In triangle $ ABC$, $ 3\sin A \plus{} 4\cos B \equal{} 6$ and $ 4\sin B \plus{} 3\cos A \equal{} 1$.  Then $ \angle{C}$ in degrees is

$ \textbf{(A)}\ 30 \qquad \textbf{(B)}\ 60 \qquad \textbf{(C)}\ 90 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 150$

Thanks in advance!

Wait are you saying 3+4=6, cause it's 7..

I don't get it

this very confusing.  are those supossed to be angle numbers?

[quote="314.math"]In triangle $ ABC$, $ 3\sin A + 4\cos B = 6$ and $ 4\sin B + 3\cos A = 1$.  Then $ \angle{C}$ in degrees is

$ \textbf{(A)}\ 30 \qquad \textbf{(B)}\ 60 \qquad \textbf{(C)}\ 90 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 150$

Thanks in advance![/quote]

[hide="Solution"]
The $ \sin, \cos$ and the $ 3,4$ immediately suggest squaring:

\begin{align*}
9\sin^2 A + 16\cos^2 B + 24 \sin A \cos B &= 36 \\
+ 9\cos^2 A + 16\sin^2 B + 24 \sin B \cos A &= 1 \\
\Longrightarrow 25 + 24(\sin A \cos B + \sin B \cos A ) &= 37
\end{align*}

Where we arrived at the third equation by summing and using Pythagorean identities. Now we recognize the sine addition identity, so $ \sin (A+B) = \sin (180 - C) = \sin C = \frac{12}{24} = \frac 12$. Thus either $ C = 30^{\circ}, 150^{\circ}$.

If $ C = 150$, then $ A+B = 30 \Longrightarrow A,B < 30$, and $ 3 \sin A + 4\cos B < 3\left(\frac 12\right) + 4(1) = 5.5 < 6$, contradiction; thus $ C = 30 \Longrightarrow \mathrm{(A)}$.

[/hide]

[quote="azjips"]$ \sin (A\plus{}B) \equal{} \sin (180 \minus{} C) \equal{} \sin C \equal{} \frac{12}{24} \equal{} \frac 12$[/quote]

How was the above determined?

Thanks!

Since it is a triangle, $ A\plus{}B\plus{}C\equal{}180$, so $ \sin(A\plus{}B)\equal{}\sin(A\plus{}B\plus{}C\minus{}C)\equal{}\sin(180\minus{}C)$. Also, from the equation $ 25\plus{}24(\sin A \cos B \plus{} \sin B \cos A)\equal{}37$, we get $ \sin A \cos B \plus{} \sin B \cos A \equal{} \sin(A\plus{}B)\equal{}\frac{1}{2}$, so $ \sin C\equal{}\frac{1}{2}$.

[quote="azjps"][quote="314.math"][quote="azj[i]i[/i]ps  :huh: "] [/quote] [/quote]

Brut3Forc3, I could use your avatar.  :P[/quote]

Oops.  :blush:  Typing Error - Sorry, azjps.

this is a classic example of a problem that can be easily solved using symmetry

[hide="Initial Thoughts"]We have 2 different clues to use squares.  We have symmetric sines and cosines and we have 3 and 4.  These are absolutely dead giveaways.[/hide]
[hide="Solution"]Squaring both equations and adding them
$ 9\sin^2 A \plus{} 24 \sin A \cos B \plus{} 16\cos^2 B \equal{} 36 \\
\plus{} 9\cos^2 A \plus{} 24 \sin B \cos A \plus{} 16\sin^2 B \equal{} 1 \\
\equal{} 25 \plus{} 24(\sin A \cos B \plus{} \sin B \cos A ) \equal{} 37$
From this we get
$ \sin A \cos B \plus{} \sin B \cos A \equal{} \frac{1}{2}$
Using our knowledge of trig and triangles,
$ \sin A \cos B \plus{} \sin B \cos A \equal{} \sin {(A\plus{}B)}$ and $ A\plus{}B\plus{}C\equal{}180$
Plugging these in we get
$ \sin {A\plus{}B}\equal{}\sin {180\minus{}C}\equal{}\frac{1}{2}$
And because
$ \sin {180\minus{}C}\equal{}\sin{C}$
We have
$ \sin{C}\equal{}\frac{1}{2}$
Therefore $ C\equal{}30 \text { or }150$.
If $ C\equal{}150$, then $ A\plus{}B\equal{}30$ which means that $ A<30 \text{ and } B<30$
$ 3 \sin A \plus{} 4\cos B < 3\left(\frac 12\right) \plus{} 4(1) \equal{} 5.5 < 6$
This means $ C \neq 150 \implies C\equal{}30 \implies \boxed{A}$[/hide]


algebra
polynomial
inequalities
trigonometry
algebra proposed

Find all families of polynomials with complex coefficients $ (f_n)$ such that $ deg f_n=n$ and $f_n,f_m$ commute for all m,n.

What do you mean by 'family'?
At least I know one such family: The Chebyshev polynomials.. :)

This is a result known as Julia-Ritt's theorem, and it has been generalised by Jacobsthal :

Any such a family of polynomial is an affine transformation (on the variable and on the polynomial) from one of the two sequences $(X^n)$ and $(T_n)$, where $T_n$ is the Tchebychev polynomials.

Pierre.

I knew that Julia has proved a much stronger result in 1922, but which is proved with very complicated methods. Anyway, this problem\theorem has a really nice, short ( :D )and super beautiful solution. I will let the pleasure of finding it. The answer is that indicated by Pierre.


I type in english, but I can read korean and japanese.

I voted.. but I am kinda in the middle..

I am Vietnamese I know some Korean Eg hi, alo, sorry , i love you ...  :D but i write English beter than  :lol:

my popa said that my first language was english rather than Korean....... :P

Of course, Korean. Because I'm a native Korean -_-;;

I'm learning two foreign languages, English and Japanese, but still not so fluent.

A funny thing is that I have spent much more time learning English than Japanese, but now I uses Japanese better.  :maybe:

둘다 잘하는데... ㅋㅋ

English for writing, speaking, reading, etc.
Of course, I graduated middle school, high school, and now attending a college in United States so I did more literature work in English. Nevertheless, I can do most of works in Korean... just don't make me translate a book or something.

ㅋㅋ.  :)

Because I'm native Koreanㅡㅡ;;

My second language is English, and the third is Russian(but not very fluent).

Grew up in USA, so English is better, but I have been in Korea for the last 7 years (I'm going back to NJ in March)
My English Reading skills are WAY better than my Korean, but I can't speak English properly because I dont use English in everyday life.

[quote="Erken"][quote="GrounD_ZerO"]and the third is Russian(but not very fluent).[/quote]

wow really?

how did you learn it?[/quote]

by myself, but still in progress, and now I stopped it tempoarily because of Physics.

I was born in Korean, but I lived in the US for half my life now :P

whee all these Korean people talking about themselves! hi unnis rang opas

I can speak Korean pretty well and understand it fairly well as well.

But I'm more comfortable writing and speaking in english.

I am fluent in both languages, apparently.

한국인이 생각보다 많네여...ㅋㅋㅋ :) And I'm way better at Korean than English.

I can do Chinese and Korean, but honestly I type a :censored: load better in English.
That above was typed in approximately twelve s econds(including this)
아놔 한국어 타자 나 진짜 느리네
이것이 20초쯤 걸렸어 :P

나는 둘다 잘하는데... ㅋㅋㅋ

[quote=Valkoorsky18]i legit can't type korean for the love of me...

even though i am half-korean, i live in the u.s. and never had a chance to learn.[/quote]

same......
ㅋㅋㅋㅋㅋ

i'm pure korean but was born in america ;-;


Ok I know this week was sprint round practice.
I'm just curious who pwned it and who got pwned.
Someone(I mean you bluestalk) post the ppl's scores here(i don't care about ppl's 'privacy', so just post).

What the...I POST THE WEEKLY DISCUSSIONS! JEEZ! EVEN IF I'M LAZY AND DON'T GET AROUND TO IT UNTIL FIVE!


calculus
integration
trigonometry
calculus computations

Compute this integral:
\[\int_{0}^\pi{\frac{{x\sin (x)}}{{1+\cos^{2}(x)}}dx}\]

[hide]Change varible $x=\pi-t$ !  :wink: [/hide]


A question: How can you participate in this olympiad? what are the request?????.

When and where is the next IMC??????

Have you visited this site ?
[url=http://www.imc-math.org/]www.imc-math.org[/url]

the exact request is "no more than 4 years of (university) mathematics study at the moment of participation". So, unlike the rules state, you can e.g. participate while you're in high school.

[quote="Peter VDD"]you can e.g. participate while you're in high school.[/quote]

!!!!!!!!!!!!!

Yes, is that so surprising?

I cannot guarantee you'd get a good result, but you are allowed. ;)

It's not surprising, as people from Romania have done this. Yet there is not point in going if you have no idea about Calculus or Linear and Superior Algebra  :roll:

Well actually I thik 4 questions could be solved elementary last year. Enough for a third prize if you could do that.


Is there a finite set $S$ of $n$ points in a plane ($n>3$), such
that for every two points $A,B\in S$ there is a point $C\in S$ s.t.
triangle $ABC$ is equilateral?

i believe no, at least in two dimension (in three take a thetraedron), since given two points such that their distance is maximum, we have to conctruct another point. By induction, you can show that there won't be two couples of points whose distance is equal and maximum


integration
calculus
trigonometry
real analysis
real analysis unsolved

It's a really nasty-well-known integral

First set $u^{2}=\tan t$, after use partial fractions...


Are there any good and free (shareware is fine) software which can draw laboratory set ups? Very urgent. Thanks.

[quote="karthik_k"]Try Chemsketch --
http://www.acdlabs.com/download/
It's amazing when it comes to anything to do with chemistry..[/quote]
i've tried that, but where can I draw laboratory setups? I can't find it..

Try the older version of the software -- It supports drawing lab setups ------- as far as the latest version is concerned , I dont know where the stuff is....... :?

hmm,i wonder if thers a program that can do equations for us???? :D


[quote="firecricket91"]Alright guys im going to stop spamming now (although half of the posts actually have a point)

to arch since you brought it up in another post, my goal was trying to find vikas's account so im saying things that would make him post (since he doesnt want to tell me or anyone else his account here for that matter)-i dont know why though since its not a big deal.

Umm yea..i guess ill stop spamming now but we can still revive the problem marathon.................[/quote]

yea i have been watching the Indiana forum for quite some time and its very very spammy

I think ill apply to be the mod of this forum.

lol i pmed an administrator, asking to be the indiana mod, even though i'm moving.  Ill still go through this forum though.

[quote="hunter34"]You don't apply to be a moderator, or at least I don't you do.[/quote]

http://www.artofproblemsolving.com/Wiki/index.php/AoPS_forums_tutorial#Becoming_a_Moderator

You don't. Though all those people in the "Spam in Mathcounts forum" topic don't seem to known that.  :roll:


A charity sells $140$ benefit tickets for a total of $2001$ Dollars. Some tickets sell for full price ( a whole dollar amount ), and the rest sell for half price. How much money is raised by the full price tickets?

A. 782 dollars

B. 986 dollars

C. 1158 dollars

D. 1219 dollars

E. 1449 dollars

[hide="Suggestions"] Find a range for the price and then set up the equation.  At this point, you can either guess-and-check or solve using the fact that you know the ticket price is an integer. [/hide]


Let be $ x,y,z>0$ such that $ xyz\equal{}1$ . Show that :

$ (xy^2\plus{}yz^2\plus{}zx^2)^2\ge \frac{x^2}{\sqrt{y}}\plus{}\frac{y^2}{\sqrt{z}}\plus{}\frac{z^2}{\sqrt{x}}$

[quote="alex2008"]Let be $ x,y,z > 0$ such that $ xyz \equal{} 1$ . Show that :

$ (xy^2 \plus{} yz^2 \plus{} zx^2)^2\ge 3( \frac {x^2}{\sqrt {y}} \plus{} \frac {y^2}{\sqrt {z}} \plus{} \frac {z^2}{\sqrt {x}})$[/quote]$ (*)$
$ (*) < \equal{} > (a^2 \plus{} b^2 \plus{} c^2)^2 \geq 3(a^3b \plus{} b^3c \plus{} c^3a)$
(with $ a \equal{} \sqrt {\frac {x}{y}}  ,...$)


geometry
3D geometry
sphere
email
logarithms
AMC

Help me with these problems please:

1. Prove that power of 2 has a decimal expansion that begins with 1999 i.e. prove there exists n such that 2^n = 1999....

2. 2001 flies are inside a cube of side 1. Prove that three of them are within a sphere of radius 1/11

Email is always better!! (kky1638@hotmail.com) :D

[quote="avolfson"]For the second one it seems that you need to find how many cubes of side length 1/11 are in the cube. This would be

\[\ (1/11)^3 \]
or something like
\[\ 1/14641 \]?

No, ignore that... you have a radius.... hmmm maybe its realated to that somehow...[/quote]

11^3 = 1331.

For the first problem:
Hint #1: [hide]You are going to use the Pigeon Hole Principle.

Hint #2: [hide]If $n = 1999abcdef\dots$, what can be said about $\log{n}$?

Hint #3: [hide]What can be said about $\log{2}$?[/hide][/hide][/hide]

Here's a more thorough chain of hints for the second one:
Hint #1: [hide]You are going to use the Pigeon Hole Principle.

Hint #2: [hide]Try to find 1000 spheres such that the union of the 1000 spheres contains the entire cube.

Hint #3: [hide]Does a cube of side length 1/10 fit inside a sphere of radius 1/11?[/hide][/hide][/hide]


Δίνεται το σύνολο Χ των 2x2 πινάκων με μιγαδικούς συντελεστές  και πίνακας Α που ικανοποιεί τη σχέση $ A^4 \equal{} A$ Βρείτε ένα υποσύνολο Y του Χ με τις ιδιότητες
1. Αν A ανήκει στο Χ υπάρχει ένας πίνακας Β που ανήκει στον Y και είναι όμοιος με τον Α
2. Αν Α ανήκει στον Y και Β ανήκει στον Y τότε ο Α και ο Β δεν είναι όμοιοι

[color=red][Moderator edit: Σε παρακαλώ μην ξαναβάλλεις πίνακες στο φόρουμ των ολυμπιάδων υπάρχει αντίστοιχο υποφόρουμ][/color]

Ypodeiksi : O pinakas A einai diagonopoiisimos. Ara, poia einai i moni sobari epilogi gia to Y ?


trigonometry
calculus
calculus computations

Show that $ s_n\equal{}\sum_{k\equal{}1}^n \frac{\sin k}{2^k}$ is a Cauchy sequence.

[quote="facis"]What is a cauchy sequence?[/quote]

[b][color=red]"That kind of stuff is easily found with Google."[/color][/b]
[url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=50490]BlueVelvet[/url] in this [url=http://www.mathlinks.ro/viewtopic.php?t=252125][i]post[/i][/url].  :wink:

[b]References[/b]

[1] [url=http://en.wikipedia.org/wiki/Cauchy_sequence]Wikipedia's take on Cauchy sequences[/url]
[2] [url=http://mathworld.wolfram.com/CauchySequence.html]MathWorld's respective entry[/url]

I'll keep it simple.  If you're working with real numbers, a Cauchy sequence is any sequence that converges (this is not true in an arbitrary metric space).

Wait a minute-what does "converges" mean? :wink:

thanks everyone.  btw, I know what convergence is.  We are doing problems liek this in my calculus class, but we just haven't used the term Cauchy sequence and aren't worrying about converging within a specific set.


anyone knowing a good converter for Windows which doesn't use perl ?

i am not sure, but doesnt MikeTex is Latex for windows?  btw you should really add Latex support to this cbb!!  the equation would come out nicer

you didn't understood what I meant  - I need a LaTeX2Gif converter - unfortunately all that I could find on the web are perl-based converters.

My host-provider doesn't have perl installed, so I need some other kind of converter.

Some years ago I used emTeX for MSDOS because of my old slow CPU,
and there was a batch file in that package called djpcx that used emTeX dvi-driver to convert dvi->pcx (which can be converted to gif easily) with pretty good quality.
BTW you can search anything related to TeX at http://www.ctan.org   ...

yes I know. but unfortunately my current provider cannot offer me support (and does not agree with installing) LaTeX (namely tetex) so I am searching for a new provider. I am waiting for an  answer for a long time now from a certain firm, but they seem very un-communicative.

try teXaide... i'm using it... it's a stand-alone aplication... but unfortunatly it doesn't have save or export...
i'll send u via email to see what's about maybe it will help you :D

no thanks. I know what I have to do, I only need some material support.

There was one that used php, just have to contact the guy that wrote it, he's not online currently.

People fail to recognize this is 12 years old.

Twelve years old? What a shocker! I can't believe it![color=#eee]'s not butter[/color]

Valentin Vornicu is the instructor of my class that is going on right now.


[hide]the sum of the angles in a triangle is 180. It has 3 angles so 180/3=[b]60[/b][/hide]

[hide] The sum of a triangle's angles are 180, 180/3 angles=60 degrees[/hide]

[quote="bob123"][quote="solafidefarms"]What is the average of the degree measures of the angles of a triangle?

Some people missed it at Nats!!![/quote]

Actually, you were supposed to give the answer in radians, and only 1 person missed it (227 people got it right, I checked the statistics sheet)[/quote]

Actually, I just checked the stat sheet, and 225 people got it right. The question was actually [quote]What is the mean value of the measures of the three interior angles of any triangle?[/quote] and the answer blank had units of degrees. They don't use radians at MC.

[quote="uberl33tmage"][hide]60[/hide][/quote]
could you show any work?

Where did you guys get the stat sheets?

If it was radians, it would be $\frac{\pi}{3}$, I dont' have the test.

[quote="solafidefarms"]One of the perqs of being at Nats :P[/quote]
WHAT!! I was at nats why didn't I get one??


algebra
function
domain
topology
real analysis
real analysis unsolved

Let $ \mu$ be a Borel measure on $ \mathbb {R}$.

Denote $ \mathbb {M} _ \mu$ as the domain of $ \mu$

Then for any $ E \in \mathbb {M} _ \mu$, we have:

$ \mu (E) = inf \{\sum _{1}^ \infty [F(b_n) - F(a_n)]: E \subset \bigcup _1^ \infty (a_n,b_n] \}$

By a theorem, I also know that:

$ \mu (E)=inf \{ \mu (U) : E \subset U,U \ open \}$

Question: If $ E \subset \mathbb {R}$, and if $ E \in \mathbb {M} _{ \mu }$ , prove that $ E=V \backslash N$ where $ V$ is a countable union of open subset and $ \mu (N) = 0$

Proof so far.

Case 1.  If $ \mu (E) < \infty$ :

Given $ \epsilon > 0$ , there exists open sets $ G_n$ with $ E \subset G_n$ , $ \mu (G_n) < \mu (E) + \frac { \epsilon }{2^n}$ since $ \mu (E)$ is the infimum of these sets.

Define $ V = \bigcap G_n$ .

Then we know the followings:

i. $ E \subset V$, implies that $ \mu (V) \geq \mu (E)$

ii. $ \mu (V) \leq \mu (G_n) < \mu (E) + \frac {\epsilon }{2^n}$
implies that $ \mu (V) \leq \mu (E)$

Using both i and ii, I then conclude that $ \mu (V \backslash E ) = 0$ , in which implies $ V \backslash E = N$ , implies that $ V \backslash N = E$ with $ \mu (N) = 0$

Case 2. $ \mu (E) = \infty$:

In here, I want to do the same trick.  So can I assume that $ \exists F \subset E \ s.t. \ 0 < \mu (F) < \infty$ ?  I know I can do that for any semi-finite measures.

Any hints?  Thank you.

[quote="tttcomrader"]
Case 2. $ \mu (E) \equal{} \infty$:

In here, I want to do the same trick.  So can I assume that $ \exists F \subset E \ s.t. \ 0 < \mu (F) < \infty$ ?  I know I can do that for any semi-finite measures.

Any hints?  Thank you.[/quote]
I don't know how such $ F$ helps you, but yes, it exists, you can take $ F \equal{} E\cap [ \minus{} n,n]$ for $ n$ large enough (so that $ \mu(F) > 0$).

Anyway, in this case it is also possible to show that for any $ \epsilon > 0$ there exists an open set $ O\supset E$ s.t. $ \mu(O\setminus E) < \epsilon$.

I'm sorry but why is $ \mu(O\setminus E) <\epsilon$ true?

Because a unit of open sets is an open set ;)
More precisely: approximate $ E\cap [n,n\plus{}1]$ properly by an open set and then take the union of those sets. I believe, it should work.


can we prove that only by using limit (that is what i've done) if not i will be grateful to you if you post the solution.

$]0,\pi]=\coprod_{k=1}^{+\infty}$ $]\frac{\pi}{k+1},\frac{\pi}{k}]$


Well, it's getting more difficult to read what you write. Could you please try to use real sentences and real spelling (in English) so we can understand what you are saying?

Recent examples:
[quote="churchilljrhigh"]
you should checek out OPus Dei boy they are touhg they wear metal clices. on their legs that sting and draw blood. adn self whip themselves.[/quote]

[quote="churchilljrhigh"]
during the vientam war the vientaemes hated south vietnam governemnt becasue the dictaor of south vietnam christian and persecuted vietense (the reason u.s lost vietnam the u.s. supported likes to support any person going agianst communists even if he is a brutal dictator)[/quote]

I don't mean this post in a derogatory way.

mmm, interesting how there's always at least one person with bad spelling/grammer at all times during AoPS history ('tis true, I kept track)...maybe we do need that policeman of Jim's :lol:

What's with the singling out?  To be totally honest I doubt there is "one" person that we should all shake the finger at to look over his/her grammar before posting.  Maybe instead of directing this topic to a specific person, which instantly brings to my mind "ouch! sorry man!", we should do a general sweep.  sorry man!

[quote="MagnusRex"]Well, it's getting more difficult to read what you write. Could you please try to use real sentences and real spelling (in English) so we can understand what you are saying?[/quote]

Allow me to agree with this request and, as a mathematician might, generalize it. The quality of the AoPS Web site is judged in part by the content contributed by forum participants, and I would urge everyone to fact-check, spell-check, and grammar-check before posting. Additionally, I would urge people to go back to their previously posted messages and EDIT the ones most in need of editing. If you are logged in with your user name and go back to read your own previous posts, you will find out you can edit them.

That's the basic principle: remember that this is a place where you can begin to establish your reputation among your possible future colleagues. An electronic forum that is, in principle, viewable by everyone in the world with Internet connectivity is hardly the place to keep up junior high social habits. Rather, it is the place to begin developing grown-up habits of communication.

Test Name: S.A.T. 9th Edition     	   Level: ADV I
  Test Date: October, 2002   	  Type: NR
  School: CHURCHILL JUNIOR HIGH    	  Grade: 8

  TITLE 	  PERCENTILE*
  READING VOC. 	96
  READING COMP. 	99
  TOTAL READING 	99
  MATH PROB SOLV 	99
  MATH PROCEDURES 	99
  TOTAL MATH 	99
 [b] PREWRITING 	92[/b]
  COMPOSING 	98
  EDITING 	96
  TOTAL LANGUAGE 	97
  SCIENCE 	98
  SOCIAL SCIENCE 	95
  LISTENING 	99
  USING INFO. 	97
  THINKING SKILLS 	99
  TOTAL BATTERY 	98

I guess I need to work on my prewriting skills. OK I will try to change.

Interesting how people who aces tests doesn't necessarily speak with correct grammar in everyday life...

You're right, I've seen at least two more people that can currently use a whole lotta of the things tokenadult wrote (although I kinda trailed off at around the first paragraph :lol:) especially :wink:

[quote="Tare"]Interesting how people who aces tests doesn't necessarily speak with correct grammar in everyday life...
[/quote]

You *did* do that on purpose eh? ;)

[quote="churchilljrhigh"]
Test Name: S.A.T. 9th Edition Level: ADV I
Test Date: October, 2002 Type: NR
School: CHURCHILL JUNIOR HIGH Grade: 8

TITLE PERCENTILE*
READING VOC. 96
READING COMP. 99
TOTAL READING 99
MATH PROB SOLV 99
MATH PROCEDURES 99
TOTAL MATH 99
PREWRITING 92
COMPOSING 98
EDITING 96
TOTAL LANGUAGE 97
SCIENCE 98
SOCIAL SCIENCE 95
LISTENING 99
USING INFO. 97
THINKING SKILLS 99
TOTAL BATTERY 98

I guess I need to work on my prewriting skills. OK I will try to change.

[/quote]

Well, ok. Good Job. I wasn't questioning your intelligence. I was asking you to be less careless.

:) I know Im one of those people as some people  :wink: on this sight should know. Churchhill just know your not alone  :). :wink: :wink:...

Hmm, it's so tempting to do spelling fixes ...

Joel, if you did that, you would be busy on Aops 24/7.

Gaah why is everyone reviving year-old topics at random all of a sudden?!

Oh, yeah.  :blush:

Just that JBL's comment was so worth commenting on. I'll be sure not to do it again.

The "real" aalindsey did not post this. It seems like someone is using her screen name and everything.

The posts before Wednesday came from 2004. But since when has churchilljrhigh been a girl?

since a while back even though posts from early on led me to think churchill was a boy, unless it's all some weird thing to confuse us


Hi,
Using DePolignac's Formula, we write the expression for the maximum exponent of 2 that divides (2n)!, and also an expression for the maximum exponent of 2 that divides (n)!, the subtaction of these two expressions ought to give the maximum exponent of 2 that divides the given expression. If one calculates it's easy to see that the result is n. Therefore, 2^n divides the given expression.

Multiply the top and bottom by $(n)!$ to get $\frac{(2n)!}{2^{n}n!}$. Now factor out $2^{n}$ from the numberator, leaving

$\frac{2^{n}n!(1*3*5*\ldots*(2n-1))}{2^{n}n!}$

This leaves the expression $1*3*\ldots*(2n-1)$

[quote="José"][quote="Farenhajt"]Or just use induction :D[/quote]

Please, can you show how??[/quote]

For $n=1$ we get $\frac{2}{2}=1$

If the fraction is integer for $n=m$, then for $n=m+1$ we get

\begin{eqnarray*}\frac{(m+2)(m+3)\dots(2m)(2m+1)(2m+2)}{2^{m+1}} &\stackrel{\text{put the last factor first}}{=}& \frac{(2m+2)\cdot(m+2)(m+3)\dots(2m)(2m+1)}{2^{m+1}}\\ &\stackrel{\text{cancel out the 2}}{=}& \frac{(m+1)\cdot(m+2)(m+3)\dots(2m)(2m+1)}{2^m}\\ &=& (2m+1)\frac{(m+1)(m+2)(m+3)\dots(2m)}{2^m}\end{eqnarray*}

The last fraction is integer by the inductive assumption.


geometry
pigeonhole principle
algebra proposed
algebra

Let $x_0,x_1,...,x_{n-1}$ be some points on the unit circle. Let also $A_1...A_n$ be a regular polygon inscribed on this circle. Fix a number $k$ (entire) between 1 and $n/2$. Prove that one can find $i,j$ such that $d(A_i,A_j)\geq d(A_1,A_k)\geq d(x_i, x_j)$. Here $d(A,B)$ is the usual distance between two points.

are the $x_i$ arranged in that order in the circle?

sorry for the stupid question above! :D
anyway here's my solution:
we may find an arc corresponding to a central angle of $\frac{2\pi}{k-1}$ containing at least k of the $x_i$ let's say $\{x_{i_1},...,x_{i_k}\}$(all these points satisfy then $d(A_1,A_k)\geq d(x_{i_r},x_{i_s})$)
[hide="proof"]let us call the $x_i$ as $M_i$acording to the order they appear on the circle. then the sum of the angles $M_i O M_{i+k-1}$ is $n \frac{2\pi}{k-1}$so at least one of them by pigeonhole is going to be bigger than$\frac{2\pi}{k-1}$[/hide]
now we have to show that among k elements of {1,2,...,n} there are two whose difference is between k-1 and n-k+1
and we are done !
but that 's easy just put the numbers 1,2,...,n in that order in the circle and notice that at least two points will subtend an angle of less than $\frac{2\pi}{k-1}$ !


Let $ x_1,x_2,...,x_n$ be $ n$ real numbers.Consider $ S_k\equal{}x_1^k\plus{}x_2^k\plus{}...\plus{}x_n^k$. We have $ S_1\equal{}S_2\equal{}...\equal{}S_{n\plus{}1}$ .Prove that $ x_i\equal{}0$ or $ x_i\equal{}1$ for all $ i\equal{}1,2,...,n$

First we can WLOG assume $ x_1,\ldots ,x_n$ are all nonzero.
Then if write $ (X \minus{} x_1)\cdots (X \minus{} x_n) \equal{} X^n \plus{} a_1 X^{n \minus{} 1} \plus{} \cdots \plus{} a_n$,
we can apply the [url=http://en.wikipedia.org/wiki/Newton%27s_identities]Newton identities[/url] to get
$ S_1 \plus{} a_1 \equal{} 0$
$ S_2 \plus{} S_1a_1 \plus{} 2a_2 \equal{} 0$
$ S_3 \plus{} S_2a_1 \plus{} S_1a_2 \plus{} 3a_3 \equal{} 0$
$ \vdots$
$ S_n \plus{} S_{n \minus{} 1}a_1 \plus{} \cdots \plus{} S_1a_{n \minus{} 1} \plus{} na_n \equal{} 0$
$ S_{n \plus{} 1} \plus{} S_{n}a_1 \plus{} \cdots \plus{} S_2a_{n \minus{} 1} \plus{} S_1a_n \equal{} 0$

Then for $ S: \equal{} S_1 \equal{} \cdots \equal{} S_{n \plus{} 1}$ we get
$ S \plus{} a_1 \equal{} 0$
$ S(1 \plus{} a_1) \plus{} 2a_2 \equal{} 0$
$ S(1 \plus{} a_1 \plus{} a_2) \plus{} 3a_3 \equal{} 0$
$ \vdots$
$ S(1 \plus{} a_1 \plus{} \cdots \plus{} a_{n \minus{} 1}) \plus{} na_n \equal{} 0$
$ S(1 \plus{} a_1 \plus{} \cdots \plus{} a_{n \minus{} 1} \plus{} a_n) \equal{} 0$

Subtracting successive equations gives
$ (S \minus{} k)a_k \plus{} (k \plus{} 1)a_{k \plus{} 1} \equal{} 0$ for all $ 1 \le k < n$ and $ (S \minus{} n)a_n \equal{} 0$.
But $ a_n \equal{} ( \minus{} 1)^nx_1\cdots x_n \not \equal{} 0$.
So $ S \equal{} n$, and therefore $ a_1 \equal{} \minus{} S \equal{} \minus{} n$ and $ a_{k \plus{} 1} \equal{} \minus{} {n \minus{} k\over k \plus{} 1}a_k$ for all $ 1 \le k < n$.
This gives $ a_k \equal{} ( \minus{} 1)^k\left(n\atop k\right)$ for $ k \equal{} 1,\ldots ,n$ by induction.
And therefore $ (X \minus{} x_1)\cdots (X \minus{} x_n) \equal{} \sum_{k \equal{} 0}^n( \minus{} 1)^k\left(n\atop k\right)X^{n \minus{} k} \equal{} (X \minus{} 1)^n$.
So $ x_1 \equal{} \cdots \equal{} x_n \equal{} 1$.


Let P equal the sum of the measures of the interior angles of a convex polygon of n + 4 sides. Let Q equal the sum of the measures of the interior angles of a convex polygon of n - 4 sides. If n is an integer greater than 6 what is the value of P - Q, in degrees?

[hide="Solution"]
In general, the sum of the measures of the interior angles of a polygon with $ n$ sides is $ 180(n\minus{}2)$ (just consider the number of triangles, each with $ 180^{\circ}$, that fit in the polygon).

Thus,
$ P\equal{}180(n\plus{}2), Q\equal{}180(n\minus{}6) \Rightarrow P\minus{}Q\equal{}\boxed{1440^{\circ}}$
[/hide]


The equation [b](x)[/b]^6-5[b](x)[/b]^4+16([b]x[/b])^2-72[b](x)[/b]+9=0 has
a) exactly 2 distinct real root
b) exactly 3 distinct real root
c) exactly 4 distinct real root
d) exactly 6 distinct real root....

Dont post a complete solution try to post a solution by logical reasoning seeing the options.

hello, do you mean the equation $ x^6-5x^4+16x^2-72x+9=0$? Why don't you use $ \text{\LaTeX}$?
Sonnhard.


calculus
integration
trigonometry
logarithms
algebra
partial fractions
calculus computations

Evaluate $ \int_0^{\frac{\pi}{4}} \frac{\tan x}{1\plus{}\sin x}\ dx$.

I'm guessing the substitution $ u\equal{}\tan (\frac{x}{2})$ would work but is not the neatest way.

$ \frac{\tan x}{1\plus{} \sin x}\equal{} \frac{\sin x}{\cos x(1\plus{} \sin x)} \equal{}\frac{\sin x(1\minus{}\sin x)}{\cos x(\cos^{2}x)}$

$ \equal{}\tan x \cdot \sec^{2} x(1\minus{}\sin x) \equal{} \tan x\sec^{2}x \minus{} \tan^{2}x \sec x$

$ \equal{}\tan x\sec^{2}x\minus{}(\sec^{3}x\minus{}\sec x)$.

Now $ 2\sec^{3}x \minus{}\sec x\equal{}\frac{d}{dx}(\tan \sec x)$

so $ \int \sec^{3} x \ dx\equal{} \frac{\tan x\sec x}{2} \plus{}\frac{1}{2}\int \sec x  dx$.

Therefore, $ \int \frac{\tan x}{1\plus{}\sin x}\ dx \equal{} \frac{1}{2}\sec^{2} x \minus{} \frac{\tan x\sec x}{2} \plus{} \frac{1}{2}\int \sec x \ dx$

$ \equal{} \frac{1}{2}\sec^{2}x \minus{} \frac{\tan x\sec x}{2} \plus{} \frac{1}{2}\ln \mid \sec x \plus{} \tan x\mid \plus{}c$.

Putting in the limits $ 0$ and $ \frac{\pi}{4}$, we get

$ 1\minus{} \frac{\sqrt{2}}{2} \plus{} \frac{1}{2} \ln (1\plus{}\sqrt{2}) \minus{} \frac{1}{2}$

$ \frac{1}{2} \minus{} \frac{\sqrt{2}}{2} \plus{} \frac{1}{2}\ln (1\plus{}\sqrt{2})$.

Thanks kunny.

Is there a neater way of doing this?

O.K. Here is my solution:

$ \int_0^{\frac {\pi}{4}} \frac {\tan x}{1 \plus{} \sin x}\ dx \equal{} \int_0^{\frac {\pi}{4}} \frac {\tan x}{\cos ^ 2 x}\cdot (1 \minus{} \sin x)\ dx$

$ \equal{} \left[\frac 12 \tan ^ 2 x \cdot (1 \minus{} \sin x)\right]_0^{\frac {\pi}{4}} \plus{} \frac 12 \int_0^{\frac {\pi}{4}} \tan ^ 2 x\cdot \cos x\ dx$

$ \equal{} \frac 12 \left (1 \minus{} \frac {1}{\sqrt {2}}\right) \plus{} \frac 12 \int_0^{\frac {\pi}{4}} \frac {\sin ^ 2 x}{\cos x}\ dx$

$ \equal{} \frac 12 \left (1 \minus{} \frac {1}{\sqrt {2}}\right) \plus{} \frac 12 \int_0^{\frac {\pi}{4}} \left(\frac {1}{\cos x} \minus{} \cos x\right)\ dx$

$ \equal{} \frac 12 \left (1 \minus{} \frac {1}{\sqrt {2}}\right) \plus{} \frac 12 \left[\ln \tan \left(\frac {x}{2} \plus{} \frac {\pi}{4}\right) \minus{} \sin x\right]_0^{\frac {\pi}{4}}$

$ \equal{} \frac 12 \left (1 \minus{} \frac {1}{\sqrt {2}}\right) \plus{} \frac 12 \left (\ln \tan \frac 38 \pi \minus{} \frac {1}{\sqrt {2}}\right)$

$ \equal{} \frac 12 \left (1 \minus{} \frac {1}{\sqrt {2}}\right) \plus{} \frac 12 \ln (1 \plus{} \sqrt {2}) \minus{} \frac {1}{2\sqrt {2}}$

$ \equal{} \boxed{\frac {1 \minus{} \sqrt {2}}{2} \plus{} \frac 12 \ln (1 \plus{} \sqrt {2})}$

I have just found another solution.

Maybe this is neater than our solutions.

Let $ \sin x\equal{}t,\ \int_0^{\frac{\pi}{4}} \frac{\tan x}{1\plus{}\sin x}\ dx\equal{}\int_0^{\frac{1}{\sqrt{2}}} \frac{t}{(1\minus{}t)(1\plus{}t)^2}\ dt$.

The rest is left for everyone. :)

$ \frac{\tan{x}}{1\plus{}\sin{x}}\equal{}\frac{1}{\cos{x}}\minus{}\frac{1}{\cos{x}(1\plus{}\sin{x})}$.

$ \therefore\int_{0}^{\frac{\pi}{4}}\frac{\tan{x}}{1\plus{}\sin{x}}dx\equal{}[\ln(\sec{x}\plus{}\tan{x})]_{0}^{\frac{\pi}{4}}\minus{}\int_{0}^{\frac{\pi}{4}}\frac{1}{\cos{x}(1\plus{}\sin{x})}dx$
$ \equal{}\ln{(1\plus{}\sqrt{2})}\minus{}\int_{0}^{\frac{\pi}{4}}\frac{\cos{x}}{(1\minus{}\sin^2{x})(1\plus{}\sin{x})}dx$.
Substitute $ \sin{x}$ to $ t$, we get
$ \int_{0}^{\frac{\pi}{4}}\frac{\cos{x}}{(1\minus{}\sin^2{x})(1\plus{}\sin{x})}dx\equal{}\int_{0}^{\frac{1}{\sqrt{2}}}\frac{1}{(1\minus{}t^2)(1\plus{}t)}dt$.

To evaluate this, let's applying the identity $ \frac{1}{(1\minus{}t^2)(1\plus{}t)}\equal{}\frac{1}{4}\left(\frac{1}{1\plus{}t}\plus{}\frac{1}{1\minus{}t}\plus{}\frac{2}{(1\plus{}t)^2}\right)$ gives:
$ \int_{0}^{\frac{1}{\sqrt{2}}}\frac{1}{(1\minus{}t^2)(1\plus{}t)}dt\equal{}\minus{}\frac{1}{4}\ln{(1\plus{}\frac{1}{\sqrt{2}})}\minus{}\frac{1}{4}\ln{(1\minus{}\frac{1}{\sqrt{2}})}\minus{}\frac{1}{2}\left(\frac{1}{1\plus{}\frac{1}{\sqrt{2}}}\minus{}1\right)\equal{}\frac{1}{2}\ln{(1\plus{}\sqrt{2})}\plus{}\frac{1}{2}(\sqrt{2}\minus{}1)$.

So, the answer is $ \boxed{\frac{1}{2}\ln{(1\plus{}\sqrt{2})}\minus{}\frac{\sqrt{2}\minus{}1}{2}}$.
My approach is as same as kunny's. kunny typed first while I was typing....

Very nice, kunny!

I've worked through this, using partial fractions, and got the same answer as above:

$ \frac {1}{4} \int_{0}^{\frac {1}{\sqrt {2}}} \frac {1}{1 \minus{} t} \plus{} \frac {1}{1 \plus{} t} \minus{} \frac {2}{(1 \plus{} t)^{2}} \ dt$

$ \equal{} \frac {1}{4} [\ln\frac {1 \plus{} t}{1 \minus{} t} \plus{} \frac {2}{1 \plus{} t}]_{0}^{\frac {1}{\sqrt {2}}}$

$ \equal{} \frac {1}{4}[\ln \frac {\sqrt {2} \plus{} 1}{\sqrt {2} \minus{} 1} \plus{} \frac {2\sqrt {2}}{\sqrt {2} \plus{} 1} \minus{} 2]$

$ \equal{} \frac {1}{4} [\ln (3 \plus{} \sqrt {2}) \plus{} 2\sqrt {2}(\sqrt {2} \minus{} 1) \minus{} 2$

$ \equal{} \frac {1}{2} \minus{} \frac {\sqrt {2}}{2} \plus{} \frac {1}{2}\ln(1 \plus{} \sqrt {2})$.

Thanks kunny (and J.Y. Choi: my computer crashed twice between my last post and yours), I enjoyed that question.


calculus
integration
trigonometry
logarithms
function
calculus computations

$ I \equal{} \int_0^{\frac{\pi}{2}} \frac{\cos\theta}{1\plus{} \cos\theta}\, d\theta$.

Using the substitution $ t\equal{}\tan\frac{1}{2}\theta$ or otherwise, show that $ I\equal{} \int_0^1 (\frac{2}{1\plus{}t^{2}} \minus{}1 )\, dt$.

Find $ I$ in terms of $ \pi$.

hello, i have forgotten to say that $ \cos(\theta)\equal{}\frac{1\minus{}t^2}{1\plus{}t^2}$ and $ \int_{0}^1\frac{2}{1\plus{}t^2}\minus{}1\,dt\equal{}\frac{\pi}{2}\minus{}1$.
Sonnhard.

There are other ways to find $ \int_{0}^\frac{\pi}{2} \frac{\cos \theta}{1\plus{}\cos \theta}\, d\theta$ besides using the substiution $ t\equal{}\tan \frac{1}{2}\theta$.

For example, $ \frac{\cos \theta}{1\plus{}\cos \theta} \equal{} \frac {2\cos^{2}\frac{1}{2}\theta \minus{}1}{2\cos^{2}\frac{1}{2}\theta}$

$ \equal{} 1\minus{} \frac{1}{2}\sec^{2}\frac{1}{2}\theta$

$ \Rightarrow I \equal{} 2\int_{0}^\frac{\pi}{4} 1\minus{} \frac{1}{2}\sec^{2}\phi\,d\phi$, where $ \phi \equal{} \frac{1}{2}\theta$.

$ \Rightarrow I \equal{} \frac{\pi}{2} \minus{} 1$, which is the same answer as Sonnhard's.

Are there any other ways of doing this integration?

There [b]must[/b] be other ways of doing this!

Try "rationalizing the denominator": $ \frac{\cos\theta}{1\plus{}\cos\theta}\equal{}\frac{\cos\theta(1\minus{}\cos\theta)}{1\minus{}\cos^2\theta}.$

What can you do with that?

[hide="plenty"]
\[ \displaystyle F(\theta) \equal{} \int\left(\frac {\cos{\theta}}{1 \plus{} \cos{\theta}}\right )\text{d}\theta
\]
After a quick application of Merryfield's hint,
\[ \displaystyle F(\theta) \equal{} \int \cot{\theta}\csc{\theta}\text{ d}\theta \minus{} \int \cot^2{\theta}\text{ d}\theta \equal{} \int \cot{\theta}\csc{\theta}\text{ d}\theta \minus{} \int \csc^2{\theta}\text{ d}\theta \plus{} \int\text{d}\theta
\]

\[ F(\theta) \equal{} \minus{} \csc{\theta} \plus{} \cot{\theta} \plus{} \theta\plus{}C
\]

\[ \displaystyle F(\pi/2) \minus{} F(0) \equal{} \frac {\pi}{2} \minus{} 1
\]
[/hide]

Thanks Kent and igiul. Nice.

But I'm bothered by $ F(0)$ as both $ \csc \theta$ and $ \cot \theta$ go to $ \infty$ as $ \theta$ goes to $ 0$.

But $ \cot\theta\minus{}\csc\theta\equal{}\frac{\cos\theta\minus{}1}{\sin\theta}$ tends to $ 0$ as $ \theta\to0.$ You do have to combine those two terms so the badness can cancel out.

Thanks Kent.

I'm guessing you differentiated top and bottom, using L'Hopital's rule (don't know how to do a circumflex accent), to get $ \frac{\minus{}\sin \theta}{\cos\theta}$ which goes to $ \frac{\minus{}0}{1} \equal{} \frac{0}{1} \equal{} 1$. Or do you have another method of evaluating the limit?

Any ideas on other ways of doing this integration?

Have you seen my $ \ln(\cos x)$ post: I'm fascinated by $ \int_{0}^{1.5} \ln(\cos x)\, dx$!

I don't make L'Hôpital my default first resort and I try to talk students out of always resorting to it - I think it gets in the way of thinking, especially thinking about size. As a matter of quick and dirty power series (and I do have the power series of the common functions memorized), I know that $ \cos\theta\minus{}1\equal{}O(\theta^2)$ as $ \theta\to 0,$ while $ \sin\theta\approx\theta.$ That's enough information to know that the limit is zero. (And it took far longer for me to write this than it took to see it - quick and dirty power series is, first and foremost, quick.)

I'll add that in this case we were integrating a function that is continuous at $ 0,$ so its antiderivative must be continuous at zero as well. That gives is the advantage in thinking about that limit of knowing that the limit must exist and be finite.

As for that other topic: it looks to me that everything worth saying has already been said there.

Thanks Kent. I'll remember that when I'm next tempted to use l'Hopital's rule without thinking.

I'd still like to see other ways of integrating this. I'm sure there are people out there like kunny et cetetra who can see other ways quite easily. I'm off on holiday until Saturday night so if there are any responses, I won't be able to thank you for a while.

jus add one and subtract 1. write 1 / (1+cos x) as 1/2* (sec (x/2))^2...

$ I \equal{} \int_0^{\frac {\pi}{2}} \frac {\cos\theta}{1 \plus{} \cos\theta}\, d\theta$

$ \equal{}\int_0^{\frac{\pi}{2}} \left(1\minus{}\frac{1}{1\plus{}\cos \theta}\right)\ d\theta$

$ \equal{}\int_0^{\frac{\pi}{2}} \left(1\minus{}\frac{1}{2\cos ^ 2 \frac{\theta}{2}}\right) d\theta$

$ \equal{}\left[\theta \minus{}\tan \frac{\theta}{2}\right]_0^{\frac{\pi}{2}}$

$ \equal{}\boxed{\frac{\pi}{2}\minus{}1}$


inequalities
function
calculus
quadratics
algebra
floor function

Helklo, ssupermath!

Quote:1) For x a real number,
find the minumum value of

I have a non-Calculus solution for this one . . . [hide]

We have: f(x) = (x - 1)(x - 4)(x - 2)(x - 3) = (x2 - 5x + 4)(x2 - 5x + 6)

= [(x2 - 5x + 5) - 1][(x2 - 5x + 5) + 1] = (x2 - 5x + 5)2 - 1

This is a quadratic function of the form: y = u2 - 1

which obviously has its minimum when u = 0.

Therefore, the minimum value of f(x) is -1.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

When does this minimum value occur? . . . when x2 - 5x + 5 = 0

. . . I'll let you work out the details.[/hide]

1)For x element of real numbers find the minumum value of
(x-1).(x-2).(x-3).(x-4)

2)Let a,b,c,d,e are elements of {0,1,2,...,9}
If db(c^2)=aeadbc then find e-a=?(answer:3)

3)What is the sum of real roots of equation
x^2-4x^3+5x^2-4x+1=0

4)For all x real numbers that satisfy inequality at follows, find the maximum value of C real number.
x^2>=C.[x].(x-[x])
[x] denotes greatest integer function

for number 3, it would be 4 because the sum of the roots is -b/a

Quote:for number 3, it would be 4 because the sum of the roots is -b/a

How can you tell that all the roots are real?

Quote:3)What is the sum of real roots of equation
x^2-4x^3+5x^2-4x+1=0

Should the first x^2 be x^4?

If so, [hide]

We have a reciprocal equation in x. Dividing by x^2 we get:

x^2 - 4x + 5 -4/x + 1/x^2 =0

(x^2+1/x^2) - 4(x+1/x) + 5 = 0.

Putting x+1/x = a, and noting that x^2+1/x^2 = (x+1/x)^2 - 2, we have:

a^2 - 2 - 4a + 5 = 0 or

a^2 - 4a + 3 =0

(a-3)(a-1)=0

a=3 or a=1, so

x+1/x = 3 or x+1/x=1, and indeed, we have the two quadratics

x^2 - 3x - 1 = 0 and x^2 - 2x -1 =0, with solutions

(3+/-sqrt(3))/2 and (1+/-sqrt(5))/2

Indeed, all four roots are real, and they sum to 4.

[hide][/hide][/hide]

You were good until the end...
The two equations are x + 1/x = 3 and x + 1/x = 1, which the first has solutions in R and the second does not. So the sum is the sum of the first one, which is 3.


If $ a\ge b>1$, what is the largest possible value of $ \log_a(a/b)\plus{}\log_b(b/a)$?

$ \textbf{(A)}\ \minus{}2 \qquad
\textbf{(B)}\ 0 \qquad
\textbf{(C)}\ 2 \qquad
\textbf{(D)}\ 3 \qquad
\textbf{(E)}\ 4$

Elemennop wrote:[hide]

why is it so easy?

[/hide]

Ummm... make sure equality can be reached.

[hide]$ \log_a a/b \plus{} \log_b b/a \equal{} \log_a a \minus{} \log_a b \plus{} \log_b b \minus{} \log_b \equal{} 2 \minus{} \log_a b \minus{} \frac{1}{\log_a b}$
Let $ x\equal{}\log_a b$.
We need the maximum value of $ 2\minus{}\left(x\plus{}\frac{1}{x}\right)$.  We do this by getting the minimum value of $ x\plus{}\frac{1}{x}$.

Let $ x\equal{}\tan\theta$.  It is obvious that $ 0<x<1$, so $ \theta$ can be acute.  Let this be so.  herefore $ x\plus{}\frac{1}{x}\equal{}\tan\theta\plus{}\cot\theta\equal{}\frac{\sin\theta}{\cos\theta}\plus{}\frac{\cos\theta}{\sin\theta}\equal{}\frac{\sin^2\theta\plus{}\cos^2\theta}{\sin\theta\cos\theta}\equal{}\frac{2}{\sin 2\theta}\geq 2$.

So the answer is $ 2\minus{}2\equal{}\boxed{0}$.[/hide]

Alternatively reduce to $2-\Bigg ( \frac{\log b}{\log a}+\frac{\log a}{\log b}\Bigg )$. Since $a \ge b > 1$, both terms in the parentheses are positive. Then by AM-GM, $\frac{\log b}{\log a}+\frac{\log a}{\log b} \ge 2$. So the entire quantity is at most $0$.

[b]Solution:[/b] Notice by basic algebra this simplifies to $2 - (x+ \frac{1}{x})$. Since $x$ is positive, this is maximized when $x=1$ so our answer is $\boxed{B}$.


calculus
integration
conics
ellipse
calculus computations

Consider the iterated integral
$ \int_{0}^{2}\int_{0}^{\frac{1}{2}\sqrt{36\minus{}9x^{2}}}\int_{5x^{2}}^{36\minus{}4x^{2}\minus{}4y^{2}}2\mbox{ dz dy dx}$.
Consider integrating first with respect to $ y$, then $ x$, then $ z$.  Set up a sum of iterated integrals, that, when evaluated, give the same result.

Repeat the previous part for integration first with respect to $ x$, then $ z$, then $ y$.

The 3-dimensional region of integration is bounded from below by the parabolic cylinder $ z \equal{} 5x^2$ and from above by the upside-down paraboloid $ z \equal{} 36 \minus{} 4x^2 \minus{} 4y^2$. These two surfaces intersect when $ 9x^2 \plus{} 4y^2 \equal{} 36$, which explains the upper limits for $ y$ and $ x$. The lower limits for $ x$ and $ y$ indicate that we are only dealing with the first octant. Try to draw this in 3D.

Now, the limits for $ y$ as the innermost variable are going to be $ 0$ and $ \frac {1}{2}\sqrt {36 \minus{} 4x^2 \minus{} z}$ (this comes from the equation of paraboloid), because the parabolic cylinder does not involve $ y$. Then you project the solid onto the $ xz$-plane - the projection is bounded by $ x \equal{} 0$, $ z \equal{} 5x^2$, and $ z \equal{} 36 \minus{} 4x^2$. This gives you the remaining limits.

For the last part you are going to have to split the integral, because the upper limit for $ x$ gets tricky.

[quote="JRav"]Consider the iterated integral
$ \int_{0}^{2}\int_{0}^{\frac {1}{2}\sqrt {36 - 9x^{2}}}\int_{5x^{2}}^{36 - 4x^{2} - 4y^{2}}2\mbox{ dz dy dx}$.
Consider integrating first with respect to $ y$, then $ x$, then $ z$.  Set up a sum of iterated integrals, that, when evaluated, give the same result.

Repeat the previous part for integration first with respect to $ x$, then $ z$, then $ y$.[/quote]
[hide]Consider $ I = \int\int\int_{E}2~dV = \int_{0}^{2}\int_{0}^{\frac {1}{2}\sqrt {36 - 9x^{2}}}\int_{5x^{2}}^{36 - 4x^{2} - 4y^{2}} 2~ dz ~dy ~dx$

Thus $ E=\{ (x,y,z) \in \mathbb{R}^3 ~ | ~~0<=x<=2, ~~0<=y<=\frac{1}{2}\sqrt{36-9x^2}, ~~5x^2<=z<=36-4x^2-4y^2 \}$

So what you're basically doing is finding two times the volume of the region in the first octant bounded below by the cylinder $ z=5x^2$ and above by the the paraboloid $ z=36-4x^2-4y^2$.  They also intersect in the first octant along the ellipse $ \frac{x^2}{4} + \frac{y^2}{9}=1$  Picture below:

[hide="Picture"][img]http://img219.imageshack.us/img219/9089/picture1xs6.png[/img][/hide]
Observing this we can also note that that:
$ E=E_{1} \cup E_{2}$

$ E_{1} = \{ (x,y,z) \in \mathbb{R}^3 ~ | ~~0<=z<=20,  ~~ 0 <= x <= \sqrt{\frac{z}{5}}, ~~ 0<=y<=\frac{1}{2}\sqrt{36-4x^2-z}\}$
$ E_{2} = \{ (x,y,z) \in \mathbb{R}^3 ~ | ~~20<=z<=36,  ~~ 0 <= x <= \frac{1}{2}\sqrt{36-z}, ~~ 0<=y<=\frac{1}{2}\sqrt{36-4x^2-z}\}$

Thus:
$ I = \int\int\int_{E} 2 ~dV$
$ I = \int\int\int_{E_{1}} 2 ~dV + \int\int\int_{E_{2}} 2 ~dV$
$ I = \int_{0}^{20}\int_{0}^{\sqrt{\frac{z}{5}}}\int_{0}^{\frac{1}{2}\sqrt{36-4x^2-z}} 2 ~dy ~dx ~dz + \int_{20}^{36}\int_{0}^{1/2\sqrt{36-z}}\int_{0}^{\frac{1}{2}\sqrt{36-4x^2-z}} 2 ~dy ~dx ~dz$

I'll leave with respect to $ x$ first up to you.[/hide]

Actually, the second part is really the one I need.

[quote="mlok"]That's too bad, because I'm tired after doing the first, unneeded part, and don't want to deal with the second.[/quote]

you cracked me up.....  :rotfl:  :D


Do ya'll have an advanced number theory class?

sweet!

NT seems to be one of the major competition topics (behind geometry) so I was curious. Thanks.

That's undecided; it will probably become more clear as we start to work on the course.  It's likely that in time we'll have 3 levels of number theory, with the two we'll add being above the one we currently offer.  We're also kicking around the idea of a number sense course for younger students.


trigonometry
integration
calculus
calculus computations

$ y' \plus{} 2xy \equal{} \sqrt {y}sinx$  ,  $ y(0)\equal{}0$

[quote="the_wonder"]$ y' \plus{} 2xy \equal{} \sqrt {y}sinx$  ,  $ y(0) \equal{} 0$[/quote]

$ \frac{dy}{dx} \plus{} 2xy \equal{} \sqrt{y} sinx$

Divide by $ 2 \sqrt{y}$

this gives

$ \frac{1}{2 \sqrt{y}} \frac{dy}{dx} \plus{} x \sqrt{y} \equal{} \frac{sinx}{2}$

Let $ \sqrt{y} \equal{} z$

Differentiate w.r.t x

$ \frac{1}{2 \sqrt{y}} \frac{dy}{dx} \equal{} \frac{dz}{dx}$

$ \frac{dz}{dx} \plus{} z.x \equal{} \frac{sinx}{2}$

multiply throughout by the Integrating Factor which is $ e^{\int x. dx} \equal{} e^{x^2/2}$

This gives

$ \frac{d}{dx}(z.e^{x^2/2} \equal{} \frac{1}{2}sinx.e^{x^2/2}$

$ z.e^{x^2/2} \equal{} \frac{1}{2} \int sinx.e^{x^2/2} . dx$

Can you proceed from here?

Yes of course.It`s easy.
Thanks for helping!!!  :lol:

hello, i have found $ y(x)\equal{}0$ as the solution.
Sonnhard.


ok so i know that i probably should get ctc permission but this is a CHESS tourney not a MATH tourney so i am going to host one to join just post that you want to join my tournament! winner gets a prize! it will be single elimination and we will have 3 prizes for 1st 2nd and 3rd!

[hide="3rd prize"]you gotta win to find out[/hide]

[hide="2nd prize"]you gotta win it to find out[/hide]
[hide="1st prize"][hide="you wanna know?"]
[hide="sure?"]you gotta win it to find out
[/hide][/hide][/hide]

Hell I'll join, I haven't played in a while but took 25th at nationals a few years ago.

ok, ill join. where and when is the tourney going to take place?

I'll join, even though I kinda suck at chess.   :D

il join :D shaggy u stole my avatar D:

... jk :P

I'm in. So what do we play over? Time control?

I'm better at blitz  :D . But best scenario is 25 minute clocks, 0 delay, and 10 minute moves? Btw, how many people are going to be in the tournament. Also, if were going to do more than 2 rounds a day I might not be able to do it.

when i pm you reply to me and when i pm you again then you are to play within 7 days of the day i send you a second pm we will have 8 players in/excluding me. i will alert you all by pm when the time comes...

so is it single elimination or something?

I will join, I guess. Just tell me where it is.
Please not on another chess server (like FICS)
Not that I have anything against then, but I paid good money for my ICC account, so I'm sticking with it.

Would everyone mind posting their USCF ratings?
Unless, of course, you're not from the USA.

Mine: 1709 (though probably lower after Seafair Open)
I [i]would[/i] have been on board one last nationals but I fell to a FREAKIN SKEWER ARRRRGGGGGHHHHHHH (against Brennan Lee). Explanation: I beat two 1900's in a row to have a perfect score by round 5, while Alec Getz drew with George Qi in round 4, so I became to only person with a perfect score.
I ended up around 46 after losing to an expert and then another 1900.
The paradox of the Gruenfeld: I always get winning positions, I always lose
The paradox of the Accelerated Dragon: I always drop a piece, I always win

[quote="serialk11r"]Oh dang you're pretty good then.
I just started last year, I think I can compare to a 1600. I beat a 1700 before, but the time control was 45 0 so dunno if yall think that's legitimate.
I played in only one tournament before, and I was in the u1000 division and somehow screwed up really bad, so my provisional rating is like 900 :P[/quote]

The time is legit. If you played at nationals everyone gets a 2 hour clock. It was really annoying because people would stall for like an hour and it made me want to hurt them.

[quote="Lawrence Wu"]i love nationals. you get to walk around between moves[/quote]

One of the players on our team just so happened to have his dad work at the hotel the tournament was set up that year. We got free hotel rooms, access to the swimming pool and discounts at all the restaurants in the hotel. It was pretty sweet.  :P

I wanna participate if there are any vacancies!!!! BTW My favourite opening used to be [i]Giuoco Piano[/i].

I join my official rating is 1836 muahahahahah won nationals few years ago muahahahahah

Wait Brennan Lee... OMG are you serious? that's the person I beat to win nationals...

I think this kinda failed. I'll just drop this for now and try to start another one in a m onth or so. Sorry. :(

[quote="shaggy75"]OK, since mathblitz was overloaded w/ school, I volunteered to take his place and organize this tourney. mathblitz agreed.

Signups will End In 3 Days. I would end it today, but I have a three-day-chess tourney starting tomorrow.
After that, I will post up the pairings for the first round.

[b]If you cannot play, please post or pm me within the next three days.[/b]

The format for the tournament will be swiss. If there is a tie for first, second, and/or third place there will be a blitz playoff game(s)

Please vote for where the tourney will be held(FICS, ICC etc.) within 5 days or i will decide.

[u][b]Current Signups(11)[/b][/u]

crazychinesemaniac
BOGTRO
B_Gasperov
Meta_Knight
klrocks
evilhamster
mathemonster
BoesFX
serialk11r
Squarerootofisoscelestria
chenhsi

If you have any comments or complaints or concerns, please pm me.[/b][/quote]

Shaggy i never agreed to this but go on. Next time i will confirm before i let someone take over.

[quote="SuperMathBoy"]I join if this is to start going again, my rating is 2094 and I won the nationals in 4th grade. Lol, I beat Ray robson in the last round. I guess it should be on fics since that is free[/quote]

YOU BEAT RAY ROBSON?! OMG *faints*

I'm 1945 now btw.

*crawls away weakly in the view of a beast*


Inner Napoleon triangle has side lenght?

As proven in http://www.cut-the-knot.org/proofs/napoleon.shtml (Proof #1), the sidelength s of the external Napoleon triangle of triangle ABC satisfies $3s^2=\frac12\left(a^2+b^2+c^2\right)+2\sqrt3\cdot F$, where F is the area of triangle ABC. Similarly, the sidelength s' of the internal Napoleon triangle satisfies $3s^{\prime}\ ^2=\frac12\left(a^2+b^2+c^2\right)-2\sqrt3\cdot F$.

  Darij


function
logarithms
algebra unsolved
algebra

Let f(x)=e^3x . Find f ' (ln(3)).

well I tried and I found the answer was 27 ...which was in the choices...However it was not the answer .

The choices were
A.81     B. 27       C.9       D.3       E.1

Please help thank you.

I don't think this belongs in the algebra olympiad section, but here it is:

\begin{eqnarray*}
f(x) &=& e^{3x}\\
f'(x) &=& 3e^{3x}\\
f'(\ln{3}) &=& 3e^{3\ln{3}}\\
&=& 3e^{\ln{27}}\\
&=& 3(27)\\
&=& 81
\end{eqnarray*}

Hope that helps. :)

what is f ' ..what is the difference between f and f ' ...I mistunderstood that one

I really think that this kind of problems should be treated in other part.  :?


probability
geometry
MATHCOUNTS
AMC
AIME
search
3D geometry

Well, try to look for the counting and probability book on this website. That's all I can give you, but the intermediate book will be out this winter.

[quote="D.P.L"]Does any one have/know where i can get some combinatorics papers?

Thanx in advance :)[/quote]
Can you specify what you mean by combinatorics papers? If you want problems, combinatorics is a great subject area, and I reccomend you start with "102 Combinatorics Problems from the Training of the USA IMO Team" (the book). If you want instruction, for the absolute basics I would reccomend the intro book and class here, if you already know mathcounts-level applications I would reccomend the intermediate class. For any higher level of competetence, just start doing AIMEs or looking at the combo section of IMO shortlists on kalva. You're bound to find something to work on.

As always, if you need answers, post your question here.

[quote="D.P.L"]This might be a bit of topic, but what is $CPIG$.

@penguin: i'm looking for combinatoric papers that are aime level.[/quote]

CPIG is Challenging Problems in Geometry. It's a great problem book, but you might want to supplement it with "Geometry Revisited", or you miss out on some important parts you need to solve the problems.

For Combo, I still reccomend "102 Combinatorics Problems from the Training of the USA IMO Team", just as a library of problems. There will be about 50 out of the 100 at your level, but looking at harder problems is very enjoyable. Try to find it at a local college library instead of buying it.

If you still need training on the basics of Combo, just ask questions.

For problems: for AIME level, either go fish around kalva's AIME archive (much more complete in the eariler years than ours) or go search through tourament of towns papers, I hear they are combo heavy.

Just to test your skills, how do you fare with the following problem:

Given a cube and six distinct colors, how many ways can you color the cube (if you exclude colorings that can be found by rotating or flipping other colorings)?

:huh:
Is this going to be counting all the ways like this...

all faces same color,
5 faces same color and 1 face another color...and so on till 2 faces...
then all faces different colors....
seems very difficult...any simpler methods??? also anywhere where i can download counting tutorials, problems & solutions... good ones...i checked the one on 3 types of probability and that was interesting...i'm looking for more and unfortunately down where i live none of aops books are available!

what do you mean by "flipping other colorings"?

well a cube is basically 3 pairs of opposite faces... you can go from there.

Err, sorry, you can only use each color once...

In that case, would it just be

[hide]30? Because there are $6!$ or 720 ways to color the cube, and each seperate cube can be rotated 24 ways?[/hide]


I was just wondering could you do proofs like this? :

Let x=0
If the equation is correct for this value, x=0 is true.

I'm trying to do a problem where I can't get a proof, so I am trying to work backwards.
Now I'm stuck here.

This is not very clear, but I'm not really good at describing this problem.
It's from a high school text book, so I'm trying to make the answer simple by using the condition above

This is redundant.  You can't say if x=0 prove x=0.

I have to prove a sequence.
I tried to make a general sequence.
I wanted to prove the general sequence works.
So I worked backwards and put
If x=0 and x=n
Then if the answer is true, the values are true.

So is this not allowed in making proofs?

Sequence? Do you mean inductory proof?

As for "working backwords"...if I understand what you wrote, it's more common (and rigorous) to say: the formula is true only if it is true for x=0. (I don't know what "true values" really express...)

You're being very unclear.  Please retype the problem for us.

The way I see it is:

You're trying to prove some equation is satisfied by $ x\equal{}0$, and then induct to the fact that the equation is true for all $ x$.

If so, then this is a method of proof known as [[Induction]].

Yeah I think its a mathematical induction question

if you say its true for x=0 how about x=1,x=2 etc etc

Just type the exact question here someone will solve it soon


Let $ ABC$ be an acute triangle and choose points $ A_1, B_1$ and $ C_1$ on sides $ BC, CA$ and $ AB$, respectively. Prove that if the quadrilaterals $ ABA_1B_1, BCB_1C_1$ and $ CAC_1A_1$ are cyclic then their circumcentres lie on the sides of $ ABC$.

we get $ \angle C_1A_1B \equal{} \angle B_1 A_1C \equal{} \angle BAC$ and cyclic. We get $ \angle C_1A_1A \equal{} \angle C_1CA \equal{} \angle C_1BB_1 \equal{} \angle B_1A_1A$, then $ AA' \perp BC$ and cyclic, from wich the thesis.


geometric series
calculus
calculus computations

inf    cos(n*pi)
sum  -----------
n=0      5^n

I'm thinking it would converge to 0 since cos(n*pi) would be alternating from 1 to -1... cancelling out terms.. and at the same time 5^n would be increasing and making the terms smaller to the point they'd be almost 0, I don't know how I can do this algebraically though.

It looks a lot easier if you rewrite $\cos(n\pi)=(-1)^{n}.$

Now we have (if I interpret your notation correctly):

$\sum_{n=0}^{\infty}\frac{(-1)^{n}}{5^{n}}.$

Is there any chance that that might be a geometric series?


I have the book challenging problems in geometry. I'm having a lot of problems doing even the simplest of problems, I'm quite sure I know all the basics from intro to geometry but these problems seem very very hard.

Then you should probably read the solutions to the problems you don't know how to do and actually work through that solution method yourself after you finish reading. Do this for a few problems and you will learn some strategies for tackling these geometry problems and then you can apply these on later problems.

wow, I think it's interesting how this thread's name is almost identical to the one below it (or above.. or whatever)

well, same here.  but when i read the solutions, they seem REALLY REALLY EASY.

it's so annoying, but u get used to them after you do a few.


real analysis
geometry
geometric transformation
real analysis unsolved

Hi,
I'm wondering why a Lebesgue measurable set E with m(E) > 0 in R provides a measurable sum set E + (-E) = {x-y/ x, y belonging to E} for the Lebesgue's measure.
I have already read some posts about this topic on this forum but these ones were only dealing with the non vacuity of its interior, not measurability.
Thanks for any help!

Forgive me if I am missing something obvious, but it seems that this is trivial. I mean, $E + (-E) = \{x - y | x, y \in E\} = \{x - k | x, k \in E\}$ ($k$ not varying) $\cup \{x - y | x, y \in E, y \ne k\}$. By translation invariance of measure, the first piece has $m(E)$, and since it's a union, we then have $m(E + (-E)) \ge m(E) > 0$.

To ludo: it is a very bad idea to start a new thread in the Solved Problems section :)

To Xevarion: your argument only shows that the inner measure is positive, which is not sufficient for measurability...

To ludo again: The difference set  of a Lebesgue measurable set of positive measure may fail to be Lebesgue measurable. By the way, positivity of measure doesn't help much: if you have a counterexample with a set of measure $0$, just unite it with a very short interval of positive length lying very far away to get a counterexample with positive measure.

Thanks for your answers! Does it mean this question is wrong ?
To the moderator: In which forum should I send similar questions/answers ? It wasn't clear to me, even with the available informations provided about topics dealt in each forum.

Sorry, for some reason I thought I was supposed to show the measure wasn't 0... I guess I can't read! Now I see that's exactly what you already heard and didn't care about.

You should post such problems in "Unsolved and Proposed Problems" (or if it's more of a calculation than a theoretical problem, post it in Computations and Tutorials).

So fedja, from your post should I gather that the set is NOT always measurable? If we take some dense uniformly distributed set of measure 0 in $[0,1]$ and unite it with a totally disconnected set of measure > 0, could that be such an example?


Let $m,\ n$ be integers such that $2\leq m\leq n$ and let $a,\ a'$ be integers which are less than or equal to $m$ and let $b,\ b'$ be integers which are less than or equal to $n$ such that $(a,b)\neq (a'\ b').$ Given a town of the rectangular shaped chessboard which is made up of $m's$ road running north and south which is called Line and $n's$ road running west and east which is called Street. Denote the intersection point of the $a$ th Line from the west and $b$ th Street from the north by $A$, and $a'$ th Line from the west and $b'$ th Street from the north by $B,$ including the edge for both cases.Find all pair of $(m,n,a,b,a', b')$ such that by passing through each crossroads of the town exactly one time, you can reach the point $B$ from the point $A$ including in the start point and goal one.

[quote="kunny"] you can reach the point $B$ from the point $A$ including in the start point and goal one.[/quote]

What does this mean?

Well, it's been a year now, so can [b]kunny[/b] please post your solution?  :D

Hey, kunny, please show the solution, or at least give a hint. Don't just not reply!  :)  :D

[hide = "Sketch"]

We claim that the answer is whenever $mn-1$ has the same parity as the taxicab distance between $(a, b)$ and $(a', b')$, except for the few scenarios listed below:

1. On a $2 \times n$ grid (WLOG it has $2$ rows), $A, B$ are vertically adjacent.
2. On a $3 \times n$ grid (WLOG it has $3$ rows), one of $A, B$ is at the leftmost or rightmost square on the second row, or if $n = 3$ one of them is on one endpoint of the second column.

It's straightforward to verify that these two scenarios indeed both fail, and that they are the only failing examples on $2 \times n$'s and $3 \times n$'s whenever they satisfy the taxicab parity condition.

Now, we will tackle cases where both dimensions are $\ge 4$ using induction.We wish to show that it always works in these cases. If both dimensions are in $\{4, 5\}$, then it's again straightforward to verify that the taxicab parity condition is sufficient and necessary (these are our base cases). Otherwise, we'll suppose that $m \ge 6.$ We can now consider removing a $2 \times n$ from the top or bottom and inducting down. The only case where this fails is when one of $A, B$ is in the top $2 \times n$ and the other is in the bottom $2 \times n$. Suppose WLOG that $A$ is in the topmost $2 \times n$, where $n \ge 4$. Then, let's pick any point $P$ in the third row with an even taxicab distance from $A$. By the $2 \times n$ case, we know that there exists a path of length $2n$ from $A$ to $P$ which covers all intersections in the topmost $2 \times n$ before reaching $P$. From here, we can easily just use induction to see that $P$ and $B$ can be connected (since the resulting grid has both dimensions $\ge 4$).

[/hide]


superior algebra
superior algebra unsolved

Let $\left( A,+, \cdot \right)$ be a ring and $a,b \in A$ such that $x^{3}+a x^{2}+bx = 0, \, \forall x \in A$.
Prove that $A$ is commutative.

If we assume that the ring is unitary, which is as far as I know usual in those contests, the problem readily reduces to $\forall x,\ x^{3}=x$ that has been discussed on the forum several times.
(I've omitted the computations.)Letting $x=1, x=-1$ gives us a system of two equations and two unknowns yielding $2a=0, 2b=-2$. Then multiply the given equation by two to get $2x^{3}-2x=0(*)$. Now replace $x$ by $x+1$ to get $3x^{2}+3x=0$, hence $3x^{2}=3x$. Use $(*)$ and the last equation to get $x^{3}=x$. QED


[quote="digitalboy1"]Let $x_1,x_2,...x_n$ and $y_1,y_2,...y_n$ be real numbers satisfying\[x_1^2+x_2^2+...+x_n^2=y_1^2+y_2^2+...+y_n^2=1\]Prove that\[2\left|1-\sum_{i=1}^n x_iy_i\right|\geq(x_1y_2-x_2y_1)^2\]Determine when equality occurs.[/quote]
\[2\left|1-\sum_{i=1}^n x_iy_i\right| = \sum_{i<j} (x_iy_j-x_jy_i)^2\geq(x_1y_2-x_2y_1)^2\]

siuhochung,maybe you have made a small mistake.I think the sum of
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/4a0d0dccb878570afd06792cc1b9a643.gif[/img] should be square :D

following siuhochung,
I think $2\left|1-\sum_{i=1}^n x_iy_i\right| = 2\sum_{i<j} (\sqrt{x_iy_j}-\sqrt {x_jy_i})^2 \geq 2(\sqrt{x_1y_2}-\sqrt{x_2y_1})^2$
then we just have to prove $2(\sqrt{x_1y_2}-\sqrt{x_2y_1})^2 \geq (x_1y_2-x_2y_1)^2 $
not hard.


geometry
3D geometry
geometric transformation
rotation
reflection
factorial

How many ways are there to put 8 beads of different colors on the vertices of a cube, if rotations of the cube (but not reflections) are considered the same?

The cube has 8 verticies, and there's 8 colors right?
I am not sure if this is right but i thought you could find 8! (as in factorial) and get...40320.
But that doesn't seem right.

[quote="thinkinavi"]What does the solution mean when it says "...there are 3 ways to rotate a cube onto itself with that vertex fixed"? I thought that a cube could be rotated 8 ways, since there are 8 vertices, so I answered 5040. How could a cube be rotated with one vertex fixed? I thought about around the fixed vertex so that it would be at right angles, but that would give 4 ways, not 3.[/quote]
[hide="Explanation"]When the solution says that it can be rotated with one vertex fixed, that means you can choose any bead on any vertex, and the cube can be rotated around the vertex just chosen.  Visualize a cube, and have a bead on one of them.  You can find a cube (a dice would work) and look at that.  Look at one of the vertices as your top view, with the vertex at the center.  It should be connected to three other vertices.  That gives three ways of rotation.  The cube can be rotated $120^\circ$ and the cube will be in the same position as the starting view (ignore the changing faces because a dice has faces that are different).  That gives a total of $3$ ways of rotation.
[/hide]

I need a little help.

[hide=What I did]I drew out each of the possible rotations for a cube and got 6. $\frac{8!}{6}=\boxed{6720}$. By the way, I counted the rotations in which you went back and forth with the cube, and when you went side to side with it. Hope that didn't confuse you guys though...[/hide]

Can someone help me figure out what went wrong?

[hide=Solution]
Each permutation of space diagonals maps to a distinct rotation. There are $4!=24$ rotations, so the answer is $8!/4!=1680$.
[/hide]

[quote=Alcumus Sol]...there are 3 ways to rotate the cube onto [b]itself[/b] with that vertex fixed...[/quote]

wait this literally makes no sense what

People have explained it above. Basically if you turn the cube on one vertex (so it kinda looks like a 3D diamond or smth) you can rotate it 3 ways and the colors will be in the same places. Thats why. So 8! / 8 ways to pick the vertex and 3 ways to rotate each turn = 8!/24 = 1680.

Hope this helps!

[hide=solution:]There are $8!$ ways of arranging the beads on the vertices, but this counts the rotations.  We need to divide by 24, since there are 24 rotations.  So the answer is $\dfrac{8!}{24}=\boxed{1680}$ [/hide]

My first try was (8*7*6*5)/4 * 4!/4 which was wrong

[quote=mathwizarddude]Could somesome please point out what's wrong with this? It seems that if I double my answer then it would be correct...
First take any of the 8 vertices and pick any one of the 8 beads (since we do not account for rotation), then for the 3 vertices that share an edge with the chosen vertices, there are $ \binom73$ ways to choose 3 out of the 7 remaining beads, then for the rest of the 4 vertices, there are $ 4!$ ways to choose the 4 beads left since each of the possible combinations for the 4 remaining positions relative to the other 4 vertices is unique (so no need to account for rotations). So the answer would be $ \binom73*4!\equal{}840$... However, if we do $ 8!/24$ we get $ 1680$... what's wrong :?:[/quote]

Someone please respond to this! I am confused as well...


analytic geometry
graphing lines
slope
calculus
derivative

what's the relationship between f(x) and f '(x)? Can somebody tell me?

http://en.wikipedia.org/wiki/Derivative

$f'$ is the slope of $f$ in the point $x$

Calculus problems should not go in this forum.

Anyway. $f''(x)$ is the second derivative of $f(x)$ — it is the derivative of the derivative.


Please read everything on this hidden text.

[hide="COPSD 2nd"]
Welcome to COPSD 2nd.

Here are the rules:

[hide="rules"]
1. You have 10 questions to do and you have to send the solutions (meaning answer + explanations on how you got them). If you get at least 7 right, you'll be sending to Round 2.
2. You have a week to do.
3. Only middle schoolers are allowed.
4. Do not ask for online help or anything like that (This is same for COP except I didn't put it there. It's on the PDF document so). Calculator is allowed.
5. PM me the answers.[/hide]

Here are the problems:

[hide="Problems"]
1. Let $m = 10! + 14!$. Find the largest prime factor of $m$.

2. For $x,y > 0$, $xy + x + y = 220$. Find $x+y$.

3. $x^2 + y^2 = 290$ and $x + y = 24$ with $x > y$. Find $x^2 - y^2$?

4. Let $r$ be a remainder when $11^{2005}$ is divided by 3. Find $r$.

5. Find the GCF of $(12,24,36,48)$.

6. If $x + \frac{1}{x} = 5$, then find the value of $x^2 + \frac{1}{x^2}$.

7. Find $n$ if $\frac{1}{16807} = 7^{5n-6}$.

8. If $z = 20$, then the result of function $k(z) = \frac{m}{n}$ where $m,n$ are in simplified form. Find the value of $m+n$.

$k(z) = \frac{z^2+4z+4}{z^2-4} \cdot \frac{z^2-9}{z^2+6z+9}$

9. $\sqrt{x} - \sqrt{y} = z$ where $z$ is a perfect square. What is the smallest possible value of $x+y$?

10. How many ways can you pick 2 girls out of 10 girls on the cafeteria?[/hide]

PDF version is also available.[/hide]

[quote]
8. If $z = 20$, then the result of function $k(z) = \frac{m}{n}$ where $m,n$ are in simplified form. Find the value of $m+n$.

$k(z) = \frac{z^2+4x+4}{z^2-4} \cdot \frac{z^2-9}{z^2+6z+9}$
[/quote]

That 4x is supposed to be 4z right?

Few addition for number 9.

$x,y,z$ are natural numbers.

Wow, this is the most problems I've seen posted before.

Maybe I guess.

I didn't look up on middle school test when I made this. All came from my head when I tried to estimate the level of student who are decent at math competition (not number 1 but decent) so that you might not get all of them in 5 seconds but it wouldn't take too much time neither. :)

COPSD are due today, FRIDAY!!

Please send me your solutions!


geometry
trigonometry
perpendicular bisector

[hide]
It seems that $?=24^{o}$ is a possible solution, but how to find it? Is there a pure geometry way to solve it instead of trig calculation?[/hide]

[hide]Suppose lower-left point is A, then clockwise B, C, D, and the point in the middle is M.
angleDAC = 36°
angleBDA = 48°
angleBAC = 12°
angleBDC = 24°
angleDCA = x

In AMD:
angleAMD
= 180° - angleDAM - angleMDA
= 96°
= angleBMC

angleBMA = angleCMD
= 180° - angleAMD
= 84°

In ABCD:
angleABC
= 360° - 24° - 48° - 12° - 36° - angleBCD
= 240° - angleBCD

Now in ABC:
180° = 12° + x + angleABC
=> 168° = x + 240° - angleBCD
=> angleBCD = 72° + x  (1)

And in BCD:
180° = 24° + x + angle BCD
=> angleBCD = 156° - x  (2)

Combine (1) and (2)
72° + x = 156° - x
=> 2x = 84°
=> x = 42°
[/hide]

Label the quadrilateral $ABCD$ counter-clockwise with the leftmost vertex $A$. Extend the rays $BC, AD$ until they meet at a point $P$. From the triangle $\triangle ABP$,

$\angle BPA = 180^o - (\angle ABP + \angle BPA) =$

$= 180^o - (48^o + 24^o + 36^o + 12^o) = 180^o - 120^o = 60^o$

Construct an equilateral triangle $\triangle ABQ$ on the segment $AB$. The triangle $\triangle ABD$ is isosceles with $DA = DB$, because the angles $\angle ABD = \angle BAD = 48^o$ are equal ($\angle BAD = 36^o + 12^o = 48^o$). Hence, the line $QD$ is the perpendicular bisector of the segment $AB$, the common base of the isosceles triangles $\triangle ABQ, \triangle ABQ$. The triangle $\triangle ABC$ is isosceles with $AB = AC$, because the angles $\angle ABC = \angle CAB = 72^o$ are equal ($\angle ABC = 48^o + 24^o = 72^o$, $\angle CAB = A80^o - (36^o + 72^o) = 72^o$).

The triangles $\triangle APC \cong \triangle APQ$ are congruent by SAS, because the side $AP$ is common, the sides $AQ = AB = AC$ are equal and the angles $\angle PAC = \angle PAQ = 12^o$ are also equal ($\angle PAQ = 60^o - (36^o + 12^o) = 12^o$). Hence, their remaining angles $\angle APQ = \angle APC = 60^o$, $\angle ACP = \angle AQP = 180^o - (60^o + 12^o) = 108^o$ and  remaining sides $PC = PQ$ are also equal. Consequently, the triangles $\triangle DPC \cong \triangle DPQ$ with the common side $DP$ are also congruent by SAS. The angles of the triangle $\triangle DPQ$ are easy to calculate:

$\angle ADQ = 180^o - \frac{\angle BDA}{2} = 180^o - \frac{180^o - 2\angle ABD}{2} =$

$= 90^o + \angle ABD = 90^o + 48^o = 138^o$

$\angle DCP = \angle DQP = \angle ADQ - \angle DPQ = 138^o - 60^o = 78^o$

$\angle CDP = \angle QDP = 180^o - \angle ADQ = 180^o - 138^o = 42^2$

As a result, the unknown angles $\angle ACD, \angle BDC$ are equal to

$\angle ACD = 180^o - [\angle DCP + (180^o - (36^o + 48^o + 24^o)] = 108^o - 78^o = 30^o$

$\angle BDC = 180^o - [\angle CDP + (180^o - (48^o + 36^o + 12^o)] = 96^o - 42^o = 54^o$

Anyway, trigonometry is a powerful geometric tool and there is nothing dirty about it.

Yetti

[quote="Stijn"]Hmmm, other result, must be some calculating error in mine, I wrote it down quite quickly, but I thought everything was ok...[/quote]
The error is here:

[quote="Stijn"]And in BCD:
180° = 24° + x + angle BCD
=> angleBCD = 156° - x  (2)
[/quote]

Actually, the solution is simpler than what I put down earlier. After finding the isosceles triangles $\triangle ABD$ with $\angle ABD = \angle BAD = 48^o,\ DA = DB$ and $\triangle BCA$ with $\angle BCA = \angle CBA = 72^o,\ AB = AC$, construct an equilateral triangle $\triangle ABQ$ on the base $AB$. Then $QD$ is the perpendicular bisector of the segment $AB$. The triangles $\triangle ADC \cong \triangle ADQ$ are congruent by SAS, because the angles $\angle DAQ = \angle DAC = 12^o$ are equal, the side $AD$ is common and the sides $AQ = AB = AC$ are also equal ($\triangle ABQ$ is equilateral with $AQ = AB$, $\triangle ABC$ is isosceles with $AB = AC$). Since $QD$ bisects the angle $\angle AQB = 60^o$, the unknown angle is equal to $\angle ACD = \angle AQD = \frac{\angle AQB}{2} = 30^o$.

[quote="shobber"]And for the trig solution, I think it would be like solving a trig equation:
$\frac{\sin{x}}{\sin{12^o}}=\frac{\sin{24^o}}{\sin{(x+72^o)}}$[/quote]
Not really - you can plug in $x = 30^o$ and the 2 sides of the equation will not be equal. Using the sine theorem for the triangles $\triangle ACD, \triangle ABD$,

$\frac{\sin{(180 - 12^o - x)}}{\sin x} = \frac{AC}{AD} = \frac{AB}{AD} = \frac{\sin{(180^o - 2 \cdot 48^o)}}{\sin{48^o}}$

$\sin{(168^o - x)} \sin{48^o} = \sin{84^o} \sin x$

This equation is easily solved without any use of numerical methods.

Using $\cos{(a - b)} - \cos{(a + b)} = 2 \sin a \sin b$ with $a = 168^o - x,\ b = 48^o$ on the left and $a = 84^o,\ b = x$ on the right:

$\cos{(120^o - x)} - \cos{(216^o - x)} = \cos{(84^o - x)} - \cos{(84^o + x)}$

Substituting $\cos{(x - 36^o)} = -\cos{(216^o - x)}$,

$\cos{(120^o - x)} = \cos{(84^o - x)} - [\cos{(84^o + x)} + \cos{(x - 36^o)}]$

Using $\cos{(a + b)} + \cos{(a - b)} = 2 \cos a \cos b$ with $a = x + 24^o,\ b = 60^o$ on the right:

$\cos{(120^o - x)} = \cos{(84^o - x)} - 2 \cos{60^o} \cos{(x + 24^o)} = \cos{(84^o - x)} - \cos{(x + 24^o)}$

Substituting $\sin{(x - 30^o)} = \cos{(120^o - x)}$ and using $\cos{(a - b)} - \cos{(a + b)} = 2 \sin a \sin b$ with $a = 54^o,\ b = x - 30^o$ on the right:

$\sin{(x - 30^o)} = 2 \sin{54^o} \sin{(x - 30^o)}$

Since $\sin{54^o} \neq \frac 1 2$, it must be $\sin{(x - 30^o)} = 0$. As a result, $x - 30^o = 0$ or $\pm 180^o$, $x = 30^o, 210^o$ or $-150^o$. Clearly, the only reasonable root for the quadrilateral $ABCD$ is $x = 30^o$.


Let $f(x)=3\cos x + 2 \sin x$, where $x$ is real.

Find maximum of $f(x)$.

And don't use Calculus. :)

well for functions in the form $asin(x)+bcos(x)$ the amplitude is $\sqrt{a^2+b^2}$
and in this case $a=2$ and $b=3$
so the maximum value is $\sqrt{4+9}=\sqrt{13}$

doing it with calculus is just plain messy even though u could

[quote="maokid7"]well for functions in the form $asin(x)+bcos(x)$ the amplitude is $\sqrt{a^2+b^2}$[/quote]

Could you please explain that?

Let $n=\sqrt{a^2+b^2}$. If $a=n\cos c$, then $b$ will be equal to $n\sin c$ by Pythageorean Theorem. Then $a\sin x+b\cos x=n(\cos c \sin x+\sin c \cos x)=n\sin (x+c)$. This makes the amplitude $n$.


function
real analysis
real analysis solved

$f(x)$ is a measurable function defined on $\mathbb{R}^1$ and it holds that \[f(x+1)=f(x) \quad \hbox{a.e.}\]
Find a function $g(x)$ such that
\[g(x)=f(x) \quad \hbox{a.e.}, \quad g(x+1)=g(x), \quad x\in \mathbb{R}^1\]

[note: a.e. means almost everywhere, maybe French books use p.p.]

oops..I realized how to solve it soon after I posted it :(  :blush:

Suppose $E = \{x | f(x+1)\neq f(x) \}$, we know $mE=0$.

Let $F = \bigcup\limits_{-n}^n (E+\{n\})$, $F$ is the union of countable many zero-measure sets, hence $mF=0$ too.

Define \[
g(x) =
\left\{
\begin{array}{ll}
f(x), & x\notin F\\
0, & x\in F
\end{array}
\right.
\]


Does any1 have any idea what the answer to this question is?

The MATHCOUNTS answer is wrong.

9. Using the digits 2, 3, 4, 7 and 8, Carlos will form five-digit
positive integers.  Only the digit 2 can be used more than once
in any of Carlos’ five-digit integers.  How many distinct five-
digit positive integers are possible?

I don't like combinatorics but this what I think (no clue if it's right):
The first digit can be any of the 5.  Since 2 can be re-used, the second digit can also be any of the 5.  The third can be 1 of 4, the fourth 1 of 3, and the fifth 1 of 2.  So that makes $ 5\cdot5\cdot4\cdot3\cdot2\equal{}600$.  What does the key say?

You're clearly overcounting;the 3rd does not have to have 4 choices.
[hide]Casework-
5 twos=1 way.
1 twos=120 ways.
2 twos=10*24=240 ways.
3 twos=10*12=120ways.
4 twos=20 ways.
Summing, gives 501.[/hide]

kinda funny, mathcounts gave the right answer but the completely wrong solution

just look at the last line

120 + 240 + 120 + 200 + 1 = 501

I guess they had an extra number.

Ugh, I remember our team got this one wrong;we did everything but forgot 22222.


analytic geometry
graphing lines
slope
geometry
rectangle
area of a triangle

In the coordinate plane, an L-shaped region is bounded by horizontal and vertical segments with vertices at (0,0), (0,3), (3,3), (3,1), (5,1), (5,0). Find the slope of the line passing through the origin that divides the area of the region in half.

A. 2/7     B. 1/3    C. 2/3    D. 3/4    E. 7/9

[hide]The area of the L-shaped region is clearly $ 9\plus{}2\equal{}11$. We need to find the line through the origin that divides the region into two pieces, each with area of $ \frac{11}{2}$. Since the $ 2$ by $ 1$ rectangle on the right has an area of $ 2$, any line passing through the rectangle clearly will not divide the area in half.
Therefore, we need to find the line passing through the $ 3$ by $ 3$ square which will break it up into two regions, of area $ \frac{11}{2}$ and $ \frac{7}{2}$, on the top and bottom, respectively.
Using the standard formula for the area of a triangle, we have $ A\equal{}\frac{bh}{2}\equal{}\frac{3h}{2}\equal{}\frac{7}{2}$, so $ h\equal{}\frac{7}{3}$. The slope is then $ \frac{h}{b}\equal{}\frac{\frac{7}{3}}{3}\equal{}\boxed{\frac{7}{9}}$.[/hide]


From past year postings in mathcounts.org:

2002
1st[tab]43
2-3[tab]42
4-6[tab]41
 7 [tab]40
8-9[tab]39
10[tab]38

2001
1st[tab]42
2[tab]41
3-4[tab]40
5[tab]39
10[tab]36 or 37
33[tab]30
67[tab]26
95-106[tab]22

Does anyone have additional past results, especially for top 10 in other years?

2000

1st  - 42
2nd  - 41
3rd  - 40
4th -5th  - 39 (one of the 39s actually won)
6th-7th  -  38
8th -  37
9th-12th  -  35
needed a 26 for top 25%

1999

1st  - 45
2nd -3rd  - 43
4th  - 41
5th-6th  -  40
7th-8th  -  39
9th  -  38
10th  -  37
needed a 27 for top 25%
There you go!  that's all I have

frost

Thank you!

Does anyone else have any results from earlier years?

It is of historical interest, and useful for those practicing past nationals to see where they stand.

Hmm ... Alison, do you remember the scores from our year?  I remember how we placed, but not how we scored :)


calculus
integration
real analysis
real analysis unsolved


Find the radius and interval of the following power series.
$ \sum_{n\equal{}1}^{\infty}\frac{n!}{n^n}x^n$
Don't know, actually, how to solve it. :(
Cannot find the $ \lim_{n\rightarrow\infty}\sqrt[n]{\frac{n^n}{n!}}$

Just use the fact that $ \lim_{n\to\infty}\frac{a\plus{}1}{a_n}\equal{}\lim_{n\to\infty}{a_n}^{\frac1n}$, where $ a_n\equal{}\frac{n!}{n^n}$.
I think you should get $ 1$.

[quote="azatkgz"]Find the radius and interval of the following power series.
$ \sum_{n \equal{} 1}^{\infty}\frac {n!}{n^n}x^n$
Don't know, actually, how to solve it. :(
Cannot find the $ \lim_{n\rightarrow\infty}\sqrt [n]{\frac {n^n}{n!}}$[/quote]

well, we know
$ R \;\equal{}\; \dfrac{1}{ \limsup_{n\rightarrow\infty}\; \bigg|\; \dfrac{n!}{n^n} \;\bigg| \;^{\dfrac{1}{n}} }$

so just evaluate the limit.

and why can't you find $ \liminf_{n\rightarrow\infty}\; \sqrt [n]{\frac {n^n}{n!}} \;\equal{}\; \lim_{n\rightarrow\infty}\; \sqrt [n]{\frac {n^n}{n!}}$ ?

the limit is simply equal to $ e$. hence, $ R \equal{} e$  :(

(i will leave it as an exercise for you to show that $ \lim_{n\rightarrow\infty}\; \frac {n}{\sqrt [n]{n!}}$ is indeed $ e$. it is quite simple. )


Define a sequence such that $a_1 = a_2 = 1$ and for all $n>2$, $a_n = a_{n-1}+2*a_{n-2}$

Is there a closed form expression for $a_n$ in terms of only $n$?

yes...there is a standard process for doing this, look in aops 2 for an explicit form for the fibbonacci sequence

[quote="Altheman"]yes...there is a standard process for doing this[/quote]

Can you post the link or the process, please???

the characteristic polynomial for the recursion: $Aa_n+Ba_{n-1}+Ca_{n-2}=0$
is $P(x)=Ax^2+Bx+C=0$, let $P(r)=P(s)=0$, then we can say that the explicit formula is
$F(n)=k*r^n+c*s^n$ where $k$ and $c$ are constants
the way to get this is kind of complicated, but it is easy to prove with induction

Rewrite the recurrence as
$a_n+a_{n-1}=2(a_{n-1}+a_{n-2})=2^2(a_{n-2}+a_{n-3})$

$=\cdots =2^{n-2}(a_2+a_1)=2^{n-1}\ \cdots [1].$

We can also rewrite as $a_n-2a_{n-1}=-(a_{n-1}-2a_{n-2})=(-1)^2(a_{n-2}-2a_{n-3})$

$=\cdots =(-1)^{n-2}(a_2-2a_1)=(-1)^{n-1}\ \cdots [2].$

Solving system equation $[1]$ and $[2],$ yielding $a_n=\frac{2^n-(-1)^n}{3}.$

This is the standard solution for Japanese junior, high school students.

Thanks very much

 :blush: I should've been able to figure that out
(btw this was part of exploring an extension of a 2003 AIME2 problem 13)


Do a search of the forum -- this question is here a half-dozen times at least.

[quote="mgao"]Well, I was just wondering, are there any free 3D geometric programs on the internet available for download?  If so, plz let me know.[/quote]

I have now a new one for me, its called Derive 5... it's nice!


[hide]Either 2x-1=-5 or 2x-1=5. In the first case, x could be -2. In the second, 3. So the solution set is {-2,3}.[/hide]

Billy

[quote="symmetry"]What is the solution set of the equation

|2x - 1| = 5?[/quote]

[hide]Since it's in absolute values, we can make this equation into 2 equations:

$2x-1=5$

and

$2x-1=-5$

Solving both, we get that $x$ can equal $3$ or $-2$

{$3$, $-2$}[/hide]

[hide]
|2x - 1| = 5
breaks down into 2x-1=5 and 2x-1=-5.
solving for x, we get x is an element of the solution set {3, -2}

[/hide]


Say that I play a game against Michael Jordan (when he was with the Bulls) and the probability that I can win a game, any game, is $0.20$. What is the smallest integer $n$ for which the following statement is true: "If Michael and I play $n$ games, then I have better than $60$% chance to win atleast one".

[hide="Answer"]This is simply the case when the chance to win no games drops below 40%. You have a $\frac{4}{5}$ chance to lose, or $.8$, so we have $(.8)^4=.4096$, so the answer will be $n=5$ games.[/hide]

Sorry if this is really wrong but
is the answer
[hide]
$n=3$ games because the probability to win $1$ game out of $n$ is $\frac{1}{n}$ but as there is $60$% chance so $\frac{\frac{3}{5}}{n}$ which gives the answer as n=3 because $\frac{\frac{3}{5}}{n}=\frac{1}{5}$
[/hide]
Is this right???? :?  :(  plese can somebody correct me????

Why is the probability to win 1 out of n equal to $\frac 1n$?  Think about that for a while and then you might see why that doesn't work.

no that was if the $60$% was not given it is given so it is $\frac{\frac{3}{5}}{n}$.

But why would it be that even if the 60% weren't given?

I really dont know but then what will it be????

Look at it this way.

The probability of winning game 1 is 0.2

The probability of winning game 2 (given that you don't win game 1) is (1-0.2) x 0.2

The probability of winning game 3 (given you didn't win either of the other two is (1-0.2)^2 x 0.2

Proceed.

The probability that you win at least one game in the first $n$ games is equal to 1 minus the probability that you don't win any games in the first $n$ (we say this because it's easier to figure the latter out).  The probability of not winning any games in the first $n$ is $\underbrace{.8\cdot .8\cdot .8\cdots .8}_n=.8^n$.  Thus the probability is $1-.8^n$.  We want this probability to be greater than $.6$, thus

\[1-.8^n>.6\Rightarrow .8^n<.4.\]

The first $n$ such that this is true is 5.  This is the answer.

Note that this is the same method as JesusFreak's.  The idea of counting the complement is very powerful in probability.

I'll note that my method, if we use geometric series, reduces to

$0.2(1+0.8+\ldots+0.8^{n-1})=0.2\left(\frac {1-0.8^n}{1-0.8}\right)=1-0.8^n$

which is exactly the above result.

Well that means that the probability that Michael Jordan will win every game must be less that 40%.  The chance of his winning any game is 80%.  Thus, the answer is [hide]$\log[.8]{.4}$, rounded up, i believe.  That is 5.  Sure enough, 4 games is 40.96%, but 5 games is 32.768 %[/hide]


AMC 10
AMC 10 A
AMC 12 A
AMC 10 B
AMC 12 B
AMC

How many people take the AMC tests yearly.

These stats are pretty easily available on the AMC website. It's around 100,000.

Check your addition.

The statistics for 2008 AMC 10 A, 2008 AMC 12 A, 2008 AMC 10 B and 2008 AMC 12 B
are at:
http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2008-12a/pdfs/2008--1012%20Grand%20Statistics.pdf

You will find that the total for all four is more than 230,000.

Steve Dunbar
MAA Director, American Mathematics Competitions


Hey startogalaxy,

can you send me our group picture with Jacob? You know my email I presume.

So for those of you who got the luxury to go to Winter Camp, how was it?

didnt it??

EDIT: Okay i was right...phew

Winter Camp was really nice. :)
so much training...

And great services from Alex. lol. kidding

btw, it'd be a great idea to post most of the unsolved tough problems we had during the week. :P

Find all functions from reals to reals, such that:
$f(x+y)+f(x-y)=2f(x)cosy$

The only solution is all functions of the form $f(x)=a\sin x+b\cos x,\ a,b\in\mathbb{R}$. :)

Very nice winter camp!!hopefully next year i can be there too... :lol:

[quote="rem"]I know, but I do not know how to prove it...[/quote]
A hint
[hide]Use subsitution, $x=0, y=t$
                                $x=\frac{\pi}{2}+t, y=\frac{\pi}{2}$
                                $y=\frac{\pi}{2}+t, x=\frac{\pi}{2}$[/hide]

oh...y'all discussin still the problem proposed by Dr Ed Doolittle?


calculus
integration
function
calculus computations

I have found this integral
$\int e^{x}(e^{2x}+\frac{x^{2}}{e^{x}}) \sqrt{1+e^{-2x}}dx$
here: http://www.dmi.unict.it/%7Eemmanuele/PDF/compiti/ingegneria/tuttiinginfAM1.pdf   page 75.

The [url=http://integrals.wolfram.com/index.jsp]Wolfram integrator[/url] gives something unintelligible (including a lot of special functions, etc.) for $\int x^{2}\sqrt{1+e^{x}}$, which is the essential part of the problem (the rest is piece of cake). Normally, this means that there is no elementary antiderivative though I'm too lazy to try to prove it formally.


I'm trying to execute a loop to place text from an array at regular intervals along a horizontal path. However, the text isn't aligning horizontally like I want it to. I want it to have the same baseline (like if it were printed in a LaTeX document), but instead when I run this:

void tablabel(string[] labs, int ht, int hset=0, int vset=0) {
	int numlabs = labs.length;
	for (int c=0; c<numlabs; ++c) {
		label(labs[c], (ht*(c+.5)+hset, vset), N);
	}
}

with an array like this...

tablabel(new string[]
	{"$ L$:",
	 "$ x$",
	 "$ \minus{}x$",
	 "$ \minus{}1$",
	 "$ 1$",
	 "$ \minus{}x$",
	 "$ \minus{}1$",
	 "$ 1$",
	 "$ x$",
	 "$ \minus{}x$",
	 "$ 1$",
	 "$ \minus{}1$",
	 "$ 1$"}, 30, -30, 0);

The terms with a minus sign are on a different horizontal baseline than those without. Does the minus sign change the height of the bounding box, or throw off the "center" of the label? I've done a lot of googling but can't find a simple explanation of the 'align=' options.

Thanks again for all your help  :)

If I understand it right, the $ \minus{}$ sign changes the bottom of the box the same way as the $ \plus{}$ sign (with the latter it is pretty apparent). The simplest remedy is to use \vphantom. Of course, you do not need to type it 10 times, just use it in the for loop:

[asy]void tablabel(string[] labs, int ht, int hset=0, int vset=0) {
int numlabs = labs.length;
for (int c=0; c<numlabs; ++c) {
label("$\vphantom{+}"+labs[c]+"$", (ht*(c+.5)+hset, vset), N);
}
}

draw((-20,3)--(100,3));

tablabel(new string[]
{"1", "-1", "x", "-x", "+x", "\times t"}, 20, -20, 0);[/asy]


linear algebra
matrix
linear algebra unsolved

EDIT: Fixed mistake in problem (in italics)

Let $ A$ be a non-singular skew symmetric matrix over a field with characteristic $ \neq2$. Show that $ A$ is [i]congruent[/i] to a block sum of matrices $ \left(\begin{array}{cc}0 & 1 \\
\minus{} 1 & 0\end{array}\right)$.

Let $ J$ denote the matrix given in the problem statement. Since $ A\equal{}\left(\begin{array}{cc}0 & a\\\minus{}a&0\end{array}\right)$ for some $ a\not\equal{}0$ holds, the claim is verified by setting $ T\equal{}\left(\begin{array}{cc}a & 0\\0&1\end{array}\right)$ and observing $ T^TJT\equal{}A$


I think I did deserve to get banned or muted. BUT! i think its very unfair that i got banned for saying something while lots of people say bad stuff or even worse things and they never get banned
. Is there anything we can do about that

I also wonder why my ban was extended to one year from one month. I mean what i said wasn't really bad. and what did i do that had it extended. Like i said lots of people have said worse stuff.

[quote="myyellowducky82"][quote="ernie"]I think it has been more enforced.[/quote]
especially w/ all the new rules.[/quote]

Yeah, and with more people swaring, flooding, etc.

[quote="jain_harshi"]There should be a way to catch only the people who swear or flood. Not everyone should be banned[/quote]

You could report.

[quote="violin321"]the new rule is so unfair. Some people (like me) don't use the forum. I never make rude remarks or anything, so why should I not be able to ask for the answer or whatever in ftw?!
oh shoot this would be the wrong thread...[/quote]

you could always review for the answer
and most of the time, if you don't make productive posts in the forum, you might not make productive replies in the chat

BTW, if you think this is the wrong thread, y'd you click submit??


Let $O$ be the intersection of the diagonals of convex quadrilateral $ABCD$. The circumcircles of $\triangle{OAD}$ and $\triangle{OBC}$ meet at $O$ and $M$. Line $OM$ meets the circumcircles of $\triangle{OAB}$ and $\triangle{OCD}$ at $T$ and $S$ respectively.

Prove that $M$ is the midpoint of $ST$.

By simply angles we can prove that triangles MDA, MBC, SDC and TBA are similar. Thus we have:
MS=BD*CM/BC=(AC*BM/CM)*(CM/BC)=AC*BM/BC=MT :D
PS: What is CGMO?

CGMO = Chinese Girls Mathmetical Olympiad. It is a contest only for girls. The problems are relatively easy.  :)

My solution:

Since $ M $ is the Miquel point of complete quadrilateral $ \{ AC, AD, BC, BD \} $ ,

so we get $ \triangle MCA \sim \triangle MBD $ and $ \frac{BM}{CM}=\frac{BD}{AC} $ . ... $ (1) $

Since $ \triangle BTM \sim \triangle BAC $ ,

so we get $ \frac{TM}{AC}=\frac{BM}{BC} $ . ... $ (2) $

Since $ \triangle CMS \sim \triangle CBD $ ,

so we get $ \frac{MS}{BD}=\frac{CM}{BC} $ . ... $ (3) $

From $ (1), (2), (3) $ we get $ TM=MS $ .

Q.E.D

Dear Mathlinkers,
you can see and more

http://jl.ayme.pagesperso-orange.fr/  vol. 16  Deux triangles adjacents..., p. 57-61.

Sincerely
Jean-Louis

Apply an inversion around $P$: then $\{P\}= A'C' \cap B'D'$, $\{M'\}= A'D' \cap B'C'$, $\{T'\}=A'B' \cap PM'$ and $\{S'\}=C'D' \cap PM'$. Since $(M',P;T',S')=-1$, it follows that $$\frac{MT}{MS}=\frac{\frac{M'T'}{T'P}}{\frac{M'S'}{S'P}}=1.$$

Inverse around $O$,then
$M'=A'D'\cap B'C',T'=OM'\cap A'B',S'=OM'\cap C'D'$.
Since inversion preserves cross ratio,we have
$$(T,S;M,\infty_{TS})=(T',S';M',O)=-1$$
Done.


Prove that $ \frac{5^{125}\minus{}1}{5^{25}\minus{}1}$ is a composite number.

[quote="kunny"]$ \frac {a^5 \minus{} 1}{a \minus{} 1} \equal{} a^4 \plus{} a^3 \plus{} a^2 \plus{} a \plus{} 1.$[/quote]

but how does this prove that the no. = $ b \times c$, b and c being +ve integers?

@ Kunny can u please explain ,i don understand :(

[quote="kunny"].
.
$ \equal{} \{a(a \plus{} 1) \plus{} 1\}\{(a \minus{} 1)a \plus{} 1\} \plus{} (a \minus{} 1)a(a \plus{} 1) \plus{} 2a.$[/quote]

well, this doesn't explain why its a composite number!

Sorry, my solution didn't explain the proof and make you guys, confuse you.

kunny

well, i don't know how to do this  :|

but i found this Q under the category "Number theory" in a book ...

[b]Thank you very much[/b] for the method, Kouichi Nakagawa  :)

I have just found the problem [url=http://www.mathlinks.ro/viewtopic.php?p=849468#849468]here, IMO SL Problem 1992[/url] :lol:


A "rising number" is a positive integer with multiple digits such that each digit is larger than the one to its left.
(ex. 578, 145 are, but 242, 144, and 321 aren't)

What is the total number of "rising numbers"?

[hide="hint"]Break it up into cases; 2-9 digits[/hide]

[hide="solution"]For any given sequence of different numbers, there will be one "rising number" that can be generated from it.  There are 9 digits to use, thus, there are

$\binom92+\binom93+\cdots+\binom99=2^{9}-9-1=502$
"rising numbers."[/hide]

That seems right.

Also, the number of falling numbers is just

$2^{10}-10-1=1013$.

Unfortunately, that is wrong Quevvy. b-flat's answer is nearly right.

b-flat got it.

Though there's a minor error in arithmetic in there:
We should have:
$\binom92+\binom93+\cdots+\binom99=2^{9}-9-1=512-9-1=\boxed{502}$

Oops, forgot that $512-9-1=502\neq501$  :blush:

I'll fix that.

[hide=" :ninja: "]Similarly, for any given sequence, there is exactly one number that can be formed, but repeated digits are allowed, so there are $\binom{19}{10}=92,378$ numbers.[/hide]

I don't really get the formulas that you used. Can someone explain them to me? I mean, how does it end up as $2^{9}-9-1$?

Well, let's line up the digits 1 to 9.

1, 2, 3, 4, 5, 6, 7, 8, 9

If we wanted a 3-digit long increasing number, we simply select any 3 of the 9 numbers, and automatically they're in increasing order. How many ways can you choose 3 digits from the 9? $\binom{9}{3}= 84$ ways.

Now, we want increasing numbers of any length, so we can take any subset of the digits from 1 to 9. There's $2^{9}= 512$ such subsets (why?). But we can't include the 0-digit subset (empty set) or a single digit set. So we subtract 1 + 9 = 10.

Ok thanks!  :blush: I just didn't get the equations. Now I get it

[quote="JavaMan"]Here's a variant:

A positive number is non-decreasing if each digit is more than or equal to the digit to its left. How many 10 digit non-decreasing numbers are there?  :)[/quote]

I believe the answer to that is $\binom{18}{10}$, because there are 10 digits and 8 dividers for the separation from 1 to 2, 2 to 3, 3 to 4, ...8 to 9.


$p$ is prime .${a_n}$ is defined $a_n=n$ for all $0=<n\leq p-1, a_n=a_n-1 +a_n-p$ for all $n\geq p$.
caculator :$a_p^3mod(p)$

Problem from Polish MO 1995, discussed before:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23201