Given a regular 19-gon ABCDEFGHIJKLMNOPQRS.  Consider the octagon ACDGIJNQ.  In this octagon, compute the value of
(a - c + d - g + i - j + n - q), where each letter is the measure (in degrees) of it's respective angle.

How much can you generalize this?

maybe it would help if we had a picture of a 19-gon and
maybe you should run a poll and see how many people know what a enneadecagon is.(i sure don't  :D )

[img]http://mathworld.wolfram.com/eimg1957.gif[/img]
http://mathworld.wolfram.com/eimg1957.gif

You'd be better off drawing it yourself.  But read the question -- you don't really have to draw the 19-gon (which is what enneadecagon means) anyhow.

[hide](a - c + d - g + i - j + n - q) = 0

I thought of this problem as how much each angle is from the base angle of the regular polygon.  To do that, I counted the number of points on the enneadecagon between the points on the octagon.(from D to G would be 3)  With that, it is easy to compute that the measure of each angle is equal to the base angle (3060/19) - (180(x-2)/19) where x = the sum of the distances to the points next to the angle on the octagon.

After I found the answer, I found that if you substitue the sum of the distances the the ajacent points on the octagon for the measure of the angle, you can plug them into your equation, and then use them in (3060/19) - (180(x-2)/19), instead of finding the measure of each individual angle.

I am working on a generalization for this, for all polygons inside regular polygons.[/hide]

[quote]If you draw a circle, you have a pretty good picture of an enneadecagon. (Thanks for teaching me a new word, Joel!)[/quote]

Yep.  If you go to the Polygon entry on mathworld, it has lots of fun names for many-sided polygons.  (You know, I'm a little rusty on my greek, but I'd give good odds that "many-sided polygons" is a redundant way of saying the same thing twice.  That last sentence wasn't nearly as clever as I had hoped it could have been.  Oh, well.)

Good job, bookworm -- generalizing to polygons inside polygons is an excellent thing to try.  If you try to do it strictly computationally (by actually calculating or adding up the angles and finding specific values), it's going to be quite hard I think.  See if you can come up with a way of showing it without computation.  Drawing many pictures I think is helpful here.


symmetry
geometry
parallelogram
ratio
geometry solved

It follows from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=tuff+nut&t=17]this[/url], which has received several solutions.

Projective soln:

Send line PRQ to infinity, maintaining symmetry about line OR. Then AD || BQ and DC || AB so ADCB is a parallelogram => O is the midpoint of BD => B and D are symmetric with respect to line OR. Transforming back, <ORB = <ORD.

[quote="Nuwe"]Does anyone know the solution of this problem?
          Let ABCD is a convex quadrilateral,line AB meet CD at P;line AD meet BC at Q.
O is the intersection of two diagonals.R on PQ such that OR is perpendicular to PQ.prove that (angleORD)=(angleORB).(it's from China)[b][/b][/quote]

If $X$ is the intersection of DB and PQ, $X$ and $O$ are harmonic conjugates, i.e. the cross-ratio of $(D,O,B,X) = -1$.  So $O$ and $X$ have the same ratio $r$ of distances to $D$ and $B$, and we want to prove that point $R$ is on the locus of points with the same ratio $r$: Appollonius' circle with diameter $OX$.  But this is clear from the perpendicularity.

Nice, but rather easy for a Chinese problem.


analytic geometry
Functional Analysis
real analysis
real analysis unsolved

Suppose that we have two norms, $ \|\cdot\|_1, \ \|\cdot\|_2$ that make $ X$ into a Banach space(in other words, $ (X,\|\cdot\|_1)$ and $ (X,\|\cdot\|_2)$ are both complete). Are these norms necessarily equivalent?

[hide="OK, I withdraw this argument."]Yes, they are equivalent. Consider the identity mapping from $ (X,\|\cdot\|_1)$ to $ (X,\|\cdot\|_2)$ The graph of this mapping is closed (being all of $ X\times X$) so the closed graph theorem says that it is continuous - that is, $ \|x\|_2\le c\|x\|_1$ for some $ c.$ The same argument with the places reversed gives the reverse inequality.[/hide]

Really?  Not that I don't believe the proof, but are Banach spaces uniquely determined up to equivalence by their cardinality?

[quote="Kent Merryfield"]Consider the identity mapping from $ (X,\|\cdot\|_1)$ to $ (X,\|\cdot\|_2)$ The graph of this mapping is closed (being all of $ X\times X$) [/quote]
 :what?: The graph is just the diagonal, not the whole product...

[quote="t0rajir0u"]If I understand the problem correctly, no.  The issue is that the two norms need not be continuous with respect to each other, so one can think of $ X$ as being two different Banach spaces related by a pathological bijection.[/quote]
Do you have a counterexample? May be some construction with linear discontinuous mapping would work..

The following construction seems to be working: let $ (X,\|\cdot\|_1)$ be a Banach space  with a norm in which $ T: X\mapsto X$ a linear discontinuous 1-1 map(for example., permuting the first and second coordinate in the basis representation).
Let $ \|x\|_2: \equal{}\|Tx\|_1$. The second norm is also complete: given a Cauchy sequence $ x_n$, we can find $ y$ s.t. $ \|Tx_n\minus{}y\|_1\to 0$. Because $ T$ is 1-1, $ \exists \ z: \ Tz\equal{}y$, and $ \|x_n\minus{}z\|_2\equal{}\|Tx_n\minus{}y\|_1\to 0$. But $ \|\cdot\|_1, \ \|\cdot\|_2$ are not equivalent due to discontinuous $ T$.

[quote="Lippilus"]... a linear discontinuous 1-1 map(for example., permuting the first and second coordinate in the basis representation).[/quote]
That looks continuous to me.

Let $ \{u_i\}_I$ be a Hamel basis for the space X. Let $ (c_n)_n$ be a countably infinite subset of $ \{u_i\}_I$ and let $ T(u_i) \equal{} T(c_n) \equal{} nc_n$ if $ u_i \equal{} c_n$ and $ T(u_i) \equal{} u_i$ otherwise. That gives us a discontinuous linear isomorphism of $ X$.

Of course swapping two basis elements does not promote discontinuity, but also note that there are no "first" or "second" coordinates. A basis for a Banach space $ X$ is necessarily uncountable.

Well, permuting 2 elements of a Hamel basis($ x_1e_1 \plus{} x_2e_2\mapsto x_1e_2 \plus{} x_2e_1$) can be discontinuous - depends on the norm. And of course we can always fix 2 elements and call them 'first' and 'second'.

Yes, you are right, sorry. You can in fact always do it by permuting two elements. Let $ \{u_i\}$ be a Hamel basis, and let $ \ell$ be a discontinuous linear functional.   The complement $ N$ of $ \ker \ell$ is one-dimensional and the projection onto it is discontinuous. If we let $ \sum a_{i_j} u_{i_j}$ be a non-zero element of $ N$ we can always choose one of the $ u_{i_j}$ and permute it with another basis element to get a discontinuous map.

But you can not just permute any two. I kind of made the ridiculous assumption that the projection onto a finite-dimensional subspace is continuous. Again, sorry about that.


inequalities
function
inequalities proposed

Let $ a,b,c$ be postive real numbers. Prove that
\[ \left( \frac {a}{a \plus{} b}\right)^2 \plus{} \left( \frac {b}{b \plus{} c}\right)^2 \plus{} \left( \frac {c}{c \plus{} a}\right)^2 \plus{} \frac {2abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \ge 1
\]
P/s. The first Cauchy Schwarz solution of mine is to use
\[ \left[\sum \frac {a^2}{(a \plus{} b)^2}\right]\left[ \sum (a\plus{}b)^2 (a \plus{} c)^2\right] \ge \left( \sum a^2 \plus{} \sum ab\right)^2
\]
:)

Sorry can_hang2007,i cant find the proof that be used C-S now,but,i found an other  nice proof,Here is it
Let $ \frac{a}{b}\equal{}x,\frac{b}{c}\equal{}y.\frac{c}{a}\equal{}z$ thus $ xyz \equal{}1$
Rewrite this ineq in the pqr form,we need prove
$ 2p^2\plus{}q^2\plus{}2pq\plus{}2p\minus{}3\plus{}2(p\plus{}q\plus{}2)\geq (p\plus{}q\plus{}2)^2$
It is true with $ p^2\minus{}2q\ge 3$ (AMGM)
We have done :P

Or   if  $ xy \plus{} yz \plus{} zx \geq x \plus{} y \plus{} z$  then the inequality is iquavilent to :
$ \frac {1}{(1 \plus{} x)^2} \plus{} \frac {1}{(1 \plus{} y)^2} \plus{} \frac {1}{(1 \plus{} z)^2} \plus{} \frac {1}{1 \plus{} xy \plus{} yz \plus{} zx} \geq 1$
And : if $ x \plus{} y \plus{} z \geq xy \plus{} yz \plus{} zx$ then
$ \frac {1}{(1 \plus{} x)^2} \plus{} \frac {1}{(1 \plus{} y)^2} \plus{} \frac {1}{(1 \plus{} z)^2} \plus{} \frac {1}{1 \plus{} x \plus{} y \plus{} z} \geq 1$

Using can_hang2007 hint I found

$ f(r)\equal{} 8r^2 \plus{}2r(1\plus{}2q^2\minus{}2q) \minus{}2q^3 \geq 0$

with $ r\equal{}abc$, $ q\equal{}ab\plus{}bc\plus{}ca$ and $ a\plus{}b\plus{}c\equal{}1$

As $ f(r)$ is a convex up funtion of $ r$

I will try to finish it using

$ A\equal{}\frac{4q\minus{}1}{9} \leq r \leq \frac{q}{9}\equal{}B$

and $ f(r) \leq max( f(A),f(B))$

I hope it is correct

Yes, your solution is very nice and simple.  :wink:

[quote="Allnames"][quote="manlio"]Using can_hang2007 hint I found

$ f(r) \equal{} 8r^2 \plus{} 2r(1 \plus{} 2q^2 \minus{} 2q) \minus{} 2q^3 \geq 0$

with $ r \equal{} abc$, $ q \equal{} ab \plus{} bc \plus{} ca$ and $ a \plus{} b \plus{} c \equal{} 1$

As $ f(r)$ is a convex up funtion of $ r$

I will try to finish it using

$ A \equal{} \frac {4q \minus{} 1}{9} \leq r \leq \frac {q}{9} \equal{} B$

and $ f(r) \leq max( f(A),f(B))$

I hope it is correct[/quote]

Sorry manilo,if we use  pqr technique,i think my solution is quite nice (and we dont need the magic hint of Canhang2007[/quote]
Dear Allnames,
I dont think so, since if we use the direct $ pqr$, it will cost a lot of calculation here.
But if we use Cauchy Schwarz and then $ pqr$, it will be simpler.


[size=184][color=darkblue]Preface[/color][/size]

Long ago, a way outgrown,
Far away, a type unknown.
Long ago, a group divides,
Far away, a species hides.
When truth invades, where aspire?
Lest be known, now known higher.
Intelligence, like color, has dimensions,
And needs, like air, for proper ascension.

Oh, so that was what it stood for. :blush:

[size=184][color=darkblue]Introduction[/color][/size]

Evolution has separated Homo sapiens sapiens into further species. To accommodate the extra need, scientists have renamed the genus "Homosapiens," from which consisted of about five species. Unfortunately, the small population of [i]Homosapiens nerdicus[/i] has been highly oppressed and discriminated against, undoubtedly due to jealousy.

But, here we are, in the middle of a humongous famine. All humans are struggling to just get by. [i]Homosapiens nerdicus[/i] is no exception, although with clever deep thinking, they have found ingenious ways to get through certain cases. This promoted more rampant hatred towards the group of high IQ, and resulted, unfortunately, into near genocide.

So here we are, with nerds dying out at a splintering pace. The nerds elected a group of fifteen to help them out of this catastrophe. This group of fifteen will now start a wondrous journey to redeem the species, and indeed, hope for the true place long deserved: the position of recognition and greatness.

[size=184][color=darkblue]Stats![/color][/size]

Yes, there will be some sort of fighting in this game.

1. monkeygirl13 [85,52,42,57]
2. westiepaw [77,55,71,82]
3. wrightboy10 [10,59,58,10]
4. Richard_Min [33,18,42,93]
5. Butler [79,83,38,62]
6. randomguy64 [46,86,82,66]
7. tennis123 [44,38,61,46]
8. jjfun1 [78,63,25,98]
9. andrewjjiang97 [70,11,65,95]
10. vahalla [52,11,62,28]
11. ksun48 [93,27,42,87]
12. AIME15 [21,83,38,18]
13. aleph0 [77,56,24,80]
14. Yoshi [77,46,20,88]
15. bluecarneal [46,43,79,48]

[a,b,c,d]
a=attack
b=speed
|a-b|=defense
10c=HP
d/2+|d-50|=points you can add to attack, speed, defense, or HP, whichever one you want =)
For example, Yoshi has 88/2+|88-50|=44+38=82 points, and could choose, for example, to +42 on HP and +40 on defense, bringing (a,s,d,h) to (77,46,71,242). Your decisions here stay for the rest of the game. PM me your decision.

Bump. Waiting for eleven more decisions. If you see them, remind them to check this thread, please.

As of now, the following people have submitted.

aleph0
bluecarneal
randomguy64
math_explorer
vahalla
westiepaw

Therefore, people not on the above list haven't.

If piegraph doesn't submit, can I replace him?  :P

or do I have to remind him...

YOU WILL NOT REPLACE ME ABCAK!
I WILL TELL CHUCK NORRIS TO SEND A ROUND-HOUSE KICK TO YOUR PRIVATE MESSAGE BOX IF YOU DO SO.

Hey! No hostility here! piegraph, do please send your submission, though.

if it isn't done, say by tonight, you can try prodding.

we are a team right? is there anything that will split us up later?

uh.. i don't get what to submit.
-
What stats you want to boost

And, the stats are unfair.
Shouldn't they add to the same thing? Like everybodys stats in total would be 200? Because right now it is really unfair :o

Read. The. Freaking. Posts.

You. Submit. The. Stats. That. You. Want. To. Boost. And. By. How. Much.

Got it?

Life is unfair.  :(  Random number generators are a part of life.

But, whatever, because I'm feeling nice, new system. Everyone has 300 points to add anywhere they want. If not everybody gets it in by this Friday, though, then hello old system.

I yell at the top of my lungs to get yourselves to gether and CHAAAAARGGGEEEEE.

Smash piegraph in the stomach with my far superior strength so his lung's air is forced out, and say "I'll do the same to you, randomguy if you two DON'T STOP!"

I run off.

Then I grab some sticks. And run back.

That takes a while. In any case, congratulations, after about 10 minutes of hard work, you won the fight (while randomguy64 is coming back). Damage statistics are coming out tomorrow.

So it turns out that here are the HP decreases from the battle. Ouch. It seems this one really hurt. And piegraph and randomguy64 suffered a bit extra from infighting. Tysq, tysq.

monkeygirl13: 75/117/100/104
westiepaw: 95/120/114/91
piegraph: 78/75/78/75 (-35,-11→32)
Richard_Min: 60/203/15/91 (-30→30)
Butler: 75/107/96/86
randomguy64: 60/102/63/175 (-19,-9→32)
tennis123: 75/117/78/97
jjfun1: 75/75/76/75
andrewjjiang97: 75/75/75/81
vahalla: 120/43/162/30
ksun48: 75/121/75/100
AIME15: 75/75/75/75
[aleph0: 70/70/80/80]
Yoshi: 15/62/23/284
bluecarneal: 85/90/65/60 (-37→48)
[hide=" "]Check their pockets![/hide]
The guards now lay dead on the ground.

Goes in and sees how aleph is...

Phire thinks his game fails, and his fail game has almost 400 posts...

That makes him pwn/awesome or something...

I keep an eye out just in case reinforcments come.

You reach a hallway with 20 rooms coming out. Good grief, you exclaim. Any psychologists out there?

How about asking the prisoner about the rooms?

He ain't here.

Although...

I run back, ask the prisoner how to get to Aleph, and run back.

He hands you a slip that his friend says has something mathematical to do with aleph0's position. You stare at the paper, puzzled. Certainly that won't help, would it?

[size=125][b][i].2[/i]2.3[/b][/size]

"Anybody know why one of the 2's are slanted and one isn't?"

[size=75][color=red]Meh. This game's dead. I guess aleph0 escapes and some celebration happens and whatnot. [/color][/size]


[i]Solve the equation in $ \mathbb{R}$

$ 3^{x^3 \plus{} x \plus{} 2 } \plus{} (x^3 \minus{} 3x \plus{} 1) . 3^{2x \minus{} 3x^3} \equal{} 3^{4x \plus{} 1}$[/i]

The equation writes $ 3^{x^3\minus{}3x\plus{}1} \plus{} (x^3\minus{}3x\plus{}1)3^{\minus{}3x^3\minus{}2x\minus{}1} \equal{} 1$. Now, if $ x^3\minus{}3x\plus{}1 < 0$, then LHS is less than $ 1$; if $ x^3\minus{}3x\plus{}1 > 0$, then LHS is more than $ 1$; while if $ x^3\minus{}3x\plus{}1 \equal{} 0$, then LHS is exactly $ 1$. So there are three solutions, in the intervals $ (\minus{}2, \minus{}1)$, $ (0, 1)$ and $ (1, 2)$.

[i]Since nobody can solve that problem , i will give a big hint ::D

      Multiply two sides with $ 3^{3x^3 \minus{}2x}$[/i]

And then, did I not state where the roots are?

[i]The problem ask to find roots , not demand to find where are  the roots ? :mad: [/i]

Dear, dear author of the post, rather than using frowning emoticons, you better read carefully the posted answers.
[quote]... while if x^3-3x+1 = 0, then LHS is exactly 1 ...[/quote]
Did I not write this? So the three real roots, for which I kindly specified the location, are the roots of this [color=blue]depressed cubic[/color], to which are you asking me to apply Cardano's formula?


AMC
USA(J)MO
USAMO
geometry
circumcircle
parallelogram
ceiling function

Rules are all the usual: 4.5 hours for each of the 2 sets of 3 problems. No calculators or computers. &c.

And now, the problems.

[hide="Set 1"]1. For what integers $x$ is $\gcd(x^{4}+3x^{3}+2x^{2}+5x+10,x^{2}+1)>1$?

2. A robot has an row of 10 boxes. A number of rocks are dropped into the rightmost box, and if in any box there are at least 7 rocks, the robot will take out 7 rocks from that box, put 5 rocks in the box immediately to the left of the original box, and discard the other 2 rocks. The robot will continue this process until none of the boxes have more than 6 rocks. Find, with proof, the largest possible number of rocks that can be put in the first (rightmost) box such that the robot does not require more than 10 (originally empty save the rightmost) boxes to put the rocks. (If the robot has more than 6 rocks in the leftmost box, he will require another box; e.g. if he has 7|2|2|2|2|2|2|2|2|2, he will require an 11th box. Also, the arrangement 7|2|2|2|2|2|2|2|2|2 is not possible.)

3. Let $\triangle ABC$ be scalene and have circumcircle $O$ with center $O$. $AO$, $BO$, $CO$ are extended to meet circle $O$ at $A'$, $B'$, $C'$. The perpendicular from $A$, $B$, $C$, to $BC$, $CA$, $AB$, intersect circle $O$ at $A''$, $B''$, $C''$, respectively. $B'B''$ intersects $C'C''$ at $X$, $C'C''$ hits $A'A''$ at $Y$, $A'A''$ intersects $B'B''$ at $Z$. Prove that $AX$, $BY$, $CZ$ are concurrent at $P$ and $\angle APA''=\angle BPB''=\angle CPC''=180^\circ$.[/hide]

[hide="Set 2"]4.  Given a $2n$-sided regular polygon, find with proof the number of parallelograms whose vertices are at intersections of the diagonals and sides of the polygon and whose sides are on diagonals or sides of the polygon.

5. An ant stands at the origin in a 4-dimensional space. Every second he may move to any of the 80 points which are adjacent to him. (Thus he may move 1, $\sqrt{2}$, $\sqrt{3}$, or even 2 units in one second.) The ant may not visit a previously-visited gridpoint, but he may cross his path. After 10 seconds, how many different paths could he have taken?

6. If $f(x)=q\left\lceil \frac{px}{q^{2}}\right\rceil$, prove $n(f^{n}(q)) \le \sum_{i=0}^{n}f^{i}(q)f^{n-i}(q)$ for all positive integral $p,q,n,x$ such that $p>q$. (Note that $f^{0}(x)=x$.)[/hide]
Or the problems may be found, most of the time, at http://billydorminy.homelinux.com/MockUSAMO2006a.pdf . (Thanks to 10000th User for typing it in a PDF.)
Please write your solutions as ID Number - Problem Number - Solution. You have until July 25 to send me your solutions.

[size=200]EDIT:[/size]
Problem 6 had a typo: $p>q$, not $q>p$. Note that the PDF does not have this change, and won't till next week. (I'm leaving on a weeklongish trip...) Thanks to rem for pointing this out. He shall get an extra point, unless of course he gets a perfect in the first place :).

Have fun!

Billy

1) Rightmost
2) Yes, the other boxes are empty.

[quote="rem"]I have a question about problem 2:
1. What do you mean by first box -- the leftmost or the rightmost box?
2. When the rocks are dropped into the rightmost box, are the rest of the boxes empty or not?[/quote]The problem is perfectly clear to me but meh.

solafidefarms, for #6, can you specify what $p,q$ is? It only says 'positive'. Yes, I know what they are but it's better to specify it. Thanks.

[EDIT]I made a pdf of the problems. Unless approved by solafidefarms, it will not be posted.

Fixed. Yes, p,q are integral.

Question: Are these problems too easy?

No no no, don't get me wrong. The problems are mildly difficult. Not extremely hard but they are fun and I like it :). solafidefarms, I'm going to send you the pdf of the problems and you decide whether to post it or not. Thanks. Now I'm going to time myself on the first set of problems. 4.5 hours here we go!!! :).

[EDIT]Good thing that I tested the attachment thing because I just realized the attachment thing is not working for me. I was going to send a pdf to solafidefarms but the thing won't add anything. The browser just dies off with a blank screen. Does anybody else have that problem with the PM's attaching a file?  :maybe:

Must not look at problems...
Must not look at problems...
Must not look at problems...

So tempting, but I must wait until I have time....

10000th user.... i'd just email it  :wink:

As in, if he is at (a,b,c,d), adjacent points are $(a_{1},b_{1},c_{1},d_{1})$ where $a_{1}\in\{a-1,a,a+1\}$ and so forth, except that you cannot stay at the same point. Thus there are 81-1 points adjacent to any given point.

[quote="10000th User"]
Good thing that I tested the attachment thing because I just realized the attachment thing is not working for me. I was going to send a pdf to solafidefarms but the thing won't add anything. The browser just dies off with a blank screen. Does anybody else have that problem with the PM's attaching a file?  :maybe:[/quote]

Yes!  I thought it was just my slow CPU.  This is a bug I guess.  Report it in the Bug Report Forum, or I will if I have time later.

There is probably a typo.

After briefly scanning the problems, I believe that #6 is solafidefarms's favorite.

EDIT: After thinking about #2 for 5 mins, I can see why solafidefarms likes that one.

[quote="JSteinhardt"]Can we write out our proofs on paper then LaTeX after our contest time ends?[/quote]Yes.

[quote="JSteinhardt"]Also what should we do about the diagram for 3?

And is it okay if instead of attaching the solution files, I give you a link to them (it will be very hard for people to guess the link, don't worry)?[/quote]

i don't see the problem with that...diagrams are important for that problem since there are a few situations that prove it, but certain things can make it a pain to reproduce the diagram (as i have learned the hard way)

[quote="JSteinhardt"]What is $f^{n}(x)$? Is it $f(f(f(...f(x))))$ n times?[/quote]

Yes, it is.

[quote="JSteinhardt"]Also what does $x$ have to do with problem 6? It says to prove it for all $x$ but $x$ never appears.[/quote]

It says to prove it for all positive integers $p$, and $q$

[quote="JSteinhardt"]What is $f^{n}(x)$? Is it $f(f(f(...f(x))))$ n times?[/quote]

This question was answered but I have to admit the wording on problem 6 was quite poor.... Didn't really notice the first time I looked at the problems, my bad.

[quote="JSteinhardt"]Hm now I'm not actually sure what format you want our stuff in.

Should we have 1 file, 2 files, or 6 files?[/quote]
This question was already answered: it does not really matter, but preferably you should submit two sets one with the first day's solutions and the other with second day's solutions. :wink:

The 31st.

Just a reminder: there are 4 days left to submit your solutions.

Unfortunatly it was.

I procrastinated in typing up my responses until 11 last night. But I submitted it on time.

problems have been posted in separate threads inside aops and are collected [url=https://artofproblemsolving.com/community/c3624993_2006_mock_usamo]here[/url]
(among others [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url], among USA Mocks/ USAMO Mocks)

Related post [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url]


What is the mean of all the positive five-digit numbers that can be formed by using each of the digits 1,2,3,4 and 5 only once in a number.

instead of adding them all together and dividing, let's think about this a little differently

take the ones place
how many numbers can go there? (5)
if we put a number, say 1 in the ones place, how many numbers can be made such that we have a one in the ones place?

well, that will just be the number of ways we can arrange all the other numbers
since we can arrange the other numbers in $ 4!$ ways, there are $ 24$ ways to arrange all the other numbers.

this method can be used to find the number of times 2, 3, 4, and 5 go into the ones place

therefore to find the number that will represent the mean in the ones place, we have
$ \frac{1\times24\plus{}2\times24\plus{}3\times24\plus{}4\times24\plus{}5\times24}{120}$

note that i got 120 because there 120 ways to arrange numbers

solving our expression, we get $ 3$

but that's just the ones place, what about all the other places??

well, we can use the exact same approach to get the other place values, and since we already know that this approach will get us an answer of 3, than there must be 3's in the rest of the place values

therefore, our answer is $ \boxed{33333}$

$\boxed{33333}$ should be the correct answer.

[hide]
Simply, the greatest number plus the least number is $66666$. The second greatest and second least is $66666$, so on. Note, of course, that the mean is $\frac{66666}{2} = \boxed{33333}$
This method can be extrapolated many of this types of problems (Example: what is the mean of all $3$ digit natural numbers with digits of $2$ and $8$, Answer= the mean of $222$ and $888$ = $555$), and vallon gives a good proof.
[/hide]

Because this is symmetric, just take the middle number for everything to get 33333


inequalities
IMO Shortlist
inequalities unsolved

[quote="vnmathboy"]Let $a_1\leq a_2\leq ...\leq a_n$ such that $a_1+a_2+...+a_n=0$. Proved:
$a_1^2+a_2^2+...+a_n^2+na_1a_n\leq 0$[/quote]

Another proof:

Since the real numbers $a_1$, $a_2$, ..., $a_n$ satisfy $a_1\leq a_2\leq ...\leq a_n$ and $a_1+a_2+...+a_n=0$, we have

$a_1\leq a_2\leq ...\leq a_{u-1}\leq a_u\leq 0\leq a_{u+1}\leq a_{u+2}\leq ...\leq a_n$

for some index u.

Now, let i be an index with $i\leq u$. Then, $a_i\leq 0$, but $a_i\geq a_1$. Hence, $a_i\cdot a_i\leq a_i\cdot a_1$. In other words, $a_i^2\leq a_1a_i$. Hence,

$a_1^2+a_2^2+...+a_u^2\leq a_1a_1+a_1a_2+...+a_1a_u=a_1\left(a_1+a_2+...+a_u\right)$.

Now, since $a_1+a_2+...+a_n=0$, we have $a_1+a_2+...+a_u=-\left(a_{u+1}+a_{u+2}+...+a_n\right)$; on the other hand, since all $a_k$ are $\leq a_n$, we have $a_{u+1}+a_{u+2}+...+a_n\leq\underbrace{a_n+a_n+...+a_n}_{n-u\text{ times}}=\left(n-u\right)a_n$. Thus, $a_1+a_2+...+a_u=-\left(a_{u+1}+a_{u+2}+...+a_n\right)\geq -\left(n-u\right)a_n$. In view of $a_1\leq 0$, this yields $a_1\left(a_1+a_2+...+a_u\right)\leq a_1\cdot\left(-\left(n-u\right)a_n\right)$. Hence,

$a_1^2+a_2^2+...+a_u^2\leq a_1\left(a_1+a_2+...+a_u\right)\leq a_1\cdot\left(-\left(n-u\right)a_n\right)=-\left(n-u\right)a_1a_n$.

On the other hand, let j be an index with j > u. Then, $a_j\geq 0$, but $a_j\leq a_n$. Hence, $a_j\cdot a_j\leq a_j\cdot a_n$, what rewrites as $a_j^2\leq a_na_j$. Thus,

$a_{u+1}^2+a_{u+2}^2+...+a_n^2\leq a_na_{u+1}+a_na_{u+2}+...+a_na_n=a_n\left(a_{u+1}+a_{u+2}+...+a_n\right)$.

But $a_1+a_2+...+a_n=0$ yields $a_{u+1}+a_{u+2}+...+a_n=-\left(a_1+a_2+...+a_u\right)$; on the other hand, as all $a_k$ are $\geq a_1$, we have $a_1+a_2+...+a_u\geq\underbrace{a_1+a_1+...+a_1}_{u\text{ times}}=ua_1$. Thus, $a_{u+1}+a_{u+2}+...+a_n=-\left(a_1+a_2+...+a_u\right)\leq -ua_1$. Since $a_n\geq 0$, this yields $a_n\left(a_{u+1}+a_{u+2}+...+a_n\right)\leq a_n\cdot\left(-ua_1\right)=-ua_1a_n$. Hence,

$a_{u+1}^2+a_{u+2}^2+...+a_n^2\leq a_n\left(a_{u+1}+a_{u+2}+...+a_n\right)\leq -ua_1a_n$.

As a consequence of the above,

$a_1^2+a_2^2+...+a_n^2=\left(a_1^2+a_2^2+...+a_u^2\right)+\left(a_{u+1}^2+a_{u+2}^2+...+a_n^2\right)$
$\leq \left(-\left(n-u\right)a_1a_n\right)+\left(-ua_1a_n\right)=-na_1a_n$,

so that $a_1^2+a_2^2+...+a_n^2+na_1a_n\leq 0$, as intended to prove.

[b]EDIT:[/b] Source of the problem: IMO Shortlist 1972, proposed by Czechoslovakia (CZS 4).

  Darij

Since $a_n \ge a_i \ge a_1$ for all $1 \le i \le n$, $(a_n - a_i)(a_i - a_1) \ge 0$, which expands as
\[ a_n a_i - a_n a_1 - a_i^2 + a_i a_1 \ge 0. \]
Summing from $i = 1$ to $n$ and using $a_1 + a_2 + \dots + a_n = 0$ gives
\[ a_1^2 + a_2^2 + \dots + a_n^2 + na_1 a_n \le 0. \]


modular arithmetic
logarithms
number theory unsolved
number theory

$n$ is a positive integer, $F_n=2^{2^{n}}+1$. Prove that for $n \geq 3$, there exists a prime factor of $F_n$ which is larger than $2^{n+2}(n+1)$.

[quote="KDS"][quote="ehsan2004"]I think it will help:

all of the divisor$>1$ of $2^{2^n}+1$ are in the form $2^{n+2}k+1$ where $k\in \mathbb{N}$  ;)[/quote]
I know this , but anyone give more detailed solution please?[/quote]

It`s really hard ! It was ChinaTST .

Hint :

Assume $F_n = \prod_{i=1}^{m} {p_i}^{\alpha _i}$ , Find an upper and a lower  bound for $\sum_{i=1}^{m} \alpha _i$.

Let 2^(2^n) = m^2. Now, try to find the smallest k, such that p is a prime divisor of m^2 + 1, and p > 2m + sqrt(2m), for every m >= k. (This problem is similar to IMO 2008 - Problem 3.)

It is well-known that all prime divisor of $F_n$ are $1 \pmod {2^{n+1}}$.
Let the prime divisors of $F_n$ be $2^{n+1}p_1+1, 2^{n+1}p_2+1 \cdots, 2^{n+1}p_t+1$.
We have $2^{2^n}+1 = \prod_{i=1}^t (2^{n+1}p_i+1)^{e_i} \equiv \prod_{i=1}^t (2^{n+1}p_ie_i+1) \equiv 1 \pmod {2^{2n+2}}$
Therefore, we have that $\sum_{i=1}^t p_ie_i \equiv 0 \pmod {2^{n+1}} \implies \sum_{i=1}^t p_ie_i  \ge 2^{n+1}$
Let $max[p_i]=p$. We have $p>\frac{2^{n+1}}{\sum_{i=1}^t e_i}$. It suffices to show that $\frac{2^{n+1}}{\sum_{i=1}^t e_i} \ge 2(n+1)$
This follows from $2^{2^n}+1 = \prod_{i=1}^t (2^{n+1}p_i+1)^{e_i} \ge \prod_{i=1}^t (2^{n+1}p_i)^{e_i} +1 \ge 2^{(n+1)\sum_{i=1}^t e_i}+1$.  $\blacksquare$

here's a very different solution  :-D
[b][color=#f00]claim(1):[/color][/b] Let P a prime  number and $p=k.2^{n+1} +1$ such that $k <2n$ then there's $m <2^n$ such that $p | 2^m +1$
[color=#600]proof:[/color]
suppose the contrary
$2^{p-1} \equiv $ $1$  $mod p$
$2^{2^{n+1}.k} \equiv $ $1$  $mod p$
$2^{2^{n}.k} \equiv $ $1$  $mod p$
.........
$2^{k} \equiv $ $1$  $mod k.2^{n+1} +1$
but $2^k <2p  \implies 2^k=p+1= k.2^{n+1} +2$
a contradiction
(there's a typo in this proof )
let $p | 2^{2^n} + 1 \implies p | 2^{2^{n+1}} - 1 \implies ord_p(2)=2^{n+1} \implies 2^{n+1}|p-1$
thus $p=2^{n+1}.k+1$
now by zsigmondi we have a prime divisor $q$ for   $2^{2^n} + 1$ that doesn't divide any $2^m +1 : m<2^n$
from the claim above we have $q \ge 2^{n+1}.(2n)+1=2^{n+n}.n+1$
and we are done :D

[quote=Ereb15]Can you use Zsigmondi at school olympiads?[/quote]

why not :blush: ?

[quote=Zhero][hide="Stronger claim"]For $n \geq 4$, the largest prime factor is larger than $2^{n+4} (n+2)$. [/hide]
[hide="Proof of stronger claim"]
[b]Lemma:[/b] If $n \geq 2$, $p$ is a prime, and $p | 2^{2^n} + 1$, then $2^{n+2} | p - 1$.
[i]Proof:[/i] Let $k$ be the order of 2 modulo $p$. Since $2^{2^n} \equiv -1 \pmod{p}$, $2^{2^{n+1}} \equiv 1 \pmod{p}$, so $k$ must be a power of 2 less than or equal to $2^{n+1}$. But if $k < 2^{n+1}$, we may square the congreunce $2^k \equiv 1 \pmod{p}$ a total of $n - \log_2 k$ times to get $2^{2^n} \equiv 1 \pmod{p}$. But $2^{2^n} \equiv -1 \pmod{p}$ and $p \neq 2$, so we must have $k = 2^{n+1}$. Because $n \geq 3$, $8 \, | \, 2^{n+1} \, | \, p - 1$, whence $\left( \frac{2}{p} \right) = 1$. Hence, $a^2 \equiv 2 \pmod{p}$ for some integer $a$. The order of $a$ must therefore be $2k$, so $2^{n+2} = 2k \, | \, p - 1$, as desired.

------------------------------------------------------------------------------------------------

The problem is clearly true when $n = 3$, so we may suppose that $n \geq 4$. Let $p_1 \leq p_2 \leq \ldots \leq p_m$ be primes such that $2^{2^n} + 1 = p_1 p_2 \ldots p_m$. By the lemma, there exist integers $a_1, a_2, \ldots, a_m$ such that $p_i = 2^{n+2} a_i + 1$ for each $i$. Hence, we have
\begin{align*} 2^{2^n} + 1 = (2^{n+2} a_1 + 1)(2^{n+2} a_2 + 1) \ldots (2^{n+2} a_m + 1) . \tag{1} \end{align*}
It follows that $2^{2^n} \geq (2^{n+2} + 1)^m > 2^{m(n+2)}$, whence
\[  m < \frac{2^n}{n+2}. \]

Since $n \geq 4$, $2n + 4 \leq 2^n$. Taking (1) modulo $2^{2n+4}$ yields
\begin{align*}
1
&\equiv 2^{2^n} + 1 \\
&= (2^{n+2} a_1 + 1)(2^{n+2} a_2 + 1) \ldots (2^{n+2} a_m + 1) \\
&= 2^{2n+4}(\mbox{some terms}) + 2^{n+2}(a_1 + a_2 + \ldots + a_m) + 1 \\
&\equiv 2^{n+2}(a_1 + a_2 + \ldots + a_m) + 1 \pmod{2^{2n+4}},
\end{align*}
whence
\[ 2^{n+2} | a_1 + a_2 + \ldots + a_m. \]
Since $a_m \geq a_{m-1} \geq \ldots \geq a_1$, we have $a_m \geq \frac{2^{n+2}}{m}$. Hence,
\begin{align*}
p_m
&= 2^{n+2} a_m + 1 \\
&\geq 2^{n+2} \cdot \frac{2^{n+2}}{m} + 1 \\
&> 2^{n+2} \cdot \frac{2^{n+2}}{\frac{2^n}{n+2}} \\
&= 2^{n+4} (n+2),
\end{align*}
as desired. [/hide]

[quote="chuyentoan"]It is really false!all of the divisor$>1$ of $2^{2^n}+1$ are in the form $2^{n+1}k+1$  but not $2^{n+2}$ where $k\in \mathbb{N}$ :wink:[/quote]
No, it is true; see my my first lemma.[/quote]
How to prove that there exists such $p_1 \leq p_2 \leq \ldots \leq p_m$? I don't understand what you mean by writing $2^{2^n}+1$ in that form? It looks like prime factorization but where are the exponents of primes?

Note that the consecutive primes can be equal, so it's just spliting the exponent


analytic geometry
Pythagorean Theorem
geometry

[hide]The distants formula is $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ so if you plug the numbers in, you get: $\sqrt{(-3-2)^2+[7-(-5)]^2} = \sqrt{25+144} = \sqrt {169} = 13$[/hide]

[quote="MCrawford"]How many units long is a line segment whose endpoints have coordinates $(-3, 7)$ and $(2, -5)$?[/quote]
[hide="solution"]
5-12-[b]13[/b]
[/hide]

Easy if you know basic geometry.
[hide="Click to reveal hidden content...if you dare!"]The difference of the x coordinates is 5, and the difference of the y coordinates is 12. So the length of the segment could be seen as the hypotenuse of a right triangle with legs 5 and 12. Use the Pythagorean theorem (a 2 + b 2 = c 2 ) to find that the answer is 13. Or just be familiar with the 5, 12, 13 Pyth. triple.[/hide]

haha that's a number sense problem...
[hide]the famouse 5-12-13 triangle![/hide]

Lucy, thank you for showing your work, and I like it the most  :D

[hide]13 with the distance formula.[/hide]

extension:

how does the distance formula work? what's the theorem called?

Differences in the y values and the x values just serve as legs of a right triangle, so its basically pythagorean theorem.

I believe the distance formula for the distance between (x1, y1) and (x2, Y2) is:

[hide]sqrt[(x1-x2) 2 +(y1-y2) 2 ][/hide]

As mentioned, it's basically the Pythagorean theorem.

[hide="answer to original problem"]
u can use a 5-12-13 triangle
so...[b]13[/b]is the answer[/hide]


i just found out about mathcounts this year...does anybody have tips on how to prepare for mathcounts next year so that i can get in the top three in chapter, go to state, and go on from there?...i'm open to recommendations for resources (e.g. books, tests, and other stuff)...thanks

There are about 100 of this topic already-search "MathCounts Prep".


ARML 2004

SF Bay A vs. Massachusetts A

Who will triumph over the other?

I want to vote for Massachusetts A to scare the SFBA kids into studying, but I don't think I will. (Grace and Darren, you had better not be reading this.)

I keep hearing about competition between SFBA and Massachusetts... Why is there such a big rivalry?

(Oh, and I voted for............ [hide]neither[/hide]:) )

There's a big rivalry because the Massachusetts kids don't want to admit that the SFBAers are smarter. It might have something to do with Gabriel Carroll and Reid Barton winning tons of contests as well.

Eh, SFBA will win.  But why isn't this SFBA A vs. TJHSST A?  I think that's the biggest contest, although TJ almost always wins.

SFBA A can beat TJ A. And what's this almost always? Before 2002, SFBA completely dominated ARML.

Why are there two options only? I find that a bit presumptious.... The MI/ICAE team will make a killing this year. Prepare yourselves for domination. [/humor].

[quote]Why are there two options only? I find that a bit presumptious.... [/quote]

No, the poll is about which of the two teams will beat the other, not which team will win ARML entirely.

And Simon, how can you be so sure that SFBA can win it all this year? They don't even really have a super-genius anymore who can lead them and do it all on the team round (unlike during their golden years, when they had Gabriel, Inna, etc...).

not sure if this matters, but MA has a lot less USAMO qualifiers than last year.

A super-genius doesn't win ARML. Having lots of people who are pretty good is much more useful. I think a team of 15 copies of me would easily beat a team consisting of Gabe, Reid, Tiankai, and 12 random people, even though the team with Gabe, Reid, and Tiankai could get the top 3 individual scores. (Of course, that isn't how ARML teams are formed, so it's a moot point.) To demonstrate my point, notice that Wisconsin hasn't been a threat for the national title recently despite having probably the most impressive top end in the country: Po-Ru, Dani, Po-Ling, and Yian. 4 people don't make a team. 15 people make a team.

It's obvious that SFBA got far worse between 2001 and 2002. Eight of the fifteen people on SFBA A 2001 were seniors (as were I think 6 on the B team). But there were also six seniors on the 2002 A team. That seemed like a big loss, but when we started practicing, it was absolutely clear that the 2003 team was far stronger than the 2002 team. I can't imagine that the 2004 team will do nearly as well as the 2003 team (whose top 5 scorers were 4 seniors and a junior), but they won't do horribly either.

Anyway, I'm nearly certain I know who will win ARML 2004, and it's not SFBA, Massachusetts, or TJ.

[quote]Anyway, I'm nearly certain I know who will win ARML 2004, and it's not SFBA, Massachusetts, or TJ.[/quote]

Who, Exeter?

By the way, is anyone else here actually going to be on the SFBA or Massachusetts team this year?

That presumably leaves it to Chicago, then?

By the way, Simon is completely right about what it takes to win ARML.  This year's NYSML exam is a good example, where by virtue of having Alison on their team, the Albany Area Math Circle was able to beat NYC A on the team round, and NYCA still won the whole thing quite handily.

Yes. I definitely expect Chicago to win this year.

No, Georgia A will pwn the rest of you phr34k5!!! (ROFL, I wish)

Who has Tiankai? SFBA? Because, unless I'm completely wrong, he's still eligible to compete, right?

[quote]Who has Tiankai? SFBA? Because, unless I'm completely wrong, he's still eligible to compete, right?[/quote]

Exeter does.

Simon, why do you say Chicago?

Chicago is the only top team that has anything like the strength it did last year.

haha, i tried out for the massachusetts arml team a few months ago and i placed 14th or so (that would've put me towards the bottom of the A team) but i cant go to arml cuz i have high school graduation that weekend.  i didnt qualify for usamo, either... my score was 200, needed 210....

a friend at school was 3rd at the arml tryouts and he didn't make usamo either (202.5) so i dont know how strong the Mass A team will be.

What? Nobody is giving Southern California a chance? Oh well, we'll be  the San Antonio Spurs of the ARML teams then - be Quietly Dangerous! (we wish  :D )

USAMO qualifiers isn't really anything to go by for ARML, but SFBA has at least 22 USAMO qualifiers, whereas Southern California has at most 10. (I can guarantee beyond a reasonable doubt that the SFBA A team will not have 15 USAMO qualifiers though.)

To give the rest of the team some credit:  they would have done quite respectably on the team (6/10, to be conservative?  I certainly didn't do all the problems) without me.  And Albany Area also did quite well on the Power Round, where I was practically useless, having tied myself up with an annoying problem.

[quote="JBL"]This year's NYSML exam is a good example, where by virtue of having Alison on their team, the Albany Area Math Circle was able to beat NYC A on the team round, and NYCA still won the whole thing quite handily.[/quote]


induction
inequalities
function
algebra unsolved
algebra

actually this is china 1995 and there is an additional condition $ n\geq 3$,anyway the solution is as follow:

let $ F_n$ be the n-th fibonacci number,where $ F_0 = 0,F_1 = 1$.then we would have $ a_i = F_ia_1$.now let $ d_2 = b_2 - b_1$ and for $ i > 2$:

$ d_i = b_i - b_{i - 1} - b_{i - 2}$

it follows that:

$ b_i = F_{i - 1}d_2 + \ldots + F_1d_i + F_ib_1$

also by induction we can show that:

$ F_1 + \ldots + F_k = F_{k + 2} - 1$ hence:

\begin{eqnarray*}\frac{b_{n - 1} + b_n}{b_1 + \ldots + b_{n - 2}} & = & \frac {F_nd_2 + \ldots + F_2d_n + F_{n + 1}b_1}{(F_{n - 1} - 1)d_2 + \ldots + (F_1 - 1)d_n + (F_n - 1)b_1} \\
& \geq & \frac {F_{n + 1}b_1}{(F_n - 1)b_1} = \frac {a_{n - 1} + a_n}{a_1 + \ldots + a_{n - 2}}\end{eqnarray*}

(in the last inequality we used the fact that if $ b,d > 0$ and $ \frac ab\leq \frac cd$ then $ \frac ab\leq \frac {a + c}{b + d}$)

now let $ s = a_1 + \ldots + a_n$.then we have showed that:

$ \frac {a_{n - 1} + a_n}{s - a_{n - 1} - a_n}\leq\frac {b_{n - 1} + b_n}{s - b_{n - 1} - b_n}$

but the function $ f(x) = \frac {x}{s - x}$ is increasing on $ [0,s]$ hence we must have:

$ a_{n - 1} + a_n\leq b_{n - 1} + b_n$.


April 1991 Mandelbrot Round Five Individual Test Question:

8. In acute triangle $ABC$, points $D$, $E$, and $F$ are located on sides $BC$, $AC$, and $AB$ so that $AD \bot BC$, $DE \bot AC$, and $DF \bot AB$. Let $R_1$ and $R_2$ be the radii of the circles around $\triangle ABC$ and $\triangle AEF$, respectively. Determine $\angle A$, in degrees, if $area(ABC) = R_1 R_2$

here's an image

[hide="and a hint"]
what is AFED?
[/hide]


algebra
polynomial
function
vector
domain
linear algebra
Ring Theory

Show that the set of polynomial function P(R) is not a finitely generated vector Space
Thank u for ur help in advance

That's trivial if you show that the family (1,X,X^2,...) is free over R ?

[quote="blahblahblah"]Do you really want to bring the more analytical properties of polynomials (that they are algebraically closed over $\mathbb{C})$ to bear on what is really a purely algebraic question?[/quote]
I didn't bring that in. I said that an $n$th degree polynomial has [i]at most[/i] $n$ roots, which is a purely algebraic fact that holds over any field. There was no need to claim that it had exactly $n$ roots, or indeed any roots at all.

[quote]Two polynomials are equal if the sequences associated with them are equal.[/quote]
That statement is completely equivalent to the statement that $\{1,x,x^2\dots\}$ is independent. What we make of that equivalence depends on how we define polynomials. There are two main ways to think of the set of polynomials:

1. a polynomial [i]is[/i] a (finitely supported) sequence; $x$ and its powers are formal symbols with no intrinsic meaning,

or

2. a polynomial is a function which can be written in the appropriate form, and $x$ is a variable whose domain is the field in question.

The proof I wrote was with definition (2) in mind; blahblahblah seems to be using definition (1), in which case there's nothing to be proved.

In either case, we have the canonical example of a vector space with an countably infinite (Hamel) basis.

Of course, you're correct - I misread your post.  But there's still something to be said for working with polynomials without actually referring to their values as functions in the context of linear algebra.  Of course, we need to use values when constructing, say, a dual basis for the vector space of polynomials of degree $\leq n$, but most of the time we don't.

edit:  It seems that we both edited our posts without reading the other's edits, which makes this a bit confusing.

A way of defining the polynomial ring $R[X]$ is as the set of finite functions from $N \rightarrow R$ (where I use the convention $0 \in N$), "finite" meaning here only finitely many non-zero values.  The operations are pointwise addition and $f\cdot g : n \mapsto \sum_{i+j=n} f(i) g(j)$.

In this formulation the functions $e_i : n \mapsto \delta_{in}$ are clearly linearly independent.

What rgep just posted is a formalization of what I meant by definition (1) in my post #6.

[quote="Kent Merryfield"]julien_santini used the word "free" for a concept that is called, in most English-language linear algebra courses, "linearly independent."

If a vector space has dimension $n$, then any set of $n+1$ vectors must be linearly dependent. (This is a major theorem in the elementary theory.) It follows that if a vector space admits an infinite independent set, then that space must be infinite dimensional.

As for the particular point, the independence of $\{1,x,x^2,\dots\}$:

Let $P(x) =a_0x+a_1x^2+\cdots+a_nx^n$ be an arbitrary linear combination of this set. Either it is the trivial linear combination (all $a_m=0$) or we may WLOG assume that $a_n\ne0.$ Assume that this is not the trivial linear combination. But the theory of polynomials now tells us that to each root of this polynomial must correspond a factor. An $n$th degree polynomial can have no more than $n$ zeros. Since the real numbers are infinite, there must be a real number that does not equal any of the roots, and hence there is an $x$ such that $P(x)\ne0.$ That means that $P$ is not the zero polynomial. So the only way to produce the zero polynomial is to have the trivial linear combination. That proves independence.

This proof used exactly two properties of $\mathbb{R}$: it is a field (used to prove that an $n$th degree polynomial can have no more than $n$ roots) and it is infinite. The same proof would work for polynomials over any infinite field.[/quote]

Thank u very much , Now it makes senses

[quote="blahblahblah"]A minor quibble:  Do you really want to bring the more analytical properties of polynomials (that they are algebraically closed over $\mathbb{C}$) to bear on what is really a purely algebraic question?

To every polynomial $p(z)=a_0+a_1z+\ldots+a_nz^n$ we may associate the sequence $\{b_i\}^{\infty}_{i=0}$ where $b_i\equiv 0$ for $i>n$ and $b_i\equiv a_i$ otherwise.  Two polynomials are equal if the sequences associated with them are equal.  We may thus conclude that $\{1,x,x^2,\ldots\}$ is linearly independent, because if $c_1\cdot 1+c_2\cdot x+\ldots=0$, then the LHS is the sequence $c_1,c_2,\ldots$ and the RHS is the sequence $0,0,0,\ldots$, so all of the $c_i$'s are zero, which is linear independence.[/quote]

RHS ? , LHS ? Thank u

In general, defining polynomials as functuions seems to me a bad idea: the polynomials $x$ and $x^p$ define the same function on $GF(p)$, different functions on $GF(p^2)$ and yet are clearly not the same polynomial.


to make the lives of people simpler.. :D

[hide="hint"]

three consecutive numbers = (x-1)(x)(x+1)

it could be written out as

$(x^2-1)(x)$

I hope it helps
 [/hide]

[quote="shinwoo"]to make the lives of people simpler.. :D

three consecutive numbers = (x-1)(x)(x+1)

it could be written out as

$(x^2-1)(x)$

I hope it helps
[/quote]

Cough. It's three consecutive [b][i][u]EVEN[/u][/i][/b] integers, not three consecutive integers. There's a difference.

There is a [b]slightly[/b] quicker way than Drunken_Math's solution...  :D

o

sorry
 :blush:

I read the problem wrong

then

it would be better with

$(x-2)x(x+2)$

[quote="shinwoo"]it would be better with

$(x-2)x(x+2)$[/quote]

Yep, and the first and third terms are $(x+2)(x-2)$. Does that factorization look familiar?

[hide]
$x\cdot(x+4)=192$
Using guess and check, you get that the answer is $\framebox{12, 14, 16}$ and, you could get -12, -14, and -16[/hide]

[quote="shinwoo"]what happened to the qudratics formula pluggies??

noone's using it?[/quote]

Show us your solution...

I'll do this for the fun of it.

We know that the middle term will be $x$ so teh three consecutive terms are $(x-2)x(x+2)$.
We know that $(x-2)(x+2)=x^2-4\rightarrow x^2=196\rightarrow x=14$.  Asa result, the consecutive numbers could be both $12,14,16$ and $-16,-14,-12$. :)

Masoud Zargar

[hide]12,14,16. I suppose the negative one would work but it doesn't seem right...[/hide]

[quote="236factorial"]Find three consecutive even integers if the product of the first and third integer is 192.[/quote]

[hide]Three integrers: $K; K+2; K+4.$

Then, $K*(K+4)=192$
${}^K{}^2+4K-192=0$
$K=12$ or $K=-16$.

Then, the numbers you`re looking for are $\boxed {12; 14; 16}$ or $\boxed {-16; -14; -12}$.[/hide]


[quote="ml2"]I don't quite understand a couple of things that AOPs vol. 1 says about russels paradox.

First if A= {1,2,A} then A contains 1 and 2. Thus A is really just {A} because otherwise we have 1 and 2 twice in the set?

Also it later states that the set {1,2} is in the set M which consists of every set thta does not contain itself. Doesn't {1,2} contain {1,2}?

Thanks[/quote]Could you quote where it says that {1,2} is the set of all sets that do not contain themselves? I find that hard to believe. Or do I misunderstand?

M is the set of all sets who do not contain themselves.

If M does not contain itself, it does contain itself.

If M does contain itself, it doesn't contain itself.

So does it contain itself.

A not-so-satisfying solution is to invent metalanguages. Imagine a ladder. Put M on a rung that is not the bottom rung. Then let M only apply to the sets on the rung(rungs) below it. Thus there is no paradox.

Another example with the ladder. Let statement A be "The sum of the angles of a triangle is pi radians." Say we wanted to prove it. Put it on the bottom rung. Then on rung two we could put the statement "Statement A is true." and call it B. Then on rung three we could put "Statement B is true."

Sets can contains themselves, or any other set, as an element of the set.

For example: a basic construction in set theory is the [i]power set[/i], which is the set of all possible subsets of a set.  To take a simple example, suppose
\[
S = \{2,3,5\}.
\]
Then the power set of S, sometimes denoted P(S), is
\[
P(S) = \{ \varnothing, \{2\}, \{3\}, \{5\}, \{2,3\}, \{2,5\}, \{3,5\}, \{2,3,5\} \}.
\]
P(S) has 8 elements.  Each element is itself a set.  In particular, $S \in P(S)$.

[quote="ml2"][quote="RC-7th"]But $A\ne \{1,2\}$.

$\{1,2\}$ does not contain itself. However, $A=\{1,2,A\}$ does contain itself.[/quote]

I understand the paradox, I just don't understand the statements about sets containing themselves.. what does it mean for a set to contain itself is what confuses me. I don't understand what set A is if it is {1,2,A}.[/quote]

As DPatrick said, you are misunderstanding about sets being elements of sets. If, for example B = {1, 2, 3}, then the set {4, B} is not equal to {1,2,3,4}, it is equal to {4, {1,2,3}}. It contains 2 elements, one of which is 4 and the other is a set containing 1, 2, 3. So {1,2} does not contain itself, since its only elements are 1 and 2 - none of its elements is (the set containing 1 and 2).


floor function
logarithms
real analysis
real analysis unsolved

What is nature of $ \sum_{n \equal{} 1}^{\infty}\frac {( \minus{} 1)^{\displaystyle{\lfloor\ln n\rfloor}}}{n}$?

A typical block of terms between sign changes has sum approximately 1 in absolute value. The series diverges by the Cauchy criterion.


Compute:
$\sum_{1\leq i\leq n}{\frac{1}{i}}+\sum_{1\leq i<j\leq n}{\frac{1}{ij}}+\sum_{1\leq i<j<k\leq n}{\frac{1}{ijk}}+...+\frac{1}{1\cdot 2\cdot 3\cdots n}$


conics
ellipse
function
integration
calculus
algebra
system of equations

In preparation for the 2009 exam, I was looking at last year's multiple choice exam. I was confused about these two questions:

18. A uniform circular ring of radius R is fixed in place. A particle is placed on the axis of the ring at a distance much greater than R and allowed to fall towards the ring under the influence of the ring’s gravity. The particle achieves a maximum speed v. The ring is replaced with one of the same (linear) mass density but radius 2R, and the experiment is repeated. What is the new maximum speed of the particle?

[hide="Partial Progress"]The circumferences of the ring doubles, so the gravitational force should double. Thus the potential energy at a given height should double, so the speed should be $ \sqrt {2}v$. But the answer is apparently $ v$. Where did I go wrong?[/hide]

25. Two satellites are launched at a distance R from a planet of negligible radius. Both satellites are launched in the tangential direction. The first satellite launches correctly at a speed v_0 and enters a circular orbit. The second satellite, however, is launched at a speed $ \frac {1}{2} v_0$. What is the minimum distance between the second satellite and the planet over the course of its orbit?

I have tried different approaches, and none seem to work. Need I use the equation of an ellipse?

for #25, the 1/2v_0 object moves in an ellipse.

Heh, there were the exact same two I also had problems on...

What? Is there some diagram given? What's "the axis of the ring"???

No diagram was given. The axis of the ring is the line normal to the plane of the ring through it's center.

For number 25, the fact that the first satellite goes into circular orbit tells you that $ v_0 \equal{} \sqrt {GM}$, where $ M$ is the mass of the planet.  Then use conservation of both energy and angular momentum to get a system of equations and plug in that expression for $ v_0$.  You actually don't need the equation for the ellipse.

As for 18, see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=190314.  You would be correct in saying that the potential energy doubles if the ring were a sphere.  Here, it doesn't exactly double - it does, of course, approximately double if the particle is very far away from the ring, in which case the ring appears as a point.  However, the particle does approach the ring arbitrarily closely, so we can't make that approximation forever.  Hence we have to integrate.  Yeah, it's a pain.

EDIT: If you want more detail for number 25, here it is.  Since energy is conserved, the satellite will be a minimum distance away from the planet when its speed is maximum.  We have the following equations:
\[ \frac{1}{2}m\left(\frac{v_0}{2}\right)^2 \minus{} \frac{GMm}{R} \equal{} \frac{1}{2}mv_\textrm{max}^2 \minus{} \frac{GMm}{r_\textrm{min}}
\]
and
\[ m\left(\frac{v_0}{2}\right)R \equal{} m v_\textrm{max} r_\textrm{min}.
\]
Eliminate $ v_\textrm{max}$ and solve for $ r_\textrm{min}$.

[hide="full solution to ring thingy?"]
The particle going through the ring with radius R has kinetic energy 1/2mv^2=E.
Let the distance the particle is from the center of the ring be d. Let the gravitational acceleration at any moment be G. Since the gravitational acceleration is a function of the distance from the center of the ring, we can say f(d)=G. Then, $ \int^{a}_0 f(d) \delta d \equal{} \frac12 v^2$.
We see that for each d, G=(product of masses)/r^2, where r is the distance from the particle to any point on the ring. When R is doubled, the product of masses is doubled. Consider the angle between the axis of the ring and any line from the particle to a point on the ring. Let that angle be theta. At the same theta, the particle in the system with ring of radius R/2 has 2 times the distance to the ring, so r^2 is quadrupled. However, the effective distance from which the same force acts is doubled (well, force changes at half the rate with regard to distance, so its the average), since the distance is arbitrarily long (we can let it be infinity, for a in the equation up there).
Therefore, we have that the kinetic energy is 2*2/4E=E, and the velocity is the same.

[/hide]

I actually just realized my problem with the ring question. It is true that the potential energy from very far away does double, but since we're setting the potential energy to be $ 0$ at $ \infty$, it's the potential energy in the center of the ring which is important. If the initial potential energy of the object for the first ring is $ \minus{}\frac{GmM}{d}$, then the value for the second ring is $ \minus{}\frac{2GmM}{d}$, where $ d$ is the vertical distance from the ring to the object. However, in both cases, the final potential energy is $ \minus{}\frac{\minus{}GmM}{R}$, and since $ d>>R$, the final is much more significant. Therefore, the maximum speed is approximately the same in each case. And we don't have to use integration  :lol:

When far away PE =0, and when at the centre PE = -KMm/r . Now M = $ \lambda r$ , so r cancels and we get fixed PE and hence fixed speed in both cases.

[quote="calc rulz"]I actually just realized my problem with the ring question. It is true that the potential energy from very far away does double, but since we're setting the potential energy to be $ 0$ at $ \infty$, it's the potential energy in the center of the ring which is important. If the initial potential energy of the object for the first ring is $ \minus{} \frac {GmM}{d}$, then the value for the second ring is $ \minus{} \frac {2GmM}{d}$, where $ d$ is the vertical distance from the ring to the object. However, in both cases, the final potential energy is $ \minus{} \frac { \minus{} GmM}{R}$, and since $ d > > R$, the final is much more significant. Therefore, the maximum speed is approximately the same in each case. And we don't have to use integration  :lol:[/quote]

An even better solution.  Darn, I should have realized that.


induction
inequalities proposed
inequalities

Let $S = \sum x$, then ineq becomes
\[
\sum\limits_{i = 1}^n {\frac{{\left( {S - x_i  - x_{i + 1} } \right)\left( {x_i  - x_{i + 1} } \right)}}{{x_i  + x_{i + 1} }}}  = \sum\limits_{i = 1}^n {\frac{{S\left( {x_i  - x_{i + 1} } \right)}}{{x_i  + x_{i + 1} }}}  - \sum\limits_{i = 1}^n {\left( {x_i  - x_{i + 1} } \right)}  = S\sum\limits_{i = 1}^n {\frac{{x_i  - x_{i + 1} }}{{x_i  + x_{i + 1} }}}
\]
So it suffices to prove that
\[
\sum_{i=1}^n \frac{x_i - x_{i+1}}{x_i + x_{i+1}} \geq 0
\]
and we'll use induction.

The original ineq only makes sense only $n \geq 3$. If $n$ = 3, then
\[
\sum_{i=1}^3 \frac{x_i - x_{i+1}}{x_i + x_{i+1}} = \frac{(x_1 - x_2)(x_1-x_3)(x_2-x_3)}{(x_1 + x_2)(x_1+x_3)(x_2+x_3)}
\]
and if it's true for $n$, then
\[
\frac{x_1 - x_2}{x_1 + x_2} + \cdots + \frac{x_{n+1} - x_1}{x_{n+1} + x_1}
=
\left( \frac{x_1 - x_2}{x_1 + x_2} + \cdots + \frac{x_{n} - x_1}{x_{n} + x_1} \right)
+
\left(  \frac{x_{1} - x_n}{x_{1} + x_n} +  \frac{x_{n} - x_{n+1}}{x_{n} + x_{n+1}} + \frac{x_{n+1} - x_1}{x_{n+1} + x_1}  \right)
\]
which is nonnegative by induction hypothesis.

(except in the last case, $x_{n+1} = x_1$.)


i see why if for
$ x\equiv a_1 \pmod{n_1}$
$ x\equiv a_2 \pmod{n_2}$
.
.
.
$ x\equiv a_k \pmod{n_k}$
there is some k which divides more than one n_i, then the system can't have a solution. For example, if we have two congruences written
$ \frac{x\minus{}a_1}{4}$ and $ \frac{x\minus{}a_2}{8}$ can't both produce whole numbers for some $ x$. But how do you show this is generally true?

[quote="Los"]there is some k which divides more than one n_i, then the system can't have a solution.[/quote]
No: if there is some $ k > 1$ that divides more than one $ n_i$ then it is possible for the system to not have a solution.  (But it is also possible that the system does have a solution.)  This is not equivalent to what you wrote.

In general, you just need to choose values for the associated $ a_i, a_j$ that are incompatible mod $ k$, e.g. $ x \equiv 1 \pmod n_i$ and $ x \equiv 0 \pmod n_j$ (where $ k | (n_i, n_j)$).

aha. so for example

$ x \equal{} 1\pmod{9}$
$ x \equal{} 4\pmod{12}$
$ x \equal{} 22\pmod{26}$

has a solution (it is $ x \equal{} 100$), even though there is some $ k \equal{} 3$ for which $ k|9$ and $ k|12$.

To see this write
$ x\equiv 4\pmod{12}\Rightarrow x\equiv 4\pmod{3} and x\equiv 4 \pmod{4}$
     and
$ x\equiv 22\pmod{26}\Rightarrow x\equiv 22\pmod{13} and x\equiv 22 \pmod{2}$,
and you pick to work with the first of each, [color=red]because they have smaller numbers[/color](?). Then one solves the system

$ x \equal{} 1\pmod{9}$
$ x\equiv 4\pmod{3}$
$ x\equiv 22\pmod{13}$

in the usual way, whose numbers modulo all of which are coprime to one another.


geometry
3D geometry
tetrahedron
analytic geometry
octahedron
symmetry
calculus

What does it mean to "embed" a tetrahedron into 4 space? In a solution to a problem I read, it labeled the vertices of a tetrahedron with ordered quadruplets (? right word) $ (1,0,0,0), (0,1,0,0), 0,0,1,0), (0,0,0,1)$. How does this work and why does it make calculations simpler?.

Also, when is it useful to use coordinates for 3d geometry problems? Given an octahedron, is there a way to label its vertices and quickly find a useful point in it such as the centroid of one of the faces of the octahedron?

I think the word is "quadruple".  To me $ \mathbb{R}^{4}$ means your 3 usual dimensions and time as the fourth, so I have no idea how you could lable a tetrahedron with four coordinates unless it somehow changed shape with time.
To answer your other question, I often find it useful to draw a picture if I'm trying to parametrize a surface or get a better picture of some shape.  Also, if you need to find some property, like area, distance, volume,... it may  help to draw the region.  Hope that helps.

You can theoretically have 4 spacial coordinates, you can theoretically have infinitely many spacial coordinates! For an informal discussion, Flatland is an interesting novella to check out.

I'm using hide-tags because this is sorta long

[hide="Regarding the usage of 4-space"]When you're faced with an abstract situation in several dimensions, it often helps to think of lower-dimension analogies.

Here, the problem is analogous to naming the vertexes of an equilateral triangle $ \{ (1,0,0), \; (0,1,0), \; (0,0,1) \}$. Did you have to put the triangle in 3-space? Of course not, you were welcome to use
 $ \left \{ (0,0), \; ( \sqrt {2}, 0), \; \left( \frac {\sqrt {2}}{2}, \frac {\sqrt {6}}{2} \right) \right \}$ instead.

It definitely makes distances easier to calculate if your numbers are all 0's and 1's. For example, you can easily find the centers of each face; they are the permutations of $ \left( \frac {1}{3}, \; \frac {1}{3}, \; \frac {1}{3}, \; 0 \right)$. The center of the whole tetrahedron is $ \left( \frac {1}{4}, \; \frac {1}{4}, \; \frac {1}{4}, \; \frac {1}{4} \right)$.

I can see how someone might be uncomfortable with the thought, but I think it's a wonderful simplification of calculations.[/hide]
[hide="When to use coordinates"]Great reasons to use Cartesian coordinates are when:
- there are lots of right angles
- the numbers are relatively simple/there is lots of symmetry
- when there are free variables in the problem, ie. you can make simplifying assumptions
- you need to optimize something and you are good with inequalities/with calculus
- you want to use vectors, dot products, cross products, etc.

Good reasons to use coordinates are when:
- you are in the mood to brute-force something
- you don't know what to do[/hide]
[hide="Coordinatizing an octahedron and such"]
The coordinates of an octahedron's vertexes are just the permutations of $ (\pm 1, 0, 0)$. It helps to consider duality here - the [color=blue][url=http://en.wikipedia.org/wiki/Dual_polyhedron]dual polyhedron[/url][/color] to the octahedron is the cube.

The centroid of any regular polygon/polyhedron/etc and of most high-symmetry polygons/polyhedrons/etc is just the arithmetic mean of the vertexes.[/hide]


What is the product of all positive odd integers less than $1000$?
(Using factorial notation where necessary).

I know the answer but I need a solution. Please show your work.

[hide="Sure"]
$1 \times 3 \times 5 \times... \times 999$ is the answer we wish to express in factorial notation.  Note that this is $999!$ but with the even numbers divided out, or $\frac{999!}{2 \times 4 \times... \times 998}$.  Furthermore, we can factor out a $2$ from each factor in the denominator, producing $\frac{999!}{2^{499} \times 1 \times 2 \times... \times 499} = \frac{999!}{2^{499} 499!} = \frac{1000!}{2^{500} 500!}$.
[/hide]

Yeah, thats what I got after as well. On the actual AMC, the number was $10000$ not $1000$ sorry :oops:


We're now focusing on getting the blogs in shape.  I'm going to maintain a bug list and a wish list/todo list here.  Please add any bugs you find here and I'll add them to the BUG list.  Also, add items you'd like to see in the blogs here, and I'll maintain a TODO list.

BUG list:
* Get the post count and views count on the list of all blogs correct
* Add the Reply Counter
* Allow remove contributors
* Still a few missing language entries in the Blog options area.
* Post/reply does not have the same style as the blog itself.

TODO list
* Bring back the shout boxes (we have all the old shouts)
* Bring back the old two templates
* Implement categories, so people can have categories for their blog entires.
* Make the Blog index page more distinct from the forums.
* Need a way to see a list of all the subjects posted in a blog (this existed in the old system)
* Email subscriptions/RSS
* Mood emoticons

Please add more bugs and wishlist items so we can get the blog system running smoothly!

On background images: I suppose you just type the address into the box, http://billydorminy.homelinux.com/bg.jpg , right? Because it doesn't seem to be working on my blog. Is the image too big? (2272x1704, 252KB) Or does it not work because it's not hosted on the artofproblemsolving.com server?

Just like before, you have to use the proper CSS syntax: namely you have to type
[code]url(http://billydorminy.homelinux.com/bg.jpg)[/code]

All items have CSS identifiers, so you can easily put your own CSS code (in the Blog Customizing section) if you are not satisfied with your look) ;)

By the way, as always, if you are posting a bug here, please let us know what browser (and version) you are using -- certain bugs will only show up in certain browsers.

This time, I think I have found a real problem with the blogs  :)

Actually, the view thing is better the way it is.

I am more concered about the number of entries being different from the actual number posted.

For example, in my blog, I have posted 9 entries, yet it only says 7 on the blog page (it says 9 inside my blog).  When I posted my 9th entry today, it did not increase the number of entries.

Also this appears to be true in almost ever blog I look at.

On my blog, one of the entries is contained in another entry. The two entries are connected, with two scrollbars.

[url=http://www.artofproblemsolving.com/Forum/weblog.php?w=223]The last two entries on this page.[/url]

I have posted another entry in my blog, yet it still says on this page:

http://www.artofproblemsolving.com/Forum/apsjour.php

that I have only 7 entries.

Also, the IMO blog has 0 entries (there are at least 3+).

A possibly related bug: The count for replies are off for some entries.

For example:
http://www.mathlinks.ro/Forum/weblog.php?w=305
The welcome topic has 4 replies, but it only shows up as 2.

Yeah, the replies counter is not fixed yet apparently. I will deal with it once I am back from the IMO.

[quote="Captain Sugar"]my post counter is messed up. before i had 15 entries, and it said i had 13 entries. now i put in another one, and it still says 13 entries. i think it's stuck on 13 or something, which would be creepy, except i'm not superstitious. :lol:  i use internet explorer something. i don't know the number, but i know it's the newest one.[/quote]

press help$\to$about to see the version

Also something is that I can't remove contributors from my list. It always says, General Error, SQL requests not achieved, etc...

[quote="b-flat"]I am more concered about the number of entries being different from the actual number posted.[/quote]

Me too--I posted three entries in my new blog and the counter is still stuck at 0.

IE 6, Windows XP

i think we should bring the reply counter back. it's annoying looking at every post to see if anyone's replied.

I have this problem: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=112086 and I don't know what to do about it, and nothing has happened about it. I'd really like to remove some contributors. :(

[quote="Xevarion"][quote="rrusczyk"]* Email subscriptions/RSS[/quote]
Is this going to happen any time in the near future? I'd like Facebook to be able to read my AoPS blog posts so that I can keep maintaining just one blog and conveniently allow my friends to navigate to it / know when I've updated.[/quote]First steps are already taken: here's the feed for the site: http://www.mathlinks.ro/Forum/feed.php (still in beta though as you can see :) ).

[quote="chess64"]Haha I made a php script to automatically create the RSS of my blog. If you want, I can make you one too...[/quote]

Could you please make me one? Maybe I could edit it to add a body text...

What about the problem that the style of the index of a blog isn't the style of the blog itself.

[quote="anirudh"][quote="chess64"]Haha I made a php script to automatically create the RSS of my blog. If you want, I can make you one too...[/quote]

Could you please make me one? Maybe I could edit it to add a body text...[/quote]You have to understand that PHP is a server language. Unless you have your own server/site/php server you can't do anything with a php script.

[quote="#H34N1"]Yes.

But I blocked a few people from my blog. :lol:

I created a .PHP script somewhat like this:

[code]/* MY CSS PORTION */
CSS GOES HERE

CSS WENT HERE
/* END CSS PORTION */

/*START IP BANNING */
<?php // Allows you to ban people from viewing your website.
$getip = $_SERVER["REMOTE_ADDR"];
$getdate = date( "l dS of F Y" );
$gettime = date( "h:i:sa (@B" );
$banned_ip = array();
$banned_ip[] = '111.111.111.110';
$banned_ip[] = '111.111.111.111';
$banned_ip[] = '111.111.111.112';
$banned_ip[] = '111.111.111.113';
$banned_ip[] = '111.111.111.114';
foreach($banned_ip as $banned) {
$ip = $_SERVER['REMOTE_ADDR'];
if($ip == $banned){
echo "It seems you have been banned from viewing this website.";
echo "
";
echo "If you think you have been banned in error please contact me.";
$fp = fopen("code/data/ip_data.dat",  "a");
fputs($fp, "**BANNED** Visit logged on $getdate at $gettime internet time) for IP: $getip
");
fputs($fp, "");
fclose($fp);
exit();
}
}
echo "Update in progress.";
echo "
";
echo "'Authorized Visit' detected and logged on $getdate at $gettime internet time) for IP: $getip";

$fp = fopen("code/data/ip_data.dat",  "a");
fputs($fp, "Authorized Visit logged on $getdate at $gettime internet time) for IP: $getip
");
fputs($fp, "");
fclose($fp);
/* END IP BAN */
?>

[/code]

I htink it worked okay. :maybe:

OH. Yes. I do have a PHP supporting website, theani.isgreat.org

It parses the PHP scripts...[/quote]
I don't think that works... it's an external linked .css file, not an inline style.

EDIT: Yeah, it doesn't work. It just assumes that it's awry CSS. See: http://www.artofproblemsolving.com/Forum//templates/ptifo/blogs/Hyperion/style/c746.css

EDIT2: "Blog Started: Today, at 4:14 am" It's been around for a month. What's going on?

[quote="Temperal"]EDIT: Yeah, it doesn't work. It just assumes that it's awry CSS. See: http://www.artofproblemsolving.com/Forum//templates/ptifo/blogs/Hyperion/style/c746.css[/quote]Of course it doesn't work. That would be a serious security hole if it would work!

I suppose..
Is there any place in the blog that you can place raw HTML?

I can't recreate my blog since I can't select one. I THINK THAT THIS IS A SAD EVENT BECAUSE I WILL NOW BLOG ELSEWHERE.

[quote="mathnerd314"]The server also thinks that I started my blog at 7:47 today (that's a plane!!! :pilot:). Probably because that's when I last accessed it.[/quote]

I have the same problem... Instead of showing the date when I posted the entry, it displays the time when I accesed the blog. It says "filed on Today, ...".
I want it to display the date and time when I submitted my entry.

All right I'll see what I can do. Seems to be a bug indeed.

[quote="Temperal"]Plus, could some form of BB Code or at least the basic html tags be allowed in the custom block thing?[/quote]

That would be great :)


Is it possible to map the open unit disk conformally on $ \{z: \ 0<|z|<1\}$? Since boundary should be mapped onto boundary, it seems to be impossible, but this problem is given in one of the first topics and it requires elementary solution.. Any ideas are appreciated.

Is a conformal map here bijective?

If so the standard argument is that by the open mapping theorem, the map is actually biconformal, and in particular a homeomorphism. Hence such a map preserves simple connectedness (although to prove this you probably need to know about induced homomorphisms of fundamental groups). Thus, such a map cannot exist since the unit disc (disk?) is simply connected, while the given set is not.

I guess not too helpful for your purposes  :oops:
Do you know about isolated singularities? If so, this also gives a direct argument. What do you know?

Thanks for your comment. Yes, i know about isolated singularities. How does it help here? :maybe:

Well, again I am supposing you mean a bijective map. So suppose the map exists, then we have an analytic inverse map $ f$ from the punctured disc to the disc. $ 0$ is an isolated singularity of f. It cannot be a pole or an essential singularity, because then points close to 0 would be mapped outside the disc. So it's removable, say $ f(0) \equal{} a$. Clearly $ |a| \leq 1$. $ |a| \equal{} 1$ contradicts the open mapping theorem, so $ |a| < 1$. If $ f'(0) \equal{} 0$ then we immediately have a contradiction to injectiveness of $ f$ on the punctured disc (do you know why?). So $ f'(0) \neq 0$, but this means that $ f$ maps an arbitrarily small neighborhood of 0 topologically with a neighborhood of $ a$. Since there's also a different point $ z$ in the unit disc which is mapped to $ a$ there is again a contradiction to the injectiveness of $ f$ on the punctured disc.

This turned out to look sort of complicated, but it's really not

The argument with singularity is nice, might be useful in future :)

Well not really, because the problem of which sets can be biconformally mapped to the unit disc has been completely solved. The solution is what I said in my first reply + riemann mapping theorem.

How about if the map is not required to be injective?

It's doable. Apply a Möbius transform that maps it to the left half-plane. Apply the exponential function.


Solve this nice ineq
If $a,b,c$ are positive prove that

$\frac{a}{(2a+b+c)(b+c)}+\frac{b}{(2b+a+c)(a+c)}+\frac{c}{(2c+a+b)(a+b)}\geq\frac{9}{8(a+b+c)}$

By the way who is the creator ?

The official solution is always appeared with a big speed :rotfl:

Could we prove it with expansion?

[quote="silouan"][quote="Tiks"]The reason not in,it is simple it a children's problem :P  :rotfl:[/quote]

It is from Romania Shortlist 2004. It is not a very easy problem[/quote]
Mabe,but I think it is very-very easy.

Here is another solution
${\sum\frac{a}{(2a+b+c)(b+c)}=\frac{1}{2}\sum[\frac{1}{b+c}-\frac{1}{2a+b+c} ] \ge \frac{1}{2}[\sum\frac{1}{b+c} - \frac{1}2}\sum\frac{1}{b+c} ] = \frac{1}{4}\sum\frac{1}{b+c} \ge \frac{9}{8(a+b+c)}$
where the first inequality follows from the well-known fact
$\sum\frac{1}{x+y}\le \frac{1}{2}\sum\frac{1}{x}$ (put $x=b+c$, $y=c+a$ and $z=a+b$)
and the second from CBS inequality.

[quote="razec upul"]Silouan, what shorlist are you talking about? :D[/quote]

It is Shortlist Romania 2004. This was told by Nickolas and mr. Stergiou in their book
page 237

I don't quite think so. I have the shorlist and it isn't such an inequality. Actually this inequality has been published for
the first time in romanian olympiad literature in 1988, if i'm not mistaken. :D  :P


geometry
geometric transformation
reflection

I assume we can discuss it, as it's on the MC site.

This was one of the two I skipped.  I got 44 in chapters.  I was wondering if anyone else got it?

http://www.mathcounts.org/competition/2008%20Chapter%20Sprint%20Round.pdf

I have no clue on this one, esp. w/o a calculator.  8 min on this would be fine, as that would be an accurate reflection of many people's leftover time.

most important step is to notice that "pi" is greater than 3.14 but less than 22/7.
Knowing that you can rewrite without the abosolute value signs:
|pi-3.14|-|pi-22/7|=pi - 3.14 - pi + 22/7=22/7 - 3.14=3 1/7 - 3 7/50= 1/7 - 7/50= (50-49)/350= [b]1/350[/b]

I don't want to rewrite how I did it, so...
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=195667

Lol Alex, I got that wrong because I forgot to simplify, i had 2/700. That question made me get a 45. :(

I'm an idiot.   :mad:  :(

[quote="TheHobo"]Lol Alex, I got that wrong because I forgot to simplify, i had 2/700. That question made me get a 45. :Sad

I'm an idiot. :Mad :Sad[/quote]

Don't feel bad...I did that too. :D

OMG! I did that too! I got a 45 all because of it.

I was running out of time and I couldn't finish #30 so I had to take an educated guess and this is what I put:

1/350

I was like, OMG OMG OMG I GOT IT RIGHT!

I still only got a 41, though. :blush:

Educated guess? Could you describe that, please?


geometry
circumcircle
inradius
inequalities
limit
inequalities proposed

We well know with triangle $ABC$ and three median $m_{a},m_{b},m_{c}$, circumradius $R$, inradius $r$ then
$m_{a}+m_{b}+m_{c}\le \frac{9}{2}R$
$m_{a}+m_{b}+m_{c}\le 4R+r$
thus with the inequality
$m_{a}+m_{b}+m_{c}\le (\frac{9}{2}-k)R+2kr,\forall k>0$
What's the best const $k$ ?
 :)

With my calculation
$k=\lim_{\triangle ABC\rightarrow equaliteral}\frac{m_{a}+m_{b}+m_{c}-\frac{9R}{2}}{R-2r}=\frac{3}{2}$
But follow me it isn't the best const because we cann't find the best const by limit.
What is your ideal ?

[quote="evarist"][color=purple]gemath i think if $k=\frac{3}{2}$ the triangle need the condition. Have you got a solution of this problem.If you have please post it[/color][/quote]
  [color=purple]  In triangle which has two angle > $\frac{\pi}{3}$  we have$p\leq \sqrt{3}(R+r)$ thus$\sqrt{3}p\leq 3(R+r)$so
$m_{a}+m_{b}+m_{c}\leq \ 3(R+r)+\frac{|a-b|+|b-c|+|c-a|}{4}$
  What a pity! Thus i belive$k=\frac{1}{2}$ is the best constant.Vacs can you show me the best constant? [/color]

There isn't any pity here  :D  :D I don't say the best constant is $\frac{3}{2}$ I only show the limit equal $\frac{3}{2}$ and I don't understand your example for this problem, if you want to say the best constant isn't $\frac{3}{2}$ then you only show to exist a triangle such that $m_{a}+m_{b}+{m}_{c}\ge 3(R+r)$ and actually I have many examples for it (by computer), you shouldn't make wordy !


There are 4 different pair of gloves each having a different color. There are 4 persons. All the right hand gloves are distributed at random, one to each person. Then, all the left hand gloves are distributed randomly in the same manner.

What is the probability that at least one person gets a matching pair?

[hide="Brute force"]
4 is small enough where we can just list all possibilities. We can assume the right hand gloves are RBYG. Then, we can systematically list all possibilities where it doesn't work:
BRGY
BYGR
BGRY
YBGR
YGRB
YGBR
GRBY
GYBR
GYRB
There are 9 possibilities out of 24, so the answer is $ 1\minus{}\frac {3}{8}\equal{}\frac{5}{8}$[/hide]

[quote="dgreenb801"][hide="Brute force"]
4 is small enough where we can just list all possibilities. We can assume the right hand gloves are RBYG. Then, we can systematically list all possibilities where it doesn't work:
BRGY
BYGR
BGRY
YBGR
YGRB
YGBR
GRBY
GYBR
GYRB
There are 9 possibilities out of 24, so the answer is $ 1 \minus{} \frac {3}{8} \equal{} \frac {5}{8}$[/hide][/quote]

The left hand gloves are distributed first and then the right hand gloves are distributed, both at random. We have to find the possibility of getting an identical pair of gloves. How are you finding that?

It is 1- the probability of no one getting a pair of the same gloves. We can assume it starts out with any order we want because for any other order, the probability would be the same.

Side note: this is just a special instance of the derangement or rencontre problem, $ 9 \equal{} D(4)$.


AMC
USA(J)MO
USAMO
Support
summer program
MathPath
Davidson Fellows

Hi, I see no one has posted here yet about the [url=http://www.ditd.org/public/article.aspx?cid=9&mid=124&tp=122]Davidson Young Scholars[/url] program. Maybe the program is still too new to have gained much attention, so I'll bring it up here for those of you who might be interested. Do any of you have any experience with the program?

i received a phamplet in the mail. the people who wins the prizes for mathematics are always imo people. (no point in registering in math unless you are at aleast USAMO level i guess).

[quote="kel227"]If people wanted to know more about it, you could ask anders. He is registered on this site and is a Davidson fellow for math. He received a $25,000 scholarship. He also received a gold medal at the IMO.[/quote]
...or Daniel Kane, another familiar name to some of us, for that matter.  He got $50K from Davidson.

The first three replies are referring to the [url=http://www.ditd.org/public/article.aspx?cid=1&mid=101&tp=2]Davidson Fellows[/url] program, which indeed gave high awards to IMO team members Daniel Kane and Anders Kaseorg, and which looks very interesting. But for the younger mathletes who aren't at the level of a MOPper yet, there is a distinct program, the [url=http://www.ditd.org/public/article.aspx?cid=9&mid=124&tp=122]Davidson Young Scholars[/url] program, which may be of interest to the younger participants on this AoPS site. I'm wondering if anyone here has heard of the Young Scholars program.

I am going to reenergize this thread because there are many new young people on AoPS who I think ought to know about the Davidson Institute for Talent Development [url=http://www.ditd.org/public/article.aspx?cid=9&mid=124&tp=122]Young Scholars Program[/url]. I get the impression that there are several AoPSers who would qualify for and enjoy being part of the Young Scholars Program.

Yea I've heard of the Davison Young Scholars. I just got accepted as one, I'm really happy!

I'm in the program since 2003, didn't go to any gatherings yet (out of the country every summer).

It's good to meet Young Scholars program participants here.

how do we apply and when? and what is it? is it like summer program??

The Davidson Young Scholars program has an [url=http://www.ditd.org/public/article.aspx?cid=129&mid=250&tp=122]application information page[/url] on its Web site that explains most of the procedures for applying. I have also heard that the program staff, reachable through various channels listed on the Web site, are happy to answer questions about the application process.

The program operates partly as a way for young people to connect with one another through online groups and occasional in-person gatherings, and partly as a parent support network through those channels. There are also staff members and professional consultants to the program who have advice on educational planning and learning opportunities. The program provides financial aid to some families based on financial need.

It's my impression that some of the young people on AoPS are advanced enough in math that they might be eligible for the Davidson Young Scholars program, which is why I bring this up.

Hi,

I'm one of the Hobbits (2002 Young Scholars), and I'm going to the Denver Gathering in June.

Since I have been to MathPath I have become even more aware than before that there are some REALLY SMART young people who hang out here on AoPS (sometimes lurking more than posting) who ought to apply to the [url=http://www.ditdservices.org/Articles.aspx?ArticleID=24&NavID=0_0]Davidson Young Scholars Program[/url]. It's one more way, besides AoPS, for math-liking young people to network with one another, and for some families the Davidson Young Scholars program may make many summer opportunities (such as those discussed in other threads on this forum) more affordable, through financial aid. Don't be bashful about [url=http://www.ditd.org/Articles.aspx?ArticleID=26]contacting the Davidson Institute headquarters[/url] if you have questions about the program, and by all means apply if you think you might fit into the program.

I've been in the program since it started, 1998-1999 I think, I was 4, they almost didn't accept me cause I was so young.

Greetings to the new and old and could-be Young Scholars who are posting in this thread.

Do i still qualify if i got a 790 in the Math portion of the SAT LAST year, but did horrible on the Verbal? Do I have to retake it this year? Also, it says i need a copy of an IQ test. Where do i take IQ tests?

You would have to ask, although I don't think SAT scores have anything to do with it, four year olds can't exactly take SAT's can they?  The application has gotten much easier over the years, the standards really arn't that high now.  I don't think most people on here would have a problem getting in.  As for IQ tests, just ask your parents and/or school so you can go take one.


articles search Princeton college topology MIT

[quote="answerseeker95"]I just read a recent news article about a guy who solved the poincare conjecture few weeks ago. How many of you know about this? Can someone please tell me more about it (the article was in Chinese and it very brief)?[/quote]

Here's a Google News search that will turn up a recent AP article (multiple instances of it) in English:

http://news.google.com/news?q=Poincare+conjecture

Hope this helps!

[quote="answerseeker95"]I just read a recent news article about a guy who solved the poincare conjecture few weeks ago. How many of you know about this? Can someone please tell me more about it (the article was in Chinese and it very brief)?[/quote]
Recent? You mean they officially declared it a legitimate proof? If so then I didn't know that.

I'm guessing most of us know of this because the Poincare Conjecture were proven months ago, there were threads about it in like September. Just search for it and you should find a thread about it.

Nope, it hasn't been verified yet.  Give it another couple of years :).

Huh, I heard in two years so I thought they'd be done next year...

Here is a slightly more informative article on Perelman's Proof of the Poincare Conjecture: http://mathworld.wolfram.com/news/2003-04-15/poincare/

Most people who understand it seem to be more or less convinced of the legitimacy of the proof. But technically, Perelman has not published the proof in any journal yet - just on the internet. In order to receive the $1 million prize he'll have to publish it in a journal and then wait 2 years from that point for it to pass scrutiny. Those are the rules set up by the Clay Institute, who sponsor the reward. No one is sure why he hasn't published the result yet. People are speculating that he's just not interested in the money.

There's an interesting article in Discover magazine this month (Jan. '04) on the nature of proof in mathematics. Proofs today are so long and complicated that guaranteeing their validity is becoming less and less certain, even to experts.

Oh, I was talking about legitimate proof. Well, I was quite mistaken, but the proofs and news certainly seem interesting.

According to some professors at my university, there is quite a bit of debate about the correctness of Perelman's proof. He might not want to publish it until he is more or less certain that it is correct. By the way, where did he publish it on the internet?

try
[url]http://www.math.lsa.umich.edu/research/ricciflow/perelman.html[/url]

it has a lot of commentary and notes on perelman's work.

Here's another good layman's article on the proof from the Daily Princetonian (the campus newspaper for Princeton University): http://www.dailyprincetonian.com/archives/2003/04/17/news/7979.shtml

Here goes a writer's somewhat funny take on Perelman's achievement: [url=http://www.reflector.com/news/newsfd/auto/feed/news/2004/01/09/1073710748.26609.2365.0197.html;COXnetJSessionID=AEK0Yx3GUTkw0JvfOx0MfYuIfIfBAFQO2YF0HfcfPUNqgkYHgxgo!-759395321?urac=n&urvf=10740394121510.5083651816148208]Funny Article[/url]

And here goes a more technical explaination of the actual Conjecture that he proved: http://en2.wikipedia.org/wiki/Thurston%27s_Geometrization_Conjecture

Here is a timeline that I constructed (as best I could) of the confusing chain of events surrounding the proof:

Nov. 11, 2002 - Grigori "Grisha" Perelman quietly publishes the first of his papers, "The entropy formula for the Ricci flow and its geometric applications" (39 pages) on the internet through arXiv.org - an archive for electronic preprints of math and physics papers. Here he (purportedly) proves Thurston's Geometrization Conjecture, which has the Poincare Conjecture as a consequence. There is very little fanfare or mention of the history or significance of these problems within the paper.

March 10, 2003 - Perelman puts up a second paper "Ricci flow with surgery on three-manifolds" (22 pages) on arXiv clearing up some of the points in the first paper. Again it passes without a great deal of attention from the popular media, but the buzz begins to spread in the mathematical world.

April 7-11, 2003 - Perelman gives a series of lectures at MIT describing his work to peers. The international buzz really picks up that he may have proven the Poincare Conjecture.

April 15, 2003 - Eric Weisstein puts up a news article on MathWorld with the headline "Poincar Conjecture Proved--This Time for Real" (following a faulty proof that was announced just a month before). An article also appears in the NY Times. Several other announcements follow.

April 16, 2003 - Perelman gives a lecture at Princeton University describing his work.

Jul. 17, 2003 - Perelman puts up a third paper "Finite extinction time for the solutions to the Ricci flow on certain three-manifolds" (7 pages) on arXiv dealing with some more points from the first two papers.

Everything remains quiet for a while as mathematicians ponder and try and understand Perelman's work. Perelman remains quiet and refuses to talk to the media.

Dec. 2003 - Discover magazine includes a discussion of Perelman's proof in an article on the 100 biggest science stories of 2003.

Jan. 2004 - Paul Elias of the Associated Press writes an article about Perelman and the proof which is picked up by several outlets, including CNN.com. Several mathematicians claim they are becoming convinced of the validity of Perelman's proof. A renewed buzz begins and people wonder if Perelman will ever publush his results in a formal journal; and if, or when, he will receive the $1 million dollar prize from the Clay Institute. Will he have to publish it in an actual journal, or is the electronic publication and peer review currently going on sufficient? Stay tuned for more....[/url]

[quote="gauss202"]Here is a timeline that I constructed (as best I could) of the confusing chain of events surrounding the proof: . . . Jan. 2004 - Paul Elias of the Associated Press writes an article about Perelman and the proof which is picked up by several outlets, including CNN.com. Several mathematicians claim they are becoming convinced of the validity of Perelman's proof. A renewed buzz begins and people wonder if Perelman will ever publush his results in a formal journal; and if, or when, he will receive the $1 million dollar prize from the Clay Institute. Will he have to publish it in an actual journal, or is the electronic publication and peer review currently going on sufficient?[/quote]

Thanks for the timeline, which is very helpful for explaining why this story is resurfacing at intervals. The aspect of the Associated Press story that I found most interesting was precisely the discussion of the Clay Institute requirement that a prize-winning proof be "published"--with the tantalizing suggestion that print publication in a peer-reviewed dead-tree journal may not be the only form of publication that would meet the prize criterion. Perelman's preprints must make some interesting reading for mathematicians searching for clues to Poincare's long-standing problem.

There is an article about the proof on CNN (the website). I'm sure it won't be too hard to find once navigated to the site itself.


Let $x$, $y$, $z$, $p$, $q$ and $r$ be positive reals s.t. $p+q+r=x^py^qz^r=1$. Prove inequality
\[\frac{p^2x^2}{qy+rz}+\frac{q^2y^2}{px+rz}+\frac{r^2z^2}{px+qy}\geq\frac{1}{2}.\]

[quote="Myth"]х?н[/quote]

What does that mean ?

:blush:  :blush:  :blush:
At least no-russians can't understand it.

[quote="orl"]Kvant volume X, problem number $Y$ required.[/quote]

Request is still up-to-date. I guess I should make a language journey to Russia for 3-4 month to refresh my knowledge.  :lol:

[quote="Myth"]Let $x$, $y$, $z$, $p$, $q$ and $r$ be positive reals s.t. $x+y+z=x^py^qz^r=1$.[/quote]

Myth, $x+y+z=x^py^qz^r=1$ doesn't make any sense at all, and neither does $xyz = 1$ as stated in the topic subject. Are you sure that all the variables are positive reals?

You are right, it was typo.
I will edit it.

Oh,Dear Myth.
Before you edit problem,I couldn't find any $x,y,z,p,q,r$ satisfy,because $x+y+z=1$.
Now it's quite clear.
$\frac{p^2x^2}{qy+rz}+\frac{q^2y^2}{px+rz}+\frac{r^2z^2}{px+qy}\ge\frac{px+qy+rz}{2}$
But use Cauchy,$px+qy+rz \ge x^py^qr^z=1$
And we are done!
Am I wrong? :)  :)


Well, i know a lot of people who remember tons of interstnig quotes (I'm one of them, actually)... Quotes by famous scientists, poets and philosophers... And ofcourse quotes from movies, TV shows... What are your favourite quotes?

[quote="bubka"]look in round table and find vihag's old topic by the same name.[/quote]

Well, I saw it, but I didn't want to revive it... Maybe some of new members wolud like to share some of their favourite quotes with us?  :)

Are you looking for serious quotes, or random stuff people say on AIM?

"I'm the most terrific liar you ever saw in your life. It's awful. If I'm on my way to the store to buy a magazine, even, and somebody asks me where I'm going, I'm liable to say I'm going to the opera."

Read my Sig :D
another good quote:
 "<(^^);..> You have just recieved the Amish Computer Virus. Since the Amish don't have computers, it is based on the honor system. So please delete all the files from your computer. Thank you for your cooperation"

 "<Lamer> The tradgey of Canada is that they could have had British culture, French cooking, and American technology. Instead they got American culture, British cooking and French technology"

"I'm not racist. Racism is a crime, and crime is for black people."

"I'd give my right arm to be ambidextrous."

"Procrastinators unite! Tomorrow..."

"All rentals are due on the due date." -Blockbuster

I have some from Black Saga and the game Halo 2:

"I can accept failure. Everyone fails at something. But I cannot accept not trying." ~Michael Jordan.

"Teamwork wins championships." or something like that, from Michael Jordan

"It blew right through us--50 cal, Rockets, didn't do a thing." ~Marine

"Retrieving the Icon is my only concern." ~The Arbiter

"It is easy human. Take the icon in your hands..AND DO AS YOU ARE TOLD!" ~Tartarus

"Grab a banshee. He's gonna know we're coming." ~ Seargant Johnson

"Me, inside your head, now." ~Cortana
"Chief, blink if you can hear me." ~Cortana

"The Demon? He is here?" ~The Arbiter

"This is not your grave...but you are welcome in it." ~Gravemind.

"Do not reveal yourselves only after the Arbiter has joined the enemy in battle. Arbiter, you too may wish to do the same. But take heed-your armor is not as new as ours." ~White Spec Ops Commander

More to come but I am forgetting at this very moment


Euler
geometry
circumcircle
geometry proposed

Let triangle ABC inscribed in circumcircle (O). The Euler line of triangle ABC respectively intersects AB, BC, CA at C', A', B'. (The Euler line of triangle ABC is the line which passes through the orthocenter and the center of circumcircle of triangle ABC).
1) Prove that the circles that take AA', BB', CC' be their diameter have the same radical axes.
2) Suppose those circles above intersects at M, N. Prove that: In two points, one lies on the (O), one lies on the Euler circle of triangle ABC.
 :D  :blush:

(For generalized problem...)

I proved $ (1)$ and half of $ (2)$

But why do those three circles meet on the Euler circle?! T. T

I cannot get it

Plz anybody show me the solution for the very last problem.. :wallbash:  :wallbash_red:

Dear Mathlinkers,
the problem posted at first comes from Victor Thébault.
Any reference?
Sincerely
Jean-Louis

[quote="ma 29"]I think we have the next problem: :D

Let triangle $ ABC$ inscribed in circumcircle $ (O)$. Let $ d$ is a line through $ O$.The line $ d$ respectively intersects $ AB, BC, CA$ at $ C', A', B'$.
1) Prove that the circles that take$ AA', BB', CC'$ be their diameter have the same radical axes.
2) Suppose those circles above intersects at $ M, N$. Prove that: In two points, one lies on the $ (O)$, one lies on the Euler circle of triangle $ ABC$.[/quote]
Mr. ma 29, vey nice. Can you show us your solution to my proposed problem?? I am hoping to see it from you. Actually, I'm "dying" to see that if there is a nice solution about this problem!! :blush:  :D

[quote="ma 29"]I think we have the next problem: :D

Let triangle $ ABC$ inscribed in circumcircle $ (O)$. Let $ d$ is a line through $ O$.The line $ d$ respectively intersects $ AB, BC, CA$ at $ C', A', B'$.
1) Prove that the circles that take$ AA', BB', CC'$ be their diameter have the same radical axes.
2) Suppose those circles above intersects at $ M, N$. Prove that: In two points, one lies on the $ (O)$, one lies on the Euler circle of triangle $ ABC$.[/quote]

See [url]http://www.mathlinks.ro/viewtopic.php?t=186883[/url], the supporting lemma, for all claims.

Pedal circle of point on $ d$  pass through constant point which lie on Euler circle of triangle $ ABC$. Circle with diameter $ AA'.BB',CC'$ are pedal circle of $ A',B',C'$ respectively.  Hence this three circle have common point on Euler circle.


superior algebra
superior algebra solved

Determine the structure of the ring
(a) $\mathbb{Z}[x]/(x^2+3,3)$,
(b) $\mathbb{Z}[x]/(x^2+3,5)$,
(c) $\mathbb{Z}[x]/(x^2-3,2x+4)$.

My answers are
(a) $\mathbb{Z}_3[\sqrt 3]$
(b) $\mathbb{Z}_5[\sqrt{-3}]$
and I have no idea about (c).

p.s. Can the answer to (a) be simplified more?


I am totally lost at how to go about solving this problem could someone please help? I read your stickies and other posts to make sure I ask a question in the correct forum if I did not I apologize and make sure I post in the correct place next time.

5x+2y=2x+1
2x-3y=3x+2
solve by addition  Thank you very much.[/u]

The two most conventional ways to solve a system of equations are substitution and elimination (in your case, 'addition' is elimination) we proceed as follows (some may have a more concise way, but this is how I did it first time through):

$5x+2y = 2x+1$
$2x-3y = 3x+2$

move all variables to one side:
$3x+2y = 1$
$-x-3y = 2$

multiply the second equation by 3, so that upon addition, we [b]eliminate[/b] the x variable:

$3x+2y = 1$
$-3x-9y = 6$

add:

$-7y = 7$

solve:

$y =-1$

then you substitute that back in to one of the original equations to find x! Done and done.

btw, this would probably be more appropriate for the [url=http://www.artofproblemsolving.com/Forum/index.php?f=300]classroom math[/url] forum :wink:

welcome to AoPS slowlearner