly for $\frac{z}{40}$, we get another semi-circle: $a^2+(b-20)^2>20^2$.

Our area is the overlapping portion of these two semicircles. With some quick calculations, $A= 1200-200\pi=200(6-\pi)\approx \boxed{572}$.[/hide]

How do you find the overlapping area of two circles?

A picture is worth a thousand words.  :D

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=604[/img]
Two times the nice blue shaded circular segment. Notice that we can always take advantage of the symmetry.  :)

Or it can be simplified - the area is three quarters of a square minus two quarters of a circle (sketch it to see it).

why is it the area overlapping? i got to $(a-20)^2 + b^2 > 20^2$ and $a^2 + (b-20)^2 > 20^2$, but wouldnt these inequalities give us everything outside of the two circles? and also, from your diagram, how would you get $1200 - 200\pi$? if anything wouldnt you take the quarter circle and subtract the triangle, then multiply by 2? so itd be like something times $\pi$ minus something else....

Use the search function, 4everwise! http://www.artofproblemsolving.com/Forum/viewtopic.php?t=73967&postorder=asc&start=39

[quote=4everwise]Let's approach this problem algebraicly. Notice that $\frac{40}{\overline{z}}=\frac{40}{a+bi}=\frac{40a+40bi}{a^2+b^2}$. Examining this form of $\overline{z}$, we see that $a>0$, $b>0$, and $a^2+b^2>40a\implies (a-b)^2>0$.[/quote]
Is it just me or shouldn't the equation be $a^2+b^2>40a+40b$ ?. Why is it not this. Can someone else also post a solution that explains 4everwise's reasoning, I am having a hard time understanding it, thanks

[hide=Alternate Solution] The key to solving this problem is notice the bounds that we have.  So for $\frac{z}{40}$, this means that it is equivalent to $\frac{a+bi}{40}$.  Trivially, since we have the inequality $$0\leq a,b \leq 40$$ which basically tells us that one of our figures is a square with a length of 40 units.  Now for $\frac{40}{\overline{z}}$, we know that it is equivalent to $\frac{40a+40bi}{a^2+b^2}$ as mentioned by $@4everwise$.  Now that we found the second bound, we know our inequality for this part of the problem is $0\leq a,b \leq \frac{a^2+b^2}{40}$ which means that we have two circles with centers of $(20,0)$ and $(0,20)$.  So basically, the answer is the area outside the two circles, but it has to be inside the square in order for it work.  So basically, all we do now is some basic computation which is basically $1200-200\pi$ so you should get $572$ as the integer that is closest to the area of $A$. [/hide]

Let $z=b+ai$. Then, $\frac{z}{40}=\frac{b+ai}{40}$, so $0 \le \frac{a}{40} \le 1 \implies 0 \le a \le 40$ and $0 \le \frac{b}{40} \le 1 \implies 0 \le b \le 40$. Similarly, , $\frac{40}{\overline{z}}=\frac{40}{a-bi}=\frac{40a+40bi}{a^2+b^2}$. Hence, $0 \le \frac{40a}{a^2+b^2} \le 1 \implies$ $0 \le 40a \le a^2+b^2 \implies$ $40a \le a^2+b^2 \implies$ $(a-20)^2+b^2 \ge 400$. By the same logic it can be shown that $a^2+(b-20)^2 \ge 400$.
Now, we simply want to find the area of the shaded region below.

[asy]
size(200);
defaultpen(fontsize(8));
path kt = arc((20,0),20,0,90)..arc((0,20),20,0,90)--(40,40)--cycle;
fill(kt,black);
draw((0,-30)--(0,50));draw((-30,0)--(50,0));
draw(circle((20,0),20));draw(circle((0,20),20));
draw((40,0)--(40,40)--(0,40));
draw((20,0)--(20,20)--(0,20));
label("W",(20,0),(0,-2));label("Y",(0,20),(-2,0));label("Z",(0,0),(-1,-1));label("X",(20,20),(-1,-1));
label("b",(50,0),(2,0));label("a",(0,50),(0,2));
label("(40,0)",(40,0),(1,-1));label("(0,40)",(0,40),(-1,1));
label("(40,40)",(40,40),(1,1));
dot((40,0)^^(0,40)^^(40,40));
[/asy]
Let's use complementary counting and find the area of non-shaded region that lies inside the square with side length $40$. We speculate that $WXYZ$ is a square (*). Then $[WXYZ]=20^2=400$ and the combined areas of the quarter circles is $100\pi+100\pi=200\pi$. So our answer is simply the integer closest to $40^2-(400+200\pi)$, which is $\boxed{572}$.

(*) As a property of the radical axis, $XZ \perp YW$. Let the intersection of $XZ$ and $YW$ be $P$ be $W$ and $\angle YZX=n$. Then, since the x-yaxises are perpendicular, $\angle XZW = 90 - n$, $\angle PWZ = n$, and $\angle ZYP = 90- n$. Because $YZ=ZY$, $n=45$. WLOG, $WP=PX=PY=PZ$, so $WXYZ$ is a square.

This is probably me just being dumb, but isn't $\frac{40}{z}=\frac{40}{a+bi}=\left(\frac{40}{a+bi}\right)\left(\frac{a-bi}{a-bi}\right)=\frac{40a-40bi}{a^2+b^2}$ and not $\frac{40a+40bi}{a^2+b^2}$?

[quote=eagles2018]\bump...Can someone help me out? What am I doing wrong?[/quote]

Original has a typo. Its $\frac{40}{\overline{z}}$, not $\frac{40}{z}$, so it should be $\frac{40}{a-bi}$.

Let $z=a+bi$. We get that $\tfrac{a+bi}{40}$ and $\tfrac{40}{a-bi}$ both have real and imaginary parts in the range $[0,1]$. Forming inequalities, we find $0 \leq \tfrac{a}{40} \leq 1 \implies 0 \leq a \leq 40$, $0 \leq \tfrac{b}{40} \leq 1 \implies 0 \leq b \leq 40$. Then, taking $\tfrac{40}{a-bi}=\tfrac{40a+40bi}{a^2+b^2}$, we have $0 \leq \tfrac{40a}{a^2+b^2} \leq 1 \implies 0 \leq 40a \leq a^2+b^2$, $0 \leq \tfrac{40b}{a^2+b^2} \leq 1 \implies 0 \leq 40b \leq a^2+b^2$. We interpret the inequalities geometrically, throwing them on the complex plane. The first two are the easiest to deal with, giving is a square with vertices $0$, $40$, $40+40i$, $40i$. For the last two inequalities, we know that they are greater than or equal to $0$ already, so we complete the square, giving us $(a-20)^2+b^2 \geq 400$ and $a^2+(b-20)^2 \geq 400$, which are circles centered at $20$ and $20i$ with radii $20$. We are looking for the area of the locus of points outside the square but inside the semicircles. Splitting up the semicircles as in #11, we get that the area is $1200-200\pi \approx \boxed{572}$.

Piggy-backing off the solution in post #11. Once you get to find the area of the shaded region in the presented diagram, rename $Z=O$ and $X=P$.

Note that the two circles intersect at the origin $O=(0,0)$ and $P=(20,20)$ which is colinear with $B=(0,40)$ and $A=(40,0)$. Hence $\triangle AOP\cong \triangle BOP$ since they are both $45-45-90$. So the answer will turn out to be half the area of the square minus the overlapping region of the circles. This is equal to $[BOA]$ minus twice the area of the lune formed by the line $y=x$ and minor arc $\widehat{OAP}$. Clearly, $[BOA]=800$. To find the twice the area of the lune, we just need to compute the twice the area of the region enclosed by the line $y=x$ and the semicircle $y=\sqrt{20^2-(x-20)^2}$ which is simply$$2\int_0^{20} \left(\sqrt{20^2-(x-20)^2}-x\right)dx=200\pi-400.$$So the final answer is$$800-(200\pi-400)=1200-200\pi\approx \boxed{572}.$$


Hey, I've been planning a Mock AMC on another site, but as I am not very creative, I've been wondering if I could borrow some of the questions from past Mock AMC's hosted here, on AoPS. I would greatly appreciate if I were allowed to use all of the questions. They are all really good and I'm hoping to use them, so all Past Mock AMC writers, just post here Y/N. Thanks.

Since no one's answered, I'll assume that everyone's cool with me using their questions. Thank you all.

Consider that people who have already taken those Mock AMCs will have an unfair advantage, and that silence does not imply consent.

Hey Nintendo can I use your Mario character?

... Since there was no response I'll assume I can.

(I'm gonna assume that the Mock AMC writers hold an implied copyright on all of their questions, and you need to get their explicit permission in order to use them)

[quote="lingomaniac88"][quote="randomdragoon"]Hey Nintendo can I use your Mario character?[/quote]

Well, it's not like you'd be making any money administering the test... I hope.  I'm not sure if Mock AMCs, AIMEs, etc. would be copyrighted by the person or people who write its questions.

Either way, I would still acknowledge the fact that Mario is a copyright of Nintendo.[/quote]
The question writers still hold a copyright. Whether it would be profitable for them at all to take a copyright infringment of this sort to court is extremely doubtful.

Ok, I see your point. I pm'ed two of my three anonymous sources, they said it was ok. I'm just waiting on the reply of mycalc. mycalc, if you're out there, pm me or respond to this thread soon. I've already pm'ed you.

correction for previous post: the user is not mycalc, it's calc rulz (don't ask how I got that mixed up.).

i don't see the point of compiling a test of problems that we have already done? (the audience of mock amc test takers have looked at most of the mock tests...)


quadratics
inequalities
inequalities proposed

[hide]
(This is wrong, see below)

WLOG, let $a \leq b \leq c \leq d$.

$(a^{2}+b^{2})(b^{2}+c^{2})(c^{2}+d^{2}) \leq 1$ is equivalent to:

$\sqrt{\frac{a^{2}+b^{2}}{2}}\sqrt{\frac{b^{2}+c^{2}}{2}}\sqrt{\frac{c^{2}+d^{2}}{2}}\sqrt{\frac{d^{2}+a^{2}}{2}}\leq \frac{1}{4}$*

Notice that the LHS is $4$ quadratic means. Thus:
$LHS \leq bcdd$**      ( as the QM(a,b)<=max(a,b) )

Now, $2 \geq b+c+\frac{d}{2}+\frac{d}{2}\geq 4\sqrt[4]{\frac{bcd^{2}}{4}}$

${\frac{1}{2}}^{4}\geq \frac{bcd^{2}}{4}$

$bcd^{2}\leq \frac{1}{4}$

Put this into **, and we get *, which is equivalent to the inequality we're trying to prove.
[/hide]

You cannot assume $a \ge b \ge c \ge d$, this is not symmetric.

Nice, Silver 1989, if we replace $\frac{a+b+c+d}{4}$ with $\frac{a+b+c+d}{2}$

The following is more interesting.
If $a,b,c,d,e$ are nonnegative real numbers such that $a+b+c+d+e=5$, then

$(a^{2}+b^{2})(b^{2}+c^{2})(c^{2}+d^{2})(d^{2}+e^{2})(e^{2}+a^{2}) \le \frac{729}{2}$.


How can I type fractions, square roots, and other symbols?

(If there is already a post about this, please tell where it is, thank you.)

Look at the FAQ post in the LaTeX subforum.

They actually moved everything to [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX]Here[/url]

I was referring to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]this[/url].


ratio
trigonometry
geometry
trig identities
Law of Sines
angle bisector

[quote="mystagename"]:(  So we have this problem

where using the triangle we have to find the ratio of p:a in terms of sin x and sin y and also the ratio of q:c in terms of sin x and sin z.

anyone know what to do?[/quote]

law of sines:

$\frac{\sin{x}}{q} = \frac{\sin{z}}{c}$ and $\frac{\sin{x}}{p} = \frac{\sin{y}}{a}$

z and y are complementary angles

$sin(z) = sin(y)$

and $\frac{q\sin{z}}{c} = \frac{p\sin{y}}{a}$

so $\frac{q}{c} = \frac{p}{a}$

[quote="mystagename"]thanks.

i have another couple of questions.

how would you develop a fromula for an angle bisector of a triangle from this?[/quote]

I assume you mean the length of the angle bisector.

in this case, we need to use Stuwart's Theorem to find a simplified version of the length of the angle bisector.

In case you dont know what's Stuwart's Theorem, it states that in $\triangle ABC$, where the side opposite $A$ has length $a$, and the side opposite $B$ has length $b$, and finally the side opposite $C$ has length $c$.  if there's a segment drawn from vertex $A$ to side $BC$, splitting it into two segments of length $m$ and $n$, where $m$ is opposite of $b$, and $n$ is opposite of $c$, then the length of this segment, denoted by $p$, satisfies:

$a(p^2 + mn) = b^2m + c^2n$

for example, let $l$ be the length of the angle bisector in your picture, and $b = q + p$, then $b(l^2 + qp) = c^2p + a^2q$

i suggest you try it out from now on by yourself to find $l$ in terms of $a$, $b$, and $c$(good practice :lol:) , but the solutions below:

[hide]

first, we separate $l^2$ in the equation given:

\begin{eqnarray*} b(l^2 + qp) &=& c^2p + a^2q\\ l^2 &=& \frac{c^2p + a^2q}{b} - qp\\ \end{eqnarray*}

using law of sines, we find $q = \frac{c\sin{x}}{\sin{z}}$ and $p = \frac{a\sin{x}}{\sin{y}}$, but since $z$ and $y$ are supplementary angles, $\sin{z} = \sin{y}$.  so let $k = \frac{\sin{x}}{\sin{y}}$, then $q = ck$ and $p = ak$.  Plug into equation above:

\begin{eqnarray*} l^2 &=& \frac{c^2ak + a^2ck}{b} - ack^2\\ l^2 &=& ac* (\frac{ck + ak}{b} - k^2)\\ \end{eqnarray*}

but $ck + ak = q + p = b$, and from this equation we get $k = \frac{b}{a+c}$, so:

\begin{eqnarray*} l^2 &=& ac* (\frac{ck + ak}{b} - k^2)\\ l^2 &=& ac* (1 - k^2)\\ l^2 &=& ac* \left(1 - \left(\frac{b}{a + c}\right)^2\right)\\ l &=& \sqrt{ac*\left (1 - \left(\frac{b}{a + c}\right)^2\right)}\\ \end{eqnarray*}

and this is the answer :lol:
[/hide]


How many real solutions does the equation $ x^{3} 3^{1/x^{3} } \plus{}\frac{1}{x^{3} } 3^{x^{3} } \equal{}6$ have?

$\textbf{(A)}\ 0  \qquad\textbf{(B)}\ 2  \qquad\textbf{(C)}\ 3  \qquad\textbf{(D)}\ \text{Infinitely many}  \qquad\textbf{(E)}\ \text{None}$

[hide]$ x$ must be positive, since otherwise LHS would be negative.

$ LHS\geqslant 2\sqrt{3^{x^3\plus{}x^{\minus{}3}}}\geqslant 2\sqrt{3^2}\equal{}6$

Hence $ x^3\equal{}x^{\minus{}3}\iff x\equal{}1$. That's the unique solution, so the answer is $ E$.[/hide]

[quote="xdanielxedgex"]LHS? what is that?[/quote]

Left-hand side.

hello, the only real solution is $ x\equal{}1$, why is the correct answer $ \boxed{E}$?
Sonnhard.

[quote="Arrange your tan"][quote="Farenhajt"][quote="Dr Sonnhard Graubner"]hello, the only real solution is $ x \equal{} 1$, why is the correct answer $ \boxed{E}$?
Sonnhard.[/quote]

"None" means "None of the previous options" (note that zero solutions is option A) - and you'll agree that "exactly one solution" fits nothing from A to D.[/quote]

"None" is not acceptable for the original poster to give as a choice.  It also means "zero"
in this context.

xeroxia, choice E) should have been something such as "None of the above" or "None of the previous."
Typing just "None" as a choice in the question, combined with what's already there is wrong.[/quote]

Assuming this was given on an official test of some kind (as questions of this form usually are), wouldn't you agree that it would be a grave error by the question inventors to offer TWO options for "zero solutions" - namely A and E? Math boils down to "read carefully" more often than not.

Besides, something may have been lost in the translation... Perhaps the best solution would be "Other".

[quote="Farenhajt"][/quote][quote="Arrange your tan"][quote="Farenhajt"][quote="Dr Sonnhard Graubner"]hello, the only real solution is $ x = 1$, why is the correct answer $ \boxed{E}$?
Sonnhard.[/quote]

"None" means "None of the previous options" (note that zero solutions is option A) - and you'll agree that "exactly one solution" fits nothing from A to D.[/quote]

"None" is not acceptable for the original poster to give as a choice.  It also means "zero"
in this context.

xeroxia, choice E) should have been something such as "None of the above" or "None of the previous."
Typing just "None" as a choice in the question, combined with what's already there is wrong.[/quote][quote="Farenhajt"]

Assuming this was given on an official test of some kind (as questions of this form usually are), [1] wouldn't you agree that it would be a grave error by the question inventors to offer TWO options for "zero solutions" - namely A and E? [2] Math boils down to "read carefully" more often than not.

[3]Besides, something may have been lost in the translation... Perhaps the best solution would be "Other".[/quote]

[1] I was communicating/implying that.
[2] Math also boils down to being careful to not give a faulty question.
When it is, such as this, the person answering is not to assume what is meant.
The presenter should be able to correct it, or throw out the question.
[3] Yes, I would chalk it up to being lost in translation, or a cut-and-paste
was incomplete.

I have written certain quiz/forum problems with the wrong quantities
(by accident), and those problems were undoable as stated, despite
my intentions for what was supposed to happen in approaching them.

How many real solutions does the equation"
x^3=y
yx3^(1/y)+(1/y)x(3^y)=6
y^2x3^(1/y)-6y+3^y=0
  36-4x3^(y+1/y)=>0
9-3^(y+1/y)>=0 => 9>=3^(y+1/y)
y+1/y<=2
(y-1)^2<=0,
so only solution y=1
thus x=1

 thanks  sonhard  for pointing out my mistake

hello, but $ x\equal{}1$ fulfills our equation.
Sonnhard.

[quote="bensile"]But "Hence x^3=x^(-3)" why can't it be x=-1? I know -1 doesn't work for this equation but why can't?[/quote]

I see no reason why it can't...


induction
function
conics
parabola
algebra unsolved
algebra

Let $\large f: N to N$ be defined recursively by $\large f(1)=0$ and:
$\large f(n) = max (f(j)+f(n-j)+j)[/tex] with [tex]\large 1\leq f\leq$ for all $\large n\geq 2$.
Determine $\large f(2000)$.

$f(2000)=10864$The solution in "Mathematical olympiads 2000-2001"but it's very long.I can't post it

$f(n) = \max_{1\leq j \leq n-1} \left(f(j)+f(n-j)+j\right)$

$f(n)=\frac{n(n-1)}2$ for all $n$.  It is not very hard to prove by induction: the function is the maximum of an upward-facing parabola, and the rightmost endpoint is always slightly higher than the leftmost endpoint and that gives us this function $f$.  If $f$ were given by the minimum instead, we would have $f(n) = n-1$.

Thanks! I has just solution for my problem


superior algebra
superior algebra unsolved

Hi everyone. I will be glad if some one will try to solve this some problems or post some idea:
Problem $ 1$:
Prove  that the order of Orthogonall group $ O_{2n\plus{}1}(q)$is equal to :
$ 2q^{(n\minus{}1)^{2}/4}\prod_{i\equal{}1}^{(n\minus{}1)/2}(q^{2i}\minus{}1)$.  you can suppose that $ q$ is like $ 2k\plus{}1$ where $ k$ is natural number, but formula also is true for $ q\equal{}2^{m}$

I think you would like to correct some typos (cf. http://en.wikipedia.org/wiki/Orthogonal_group#Over_finite_fields ). Your equation does not make sense for even $ n$.

  darij

Some of your terms aren't integers then.

[quote="darij grinberg"]I think you would like to correct some typos (cf. http://en.wikipedia.org/wiki/Orthogonal_group#Over_finite_fields ).

  darij[/quote]
yes you are right. thanks  :) .
I need a proof ofr formula which is on wikipedia.

Does anyone know where can I see some article about computing order of this group? :huh: