geometry MATHCOUNTS AMC AIME USA(J)MO USAMO calculus

Grade: 8
Age: 13
Math Class Currently in: Geometry
Weakest Math Area: Anything Alg II and above
Favorite Math Area: Geometry
Chapter Mathcounts (tell ranking, score): Never did
State Mathcounts(same as above): Never did
National Mathcounts(same as above): Never did
AMC 10 Highest Score: n/a
AMC 12 Highest Score: n/a
AIME Highest Score: n/a
USAMO Highest Score: n/a
A Notable Achievement(s) so far this year (list): n/a? lol first in-state competition is in two weeks, and all I've done to this point are math leagues this year.

mine doesn't have much, but this is my 1st year ,and most of th e tournaments are next semester.

Grade: 9
Age: 14
Math Class Currently in: Math 24
Weakest Math Area: Anything > 24
Favorite Math Area: n/a
Chapter Mathcounts (tell ranking, score): 6th, 2nd, dunno scores
State Mathcounts(same as above): 8th, 3rd, dunno scores
National Mathcounts(same as above): 56th, 30
AMC 10 Highest Score: n/a
AMC 12 Highest Score: n/a
AIME Highest Score: n/a
USAMO Highest Score: n/a
A Notable Achievement(s) so far this year (list): Doing well in GNHML????

Grade: 11
Age: 15
Math Class Currently in: Physics C: Mechanics (bleh)
Weakest Math Area: Everything, right now.
Favorite Math Area: Calculus, probably.
I don't know what mathcounts is.
AMC 10 Highest Score: 112.5
AMC 12 Highest Score: 112.5 (ironic, isn't it?)
AIME Highest Score: I missed everything.
USAMO Highest Score: n/a
A Notable Achievement(s) so far this year (list): I managed to forget sine and cosine of sixty degrees... Negative, yet notable.

Doesn't this count as spam though?

Grade: 9
Age: prime
Math Class Currently in: not sure really
Weakest Math Area: any calc, trig
Favorite Math Area: Linear Alg
Chapter Mathcounts (tell ranking, score): Never did
State Mathcounts(same as above): Never did
National Mathcounts(same as above): Never did
AMC 10 Highest Score: n/a
AMC 12 Highest Score: n/a
AIME Highest Score: n/a
USAMO Highest Score: n/a

Grade: 9
Age: one less than the number of sides on a quindecagon
Math Class: APH (Precalc)
Weakest Math Area: Combinatorics/Functional Equations
Favorite Math Area: probably Number Theory
Chapter Mathcounts:17-18 (place), can't remember score
State/National Mathcounts:n/a
AMC 10: 106
AMC 12: N/A
ARML: 1  :blush:
Notable achievements this year: 22 for first round USAMTS
24/27 on recent local math league :)
Promising results on a practice AMC 12 I took this year (123 ish) :D
Mock AIME 3 for 2007: 4 officially, probably could have gotten 7-8

i hope i will qualify for ARML,.... someday  :blush:

Grade: 11
Age: 17
Math Class Currently in: BC calc (took stats last year)
Weakest Math Area: Geometry, inequalities
Favorite Math Area: Combinatorics
Chapter Mathcounts (tell ranking, score): 6th, 2nd, 3rd
State Mathcounts(same as above): 7th, 2nd, 3rd [our chapter is beastly]
National Mathcounts(same as above): did not qualify, 36th, 26th
AMC 10 Highest Score: 150
AMC 12 Highest Score: 129
AIME Highest Score: 11
USAMO Highest Score: 15
A Notable Achievement(s) so far this year (list): Our alpha [i]team[/i] got a perfect score on the Greater Boston Math League meet #2, perfect individual score on WPI, telling my teachers the wrong date that I was going to be out of class to take the MA Olympiad

Grade: 8
Age: 13
Current Math Class: Algebra II
Weakest Math Area: Non-trig geometry, complex numbers
Favorite Math Area: Combinatorics
Chapter MC: 12th, 2nd, TBD
State MC: 29th, 1st, TBD
National MC: Didn't qualify, 112th (don't ask), TBD
AMC 10 Highest Score (Practice): 150
AMC 12 Highest Score (Practice): 123
AIME Highest Score (Practice): 9
USAMO Highest Score: N/A

Other stuff...

2006 ARML, unofficially, 6
MAML Level 1: 109
Math League (Middle): 39,40,TBD
Math League (High): 4 on first one  :(
USAMTS Round 1: 16  :mad:

Grade: 9
Age: 14
Math Class Currently in: analysis(they just call it analysis. its really just precalc+calcA :dry: )
Weakest Math Area: combinatorics
Favorite Math Area: prolly number theory or trig
Chapter Mathcounts (tell ranking, score): my team got 2nd. Let's just say i sucked as an individual....
State Mathcounts(same as above): I....got brutally killed.
National Mathcounts(same as above): n/a
AMC 10 Highest Score: 124(hoping for a 150 this year)
AMC 12 Highest Score: n/a, hoping for 145+ this year
AIME Highest Score: 5, 8 on practice which should have been 10 if not for dumb@ss mistakes. aiming for 10+ this year
USAMO Highest Score: aiming for 10+....assuming i qualify
ARML:4 last year. Not going this year if I make MOP.
Mandelbrot:8 on nats this year(wow what a gay score.hoping to improve that)
santa clara math contest:I'm sure I got them all right(I mean 3hours for 7 trivial problems? wtf?). Not sure if my tiebreaker soln was elegant enough for a good ranking overall...
9th grade CML last year:5/5/5/6/5(yes its bad.(its out of 6 for each round))
math league: I pretty much didnt try on this one. Just took it for ec in math class.
USAMTS: didn't do. Might go for a medal next year. dunno.
iTest: don't remember exactly, but i think my team got 62.
A Notable Achievement(s) so far this year (list):
Hmm I managed to convince myself that $\tan\varphi=\frac{\cos\varphi}{\sin\varphi}$ on my school test. That killed my grade.
Um, 41/42 on w00t problem set?

Grade: 12
Age: 18 (The oldest person to post here, besides Kent Merryfield.)
Math Class Currently in: Analysis (Integrals) and Algebra (Groups, Rings, Fields) [in Romania everybody has to study the same thing].
Weakest Math Area: Geometry (especially solid geometry).
Favorite Math Area: Don't really have one. I'm mostly interested in new stuff.
Romanian Mathematical Olympiad: I reached the last phase in the 10th grade. I also participated at two TSTs that year.
Notable achievements so far this year (list): Keine.

Grade: 11
Age: 17
Math Class Currently in: Calculus this semester, AP Calc AB next
Weakest Math Area: Geo
Favorite Math Area: Huh??
Chapter Mathcounts (tell ranking, score): 3rd
State Mathcounts(same as above): 29th
National Mathcounts(same as above): i beat the winner that year
AMC 10 Highest Score: don't remember, it's been a while
AMC 12 Highest Score: 101.5  :blush:
AIME Highest Score: 2 lol it was so hard i fell asleep
USAMO Highest Score: same as National MC, beat the winner that year
A Notable Achievement(s) so far this year (list): being lazy and now i have B in calc, sleeping in all my classes, Governor's Honors Program: Math Major, Comm Arts minor :lol: , and totally transformed our school's sucky math team.

Grade: 1. semester university.
Age: 19.

Classes:
formally there (thus not really, occuring to write the tests ;) ): Linear Algebra 1, Algebra 1, Stochastics (partially there).
not really interesting (until now, hope they get better): representations of finite groups, logic 1, analytic number theory.
interesting: topology 1 (=algebraic topology).
Seminars: Expander Graphs, Central Simple Algebras and Galois Cohomology.

German Math. Olympiad: 2., 3., 3., 2. price.
German Math. Competition (BWM): 3., 3., 1., 1., 1. price (second/third round).
IMO: silver, bronce.

Weak: Geometry.
Good: Number Theory.
Dislike: Combinatorics, some types of Geometry.
Like: Number Theory, (Superior) Algebra, Complex Analysis.

Archievements: nothing.

Grade: 12
Age: 17
Math Class Currently in: None
Weakest Math Area: NT
Favorite Math Area: Combinatorics
Chapter Mathcounts (tell ranking, score): 1st
State Mathcounts(same as above): 6th
National Mathcounts(same as above): n/a
AMC 10 Highest Score: 150
AMC 12 Highest Score: 118
AIME Highest Score: 12
USAMO Highest Score: 10
A Notable Achievement(s) so far this year (list): Organizing teams for AHSIMC/PUMaC/HMMT

Grade: 7th
Age: 12
Math Class Currently in: Geometry
Weakest Math Area: Counting/Probability
Favorite Math Area: Algebra, Geometry
AMC 10 Highest Score: 136.5
AMC 12 Highest Score: 118.5
AIME Highest Score: 8, give or take 1 (not sure yet =P)
USAMO Highest Score: I wish...
Chatper Mathcounts: 1st
State Mathcounts: 7th (screwed up real bad this year =D)
A Notable Achievement(s) so far this year (list): 800 on SAT Math section, 2140 overall, w00t!!! Qualified for Texas ARML

Grade: 8
Age: 13
Math Class Currently in: Honors Algebra II
Weakest Math Area: Combinatorics, Geometry
Favorite Math Area: Algebra
Chapter Mathcounts (tell ranking, score): 1st, 4th
State Mathcounts(same as above):..not great  :(
National Mathcounts(same as above): Never did
AMC 10 Highest Score: 132
AMC 12 Highest Score: 120
AIME Highest Score: 6
USAMO Highest Score: n/a
A Notable Achievement(s) so far this year (list): nm

Grade: 8
Age: 13
Math Class Currently in: N/A - study-hall independant study, so around BC calc
Weakest Math Area: Geometry
Favorite Math Area: ?
Chapter Mathcounts: 6th grade: 31, no idea of place
7th: 46, 1st written, 1st countdown
8th: 45, 1st written 1st countdown

State Mathcounts: 6th - N/A
7th - 3rd written 4th countdown  39 points
8th - 1st written 2nd countdown can't tell score yet

National Mathcounts: Lets say I got 3 on the target in 7th grade so I got 70th.

AMC 10 Highest Score: 137 (7th grade AMC10A 2006)
AMC 12 Highest Score: 141 (8th grade AMC12B 2007)
AIME Highest Score: 11 (8th grade AIME I 2007)
USAMO Highest Score: N/A (coming soon this year, probably 0)

A Notable Achievement(s) so far this year (list): that AIME 11 is fun.

[quote="Kent Merryfield"]Age: 53 (OK, it's been a long time since I was a student.)

Current Math Class: this semester I'm teaching Calculus II, Probability (the calculus-prerequisite version), and "Honors Seminar in Problem Solving" (that is, Putnam preparation.) Next semester it will be Calculus III and Linear Algebra.

Highest AHSME score: ~88, in 1970 (the scoring system was different). 3-time school AHSME leader.

AIME: didn't exist. (Had it existed, I would probably have been at least a 2-time qualifier. But there's no way I would have gotten anywhere close to the USAMO had it existed.)

Highest Putnam score: Positive. I don't remember any more detail than that.

Strongest math area: analysis. (Hey, it's what my Ph.D. is in.)[/quote]
Kent, I think following assertion is true :

''Age of for every member of Mathlinks.ro ( not is Kent) small than age of Kent'' :P

Grade: 11
Age: 15
Math Class Currently in: Geometry
Weakest Math Area: Combinatorics
Favorite Math Area: Number Theory
Chapter Mathcounts: 6th grade - 3rd, 7th grade - 1st, 8th grade - 1st
State Mathcounts: 6th grade - 13.5th, 7th grade - 4th, 8th grade - 3rd
National Mathcounts: 7th grade - 64th, 8th grade - 6th
AMC 10 Highest Score: 117.0
AMC 12 Highest Score: 144.0
AIME Highest Score: 16
USAMO Highest Score: 22
A Notable Achievement(s) so far this year (list): none

Um, a bunch of Alex's stuff is untruths. Like... 6th grade, 10th place in States. He was in PA, so I know.

Alex, why did you leave PA? Sigh...

And a 22 on the USAMO? I'm pretty sure he got higher than that. Didn't you? After all, you were a USAMO winner.

And no special achievements. That's funny.

Grade: 7
Age: 12
Current math class: pre-algebra honors
weakest math area: geometry
strongest math area: linear equations
6th grade: 18th chapter(no state)
7th grade: 3rd school, 4th chapter, 34th state
Highest amc 8 score : 21(first and only time)
Highest amc 10 score: 117(i suck at math)
Highest amc 12 score: n/a
Highest AIME score: 2 :blush:
Highest USAMO score: n/a

Grade: 6
Age:11
Current math class: Algebra 1
weakest math area: geometry, probability
strongest math area: Addition (and  linear equations in 2 variables)
6th grade: 8th chapter, 22nd State
Highest Amc 8 score : 23(This year)
Highest Amc 10 score: 109.5(Awful, but highest 6th grade score in MI)
Highest Amc 12 score: n/a
Highest AIME score: n/a
Highest USAMO score: I wish...
Notable Achievement: Perfect 40/40 on 6th Grade MML.

Updated:
Grade: 9
Age: 14
Math Class Currently in: Precalculus Honors
Weakest Math Area: Geometry.
Favorite Math Area: Algebra/Combinatorics
Chapter Mathcounts: 4th (41 in 7th grade)
State Mathcounts: 20-something-th. (7th grade; I sucked in 8th)
National Mathcounts: Nope.
AMC 10 Highest Score: 138 (8th)
AMC 12 Highest Score: 112.5 (9th)
AIME Highest Score: [b]5[/b] (9th)
USAMO Highest Score: Nope.
A Notable Achievement(s) so far this year (list):
5x better on the AIME this year? And not reading #2 on the AIME I carefully.

Grade: 7
Age: 11
Math class currently in: Interm. Algebra + Num. Theory
weakest math area: combinatorics
strongest math area: I don't know whether to choose geometry, algebra, or number theory...
Chapter Mathcounts highest ranking/score: 2nd/41 this year
State Mathcounts highest ranking/score: 2nd/36 this year
Highest AMC 8 score: 21 this year
Highest AMC 10 score: Won't take it in my lifetime
Highest AMC 12 score: 111
Highest AIME score: 3 :(
Notable Achivements this year: Got first state mathcounts countdown(yay), highest scoring middleschooler on AMC 12 A in the state this year, highest scoring 7th grader on AMC 12 B in the State this year

Grade: 11
Age: 15
Math Class Currently in: Learning more geometry
weakest math area: Calc
favorite math area: Combinatorics :D
Achievements:
Silver medal in the "Olimpiada de Mayo" 2009, Level 2
Best in my class
Notable Achievement so far this year: I'm going to participate in the XV Centroamerican an Caribbean Math Olympiad, Puerto Rico 2010 :D (I'll try to be the best ;))

... I know it's not a lot, but is all I've got so far, but i'm trying to be better every year :)

Please, Please, don't revive old threads.
Just a note.

This is a REALLY old thread. Geez.

Grade: 9 / 12 (9th for English and Gym, 12th for all the rest)
Age: 13

Math Class Currently In: In school - Pre-Calc Honors (bleh...I just moved....)
                                        Out of school - Writing a thesis and working with a univ. professor
Weakest in: Idk.
Favorite: Calc, algebra, comb.
Achievements: Top-scoring female AMC'er in tri-state area
USAMO qualifier :)
6-time national champ in Continental Math League

Grade: 7
Age: 12
Math Class Currently in: Algebra 1
Weakest Math Area: Probability (more specifically, Combinatorics)
Favorite Math Area: Geometry
AMC 10 Highest Score:  [Taking next year]
AMC 12 Highest Score: [Taking next year]
AIME Highest Score: [N/A]
USAMO Highest Score: [N/A]
A Notable Achievement(s) so far this year (list): Haha, don't make me laugh.  :D I did more last year than this year...which is kinda sad... :maybe:


function
real analysis
real analysis unsolved

If $f(x)$ and $g(x)$ are non-differentiable functions then show that $f(x)g(x)$ can be a differentiable function.

[quote="Kent Merryfield"]Are we talking about a single point? Here's a cheap example for one point: $f(x)=g(x)=|x|.$

If you want more than that, I'd suggest playing around with lacunary trigonometric series, although I haven't looked hard enough to see a clear example yet.[/quote]

Actually I am talking about on $\mathbb{R}$.

Denote $f$ for [url=http://en.wikipedia.org/wiki/Weierstrass_function]this[/url]. Take some $E \subset R^{1}$. Multiply $f$ once with $\chi_{E}$ and once with $\chi_{E^{c}}$. Then the product is the zero function.

But this is not original :)

You don't need Weierstrass or trigonometric series. Set $f(x)=2$ for rationals and $f(x)=1/2$ for irrationals and $g(x)=1/2$ for rationals and $g(x)=2$ for irrationals.

Here is my tricky answer  :rotfl:

[hide]
Let $f(x)$ be the Dirichlet function defined being 1 for rational and 0 for irrational.
Let $g(s)$ be the Dirichlet function defined being 1 for irrational and 1 for rational.

Now this also shows that $f(x)+g(x)$ is differenciable. Note only $f(x)g(x)$.
[/hide]

EDIT: puuhikki is similar.


Let be $ a,x_1>1$ .

a)Prove that $ \prod_{cyc}\sqrt[n]{\mathrm{log}_a\ x_1\plus{}\frac{\mathrm{log}_a x_2}{\mathrm{log}_a^2x_3}}\ge 2$

b)If $ x_1x_2...x_n\equal{}a$ show that $ \sum_{cyc}\sqrt[3]{\mathrm{log}_ax_1\cdot \mathrm{log}_ax_2\cdot \mathrm{log}_ax_3}\le 1$

[quote="Pain rinnegan"]Let be $ a,x_1 > 1$ .

a)Prove that $ \prod_{cyc}\sqrt [n]{\mathrm{log}_a\ x_1 \plus{} \frac {\mathrm{log}_a x_2}{\mathrm{log}_a^2x_3}}\ge 2$

b)If $ x_1x_2...x_n \equal{} a$ show that $ \sum_{cyc}\sqrt [3]{\mathrm{log}_ax_1\cdot \mathrm{log}_ax_2\cdot \mathrm{log}_ax_3}\le 1$[/quote]

I think you mean $ a,x_i >1$ or $ b)$ can't hold.

If so, then

a)Let $ \mathrm{log}_{a}x_{i}\equal{}k_i$,
$ L.H.S.\equal{}\prod_{cyc}\sqrt[n]{k_1\plus{}\frac{k_2}{(k_3)^2}} \ge \prod_{cyc}\sqrt[n]{2 \sqrt{\frac{k_1k_2}{(k_3)^2}}}\equal{}2$

b)$ k_1\plus{}k_2\plus{}k_3\plus{}....\plus{}k_n\equal{}1$, so

$ L.H.S.\equal{}\sum_{cyc}\sqrt[3]{k_1 k_2 k_3} \le \sum_{cyc} \frac{k_1\plus{}k_2\plus{}k_3}{3}\equal{}1$


Will the score come out before AIME 2?  Just curious. It is the first time I take AIME 1 (I took AIME 2 last year).

I am afraid that AIME1 scores will be a little late.

Last year, they came out after they announced who passed the AIME.  They wanted to get all the AIME scores in (I and II) before announcing.  I do not know if this is the same case this year.  If you want, I will ask my teacher tomorrow to see if she knows.


[hide="1"]
can be rewritten as:

$(xy - 1)^2 - (x + y)^2$

though I don't know if that helps. I just expanded and rearranged.
[/hide]

Number 1[hide](xy+x+y-1)(xy-x-y-1)[/hide]

Number 2[hide](xy-1)(xy+x+y+1)[/hide]

Number 3[hide][(a+1)x+1](ax^2+x-1)[/hide]

[quote="jhcreinhardt"]
Number 2[hide](xy-1)(xy+x+y+1)[/hide][/quote]

Note that this can be further factored $(xy - 1)(x + 1)(y + 1)$.

[hide="3"]
$a^2x^3+ax^3+2ax^2+x^2-ax-1$
$a^2x^3+ax(x^2+2x-1)+(x^2-1)$
Apply quadratic formula to get answer
[/hide]


[quote="opple"]does anyone have the result for this year's CAP competition?[/quote]
What's CAP?

Physics equivalent of COMC.  But of course, it's not really equivalent, since it's not as prestigious as the math contests  ;)

Here're some results for 2005: http://www.physics.umanitoba.ca/uber_news/National_Ranking.html

Actually, if a comparison were to be made, then CAP is greater than COMC: CAP has good money prizes, and it plays a crucial role in determining which 15 people go to National Olympiad Finals (NOF). And it's a lot harder too - the top score this year is under 70%.

Also, note that Peng isn't actually in my school; he just wrote it there.

[url=http://www.cap.ca/edu/HSPrize/2005HSresults.html#New%20Brunswick:]Official Contest Result[/url] is out


special factorizations
number theory unsolved
number theory

$ (a)$ Find all naturals $ n$ such that $ n^2 \plus{} 2008n$ is a perfect square.
$ (b)$ If $ x_1,x_2,...,x_t$ ar all the solutions to the equation at point $ (a)$, then find the residue of $ x_1 \plus{} x_2 \plus{} ... \plus{} x_t$ when dividing by $ 2008$.

Part a can be factored by completing the square:
(n+1004)^2-k^2=1004^2
(n+k+1004)(n-k+1004)=1004^2
The sum of the factors is even so they are both even. Giving a two to each, we are left with $ 5002^2\equal{}2^2 \cdot 2501^2$.
The left factor is also greater than the right. So we try out the pairs
$ (2(2501^2)(4),2(1))
(2(2501^2)(2),2(2))
(2(2501^2),2(4))
(2(2501)(2^2),2(2501))$
Solving all four systems, we get:
(25019001,25020003)
and three others, which I don't feel like finding.


AMC AIME videos USA(J)MO USAMO MATHCOUNTS

[quote="PenguinIntegral"][/quote][quote="calc rulz"][quote="PenguinIntegral"][quote="MysticTerminator"]uh well ok so it wasn't like absolute pwnage (only 3 people did that), but like 1 out of the 2 American gold medals isn't bad

and yeah, they require different skill sets (I've found Olympiad geo and AIME/HMMT geo to be pretty different)

but you're in like 8th grade, so you'll live[/quote]
Ok so let me give you a scenario:

You're in 8th grade. You have basically (in terms of learning material and techniques) mastered all pf AoPs 2 except for the some of the obscure Geometry/Trig/other obscure stuff you don't care about. You can get double digits on the AIME if you don't make stupid errors like assuming 3*8 is 22. It is the start of September. You wish to pwn the USAMO. You have access to every Iron Maiden song and "The Muppets" video ever made, in addition some common materials (CPIG, kalva, the USAMO book, AoPS2, PSS, and Acops, though you think Acops sucks) and could aquire more if they didn't cost like a gazillion dollars.

What would be your strategy for maximizing USAMO pwnage that doesn't involve kidnapping Richard Rusczyk?[/quote]

WOOT, definitely.

And of course, practice as many olympiad problems as you can.

You may also want to check out PSS, I've heard it's good.[/quote][quote="PenguinIntegral"]
I think not costing a gazillion dollars was one of the criteria.[/quote]

Uh, get a job?

[quote="MysticTerminator"][quote="PenguinIntegral"][quote="MysticTerminator"]uh well ok so it wasn't like absolute pwnage (only 3 people did that), but like 1 out of the 2 American gold medals isn't bad

and yeah, they require different skill sets (I've found Olympiad geo and AIME/HMMT geo to be pretty different)

but you're in like 8th grade, so you'll live[/quote]
Ok so let me give you a scenario:

You're in 8th grade. You have basically (in terms of learning material and techniques) mastered all pf AoPs 2 except for the some of the obscure Geometry/Trig/other obscure stuff you don't care about. You can get double digits on the AIME if you don't make stupid errors like assuming 3*8 is 22. It is the start of September. You wish to pwn the USAMO. You have access to every Iron Maiden song and "The Muppets" video ever made, in addition some common materials (CPIG, kalva, the USAMO book, AoPS2, PSS, and Acops, though you think Acops sucks) and could aquire more if they didn't cost like a gazillion dollars.

What would be your strategy for maximizing USAMO pwnage that doesn't involve kidnapping Richard Rusczyk?[/quote]

ok well I think in your middle school so first off definitely don't fall into the MATHCOUNTS TRAP because it is silly and will just waste your time
[/quote]
LIES!!!! The extra .213 second I shave off my countdown time will help me throughout my entire IMO career. You crazy man

[quote="MysticTerminator"]
then work olympiad problems
[/quote]
Like which ones ? Cause like the Chinese ones are like 10x harder than the IMO itself

And my little brother could do the first couple Canada MOs. And that doesn't say much for my little brother
[quote="MysticTerminator"]
books will be helpful at first (work through engel)
[/quote]
What books? Like "Run to the Hills: Iron Maiden, the Authorized Biography"? Because we all know Iron Maiden rocks

[quote="Iversonfan2007"][quote="MysticTerminator"]

ok well I think in your middle school so first off definitely don't fall into the MATHCOUNTS TRAP because it is silly and will just waste your time
[/quote]

sorry i don't understand this please explain! :rotfl:  :P  :wink:[/quote]

[quote="Me"]
LIES!!!! The extra .213 second I shave off my countdown time will help me throughout my entire IMO career. [/quote]

[quote="bpms"][quote="Iversonfan2007"][quote="MysticTerminator"]

ok well I think in your middle school so first off definitely don't fall into the MATHCOUNTS TRAP because it is silly and will just waste your time
[/quote]

sorry i don't understand this please explain! :rotfl:  :P  :wink:[/quote]
Excuse me what?[/quote]

inside joke :wink:

[quote="Iversonfan2007"][/quote][quote="bpms"][quote="Iversonfan2007"][quote="MysticTerminator"]

ok well I think in your middle school so first off definitely don't fall into the MATHCOUNTS TRAP because it is silly and will just waste your time
[/quote]

sorry i don't understand this please explain! :rotfl:  :P  :wink:[/quote]
Excuse me what?[/quote][quote="Iversonfan2007"]

inside joke :wink:[/quote]
I was reffering to his comment. Is that the inside joke or is your comment? :confused:

[quote="MysticTerminator"]what no mathcounts is not an inside joke

it is a vile competition that wastes time of those who could be doing much better things[/quote]
I know what mathcounts is and I would appreciate it GREATLY if you didn't insult it. Mathcounts is my favorite math competition.

[quote="chess64"][quote="Karth"]Since olymipad problems take quite a bit of time to do, how should I do these? Depending upon how much time I have, how many problems should I do per day? On an average, how much time does an olympiad problem take to do?[/quote]

:|

did you not just read the post above yours...

:huuh:[/quote]

Yes, obviously time varies... but on an AVERAGE, how much time should we spend to solve it? 1 hour? 1 day?  :)


Two noncongruent triangles have the same area but different perimeters. Triangle A has sides lengths of 8, 8 and 6 cm. Triangle B has two sides with length of 8 cm. Find the length of the third side length of Triangle B. Round your answer to the nearest hundredth of a cm.

somehow i keep getting wrong answers... please explain

you posted it [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=113012]here[/url] too

oh ok, i'll let these posts die,... i thougth i didn't post it/..


Hey slamchess havent seen u in a while.

im sure the above hint helped u a lot.

[hide] the formula for this is: total number = a + b + c - ab - ac - bc + abc + n, where a is the number of people that take the first class, b is the number that take the second class, c is the number that take the 3rd class, ab is the number that take the first and second class, ac is the number that take the 3rd and first class, bc is the number that take the 2nd and 3rd class, abc is the number that take all three, and n is the number that take none.

[u]note[/u]: the #a includes ALL the people that take the first class -- even if they take other classes. same for b, c, ab, ac, and bc.

For this problem:

400 = a + b + c - ab - ac - bc + abc + n
400 = 117 + 109 + 114 - 70 - 34 -29 + abc + 164.

Add, subtract, etc. and you get $29$ [/hide]

[quote="Naga"][quote="slamchess"]I have never figured out how to do these type of problems[/quote]

use Venn diagrams.[/quote]

Definitely.  Try working from the inside out and use a variable if necessary.

[quote="Naga"][quote="slamchess"]I have never figured out how to do these type of problems[/quote]

use Venn diagrams.[/quote]

I dare you to make a Venn Diagram with four variables.  :P

At the competition, do not make a Venn Diagram.  However, you can use a Venn Diagram to see why Fanatic did what he did.  At the contest, since Sprint Round is mostly about how quick you can do a problem, you should use Fanatic's example of PIE

[quote="vishalarul"]I dare you to make a Venn Diagram with four variables.  :P[/quote]
Yeah, I've always wondered how one of those would look.

http://www.combinatorics.org/Surveys/ds5/VennWhatEJC.html

Just google it, you will find lots


Consider the integer sequence defined by the recurrence relation

Z(n) = Z(n-1)*Z(n-2) +1

where Z(1) = Z(2) = 1

The first few terms are

1, 1, 2, 3, 7 ...

Prove that Z(n) | Z(mn)

 Z(n) means Z sub n, not z*n

Spoiler:

[hide] if you want the solution, go to my website and go to the update of 02/10/04

[/hide]

a | b means a divides b
so
Prove that:
Z(mn)=K *Z(n) where K is an integer.

Okay, here we go:

Consider the list {z(1), z(2), ...} mod z(n).  We get {1, 1, 2, 3, ..., 0, ...} where 0 is the nth term.  (This is because z(n) :equiv: 0 (mod z(n)).)

Now, z(n + 1) = z(n)*z(n - 1) + 1 :equiv: 0*z(n - 1) + 1 :equiv: 1 (mod z(n)).
z(n + 2) = z(n + 1)*z(n) + 1 :equiv: 1*0 + 1 :equiv: 1 (mod z(n)).

Thus, z(n + 1) :equiv: z(n + 2) :equiv: 1 (mod z(n)).

However, our sequence is uniquely determined by any two consecutive values.  We already know what the sequence following two consecutive values of 1 looks like -- and it has a zero in the nth place.  Or in other words, mod z(n), our sequence is periodic of period n.  Thus, z(mn) :equiv: 0 (mod z(n)), and z(n) | z(mn) for all positive integers m.


To measure the height of a mountain, Peter takes two sightings of the peak at a distance of 900 meters apart on a direct line to the mountain. The first observation results in an angle of elevation of 47 degrees and the second results in an angle of elevation of 35 degrees. If the transit is 2 meters high, what is the height h of the mountain?

[hide]
Use basic trigonometry. You know, that thing where the function of an angle equals the ratio of sides in a triangle. You might want to read up on that.
[/hide]

Do the sightings happen on the same side of the mountain? (could be only on one side or one on each side.)


[quote="The Sigma"]I currently both Volume 1 and 2 of the Art of Problem solving books and all the introductory books, and let me tell you that I am very pleased with the books. I have two questions:
Does anyone know of a good calculus book that teaches as well as that of the original AOPS books?
Also, does anyone know of any physics books that train for the physics olympiads just as the AOPS texts train for the math competitions?
Anyones input would be greatly appreciated.[/quote]
here a topic on physics resources, hope you find it helpful: :)
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=books+physics&t=33967


Any clinicians out there...?

The QT interval of an ECG is known to lengthen when plasma postassium falls. This is due to delayed repolarization. Repolarization is an event driven by voltage gated potassium channels. But if plasma potassium is lower then the electro-chemical gradient for postassium flowing out of the cardiac myocyte will be higher. Surely then the cell will repolarize faster and the QT interval will be shortened, not lengthened!!

I have googled this to death, but no clarification. Can anyone help? Please?

Thanks,

Terence

This isn't my area of study, but a quick pubmed search yielded a few articles that look interesting.  (For future reference, you might want to use http://www.pubmed.com for bio related literature searches as opposed to a generic search engine like google).

You might try this guy: http://www.ncbi.nlm.nih.gov/entrez/query.fcgi?cmd=Retrieve&db=pubmed&dopt=Abstract&list_uids=16101630&query_hl=1&itool=pubmed_docsum
Action potential duration and QT interval during pinacidil infusion in isolated rabbit hearts.

Or this one http://www.jbc.org/cgi/content/full/278/12/10417
 Impairment of Human Ether-à-Go-Go-related Gene (HERG) K+ Channel Function by Hypoglycemia and Hyperglycemia

Or this http://circres.ahajournals.org/cgi/content/full/89/1/33
Molecular Interactions Between Two Long-QT Syndrome Gene Products, HERG and KCNE2, Rationalized by In Vitro and In Silico Analysis

Hope that's somewhat helpful.


superior algebra
superior algebra unsolved

Calculate the order of the following orthogonal groups over finite field |O(3, F5)|, |O(3, F7)|

I found a formula in wiki, http://en.wikipedia.org/wiki/Orthogonal_group, but I don't understand how to proof that formula.

We had this question at http://www.mathlinks.ro/Forum/viewtopic.php?t=180202 , and I'll post at least a sketch of a proof as soon as I succeed to find one. As for now, I just know that it is enough to find the number of invertible symmetric $ n\times n$ matrices over the field, but this doesn't seem to be easier.

  darij


Let $ (O_1), (O_2), (O_3)$ be three circles one by one outside of each other. Let $ t,t'$ be the inner common tangents of $ (O_1)$ and $ (O_2)$, $ p,p'$ be the inner common tangents of $ (O_2)$ and $ (O_3)$, $ q,q'$ be the inner common tangents of $ (O_3)$ and $ (O_1)$. Prove that $ t,p,q$ concurs if and only if $ t',p',q'$ concurs.

Dear Mathlinkers,
this result comes from Mannhein.
Any reference?
Sincerely
Jean-Louis


function
modular arithmetic
algebra unsolved
algebra

[quote="cnyd"]Prove that there is no bijective function,$ f: \mathbb{N}\mapsto\mathbb{N}_{0}$
$ f(mn) \equal{} f(m) \plus{} f(n) \plus{} 3f(m)f(n)$[/quote]
Assume that such bijective function $ f$ exists.

Let $ g(x)\equal{}3f(x)\plus{}1$, then we have $ g(mn)\equal{}g(m)g(n)$, and clearly $ g(1)\equal{}1$.

Note that $ g$ is injective and for all $ k \in \mathbb{N}_{0}$ we can find $ n$ so that $ g(n)\equal{}3k\plus{}1$

And as $ 10\equal{}2 \times 5$ and $ 2,5 \neq 1 \pmod{3}$ , we know $ 10\equal{}g(p_1)$ for some prime $ p_1$ .
Similarly $ 187\equal{}g(p_2) ,  22\equal{}g(p_3) , 85\equal{}g(p_4)$ for some primes $ p_2,p_3,p_4$.

Now $ g(p_1p_2)\equal{}10 \times 187\equal{}22 \times 85\equal{}g(p_3p_4)$ , contradiction as $ g$ is injective and $ p_1,p_2,p_3,p_4$ pairwisely different.


advanced fields
advanced fields unsolved

I am having trouble understanding Cauchy sequences.
Can someone explain them and give a couple of concrete examples?

$(\forall\varepsilon > 0) (\exists N\in N) / (\forall(n,p)\in N^{2}) (N\leq n \leq p \Rightarrow \vert u_n-u_p\vert \leq \varepsilon )$
$\sum_{k=0}^{n}\frac{1}{k!}$ is a cauchy sequence


16.) The Rangers won 5 of their first 6 games. How many of their next 30 games must the Rangers win to have twice as many wins as losses?

Note: Please give the fastest/most reasonable way to do the problem.

[hide]
They've played 6 games already and are going to play 30 more. That means 36 games total. To have twice as many wins as losses, they need to win 24 of them. Since they've already won 5, they only need to win 19 of their next 30 games which is your answer.[/hide]

whew.....i thought i was wrong... :D  thanks

[hide]This is probably a slow and bad idea, but....

Well, lets say that the Rangers won 1 and lost one of their next 3 games.  They would have a 6-3 record.  This gives them 27 games left.  Then, for each of the next games, they have to win 2 for every three games to win twice as much and keep that average.  $\displaystyle{\27\cdot\frac{2}{3}=18}$  then, $\displaystyle{\118+1=\boxed{\119}}$[/hide]

[hide]They play 36 games in total, and out of those they need to win 24 times total. They've already won 5 times, so they only need to win 24-5=19 more games.[/hide]


quadratics function algebra polynomial trigonometry

Suppose that $P(x)$ is a quadratic function (polynomial of degree two) such that $P(0)=\cos^{3}{40^\circ}, P(1)=(\cos{40^\circ})(\sin^{2}{40^\circ})$, and $P(2)=0$. Evaluate $P(3)$ numerically.

Alternatively, since it is quadratic we can use differences: $P(0)-P(1)=\Delta_{1}, P(1)-P(2)= \Delta_{2}, \Delta_{1}-\Delta_{2}=K$ where K is a constant. $\Delta_{1}=2\cos^{3}40^\circ-\cos 40^\circ, \Delta_{2}=\cos 40^\circ-\cos^{3}40^\circ$
$\Delta_{1}-\Delta_{2}=3\cos^{3}40^\circ-2\cos 40^\circ=\Delta_{2}-\Delta_{3}\implies \Delta_{3}=3\cos40^\circ-4\cos^{3}40^\circ$
$\Delta_{3}=P(2)-P(3)=-P(3) \implies P(3)$
$=4\cos^{3}40^\circ-3\cos40^\circ=\cos 3*40^\circ=\cos 120^\circ=-\frac12$


Find all sets A of positive integers, A has at least two elements such that if a,b belong to A, a<b then a^2/(b-a) belong to A, too.  :D

The method of the extremal element works wonders. Take $ a \equal{} \min A$; then for any $ b \in A \setminus \{a\}$ we need $ a^2/(b \minus{} a) \geq a$, so $ a < b \leq 2a$.
So now we also need $ a^2/(b \minus{} a) \leq 2a$, so $ 3a \leq 2b$.

Asume $ A$ contains three elements $ a < b < c$. Then $ b^2 > a^2$ and $ c \minus{} b \leq 2a \minus{} 3a/2 \equal{} a/2$, so $ b^2/(c \minus{} b) > a^2/(a/2) \equal{} 2a$, contradiction.
Therefore $ A$ contains exactly two elements $ a < b$, so $ a^2/(b \minus{} a) \equal{} a$, i.e. $ b \equal{} 2a$,
or $ a^2/(b \minus{} a) \equal{} b$, i.e. $ b^2 \minus{} ab \minus{} a^2 \equal{} 0$, but this yields $ b \equal{} a(1 \pm \sqrt {5})/2 \not \in \mathbb{Z}$.

Finally, the only such sets are $ A \equal{} \{a,2a\}$, for any positive integer $ a$.

see it here:(baltic way 2008)
http://www.mathlinks.ro/viewtopic.php?p=1327937#1327937


Let $ x_1,x_2,\cdots,x_n$ be positive numbers. Prove that,
\[ \sum_{i \equal{} 1}^n \frac {x_i^2}{x_i^2 \plus{} x_{i \plus{} 1}x_{i \plus{} 2}} \leq n \minus{} 1
\]
[Note that, $ n \plus{} i\equiv i$]

[i]Edit: Sorry, it is ISL 1985/18, not 19[/i]

[quote="Moonmathpi496"]Let $ x_1,x_2,\cdots,x_n$ be positive numbers. Prove that,
\[ \sum_{i \equal{} 1}^n \frac {x_i^2}{x_i^2 \plus{} x_{i \plus{} 1}x_{i \plus{} 2}} \leq n \minus{} 1
\]
[Note that,$ n \plus{} 1\equiv1$ and $ n \plus{} 2\equiv 2$][/quote]
If $ n\equal{}2$, $ \sum_{i \equal{} 1}^n \frac {x_i^2}{x_i^2 \plus{} x_{i \plus{} 1}x_{i \plus{} 2}} \equal{}1\le 1$. When $ n\ge3$,
$ \Leftrightarrow \sum_{i\equal{}1}^n\frac{x_{i\plus{}1}x_{i\plus{}2}}{x_i^2 \plus{} x_{i \plus{} 1}x_{i \plus{} 2}}\ge 1$
$ \Leftrightarrow \sum_{i\equal{}1}^n\frac{1}{1\plus{}\frac{x_i^2}{x_{i\plus{}1}x_{i\plus{}2}}}\ge1$
Let $ \frac{x_i^2}{x_{i\plus{}1}x_{i\plus{}2}}\equal{}p_i\Rightarrow p_1 p_2 \cdots p_n\equal{}1$
we can assume $ p_i\equal{}\frac{q_i}{q_i\plus{}1},q_{n\plus{}1}\equal{}q_1$, We need to prove
$ \sum_{i\equal{}1}^n\frac{q_i}{q_i\plus{}q_{i\plus{}1}}\ge1$, by cauchy, we have
$ (\sum_{i\equal{}1}^n\frac{q_i}{q_i\plus{}q_{i\plus{}1}})\ge \frac{(\sum q_i)^2}{\sum_{i\equal{}1}^n q_i(q_i\plus{}q_{i\plus{}1})}\ge1$
Last inequality is obvious. Equation is only true when $ n\equal{}2$.

[quote="math_07"][quote="NextPeace"]When $ n\ge3$,
$ \Leftrightarrow \sum_{i \equal{} 1}^n\frac {x_{i \plus{} 1}x_{i \plus{} 2}}{x_i^2 \plus{} x_{i \plus{} 1}x_{i \plus{} 2}}\ge 1$[/quote]
Sorry for my stupid question, but how did you get that? :ninja:  (Could you please explain?)[/quote]

$ \sum_{i \equal{} 1}^n \frac {x_i^2}{x_i^2 \plus{} x_{i \plus{} 1}x_{i \plus{} 2}} \leq n \minus{} 1$
$ \Leftrightarrow \sum_{i \equal{} 1}^n (\frac {x_i^2}{x_i^2 \plus{} x_{i \plus{} 1}x_{i \plus{} 2}}\minus{}1) \leq \minus{}1$
$ \Leftrightarrow \sum_{i \equal{} 1}^n \minus{}\frac {x_{i \plus{} 1}x_{i \plus{} 2}}{x_i^2 \plus{} x_{i \plus{} 1}x_{i \plus{} 2}}\leq \minus{}1$
$ \Leftrightarrow \sum_{i \equal{} 1}^n\frac {x_{i \plus{} 1}x_{i \plus{} 2}}{x_i^2 \plus{} x_{i \plus{} 1}x_{i \plus{} 2}}\ge 1$


Is it possible to keep the "view your posts" link in the main page of mathlinks new mainpage, otherwise one
has to go through another page. Am I right or is something missing?
Please reply.

Well, the point is to eventually configure the whole of MathLinks to that layout, but are you saying you want that link to be in the layout?

Yes, or at least something similar that is accessible.

Noted (it will be present). However remember that this is a beta layout so some stuff will change until released.


algebra
polynomial
real analysis
real analysis unsolved

hi,
consider the sequence of polynoms $ (P_{n})_{n \geq 0}$ with
\[ P_{n \plus{} 1}(x) \equal{} \minus{} 2xP'_{n}(x) \plus{} P_{n}(x), P_{0}(x) \equal{} 1.
\]
Find $ P_{n}(0)$.

From the relation given:

$ P_{n\plus{}1}(0) \equal{} \minus{}2 \cdot 0 \cdot P_n'(0) \plus{} P_n(0) \equal{} P_n(0)$. From $ P_0(x) \equal{} 1$ we get $ P_0(0) \equal{} 1$, so $ \boxed{\forall n: P_n(0) \equal{} 1}$.

Somehow it doesn't seem like this is what you meant.


geometry
circumcircle
geometric transformation
homothety
cyclic quadrilateral
geometry solved

Hi all,

   In the following configuration:

      3 circles of equal radii touching each the other two,  the 4th big circle
       touches all 3 internally,

   prove that  tangents from a point on the big circle to the 3 equal
   circles are  tC1 +tC2 = tC3.

   Regards,

   Pestich.

The three smaller circles need not touch each other: it's enough for them to have equal radii. We use Casey's theorem on the system formed by the three small circles and the point chosen on the large circle and we're done. It's perfectly analogous to the proof of the fact that if $X$ is a point on the arc $BC$ which doesn't contain $A$ of the circumcircle of the equilateral triangle $ABC$, then $XA=XB+XC$.

I had the idea with using Casey's theorem, too, but actually Casey's theorem is quite a hard-to-prove result (BTW I have not seen any proof of the converse - it should be rather difficult). Here is a solution of Pestich's problem without Casey.

As always, I must rename everything:

[i][b]Problem.[/b] Consider three "little" circles a, b, c with centers A, B, C and equal radii r externally tangent to each other. Let a "big" circle p with center O touch all three circles a, b, c internally. Let P be a point on the circle p. One of the two tangents from P to the circle a touches a at the point X; similarly define the points Y and Z by drawing tangents from P to the circles b and c. Then prove that one of the lengths PX, PY, PZ equals the sum of the two others.[/i]

So the solution goes:

Since BC = r + r = 2r, and similarly CA = 2r and AB = 2r, the triangle ABC is equilateral. Let the circle p touch the circles a, b, c at A', B', C'. Then we have AA' = r, BB' = r, CC' = r, OA' = R, OB' = R, OC' = R, where R is the radius of p. Now, if we call $\displaystyle k=\frac{R-r}{R}$, then we have $\displaystyle k=\frac{OA}{OA^{\prime }}=\frac{OB}{OB^{\prime }}=\frac{OC}{OC^{\prime }}$. [Honestly, $k=4-2\sqrt{3}$, but this won't bother us.] Therefore, the triangle ABC is the image of the triangle A'B'C' in the homothety with center O and factor k. Since the triangle ABC is equilateral, its preimage A'B'C' must be equilateral, too. This means B'C' = C'A' = A'B'. Now, the point P lies on the circumcircle of this triangle A'B'C'; assume, without loss of generality, that the point P lies on the arc C'A'. Then, the quadrilateral A'B'C'P is a convex cyclic quadrilateral, and by the Ptolemy theorem we have $PC^{\prime }\cdot A^{\prime }B^{\prime }+PA^{\prime }\cdot B^{\prime }C^{\prime }=PB^{\prime }\cdot C^{\prime }A^{\prime }$. Dividing by B'C' = C'A' = A'B', we get PC' + PA' = PB'.

If the line PA' meets the circle a at the point A" (apart from A'), then, by the theorem on the intersecting secant and tangent, we have $PX^{2}=PA^{\prime }\cdot PA^{\prime \prime }$. But since the point A' is the point of tangency of the circles a and p, there exists a homothety with center A' mapping the circle a to the circle p; this homothety maps the center A of a to the center O of p, and this homothety maps the point A" lying on a to a point (different from A') lying on p and also lying on the line A'A", i. e. to the point P. Therefore, OP || AA". Hence, by Thales, $\displaystyle \frac{PA^{\prime \prime }}{PA^{\prime }}=\frac{OA}{OA^{\prime }}=k$, so $PA^{\prime \prime }=k\cdot PA^{\prime }$, and $PX^{2}=PA^{\prime }\cdot PA^{\prime \prime }=k\cdot PA^{\prime }\;^{2}$, leading to $PX=\sqrt{k}\cdot PA^{\prime }$. Similarly $PY=\sqrt{k}\cdot PB^{\prime }$ and $PZ=\sqrt{k}\cdot PC^{\prime }$, so that PC' + PA' = PB' implies PZ + PX = PY. This solves the problem.

  Darij


Find all pairs of integers so that $1+5*2^{m}=n^{2}$ .

[hide] $(n+1)(n-1) = 5*2^{n}$.

So one of ${n+1, n-1}$ is a power of 2, the other is 5 times a power of 2. L

Case 1:
$n+1 = 2^{m}\\ \\ n-1 = 5 (2^{j}) \\ \\ 2^{m}-5(2^{j}) = 2$ Now obviously $2^{j}< 2^{m}$ so:

$2^{j}(2^{m-j}-5) = 2$. Thus j = 0 or 1, neither of which yields solutions.

Case 2:
$n+1 = 5(2^{j}) \\ \\ n-1 = 2^{m}$ Proceeding as before,

$2^{j}(5-2^{m-j}) = 2$. j = 0 yields no solutions. For j=1, m=3 yields a solution. Thus, the only solution is $1+5*2^{4}=3^{2}$. [/hide]

I did it the Farenhajt way, but that's a great solution Phelpedo. Thank you.

I did it the way Phelpedo did it on my second try.

Here is my first try that came to mind, just wanted to put it done to get rid of this stupid thought from my mind:
[hide]
5 x 2^m = 10, 20 + 20(m-1)

It is easy to see that 10 +1 is not a square, so disregard that.

Then, replace 5 x 2^m with 20 + 20 (m-2)

Now it is a very easy problem.

21 + 20m - 20 = n^2

n = (20m + 1)^ (1/2)

Then we can use all of the math junk :!:  :lol:  :idea:

The immediately perceptible solution is when m = 4, n=9.
From then on, n must be a number with a unit digit of 1 or 9.

We can graph this function and notice that there are no other solutions.
[/hide]

Please don't make fun of me... :(


I have two hats.  In one hat are balls numbered 1 through 15.  In the other hat are balls numbered 16 through 25.  I first choose a hat, then from that hat, I choose 3 balls, without replacing the balls between selections.  How many different ordered selections of 3 balls are possible?

If you pick hat 1, there are 15*14*13 choices.
If you pick hat 2, there are 10*9*8 choices.

Add em up!  We get 3450 as the total answer.

I got this answer but also understood that he had two choices of hat to choose from so 2(2 choices of hat)* 3450 (as computed above by AIME15) = 6900 total choices.  Why not multiply 3450 by 2?

AIME15's solution is separating the count into two cases. If we pick hat 1, there's no reason to choose hat 2 or include its count in the total count. Same for hat 1 if we choose hat 2. What is your reasoning for multiplying by 2?

Because once a hat is chosen, only certain balls are available.

Also who has more than one hat? A hat is a lifelong commitment.

[quote=OlympusHero]We know that there are 25 balls in total, so why not do 25*24*23?[/quote]

All three balls must be from the same hat.


School is really fun, although I don't like most of the students.

Without school, everyone would be at home with nothing to do.

don't like students.

don't like curriculums.

don't like teachers.

don't like the facilities.

don't like the environment (except that it is very very safe)

but, ultimatley, there's no one who would actually spend good deal of money to study math deep enough.

Other than friends or math class, I don't like school especially the people who make weird comments (with lots of sarcasm) and "grown-up" talk. The classes are okay, except for the homework. :D

I don't like the poll. Granted school isnt something I would love, I wouldn't hate it either. I somewhat like it due to interesting projects.

Yeah, it's not as if you either hate school or you love it.

what do you mean? are you a truant? HOLY ____!

i like school. mostly cuz i see my friends if i never saw my friends then i would hate school =]

School is great if you know the right people  :P which is true with everything else in life, so get used to it

I love school, cause I meet with all my friends. Also, I like learning new things.  :o

School is the worlds biggest waste of time in my opinion. Life is short and childhood is shorter.

If there wasn't school, I'd get bored very fast...another plus side is of course seeing all my friends  :)

haha i need to start focusing. all i do in english is make faces at my friend, who makes faces back. yeah... we're bored :D i sleep in math (lean against the wall so i still look kinda alert) yeah. i need to wake up.  :wink: i'll go on a sugar rush!! :wow: haahaa no...

Lol me and my friend never pay attention in French. We got a detention for playing taht game where you use your pencil and tap inbetween your fingers rapidly againist the table and try not to hit a finger. I heard your suppose to play it with a knife. :wink:

i used to love school when i was in los alamos because the teachers didn't make you memorize things they actually showed proofs. not only in mathematics, but in history(!) and science :wow:. But, when i say that they don't have proof for what they say the teachers get mad at me  :mad:  :mad:

hmm

you could say my school is just bunch of people who are excellent at making you go to a good college. (with exception of 1 english teacher; but he quit to open a car repair shop)

proof stuff is true. now i don't take math at my school, so i go to nearby college, and then the professor is supposed to be a math geek, but still does no proofs. I fell asleep everyday in that class.

[quote="(^_^)"]no, math is the biggest waste of time[/quote]
 :o  :o  :o

My school is safe, clean, and filled with good teachers teaching the same old crap year after year. I hate school. I love academics, but I'd rather study trigonometry than triangle parts (what is a median? Hmm... that's incredibly hard!)
This year I'm getting too lethargic to draw doodles in class. Last year I probably filled 20 pages with doodling in various composition books, my student planner and notebook paper.

Just sneak a practice AIME into your notebook.

i still hate school. My school is not the safest place in the world. Its stupid and boring. They won't lemme take Calculus AB because they say i haven't gotten all the credits so i'm stuck with pre-algebra. Social Studies teacher is a *bleep*. Science teacher isn't even certified to teach science. Language Arts teacher is short tempered and makes us memorize A LOT. P.e. is stupid. Health is filled with the worst people in the school.

[quote="Poincare"]I HATE SCHOOL!!!!!!!!! i just can't stand it i don't really know why.....[/quote]
I feel the same way here.

[quote="TZF"][b]@ Poincare:[/b]

They won't lemme take Calculus AB because they say i haven't gotten all the credits.
[i][color=blue]Petition. Go negotiate with the math department head. If you really have to, get your parents involved. Don't let them convince you that school policy must be enforced in all cases.[/color][/i]

Social Studies teacher is a *bleep*.
[color=blue][i]Get used to it. The world is full of ƒ~ers.[/i][/color]

Science teacher isn't even certified to teach science.
[color=blue][i]Get used to it. The world is full of stupid people.[/i][/color]

Language Arts teacher is short tempered and makes us memorize A LOT.
[color=blue]Take the opportunity to experiment with your writing and try to develop a style. English is nice because it's a class in which you can move ahead and keep learning without neglecting what's being taught. Oh, and also... get used to it! You'll have to memorize a lot of things in life.[/color]

P.e. is stupid.
[color=blue][i]It's brainless fun. What's wrong with that?[/i][/color]

Health is filled with the worst people in the school.
[color=blue][i]Doesn't everyone have to take it? When someone tries to give you a cigarette later in life, you will totally remember the gross pictures of black lungs and the movies with the people with cancer talking through holes in their necks.[/i][/color][/quote]
for the pe part, maybe he's just not that athletic so he doesn't really like it?
that situation applies to me too

Well, I feel like such a jerk now...
I complain about my school when it's probably one of the best public schools (at least in the top 15%) around...
P.E. isn't that intensive for me. I can probably do a 9:30 mile (I know, I suck) now.

Uggh

1) one of the best is like nowhere near top 15%?
2) My school is awesome. Though I do agree somewhat (mathwise) with Twain: "I never let school get in the way of my education."

[quote="1=2"]I can't run(I mean run and not stop) for a whole mile and not die.[/quote]
Me neither. Even worse, I can't run half a mile that way.

[quote="1=2"]I can't run(I mean run and not stop) for a whole mile and not die.[/quote]

Me too, I run 1.2 miles in 14min. Our P.E. department makes us run 1.2 mi.


trigonometry
complex numbers
geometric series

[quote="yuch3n"]Hey guys, I'd appreciate input and help on how to prove that:

[img]http://upload.wikimedia.org/math/4/e/c/4ec07ced5be261ded5f7f081bb74f1fe.png[/img]

and

[img]http://upload.wikimedia.org/math/0/f/2/0f25aafc310a3545afa1683df1ee0652.png[/img]

Any help is appreciated![/quote]

USSR Olympiad?

[hide="Hint"]
Think in terms of complex numbers
[/hide]

Another interesting technique is to multiply by $ \sin\frac{\alpha}{2}$ and apply product-to-sum formulas.

[quote="MellowMelon"]The place where I saw this problem was the USSR Olympiad Problem Book. t0rajir0u's idea is the solution given in the back of the book, as well as the one I found when solving it.[/quote]

eh? I think my complex number solution coupled with expressing this as a geometric series is clearly the solution given in the book...

Thanks for the all the help, but I actually proved it both ways discussed in the board before viewing. The product-sum method is a lot longer than the geometric-sum method.


I only know these:

$ \gneq$ \gneq

$ \lneq$ \lneq

$ \gneqq$ \gneqq

$ \lneqq$ \lneqq

[quote="aufha"]When I type this: > in LaTeX, it appears as a question mark in the pdf. How do I type >? (without the "or equal to part")[/quote]You can just use the > on the keyboard in both text and maths modes. If this doesn't work please post a minimal example including the preamble which shows the problem.


geometry
parallelogram
incenter
geometry proposed

From the parallelogram $ABCD$ we have four points:
$A^{'}$ : incenter(center of incircle) of triangle $DAB$
$B^{'}$ : incenter of triangle $ABC$
$C^{'}$ : incenter of triangle $BCD$
$D^{'}$ : incenter of triangle $CDA$
and also we have the lentgh of only one of the segments $A^{'}B^{'}$,$B^{'}C^{'}$,$C^{'}D^{'}$,$D^{'}A^{'}$. Draw $ABCD$.

My opinion is that we can draw an infinite number of parallelograms $ABCD$ such that the obtained segment $A'B'$ has a given fixed length.

So, I think that there is not a unique solution


algebra
polynomial
superior algebra
superior algebra unsolved

Show that $ x^5 \minus{} 27x^3 \plus{} 15x \plus{} 6$ is irreducible over $ \mathbb{Q}(\sqrt{2}, \sqrt{3})$

Describe the reticle of subfields of a Galois extension of order 595. (I tried to use the Sylow theorems, but I didn't get everything)

What is a "reticle"¿

For the polynomial, show that it needs to show irreducibility over $ \mathbb Q$.


I recently learned about TR's trust-busting in American History class and thought, "wow, what a great guy." I wonder what AoPS thinks about our nation's leaders. Which presidents do you think are the most effective, best leaders, most popular...?

I think the best president's were the Roosevelts.
However, I think Franklin D. Was a little better.
Here's why:
The Great Depression.
He single handedly helped us get out of the depression. He did the new deal. People loved him, and this was not only because he was the president; they hated Herbert Hoover
WWII
He led our country through the greatest war in history. At first, he kept us out until we were attacked, then we came in and the axis was doomed. He was incredibly popular; he was reelected 3 times.

Theodore Roosevelt:
The start of the progressive movement.
Roosevelt basically started the pregressive movement. During that time period, there was the stamp act, the food and drug act, and he was the first president to side witht the people. He was a war hero as well. He was a trustbuster, and was the first to really recognize the problems that our work force had and the problems with factories.

Franklin D. Roosevelt could have prevented the attack it is proven he recieved inforation on it so that'swhy I say Theodore.

We should make a poll containing at least a couple of the great presidents.

I say the best is president George W. Bush. Whenever you need a retard to laugh at, he's there. I mean, how often do you actually get the chance to [b]see[/b] someone say "nukyuler" an television? Think about it. And that joke about Iraq having weapons of mass destruction, man, that one cracks me up -- the best part is, nearly all of america fell for it!

On top of that, he represents the majority of the american people very well -- he's a greedy racist irresponsible idiot (not to say that I'm not one myself).

---

Disclaimer: I am lying. Right now. This whole paragraph is a lie.

On a fine summer day, a teacher was arrested for boarding a plane with a protractor, compass, ruler, a graphing calculator and a calculus book. She was claimed to have weapons of math instruction.

You can continue from there.

my mom got an email like that awhile ago.


calculus
analytic geometry
graphing lines
slope

Find the inverse of f(x)=e^(2x)+2e^(x)+1, then find the equation of the tangent to the inverse at x=4

[quote="knexpert"]The latter part of this problem is calculus, so it shouldn't be in this forum.[/quote]

my bad, and that is correct


Through the focus of $y^2=2px$($p>0$) draw a line, which intersets the parabola at $A: (x_1,y_1)$ and $B: (x_2, y_2)$. Find $\frac{y_1y_2}{x_1x_2}$.

[hide]
The focus of this parabola is $\frac{1}{4c} = \frac p 2$ where c is the leading coefficient (of y in this case).  The point itself is then $(\frac p 2, 0)$.  We can make a vertical line to make it easier, and it has intersections $y^2 = p^2 \implies y = \pm p$.  So the product of the y's is $-p^2$ and the product of the x's is $\frac{p^2}{4}$, so the ratio is $-4$
[/hide]

Shouldn't you prove that the quotient is the same for all the lines passing through the focus?

Hm, I just assumed it would be constant for all lines...
[hide]
The line can be expressed $y = m(x - \frac p2)$ where $m$ is the slope.  Plugging into $y^2 = 2px$, we get $m^2(4x^2 - 4xp + p^2) = 8px \implies 4m^2x^2 - (4pm^2 + 8p)x + m^2p^2 = 0$.  So the product of the x's is $\frac ca = \frac{p^2}{4}$.  If the roots are $x_1, x_2$, then $y^2 = 2px_1, 2px_2$.  So the product of the y's is $\sqrt{2px_1} \sqrt{2px_2} = \sqrt{4p^2}\sqrt{x_1 x_2} = 2p \cdot \frac p2 = p^2$.  However, since the vertex of the parabola is $(0,0)$, and one branch opens down, one solution to the equation, as long as $m>0$, must have a negative y value.  So the product of the y's must be negative, and since we didn't account for sign when we took the square root before, the product is then $-p^2$.  Thus the ratio of the products is $\frac{-p^2}{ \frac{p^2}{4}} = -4$
[/hide]


geometry
circumcircle
trapezoid
parallelogram
geometric transformation
reflection
Euler

Given triangle $ABC$.Through an arbitrary point $P$ on circumcircle $(O)$ of $ABC$ drawn ines parallel to $BC,CA,AB$ and $A',B',C'$ are the second intersection of these line with the circumcircle, and $A'',B'',C''$ are symetric point of $A,B,C$ with lines $B'C',C'A',A'B'$ prove that $OO''=PH$ with $O''$ is cricumcenter of $A''B''C''$ and $H$ os orthocenter of $ABC$.

WLOG, let P be of the arc AB of (O) opposite to C. $\angle C'A'B' = 180^\circ-\angle C'PB' = \angle CAB,$ etc., $\triangle A'B'C' \cong \triangle ABC$ (similar with common circumcircle (O)). A'B' = AB, AA'B'B is a cyclic isosceles trapezoid with $AA' \parallel BB'$ and with equal diagonals $AB' = A'B.$ $\triangle AB'A'', \triangle BA'B''$ are both isosceles with $A''B' = AB' = BA' = B''A'.$

Let AA'', BB'' meet at Q. $\angle BQA = 180^\circ-\angle B'C'A' = 180^\circ-\angle BCA,$ $Q \in (O).$ Similarly, the intersection of BB'', CC'' is on (O) and since BB'' meets (O) at B and exactly one other point, this other point is Q and AA'', BB'', CC'' concur at $Q \in (O).$

Since $\angle A''B'C' = \angle C'B'A = \angle C'QA$ and $AA'' \perp B'C',$ it follows that $C'Q \perp B'A''.$ Since $\widehat{(BB'', B''A')}= \angle A'B''Q = \angle A'BQ = 180^\circ-\angle A'C'Q = \widehat{(QC', C'A')}$ and $QC' \perp B'A'',$ $BQ \equiv BB'' \perp C'A',$ it follows that $\widehat{(B'A'', BB'')}= \widehat{(QC', C'A')}= \widehat{(BB'', B''A')},$ so that $A''B' \parallel B''A'$ (in addition to A''B' = B''A' shown previously), A'B'A''B'' is a parallelogram and $B''A'' \parallel A'B',$ $B''A'' = A'B' = AB.$ Similarly, $C''B'' \parallel B'C',$ $C''B'' = B'C' = BC$ and $A''C'' \parallel C'A',$ $A''C'' = C'A' = CA.$ Consequently, $\triangle A''B''C'' \cong \triangle A'B'C'$ are centrally similar (congruent) with coefficient -1.

Since $C''Q \equiv CC'' \perp A'B' \parallel B''A'',$ C''Q is the C''-altitude of the $\triangle A''B''C''$ and similarly, A''Q, B''Q are its A''-, B''-altitudes, Q its orthocenter. Let H' be orthocenter of the $\triangle A'B'C'.$ Then OH'O''Q is a parallelogram and H'O'' = OG = OP = R (circumradius of $(O) \cong (O'')$, O''G = OH' = OH.

Let perpendiculars from $P \in (O)$ to BC, CA, AB meet (O) again at D, E, F. Since $PD \perp PA',$ A'D is a diameter of (O) and similarly, B'E, C'F are diameters of (O). Thus the $\triangle DEF$ is a reflection of the $\triangle A'B'C'$ in O. If K, H' are orthocenters of these 2 triangles, then K is also a reflection of H' in O, i.e., these 2 triangles have a common Euler line KOH' and $\angle POK+\angle POH' = 180^\circ.$ The $\triangle ABC$ is obtained from the $\triangle A'B'C'$ by dropping perpendiculars from $Q \in (O)$ to its sides B'C', C'A', A'B', meeting (O) again at A, B, C. On the other hand, the $\triangle A'B'C'$ is obtained from the $\triangle ABC$ by drawing parallels from $P \in (O)$ with its sides BC, CA, AB, meeting (O) again at A', B', C'. Therefore, $\angle POH+\angle QOH' = 180^\circ.$ Since OH'O''Q is a parallelogram, $\angle OQO''+\angle QOH' = \angle OH'O''+\angle QOH' = 180^\circ.$ As a result, the $\triangle PHO \cong \triangle OO''Q \cong \triangle O''OH'$ are congruent by SAS and $OO'' = PH.$


Putnam
number theory proposed
number theory

Let $a$ a base. Prove that any number $n>0$ has a multiple which uses all digits in base $a$.

Here is a crude attempt to construct a such multiple:

Denote the a digits $N_{i} \quad i=0..(a-1)$

First find $p$ so that $a^{p}\geq n$

Now it's clear that for each digit $N_{i}$ there is a multiple $m_{i}$ of $n$
such that $d_{p+1}=N_{i}$ where $d_{p+1}$ is the (p+1)th digit in $m_{i}$ in base a.

Now construct the multiple as
$m=m_{0}+m_{1}\cdot a^{(p+1)} +m_{2}\cdot a^{2(p+1)} +\cdots m_{a-1}\cdot a^{(a-1)(p+1)}$
and it's clear that it has the desired property.

I had other idea. The sum of the inverses of all numbers which do not contain a fixed digit is finite. Thus the sum of the inverses of all numbers which do not contain all digits is finite. But the sum of the inverses of all multiples of a number diverges.

Harazi, that is an absolutely beautiful idea.  Anyways, the plodding solution is not difficult.  It suffices to prove that for each $i\in\{1,2,\ldots,a-1\}$, there exists $k\in\mathbb{N}$ such that $kn$ has leading digit $i$, that is, that $ia^j\leq kn<(i+1)a^j$ for some $j\in\mathbb{N}$.  But existence is trvial, because for large enough $j$, we have $a^j>n$, so that there must be a $k$ with the desired property.  Now we just construct our multiple in 'blocks' and we are done.


[quote="metafor"]Consider a triangle $ ABC$, with $ \angle A \equal{} 90^{\circ}$, and $ AC > AB$. Let $ D$ be a point in $ AC$ such that $ \angle ACB \equal{} \angle ABD$. Draw an altitude $ DE$ in triangle $ BCD$. If $ AC \equal{} BD \plus{} DE$, find $ \angle ABC$ and $ \angle ACB$.[/quote]

$ \angle ACB\equal{}\angle ABD\equal{}\beta$ -----> $ \angle DBE\equal{}90^{o}\minus{}2\beta$ ----->

$ \bigtriangleup BDE$ ----> $ sin(90^{o}\minus{}2\beta)\equal{}\frac{DE}{BD}\equal{}cos2\beta$ ----> $ DE\equal{}BDcos2\beta$

$ \bigtriangleup ABD$ ----> $ cos\beta\equal{}\frac{AB}{BD}$------> $ BD\equal{}\frac{AB}{cos\beta}$------> $ DE\equal{}\frac{ABcos2\beta}{cos\beta}$

$ AC\equal{}BD\plus{}DE\equal{}BD\equal{}\frac{AB}{cos\beta}\plus{}\frac{ABcos2\beta}{cos\beta}\equal{}\frac{AB(1\plus{}cos2\beta)}{cos\beta}\equal{}\frac{AB(1\plus{}2cos^{2}\beta\minus{}1)}{cos\beta}\equal{}2ABcos\beta$ ---->

 $ \frac{AC}{AB}\equal{}2cos\beta$

$ \bigtriangleup ABC$ ---------------> $ cot\beta\equal{}\frac{AC}{AB}$ -----> $ \frac{AC}{AB}\equal{}2cos\beta\equal{}cot\beta$ -----> $ sin\beta\equal{}\frac{1}{2}$ ----> $ \beta\equal{}30^{o}$

[quote="metafor"] [color=darkred]Consider $ \triangle ABC$ with $ A \equal{} 90^{\circ}$ and $ b > c$ . Denote $ \left\|\begin{array}{c}
D\in (AC)\ ,\ m(\widehat {ABD})\equal{}C\\\\
E\in (BC)\ ,\ DE\perp BC\end{array}\right\|$ . Prove that $ AC \equal{} BD \plus{} DE\ \Longleftrightarrow\ a\equal{}2c$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote $ F\in (BC)$ for which $ AF\perp BC$ . Observe that $ b^2\plus{}c^2\equal{}a^2$ and $ AF\equal{}\frac {bc}{a}$ . From $ \triangle ABD\ \sim\ \triangle ACB$ obtain

$ \frac cb\equal{}\frac {BD}{a}\equal{}\frac {DA}{c}$ , i.e. $ BD\equal{}\frac {ac}{b}$ and $ DA\equal{}\frac {c^2}{b}$ . Thus, $ DC\equal{}AC\minus{}AD$ , i.e. $ DC\equal{}\frac {b^2\minus{}c^2}{b}$ . Since $ DE\parallel AF$ obtain

$ \frac {CD}{CA}\equal{}\frac {DE}{AF}$ $ \implies$ $ \frac {b^2\minus{}c^2}{b^2}\equal{}\frac {DE}{\frac {bc}{a}}$ , i.e. $ DE\equal{}\frac {c(b^2\minus{}c^2)}{ab}$ . Therefore, $ \boxed {\ AC\equal{}BD\plus{}DE\ }$ $ \Longleftrightarrow$ $ b\equal{}\frac {ac}{b}\plus{}\frac {c(b^2\minus{}c^2)}{ab}$ $ \Longleftrightarrow$

$b^2(a\minus{}c)\equal{}c(a^2\minus{}c^2)$ $ \Longleftrightarrow$ $ \boxed{b^2\equal{}c(a\plus{}c)}\ (*)$ $ \Longleftrightarrow$ $ a^2\minus{}c^2\equal{}c(a\plus{}c)$ $ \Longleftrightarrow$ $ a\minus{}c\equal{}c$ $ \Longleftrightarrow$ $a=2c\iff$ $B=60^{\circ}$ .

[b][u]Remark[/u].[/b] From the relation $(*)$ , i.e.  $b^2=c(a+c)$ obtain shorterly that $B=2C$ , i.e. $B=60^{\circ}$ .[/color]

So-called ... middling. And yet I like it.

$AB$ is tangent to the circle $\odot (BCD)$, hence $AB^2=AD\cdot AC \ (\ 1\ ), BADE$ is cyclic, consequently $\widehat{AED}=\widehat{ABD}=\widehat{BCA}$ and $AE$ is tangent to the circle $\odot (CDE)\implies AE^2=AD\cdot AC\ (\ 2\ )$.
From $(1)$ and $(2)\implies AE=AB \iff\angle CDE=\angle ADB\ (\ 3\ )$.
Extend $(BD$ with $DF=DE$; from the given relation we get $BF=AC$ and, with $(3)$ we infer $\triangle DFC\equiv\triangle DEC$, hence $CF=CE$ and $DFCE$ is cyclic and $\widehat{DFE}=\widehat{DCE}$.
We conclude that $\triangle BEF\equiv\triangle AEC$ (a.s.a), hence $EF=CE$, or triangle $CEF$ is equilateral and the result follows.

Best regards,
sunken rock

well i don't think i have a good solution like that of zaya_yc...!!
But yea this is one..!!

[img]http://i48.tinypic.com/6fm9hl.jpg[/img]

$since \ \angle CDB = 90^\circ + \theta \ and \ \angle CDE = 90^\circ - \theta \ so \ \angle EDB = 2\theta$

we draw DM || AB so $\angle EDM = \angle BDM = \theta$

By Thales' theorem,

$\frac {DM}{AB} = \frac {CD}{AC}$

$so, \ \frac {q sec\theta}{p cos\theta} = 1 - \frac {p sin\theta}{(p+q)}$

$or, \ \frac {q}{p cos^2\theta} + \frac {p sin\theta}{(p+q)} = 1$

$\implies pq + q^2 + p^2cos^2\theta.sin\theta = p^2cos^2\theta + pq cos^2\theta$

$\implies pqsin^2\theta + p^2cos^2\theta(sin\theta - 1) + q^2 = 0 ----- (i)$

$Now, \ clearly \ \triangle CDE \sim \triangle BDA$

$so, \ \frac {BD}{AD} = \frac {CD}{DE}$

$=> \frac {p}{p sin\theta} = \frac {p + q - psin\theta}{q}$

$\implies pq = p^2 sin\theta + pq sin\theta - p^2 sin^2\theta$

$\implies pq (1 - sin\theta) = p^2 sin\theta(1 - sin\theta)$

$=> q = psin\theta ------ (ii)$

putting q from (ii) in (i) we get,

$p^2 sin^3 \theta + p^2 cos^2\theta (sin\theta - 1) + p^2 sin^2 \theta = 0$

$\implies sin^3\theta + sin\theta - sin^3 \theta - cos^2\theta + sin^2 \theta = 0$

$\implies 2 sin^2\theta + sin\theta - 1 = 0$

$\implies 2 sin^2\theta + 2sin\theta - sin\theta - 1 = 0$

$\implies (sin\theta + 1)(2 sin\theta - 1) = 0$

$rejecting\ sin\theta = -1 \ we \ get, \ sin\theta = 1/2 \ or, \ \theta = \pi/6$


Could somebody please show me how to solve this:
S0=0
S1=1
S2=4
S3=9
Sn=2(Sn-1) -Sn-4
what is S30?


Hello,

Who can help my with the next I can integrate this integral normaly if you use just you tabel with standard integrals its not so difficult but if you will do it with al realy goniometric substution.
Where do I miss?
Who can I modify my solution or did I make a seriuos mistake?

[img]http://img447.imageshack.us/img447/1863/intergral2el.jpg[/img]

Thanks a lot.

Nice handwriting, but this is in the wrong forum.

A moderator will move this, you may want to check the Calculus forums.

can someone remove this message I have post it in the right forum

im trying to but i keep getting this:

The target forum you choose is a category or a link, so can not contain topics.

help fellow moderators

are you trying to find the thing in the top left corner?


calculus
integration
trigonometry
calculus computations

$ \int\limits_{t \plus{} 2\pi}^{t \plus{} \frac {5\pi}{2}}{[sin^{ \minus{} 1}(cosx) \plus{} cos^{ \minus{} 1}(cosx)]}dx$

I just dont know how to start this problem.
Thanks for any hints.

If your $ \sin^{ \minus{} 1}$ (resp. $ \cos^{\minus{}1}$) has value range $ \left [ \minus{} \frac {\pi}{2},\frac {\pi}{2}\right]$ (resp. $ [0,\pi]$), then you only have to take care of the "signs", i.e. $ \sin (x\plus{}\pi)\equal{}\minus{}\sin x$ resp. $ \cos(x\plus{}\pi)\equal{}\minus{}\cos x$, and you have to use $ \cos(\frac {\pi}{2} \minus{} x) \equal{} \sin x$ in order to tackle $ \sin^{ \minus{} 1}\cos$

[quote]$ \int\limits_{t \plus{} 2\pi}^{t \plus{} \frac {5\pi}{2}}{[sin^{ \minus{} 1}(cosx) \plus{} cos^{ \minus{} 1}(cosx)]}dx$[/quote]
is $ sin^{\minus{}1}(cosx)\equal{}\frac{1}{\sqrt{1\minus{}x^{2}}}$ and $ cos^{\minus{}1}(cosx)\equal{}x$ ?

$ \arcsin ({\cos x}) \plus{} \arccos ({\cos x}) \equal{} \frac {\pi}{2}$

the integral becomes

$ \int \frac {\pi}{2} dx$ :wink:


geometry
rectangle
number theory proposed
number theory

A rectangle is divided into $200*3$ unit squares .Prove that the number of way of splitting this rectangle into rectangle of size $1*2$  is divisible by $3$

Let's not post these hundreds of times, guys :).

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=60140[/url]


Prove that prime divisors of $a^{p-1}+a^{p-2}+...+a+1$ where $p$ is prime number($a\in \mathbb{Z}$) are congruent to $0$ or $1$ by modyle $p$.


:gleam:  :welcome: [b]Jack and Jim are next door neighbors as well as friends. For their science project, they want to connect their bedrooms with an intercom kit. The windows in their rooms were both 15 ft off the ground and the houses were 32 ft apart. The wire was going to be buried in a trench 1 ft underground. The boys want to be able to install the intercoms on any wall of their rooms, each of which is 12 ft square. How much wire will be needed for this project if the boys want to spend as little money on wire as possible?[/b] :pilot:

Hmmm...judging that not even Jesusfreak or 236factorial have answered so far, mebbe i should change the title to "a hard intercom problem".  :D  :rotfl:

First of all, it's complete spam to post something like that. Second, as I'm not even going into 9th grade but 10th grade, I probably shouldn't be answering these problems at all, but rather spending more time in the high school forums. I can't tell you why 236! hasn't answered, but I'm sure he has his reasons.

[quote="236factorial"]Is the answer rational? Does the wire have to go out the window? If so, how can it also be plugged into a wall?[/quote]
it has to go out the window, and plugged into any of the walls inside.

ill post the solution to this problem, if no one gets it ;) , Thursday.

I get $24 + 2\sqrt{481}$ If that's right, I'll post a solution.

The answer is rational...at least im pretty sure it is. even im not really sure of mi answer... :blush:
i kinda drew a diagram and meditated upon the words [b]"little money on wire as possible"[/b]. Hmmmm... :D

wait, so you don't know the answer either? great.  You should know the solution to your problem when you post it unless you need help with it.  If you do need help with it, then don't say you know the answer :!:

Does any wall mean that I can choose which wall? And which walls are you choosing for measuring the distance between houses? Can the wire go through the house itself? ARE THE ROOMS NEXT TO THE WINDOWS?? You can't assume anything here! Maybe the wire costs less the more you buy!! Maybe they're 32 feet apart but there's a big HEDGE between the houses. Maybe there's an underground cable! AND DON'T DOUBLE POST!!!

Ahhhhhh, I feel much better now. ;)

i sed i do know the answer, but am not perfectly sure. i think it is rational, and i think it is positive. ill post it as soon as i get bak 2 usa.

Of course it is positive. It's a distance.

im getting the wires in a parabola, but I forget how to measure the distance of points along the parabola in an equation.... :?:  :huh:

maybe it cant be measured, so the problem is not possible! so that means.....they could try using cell phones! :D

It can be done with Calculus... but that's definitely not appropriate for MC.


probability
floor function
geometric series
algebra
binomial theorem
probability and stats

How can we calculate the following probabilities:
1.We rolled 3 dices and the sum is 9
2.We rolled 3 dices and the sum is 10

Thanks in advance.

This can also be answered by recursion - building a recursive table. Let $ p(k,n)$ be the probability of rolling a sum of $ n$ with $ k$ dice, and let $ b(k,n)\equal{}6^kp(k,n).$ Then
\[ p(k,n)\equal{}\frac16\sum_{j\equal{}1}^6p(k\minus{}1,n\minus{}j)\]
from which we deduce that
\[ b(k,n)\equal{}\sum_{j\equal{}1}^6 b(k\minus{}1,n\minus{}j)\]
It's pretty easy to build a table for $ b$ by hand and even easier with a spreadsheet. Here are the first three columns of that table:
\[ \begin{tabular}{cccc}{\it n}&1 die&2 dice&3 dice\\1&1&&\\2&1&1&\\3&1&2&1\\ 4&1&3&3\\ 5&1&4&6\\ 6&1&5&10\\ 7&&6&15\\ 8&&5&21\\ 9&&4&25\\ 10&&3&27\\ 11&&2&27\\ 12&&1&25\\ 13&&&21\\ 14&&&15\\ 15&&&10\\ 16 &&&6\\ 17&&&3\\ 18&&&1\end{tabular}\]
So $ b(3,9)\equal{}25$ and $ b(3,10)\equal{}27,$ or $ p(3,9)\equal{}\frac{25}{216}$ and $ p(3,10)\equal{}\frac{27}{216}.$

The short (and dirty) way is to list out all of the possible combinations of numbers 1-6 that add to 9 or 10, then divide by the total number of combinations. This can be done somewhat systematically for small numbers of dice, like say 2 or 3, but can become unwieldy for a large number of die.

The more general way of solving this involves the use of generating functions. Suppose we are tossing $ n$ die and want to obtain a sum $ c$. Let $ x_1, x_2, x_3, \ldots x_n$ be the number that we get on the nth die respectively. We want to know the number of possible combinations with $ x_1 \plus{} x_2 \plus{} x_3 \plus{} \ldots \plus{} x_n \equal{} c$, for some number $ c$ (in your case $ c \equal{} 9$ or $ 10$).

If you look at the product $ (z \plus{} z^2 \plus{} z^3 \plus{} \ldots \plus{} z^6)^n$, then the coefficient of $ z^c$ in the resulting product will tell you how many combinations of exponents add to $ c$. This will be the number of combinations of numbers 1 to 6 that add to c. To calculate this coefficient without having to multiply the product all out, we can use generating functions.

$ (z \plus{} z^2 \plus{} z^3 \plus{} \ldots \plus{} z^6)^n \equal{} \left(\frac {z \minus{} z^7}{1 \minus{} z}\right)^n$ (By sum of a geometric series)
$ \equal{} z^n(1 \minus{} z^6)^n(1 \minus{} z)^{ \minus{} n}$
$ \equal{} z^n \left(\sum_{k \equal{} 0}^n ( \minus{} 1)^k \binom{n}{k} z^{6k}\right) \left(\sum_{k \equal{} 0}^\infty \binom{n \plus{} k \minus{} 1}{k} z^k\right)$ (By the regular Binomial Theorem and the General Binomial Theorem)

A general formula can be worked out from this point, but you can basically find the coefficient of $ z^c$ in the product by matching powers of the two series that add to $ c \minus{} n$.

For an example of this method see p. 161 of this [url=http://books.google.com/books?id=ckklgif1M68C&pg=PP1&dq=Venkatarama+Krishnan#PPA161,M1]book[/url].

[quote="gauss202"]If you look at the product $ (z \plus{} z^2 \plus{} z^3 \plus{} \ldots \plus{} z^6)^n$, ...[/quote]
The recursion in my previous post can be seen as one way to multiply out that polynomial.

(Just for completeness) I think the general formula for the number of ways to roll a sum of $ c$ with n die would be $ \sum_{k \equal{} 0}^{\lfloor\frac {c \minus{} n}{6}\rfloor} ( \minus{} 1)^k \binom{n}{k} \binom{c \minus{} 6k \minus{} 1}{n \minus{} 1}$


The value of $ 85^{9} \minus{} 21^{9} \plus{} 6^{9}$ is  divisible by an integer between 2000 and 3000. Find this integer.

Any general tips to start this off?

EDIT: Thanks for pointing that out :D

Are you sure it is an integer in the thousands? $ 85^9$ is quite a bit larger than $ 21^9$.

[hide]
$ 85^{9} \minus{} 21^{9} \plus{} 6^{9} \equal{} (85 \minus{} 21)(\cdots) \plus{} (2\cdot 3)^9 \equal{} 2^6(\cdots \plus{} 2^33^9)$, so $ 64$ is a factor of our answer.

Now we just try some small prime numbers. $ 3$ doesn't work, $ 85^{9} \minus{} 21^{9} \plus{} 6^{9} \equiv 0 \minus{} 1 \plus{} 1 \equiv 0 \pmod{5}$ and $ 85^{9} \minus{} 21^{9} \plus{} 6^{9} \equiv 1 \plus{} 0 \minus{} 1 \equiv 0 \pmod{7}$, so $ 2^6 \cdot 5 \cdot 7 \equal{} \boxed{2240}$ divides into our number. And yay, it fits in our range.

There's not really a better method than just trying small numbers.
[/hide]

Hmm... I'd factor (like you guys have done).. but azpjs, it does not factor like that:

$ 85^9\minus{}21^9\plus{}6^9 \equal{} (85\minus{}21)(85^9\plus{}21\cdot85^8\plus{}21^2\cdot85^7\plus{}21^3\cdot85^6\plus{}21^4\cdot85^5\plus{}21^5\cdot85^4\plus{}21^6*85^3\plus{}21^7\cdot85^2\plus{}21^8\cdot85\plus{}21^9) \plus{} 2^9\cdot3^9$
$ \equal{}64(85^9\plus{}21\cdot85^8\plus{}21^2\cdot85^7\plus{}21^3\cdot85^6\plus{}21^4\cdot85^5\plus{}21^5\cdot85^4\plus{}21^6*85^3\plus{}21^7\cdot85^2\plus{}21^8\cdot85\plus{}21^9 \plus{} 2^3\cdot3^9)$

This is easy to examine mod 3, 5, 7, and 17 since many terms have these as factors.  mod 3, we have $ 64(85^9)$ which does not reduce to $ 0$.  Mod 5 we have $ 64(21^9\plus{}2^3\cdot3^9) \equal{} 64 \cdot 3^9(7^9\plus{}2^3)$ and $ 7^9\plus{}2^3 \equal{} (7^3\plus{}2)(7^6\minus{}7^3\cdot2 \plus{} 2^2)$.  Since $ 7^3\plus{}2\equal{}49*7\plus{}2\equal{}280\plus{}63\plus{}2\equal{}345$ which is a multiple of 5, the number is divisible by 5.

Mod 7 we have $ 64(85^9\plus{}2^3\cdot3^9) \equal{} 64(85^3\plus{}2*3^3)(85^6\minus{}85^3*2*3^3\plus{}2^2*3^6)$.  $ 3^6 mod 7 \equal{} 729 mod 7 \equal{} 1 mod 7$, $ 85^6 \equal{} 15^6 mod 7 \equal{} 1 mod 7$, $ 3^3 \equal{} 27 \equal{} 6 mod 7$ and $ 85^3 \equal{} 1 mod 7$ hence the second factor is equivalent to $ 1\minus{}1*2*6\plus{}4*1 \equal{} \minus{}7 \equal{} 0 mod 7$, so our number is divisible by 7. (To find a sum mod x, we can subtract x as many times as necessary from each factor, and use the results in place of the original numbers)

Hence we have $ 64*5*7 \equal{} 320*7\equal{}2240$.

Looking back, I could have factored it as $ 85^9\minus{}21^9\plus{}6^9 \equal{} (85^3\minus{}21^3)(85^6\plus{}85^3*21^3 \plus{} 21^6) \plus{} 6^9$ and $ 85^3\minus{}21^3 \equal{} (85\minus{}21)(85^2\plus{}85*21\plus{}21^2) \equal{} 64*(6400\plus{}800\plus{}25\plus{}1600\plus{}100\plus{}80\plus{}5\plus{}441) \equal{} 64(9451)$ so combining $ 6^9$in we have $ 64((9451)(85^6\plus{}85^3*21^3\plus{}21^6)\plus{}2^3*3^9)$

hmm.. perhaps could yield something...

Did you actually read azjps's post facis? In your post you just did the exact same factorization he did.

gauss202, I did.  If you look closely, he accidentally treated the exponents of 9 on the 85 and 21 as 3s and has the incorrect sign in the factorization: he has incorrectly written that $ 85^{9} \minus{} 21^{9} \plus{} 6^{9} \equal{} (85 \minus{} 21)(85^2 \minus{} 85 \cdot 21 \plus{} 21^2) \plus{} (2\cdot 3)^9$.  What I did and what he did are different.  Essentially, though, my work is just what he would have had if he had done it correctly.

Yeah, that was a typo.  :oops: Either way though, its clear that $ 2^6$ divides into the quantity, and the rest doesn't change.

Edit: Wait, that was a lot of typos ... argh.


I'm still pretty fresh at LATEX. Is there a USAMTS header template for latex? I just couldn't put the box of year, round, problem on the up-right corner.

I'm not sure I understand your question.  If you download the template, and compile normally, and it doesn't work, you need to save the fancyhdr.sty file into the folder where your USAMTS template for the solutions is.  To download the fancyhdr.sty file, just right-click and "save as" below.

[url=http://www.artofproblemsolving.com/LaTeX/Images/fancyhdr.sty]fancyhdr.sty[/url]

Or you can just install the fancyhdr package using Miktex Package Manager and then you don't need that file.


If $ \frac{x(y\plus{}z\minus{}x)}{log x}\equal{}\frac{y(x\plus{}z\minus{}y)}{log y}\equal{}\frac{z(x\plus{}y\minus{}z)}{log z}$, prove that $ x^y.y^x\equal{}z^y.y^z\equal{}z^x.x^z$

[hide]Since x,y, or z can't be zero, we can safely take the recipricols of the given equation to get
ln(x)/x(y+z-x)=ln(y)/y(x+z-y)=ln(z)/z(x+y-z)
Multiplying the whole thing by 2xyz gives:
2zy*ln(x)/(y+z-x)=2xz*ln(y)/(x+z-y)=2yx*ln(z)/(x+y-z)
Call this result A.

Take from the given equation (by cross multiplication)
xln(y)*(y+z-x)=yln(x)*(x+y-z)
xln(y)/yln(x)=(x+z-y)/(y+z-x)
Call this result B

Repeat this with the two other combinations to get

C: xln(z)/zln(x)=(x+y-z)/(y+z-x)
D: yln(z)/zln(y)=(x+y-z)/(x+z-y)

Combine
B+C=> zln(y)+yln(z)=2zy*ln(x)/(y+z-x)
1/B+D=> xln(z)+zln(x)=2xz*ln(y)/(x+z-y)
1/C+1/D=> yln(x)+xln(y)=2xy*ln(z)/(x+y-z)

From result A we know that
2zy*ln(x)/(y+z-x)=2xz*ln(y)/(x+z-y)=2yx*ln(z)/(x+y-z)
Thus,
zln(y)+yln(z)=xln(z)+zln(x)=yln(x)+xln(y)
ln(y^z*z^y)=ln(z^x*x^z)=ln(x^y*y^x)
y^z*z^y=z^x*x^z=x^y*y^x
As desired.[/hide]


In the sequence $1, 0, 1, 0, 1, 0, 3, 5, \cdots$, each member after the sixth one is equal to the last digit of the sum of the six members just preceeding it. Prove that in this sequence one cannot find the following group of six consecutive members: \[0, 1, 0, 1, 0, 1\]

Actually every six consecutive terms(after the second term)couldn’t be in  this type:$ a,b,a,b,a,b$
Note that $ x_i$ is  unic number that $ 0\leq{x_i}\leq9$ and $ x_i = 2x_{i + 6} - x_{i + 7} \mod 10$ because:
$ x_i + x_{i + 1} + x_{i + 2} + x_{i + 3} + x_{i + 4} + x_{i + 5} = x_{i + 6} \mod 10 \\
x_{i + 1} + x_{i + 2} + x_{i + 3} + x_{i + 4} + x_{i + 5} + x_{i + 6} = x_{i + 7} \mod 10 \\
\Longrightarrow{ x_i = 2x_{i + 6} - x_{i + 7} \mod 10}$
Therefore if $ x_i,x_{i + 1},x_{i + 2},x_{i + 3},x_{i + 4},x_{i + 5},a,b,a,b,a,b$ appearance in the sequence,then:
$ x_i = x_{i + 2} = x_{i + 4} = 2a - b \mod 10 \\
x_{i + 1} = x_{i + 3} = x_{i + 5} = 2b - a \mod 10$
therefore the six concecutive terms which appearance before $ a,b,a,b,a,b$ is also in this type:$ c,d,c,d,c,d$ and  the six concecutive terms which appearance before $ c,d,c,d,c,d$
is also in the same type and so on … contradiction!(note that  the start of sequence is:
$ 1,0,1,0,1,0,3,5,0,9,8,5,...$

Actually, I think the above proof is flawed, because some of the indices are off by one.  Working backwards from 0,1,0,1,0,1, I get the sequence would have to be $ \ldots, 2, 9, 2, 9, 9, 0, 1, 0, 1, 0, 1$, which doesn't match the pattern claimed.

Here's my solution:

Define the function $ Q(i) \equal{} 2 x_i \plus{} 4 x_{i\plus{}1} \plus{} 6 x_{i\plus{}2} \plus{} 8 x_{i\plus{}3} \plus{} 2 x_{i\plus{}5} \mod 10$.  From the beginning of the sequence, compute $ Q(1) \equal{} 8$.

Next, I claim that $ Q(i\plus{}1) \equal{} Q(i)$.  To see this, we have:
$ Q(i\plus{}1) \equal{} 2 x_{i\plus{}1} \plus{} 4 x_{i\plus{}2} \plus{} 6 x_{i\plus{}3} \plus{} 8 x_{i\plus{}4} \plus{} 2 x_{i\plus{}6} \mod 10$
$ \equal{} 2 x_{i\plus{}1} \plus{} 4 x_{i\plus{}2} \plus{} 6 x_{i\plus{}3} \plus{} 8 x_{i\plus{}4} \plus{} 2 (x_i \plus{} x_{i\plus{}1} \plus{} x_{i\plus{}2} \plus{} x_{i\plus{}3} \plus{} x_{i\plus{}4} \plus{} x_{i\plus{}5})$ $ \mod 10$
$ \equal{}  2 x_i \plus{} 4 x_{i\plus{}1} \plus{} 6 x_{i\plus{}2} \plus{} 8 x_{i\plus{}3} \plus{} 10 x_{i\plus{}4} \plus{} 2 x_{i\plus{}5} \mod 10$
$ \equal{}  2 x_i \plus{} 4 x_{i\plus{}1} \plus{} 6 x_{i\plus{}2} \plus{} 8 x_{i\plus{}3} \plus{} 2 x_{i\plus{}5} \mod 10$
$ \equal{} Q(i)$

We must therefore have $ Q(i) \equal{} 8$ for all $ i$.

Now suppose that the sequence contained the terms "0, 1, 0, 1, 0, 1"; that is, there is some $ k$ so that  $ (x_k, x_{k\plus{}1}, x_{k\plus{}2}, x_{k\plus{}3}, x_{k\plus{}4}, x_{k\plus{}5}) \equal{} (0,1,0,1,0,1)$.
Then we compute $ Q(k) \equal{} 4$. But this is impossible; we conclude that "0, 1, 0, 1, 0, 1" never appears.

[quote="juggle5"]Actually, I think the above proof is flawed, because some of the indices are off by one.  Working backwards from 0,1,0,1,0,1, I get the sequence would have to be $ \ldots, 2, 9, 2, 9, 9, 0, 1, 0, 1, 0, 1$, which doesn't match the pattern claimed.
[/quote]
yes...you are right...sorry :oops:


hey
using my new found knowledge of barycentrics i found that
for a triangle ABC , the lemoine point lies on the line joining the mid point of the altitude AD (to BC)  and the midpoint of BC.
In case this thing is correct , cud ne1 suggest a synthetic proof (projectiv geo mayb)
best regards
(PS does ne1 have ideas abt free software allowing u to draw geometrical diagrams to verify facts)

I don't think we need projective geometry. First try to prove that the antiparallel lines to the sides of the triangle drawn through K (Lemoine's point, I have no idea why they call it K :D) are bisected by K. Then we can see that 2 such lines are the diagonals of a rectangle which is inscribed in ABC, so K is the center of such a rectangle. The third part is proving that the locus of the center of a rectangle inscribed in ABC (that means it has 2 vertices on a side of ABC, and the other 2 vertices on the remaining 2 sides, so we have three cases) is formed by the reunion of those segments you found, which link the midpoint of a side to the midpoint of the corresponding height.


number theory
superior algebra
superior algebra theorems

Hi all

I have been trying to learn some class field theory as well as develop my knowledge of number fields through various texts. Initially I worked through Marcus' Number Fields, which was fantastic, because every chapter ended with 40-something problems that really helped me understand the material.

Now I have moved on to a bit more advanced material, particularly CFT which is hardly touched upon in Marcus, and I have been reading Janusz, Algebraic Number Fields, and intend to go over Lang's Algebraic Number Theory as well. Unfortunately, neither of these texts have much in the way of problems, and I am finding that I am getting overwhelmed with text, and never getting the opportunity to really learn the material the way I did with Marcus. I usually end up reading 50 pages of material over 2 days and coming to the end of the chapter and not remembering anything other then what I read in the last few minutes.

Anyways, I was wondering if anyone out there can suggest a text that covers the same material as the 2 aforementioned texts (Or at least the topics in CFT covered by them since that is really what I am interested in learning), but has a lot of problems as well to help alleviate some of my issues with long blocks of text. Particularly a text that is focused on proofs via local properties, since Marcus did not have that at all and I would really like to get a feel for localization.

Thank you for your help.

Hi,

I also met a problem like you. I read and then forgot as if I did not read any thing, because of lack of doing exercises after each part. I've just finished one basic course in Algebraic Number Theory, so I dont know about what u are studying now, but I saw the book [b]Algebraic Number Theory [/b]of [b]Jurgen Neukirch[/b] has about 6-8 exercises after each part, so It will help in practicing. And I saw it has also the part in CFT. I think It may help.


Given a circle with center O,and radius R.A and B are 2 points outside the (O).OA=2R.Determine a point M on (O) such that (MA+2MB) is a minimum.

[quote="Altheman"][quote="cincodemayo5590"][hide]
Since there is no restriction as to where point $B$ needs to be placed, we can just place the point infinitesimally close to wherever point $M$ is placed so that the distance approaches zero. Then, the shortest distance from $M$ to $A$ will occur when $M$ is on $\overline{OA}$ (do you see why?). This means that the sum can approach but never attain the minimum of $\boxed R$.[/hide][/quote]

B is fixed.... :wink:

i don't believe there is a nice solution for this problem...what is the source? a solution can be attained through calculus, but the algebra is horrid...[/quote]

[hide="only for Alex to read"] ONO!! YOUR MOM IS FIXED!! [/hide]

But ya, if B is a fixed point then i see only the solution to which alex hinted, and it is pretty nasty...


Who do you think will win the national championship?

[quote="mustafa"]Ohio State[/quote]
of course. OSU pwns

:whistling: Lets agree to disagree :whistling:

OSU rocks!

I ve been a fan for the last 6 years!!!!

Go OSU!!!!!! #1!

[quote="azndemon43"][quote="junggi"][quote="azndemon43"]no, no, you guys got this all wrong

USC ALL THE WAY

TROJANS OWN ALL

but they played terrible last saturday, i will admit[/quote]
whoawhoa.
NO ONE, I repeat, NO ONE can be better than OSU. :mad:[/quote]

oh really, which team was it that almost pulled a three- peat

OSU doesnt stand a chance :D[/quote]
honestly I have no idea b/c i never watch football
I just like OSU b/c THEY HAVE THE BEST CHINESE FOOD EVER IN THEIR CAF.

[quote="mysmartmouth"]I would like to point something out here.  Remember the year OSU was the national champ?  They were 13-0.  However, they had one loss that year.  Not that season.  That year.  They lost to the University of South Carolina on January 1st at the Outback Bowl the previous year.  Go Gamecocks![/quote]
no, not SoCarolina, were talkin' SoCal.

Just look at my avatar

Oh, well, if this is about good food, you have swayed me.

Good food > football

Tennessee rocked Georgia!

I wish Florida could lose a game somehow so that Tennessee would make it to the SEC championship.

[quote="mustafa"]Food is much more important than football[/quote]
so? OSU ROCKS.
edit lol i read that wrong

[quote="baseballfreak2"]CAL all the way!!! I think they CAN beat USC this year...[/quote]

cal's already lost a game.
I give them 15% chance to beat USC

[quote="mysmartmouth"][quote="azndemon43"]
no, not SoCarolina, were talkin' SoCal.

[/quote]

I know.  But by the transitive property, ( :rotfl: ) SoCar > OSU > SoCal, so SoCar > SoCal.

(I was just adding something random to the above discussion...nvm)[/quote]
serious flaw there.
$OSU>everything else>socar$
so your logic doesnt hold


ratio
geometry
AMC
AIME
power of a point

In a circle, chords AB and CD intersect at R. If AR:BR = 1:4 and CR:DR = 4:9, find the ration AB:CD.

Difficult:

Let S be the set of lattice points in the region defined by $0 \le x \le 3$ and $0 \le y \le 3$. Triangles are formed by choosing three non-collinear members of S as vertices. Two triangles are distinct if they share no more than two vertices. Compute the number of distinct triangles with an area of 3/2. (Make sure you count right!)

Proof:

A sequence of positive integers $a_1, a_2, a_3 \ldots$ is defined as follows:

$a_{n+1} = a_n - S_n$, where $S_n$ is the sum of the digits of $a_n$.

If $ 100 < a_1 < 1000$, show that the sequence must contain the number 72.

Power of a Point states:

Given a point P and a line through P which intersects some circle in two points A and B, the product (PA)(PB) is the same for any choice of the line.

There are several formulas based on this theorem.

The one in use is  two chords through a point.

Given chords AB and CD which intersect at R, (AR)(BR) = (CR)(DR).

[quote="dangvh"][quote="mna851"]The first one is very easy

[hide="Hint"]Power of a Point[/hide][/quote]

I'm probably just stupid but could you tell me what "power of a point" is?[/quote]

http://www.cut-the-knot.org/pythagoras/PPower.shtml

Power of a point will show up in a lot of places (like one of the last ones on the AIME this year used that as an intermediate step) so it's one of those theorems you would want to get used to using.

Proof
[hide]$a_1$ is a three-digit number that can be expressed in the form $100H + 10T + U$.

When we subtract the sum of the digits we have

$100H + 10T + U - H - T- U = 99H+9T$

which is a multiple of 9. Since the digits of all multiples of nine sum to a multiple of nine, only multiples of 9 will be subtracted from $a_1$. The maximum value of the sum of the digits of a two digit number is 18. That number is 99, which is greater than 72 and thus we will only subtract 9 from every other two digit number less than 99 and hence cannot miss 72.[/hide]

[quote="dakyru"][hide="Answer to first one."]
$AR = x$, $BR = 4x$, $AB = x + 4x = 5x$
$CR = 4y$, $DR = 9y$, $CD = 4y + 9y = 13y$
$4x^2 = 36y^2$
$x^2 = 9y^2$
$x = \sqrt{9y^2} = 3y$
$AB:CD = 5x:13y = 5(3y):13y = 15:13$[/hide]
That's what I got.[/quote]
Yeah, that's right--I made a stupid mistake


Periodical sequences $ (a_n), (b_n), (c_n)$ and $ (d_n)$ satisfy the following conditions: $ a_{n \plus{} 1} \equal{} a_n \plus{} b_n, b_{n \plus{} 1} \equal{} b_n \plus{} c_n,c_{n \plus{} 1} \equal{} c_n \plus{} d_n, d_{n \plus{} 1} \equal{} d_n \plus{} a_n$ for $ n\geq 1$.
Prove that $ a_2 \equal{} b_2 \equal{} c_2 \equal{} d_2 \equal{} 0$.

First note that
\[ a_{n\plus{}1}\plus{}b_{n\plus{}1}\plus{}c_{n\plus{}1}\plus{}d_{n\plus{}1}\equal{}2(a_n\plus{}b_n\plus{}c_n\plus{}d_n)\]
\[ \Rightarrow a_n\plus{}b_n\plus{}c_n\plus{}d_n\equal{}2^{n\minus{}1}(a_1\plus{}b_1\plus{}c_1\plus{}d_1)\]
If the sequences are going to be periodic, there must be a common period (any common multiple of the periods of the four sequences) and this forces $ a_n\plus{}b_n\plus{}c_n\plus{}d_n\equal{}0$ for all $ n$.

Because of $ a_n\plus{}b_n\plus{}c_n\plus{}d_n\equal{}0$, we require $ c_n\equal{}\minus{}a_n,d_n\equal{}\minus{}b_n$ for $ n\geq 2$. This means, for $ n\geq 2$,
\[ a_{n\plus{}1}\equal{}b_n\plus{}a_n,b_{n\plus{}1}\equal{}b_n\minus{}a_n\]
In particular,
\[ a_{n\plus{}1}^2\plus{}b_{n\plus{}1}^2\equal{}2(a_n^2\plus{}b_n^2)\Rightarrow a_n^2\plus{}b_n^2\equal{}2^{n\minus{}2}(a_2^2\plus{}b_2^2)\]
Using periodicity again, we need $ a_2^2\plus{}b_2^2\equal{}0\Rightarrow a_2\equal{}b_2\equal{}0$. Since $ c_2\equal{}\minus{}a_2,d_2\equal{}\minus{}b_2$, all are 0.


Prove that a normed space $ E$ is euclidian if and only if its norm verifies Ptolemy's inequality.

This is a very classical problem. For example, you can find a proof in the book ""Analyse" by Xavier Gourdon :)


induction
number theory unsolved
number theory

:arrow: Let $ a,b,c \in Z^*$ and$ a\plus{}b\plus{}c$=0
   prove that:$ a^n\plus{}b^n\plus{}c^n$ $ \equiv$0 ($ mod$ ($ a^4\plus{}b^4\plus{}c^4$)) if $ n \equiv$1($ mod$3).  :ninja:

I proved this problem by induction  :lol: but it is very long  :(

Let $ A_n \equal{} a^n \plus{} b^n \plus{} c^n$, then $ A_0 \equal{} 3,A_1 \equal{} 0,A_2 \equal{} 2P,P \equal{} a^2 \plus{} ab \plus{} b^2,$ and
$ A_{n \plus{} 3} \equal{} PA_{n \plus{} 1} \minus{} 2QA_n,Q \equal{} \frac{ab(a \plus{} b)}{2}\in Z.$
It give $ A_3 \equal{} \minus{} 6Q,A_4 \equal{} 2P^2, A_5 \equal{} \minus{} 8PQ,..$
Obviosly $ A_n$ is polinom with degre n $ A_n \equal{} f(P,Q)\equiv B_nPQ^k \plus{} C_nQ^l\mod A_4 \equal{} 2P^2,$ were degre calculate $ degP \equal{} 2,deg Q \equal{} 3$. Therefore $ B_n \equal{} 0 \equal{} C_n$, when $ n\equiv 1\mod 3$.
I prove only $ 2|A_4, \frac{2A_{3n\plus{}1}}{A_4}\in Z$.
If $ A_{3n}\equal{}C_nQ^n\plus{}C_{n1}Q^{n\minus{}2}P^2\plus{}...$
$ A_{3n\plus{}1}\equal{}2P^2D_nQ^{n\minus{}1}\plus{}...$
$ A_{3n\plus{}2}\equal{}2B_nQ_n\plus{}...$
$ C_{n\plus{}1}\equal{}\minus{}2C_n,B_{n\plus{}1}\equal{}\minus{}(C_n\plus{}B_n),D_{n\plus{}1}\equal{}B_n\in Z$.

why you have  $ B_n$=0=$ C_n$ when n$ \equiv$1($ mod$ 3)

[quote="hoangclub"]I think this statement is false.[/quote]
$ A_n\equal{}R_n(P,2Q)$, were $ R_n(x,y)\in Z[x,y]$ and $ deg(R_n)\equal{}n, deg_(P)\equal{}2,deg(Q)\equal{}3)$.
Therefore $ R_{3k\plus{}1}\equal{}a_0P^2(2Q)^{k\minus{}1}\plus{}a_1P^5(2Q)^{k\minus{}1}\plus{}...$.
It proved, that $ 2P^2|A_{6k\plus{}1}$.
$ A_{6k\plus{}4}\equal{}PA_{6k\plus{}2}\minus{}2QA_{6k\plus{}3}\equal{}P^2A_{6k}\minus{}2QPA_{6k\minus{}1}\minus{}2QA_{6K\plus{}1}.$
Therefore it is true, if $ 2|A_{6k}\equiv A_{6k\minus{}2}\mod 2 ...\equiv A_2\mod 2\equiv 0$.


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Given that $2x^2+3xy+2y^2=1$.
Find the minimum value of $f(x,y)=x+y+xy$.

I want to solve this problem by Lagrange Multiplier but I am not familiar with that.
Please help.

This is Hosei University/Law entrance exam 1978 in Japan, but Japanese high school students don't know Lagrange's multiplier.

The idea is that the gradients of the function and its restrain shult be scalar multiples.
So:

$\nabla{f} = \lambda\nabla{g}$

now :

$\nabla{f}= (\frac{\delta{f}}{\delta{x}},\frac{\delta{f}}{\delta{y}})$
in the same way find $\nabla{g}$ and you get an equation in two unknowns so find the reult. Hope it helps.

This problem can be solved by other methods, for example:
Method 1
With a suitable choice of rotation, we can transform $2x^2+3xy+2y^2=1$ into a standard form without the xy term. Then apply the usual parametric form $x=r \cos\theta,y=r \sin \theta$.
Substitute that into f(x,y) and deal with a minimzation problem for 1 variable.
Method 2
$2x^2+3xy+2y^2=1 \implies 2(x+y)^2=1+xy \\ \implies 2(x+y)^2 =f(x,y)-(x+y)+1$.
Re-arrange terms: $2(x+y)^2+(x+y)-[1+f(x,y)]=0$.
Regarding this equation as an equation in (x+y).
Then $discriminant \geq 0$.
$1+8[1+f(x,y)]\geq 0 \implies f(x,y)\geq -\frac 9 8$.
Therefore, the minimum value of f(x,y) is $-\frac 98$.

I want to learn how to apply Lagrange Multiplier too.  :)

Let $g(x,y) = 2x^2 + 3xy + 2y^2 = 1$.

First, you must get rid of the boundary cases. (Here you will get a minimum, I think. I haven't checked all the computations)

Next, use the Lagrange multiplier $\lambda$:

$\displaystyle \frac{\partial f}{\partial x} = \lambda \frac{\partial g}{\partial x}$; $\displaystyle \frac{\partial f}{\partial y} = \lambda \frac{\partial g}{\partial y}$.

$\displaystyle \, \, \Longleftrightarrow \, \, \, \, \, y+1 =\lambda \left( 4x + 3y \right)$; $\displaystyle x+1 =\lambda \left( 4y + 3x \right)$.

Subtracting we get $\displaystyle y-x = \lambda \left( x - y \right) \, \, \, \Longleftrightarrow \, \, \, \left( \lambda + 1 \right) \left( x-y \right) = 0$.

[i]Case 1.[/i] $x=y$;

We have $2x^2 + 3x^2 + 2x^2 = 7x^2 = 1$, so $\displaystyle x = \frac{1}{\sqrt 7}$, so $f(x,x) = \frac{2}{\sqrt 7} + \frac{1}{7}$.

[i]Case 2.[/i] $\lambda = -1$.

We get that $4x+4y+1=0$. Combining this with $g(x,y)=1$ we get some $x,y$ and then we compute $f(x,y)$.

Lagrange Multipliers are very algorithmical, so it's easy to get the hang of it, but in most problems they are useless.

By the way, I'd take the non-calculus solution any day.

Very nice method in case of this type of problems. Thanx for writing up the whole solution!

What's values of the pair $(x+y,xy)$ for minimum value?

In Romania this problem is for the grade $9$
(only with the second degree trinomial).[hide="Here is a simple solution."][color=darkblue][b]Proof.[/b] With the substitutions $x+y=u$, $xy=v$, the our problem
 becomes : $\min \{\ u+v\ |\ 2u^2=v+1\ ,\ u^2\ge 4v\ ,\ u\in R\ ,\ v\in R\ \}$, i.e.
$\min_{u\in R} \left(\ 2u^2+u-1\right)=\boxed {\ -\frac 98\ }\ .$ The co-ordinates of the minimum point
$\left(\ x_{m}\ ,\ y_{m}\ \right)$ are the roots of the equations $8t^2+2t-7=0$[/color]
$\left(u_{m}=-\frac 14\ ,\ v_{m}=-\frac 78\right )\ .$[/hide]

My approach is the same as Virgil Niciula, but we should limit the range of $u.$

Indeed, that's the best solution, but note that it's been already posted.

No. Do you think that $x+y\in \mathbb{R}\Longrightarrow x\in \mathbb{R}$ and $y\in \mathbb{R}?$

If $x\in R$, $y\in R$ for which $x+y=u$ and $xy=v$ then :

$\boxed {\ x\in R\ ,\ y\in R\ ,\ 2x^2+3xy+2y^2=1\ }$ $\Longleftrightarrow$

$\boxed {\ u\in R\ ,\ v\in R\ ,\ 2u^2=v+1\ ,\ u^2\ge 4v\ }\ .$

[b]Remark.[/b] $|u|\le \frac{2\sqrt 7}{7}\ ,\ v\le \frac 17\ .$

[quote="Virgil Nicula"]If $x\in R$, $y\in R$ for which $x+y=u$ and $xy=v$ then :

$\boxed {\ x\in R\ ,\ y\in R\ ,\ 2x^2+3xy+2y^2=1\ }$ $\Longleftrightarrow$

$\boxed {\ u\in R\ ,\ v\in R\ ,\ 2u^2=v+1\ ,\ u^2\ge 4v\ }\ .$[/quote]

This is what I wanted to say, thank you Virgil Nicula.

kunny