Could anyone tell me how I could write contests that are not held in my school?

Find a school where you can write that contest and register for the contest there and write it there. It's really simple to do unless that school is far away.

[quote="siva.eas"]Will I be an official contestant or an unofficial contestant if i wrote it in another school?[/quote]

Official.  I've done this before, first e-mail/talk to the teacher in charge of contests at that other school and ok it with them first.

Could anyone tell me what schools in Toronto hold the Canadian Open Mathematics Challenge.

[url=http://www.cemc.uwaterloo.ca/english/contests/past_result/2006/comc_results.pdf]2005 COMC results[/url]

u can see what schools in toronto hold COMC, just scroll down to provincial medallist section, specifically Ontario Metro.

[quote="Inspired By Nature"][quote="Inspired By Nature"][quote="rem"]Sometimes for the name of the school you have to write down the name of the school you are writing the contest in and not your own school.[/quote]

So if you have to write down the name of the school that you are writing the contest from, I have the following questions:

1) How would the Canadian Olympiad Committee ppl know that it is the same person who wrote all these other contests from school A, and this specific contest from school B? :?
2) Say, in the case of AMC-12, you write it from another school and end up qualifying for the AIME, then would you also take the AIME at school B?[/quote]

I guess my questions(above) kinda got lost in all the posts. If you know the answers, I would greatly appreciate a response.[/quote]

1) Their database is not sorted only by school... they know your name and can match if need be.  They're not stupid about that.
2) Maybe.  Does it matter?  You'll get to write it one way or another.

[quote="Inspired By Nature"]
2) Say, in the case of AMC-12, you write it from another school and end up qualifying for the AIME, then would you also take the AIME at school B?[/quote]

You need to write AIME at the same place where you wrote the AMC, because your question paper and bubble sheet are sent to the teacher who administrated AMC.

This is extremely late, but i am a USAMO qualifer and my school is not holding the USAMO.
I was wondering if i could take it at another school. Could those of you who have qualify for USAMO and live in Toronto, ask your USAMO protor if i may write it at your school, or may u give me their contact info (email).
Thanks alot, I would appreciate it!

[quote="p.c"]This is extremely late, but i am a USAMO qualifer and my school is not holding the USAMO. [/quote]

Oh-oh, that is indeed late. You have to fax the USAMO people some forms before April 14 (the forms are in the USAMO teacher manual you can find on their site), which is tomorrow, while the long weekend is ahead, and the contest is right after it.

If I'm not mistaken, if you're invited to the USAMO your school is obliged to assign a teacher to supervise you, even if you are the only one writing. In my school (Vaughan Road Academy), USAMO was probably never held before, but it was organized for me and Richard Peng to write it. I don't have my proctor's email, though, since he's really secretive.

So if you can contact a math teacher from you school during the weekend, then maybe you can complete the forms and send them during the weekend (e.g. you complete your part of the form, mail it to teacher, teacher signs and faxes). You could also ask Mr. Bill Sands (sands@math.ucalgary.ca) to help you - he was going to supervise USAMO for some Canadian students.

Good luck, and hope you get to write the contest officially.

[quote="YawningLion"]
Oh-oh, that is indeed late. You have to fax the USAMO people some forms before April 14 (the forms are in the USAMO teacher manual you can find on their site), which is tomorrow, while the long weekend is ahead, and the contest is right after it.

If I'm not mistaken, if you're invited to the USAMO your school is obliged to assign a teacher to supervise you, even if you are the only one writing. In my school (Vaughan Road Academy), USAMO was probably never held before, but it was organized for me and Richard Peng to write it. I don't have my proctor's email, though, since he's really secretive.

So if you can contact a math teacher from you school during the weekend, then maybe you can complete the forms and send them during the weekend (e.g. you complete your part of the form, mail it to teacher, teacher signs and faxes). You could also ask Mr. Bill Sands (sands@math.ucalgary.ca) to help you - he was going to supervise USAMO for some Canadian students.

Good luck, and hope you get to write the contest officially.[/quote]

Thanks for the help! I was lucky to be able to contact and to write the USAMO at University of Toronto Schools. I thought I would be unable to find someone with the long weekend and schools being closed.

Yes, in my school I am probably the first to have qualified for the USAMO. Therefore, tt is hard  to find 9 hours of supervision for just 1 person. The teacher would probably to extremely bored to sit with me for 9 hours.

Good luck to everyone writing the USAMO!

[quote="lightrhee"][quote="p.c"]Yes, in my school I am probably the first to have qualified for the USAMO. Therefore, tt is hard  to find 9 hours of supervision for just 1 person. The teacher would probably to extremely bored to sit with me for 9 hours.[/quote]

In my school, my teacher lock(?) me up in a small, empty room next to the music room. He checks me once or twice during the contest. :rotfl:[/quote]

For me...about the same. Bag in locker, and me in a cupboard(?), and the teacher(s) may or may not be with me all the time. Oh well...not that there's a whole lot I can do to cheat anyway... :lol:  :rotfl:


The ten-letter code BEST OF LUCK represents the ten digits $ 0\minus{}9$ in order.  What $ 4$-digit number is represented by the code word CLUE?

$ \textbf{(A)}\ 8671\qquad
\textbf{(B)}\ 8672\qquad
\textbf{(C)}\ 9781\qquad
\textbf{(D)}\ 9782\qquad
\textbf{(E)}\ 9872$

C=8, L=6, U=7, and E=1. So the number is 8671. Thus the answer is A.

[hide]

$ C$ will equal $ 8$, $ L$ will equal $ 6$, $ U$ will equal $ 7$, and $ E$ will equal $ 1$ for an answer of $ \boxed{\textbf{(A)}\ 8671}$[/hide]


Does anyone hear like " The Killers" ?  I think there music is brilliant, they muster a hard-rock tune while styling a pleasant 80's touch, this combined with a superb lead singer whose voice I think is truly amazing (yes, im a guppie for english voices), has allowed them to produce brilliant songs, Mr.Brightside, Jenny is a friend of mine, for instance, and I hope they continue making me wear out my sony surround-sound speakers.  :D

I don't really like their music that much.

I've heard their music sometimes... "Somebody told me" is my favorite by them.

[quote="mathclass"]Mm, I don't really understand the lyrics sometimes but I still like them.[/quote]

  That is the exact same case with me.  I do understand the lyrics in the slighest, the only correlation i can make is the songs seem to be related to love and relationship problems.  My favourite would be Mr.Brightside and Jenny was a friend of mine.

Mr. Brightside is ok in my opinion. I listen to the band occasionally but I usually skip them to get to other songs.

[quote="G-UNIT"]
  That is the exact same case with me.  I do understand the lyrics in the slighest, the only correlation i can make is the songs seem to be related to love and relationship problems.  My favourite would be Mr.Brightside and Jenny was a friend of mine.[/quote]

I understood Mr. Brightside after I read the lyrics. It was, the guy's girlfriend ran off with a rich old guy... I think?

But whatever, I just like the tunes.

They're good, but I wouldn't call them hard rock.

I have their CD, I usually like them when I'm in the mood for them. My fav is Somebody Told Me, because it makes no sense. Did they do a song concerning a hypnotist? If so, I love that song. And G-UNIT, I favor the english accent too!  :D  I like MUSE which some of their albums are more hard rock than others, but really cool. They guy has a really cool voice and has the talent of queen, he can hit the high notes better than I can and i'm soprano (a low one though...). Absolution is their newest album that I love! But it has more of a classical sound to some of their songs, he plays countless instruments so he plays some piano in tracks: 02, 03, 09, 10 (a LOT!), and 14. I'm not talking keyboard stuff, i'm talking downright concert pianist!!!! It's awesome! But if you favor the hard rock sounds, go for Hullabaloo, if you can find it. I think that that can only be found in a few music stores outside of england, they're not that famous yet. But anyway, the Killers are pretty cool.

im not really a big fan of crazy music, I like listening to Grieg and his composition; the Peer Gynt Suite.  I can play the original piano version of Hall of the mountain king! :) I have been playing for 8 years

OH I love King of the Mountain hall! ( I thought it was King of the Mountain hall, is it the other way around? Must check my music...)  I play that too, it's so fun! I've been playing piano for 6-8 years also, but I think I took a year off or  something like that. I was bored to death with the kids books, but when I got the adult book I couldn't stop! I really like Prelude IV by Chopin, and Alla Turca by mozart, of the ones that I play. The only piece that I ever liked by Gershwin was Rhapsody in Blue. I like the Nocturnes by Chopin too, and the first prelude. The heroic and military polonaises are awesome! My dad was so funny, he said if I can master the military he would buy me a baby grand piano. I'm thinking, yeah I really want to go do that, not that I think that I can yet.  I can play the first measure! That's it!  :roll:  Give me a year or two... And the heroic???  :|

i like green day better than the killers. :)

Ohhhh yeah! greenday rocks... I definately prefer them over the killers.

G-Unit, I want your screen name!  :)
I like The Killers. Though I have absolutely no idea what Somebody Told Me is about, it's still fun to listen to. Mr. Brightside, I used to enjoy listening to as well. Good song, but I played it too much on my MP3 player haha.

Yeah, I played Somebody told me too many times on my sister's MP3 player, being that I don't have one... I also like Evanescence!


calculus
integration
trigonometry
logarithms
calculus computations

[quote]Today's calculation of Integral 444
created by kunny
Evaluate $ \int_0^{\frac {\pi}{6}} \frac {\sin x \plus{} \cos x}{1 \minus{} \sin 2x}\ln\ (2 \plus{} \sin 2x)\ dx.$

[/quote]
is the answer

$ \equal{}\frac{2}{\sqrt{3}\minus{}1}\ln\frac{4\plus{}\sqrt{3}}{4} \minus{} \frac{1}{\sqrt{3}}\ ln\frac{2\plus{}\sqrt{3}}{5\minus{}\sqrt{3}}$

As the numerator contains put $ \ sin x \plus{} cos x$so put $ \ sin x \minus{} cos x \equal{} t$as $ \sin x\minus{}\cos x \equal{} t\ \Rightarrow\ (\sin x\plus{}\cos x ) dx \equal{} dt$

therfore the integral reduces to I = $ \int_{\minus{}1}^{\frac{1\minus{}\sqrt{3}}{2}}{\ln\frac{3\minus{}t^{2}}{t^{2}}}dx$

now integrate using parts ..


trigonometry logarithms function geometry circumcircle algebra polynomial

Let's compile a list of formulas that will come in handy during the AMC-10 and AMC 12. Try not to memorize these formulas; the point is to derive them; (unless of course, they're hard to prove :( ) . My hope is that e can have a sticky. Examples include, fermat's theorem, heron's formula, sum of geometric sequences, trig formulas, combanitorics, etc. Please bold the formulas, and try not to go past AMC level. Also, can mods Tex some of the messier formulas
I'll start off:

[b]Important trig formulas using squares:[/b]
sin 2 (x) + cos 2 (x) = 1,
tan 2 (x) + 1 = sec 2 (x),
1 + cot 2 (x) = csc 2 (x)

[b]Important double angle formulas:[/b]
sin(2x) = 2 sin(x)cos(x)
cos(2x) = cos2(x) - sin2(x)
            = 2 cos2(x) - 1
            = 1 - 2 sin2(x)

[b]Important angle addition formulas:[/b]
sin(x+y) = sin(x)cos(y) + cos(x)sin(y)
cos(x+y) = cos(x)cos(y) - sin(x)sin(y)
tan(x+y) = [tan(x)+tan(y)]/[1-tan(x)tan(y)]

[b]The Law of Cosines[/b]
c 2  = a 2  + b 2  - 2ab cos(C).

[b]The Law of Sines[/b]
A/sin(A) =B/sin(B) = C/sin(C)

For AMC-10, these are fairly ones you really need.

Pythagorean Theorem

$a^2+b^2 = c^2$

You can prove it by yourself on that one.

Heron's Formula

$s = (a+b+c)/2\\
A = \sqrt {s(s-a)(s-b)(s-c)}$

Again, theorems ARE NOT big deals. Exception of Pythagorean Theorem, you really don't need anything for AMC-10.

For AMC-12, you might want to go over laws of cosines and sines and perhaps logarithrm but other than, I don't think you really need them.

on the amc 12, you dont want to be stuck deriving or proving the trig or pythagorean theorem, it takes to long.  So its best to just memorize them.  Good luck

-Matthew

It is far more important to understand the processes to derive formulas than the formulas themselves.  You can derive many, many trig formulas with just knowing a few basic identities.  I suggest becoming acquainted with the relationships between the basic trig functions:  $\tan \theta=\frac{\sin \theta}{\cos \theta}$, $\csc \theta=\frac{1}{\sin \theta}$ , and so on.  If you know $\sin^2 \theta + \cos^2 \theta =1$ the 2 other pythagoreans come if you just divide by $\sin^2$ or $\cos^2$.  The double angles are good to know..you can derive half angle from these.  Becoming familiar with the relationships like $\sin(-\theta) = -\sin \theta$, $\sin(180-\theta)= \sin(\theta)$, etc.

For problems with cevians consider using the Law of Cosines since you will have supplementary angles (the cosines will cancel out if you add since $\cos(180-\theta)=-\cos \theta$.)

The [b]angle bisector theorem[/b] is useful:

If you have $\triangle ABC$ and you draw the angle bisector from $A$ to side $BC$ and call the intersection $D$ then:

\[(AC)(BD)=(AB)(CD).\]

[b]Stewart's Theorem[/b] isn't the most important of theorems to have on AMC but it might come in handy-it can simplify some problems that would otherwise take a bunch of intermediate steps to solve.  So once again we have $\triangle ABC$ with cevian $AD$ (this time it is just any old cevian....not necessarily the angle bisector) and Stewart's says:

\[(AB)^2(CD)+(AC)^2(BD)=(BD)(CD)(BC)+(AD)^2(BC).\]

Note: if you are trying to improving your AMC score bw not and tuesday...do NOT try memorizing this if you haven't seen it before (unless you have some strange desire to).  I would, however, try proving it.

I'll try to post a few more formulas (or rather general problems) later.

i know that it is important to derive the formulas on your own,  i do it all the time! But i dont want to be messying around with it when im taking the test.  You want to know rather quickly about the intersection of the medians, trig formulas, and others.

Anyway, trig identities are important.

I agree with you.  It helps a lot knowing the formula off the top of your head.  I'm just saying that it is tragic for one to just memorize a formula and not have a clue how and why it works...I can understand forgetting a few parts of the derivation, however.

I love Stewart, but I haven't used it a lot on the AMC.  If you want to read more about Stewart, go to chapter 3 of volume 2.

It's good to know certain [b]factoring tricks[/b], like

$a^3 - b^3 = (a-b)(a^2 + ab + b^2)$
$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$
$a+b+ab+1 = (a+1)(b+1)$

Also, note that in the [b]law of sines[/b]:

$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} = 2R$, where R is the circumradius of the triangle (I think; please edit if this is wrong)

So you can plug into $area = \frac{absinC}{2}$ to get $\frac{abc}{4R}$

It's also good to be familiar with the following (and how to use them):

\[a^2+b^2=(a+b)^2-2ab\]

\[a^3+b^3=(a+b)^3-3ab(a+b)\]

and so on.  You can look up Newton's Sums for more info.

It's also good to know the relationship between the coefficient of a polynomial and the roots:

If the $r_i$ are the roots then:

\[x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=x^n-(r_1+r_2+\cdots r_n)x^{n-1}+(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)x^{n-2}+\cdots+(-1)^nr_1r_2\cdots r_n.\]

I'm just wondering if Cauchy-Schwarz is ever helpful in AMC 12?

It is hardly ever useful and definitely not required for any problems.  AM-GM might help out on a hard problem.  I wouldn't count on seeing a problem using either of them on this year's AMC though.

I don't think anyone has SEEN the AMC12 yet (for if they have, they'd be cheating) so nobody knows the difficulty prior to actually taking the test. Besides, no discussion until Thursday so yeah, even after we actually take it you still won't know for 2 days how other people compare the difficulty.

[quote="DPopov"]Have any heard on the difficulty on this year's AMC 12?[/quote]

Even the people who write the test don't know about its difficulty any other way than empirically, and until all the answer sheets are submitted and scored, no one has statistics that compare this year's self-selected group of test-takers to last year's self-selected group of test takers. As always, what's easy for you may be hard for me, or (more likely  :)  ) the other way around.

True, though there was some rumor that they were going to make the tests harder this year ;)

Where did the rumor of harder tests come from? Not that I'm complaining, I think I'd like a hard test better.

[quote="probability1.01"]It shouldn't matter though--they select AIMEers by percentage rank (primarily).  This year, if it's harder, it might squeeze out the borderline people, but if you want to qualify, your goal is not to be one of them in the first place :)[/quote]

LOL. Dude easy for you to say.

You got a 10 or 11 on the AIME in 8th...

Or was it seventh? Alex...

welll, I guess it is also important to at least be able to derive inraidus and circumradius formulae... also: Pick's Theorem is helpful: A=B/2 + I -1 (for area of figure with lattice pts.)
trig identities are a must (for 12) not 10... also, maybe going over trig values (tan45 = 1, sin45=sqrt2/2, etc...) though I doubt many of you need it...
maybe the formula for adding combinations: nC0 +nC1+nC2+.... nCn = 2^n

Property of inclusion-exclusion is important.

Block walking.

3-D to 2-D interpretation.

This may have already been said:
area of a triangle=
$ab\sin\theta$

and the perimeter and area of any regular polygon is, resp.,
$2vr$
$v \cos \frac{180}{n} r^{2}$
where $r$ is its radius, $n$ is the # of sides, and $v=n \sin \frac{180}{n}$

area of any quadrilateral is: $\sqrt{(s-a)(s-b)(s-c)(s-d)}$ $s=$ semiperimeter

Umm...your area for the area of a quadrilateral only applies for cyclic quadrilaterals In(dx/dy).

But the true formula is...

$(ABCD) = \displaystyle \sqrt{(s-a)(s-b)(s-c)(s-d) - \cos^2 \frac{B+D}{2}}$, where s is the semiperimeter and a,b,c,d are side lengths of quadrilateral ABCD.

There's a proof of it somewhere on this site.


geometry
analytic geometry
conics
geometry proposed

Let $M$ be the midpoint of the side $BC$ of $\triangle  ABC.$ $\tau$ is  a line on its plane which  intersects sides $ AB,$ $ AC$ at $ X$ and $ Y,$ respectively. Let $ S_1$ and $ S_2$ be the areas of  $\triangle MBX$ and $ \triangle MCY.$ Show that

$ \frac {6}{[\triangle ABC]} \equal{} \frac {1}{S_1} \plus{} \frac {1}{S_2} \Longleftrightarrow  \tau$  is tangent to the Steiner inellipse of $\triangle ABC$

Let us use barycentric coordinates WRT $\triangle ABC.$ $px \plus{} qy \plus{} rz \equal{} 0$ is the barycentric equation of $\tau.$ Then $\tau$ cuts $ AB$ and $ AC$ at $ X:(q: \minus{} p: 0)$ and $ Y : (r: 0: \minus{} p)$

$ \frac {[\triangle XBM]}{[\triangle ABC]} \equal{} \frac {q}{2(q \minus{} p)} \ \ , \ \ \frac {[\triangle YCM]}{[\triangle ABC]} \equal{} \frac {r}{2(r \minus{} p)}$

$ \Longrightarrow \ \frac {[\triangle XBM] + [\triangle YMC]}{[\triangle XBM] \cdot [\triangle YCM]} \equal{} \frac {4qr+2pr-2pq}{qr}[\triangle ABC]$

$\Longrightarrow \ \frac {4qr-2pr-2pq}{qr} \equal{} 6 \ \Longrightarrow \ pq \plus{} pr \plus{} qr \equal{} 0 $

$\tau$ satisfies the wanted condition $\Longleftrightarrow$ $ pq \plus{} pr \plus{} qr \equal{} 0,$ which indeed means that $\tau$ is tangent to Steiner inellipse $\mathcal{E} \equiv x^2 \plus{} y^2 \plus{} z^2 \minus{} 2xy \minus{} 2xz \minus{} 2yz \equal{} 0,$ since $\tau \equiv [p,q,r]$ is tangent to $\mathcal{E}$ $\Longleftrightarrow$ Point $(p:q:r)$ lies on the dual conic $ \mathcal{D} \equiv xy \plus{} xz \plus{} yz \equal{} 0$ of $\mathcal{E}.$


I'm gnna take the usabo and usapho and maybe the chem one, too. what textbooks should i use and, especially WHERE CAN I FIND PRACTICE TESTS?!?!?! (do they release any)

I've heard USABO suggests the Campbell Biology textbook, I've never taken any of these tests so I don't really know.

lol yeah, they do. im interested in trying to find practice tests, tho. does anyone have any links or know where to buy them?


No, I'm not referring to the clause "math rocks," but a kind of rock music.  Its name derived from the complex rhythms and beats that it seems only a mathematician can derive them.
A famous post-21st century example would be Foals, whose latest single Cassius illustrate some of the stereotypical dissonant cords of math rock (and somewhat the complex rhythm, though I think only at the end)
http://www.youtube.com/watch?v=A3lWbPEOJp0
And one of the best math rocks songs in my opinion is steady as she goes by shellac, which can be found here
http://www.youtube.com/watch?v=wAf_TMirQSI
the drummer displays a lot of rhythm variety midway through the performance.  (listen carefully to his bass and you'll hear a lot more)
Just like to point out this interesting music genre unknown to most people (I think).

This is quite intriguing.

At first, I thought the post would be about those horrible "math rock" music that my teacher forced us to listen to in order to learn the multiplication table and such. Does anyone know what I'm talking about?

another interesting to note is that, the math rock band Foals actually released a single called "mathletics"
http://www.youtube.com/watch?v=WjNNWvz3JT4&feature=related

Of Course, I would know about this. The band The Dillinger Escape Plan is considered mathcore, because it is extremely complex and technical metalcore. I can connect this with math rock.


vector Stanford college AMC AIME geometry Gmaas

OK, here's how it works. Someone posts an acronym that they just made up, and the next poster makes up a new meaning for that acronym and posts a new acronym of their design. I'll start.

ILAP!

[hide="Do you wanna see my meaning?"][hide="Are you sure?"][/hide]
[hide="Completely? Well, OK."]I Like Apple Pie![/hide][/hide]

Let's go with this one.

[hide="WTH"]Will the Human. What else?[/hide]
EDIT: I'm sorry I didn't see aleph0's post in the other thread with this same name.

WTH = Weird Though Helpful

Mine is: AMAP.

[hide="my definition"] As Much As Possible! [/hide]

Articuno may appear purple

...I wish...:(

IWLYMIYWR

I would love you more if you were rich. xD

you were supposed to hide it.

I would like your mom in your white room.

TRISSMITP

[hide="in here"]The rain in Spain stays mainly in the plains.[/hide]

The rabbits in salads slay many identified tablet PCs.

[hide="IYCTHTTYACIYPN"]
If you click this hide tag then you are cheating if you post next.
[/hide]

If you care that hats talk to you and cinemas ink you, pretzels nod.
[hide="TWTFAIHED"]That Was The Funniest Acronym I Have Ever Done![/hide]

Twits with tongues fail abominable in heating edible dinghies.

[hide="DADAFD"]Doe, a deer, a female deer.[/hide]

Dark archon defected a fine dragoon!

(Starcraft joke...)
[hide="HHLBBCNOFN"]Hydrogen, Helium, Lithium, Beryllium, Boron, Carbon, Nitrogen, Oxygen, Fluorine, Neon[/hide]

He had little bacon before; concussions not off far now.

[hide="FTOTOSSNVLHCSTHNMFUTAPWACFR"]Facing tens of thousands of supporters Sunday night, Venezuelan leader Hugo Chavez said that his new mandate for unlimited terms as president was a call for revolution.[/hide]

LOL

FTOTOSSNVLHCSTHNMFUTAPWACFR
For those others that owe someone such new vicinty LHC, some takeovers hold new mutations for usual toads appearing pink while all cars fail recklessly.

I fail.

[hide="IABOHTGTSAY"]
I Am Bored Of Having To Go To School Aren't You?[/hide]

I am Borat Ogart. Here the gates to Saudi Arabia yawn.

[hide="FAIL"]For all in Lancaster![/hide]

Fantisize about incarcerated llama's?

[hide="GOD"] Gnomes owning dwarves[/hide]

Glorified omelets die.

[hide="IBIWLTB"]I bet I will lose this bet.[/hide]

[quote="asterday1"]True, when everyone finds a ten foot acronym, the first action that is actuated-a sensible, perfect failure- actually triggers initialization of greatness, a long shout having hyper active timing, inside every perfect initialization of coolness, EVERY perfect initialization of coolness. For, saying in full aligned madness, can any imperfect full soul actuate this inside realization of light? Well, saying THAT takes completely unnecessary-wait, says Forte-time.

[hide="BEATTHAT"]
Be everything and anything that totally has actual time.

[/hide][/quote]

Beating Everyone Around That Thing Has Annoying Traps (???)

[hide="TWBVLSBPATTIWCOYATPAMDMSATIFAKFAKITAKFASASDIBAIMEFMBHETMATKAKKAWMDFAKFAJPIABAIMAIMEWIOTOMATIGRLAMNIBOGTROAWYATCNPWATIGALSIGIBLOHSBAKKAJOMAYSBEATIRLWBBOGTRO"]This Will Become Very Long So Be Prepared And This Time I Will Completely Own You And The Post Above Me Didn't Make Sense And This Is Failing And Kelvin Fails And Kelvin Is Ticklish And Kelvin Failed At Stanford And So Did I But AIME15 Failed More Because He Lost To Me And Tickle Kelvin And Kill Kelvin And World Math Day Fails And Kelvin Fails And JP Is A Beast And I Made AIME Which Is On The 17th Of March And This Is Getting Really Long And My Name Is BOGTRO And Whats Yours And This Contains No Puncuation Whatsoever And This Is Getting Annoyingly Long So I Guess I Better Leave Off Here So Bye And Kill Kelvin And Jp Owns Me And You So Bye Everyone And This Is Really Long Written By BOGTRO[/hide]

The way before viewing light space beats partying a totally totally ignorant way come on you already tripped past a meter denying mods do mean shitake a too if fat a klondike fat albert klondike is the absolute fattest albert side angle side does it belong american invitational mathematics examination for much bets however Extra terrestrial momma a trip klondike a klondike klondike a wofford man dang fo' absolute klondike fo' absolute Jaehoon please I am bored aol instant messaging american invitational mathematics examination why i ought to oh my a tiger in grass rules lamination a mini nut  is bogtro a while you at to catch no pokemon what a tigger i good a lip sync is good is bad lol oh hey sure bye a klondike klondike joe oh my amazing yes sir beat eat at to is really long what bomb bogtro = i dont even know

[hide="HMFA"]Hello, my fellow Americans.[/hide]

hi my friend andy

[hide="WDYTTM"]What do you think this means??[/hide]

Why do you talk to me?

Hm...
[hide="IASOCEBISRLCHIHBPIAGIPCAIAATAMIGIA"]I am stumped on chemical equations but I still really love chemistry. However, I hate biology. Physics is awesome. Geometry is pretty cool, as is Algebra and Trigonometry. Actually, math in general is awesome.[/hide]

[quote="bubka"]IASOCEBISRLCHIHBPIAGIPCAIAATAMIGIA

Intersecting a succintly obnoxious covertly eccentric bisector in some relatively less crucial hence insufficiently hidden bifocal points is a grossly impudent parabolic curve, attired in an aboriginal turban and mimicking isomorphic groups in algebra.

(gosh, that was huge, I'll keep mine smaller)

[hide="I"]I[/hide][/quote]

Impedimenta.

[hide="ILAATFWF"]I Laugh At All The Failures Who Fail.[/hide]

I laughed at a twenty-foot water fountain.
 :huh:

[hide="HTWWCYTTTCABFPOSAINRTFTELAIFYPAIYIWDYTWTMPSAATAELSTRYTIUYDSLUTLWHOTTITEPIDKAIDCBATRLSIGEMRLALALALBTSNSWITSTRIDSNCPMOTAPBNIWARHSCYCASLTOTWT"]Hello, these words will commence your torture, trying to construct a back fitting phrase (or sentence, as I now realize) to fit this extraordinarily long acronym; I feel your pain as I (yes, I, who did you think was typing, my personal secretary???) am also typing an equally long sentence that rivals yours; that is, unless you're doing stuff like using twelve letter words half of the time - is that even possible... I don't know, and I don't care - but anyhow this ridiculously long sentence is getting even more ridiculously long, and longer, and longer, and longer, but there seems no stop - who invented this sentence; that really is disappointing seeing no close punctuation mark (.) OH THERE'S A PERIOD - but no, it was a red herring - so can you construct a sentence like this, only time will tell.[/hide]

High time with wally can you teach the truth could a bear fight people or see an insect new rhino the fat tim elephant lady anna if finding your prizes and in your igloo with days yours truly with teeth maybe pork said another ambulance the angry egg likes silver the rookie yelled time I understand your dialect so let us try little wally heart of the tie in texas eat pie in da kind and innocent district columbia be a too ripped lagging sore in game end my religion land and lubber and lobsters and lord live behind the south north sick window in the sweet taffy riding in dust seeing no cool people making over the ants peanut butter new igloo with a rhino he saw cool years called a sorry little toddler octopus...[b]THAT WAS TERRIBLE[/b]

seriously, why would you make one that long?

[hide="ILTEPAMH"]I like to eat pie at my house[/hide]

i love to egg pakistani-american mega houses

[hide="TSIPIBJSIMGHSCFMMITAYBTOSLSOTBSIMBAFYBIWLAFMAITOAOAHWPMADYJLIWYDTWSPMOYATSLHMUEWTMSIIAWMWIJSHTOAAMTWCOTTOMHJPATCOOMBOTS"]this sentence is partially inspired by james4l so i must give him some credit for making me inclined to annoy you by typing out such long sentences; on the bright side it may be annoying for you but it was less annoying for me as i type on average over a hundred words per minute and dont you just love it when you decide to waste such precious minutes of your all-too-short life here matching up every word to make sure it is accurate with mine while i just sit here taking only about a minute typing whatever comes off the top of my head, just pouring all the contents out of my brain onto this screen?[/hide]

ok...just stop with the long acronyms and people will start posting again.  Long acronyms ruin the game and they are not fun at all.

NOVRYFWBTGDTKOTBRNJWOTT
[hide] Number one victory royale yeah fortnite we 'bout to get down ten kills on the board right now just wiped out tomato town

No cap I had to do it because my little bro's friends are obsessed with that song


Let $A$ be a subset of the real line, we denote by $A*$ the set that is the union of $A$ and its limit points.
Give an example in which $A$ is an open set, and the sets $A \cap B*$,$A*\cap B*$ and $(A \cap B) *$ are all distinct. Give an example in which $A$ is not open and $A \cap B*$ is not a subset of $(A \cap B) *$.

I think $B$ is also a subset of the real number line.
Is the first example possible?  (Does the limit points include points in the interior of the set?)

$B$ is any subset of the real line. Either open, closed, or none.

Yes, the limit points include the points in the interior of the set.

2) Let $A=[0,1]$ and $B=(1,2)$.
Then, $A \cap B*=1$, and $(A \cap B)*$ does not exist.


Ενας ενδομορφισμός $ f: E\minus{}>E$ ενός ευκλείδιου χώρου $ E$ ονομάζεται [b]ομοιότητα[/b] με λόγο [b]k[/b] ανήκει [b]R[/b] (πραγματικοί αριθμοί) αν για κάθε $ X$ανήκει στο [b]Ε [/b]ισχύει

$ \parallel{}f(X)\parallel{}\equal{}k\parallel{}X\parallel{}$

Nα δείξετε ότι:

α)  $ f(X). f(Y)\equal{}k^2(X.Y)$ ,όπου $ (X.Y)$ το εσωτερικό γινόμενο των διανυσμάτων $ X,Y$

β) Αν το $ X$ είναι κάθετο στο $ Y$ τότε $ f(X)$ κάθετο στο  $ f(Y)$

γ)  Ένας (μη μηδενικός ) ενδομορφισμός $ f: E\minus{}>E$ είναι μια ομοιότητα αν διατηρεί την ορθογωνιότητα

Hmmm, απλό αλλά θεμελιώδες :)

[hide]Χρησιμοποίησε την polarization identity (btw, ξέρει κανείς πώς τη λέμε στα ελληνικά?) και το γεγονός ότι η $ f$ είναι γραμμική:
$ 2\langle X, Y\rangle \equal{} \|X \plus{} Y\|^2 \minus{} \|X\|^2 \minus{} \|Y\|^2$
[/hide]

Cheerio,

Durandal 1707


You have a quadrilateral.
You pick four points on different sides and connect it to form another quadrilateral.
Show that new quadrilateral engulfs the center of the old quadrilateral.

Does it matter whether or not the quadrilateral is convex?

What is the center of a general quadrilateral?  Do you mean the intersection of the diagonals?

If the quadrilateral is concave, then how can you have an intersection of the diagonals?

The original quadrilateral is shown in blue.

One diagonal, and extension of the other, are shown in red.

The quadrilateral constructed by picking 4 random points on the 4 sides is shown in brown.

Where is the centre?

Assume he means convex and that the center is the intersection of the diagonals and you can do it without too much trouble (hence its home in Getting Started).


a) Show that in a triangle $ \bigtriangleup MNP$, whose sides are less then $ 2$, the lenght of the height coresponding to $ MN$ is less than $ \sqrt{4\minus{}\frac{MN^2}{4}}$.
b)In a tetrahedron $ ABCD$, at least $ 5$ sides are $ \leq 1$. Show that the volume of $ ABCD$ is less than $ 1$.

[hide="Part A"]
Let the foot of the altitude from $ P$ to $ \overline{MN}$ be $ Q$. WLOG we can say that $ MQ\ge NQ$, or it would be the same thing flipped. Therefore, $ 2MQ\ge MN$, or $ 4\minus{}(MQ)^2\le 4\minus{}\frac{(MN)^2}{4}$. $ MP\le 2$, so $ (MP)^2\le4$. Now we have that $ (MP)^2\minus{}(MQ)^2\equal{}(PQ)^2\le 4\minus{}(MQ)^2\le 4\minus{}\frac{(MN)^2}{4}$. Since both sides are positive and $ \sqrt{x}$ monotonically increases,

$ \boxed{PQ\le \sqrt{4\minus{}\frac{(MN)^2}{4}}}$.
[/hide]


guys,i heard of sumthing called the kantorovich inequality.could anyone here tell me more about it?
thanx

If I remember corectly, it says that if a_1,...,a_n are in [a,b] then
 ( \sum a_i)*( \sum 1/a_i) \leq n 2 2*( a+b) 2 /4ab or something like this.

( \sum xi*ti)( \sum ti/xi) \leq ((a+b) 2 /4ab)*( \sum ti) 2


P and Q are prime numbers not 3 or 5

Prove that P^4 - Q^4 is a multiple of 15.

Well

$q^4-p^4=1-1=0 \mod {5}$

because of fermats little theorem

$q^4-p^4=q^2-p^2=1-1 \mod {3}$

Again fermats theorem. And we are done


Yeah, whoops I did this wrong the first time....840 was kind of a ridiculous answer.

To separate the four items into three bags, only two divisors are necessary; if X is an item and l is a divisor:

X l X l XX     - only two divisors necessary to split the four items into three bags.

Now, there are many possibilities, but you can treat them as a word, with some of the same letters.  There are six total letters, but not all letters are different.  Thus:

6!/(4!2!)

Lets look at it look this. You have 4 items

_A_B_C_D_

Let's divide these into groups so to speak. All things in the left most group go in the 1st bag, all things in middle go in 2nd,etc. Since there are 3 bags, we need two dividers!

Therefore we have 5 places to place the first divider

Now let's think. When we place the divider, we will now have 6 places to put the 2nd divider. Can you see why?

Thus we have 6 places to put the 2nd divider. So we have 30 ways. But wait! The 2 dividers are indistinct, so we must divide by 2! (can you see why?). Therefore there are $ \frac{6\cdot5}{2\cdot1}\equal{}15$

It's no coincedence I wrote the last step that way, because that is the same as $ \binom{6}{2}$ In fact, for all problems of this type, it is $ \binom{d\plus{}i\minus{}1}{i}$ where $ d$ is the number of distinct objects, and $ i$ is the number of indistinct.

840 is most certainly not correct. Even if all were distinct, the case which would produce the most possibilities, it would be 4^3=64 so our answer is less than that. Before doing it helps to think about what the answer might be. Is 840 a realistic answer? If not, then figure out where you went wrong.

[quote="mewto55555"]Lets look at it look this. You have 4 items

_A_B_C_D_

Let's divide these into groups so to speak. All things in the left most group go in the 1st bag, all things in middle go in 2nd,etc. Since there are 3 bags, we need two dividers!

Therefore we have 5 places to place the first divider

Now let's think. When we place the divider, we will now have 6 places to put the 2nd divider. Can you see why?

Thus we have 6 places to put the 2nd divider. So we have 30 ways. But wait! The 2 dividers are indistinct, so we must divide by 2! (can you see why?). Therefore there are $ \frac {6\cdot5}{2\cdot1} \equal{} 15$

It's no coincedence I wrote the last step that way, because that is the same as $ \binom{6}{2}$ In fact, for all problems of this type, it is $ \binom{d \plus{} i \minus{} 1}{i}$ where $ d$ is the number of distinct objects, and $ i$ is the number of indistinct.[/quote]

That's for indistinguishable to distinguishable.

This problem is distributing distinguishable(4 different items) to indistinguishable(3 grocery bags)

Casework:
4-0-0: 1 way
3-1-0: 4 ways
2-1-1: 6 ways

add up gives 11 total ways

You forgot 2-2-0, which has $ 3$ ways. So $ 14$ total.

Sigh...this reminds me that I forgot 1+1+4=6 before...


Let $ x$, $ y$ and $ z$ be positive reals such that $ xyz \equal{} 1$. Prove that
\[ \prod{(x^4 \plus{} x^2y^4 \plus{} 1)}\ge\prod{(x^2 \plus{} xy^2 \plus{} 1)}
\]
:)

[quote="ElChapin"]jo jo nice solution, but the problem got easier when I introduced the three variables...

The orginin of this problem was the inequality

$ x^4 \plus{} x^2y^4 \plus{} 1\ge xy(x^2 \plus{} xy^2 \plus{} 1)$ for positve $ x$ and $ y$

:)[/quote]
Of course it is harder! It is wrong! :)

Let $ x\equal{}y\equal{}1\plus{}\epsilon$ where $ \epsilon$ is very small, and then you'll get $ LHS > RHS$.
If you let $ x\equal{}y$ we have to prove $ f(x) \equal{} x^6 \minus{} x^5 \minus{} x^2 \plus{} 1 \ge 0$. But $ f'(x) \equal{} 6x^5 \minus{} 5x^4 \minus{} 2x$, so $ f'(1) < 0$, and thus it is decreasing in that point and $ f(1\plus{}\epsilon) < f(1) \equal{} 0$ where $ \epsilon$ is small.

mmm that happens when you forget to square a term in some calculations jajaja you end creating an inequality which happens to be true but your method isn't... I'll post the real inequality soon... sorry


Source: Massachusetts Mathematics League March 2003

Two railroad workers are together in a 0.8 mile tunnel. One jogs east and the other west in order to be out of the tunnel before the Express comes through at 40 mph. Each man reaches his respective end in three minutes. The man jogging east reaches the east entrance just as the train enters, and the front of the train is 0.025 miles beyond the other man when it is 0.2 miles beyond the west end of the tunnel. At what rate did each man jog?

[hide="solution"]Since the train is traveling 40mph, it takes it 1.5 minutes to travel to where it is .2 miles out of the tunnel.  Both men, reached their end of the tunnel at the instant the train entered, and the train is .025 miles ahead of the man who ran west when it is .2 miles out of the tunnel.  Therefore, the man who ran west ran $.2-.025=.175$ miles in 1.5 minutes.  Thus, his speed was $\frac{.175}{1.5}=\frac{7}{60}$ miles per minute, or $7$ mph.  This means that he ran $7*\frac{3}{60}=.35$ miles in the 3 minutes while he was running to the end of the tunnel.  So the man running east ran .45 miles in 3 minutes, giving him a speed of $\frac{.45}{3}*60=9$ mph.

So the man running west ran $\boxed{7mph}$, and the man running east ran $\boxed{9mph}$[/hide]


given:
a)  tan(t/2)=sin(t)/(1+cos(t))
b)  atan(x)=2atan(x/(1+sqrt(1+x^2)))

explain how does the second comes from the first

a):
[hide]
$\tan(\frac{\theta}{2}) = \frac{\sin(\frac{\theta}{2})}{\cos(\frac{\theta}{2}$

$=\frac{\sqrt{\frac{1-\cos\theta}{2}}}{\sqrt{\frac{1=\cos\theta}}{2}$

$=\frac{\sqrt{1-\cos\theta}}{\sqrt{1+\cos\theta}}$

multiply by $\frac{\sqrt{1+\cos\theta}}{\sqrt{1+\cos\theta}}$
$=\frac{\sqrt{1-\cos^{2}\theta}}{1+\cos\theta}$
$=\frac{\sin\theta}{1+\cos\theta}$

EDIT: Ugh, solving stuff directly in latex confuses me... for the unparseable parts basically you interpret tan as sin/cos and then use half angle formulas. It simplifies to the first visible expression in my post.
[/hide]


Since $a_{0}=b_{0}$ and $b_{n}$ is growing much faster than $a_{n}$ we know that $b_{2006}$ will be greater, or am I missing some major point?

You're missing some major point. $b_{n}$ is not growing "much faster than" $a_{n}.$

The essential feature of the rate of growth of both is the squaring; you saw that adding $5$ will eventually be adding something very small compared to the size of the main term; what you missed was that adding $2^{n}$ is [i]also[/i] adding something very small compared to the size of the main term.

$a_{n}$ gets a "head start" over $b_{n}$ and $b_{n}$ never catches up. $a_{2006}>b_{2006}.$

Admittedly, this all requires a more careful argument that what I just gave - but this is the essence of it.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ k$ $ \subset$ $ K$ be a Galois extension.

-For $ k$ $ \subset$ $ L$ $ \subset$ $ K$, show that $ G(K/L)$ is solvable if $ G(K/k)$ is solvable.

-For $ k$ $ \subset$ $ L$ $ \subset$ $ K$ with $ k$ $ \subset$ $ L$ normal show that $ G(L/k)$ and $ G(K/L)$ are solvable if and only if $ G(K/k)$ is solvable.

-For $ k$ $ \subset$ $ L$ with $ K$ and $ L$ in a common field, show that $ G(KL/L)$ is solvable if $ G(K/k)$ is solvable.

1) For Galois Correspondance $ Gal(K/L) \leq Gal(K/k)$ hence if the latter is solvable, also its subgroups are.

2) Again Galois Correspondance says that $ Gal(K/L) \triangleleft Gal(K/k)$ and $ Gal(L/k) \simeq \frac {Gal(K/k)}{Gal(K/L)}$. Now focus on groups and use that the extensions and factors of solvable groups are solvable.

3) dunno :)


MATHCOUNTS
geometry
3D geometry
frustum
ratio

A water container with a circular bottom tapers linearly to a circular top with a radius one-halp that of the base.  A mark halfway up the side says "200 gallons."  How many gallons, to the nearest tenth, does the water container hold?

i would just like to note to not get discouraged if youre not able to solve these problems, these are mathcounts national probs, and there is a long time to just chapter mathcounts, so keep on truckin.

calculator allowed

[hide="My long boring solution"]
We know that

V=h/3 * (B + sqrt(Bb) + b) for any frustum for h=height, B and b = areas of bases

Let R be the radius of the bottom base.

Then we see that R/2 is the radius of the top base, so the total volume of the frustum is:

h/3 * (R^2 + sqrt(R^2 / 4 * R^2) + R^2 / 4) = h/3 * (2R^2)

But since the radius of the halfway base is 3R/4, 200 gallons is

h/6 * (R^2 + sqrt(R^2 * 9R^2 / 16) + 9R^2 / 16) = h/6 * (3R^2 / 2)

Therefore the ratio of 200 gallons to the total volume is

(h/6 * [3R^2]/2) / (h/3 * 2R^2)

= 3/8

So the total volume is 200 * 8/3 which equals 533 1/3, or about 533.3

Anyone still there? Hello?

[/hide]

[color=red][size=150][i][u][b]PLEASE STOP POSTING IN TOPICS FROM 2003 AND 2004[/b][/u][/i][/size][/color]  :mad:  :mad:  :mad:  :mad:

THIS CAUSES THE FORUM TO BE LITTERED WITH OLD TOPICS, AS IT IS RIGHT NOW! THE IMPORTANT TOPICS, LIKE THE MARATHON, AND SIRORANGE'S CHALLENGE, ARE EITHER ON THE LAST PAGE, OR ARE AT THE BOTTOM OF THIS ONE! IF YOU WANT PROBLEMS TO SOLVE, YOU CAN GO TO MARATHON, OR SIRORANGE'S CHALLENGE! PLEASE STOP!

IT CAN BE CONSIDERED SPAMMING!

(Yes, that was meant to be shouted. :P) :cursing:

And it's not just you mathnerd314, sorry to do this to you, even though you only did it once or twice (that i've noticed). Some other people are doing it a lot.  :roll:

when you post old problems it moves the good new problems to page two and nobody can see them!!

:oops_sign: . Didn't see the "2003".  :oops:  :blush:  :surrender:

Now that I think of it, I had the wrong answer anyway  :(  :(

What do you mean you didn't see the 2003?  To be able to even see the topic, you wouldve had to have gone to like page 19 or something

If the problem has no solution posted, it is perfectly fine to answer and bump it up to the front page. If it has been solved before, don't answer it unless you have a novel new twist on the solution. It's not that the good new problems go to the second page - these are problems that don't have solutions either.


inequalities
calculus
derivative
inequalities open

hello
can anyone explaine me "the entirely mixing variable method" .
I have a paper wich peaks about but i didn't andertsood the main idea !!

What is the problem that you come up with? I am the author of this article in Math Reflections. Tell me if I can explain.

goodmorning everybody
@ hungkhtn : i didn't anderstood why we may assume that c=0 if one side is increased or decreased and the other doesn't change when we decrease a , b and c by a number wich is not bigger than min(a,b,c) !!
i think that the problem come from my bad english  :blush:

We decrease all variables. Of course, both sides may be changed.

Sometime, one side remains the same.

For example, show me how could you use this method to prove the simple sample: for $ a,b,c\ge 0$

\[ a(a\minus{}b)^2\plus{}b(b\minus{}c)^2\plus{}c(c\minus{}a)^2\ge (a\minus{}b)(b\minus{}c)(c\minus{}a).\]

it is easy to understand main idea of E.M.V. I think so  :)
When you solve any inequality, if you see have at least one varible equal to 0 and remain variable are not same values so you should using E.M.V.
I will explain clearly. You want to solve the following inequality:
\[ f(a,b,c)\ge 0
\]
The first, you should transform
\[ f(a,b,c)\ge 0 \iff \sum S_a(b \minus{} c)^2 \plus{} k(a \minus{} b)(b \minus{} c)(c \minus{} a)\ge 0
\]
Using E.M.V we will prove that:
\[ f(a,b,c)\ge f(a \minus{} c,b \minus{} c,0)\ge 0
\]
Of course, as you saw, in hungkhtn's article he wrote that:
\[ f(a,b,c)\ge f(a\minus{}b,b\minus{}c,c\minus{}a)\ge 0\]
\[ f(a,b,c)\ge f(a \minus{} c,b \minus{} c,0)\ge 0
\]
After it was solved, an inequality is equivalent to only 2 variables or 1 variable. After we are reduced varialbe then we should use derivative method to solve it. The rest part is always quite easy.
If you am not stupid explainer, I think one of above explanation can help you understand a little about E.M.V


I've been kind of teaching myself the AP Physics C syllabus after having completely a first year physics course. I looked through a few of the FRQs on the collegeboard website but none of the past exams seem to have questions concerning only kinematics. I've looked through a few review books and gone through those too but I feel like I just need some more practice.

Also, how in depth does Physics C go into vectors? I mean I understand basic resolution of vectors into components, addition, subtraction, etc. but I really don't know much about cross and dot products or vector-valued functions and such.

Well chances are, you're not going to be able to find a collection of pure kinematic AP problems.  If you do problems from a wide collection of review books, that should be sufficient practice.  Or, you can pick up a college textbook (Halliday/Resnick is good) and do some of those problems.  I think those are good practice.


Let A be a set of n+1 complex numbers. Prove that there exists a constant k such that for any monic polynomial f of degree n at least one of the numbers $ f(z)$ with z in A has modulus at least k.
  Aditional question (because I didn't like their solution: too complicated): give a bound for k ( if you can, the best-this one I didn't manage) in terms of the elements of A.

I didn't find any bounds because I didn't want to go through with the computations (even though they're not hard at all :)), but here's what I thought we could do for the first part of the problem:

We have to show that when $x_1,x_2,\ldots,x_n$ move around in $\mathbb C$, the expressions $P_i=\prod_{j=1}^n|z_i-x_j|$ can't all become as small as we want at the same time. Take $\varepsilon$ small enough that the disks centered at $z_i$ and having radius $\varepsilon$ are all disjoint. If we assume that all $P_i$ can be as small as we want at the same time, then we will be able to find $(x_j)$ s.t. $P_i<\varepsilon^n,\ \forall i=\overline{1,n+1}$. This means that in each $P_i$, at least one factor must be smaller than $\varepsilon$, meaning that for each $i=\overline{1,n+1}$ we must find a $j\in\{1,2,\ldots,n\}$ s.t. $|z_i-x_j|<\varepsilon$. Because of our choice of $\varepsilon$, these $j$'s must all be distinct, and we have obtained a contradiction: we have an injection from $\overline{1,n+1}$ to $\overline{1,n}$.

Is it Ok?

Nice and ok for me, but we have an aditional question for 1000000 dolars  ;) . And it is not based on any stupid computation!

[quote="harazi"]...we have an aditional question for 1000000 dolars  ;) . And it is not based on any stupid computation![/quote]

Are thinking about the mathematics problem of Clay Institute
7 problems solved =   7 000 000 Dollars

Please, Moubi, do not touch those problems, the money are mine!


[quote="ridgers"]Find all the natural pairs a,b such that $ ab^{2} \plus{} b \plus{} 7$ is divisible by $ ab^{2} \plus{} a \plus{} b$

[hide]I found that only a=7 and b every natural number is a solution. Is it right?[/hide][/quote]

$ (a,b)\equal{}(1,1),(2,1),(7,n)$

[quote="SUPERMAN2"]See [hide="here"]http://www.mathlinks.ro/Forum/viewtopic.php?p=124428#p124428[/hide][/quote]

This is not the same problem. There is $ ab^2$ versus $ a^2b$ in your link and there is twice $ ab^2$ in the current problem, even after the edit made by OP.

[quote="ridgers"]Find all the natural pairs a,b such that $ ab^{2} \plus{} a \plus{} b$ is divisible by $ ab^{2} \plus{} b \plus{} 7$

[hide]I found that only a=7 and b every natural number is a solution. Is it right?[/hide]

Edit: Sorry pco but the question was wrong, now it is correct[/quote]

clearly, $ a \minus{} 7$ has to be divisible by $ ab^2 \plus{} b \plus{} 7$, which is not feasible, as long as $ a\neq 7$. (That is because $ (ab^2 \plus{} a \plus{} b) \minus{} (ab^2 \plus{} b \plus{} 7)$ is divisible by $ ab^2\plus{}b\plus{}7$)

if $ a \equal{} 7$, any natural $ b$ fits the statement.

[quote="ridgers"]Find all the natural pairs a,b such that $ ab^{2} \plus{} a \plus{} b$ is divisible by $ ab^{2} \plus{} b \plus{} 7$

[hide]I found that only a=7 and b every natural number is a solution. Is it right?[/hide]

Edit: Sorry pco but the question was wrong, now it is correct[/quote]

With this new problem, the only solution is $ (a,b)\equal{}(7,n)$ :

Just write $ ab^2\plus{}a\plus{}b\equal{}k(ab^2\plus{}b\plus{}7)$

$ k\equal{}1$ implies solution $ (a,b)\equal{}(7,n)$
$ k>1$ implies $ a\minus{}7\equal{}(k\minus{}1)(ab^2\plus{}b\plus{}7)$ and so LHS$ <a$ while RHS$ >a$ so no solution.

My solution is same as Erken's one. Also pco has nice solution


[quote="jhredsox"][quote="anirudh"] $\triangle ABD\sim\triangle DBC\sim\triangle ABC$.[/quote]

what does this mean?
-jorian[/quote]

It means similar. I think. :lol:

o, thats why the writing for it had $..../sim/.......$
-jorian

no, not gauss goldendomer, fermat!

[quote="anirudh"][quote="ragnarok23"][quote="anirudh"][img]http://i96.photobucket.com/albums/l167/adasarathy/scan0002.jpg[/img]

I just need a clarification. I know the answer is $\frac{6}{9}=\frac{2}{3}$. However, the problem is that the solution assumes that the triangles $\triangle ABD\sim\triangle DBC\sim\triangle ABC$.[/quote]
Which is always true.....[/quote]

No. That's what I was defending. I just need a proof.  :([/quote]
Let ACB=x and BAC=y. Thus, since ACB and BAC are complementary, DBC=y and BAC=x. By AA, DAB~DBC.

So you want us to prove $\triangle ABD\sim\triangle DBC\sim\triangle ABC$. Here is the proof.

[hide="proof"]

$\angle{DAB}=\angle{CAB}$

$\angle{ADB}=\angle{ABC}$

Since those two angles are equal(and they are corresponding angles in the two triangles), then

$\angle{ABD}=\angle{ACB}$

Therefore,

$\triangle ADB \sim \triangle ABC$

All we need to do now is prove that

$\triangle ABC \sim \triangle BDC$

$\angle{ACB}=\angle{DCB}$

$\angle{BDC}=\angle{ABC}$

We use the same logic we used for the last part of the proof, and get

$\triangle ABC \sim \triangle BDC$

$\therefore \triangle ABD \sim \triangle DBC \sim \triangle ABC$

[/hide]

[hide]
Angle A in the triangle DAB is the same as angle A in triangle BAC.
Both triangle DAB and BAC have a right angle. If we rotate triangle DAB, then it has two of the same angles as BAC in the right places.
By AA similarity, those two triangles are congruent.
As for triangle DBC and triangle BAC, they both have a right angle and C in the right places.
By AA similarity again, those two triangles are congruent.
Since DAB and DBC are similar to BAC, all three triangles are similar.[/hide]


algebra
polynomial
function
Ramsey Theory
combinatorics proposed
combinatorics

What is your favorite combinatorics problem ever?  Please 1 problem per person.

[b]Problem)[/b]  Show there exists two disjoint sets of positive integers $A=\{a_{1},a_{2},\ldots,a_{n}\}$ and $B=\{b_{1},b_{2},\ldots,b_{n}\}$ such that \[\{a_{i}+a_{j}~|~i<j\}=\{b_{i}+b_{j}~|~i<j\}\] (repetition included) if and only if $n$ is a power of 2. Example: $A=\{1,4,6,7\}$ and $B=\{2,3,5,8\}$.

[b]Solution)[/b] [hide]Let $A(x)=\sum_{k\in A}x^{k}$ and $B(x)=\sum_{l\in B}x^{l}$ and let $P(x)=A(x)+B(x)$ and $Q(x)=A(x)-B(x)$.  Note that \[A^{2}(x)-A(x^{2})=2\sum_{i<j}x^{a_{i}+a_{j}}=2\sum_{i<j}x^{b_{i}+b_{j}}=B^{2}(x)-B(x^{2})\] so \[P(x)Q(x)=A^{2}(x)-B^{2}(x)=A(x^{2})-B(x^{2})=Q(x^{2})\] Thus $P(1)Q(1)=Q(1)$.  Since $P(1)=2n$, it follows that $Q(1)=0$. Suppose that $n$ is not a power of 2.  A little experimentation suggests that $Q(1)=Q'(1)=Q''(1)=\cdots=0$.  We prove this with Strong Induction.  If $Q(1)=\cdots=Q^{(k-1)}(1)=0$ then equating \[\left.\frac{d^{k}}{dx^{k}}P(x)Q(x)\right|_{x=1}=2nQ^{(k)}(1)~~~~~~~~\mbox{and}~~~~~~~~\left.\frac{d^{k}}{dx^{k}}Q(x^{2})\right|_{x=1}=2^{k}Q^{(k)}(1)\] shows that $Q^{(k)}(1)=0$ (since we assumed $n$ is not a power of 2). This completes the proof that $Q^{(k)}(1)=0$ for all $k$.  Then $Q$ is the zero polynomial so $A=B$, a contradiction.  This establishes the only if.

To establish the if, let $A$ be the set of integers less than $2n$ whose binary expansions have a even number of 1's, and let $B$ be the odds.  For each pair $(x, y)$ of unequal integers, let $f(x, y) = (x', y')$ be defined as follows: locate the rightmost bits where the binary expansions of $x$ and $y$ differ, and switch these bits. Notice that (i) $x+y = x'+y'$, (ii) $f$ is an involution, and (iii) if $x, y \in A$ then $x', y' \in B$ and vise-versa. It follows that this choice of $A$ and $B$ satisfies the given condition for sets of non-negative integers.  Add 1 to each of their elements to make them satisfy the given condition for sets of positive integers.[/hide]

This problem is still (notoriously) unsolved, but it is still my favorite:

Define the Ramsey number $R(k)$ to be the largest $n$ such that there exists a graph on $n$ vertices such that

1. The graph does not contain a complete subgraph on $k$ vertices (a set of $k$ vertices such that every pair of vertices in the set is connected).
2. The graph does not contain an empty subgraph on $k$ vertices (a set of $k$ vertices such that no pair of vertices in the set is connected).

Erdös proved in the 1940's that for large enough $k$, $R(k)>(1.4)^{k}$.   In fact proved something much stronger: He showed that if you took a collection of $(1.4)^{k}$ vertices and picked a graph at random on them (by flipping a coin to decide whether to include each edge, for example, or by taking all the possible graphs on that many vertices and picking one uniformly at random), then it will (again for large enough $k$) be more than 99.99% certain to satisfy both of the above properties.  However, he was never able to solve the following problem:

[b]Give an example of a graph on at least $(1.001)^{k}$ vertices which has both of the above properties.  [/b]

Of course, this is easy to do for very small $k$, but the catch is to come up with a construction which will work for arbitrary $k$, or at least for infinitely many $k$.

I like this problem just because it  has interesting implications as to what a "typical graph" is like: If you asked me to describe properties of a "typical graph", the two properties above would have to be included...after all, almost every graph satisfies them.  On the other hand, if you asked me to describe a graph and then to describe its properties, I could not give an example of a graph with both of them.  In other words, every (large) graph mathematicians can currently describe to you is abnormal.

An [i]increasing sequence[/i] of integers is said to be alternating if it starts with an odd term, the second term is even, the third term is odd, the fourth is even, and so on. The empty sequence is counted as alternating.
Let $A(n)$ denote the number of alternating sequences involving only the integers ${1,2, \dots n}$. Find $A(20)$

[b](British MO 1994)[/b]

Here is a solution to tim1234133's problem:[hide]\begin{eqnarray*}A(n)&=&[x^{n}]\sum_{k=0}^{\infty}\frac{1}{1-x}\left(\frac{x}{1-x^{2}}\right)^{k}\\ \ &=&[x^{n}]\frac{1+x}{1-x-x^{2}}\\ \ &=&F_{n+1}+F_{n}\\ \ &=&F_{n+2}\end{eqnarray*}

This solution uses the known generating function for Fibonacci numbers.

[i]Remark: Some people might prefer a nice recurrence solution instead.[/i][/hide]


ARML email probability geometry MATHCOUNTS

For new members, here are some sites that will help for developing strong problem solving skills:

http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm

http://www.kalva.demon.co.uk/aime.html

http://www.math.sc.edu/~filaseta/contests.html

Note that the sites from top to bottom go from easiest to hardest problems.

Here's a site which has links to many math contests around the nation:
http://www.arml.com/state_and_regional.htm
I suggest starting with the local contests here in SC and then moving on to others such as the Michigan State Contest and the UGA contest (try finding contests with solutions)

Taking time to do these contests will greatly increase your math skills and you should take time to do it. If everyone on the SC team does, we will have a great chance of placing at ARML :lol:

Oh, almost forgot, take time to review the solutions and problems we have posted on this site, primarily in the ARML marathon, and for the power contest, take a look at Joe's Proof thread

It seems that we are really trying to recruit/involve new members, which is a good thing!  Also, I (and Joe, I think) have a large collection of problems on disks and on my hard drive.  Email or PM if you need anything from previous years.  Also, books you might should buy or (check out from your library) are:

Art of Problem Solving Vols. 1 & 2
Introduction to Counting & Probability
Introduction to Geometry

Some may be too easy, you'll have to see.  But I own all of these, and I think they are GREAT!  I know you people are above MathCounts, but you can find some GOOD problems in this book as well:

MathCounts 100 All-Time Greatest Problems.

Feel free to add more!

[quote="mysmartmouth"]

Feel free to add more![/quote]

Contest Problem Books 1,2,3, and 4...which contain all the AHSMEs from 1950-1982, all the other AHSMEs and AMCs can be located not only at the Contest Resources on AOPS but also that website I listed w/ the AMCs.

The ARML book which contains the ARMLs from 1995-2003 (the green one) and the ARML NYSML book which has the ARMLs from 1989-1993 (which sadly I don't have yet :( )

The last 2 ARML books Shobhit mentioned prove very hard to acquire.  Outside of the actual ARML site, you have little chance of finding one for sale.  Your best bet is libraries, or perhaps Froogle.

[quote="mysmartmouth"]

Art of Problem Solving Vols. 1 & 2
![/quote]

You really need to buy Volume 2 ;)


Consider ABCD be a cyclic quadrilateral and O is its circumcircle.  AC meets BD in E and AD meets BC in F.
Define P(X) be the power of point X to the circle O.  Prove that :
P(E) + P(F) = EF^2

dooste azizam, avval niz be ye seri ghazyeha hast keh moteassefaneh
vaght nemikonam benvisameshoon.mitooni inja chand tashoono bebini:
http://www.mathlinks.ro/Forum/viewtopic.php?t=51865
masale:farz konim $ABCD$ charzellie mohatie,$AC,BD$, hamo too
$E$, va $AB,CD$, hamo too $F$, va $AD,BC$, hamo too $G$ ghat
mikonan.sabet konid $P(F)+P(E)=EF^{2}$
ghazie:az linke bala natijeh migirim keh $OE$, bar $FG$, amoodeh.
khob farz konim hokme masale bar gharar bashe:
$P(F)=FO^{2}-R^{2}$($R$, shoae dayere mohiti)
$P(E)=R^{2}-EO^{2}$(-(dar nazar begir keh ghovat ro tebghe tarif,dar
khareje
dayere mosbat,va dar rooye dayere sefr, va dar dakhele dayere manfi
dar nazar migirim)
khob, dota rabeteye bala ro jam konim bebnim chi mishe!
$FO^{2}-R^{2}+R^{2}-EO^{2}=FE^{2}$
va dar natijeh bad az sadeh kardan khahim dasht:
$FO^{2}-EO^{2}=FE^{2}$
va in yani mosalase $OEF$, baiad ghaeme bashe.
ama midoonestim keh $OE$, ba $FG$ amoodeh.
va nemitooneh dar eine haal ham be $FG$va ham be $FE$ amood bashe.
pas behtareh keh masalaro sooratesho taghir bedi.
khodet ham ye tajdide nazar bokon, bebin dorost migam ya na :wink:

aslbatteh ye eshtebah koochik pish oomadeh,

va man ghalat boodane rabeteye$P(E)-P(F)=EF^{2}$ro sabet kardam.

negaran nabashin masale kamelan doroste.

etefaghan halate koli ham dareh.

inkeh, majmooe ghovathaye do noghteye $E,F$, nesbat beh dayereye$O$

,agar in do noghteh nesbat be ham mozdavaj bashan mishe$EF^{2}$.

do noghteye mozdavaj ham noghati hastand keh ghotbie harkodoom

nesbat be dayere, noghteye digar migzarad.

dar morede in masale ham , hamoontor keh too linke bala, tozih dade

boodam,$4$zelie $ABCD$,yek $4$zelie kamele, va $E,F$ do noghteye

mozdavajand. :wink:

[quote="makmib"]bebakhshid $!$
man dar soorate soal be eshtebah neveshtam $AC$ meets $BD$ in $E$,
soorate asli in ast ke $AB$ meets $CD$ in $E$.
albatte fekr konam ba vojoode in eshtebah soal dorost bashad.[/quote]

bale kamelan doroste,

kolan masaleye ghashangie,ageh ino hal kardi, mitooni badan roo halate

kolisham fekr koni :wink:


limit
trigonometry
calculus
calculus computations

Please someone answer these questions from some of my assignments. I need them to be answered as a guide for my other question hope you can help me.. I know how to answer but Im not good in factoring and I always have zero as answer.  :(

Evaluate

[img]http://img338.imageshack.us/img338/7890/problem1.jpg[/img]

Differentiate

[img]http://img15.imageshack.us/img15/170/problem2o.jpg[/img]

1. I can't understand what that other x in the numerator is.
2. Try multiplying the numerator and denominator by $ (2\plus{}\sqrt{x\plus{}5})$.
3. Pull out the 2.  $ \lim_{x\to 0}\frac{\sin{5x}}{x}$ is well-known to be 5.
4. Look at the degree of the numerator and denominator.

1. Either multiply everything out and use the power rule, or just do a straightforward product rule.  All of the individual terms are easy to differentiate.
2. Use the quotient rule.  All of the individual terms are easy to differentiate.

Can you help me for its answer . Just the differentiate. I got already how to evaluate.


find all functions f defined from IR to IR such as :
           (f(x)-f(y))(x-y)=xy(f(x)+f(y))

Er, what's IR?  I'm going to assume you mean reals.

$x = y \implies 2f(x) x^2 = 0 \implies f(x) = 0$ for all $x$.  Err... yeah.  I'm confused.


What is the smallest positive integer c for which sqrt(a)+sqrt(b)=sqrt(c) has a solution in unequal positive integers?

Just take the lowest two perfect squares higher than 0 (because if you use 0 the integers repeat) and you get 1 and 4.  Sqrt of those gives you 2 and 1, which make 3, squared you get 9.  Therefore c=9.

A little too easy if you ask me.  I wonder why no one did it before me.

they probably thought that it was a time and went looking for more of a challenge. Maybe I should've posted this in the basics forum.


linear algebra
matrix
algebra
polynomial
probability
expected value
probability and stats

Matrix $ A\in M_n(R)$, is filled in a random way with numbers from set $ \{0,1\}$.
Calculate the expected value of $ (\det A)^2$.

Remark:elements of $ A$ are random variables with values $ \{0,1\}$, and the probability's are $ \mathbb{P}(X\equal{}1)\equal{}\mathbb{P}(X\equal{}0)\equal{}\frac{1}{2}$.

It comes out to be $ \frac{(n\plus{}1)!}{2^{2n}}$.

The idea: Use the permutation definition of the determinant. If $ f_{\sigma}$ is the random variable which is the product of the $ (k,\sigma(k))$ entries in $ A$, $ \det A\equal{}\sum_{\sigma}\text{sgn}(\sigma)\cdot f_{\sigma}$, and $ E((\det A)^2)\equal{}\sum_{\sigma}\sum_{\phi}\text{sgn}(\sigma)\text{sgn}(\phi)\cdot \text{Cov}(f_{\sigma},f_{\phi})$. That covariance is easily calculated: it's $ 2^{\minus{}2n\plus{}s}$, where $ s\equal{}\#\{k: \sigma(k)\equal{}\phi(k)\}$. We can also write this as the number of fixes points $ s(\omega)$ of $ \omega\equal{}\sigma^{\minus{}1}\circ\phi$.
Rearranging the inner sum,
$ E((\det A)^2)\equal{}\sum_{\sigma}\sum_{\omega}\text{sgn}(\omega)\cdot 2^{\minus{}2n\plus{}s(\omega)}\equal{}\frac{n!}{2^{2n}}\sum_{\omega}\text{sgn}(\omega)\cdot 2^{s(\omega)}$

How do we calculate that latter sum? It's the determinant of $ \begin{bmatrix}2&1&2&\cdots\\1&2&1&\cdots\\1&1&2&\cdots\\ \vdots&\vdots&\vdots&\ddots\end{bmatrix}\equal{}J\plus{}I$ ($ J$ is the matrix of all 1s). Since $ J$ has rank 1, only the trace matters and we can easily calculate its characteristic polynomial: $ \det(\lambda I\plus{}J)\equal{}\lambda^n\plus{}n\lambda^{n\minus{}1}$. The specific determinant we're looking for is $ n\plus{}1$, and we have the formula.

This generalizes: if each entry is filled with probability $ p$ (independently), the expected value of the square of the determinant is $ p^{2n}\cdot n!\cdot\left(\frac1p\minus{}1\right)^{n\minus{}1}\left(\frac1p\minus{}1\plus{}n\right)$


geometry
parallelogram
rhombus
geometry proposed

It is known that $\triangle ABC$ is an acute triangle. Let $C'$ be the foott of the perpendicular from $C$ to $AB$, and $D$, $E$ two distinct points on $CC'$. The feet of the perpendiculars from $D$ to $AC$ and $BC$ are $F$ and $G$, respectively. Show that if $DGEF$ is a parallelogram then $ABC$ is isosceles.

From $DF \perp AC$ we get $EG \perp AC$, similarly  $EF \perp AB$, so $CE \perp FG$, that is $DE \perp FG$, so DFEG is a diamond.
Then we soon get CC' is also a bisector of $\angle BCA$, that completes the proof.□

[quote="k2c901_1"]Assuming you meant "rhombus" when you said "diamond", I think you're pretty much correct.[/quote]
Thanks for teaching me an English word of maths.


function
induction
algebra proposed
algebra

find all serjective functions $f:N_0\rightarrow N_0$with the propertythat for all $n\geq 0,f(n)\geq n+(-1)^n$

Compared with the average level of the other problems you post, we could say that this one is really trivial.... :D

Let $f$ be such a function.
For $n \geq 2$ we have $f(n) \geq 1$, and $f(0) \geq 1$.  Since $f$ is surjective we must then have $f(1) = 0$.
But $f(n) \geq 2$ for $n \geq 3$ and $f(2) \geq 2+1 = 3$. As above, we deduce that $f(0) = 1$.
Now, following the same reasoning as above, an easy induction leads to $f(2n) = 2n+1$ and $f(2n+1) = 2n$,
that is $f(n) = n + (-1)^n$.
Convesely, it is easy to verify that this function is a surjective one which satisfies the statement of the problem.

Pierre.

ya! but some problems are needed to be solved on bed and make you sleep better :D


There were 10 boys and 13 girls in Ms. Floop's homeroom class. The local newspaper had sent a photographer to take some pictures of the students at work in class. Assuming the photographer chooses students at random to be photographed, what is the probability that the first student photographed will be a girl?

[hide]there are 23 students and 13 girls so 13/23[/hide]

[hide]13/23 because there are 13 girls out of 23 students.[/hide]


[quote="MCrawford"]What is the greatest odd integer that is a factor of $5!$?[/quote]

[hide="Answer"] $15$[/hide]

[hide] even * even= even,

[even * odd] or [odd *even]=even

odd * odd = odd,

  So basically we have to find the product of the two largest odd numbers.

 In this case its 5 and 3. And their product is 15. [/hide]

[quote="solafidefarms"]Actually, we need to multiply ALL the odd numbers together, not just 2.[/quote]

Yeah thats what i did.

But in this case, the answer is exactly the same.

Units digit is obviously 0. As for the other part, it's a little tedious but not too hard.

[quote="chess64"][quote="MCrawford"]What is the greatest odd integer that is a factor of $5!$?[/quote]

[hide]$5!=5 \cdot 4 \cdot 3 \cdot 2$. Taking away the even numbers, we see $5 \cdot 3 = \textbf { 15.}$[/hide]

A little bit harder:

$\text {What is the greatest odd integer that is a factor of } 15! \text {, and what is the units digit of } 15! \text { ?}$[/quote]

[hide="Answers"]Units Digit = 0
Largest Odd Factor: 2027025[/hide]

[hide=Solution of the original one]
We can prime factorize $5! = 120$. We get $2^3 \cdot 3 \cdot 5$. To get an odd number, we must multiply 2 or more odd numbers. In the prime factorization of 120, we have 2 odd numbers: 3 and 5. Multiplying them together gives $3 \cdot 5 = \boxed{15}$[/hide]


AMC
AIME
geometry
USA(J)MO
USAMO
geometric transformation
homothety

Well... I'm not so sure whether practicing hard from the questions or actually learning the information from books is more useful. I would lean towards actually taking a year or so, looking in detail at some number theory books, geometry books, etc... and then when you take the AIME, you will find so many things have naturally become easier. For those of you with >7 on AIME, do you think this strategy will work as opposed to brute force problems?

Advice is appreciated.

I fully agree - by reading math, you will learn some interesting math, but by doing math, you will improve your ability to do math.

My first two times taking the AIME, I scored a 2 and a 4.  I think that's the first time I ever "bombed out" (by my standards) on a math competition.  My junior year, I practiced hard and worked old exams.  I even started looking at some USAMO problems.  That year, I got a 9 on the AIME.  It felt great to conquer an obstacle that had stood in my way for two years.

Best of luck to you.  By the way, another good way to improve your skills is to cruise this forum - there are always people around posting nice problems.  I sure wish I'd had something like this when I was in high school!

Oh, and by the way, if you haven't already, sign up for NOML (Cody Bowl) - my problems are the same format as the AIME, the pacing is similar, and the hard problems on the tests are similar difficulty to the AIME.

The AIME question style is so distinctive I think you have to practice for a year on old AIME exams to have a good chance at qualifying for USAMO from the AIME.

It is also a good idea to review the math "you already know," because the AIME demands a thorough knowledge of precalculus mathematics. The AoPS online course are good for that, as are the AoPS books, and so are [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22050]other books[/url].

My eighth grade year I had no idea what was going on in AMC and AIME. I just ended up picking up a bunch of old contests for AMC and practicing. I also bought 96-2000 AIMEs from the MAA, and also practicing on that year's first AIME (as I took the second one).

All it was was practice, but I got an 8 that year, 2 above the requisite score. That's really all it takes, practicing old contests. However, when you practice, make sure you take the FULL 3 HOURS, and do NOT split up the time over days because getting used to the fatigue after working for hours is just as important as knowing how to do many problems. Additionally, I suggest that even after the 3 hr time period, you mark down the problems you'd done, and then continue to try some of the harder problems you didn't get to, for at least another hour and a half (this doesn't have to be on the same day necessarily).

Finally, you may check your answers. Good luck.

I agree with what everyone said about practicing old exams, but I still think that practicing the questions first demands adequate knowledge of many of the concepts such as advanced geometry, combinatorics, and NT. If I've read through all the comb. and geometry in AOPS Volume 2, would it be more beneficial to practice problems, or to continue to skim over inversions, homothecy (spelling?), block walking, etc, diophantine equations, etc...? Thanks.

The amount of concepts you need is not as numerous as you might think.  AIME problems are usually just hard extensions of basic concepts.  You don't need advanced geometry like homothecy and inversion to do the problems.  While some things like Ceva and Menelaus might facilitate some problems all you really need is a good base in Euclidian geometry (i.e. able to angle chase, use congruent/similar triangles, know how to use area even when it isn't directly mentioned in a problem etc.) and trig for some problems.

Most people who make it to AIME already know all the theorems needed to solve the problems.

I can almost guarantee that 'skimming' a topic will not increase your AIME score one bit.  You need a complete understanding of a topic to answer most of the AIME questions.  That being said, at this point in time, I suggest practicing problems more than learning more material.  There is also enough time for you to learn a few more topics.  I wouldn't try to get everything at once (especially with only 1 month) so try focusing on a few topics and be sure to attempt problems.

Of course, that is just my thought on it.  I'm not a master of the AIME so maybe I am completely wrong...

since this is my first year taking AIME, i am going to be solving past AIMES (of course, after mathcounts)

I don't think I had ever heard of inversion until I started graduate school in mathematics.  I have no clue what homothecy is.  These things are possibly useful on the IMO, and maybe to a lesser extent the USAMO.  Sometimes you'll see some exotic mathematics (particularly geometry) on the IMO because the problems are contributed by people from very different mathematical cultures.

For the AIME, here are the geometric concepts which in my opinion are underemphasized in most standard high school curricula but are very useful on the exam.  I hope others will add to this list as they see fit.

* angle bisector theorem
* mass-point geometry (I have never actually used this, I always use...)
* translating base ratios / height ratios / similarity into area ratios
* setting coordinate systems to make a problem easier
* properties (such as area ratios) that are preserved under affine transformations (an affine transformation is basically a way of stretching, shrinking, skewing, rotating or moving around a figure)
* law of cosines (this one actually seems to get adequate play in HS, but I can't emphasize this enough because it's handy), ESPECIALLY using law of cosines to get the cosine of an angle
* the determinant formula for area of a triangle or parallelogram in the coordinate plane, or for volume of a pyramid in coordinate space
* being able to visualize things in three-dimensions is a terrific skill to have; I remember once seeing a problem #15 on the AIME that was a solid geometry problem, that I thought was overvalued and should have been in the #10 spot at latest.
* Heron's formula (occasionally useful, but not as often as...)
* the formula A = 1/2 ab sin C

The trend in recent AIMEs, in my opinion, is to rate a geometry problem harder than most other problems that require the same amount of ingenuity.  Of course, all these things are in the eye of the beholder, but the teaching of geometry in most high schools these days is so abysmal that students who are capable of doing REAL geometry (i.e. problem solving, not trivial vocabulary exercises and mind-numbingly easy proofs) suffer quite a bit.

IMHO, the best way to improve your score on AIME quickly is learning trig.  Lots of trig.
It's often directly tested and can solve almost any geometry problem.  My freshman year, I did all the problems from a Chinese prep book for the trig portion of the college entrance exam -- (sum to product, product to sum, sin/cos/tan in terms of tan(theta/2)) -- this raised my score by 2 the first year and then 3 more the next (I say that this was solely due to trig because of the many careless errors i made).
Of course, if you really want to improve your AIME score, there's no real alternative to learning the math.  :D

Well I come from the "practice old AIMEs" faction :D.

Seriously though, I took the AIME II last year with only basic knowledge of the typical precalc stuff (self-taught).  I didn't know what the angle bisector theorem, Ceva's theorem, Menelaus' theorem, Ptolemy's theorem, and a host of others were, and for the most part, I still don't :blush:.  Everything I knew was basically expected on the AMC.

The key was to practice lots of AIMEs and get used to problem-solving mode.  Fatigue from problem solving wasn't a problem for me, but if it is for you, you should work on that too.  You should definitely assign variables to everything given in the problem and play around with them a bit before trying to solve, unless you already know how to do the problem.  The basic way to solve every problem is the same: you are looking for a value, and you are given some values.  You must relate the value you are looking for to the values that you are given.  I find that it is often helpful to visualize things (even if they are not visual type things at all) and move them around in my mind to make the problem more tangible.

[quote="jmerry"]I improved dramatically (increased my score by 7) on the AIME between my 10th and 11th grade years. What did it? I never practiced an AIME, and didn't know where to find the questions anyway.
The only explanation I can come up with is that it was what I learned at MOP, even as a dazed rookie who could barely keep up. I spent several months after that just digesting what I'd picked up and thinking about hard problems.

Of course, this can't really be taken as advice- I was already in the USAMO, even honorable mention despite my minimal qualifying index. It was a real improvement on the AIME- my scores in order were 6,7,14,13.

Depth is always good. I didn't really pick up anything new at the AIME level, just a deeper understanding of how to solve hard problems.[/quote]

So the best way to pass the AIME is to pass the USAMO first?  :P

That means there's no best way to pass the AIME! We've reached a conclusion, folks.

Alot of help comes from doing past AIMEs; do some of the tests by actually sitting down for 3 straight hours and doing it (the mental exhaustion at the end can only be simulated this way), and doing other problems whenever you have time.

Attempt to do every single problem, and once you've absolutely exhausted every single option you can think of for doing it, do it some more. Then look at the solution, do the solution a step at a time, and make sure you understand the fundamentals behind it.

On top of that, it's very important to be able to understand many fundamental concepts. Only a basic understand of geometry will get you no where; what you are striving for is being able to manipulate any geometric figure you see to what you want it to be. That's the understanding you need to be at to efficiently and effectively do the AIME.


Find the determinant of the $ n$x$ n$ matrix with entries $ a_{ii}\equal{}i\plus{}4$, and $ a_{ij}\equal{}1$ for $ i\neq j$.

[b]Edit:[/b] You're absolutely right! What was I thinking  :maybe:

[hide="Original Post"]Did you have a particular application in mind?

[hide="Hint"]
We can easily generalize the diagonal method:
[img]http://upload.wikimedia.org/wikipedia/en/d/d2/Determinant_3x3_Example_Barking_Mad_1.jpg[/img]
[/hide]

[hide="Answer"]
A. Four our \ diagonals, the first will be $ \prod_{i = 1}^{n}i+4 =\frac{(n+4)!}{24}$, and the rest will be $ \prod_{i = 1}^{n}1 = 1$.
Adding these gives $ \frac{(n+4)!}{24}+n-1$

B. For our / diagonals, we have to be careful; there are two cases.

1. For odd n, each / diagonal will have $ n-1$ factors of $ 1$ and $ 1$ factor of some $ i+4$, each distinct. Thus, the product of each / diagonal is some distinct $ i+4$, and the sum of these is $ \sum_{i = 1}^{n}i+4 =\frac{(n+4)(n+5)}{2}-(1+2+3+4) =\frac{1}{2}n^{2}+\frac{9}{2}n$

2. For even n, the diagonals alternate.
2a. Every alternate / diagonal will be all 1's. Summing over the $ n/2$ of these gives $ \frac{n}{2}$
2b. Every other / diagonal will have $ n-2$ factors of $ 1$, and a pair of $ i+4$'s, separated by $ n/2$. We then want to find $ \sum_{i = 1}^{n/2}(i+4)( i+\frac{n}{2}+4)$. Using the sum of squares formula $ \frac{n(n+2)(2n+1)}{3}$, we find that this is equal to $ \frac{5}{48}n^{3}+\frac{9}{4}n^{2}+\frac{121}{12}n$.
2c. Adding the previous results gives $ \frac{5}{48}n^{3}+\frac{9}{4}n^{2}+\frac{127}{12}n$

The determinant for odd $ n$ is the result of (A) minus the result of (B1):
$ D_{n}=\frac{(n+4)!}{24}-\frac{1}{2}n^{2}-\frac{7}{2}n-1$

The determinant for even $ n$ is the result of (A) minus the result of (B2c):
$ D_{n}=\frac{(n+4)!}{24}-\frac{5}{48}n^{3}-\frac{9}{4}n^{2}-\frac{115}{12}n-1$
[/hide][/hide]

No, I don't think the diagonal method generalizes that way. If you use that method, you have $ n!$ (not $ 2n$) ways of selecting an element from each row such that all columns are picked. Then you have to multiply all of those elements, and also multiply by negative one if you need an odd number of transpositions to get that permutation (now we're on the edge of group theory).

The reason that method works for $ 3\times 3$ matrices is because it gets all of the $ 3!$ permutations. This is not true for higher dimensions.

The "diagnol method" generalizes under a rule called "Leibniz's Rule" I believe, (or something like that).

[quote="cancer"]Find the determinant of the $ n$x$ n$ matrix with entries $ a_{ii}= i+4$, and $ a_{ij}= 1$ for $ i\neq j$.[/quote]I have an idea but I don't think i can prove it formally.. so i'll try to explain with an example:[hide="Long example"]\begin{eqnarray*}\det\left(\begin{array}{cccc}5&1&1&1\\1&6&1&1\\1&1&7&1\\1&1&1&8\end{array}\right)&=&\det\left(\begin{array}{cccc}5&1&1&1\\1&6&1&1\\1&1&7&1\\0&0&0&7\end{array}\right)+\det\left(\begin{array}{cccc}5&1&1&1\\1&6&1&1\\1&1&7&1\\1&1&1&1\end{array}\right)\\ &=&7\det\left(\begin{array}{ccc}5&1&1\\1&6&1\\1&1&7\end{array}\right)+\det\left(\begin{array}{cccc}5&1&1&1\\1&6&1&1\\1&1&7&1\\1&1&1&1\end{array}\right)\\ &=&7\det\left(\begin{array}{ccc}5&1&1\\1&6&1\\1&1&7\end{array}\right)+\det\left(\begin{array}{cccc}5&1&1&1\\1&6&1&1\\0&0&6&0\\1&1&1&1\end{array}\right)+\det\left(\begin{array}{cccc}5&1&1&1\\1&6&1&1\\1&1&1&1\\1&1&1&1\end{array}\right)\\ &=&7\det\left(\begin{array}{ccc}5&1&1\\1&6&1\\1&1&7\end{array}\right)+\det\left(\begin{array}{cccc}5&1&1&1\\1&6&1&1\\0&0&6&0\\1&1&1&1\end{array}\right)\\ &=&7\det\left(\begin{array}{ccc}5&1&1\\1&6&1\\1&1&7\end{array}\right)+\det\left(\begin{array}{cccc}5&1&1&1\\1&6&1&1\\1&1&1&1\\0&0&0&6\end{array}\right)\\ &=&7\det\left(\begin{array}{ccc}5&1&1\\1&6&1\\1&1&7\end{array}\right)+6\det\left(\begin{array}{ccc}5&1&1\\1&6&1\\1&1&1\end{array}\right)\\ &=&7\det\left(\begin{array}{ccc}5&1&1\\1&6&1\\1&1&7\end{array}\right)+6(5)\det\left(\begin{array}{cc}5&1\\1&1\end{array}\right)\\ &=&7\det\left(\begin{array}{ccc}5&1&1\\1&6&1\\1&1&7\end{array}\right)+6(5)(4)\end{eqnarray*}[/hide]Line 1 follows from the fact that if you have two matrices that are exactly the same except for one corresponding line, then the determinant of their sum is equal to the determinant of one matrix with the two different lines' sum.
Line 4 follows from the fact that if you have two of the same row in your matrix, the determinant is 0.
Line 5 follows from doing a row swap once and a column swap once, so the determinant is unchanged. (true for any $ n$)
Lines 6 and 7 follow from doing the splitting and simplifying steps.

So if you define the nth matrix satisfying the requirements as $ A_{n}$, then $ \det(A_{n})=\frac{(n+2)!}{3!}+(n+3)\det(A_{n-1})$.

[quote="MellowMelon"]The way I had thought this was supposed to be done was subtracting the first row from all of the others (an elementary row operation that does not change determinant). This gives you a main diagonal that goes $ 5, 5, 6, 7, 8,\ldots n\plus{}3$, a first column of $ 5,\minus{}4,\minus{}4,\minus{}4,\ldots$, a first row of $ 5, 1, 1, 1,\ldots$, and every oth element $ 0$. If you use Leibniz's rule now, the $ 0$s make it so you only have to compute the product of $ n$ diagonals.

Only thing is, I'm not perfectly well versed with the rule of when to multiply by negative one and when not to. I'll have to look it up some time, but does anyone want to finish it out?[/quote]

Using this (correct) method, I found the same as minsoens: $ D_{n}\equal{} (n\plus{}3)D_{n\minus{}1}\plus{}\frac{1}{6}(n\plus{}2)!$, where $ D_{n}$ is the determinant of the $ n\times n$-matrix. Therefore (for $ n\geq 3$),\[ \begin{aligned}\frac{D_{n}}{(n\plus{}3)!}& \equal{}\frac{D_{n\minus{}1}}{(n\plus{}2)!}\plus{}\frac{1}{6(n\plus{}3)}\equal{}\frac{D_{n\minus{}2}}{(n\plus{}1)!}\plus{}\frac{1}{6(n\plus{}2)}\plus{}\frac{1}{6(n\plus{}3)}\\ & \equal{} ... \equal{}\frac{D_{2}}{5!}\plus{}\frac{1}{6}\sum_{k\equal{}6}^{n\plus{}3}\frac{1}{k}\\ &\Rightarrow\ D_{n}\equal{} (n\plus{}3)!\left(\frac{29}{120}\plus{}\frac{1}{6}\sum_{k\equal{}6}^{n\plus{}3}\frac{1}{k}\right)\end{aligned}\]

In the following post, I had shown how to evaluate such determinants using a formula for the determinant of the sum of two matrices:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=591700#591700[/url]

To be specific, let $ J$ be the all-ones matrix.  Let $ D$ be the diagonal matrix with diagonal entries 4, 5, ..., $ n\plus{}3$.  Our problem is to evaluate the determinant of $ D\plus{}J$.  Using the formula, we just need the cases when zero or one rows are taken from $ J$, and the remaining rows are taken from $ D$.  We get the sum
\[ 4 (5)\cdots (n\plus{}3)\left(1\plus{}\sum_{k\equal{}4}^{n\plus{}3}\frac{1}{k}\right).\]
That answer is equivalent to Kurt Godel's.

[quote="MellowMelon"] If you use Leibniz's rule now, the $ 0$s make it so you only have to compute the product of $ n$ diagonals.[/quote]

What does the Leibniz's Rule say?

[quote="MellowMelon"]I didn't end up finding any references to it, so I'm going have to ask sylow_theory exactly what Leibniz's rule is. Basically, what I meant to say was the method listed in my post in this thread (with permutations), so that may or may not be it.[/quote]

Well, I don't really see it exactly. Can you state your method?


Give $ x>0;y>0,x \neq y $ and $m,n$ are positive integers.
Prove that:
[tex] x^2+y^2 > (\frac{16}{9})^{\frac{1}{m+n}}[/tex] if
[tex] x^n+\frac{1}{x^m}=y^n+\frac{1}{y^m}[/tex] :)

old problem.....
over a year unsolved....
can you give a try??...
 :)


This is an odd probability question I was given.  My teacher forgot the answer to it and can't remember how it is supposed to be done.

But here it is for you guys:

[hide="What I think is the Answer"]If I have a regular deck of playing cards (there are 52 cards) and I drop the top card, what is the probability that the next card that I draw is a king?

What are the cases that can come up since we have removed the top card?

1a. The probability of the card dropped being a king is $\frac{4}{52} \Rightarrow \frac{1}{13}$.
2a. The probability of the card dropped not being a king is $\frac{48}{52} \Rightarrow \frac{12}{13}$

Let's expand on these two events

1b. In the first case, what is the probability that the next card is a king? $\frac{3}{51} \Rightarrow \frac{1}{17}$.
2b. In the second case, what is the probability that the next card is a king? $\frac{4}{51}$.

Multiply these two subcases by their parent cases and add.

$\frac{1}{13} \cdot \frac{1}{17}+\frac{12}{13} \cdot \frac{4}{51} \Rightarrow \frac{17}{221}$[/hide]

If this is wrong... Let me know...

[hide="Perhaps A Faulty Method Of Thinking"]When the cards are first put in order, the probability that the card second from top is a king is 1/13. However, if after we took off the top card we reshuffled the cards, then I agree with your answer.[/hide]

Actually, both have the same answer, I think you added wrong.

[quote="Gyan"]You do notice that $(17/221)$ is same as $(1/13) $   The probability that the second (or for that matter 27th card ..or any other card)  being a king (or a queen, for that matter)  is simply $(1/13)$ :)

BTW - If I am not mistaken in Mathcounts, you may  have to put the answer in the simplest form, so  it might be a good idea to see if you can simplify the fraction.....:) Have fun.,,[/quote]

Ha! that was stupid.  I can't believe I went through all that work for nothing, and I even forgot to reduce...

[b]I still don't know why you multiply the parents and the subcases![/b]


1. consider 10 ppol sitting around a circular table. how many different ways can they change seats so that each person has a different neighbor to the right?

2. in how many ways can we place r red balls and w white balls in n boxes so that each box contains at least one ball of each color?

[hide="Number 2"]
Stick one of each ball into each box so that there are $ r\minus{}n$ reds and $ w\minus{}n$ white balls left.
Then from balls and urns it's just $ \binom{r\minus{}n\plus{}w\minus{}n\plus{}n\minus{}1}{n\minus{}1}\equal{}\boxed{\binom{r\plus{}w\minus{}n\minus{}1}{n\minus{}1}}$
[/hide]

[hide]I'm not 100% sure that this will work out, but you might want to try using recursion on #1[/hide]


So I have been wondering, what exactly is the difference between Financial Engineering, Financial Mathematics and Computational Finance?

I am strongly considering pursuing a PhD in Financial Math at some school that offers it, and I'd like to know if I will be less preferred than someone with a Financial Engineering degree for a quant job, or some nonsense like that. Is the demand for people from these degrees going to decrease due to the economic downturn?

I'm aware that the most important skills I can have are good programming skills and a keen brain, but I figure taking the right classes really can't hurt.

The current schools on my short list that I'm confident are not shams (yeah, I'm looking at you, Fordham): Carnegie Mellon and Stanford (though they don't have a PhD program)

If anyone knows anything about the real reputation about any other programs, I would certainly love more insight. Since I live in Tallahassee, Florida State is naturally my backup school, and they surprisingly have a PhD program in FinMath.

Other related questions: I can finish both my pure math degree and my actuarial science degrees in two years (i.e. Summer 2010). Should I just apply to grad school, work for a year doing some sort of entry level actuary work, take more classes at the graduate level at Florida State without really pursuing a graduate degree, try and do some research with some of the FinMath professors at FSU or [sarcasm] work on my novel[/sarcasm]?

Options 2, 3, and 4 and be done at the same time, or at least 2 and 3.

you do not need to have PhD in fin. math to [b]work [/b]as a quant. In fact, most fin. math PhDs stay in academia.
Do PhD in some real science instead (math/physics).

and, regarding your main question, no, there is not much difference between quant finance / fin engineering / fin math.

...and, in general, it is an awful time now to think about career as a quant. the glory days of investment banking are over  :roll:

[quote="Different"]...
...and, in general, it is an awful time now to think about career as a quant. the glory days of investment banking are over  :roll:[/quote]

Of course this year has been a bad year all around for the economy, but I'm curious why you'd think that being a quant is suddenly a bad long term career choice.  Is Moore's Law no longer in force or something?  As long as Moore's Law is in play, there is going to be a need to make quantitative trading decisions in shorter and shorter time frames--already far too short for direct human involvement in a lot of trading decisions.

Needless to say with the economy as a whole having taken a major hit that has affected quant roles too, but I would expect quant positions to be one of the first things to bounce back.  Just as security type roles experienced a major boom in the 2000's following an initial post-9/11 shakeout, I would expect financial risk management to be THE big career of the 2010's.  People are going to want more financial security and it is not clear to me how this can be done--in a hi tech world--without quants.

Now might be about the right time to start earning some credentials in this area.


[quote="cubedude"]though it could be better if not for all the learning and online classes. [/quote]

I don't think that you quite grasp the reason for this website :rotfl:

This is for MATH.

[quote="junggi"][quote="cubedude"]what do you people think about this site?? it is not bad.though it could be better if not for all the learning and online classes. i originaly got here by a rubix cube step by step help site... and i somehow got to this. just respond to me about what you think of the site.[/quote]
could someone PLEASE lock this guys topics.
they are pretty useless and offending.[/quote]

yeah, i beleive you must leave.

[size=200]Now[/size]

this is a website for MATH people.

I LOVE THIS SITE! All said. But I don't think it is a replacement for school if that was what you meant. Where is this cube help site? i am still puzzling over the helpsite i use...?

LMAO that's like criticizing Waffle House because it's not a good place to go and eat dinner.

EWWW I HAte waffle house's dinners! it is SOOOOOOOOOO greasy!
OK I just had to say that.


In the triangle $ ABC$, the bisectors of $ \angle A$ and $ \angle B$ meet at the incentre $ I$, the extension of $ AI$ meets the circumcircle of triangle $ ABC$ at $ D$. Let $ P$ be the foot of the perpendicular from $ B$ onto $ AD$, and $ Q$ a point on the extension of $ AD$ such that $ ID \equal{} DQ$. Determine the value of $ (BQ \times IB)/(BP \times ID)$.

we have $ \widehat{BID}\equal{}\widehat{DBI}\equal{}\frac{1}{2}(\widehat{A}\plus{}\widehat{B})$ so $ DQ\equal{}ID\equal{}BD\equal{}DC$ therefore $ D$ is the circumcenter of quadrilateral $ BQCI$ and so $ \widehat{IBD}\equal{}90$ so $ BQ.ID\equal{}BP.IQ\equal{}2BP.ID$,hence so the ratio equal to $ 2$.

Nice. I just used an equilateral triangle in the contest  :D


[i] Από το εξαιρετικό βιβλίο του Sharygin σας προτείνω την παρακάτω άσκηση :[/i]

[b] ΑΣΚΗΣΗ[/b]

 Από το ορθόκεντρο  $ H$ ενός οξυγωνίου  τριγώνου  $ ABC$  φέρνουμε τις προβολές  $ K$  και   $ L$ πάνω στην εσωτερική και την εξωτερική διχοτόμο της γωνίας $ B$.Να αποδειχθεί  ότι η  ευθεία $ KL$  περνάει από το μέσο της πλευράς $ AC$.

 [u]Μπάμπης[/u]

τι ειναι η EMO ???

Είναι η κίνηση Emotional  :rotfl: Βασικά πιστεύω ότι μετά από μια τέτοια άσκηση γίνεσαι πραγματικά over-emotional που δεν την έλυσες. :lol: Κανονικά είναι International Mathematical Olymbiad.

[quote="kritikos"]τι ειναι η EMO ???[/quote]

Επειδή κυκλοφορούν πολλά ανεκδοτάκια για τους emo θα μπορούσε να πει κανείς ότι ΕΜΟ είναι ο διαγωνισμός στον οποίον συμμετέχουν emo μαθηματικοί :rotfl:  Πάντως τέτοιο διαγώνισμο δεν έχω ξανακούσει...μάλλον ο κ. Στεργίου θα εννοεί IMO ή BMO (καλά...αν πραγματικά υπάρχει τέτοιος διαγωνισμός πάντως θα είναι πολύ επίκαιρος :P ).

Σε ότι αφορά το πρόβλημα τώρα έστω $ O$ το σημείο τομής $ BH, KL$ kai $ M$ το μέσο $ AC$ και $ D$ το σημείο τομής $ BH, AC$. Τότε επειδή $ BHKL$ ορθογώνιο θα ισχύει $ BO\equal{}OH$ και $ \angle{DOL}\equal{}2OBL\equal{}2(\frac{B}{2}\minus{}90\plus{}A)\equal{}B\plus{}2A\minus{}180\equal{}A\minus{}C$ (έχω υποθέσει χωρίς βλάβη της γενικότητας ότι $ A>C$).
Αρκεί να αποδείξω τώρα ότι $ \angle{DOM}\equal{}A\minus{}C$ το οποίο όμως αποδεικνύεται εύκολα λόγω του κύκλου των 9 σημείων αφού (αν θεωρήσουμε $ P$ το σημείο τομής $ AB, CH$) $ \angle{DOM}\equal{}DPM\equal{}APM\minus{}DPA\equal{}A\minus{}C$

Q.E.D

Γιατί να μην εννοεί Εθνική Μαθηματική Ολυμπιάδα? :D

[quote="kritikos"]τι ειναι η EMO ???[/quote]

Το βρήκατε ήδη , αλλά ας το επιβεβαιώσω : Είναι η εθνική μας μαθηματική ολυμπιάδα.

 Την λέμε και έτσι καμιά φορά .

  Μου άρεσε πάντως η ... φαντασία σας με τους  ΕΜΟ! :)  !

[u] Μπάμπης[/u]

Δημήτρη και Μανιέ προβλέπω να σας πετυχαίνουν πουθενά καμιά σπείρα από αυτούς και να μην ξέρετε από πού να φύγετε... :lol:  :D


There are 16 blank cards in a cube. A man walks over the cube and randomly decides to paint some red.
the probability that 1 card is painted, compared to 2 cards, and 3 cards etc is 1:2:3:4.....:14:15:16
After he is done I draw 4 cards without replacement; what is the probability that at least 2 of them are red?

I tried using brute force, but that didn't seem to work.

I did this question by finding a general formula for each n number of cards he painted and plugging it into a calculator.
Here is my equation. (x\y) means x choose y
n{(n\2)[(16-n)\2]+(n\3)(16-n)+(n\4)}/[(16\4)*136]
When you do all that you should get:
204204/247520=33/40
[b][i]33/40[/i][/b]