AMC AIME Duke college calculus Harvard integration

I was wondering how bad getting a 3.3 in Complex Variables in high school is when applying to colleges as a prospective math major. You see, due to troubles with sequences and series, procrastination, and prepping for AIME, I got a 64 on my 3rd exam, which brought my final grade down to an 88 (I got a 96 on the final, by the way). Had it not been for that test, I would have gotten a 4.0. So I was wondering, do colleges see 3.3 as a "bad" grade for someone who tries to focus on math in their application? By the way, I did get an A in Calc III last semester.
Also, at more competetive scholarship-offering schools (Caltech, Michigan, Duke) , do 1 bad grade hurt your chances for a scholarship by a lot?

By getting a 3.3 I assume you mean getting the equivalent of a B in the class...

Well if everything else is 4.0 then your cumulative should still be 3.9+ and I think that's fine... what is Complex Variables anyway? Is it calculus? Or beyond? I've never heard of it as a class...

Complex Variable classes can be at any level, but I think it mostly comes after the calculus sequence.... unless the class goes very slow and covers very little material. One class never makes or breaks you, I wouldn't sweat about it.

What does it mean to have 3.3?  Wouldn't a high B be like 3.8 or something?

GPA is based on the following scale.
4.0=A
3.666=A-
3.333=B+
3.0=B
and so on

so a 3.3 would be about a B+

if you are worrying about getting into a good college, this shouldn't matter, as most people don't take complex variables in high school! you should be in a very competitive spot with all the classes you've taken and the fact that you have better grades otherwise
i don't know about scholarships though, i would assume they may only consider the GPA and not the classed you took

it definitely can't help but it shouldn't hurt that much

[quote="polymorphic"]3.3 is a B+.
I would guess that "Complex Variables" is something similar to Complex analysis.[/quote]

Names are just a bunch of blahblahblah.

On a side note, having taken and received an A in calc III should be a great benefit to you.

notsosilly, you're just wrong.

Spoon, "complex analysis" is offered as the title of an undergraduate course at many schools, Harvard and NYU among them.

A "complex variables" course would presumably have as prerequsites at the very least a full calculus sequence, including multivariable calculus. Highlights of the course would include Laurent series, the residue theorem, conformal mapping. Even if this particular course is on the computational side, these are among the things being calculated.

I don't really know what goes on in the admissions committees of highly selective universities, but at my own university, I am quite familiar with high school students taking college courses. (We have a program ourselves for that with a very attractive cost.) Mostly these students take the next few courses after AP Calculus BC: Calculus III, Linear Algebra, Differential Equations, Probability; some go on to include courses beyond that (including, possibly, the complex variables course, among other choices.)

The students that do this are, after all, highly able students who skipped over some level or levels of math in school. In some cases, they may be students driven by ambitious parents. And mostly they make very good grades in these courses - but they don't always make A's. Some things to remember about that:

1. College grades in general run lower than high school grades. Although there is grade inflation at all levels, college grades are, on average, less inflated.

2. Such students are attempting something quite ambitious - something that few if any of their high school contemporaries try, competing directly with older students. After all, you could have played it safe and not skipped over those earlier courses, and you'd be taking AP Calculus in your senior year at your home high school.

3. Admissions committees may sometimes work in mysterious ways, but mostly they're not stupid or blind. It's not that hard to figure out #1 and #2.

So - what do you do if you make a B in that advanced college course you took in high school?

* Smile. It's not the end of the world. That is a quality grade.
* Claim that course and submit that transcript with your applications. Proudly. It [i]will[/i] be a net plus for you.
* Tell your mother or father - the one who's about to call up that college instructor demanding a meeting - to chill out. If you do want to talk to the instructor, do it yourself.

[What the relative importance of is or should be for advanced curriculum versus problem solving ability as demonstrated through competitions - that's a debate for another time and place.]

Another piece of advice for for high school students who want to take an upper division course within the core of a math major - a real analysis course, perhaps, or an abstract algebra course:

Assume that where you're going you'll probably want to take this again anyway, to get it in your eventual college's context. Doing it now is mostly to keep your mind going and to give you a better general background. Once you're at a certain point, it's no longer that important exactly what you take. The "wrong" course with the right instructor may be better than the right course with the wrong instructor.

some advice, which is probably more pertinent than anything particular about complex variables or college admission.

1. It is tasteless to solicit comments on one's personal [i]embaras de richesse[/i], such as the receipt of any grade between A and Z in complex variables prior to university. Note that you are fortunate to be able to take such a class at this stage in lieu of calculus or advanced jump-rope or whatever drivel is typically offered, and leave it at that.
By the same token, publically bemoaning the absence of courses on algebraic topology at your local magnet kindergarten, or the lack of peers at your level, is much like complaining about the caviar in your neighborhood grocery store, and will inspire similar (and often silent) reactions.

2. If you (original poster) can't figure out the answer to the question on your own, and really need to know, you would do better to ask it in private, or to ask it in some abstract form that can be discussed in public without appearing as a personal solicitation of congratulatory assurance.

[quote="Spoon"]The only "Complex Analysis" I know of it at the advanced graduate level. I highly highly doubt a high school student took anything remotely close to anything with "analysis" in the material. [/quote]

Complex analysis is not fundamentally different from calculus in that it can be taught as a mostly computational undergraduate subject.  It is taught that way in engineering curricula. Likewise for linear algebra, differential equations, multivariable calculus, Fourier series, and numerical analysis.  The rigorous mathematical version of these subjects is a different story, but the half-baked computational version is not far in spirit or pedagogy from high school calculus.  It is omitted at high school level for lack of time and general interest, not for reasons of difficulty or abstraction.   Churchill's book is/was the standard text for complex analysis in that style.

It was complex variables, yes.
We only had a few basic proofs to do on tests, but we did recreate all proofs in the book in the lectures (Most people just zoned out while the professor was explaining the proofs :lol: ).

Tut tut :P you should have been [u]paying attention[/u]!!! But yeah, you should be extremely happy with the 3.33, cause you're a quality student, whereas here in Shanghai we have no options after AP Calculus (that's where I am at now) therefore I'll be trying the EPGY courses online.

[quote="blahblahblah"]It generally refers to a course in which you cover the basic theory of complex numbers, the cauchy-riemann conditions, contour integration, taylor/laurent series, and finally the residue theorem.  It's mostly glorified multivariable calculus, at that level.[/quote]

I do not consider this to be Complex "Analysis". These topics would go under (what I consider) what should be titled "Complex Variables" and it IS most definitely an undergraduate course.

By complex analysis, I refer to stuff along the lines of:

Complex integration. Cauchy's theorem and its corollaries. Taylor series and the implicit function theorem in complex form. Conformality and the Riemann-Caratheodory mapping theorem. Theorems of Bloch, Schottky, and the big and little theorems of Picard. Harmonicity and Dirichlet's problems.

And this assumes they are covered in a completely rigorous manar.

In any event, I'm sure you could find different titles at different levels with different textbooks covering different material in lots of places. It makes no difference really. Any complex course taken as a high schooler is impressive, and, as someone already said, you should probably be able to answer your question yourself. Your post appears more as arrogance than a modest question. (Which I think with growing competition among members in this forum is becoming more and more rampant.)

No, I'm definitely not trying to be arrogant.
I am sorry for starting this thread in the first place. Perhaps I overreacted when i saw my grade. ;)

As moderator here, let me make clear that the original post that opened this thread asks a legitimate question, a question of interest to more than one person and very on-topic, and I'm glad you asked.

Kent Merryfield's reply in post #13 is very useful for thinking about the issue of taking advanced math courses while still in secondary schooling.

I like pretty nearly all posts by Fleeting Guest and appreciate his joining the discussion, but post #14 sounds too much like a personal slam. The original post was an informational question. A busy high school student in a highly competitive college admissions environment does have to make choices about time allocation. What I generally advise young people is to dare to be themselves: do what is personally important, and spend time on what produces the most personal learning and development. It is well apparent that the AoPS staff heartily approves math competitions as a developmental activity   :)  and it is also well known that many college applicants to top colleges take "college-level" courses in math while still of high school age. I personally advise my son to take a both-and approach to preparing for higher-level math study, but ultimately HE has to decide what use of his limited time is most satisfying and challenging to him, whether or not it is impressive to a college admission officer.

Spoon, does it really matter if the class is called Complex Variables or Complex Analysis? Of course a graduate level class should be more advanced than an undergraduate class, and, indeed, most universities with graduate programs in mathematics do offer an undergraduate class and a graduate class.

No, it doesn't. Which is why I said in my post "I'm sure you could find different titles at different levels with different textbooks covering different material in lots of places. It makes no difference really." Your statement is exactly what I meant. I just felt I should justify myself because many people commented.

It's also kind of an odd statement that complex analysis can only be a course offered at the graduate (advanced graduate?) level.

As long as the spirit behind the course is to prove real analysis-like results in $\mathbb{C}$ (or $\mathbb{C}\cup\{\infty\}$  :D  ), I don't think it's necessary to quibble with whether it is an undergraduate course or a graduate level course.  When I took my complex analysis course last semester (yes, it was an undergraduate course) we used, for example, real analysis results (equicontinuity, for example) to prove results from complex analysis, as well as the open mapping theorem and a proof of Liouville that depended centrally on the one-point compactification of the complex plane.  I tend to think that these are definitely results which belong to analysis, and hence are very much complex analysis.

We've establsihed that... I was wrong. Do you have a NEW point to contribute?

No, I don't have a 'NEW' point to contribute, but it would be nice to know why you find the need to be so aggressive in your posting.  I stated the conditions under which I think a course should be called 'complex analysis'.  That is hardly off topic given the thread title.

[quote="Spoon"]Because you called my statement odd and then completely misrepresented what I said. My statement was "None that I know of." Nowhere did I say it could only be this or that. At least read what people say before posting that it's "odd".[/quote]

Well, since complex analysis is a well-defined term (that is, if you say I am interested in complex analysis to a math professor, they are not going to start talking about graph theory with you), when you say 'the only complex analysis I know of is at the advanced graduate level', you imply that complex analysis is primarily a course for grad students.  That is, of course, patently untrue, and you should be careful of giving people advice based on topics which you may not have a strong grasp of.  This is not to say that I am great at complex analysis, but at least I have taken a course on it (and courses in real analysis as well).

I then outlined some topics typically discussed in a first course in complex analysis (or complex variables.  or 'functions of a complex variable', etc.).  You then proceeded to tell me exactly what you consider to be complex analysis.  However, probably because you are unfamiliar with the topic, you listed, for example, cauchy's theorem and it's corollaries, which are taught in any first course on complex variables (there is a case to be made that this is the central result of the course).  Similarly for conformal mapping, taylor series, and the inverse function theorem.

You accused the original poster of arrogance, but isn't it a bit arrogant to assume that you have the necessary expertise to say exactly what is and what isn't a complex analysis course?  In any event, I don't really want to debate this, because given the choice I will study complex analysis rather than argue its exact nature.

Re: posts #21 through #27:

Thanks for the discussions of definitions. I think the definitional argument here has been well explored on all sides. (In other words, let's move on to other issues if there are still topical issues to explore in this thread.)

More generally in this thread, PLEASE take care not to accuse other people of arrogance (and also please don't assume that people who are correcting you factually are slamming you personally). Then we will all have more productive discussions here. Even highly math-capable people fare better in life with better social skills.

tokenadult raises some interesting points on the subject of etiquette.

[i]I beg your pardon, but I'm going to adopt a zero-tolerance policy here when it comes to metadiscussion of how the forum is moderated. But you are very welcome to continue to participate here as you have in the past.[/i] -- moderator edit by tokenadult

[quote="fleeting_guest"]Post no. 14, which I wrote, varied between harshly sarcastic and humorously sarcastic; people may disagree about whether that rises to the level of a "slam", but it was deliberately written so as to avoid being [i]personal[/i].[/quote]

Excuse me, but explain to me how a post entirely in the second person (and thus unquestionably directed at an individual) which you [i]begin[/i] by using the word "tasteless" is not personal?  You last long and rather pretentious post aside, that is quite clearly a personal comment.  In addition, tokenadult's post very conspicuously did not adopt a similar tone -- indeed, he even went so far as to open with a complement, in contrast with your own post.  As a moderator, it is his responsibility to maintain a civil level of discourse on this forum.  He consistently maintains civility in his own posts.  You have demonstrated an inability to do so.

I will note for the record that I am quite aware that this post contains quite personal remarks which might fairly be construed as personal attacks.  But, frankly, much of what you say deserves nothing better; you've already demonstrated that you respond to civility with rudeness.  I'd also like to say that your comment which began all this nonsense had some merit.  Nevertheless, your tone in that post and in your further posts has been totally inappropriate.

Two things.

First of all, I think fleeting guest's earlier comment on congratulatory reassurance is about right.  Though I would be extremely annoyed by a B+ in any math course at my university, which manages to have a math department of reasonably high quality, I would, of course, draw the line at communicating my personal opinions on grades to someone else.  Since most people in the know (save for the sophomoric know-it-alls that college admissions boards seem to have engendered in large quantities) would adopt a similar line of thought upon such questioning and respond that that is a good mark, and that any complex analysis class is better than no complex analysis class, we're pretty much left with congratulatory reassurance.

Secondly, I didn't appreciate tokenadult's post about "Even highly math-capable people fare better in life with better social skills.", which was at least tangentially directed at me.  I am not a brilliant mathematician, nor am I an excellent self promoter (although that skill is at least easy to cultivate), yet I managed to be accepted to one of, by anyone's standards, the three best universities in the United States as an international student, to be named the single top applicant at the university I go to (out of thousands of applicants every year), and also win national and provincial awards for community work and all-around excellence, respectively.  Although listing such things is inevitably more than a bit gratuitous, my social skills have gotten me to where I am today, not hindered me - and I certainly resent having such a label stuck on me for making a reasonable characterization.

On a marginally personal level, you (Joel) consistently respond to posts in F&G concerning political conservatism and religious fundamentalism with sarcasm, so I hardly see someone else responding likewise to blatant self-aggrandizement to be any worse.

Err, that was three things.

[quote="blahblahblah"]Secondly, I didn't appreciate tokenadult's post about "Even highly math-capable people fare better in life with better social skills.", which was at least tangentially directed at me.[/quote]

Well, sorry about that. I did reference the entire thread, didn't I, when I mentioned that we should all be a bit less prickly here and all slow to take offense? So my comment was very tangential indeed as a post directed to any one person. Actually, it appears to me that your participation, blahblahblah, in this thread and in many other threads on this forum has been just fine, certainly quite interesting to me personally, and I don't have any doubts about your social skills.

Returning to the original poster's concerns, so that this thread doesn't go too far into metadiscussion, I would simply say that of course the IDEAL thing is to dominate in both the math contest and in the college math class. But few young people can hit on all cylinders all the time. Everyone gets tired or too busy some of the time. So don't worry about your grade in the course too much. Enjoy the most challenging classes you take for the intellectual stimulation they provide, enjoy math contests if they advance your learning, and feel free to express your real inner self when you apply to college.

Good luck to the college applicants in high school class of 2006 and best wishes to the current college students, recent alumni, and college teachers who add so much useful content to this forum.

[quote="blahblahblah"]On a marginally personal level, you (Joel) consistently respond to posts in F&G concerning political conservatism and religious fundamentalism with sarcasm...[/quote]

You're absolutely correct.   (Indeed, I even noted in my previous post that it was an example of the same behavior.)  However, the circumstantial difference between a conversation in which there is a debate or an inherent clash of viewpoints and a conversation in which someone has asked an informational question is very great.  My personal insults are almost exclusively directed at the quality of an individual's argument, not at the "tastelessness" of his question-asking style.  I agree that self-aggrandizement is annoying and that these fora often suffer from it.  However, I see very little evidence of it in this case: 1234567890 is a high school student, presumably ready to apply to competitive colleges.  At my high school, the average GPA was approximately 91/100 -- an A-, say.  That means a B+ (88) would have brought down the GPA of more than half of the school.  If I were unfamiliar with the differences between standard college and high school grading scales, the worry of the initial poster would seem very valid.  His rationalization of the grade is typical behavior for anyone who got a grade they didn't like and doesn't have much to do with his question, but it hardly merits the description "tasteless."

As for tokenadult, the statement that math people would benefit from improved verbal and interpersonal skills is one that he makes frequently -- look through old posts in this forum about the SAT.  As another person who views myself as a personable mathy individual (and also as someone who parallels your accomplishments reasonably closely), I see no reason to take that as a comment about me -- indeed, as a personable mathy person, I agree with him that many mathy people could shift some of their focus away from mathiness towards other skills and be much more interesting, likeable people as a result.

This thread seems like it's about ready to be locked or split or something.  Although in debating fairness I suppose I shouldn't advocate that lest fleeting_guest feel I have snookered him out of the last word.

I wasn't accusing tokenadult of sniping at me - certainly, he is a forum member who I am inclined to give the benefit of the doubt to always based on what his posting, as a whole, has represented.  But the comment about arrogance related directly to me, and its juxtaposition with the bit about 'better social skills' certainly created the general impression of a rebuke.  Anyways, water under the bridge.

I'll reserve further comment in the interest of maintaining a semblance of focus in this thread.


Lte us start a treething fan club!!!!!!
My secret treething fan club name is stan!!!!

sweet.  and mine will be cartman.  somebody be kyle next

asmallgoat go to wikipedia and type in mathcounts

whatever. I guess I'm kyle anyway, but I don't want to join no club. I can be a honorary guinea pig

this is private post but im youre bigggest fan :blush:  :blush:  :love:

This isn't the slightest bit weird... :huh:

if you will join please post this in english!!!!!!!! The club is named TIC for treething is cool!!!!!!!

I can TOXIC and then tank using superior DEF and SP.DEF.

yeaah... i'm pretty sure this guy isnt me. if any admin wants to run an ip address check, they can.

Enoch: Oh, atheist temple only got an A on Oi Punk?
He's not that good at stepmania then

treething this guy isn't you-some people already know that from ip
in fact i know that from knowing this person

[quote="Treething"]Who are you?[/quote]

Yay now you can earn treething bucks!!!! You get one tb (treething buck) every time you get someone to join the club. you get banned if you send spam to treething. you get 100 bucks if you REALLY converse with treething about a serious topic!!!!! you get 999999999 if he allows you to look at his blog.

Oh yeah it's also cool to get treethings avatar, his signature, or/and quote him.

      :weightlift:  :showoff:  :weightlift:
      [b]Earn your Treething Bucks![/b]

[quote="mustafa"][quote="bpms"][quote="randomdragoon"]I'll use SWAGGER, then PSYCH UP.
Then I'll hit you with a FAINT ATTACK for the win.[/quote]
I'll counter with SUBSTITUTE then FOCUS PUNCH from my Tyranitar![/quote]

But wait! Hitmonlee comes out, and what will he do? KICK! Critical hit, and Tyranitar is down!

[/Pokemon Stadium voice][/quote]
But uh oh, it only broke his substitute!

A challenger appears! Skarmory, and he uses SPIKES three times and then uses WHIRLWIND like there's no tomorrow!

[quote="mustafa"][quote="bpms"][quote="randomdragoon"]I'll use SWAGGER, then PSYCH UP.
Then I'll hit you with a FAINT ATTACK for the win.[/quote]
I'll counter with SUBSTITUTE then FOCUS PUNCH from my Tyranitar![/quote]

But wait! Hitmonlee comes out, and what will he do? KICK! Critical hit, and Tyranitar is down!

[/Pokemon Stadium voice][/quote]

if i recall correctly breloom can go up and use his dual resistance to earthquake and rock slide and use mach punch to break the substitute, take one hit, and use brick break for the kill (or another mach punch if tyranitar's under 60% health)

netbattle ftw

[quote="BoesFX"][quote="randomdragoon"]I'll use SWAGGER, then PSYCH UP.
Then I'll hit you with a FAINT ATTACK for the win.[/quote]

Nah, it doesn't do any good, FAINT ATTACK is counted as special attack and SWAGGER+PSYCH UP only increase physical attack. You should try another attack like RETURN or another physical attack. :wink:[/quote]
Nope, dark is physical attack (at least up to gen. III, I don't know what happens in the unreleased gen.IV)

I got myself to join the club. I demand 10 treething bucks, or I'll boycott.

[quote="randomdragoon"][quote="Treething"]nope, dark is special.[/quote]
Oh doh, I was thinking about SHADOW BALL. I keep getting them mixed up.

Well in that case I'll just TOXIC, SUBSTITUTE and REST my way to oblivion[/quote]\

I HACK THE GAMES AND MAKE ALL OF YOUR POKEMON LEVEL 2 (are there even level 1's?)

and I will construct a Warp Gate, a Cibernetics ore and make some Protoss Dragoons.

oh, damn, I need more Protoss Pylows

[quote="randomdragoon"]Sure there are level 1s (by hacks or the MISSINGNO glitch) but ... using PSYWAVE with a lvl 1 pokemon will freeze your game.

Also:

My STARMIE's move list:
Surf
Psychic
Thunderbolt
Recover

Beat that[/quote]

Starmie @ Leftovers ** Divination
Trait: Natural Cure
EVs: 220 HP / 36 Spd / 252 SAtk
Timid Nature (+Spd, -Atk)
- Rapid Spin
- Recover
- Surf
- Thunderbolt

Treething quoted me, do I get Treething bucks now?

[quote="Fabiomr1"]and I will construct a Warp Gate, a Cibernetics ore and make some Protoss Dragoons.

oh, -, I need more Protoss Pylows[/quote]

Zergling Rush ...  :ninja:

BTW,
[quote="Treething"]

Starmie @ Leftovers ** Divination
Trait: Natural Cure
EVs: 220 HP / 36 Spd / 252 SAtk
Timid Nature (+Spd, -Atk)
- Rapid Spin
- Recover
- Surf
- Thunderbolt[/quote]

Blissey lv 100 @ Leftovers
Trait: Natural Cure
EV: 252 HP / 252 Def / 6 Sp. Def
Bold Nature (+Def -Atk)
- Seismic Toss
- Toxic
- Softboiled
- Aromatherapy

Comment: Tank War :D

I'm not even sure what we're talking about here

Starfox64 personal high score

Total hits 1657
----
Corneria: 197
Meteo: 311
Katina: 177
Sector X: 211
Macbeth: 156 (bad score on this level, usually do better)
Area 6: 379
Venom: 227
---
Lives remaining at end: 20
Peppy: 7
Slippy: 7
Falco: 7

aroma the rapist i see

[code]
Slaking (M) @ Choice Band ** Apocalypse
Trait: Truant
EVs: 4 HP / 252 Atk / 252 Spd
Adamant Nature (+Atk, -SAtk)
- Earthquake
- Fury Swipes
- Return
- Shadow Ball

Skarmory (M) @ Leftovers ** Bleeding Skies
Trait: Sturdy
EVs: 252 HP / 16 Atk / 144 Spd / 96 SDef
Impish Nature (+Def, -SAtk)
- Drill Peck
- Roar
- Spikes
- Taunt

Vaporeon (F) @ Leftovers ** Repent (Whore)
Trait: Water Absorb
EVs: 204 HP / 216 Def / 88 SDef
Bold Nature (+Def, -Atk)
- Protect
- Surf
- Toxic
- Wish

Metagross @ Leftovers ** Contempt
Trait: Clear Body
EVs: 252 Atk / 108 Def / 104 SAtk / 44 SDef
Brave Nature (+Atk, -Spd)
- Explosion
- Ice Punch
- Meteor Mash
- Psychic

Celebi @ Leftovers ** Resurrection
Trait: Natural Cure
EVs: 156 HP / 56 Def / 104 Spd / 120 SAtk / 72 SDef
Bold Nature (+Def, -Atk)
- Heal Bell
- Leech Seed
- Psychic
- Recover

Breloom (F) @ Leftovers ** Ascendancy
Trait: Effect Spore
EVs: 56 HP / 200 Atk / 252 Spd
Jolly Nature (+Spd, -SAtk)
- Focus Punch
- Mach Punch
- Sky Uppercut
- Spore
[/code]


Nice job Neal Wu!!  Only a seventh grader.  He's awesome at countdown!

I am from IN (not on team, got 5th at state) so it really stinks that Texas beat us on a tie breaker.  especially since they had only 1 in the top 12 compared to our three.Compare the scores, IN:36,35,32,25 8 team score:48.  TX:35,29,28,28 9 on team score 48.   The should both get laptops! (CA only got 43)

Awesome job by our Benjamin brothers.  Nathan the 7th grader got 2nd written!!  He will easily win written round next year.  And David, the top 7th grader last year, tied with Mark Zhang and Patricia Li for 3rd written but was bumped to 5th.  Also, Pardha got 10th written. Yixin tounded out the IN team.

I told you Mark Zhang wouldn't win! But I sure didn't know Neal Wu would.

Did you know that Sergei Bernstein and David and Nathan Benjamin are realted?  Half half cousins or something like that.

IN killed MA in lazer tag at Joe DUmar's!

[quote="nat mc"]

Did you know that Sergei Bernstein and David and Nathan Benjamin are realted?  Half half cousins or something like that.

[/quote]
Seriously?
I thought that target was reaaly hard (i got 1 :blush:  :blush: )
sprint was average
so was masters
countdown and team were hard

Sure, the tests were pretty stiknin' hard (stupid target!) but the whole experience was amazing! I mean - the whole elegance surrounding the awards banquet, the sheer excitement of the countdown round - and that screen was so much cooler than the cheap projection thingy at states - I feel at the top of my game because of the professionalism surround nationals MathCounts. I am so much better for having gone and having had the experience, not just including the test-taking.

[quote="sirorange"]Ok, Indiana was pwned at laser tag, get it through your head![/quote]
yeah yeah whatever lol

The lazer tag sucked. The minigolf was pretty fun though.

Okay, I posted this somewhere else, but here, my opinions

MY OPINIONS:
[size=200]
My Opinions[/size]

The sprint was, in my opinion, a bit easier than sprints of past years.

The target was, in my opinion, about the same as last year.

The team round was exceptionally hard this year.

Fun Round - fun. Two of my teammates got mentioned in this round.

Countdown Round - Harder than usual - many times neither competitor got the right answer. Bob, Kevin, and I were keeping score of how many times we can get the right answer before the compeitors did. Bob got 17. Kevin got 11. I got 7. I suck at Countdown.

Masters Round - I've seen some Masters Round problems from past years and this one seemed to require a lot more thinking than usual. Several people didn't finish in time. Neal should have won this round.

I got 35 trading pins.

[quote="Dnas"][quote="biffanddoc"][quote="nat mc"]

Did you know that Sergei Bernstein and David and Nathan Benjamin are realted?  Half half cousins or something like that.

[/quote]
Seriously?[/quote]
Well, half half cousins is the simplest way I could figure out how to do it.

The more specific version is: he is my half-brother's mother's second husband's brother's son.
In other words, his uncle married my dad's first wife.

So, technically, 3 members of the family made it to countdown :)

[quote]Ok, Indiana was pwned at laser tag, get it through your head![/quote]
No... we won :P

Probably no way to tell because we didn't add the scores and some of the guns didn't work right part of the time.
w/e... it was fun.[/quote]

Wait... are you David or Nathan Benjamin?

1 2 and 5 from the same family is pretty good.  Sergei should move to West Lafayette.

NoSoupForYou: What score and rank did you get?

Score: 16/5
Rank: 51st (dropped from 41st because of tiebreakers  :( )

[quote="chess64"][/quote][quote="Dnas"][quote="biffanddoc"][quote="nat mc"]

Did you know that Sergei Bernstein and David and Nathan Benjamin are realted?  Half half cousins or something like that.

[/quote]
Seriously?[/quote]
Well, half half cousins is the simplest way I could figure out how to do it.

The more specific version is: he is my half-brother's mother's second husband's brother's son.
In other words, his uncle married my dad's first wife.

So, technically, 3 members of the family made it to countdown :)

[quote]Ok, Indiana was pwned at laser tag, get it through your head![/quote]
No... we won :P

Probably no way to tell because we didn't add the scores and some of the guns didn't work right part of the time.
w/e... it was fun.[/quote][quote="chess64"]

Wait... are you David or Nathan Benjamin?[/quote]
That's David

[quote="Phelpedo"]I think the test was a lot harder. Didn't you think that your scores were lower than your practice test scores?[/quote]

No, they're actually about the same.

Ah, good gene pool. Another other members of the family do math?

[quote]
Funniest moment = watching my shoe almost hit someone in the back when it fell 20+ feet [/quote]
Huh? Please explain  :D

Well, I was on the obstacle course, and my shoe fell off on the high level. At Joe Dumar's. Someone was bending down to pick up a couple tickets he dropped, and the shoe missed him by about an inch.

[quote="nat mc"][quote="Treething"]Bah. I got my AoPS book signed by Mr. Ruszcyk. I seriously expected him to be in his 50s. That person at the shark slide wouldn't let me go again because I jumped down the slide. And I lost my shoe on the upper obstacle course, and I almost landed on someone after jumping from the top of the hard rock climbing wall, the middle one. It was fun overall.

Except when one of my teammates got lost in the stairwell and my coach got very angry :P .[/quote]
Your last statement made me think of something very funny. Ok here goes,

After the banquet, all the elevators were full and Pardha decided that we would take the stairs. So when we got in the stair well and went up to the 5th floor, it said no exit and was locked.  So, we went back down to the 4th, and it also said no exit and was locked.  WE looked down the stairs and there was an exit sign pointing down.  So we went down and down and down.  I was carrying NAthans trophies so he and I were way back and almost lost them.  Then we were down as far as you could go and there was one more set of stairs that went back up a little ways.  At the top, there was a door.  It was unlocked.  WE wisely checked to see if it would open from the inside in case there was no exit.  It was locked on the inside.  SO we told Nathan to hold the door.   There was a very long hall way that seemed to go on forever.  At the end there was a fire alarm thing and a red door.  It looked like an emergency exit that would set off the alarm.  David got worried that we would get in trouble if it went off.  Meanwhile, Nathan couldn't see us so he yelledf something.  I thought he said "Should I come"  so I yelled "NO" but what he really said was" Should I keep holding the door".  So he came, and we were trapped.  We couldn't go back up, so we had to open the door.  An alarm did not go off.  But again, the door wouldn't open from the inside, so we had to hold it again.  WE were in a place that seemed like a parking garage.  Across the garage floor, which seemed to be covered in wet paint, there was a door but we didn't know where it led.  Pardha ran across with Nolan on the wet paint and opened the door and told us toi come.  It had led us outside, but we didn't know where.  There was a woman at the parking garage and Pardha asked where the entrance was.  We ran to the entrance.  Nathan and I were relieved and were laughing about it.  But when we got in the hotel, David thought that Pardha was going to get arrested and that the hotel had a curfew. Nathan and Pardha and I all made fun of him.

It was so hilarious.[/quote]

You were at Nationals? And where was Yixin in this?

He was with us too.  He was also laughing with Nathan and I

[quote="Phelpedo"][quote="nat mc"]....stuff about stairs....[/quote]

You were at Nationals? And where was Yixin in this?[/quote]
Phil was among the 20-some guests we brought from Indiana. :)

Know what? I didn't sleep at all the night before the test. :? And I was more sleepy on Saturday morning than anytime Friday or Saturday.

The test was pretty hard, but I had fun.

:lol: I met so many people! Very smart people too. Especially compared to me...

Billy

[quote="Treething"][quote="Secret Asian"]The good and bad of 2005 nats? Let's start with good news first because that's the fun stuff.

Good:  Had fun became even more insane than I usually am.

Bad:  Er....completely screwed up the test. Don't even ask how I did. I usually do pretty well on practice tests..but the mixture of 2 hours of sleep plus what I thought was a difficult test screwed me up very very badly.

Something Weird:  Some kid from New Jersey thought I was a girl.  I was sitting in a chair waiting for the coach to come so we could go to the GM museum and I was randomly taking pictures of every little thing I saw and the guy from NJ came over to me and asked if I was Valerie Chia.  Very strange.[/quote]

HAHAHAHA!

That must've been Jason xD .

Well, if you're Miles, I asked you on the bus what gender you were. But I doubt you are.[/quote]

Who's Miles?  But seriously...I  do not look like Valerie at all...you can take a look at the yearbook.  I'm from Utah, Konlin Shen.

[quote="AznChristmas"][quote="Secret Asian"]Who's Miles?  But seriously...I  do not look like Valerie at all...you can take a look at the yearbook.  I'm from Utah, Konlin Shen.[/quote]

No offense or nothing, but you're make for a pretty ugly girl.

You don't look anything like a girl. That's strange.[/quote]

Of course I'd make an ugly girl.  That's why I'm a guy.  It was very strange.  Although everyone else on my team thought it was hilarious....the one question I have to ask though is how?   How did that guy mistake me?

I don't know, maybe he was blind, maybe you imagined him, maybe the world will come to an end next Wednesday...
Anything can happen.  :D


function
parameterization
calculus
calculus computations

Express the following parametrization in the form $ z\equal{}f(x,y)$.
$ \textbf{X}(s,t)\equal{}(s^{2}\minus{}t^{2},s\plus{}t,s^{2}\plus{}3t)$
Thank you.

Let $ x\equal{}s^2\minus{}t^2\equal{}(s\minus{}t)(s\plus{}t)$ and $ y\equal{}s\plus{}t$. Then, $ s\minus{}t\equal{}\frac{x}{y}$. Solve the system of equation and get $ s\equal{}\frac12\left(y\plus{}\frac{x}{y}\right)$ and $ t\equal{}\frac12\left(y\minus{}\frac{x}{y}\right)$. Therefore,

$ z\equal{}s^2\plus{}3t\equal{}\left(\frac12\left(y\plus{}\frac{x}{y}\right)\right)^2\plus{}\frac32\left(y\minus{}\frac{x}{y}\right)$

which simplifies to $ z\equal{}\frac14y^2\plus{}\frac32y\plus{}\frac12x\minus{}\frac{3x}{2y}\plus{}\frac{x^2}{4y^2}$

Certainly more complicated than I anticipated. :lol:


quadratics
calculus
integration
number theory
relatively prime
algebra
quadratic formula

[quote="Ish"]Problem:

Find all pairs of positive integers (a,b) such that a and b are relatively prime and a/b + 19b/a is an integer.
Relatively Prime means no common factors other than the number 1.[/quote]
[hide] Say that $\frac{a}{b}+\frac{19b}{a}=k$.  We know that $\frac{a}{b}+\frac{19b}{a}=\frac{a^{2}+19b^{2}}{ab}\Rightarrow \frac{a^{2}+19b^{2}}{ab}=k\Rightarrow a^{2}+19b^{2}-kab=0$.

Now, applying the quadratic formula, we find that $x=\frac{kb\pm\sqrt{b^{2}k^{2}-76b^{2}}}{2}=\frac{b(k\pm\sqrt{k^{2}-76})}{2}$.

Thus, $k^{2}-76$ must be a perfect square, otherwise x would be irrational.  So, say that $k^{2}-76=y^{2}\Rightarrow k^{2}-y^{2}=76\Rightarrow (k+y)(k-y)=76$.  We know that k must be an integer, so $y$ must be an integer.

The only system of equations that makes both k and y integral is $\text{k+y=38, k-y=2}\Rightarrow k+y+k-y=40\Rightarrow 2k=40\Rightarrow k=20$.  So, since $k=20$, we have that $a^{2}-20ab+b^{2}=0\Rightarrow (a-b)(a-19b)=0\Rightarrow \text{a=b, a=19b}$

Now, since $a$ and $b$ are relatively prime, we know that $b$ must equal one in either case or else $a$ and $b$ will share the factor of $b$.  So, $a=1$ or $a=19(1)=19$.  Thus, our pairs of $\text{(a, b)}$ are $\boxed{\text{(19, 1) and (1, 1)}}$ [/hide]

[quote="Ish"]Good answer.
That's what I had .
Is there anything more elegant and shorter?  :([/quote]
[hide="A Little Bit Shorter"]
When you have $a^{2}+19b^{2}-kab$, we can say that we must find an integer $k$ such that $a^{2}+19b^{2}-kab$ is factorable into $(a-mb)(a-nb)$.  From there we can multiply out so we have that $a^{2}-(m+n)ab+mnb^{2}=a^{2}-kab+19b^{2}\Rightarrow kab=(m+n)ab\Rightarrow m+n=k$ and $mnb^{2}=19b^{2}\Rightarrow mn=19$.  Now, since $m+n$ and $mn$ are integral, we know that $m$ and $n$ are integral, so the only possibilities for $(m,n)$ are $(1, 19)$ and $(19, 1)$.  Thus the factorization is $(a-b)(a-19b)$ so we have that $(a,b)$ is $(1,1)$ and $(19,1)$
[/hide]

lol nice! That is more elegant in my book!


Bold indicates a vector:

Find the component form of  [b]v[/b]

Given:

$ \parallel{}$[b]u[/b]$ \parallel{} \equal{} 1$, $ \Theta \equal{} 45$; and ||[b]u[/b] + [b]v[/b]||$ \equal{} \sqrt{2}$, $ \Theta \equal{} 90$

Where $ \Theta$ is in degrees.

The solution becomes obvious if you solve it geometrically, but how would you solve it algebraically?

$ u \equal{} (\frac{1}{\sqrt{2}}i \plus{} \frac{1}{\sqrt{2}}j)$

$ u\plus{}v\equal{} \sqrt{2}j$

$ v\equal{}(\frac{\minus{}\sqrt{2}}{2}i \plus{} \frac{\sqrt{2}}{2}j$)

Therefore the length of $ v$ is $ 1$ and the angle is $ 135$ degrees or $ \frac{3\pi}{4}$ radians.


calculus
derivative
function
geometry
rectangle
integration
calculus computations

1. for second order mixed differentiation, the order in which the derivatives are found doesnt matter. Why?

2. to find the gradient we use partial functions for more than 1 variable instead of total, Why it doesnt affect the result?

[b]Lemma:[/b] Let $ f_{xy}$ and $ f_{yx}$ be continuous on the rectangle $ R\equal{}[a,b]\times [c,d]$.  Then $ \iint_{R}f_{xy}\,dA\equal{}\iint_{R}f_{yx}\,dA$.

[b]Theorem:[/b] If $ f_{xy}$ and $ f_{yx}$ are continuous, then they are equal.
[i]Proof:[/i]Suppose not, so that at some point $ (a,b)$, $ f_{xy}(a,b)\minus{}f_{yx}(a,b)\equal{}h>0$.  By continuity of $ f_{xy}$ and $ f_{yx}$, there exists some small rectangle $ \Delta x\times\Delta y$ centered at $ (a,b)$ on which $ f_{xy}(x,y)\minus{}f_{yx}(x,y)\geq\frac{h}{2}$, so that $ \iint_{R}f_{xy}\minus{}f_{yx}\,dA\geq\iint_{R}\frac{h}{2}\,dA\equal{}\frac{h}{2}(\Delta x)(\Delta y)>0$.  But this contradicts the lemma, since $ f_{xy}$ and $ f_{yx}$ have equal integrals over the rectangle, meaning $ \iint_{R}f_{xy}\minus{}f_{yx}\,dA\equal{}\iint_{R}f_{xy}\,dA\minus{}\iint_{R}f_{yx}\,dA\equal{}0$.


linear algebra
matrix
vector
real analysis
real analysis unsolved

For each integer \[n \ge 1 \] let
\[S_{n}\] be the vector space of symmetric real n x n matrices, and \[S_{n}^{+}\subset S_{n}\] be the set of positive-definite symmetric matrices.

Prove that \[S_{n}^{+}\] is an open subset of \[S_{n}\] that for each
\[A \in S_{n}\]
the matrix exponential \[e^{A}\] lies in \[S_{n}^{+}\] and that the resulting map \[A \mapsto e^{A}\] is a \[C^\infty \] bijection \[S_{n}\to S_{n}^{+}\] whose inverse also is C-infinity.

Prove analogous statements for Hermitian matrices.

Reformatted for readability:
---
For each integer $n \ge 1$ let $S_{n}$ be the vector space of symmetric real $n\times n$ matrices, and $S_{n}^{+}\subset S_{n}$ be the set of positive-definite symmetric matrices.
Prove that $S_{n}^{+}$ is an open subset of $S_{n}$, that for each $A \in S_{n}$ the matrix exponential $e^{A}$ lies in $S_{n}^{+}$ and that the resulting map $A \mapsto e^{A}$ is a $C^{\infty}$ bijection $S_{n}\to S_{n}^{+}$ whose inverse also is $C^{\infty}$.
---

1) $A\in S_{n}$ is positive definite iff there is $c>0$ such that $Ax\cdot x\ge c|x|^{2}$ for all $x$. If $B\in S_{n}$ is such that the norm of $A-B$ is less than $c/2$, then $Bx\cdot x\ge \frac{c}{2}|x^{2}|$.
2) Use the identities $\exp(A)^{T}=\exp(A^{T})$ and $\exp(A)=\exp(A/2)\exp(A/2)$.

How do we use the exponential identities in connection with the C-infinity bijection mapping?

The identities are to prove that $\exp$ takes its values in $S_{n}^{+}$. The $C^{\infty}$ smoothness of $\exp$ follows from the fact that it is represented by a (multivariable) Taylor series. I don't know what to do about the inverse.


The November 5 SAT was the first SAT I took and I got:

Math: 770
Verbal: 560
Writing: 560
Essay: 9

Should I be worried about this or is it okay if I improve a little bit simce it's my first time? Also, I was wondering, do colleges look at the highest scores from each subject from numerous SAT scores?

Supposedly, they will look at your best score in each category from different tests, but I'm not too sure about that... I guess it makes sense, though.

as far as I know they only look at your highest scores, but some were saying they also consider the number of times you took the test, so my honest opinin is to better prepare before you take it next time ;)


from some old problem of the week thing...

$\frac{a+b+c}{3}=P_1$

$\frac{a+b}{2}=P_2$

$\frac{b+c}{2}=P_3$

$\frac{a+c}{2}=P_4$

Find $a, b, \text{and}, c$ where $a, b, \text{and}, c$ are [b]distinct two digit primes[/b], and $P_n$ represents another prime number (which may or may not be distinct).

[hide="an idea"]
All 2-digit primes are congruent to $1$ or $5$ in mod $6$.
From the first condition, it is clear that all 3 primes $a$, $b$ and $c$ must have the same parity (all 3 are in the form $6k+1$, or all 3 are in the form $6k+5$).

But I'm not sure how to go from here...
[/hide]

[hide="Hint"]
Five primes in arithmetic progression.  ;)
[/hide]


Find all integers $ a$, $ b$ and $ c$ that satisfy equation:

$ 2(a\plus{}b\plus{}c)^{3}\plus{}(a\plus{}b)(b\plus{}c)(c\plus{}a)\plus{}2abc\equal{}2$

See here :
 http://www.mathlinks.ro/Forum/viewtopic.php?t=125143
 http://www.mathlinks.ro/viewtopic.php?t=127950   :wink:


Find the smallest $n$ so that no matter how $n$ points are placed in a unit square, there exists a pair of points separated by a distance no greater than $\frac{\sqrt{2}}{3}$.

[hide] Divide the square into 9 smaller squares.  The diagonal of the smaller squares is $\frac{\sqrt{2}}{3}$, so we need to have multiple points in one square.  By Pigeonhole we need 10 points to be sure of this.[/hide]


There are 3 bags which are known to contain 2 white and 3 black , 4 white and 1 black and 3 white and 7 black balls respectively . A black ball was drawn at random from one of the bags.Then what is the chance that it was drawn from the bag containing the most balls???

Since there are 11 black balls total and we know that we drew one of them, there's a $\frac{7}{11}$ chance that we drew it from the bag with 10 balls (the most).

I think we have to use Baye's Theorem in this.


use cramers rule to solve the following system of linear equations
x+2y=7
3x-9=-9
for this equation I got  x=10/11 and y= 3

Using Cramers rule solve
2x-y=4
3x+y-z=10
y+z=3
for this equation I got lost and didnt understand how to do it

Use determinants to determine the number of solutions for each of the following systems.
(a) 3x-y=-6
         2x=3y+2
for this I got (7,24)
(b) 3x-y=7
      6x-2y=10
for this I got no solution
Can someone please tell me if my answers are right it would be very helpful.

just do a 3 x 3 martrices.[url=http://fclass.vaniercollege.qc.ca/web/mathematics/comp/mapla_2.htm]fclass.vaniercollege.qc.ca/web/mathematics/comp/mapla_2.htm[/url]

PM me if you need more help

Check the first system, I think you left out a $y$.  Problem 2a is wrong, just plug the numbers back in.  Problem 2b is simple, just multiply the first equation by 2 and it's obvious.

This sounds a lot like homework... Please don't post here for homework help


geometry
circumcircle
trigonometry
angle bisector
perpendicular bisector
geometry unsolved

In a triangle $ \triangle ABC, BC>AB$, and the angle bisector of $ \angle ABC$ meets AC at D. The perpendicular from C to BD meets BD at P, and the perpendicular from A to BP is Q. Let M be the midpoint of AC, E be the midpoint of BC. Let the circumcircle of PQM intersect AC at point H (not equal to M). Prove that O, H, E, and M are cyclic.

Dear Mathlinkers,
I suppose that O is the center of PQM.
This circle is attribuate to a french geometer Calabre.
I need this situation in order to prove synthetically the Feuerbach's theorem.
See: http://perso.orange.fr/jl.ayme   vol. 1  Le théorème de Feuerbach  page 2...(Conséquence du lemme 1)
Sincerely.
Jean-Louis

1.PE∥AB,ME∥AB  P,M,E are collinear
2.let AB/BC=k, we have that QD/DP=AD/DC=AB/BC=k
QD=k*DP  BQ/BP=AB/AC=k   BQ =k* BP
=> QD/ BQ= DP/ BP
MC/MD=MA/MD=1+AD/DM=1+BD/DP=BP/DP
=> MD/MC=DP/BP
=> QD/ BQ= MD/MC => MQ∥BC
3.MQ=MP  ∠QHM=∠QPM=∠PQM=∠PBC=∠PBA  => B,A,H,Q are concyclic
  => BH⊥AC
4. ∠EHO+∠OME=∠EPO+∠OME=∠OMP+∠OME=180
Done!

Another solution~

We shall first prove that $ H$ is the foot of perpendicular from $ B$ to $ AC$. (which is somehow inspired by jayme's link~)
First extend $ AQ$ and intersect $ BC$ at $ X$.
Since $ BQ$ bisects $ \angle ABX$ and $ BQ \bot AX$, so $ Q$ is the mid-pt of $ AX$.
Since $ M$ is also the mid-pt of $ AC$, so $ QM//BC$.
Thus it is not hard to prove that $ H, B, C, P$ concyclic.
So $ \angle BHC \equal{} \angle BPC \equal{} 90$ degrees.

As $ E$ is the mid-pt of $ BC$, So $ \angle BEH \equal{} 2\angle BCH$
$ \angle CEM \equal{} \angle ABC$ since $ EH//AB$.
So $ \angle HEM \equal{} 180 \minus{} \angle HEB \minus{} \angle MEC \equal{} 180 \minus{} 2\angle BCA \minus{} \angle ABC$

On the other hand, since $ M$ is the mid-pt of AC, So the foot of perpendicular of M to PQ is the mid-pt of $ PQ$, denote as $ Y$, i.e. $ MY$ is the perpendicular bisector of PQ.
As $ OY$ is also perpendicular bisector of PQ, so $ O,Y, M$ collinear.

As $ B, H, Y, M$ concyclic ($ \angle BHM \equal{} \angle BYM \equal{} 90$, from some angle tracing we can get:
$ \angle HOM \equal{} 180 \minus{} 2\angle OMH \equal{} 180 \minus{} 2\angle YMH \equal{} 180 \minus{} 2\angle HBY \equal{} 180 \minus{} 2(0.5\angle ABC \minus{} (90 \minus{} \angle BAC)) \equal{} 360 \minus{} \angle ABC \minus{} 2\angle BAC \equal{} \minus{} \angle ABC \plus{} 2 (\angle ABC \plus{} \angle ACB) \equal{} 2\angle ACB \plus{} \angle ABC$.
So $ \angle HOM \plus{} \angle HEM \equal{} 180$
So $ O, H, E, M$ concyclic

I should've been in this competition... but then i prefer preparing for my public exam to sitting for this. But anyway it's the same coz i can solve the problems at home~

Hope anybody can post the rest of the questions.

[size=117][color=darkblue] [b]I know an old problem which is similarly with this proposed problem.[/b] [/color][/size]

[quote="Virgil Nicula"] [color=darkred]Let $ ABC$ be an acute and nonisosceles triangle  . The bisector of $ \widehat {BAC}$ meets $ BC$ at $ D$ . Denote the

midpoints $ M$ , $ N$ , $ P$ of $ [BC]$ , $ [CA]$ , $ [AB]$ respectively and the projections $ \left\|\begin{array}{c}
H\in BC\ ,\ AH\perp BC\\\
R\in AD\ ,\ BR\perp AD\\\
S\in AD\ ,\ CS\perp AD\end{array}\right\|$ .

Prove that $ MRHS$ is cyclically and $ S\in MN$ , $ R\in MP$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] The quadrilateral $ ABHR$ is cyclically $ \implies$ $ \widehat {MHR}\equiv \widehat {BAR}$ . It is well-known that $ AD\equal{}\frac {2bc\cos\frac A2}{b\plus{}c}$

and $ BD\equal{}\frac {ac}{b\plus{}c}$ . Observe that $ AS\equal{}b\cos \frac A2$ and $ \frac {BD}{BM}\equal{}\frac {2c}{b\plus{}c}\equal{}\frac {AD}{AS}$ , i.e. $ MS\parallel AB$ . Thus, $ \boxed {\ S\in MN\ }$ and

$ \widehat {MSR}\equiv \widehat {BAR}$ . In conclusion, $ \widehat {MHR}\equiv\widehat {MSR}$ , i.e. [b]the quadrilateral $ MRHS$ is cyclically[/b]. The quadrilateral $ ACSH$

is cyclically $ \implies$ $ \widehat {MHS}\equiv\widehat {CAS}$ . Therefore, $ \widehat {MRS}\equiv \widehat {MHS}$ $ \implies$ $ \widehat {MRS}\equiv \widehat {CAS}$ , i.e. $ MR\parallel AC$ . Thus, $ \boxed {\ R\in MP\ }$ .[/color]


geometry
3D geometry
counting
distinguishability
geometric transformation
rotation
AMC

So I've basically gone through the group theory chapter of Topics in Algebra by Herstein and done most of the problems, and I have a good understanding of topics like homomorphisms, permutation groups, conjugacy classes, Sylow, etc. However, I realized that I have no clue how one might apply these ideas to solve elementary combinatorics problems.

So, does anyone know any resources that would give examples of applications of group theory to combinatorics problems? Or posts on AoPS? Or an example that you could post here? Anything would be appreciated. Thank you very much!

A Couple Notes:
I recently discovered how to use Burnside's Lemma to solve nice problems, for example, find the number of colorings of a cube that are indistinguishable wrt rotations, or colorings of a necklace that are indistinguishable wrt flipping and rotation. I found that particularly neat. I also liked the use of group theory in USAMO 2008 #6. But I don't know much more than that.

In the context of USAMO 2008 #6 the relevant aspect of "group theory" is really linear algebra (there, over $ \mathbb{F}_2$, which is common in combinatorics) - see the examples given [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=220351]here[/url].  The correspondence between groups where every element has order $ 2$ and vector spaces over $ \mathbb{F}_2$ has a particularly nice interpretation in the notion of a [url=http://en.wikipedia.org/wiki/Boolean_ring]Boolean ring[/url].  Speaking of which,

[b]Exercise:[/b]  Let $ G$ be a finitely-generated group where every element has order $ 2$.  Prove that $ G$ is finite and abelian and therefore isomorphic to a finite-dimensional vector space over $ \mathbb{F}_2$.


geometry circumcircle rectangle TST parabola trapezoid symmetry

Let $ BO_B\cap CO_C \equal{} Y$. We have $ \angle CBO_B \equal{} 90^{\circ} \minus{} \angle DAB$ and $ BCO_C \equal{} 90^{\circ} \minus{} \angle DAC$ hence $ \angle BYC \equal{} 180^{\circ} \minus{} (90^{\circ} \minus{} \angle DAB) \minus{} (BCO_C \equal{} 90^{\circ} \minus{} \angle DAC \equal{} \angle BAC$ which means that point $ Y$ lies on the circumcircle of triangle $ ABC$ (1). A simple angle chase shows that $ \angle BAO_B \equal{} \angle CAO_C$ which implies $ \angle O_BAO_C \equal{} \angle BAC$ and point $ Y$ lies on the circumcircle of triangle $ AO_BO_C$ (2). From (1) and (2) we have $ AY$ - radical axis of circles $ (ABC)$ and $ (AO_BO_C)$, $ O_BO_C$ - radical axis of circles $ (BCO_CO_B)$ and $ (AO_BO_C)$ and $ BC$ - radical axis of $ (ABC)$ and $ (BCO_CO_B)$ and these three lines are concurrent in point $ R$. Let $ AH\cap BC \equal{} T$ and $ O_BO_C\cap AD \equal{} Q$. We know that $ AD$ is radical axis of circles $ (ABD)$ and $ (ACD)$ hence $ O_BO_C\perp QR$ and $ AT\perp RT$ which implies that $ RTQA$ is cyclic quadrilateral and $ \angle DAH \equal{} \angle RAT \equal{} \angle DRQ$ (3). We know that $ AX\perp AY$ (4) (it was posted on the forum, I'll write proof later). A simple angle chase shows that $ RBO_BA$ is cyclic quadrilateral and $ BO_B \equal{} AO_B$ hence $ QR$ is angle bisector of $ \angle DRA$ which combined with $ AD\perp QR$ implies $ AR \equal{} DR$ and $ \angle RDA \equal{} \angle DAR$ (5). Let $ AX\cap QR \equal{} S$. Using (4) and (5) we get $ \angle RDA \equal{} \angle DAR \equal{} \angle RSA$ (because $ \triangle RQA$ and $ \triangle RAS$ are similar) hence $ RDSA$ is cyclic. Using this and (3) we get $ \angle DAX \equal{} \angle DAS \equal{} \angle DRS \equal{} \angle TRQ \equal{} \angle TAQ \equal{} \angle HAD$ qed.

We can also prove that $ BO_C$, $ CO_B$ and $ AX$ are concurrent.

Let $ F \in BC$ be foot of $ AH.$ Let $ A', B', C', D', F'$ be midpoints of $ BC, CA, AB, AD, AF.$ $ O_BO_C$ is perpendicular bisector of $ AD$ $ \Longrightarrow$ $ \angle AO_BO_C \equal{} \angle ABD \equal{} \angle ABC$ and $ \angle AO_CO_B \equal{} \angle ACD \equal{} \angle ACB$ $ \Longrightarrow$ $ \triangle AO_BO_C \sim \triangle ABC \sim \triangle AC'B'$ for any $ D \in BC.$ Let $ K, M$ be midpoints of $ O_BO_C, C'B'$ and $ AD'KL, AF'MN$ rectangles; $ KL, MN$ are perpendicular bisectors of $ O_BO_C, C'B',$ respectively. Since $ D', F'$ are A-altitude feet of $ \triangle AO_BO_C \sim \triangle AC'B',$ these 2 rectangles are similar $ \Longrightarrow$ $ \triangle ALN \sim \triangle AD'F'$ are similar by SAS and $ LN \perp (AN \parallel B'C').$ Consequently, foot $ L$ of perpendicular $ AL$ to $ KL$ from $ A$ is on midparallel $ MN$ of $ AH \parallel OA'.$ This means that the perpendicular bisector $ KL$ of $ O_BO_C$ is tangent to a parabola with focus $ A,$ vertex tangent $ MN$ and directrix $ OA'.$ If $ KL$ cuts the directrix $ OA'$ at $ X,$ the parabola tangent $ KL$ bisects the angle $ \angle OXA.$ Since $ AD \parallel KL$ $ \Longrightarrow$ $ AD$ bisects the angle $ \angle XAH.$
This is true regardless of $ BCO_CO_B$ being cyclic or not. But if it happens to be cyclic, the intersection $ X$ of the perpendicular bisectors $ OA', KL$ of $ BC, O_BO_C$ is its circumcenter.

I will post the solution  later  in mind

Denote X’ is the intersection of B’C’ and ObOc
  ⊿ABOb∽⊿ACOc  => ∠ObBC’=∠OcCB’ => C’OcB’Ob is an isosceles trapezoid
  => ObOc=B’C’
  ⊿AC’B’∽⊿ACB   ⊿AOcOb∽⊿ACB  =>
  B’C’/BC=AB’/AC= ObOc/BC=AOc/AC  => AB’=AOc
 Analogously, we have AOb=AC’
 AX’ is the bisector of ∠C’AOc
On the other hand, we have AX  is the bisector of ∠C’AOc
Thus A,X,X’ are collinear
By angle chasing  ∠A’AB+∠C’AC=90 which indicates A’,O,X’’are collinear
Therefore ∠OX’’A=∠OAX=∠ABA’=∠DAH                 (1)
Notice ∠OAX=∠X’AC’-∠OAB=∠X’Oc-∠OcAD=∠DAX       (2)
Form (1)(2) we get the result

[quote="plane geometry"]Denote $ X'$ is the intersection of $ B'C'$ and $ O_bO_c$
  $ \triangle ABO_b\sim\triangle ACOc \Rightarrow \angle O_bBC' = \angle O_cCB' \Rightarrow C'O_cB'O_b$is an isosceles trapezoid
  $ \Rightarrow O_bO_c = B'C'$
 ${ \triangle AC'B'\sim \triangle ACB \triangle AO_cO_b\sim \triangle ACB \Rightarrow \frac {B'C'}{BC} = \frac {AB}{AC} = \frac {O_bO_c}{BC} = \frac {AO_c}{AC } } \Rightarrow AB = AO_c$
 Analogously, we have $ AO_b = AC'$
 $ AX'$ is the bisector of $ \angle C'AO_c$
On the other hand, we have $ AX$  is the bisector of $ \angle C'AO_c$
Thus $ A,X,X'$ are collinear
By angle chasing  $ \angle A'AB + \angle C'AC = 90^0$ which indicates $ A',O,X''$ are collinear
Therefore$ \angle OX''A + \angle OAX + \angle ABA' + \angle DAH (1)$
Notice $ \angle OAX = \angle X'AC' - \angle OAB = \angle X'AO_c - \angle O_cAD = \angle DAX (2)$
From$ (1)(2)$ we get the result[/quote]
Very nice solution! Dear plane geometry

[quote="MellowMelon"]Let $ ABC$ be an acute triangle.  Point $ D$ lies on side $ BC$.  Let $ O_B, O_C$ be the circumcenters of triangles $ ABD$ and $ ACD$, respectively.  Suppose that the points $ B, C, O_B, O_C$ lies on a circle centered at $ X$.  Let $ H$ be the orthocenter of triangle $ ABC$.  Prove that $ \angle{DAX} \equal{} \angle{DAH}$.

[i]Zuming Feng.[/i][/quote]
Dear mathlinkers! How can we construct $ D$ on $ BC$ such that $ O_B,O_C,B,C$ are on a circle where $ O_B, O_C$ is the circumcenters of triangles $ ABD$ and $ ACD$, respectively?

[quote="mathVNpro"]Dear mathlinkers! How can we construct $ D$ on $ BC$ such that $ O_B,O_C,B,C$ are on a circle where $ O_B, O_C$ is the circumcenters of triangles $ ABD$ and $ ACD,$ respectively?[/quote]
The point $ D$ is not constructible (in general). Calculate the angles $ \angle CAD$ and $ \angle BAD$ in terms of the angles of $ \triangle ABC,$ then realize that the construction is equivalent to the angle trisection.

[quote="Luis González"]...the construction is equivalent to the angle trisection.[/quote]
Calulate angle $ \theta \equal{} \angle DAH$:
$ \angle O_BBC \equal{} \theta \plus{} \angle ABC$ and $ \angle BCO_C \equal{} \angle BCA \minus{} \theta ,$
$ \angle O_CO_BB \equal{} \pi \minus{} (\angle ABC \plus{} 2 \theta )$ and $ \angle CO_CO_B \equal{} \pi \minus{} (\angle BCA \minus{} 2 \theta).$
$ BCO_CO_B$ is cyclic $ \Longleftrightarrow$ $ \angle O_BBC \plus{} \angle CO_CO_B \equal{} \pi$ $ \Longleftrightarrow$ $ \theta \equal{} \frac {_1}{^3}(\angle BCA \minus{} \angle ABC).$

Even though $ D$ is not constructible for a given $ \triangle ABC,$ you can still construct some $ \triangle ABC$ with $ BCO_CO_B$ cyclic, given $ \triangle ABD.$ (If $ F \in BD$ is A-altitude foot, pick $ D,$ so that $ \angle ABD \plus{} 3 \angle DAF < \frac{_{\pi}}{^2}.$)

We know that $ AX$ is the bisector of $ \angle BAO_c$ and $ \Delta AO_bO_c\sim \Delta ABC$
On the other side, denote $ T$ the intersection of $ BO_c$ and $ CO_b$.
We have $ \frac {O_bT}{O_cT} \equal{} \frac {BO_b}{CO_c}$
$ \Rightarrow \frac {O_bT}{TC} \equal{} \frac {R_b}{R_c}.\frac {O_cT}{TC} \equal{} \frac {R_b}{R_c}.\frac {O_bO_c}{BC}\equal{} \frac {R_b}{AC} \equal{} \frac {AO_b}{AC}$
So $ AT$ is the bisector of $ \angle O_bAC$ or $ \angle BAO_c$
We obtain $ AX, BO_b, CO_c$ are concurrent.
Applying Tri-Ceva theorem for triangle $ O_bO_cX$ we have:
$ \frac {sin \angle O_cXT}{sin \angle O_bXT}.\frac {sin \angle XO_bT}{sin \angle O_cO_bT}.\frac {sin \angle O_bO_cT}{sin \angle XO_cT} \equal{} 1$
Note that $ XO_b \equal{} XO_c$ so $ \frac {sin \angle O_cXT}{sin \angle O_bXT} \equal{} \frac {PO_c}{PO_b}$
And $ \frac {sin \angle XO_bT}{sin \angle O_cO_bT}.\frac {sin \angle O_bO_cT}{sin \angle XO_cT} \equal{} \frac {sin \angle TCB}{sin \angle TBC}.\frac {cos \angle O_bBD}{cos \angle O_cCD} \equal{} \frac {TB}{TC}.\frac {cos \angle O_bBD}{cos \angle O_cCD}$
$ \equal{} \frac {R_b}{R_c}.\frac {cos \angle O_bBD}{cos \angle O_cCD} \equal{} \frac {BD}{CD}$
Therefore $ \frac {PO_c}{PO_b} \equal{} \frac {CD}{BD}$
We get $ \Delta ABD\sim \Delta AO_bP$ then $ \angle BAD \equal{} \angle O_bAP \Rightarrow \angle DAX \equal{} \angle BAO_b \equal{} \angle DAH$

[quote]Let $ ABC$ be an acute triangle.  Point $ D$ lies on side $ BC$.  Let $ O_B, O_C$ be the circumcenters of triangles $ ABD$ and $ ACD$, respectively.  Suppose that the points $ B, C, O_B, O_C$ lies on a circle centered at $ X$.  Let $ H$ be the orthocenter of triangle $ ABC$.  Prove that $ \angle{DAX} \equal{} \angle{DAH}$.

[i]Zuming Feng.[/i][/quote]

We have $ \angle AO_BB \equal{} 2(180^{\circ} \minus{} \angle ADB) \equal{} 2\angle ADC \equal{} \angle AO_CC$. In the other hand, since $ \triangle AO_BB$ and $ \triangle AO_CC$ are all issoceles triangles respectively with respect to vertexes $ O_B,O_C$. Hence it exists a spiral similarity through center $ A$, let denote this transformation by $ f_A$ such that $ f_A:$ $ \triangle AO_BB\mapsto \triangle AO_CC$ $ (*)$. Thus $ f_A:$ $ O_B\mapsto O_C$ and $ B\mapsto C$. Therefore, it also exists a spiral similarity through $ A$, let denote this by $ g_a$ such that $ g_A:$ $ O_b\mapsto B$, $ O_c\mapsto C$ $ \Longrightarrow$ $ O_BO_C\mapsto BC$, which implies that $ \triangle AO_BO_C\mapsto \triangle ABC$ $ (**)$. From $ (*)$ and $ (**)$, we have $ (O_BO_C,BC) \equal{} (AO_c,AC)$ and $ (O_BB,O_CC) \equal{} (AB,AC)$. Thus, if $ P\equiv BO_B\cap CO_C$ and $ Q\equiv O_BO_C\cap BC$ then we have $ \{A,O_C,C,Q\}$ and $ \{P,B,C,A\}$ are two sets of concyclic points, which leads to the fact that $ A$ is the [i]Miquel point[/i] with respect to the complete quadrilateral $ (PB,PC,QO_B,QB)$. Note that $ OO_BO_CC$ is concyclic $ \Longrightarrow$ $ A,P,Q$ are collinear. Further, if $ G\equiv CO_B\cap BO_C$ then $ PQ$ is the polar of $ G$ with respect to $ (O_A)\equiv (BO_BO_CC)$, $ O,A,G$ are collinear and $ \overline {OGA}\perp PQ$ at $ A$ $ \Longrightarrow$ $ \angle PAO_A$ $ \equal{} \angle O_AAQ$ $ \equal{} 90^{\circ}$.

In the other hand, we have $ \angle DAH_A \equal{} \angle DAC \minus{} \angle H_AAC \equal{} \angle DAC \minus{} (90^{\circ} \minus{} \angle ACD)$. Also, $ \angle DAO_A \equal{} \angle PAD \minus{} \angle PAO_A$ $ \equal{} \angle PAD \minus{} 90^{\circ}$. Therefore, in order to prove that $ \angle DAH_A \equal{} \angle DAO_A$, we need to prove that $ \angle PAD \equal{} \angle DAC \plus{} \angle ACD$ $ \equal{} 180^{\circ} \minus{} \angle ADQ$ $ \equal{} 180^{\circ} \minus{} \angle QAD$, which equivalent to $ \angle ADQ \equal{} \angle QAD$, which is obviously true since $ \overline {O_BO_CQ}$ bisects $ AD$ and $ Q\equiv PA\cap O_BO_C$. Our proof is completed then. $ \square$

WLOG $B\ge C$.

First, by trivial angle chasing, $\angle{BO_B O_C}=B+2\angle{BAD}$ while $\angle{O_B BC}=90^\circ-\angle{BAD}$, so because $BO_B O_C C$ is cyclic, $\angle{BAD}=90^\circ-\frac{2B+C}{3}$ and $\angle{CAD}=90^\circ-\frac{B+2C}{3}$. We find that $\angle{DAH}=\angle{DAB}-\angle{HAB}=\frac{B-C}{3}$, so it remains to show that $\angle{DAX}=\frac{B-C}{3}$. (*)

Note that $\angle{O_B XC}=2\angle{O_B BC}=\frac{4B+2C}{3}$ and $\angle{O_C XB}=2\angle{O_C CB}=\frac{2B+4C}{3}$.

However, $\angle{O_B AC}=\angle{O_B AD}+\angle{CAD}=180^\circ-\frac{4B+2C}{3}$, so by symmetry, $ABXO_C$ and $ACXO_B$ are cyclic.

Finally,
\begin{align*}
\angle{DAX}+\angle{DAB} = \angle{XAB} &= \angle{XO_C B} \\
&= \angle{BO_B O_C}-90^\circ = 90^\circ-\frac{B+2C}{3} = \angle{DAB}+\frac{B-C}{3}, \\
\end{align*}so we're done by (*).

Suppose radius of $\odot{BAD}=r_1$ and of $\odot{CAD}=r_2$ and $\angle{DAB}=\theta$ now so $AD=2r_1Sin B=2r_2 Sinc$ also $\angle{O_BDO_C}=\pi-(B+C)=A$ and that implies $O_BDO_C$ is similar to $ABC$ and so $\angle{O_CO_BD}=B,\angle{O_BO_CD}=C$ now also note $B+2\theta+C+\frac{B+\theta}{2}=\pi \implies \theta=\frac{\pi}{2}-\frac {2B+C}{3}$.Now suppose $\angle{XBC}=\alpha$ then $aSin(\alpha)=r_1Sin(\frac {2B+C}{3}-\alpha)$ also $c=2r_1Cos(\frac{B-C}{3})$ , using this we get $\frac{Sin(B-\alpha)}{Sin(C-\alpha)}=\frac{Cos(A-B)}{Cos(A-C)}$ now apply ceva with $AX,BX,CX$ and note $\frac {Sin\angle{XAC}}{Sin\angle{XAB}}=\frac{Cos(A-B)}{Cos(A-C)}=\frac{Cos(\frac{C+2B}{3})}{Cos(\frac{B+2C}{3})}$ and so obviously $\angle{XAB}=\frac{\pi}{2}-\frac{B+2C}{3},\angle{XAC}=\frac{\pi}{2}-\frac{2B+C}{3}$ ,and hence $\angle{XAD}=\frac{B-C}{3}$ and which is indeed $\angle{DAH}$ ,so done!

Yes; see [url=http://derpy.me/cyclicquad]derpy.me/cyclicquad[/url] (Fact 12; thanks to Zhero for making this link)

Edit: made link clearer; also http://yufeizhao.com/olympiad/cyclic_quad.pdf

[color=#00CCA7]Linked the other one as well. -- mod[/color]

The link isn't working .Paste it in URL Form

Note that $AO_BO_C \sim ABC$ by simple angle chasing, so let $Y = O_BO_C \cap BC$ and $Z = AD \cap O_BO_C$. Notice that by miquel's theorem, we have $Y \in ABO_B$ and $ACO_C$ so since $O_BO_CBC$ are concyclic, we have by radical axis theorem that $BO_B, CO_C, AY$ concur, so $A$ is miquel point of $O_BO_CCB$. Hence, $XA \perp AY$. If we let the midpoint of $O_BO_C$ be $M$ then $\angle DAX = \angle AOM = \angle AYM$ and clearly $\angle HAD = \angle MYD$ (Let $AH \cap BC = E$, then $AYEM$ are concyclic). So, we must show $YM$ bisects $\angle AYE$, or $AM = MQ$ which follows since $M$ is the circumcentre of $AED$.

Sorry for a little glitch in the diagram ($AHG \perp BC, G = AH\cap BC$).
Let $BO_B \cap CO_C = Y$.
We have $AO_BO_CY, ABCY, BO_BO_CC$ concylic by a little angle chasing, and thus $A$ is the miquel point of cyclic quadrilateral $BO_BO_CC$ (by http://yufeizhao.com/olympiad/cyclic_quad.pdf )
So, we have $XA \perp YA$.
Now, let $BC \cap O_CO_B = I$. Then, $AO_BBI$ is cyclic, and $AO_B = BO_B$ implies line $IO_BO_C$ is the bisector of $\angle AIB$.
As $AH \perp IB, AD \perp IO_B$ and $AX \perp IA$, $AD$ is the bisector of $\angle HAD$, and thus $\angle DAX = \angle  DAH$.

Let $J$ be the circumcenter of $BO_BO_CC$ and $O$ be the circumcenter of $\triangle{ABC}$.Note that $\angle{AO_BO_C}=\angle{AOO_C}=B$ so $A,O_B,O,O_C$ lie on a circle.Also by some angle chasing we obtain that $BO_BO_CC$ is cyclic yeilds that $\angle{ADC}=90+\frac{B-C}{3}$.Next note that $BO_B \cap CO_C=K$ lie on the circumcircle of $ABC$ as well as $AO_BO_C$.Let $r_1,r_2,R$ denote the circumradii of $AO_BO_C,BO_BO_CC$ and $ABC$ respectively.

$\angle{BOJ}=\angle{BKO_C}=A$ and $\angle{OBJ}=\angle{OBC}-\angle{JBC}=\angle{BO_CC}-\angle{BKO_C}=\angle{KBO_C}$.Thus $\triangle{OBJ} \sim \triangle{KBO_C} \implies \frac{OJ}{R}=\frac{KO_C}{KB}=\frac{O_BO_C}{BC}=\frac{r_1}{R} \implies OJ=r_1=\frac{R}{2sin\angle{ADC}}=\frac{R}{2cos\frac{(B-C)}{3}}$.
Extend $AD$ and $OJ$ so that they meet at $X$.We now get

$\frac{AO}{AX}=\frac{OJ}{JX}$
$\Leftrightarrow \frac{OX}{OJ}=\frac{AX}{AO}+1$
$\Leftrightarrow \frac{2cos\frac{(B-C)}{3} \cdot R}{\frac{R}{2cos\frac{(B-C)}{3}}}=\frac{sin(B-C)}{sin\frac{(B-C)}{3}}+1$
$\Leftrightarrow 4cos^2\frac{(B-C)}{3}=4-4sin^2\frac{(B-C)}{3}$
$\Leftrightarrow cos^2\frac{(B-C)}{3}+sin^2\frac{(B-C)}{3}=1$

which is an obvious identity.Therefore $AJ$ bisects $\angle{DAO}$ and consequently $\angle{DAJ}=\angle{DAH}=\frac{B-C}{3}$.

Notice that $\triangle AO_BB$ and $\triangle AO_CC$ are both isoceles. Furthermore, since $O_B$ and $O_C$ are circumcenters, we know that $\measuredangle AO_BB = 2\measuredangle ADB = 2\measuredangle ADC = \measuredangle AO_CC$ where the angles are directed. It follows that $\triangle AO_BB \sim \triangle AO_CC$ with similar orientation. Therefore, $A$ is the center of spiral similarity that sends $\overline{O_BB}$ to $\overline{O_CC}$, so $A$ is the Miquel point of cyclic quadrilateral $BO_BO_CC.$ Hence, it is well-known (and can be proven using harmonic divisions) that $A$ is the inverse of $Y \equiv BO_C \cap CO_B$ with respect to $\odot (BO_BO_CC).$ Then since $C, Y, O_B$ are collinear, under inversion in $\odot (BO_BO_CC)$, it follows that $X, O_B, A, C$ are concyclic. Then because $O_B$ is the circumcenter of $\triangle ABD$ and the points $X, O_B, A, C$ are concyclic, we have \[\measuredangle DAX = \measuredangle O_BAX - \measuredangle O_BAD = \measuredangle O_BCX - \left(90^{\circ} - \measuredangle DBA\right) = \measuredangle O_BCX + \measuredangle DBA - 90^{\circ}.\] Meanwhile, since $AH \perp BC$, we have $\measuredangle HAD = 90^{\circ} - \measuredangle ADB.$ Therefore, \[\measuredangle DAX = \measuredangle HAD \iff \measuredangle O_BCX + \measuredangle DBA + \measuredangle ADB= 180^{\circ} = 0.\] Since the sum of the directed angles in $\triangle ABD$ is equal to $0$, we need only show that $\measuredangle O_BCX = \measuredangle BAD.$ But because $X$ is the circumcenter of $\odot(BO_BO_CC)$, we have \[\measuredangle O_BCX = 90^{\circ} - \measuredangle CBO_B = 90^{\circ} - \measuredangle DBO_B = \measuredangle BAD.\] $\square$

Let $W$ be the circumcircle of triangle $ABC$.and let $W_1$be the circumcircle of triangle $AO_bO_c$.and let $W_2$ be the circumcircle of $BO_bO_cC$.and let $T=W \cap W_1$ By the known lemma we know that lines $AT,O_bO_c,BC$ are concurrent. Let $M$ be their interestion.
\[\measuredangle O_bBA = \measuredangle O_cCA\].
So we find that $T=O_bB \cap O_cC$ so we find that $A$ is the Miquel point of $BC(OB)(OC)$ .
By the known lemma if $K$ be the circumcenter of $BCO_bO_c$ we find that
$MA \perp KA$
By angle casing we find that $MAO_cC$ is cyclice.
Now we know that angles$KAD=AMO_c=ACO_c=ABO_b=DAH$

We have that $\angle AO_BO_C = \frac{1}{2}\angle AO_BD = \angle B$ and likewise $\angle AO_CO_B=\angle C$. So $A$ is the center of spiral similarity sending $AO_BO_C$ to $ABC$ and is hence the Miquel point of $BCO_CO_B$. It is well known that $AO_BXC$ is cyclic. So $$\angle XAC = \angle XO_BC = 90-\angle O_BBC = \angle BAD \implies \angle XAD = \angle DAC - \angle BAD.$$ Also, $\angle BAH = 90 - \angle B = 2\angle BAD - \angle DAC$, which follows from $\angle BO_BC = \angle O_CCB$. Therefore, $\angle DAH = \angle DAC - \angle BAD$ and the result follows. $\Box$

WLOG, let $AC>AB$. Let $L$ be the midpoint of $\overline{AD},$ and set $Y=\overline{BO_B} \cap \overline{CO_C}, Z=\overline{BC} \cap \overline{O_BO_C}.$

Note that $\triangle AO_BB \sim \triangle AO_CC,$ so $A$ is the spiral center for $\overline{O_BO_C} \mapsto \overline{BC}.$ Consequently, $\overline{YZ} \perp \overline{AX}$. From $\angle ALZ=90^{\circ}$ we conclude that $\overline{AX}$ is tangent to the circle $\odot (AZL).$ Finally, we have $$\angle DAH=\angle DZL=\angle AZL=\angle DAX,$$ establishing the result.

[quote=BBAI][quote="v_Enhance"]Without loss of generality $AC > AB$.  It is easy to verify via angle chasing that $\angle AO_BB = AO_CC$.  Since $O_BO_CCB$ is cyclic, it follows that $A$ is the Miquel point of $O_BO_CCB$.  Therefore, $AO_CXB$ is cyclic.

Set $x = \angle BAD$, $y = \angle CAD$.  Then \[
	\angle BO_BO_C = \angle BO_BD + \angle D_OBC = 2x+B \implies \angle BXC = 360 - 4x - 2B \implies \angle BAX = \angle BO_CX = 2x+B-90.
\] On the other hand, $\angle BAH = 90 - B$. From here it is easy to derive that $\angle HAD = x+B-90 = \angle XAD$, as desired.[/quote]
How $AO_cXB$ is cyclic? Is it a property..[/quote]

Yes it is a property. For cyclic quadrilateral $ABCD$, circles $(AOC)$ and $(BOD)$ concur at the Miquel Point of the quadrilateral, where $O$ is the circumcenter of $(ABCD)$.

[quote=anantmudgal09]WLOG, let $AC>AB$. Let $L$ be the midpoint of $\overline{AD},$ and set $Y=\overline{BO_B} \cap \overline{CO_C}, Z=\overline{BC} \cap \overline{O_BO_C}.$

Note that $\triangle AO_BB \sim \triangle AO_CC,$ so $A$ is the spiral center for $\overline{O_BO_C} \rightarrow  \overline{BC}.$ Consequently, $\overline{YZ} \perp \overline{AX}$. From $\angle ALZ=90^{\circ}$ we conclude that $\overline{AX}$ is tangent to the circle $\odot (AZL).$ Finally, we have $$\angle DAH=\angle DZL=\angle AZL=\angle DAX,$$ establishing the result.[/quote]

This is a great solution. For anyone stuck on the angle chase(it makes a few jumps) here's a more detailed version.

Let $F$ be the foot of the perpendicular from $A$ to $BC$. Observe that since $angle{AFZ}=90$, $F$ lies on $(AZL)$. Now we have $$\angle{DAH}=\angle{LAF}=\angle{LZD}=\angle{DZL}=\angle{AZL}=\angle{DAX}$$.

Really Cute and Easy Problem. :D
[quote=USATST 2009 P2]Let $ ABC$ be an acute triangle.  Point $ D$ lies on side $ BC$.  Let $ O_B, O_C$ be the circumcenters of triangles $ ABD$ and $ ACD$, respectively.  Suppose that the points $ B, C, O_B, O_C$ lies on a circle centered at $ X$.  Let $ H$ be the orthocenter of triangle $ ABC$.  Prove that $ \angle{DAX} \equal{} \angle{DAH}$.

[i]Zuming Feng.[/i][/quote]

$\angle AO_BB=2\angle ADB=2(180^\circ-\angle ADC)=\angle AO_CC$, and as $AO_BB$ and $AO_CC$ are isosceles. Hence, there $\exists$ a Spiral Similarity$(\sigma)$  centered at $A$. $\sigma:BO_B\mapsto CO_C\implies\sigma:O_BO_C\mapsto BC$. So, $L=BO_B\cap CO_C=\{\odot(AO_BO_C)\}\cap\{\odot(ABC)\}$. So, by taking Paiwise Radical Axis of $\{\odot(ABC),\odot(BO_BCO_C),\odot(AO_BO_C)\}$ we get that $\{AL,O_BO_C,BC\}$ are concurrent at a point $K$. Now notice that $A$ is the Miquel Point of the cyclic quad $\odot(BO_BO_CC)$. Hence, $\{AX,BO_C,CO_B\}$ are concurrent and also $AX\perp AK$. So, $\angle DAH=\angle DKO_C=\angle O_CKA=\angle DAX\qquad\blacksquare$

woah cool problem

Note that $A$ is the Miquel point of $BO_BO_CC$. Let $M$ be the midpoint of $AD$ and $T$ be the intersection of $O_BO_C$ and $BC$. Let $K$ be the foot of the perpendicular from $A$ to $BC$. Observe that since $M$ is the midpoint of a chord in both $(ABD)$ and $(ACD)$, we have that $\angle TMA = \angle O_BMA = 90$, so $M$ lies on $(AKT)$. Furthermore, note that by a well-known property of Miquel points, $\angle XAT = 90$, so $AX$ is a tangent to $(AMKT)$. Now, we angle chase to finish: $$\angle XAD = \angle XAM = \angle ATM = \angle AKM = \angle KAM = \angle DAH.$$

By the USAMO 2013/6 lemma, $\triangle AO_BO_C\sim\triangle ABC$. Thus $A$ is the Miquel point of $O_BO_CCB$. Let $K=O_BO_C\cap BC$. We need to show that lines $AH,AX$ are isogonal in $\triangle ADK$. Observe that $AX$ is perpendicular to $AK$ by properties of Miquel points and $AH$ is perpendicular to line $DK$. Thus $\angle XAK=90^\circ,\angle HAK=90^\circ-\angle AKD$. We need to show $\angle DAK=90^\circ-\tfrac 12\angle AKD$. But this is immediate by cyclic quad $(KCO_CA)$ from the Miquel configuration and properties of the circumcenter of $\triangle ACD$, thus we are done.

We solve the configuration where $ABD$ is acute and $ACD$ is obtuse; other cases are analogous. Thus angle $B > C$. By angle chasing we get $AO_B O_C, ABC$ are directly similar. Therefore $A$ is the center of the spiral similarity sending $O_B O_C$ to $BC$, thus it is the Miguel Point of $BO_B O_C C$. Let $E = BO_B \cap CO_C$, so $AEO_C O_B, AECB$ are cyclic. Let $\angle ADB = \theta$, so it suffices to show $\angle DAX = 90-\theta$. Remark $AO_B O_C, DO_B O_C$ are congruent. Now angle chasing we get $\angle EBC = B+\theta-90, \angle O_B O_C C = 2\theta - C$. So $\theta = 90.+ \frac{C-B}{3}$. Now this yields $\angle O_B AE = \angle CO_C O_B = 180 - \frac{C}{3}-\frac{2B}{3}$. By Brokard $\angle XAE = 90$, so $\angle O_B AX = 90 - \frac{C}{3}-\frac{2B}{3}$. We can readily find $\angle BAO_B = \frac{B-C}{3}$. So $\angle BAX = 90 - \frac{B}{3}-\frac{2C}{3}=270-B-2\theta$. Hence $\angle DAX = \angle BAX - (180-B-\theta)=90-\theta$. This completes the proof.

Note that $AO_B=BO_B$, $AO_C=CO_C$, and $\measuredangle AO_BB=2\measuredangle ADB=2\measuredangle ADC=\measuredangle AO_CC$. Hence $\triangle AO_BB \sim \triangle AO_CC$, which means that $A$ is the Miquel Point of $BCO_CO_B$.

Let $P$ be the intersection of $BC$ and $O_BO_C$ and $Q$ be the foot of the perpendicular from $A$ to $BC$. Since $AD$ is the radical axis of $(ABD)$ and $(ACD)$, $\overline{O_CO_BP}$ is the perpendicular bisector of $AD$. It follows that $\triangle ADP$ is isosceles and $\measuredangle DAP=\measuredangle PDA$. Moreover, $XA$ and $AP$ are perpendicular since the points $ B, C, O_B, O_C$ lies on a circle. Now, $\measuredangle DAX=90-\measuredangle PAD=90-\measuredangle ADP=\measuredangle QAD=\measuredangle HAD$.

Let $E = BO_B \cap CO_C$, $T = BC \cap O_BO_C$, and $F$ be the foot of the $A$-altitude.

It's well-known that the Miquel Point of complete quadrilateral $BCO_CO_B$ is $A$. Since $BCO_CO_B$ is cyclic, we know $A \in ET$ and $XA \perp ET$.

[b][color=#f00]Claim:[/color][/b] $TAD$ is isosceles.

[i]Proof.[/i] Notice $O_BO_C$, which is the perpendicular bisector of $AD$, passes through $T$. $\square$

Now, $$\angle DAX = 90^{\circ} - \angle TAD = 90^{\circ} - \angle TDA = 90^{\circ} - \angle FDA = \angle DAF = \angle DAH$$ as desired. $\blacksquare$

[b]Remarks:[/b] I made all the "obvious" observations before getting stuck. In the end, using the fact that $BCO_CO_B$ is cyclic effectively was key. (Instead of angle chasing or Radical Axes, we need to recall the relationship between Brocard's and Miquel Points of cyclic quadrilaterals.)

[asy]import geometry; size(10cm); pen c1=orange+white+white; pen c2=orange+white; pen c3=brown+red+white; pair A,B,C,D,H,O_B,O_C,E,F,X; A=(-2,5); B=(-3,-3); C=(7,-3); D=(0,-3); H=orthocenter(A,B,C); O_B=circumcenter(A,B,D); O_C=circumcenter(A,C,D); E=extension(B,O_B,C,O_C); F=extension(B,C,O_B,O_C); X=circumcenter(B,C,O_B); fill(B--O_B--O_C--C--cycle,c1); draw(B--O_B--O_C--C--cycle,c2); draw(B--F,c2); draw(O_B--F,c2); draw(O_B--E,c2); draw(O_C--E,c2); draw(A--H); draw(A--O_B); draw(A--D); draw(A--X); draw(A--O_C); draw(E--F,dotted); draw(circle(B,C,O_B),c2); draw(circle(A,B,D)); draw(circle(A,C,D)); draw(A--B,c3); draw(A--C,c3); draw(arc(intersectionpoint(bisector(A,C),bisector(A,F)),C,F),dotted); dot(A^^B^^C^^D^^H^^O_B^^O_C^^E^^F^^X); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,S); label("$H$",H,0.6S); label("$O_B$",O_B,SE); label("$O_C$",O_C,NE); label("$E$",E,NE); label("$F$",F,SW); label("$X$",X,NE);[/asy]

WLOG let $AB<AC.$ Let $E=\overline{BO_B}\cap\overline{CO_C}$ and $F=\overline{BC}\cap\overline{O_BO_C}.$ We see that $$\measuredangle AO_BB=2\measuredangle ADB=2\measuredangle ADC=\measuredangle AO_CC,$$ $AO_B=BO_B,$ and $AO_C=CO_C,$ so $\triangle AO_BO_C\sim\triangle ABC.$ Hence, the Miquel point of $BO_BO_CC$ is $A,$ and since $BO_BO_CC$ is cyclic, $A$ is the foot from $X$ to $\overline{EF}.$ Notice that \begin{align*}90-\angle DAX&=\angle FAX\\&=180-\angle DFA-\angle ADF\\&=\angle AO_CC+\angle CDA-180\\&=2(180-\angle CDA)+\angle CDA-180\\&=180-\angle CDA\\&=\angle ADB\\&=90-\angle HAD.\end{align*} $\square$

[asy]
import olympiad;
size(230);
/** Coordinates from Geogebra **/
pair A = (-0.36 , 4.12);
pair B = (-3 , -2);
pair C = (4.64 , -1.82);
draw(A--B--C--cycle , green);
pair D = (0.27 , -1.92);
draw(A--D);
draw(circumcircle(A,B,D) , purple);
pair O_B = circumcenter(A,B,D);
draw(circumcircle(A,C,D) , heavycyan);
pair O_C = circumcenter(A,C,D);
draw(circumcircle(B , O_C , C) , orange);
pair X = circumcenter(B , O_C , C);
pair H = orthocenter(A,B,C);
draw(circumcircle(A,B,C) , blue);
pair O = circumcenter(A,B,C);

/** Dots and Labels **/
dot("$A$" , A , N);
dot("$B$" , B , SW);
dot("$C$" , C , SE);
dot("$D$" , D , S);
dot("$X$" , X , S);
dot("$H$" , H , S);
dot("$O$" , O , NE);
dot("$O_B$" , O_B , NW);
dot("$O_C$" , O_C , NE);
[/asy]

Simple angle chasing gives us $AO_BO_C \sim ABC$. Now , let $L = O_BO_C \cap BC$ and $M = AD \cap O_BO_C$. Miquels Theorem gives us $Y \in ABO_B$ and $ACO_C$.
[b]Claim:[/b] $BO_B , CO_C$ and $AL$ concur.
[i]Proof:[/i] Since we know that $O_BO_CBC$ are concyclic , we can use the radical axis theorem which completes the proof. This also implies that $A$ is the miquel point of $O_BO_CCB$ $\blacksquare$
 Also $XA \perp AL$.
Let $G$ be the midpoint of $O_BO_C$ , then we have the following $$\angle DAX = \angle AOM = \angle ALM$$ Also $\angle HAD = \angle GLD$. What remains to show is that $LG$ bisects $\angle ALE$ which is true since $G$ is the circumcenter of $\triangle AED$.
$\blacksquare$

Let $AB<AC$, $F=BC\cap O_BO_C$. Notice that the point $A$ is the miquel point of the cyclic quadrilateral $BCO_CO_B$. So $XA\perp AF$. Since the line $O_BO_C$ is the perpendicular bisector of $AD$ therefore $\triangle AFD$ is isosceles. So $$\angle XAD=90^\circ-\angle DAF=90^\circ-\angle ADF=\angle DAH$$.

This solution is longer than most other solutions, but hopefully someone will find it helpful. The intuition of simplifying geometry problems works really well here.

Angle chase to obtain that we actually want to prove that $\overline{AD}$ and $\overline{AX}$ are isogonal. Take a $\sqrt{bc}$ inversion to get the following problem:

[quote=I lost the game]Let $D$ be a point on the circumcircle of triangle $ABC$. Let $A_1$ and $A_2$ be the reflections of $A$ over $\overline{BD}$ and $\overline{CD}$, respectively. Prove that if $A_1BCA_2$ has circumcenter $O$, then $\overline{AD}$ and $\overline{AO}$ are isogonal.[/quote]

The crux of the simplifications is this: in fact, we can drop the condition that $A_1BCA_2$ is cyclic, and redefine $O$ as the intersection of the perpendicular bisectors of $\overline{BC}$ and $\overline{A_1A_2}$. Once we see that, the following two claims aren't hard to find.

[b][color=#07f]Claim 1:[/color][/b] The perpendicular bisector of $\overline{A_1A_2}$ passes through the reflection $A'$ of $A$ over the perpendicular bisector of $\overline{BC}$.
[i]Proof:[/i] Notice that $DA=DA_1=DA_2$, so $D$ is on the perpendicular bisector of $\overline{A_1A_2}$. Then, proceed with angle chasing.

[b][color=#07f]Claim 2:[/color][/b] The line isogonal to $\overline{AD}$ and the perpendicular bisector of $\overline{A_1A_2}$ are symmetric about the perpendicular bisector of $\overline{BC}$.
[i]Proof:[/i] Notice that one line passes through $A$, and the other passes through $A'$. Then, proceed with angle chasing.

By claim 2, the line isogonal to $\overline{AD}$ and the perpendicular bisector of $\overline{A_1A_2}$ must intersect on the perpendicular bisector of $\overline{BC}$, so we are done.

Let $P = O_BO_C \cap BC$, which lies on the perpendicular bisector of $AD$. Then $A$ is the Miquel point of cyclic quadrilateral $O_BO_CCB$, so
\[\measuredangle DAH = \measuredangle O_BPD = \measuredangle APO_B = \measuredangle XAD. \quad \blacksquare\]


quadratics
search
LaTeX
algebra
polynomial
ARML
quadratic formula

The quadratic formula is pretty long. Is there a formula for the roots of a equation in the form $ax^3 + bx^2 + cx + d = 0$? It's even nastier I think...

You can search for "cubic formula" in this forum or Google. I found a copy that's in LaTeX and it's attached.

This is just one root. I couldn't find GIFs of the formulas for the others. However, if you want all three roots and are willing to sort through countless non-LaTeX expressions, I can give that to you too.

EDIT: Oh yeah, other people have asked about this one too, so you can find their responses under other threads. Someone's asked "Dr. Math", and he gave the one that isn't in LaTeX but has all three roots that I can show you if you REALLY want. And there is actually a formula for ax^4 + bx^3 + cx^2 + dx + e, but it is predictably even nastier. But some guy proved that there is no formula for fifth-degree equations.

But that's only when the coefficient of $x^3$ is 1, right?
Is there a easier way to solve these types of equations? How do you factor these types of equations?

[quote="nat mc"]You can easily make the coefficient 1 by dividing by that number that is the coefficient of x^3[/quote]

You have a point there :P

Lol, thats what you do when you prove the quadratic formula

Once you get one root of the equation, you divide $ax^3+bx^2+cx+d$ by $x-r$ with r being the root. This will get you $ax^2+(b+ar)x+(c+br+ar^2)$ with remainder $ar^3+br^2+cr+d$ which equals 0. If you don't see why then you should look at the original equation again. Thus, you've created a quadratic which can easily be solved.

That is what I have my TI-84 to do, although I did the polynomial division by hand (it's hard to program).

edit: oh, and the first formula posted does require complex numbers, even if it doesn't appear to. It's obvious in the second.

btw, the quadratic formula is a short as heck. compare $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ to the ones posted for the cubic.

If the stuff under the other radical is also complex, then you might get real solutions.

that is true, but you still need the complex numbers in the intermediate calculations.

Just to add a bit more content to this post, once you've found $r$, the first root, the other two are the roots of $ax^2+(b+ar)x+(c+br+ar^2)$ which are found by:

$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\frac{-b-ar\pm\sqrt{b^2 + 2abr + a^2r^2 - 4ac - 4abr - 4a^2r^2}}{2a}$ which is all that you can simplify without plugging in.

Hmmm.... I say just type it into your calculator  :P You can use the Equation Solver, the PolySmlt App, or even just graph it!

In my book, the quadratic formula proof is first dividing the equation by a. Then, it starts to complete the square, factor, square root, and other things. Does anyone here supply a easy way to prove the quadratic formula?

I can post the proof in my textbook, Advanced Algebra, if anyone is curious.

[quote="mathnerd314"]My calculator is old and can't do complex numbers or apps.  :(

Proof:
ax^2 + bx + c = 0 Given
4*a^2*x^2 + 4abx + 4ac = 0 Multiply by 4a
4*a^2*x^2 + 4abx + b^2 + 4ac = b^2 Add b^2
4*a^2*x^2 + 4abx + b^2 = b^2 - 4ac Subtract 4ac
(2ax + b)^2 = b^2 - 4ac Factor left
2ax + b = plus or minus sqrt(b^2 - 4ac) Root
2ax = -b plus or minus sqrt(b^2 - 4ac) Subtract -b
x= (-b plus or minus sqrt(b^2 - 4ac)) / 2a Divide by 2a, QED[/quote]

AKA...

$ax^2+bx+c=0$
$4a^2x^2+4abx+4ac=0$
$4a^2x^2+4abx+b^2+4ac=b^2$
$4a^2x^2+4abx+b^2=b^2-4ac$
$(2ax+b)^2=b^2-4ac$
$2ax+b=\pm\sqrt{b^2-4ac}$
$2ax=-b\pm\sqrt{b^2-4ac}$
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

[quote="nat mc"]but what about like ARML where you can't use calculators.[/quote]

It is highly, highly unlikely you will have to solve an [i]arbitrary[/i] cubic at ARML or any competition for that matter.  Either the cubic will have one root that is easy to find (like $x=1$ for example) or you won't actually have to solve the cubic explicitly.

[quote="nat mc"][quote="mathnerd314"]My calculator is old and can't do complex numbers or apps.  :(

Proof:
ax^2 + bx + c = 0 Given
4*a^2*x^2 + 4abx + 4ac = 0 Multiply by 4a
4*a^2*x^2 + 4abx + b^2 + 4ac = b^2 Add b^2
4*a^2*x^2 + 4abx + b^2 = b^2 - 4ac Subtract 4ac
(2ax + b)^2 = b^2 - 4ac Factor left
2ax + b = plus or minus sqrt(b^2 - 4ac) Root
2ax = -b plus or minus sqrt(b^2 - 4ac) Subtract -b
x= (-b plus or minus sqrt(b^2 - 4ac)) / 2a Divide by 2a, QED[/quote]
How about:

$ax^2+bx+c=0$

$x^2+\frac{b}{a}x+\frac{c}{a}=0$

$\left(x+\frac{b}{2a}\right)^2=-\frac{c}{a}+\frac{b^2}{4a^2}$

$\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a}$

$x+\frac{b}{2a}=\frac{\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$[/quote]

That involves completing the square, right? that's the one my book uses.

yea...that's the one i use. I've been trying to derive the cubic formula but I can't seem to get anywhere. I got to the deflated cubic but how do you turn $y^3+dy+e=0$ into a quadratic?

I've been wondering about that one too.  It seems that it should be much less complicated than the full cubic formula.  You could just plug in 0 into that formula


group theory
abstract algebra
superior algebra
superior algebra unsolved

Show that the set of elements of finite order in $ \mathbb{R}/ \mathbb{Z}$ is the subgroup $ \mathbb{Q}/ \mathbb{Z}$.

well the elements of $ \mathbb{R}/\mathbb{Z}$ are the cosets $ r \plus{} \mathbb{Z}$, $ r\in \mathbb{R}$. if $ r\not\in \mathbb{Q}$, then for all natural $ n$, $ rn\not\in \mathbb{Z}$, so $ r \plus{} \mathbb{Z}$ does not have finite order. on the other hand, if $ r\in \mathbb{Q}$, $ r \equal{} \frac {m}{n}$ for integers $ m$ and $ n$, so $ nr\in \mathbb{Z}$, and thus $ r \plus{} \mathbb{Z}$ has finite order.

i.e.  in ring $ R, \; \mathbb{Z} : \mathbb{Z} \equal{} R\cap\mathbb{Q}$


geometry
perimeter
search
geometry solved

Let M be the Gergonne point of triangle ABC, and let AM intersect BC at X.
Prove that:
$\frac{AM}{MX}=\frac{a(p-a)}{(p-b)(p-c)}$.

[color=red][Moderator edit: "Gergonne" is written with two "n". What's much more important: Please define your notations. In this particular case, this means writing explicitely that $p=\frac{a+b+c}{2}$.][/color]

This is a simple consequence of Van Aubel's (Van Obel's? Obel's? Aubel's?) Theorem, which in turn is a simple consequence of center-of-mass techniques. By the way, in the future, define your terms, because otherwise it gets sort of confusing. Americans (and other Westerners, I think) use $p$ for the perimeter and $s$ for the semiperimeter while Russians use $p$ for semiperimeter. Here $p$ is evidently the semiperimeter.

Van Aubel's Theorem: If $ABC$ is a triangle with cevians $AD, BE, CF$ concurring at point $P$, then $\frac{AP}{PD} = \frac{AF}{FB} + \frac{AE}{EC}$.

I won't post the proof here, but you can do a search (or work it out; it's not difficult). Applying the above theorem to this problem, we see $\frac{AM}{MX} = \frac{p-a}{p-b} + \frac{p-a}{p-c} = \frac{\left((p-b) + (p-c)\right) \cdot (p-a)}{(p-b)(p-c)} = \frac{a(p-a)}{(p-b)(p-c)}$.

[quote="MysticTerminator"]
Van Aubel's Theorem: If $ABC$ is a triangle with cevians $AD, BE, CF$ concurring at point $P$, then $\frac{AP}{PD} = \frac{AF}{FB} + \frac{AE}{EC}$.
[/quote]
Cool theorem. Never heard of it.......

$\frac{AF}{FB}=\frac{\triangle{APC}}{\triangle{BPC}}$

$\frac{AE}{EC}=\frac{\triangle{ABP}}{\triangle{BPC}}$

$\frac{AF}{FB}+\frac{AE}{EC}=\frac{\triangle{ABP}+\triangle{APC}}{\triangle{BPC}}=\frac{S_{ABPC}}{\triangle{BPC}}=\frac{AP}{PD}$

[quote="shobber"][quote="MysticTerminator"]
Van Aubel's Theorem: If $ABC$ is a triangle with cevians $AD, BE, CF$ concurring at point $P$, then $\frac{AP}{PD} = \frac{AF}{FB} + \frac{AE}{EC}$.
[/quote]
Cool theorem. Never heard of it.......[/quote]

It is one of the theorems one really should know, although they don't occur in the canonical IMO training literature. It can be applied in various cases, e. g. in [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=van+aubel&t=5596]my solution of a Mexican MO problem[/url], or in the study of Heronian triangles ([url=http://forumgeom.fau.edu/FG2001volume1/FG200104index.html]K. R. S. Sastry, [i]Heron Triangles: A Gergonne-Cevian-and-Median Perspective[/i], Forum Geometricorum, 1 (2001) pp. 17-24[/url]; by the way, this paper gives exactly  your proof of the van Aubel theorem). See also http://cut-the-knot.com/triangle/Gergonne.shtml , where it is referred to as "van Obel theorem"; however, the right spelling is van Aubel (see http://www.pandd.demon.nl/lemoine/vanaubel.htm , where, however, another theorem of van Aubel is discussed).

Anyway, let me give a complete solution of Mathx's problem using the van Aubel theorem:

[color=blue][b]Problem.[/b] The Gergonne point G of a triangle ABC is defined as the point of intersection of the lines AD, BE, CF, where the points D, E, F are the points of tangency of the incircle of triangle ABC with the sides BC, CA, AB. Prove that $\frac{AG}{GD}=\frac{a\left(s-a\right)}{\left(s-b\right)\left(s-c\right)}$, where $s=\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC.[/color]

[i]Solution.[/i] It is well-known that, for the points of tangency D, E, F of the incircle of triangle ABC with the sides BC, CA, AB, we have AE = s - a, EC = s - c, AF = s - a and FB = s - b. Hence, for the point of concurrence G of the lines AD, BE, CF, the van Aubel theorem yields

$\frac{AG}{GD}=\frac{AF}{FB} + \frac{AE}{EC}=\frac{s-a}{s-b}+\frac{s-a}{s-c}=\frac{\left(\left(s-b\right)+\left(s-c\right)\right)\left(s-a\right)}{\left(s-b\right)\left(s-c\right)}$.

Now, since (s - b) + (s - c) = 2s - (b + c) = (a + b + c) - (b + c) = a, this becomes $\frac{AG}{GD}=\frac{a\left(s-a\right)}{\left(s-b\right)\left(s-c\right)}$, and the problem is solved.

  Darij


geometry
trapezoid
ratio
similar triangles

In trapezoid ABCD we have DC//AB. AC and BD meet at O. M and N are on AD and BC and MN//AB. MN meets AC at G and meets BD at H.
Prove MG=HN

There is a theorem which states "Parallel lines are cut by intersecting lines into lengths proportional to their distance from the intersection point of the two lines".  Or something along those lines.  Now, you see, the problem is pretty easy.  But I'll do it.
We know that $AGM~ACD$ and $BNH~BCD$.  We also know that $\frac{AG}{AC}=\frac{BH}{BD}$.  Thus, $\frac{AGM}{ACD}=\frac{BNH}{BCD}$ (those are supposed to be triangles, and the ratio of their side lengths).  Then we know that, by CPSTS (Corresponding parts of similar triangles are similar, there's also a CPCTC), $\frac{GM}{CD}=\frac{HN}{CD}$.  The conclusion follows.

I was thinking about area ratios and my own proof made me post it here. Then when I was re-reading the topic, I found out that it was nearly trivial......
Anyway, your solution is the very neat.

I would this it's common for a problem like this...
Anyways, I was at first thinking of proving that those two angles DAC and DBC are congruent by inscribing it in a circle, and then from there proving the triangles congruent, etc...and i thought the trapezoid is cyclic because two pairs of opposite angles are supplementary...and then i realized that opposite angles have to be congruent for that... :D


Is it true that for given two polynomials $ f(x)$ and $ g(x)$ $ (deg(f(x)) \equal{} deg(g(x)) \equal{} n; f(x),g(x)\in \mathbb{Q})$ that
1)if $ f(x_i) \equal{} g(x_i)$ for $ i \equal{} 1,2...,n$ we have $ f(x) \equal{} g(x)$ for all x;
2) if $ f(x_i)\equiv g(x_i)$ $ (mod$ $ p)$ for $ i \equal{} 1,2...,n$ ($ p$ is a arbitrary prime) we have $ f(x_i)\equiv g(x_i)$ for all x;
$ x_i\in \mathbb{Q}$

$ 1) \Rightarrow f(x)\equal{}a(x\minus{}x_1)(x\minus{}x_2)...(x\minus{}x_n)$ and $ g(x)\equal{}b(x\minus{}x_1)(x\minus{}x_2)...(x\minus{}x_n) \Rightarrow$

$ \Rightarrow$ if $ a \not\equal{}b$, $ a \not\equal{}0$ and $ b \not\equal{}0$ then $ f(x)\not\equal{} g(x)$, where $ x\not\equal{}x_i$

its not true :put $ g(x)\equal{}f(x)\plus{}\prod(x\minus{}x_{i})$
i think you mean:
$ f(x_{i}) \equal{} g(x_{i})$ for$ i \equal{} 1,2....n \plus{} 1$
or some condition like that both of them are monic.


Hello,

I suppose that some poeple here may be looking for grad school options or just pieces of advise. It would be nice if there were some kind of forum like that either exactly here, or at least in the site. For example, if someone is planning to study math, say dinamical systems (just an example, it could be Theoretical Computations, Coding Theory, Cryptography, Topology, or any other are as well) but doesn't know where is a good place for doing it, may be some of us could help.

Any way, it is just a piece of advise of a grad student that went though a lot of problems to find its school.

Best,

Questions about graduate school programs are very welcome here on the College Forum. You should consider this forum to be about all aspects of choosing and successfully applying to programs after free compulsory education (primary and secondary education).

A good book about your concerns is

[url=http://www.amazon.com/exec/obidos/tg/detail/-/082183455X/learninfreed]A Mathematician's Survival Guide: Graduate School and Early Career Development[/url]

by Steven G. Krantz.

Hope this helps! Welcome to the forum.

[quote]My advisor got his PhD from University of Michigan, and he keeps telling me that that's the place to go for analytic number theory.[/quote]

This guy is the reason why.
[url=http://www.ams.org/notices/199603/comm-morgan.pdf#search='Soundararajan%20math%20putnam']Kannan Soundararajan[/url]

He is among the top 3 or 4 analytic number theorists in the U.S. Even though he's really young, he's authored around 30-40 research papers in the last 8 years; 4 very significant. We AoPser's know that someone has got to be good if they beat out Kiran Kedalaya in the Frank and Brennie Morgan Prize.  Prof. Soundararajan teaches all the major analytic number theory classes.