Good luck guys and girl! kick some asses! make our country proud!!

Since this is an experienced, power house team, how much ass should I expect to be kicked? I think you guys should make it your goal to get mentioned on the six oclock news.

"Breaking news: Canadian Mathies pwn  SO HARD"   :D

Yay GO CANADA!!!!!!!!! :moose:

I know you'll all make us proud!!!!!!

Yes, you have to make some good stuff happen because next year's team would really not be that experienced. So the next time  Canada has a shot at a top-10 ranking or something around a 180 score might be years(Perhaps 2 or 3) from now. Anyway, best of luck, and make us all proud.   :)  :lol:

no pressure guys... whatever the results r, we're proud of u! just enjoy urselves!

may the wisdom of Canmoo be with you! :moose:

Good luck guys! Canada is proud of you. :omighty:

[quote="J-I-M-B-O"] kick some asses![/quote]

!??!?!?

[quote="Singular"]did someone call pizza hut?[/quote]

becuase someone is gonna get PSOWND!

Since tommorow is Day 1 of IMO 2006, good luck team, and GO CANADA! :lol:
Note to everyone else: Support your team by wishing them luck!!!!

Team canada: May your complements be approximated by zero. :D

Can't wait to get on campus with the students for my second attempt of capturing the moose  :evilgrin: this time I have infiltrated a spy in the Canadian team :)

lmao, the real life equivilant of mafia, at imo 2006! This is why they should have brought alex wice: he plays mafia for like an hour a day on 2+2, he would sniff out VV's spy in a second.

GOOD LUCK TOMORROW! You guys should own! :lol:

yay canada []*[] pawn those communist countries  :P  (including U.S.)

Congratulations Team Canada.  This is a definite improvement from last year.  We're proud of you!

congrats Team Canada!
Canmoo is the coolest mascot ever!
http://imo2006.dmfa.si/images/20060711/slika18.jpg

[quote="rem"]I wonder how many times Canmoo got moosenapped. :lol:[/quote]
3 times. Swedish team, Isreali team, and USA team successfully managed to moosenap Canmoo. British team was trying to steal the moose while Isrealis had him, so they failed. :rotfl:

Hi everyone!  It's great to see that we did so well.  No gold is kinda a bummer, but it looks like IMO was pretty hard this year.  I do think #1 was a bit too easy for IMO, yes?

#3 is much nicer than #3 last year, and #6 is a really nice geometric inequality.  Were there any special prizes given out for great solutions?

That's funny about the moose.  You'll all have to post your pictures when you get back.  IMO creates the best of memories! :D

Good job guys!!!!

Sorry for the late post.. I haven't been checking the internet lately, and now, I'm mass-posting stuff!!

But I'm so proud of you guys!!!!!!!!!!!!!!!!!!!!!

[quote="TheDreamer"][quote="rem"][quote="lightrhee"] British team was trying to steal the moose while Isrealis had him, so they failed. :rotfl:[/quote]
That's hilarious :rotfl:[/quote]

:lagh

And looks like I'm the last one to congratulate the team TT..but AMAZING job, i'm sure you guys owned soccer too, and GOOD LUCK AT IOI!!!!!!!!!!!!!!!!!!!!!!!!!  BTw, wouldn't the IOI training coincide with the IMO training?[/quote]
I'm pretty sure that Canadian IOI team doesn't have training camp.

[quote="lightrhee"][/quote][quote="TheDreamer"][quote="rem"][quote="lightrhee"] British team was trying to steal the moose while Isrealis had him, so they failed. :rotfl:[/quote]
That's hilarious :rotfl:[/quote]

:lagh

And looks like I'm the last one to congratulate the team TT..but AMAZING job, i'm sure you guys owned soccer too, and GOOD LUCK AT IOI!!!!!!!!!!!!!!!!!!!!!!!!!  BTw, wouldn't the IOI training coincide with the IMO training?[/quote][quote="lightrhee"]
I'm pretty sure that Canadian IOI team doesn't have training camp.[/quote]
They didn't have training for the IOI team last year. Instead, they sent us to the USA Invitational Computing Olympiad. I'm not sure about this year though.

Some questions:

What is IOI?

Which were your results finally at football (soccer)? (all matches)

Was there any special prize for a nice solution?

Thanks.

IOI is the international programming competition for high school students.

Thanks rem for replying. I found what I was looking for. What about the other 2 questions??? (Maybe to the Canada competitors at IMO)

[quote="José"]Which were your results finally at football (soccer)? (all matches)
[/quote]
Canada 1:0 Netherlands (penalty shootout)
Canada 0:1 I don't remember which country we played (penalty shootout)
Playing time was very short, so most of the games went to penalty shootout with 0:0.

[quote="José"]
Was there any special prize for a nice solution?
[/quote]
No, not this year.


Could some one explain with an example what component interval menas:  acc to apostol,
Let $ S$ be an open subset of $ \mathbb{R}^{1}$ . An open interval $ I$ (may be either finite or infinite) is called a component interval of $ S$ if $ I \subseteq S$ and if there is no open interval $ I \subseteq J \subseteq S$.
PLz give an example or two.

Theorem: An open subset of $ \mathbb{R}$ is a disjoint union of open intervals.

This definition is part of the proof of that theorem, and isn't very interesting otherwise.

Indeed the proof of this theorem easily uses the idea of component interval.


search
function
linear algebra
linear algebra unsolved

I was reading http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=436295929&t=138328, I used the search function but couldn't find the following question on this forum.

Let $ A \in \mathcal{M}_{n}(\mathbb F)$ and $ C(A)$ be a subspace of $ \mathcal{M}_{n}(\mathbb F)$ consisting of matrices commuting with $ A$. In http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=436295929&t=138328, it is shown that $ \dim C(A) \geqslant n$.

Question: Find a sharp upper bound of $ \dim C(A)$ when $ A \neq \lambda I_n$.

$ (n\minus{}1)^2\plus{}1$ is the upper bound. It is obtained with a projector on a hyperplane.
If $ u_1\plus{}\cdots\plus{}u_k\equal{}n$ then $ u_1^2\plus{}\cdots\plus{}u_k^2\leq(n\minus{}u_k)^2\plus{}u_k^2{\leq}{(n\minus{}1)^2\plus{}1}$.

There is even an explicit formula for the dimension of the commutant. However, just the upper bound follows also trivially from Burnside's theorem, because the commutant is obviously a nontrivial subalgebra of $ M_n(k)$, thus it has a stable subspace.

Are you referring to this explicit formula, or do you know another one?
$ \dim Com(A)\equal{}\sum_{i\equal{}1}^r (2r\minus{}2i\plus{}1) \deg p_i$

Hi Gabriel,
how do you find the upper bound with the Burnside's theorem.
It's unclear for me.

For Jcmath: yes, I refer to that formula.
  To Loup blanc: indeed, this doesn't work, the upper bound I get is $ n^2\minus{}n\plus{}1$ and not $ (n\minus{}1)^2\plus{}1$. Sorry, it seems like I should start writing things on paper before posting them...


factorial
combinatorics unsolved
combinatorics

Use the numbers 3, 3, 7, 7 with basic operations (+, -, *, /) and braces to get 24 as a result.

24=7.(3/7+3) ... I don't know are there more ways, but If I have enough spare time I may write a program :).

P.S. Don't say to Arne I still don't know Latex...

This is one of the more well-known sets of four numbers that are difficult to get to 24. The solution above is unique, I believe.

Another one that is well-known is $ 3\ 3\ 8\ 8$. It uses a similar idea.

I would consider this more of a puzzle than a math problem. :P

I'm wondering what is the criteria of "difficulty"?
Difficulty in math is very interesting term. One thing that is difficult to me may not be difficult to you and reverse. "As I said I never repeated myself" ... but I think - difficulty in problems hides in the difference about people's way of thinking, their previous experience and ... facts they knows.
Second question: is there a common methodology for determining count of solutions of this kind of problems (maybe there exists some kind of formula)?

I think such kind of problem will be more difficult and interesting if we say that ... for example ... exponents, factorials, mod are allowed or sin, cos, log etc. ... and other numbers are used :). In this case I believe there will be very hard to invent some formulae or to invent some instrumentarium...


calculus
integration
analytic geometry
trigonometry
geometry
trapezoid
real analysis

Compute
$ \iint_{D}\left( y\plus{}\frac{y}{\sqrt{x^{2}\plus{}y^{2}}}\right)\, d(x,y)$
where D is a circle: $ (x\minus{}2)^{2}\plus{}(y\minus{}3)^{2}\leq 1$

I'll be satisfied with even hints only  :)

Look like this

[img]http://img228.imageshack.us/img228/261/clipboard02aa7.jpg[/img]

OK?

That can't possibly be right. The integral of $ y$ over the disk is $ 3\pi$, and the integral of the other part is positive.

I don't think polar coordinates work well here with the different centers- I'll try Cartesian for the hard part.
$ \int_{1}^{3}\int_{3\minus{}\sqrt{1\minus{}(x\minus{}2)^{2}}}^{3\plus{}\sqrt{1\minus{}(x\minus{}2)^{2}}}\frac{y}{\sqrt{x^{2}\plus{}y^{2}}}\,dy\,dx \equal{}\int_{1}^{3}\left[\sqrt{x^{2}\plus{}y^{2}}\right]_{y \equal{} 3\minus{}\sqrt{1\minus{}(x\minus{}2)^{2}}}^{y \equal{} 3\plus{}\sqrt{1\minus{}(x\minus{}2)^{2}}}\,dx$
$ \equal{}\int_{1}^{3}\sqrt{x^{2}\plus{}9\plus{}1\minus{}(x\minus{}2)^{2}\plus{}6\sqrt{1\minus{}(x\minus{}2)^{2}}}\minus{}\sqrt{x^{2}\plus{}9\plus{}1\minus{}(x\minus{}2)^{2}\minus{}6\sqrt{1\minus{}(x\minus{}2)^{2}}}\,dx$
Substituting $ t \equal{} x\minus{}2$, this becomes
$ \int_{\minus{}1}^{1}\sqrt{4t\plus{}14\plus{}6\sqrt{1\minus{}t^{2}}}\minus{}\sqrt{4t\plus{}14\minus{}6\sqrt{1\minus{}t^{2}}}\,dt$

Still not there, but it's at least a little closer.
To estimate the value, I made the trig substitution $ t \equal{}\sin\theta$ to get
$ \int_{\minus{}\pi/2}^{\pi/2}\cos\theta\cdot\left(\sqrt{14\plus{}4\sin\theta\plus{}6\cos\theta}\minus{}\sqrt{14\plus{}4\sin\theta\minus{}6\cos\theta}\right)\,d\theta$
and integrated it numerically in a spreadsheet; the value is approximately $ 2.58858114$.

(The trig substitution was done for numerical reasons; with it, I get 10-digit accuracy using an 8-interval trapezoid rule)

Thanks for the reply.
So it seems that this integral is not elementary :|

I can't say for sure that it's not elementary; there are a lot of tricks I haven't tried.
On the other hand, I don't see a good reason for it to be elementary.

[quote="Moubinool"]does maple 11 take care about the det(matrix jacobien) for the change variables
x=2+r.cos(t), y =3+r.sin(t)  ?[/quote]

Jacobian equals to $ r$, check the definition of $ f$ to see that.

[quote="petko"][b]bookworm_vn[/b] when using substitution as you used the angle doesn't change form 0 to $ 2\pi$ and also radius not from 0 to 1 - unfortunately it's not so easy :|[/quote]

Oh, I see... in fact $ 0\leq r\leq 1$ and $ 0\leq t\leq 2\i$ are totally wrong, I will check my answer very soon.

I don't think that's the problem; you parametrized the disk just fine. I think the problem is that Maple couldn't integrate the second term symbolically, and just dropped it without telling you.


the point of this thread is to think of a person, it will be best if they are famous, so everyone would know them, and describe a thing they are doing, and the next person should write what the person is thinking about.

i'll start

Bill Clinton observing the audience of Hillary Clinton's campaign speech.

Someone's getting lucky tonight  :wink: . Too bad it's not me  :( .

Andrew Wiles looking at a Counter-Example of Fermat's Last Theorem.

"What counterexample?  There's no counterexample."

Michael Jackson listening to the Jackson 5.

"AWWW yeah, put it on the down low man, I got the sweet goods and the moves to prove it, baby!"

George W. Bush right before he speaks on national television.

"Ok...time to ignore the new development in Iraq and concentrate on how I can sucker money out of America into my own bank."

Mr. R right before IMO.

Why can't IMO be more challengin?

Tom Crusie about to meet Tom Hank

"I hope he drops the soap in the shower"

Quagmire from Family Guy discovering he has AIDS

..................

Fermat (is he comes back to life) and reads Andrew Wiles' proof of Fermat's Last Theorem.

why should he get the credit? its my theorem. it even has my name on it!!!!

Donald Trump thinking about his latest architectural hotel in the middle of California National Park

I told them gold! Pure gold! But no, they had to put cheap titanium.

Andy Roddick before facing Roger Federer in the Wimbledon Final.

I wander if the Ball Boy has slipped him that poisonous frink yet....... Oh yeah, let's play!

Terence Tao is about to receive his Olympiad Gold Medal at the age of 13.

RinnngRinggg...Yes...hello?...oh yes wire that one million dollars to the judging council...what....oh okay.

Bill Gates watching John Elway's last game.

I can't believe I lost the title of World's Richest Person to a furniture maker  :mad: .

Friedel and Craft after accidentally discovering Friedel-Craft Reactions.

WTF?

Athiests getting beat up by priests!

Atheist: God shall punish u!
Priest: you dont believe in God, remember?
Atheist: oh yeah...

Hank Aaron witnessing a pitcher break his all-time HR record

That rotten @ss muthaf^&*%a is such an effin cheater. Ah well, I'll congradulate him anyway, cuz I got a huge arse bribe!

JK of course  :lol:

random pedestrian listening to his mp3 player

dam that chick across the street is hawt :lol:

An emo guy about to cut himself for the first time

stupid plastic knives don't work

Guy in the commercials who says "Can you hear me now?"

Man that guy on the other line must go nuts hearing me repeat the same 5 lines over 8 hours a day.

Any person ($ \in\mathbb{S}$ where $ \mathbb{S}$ is the set containing the Earth's population) who's about to skydive.

A kid being beaten up by his little sister o.O

Dude this is pathetic. I've gotta hit the gym.

A basketball being bounced on the ground (wow this is getting ridiculous).

ouch, ouch, ouch, lucky baseballs.

Barry Bond breaks the hr record.

Cha-Ching!! $ \$$ $ \$$ $ \$$

You eating sushi. :wink:

This is so much more addicting that beer...

You spending 5 hours straight on AoPS forums.

Ahhh so much latex and random symbols arbitrarily placed together in such great harmony! So many problems I have yet to solve but must conquer. *falls down from lack of excercise*.

Ronald Reagan on an online chatroom.

Neil Armstrong, receiving the Nobel just for saying "I didn't do it."

I'd like to say, good luck mr.gorscky.

search it up on google or u wont get it

i'll show them when i grow up and become a professional boxer/wrestler/kung fu master.

Jackie Chan and Bruce Li and Arnold Schwazenager and  Chuck Norris having a four way match.

[quote="youdonotknowme2"]Jackie Chan and Bruce Li and Arnold Schwazenager and  Chuck Norris having a four way match.[/quote]

Jackie Chan beats Chuck Norris, Bruce Li Falls to the Might of Arnie.

Jackie vs. Arnie obviously goes out with Arnie.

Arnold Schwazenager being praised with glory by Chucky.

They made my head too big in the Simpson's Movie.

Superman getting his red underpants out of the wash.

Do I seriously have to wear these silly things over my superman outfit? Makes me look queer.

Kid being yelled at because he's doing math instead of studying to learn how to drive (actually that's me lol).

YOU WOULD RATHER DO MATH THAN TO DRIVE!!!!!!!!!
i would give almost anything to drive.

who cares if i can drive or not, when i be the best mathmatician in the world, i will hire 1 no 2 person to be my personal driver, if i can get people to donate money to me.

a kid getting yelled at for not doing math

For the last time, it doesn't matter, it is just Geometry. :P

Harry potter on drugs.

I am invincible! The power of the Elder wand lies with me and me along, oh yeah! Now that I have 3 kids maybe I should consider another one, *nudge, nudge*, *wink, wink* at Ginny.

You wolf! No wonder you were banished!

Rodney Yee doing pilates.

This is nothing compared to yoga.

Avogadro taking credits for Chuck Norris' Number ($ 6.02\times 10^{23}$).

Chuck Norris thinking 'i'll show him! i'll roundhouse kick him so hard, he'll become a 5-year old again!'

Homer Simpson teaching Spider Pig how to drive

Wow, i can't believe the pig learn to drive faster than i learned how to spell hi. wait, how do you spell hi, starts with a um..h, and end with a o, that's it!!

Shaq missing the free throw for the 10th time.

WTF! They should just ban free throws so I'm the best basketball player ever, man! But NO, stupid NBA board just HAS to have free throws. Doesn't care about it's players, just wants free throws. Well, better get going and get my 15 millions dollar paycheck. They're not giving me enough for missing my freethrows! There should be compensation!

Santa Claus the night before Christmas learning that the elves stole his reindeer to go the the local pub.

dang it, those short f*c%en as&*es, i knew i should have used those powder after i fire those f*** elves. I'm hiring dwarfes after this.

Tiger wood getting beat up by Drew Carey

I couldn't run anywhere, he was too fat. That is when he rolled over me, paralyzing me neck down.

Ubemaya finding out that Meggy-Meg loves him.


function
limit
calculus
calculus computations

Let $ f(x) \equal{} 0$ if $ x$ is irrational and let $ f(\frac {p}{q}) \equal{} \frac {1}{q}$ if $ \frac {p}{q}$ is in lowest terms. What is the function $ g$ defined by $ g(x) \equal{} \lim_{y\to\ x}f(y)$?

Basically, just find the limit mentioned above...
I'm not sure where to start.

After thought, my answer would be the same as the original function, except plug in y for x.

Anybody can help me on this??

[quote="rofler"]Because you need ever increasing denominators to get better and better approximations, the answer is 0.

Think about it this way. Fractions with n digit denominators can only approximate so well. Therefore, if the closest approximation with n digit denominator is a, then if y is between x and a, it needs a bigger denominator. Continuing this way, we see that the denominator is unbounded.

Q.E.D.[/quote]

Here's my answer after even some more thought...
the limit goes to 0 if x is irrational and DNE(Does Not Exist) is x is rational.

Here's my reasoning:
If x is irrational, it requires larger and larger denominators to approximate it well. So, the function evaluates closer and closer to 0. But, if x is rational, there will always be irrational values close-by that evaluate to 0 and also the rational value which evaluates to some positive number. A limit cannot converge to two values.

Does it sound legit to anyone?

iamhe: when $ x$ is rational, you should discount $ f(x)$ from the limit -- we care about the behavior of $ g$ near $ x$, but not necessarily at $ x$. So the fact that $ f(x) > 0$ does not affect the limiting behavior.


I have $ 3$ positive whole numbers $ x,y,z$. Every turn, I choose one of these numbers and change it to $ x\plus{}y\plus{}z$
I start the game with $ 1,2,3$. Find all positive whole numbers $ a$ and $ d$ so that I finish the game with $ a\minus{}d$, $ a$, and $ a\plus{}d$

$ a\equal{}2, d\equal{}1$ is the only solution.

So clearly $ a\plus{}d\equal{}a\plus{}(a\minus{}d)$, so $ a\equal{}2d$.  Then we think: what number came before $ a\plus{}d$?  Whatever number it was, it created $ a$ when added to $ a\minus{}d$.  So it must have been $ d$.  But clearly it's impossible for us to have two of the numbers be equal.  Therefore, there could never have been a step preceding $ a\minus{}d, a, a\plus{}d$.  So we're done.


If the equation x^3 + ax^2 + bx + c =0, has roots p, q and r,

   (i) find a, b and c in terms of p, q and r.
   (ii) determine p^2q^2 + q^2r^2 + r^2p^2 in terms of p, q and r.

[hide="hint"]if a polynomial $x^3+ax^2+bx+c$ has roots $r$, $s$, $t$, it can be written as $(x-r)(x-s)(x-t)$, do you see why?[/hide]

Here is the problem typed nicely:

If the equation $x^3 + ax^2 + bx + c =0,$ has roots $p$, $q$ and $r$,

(i) find $a$, $b$ and $c$ in terms of $p$, $q$ and $r$.
(ii) determine $p^2q^2 + q^2r^2 + r^2p^2$ in terms of $p$, $q$ and $r$.

Question #1 (Good idea to memorize this answer... it becomes useful in many algebraic problems where you don't know roots but have to manipulate them)
[hide]$(x-p)(x-q)(x-r)$
$x^3-px^2-qx^2-rx^2+pqx+prx+qrx-pqr$
$x^3-(p+q+r)x^2+(pq+pr+qr)x-pqr$
$a=-(p+q+r)$
$b=(pq+pr+qr)$
$c=-(pqr)$[/hide]
I am busy tonight so I will do the 2nd problem if I have time later...
EDIT: Uhm what does the 2nd problem ask cuz right now the answer is what is asked.

[quote="guile"](ii) determine p^2q^2 + q^2r^2 + r^2p^2 in terms of p, q and r.[/quote]The problem is already solved, determine that expression in terms of p,q,r? But it's already in terms of p,q,r  :!:  :rotfl:


integration
calculus
real analysis
function
calculus computations

Let {an} be a strictly decreasing sequemce in [0,1] such that lim n->inf an=0
Define f:[0,1]-> R by f(x)= 1 if x= an for some n
                                 f(x)= 0 if x$\neq$an for some n

Show that f is integrable
Show that $\int_{0}^{1}$f = 0

[quote="ABM"] f(x)= 0 if x$\neq$an for some n[/quote]

You surely mean "for every $n$". $f=0$ a.e whence $f$ is Riemann-integrable and its Riemann integral coincides with its Lebesgue integral, meaning $\int_{[0,1]} f dm =1$.

Careful with that- you need to say more than "0 almost everywhere"  to say that a function is Riemann integrable.

Obviously, the integral should be 0 (all lower sums are zero); we just need to find partitions that make the upper sum small. For a constant $\epsilon$, find $N$ such that $a_n<\epsilon$ if $n>N$. Then the partition with cuts at $\epsilon$ and $x_k\pm\frac{\epsilon}N$ for $k\le N$ has upper sum at most $\epsilon\cdot 1+N\cdot\frac{2\epsilon}{N}\cdot 1=3\epsilon$. This can be made as small as we like, and therefore $f$ is integrable with integral 0.

Yeah, I've already corrected that in the other post with the same title  :D


vector
linear algebra
linear algebra unsolved

Hi Guys
PLS CAN YOU CHECK FOR ME THE BASIS
REAL MATRICE 3*3   WITH TRACE =0 (TRACE IS SUM OF DIAGONAL)

Thank u in advance

Call the set of matrices which satisfy $\text{tr}(A)=0$ and $A\in M_3(\mathbb{R})$ $S$.

The matrices $[\delta_{ab}]$, where $a\not=b$, are linearly independent and have trace zero.

Now consider the $3$ matrices which are zero off diagonal, and have diagonal entries $(1,-1,0)$, $(1,0,-1)$, $(0,1,-1)$.  These are clearly independent when adjoined to the above set of matrices, and all have trace zero.

Now $\dim(M_3(\mathbb{R}))=9$.  Therefore $\dim(S)\leq 9$, since $S\subseteq M_3(\mathbb{R})$.  The matrices I have given you are linearly dependent and belong to $S$, so their dimension is also $9$.  But this means that they are a basis.  (and also that they span $S$!)

Thanks for the correction :blush: .  I don't know what I was thinking - if you set $a+b+c=0$, then once you choose $a$ and $b$, $c$ is uniquely determined.

[quote="Kent Merryfield"]blahblahblah -

Those three diagonal matrices are dependent. If you want a basis for $S,$ take any two of them but not the third.

$\dim M_3(\mathbb{R})=9,$ but $S$ is the null space of a nontrivial linear operator from $M_3(\mathbb{R})$ to $\mathbb{R}.$ Hence $\dim S=8.$ We're looking for a basis with 8 elements - the 6 you had for the off-diagonal entries, and two diagonal matrices.

sparky_tx -

The question can never be "find [b]the[/b] basis of ... " Every vector space or subspace of dimension at least 1 has infinitiely many bases; if the dimension is at least 2, we have a vast profusion of choices for an acceptable basis. The best you can hope for is to answer the question "find [b]a[/b] basis of ... "[/quote]

Thank u , how about Real Matrice 3*3  WHAT IS bASIS
Mr Kent Merryfield can u explain why trace case since 3*3 there are many basis , just pls bare up with me , I am new with linear Algebra


superior algebra
superior algebra unsolved

for p=3 mod 4, $ a \in \{2,...,p\minus{}2\}$ we have that $ \minus{}a^2$ is a generator of $ (Z_p)^{\times}$

What exactly is this problem about¿
It is surely wrong for arbitrary $ a$, and the existence of such an $ a$ isn't anything difficult, either.
Please clarify!

$ p$ prime and $ (p \minus{} 1)/2$ prime too, sorry, and I meant for every $ a$.. So:

Let $ p$ be a prime number such that $ \frac {p \minus{} 1}{2}$ is prime too. For every $ a \in \{2,...,p \minus{} 2\}$ then $ \minus{} a^2$ is a generator for $ (Z_p)^\times$

It's true if $ q\equal{}{{p\minus{}1}\over 2}$ is an [i]odd[/i] prime, in which case $ p\equiv 3\ (mod\ 4)$ is an unnecessary assumption.  Proofreading your posts can save a lot of people a lot of time.

Since $ p\minus{}1\equal{}2q$ is the product of two primes, it suffices to show that $ \minus{}a^2$ does not have order $ 1$, $ 2$, or $ q$.  The first is clear, and $ (\minus{}a^2)^q\equiv(\minus{}1)^q a^{p\minus{}1}\equiv \minus{}1\ (mod\ p)$.

So we just need to show that $ (\minus{}a^2)^2\equiv a^4\not\equiv 1\ (mod\ p)$.  This follows from the assumptions and the fact that there are no square roots of $ \minus{}1\ (mod\ p)$.


So unfortunately I'd been exposed to programming a few years ago, but completely forgot it all over the last few years (I have basically no idea how to code right now).

So.....
1. What's the best way to learn Java from scratch, learning both Java syntax and general programming techniques?
2. Are there any online tutorials that are like the USACO training thingie, but for complete beginners?

I just used introductory Java books from the library, and they worked pretty well for me.

I dunno about a training website, but USACO has a series of competitions for people who are just learning how to program, and you can find questions used on those contests at http://ace.delos.com/ioigate, just look at the LEVEL1 contests.

Also, C++ is probably a better language to try to learn, at least IMO.

I think the best way to learn is to start by figuring out what a "shell" should look like just to get a basic program running, and the console I/O commands, then start learning about logic-based programming.

In JAVA, here's what your code might look like just to do something random:

//Following the double-forward slash, everything on that line will not be read by the computer

import java.io.*;
import java.util.*; //tells the program to pull up all of the commands defined in these two places

public class ClassName { //think of a class as a basic organizational object that you can stick all of your variables and function definitions in
  public static void main(String[] args) { //this is the thing that gets executed
  System.out.println("Dude, this program just did something.");  //prints a line of output. If you don't want a newline at the end, just do System.out.print
  //the "" cause stuff to be interpreted as a string of characters rather than as code
  System.exit(0);
  }
} //brackets act as your basic object to denote the beginning and end of stuff

In C++, here's what you do

#include <fstream>
#include <iostream>

using namespace std;

//the above code does basically the same stuff as the import commands in JAVA

//we don't need classes in C++

int main() { //this is the thing that gets executed
  cout << "Dude, this program just did something." << endl;  //prints a line of output. If you don't want a newline at the end, don't use endl
  //the "" cause stuff to be interpreted as a string of characters rather than as code
  system("PAUSE"); //makes the program pause before terminating, though LINUX doens't support this; you can use getchar() instead, I believe
  return 0;
  }
} //brackets act as your basic object to denote the beginning and end of stuff

Check out http://tfprog.com/wiki/doku.php?id=your_first_language . Gives really good advice on book choosing, what language to learn, etc.

I think C++ is difficult for beginners, which I guess could be a good thing if you're up to the challenge. I started off with Java, but would recommended an even simpler (more elegant) language such as Python, of which there are numerous books online.

I learned this random language that does nothing called TCL/TK first. That was only for a year (happily). The next year I started learning C++. At the end of the camp, I knew enough to be able to write programs that worked. I kept learning C++ until about two years ago, when my sister learned Java at school. I bought myself a teach yourself java book and learned enough. I stopped at the networking stuff, but I think I might start learning that again. I now use C++ because USACO showed me how great the time differences are.

Personally, I also think that C++ is the better language. Java is very useful for easily creating GUIs, which is what we do in school, but C++ is probably better for implementing algorithms. As for learning python,  go for it, but after you learn either C++ or java, you're probably going to only use python as a calculator.

You should not forget about C and Pascal -- these may not be all that useful languages compared to C++, .Net or Java but you'll learn much more about how your memory works, what are pointers, what are virtual functions and so on. Also, they may be easier to start with if you have no idea what are classes or static members.

I started learning Java recently, and I'm using a book, as well as looking at online sources. The book I'm using, Java Methods, by Maria and Gary Litvin, seems to be pretty good, from my limited viewpoint  :lol:

I have that same book, and it is pretty good. There's no substitute for learning the syntax and then jumping into USACO training though :).

Yeah, I had registered an account to use the USACO training pages, and looked at the first problem, when I realized:

Crap, I don't know how to use file I/O  :rotfl:

I looked ahead at the chapter in Java Methods on input and output stuff, but was confused by that, as well as the online Sun tutorial. But that was a while ago, I'll probably take another look soon.

Doesn't it explain how to do FILE I/O on the training pages?

Yeah, they did explain how to do file I/O on the Training Pages. I'll post it here in case you didn't see it:

[quote]The simplest programming challenge is named 'test' and requires you to read a pair of small integers from a single input line in the file 'test.in' and print their sum to the file 'test.out'.

Below is a simple solution in the 'C' programming language. Note the use of 'exit (0);', which is usually required to exit properly.
[hide="Solution in C"]
[code]
/*
ID: your_id_here
LANG: C
TASK: test
*/
#include <stdio.h>
main () {
    FILE *fin  = fopen ("test.in", "r");
    FILE *fout = fopen ("test.out", "w");
    int a, b;
    fscanf (fin, "%d %d", &a, &b);	/* the two input integers */
    fprintf (fout, "%d\n", a+b);
    exit (0);
}
[/code][/hide]

Below is a simple solution in the C++ programming language. Note the use of 'return (0);', which is usually required to exit properly.
[hide="Solution in C++"][code]/*
ID: your_id_here
PROG: test
LANG: C++
*/
#include <iostream>
#include <fstream>
#include <string>

using namespace std;

int main() {
    ofstream fout ("test.out");
    ifstream fin ("test.in");
    int a, b;
    fin >> a >> b;
    fout << a+b << endl;
    return 0;
}[/code][/hide]

Below is a simple solution in the PASCAL programming language:
[hide][code]{
ID: your_id_here
PROG: test
LANG: PASCAL
}
Program Test;
Var fin, fout: text;
    a, b: word;
Begin
    Assign(fin, 'test.in'); Reset(fin);
    Assign(fout, 'test.out'); Rewrite(fout);
    Readln(fin, a, b);
    Writeln(fout, a+b);
    Close(fout);
End.[/code][/hide]

And here is the same program, this time in JAVA. Note that the program presumes the file opens will succeed, which they will:
[hide][code]/*
ID: your_id_here
LANG: JAVA
TASK: test
*/
import java.io.*;
import java.util.*;

class test {
  public static void main (String [] args) throws IOException {
    // Use BufferedReader rather than RandomAccessFile; it's much faster
    BufferedReader f = new BufferedReader(new FileReader("test.in"));
                                                  // input file name goes above
    PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("test.out")));

    String s = f.readLine();                      // gets entire line
    int i1 = Integer.parseInt(s.split(" ")[0]);   // first piece becomes an integer
    int i2 = Integer.parseInt(s.split(" ")[1]);   // second piece becomes an integer
    out.write( (i1+i2) +"\n");                    // output result
    out.close();                                  // close the output file
    System.exit(0);                               // don't omit this!
  }
}[/code][/hide][/quote]

:wallbash:

Should have seen that. Thanks for posting it.

Ruby and python are awesome. I wish you could use them on USACO...

Well even if they let you, you'd have to deal with the fact that stuff in python randomly adds an order of magnitude to its execution for no apparent reason, plus the fact that it runs via an interpreter rather than a compiler makes your speed about 10 times slower than C++, so you wouldn't be solving many problems...

Also the IOI doesn't allow it, though it doesn't allow JAVA either.

Does anyone know if the IOI is planning on allowing java at some point? If not, it's just one more reason to also learn C++.

Well, the [url=http://www.ioinformatics.org/Newsletters5/ioi-newsletter6.html#Java]2006 IOI Newsletter[/url] has some information about using Java in the IOI (and why they decided not to allow it):

[quote="IOI Newsletter"]Java in the IOI

by Jyrki Nummenmaa and Rob Kolstad, March 19, 2006

The ISC and the ITWG have continued to follow up on developments in Java programming environments to assess the possibilities to incorporate Java into the IOI.

It appears that the Sun's Java 1.5 SDK is the only affordable Java system coming close to meeting IOI's need. The following evaluation is based on this choice and primarily on the statistical data from the USACO contests, including some other smaller-scale testing. USACO has run 5 monthly contests using this compiler, just under 5000 entries, of which 6% is Java.

The following summarizes these findings. Two major barriers prohibit the use of Java in the IOI in the present setup.

Firstly, the execution times are larger by a magnitude and dependent on how the solution is implemented. The overhead of starting a Java solution includes 210 ms overhead of startup overhead and 500 ms memory initialization overhead. The memory allocation seems to be highly CPU-intensive and the impact of Java garbage collection is not known well enough at the moment.

Only one particular style of Java file input runs close to fast enough for IOI-style input files of reasonable size, and even then there seems to be an input time overhead.

In general, assuming that x is the execution time of a conceptually comparable C/C++ program, the execution time of a Java program can be expressed as Ax+B, where A typically is around 2 and B could be nearly 1 second. However, A and B can vary significantly depending of the actions of the program (meaning not just the task but also the implemented solution strategy).

The second problem is memory consumption. It is very diffcult to estimate and thereby reasonably limit the memory consumption in the JVM environment. Also, this depends on the libraries loaded into the JVM, as they also consume the same memory, and there, of course, could be different choices for the libraries used in the program.

There are some proposals for overcoming these difficulties and the ISC and ITWG will continue to explore possibilities to solve these issues in the hope that they will be able to recommended the use of Java in the future.

However, at the moment we do not see Java as a fair choice for a programming language in the IOI. At the March review meeting it was decided not to allow Java at IOI 2006.[/quote]

So basically they're saying,

"Java runs too slow."

I don't see why there has to be such strict time limits; the ICPC style of competition works perfectly fine with java. Is it that hard designing problems which, instead of only allowing solutions which run in less than x seconds, with x+0.5 seconds not allowed; make it so that programs that run in O(n) should work, and O(n log n) shouldn't, for example? ICPC style makes much more sense to me.

The difference is that at IOI, you can get partial points for partial solutions.

It's probably quite possible to choose time limits such that the correct solution passes both in C++ and Java and anything worse than that won't pass in either language, but if you can get points for passing smaller test cases with slow algorithm, people using Java are at disadvantage here.

[quote="joml88"]If you want to get a really good introduction to computer science, you can always self-study [url=http://ocw.mit.edu/OcwWeb/Electrical-Engineering-and-Computer-Science/6-001Spring-2005/CourseHome/index.htm]6.001[/url] on MIT's Open Courseware.  They use Scheme.  I'm not sure exactly what the class is like, but it's the core beginning class in Course 6: Electrical Engineering and Computer Science at MIT.

If you want some more applied and less theoretical, there's 1.00 (either look [url=http://ocw.mit.edu/OcwWeb/Civil-and-Environmental-Engineering/1-00Fall-2005/CourseHome/index.htm]here[/url] or [url=http://ocw.mit.edu/OcwWeb/Civil-and-Environmental-Engineering/1-00Spring-2005/CourseHome/index.htm]here[/url]...the first is Fall 05 and the second is Spring 05 so the content should be similar in both).  1.00 is more of an intro to java and it focuses more on learning to hack instead of learning more formal and organized programs.  It's in the Civil and Environmental Engineering department.  I'll be taking it this semester.  So if you do end up self studying it feel free to let me know :D  You could even try following along with my class if you want the structure.  You probably wouldn't be able to make the lectures though :D[/quote]

http://www.amk.ca/diary/2006/11/mit_to_try_python_for_introduc.html

:D

Hi

The way i learned java was in my opinion very efficient.

Try http://www.bluej.org

Buy that book and use the program from the site. I think its the best way to learn java.

I don't think that's quite the book for absolute beginners.


Determine the infimum of the following set:

$\inf \left\{ \sum_{i=1}^\infty x_i^2: x_i \in \mathbb {R}, \, \sum_{i=1}^\infty x_i \geq 1\right\}$

Do you mean $\inf \left\{ \sum_{i=1}^\infty x_i^2: x_i \in \mathbb {R}, \, \sum_{i=1}^\infty x_i \geq 1\right\}$?

That infimum is zero. Try $x_i = \frac{1}{n}$ for $i \le n$, $x_i = 0$ for $i > n$.


inequalities
algebra
polynomial
complex analysis
complex analysis unsolved

Let $ z_1$, $ z_2$,..., $ z_n$ be $ n$ given complex numbers. By considering the polynomial $ (z \minus{} z_1)...(z \minus{} z_n)$, prove that there exists a complex number $ z_0$ such that $ |z_0| \equal{} 1$ and
\[ |z_0 \minus{} z_1| |z_0 \minus{} z_2|\cdots|z_0 \minus{} z_n| \geq 1.\]
Any hints are welcome, thank you very much.

Hint:
Prove the statement by contradiction.Apply Rouche theorem :wink:


Pat is to select six cookies from a tray containing only chocolate chip,
oatmeal, and peanut butter cookies. There are at least six of each
of these three kinds of cookies on the tray. How many different
assortments of six cookies can be selected?

The numbers of the types of cookies must sum to 6.  So the set of number of cookies must be:

$ (0,0,6)\\(0,1,5)\\(0,2,4)\\(0,3,3)\\(1,1,4)\\(1,2,3)$
or
$ (2,2,2)$.

There are $ \frac{3!}{2}\equal{}3$ orders for the sets $ (0,0,6), (0,3,3)$ and $ (1,1,4)$.

There are $ 3!\equal{}6$ orders for the sets $ (0,1,5), (0,2,4),$ and $ (1,2,3)$.

There are $ \frac{3!}{3!}\equal{}1$ order for the set $ (2,2,2)$.

Thus, the answer is $ 3 \cdot 3 \plus{} 3 \cdot 6 \plus{} 1 \cdot 1\equal{}\boxed{28}$.

[hide="Alternative Solution"]Consider 6 cookies lined up in a row:
\[
\bullet\; \bullet\; \bullet\; \bullet\; \bullet\; \bullet
\]
We need to separate the cookies into 3 groups; in other words, we have to place 2 dividers. For example, we could place the dividers like this:
\[
\bullet| \bullet\; \bullet\; \bullet\; \bullet\; \bullet|
\]
There are 8 total slots, so we can place the 2 dividers in $\binom{8}{2}=28$ ways. Each arrangement of dividers corresponds with one assortment of cookies, so the answer is $\boxed{28}$.[/hide]

[hide="Another one"]This problem is equivalent to placing 6 indistinguishable balls in 3 distinguishable urns. Applying balls and urns, the answer is $\binom{6+3-1}{3-1}=\boxed{28}$ [/hide]

$\vdots | \vdots | \vdots | \vdots | \vdots | \vdots | \vdots$ so there is 7 ways of putting a bar? i dont get where the 8 ways of putting the bar comes?

I don't get why the answer isn't 729. For the first cookie, you have three choices. For the next one, another three choices, etc.

It's not 729 because not all the cookies are different.  Remember, this would be like placing 6 indistinguishable balls in 3 distinguishable urns(credits to Iamteehee).  You would be over counting quite a lot of possibilities.

[font=TIMES NEW ROMAN][hide=Logical order of listing]
c=chocolate chip
o=oatmeal
p=peanut butter
-----
6c0o0p

5c1o0p
5c0o1p

4c2o0p
4c1o1p
4c0o2p

3c3o0p
3c2o1p
3c1o2p
3c0o3p

2c4o0p
2c3o1p
2c2o2p
2c1o3p
2c0o4p

1c5o0p
1c4o1p
1c3o2p
1c2o3p
1c1o4p
1c0o5p

0c6o0p
0c5o1p
0c4o2p
0c3o3p
0c2o4p
0c1o5p
0c0o6p $$\definecolor{color}{rgb}{0.7, 0.9, 0.9} \scriptsize{\colorbox{color}{\color{black}{A pifinity solution}}}$$

[quote=Mr.Ozil]i guess if u think about it, you have seven places to put the first bar, then 8 places after u put the first bar, then divide by 2 for over counting[/quote]

wait, how could you have 8 places after you put the first bar... where does 8 come from? Im having the same confusion as tfalkenberg...

How are there 8 places to put the first bar? I'm actually confused;
$1 \bullet 2\bullet 3\bullet 4\bullet 5\bullet 6\bullet 7$
These are all the possible placement of the "sticks" right? so then isn't it $\binom {7}{2}?$

There are 8 slots because we can place dividers this way:
\[
\bullet| | \bullet\; \bullet\; \bullet\; \bullet\; \bullet
\]


trigonometry
geometry
3D geometry
sphere
USAMTS
AMC
USA(J)MO

How'd you do? As well as you thought? (The leaderboard and at least my score are updated for round 2)

I too am waiting for that problem 5 rubric, I got [b]2[/b] for a complete and correct solution, or so I think. . .

Sounds like a goodly number of complaints shall most probably be filed on 5.

[quote="solafidefarms"]But 2,3, or 4 points off for no errors but brute force?![/quote]
Brute force solutions warrant full credit (unless some part of the problem indicates otherwise, such as "no computer use", or something like that).  You just have to be more careful when writing b-f solutions, as the graders will be picky and take points off for much smaller mistakes.

[quote="zanttrang"][quote="beta"]It seems if you used too much trig/calculator, your score will be lowered?

Here's a quote:

[quote="zabelman"]A quick note on brute force calculations: graders hate this type of solution almost as much as I do, so be careful. If you make a typo in one of the lines, they could potentially knock off a point out of spite. But
more importantly, if you fail to address a minor but necessary detail, they probably will knock off a point or
two for lack of total rigor (and out of spite).[/quote][/quote]
Trig isn't considered brute-force unless you do an obscene amount of it, and I was definitely able to solve the problem using a very reasonable amount of trig.  Granted I didn't use the overlapping spheres interpretation that was given in the math jam, but my interpretation is just as valid and it would be unprecedented to take points off for a different but completely correct solution.  They're holding up on posting the problem 5 rubric just to spite me, I can tell....  ;)[/quote]

Well I verified someone's solution(who is about to protest once the rubrik is posted), and his solution had a lot of calculations with trig, and a calculator is definitely necessary to evaluate the answer. Maybe yours requires less calculations with the calculator.

[quote="solafidefarms"]But 2,3, or 4 points off for no errors but brute force?![/quote]

Well it seems for any kind of error(even typo or some really trivial error) they have the reason to take off a point.

[quote="beta"][/quote][quote="zanttrang"][quote="beta"]It seems if you used too much trig/calculator, your score will be lowered?

Here's a quote:

[quote="zabelman"]A quick note on brute force calculations: graders hate this type of solution almost as much as I do, so be careful. If you make a typo in one of the lines, they could potentially knock off a point out of spite. But
more importantly, if you fail to address a minor but necessary detail, they probably will knock off a point or
two for lack of total rigor (and out of spite).[/quote][/quote]
Trig isn't considered brute-force unless you do an obscene amount of it, and I was definitely able to solve the problem using a very reasonable amount of trig.  Granted I didn't use the overlapping spheres interpretation that was given in the math jam, but my interpretation is just as valid and it would be unprecedented to take points off for a different but completely correct solution.  They're holding up on posting the problem 5 rubric just to spite me, I can tell....  ;)[/quote][quote="beta"]

Well I verified someone's solution(who is about to protest once the rubrik is posted), and his solution had a lot of calculations with trig, and a calculator is definitely necessary to evaluate the answer. Maybe yours requires less calculations with the calculator.[/quote]

Brute force calculations with integers are just fine. However, calculations with decimals usually have some rounding error. The possibility of rounding error means that such calculations do not count as a proof. On the other hand, figuring out the trigonometry in order to accomplish those calculations in Problem 5 is a worthy feat.

I do not know the rubric for Problem 5. I graded Problem 4.

Erin Schram
USAMTS grader

[quote="zanttrang"][quote="beta"]
Well I verified someone's solution(who is about to protest once the rubrik is posted), and his solution had a lot of calculations with trig, and a calculator is definitely necessary to evaluate the answer. Maybe yours requires less calculations with the calculator.[/quote]
Take a look at mine here and tell me what you think (it's my full solution set):
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=375126#p375126

I remember the solution you're referring to, and that might warrant points off, because he used approximations throughout and thus I don't think the answer was rigorous.  Mine, however, uses exact results throughout, with absolutely no calculator use necessary.  The grader's only comment was "we were looking for a geometrical (sic) based proof/solution."  I'm not really sure how that criticizes what I've done...[/quote]

nvm, it looks good.

[quote="Erin J. Schram"]
Brute force calculations with integers are just fine. However, calculations with decimals usually have some rounding error. The possibility of rounding error means that such calculations do not count as a proof. On the other hand, figuring out the trigonometry in order to accomplish those calculations in Problem 5 is a worthy feat.

I do not know the rubric for Problem 5. I graded Problem 4.
[/quote]

Yeah I agree, that's why I don't think the person deserve a 5. However that person probably does deserve more than 1(probably a 3, depending on the rubrik).

Edit: nevermind, you addressed your own concern.

Zanttrang, just wondering what was your comment?

[quote="zanttrang"]I'll hold off commenting until the grading rubric for #5 is issued and my dispute can be resolved (I got a 3 for a complete and correct solution???)[/quote]

Same here.  The comment said my solution was correct but they were looking for a non-calculator approach.  HOWEVER, I'm very sure that my solution clearly did not make use of a calculator or any decimal approximations.  It doesn't even seem that the grader read my solution fully...

[quote="JesusFreak197"]I don't think I'm going to continue taking USAMTS. It takes too much time and isn't very enjoyable for me, and I don't think the graders always do a very good job (not to be hard on them, I know they have a lot to do), plus I really don't like the rules for complaints. I can't complain when my score for a problem is one point off? Umm... my #1 had a good explanation except that I didn't see that the middle 2x2 has some cases where it cannot work, and yet I got a 3. By the rubric, I should have a 4, yet I can't complain. :huh:[/quote]
I typed up the rubric for problem 1, so I'll try to explain the reason for the rule about not complaining about a one-point difference.

Problems 1 and 4 were graded at the NSA. We graders work on a consensus system: we discuss the first few solutions we see (selected randomly) and design a general rubric based on those. Whenever a solution that is different from previous solutions appears, we decide talk at our table to decide where it fits in the rubric. The grading comes before the rubric.

The graders for Problem 1 started at one table, but eventually, enough new graders arrived to have eight graders split between two tables. At that point, it is possible that the two tables started to diverge in their rubrics, unless they were careful to communicate between tables. And the graders for Problem 4 finished first (the solutions to Problem 1 were longer on average, so took more time to read), so some of them switched over to grading Problem 1 at the end, which means many Problem 1 solutions were graded by people inexperienced in that problem.

Near the end of the grading, I walked over to the original Problem 1 table and asked them to explain their rubric to me. After the grading, I typed up my notes on the Problem 1 rubrc in a neat list and e-mailed it to Richard Rusczyk at the Art of Problem Solving. What is posted as the grading rubric is what I sent him. My notes were a little inadequate: the graders had a clear idea what constituted a "clash" but I had not asked for a detailed explanation.

The grading rubrics are only an approximation of the actual rules that the graders use. Even with our notes on the rubric in front of us, less experienced graders often ask the opinion of other graders on how to grade a borderline case. We graders also disagree with each other sometimes. Our rules for disagreements at the grader table is that the judgment of the grader stands unless we can demonstrate that he or she misunderstood what the student wrote.

So being one point below the rubric suggests could have several causes:
(1) The grader missed something in reading the solution and mistakenly gave it the wrong grade.
(2) The grader was harsher than most graders (perhaps someone who did not know the rubric well);
(3) The table grading that solution had a harsher rubric than the table I got the rubric from;
(4) The solution had some style problems that made the grader decide unfavorably in a borderline case;
(5) The solution was graded as all the graders graded that case, but I missed listing that case in the rubric;
(6) The solution was graded exactly as the rubric suggests, but the student misunderstood how the rubric applies to his or her problem.

Only case (1) demands giving the problem a different grade. Case (2) could be given a different grade. All the others should not have the grade changed. Richard Rusczyk at the Art of Problem Solving would be in a terrible strain to tell one case from the others when he has no opportunity to talk to the graders personally. We have the one-point rule so that he does not have to try.

We try our best, but we are only human.

Erin Schram
USAMTS grader

I'm kinda upset about my score. I got {5,[b]5[/b],[b]5[/b],4,5}. I'm upset because I have no idea what I did wrong on #4. I thought it was perfect, perhaps even commendation-worthy. The comments stated that I "missed a case," which I didn't. And I'm wondering why my #5 wasn't commended when it was more or less the same solution as given in the Math Jam. Have a look.

I really worked hard on Round 2 for a 25. I think people should be allowed to appeal a score of 4, because for a lot of people the difference between 4 and 5 can be the difference between perfect and [i]just not good enough[/i].

TheAmazingOne,

Your single formula looks correct, but one could've easily mistaken it as the one form n = 3^k - 2.  However, the other answer was n = 2*3^k - 2 or something like that.

Other than that, a quick skim tells me that your solution is pretty good.  I was too lazy to actually read the whole thing though. :D  :rotfl:

I got the same comment as TheAmazingOne. I had the same solution too. I got a three. I would put it up, but I handwrote them. Oh well.

[quote="TheAmazingOne"]Yeah, and my single formula makes more sense than having two formulas. My single formula follows directly from a [i]generalization[/i] of the problem, which I put in there to try to get a commendation. I really think I deserved a [b]5[/b] on that, but instead of my generalization getting commended, the grader misunderstood the generalization and took off a point.[/quote]
Your solution is correct but unintelligible. I would have given it a five with no commendation.

Generalizations often earn commendations, but in this case, your generalization made your proof much less clear. You wrote in a stream of consciousness style that made your flow of logic hard to follow. You used phrases such as, "$a$ is of the form $zb$," without describing what form means in this context (it would have been better to say, "$a$ is $zb$ where $b$ is an integer").

You were hard to understand, and by working with a generalization, you left the grader without the familiar points of reference for tracking your proof. He must have decided that your proof was incomplete because it lacked statements common to most $z=3$ proofs. But it wasn't incomplete.

Your solution is a good argument that the "no complaints about a one-point error in grading" rule should be suspended for a complete and correct solution that earned only four points. Unfortunately I do not have the authority to suspend that rule. However, since I am full of the Christmas spirit today, I recommend that you send a request to the Art of Problem Solving asking to have your score reconsidered, and tell them that the head grader for Problem 4/2/17, Erin Schram, says that your solution was misgraded.

Erin Schram
USAMTS grader
Merry Christmas

{[b]5[/b] 5 1 3 - } = 14

Don't know how I got a commendation on #1 :huh: Oh well :D

[quote="banblink182"][quote="Erin J. Schram"]
Brute force calculations with integers are just fine. However, calculations with decimals usually have some rounding error. The possibility of rounding error means that such calculations do not count as a proof.
Erin Schram
USAMTS grader[/quote]

Not to be too argumentative, but if I "stored" an irrational number as a letter using the TI-83, and then manipulated the value from there, there wouldn't be any rounding errors. Though the calculator represents the letter as a decimal approximation, the actual value is stored in the calculator.

For example, if I saved sqrt(3) as A, then even though the calculator approximates A as 1.732050808, if you type in $A^2$, you will get 3, and not 3.000000001 (which is the square of the rational number 1.732050808).[/quote]
In that case, you'd be getting lucky - the approximation would be good enough that it would give you the right answer when doing a computation.  However, that is never proof of a statement.  Squaring a 15-digit approximation of the square-root of 3, getting 3 on your calculator, and declaring that the square of the square root of 3 is therefore 3 is not a proof of this fact.

[quote="banblink182"][quote="Erin J. Schram"]
Brute force calculations with integers are just fine. However, calculations with decimals usually have some rounding error. The possibility of rounding error means that such calculations do not count as a proof.
Erin Schram
USAMTS grader[/quote]

Not to be too argumentative, but if I "stored" an irrational number as a letter using the TI-83, and then manipulated the value from there, there wouldn't be any rounding errors. Though the calculator represents the letter as a decimal approximation, the actual value is stored in the calculator.[/quote]That's not true.  The calculator stores a handful more digits than it displays, but it doesn't (can't) store the actual value.

For example, if I ask my calculator (TI-84) for the billionth root of 1.01, it tells me the answer is 1 (exactly).  When I then raise that result to the billionth power, it tells me the answer is 1.010050167, not 1.01.

[quote="1234567890"][quote="Philip_Leszczynski"]5^5 !!  :lol: I really wasn't expecting it...[/quote]

3125?[/quote]

Whoa...  I must've reeeeeeeally blew this round.  :P  :lol:

4, 5, 5, 4, 3:   21 total..
#4: I totally forgot the multiple of 2 case, but my proof covered it, so i got 4 points.
#1: I suppose my solution was correct...but VERY hard to read, and probably unclear too.  Hence, a 4.
#5: I used a brute force sort of method, and finished it with a calculator.  Hence, a 3.  I suppose i could have found a way to simplify that huge ugly trig expression, but that would have been a problem itself.
#2 and #3: Were fine.  I probably could have cut down on the wordiness for #3.

For Round 3, I'm going over my solutions as much as possible while I have time to make sure they're as clear and thorough as possible.

[quote="K81o7"]
#5: I used a brute force sort of method, and finished it with a calculator.  Hence, a 3.  I suppose i could have found a way to simplify that huge ugly trig expression, but that would have been a problem itself.[/quote]

The problem was relatively trivial with a calculator.  I spent god knows how long figuring out a way to make the trig work out nicely.

5 5 1 5 5 =21 (no commendations)
anyway, im pretty disappointed because right off the bat, i saw the answer for 3, but later wrote it off as wrong and took a completely different viewpoint on it

edit: removed post in comment to something that was in the previous page


function
irrational number
real analysis
real analysis unsolved

I need to show if the following is true or false.

If the function $ f: (0,1)\to R$ is continuous in every irrational number x then f is continuous at every number.

How about existence of functions discontinuous at just one point $ \frac 12$? ;)

... or even every rational. There's a famous example for that.

I have no idea what you're talking about. I've never heard that name.


Alright, I'm sure there's plenty of Final Fantasy fans here...there just needs to be...

Which is your favorite?

The reason I have I through V in one category is because they're old and somewhat similar. Very few ppl play them anyways

[quote="dts"]Come now, you left some out...

What about Final Fantasy Legend I through III (Game Boy)?  Final Fantasy Tactics Advance (GBA)?  Final Fantasy Crystal Chronicles (GameCube)?

My vote goes to Tactics Advance, even though it's not on the list.

Although, I must say, The Legend of Zelda is even better than Final Fantasy (just stating my opinion)...[/quote]

Hmm my problem was running out of poll options...lol sorry. Or else I would've added some expansions and the other ones (as you pointed out) I left out.

And yea, I think Zelda is quite good. But FF is known for their awesome storylines.

Go Garnet til Alexandros XVII!!!! YEAH!!!


A swimming pool is 12 meters long, 6 meters wide, 1 meter deep at the shallow end, and 3 meters deep at the deep end. Water is being pumped into the pool at 1/4 cubic meter per minute, and there is 1 meter of water at the deep end.
a)What percent of the pool is filled?
b) At what rate is the water level rising?

[quote="JoeBlow"]The statement of this problem is a bit ambiguous.  Does the depth change linearly (not in most pools I've seen) or in some other way?  For that matter, along which axis is the depth changing?  Along the 12 meter 'length', along the 6 meter 'width', or maybe some other direction altogether?

But I guess this is from a textbook, and you probably have a picture to go along with it, so I'm going to assume it's changing linearly along the 12 meter length.  The total volume capacity of the pool is $ 144\text{m}^3$.  When the height of the water is $ 0\leq h\leq 2$, the volume of water in the pool is $ V(h) \equal{} 36\text{m}^2\cdot h$, so when the height is $ 1\text{m}$ the volume of water is $ 36\text{m}^3$, which is (a) $ \frac {1}{4}$ of the pool's capacity. (b) $ \frac {dh}{dt} \equal{} \frac {1}{36\text{m}^2}\frac {dV}{dt} \equal{} \frac {1}{144}\frac {\text{m}}{\text{min}}$.[/quote]
I think your result in a) is not correct. The volume should be 18 instead of 36 since the base is a trapezoid. Again, sorry for not scanning the picture. Indeed, you're right on b) and your assumption that the rate of water rise changes linearly. :wink:


You can't just use a rearrangement of terms- because the right hand side doesn't converge absolutely. This also requires a clever justification for any such manipulation.


MATHCOUNTS LaTeX ratio geometry trigonometry

[u][b][size=200]DUE TO MY OWN CONFLICTS THE COUNTDOWN ROUND IS POSTPONED UNTIL TUESDAY NIGHT AT 700 EST.[/size][/b][/u]
I am almost done with the first  mock national MATHCOUNTS of 2006.
It will include the regular sprint and target rounds.
If we get enough people, then we can do a team round as well, with team members communicating over AIM chatroom, and you can even choose cool team names. :w00tb:
If we have enough for a team round, I will randomly assign the teams, except for people of the same state will always be together (just to be sorta realistic and for practice with people you will work with at nationals)

We will also have a countdown round, with the top 12 participating, if there are that many. If there are 12 people then it will be standard national MATHCOUNTS style.
If there are not 12, then it will be standard ladder format.

There also can be special awards to winners (I dunno what though), maybe e-certificates :D or super smilies :w00tb: :wow: :first: :winner_second: :winner_third:

The answers will be due by [b]TUESDAY[/b] at [b]6:00EST[/b], so I can grade for an hour and then do the countdown at [b]7:00 EST[/b]

TEAMS:
Team 1: Zenith Xan, asdf4, inyoung, Goldendomer
Team 2: Slamchess, GoColts0626, Ignite168, captain sugar
Team 3: Mathgenius^ln(x), kyyuanmathcount, pianoforte, klebian
Team 4: mysmartmouth, chinesechess4all, philB, mathwhiz101
[b][u]
Team names/captains are due by 1:19AM Saturday[/u][/b]

[b]Post here if you are interested.[/b]
It will be posted later today if there is enough interest.

[size=150][b]ANSWERS RECEIVED FROM:
mathgenius^ln(x)
asdf4
goldendomer
kyyuanmathcount
inyoung
ignite168
klebian
pianoforte
philB
ehehehe
easyaspi
fanatic
team 73H 1337 Fruit Mathamaniacs
the team with asdf4[/b][/size]

Team round more details:

If we have one,

Each team will choose their own captain.
If no agreement is made within 24 hours of the team being made, up until 3 hours until the answers are due, I will randomly choose the team captain from those interested.

The team captain is responsible for submitting the final answers for the team round, and submitting the team name.
If no agreement can be made on team name, the above procedures will be used.

TIEBREAKERS:
I will use special tiebreaking rules:

They are as follows:

1. Number of sprint round problems answered correctly.
2. Number of sprint round problems answered correctly 21-30.
3. Comparison of problems in that subset (in random order)
4. Number of sprint round problems answered correctly 20-11
5. Comparison of problems in that subset (in random order)
6. Number of target round problems answered correctly 4-8
7. Comparison of problems in that subset (in random order)
8. Comparison of sprint round problems 1-10 (in random order)
9. If there is still a tie, we will have a 1 question countdown round at the beginning of the countdown round to determine who has a higher rank

Also, you should post your state so I can pair up team members.
(zmastr you don't need to beacuse I know that you are from virginia)

Count me in too.  [size=150][b]SC[/b][/size]

I'm in, Texas.

By the way, I'd recommend less than 4-people teams, unless you want 3 teams competing.
3-member teams would probably be best.

I wanna, if I can make it, Virginia
and although this is unrelated, Zmaster, what chapter are you from? Are you like team captain of your school? Maybe I'll see you at states this saturday

wow already 8 participants. and 3 virginians :w00tb: problems going up in an hour :)

[quote="Zmastr"]I've been doing mathcounts for 3 years, so this year I'm captain  :D  The rest of my team just started doing it this year :lol:   I'm from Tidewater Region, Va Beach.  Where are you from and what's your name?  I might know you.  Are you: Daniel Li, Eric Bombargdner, ChanSeok Oh, or Brian Hamrick?[/quote]

ChanSeok and Eric have graduated Middle School already  :P (I assume ChanSeok has, as I haven't heard anything about him)

[quote=WannabeCharmander]I don't know if 2017 is too late to ask, but can you provide the solutions to sprint and target?[/quote]

I doubt the creator of this contest is still on the middle school forum. Or even on AoPS. You might want to PM him/her though, and see if they, by some slight chance, still have the answers...

The last post he posted was in 2006, so he is for sure gone/.

[quote=fatant]His last post was 11 years ago, so he's buh bye. Wait, this thread was posted when I was still learning counting from 1-10... time goes by fast. :what?:[/quote]

LOL same.

Also, please don't revive threads from 2006 :P

This was so big of a bump I don't even mind

Unless he was already an adult when he started AoPS.

[quote=geoman523]wait this guy is probably married with 2 children by now[/quote]

Well he would be somewhere in his mid-twenties so :P

Yea I did the whole test (sprint and target) and now I don't even know whether my answers are correct -_-
Epiphany

How is 11 a national mathcounts question?

[quote=WannabeCharmander]I don't know if 2017 is too late to ask, but can you provide the solutions to sprint and target?[/quote]

WannabeCharmander wanted solutions 11 years later


Let $S$ be the set of positive integers in which the greatest and smallest elements are relatively prime. For natural $n$, let $S_n$ denote the set of natural numbers which can be represented as sum of at most $n$ elements (not necessarily different) from $S$. Let $a$ be greatest element from $S$. Prove that there are positive integer $k$ and integers $b$ such that $|S_n| = a \cdot n + b$ for all $ n > k $.

We can easily prove that for sufficient great n, any number less then n-a(a-1)
belongs to Sn. So consider the numbers from n-a(a-1) to n and let qn be the number of elements from them not belonging to Sn. We see that if k belong to Sn, then
k+a belongs to Sn+1. So q(n+1)<=q(n). So q(n) will become eventually constant, which proves we can take b=-q(n)

[quote="iura"]We can easily prove that for sufficient great n, any number less then n-a(a-1) belongs to Sn.[/quote]
Are you sure? :)

[quote="iura"] So consider the numbers from n-a(a-1) to n ...[/quote]
Why "to n"? Maybe "to a*n", because a*n can be represented as sum a+a+..+a (n times). Please verify your solution.

Well, I had the idea to prove that the sequence $ an-|S_n| $ is decreasing, but that comes to proving that for large n all numbers $ na+1,na+2,...,na+a$ are sums of at most n+1 elements from the set. But I couldn't prove this last assertion. I would really like to see a complete solution to this beautiful problem.

I think you realized that your assertion is false.

Yes, Myth, I realized and that is why I'm asking for a solution. I just cannot solve this beautiful problem.

I made corrections in my proof
(1) If an element x doenst belong to S(n) for a sufficiently large n, then it does not belong to any S(n)
(2) There is an element m s.t. every element >m belongs to S(n) for some n.
This follows from the Frobenius coin problem for case of 2 numbers.
In fact we can take m>(a-1)(r-1)/2 (as I remember). However, we take m=a(a-1)
(3) For sufficient large n, any element m<x<an-a(a-1) belongs to S(n).
Proof: We prove an-x=ka+lr with k+l<n.r=a-smallest element. (Then taking
(n-k-l)*a  and l*smallest gives us x.). Since an-x>a(a-1)>m it can be written as pa+qr by result of (2). replacing eventually a of r with r of a we get a representation
pa+qr where q<=(a-1). Now we can easily prove that if x>m then p+q<=n so we get our desired representation.
(4)It follows from (1) and (3) that some elements (small ones) will never enter S(n),
and some elements between an-a(a-1) and an will not enter also.
Let q(n) be the number of such elements.
As if x belongs to S(n), then x+a belongs to S(n+1) we get q(n)>=q(n+1).
Hence q(n) eventually will become constant and we can take b=-q(n)-{number of small elements not entering S(n)}

Notice that for sufficiently large $n$, all numbers less than $an-n^2$ are either expressible as a sum of $\le n$ elements of $S$ or not expressible as a sum of any number of elements of $S.$ Hence, it suffices only to show that for $n$ large enough, the number of elements in $\{an-a^2+1, an-a^2+2, \cdots, an\}$ which are expressible as a sum of at most $n$ elements of $S$ is constant.

For large $n$, this is simply the number of elements of $\{0, 1, \cdots, a^2\}$ which can be expressed as a sum of elements of the set $\{a\} \cup \{x| a-x \in S\}.$

As this is independent of $n$, we're done.

$\square$


Some Chinese probs, hopefully typo-proof.

1. $ x^2 \plus{} y^2 \minus{} 4x \minus{} 2y \plus{} 5 \equal{} 0$. Find the value of $ \frac {(\sqrt {x} \minus{} \sqrt {y})^2 \plus{} 4\sqrt {xy}}{x \plus{} \sqrt {xy}}$.
2. $ a \plus{} b \plus{} c \equal{} 0$. Find the value of $ \frac {a^2}{2a^2 \plus{} bc} \plus{} \frac {b^2}{2b^2 \plus{} ac} \plus{} \frac {c^2}{2c^2 \plus{} ab}$.
3. $ \frac {1}{a} \plus{} \frac {1}{b} \equal{} \frac {9}{2a \plus{} 2b}$. Find the value of $ \frac {a}{b} \plus{} \frac {b}{a}$.
4. Find the value of $ \sqrt {\sqrt {180} \plus{} \sqrt {196}}$.
5. Rather stupid problem: $ b \plus{} \frac {1}{b} \equal{} 2$. Find the value of $ b^{1500} \plus{} \frac {1}{b^{1500}}$.  :)
6. $ a^2 \plus{} b^2 \minus{} 6a \minus{} 4b \equal{} \minus{} 13$. Find the value of $ \sqrt {\frac {a}{b}} \plus{} \sqrt {\frac {b}{a}}$.

EDIT: Typo-proofed.

5. That's a quadratic with a double root at $ b \equal{} 1$, so $ b \equal{} 1 \rightarrow b^{1500} \plus{} \frac {1}{b^{1500}} \equal{} 2$

1. $ \frac{x\plus{}y\plus{}2\sqrt{xy}}{x\plus{}\sqrt{xy}}\equal{}1\plus{}\frac{y\plus{}\sqrt{xy}}{x\plus{}\sqrt{xy}}\equal{}1\plus{}\frac{\sqrt{y}}{\sqrt{x}}$

The condition can be rewritten as $ (x\minus{}2)^2\plus{}(y\minus{}1)^2\equal{}0$ so x=2,y=1.

4. $ 3 \plus{} \sqrt {5}$. It's square is 14 + $ 6\sqrt {5}$, or $ \sqrt {180} \plus{} \sqrt {196}$.

3. multiplying both sides by a+b, we get $ 1 \plus{} \frac {b}{a} \plus{} 1 \plus{} \frac {a}{b} \equal{} \frac {9}{2}$, or  $ \frac {a}{b} \plus{} \frac {b}{a} \equal{} \frac {5}{2}$

6. rearranging get $ (a\minus{}3)^2 \plus{} (b\minus{}2)^2 \equal{} 0$, so a = 3 and b = 2, so the answer is $ \sqrt{\frac{3}{2}} \plus{} \sqrt{\frac{2}{3}}$

for two, choose convenient values so let them be $ 1, 0, \minus{}1$ then we get 1.

plugging in values is definately the best way to do number 2.
[hide="#5"]Rearranging we have $ (a \minus{} 3) \plus{} (b \minus{} 2)^2 \equal{} 0 \implies a \equal{} 3, b \equal{} 2.$ So the answer is $ \sqrt {\frac {3}{2}} \plus{} \sqrt {\frac {2}{3}} \implies \frac {\sqrt {3}}{\sqrt {2}} \plus{} \frac {\sqrt {2}}{\sqrt {3}}$. Rationalizing the denominators we have $ \frac {5}{\sqrt {6}} \implies \frac {5\sqrt {6}}{6}.$[/hide]

About problem 2:
Hrm, I don't know where this idea that you can just "pick convenient values and plug them in" came from.  Quite frankly it's pretty ridiculous:

Example:
a+b=0  What is the value of a-b?

Since a=-b, $ a\minus{}b\equal{}\boxed{2a}\equal{}\boxed{\minus{}2b}$, depending on how you want to look at it.  If you picked any numbers for a or b, you would definitely be wrong.

I don't know for certain yet, but this problem appears to be some cute identity which really DOES boil down to 1, although you have no way of knowing that by just looking at it.

If it's the case that someone taught you you could "pick convenient values and plug them in" to reduce an expression, I am really sad they have misled you so badly.

[quote="RightAnnihilator"]
If it's the case that someone taught you you could "pick convenient values and plug them in" to reduce an expression, I am really sad they have misled you so badly.[/quote]

Since the problem asks you to evaluate some expression in $ a,b,c$ given some condition on $ a,b,c$, then the former expression must have the same value for [i]any choice[/i] of $ a,b,c$ that satisfies the latter constraints, or else the problem could not be solved, and it would not be asked (at least not in a contest setting). Therefore, you will find the right answer by choosing convenient values, though it would not be considered a correct solution in a contest setting.

Do not be so quick to dismiss what someone has to say.

Well, you [i]could[/i] pick random values and plug them in, and it's very likely to be faster, too (especially in competitions). But that would defeat the purpose of the problem, wouldn't it?  :wink:

BTW, the first thing to do for #2 is [hide="this"]convert a in terms of b and c[/hide].

[quote="Textangle"]
2. $ a \plus{} b \plus{} c \equal{} 0$. Find the value of $ \frac {a^2}{2a^2 \plus{} bc} \plus{} \frac {b^2}{2b^2 \plus{} ac} \plus{} \frac {c^2}{2c^2 \plus{} ab}$.
[/quote]

$ 2a^2 \plus{} bc \equal{} 2a^2 \minus{} b(a \plus{} b) \equal{} (a \minus{} b)(2a \plus{} b) \equal{} \minus{} (a \minus{} b)(c \minus{} a)$

$ \Longrightarrow \frac {a^2}{2a^2 \plus{} bc} \equal{} \minus{} \frac {a^2}{(c \minus{} a)(a \minus{} b)}$ e.t.c.

$ \therefore \frac {a^2}{2a^2 \plus{} bc} \plus{} \frac {b^2}{2b^2 \plus{} ac} \plus{} \frac {c^2}{2c^2 \plus{} ab}$

$ \equal{} \minus{} \frac {a^2}{(c \minus{} a)(a \minus{} b)} \minus{} \frac {b^2}{(a \minus{} b)(b \minus{} c)} \minus{} \frac {c^2}{(b \minus{} c)(c \minus{} a)}$

$ \equal{} \minus{} \frac {a^2(b \minus{} c) \plus{} b^2(c \minus{} a) \plus{} c^2(a \minus{} b)}{(a \minus{} b)(b \minus{} c)(c \minus{} a)}$

The numerator factors out to $ \minus{} (a \minus{} b)(b \minus{} c)(c \minus{} a)$, yielding the answer $ 1$.

[quote="jvenezuela716"]and it would not be asked[/quote]  Does it also follow that when a problem says, "Prove the following statement" you can say, "This question would not be asked unless the proposer had a correct proof, therefore the statement is true, QED."?  Plugging in values is a useful technique for figuring out an answer -- it's completely worthless as a demonstration of an identity.

thats what i was saying,, it is a useful technique when you know there must be a numerical answer.

[quote="JBL"]Does it also follow that when a problem says, "Prove the following statement" you can say, "This question would not be asked unless the proposer had a correct proof, therefore the statement is true, QED."?  Plugging in values is a useful technique for figuring out an answer -- it's completely worthless as a demonstration of an identity.[/quote]

[quote="jvenezuela716"]Therefore, you will find the right answer by choosing convenient values, though it would not be considered a correct solution in a contest setting.[/quote]

@JBL: I'm not quite sure what the intent of your post is, but I will optimistically and naïvely assume that you simply missed the penultimate sentence of my post; my post simply highlighted, as yours did, that using convenient values can help guide a correct solution, not replace one. I find it easier to manipulate identities/expressions when I know what the answer should be at the end; there's no point avoiding the use of substituting values simply because you think it's "unclean" or "inelegant". Sorry if you misinterpreted.

[quote="jvenezuela716"] and it would not be asked (at least not in a contest setting). Therefore, you will find the right answer by choosing convenient values, though it would not be considered a correct solution in a contest setting.

Do not be so quick to dismiss what someone has to say.[/quote]

Ok, I didn't mean to sound quite that angry :)

On principle, it is bad math, no matter how good of a contest strategy it is.  As a mathematician, I'm pretty annoyed.  As someone who will never have anything to do with such a competition, I could not complain about it being applied there.

If it's safe that in competitions a graph of a function of three variables intersected with a plane is always going to be constant, and that's going to win you points, go ahead and do it, you know better than me.

I simply bothers me that problems like this are giving students a lot of goofy ideas about how to do algebra.  (A lot of horrible college algebra students have driven me nuts!)


trigonometry
inequalities
algebra proposed
algebra

Michael Niland, from your problem follows, that $\sqrt{12}\cos10^{\circ}-1$ is a root of the equation:
$\sqrt{8-\sqrt{8-\sqrt{8+x}}}=x.$ ;)

Reply to Michael Rozenberg
Thank you for this interesting further observation.Also thank you for all the nice inequalities you create.

Michael N.


in quadrilateral $ ABCD$  $ S(ADC)\equal{}S(ABC)$. $ E$ is the intersection of $ AC$ and $ BD$.draw some parallel from $ E$ to $ AD,DC,CB,BA$ that intersect $ AB,BC,CD,DA$ in $ K,L,M,N$ find $ S(KLMN)$ if $ S(ABCD)\equal{}S$

Obviously, K, L, M, N are midpoints of AB, BC, CD, AD, hence $ KL \parallel{} NM \parallel{} AC$, $ KL \equal{} \frac{AC}{2}$ and  similarly for LM and KN, consequently $ S[KLMN] \equal{} \frac{S}{2}$.

Kindly, do not post such easy problems on this forum, thanks!

Best regards,
sunken rock

yes,you are right.i just make a mistake between two problems that writes near in my notes. sorry.now solve this problem.in $ ABCDE$ we have $ AB$ is perpendicular to $ BC$ and $ AE$ to $ ED$ $ F$ is on $ CD$ such that $ (CF)(ED)\equal{}(BC)(FD)$ prove that $ BAC\equal{}BEF$ (angles)


Has anyone ever used this series of textbooks?
[url]http://www.cpm.org/[/url]
I have been subjected to these for my high school math career and I hated them the whole time.  Just wondering if anyone else has had to use these and what they felt about them?

From your opinion on those books, I'm glad I haven't.  I used a variety of books, and most of them were excellent, in my opinion.

Nope, but my glenncoe books were very boring as well.  Ofcourse, that was middleschool.  I have yet to see what awaits me in highschool.
Though, Ripper1 I really wouldn't be all that concerned about what books they use in your highschool, because normaly you already know everything they teach you in class, unless your school has a great exceleration program, in which case I envy you.

Well, I am going to be a senior next year after going through the "accelerated" program.  When I saw my Calculus [b]TEXTBOOK[/b], I almost cried.  I think this program is for kids in California only so those in California might share my pain but I doubt it, what can you expect from my high school and district which are one of the worst (at least I think that, and it shows in test scores as well).

Califonia's 49th in education. I understand your pain.

Hey! Be fair! California is 47th in the nation for education. See, we aren't THAT bad. And we have nearly twice as many USAMO qualifiers as any other state. (Ok, we also have almost twice as many people as any other state.)

Seriously, if you don't go to a California public school, I guess it would be hard to imagine how awful it is. And they plan to make it worse. What a brilliant plan. I think CA was #1 in the nation 25 or 30 years ago. Too bad we lost that.

47th. Wow. Big improvement. And with budget cuts for education etc etc, CA'll plummet to 49th anyway...

In view of the fact that mathematics taught at American schools is way too incomprehensive, I dont' think ranking the states is very meaningful.

It definatly is, I have proof, I was tought with CPM! :D

how much better is CPM than the integrated math series?

(Yes, I was educated in middle school with the Integrated Math I and II. It was quite ghetto. Fortunately, they had better books in high school.)

I had the Integrated series too... I think all other math textbooks have to be better than them almost by default.


Support
category theory
abstract algebra
topology
advanced fields
advanced fields unsolved

Hey guys, I have a couple of questions on sheaf cohomology that I was trying to work out.  Maybe you could help me:

If I have a sheaf with $ \textbf{Supp } \mathcal{F} \equal{} V$ open in my scheme $ X$, then I want to show $ H^{k} \left(X,\mathcal{F}\right) \cong H^{k}\left(V,\mathcal{F}|_{V}\right)$ for all $ k$.

A related question is, if $ \left(\textbf{Supp } \mathcal{F}\right) \cap Z \equal{} \emptyset$ for some closed $ Z$ in $ X$, then is the cohomology with supports $ H^{k}_{Z}\left(X,\mathcal{F}\right) \equal{} 0$? (Cohomology with supports is the right derived functor of $ \Gamma _{Z}$ - those global sections with support in $ Z$.)

Thanks!

For the second one, would this work? :

For each $ x \in X$, put $ \mathcal{F}_x \hookrightarrow \mathcal{I}_x$ an injective abelian group.  Then, using the inclusion $ i_x : \left\{x\right\} \hookrightarrow X$, write:
\[ \mathcal{F} \hookrightarrow \prod i_{x,*} \mathcal{I}_x : \equal{} \mathcal{J}_1
\]
which (I believe) has the same support as $ \mathcal{F}$.  It is also injective.  Then take the cokernel $ K$ (which will have support contained in that of $ \mathcal{F}$).  Repeat this to get a second injective $ \mathcal{J}_2$.  Then $ \mathcal{J}_1 \rightarrow \mathcal{J}_2$ and we eventually get an injective resolution of $ \mathcal{F}$ each term having no support in $ Z$?

The second question implies the first by a long exact sequence for cohomology with supports, so I would be happy if the above works.  Of course, if somebody could find a proof of the first statement without going into cohomology with supports (perhaps just by using some adjoints of functors) I would be even happier :).

Well!  I've found that both statements are false haha.  Here's an example:

Let $ U$ denote the open subscheme of $ \mathbb{A}^{2} \equal{}: X$ obtained by removing three hyperplanes in general position.  Let $ j : U \hookrightarrow X$ be the inclusion and take $ \mathbb{Z}_U : \equal{} j_{!}\left(\mathbb{Z}|_{U}\right)$ a sheaf on $ X$.  Then $ \mathbb{Z}_U$ has only support on $ U$ and $ H^{1}\left(U,\mathbb{Z}_U |_U\right) \equal{} H^{2}\left(U,\mathbb{Z}_U |_U\right) \equal{} 0$ as $ U$ is irreducible and $ \mathbb{Z}_U |_U$ is a constant sheaf.  But, I can also show that $ H^{2}\left(X,\mathbb{Z}_U\right) \neq 0$.  That kills the first statement.  As for the second, we have a long exact sequence of cohomology with supports, a section of which says: \[ H^{1}\left(U,\mathbb{Z}_U |_U\right) \rightarrow H^{2}_{X \minus{} U}\left(X,\mathbb{Z}_U\right) \rightarrow H^{2}\left(X,\mathbb{Z}_U\right) \rightarrow H^{2}\left(U,\mathbb{Z}_U |_U\right)\]  which is unfortunate. :P

Man!  Nonzero cohomology with supports where I have no support?!  The Zariski topology has some issues... I wonder if Etale cohomology would fix this :o


A tangent quadrilateral $ ABCD$ is given such that circle touches $ AB$,$ BC$,$ CD$ and $ DA$ in points $ M$, $ N$, $ P$, $ Q$. If $ R$, $ S$, $ T$, $ U$ are centers of incircles of triangles $ AMQ$, $ BMN$, $ CNP$ and $ DPQ$ respectively, prove that diagonals of $ RSTU$ are perpendicular.
 [hide]Denote $ O$ as the center of $ MNPQ$ and than do with angle chasing[/hide]

It seems the problem is easy
Oc is the bisector of angle C , T is the intersection of circle O and oc . It is simple to proof NT is
the bisector of angle CNP. So T lies on the circle.
We can see STUR is a cyclical  quadrilateral.
Arc SMR + Arc TPU =180 So angle SHR=90
Thus we are done


How do I create a number line in latex?

For example, one from -20 to 20, with the intervals x<-17 x>7 shaded?

You can use the picture environment but it will probably be simpler to use Asymptote, so ask again in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=519]Asymptote forum[/url].


calculus
integration
derivative
Putnam
real analysis
real analysis theorems

I just looked up the Leibniz Integral Rule (i.e. differentiation under the integral) and I don't understand what it could be used for. Can someone post an example of an integral that can't be done using the "usual" methods but is pretty easy using the Leibniz Integral Rule? Thank you!

Here's how I see it:
$ \displaystyle \frac{\partial}{\partial u} \int f(x,u) dx \equal{} \int \frac{\partial}{\partial   u} f(x,u) dx$

For this to have any chance of working, you want it to be a definite integral, and over a special interval as well. It's really very easy to find these on the forum; one of the better examples is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64436]Putnam 2005 A5[/url].


[hide]First, prime factorize.
4^3*5^4*6^2=
(2^2)^3*5^4*(2*3)^2=
2*6*5^4*2^2*3^2=
2^8*3^2*5^4

All factors of this product must be in the form 2^a*3^b*5^c where a is from 0 to 8, b is from 0 to 2, and c is from 0 to 4.  That gives us 9*3*5=9*15=135 factors.

You can see how to solve other similar problems.
[/hide]

I actually know the answer to this question :lol: .
[hide]First, prime factorize, and you get 2^8*3^2*5^4.  Then, you apply my trademark method by doing this:  Add the powers (8 from 2^8, 2 from 3^2, and 4 from 5^4), then multiply 8*2, 2*4, 8*4, and 2*8*4, and add them all together (including the 2+8+4).  Then you should get 70.[/hide]   :D

[quote="Math Geek"]I actually know the answer to this question :lol: .
First, prime factorize, and you get 2^8*3^2*5^4.  Then, you apply my trademark method by doing this:  Add the powers (8 from 2^8, 2 from 3^2, and 4 from 5^4), then multiply 8*2, 2*4, 8*4, and 2*8*4, and add them all together (including the 2+8+4).  Then you should get 70.   :D[/quote]

uhh..I think the answer is [b]135[/b]

If you do your way (which I dont understand that clearly), you should have 8+2+4+8*2+2*4+8*4+8*2*4=134, then you ahve to add one for  the number 1, so 134+1=135, which is just 3*5*9=135

I agree with Art of Owna.
Here is my work
[hide]
The prime factorization of $4^{3}\times5^{4}\times6^{2}$ is $2^{8}\times3^{2}\times5^{4}$
From there, I used one of "Mr. Frost's tricks and tips" and did $9\times5\times3$ to get an answer of $135$[/hide]

[quote="rd5493"]I just wanted to see if anyone could help me answer the following questions:

[list]
[*]What is the answer?
[*]How do you solve it?
[*]How do you solve problems like this one?[/list]

And the problem is:

[b]How many distinct factors are there in the product $4^{3}*5^{4}*6^{2}$?[/b][/quote]
[hide]
a) The answer is 135.
b) Prime factorize, add 1 to each of the exponents, then multiply (not very clear)
c) The general formula: A number whose prime factorization is $a^{b}*c^{d}*e^{f}*....$ has $(b+1)(d+1)(f+1).....$ factors.

extra) The reason this works is that for each prime $p$, you can choose 0, 1, 2, 3, ...., e factors of p.  There are e+1 of those numbers.  This is where the formula comes from.[/hide]

[quote="mchoi815"][hide]The formula for number of factors is p being primes in the prime factorizations.  p(1)^a(1) * p(2)^a(2) * ... * p(n)^a(n) is the factorization.  (a(1)+1)*(a(2)+1)*...*(a(n)+1).  The sum of factors is (p(1)^a(1)+p(1)^(a(1)-1)+...+p(1)+1) * (p(2)^a(2)+p(2)^(a(2)-1)+...+p(2)+1) * ... * (p(n)^a(n)+p(n)^(a(n)-1)+...+p(n)+1)[/hide][/quote]

um wow that's completely impossible to read; help yourself and others by using $\LaTeX$.

Here:

[quote="mchoi815"][hide]The formula for number of factors is p being primes in the prime factorizations. $p_{1}^{a{1}}\cdot p_{2}^{a_{2}}\cdots p_{n}^{a_{n}}$ is the factorization. $(a_{1}+1)(a_{2}+1)\cdots(a_{n}+1)$.  The sum of factors is $(p_{1}^{a_{1}}+p_{1}^{a_{1}-1}+\cdots+p_{1}+1)(p_{2}^{a_{2}}+p_{2}^{a_{2}-1}+...+p_{2}+1)\cdots(p_{n}^{a_{n}}+p_{n}^{a_{n}-1}+...+p_{n}+1)$.[/hide][/quote]

On the bright side, you are correct.  :)

Here's, in a nutshell, why your formulas hold (always important to know why stuff is true):

[hide]Let's say we wanted to construct one of the factors of a number, say $N$. Well, if $N$ has mchoi's prime factorization, we proceed as follows: choose which of $p_{1}^{k}$ we will multiply in our factor, where $k\le a_{1}$. Then do the same for all the $p_{i}$'s. Each has $a_{1}+1$ choices, namely $k=0,1,2,\ldots,a_{1}$, hence resulting in the formula.

As for sum of factors, consider what happens when we multiply out that whole thing, each factor will be generated once.

Those aren't exactly completely rigorous proofs, but they do allow one to understand why those hold, which is what is important here.[/hide]


:maybe:  I have proved something so strange...

Given a bicentric quadrilateral, with baycenter $G$ and incenter $I$, then $I=G$.

[quote="Amir.S"]what is baycenter?[/quote]

The intersecing points of the lines joining the mid points of opossite sides

[quote="Virgil Nicula"][quote="hucht"]:maybe:  I have proved something so strange...

Given a [b]bicentric quadrilateral[/b], with baycenter $G$ and incenter $I$, then $I=G$.[/quote]

Please, what is "[b]bicentric quadrilateral[/b]" ? Means it [b]inscribed [u]and[/u] circumscribed quadrilateral[/b] ? In this case $AC=BD$, I think. Thanks.[/quote]

I made a mistake... my skecht pad show me thinks really strange... if you draw this (even in a very very very big picture) you wilkl get confuse with some points (for exameple the point of intersections of the lines joinings oposite incenters of the trianghles formed by a vertex an two points of tangency realtiva to this vertex.


calculus
integration
trigonometry
Putnam
function
factorial
calculus computations

Evaluate $ \int_0^\infty \left(x\minus{}\frac{x^3}{2}\plus{}\frac{x^5}{2\cdot 4}\minus{}\frac{x^7}{2\cdot 4\cdot 6}\plus{}...\right)\left(1\plus{}\frac{x^2}{2^2}\plus{}\frac{x^4}{2^2 \cdot 4^2} \plus{} \frac{x^6}{2^2 \cdot 4^2 \cdot 6^2}\plus{}...\right) dx$.

I tried doing this with the taylor series for $ \sin x$, $ \cos x$, and $ e^x$, but was unsuccessful. Any hints/solutions would be appreciated.

This is actually an old Putnam problem and has appeared here a number of times.

The function in the left hand parentheses can be specifically identified - look at the factorial denominators, factor out an $ x,$ and see what you can match it up with.

Once that has been done, distribute and integrate term by term.

This was [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=113208]Putnam 1997 A3[/url]. Follow the links for the old threads.


geometry
circumcircle
incenter
power of a point
radical axis
geometry proposed

$ABCD$ is cyclic and circumscribed quadrilateral. $O$, $J$ are circumcenter and incenter respectively. Intersection of external tangents of $\angle{A}$ and $\angle{C}$ is $P$ , $\angle{B}$ and $\angle{D}$ is $Q$. Prove that $OJ$ is perpendicular to $PQ$

let $AC \cap BD=K$ then we know that $PQ$ is the polar of $K$ so $KO$ is perpendicular to $PQ$ so it remains to show that $K,J,O$ are colinear which can be proved by pascal:
[hide]
draw $AJ,BJ,CJ,DJ$ to intersect the circumcircle at points $A',B',C',D'$ for the second time,then we get that $D',O,B'$ are colinear,samewise are $A',O,C'$ (its easy to show with some angle chasing),now consider the hexagon $A'C'DD'B'A$ according to pascal we get that the intersection point of $A'C',B'D'$,the intersection of $AA',DD'$ and the intersection of $DC',AB'$ are colinear so we get that the intersection point of $DC',AB'$ and $O$ and $J$ are colinear...
also according to pascal in hexagon $BB'ADC'C$ we get that the intersection point of $BB',CC'$,the intersection point of $BD,AC$ and the intersection point of $AB',DC'$ are colinear...
so the points $O,K$ lie on the same line that connects $J$ and the intersection of $AB',DC'$ so we get that $J,K,O$ are colinear and we are done...[/hide]

Let $p=(P,PA)$ and $q=(Q,QB)$.

[size=150]Lemma[/size] $J\in p$

[size=150][hide="Proof of the Lemma"][/size]
[color=brown]See also [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=357804#357804]yetti's solution[/url]

One of the angles $B,D$ is acute (they are not right angles because $AP\not \parallel CP$)
Suppose that $B$ is acute.

$\angle CJA = 360-(\angle AJB+\angle CJB)$
$= 360-(180-\frac{A}{2}-\frac{B}{2}+180-\frac{C}{2}-\frac{B}{2})$
$=\frac{A+C}{2}+B$

Hence $\angle CJA= 90+B<180$

$\widehat{ABC}<90\Rightarrow \widehat{AOC}<180$ hence the point $P$ is on the opposite side of $B$ wrt $AC$ with $\angle APC = 180-\angle AOC = 180-2B$

$\frac{180-2B}{2}= 90-B$ hence $J$ belongs to the minor arc $AC$ of the circle $p=(P,PA)$

(If $B$ is obtuse then $D$ is acute and the proof is the same - just replace [b]B[/b] by [b]D[/b])[/color]
$\hline$[/hide]

According to the lemma, we have $J\in p$ and also $J\in q$

Let $K$ be the other intersection point between $p,q$

$OA$ is tangent to $p$ and $OB$ is tangent to $q$. So $O$ is on the radical axis $JK$ of $p,q$

Hence $OJ\perp PQ$


Show that there is some $ x\in(0,\frac {\pi}{2})$ so that $ cos(x) \equal{} x$.

how do i even begin?

This is a ludicrously misnamed topic. It's a simple application of the intermediate value theorem.


Given 2n coplanar points, show that it is possible to draw n segments having their endpoints among the given points, such that no 2 of these n segments have a common point.

This can be solved using the extremal principle. Pair the points randomly. If there are two segments AB and CD wich intersect in a point O then by switching to segments AC and BD the sum of lengths of the segments in the pairs has decreased (because of the triangle inequality). Keep repeating this until no segments intersect, becose the number of sum of lengths in pairing the 2n points is finite the algorithm must end so we must have a pairing for wich no two segments intersect.


algebra
polynomial
function
algebra solved

Prove that  $\large{x^n+5x^{n-1}+3\,\,\,|\,\,\,n\,\in{\mathbb{N_{+}}}}$ is irreducible.

Take $f(x)=x^n+ 5 \cdot x^{n-1}$, $g(x)=3$ into Rouche's theorem, used on the unit circle.
Since $|f(x)|=|x^n+5 \cdot x^{n-1}| = |x+5| \geq 4 > 3 = |g(x)|$ for all $x$ with $|x|=1$, the function $f(x)+g(x)$ has as many comples roots with modulus $<1$ as the function $f(x)=x^n+5 \cdot x^{n-1}$, having $n-1$ roots there ($0$ counted $n-1$ times).
Now assume that $x^n+5 \cdot x^{n-1}+3$ can be factored as $a(x)b(x)$ with integer coefficients, then the constant coefficients would be $\pm 1 , p$, but theres only one root of modulus $>1$, contradicting the fact that the constant coefficient is (up to sign) the product of the roots.

Another solution: by extended Eisenstein's critterion above polynomial has to be a product of a polynomials which one is of degree at least $n-1$. Having degree $n-1$ would be a contradiction cause above polynomial does not have a integer root. Hence it is irreducible.

(This is some shortlist problem - though I don't remember year)


[quote="GameBot"]How many different squares of any size are there on the 3x3 board below?[/quote]

No board, but

9 (1x1 squares) + 4 (2x2 squares) + 1 (3x3 squares) + 4 (sqrt2xsqrt2 squares)=18.

I'm assuming the board is made of a $ 3*3$ square with points connected, not a $ 3*3$ coordinate plane

Okay, say this was the board:

[code]O   O   O

O   O   O

O   O   O[/code]

Then, wouldn't you only have one $ \sqrt{2} \times \sqrt{2}$ square?

[code]O   O   O
  /   \
O<  O  >O
  \   /
O   O   O[/code]

That isn't the board. Its:

OOOO
OOOO
OOOO
OOOO

i.e. 3x3, but 4x4 dots.


geometry
geometric transformation
rotation
real analysis
real analysis unsolved

a unit ball in R^3 can be decomposed into 5 pieces such that those 5 pieces can be ressembled via translations and rotations  to form 2 complete and disjoint unit balls . :o

http://en.wikipedia.org/wiki/Banach-Tarski_paradox

The thing that people don't understand about this paradox is that the "pieces" involved are rather... convoluted.

It makes a good advertisement for the concept of (non-)amenability.


calculus
integration
logarithms
trigonometry
calculus computations

Evaluate the following integrals

1) $ \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\bigg(\frac{sin^{2}3x}{sin^{2}x}\minus{}\frac{cos^{2}3x}{cos^{2}x}\bigg)dx$

2) $ \int_{0}^{1}\frac{dx}{2\plus{}\sqrt{1\minus{}x}\plus{}\sqrt{1\plus{}x}}$

3) $ \int_{\minus{}1}^{0}x\big(e^{2x}\plus{}\sqrt[3]{x\plus{}1}\big)dx$

4) $ \int_{0}^{\frac{\pi}{4}}\big(tanx\plus{}e^{sinx}cosx\big)dx$

5) $ \int_{0}^{2}\frac{x^{4}\minus{}x\plus{}1}{x^{2}\plus{}4}dx$

Good Luck !  :)  :)  :)

[hide="5)"]$ \int_{0}^{2}\frac {x^{4} \minus{} x \plus{} 1}{x^{2} \plus{} 4}dx \\
\equal{} \int_{0}^{2} \frac {(x^2 \plus{} 4)(x^2 \minus{} 4) \minus{} x \plus{} 17}{x^2 \plus{} 4}dx \\
\equal{} \int_{0}^{2} (x^2 \minus{} 4)dx \minus{} \int_{0}^{2} \frac {x}{x^2 \plus{} 4}dx \plus{} \int_{0}^{2} \frac {17}{x^2 \plus{} 4} dx \\
\equal{} \left(\frac {x^3}{3} \minus{} 4x\right)|_0^2 \minus{} \frac {1}{2} (\ln (x^2 \plus{} 4))|_0^2 \plus{} \frac {17}{2} arc \tan \left(\frac {x}{2}\right) |_0^1 \\ \equal{}\frac{\minus{}11}{3}\plus{}\frac{17}{2} arc \tan \left(\frac{1}{2}\right)$
[/hide]

I'm getting a weird answer,Please let me know if there is any mistake in this solution!

Edit:I seem to have messed up with the limits.

your all answers seem correct picture74, i didn't solve them, but your solutions are clear and perfect  :wink:

At picture74.

I think the answer to the first one should be  $ 2\sqrt {3} - 4$

2) [hide]2 )

$ \int_{0}^{1}\frac {dx}{2 + \sqrt {1 - x} + \sqrt {1 + x}}$

Put $ x = \cos 2\theta , dx = - 2 \sin \2theta d\theta$

Hence the integral becomes

$ \int_{0}^{\frac {\pi}{4}}\frac {2\sin 2\theta d\theta}{2 + \sqrt {2}\sin \theta + \sqrt {2}\cos \theta }$

$ = \int_{0}^{\frac {\pi}{4}}\frac {\sin 2\theta d\theta}{1 + \cos ( \frac {\pi}{4} - \theta ) }$

applying properties of definite integrals the integral becomes

$ \int_{0}^{\frac {\pi}{4}}\frac {\cos 2\theta d\theta}{1 + \cos \theta }$

For evaluating this put $ y = tan \frac {\theta}{2}$  and proceed .[/hide]


Let x be a permutation of elements in [n] and let x^-1 denote its inverse.
Let inv(x) denote the number of inversions in x.
How do I prove that inv(x)=inv(x^-1). That is the number of inversions in a permutation x is equal to the number of inversions in its inverse.


Let 0<=a<=b<=c<=d<=e and a+b+c+d+e=1, Show that ad+dc+cb+be+ea<=1/5  :wink:

By Chebyshev: $(a+b+c+d+e)((e+d)+(e+c)+(b+d)+(c+a)+(a+b)) \geq 5(ae+ad+be+bc+cb+cd+dc+da+ea+eb) = 10(ae+ad+be+bc+cd)$ then
$2(a+b+c+d+e)^{2}\geq 10(ad+dc+cb+be+ea)$
$1\geq 5(ad+dc+cb+be+ea)$, then $ad+dc+cb+be+ea\leq \frac{1}{5}$  :P
$really_{}easy$

By Chebyshev: $(a+b+c+d+e)((e+d)+(e+c)+(b+d)+(c+a)+(a+b)) \geq 5(ae+ad+be+bc+cb+cd+dc+da+ea+eb) = 10(ae+ad+be+bc+cd)$ then
$2(a+b+c+d+e)^{2}\geq 10(ad+dc+cb+be+ea)$
$1\geq 5(ad+dc+cb+be+ea)$, then $ad+dc+cb+be+ea\leq \frac{1}{5}$  :P
$very-simple$