AMC
USA(J)MO
USAMO
AIME
AMC 12
AMC 10
ARML

Hi, so... no one's posted an index poll yet, so, here is one. Please vote for whichever range in which your index falls.

USAMO Index = AMC 12 score + 10*AIME score
(please only vote if you took the AMC 12!)

what i'm worried about is if too many grade <=10 get the floor and they will have to put an index. my amc12 score is less than the amc10 qualifying score :noo:

I think that below 190 should be opened up a bit more, as the secondary index for sophomores and below could end up below that, and that above 215 is almost guaranteed in. Thanks for making a poll, though  :)  BTW, i got a 197, hopefully should make it as a soph with 117/8.

I don't think my score (AMC 10: 144 plus AIME I: 6=204) made it.  Does anyone wanna give me a confidence booster?

You just might make it if there was a 150/6... it's very possible...

Wait, but if there was a 150/6, wouldn't that be higher than mine and lessen my chance of making it?  Do you mean if I got 150, then I might have made it?  I'm still confused and depressed...

Oi, you took the AMC 10. Doesn't count for USAMO index in the poll.

$ 112.5 \text{ AMC 12A} \plus{} 10(9 \text{ AIME}) \equal{} 202.5$

But I'm a sophomore, so it's all good.  :P

[quote="TheHeinrich"][quote="Ignite168"]PREDICTION: Evan O'Dorney, because he is like Jesus, will set the floor so someone should ask him what he got.[/quote]

i think Evan O'Dorney is in university already...[/quote]
Hmmm I don't think he's taking enough college classes for it to count as being in college :)
Wow these Evan O'Dorney discussions get pretty popular...

[quote="TheHeinrich"]i think Evan O'Dorney is in university already...[/quote]
He quashed this rumor at MOP last year. News misreported apparently, not that there's any surprise there. Unless something's changed between then and now he's still doing the AMC line of contests.

remember that 4th grade kid who made usamo last year?
i heard that he has 12 on the AIME!

Haha, it was funny to watch Evan O Dorney arguing with the proctors during Countdown for the top 16 at CA State Mathcounts last year :D

Heard he got like a 21 on USAMO as an 8th grader.
And that fourth grader lives in Canada :wink:

I think the qualification system is changing for this year. Wikipedia says
1. The goal is to select about 500 of the top scorers from this years’s AIME and AMC 12A, AMC 12B, AMC 10A and AMC 10B contests to participate in the USAMO.

2. Selection will be based on the USAMO index which is defined as 10 times the student’s AIME score plus the student’s score on the AMC 12 or the AMC 10.

3. The first selection will be the approximately 330 highest USAMO indices of students taking the AMC 12A or AMC 12B contest.

4. The lowest AIME score among those 330 first selected will determine a floor value. The second selection of approximately 160 USAMO participants will be among students in the 10th grade and below who received an AIME score at least as high as the floor value. If there are more than 160 young students with a score above the floor value, then approximately 160 students will be selected from this group by using the USAMO index.

5 The student with the highest USAMO index from each state, territory, or U.S. possession not already represented in the selection of the first and second groups will be invited to take the USAMO.

6. To adjust for variations in contest difficulty, the number of students selected from A & B contests will be proportional to the number of students who took the A & B Contests.

Of course, if you actually read the selection system for both years, you'll know that they're the exact same algorithm, just with more people qualifying for 2008.

[quote="Ignite168"][quote="serialk11r"][quote="Ignite168"]PREDICTION: Evan O'Dorney, because he is like Jesus, will set the floor so someone should ask him what he got.[/quote]
What? Evan O'Dorney is a genius, he can't possibly have gotten less than 14 :P Besides pretty much no one on AoPS has contact with him (which is a problem, cuz we sorta need his beastliness for our ARML team, which I'm joining this year)[/quote]

Almost no one is in contact with Jesus either :( which is a problem, as we need him to forgive our sins.[/quote]
Try starting a [b]personal relationship[/b] with Jesus and God. Read the Bible daily, and you'll get to know Jesus better.
Romans 10:9-10 shows exactly how to start it with having all your sins forgiven.

[quote="stuck"][quote="jvenezuela716"]103.5 + 10*4: I hadn't taken any AMCs since eighth grade when i took AMC 10, and now I'm a junior. I just barely made it into the AIMEs, and my preparation was minimal. But I'm starting a whole new studying regime for the rest of this year and the beginning of next. 2009 is my last chance![/quote]

Hey! A kindred spirit!

My USAMO index was ~140... Dang, I've got some major work to do. It will be worth it though, I aim to be my school's first ever USAMO qualifier.[/quote]Haha. I'm hoping I'm gonna be my school's first USAMO qualifier as well...nobody's gotten in from my school yet.

[quote="serialk11r"]This is good, remember that even if you fail, hard work can usually get you to improve a lot and achieve your goals (okay maybe not like IMO or blue MOP for most people...). Last year I got 111 on AMC10A :( This year I know how to do all the problems :D But still failed :(
Last year I got 30s for mathcounts chapter and state (eh who cares about mathcounts but for comparison's sake), then I worked really hard on math for a while, happened to go to a middle school, picked up a chapter sprint, got all the questions correct in 15 minutes :D[/quote]Nobody from my school qualifying is basically my motivation to get into USAMO.

[quote="MaThWhIz2004"]Um. I edited out the question mark from the title, but I'm not sure why the poll looks weird now.  :maybe:  :([/quote]Looks like all the entries got duplicated...that's strange...

Hey Yo... I got a 115.5 for mah AMC 12 and 9 for AIME... Am I up for the USAMO... :?:  Am a 12th Grader... :!:  Wad ya'll think...?

[quote="The Zuton Force"]$ (141,7) \implies 211$

Without silly arithmetic errors on AIME, could've been 291  :([/quote]
Dude thank you you are my savior, if you make USAMO then you just helped out your favorite freshman friend! (me) okay no that was sorta spammy. I was pretty surprised at your 7, lets just say that. I actually checked your post multiple times to make sure that you made it lol.

[quote="the future"]the funny thing is for some people, they want a floor of 8 instead of 7.

for someone that got like a 117 on the AMC 10/8 on AIME,

they will probably want the floor to be 8 instead of 7...since a 7 means higher index (since a 130/7 will beat a 117/8),

8 will probably mean no index-or a very low index-only someone that got a 100/8 will be bumped- (since the 130/7 will already be bumped out since they did not get an 8)[/quote]

Actually, for 117/8 people like me, if the floor is 7, then the secondary index will probably not be 200, as the secondary index for last year was 181.5 for the floor of 6.  I'll probably still make if even if the floor is 7, because the secondary cutoff score will be like 190-195. Nevertheless, I'd like to remove all doubt and have the floor be an 8.  :lol:

Hmmm, not quite as high as I was hoping, but it's still a record for my school...

106.5 AMC 12B + 10*7 AIME = 176.5

anyone else think it's bizarre every other range in the poll has 0 people reporting with that range?

Can a moderator remove the extra options (the repetitive ones that have 0 votes in them)?

I have a 111 AMC 10 and a 108 AMC 12, and an AIME score of 7.

If the floor is 7, how likely is it that there will be index as well for 10th graders and below?

(I would have no chance then, with these AMC scores)

It's possible that there will be no secondary cutoff if the floor is 7... last year the secondary cutoff only assured that the person got at least a 121.5 on the AMC 10, and that was for the floor of 6.  You could get lucky, and they might admit in the upper-500's without a secondary cutoff, but it's not likely.

I give you a 1 in 5 chance.

Looks like a lot of people got pretty high indexes (indices)? this year. :o Hopefully the floor will be low.

i got 123 on the AMC 10 and somewhere between a 6 and 9 on the AIME (not sure, don't remember all my answers) 123+(worst case) 6*10=183.......but i'm an 8th grader so...well...i dunno... :?:  :(


geometry
3D geometry
tetrahedron
pigeonhole principle
area of a triangle

Given a square side 1 and 9 points inside it.Show that each 3 points are the vertices of a triangle with area <1/8.If anyone has the solution i'd pray him or her to send a solution with as much details as possible.

I'm confused about this problem statement. Are you trying to say that every one of the [tex]\binom{9}{3}[/tex] choices of points form a triangle with area less than 1/8? This seems false. Am I just misinterpreting the statement?

some questions:

is the side length of the square one?
are the nine points arbitrarily placed inside the square?

edit: looks like tetrahedron beat me to it

Im pretty sure the problem goes something like this: Given nine points inside a unit square, show that SOME three points form a triangle with area <=1/8.

Hint :

[hide]

split the square in 4 smaller ones. One of them contains three points. You can change the triangle so that the three points are one the side of the square (translating one point // other side of the triangle).[/hide]

Well it's as Zabelman wrote.In fact i have translated it and i see i haven't done this very well.but the are is <1/8 and not <=1/8.
The side is 1 and the points are arbitrary ones.

[hide]

Split the square up into 4 1/2 by 1/2 squares.  By the pigeonhole principal, 3 points must end up in the same square.  The maximum area of a triangle inside one of the 1/2 by 1/2 squares is (1/2 * 1/2)/2 = 1/8.  Thus, we can always find a triangle with an area less than or equal to 1/8.

[/hide]

The area of one of these triangles doesn't have to be <1/8, just <=1/8.  The picture below shows an example.


linear algebra
matrix
algebra
polynomial

Let $ A \in \mathcal{M}_2 \left(\mathbb{Q}\right)$, such that $ \det(A^2\minus{}2I_2)\equal{}0$. Prove that $ A^2\equal{}2I_2$.

From $ \det (A^2\minus{}2I)\equal{}0$ it follows that $ \det (A\minus{}\sqrt{2}I)\equal{}0$ or $ \det(A\plus{}\sqrt{2}I)\equal{}0$.Because $ \det(A\minus{}xI)\equal{}x^2\minus{}tr(A)x\plus{}\det A$, a polynomial with rational coefficients.If $ \sqrt{2}$  is a root of this equation then $ \minus{}\sqrt{2}$ must be.And similar, if $ \minus{}\sqrt{2}$ is a root then $ \sqrt{2}$ must be.In any of these two cases we get that $ tr(A)\equal{}0$.So from Cayley Hamilton we get that $ A^2\plus{}\det A I\equal{}O_2$, but $ \det A\equal{} \minus{}\sqrt{2} \cdot \sqrt{2}$.So the conclusion follows.


[quote="The Nguyen"]$ a,b,c > 0$. Prove that $ \frac {a^2b \plus{} b^2c \plus{} c^2a}{3(a^3 \plus{} b^3 \plus{} c^3)} \plus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\ge\frac {4}{3}$[/quote]
It's posted in here:
http://www.mathlinks.ro/viewtopic.php?t=325532


[quote="treegoner"]Let $0<z \leq y \leq x \leq 3$ and $\frac{3}{xy} + \frac{2}{y^2} \geq 1$, $\frac{18}{x^2y} + \frac{4}{y^2z} + \frac{3}{z^2x} \geq 3$
  Find the maximum value of $\frac{80}{27} x^3 + \frac{9}{4} y^3 + \frac{1}{2}xyz$[/quote]

So you are the lost.


let $ a$,$ b$,$ c$ $ \geq 0$ , prove that :

$ \frac {ab}{c^2 \plus{} ca} \plus{} \frac {bc}{a^2 \plus{} ab} \plus{} \frac {ca}{b^2 \plus{} bc}  \geq \frac {a}{a \plus{} c} \plus{} \frac {b}{b \plus{} a} \plus{} \frac {c}{c \plus{} b}$

[quote="y-a-s-s-i-n-e"]let $ x$,$ y$,$ z$ $ \geq 0$ , prove that :

$ \frac {ab}{c^2 \plus{} ca} \plus{} \frac {bc}{a^2 \plus{} ab} \plus{} \frac {ca}{b^2 \plus{} bc} \plus{} \geq \frac {a}{a \plus{} c} \plus{} \frac {b}{b \plus{} a} \plus{} \frac {c}{c \plus{} b}$[/quote]
Maybe $ a,$ $ b$ and $ c$ are positive numbers?
If so that your inequality is equivalent to $ \sum_{cyc}(a^4c^2\minus{}a^3c^2b)\plus{}\sum_{cyc}(a^3b^3\minus{}a^2b^2c^2)\geq0,$
which is true by AM-GM.

yes I edit ...

thank's , is there a simple solution?

Inequality is equivalent with:

  $ \sum {\frac{a(b\minus{}c)}{c(a\plus{}c)}}\geq0$ or  $ \sum {a^2b(bc)(b\plus{}c)(a\plus{}b)}\geq0$ or$ \sum_{cyc} (a^3b^3\plus{}a^2b^4\minus{}a^3bc^2\minus{}a^2b^2c^2)\geq0$

  wich is true because

 $ \sum {a^3b^3}\geq 3a^2b^2c^2$ and $ \sum {(ab^2)^2}\geq \sum {(ab^2)(bc^2)}$

HI
the original inequality is
let $ x$ ,$ y$ , $ z$ $ \geq 0$ and $ xyz\equal{}1$ , prove that :

$ \frac {x\minus{}1}{y\plus{}1} \plus{} \frac {y\minus{}1}{z\plus{}1} \plus{} \frac {z\minus{}1}{x\plus{}1}  \geq 0$


group theory
abstract algebra
modular arithmetic
superior algebra

Prove that $ S_{4}$ is not integrable.(There is no $ G$ such that $ G'\equal{}S_{4}$)

Assume $ S_4\le G$ with $ S_4\equal{}G'$.
$ G$ operates on $ S_4$ by conjugation, which gives a hom $ \alpha : G\to \text{Aut}(S_4)$.
Then $ \ker \alpha\cap S_4\equal{}Z(S_4)\equal{}1$ and $ S_4\cong\alpha(S_4)$ has order $ 4!$.
But $ \text{Aut}(S_4)$ permutes the $ 4$ subgroups of order $ 3$ of $ S_4$,
which gives a hom $ \beta: \text{Aut}(S_4)\to S_4$,

For $ \sigma\in\ker\beta$ we have
$ \sigma(P)\equal{}P$ for all subgroups $ P$ of order $ 3$ of $ S_4$.
Then for any $ x\in S_4$ and $ P$ subgroup of order $ 3$ we have
$ xPx^{\minus{}1}\equal{}\sigma(xPx^{\minus{}1})\equal{}\sigma(x)\sigma(P)\sigma(x)^{\minus{}1}\equal{}\sigma(x)P\sigma(x)^{\minus{}1}$
$ \Rightarrow Px^{\minus{}1}\sigma(x)\equal{}x^{\minus{}1}\sigma(x)P\Rightarrow x^{\minus{}1}\sigma(x)\in N_{S_4}(P)$.
But by Sylow $ P\not\unlhd S_4$ and $ |S_4: N_{S_4}(P)|\equiv 1\pmod{3}$, so $ |S_4: N_{S_4}(P)|\ge 4$ and $ |N_{S_4}(P)|\le 6$.
This gives $ x^{\minus{}1}\sigma(x)\in N_{S_4}(\langle (123)\rangle)\cap N_{S_4}(\langle (124)\rangle)\equal{}\langle (12)\rangle$ and analog $ x^{\minus{}1}\sigma(x)\in\langle (34)\rangle$.
So $ x^{\minus{}1}\sigma(x)\equal{}1$ and $ \sigma(x)\equal{}x$ for all $ x\in S_4$.

This gives $ \beta$ injective and $ \text{Aut}(S_4)$ of order $ \le 4!$.
But then $ \alpha(S_4)\equal{}\text{Aut}(S_4)$ and $ \alpha(S_4)\equal{}\alpha(G')\equal{}\alpha(G)'\equal{}\alpha(S_4)'\equal{}\alpha(S_4')\equal{}\alpha(A_4)$.
Then for all $ x\in S_4\setminus A_4$ we find some $ y\in A_4$ with
$ \alpha(x)\equal{}\alpha(y)\Rightarrow \alpha(xy^{\minus{}1})\equal{}1\Rightarrow xy^{\minus{}1}\in\ker \alpha\cap S_4\equal{}1\Rightarrow x\equal{}y\in A_4$.
Contradiction!


superior algebra
superior algebra unsolved

[quote="berti"]Whats the minimal number of generators of $ A_n$?[/quote]

Every elements of $ A_n$ is equal to the product of 3-cycle...

This gives only an upper bound on the number of generators.


modular arithmetic
graph theory
combinatorics proposed
combinatorics

There are $n>2$ lines given in the plane. No two of the lines are parallel and no three of them intersect at one point. Every point of intersection of these lines is labelled with a natural number between $1$ and $n-1$. Prove that, if and only if $n$ is even, it is possible to assign the labels in such a way that every line has all the numbers from $1$ to $n-1$ at its points of intersection with the other $n-1$ lines.

Let's change the language. We call the lines vertices of a graph, and the labels of the intersection points colors of the edges between the vertices.

The edges colored with a certain color form a perfect matching in our graph, so the number of vertices must be even (i.e. $n$ is even). The fact that if $n$ is even we can always do the labeling is a different wording for the fact that the chromatic index of $K_{2n}$ (the complete graph on $2n$ vertices) is $2n-1$.

In order to find a proper coloring of the edges of $K_{2n}$, we first find one for $K_{2n-1}$: Call the vertices of $K_{2n-1}$ $v_0,\ldots,v_{2n-2}$, and color the edge $v_iv_j$ with color $i+j\pmod{2n-1}$. Each vertex $v_i,\ i\in\overline{0,2n-2}$ is missing a color $c_i$, and the colors $c_i$ are all different (this is where we use the fact that $2n-1$ is odd, which implies that all of the $2i\pmod{2n-1}$ are distinct). Now add the $2n$'th vertex $v_{2n-1}$, and color the edges $v_{2n-1}v_i$ with color $c_i$.


The Mock AMC F is posted [url=http://mathjunk.googlepages.com/MockAMCFProblems.pdf]here[/url] if you cannot access it below, for any reason.

[b]Current version number is $\boxed{2.0}$.  If you do not have this version, please delete your old version and re-download.[/b]

Errata and clarifications will be fixed in subsequent versions, however they will also be posted here.  Any questions regarding test questions should be PMed to me, not posted here.

You have 75 minutes to do the test, and only calculators are allowed.  No cheating!  [b]It is due January 31[/b].  I will post results February 1st.
[b]
PLEASE TELL ME WHAT GRADE YOU ARE IN SO YOU CAN WIN A PRIZE WITHIN YOUR GRADE LEVEL![/b]

Final Word On Difficulty: According to Iversonfan2007, a local math teacher has looked over the test, and said it is harder than a normal AMC 12.  So these high scores are pretty impressive.

Errata/Clarifications:
[size=134][b]#20 - PROBLEM REPLACED[/b][/size]

[u][b]List of Participants[/b][/u]
20 PARTICIPANTS SO FAR!

How hard is this comparatively with an ordinary amc?

on 17: after u remove the digits can u rearrange the digits to make whatever the number u want??

[quote="mysmartmouth"]I would call this a regular AMC 12.  Altheman calls it easy, and some people have called it hard.  But it's definitely harder than an AMC 10.  So the goal to qualify for AIME is 100.[/quote]
wait, is there going to be a mock aime by you following this up?

Oh, no.  Even if there was, we agreed earlier that people shouldn't have to qualify for Mock AIMEs.  I was just saying if this was a real AMC, you wold be aiming for 100 (or higher).

lol, that's one really old ARML problem. This test seemed more difficult than the average AMC12, great job!

[quote="joml88"][quote="Iversonfan2007"]what the h ell i didn't sign up for this w/e[/quote]

Sean owns you, bub.[/quote]

You IMed me and told me you were taking it....good call.

u have EFuzzy twice up on the list, one at #3 and one at #26

Haha... wow you're really worked up about this man.

*whipsers to mysmart*
I think he's just miffed at not getting a 150.

[quote="MysticTerminator"]no I just curse at sean a lot

it is one of the golden rules

and "wtf" is just about the only thing that works on aops[/quote]

Yes that is very true.  He sent me his answers like this:

Arnav: *****************
Arnav: *****************
Arnav: A
Arnav: *****************
Arnav: *****************
Arnav: C

But anyway I won't count it.

[quote="mysmartmouth"][quote="MysticTerminator"]lol

well some of the problems deserved it huh

like that one where I needed to get paper

I was pretty pissed[/quote]

Well now you're just showing off....I needed paper for #1...which 2 of the top 5 people (so far) missed, by the way.[/quote]

well thats how he rolls...btw...arnav...how about "showing off" by taking my mock aime

I think I will

in a bit though

what was #1 btw

I remember you were all like OMG #[] is SO TRICKY

I'm like

wtf dude

([] inserted so that NOBODY KNOWS WHICH ONE IT IS)

[quote="bpms"][quote="anirudh"]I'm doing it.[/quote]
[quote="mysmartmouth"]The Mock AMC F has been posted:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=729827#729827

If you want to sign up, post in THIS thread.

Otherwise, please continue further discussion in the NEW thread.[/quote][/quote]

dude what

[quote="bpms"]So MysticTerminator and Elemenop got 150s?[/quote]

144s ;), although arnav said he didn't miss a question...which is very believable

I am just about finished compiling results.  I will post them around 7 PM EST.  If you have not yet turned in answers, you may do so until 7 PM EST. (It is almost 6 PM EST).

[quote="aahmadi"]I'm starting right now, ill have answers submitted b4 7:30 est, im a junior, i highly doubt ill win prize, havent done hardly ne math in the past month ana half cuz of work...[/quote]

That's fine with me, I'll let you do it, but you're delaying the results, which will be posted as soon as I get your answers and average them in.

OK, results posted here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=744285#744285


geometry
circumcircle
trigonometry
perpendicular bisector
geometry solved

Let $ABCD$ a convex quardilateral with $AB=BD$  and angle$BAC=30$.
IF angle$ADC=150$ prove that  angle$BCA=ACD$angle

it was a typo on my book. I found it and I fixed the proble, sorry :blush:
Now try to solve it

Hmm.. are you sure about your problem?  :?

Silouan,
 I think that this was proposed from Greece for some competition , but when ? I had tried it for a little but I had not solved it.

 [u]babis[/u]

Hmm ok, but, are you sure there isn't a typo in Silouan's post?
Else, could you show me a pic please? (because I can't draw it  :blush: )

It is a problem I found in Internet which is JBMO shortlist .Here I found an official solution but I think it is wrong. If yu find something tell me please.

Ok, I will. (are you sure it isn't "50" instead of "150"? sorry if I'm annoying lol)

Do you have the link? :)

[quote="mathmanman"]Ok, I will. (are you sure it isn't "50" instead of "150"? sorry if I'm annoying lol)
[/quote]

Maybe. if you find a solution with 50 instead of 150 post iit please.

@stergiu  : I also find this problem in the book you gave to me, page 93.
Also here the solution is wrong i think.

Thans Silouan!

  I ' ll try to see it again.

  [u]Babis[/u]

mr Stergiu what is your opinion ? Did you see the problem ?

[u]The idea of the solution ( in the book) is this:[/u]

  Let  E the symmetric of B wrt line AC. Then  triangle  ABE is equilateral and angle E = 60. But  angle D = 150 , so the circle with center B and radius BE passes through D.( we have ext. angle D = 30 = 1/2 angle B ). But where do we need that AB = CD ?

Babis

I found the correct version of the problem. There is a typo in the book
[quote="silouan"]Let $ABCD$ a convex quardilateral with $AB=BD$  and angle$BAC=30$.
IF angle$ADC=150$ prove that  angle$BCA=ACD$angle[/quote]

In fact now it is true and the solution of the book is correct.
Now try to solve it with many ways!

Now as the problem is corrected, let me post my solution.

[color=blue][b]Problem.[/b] Let ABCD be a convex quadrilateral with AB = BD, < BAC = 30° and < ADC = 150°. Prove that < BCA = < ACD.[/color]

[i]Solution.[/i] Denote by k the circumcircle of triangle ACD. Let this circle k have the center O and meet the line AB at a point U (apart from A). Then, the quadrilateral AUCD is cyclic, so that < AUC = 180° - < ADC = 180° - 150° = 30°. Combining this with < UAC = 30° (this is just another way to state < BAC = 30°), we see that the triangle ACU is isosceles with the base AU and the base angle 30°. By the sum of angles in this triangle, we get < ACU = 180° - < UAC - < AUC = 180° - 30° - 30° = 120°.

Since the point O is the center of the circle k, by the central angle theorem, the angle < AOU must be twice the acute chordal angle of the chord AU in the circle k. Now, < ACU = 120° is obviously the obtuse chordal angle of the chord AU in the circle k. The acute chordal angle of this chord must therefore be 180° - 120° = 60°. Hence, the angle < AOU, being twice this acute chordal angle, must equal $2\cdot 60^{\circ}=120^{\circ}$.

So we have found that < AOU = 120°. Since O is the center of the circle k, while A and U are two points on this circle, we have AO = UO, so that the triangle AOU is isosceles with the base AU, and it base angle must be $\measuredangle UAO=\measuredangle AUO=\frac{180^{\circ}-\measuredangle AOU}{2}=\frac{180^{\circ}-120^{\circ}}{2}=30^{\circ}$.

We have shown that both of the triangles ACU and AOU are isosceles triangles with the base angle 30°. Hence, they are similar; since they have a common base AU, they are even congruent. Thus, the points C and O are symmetric to each other with respect to the line AU. Consequently, < BCA = < BOA.

Now, the points O and B both lie on the perpendicular bisector of the segment AD (the point O since it is the center of the circle k and the center of a circle always lies on the perpendicular bisector of a chord, and the point B because AB = BD). In other words, the line OB is the perpendicular bisector of the segment AD. Hence, the points A and D are symmetric to each other with respect to the line OB. Hence, < AOB = < DOB, so the line OB bisects the angle AOD, and $\measuredangle BOA=\frac{\measuredangle DOA}{2}$. But, again by the central angle theorem, for the center O of the circle k we get $\measuredangle DOA=2\cdot\measuredangle ACD$. Hence, $\measuredangle BCA=\measuredangle BOA=\frac{\measuredangle DOA}{2}=\frac{2\cdot\measuredangle ACD}{2}=\measuredangle ACD$. And we are done.

  Darij

i've found a trig solution! not the nicest, but it works!

let $\angle ACB=\alpha, \angle ACD=\beta$, then a simple angle-chase yields
$\angle BAD=\angle BDA=60-\beta, \angle CAD=30-\beta$

then by the sine law

i) in $\triangle ACD$ we have that $\frac{AD}{AC}=\frac{\sin\beta}{\sin 150}=2\sin\beta$,

ii) in $\triangle ABD$ we have that $\frac{AD}{AB}=\frac{\sin(60+2\beta)}{\sin(60-\beta)}=\frac{\sin(120-2\beta)}{\sin(60-\beta)}=2\cos(60-\beta)=2\sin(30+\beta)$

iii) in $\triangle ABC$ we have that $\frac{AB}{AC}=\frac{\sin\alpha}{\sin(30+\alpha)}$

now, combining i) and ii) yields
$\frac{AB}{AC}=\frac{\sin\beta}{\sin(30+\beta)}$,
which with iii) gives
$\frac{\sin\beta}{\sin(30+\beta)}=\frac{\sin\alpha}{\sin(30+\alpha)}$

simple manipulation shows, it's equivalent to $\cos(30-\alpha+\beta)=\cos(30+\alpha-\beta) \Rightarrow \sin(\alpha-\beta)=\sin(-\alpha+\beta) \\\Rightarrow \sin(\alpha-\beta)=0 \Rightarrow \alpha-\beta=180k \Rightarrow \alpha=\beta$

Additional: Asceratin the value of
$\widehat{ABD}-\widehat{BCD}$.

This is my Christmas present for everybody on this site!

Best regards,
sunken rock


linear algebra
matrix
algebra
polynomial
complex numbers

I'm sorry for this noob question, but I'm not good in matrix so here it goes :

If $ S$ is a square matrix, and $ T$ is its complex conjugate, prove that $ det(ST)$ is a real number.

By Schur's Lemma, $ S\equal{}P^{\minus{}1}UP,$ where $ P$ is upper triangular. Then $ \overline{S}\equal{}\overline{P^{\minus{}1}}\,\overline{U}\,\overline{P}$ and

$ \det(S\overline{S})\equal{}\det{U}\det{\overline{U}}\equal{}\det\left(U\overline{U}\right).$

But $ U\overline{U}$ is itself upper triangular and its diagonal elements are the products of the the diagonal elements of $ U$ and $ \overline{U}$ and hence real.

This is really just another way of saying that $ \det(S)\equal{}\lambda_1\lambda_2\cdots\lambda_n,$ where the $ \lambda_j$ are the eigenvalues of $ S,$ and $ \det\left(\overline{S}\right)\equal{}\overline{\lambda_1}\,\overline{\lambda_2}\cdots\overline{\lambda_n}\equal{}\overline{\det(S)}.$ The product of those is real and positive.

Too complicate Kent.
$ det(S)\equal{}P(s_{ij})$ where $ P$ is a polynomial with REAL coefficients.
$ \det(S\overline{S})\equal{}det(S)det(\overline{S})\equal{}P(s_{ij})P(\overline{s_{ij}})\equal{}P(s_{ij})\overline{P(s_{ij})}\equal{}|P(s_{ij})|^2\equal{}|det(S)|^2\geq{0}$.

Agreed. What I did was too complicated and yours is the quickest way.


I just realized by looking at it just now. You know how under the fram in the AoPS website, there is always an animated proof? like under the toolbar thing with links like Resources, LaTex, Community etc. they always have some animated thing playing. any idea how they make those things? with flash?


so he missed the test this year. I wonder if it is possible to take it elsewhere if his school will still not offer it next year? Can he take 10A or 12B at different school next year? Thanks for advice.

well he cannot take 10A seeing as the test dates for 10A are over and the answers are out, but you could check with the nearest university if they offer the B tests.

If your school offers the B date, you really shouldn't worry. Most schools offer only one date, which is either A or B depending on which is more convenient for the school
If your school offers neither date, lobby them to do it next year. In the meantime, [url=http://www.unl.edu/amc/b-registration/b1-archive/2007-2008/CU2008/2008-CU-list.shtml]this list[/url] of non-HS sites that offer the B date may help.

See also Frequently Asked Question #4 at:
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions


calculus integration logarithms trigonometry geometry rectangle limit

last question for today...  :D i asked this integral for HW yesterday,
it isn't as hard as you think, just requires a [u]slight trick[/u] to simplify it...

$ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\;\;\ln\ln\tan x\;\;\textbf dx\;\;\equal{}\;\;\boxed{\frac{\pi}{2}\;\ln\left[\;\sqrt{2\pi}\cdot\frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\;\right]}$

[ hint: sometimes, a [b]1D [/b]integral can be solved rather easily only
after transforming it to a [b]2D [/b]integral...  :|  ]

Define $ f(z) \equal{} \frac {2\alpha}{(\alpha^2 \plus{} z^2)\cosh \pi z}$ then $ 2\pi i\, \text{res} f\Big|_{z \equal{} i\alpha} \equal{} \frac {2\pi}{\cos \alpha\pi} \equal{} \pi \cot \left[\left(\frac14 \plus{} \frac {\alpha}{2}\right)\pi\right] \minus{} \pi \cot\left[ \left(\frac34 \plus{} \frac {\alpha}{2}\right)\pi\right]$

and $ 2\pi i\sum_{k \equal{} 0}^\infty \text{res} f\Big|_{z \equal{} i\frac {2k \plus{} 1}{2}} \equal{} \sum_{k \equal{} 0}^\infty ( \minus{} 1)^k \left[\frac {4}{2k \plus{} 1 \plus{} 2\alpha} \minus{} \frac {4}{2k \plus{} 1 \minus{} 2\alpha}\right]$

$ \equal{} \left[\psi\left(\frac34 \plus{} \frac {\alpha}{2}\right) \minus{} \psi\left(\frac14 \plus{} \frac {\alpha}{2}\right)\right] \minus{} \left[\psi\left(\frac34 \minus{} \frac {\alpha}{2}\right) \minus{} \psi\left(\frac14 \minus{} \frac {\alpha}{2}\right)\right]$

whereat $ \psi\left(\frac34 \minus{} \frac {\alpha}{2}\right) \equal{} \psi\left(1 \minus{} \left(\frac14 \plus{} \frac {\alpha}{2}\right)\right) \equal{} \psi\left(\frac14 \plus{} \frac {\alpha}{2}\right) \plus{} \pi \cot \left[\left(\frac14 \plus{} \frac {\alpha}{2}\right)\pi\right]$

and $ \psi\left(\frac14 \minus{} \frac {\alpha}{2}\right) \equal{} \psi\left(1 \minus{} \left(\frac34 \plus{} \frac {\alpha}{2}\right)\right) \equal{} \psi\left(\frac34 \plus{} \frac {\alpha}{2}\right) \plus{} \pi \cot \left[\left(\frac34 \plus{} \frac {\alpha}{2}\right)\pi\right]$ .

Let $ \gamma_k$ be the rectangle shaped contour path with corners $ \minus{} k\pi,k\pi,k\pi \plus{} ik\pi, \minus{} k\pi \plus{} ik\pi$ then

$ \int_{ \minus{} \infty}^\infty f \, dx \equal{} \lim_{k\to\infty} \oint_{\gamma_k} f\, dz \equal{} 2\pi i \, \sum_{\text{Im} > 0} \text{res} f \equal{} 2 \left[\psi\left(\frac34 \plus{} \frac {\alpha}{2}\right) \minus{} \psi\left(\frac14 \plus{} \frac {\alpha}{2}\right)\right]$ .

Now integrate with respect to $ \alpha$ :

$ \underbrace{\int_{ \minus{} \infty}^\infty \frac {\ln(\alpha^2 \plus{} x^2)}{\cosh \pi x}\, dx}_{ \equal{} : U(\alpha)} \equal{} \underbrace{4\left[\ln\Gamma\left(\frac34 \plus{} \frac {\alpha}{2}\right) \minus{} \ln\Gamma\left(\frac14 \plus{} \frac {\alpha}{2}\right)\right]}_{ \equal{} : V(\alpha)} \plus{} C$ .

Because of $ U(\alpha) \minus{} \ln(\alpha^2) \equal{} U(\alpha) \minus{} \int_{ \minus{} \infty}^\infty \frac {\ln(\alpha^2)}{\cosh \pi x} \, dx \equal{} \int_{ \minus{} \infty}^\infty \frac {\ln\left(1 \plus{} \frac {x^2}{\alpha^2}\right)}{\cosh \pi x}\, dx\to 0$ for $ \alpha\to\infty$

and $ V(\alpha) \minus{} \ln(\alpha^2)\to \minus{} 4\ln\sqrt {2}$ the constant $ C$ must be $ 4\ln \sqrt {2}$ .

Hence the RHS can be written as $ 4\ln \left[\sqrt {2} \,\frac {\Gamma\left(\frac34 \plus{} \frac {\alpha}{2}\right)}{\Gamma\left(\frac14 \plus{} \frac {\alpha}{2}\right)}\right]$ .

For $ \alpha\to 0^ \plus{}$ we obtain $ \int_0^\infty \frac {\ln x}{\cosh \pi x}\, dx \equal{} \frac14 \int_{ \minus{} \infty}^\infty \frac {\ln(x^2)}{\cosh \pi x}\, dx \equal{} \ln\left[\sqrt {2}\, \frac {\Gamma\left(\frac34\right)}{\Gamma\left(\frac14\right)}\right]$ .

And this is $ \int_0^\infty \frac {\ln x}{\cosh x}\, \frac {dx}{\pi} \minus{} \underbrace{\int_0^\infty \frac {\ln \pi}{\cosh x}\, \frac {dx}{\pi}}_{\frac12 \ln \pi \equal{} \ln \sqrt {\pi}}$ after substitution $ x\mapsto \frac {x}{\pi}$ .

So $ \int_0^\infty \frac {\ln x}{\cosh x}\, dx \equal{} \pi \ln\left[\sqrt {2\pi}\, \frac {\Gamma\left(\frac34\right)}{\Gamma\left(\frac14\right)}\right]$ and [url=http://mathworld.wolfram.com/VardisIntegral.html]Vardi's integral[/url] $ \int_{\pi/4}^{\pi/2} \ln\ln\tan x\, dx$

is $ \frac12 \int_0^\infty \frac {\ln x}{\cosh x}\, dx$ after substitution $ x\mapsto \arctan(e^x)$ .

How did you compute the integral with transforming it in a [b]2D[/b] integral?

this was already posted here http://www.mathlinks.ro/viewtopic.php?p=1058414#1058414
i was thinking about serie expansion for $ \ln \ln \tan (x)$ when $ 0 < |x| < \frac {\pi}{2}$ I can find $ \ln \tan (z) \equal{} \sum a_{n}z^{2}$ but I can't for $ \ln \ln \tan (z)$ :(
interesting integral...


These seem like fun problems that will challenge/torment my mind. Thanks.

With February my busiest month and with AMC's right on top of this, I think I'll have lots of fun trying to solve these on top of all other things.

lol i agree i am going to be SO busy

the problems are very interesting, though i am quite stumped on a few.

Huh...being honest, I thought it easier than previous rounds...well in certain ways.

[quote="ehehheehee"]uh, being honest, you being so good can make me feel bad.[/quote]

Sorry...

But I'm comparing it to previous rounds...how do you know what i thought of those?

You got a 50 on the first two rounds  :lol:

Edit: I think that this has gotten a little far about the difficulty of the round and I could get in trouble, so I am going to shut up now.

[quote="ehehheehee"]Edit: I think that this has gotten a little far about the difficulty of the round and I could get in trouble, so I am going to shut up now.[/quote]

Good call.  In any case, good luck to everyone on round 4!

Once again, please do not discuss your perceived difficulty of the round while it is in progress.  Thank you!


Let M be a positive integer and consider the set
$S=${${n\in\mathbb{N}:M^2 \le n<(M+1)^2}$}
Prove that the products of the form ab with $a,b \in S$ are all distinct.

Suppose $ab=cd$. Due to Four factors lemma we know that $a=xy$, $b=uv$, $c=xu$, $d=yv$.
There are few cases:
1) $x\leq m\leq y$ and $u\leq m\leq v$;
2) $x\leq m\leq y$ and $v\leq m\leq v$.
It is quite easy to analyze them.
1) if $x<m$ and $u\leq m$, then $c<m^2$ -- contradiction. Thus $x,u\geq m$. And it is known that it is impossible $x,y\geq m+1$ ( ;) ). And so on...
I have no time to finish.

This is exercise #63, with solution, here (don't tell me about french) :
http://www.animath.fr/cours/arithm.pdf

Pierre.


I just finished both of Collegeboard's AP Computer Science courses, so for next year I am going to be taking Computer Science courses at an university because my High School ran out; since I do not have any friends that go there, I was wondering if you nice people at AoPS could offer me some advice on which one to pick; I was offered the option of choosing any one of the following:

http://www.cs.princeton.edu/courses/archive/spring09/cos126/info.html
http://www.cs.princeton.edu/courses/archive/spring09/cos217/info.html
http://www.cs.princeton.edu/courses/archive/spring09/cos226/info.html

The stuff I know so far:
-JAVA
-C and C++ (Lol, C++)
-Visual Basic and TI-BASIC (LOL)
-Really basic knowledge of data structure crap, algorithmic design, heuristics, computational complexity and game theory

What should I do? I am really not sure if I would be able to handle COS 217 or 226, even though it seems like I know the stuff there; if anyone here can offer me some tips and advice on which course to take and how to survive college, I would be really grateful for the input! Thank you.

-The Bankrupt Aristocrat

Well, I might be (obviously) biased, but I think algorithms / data structures are really interesting. :)
(In particular, they are the probably the area of computer science that is the most similar to math.)

Anyway, I'd recommend looking at lectures and assignments from previous years of these courses (which seem to be available online) and seeing what looks interesting to you - this will probably be the most helpful in making your decision.

126 seems like a standard intro course. AP CS is alright, but definitely not the best CS foundation you can get. If you feel like your grasp of basic programming concepts is still weak, definitely take this, as having a good foundation is very important once you get into higher level classes.

217 and 226 look like introductory systems and theory courses, respectively. A good CS major should know both, so I would go ahead and pick whichever seems more interesting to learn about first, as you will probably encounter both at some point.

If you've never experienced college courses before, it'll be a bit challenging, but not that much of a problem. Just be prepared to go through a lot of material very quickly; study hard and you're good to go :) (of course, the studying hard part is what makes it such a challenge).

The Alg. & Data Structures should be a good foundation course.


Let $(x_{n})$ and $(y_{n})$ be two sequences which satisfy
$x_{1}=a, y_{1}=b$
$x_{n}=\frac{x_{n-1}+y_{n-1}}{2} y_{n}=\sqrt{x_{n}y_{n-1}}$
Find the limitation of $(x_{n})$ and $(y_{n})$?
Easy?

Is the term for $y_n$ correct¿
If it is $y_n=\sqrt{x_{n-1}y_{n-1}}$ instead, this is the arithmetic-geometric mean which converges very fast, but it's exact value is an elliptic integral. See http://mathworld.wolfram.com/Arithmetic-GeometricMean.html .


probability
geometry
calculus
absolute value

Two positive real numbers $x$ and $y$ are randomly chosen. A real number $z$ is chosen at ranndom between $x$ and $y$.  what is the probability that if $z$ is closer to $x$ than $y$, that $z^2$ is closer to $x^2$ than $y^2$, or if $z$ is closer to $y$ than $x$, that $z^2$ is closer to $y^2$ than $x^2$

Since I'm going to Germany today, I'll tell superman the answer and he'll tell you if you got it right.

So superman, don't post the answer

but i don't know how to do. it...i just know the answer...

Here's what I make of it :
[hide]For convenience, let us call $y$ the greater number and $x$ the smaller.

Case 1 :  IF $z$ IS CLOSER TO $x$ :

$z^2$ will be closer to $x^2$ :
If we call the positive difference between $x$ and $y$ $m$, then we can call $y^2$
$(x+m)^{2}=x^2+2mx+m^2$
Halfway between would be $x+\frac{1}{2}m$.  Obviously, $x^2+mx+\frac{1}{4}m^2$ is closer to $x^2$, because $x^2+mx+\frac{1}{2}m^2$ would be half-way and $m^2$ must be positive.
Q.E.D.

Case 2 :  IF $z$ IS CLOSER TO $y$ :

We have a more complicated situation.  We must find the place where $z^2$ is half-way between $x^2$ and $y^2$.  This would be where
$z^2=x^2+\frac{1}{2}(y^2-x^2)$

$z^2=\frac{x^2+y^2}{2}$

$z=\frac{\sqrt{2x^2+2y^2}}{2}$
We need the stretch of numbers that will be greater than $z$, since when $n$ is positive, as $n$ increases, $n^2$ increases.  Its length would be
$y-\frac{\sqrt{2x^2+2y^2}}{2}$
or
$\frac{2y-\sqrt{2x^2+2y^2}}{2}$
The portion of the area closer to $y$ that this would occupy is
$\frac{\left(\frac{2y-\sqrt{2x^2+2y^2}}{2}\right)}{\frac{y-x}{2}}$,
or simply
$\frac{2y-\sqrt{2x^2+2y^2}}{y-x}$
Q.E.D.

GENERAL CASE

We need to take the average between the probabilities for the two cases, since they are equally likely.  Put as a common fraction (since we want a probablility), that is

$\frac{-x+3y-\sqrt{2x^2+2y^2}}{-2x+2y}$,

where $y>x$.

Q.E.D.

Of course, if you want an answer not in terms of $x$ and $y$, that's a different matter.  As a matter of fact, I don't know if one would be able to do it without any knowledge of calculus.  But then, I don't know calculus, so I can't say.[/hide]

There.  I think I fixed it.  How embarrassing.  (Last time, I put $\frac{|y-x|+\sqrt{2x^2+2y^2}}{|y-x|}$.)

But x, y, z are all randomly chosen, so you can't put your answer in terms of them.

As a wild guess, one half?!?!?!?!?!?!?!?!?! Just a hunch.

[hide]I stand by my answer of $\frac{-x+3y-\sqrt{2x^2+2y^2}}{-2x+2y}$ where $y>x$.  Another way to get it (look to my first post here for further explanation) :
All the places where $z$ will satisfy the equation are everywhere between $x$ and $y$ except the area between $\frac{x+y}{2}$ and $\frac{\sqrt{2x^2+2y^2}}{2}$.  The length of this segment on the number line would be $\frac{-x-y+\sqrt{2x^2+2y^2}}{2}$.  Therefore, the sum of the lengths of the segments that satisfy the problem is $(y-x)-\frac{-x-y+\sqrt{2x^2+2y^2}}{2}$, or $\frac{-x+3y-\sqrt{2x^2+2y^2}}{2}$.  Put that over the length of the segment on the number line between $x$ and $y$, and we have

$\frac{-x+3y-\sqrt{2x^2+2y^2}}{2x-2y}$

N.B.  I assume that $y>x$ unless noted otherwise.

If we do not want it in terms of $x$ and $y$, it is definitely not $\frac{1}{2}$, because the probablility is greater than $\frac{1}{2}$, as I explained above.  Here is what I have so far :  When the difference between $x$ and $y$ is 0, the probablility is undefined ($\frac{0}{0}$).  However, as the difference approaches 0, the probablility approaches $\frac{1}{2}$.  Furthermore, as far as I can tell, as the difference approaches $\infty$, the probablity approaches $1$.  Therefore the answer will be somewhere in between, probably at $\frac{3}{4}$ or $\frac{\sqrt{2}}{2}$.[/hide]

Whoooooooooooooaaaaaaaaaaaaaaaaaaa...  :what?:

As a revised WILD GUESS, I would propose something crazy like 5/8, which is equal to (1/4)^2 + (3/4)^2. But I'm pretty sure that's wrong too. :huuh:

[hide]

I get $\frac{|2x-2y|+\sqrt{2x^2+2y^2}-x-y}{|2x-2y|}$ This was worked by assuming $y>x$ but it shouldn't matter when you have the absolute values in there. Here's how I did it.

Similar to the other person's solution, I found that it had to be:

$\frac{\sqrt{2x^2+2y^2}}2 < z < \frac{x+y}2$ for $z^2$ to be closer to $x^2$ when $z$ was closer to $y$.

The distance between these points is $\frac{x+y-\sqrt{2x^2+2y^2}}2$ and the total length of the segment is $|x-y|$ so you divide by the total length to get the probability that it will be what we don't want. Thus, the probability is $1-\frac{x+y-\sqrt{2x^2+2y^2}}{|2x-2y|}$ or $\frac{|2x-2y|+\sqrt{2x^2+2y^2}-x-y}{|2x-2y|}$
[/hide]

Your way is better, because the end solution does not assume that $y>x$.  However, if you are willing to assume that, you can remove the absolute value signs and the answers are the same.

im pretty sure thats right...nat mc says

[hide]$\displaystyle \frac{2|x-y|+x+y-\sqrt{2x^2+2y^2}}{2|x-y|}$

which is the same only without the 2 distributed through the absolute value thing[/hide]

I think he actually messed up....

according to what I had, he added 1 to the probability instead of subtracting it from 1. Also, I'm pretty sure you can distribute with absolute value, and the distributive form is simpler :)

EDIT: oops...actually both answers are wrong. the correct answer should be:

$\frac{|2x-2y|-|\sqrt{2x^2+2y^2}-x-y|}{|2x-2y|}$ however since we established that $\frac{x+y}2 > \sqrt{2x^2+2y^2}$. my answer was correct :D (he did it in the wrong order) The length of the segment should be $\frac{x+y-\sqrt{2x^2+2y^2}}2$ which, when negated, will be $\frac{-x-y+\sqrt{2x^2+2y^2}}2$ which means that my answer was correct.

EDIT2: oops :blush: actually I had it in the wrong order because in actuality $\frac{x+y}2 < \sqrt{2x^2+2y^2}$ so I had it in the wrong order, but the other valid answer is: $\frac{|2x-2y|-|\sqrt{2x^2+2y^2}-x-y|}{|2x-2y|}$

EDIT3: but still, he didn't distribute so that makes up for my careless error. :P

Hamster 1800, you don't need the absolute value signs around $\sqrt{2x^2+2y^2}-x-y$, because, as you pointed out, the statement $\frac{x+y}{2}<\frac{\sqrt{2x^2+2y^2}}{2}$ is always true, so the expression will always be positive.
But you are right :  he should have distributed, in my opinion.

but he made the problem...it probably got it out of some math book

I'M BACK!!!

Here is my explanation:

z cannot be between the average of x and y, and the square root of the average of their squares.
Since:

$\sqrt{\frac{x^2+y^2}{2}}\geq\frac{x+y}{2}$

So the answer is

$1-\frac{(\sqrt{\frac{x^2+y^2}{2}}-\frac{x+y}{2})}{|x-y|}$

$=$

${\frac{2|x-y|+x+y-\sqrt{2x^2+2y^2}}{2|x-y|}}$

However, each of the $2|x-y|$s should be distributed to $|2x-2y|$ to simplify.


Find an integer $m>0$ such that there exists a function $f: \mathbb R^{+} \to \mathbb R^{+}$ which satisfies
\[ f(xf(y)) = x^2y^m, \quad \forall x,y>0.\]

[quote="peter117"]Find $ m > 0$ such that
$ \exists f: R^ \plus{} \to R^ \plus{}$ which satisfy $ f(xf(y)) \equal{} x^2y^m$ $ \forall x,y > 0$[/quote]

Let $ P(x,y)$ be the assertion $ f(xf(y)) \equal{} x^2y^m$

$ P(1,1)$ $ \implies$ $ f(f(1))\equal{}1$
$ P(x,f(1))$ $ \implies$ $ f(x)\equal{}x^2f(1)^m$

Plugging $ f(x)\equal{}ax^2$ in the original equation, we get $ a^3x^2y^4\equal{}x^2y^m$ and so $ a\equal{}1$ and $ \boxed{m\equal{}4}$ and the solution $ f(x)\equal{}x^2$


A vendor has six jugs of punch with capacities of 13, 15, 16,
21, 21 and 22 gallons. One jug, which he keeps for himself, is
full of blue punch. The other five jugs are
full of red punch. Without opening any of
the jugs, he sells 1/3 of the red punch to
Randy and the remaining 2/3 of the red
punch to Lisa. How many gallons of the
blue punch did he keep for himself?

the jug thats taken must leave a multiple of 3 that can be made with any combination of the remaining jugs

13, 15, 16, 21, 21 and 22 are the numbers

take away the 13: 95 gallons left, not multiple of 3
take away the 15: 93 gallons left, 1/3 is 31, cant be made without opening the jugs
take awa the 16: 92 left, not multiple of 3
take away one of teh 21's: 87 left, 1/3 is 29, can be made with 16+13: so teh answer is 21


isn't it fun? i just submitted my homework for hogwarts and got a 3 my teacher told me that calculus has nothing to do with potions. isn't it so fun to spam on RPG forums and such? :lol:

it is fun but ur more annoying so...

I like FPSs more

[quote="funcia"]i just...got a 3[/quote]

Hey, on an AP exam, that's not bad.

hey look lots of people don't care to give a straight answer! :(

[quote="junggi"]huh what does rpg and hogwarts have to do with each other[/quote]
really, what DO they ahve to do with each other?


posted here http://www.mathlinks.ro/Forum/viewtopic.php?t=4830  :wink:


Suppose $ f: R^{+}\rightarrow R^{+}$ is a decreasing continuous function such that for all $ x,y \in R^{+}$:
\[ f(x+y)+f(f(x)+f(y))=f(f(x+f(y))+f(y)+f(x))\]
Prove that $ f(f(x))=x$.

here we take $ 0\not\in R^{+}$, because if $ 0\in R^{+}$, $ f\equiv 0$ is also solution and we have $ fof\neq Id_{R^{+}}$

then we  have $ f(f(x+f(y))+f(x)+f(y))>f(f(x)+f(y))$ (because $ f(x)+f(y)>0$)
$ f$ decreasing gives $ f(x)+f(y)+f(x+f(y))\le f(x)+f(y)$
then $ f(x+f(y))\not\in R^{+}$ so there are not solution for this problem.

[quote="aviateurpilot"] so there are not solution for this problem.[/quote]
Sorry.. my mistake! The correct is:
Suppose $ f: R^{+}\rightarrow R^{+}$ is a decreasing continuous function such that for all $ x,y \in R^{+}$:
\[ f(x+y)+f(f(x)+f(y))=f(f(x+f(y)))+f(y+f(x)) \]
Prove that $ f(f(x))=x$.

$ f(x+y)+f(f(x)+f(y))=f(f(x+f(y)))+f(y+f(x))$
$ f(x+y)+f(f(x)+f(y))=f(f(y+f(x)))+f(x+f(y))$
then $ f(f(y+f(x)))+f(x+f(y))=f(f(x+f(y)))+f(y+f(x))$
so $ g(y+f(x))=g(x+f(y))$ where $ g(x)=fof(x)-x$
it's easly to prove that $ \exists a\in\R^{+}: \ g(a)=a$
$ \forall x\in R^{+}: \ g(f(a-f(a+x))+a+x)=g(a-f(a+x)+f(a+x))=g(a)=0$
and it's easly to prove that $ \exists c\in R^{+}: \ [c,+\infty[\subset \{f(a-f(a+x)+a+x)|\ x\in\R^{+}\}$
then $ \forall x\ge c: \ fof(x)=x$
i will continue after...


As we all know, tomorrow we leave for nationals in Arlington, VA.

This is a thread for last minute comments, etc.

GO DELAWARE

thats all i have to say

GO KANSAS

we must get atleast 54th place...

good luck everyone, I hope we all have fun.

Though I am not going, I would like to wish all the AoPSers good luck and godspeed. Lets hope that an AoPSer wins so we can ask him/her how he/she did it. ;)

well i get to take a pimped out train. >.<

if i get top 4 individuals, who wants to take my spot going to the white house?

stupid floridians... if only you could count we wouldnt have this man in office and i could meet al gore intstead but thats not much better since he would think i was manbearpig :D  :D (watch south park for that one)

I'm leaving for the airport right now...

Please do not use any word ending in -izzle in this forum, nor the other word tz used. Right now, before MC, this is a bad way to relieve your stress. Sleep. You won't sleep enough at Nats.

good luck everyone!!!
(i didn't make nats)

[quote="solafidefarms"]Please do not use any word ending in -izzle in this forum, nor the other word tz used.[/quote]

What about "drizzle"?

[quote="solafidafarms"]Right now, before MC, this is a bad way to relieve your stress. Sleep. You won't sleep enough at Nats.[/quote]

I'm proof of thattttttt... :stretcher: :stretcher:

Hope everyone had fun at nationals.


I am! I have never seen this test, and it will be interesting to see if I can do any of them.  :lol: I have been practicing geometry lately, none AMC 8 stuff because I didn't want to.  :roll:

Judging on the stuff you post in HSB, you will do fine on the AMC 8. It's much easier that everyone thinks.

I've taken quite a few practices, full scored all of them, hopefully I can avoud stupid mistakes next week...

[quote="bpms"]I haven't seen an old AMC 8.
Could someone email me one or give me a link to one?
My email is glogmalblood@gmail.com
...
...
...[/quote]

Alright bpms. I'll PM you 2001.


function
topology
advanced fields
advanced fields unsolved

I want to compute the fundamental group of P^n for n greater than or equal to 2. Here is my plan:

I want to use the fact that the S^n is a universal cover of P^n and that theorem that says that the group of covering transformations of a universal cover is isomorphic to the space that it covers.

My book also indicates that this fundamental group is generated by the composition of the functions pg: I ->S^n is any continuous map where the initial point and the end point are antipodal and p is the covering map.

Its clear that the unique lifting theorem will come in somehow.

Any insight into that?

There is a bijection from the fundamental group at x to $ \pi^{\minus{}1} ( x )$, where $ pi$ is the natural map from $ S^n$ to $ P^n$. So the group is just $ Z_2$


Simplify:

                  1/2(x-2y)-2/3(3x-y)

Thank you for your help in advance!!!

i'm assuming it's:
$\frac {1}{2(x-2y)}-\frac{2}{3(3x-y)}$
take the common denominator then you get somthing like this,
$\frac {5x-11y}{(3x-y)^2+(3x-6y)^2-25y^2}$

hexaditidom,
sorry this is a bit off topic but in your signature wouldn't the L.H.L be -infinty which would mean lim D.N.E
same for the second limit?

His signature is a math joke.  The value of the first limit is infinity, which looks like a sideways 8, and 8 is what x approaches in that problem.  So therefore when x approaches 5 in a similar limit, the value is a sideways 5.

I understood that
but does the first limit exist? I dont think it is infinity,
the L.H.L does not equal R.H.L

no the first limi does not exist, it is $\infty$

Since amirhtlusa already simplified one possibility, I'll do the other:
$\displaystyle \frac{1}{2}(x-2y) - \frac{2}{3}(3x-y)=-\frac{3x}{2}-\frac{y}{3}$

Limit does not exist and limit approaches $\infty$ are two very different things, by the way.

in this case, they are the same...!!!because there is a vertical assymptote at $x=8$.

His signature is not mathematically incorrect.  See

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=215486&highlight=#215486

 for details.


geometry Support trigonometry function probability MATHCOUNTS AMC

Hello, I will be programing a math training program for everyone to train with. This will give you sets of problems with solutions. The main area I need help in is finding problems. I need the support of the AoPS community to submit
a large quantity of problems to aid in this programs creation.

You can submit problems by putting them into the proper form and pming them to me. Please, submit problems in bulk. I don't want my inbox filling up.
(If you can program and want to help me write the parser, pm me also)

Problem submission form is as follows:
There are two kinds of problems:multiple choice and proof.
Sorry, we will not be able to use latex. Use '/' for fractions, '^; for exponents, trig functions like sin(x), and logs like log(4,5) where the first is the base.
The submission form for a multiple choice problem is as  follows:

(explanation follows)
[quote]
Name:
Type:
Difficulty:
Problem:
A:
B:
C:
D:
E:
Answer:
[/quote]
You must follow this format exactly, just copy and paste it.
Name is the name, type can be Algebra, Geometry, Counting, Probablility, or Number Theory. Difficulty is mathcounts, intermidiate(AMC, everything in the intermidate problems resources on the site, and some AIME) and Olympiad.
Problem is the actual problem text, again, do not use latex.
The choices and self explanitory. The Answers gives a correct choice [b]and explanation[/b].

Proof questions follow the Format:
[quote]
Name:
Type:
Difficulty:
Problem:
Correct Proof:
[/quote]
With everything the same as above expect 'corrent proof' should be thourogh and self explanatory.

Please take the time to transcribe a few problems into this form and PM me.
[b]If you have question or want to help with the project, post below[/b]
[size=125]Thank you for your support![/size]

Quite interesting. Perhaps it should be placed to be read in the problem-solving portion of this forum.

Moved from F&G.

You could just sort of steal the problems from the MMPC I guess, they have a ton.

[quote="chess64"]What computer language will it be in?[/quote]
Python.


calculus
integration
function
real analysis
derivative
calculus computations

This is a quick problem:
Let $F$ be the Cantor function.
Find

$\int_{0}^{1}F(x)dx=$

Consider the following sequence of piecewise linear continuous functions. We can completely describe such a function by naming its vertices; we understand that these vertices are to be connected by line segments. For each function below, I will simply list the vertices. In order to understand what is going on, I strongly suggest that you draw your own pictures of the graph.

$F_{0}(x): (0,0),(1,1)$
$F_{1}(x): (0,0),(\frac13,\frac12),(\frac23,\frac12) ,(1,1)$
$F_{2}(x): (0,0),(\frac19,\frac14), (\frac29,\frac14),(\frac13,\frac12), (\frac23,\frac12) ,(\frac79,\frac34), (\frac89,\frac34),(1,1)$

To go from $F_{2}(x)$ to $F_{3}(x),$ add in 8 new vertices. These vertices include

$(\frac1{27},\frac18),(\frac2{27},\frac18), (\frac7{27},\frac38) \cdots, (\frac{25}{27},\frac78),(\frac{26}{27}, \frac89)$

Now take the limit as $n\to\infty$ of $F_{n}(x).$ This sequence of continuous functions converges uniformly, so its limit is continuous. Call that limit $F(x),$ also known as the Cantor function, the Cantor-Lebesgue function, or the "devil's staircase." The function is constant on each interval in the complement of the Cantor set, which means that on the complement of the Cantor set, it is differentiable and its derivative is zero. Its derivative does not exist at any point on the Cantor set, but since the Cantor set has measure zero, $F$ is differentiable almost everywhere.

There are several other possible characterizations of $F,$ including some which use the binary and ternary expansions of real numbers.

A good sketch of the graph is [url=http://mathworld.wolfram.com/CantorFunction.html]here[/url]. It suggests a solution: [hide]the graph divides the square $[0,1]\times [0,1]$ into equal parts.[/hide]


(4*10-4 )-1
I need to put the simplied answer in scientific notation. Please help me, I am confused about how to put something in a scientific notation if the first digit is negative. is it possible?

[hide]first do whats inside the parentheses:
(4/10,000)^-1
-1 means just flip it over and raise to 1:
(10,000/4)^1 --> 10,000/4 --> 2,500
Convert to scientific notation:
2.5*10^3[/hide]


vector
algorithm
linear algebra
matrix
geometry unsolved
geometry

Prove that if $\vec{r} \cdot ( \vec{a} \times \vec{b} ) = 0$ , then $\vec{r} = \lambda \vec{a} + \mu \vec{b}$ where $\lambda$ and $\mu$ are scalars. I'm not looking for a proof based on a geometric interpretation - that is really simple! I am looking for a proof that uses Algebra and some clever manipulations to get the result from the fact that scalar dot product of the three vectors is zero.

Since $\vec{r} \cdot ( \vec{a} \times \vec{b} )=\left|\begin{array}{ccc}r_1&a_1&b_1 \\r_2&a_2&b_2 \\r_3&a_3&b_3 \end{array} \right|$ and the determinant is zero iff one of the columns is a linear combination of the others,  the answer is clear :)

I dont think your condition has an 'iff' -- its obvious that the determinant is zero if one column IS a linear combination of others. What about the converse which is my problem? How do you prove that?!?!?

I think it just follows from the algorithm of solving sest of  3 equations using matrix....
See http://mathworld.wolfram.com/CramersRule.html

Don't worry! I got my question answered in the Matrix Algebra section!


greatest common divisor
algebra unsolved
algebra

Let $ a_1.a_2,a_3,....   a_n$  be positive integers such that :
         i) $ a_i < 2008$    for    $ i \equal{}1,2,3...,n$
         ii) The greatest common divisor of any pair of number is greater than 2008
                Prove that
           $ \sum\frac{1}{a_i}  \leq 2$

[quote="gagarin"]Let $ a_1.a_2,a_3,.... a_n$  be positive integers such that :
         i) $ a_i < 2008$    for    $ i \equal{} 1,2,3...,n$
         ii) The greatest common divisor of any pair of number is greater than 2008
                Prove that
           $ \sum\frac {1}{a_i} \leq 2$[/quote]
I don't undestand, if $ 0<a_i<2008$, then $ gcd(a_i,a_j)\le min(a_i,a_j)<2008$.
May be $ lcm(a_i,a_j)>2008, i\not \equal{}j$.


For non-negative real numbers $x_1, x_2, \ldots, x_n$ which satisfy $x_1 + x_2 + \cdots + x_n = 1$, find the largest possible value of $\sum_{j = 1}^{n} (x_j^{4} - x_j^{5})$.

according to mathematical reflections it is due to vasc
majorization and am gm are keys and :P this is closely related to imo 99a2,hope you get ,if try.

[quote="ZHANGWENZHONGKK"]It must be 1/16 ,when two of them are 1/2.[/quote]
Obviosly, maximum when $ x_i\equal{}1,i<5, x_j\equal{}0,j\not \equal{}0$ and $ max\equal{}1$.

[quote="Rust"][quote="ZHANGWENZHONGKK"]It must be 1/16 ,when two of them are 1/2.[/quote]
Obviosly, maximum when $ x_i \equal{} 1,i < 5, x_j \equal{} 0,j\not \equal{} 0$ and $ max \equal{} 1$.[/quote] What?

Can anyone post a solution please?I really like this problem.

[quote="ZHANGWENZHONGKK"]It must be 1/16 ,when two of them are 1/2.[/quote]
$ (x_1,x_2) \equal{} \left(\frac{4}{5},\frac{1}{5}\right) \Rightarrow (x^4_1 \minus{} x^5_1)\plus{}(x^4_2 \minus{} x^5_2) \equal{} \frac{52}{625} \equal{} 0.0832 > \frac{1}{16} \equal{} 0.0625$.

Here is a sketch of my method.  The function $ f(x)\equal{}x^4\minus{}x^5$ is convex in some intervals and concave in others.  Let's say $ x_1,x_2,...,x_i$ are in the interval in which it is convex and $ x_n,x_{n\minus{}1},...,x_{i\plus{}1}$ are in the interval in which it is concave.  Then it follows that the maximum occurs when $ x_n\equal{}x_{n\minus{}1}\equal{}\cdots\equal{}x_{i\plus{}1}$ and $ x_2\equal{}x_3\equal{}\cdots\equal{}x_i\equal{}0$ and $ x_1\equal{}1\minus{}(n\minus{}i)x_n$.  Now we are simply maximizing $ (n\minus{}i)(x_n^4\minus{}x_n^5)\plus{}(1\minus{}(n\minus{}i)x_n)^4\minus{}(1\minus{}(n\minus{}i)x_n)^5$, which should be fairly straightforward.

In the end I got the maximum when all $ x_i$ are equal.

maximum, when $ x_1\equal{}\frac{3\plus{}\sqrt 3}{6},x_2\equal{}\frac{3\minus{}\sqrt 3}{6}, x_3\equal{}...\equal{}x_n\equal{}0$, Then $ f_max\equal{}\frac{1}{12}\equal{}0.83333333...$.

Yes, sorry.  My last few logic steps were mistaken.  One finds that the maximum when looking at $ n\minus{}i$ occurs when $ n\minus{}i$ is maximal.  Since $ \sum x_i\equal{}1$ and since $ f$ is concave on $ x\in(\frac{3}{5},\infty)$, it follows that the maximum of $ n\minus{}i$ is $ 1$.  Now we have a quintic of which we need to find the maximum.  The derivative turns out to be $ \minus{}(\minus{}1 \plus{} 2 x) (1 \minus{} 6 x \plus{} 6 x^2)$, and so the maximum is what Rust said.

WOW I'm literally surprised that this thread hasn't got a full solution for [b]13 years[/b]...  :D

The main idea is [u][b][color=#9a00ff]SMOOTHING[/color][/b][/u]. Let's follow sir CANBANKAN's technique: check small cases and then generalize.
So we first check $n = 2,3$. It's easy to show that there are at most two $0$s among $x_1, \cdots, x_n$ when $\sum_{j=1}^n(x_j^4-x_j^5)$.
Therefore, consider a smooth $(x,y)\to (x+y,0)$. We wish to show
$$(x+y)^4-(x+y)^5>x^4-x^5+y^4-y^5\iff 4x^2+4y^2+6xy>5x^3+5y^3+10x^2y+10xy^2$$

The LHS of the above inequality $ = \frac 72(x^2+y^2)+\frac 12(x^2+y^2)+6xy\geqslant \frac 72(x^2+y^2)+xy+6xy = \frac 72(x+y)^2$, while the RHS $\leqslant 5x^3+5y^3+15x^2y+15xy^2 = 5(x+y)^3$. Henceforth if we make sure $x+y<\frac {7}{10}$ the above inequality holds. So we have the following key claim:

[b]Claim.[/b] If $x+y<\frac {7}{10}$, then $(x+y)^4-(x+y)^5>x^4-x^5+y^4-y^5$.

Denote $k$ by the number of non-negative variants among $x_1, \cdots, x_n$. WLOG suppose
$$x_1\geqslant x_2\geqslant \cdots\geqslant x_k\geqslant 0, x_{k+1} = x_{k+2} = \cdots = x_n =0.$$
If $k\geqslant 3$, denote
$$x_i' = x_i(i = 1,2,\cdots, k-2), x_{k-1}' = x_{k-1}+x_k, x_k' = x_{k+1}' = \cdots = x_n' = 0.$$ Since $x_{k-1}+x_k\leqslant \frac 2n\leqslant \frac 23 <\frac {7}{10}$, by the Claim we have $\sum_{j=1}^n(x_j'^4-x_j'^5)>\sum_{j=1}^n(x_j^4-x_j^5)$.

As long as $k\geqslant 3$ the above smoothing process can be done. After at most $n-2$ smoothing, we can turn all the $x_3, \cdots, x_n$ into $0$ while $S$ doesn't decrease.

Denote $x_1 = a, x_2 = b$ at this very moment respectively, for convenience. Now $a+b=1$ and
$$S = a^4b+ab^4 = ab(a^3+b^3) = ab(a+b)(a^2+b^2-ab) = ab(1-3ab) = \frac 13(3ab)(1-3ab)\leqslant \frac 13\cdot\frac 14 = \frac {1}{12}.\qquad\blacksquare$$

Equality holds when $a = \frac{3+\sqrt{3}}{6}, b = \frac{3-\sqrt{3}}{6}$.  :yoda:

[size=75]Edit: Changed the tags.[/size]

Here's another smoothing solution, I suppose.

Let $P(x_1, x_2, \cdots, x_n)$ denote the sum for variables $x_1, x_2, \cdots, x_n$. Fix $x_3, x_4, \cdots, x_n$, and note that
\begin{align*}
x_1^4 - x_1 ^5 + x_2^4 - x_2^5 &\leq (x_1+x_2)^4 - (x_1+x_2)^5 \\
\iff 5(x_1+x_2)(x_1^2+x_1x_2+x_2^2) &\leq 4x_1^2+6x_1x_2+4x_2^2,
\end{align*}
which is true if $x_1 + x_2 \leq \frac 45$. So $$P(x_1, x_2, \cdots, x_n) \leq P(0, x_1+x_2, x_3, \cdots, x_n) \leq \cdots \leq P(x, y, 0, 0, 0, \cdots, 0).$$ In this case, let $x = \frac 12 + t$, $x_2 = \frac 12 - t$. With some algebra we may obtain the expression equals $$-3t^4 + \frac 12 t^2 + \frac 1{16} \leq \boxed{\frac 1{12}}$$ in this case. Equality obviously holds.


I was wondering how do you have to qualify for the ARML? From which competition. Also does anybody know what awards can you get from math league contests?

That's a hard question.  It depends on where you live and where you go to school.  As I understand each state normally sends 2 teams of 15 ppl.  So you have to be in the top 30 (or the top 30 that can go) in your state.  Different states have different ways of deciding who should be selected.  Also I believe that some of the really good schools send teams.

Don't know about math league though.

A state or region or school can send as many teams as it can fill. SFBA has been sending 5 teams for a few years.

I'd suggest contacting the arml people.  arml.com should give some info how.  Tell them where you live and they can point you to a coach that puts together a team from your region.

Any 15 pre-college people can form a team. The number of teams from a region, boundaries of the "region", and selection process are totally up to the individuals who take it upon themselves to organize teams. In Texas, there are 3 state-based teams, (increased from 2 several years ago when they switched to a bus company that could provide bigger busses). NYC, Chicago, and Bay Area all send varying numbers of city-based teams.  Thos Jefferson sends a school based team. Some teams are from multi-state regions or from just one part of a state. The ARML site has links to some of the team web-pages. You can start there and see if there is a team for you area which has already been organized. Here's the link:
http://www.arml.com/teams.htm

[quote="hello"]Also does anybody know what awards can you get from math league contests?[/quote]

Very nice plaques, plus the honor of being immortalized on the webpage.

That's true -- in 2002 someone on my team shouldn't have been, and the other team complained, and in 2003 he wasn't allowed to be on our team.

[quote="JBL"]Actually, it's not quite "any 15 people." I think you have to draw students from within well-defined, non-overlapping boundaries. So you couldn't, say, start a rival Michigan team.[/quote]

I understand your point, but there isn't a "Michigan" Team right now.

Okay, replace Michigan with Texas.  Or anywhere else that has a team.

For several years, Michigan has been sending at least two teams to the Iowa City ARML site.  I would expect them to be coming again - unless you have up to date info.

Their website is:  http://www.albion.edu/math/arml/

I would expect them too, but this year they don't have a team. Believe me, I know, I had a fairly good score on the MMPC (the ARML qualifying exam) and this year there is no state team. A local math club (ICAE) will send a team. This is the closest to an MI team there is.

Bummer about Michigan. Are you going with the local math club team? I hope so.

I suspect that whomever previously organized Michigan's state-based team decided not to do it this year for some reason, and no one else picked it up. One individual who takes an interest really makes a huge difference. Too late for this year, but maybe you could find an adult sponsor next year who would be willing to round up 15 Michigan kids who are not on the local math club's team and form a state team. Or maybe the math club would be willing to let kids from all over the state compete with them and send 2-3 teams. In Texas there are 45 kids from all over the state on the "Texas ARML team". Once they get to Iowa, they are divided into gold, silver, bronze teams based on a try-out exam. If a particular school, city, or club in Texas wanted to send their own team, I think they could without affecting the basic structure for anyone else in the state. You're right about my "any 15 people" generalization being in error. I don't think the Texas team could include stray students from Michigan, for example.


Okay, here is the problem...

[quote]Given a triangle ABC, let P,Q,R be the feet ot eh perpendiculars from a point S tot eh sides BC, CA, AB respectively. L,M,N are the midpoints, respectively, of these sides. Prove that...

(LP)(BC) + (MQ)(CA) + (NR)(AB) = 0[/quote]

(Yes, the book does say that it equals zero. I am really really confused at this part. Are they perhaps suggesting that there are negative lengths?)

Here (referring to the image) the points are labeled a bit differently.
Here is teh same question...

Given a triangle ABC, let I,H,J be the feet to the perpendiculars from a point D to the sides BC, CA, AB respectively. F,E,G are the midpoints, respectively, of these sides. Prove that...

(FI)(BC) + (EH)(CA) + (GJ)(AB) = 0

[b]EDIT: AIME_is_hard suggested that the lengths are directed segments, which is probably what the book meant since it did talk a little (but only a little) about directed segments. BUT here is my next set of questions...

Then I have another question (whose answer I can't seem to find online)...
When we consider directed segments, like GJ and EH (in the figure) can both of them be considered positive and FI negative? i.e. the question is - if they're pointing in the same general direction but are not parallel, can we assign both GJ and EH a positive sign even though they are not parallel?

Now comes the really tricky part...

To find the length of a directed segment, do we have to use the cosine of the angle between the vector and the x-axis (for performing operations such as addition and multiplication) for the x-component and sine for the y-component? This stems from the question: What happens if one component is positive and the other is negative (for example if we assume the positive x-direction to be positive and the positive y-direction to be positive, what happens if the vector/ directed line segment starts at the origin and continues through quadrant IV, i.e. it points downward)?[/b]

Yeah the lengths are positive/negative depending on the direction.

For example, AB= -1 * BA

ohhh... wait, so they're talking about directed segments?

Then I have another question (that I can't seem to find online)...
When we consider directed segments, like GJ and EH (in the figure) can both of them be considered positive and FI negative? i.e. the question is - if they're pointing in the same [b]general[/b] direction but are not parallel, can we assign both GJ and EH a positive sign even though they are not parallel?

Now comes the really tricky part...

To find the length of a directed segment, do we have to use the cosine of the angle between the vector and the x-axis (for performing operations such as addition and multiplication) for the x-component and sine for the y-component? This stems from the question: What happens if one component is positive and the other is negative (for example if we assume the positive x-direction to be positive and the positive y-direction to be positive, what happens if the vector/ directed line segment starts at the origin and continues through quadrant IV, i.e. it points downward)?

Actually, a length cannot be negative.
If I travel 500 miles to Chicago, and then I travel back home, I have not traveled -500 miles. It's the same both ways. Or at least that's how it's supposed be..

Thanks for the speedy reply, but AIME_is_hard suggested that the lengths are directed segments, which is probably what the book means because they do talk a little bit about directed segments (although very little).

Yes, they are talking about directed segments. Since $ E$ is to the left of $ H$ and $ F$ is to the left of $ I$ (when you rotate the triangle so that side BC is the base) but G is to the right of J,
$ (FI)(BC) \plus{} (EH)(CA) \plus{} (GJ)(AB) \equal{} 0$
is equivalent to
$ (FI)(BC) \plus{} (EH)(CA)\equal{}(GJ)(AB)$
if we are only talking about the lengths of the sides, because in the original equation, they're considering (GJ)(AB) to be negative.

Thanks for the quick reply!

Since I don't know a whole lot about directed segments, I'm going to re-post my questions here (and number them) so that people can respond each of them (I'm sorry, dgreenb801, but I don't know which question your response answers)

1)
When we consider directed segments, like GJ and EH (in the figure) can both of them be considered positive and FI negative? i.e. the question is - if they're pointing in the same general direction but are not parallel, can we assign both GJ and EH a positive sign even though they are not parallel?

Now comes the really tricky part...

2)
 To find the length of a directed segment, do we have to use the cosine of the angle between the vector and the x-axis (for performing operations such as addition and multiplication) for the x-component and sine for the y-component? This stems from the question: What happens if one component is positive and the other is negative (for example if we assume the positive x-direction to be positive and the positive y-direction to be positive, what happens if the vector/ directed line segment starts at the origin and continues through quadrant IV, i.e. it points downward)?


induction
number theory proposed
number theory

Define a sequence $a_n$ by $a_1=a_2=a_3=1$ and $a_{n+1} = \frac{a_n^2 + a_{n-1}^2}{a_{n-2}}$ for $n \geq 3$.
Show that all elements $a_n$ are integers.

Denote $b_n = a_{n+3}+a_n$ - from $a_{n+2}a_{n-1}-a_{n-2}a_{n+1} = a_{n+1} ^2 -a_{n-1} ^2$ we get: $a_nb_n=a_{n+2}b_{n-1}$ which leads to:

$a_n...a_rb_n=...=a_{n+2}a_{n+1}...a_{r+2}b_{r-1}$ (for r>1, by using above few times) so also
$a_{n+3}+a_n=3a_{n+2}a_{n+1}$ and by induction it follows that $a_n$ is integer for every $n$.


function
inequalities unsolved
inequalities

let $ x^2\plus{}y^2 \leq {\pi}$ prove that: $ cosx\plus{}cosy \leq 1\plus{}cosxy$

cos  is an even function so we  consider only the case   0<= x   , 0<= y  .  Let’s  suppose:
0 <=  x  <=  1 <= Pi^0.5     or   0 <=  y  <=  1 <= Pi^0.5
If       0 <=  x  <=  1   then    0 <=  xy  <=  y  <= Pi^0.5  <= Pi
On interval  [0,pi  ]      cos is an decreasing function so:
 cos (y) <= cos(xy)     then   cos(x)+cos (y) <= cos(xy) + cos(x)   <=   Cos(xy) +1
The case     1 <=  x  <=  Pi^0.5   and  1 <=  y  <=  Pi^0.5   is impossible because  :
  Pi^0.5  <2 <=  x ^2+y^2


Given a,b,c>0 and abc=4. Prove that :
$ \frac{a^3}{\sqrt{(1\plus{}a^4\sqrt{a})(1\plus{}b^4\sqrt{b})}}\plus{}\frac{b^3}{\sqrt{(1\plus{}b^4\sqrt{b})(1\plus{}c^4\sqrt{c})}}\plus{}\frac{c^3}{\sqrt{(1\plus{}c^4\sqrt{c})(1\plus{}a^4\sqrt{a})}}>\equal{}\frac{4}{3}$

Can you tell me when the equality holds?

$ a\equal{}b\equal{}c\equal{}\sqrt[3]{4}$, i guessed.

this is APMO, just set $ a^3\equal{}x^2$,then you will get $ xyz\equal{}8$ and equivalent problem..


integration
calculus
function
algebra
partial fractions
real analysis
real analysis unsolved

Well, it is easy, but you need to get a bit more practice in this kind of integrals, I guess.

[hide]Write the subintegral function in form:

$ \frac {x^2 \plus{} 5x}{(x \minus{} 4)(x^2 \plus{} 9)} \equal{} \frac {A}{x \minus{} 4} \plus{} \frac {Bx \plus{} C}{x^2 \plus{} 9}$ ($ A,B,C$ are real coefficients), sum the right hand side and find the value of undetermined coefficients $ A,B,C$ and you get two integrals "from the tables". The method shown here is the most common method in dealing with this kind of integrals.

Material for further reading:[url]http://www.analyzemath.com/calculus/Integrals/partial_fractions.html[/url]
[url]http://www.analyzemath.com/calculus/Integrals/integral_partial.html[/url]

Of course, you can find a better explanation of the Partial fractions method in your calculus textbook.

[/hide]


geometry
geometric transformation
inequalities
function
algebra
polynomial
angle bisector

does any body have Austrian-Polish 2004-2005 problems.

Hi Mathx!

You can find the problems of the Austrian-Polish-Competition on the Homepage of the Austrian Mathematical Olympiad ([url=http://www.oemo.at]www.oemo.at[/url]), they are only available in German there though.

Maybe I'll translate them soon.....

//Edit: I've just seen that you're from Germany, so translation will not be neccessary  :lol:

Can you please translate them if you have time? So we can share the problems togather.

Of course I can, just excuse my poor English :)

[b][size=200]27th Austrian-Polish Mathematical Competition 2004[/size][/b]

[size=150]Single Competition, Day 1, June 24th 2004 [/size]

1) Let $S(n)$ be the sum of digits for any positive integer n (in decimal notation).
Let $N=\displaystyle\sum_{k=10^{2003}}^{10{^{2004}-1}} S(k)$. Determine $S(N)$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=26789]Austra-Poland 2004, single competition, problem 2[/url]

 In a triangle ABC let D be the point where the angle bisector of $\gamma$ (angle at C) intersects AB. Let F be the area of the triangle ABC

Prove, that the following inequality holds:
\[2\cdot F \cdot ( \frac{1}{AD} - \frac{1}{BD}) \leq AB\]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=26790]Austria-Poland 2004, single competition, problem 3[/url]

Solve the following system of equation in the real numbers:

\[a - \sqrt{1-b^2} + \sqrt{1-c^2} = d\]
\[b - \sqrt{1-c^2} + \sqrt{1-d^2} = a\]
\[c - \sqrt{1-d^2} + \sqrt{1-a^2} = b\]
\[d - \sqrt{1-a^2} + \sqrt{1-b^2} = c\]
(All square roots are nonnegative)

Time: 4,5 hours
A maximum of 8 points can be reached at each problem

[i]Edited by orl.[/i]

[b][size=200]27th Austrian-Polish Mathematical Competition 2004[/size][/b]

[size=150]Single Competition, Day 2, June 25th 2004 [/size]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=26534]Austria-Poland 2004, single competition, problem 4[/url]

Determine all positive integers n, so that $n^{10} + n^{5} + 1$ is a prime number.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=26531]Austria-Poland 2004, single competition, problem 5[/url]

Determine all positive integers n, so that the system of equations
\[ x_1 + x_2 + ... + x_n = 27\] \[ x_1 \cdot x_2 \cdot ... \cdot x_n = (\frac{3}{2})^{24}\]
has positive real solutions.

6)For $n=2^m$ (m is a positive integer) consider the set $M(n) = \{ 1,2,...,n\}$ of natural numbers.
Prove: There exists an order $a_1, a_2, ..., a_n$ of the elements of M(n), so that for all $1\leq i < j < k \leq n$ holds: $a_j - a_i \neq a_k - a_j$.

[i]Edited by orl.[/i]

[b][size=200]27th Austrian-Polish Mathematical Competition 2004[/size][/b]

[size=150]Team Competition, June 26th 2004 [/size]

7) Determine all functions f, defined over the positive integers with integers as function values, which fulfill the following condition:
For all pairs (x,y) of relatively prime positive integers holds:
\[f(x+y) = f(x+1) + f(y+1)\]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=26792]Austria-Poland 2004, team competition, problem 8[/url]

A) Prove: For $n=4$ or $ n \geq 6$ each triangle ABC can be seperated into n triangles which are similar (not neccessarily congruent) to ABC.
B) Prove: An equilateral triangle can neither be seperated into 3 nor into 5 equilateral triangles.
C) Does there exist a triangle ABC, that can both be seperated into 3 and into 5 triangles (analogue to part A)? (Either specify such a triangle and such seperations or prove that no such triangle exist).

9) Given are the double-sided sequences
\[< ..., a_{-2}, a_{-1}, a_0, a_1, a_2, ...>, < ..., b_{-2}, b_{-1}, b_0, b_1, b_2, ...>, < ..., c_{-2}, c_{-1}, c_0, c_1, c_2, ...>\]
of positive real numbers. For each integer n the following inequalities holds:
\[a_n \geq \frac{1}{2} (b_{n+1} + c_{n-1})\]
\[b_n \geq \frac{1}{2} (c_{n+1} + a_{n-1})\]
\[c_n \geq \frac{1}{2} (a_{n+1} + b_{n-1})\]
Determine $a_{2005}$, $b_{2005}$, $c_{2005}$, if $a_0 = 26, b_0 = 6, c_0 = 2004$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=26535]Austria-Poland 2004, team competition, problem 10[/url]

For each polynomal Q(x) let M(Q) be the set of those nonnegative integers x so that $0 < Q(x) < 2004$. We consider polynomals $P_n(x)$ of the form $P_n(x) = x^n + a_1x^{n-1}+...+a_{n-1}x+1$, with the coefficients $a_i = \pm 1$ for $i = 1,2,...,n-1$.
For each $n=3^k\ \ (k > 0)$ determine:
A) $m_n$ - the maximum of the numbers of elements in $M(P_n)$ over all such polynomals $P_n(x)$;
B) all polynomals $P_n(x)$, such that $|M(P_n)| = m_n$.

Time: 4,5 hours
8 points at each problem

[i]Edited by orl.[/i]

Thank you so much!  :D  And your English is exellent!

It has just one Geometry Problem  :mad:  :(  :mad:

THX@shobber! Now the problems of this year

[b][size=200]28th Austrian-Polish Mathematical Competition 2005[/size][/b]

[size=150]Single Competition, Day 1, June 30th 2005 [/size]

1) For a convex n-gon $P_n$, we say that a convex quadrangle Q is a diagonal-quadrangle of $P_n$, if its vertices are vertices of $P_n$ and it's sides are diagonales of $P_n$.
Let $d_n$ be the number of diagonal-quadrangles of a convex n-gon. Determine $d_n$ for all $n\geq 8$.

2) Determine all polynomals with integers as coefficients, so that for all real numbers x \[P(P(P(P(P(x)))))=x^{28}\cdot P(P(x))\] holds.

3) Let $a_0, a_1, a_2, ... , a_n$ be real numbers, which fulfill the following two conditions:
a) $0 = a_0 \leq a_1 \leq a_2 \leq ... \leq a_n$.
b) For all $0 \leq i < j \leq n$ holds: $a_j - a_i \leq j-i$.
Prove: $( \displaystyle \sum_{i=0}^n a_i )^2 \geq \sum_{i=0}^n a_i^3$

@amir2: Well, I think only little Austrians will complain about that because geometry is far away from beeing the Austrians strenght (which is also result of a continuously massive lesson- and topic- cutting at Austrian schools (the leader of the Austrian Mathematical Olympiad once said: "If that goes on like that we will soon be only teaching stuff like 'Draw a triangle and count the vertices'") but I dont want to become political)

[b][size=200]28th Austrian-Polish Mathematical Competition 2005[/size][/b]

[size=150]Single Competition, Day 2, July 1st 2005 [/size]

4) Determine the smallest natural number $a\geq 2$ with following properties:
There exist a prime number p and a natural number $b\geq 2$ so that \[\frac{a^p - a}{p}=b^2\]

5) Given is a convex quadrangle ABCD with AB=CD.
Outside of the quadrangle we draw the triangles ABE and CDF, so that
$\angle{ABE} = \angle{DCF}$ and $\angle{BAE}=\angle{FDC}$.
Prove that the centers of AD, BC and EF are collinear.

6)Determine all monotone functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$, so that for all $x, y \in \mathbb{Z}$
\[f(x^{2005} + y^{2005}) = (f(x))^{2005} + (f(y))^{2005}\]

[b][size=200]28th Austrian-Polish Mathematical Competition 2005[/size][/b]

[size=150]Team Competition, July 2nd 2005 [/size]

7) For each natural number $n\geq 2$ solve the following system of equations in the integers $x_1, x_2, ..., x_n$.
\[(n^2-n)x_1 + x_2\cdot x_3\cdot ... \cdot x_n\cdot ( \displaystyle \sum_{j=1}^n x_j^2) = n^3 - n^2\]
\[(n^2-n)x_2 + x_3\cdot ... \cdot x_n \cdot x_1\cdot ( \displaystyle \sum_{j=1}^n x_j^2) = n^3 - n^2\]
\[...\]
\[(n^2-n)x_i +  x_i\cdot ... \cdot x_n\cdot x_1\cdot ... \cdot x_{i-1} \cdot ( \displaystyle \sum_{j=1}^n x_j^2) = n^3 - n^2\]
\[...\]
\[(n^2-n)x_n + x_1\cdot x_2\cdot ... \cdot x_{n-1} \cdot ( \displaystyle \sum_{j=1}^n x_j^2) = n^3 - n^2\]

8) We define the sets $R_{mn} = \{ (x,y) \mid 0 \leq x\leq m, 0 \leq y \leq n, x\ and\ y\ are\ integers \}$.
We consider the functions $f: R_{mn} \rightarrow \{-1, 0, 1\}$ with the following property:
For each quadruple of points $A_i(x_i, y_i) \in R_{mn} \ \ (i=1,2,3,4)$, which are vertices of a square with side length $0 < s < 3$, the sum
\[f(x_1, y_1) + f(x_2, y_2) + f(x_3, y_3) + f(x_4, y_4) = 0\]
For each pair (m, n) of positive integers determine F(m,n), the number of such functions on $R_{mn}$

9) We consider the equation $x^3 + y^3 + z^3 = 2$.
a) Prove: It has infinitely many integer solutions x,y,z
b) Determine all integer solutions x, y, z with $|x|, |y|, |z| \leq 28$.

10) Determine  all pairs (k,n) of nonnegative integers with the following property:
For all positive real numbers x and y the following inequality holds:
\[1+ \frac{y^n}{x^k} \geq \frac{(1+y)^n}{(1+x)^k}\]

Have fun!
Yimin

[quote="Yimin Ge"]
9) We consider the equation $x^2 + y^2 + z^2 = 2$.
a) Prove: It has infinitely many integer solutions x,y,z
[/quote]

Are you sure? :P

[quote="Yimin Ge"]
3) Let $a_0, a_1, a_2, ... , a_n$ be real numbers, which fulfill the following two conditions:
a) $0 = a_0 \leq a_1 \leq a_2 \leq ... \leq a_n$.
b) For all $0 \leq i < j \leq n$ holds: $a_j - a_i = j-i$.
Prove: $( \displaystyle \sum_{i=0}^n a_i )^2 \geq \sum_{i=0}^n a_i^3$[/quote]

Actually it should be $a_j - a_i \leq j-i$ in b)  :)

@Arne and Megus: I'm terribly sorry (well, somehow also a bit embarrassed), I was a bit in a hurry. I corrected it.

It would be nice if the problems could be posted as single threads. I added some links of problems already posted.  :)


calculus
integration
trigonometry
logarithms
real analysis
real analysis unsolved

The integral equals 0.
We have
$ I\equal{}\int_0^{\pi/2}\dfrac{\tan x\ln\tan x}{\sec^2x}\ \mathrm dx\equal{}\dfrac{1}{2}\int_0^{\pi/2}\sin2x\ln\tan x\ \mathrm dx$
Applying the property: $ \int_a^bf(x)\ \mathrm dx\equal{}\int_a^bf(a\plus{}b\minus{}x)\ \mathrm dx$, we have
$ I\equal{}\dfrac{1}{2}\int_0^{\pi/2}\sin2x\ln\cot x\ \mathrm dx$
Hence, we get
$ 2I\equal{}\dfrac{1}{2}\int_0^{\pi/2}\sin2x(\ln\tan x\plus{}\ln\cot x)\ \mathrm dx\equal{}\dfrac{1}{2}\int_0^{\pi/2}\sin2x\ln(\tan x\cot x)\ \mathrm dx$

$ \equal{}\dfrac{1}{2}\int_0^{\pi/2}\sin2x\ln1\ \mathrm dx\equal{}\boxed{0}$

[quote="eureka-123"]$ \int_{0}^{\frac {\pi }{2}}\frac {tanx.log(tanx))}{sec^{2}x}dx$[/quote]

Use substitution $ x\equal{}y\minus{}(\pi/2)$, then your integral become
$ \int_0^{\frac{\pi}{2}}\frac{\sec^2 y\log(\tan y)}{\tan y}dy$

Now use substitution $ z\equal{}\log(\tan y)$.


A square piece of paper with side length 12 inches has its corners labeled $A$, $B$, $C$, and $D$. Corner $A$ is folded down onto $CD$ so that it lies at the midpoint. What is the area (in square inches) of the small triangular piece containing corner $B$ where there is no overlap?

Sorry if the wording is wierd. I had to change it so it could be understood without the diagram. I finally did solve it (after about 30 minutes of work). I was wondering if there is an easy solution. :lol: Hope you enjoy the problem.

[quote="Ignite168"][hide]I don't know if I am understanding it correctly but it corrently looks like it would be a right triangle with sides 6 and 12. Just multiply and divide by 2 to get $\boxed {36}$?[/hide][/quote]
Sorry, no. Try it with a piece of paper. If you fold a corner to the middle of an opposite side, there will be a small triangle where it doesn't overlap. (There are a total of 4 areas: 1 big trapezoid in the middle where it overlaps and 3 triangles that don't overlap.) This area is the smallest of the three.

I took a square cardboard thing...

[hide]The medium triangle is

(12-x)^2=x^2+36

so x=4.5

the meduim triangle is 6 by 4.5

the large triangle is similar so it is 6 by 8.

the small triangle is similar as well and one leg is 2 (12-2).

The other leg is 1.5 because it is similar.

so 1.5?[/hide]

Wow!! That was an easy question. I can't believe I didn't see similar triangles. Thanks ehehheehee. This is why I got a 36 at the states.  :roll:


In the adjoining figure, the triangle $ABC$ is a right triangle with $\angle BCA=90^\circ$. Median $CM$ is perpendicular to median $BN$, and side $BC=s$. The length of $BN$ is

[asy]
size(200);
defaultpen(linewidth(0.7)+fontsize(10));real r=54.72;
pair B=origin, C=dir(r), A=intersectionpoint(B--(9,0), C--C+4*dir(r-90)), M=midpoint(B--A), N=midpoint(A--C), P=intersectionpoint(B--N, C--M);
draw(M--C--A--B--C^^B--N);
pair point=P;
markscalefactor=0.005;
draw(rightanglemark(C,P,B));
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$M$", M, S);
label("$N$", N, dir(C--A)*dir(90));
label("$s$", B--C, NW);[/asy]

$\textbf {(A) } s\sqrt 2 \qquad \textbf {(B) } \frac 32s\sqrt2 \qquad \textbf {(C) } 2s\sqrt2 \qquad \textbf {(D) } \frac 12s\sqrt5 \qquad \textbf {(E) } \frac 12s\sqrt6$

[quote="Altheman"][hide]using the centroid divides medians in the ratio of 2 to 3, and altitude on hypotenuse theorems, we get $s^2=\frac{2}{3}(BN)^2$, $BN=\frac{\sqrt{6}}{2}s \implies \boxed{E}$[/hide][/quote]What's the altitude on projection theorem??  :huh:

[hide="Solution"]Let the median intersection be $P$, and $BP=2NP=2x$.

Using similar triangles, $\frac{s}{2x}=\frac{3x}{s}$. Solving, we get the answer.[/hide]

[quote="Altheman"]the altitude on hypotenuse theorems are just a thing that my school geometry book had, all it's just using the similar triangles created by the 2 right angles, (the right angle of the whole triangle, and the altitude), the same thing you did, just it is the way that i remember it[/quote]:coolspeak:


prove that for any finite subset A of P (the set of prime numbers),  sum (x \in A) sqrt(x) is not a rational number.

One more time...

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=rational&t=4496

Pierre.


bueno una desigualdad  (especialmente para amilcar  :D )

sean a y b reales ademas a es diferente de cero

demostara que

a¨2 + b¨2 + 1/a¨2 + $b/a$ $\geq $ raiz de 3

disculpen por no saber latex

Bueno marco, aca va la solucion. :D a y si conegita tiene razon lo k  nos pide demostrar es que $a^{2}+b^{2}+1/a^{2}+b/a \geq 3^{1/2}$
bueno nos dan de dato que $a,b$ pertencen a los reales y a diferente de cero.
->$S =a^{2}+b^{2}+1/a^{2}+b/a$
$a^{2}+b^{2}+1/a^{2}+b/a=b^{2}+2b(1/2a)+1/4a^{2}+3/4a^{2}+a^{2}=(b+1/2a)^{2}+3/4a^{2}+a^{2}....(1)$ ; ahora apliquemos $MA\geq MG$, a los terminos $3/4a^{2}, a^{2}$ como $a^{2}$ es real positivo, esto es posible.
$3/4a^{2}+a^{2}\geq 3^{1/2}...(2)$
ahora $(b+1/2a)^{2}\geq 0...(3)$
AHORA SUMANANDO (2) Y (3), TENEMOS (1) QUE ES IGUAL A S.
ENTONCES $S\geq 3^{1/2}$.
Y ES LO QUE QUERIAMOS DEMOSTRAR,
YA ESTA MARCO  :D

Buena demostración, aunque creo que en vez de escribir $2/a$ quisiste escribir $\frac{1}{2a}$, esos errores pueden costar una medalla :D .

:blush: :rotfl:    tiene razon prof, lo corregire,  :D
salu2


modular arithmetic
quadratics
algebra
quadratic formula

Two thousand points are given on a circle.  Label one of the points 1.  From this point, count 2 points in the clockwise direction and label this point 2.  From the point labeled 2, count 3 points in the clockwise direction and label this point 3.  (See figure.)  Continue this process until the labels $1, 2, 3, \dots, 1993$ are all used.  Some of the points on the circle will have more than one label and some points will not have a label.  What is the smallest integer that labels the same point as 1993?

[asy]
int x=101, y=3*floor(x/4);
draw(Arc(origin, 1, 360*(y-3)/x, 360*(y+4)/x));
int i;
for(i=y-2; i<y+4; i=i+1) {
dot(dir(360*i/x));
}
label("3", dir(360*(y-2)/x), dir(360*(y-2)/x));
label("2", dir(360*(y+1)/x), dir(360*(y+1)/x));
label("1", dir(360*(y+3)/x), dir(360*(y+3)/x));[/asy]

[hide]
For a number $n$ to be on the same point as 1993, the sum $S_n=\displaystyle\frac{n(n+1)}{2}$ has to be equal to $\displaystyle\frac{1993(1994)}{2}$ in mod 2000.

$S_n=\displaystyle\frac{n(n+1)}{2}\equiv\displaystyle\frac{1993(1994)}{2} \mod 2000$
$\displaystyle\frac{1993(1994)}{2}=1987021 \equiv 1021 \mod 2000$
$\displaystyle\frac{n(n+1)}{2}=2000k+1021$
Quadratic formula on:
$n^2+n-4000k-2042=0$
$n=\displaystyle\frac{-1\pm\sqrt{1-4(-4000k-2042)}}{2}$
So $16000k+8169$ has to be a perfect square.
Since 8169 is not a perfect square, $k>0$
I test the values for $\sqrt{16000k+8169}$ that have a ones digit of 3 or 7, because $3^2=9$ and $7^2=49$...
...
...
...
hmm...
...
...
...
THANKFULLY, at $k=3$, $\sqrt{16000k+8169}=237$
So $n=\boxed{118}$

Sorry about being such a noob in number theory.. the method was the first thing that came to mind, which is probably not a very good thing

[/hide]

EDIT: oh.. i see... nice job 4everwise

[quote="4everwise"][hide]The label $1993$ will occur on the $\frac12(1993)(1994) \pmod{2000}$th point around the circle. (Starting from 1) A number $n$ will only occupy the same point on the circle if $\frac12(n)(n+1)\equiv \frac12(1993)(1994) \pmod{2000}$.

Simplifying this expression, we see that $(1993)(1994)-(n)(n+1)=(1993-n)(1994+n)\equiv 0\pmod{2000}$. Therefore, one of $1993-n$ or $1994+n$ is odd, and each of them must be a multiple of $125$ or $16$.

For $1993-n$ to be a multiple of $125$ and $1994+n$ to be a multiple of $16$, $n \equiv 118 \pmod {125}$ and $n\equiv 6 \pmod {16}$. The smallest $n$ for this case is $118$.

In order for $1993-n$ to be a multiple of $16$ and $1994+n$ to be a multiple of $125$, $n\equiv 9\pmod{16}$ and $n\equiv 6\pmod{125}$. The smallest $n$ for this case is larger than $118$, so $\boxed{118}$ is our answer. [/hide][/quote]
to simlify the problem further, once we have achieved $(1993)(1994)-(n)(n+1)=(1993-n)(1994+n)\equiv 0\pmod{2000}$ (it is actually mod 4000, but doesn't matter) we can replace $1993$ and $1994$ by $-6$ and $-7$ and find the following: $(n+7)(n-6)\equiv 0\pmod{2000}$ which makes the problem ridiculously simple (first guess is right, if you sub $125$ in for $n+7$)

[quote="4everwise"] $ (1993 \minus{} n)(1994 \plus{} n)\equiv 0\pmod{2000}$. Therefore, one of $ 1993 \minus{} n$ or $ 1994 \plus{} n$ is odd, and each of them must be a multiple of $ 125$ or $ 16$. [/quote]
Are you sure the implication here is correct? One of the factors is odd doesn't imply it must consist of all the odd prime factors of 2000 since it could be only 5 instead and the other even factor would be $ 25*16$ which is still even...

[hide="Solution"]

From the point $1$, the distance to each must be the same $\pmod{2000}$.

Therefore, $2+3+\cdots 1993=2+3+\cdots+x$ letting the smallest be $x$.

Therefore, $(x+1)+\cdots+1993\equiv 0\pmod{2000}$

Therefore, $(1993-x)(\frac{x+1994}{2})\equiv 0\pmod{2000}$ or $(1993-x)(x+1994)\equiv 0\pmod{4000}$.

Note that $2$ and $5$ can't divide $\gcd(1993-x, x+1994)$ using basic Euclid's, therefore, $1993-x\equiv 0\pmod{25}$ and $x+1994\equiv 0\pmod{16}$ or $1993-x\equiv 0\pmod{16}$ and $x+1994\equiv 0\pmod{25}$ (unless one of them is $0\pmod{4000}$ which will be obviously not true after doing computations).

The first gives $x\equiv 1993\pmod{25}\equiv 18\pmod{25}$ and $x\equiv 6\pmod{25}$.  The second gives $x\equiv 9\pmod{16}, x\equiv 6\pmod{25}$.

The first gives $x=118$ (after doing some simple cases).  For the second, we get a cycle of:  $6, 31, 56, 81, 106, 131$ which mod 16 gives us $6, -1, 8, 1, 10, 3$ therefore we are done and the answer is $\boxed{118}$.

[/hide]

Sorry to revive, but can someone explain why 6 does not work? I know it has to do with the legality of division in mods. When is it legal to divide in modular equations?

Technically, I guess you can divide what ever you want, it just depends on what you are dividing by. That is, if the dividend is relatively prime to the mod, we have to divide the mod. When you have $$3x\equiv 6 \pmod{9}\implies 3x=6+9n$$ for integers $n$. We divide by $3$, and get $$x=2+3n \implies x \equiv 2 \pmod3$$. If the dividend is relatively prime to the mod, we don't have to do anything.

[quote=4everwise][hide]The label $1993$ will occur on the $\frac12(1993)(1994) \pmod{2000}$th point around the circle. (Starting from 1) A number $n$ will only occupy the same point on the circle if $\frac12(n)(n+1)\equiv \frac12(1993)(1994) \pmod{2000}$.

Simplifying this expression, we see that $(1993)(1994)-(n)(n+1)=(1993-n)(1994+n)\equiv 0\pmod{2000}$. Therefore, one of $1993-n$ or $1994+n$ is odd, and each of them must be a multiple of $125$ or $16$.

For $1993-n$ to be a multiple of $125$ and $1994+n$ to be a multiple of $16$, $n \equiv 118 \pmod {125}$ and $n\equiv 6 \pmod {16}$. The smallest $n$ for this case is $118$.

In order for $1993-n$ to be a multiple of $16$ and $1994+n$ to be a multiple of $125$, $n\equiv 9\pmod{16}$ and $n\equiv 6\pmod{125}$. The smallest $n$ for this case is larger than $118$, so $\boxed{118}$ is our answer. [/hide][/quote]

Why is it not $\frac{n^2 + n}{2}  - \frac{1993(1994)}{2} \equiv 0 \pmod{2000}$ instead of the other way around ($\frac{1993(1994)}{2} - \frac{n^2 + n}{2} \equiv 0 \pmod{2000}$)?

I used 0 based indexing, so why doesn't this work? Thanks!

Easy to see that each value has index $\frac{n^2+n}{2}-1$ (0 based indexing). So we want $$\frac{n^2+n}{2}-1\equiv \frac{1993^2+1993}{2}=\frac{(-7)^2-7}{2} -1 = 20\pmod{2000}.$$ Clearly, $6$ is the smallest.

Again, thanks for any help!

I used expansion to find that (1993)(1994) is 42 (mod 2000) which is the same as (6)(7) so clearly the smallest is six.  It means there's a problem with the solution giving 118.

im assuming you did $1993 \cdot 1994 / 2$ and then took mod 2000. However, you actually have to take mod 4000 because of the "/2" part


[quote="ridgers"]2n+9=k2 than find all natural values of n .. 2n means 2 to the power of an and k2 means k square[/quote]
[hide="Solution"]
Notice that $ 2^n \equal{} k^2 \minus{} 9 \equal{} (k \plus{} 3)(k \minus{} 3)$, so $ k \plus{} 3$ and $ k \minus{} 3$ are powers of $ 2$. Observe that $ k \plus{} 3$ and $ k \minus{} 3$ are both of the same parity, and thus they both must be even.  Hence, $ k \plus{} 3 \equal{} 2^x$ and $ k \minus{} 3 \equal{} 2^{n \minus{} x}$ for an integer $ x$ such that $ x < \frac {n}{2}$ .  The only solution to this is $ k \equal{} 5\implies \boxed{n \equal{} 4}$.  The reason why this is the only solution is as follows.  We can check that if $ n$ was any other value between $ 1$ and $ 5$, then it wouldn't work.  Now, if $ n\ge 6$, then we see that $ 2^{n \minus{} x} \minus{} 2^{x}\ge 2^{c(\frac {n \plus{} 1}{2})} \minus{} 2^{c(\frac {n \minus{} 1}{2})} > 2^{c(\frac {n\minus{}1}{2})}\ge 8$, where $ c(x)$ means the ceiling function of $ x$.  So $ 2^{n \minus{} x} \minus{} 2^{x}\ne 6$ when $ n\ge 5$.  [/hide]
EDIT: My edits were minor edits to the solution (proofreadings), but [b]ridgers[/b] posted quickly after me.


I've seen the relationship between $ \cot^{\minus{}1}$ and $ \tan^{\minus{}1}$ defined in 2 different ways:
$ \cot^{\minus{}1}x\equal{}\frac{\pi}{2}\minus{}\tan^{\minus{}1}x$
or
$ \cot^{\minus{}1}x\equal{}\tan^{\minus{}1}\left(\frac{1}{x}\right)$.

Which one is more common? Is one of them "better"?

Neither of them are "better." They are just different ways to express the same thing, and are used for different purposes.

What different purposes are they used for?

It's hard to say in general, but based on the context of the problem, one might be easier than the other. (Using one of them might make it easy to cancel terms, or something.)

An example using a different formula:

\[ y=\frac{cos{{(}2\theta{)}}}{cos^2\theta-sin^2\theta}\]

We know that $ \cos{{(}2\theta{)}}=cos^2\theta-sin^2\theta=2cos^2\theta-1=1-2sin^2\theta$

Substituting any of these into the numerator would do, but the first immediately cancels with the denominator to yield $ y=1$.

That's why one of the expressions might be more useful in one situation than another.


superior algebra
superior algebra solved

Prove that for any $n\geq 2$ you can find a noncomutative group of order $n^{3}$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=97954 (for primes at least; for composite $n$ I think we can use direct products)