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MATHCOUNTS
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AMC 8
FTW
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I've finally decided, to start posting on this AoPS thing. As to who I am, you can go on http://www.xtremesystems.org/forums look for user serialk11r and search through my 4000 posts to find the one where I revealed it. Nah here's a shortcut: I failed at mathcounts last year, 36th in the state. Yes I'm the lone JLS person in the top 25% at state out of the 3 individuals who made state (which wasn't celebrated to my disappointment :mad:  jk, I'm failing at math so I don't deserve it). I just moved to Palo Alto last year, so I don't really have JLS pride anyways :P
Sorry to Lynnelle for not joining up 1 year ago, I was sorta busy with that other forum as you can see  :oops:
Good to see some familiar people :)

PS dang at 3 pm I managed to make a ton of errors when typing that simple message.

Ahaha. I fail more than you.

At JLS I am known as a self-proclaimed failure... by some people.

Did this make sense?  :huh:

[quote="bowei"]Are you another of Ubemaya's accounts?

(It does if you are Ubemaya)
 :D[/quote]
No, I'm not Ubemaya(albert), I know him though sorta...I wish I could "suck" as much as he does at math :) I go to Gunn, I'm a freshman. The palo alto people know who I am. I joined cuz I finally decided to take a look here, one year after Lynnelle tried to force me and the rest of the JLS math club (which is dead) to, and I was like wow this forum is dead, and I love forums!!! so I decided I'm gonna try to revive it, get some more math pro people around here to join, get older people to post more.

[quote="bookaholic"]Yay, hi! JLS MathCounts isn't dead, it's just its online presence that I've given up on. JLS MathCounts is pretty awesome right now--it won't start dying until around March/April. So John joined too? What's his username?[/quote]
John didn't join yet, he'll be making a little intro post too :) He forgot his username you see...
I should go visit JLS sometime, see how math club is going. I should also visit jordan math club, only problem is I don't know when it is. If it was the same time as last year, then Gunn math club will be in the way.

[quote="bookaholic"]I think Jordan math club is Friday after school. If you ever visit JLS math club, we're now Thursday after school in room 350.[/quote]
Hmmm that means I'll have to miss chess to go to JLS and I'll have to miss math club to visit jordan :(

[quote="serialk11r"][quote="John the Scavenger"]Yoyo, everyone.

The Jordan math club just took the test to arrange Mathcounts teams.  Compared to us, it was epic fail.  I believe the highest scores on the sprints were three 21's and a no-name 19 (gaaaaaaaah!).  Oh well... we've got some promising 6th grade noobs, though, and I think we can train them to own Mathcounts in a couple of years.[/quote]
Boylie, nice to see you!
Hahahahaha so it turns out JLS is STILL sucking worse than Jordan.[/quote]
1. Hi John!
2. O rly? Explain.

well JLS rocked in '06 so it doesn't matter and appararently 1992

Apparently Steven SUDDENLY turned super beast at mathcounts...so I guess JLS has a chance. But on thursday I went to JLS and to my horror, they were taking LAST YEAR's school round.

[quote="bookaholic"]W...T...F? Are you serious? Okay, I'm emailing Mr. Peterson.[/quote]
Of course I'm serious, you can ask Steven too.
If JLS isn't able to participate that would really suck...seeing that Steven has improved so much.

Oh, that's not a problem--MathCounts doesn't care how schools choose team members. I'm more annoyed about 1) the unfair advantage people who remember problems better got and 2) the sheer silliness of it.

[quote="CheeseIsGood"]I did not take the AMC 8 last year.

Where did you get the number 17 from?[/quote]
LOL okay okay okay sorry I thought you took it and failed miserably...
Seriously though, on FTW you had INSANE speed and last year you were no where near that fast.

Welcome to the (100% 1337) bay area forums.  Enjoy your stay, reception is to your right.

[quote="xscapezaer"]Which way is right on a forum?[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48329

[quote="serialk11r"][quote="CheeseIsGood"][quote="serialk11r"]Apparently Steven SUDDENLY turned super beast at mathcounts...[/quote]

No. Definitely not.

I still suck. You can't come to a conclusion just based on how I did on AoPS FTW. I'm doing pretty horribly now though. My rating died in the last week.[/quote]
uh, I think I can.
Also, please explain how you got a perfect AMC8 score, when you got like 17 last year or something like that.
I think you'll totally destroy everyone ;) maybe not the redwood and harker kids but whatever.[/quote]

Redwood sucks this year. Which is surprising, considering that I'm not on the team anymore.

[quote="Ubemaya"][/quote][quote="serialk11r"][quote="CheeseIsGood"][quote="serialk11r"]Apparently Steven SUDDENLY turned super beast at mathcounts...[/quote]

No. Definitely not.

I still suck. You can't come to a conclusion just based on how I did on AoPS FTW. I'm doing pretty horribly now though. My rating died in the last week.[/quote]
uh, I think I can.
Also, please explain how you got a perfect AMC8 score, when you got like 17 last year or something like that.
I think you'll totally destroy everyone ;) maybe not the redwood and harker kids but whatever.[/quote][quote="Ubemaya"]

Redwood sucks this year. Which is surprising, considering that I'm not on the team anymore.[/quote]

If you think that you suck, then you saying Redwood sucks means they will kill everyone, am I right?

[quote="Jack Sparrow"]
You're not making any sense.
[/quote]

[quote="bookaholic"]Um. Do you guys know what JLS's AMC-8 team score ended up being? Like, what were the top few scores, other than Steven's? (Congratulations on the perfect, by the way!)[/quote]
No... Lemme guess though, 17, 17, 16, 15.....? It doesn't matter if everyone else at JLS sucks (that's the case if I remember correctly...no offense to some of the other kids if they ever see this), if Steven beasts mathcounts like I expect him to, then he can go to state as an individual easily, and I'm hoping top 16 in state :D you'll be pretty good at countdown. This year chapter will be really easy to beast since Jordan has no one left, Terman has no one left, Bowditch's 2 good people left, eh...no other schools are worth mentioning.
Last year I was a total failure. I managed to get a 20...I only finished 20 problems though...my time management on tests is pretty bad I must say, but usually tests are easy enough for that not to matter. Argh another example of where school is destroying my math...School math is messing up my thinking processes because they teach the most retarded ways to do problems.

[quote="CheeseIsGood"]I know that there was a 19. The scores haven't been announced yet.... I got my score by asking Mr. Peterson.[/quote]
dannggg that's pretty low...
last year was 22, 21, 20, then a whole bunch of crappy scores. :(
If I wasn't being stupid JLS could've made the honor whatever list thingy.


Let $x_1, x_2,\cdots,x_n$  be real numbers such that $0\le x_i\le1$ for $i=1,2,\cdots,n.$ Prove that
\[
\frac 1 n\sum_{i=1}^n x_i^2-(\frac 1 n\sum_{i=1}^n x_i)^2\le\{\begin{array}{ll}
\frac 1 4 & \textrm{if  n is even }\\
\frac 1 4-\frac 1{4n^2}& \textrm{if n is odd }
\end{array}
\]

[i]Edited by Myth[/i]

Indeed, LHS is a convex function on each $x_i$, so it attains maximum for $x_i\in\{0,1\}$. The result follows.


Εμένα μου άρεσαν τουλάχιστον.... :D  :D

1) Ν.δ.ο το άθροισμα των συντελεστών του πολyωνύμου

[b]P(x) = Q(x)*(x-1) + 1 [/b]αποτελεί λύση της εξίσωσης [b]P(P(x))=x[/b]

2) Να βρεθεί το άθροισμα των συντελεστών του πολυωνύμου P(x) αν το P(x) έχει την εξής ιδιότητα

[b]P(x+1) = P(x) +2007 - a[/b]

όπου a ο σταθερός όρος του P(x)

Πολύ ωραία άσκηση (και λύση).
Μπράβο!  :wink:

Ωραίος Γιάννη...Και η δεύτερη κάπως έτσι λύνεται....

Εφτιαξα και μια άλλη που μου άρεσε....Είναι λίγο πιο δύσκολη από τις προηγούμενες.

Εστω πολυώνυμο [b]P(x)[/b] με (μόνο) θετικές ρίζες [b]X1,X2,....Xn[/b]

τέτοιες ώστε [b]X1*X2*X3*.....Xn = 1.[/b]

Ν.δ.ο [b]P(1)<=0[/b]

 :D

Fotis είσαι σίγουρος για την τελευταία άσκηση;
Το πολυώνυμο [b]P(x)=(x-1/6)(x-2)(x-3)[/b] ικανοποιεί τα δεδομένα του προβλήματος ,αλλά [b]P(1)=5/3 > 0[/b] .

Για την 2 θέτουμε όπου [b]x[/b] το [b]0[/b] και βγαίνει [b]P(1)=2007[/b].

Σκέφτηκα κι εγώ μια σχετική(?) άσκηση και τη βάζω.
'Εστω [b]P(x)[/b] ένα πολυώνυμο με ακέραιους συντελεστές. Αν ισχύει [b]P(P(P(1)))=1[/b] να βρεθεί το άθροισμα των συντελεστών του [b]P[/b].

Δίκιο έχεις Stedes...Τώρα που ξανακοιτάω τη λύση μου βρίσκω λάθος.... :maybe:  :blush:

[quote="stedes"]
'Εστω [b]P(x)[/b] ένα πολυώνυμο με ακέραιους συντελεστές. Αν ισχύει [b]P(P(P(1)))=1[/b] να βρεθεί το άθροισμα των συντελεστών του [b]P[/b].[/quote]

=1..... :wink:

 :D

Ok, για τη 2 έχουμε:

Άθροισμα συντελεστών δίνεται για $P(1)$. Επίσης, προκύπτει εύκολα ότι $x=0$. Όμως $P(x)=a$ από υπόθεση. Άρα $P(1)=2007\ldots$

[quote="bouzoukman"]Με την ίδια λογική με τα παραπάνω και το γινόμενο των ριζών ενός πολυωνύμου ν βαθμού ακόμη και να έχει μιγαδικές ρίζες θα ισούται με
\[a_{0}\]
αν ν=2κ και με
\[-a_{0}\]
αν ν=2κ+1.
Πως βάζουμε αυτά που γράφουμε σε Latex να είναι συνέχεια του κειμένου??
Δεν μπορω να καταλάβω...... :wallbash: :wink:[/quote]
an katalava kala bouzoukman legontas sinexia tu kimenu malon enois ayto...

isute me $a_{0}$ an n=2k i me $-a_{0}$ an n=peritos..

an enois ayto,tote kane quote ayto to post pu diavazis tora gia na dis pos to exo grayi... :wink:

 :D

Thanks Rakar!!!!!!!
Καλά κατάλαβες,το βρήκα αυτό που ήθελα..
Ισχύει αυτό που είπα για το γινόμενο των ριζών ενός πολυωνύμου??
Διότι [b]P(χ)=(χ-$x_{1}$)..(x-$x_{n}$)[/b] οπότε [b]P(0)=(-$x_{1}$)..(-$x_{n}$)[/b].
Άρα για n=2κ έχουμε ότι
[b]P(0)=$x_{1}$..$x_{n}$[/b]

ενώ για n=2κ+1
[b]P(0)=-($x_{1}$..$x_{n}$)[/b].
Όμως P(0)=$a_{0}$ οπότε εύκολα βγαίνουν αυτά που έλεγα παραπάνω...

@bouzoukman:
ontos etsi ine...ala ayto mporis na to pis +amesos ap tus tipus tu vieta afu
$r_{1}r_{2}...r_{n}=(-1)^{n}\frac{a_{0}}{a_{n}}$

@stedes:
i lisi pu ixa vri =1....ida oti kolai kapu,ala parolayta pali nomizo oti =1...8a to ksanayakso...

@thatin:
des+ta pm su...

 :D

edit:iparxi mipos kanenas pu na exi enan [b]olokliromeno[/b] odigo xrisis tu mathematica?an kapios mpori na voi8isi, na mu stili pm...
thanks..


algebra
polynomial
number theory
prime numbers
algebra unsolved

Prove that if $a^p+b^p$ is integer for infinitely many primes $p$, then $a$ and $b$ are integers.

I can only prove that $a$ and $b$ are algebraic numbers.

Also, check problem S5 [url=http://reflections.awesomemath.org/2006_1/2006_1_problems.pdf]here[/url]

Oh.. I see there were stonger conditions. I think I have a proof.

I can get now, that there are two polynomials of  degree  4 which vanishes: one in a, and the other in b, and which have the coeff for $X$ zero.
 [hide="hint"]$b^6=(k+a^2)^3=(p+a^2)^3$ ,with k,p integers[/hide]
hence $a^4-b^4=(a^2-b^2)(a^2+b^2) \in \mathbb{Z}$. And since $a^2-b^2\in\mathbb{Z}$ we get that $a^2+b^2\in \mathbb{Q}$ then $2b^2\in \mathbb{Q}\Rightarrow b^2,b^4\in\mathbb{Q}\Rightarrow$ since there is a polynomial of degree 4 with coeff for X 0 whih vanishes in b  $b^3\in \mathbb{Q}\Rightarrow b\in \mathbb{Q}$.

Hence $a,b$ are rational numbers. Write them as $\frac m n$ and $\frac k n$ .Assume that $n\neq 1$ and fix an $a$. We have now that $m^q\equiv k^q (\mod n^a)$ for all primes $q>a$. Since prime numbers take all residues mod $\phi(n^a)$ we get that $m^q\equiv k^q (\mod n^a)\forall q\in \{1,\dots, \phi(n)\}\Rightarrow m\equiv k (\mod n^a)$. since a was chosen arbitrarly we have that $n^a\cdot m-n \forall a$, hence $m=n$ which leads us to $a=b$. Contradiction. Hence $n=1$, and so, a,b are integers.

Am I wrong somewhere?

[quote="xirti"]Oh.. I see there were stronger conditions. [/quote]
And also, $a^p-b^p$ instead of $a^p+b^p$.
Please, don't post any solutions until March 1st.


Chuck Norris can roundhouse kick someone in 15 nanoseconds, exactly. Him working together with his best friend, Norris Chuck, can roundhouse kick someone in 10 nanoseconds. How long would it take Norris Chuck to roundhouse kick someone?

Yay for roundhouse kicks! And somehow I have this feeling that someone will post some kind of Chuck Norris joke here...  :huh:

How can two people land a roundhouse kick faster if both of them kick slower than that rate?
Unless they like help them or something with the technique lol.

Let's convert this to:

Chuck Norris can mow his lawn in 15 nanoseconds.  Working with Norris Chuck, they can mow it in 10 nanoseconds.    How long does it take for Norris Chuck to mow the lawn by himself?

$ \frac{1}{15}\plus{}\frac{1}{x}\equal{}\frac{1}{10} \implies \frac{1}{x}\equal{}\frac{1}{30} \implies x\equal{}\boxed{30}$ nanoseconds.

Gah I made this problem earlier, but mixed up the numbers! Nooooo! :mad:

EDIT: AIME15, that was the intended answer.  :roll:

[quote="Cheese12345"]chuck norris's kick is so fast u cant time it lol  :P[/quote]
1) This is spam
2) Bruce Lee pwns Chuck Norris

1. Stop spamming, especially if it's off-topic.
2. This is also spam, but all the posts on this page have been rated.
3. In general, for these problems, if person A does an action in $ n$ hours, and person B does an action in $ k$ hours, and the amount of time it takes for them to do it together takes $ h$ hours, we can say $ \frac{1}{n}\plus{}\frac{1}{k}\equal{}\frac{1}{h}$.  This post is no longer spam.


function
limit
real analysis
real analysis unsolved

Can we show that $\mathcal{C}[0,1]$ is complete with the norm $\|f\|=\max |f(x)|$ by using this argument.
$C[0,1]\subset L^\infty[0,1]$ by using the norm $\|f\|_\infty$. It is known that $L^\infty$ is complete. So we just need to show that $C[0,1]$ is closed. But convergence in sup norm (max norm) in this space means convergent uniformly, so every convergent sequence converges to continuous function. So $C[0,1]$ is closed and we are done.

thanks in advance

An argument that's simpler ($L^\infty$ is a pretty ugly object):
A Cauchy sequence in the uniform norm is pointwise Cauchy, so it converges pointwise. This convergence is clearly uniform, and the uniform limit of continuous functions is continuous.

[quote="Kent Merryfield"]It's not wrong, but it is very odd. We have to ask, how did we prove the completeness of $L^{\infty}$ in the first place? If we were to look at the proof of that, we would find arguments that, much simplified, could be used in the case of $\mathcal{C}.$

The first problem in any completeness proof is to produce a candidate: an object that might be the sought-for limit.

In this case: assume that $(f_n)$ is a uniformly Cauchy seqence of functions. Then it follows that for each $x,\,(f_n(x))$ is a Cauchy sequence of real (or maybe complex) numbers. By the completeness of $\mathbb{R}$ (or $\mathbb{C}$), $\lim f_n(x)$ exists. Define $f(x)$ as that limit.

We then have our candidate. The two remaining steps are to show that $f_n$ converges uniformly to $f$ and that $f$ (as the uniform limit of continuous functions) is itself continuous and hence a member of $\mathcal{C}.$

(OK, jmerry just said the same thing in many fewer words. I'll finish the post anyway.)

The proof that $L^{\infty}$ is complete is harder that that; in particular, we have to work harder to produce the candidate, because we can't claim that $(f_n(x))$ is a Cauchy sequence of numbers for every $x.$[/quote]

I know that the usual procedure to show completeness of a space. I used the above argument because the problem is appear in a section about $L^p$ space (Royden book). So I just suspect that he expect to use properties of $L^p$ space. Thanks a lot for all commentts (yassin, jmerry, and Dr. Kent)

[quote="yassinus"]but if we dont use completeness of L°° , we will use the same argument that prove that L°° is a Banach to show the other space is complet ( i say almost the same technic are used ...)[/quote]

Royden claim that completeness of $L^\infty$ is easy, so he left it as exercises
One of the exercises, there is a problem asking to prove that convergence in $L^\infty$ is equivalent to uniform convergence outside a set of measure zero.

Using this exercise to prove completeness of $L^\infty$ I proceed as the following:
$(f_n)$ is Cauchy in $L^\infty$ implies $f_n$ Cauchy (uniformly) outside a set of measure zero  $A$. So by redefining $f_n$ and set $f_n(x)=0$ in $A$, we have a new $f_n$ that Cauchy in $\mathbb{R}$. By completeness of $\mathbb{R}$ we have the candidate $f$ for our limit.  MOreover $f_n\to f$ uniformly on $A$, so $f_n\to f$ in $L^\infty$ by the above exercise.
Hopefully its ok


inequalities
trigonometry
inequalities proposed

Let $ x_i>0, y_i\in \Bbb{R}$ such that $ \sum_{i\equal{}1}^n x_i\equal{}\sum_{i\equal{}1}^n y_i\equal{}\pi$ prove that
\[ \sum_{i\equal{}1}^n\frac{\cos y_i}{\sin x_i}\le \sum_{i\equal{}1}^n\cot x_i\]

Any idea?  :wink: actually I don't have solution? Can anyone move it to unsolve problems?


linear algebra
matrix
algebra
polynomial

$ c$ is a constant real and $ \{a_n\}$ is a sequence defined as

$ a_1\equal{}a\in \mathbb R$,　$ a_{n\plus{}1}\equal{}\frac{ca_n\minus{}1}{a_n}  (n\equal{}1,2,\cdots )$.

If the sequence $ \{a_n\}$ has just three distinct values,then determine the possible value of $ c$ ,and find the range of $ a$ for each $ c$.

[hide]Let $ f(x) \equal{} \frac {cx \minus{} 1}{x}$. Our sequence can be defined as $ a_{n \plus{} 1} \equal{} f(a_n)$

Observe that the sequence must be periodic -- therefore, $ a_{n \plus{} 3} \equal{} a_n \implies f(f(f(a_n))) \equal{} a_n$

We solve the equation $ f(f(f(a))) \equal{} a$ for $ c$. The LHS simplifies to $ \minus{} \frac {1 \minus{} c^2 \minus{} 2 c a \plus{} c^3 a}{c \plus{} a \minus{} c^2 a} \equal{} a$.

Clearing the denominators and factoring, we get that $ c \equal{} 1$ or $ c \equal{} \minus{} 1$ or $ c \equal{} \frac {1 \plus{} a^2}{a} \equal{} a \plus{} \frac {1}{a}$

We know this gives at most $ 3$ distinct values, but we need to have at least $ 3$ distinct values. We need to eliminate the cases in which $ f(a) \equal{} a$. Solving this equation for $ c$ gives $ c \equal{} a \plus{} \frac {1}{a}$, so we have to strike out that entire class of solutions.

Finally, we have to make sure that we're never dividing by $ 0$. For $ c\equal{}1$ and $ c\equal{}\minus{}1$, we have to make sure none of $ a$, $ f(a)$, $ f(f(a))$ are zero.

$ \blacktriangleright$ For $ c\equal{}1$, $ a \neq 0$, $ f(a) \neq 0 \implies a \neq 1$, and $ f(f(a)) \neq 0$, which is always satisfied.
$ \blacktriangleright$ For $ c\equal{}\minus{}1$, $ a \neq 0$, $ f(a) \neq 0 \implies a \neq \minus{}1$, and $ f(f(a)) \neq 0$, which is always satisfied.

The final solution set is therefore:

$ c  \in \{ \minus{}1, \; 1 \} \; \& \; a \not \in \{ 0, c \}$

[/hide]

[hide="Alternately"] The third power of $ \mathbf{M} \equal{} \left[ \begin{array}{cc} c & \minus{}1 \\ 1 & 0 \end{array} \right]$ must be a multiple of the identity matrix, so its satisfies $ \mathbf{M}^3 \equal{} k \mathbf{I}_2$ (but no smaller power should be the identity matrix) and its characteristic polynomial should share roots with it.  That polynomial is

$ \lambda(\lambda \minus{} c) \plus{} 1 \equal{} \lambda^2 \minus{} c \lambda \plus{} 1$.

This is clearly possible when $ c \equal{} \pm 1$.  Otherwise, there are either repeated roots when $ c \equal{} \pm 2$ (which would make the period $ 1$ and not $ 3$) or there are two roots with different absolute values. [/hide]


summer program
Mathcamp
AwesomeMath
PROMYS
HCSSiM

Does anyone know any GREAT math summer camps that they would recommend?  :)

[quote="worthawholebean"]I know of [url=http://cty.jhu.edu/]CTY[/url], [url=http://www.mathcamp.org]MathCamp[/url], and [url=http://www.awesomemath.org]AwesomeMath[/url]. But there may be more I haven't heard of (as well as the MOP, but qualification for that is extremely, extremely difficult).[/quote]
I think mathcamp is the favored of those three.
Try promys as well.

There is quite a comprehensive list in the resources section of this site, and it had links too, i would look there to start.

Ah. thank you. i will check all of those out.

It also depends on what state you live in. If you live in Texas, I would recommend the math camp at Texas State University.


probability AMC AIME geometry function quadratics algebra \/closed

Hi,

I am debating on whether to take the Intermediate Algebra or the Intermediate Counting and Probability online course (I am only going to take one of them). I am not familiar with the concepts that are going to taught in both those classes. So, I am basically aiming to take the course that proved to be comparatively difficult between Intermediate Algebra, and Intermediate Counting and Probability for the majority (it is true that everybody has a different perspective on what is difficult, but I am just looking for a general opinion).

My level right now: I am almost done with Introduction to Counting and Probability (2 more chapters to go.) I am also almost done with Introduction to Algebra (4 more chapters to go.) I have only done like 2 chapters from Introduction to Number Theory (I'm planning to do more of it when I am done with Intro. to Algebra and Intro. to Counting and Probability), and I am taking the online course on Introduction to Geometry.

Thank you very much for the help.

I would recommend you to take Intermediate Algebra.

I myself have gotten Intermediate Algebra a few days back and even Chapter 1 challenge problems proved to be difficult for me (though you're probably way smarter than me). I would assume the course would be around the same level as the book.

On the other hand, I was able to do about half of Chapter 11 problems from Intermediate Counting and Probability (I know this isn't much) without any lessons whatsoever.

Looking at it from a competition perspective, I see that you're still in middle school (I'm not a stalker). For that age, I would recommend Intermediate Algebra as the AMCs, AIME, etc. focus much more on Algebra than on probability. Surely, taking the Algebra course would steeply improve your score on such competitions.

Next, we look at the size of the book (though it is the course you are talking about). As Intermediate Algebra is a much bigger book, definitely you will do more problems and learn more material by taking that course.

Finally, it comes to your preference. No matter what others say, you should always take whichever branch of mathematics that interests you more regardless of the pros and cons.

(I hade the same dilemma when it came to the books)

I think definitely Intermediate Algebra.  You will need the material for Intermediate Algebra a lot more and earlier than Intermediate C & P in my opinion.

You should work through some/most of Intro NT and work through Intro to Geometry with the book as a first priority.  The fundamentals taught in the Intro series are probably more important for you (in terms of succeeding in contests) right now.

As for your question, the Intermediate Algebra class probably suits you better (this is the same advice I gave to pytheagle).

Yes, the Intermediate Algebra book/Algebra 3 class is the right next step for you.

Thank you very much for the replies. Here is the course outline for Algebra 3:

Tentative Class Outline

Week 1: Equations+Functions Review
Week 2: Equations+Functions Review
Week 3: Complex Numbers
Week 4: Quadratics
Week 5: Advanced Quadratics and Generalizations
Week 6: Higher-Degree Polynomials 1
Week 7: Higher-Degree Polynomials 2
Week 8: Higher-Degree Polynomials 3
Week 9: Multi-Variable Polynomials
Week 10: Polynomial Problem Solving
Week 11: Factorizations
Week 12: Sequences & Series
Week 13: Advanced Sequences & Series
Week 14: Identities & Induction
Week 15: Exponents & Logarithms
Week 16: Special Functions
Week 17: More Special Functions
Week 18: Functional Equations
Week 19: Binomial Expansion
Week 20: Inequalities
Week 21: Advanced Inequalities
Week 22: Symmetry in Algebra
Week 23: Substitution & Systems of Equations
Week 24: Advanced Problems

Will this cover a majority of the algebra that AIME has before next year's AIME which I think will be sometime in March 2010? The first day of the course is 7th Oct, 2009, and the last day is April 7th 2010.

Thank you.

It will cover a vast majority of the non-trig algebra you'll see on the AIME.

Got it! Thank you very much for the replies, I really appreciate it! :)

limac, just curious what grade are you going to?


vector
geometry
parallelogram
linear algebra
matrix
complex numbers

Algebraically, write $a=\langle a_{1},a_{2}\rangle$ and similarly for $b$. Each vector equation (in 2 dimensions) can be written as 2 scalar equations. This gives four equations with four unknowns (this is easier than it sounds :) ). $-1$ is probably $-i$.

 Geometrically: if $a$ and $b$ are sides of a parallelogram, then $a+b$ and $a-b$ are the diagonals.

[quote]Each vector equation (in 2 dimensions) can be written as 2 scalar equations. This gives four equations with four unknowns [/quote]
Well, that $4\times 4$ matrix can be effeciently block-structured as a $2\times 2$ with diagonal blocks.

Just write:

$\vec{a}+\vec{b}=\vec{u}$
$\vec{a}-\vec{b}=\vec{v}$

Add these to get $2\vec{a}=\vec{u}+\vec{v}$ or $\vec{a}=\frac12\left(\vec{u}+\vec{v}\right).$

Subtract to get $2\vec{b}=\vec{u}-\vec{v}$ or $\vec{b}=\frac12\left(\vec{u}-\vec{v}\right).$

[quote="rockin_woody"]Ok. Could somone please answer this, or give me a link to a web site that can help me...

if a + b = 4j - 1 and a - b = i + 3j find a and b algebraically.

Then, i need to be able to draw a diagram to show how to find them geometrically. Any ideas?

Thanks[/quote]

u find answer on complex plane? j looks like sqrt(-1), i it's 1 (hard to believe in four variables :lol: )???? If so, really sum equations algebraically: shortly a=[(4j-1)+(1+3j)]/2=7j/2.

To sum the same geometrically: draw (4j-1) and (1+3j) as vectors on complex plane with basis (1,j), build parallelogram on them, take 2a as diagonal. Find center of diagonal known in geometry methods. The same ideas with b (bring - to coefficients). If i is unknown constant, it changes nothing especially, it will be presented at formulaes (if i and j it's unknown constants u can consider basis (i,j) with the same geometrical pictures).

Vector notations could be used, but u'llve no 3D?
To vectorize a and b like complex numbers (and corresponding vectors) u have to decide, what is it i,j?? Can be written also simply v=(a,b) then  $A$=[1 1; 1 -1], so $A$v=d is usual system, d consists right part. In any case it'll be 2D. Why u call it 3D vector algebra question? Where is the reson? Of course, complex plane could be observed and in 3D space but it looks like too complicated approuch to this sample.


WAP comes from Wireless Access Protocol. It's a sort of WWW for mobile phones.

Here are some FAQs

[b]1) How can I access this WAP version of the site? [/b]
You must first posses a mobile device such as a mobile phone (recommended), pda or i-pod. If you have a mobile phone, please contact your mobile operator to obtain details about connection to WAP, settings and pricings.

[b]2) What is GRPS? What is the difference between GPRS and CSD? [/b]
GPRS is a techology that allows instant connections to the WAP network. Most recent phones poses this tech. It can reach high speeds. Usually charged by the amount of traffic (kB) made.

The connection on CSD support is a dial-up connection type made by dialling a number supplied by your mobile network. Generally slow. Usually charged by the amount of time (s) the phone is connected.

The difference between the two are most similar with the difference between cable internet and dial-up.

[b]3) What is the purpose of this WAP access method?[/b]
When you are away from your computer, or unable to reach internet, you can still view posts, reply, quote, edit your posts, send and receive private messages, and much more. Surfing unlimited.

[b]4) What other WAP sites are there in the world?[/b]
The best WAP site I know, besides AoPS - MathLinks ;), is http://tagtag.com

[b]5) I cannot see anything in viewforum list. I receive a network error. [/b]
Please check if you have the appropriate settings correct with your mobile operator. Or you can try accessing the tagtag site to see if your WAP phone works correctly.

[b]6) At some people their mobile device from which they posted appears written in the post. At mine it just says "mobile device". Why?[/b]
Because your phone is not (yet) in our database, so the WAPgate doesn't know "who" your phone is. But it learns quickly, so just have patience! :)

[b]7) Does this WAP mod make coffee? [/b]
I am sorry, but this version of the AoPS - MathLinks WAPgate does not support coffee machines. It will however be a feature that might be implemented in the next versions.

If it something does not work correctly, please report this in the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16274]bug / support thread.[/url]

[to be completed in the future as questions arrise].

[quote="Valentin Vornicu"][b]1) How can I access this WAP version of the site? [/b]
You must first posses a mobile device such as a mobile phone (recommended), pda or i-pod. If you have a mobile phone, please contact your mobile operator to obtain details about connection to WAP, settings and pricings.
[/quote]

iPods don't have internet... yet ;)

[quote="nr1337"] iPods don't have internet... yet ;) [/quote] I was looking into the future there  ;-)

[i]Posted from a Sony Ericsson K700i [/i]


Farz konid $ a,b,c\ge 0$ va $ a^2\plus{}b^2\plus{}c^2\equal{}3$ sabet konid

\[ a^2b\plus{}b^2c\plus{}c^2a\le 3\]

Hale khobi bod sumita :wink:
va yek soale jadid

Farz konid $ a,b,c,d\ge 0$ va $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}4$ sabet konid

\[ a^2b\plus{}b^2c\plus{}c^2d\plus{}d^2a\le 4\]

Khob bod :)

Farz konid $ a,b,c,d,e\ge 0$ va $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}e^2\equal{}5$ maximome ebarate zir ra bedast avarid

\[ a^2b\plus{}b^2c\plus{}c^2d\plus{}d^2e\plus{}e^2a\]

Aghaye M.A age soalaye shortliste $ 2007$ ro dide bashin soale $ 6$ e jabr daghighan yechizi too hamin mayehast .
ba tavvajoh be oon chizi ke oon too neveshte shode mishe sabet kard ke maximum ye adadi bozorgtar az $ 5.04568$ hast va kamtar az $ 5.13065$ hast .
oon meghdare $ 5.04568$ ke goftam be ezaye adade zir be dast miad :
$ a\equal{}1.31324$ , $ b\equal{}1.51404$ , $ c\equal{}0.9445$ , $ d\equal{}0.3005$ , $ e\equal{}0.02973$


Is it possible to configure a webserver like Apache to handle latex for testing locally? If so can someone please give me the directions or a link to the directions?

I am using Apache 2.2 and have configured PHP 5.26 on it
OS = Windows

I'm not sure what the problem would be. Install MiKTeX or TeXLive to a directory without spaces (usually c:\texmf), make sure that the path environment includes the bin directory and that's it.
[quote="mathemagician1729"]Is it possible to configure a webserver like Apache to handle latex for testing locally?[/quote]Don't you mean remotely since otherwise what would Apache have to do with it?
To access LaTeX from your browser you write pages using PHP's exec command to produce something like the [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_TeXer.php]TeXer[/url] for small images or a complete docment compiler like [url=http://nirvana.informatik.uni-halle.de/~thuering/php/latex-online/latex.php?sprachauswahl=2&aufruf=22103]LaTeX-Online-Compiler[/url].


Due to suggestions for a Mathcounts type competition in the Getting Started forum I am thinking about hosting a Mock Mathcounts Math Chat.  Basically what would happen is that I will post the problem in the chat room and then the first 3 or so ppl to correctly answer it get a point.  It will be somewhere along those lines.  Anyone who is interested please post here(or in the getting started forum where they are talking about this).  It should run around an hour.  The time will probably be on Friday or the weekends of one of the next two weeks(not this weekend but the two after it).

Also, if there happens to be too many people I might try to open two chat rooms to keep the lag down... does anyone know if this would work?

I'd like to do it. You could have a chatroom with easy problems like state level, and one with national level problems.

[quote="RC-7th"][quote="h_s_potter2002"]I don't think we should do that untill more people want to participate.[/quote]

I don't think we'll need to worry about lag if this many people participate.[/quote]

I think what h_s_potter2002 means is that if there are only 5 or 6 people participating in the chats in total, it wouldn't be very convenient to have multiple chats up at once.

Great idea Joml! [color=green][b]stickificated[/b][/color]

4 people doesn't seem a like a lot to me.

I would be interested. I understand what the others were saying about how it would be silly to have 2 with so few people, but the idea of two at different levels would be convenient. It would give people a variety and would allow people who are past states to do something valuable with their time.

For those attending, please post if you would like this to be on a certain day or at a certain time.  I would like to have it so that as many people that want to attend can.  How is Friday?  If not Friday it would have to be either Saturday or Sunday due to school.  For the time I can start it anytime from 6 at the earliest to 7:30 at the latest.  It should run for about an hour.  I think that 7 PM EST will probably suit me best.  I can arrange it at another time if there are complaints, however.  As for the setup, I am going to have only one chatroom(unless a massive amount of ppl just decide to do it in the near future  ;) ).  I am thinking that the first 30 minutes will be about Local/State level and the last 30 minutes will be National level.  I will clear the scoreboard after the Local/State level is done.  I encourage ppl participating to do both levels.

I can make it on Friday and Sunday, but I probably won't be able to make it on Saturday.

Friday evening at like 6 or 6:30 would be fine with me.  Saturday and Sunday is ok too, but I prefer Friday.

I am going to make it Friday at 6:15 EST since I have cross country and I am not sure when that gets over.  O yeah, could those who are participating please notify me of you AIM sn.  You can either post it here, IM  me(BaseballJoe2002), pm me, or email me at jlaurendi@gmail.com

Anyone can still come.  If you know how to invite yourself in a chatroom, we will be using Math Chat.  Those who aren't in middle school can still come(But if your at an extremely high level I would ask that you don't steal all the fun from everyone).  Also, I am thinking that the best way to conduct this is for me to post the problem in the chat room and then for you to IM me your answer.  This way we can give points to the first three to correctly answer the question.

Joml88 might not make it the first few minutes, so he gave me the problems, and I can host it. If people want to start coming to the chat room.

If anyone else wants to do this we are starting right now.  IM me if you want in BaseballJoe2002

Can we have the scores from the Mathchat now?  I couldn't be there for the whole thing.

Ok here are the scores that I have.  I am putting the AIM sn's on here because I don't know all of their AoPS names.

AoPS Mathlinks  20

lonelytraveler8  6

hspotter2002  7

spillsdaughter  1

[edit]koopakingz0  4

KillinginSleep 0

The screename u forgot is koopakingz0.  And it's me, dinoman.

this kinda happened a long time ago and i dont think there has been one since
it would be great if we could do another one

Maybe... I could do one... on Friday... but then again... im often busy.

[quote="white_horse_king88"]Hmm... sounds good. I'll comoderate then... if I can make it. I'll notice people in the beginner's forum too.[/quote]

If you're one of the people who care, that's notify, not notice.

[quote="Treething"]I could host one...

Of course, I would have to cut through tons of parental red tape... 10 PM bedtime, 1 hour for piano, 30 minutes for cello, tons of schoolwork to study, independent study, etc.

I think I could host one on Monday at 4:30 PM, right after MathCounts practice...[/quote]

4:30 pm in which time zone?

I meant 7:30 east coast. Sorry.

AIM SN: Gupti Aga

I'm going to un-sticky this for now, since there's a lack of interest for the moment. If a new math-chat is set up or scheduled, I will consider stickying it again.

my team practices every Saturday from 9:30 - 2:30 if you ever wanted to run one of these on a Saturday during those times, I could have 8-12 kids all do it online together as long as the school's server doesn't crash :lol:

Let me know,

Josh Frost

I'm thinking of holding another math chat this weekend.

If interested, please post.

P.S. If you're not sure I'm going to be the good math chat leader, trust me, I started this originally.

if you can give a firm time, then I'll tell you if I(we) can make it,

Josh Frost

i'm interested
if its gonna be on saturday it would be after like 12:30 to work for me

I can't do on saturday. I would be back really late.

I'm thinking of like 2:00 EST (dunno that in other period) and like for 30 mins (for ones that I'm going to be on)

What we might do is since it's only 30 mins, we can like gather some problems and talk about it.

I'll bring some from AHSME.

I can't do it on Friday between 3:45 and 6:30.

Saturday is fine, though. All I need to do is convince my parents to go online for an educational event, which shouldn't be too hard.


Does there exist a sign to present the first coeffients of the polynomail $P(x)$?(My english  is horrible, I hope you can understand me)

Sorry, but I don't understand what you ask for.

[quote="modeler"]Do you mean $\lim_{x \rightarrow \infty}\frac{P(x)}{x^{n}}$ where $n$ is the degree of $P(x)$?[/quote]
Yes,this is what I want,But do there anther sign for it (like $deg P(x)$ means the degree.

My teacher used to write things like $coef_{0}\{P(x)\}$ or $coef_{n}\{P(x)\}$ and everybody just assumed it was standard notation. It seems international too.

me neither, I think it's just called leading coefficient. :P


geometry
3D geometry
trigonometry
trig identities
Law of Cosines

Let $ ABC$ be the axial section of a cone with diametre $ AB\equal{}3 cm$ and generatrix $ AC\equal{}6 cm$. Find the shortest path to reach the midpoint of $ BC$ from point $ A$.

[hide]The generatrix is the slant height, so we have that the diameter is $ 3$ and the slant height is $ 6$.
Unfolding the cone, we have a sector of a circle with a radius of the slant height, which is $ 6$. Then, the arc intercepted is the circumference of the base of the cone, which is $ 3\pi$. Since the circumference of the circle with radius $ 6$ is $ 2\cdot6\pi\equal{}12\pi$, the sector is a quarter of a circle. Drawing a diagram, and using the Law of Cosines, we have:
\[ x^2 \equal{} 6^2\plus{}3^2\minus{}2\cdot 3\cdot 6\cdot \cos \frac{\pi}{4} \\
\equal{} 45\minus{}36\cdot \frac{\sqrt{2}}{2} \\
\equal{}\boxed{\sqrt{45\minus{}18\sqrt{2}}}\][/hide]

[quote="123456789"][hide]The generatrix is the slant height, so we have that the diameter is $ 3$ and the slant height is $ 6$.
Unfolding the cone, we have a sector of a circle with a radius of the slant height, which is $ 6$. Then, the arc intercepted is the circumference of the base of the cone, which is $ 3\pi$. Since the circumference of the circle with radius $ 6$ is $ 2\cdot6\pi \equal{} 12\pi$, the sector is a quarter of a circle. Drawing a diagram, and using the Law of Cosines, we have:
\[ x^2 \equal{} 6^2 \plus{} 3^2 \minus{} 2\cdot 3\cdot 6\cdot \cos \frac {\pi}{4} \\
\equal{} 45 \minus{} 36\cdot \frac {\sqrt {2}}{2} \\
\equal{} \boxed{\sqrt {45 \minus{} 18\sqrt {2}}}
\]
[/hide][/quote]

Almost, except $ \frac {\pi}{4}$ is an eighth of a circle.  Well, since he's not online, I'll correct it right now.

[hide]$ x^2 \equal{} 6^2 \plus{} 3^2 \minus{} 2\cdot 3\cdot 6\cdot \cos \frac {\pi}{2} \equal{} 36 \plus{} 9 \minus{} 0 \equal{} 45$, so $ x \equal{} \sqrt{45}$.[/hide]

[quote="syckls"][quote="123456789"][hide]The generatrix is the slant height, so we have that the diameter is $ 3$ and the slant height is $ 6$.
Unfolding the cone, we have a sector of a circle with a radius of the slant height, which is $ 6$. Then, the arc intercepted is the circumference of the base of the cone, which is $ 3\pi$. Since the circumference of the circle with radius $ 6$ is $ 2\cdot6\pi \equal{} 12\pi$, the sector is a quarter of a circle. Drawing a diagram, and using the Law of Cosines, we have:
\[ x^2 \equal{} 6^2 \plus{} 3^2 \minus{} 2\cdot 3\cdot 6\cdot \cos \frac {\pi}{4} \\
\equal{} 45 \minus{} 36\cdot \frac {\sqrt {2}}{2} \\
\equal{} \boxed{\sqrt {45 \minus{} 18\sqrt {2}}}
\]
[/hide][/quote]

Almost, except $ \frac {\pi}{4}$ is an eighth of a circle.  Well, since he's not online, I'll correct it right now.

[hide]$ x^2 \equal{} 6^2 \plus{} 3^2 \minus{} 2\cdot 3\cdot 6\cdot \cos \frac {\pi}{2} \equal{} 36 \plus{} 9 \minus{} 0 \equal{} 45$, so $ x \equal{} \sqrt {45}$.[/hide][/quote]But doesn't $ BC$ bisect the sector?

[quote="syckls"][quote="123456789"]But doesn't $ BC$ bisect the sector?[/quote]

Doh, you're right.  I had it right the first time I worked out this problem, but then I forgot that part.[/quote]I almost made that mistake at first too.

But if the circumference is $ 6\pi$ and $ BC$ bisects it , doesn't that mean that there is indeed a quarter of circle and the answer is $ \sqrt{45}$? :maybe:

That was the way i solved the problem too but my book says the answer is $ \sqrt{23}$, that's way i posted it...

[quote="sunnykim"]That was the way i solved the problem too but my book says the answer is $ \sqrt {23}$, that's way i posted it...[/quote]I'm pretty sure that's wrong, unless we're all misinterpreting the question...

General principle of forum posting: there is [i]never[/i] a good reason to block-quote the entire post preceding yours.

I could read the problem in two ways: the first is that the path is supposed to lie on the surface of the cone, in which case 123456789 got it.  The second is that the path is supposed to lie in space.  Neither gives you $ \sqrt{23}$ -- textbooks have been known to contain the occasional error  :wink:

[quote="mon"]But if the circumference is $ 6\pi$ and $ BC$ bisects it , doesn't that mean that there is indeed a quarter of circle and the answer is $ \sqrt {45}$? :maybe:[/quote]

Ok..but finally the answer is $ \sqrt {45 \minus{} 18\sqrt {2}}$ or $ \sqrt{45}$? :?:


inequalities
geometry
trapezoid
geometry proposed

We have the trapezoid ABCD with m(<A)=m(<D)=90.Prove that AC*BD>=2AB*CD.

What do you mean with m(<A)=m(<D)=90? Maybe that <DAB=<ADC? Because if I'm right I think that the inequality is wrong :?

just take AB=2,CD=3,AD=1.then AC=sqrt(10) and BD=sqrt(5) and you will see your inequality is not true.(or i am in a hurry and i did not do the computations right :P )

I don't know what is wrong, but this is the conclusion.


I'm new here . How can i write equations , fractions ,... , as all users do here ?

Start with the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]FAQ[/url], particularly the first two questions.

Thanks , this evening i'll learn all those  :D


logarithms calculus derivative trigonometry function limit inequalities

[hide]$2^{\sqrt{3}} \text{ or } 3^{\sqrt{2}}= 2^{\log_2{3}\cdot\sqrt{2}}$,

Now we have to determine which is bigger $\sqrt{3}\text{ or }\log_2{3}\cdot\sqrt{2} \implies 3\text{ or }2(\log_2{3})^2 \implies \log_2{8} < \log_2{3}\cdot\log_2{9}$

Thus the answer is $2^{\sqrt{3}} < 3^{\sqrt{2}}$.[/hide]

[quote="José"][quote="Centy"]Determine without a calculator which is bigger: $2^{\sqrt{3}} \text{ or } 3^{\sqrt{2}}$.[/quote]

[hide="is this correct?"]$x> y \Longrightarrow 3^x> 3^y$
$3^x=4 \Longrightarrow x=\frac{log 4}{log 3}$
$\sqrt{2}> \frac{log4}{log 3} \Longrightarrow 3^{\sqrt{2}}> 4$

$x>y \Longrightarrow 2^x>2^y$
$2^x=4 \Longrightarrow x=2$
$2>\sqrt{3} \Longrightarrow 4>2^{\sqrt{3}}$

Then, $3^{\sqrt{2}}>2^{\sqrt{3}}$ [/hide][/quote]José, why is this true: $\sqrt{2}> \frac{log4}{log 3}$? Remember no calculators...

[quote="José"]$\sqrt{2}>\frac{7}{5}$
$3^{\frac{7}{5}}>4$[/quote]
OK, and how do you know $\frac75\approx\frac{\log 4}{\log 3}$? Show me something that you have not used a calculator please.

[hide="A slightly different solution"]
If $3^{\sqrt{2}}>2^{\sqrt{3}}$, then $3^{\sqrt{2}\cdot \sqrt{3}} >2^{\sqrt{3}\cdot \sqrt{3}}=8$

Since $3^{\sqrt{6}}>3^{\sqrt{4}}=9>8$, the assumption was true, so $3^{\sqrt{2}}>2^{\sqrt{3}}$
[/hide]

Jose, your solution seems to use calculator work over the exact values of $\log 4$ and $\log 3$.

The solutions which simply use $x < y \Rightarrow x^k < y^k$ if $x, y, k > 1$ are far nicer.

[quote="Kalle"]Which is bigger,

$\frac{1}{e}$ or $\frac{\ln 3}{3}$ ?[/quote]I did it 'backwards' and keeping the binary relation in question:

[hide="a not-so-neat solution"]$\frac1e\stackrel{?}{<<}\frac{\ln3}3$

$e\stackrel{?}{>>}\log_3e^3$

$3^e\stackrel{?}{>>}e^3$

$3^\frac13\stackrel{?}{>>}e^\frac1e$

Using the fact that the maximum of $f(x)=x^\frac1x$ occurs when $x=e$, then $\stackrel{?}{>>}\ =\ <$ and so $\frac1e$ is bigger.[/hide]

PS: this doesn't seem like a Getting Started level type of problem

[quote="LordoftheMorons"]Squaring each side, we see that $8<9$. Thus, $2^{\sqrt{3}}<3^{\sqrt{2}}$.[/quote]

You cant manipulate them like that.

$(2^{\sqrt{3}})^2 = 2^{2\sqrt{3}}$

[quote="stancioiu sorin"]About this problem, which is bigger
$\ 1000*1000...*1000(1000 \text{ terms}$ or $\ 1*3*5*...*1999$?[/quote]

[hide]Notice, that $S=1000+1000+...+1000 (1000 \text{ terms})=1+3+5+...+1999$. By AM-GM we know that the product of $n$ values assumes it's biggest value when $a_1=a_2=...=a_n$. Thus, $1000*1000...*1000 (1000 \text{ terms}) > 1*3*5*...*1999$.[/hide]

Yes, the culprit is me. I did it again, posting hard problems in Getting Started. So yeah, sorry guys that this got moved. I removed all off-difficulty problems (all the previous bonus problems are removed so please don't ask me for the solutions), and I will not post anymore off-difficulty problems. Better yet, this is not a marathon for me, so I'm out. See ya.

Well, let's continue. Anybodycan post the next one.......

[hide]$\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$
$x=\sqrt{1+x}$
$x= \phi=\frac{1 + \sqrt 5} 2$

$\frac{1}{1+\frac1{1+\frac1{1+\frac1{\vdots}}}}$
$y=\frac{1}{1+y}$
$0=y^2+y-1$
$y=\frac{-1+\sqrt 5}{2}$

By adding them we get $\sqrt 5$

$e<\pi$, $\frac e {\pi}<1,\ln{\frac{e}{\pi}}<1$
$\sqrt 5>2$, so
$\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}-\frac{1}{1+\frac1{1+\frac1{1+\frac1{\vdots}}}}> \ln {\frac{e}{\pi}}$[/hide]

Anyone else can post the next problem....

[quote="Hamster1800"]I'll post a relatively simple one to determine who posts next.

$\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}-\frac{1}{1+\frac1{1+\frac1{1+\frac1{\vdots}}}}$ or $\ln{\frac{e}{\pi}}$

It's actually very simple ;)[/quote]

You could just say $\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$ is greater than 1 while $\frac{1}{1+\frac1{1+\frac1{1+\frac1{\vdots}}}}$ is smaller than 1.

So the LHS is positive.

$\ln{\frac{e}{\pi}}$ is obviously negative.

Thus the LHS is larger.

is this right?

[hide]

$\frac{600!300!100!}{400!400!200!} = \frac{401 \cdot 402 \cdot \dots \cdot 599 \cdot 600}{301 \cdot 302 \dots 399 \cdot 400 \cdot 101 \cdot 102 \dots 199 \cdot 200}$

and both the numerator and the denominator are the poduct of 200 integers. Furthermore, all those in the numerator are greater than those in the denominator, so the numerator  ( $600!300!100!$ ) is greater. [/hide]

I'll just post another one.

$\left(e^{100}\right)\left(100!\right)$ or $100^{100}$

Although I don't quite follow you, you got the right answer, so I'll post another one.

Let $a$ be the sum of the absolute values of the solutions of

$3x^4 + 24x^3 + 72x^2+96x+\frac{671}{27} = 0$

Let $b$ be the sum of the absolute values of the solutions of

$\left(25^{25}\right)\left( x^{23}\right) - \frac{10^{3^3}}{2^{2^2}} = 0$

Which is greater, $a$ or $b$?

Only real solutions or are complex solutions included also?

Complex solutions included in both cases.


A regular hexagon is inscribed in a circle of radius $ 10$ inches.  Its area is:

$ \textbf{(A)}\ 150\sqrt{3} \text{ sq. in.} \qquad
\textbf{(B)}\ \text{150 sq. in.} \qquad
\textbf{(C)}\ 25\sqrt{3} \text{ sq. in.} \qquad
\textbf{(D)}\ \text{600 sq. in.} \qquad
\textbf{(E)}\  300\sqrt{3} \text{ sq. in.}$

[hide="Click for solution"]
$ A\equal{}\frac{1}{2}ap\equal{}\frac{1}{2} \cdot 5\sqrt{3} \cdot 60\equal{}150\sqrt{3}$, or $ \boxed{\textbf{(A)}}$.
[/hide]


function
quadratics
calculus
derivative
linear algebra
matrix
algebra

If $ f \equal{} f(x_1,x_2,...,x_n)$, $ \frac {{\partial}^2 f}{\partial x_i^2} \geq 0$ for $ i \equal{} 1,2,...,n$ and $ \sum_{i \equal{} 1}^{n}{x_i} \equal{} 1$, than prove or disprove that $ f_{max} \equal{} f(1,0,0,...,0)$ or any permutation of set $ (1,0,...,0)$ -f reaches maximum for an permutation of a set (1,0,...,0).

* $ f_{max}$ means maximum value of a function f.

Obviously false. Take $ n\equal{}2$ and $ f(x)\equal{}x_1x_2$. You need the second differential to be a non-negative quadratic form for this, not just pure partial derivatives being non-negative.

[quote="fedja"]Obviously false. Take $ n \equal{} 2$ and $ f(x) \equal{} x_1x_2$. You [b]need the second differential to be a non-negative quadratic form[/b] for this, not just pure partial derivatives being non-negative.[/quote]

That means if this bold is true, than the question I've asked is true?

Thanks!

I don't think so.

Unless I'm seriously mistaken (and note that it's highly probable, as multivariate calculus still doesn't sit quite right with me :)), take $ n \equal{} 2$ and $ f(x,y) \equal{} x^2 \plus{} y^2$. Obviously, this has no maximum, even taking $ x \plus{} y \equal{} 1$ into account.  However, the Hessian of $ f$ (if that's what fedja was referring to) is $ H_f \equal{} \left( \begin{array}{cc} 2 & 0 \\
0 & 2 \end{array} \right)$, which is obviously a positive-definite matrix (meaning the second differential is a positive quadratic form) and it also satisfies your original condition.

(By the way, I fail to see the reasoning behind the idea that $ (1,0,0,\ldots,0)$ is a maximum - besides being a member of canonical basis for $ \mathbb{R}^n$, I don't see what's so special about it.)

Having a positive-definite Hessian does indeed, if I'm not mistaken, guarantee that there are no local maximums in the interior of the domain, which is kind of close to what you seem to be looking for. However, I guess the domain here would be $ \mathbb{R}^n$, as the set defined with $ \sum_{i \equal{} 1}^n x_i \equal{} 1$ isn't an open set (nor it even has an interior, if I'm interpreting this correctly) in $ \mathbb{R}^n$, so differentiating the function on that set might be problematic.

So, if $ \mathbb{R}^n$ is indeed the domain of $ f$, it contains no maximums whatsoever. Please note that doesn't mean we can say anything about the existence or locations of possible maximums of $ f$ on its restriction defined by $ \sum_{i \equal{} 1}^n x_i \equal{} 1$ - again, since this is not an open set, differentiation might not be a way to go here.

If you were to modify the condition into $ \sum_{i \equal{} 1}^n |x_i|\leq 1$, the argument which I described would, I believe, guarantee that the maximum is attained somewhere on the edge (where $ \sum_{i \equal{} 1}^n |x_i| \equal{} 1$), but that still probably isn't what you were looking for.

Urgh, I wrote way too much than I wanted to :D (And much of it is probably wrong :blush:... my luck these days :))

You are formally right, of course, but I'm ready to bet \$1000 against 1c that OP just forgot to write the condition $ x_j\ge 0$ in his definition of the domain over which the maximum should be taken ;).

:lol: Yeah, you're probably right.

In that case, then, the short answer is yes - I think it follows easily from the convexity of the function, which is guaranteed if $ H_f\geq 0$ (in this context, that means $ H_f$ is positive-semidefinite).

That's true that I forgot to write that $ x_1,...,x_n$ are positive numbers.
Thank you!  :lol:


[quote="drunner2007"]Given three standard six-sided dice, what is the probability that the sum of the numbers on each dice is an odd number?[/quote]
[hide]There is odd-odd-odd and odd-even-even.
1) a) If no odds are the same, there are 6 ways to arrange, $6(1/2)(1/3)(1/6)=1/6$
b) If two odds are the same, there are 3 ways to arrange, $3(1/2)(1/2)(1/3)=1/4$
c) If all odds are the same, there is one way to arrange, $(1/2)(1/6)(1/6)=1/72$
2) a) If the evens are different, there are 6 ways to arrange, $6(1/2)(1/2)(1/3)=1/2$
b) If the evens are the same, there are 3 ways to arrange, $3(1/2)(1/2)(1/6)=1/8$
But then the probability is more than 1...can someone tell me what I did wrong?[/hide]

[hide] is it 1/2?

27 ways to get o-o-o
81 ways to get o-e-e or permutation.[/hide]
I'm probably missing something though.[/hide]

[quote="drunner2007"]Given three standard six-sided dice, what is the probability that the sum of the numbers on each dice is an odd number?[/quote]

What do you mean by sum of the numbers on each dice?  Is it like, the sum of the top faces or the sum of all of the faces?  It is not clear to me.

i mean sum on all of the [i]top[/i] faces. i'm not 100 percent sure of the answer, but i got jclarke's answer using the exact same reasoning after looking at this problem for a while.

i'm not sure about ch1n353ch3s54a1l 's answer

if i can make sure that my answer correct, I'll probably use this on the tournament that I'm writing.

[hide="Possibility"]Well since the low sum is 3 and the high is 18, there is an even number of sums. The median sum is 10.5 and there is a correspondence between the terms in the lower half and upper half, so the 10 and 11 have the same probability, 9 and 12, etc. down to 3 and 18 so the probability is ...1/2...?
[/hide]

Think of it as three coins. One side of a coin is "odd," and the other is "even." When you toss these coins you must end up with an odd number of odds. There is an equal probability, however, of getting an odd number of evens and thus an even number of odds. So the probablity of getting an odd number of odds is 1/2.

[hide="question"]I also thing it's a half, but does the calculate is right: the first  throwing doesn't matter, and then $2\cdot(\frac{1}{2})^2=\frac{1}{2}$?[/hide]

yeah, that seem's right.

wow, that's the best solution yet.

[quote="drunner2007"]wow, that's the best solution yet.[/quote]

How about this one:

The probability distribution doesn't change with each added die.  :)

[hide]there are 6^3=216 ways to get a number with combinations, there are 27 ways to get odd-odd-odd and 81 ways to get odd-even-even, this is just permutations, add them and divide by 216, you get 108/216=[b]1/2[/b][/hide]


geometry
cyclic quadrilateral
perpendicular bisector
geometric transformation
geometry solved
USA TST
steiner line

Let $ ABCD$ be a cyclic quadrilateral and let $ E$ and $ F$ be the feet of perpendiculars from the intersection of diagonals $ AC$ and $ BD$ to $ AB$ and $ CD$, respectively. Prove that $ EF$ is perpendicular to the line through the midpoints of $ AD$ and $ BC$.

Posted before, even quite recently (few months ago only!), if I am not wrong.

Best regards,
sunken rock

It has now been corrected  :lol:

For reference, also posted here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=296040
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=12098

Another way is possible also.

You can use just a bit of complex numbers.  Let $P$ be the origin, and $E$ be at $e$, $F$ at $f$.  We have that for some $k$, $A=e+ike$ and $D=f-ikf$ because of the perpendicularity and $PFD$ is similar to $PEA$.

Let the midpoint of $AD$ be $M$, which is half of $e+f+ike-ikf$.  Now, $ME$ is half $-e+f+ike-ikf$ and $MF$ is half $e-f+ike-ikf$.  These are $(e-f)(-1+ik)$ and $(e-f)(1+ik)$ respectively, and they have the same magnitude because $-1+ik$ and $1+ik$ have the same magnitude.

Therefore, $ME=MF$, and the midpoint of $AD$ is on the perpendicular bisector of $EF$.  Similar holds for the midpoint of $BC$ so we are done.

[quote="jgnr"]This problem is missing from the [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=35&year=2000&]resource page[/url].

[hide="solution"]Let $AC$ and $BD$ meet at $P$, the midpoints of $AD,BC,DP,AP$ are $M,N,R,S$, respectively. We have $FR=\frac12DP=MS$, $ES=\frac12AP=MR$, \[\angle MRF=\angle MRD+\angle DRF=\angle ASM+2\angle DPF=\angle ASM+2(90^{\circ}-\angle PDF)=\angle ASM+2(90^{\circ}-\angle PAE)=\angle ASM+2\angle APE=\angle ASM+\angle ASE=\angle MSE.\] Hence $\triangle MRF\cong\triangle MSE$, which implies $ME=MF$. Similarly, we get $NE=NF$. Therefore $MENF$ is a kite, whence $EF\perp MN$.[/hide][/quote]
Dayummm this is a nice solution. Although, what is the motivation for constructing the midpoints of S and R besides the right angles ad PF and PE?

Does anyone have a proof using spiral similarity?

^^^Here is a solution that is absolutely effortless. So, note that the line adjoining the midpoints is antiparallel to the line $XY$ where $X, Y = AC \cap BD, AB \cap CD$. But clearly, $XY$ is anti-parallel to the line perpendicular to $EF$, so the lines adjoining the midpoints must be perpendicular to $EF$ so done.

@mathisfun7 yes I found a spiral similarity solution last night! Let M be the midpoint of AD, N of BC, R of DP, and S of CP. I will prove that F is the center of the spiral smiilarity bringing quad. FRPS to FMEN. Since clearly RS||CD perpendicular to FP then this will solve the proof.
Denote by X <BAC=<BDC
(According to my diagram, yours might be flipped) <FRM=<FRD+<DRM=2(90-X)+<DPA=2(90-X)+<DPA, but <FPE=<FPE+<EPA+<EPA=2(90-X)+<DPA and you can say the same thing for <FSN.

Now, we have FR/RM=(DP/2)/(AP/2)=DP/AP, FP/PE=(DPsinx)/(APsinx)=DP/AP and the same for FS/SN. So then FRM is similar to FPE is similar to FSN so F is the center of that spiral similarity, thus meaning that MN is perpendicular to FE.

Dear Mathlinkers,
just a remark... EF is parallel to the Steiner's line of ABCD...
Sincerely
Jean-Louis

We proceed with complex numbers. Denote $P \equiv AC \cap BD$, and let $M, N$ be the midpoints of $\overline{AD}, \overline{BC}$, respectively. Then $2m = a + d, \; 2n = b + c$, and from the projection formula, we find $2e = a + b + p - ab\overline{p}, \; 2f = c + d + p - cd\overline{p}.$ To prove that $EF \perp MN$, we need only show that $\tfrac{e - f}{m - n} \in i\mathbb{R}.$ To see this, let \[z = \frac{e - f}{m - n} = \frac{a + b - c - d + (cd - ab)\overline{p}}{a + d - b - c}.\] From the intersection of chords formula, we have \[p = \frac{ac(b + d) - bd(a + c)}{ac - bd} \implies \overline{p} =  \frac{\frac{1}{ac}\left(\frac{1}{b} + \frac{1}{d}\right) - \frac{1}{bd}\left(\frac{1}{a} + \frac{1}{c}\right)}{\frac{1}{ac} - \frac{1}{bd}} = \frac{b + d - a - c}{bd - ac}.\] Hence, \[z = \frac{a + b - c - d + (cd - ab)\left(\frac{b + d - a - c}{bd - ac}\right)}{a + d - b - c} = \frac{-a^2c + b^2d + ac^2 - bd^2 - ab^2 + cd^2 + a^2b - c^2d}{(bd - ac)(a + d - b - c)},\] where we have multiplied through by $bd - ac$, and expanded the numerator. This factors as \[z = \frac{(a + d - b - c)(ab + cd - ac - bd)}{(bd - ac)(a + d - b - c)} = \frac{ab + cd - ac - bd}{bd - ac}.\] It remains to show that $\overline{z} = -z$, which is just a straightforward computation..

Denote $P \equiv AC \cap BD$ and $X \equiv AB \cap CD.$ Let $Q$ be the reflection of $P$ in the midpoint of $\overline{BC}.$

We claim that $\triangle XBQ \sim \triangle XDP.$ Indeed, note that $BPCQ$ is a parallelogram because its diagonals bisect one another. Hence, \[\measuredangle XBQ = \measuredangle XAC = \measuredangle BAC = \measuredangle BDC = \measuredangle PDX,\] where the angles are directed. Now, let $A'$ be the reflection of $A$ in $E$, and note that $X, D, P, A'$ are concyclic because \[\measuredangle XDP = \measuredangle CDB = \measuredangle CAB = \measuredangle PAA' = \measuredangle XA'P.\] It follows that $\triangle BA'P \sim \triangle BDX$, and therefore $XD : XB = PA' : PB = PA : PB.$ Because $\triangle PAB \sim \triangle PDC$, we obtain $XD : XB = PD : PC = PD : BQ.$ Thus by side-angle-side similarity, we have $\triangle XBQ \sim \triangle XDP$ as desired. It follows that $XP$ and $XQ$ are isogonal w.r.t. $\angle BXC.$ Therefore, it is well-known that $XQ \perp EF.$ Then note that a homothety with center $P$ and ratio $\tfrac{1}{2}$ sends $XQ$ to the [url=http://forumgeom.fau.edu/FG2012volume12/FG201212.pdf]Newton-Gauss line[/url] of quadrilateral $XAPD.$ The desired result follows. $\square$

Let the midpoint of $BC \equiv M$ and the midpoint of $AD \equiv N$. Let $BD \cap AC \equiv P$. Denote the reflections of $P$ about $E$ and $F$ be $E'$ and $F'$ respectively. It is clear that $\triangle PCD \sim PBA$, so by definition of reflection $F'DPC$ and $PAE'B$ are kites similar to one another. Then since $M$ and $N$ are midpoints of $BC$ and $AD$, and since $E$ and $F$ are midpoints of $PE'$ and $F'P$, we have that $NFME$ is a kite similar to $F'DPC$ and $PAE'B$ by MGT, so $EF \perp MN$ as desired.

[hide=Different solution?]
Let $G$ be the intersection of $AC$ and $BD$, and let $H$ be the intersection of $AB$ and $CD$. Suppose that $M,N,P$ are the midpoints of $AD,BC,GH$ respectively. Then $M,N,P$ collinear by the Newton-Gauss Line, so $PE=PF$.

Now consider the Miquel Point of $ABCD$. Clearly it is the center of the spiral similarity sending $AD$ to $BC$, so thus by the Gliding Principle it also sends $AD$ to $EF$. Thus by the Gliding Principle again it follows that the midpoint of $EF$ is collinear with the midpoints of $AD$ and $BC$; letting $Q$ be the midpoint of $EF$, we know that $QE = QF$. Thus $PQ$ is the perpendicular bisector of $EF$.

However, we already know that $M$ and $N$ lie on $PQ$ so we are done.

Let $G=AC\cap BD, H\in AB: \overline{HE}=\overline{EB}, I\in CD: \overline{IF}=\overline{FC}, M$ is the midpoint of $BC$
We have $\Delta GHC=\Delta GBI\Rightarrow ME=MF\Rightarrow Q.E.D$

The reflections of $A, D$ about $F, E$ are $S, T$ respectively. We have $\Delta APT=\Delta SPD\rightarrow SD=AT$. Let $M, N$ are midpoint of $AD, BC$  respecttively so $MF=DS/2=AT/2=ME$
Analogously,$NE=NF\rightarrow MN$ is perpencular bisector of EF

Let the diagonals meet at $P$ and $H_1,H_2$ be the orthocenters of $PDC,PAB$\\
it is not difficult to show $H_1H_2\parallel EF$ (just prove $\frac{PH_1}{PF}=\frac{PH_2}{PE}$)\\
so $EF$ is parallel to the Steiner line which is perpendicular to the Gauss line which passes throw the midpoints of $AD,BC$

Let $AB\cap CD = P$, $AC\cap BD = K$ and $X, Y$ be the midpoints of $AD, BC$. Since $XY$ is Newton-Gauss line of $ABDC$, it pass through midpoint of $PK$, which is the center of $\odot(PKEF)$. Moreover, since $\triangle AKB\cup E\sim\triangle DKC\cup F$, we get $AE : EB = DF : FC$. Thus by E.R.I.Q. lemma, $XY$ also pass through midpoint of $EF$. Hence $XY$ is perpendicular bisector of $EF$ so we are done.

Here's a nice Elementary Proof:-

Let $AC\cap BD=K$ and let $X,Y$ be the midpoints of $KC,KB$ and $P,Q$ are the midpoints of $KA,KD$.

Hence, $$\begin{cases} [XZ=KY=EY] , [FX=XK=ZY], [\angle FXZ=\angle FXK+\angle KXZ=2\angle DCA+\angle KYZ=\angle EYK=\angle KYZ=\angle EYZ]\implies \triangle FXZ\cong\triangle ZYE\implies \boxed{FZ=ZE} \\
 \triangle EPM\cong\triangle FQM\implies \boxed{ME=MF} \end{cases}$$

So, $MENF$ is a kite $\implies EF\perp MN$.

[asy]
unitsize(1.2inches);
pair A=dir(145);
pair D=dir(200);
pair C=dir(-20);
pair B=dir(80);
pair Q=extension(A,B,C,D);
pair P=extension(A,C,B,D);
pair R=extension(A,D,B,C);
pair E=foot(P,A,B);
pair F=foot(P,C,D);
pair M=foot(0,A,D);
pair N=foot(0,B,C);

draw(circumcircle(A,B,C));
draw(Q--B--C--cycle);
draw(D--R--B);
draw(A--C,linewidth(0.3));
draw(B--D,linewidth(0.3));
draw(P--E,linewidth(0.4));
draw(P--F,linewidth(0.4));
draw(0*dir(0)--M,linewidth(0.4));
draw(0*dir(0)--N,linewidth(0.4));

dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$D$",D,dir(240));
dot("$P$",P,dir(15));
dot("$Q$",Q,dir(Q));
dot("$R$",R,dir(R));
dot("$E$",E,dir(75));
dot("$F$",F,dir(-90));
dot("$M$",M,dir(45));
dot("$N$",N,dir(N));
dot("$O$",0*dir(0),dir(-90));
[/asy]

Let $O$ be the circumcenter of $ABCD$, let $R=AD\cap BC$, and let $Q=AB\cap CD$. Brokard's theorem tells us that $OR\perp PQ$. The rest is essentially angle chasing, but we'll phrase the proof more structurally.

The first claim is that the angle bisectors of $\angle EPF$ and $\angle MON$ are perpendicular. The angle bisector of $\angle EPF$ is parallel to that of $\angle BQC$, and the angle bisector of $\angle MON$ is parallel to that of $\angle DBC$. Let $I$ be the incenter of $\triangle BQC$, and let $J$ be the incenter of  $\triangle DRC$. Note that $\angle QIC=\pi/2+B/2$  and $\angle RJC=\pi/2+D/2$. Thus, $\angle QIC+\angle RJC=3\pi/2$, and since $C,I,J$ collinear, this shows the claim.

Now, we know that the line through $P$ perpendicular to $EF$ is the reflection of $PQ$ about the angle bisector of $\angle EPF$, and the line through $O$ perpendicular to $MN$ is the reflection of $OR$ in the angle bisector of $\angle MON$. By the above claim, the reflections are the same (up to translation), and since we have $PQ\perp OR$, we're done.

Let $P=AC \cap BD$, $R=AD \cap BC$, $Q=AB \cap CD$.

The crucial observation is the following:

If points $X$ and $Y$ satisfy $\angle AQX = \angle CQY$ and $\angle ARX = \angle CRY$, then $\angle QXR + \angle QYR = 180^{\circ}$.

The proof is a simple angle chase. Now, note that if the line through $Q$ perpendicular to $EF$ and the line through $R$ perpendicular to the midpoints meet at $Y$, then we can take $X$ to be the intersection of $QO$ and $PR$. We get $\angle QXR=90^{\circ}$ by brokard, and the crucial observation finishes the problem

$\vec{MN} = \frac{\vec{AB}+\vec{DC}}{2}\\
\triangle ABX \sim \triangle CDX \implies \frac{XE}{AB}=\frac{XF}{CD} = k$\\
$\vec{EF} = \vec{EX} + \vec{XF} = ik\vec{AB} + ik\vec{DC} = ik(\vec{AB} + \vec{DC}) = 2ik\vec{MN} \implies MN\perp EF$

Here's a different approch:

Let $X=AB \cap CD$
$G=BD \cap AC$
 $M,N$ be the midpoints of $AD,BC$
$M'=MG \cap BC , N'=NG \cap AD$
and $O$ be the circumcetner of $(ABCD)$
[b][u][color=#f00]claim(1):[/color][/u][/b] $M'N' || GX$
[b][u][color=#00f]proof:[/color][/u][/b]
take a projective transformation that preserves $(ABCD)$ and move $G \Rightarrow O$ then $ABCD$ will be a square and its easy to see that's true
$\blacksquare$
Let $\ell$ be the internal angle bisector of $\angle AXD$ and define $f(z)$ the image of $z$ over the reflection about $\ell$
[b][u][color=#f00]claim(2):[/color][/u][/b] $f(M'N') || MN$
[b][u][color=#00f]proof:[/color][/u]
[/b]clearly $f(M') \in XN$ and $f(N') \in XM$ so it suffices to show $\frac{XM'}{XN} =\frac{XN'}{XM}$ which is folllow from the law of sines (I'm too lazy to write it down  :whistling:  )
$\blacksquare$
so $MN || f(m'n') || f(XG)$
but $f(XG) \perp EF$ (since $f(XG)$ is the isogonal of $XG$ wrt $\angle AXD$
$\implies MN \perp EF$
and we win :D

Let $\overline{MN}$ and $\overline{EF}$ concur at a point $Q$. Let $L$ be the miquel point of cyclic quadrilateral $(ABCD)$, and suppose that the lines $\overline{AB}, \overline{CD}$ and $\overline{AD}, \overline{BC}$ intersect at points $X$ and $Y$, respectively.

Note that $EFXL$ is cyclic with diameter $\overline{EX}$, so $L$ is the miquel point of $AEFD$. Similarly it is also the miquel point of $EBCF$. Note that $Q$ is the midpoint of $\overline{EF}$, so putting it all together we find there is a spiral similarity around $L$ sending $\overline{AMD} \to \overline{EQF} \to \overline{BNC}$. Now we angle chase to finish:

$$\angle EQL + \angle LQN = \angle BNL + \angle LEB = \angle BXL + \angle LEB = \angle EXL + \angle LEB = 90^\circ.$$

Oops...

Let $M,N$ be the midpoints of $\overline{AD},\overline{BC}$ respectively, and let $X,S,T$ be $\overline{AC}\cap\overline{BD},\overline{AB}\cap\overline{CD},\overline{AD}\cap\overline{BC}$ respectively. Additionally, let $G$ be the Miquel point of quadrilateral $ABCD.$ \\\\By Brokard, $\angle XGS=90^\circ,$ so $X,E,F,G,S$ are concyclic. Moreover, $\angle ONT=\angle OMT=90^\circ,$ so $O,M,N,T$ are concyclic.\\\\ Now observe that
\begin{align*}
    \angle FEB+\angle BNM+\angle ABC
    &= \angle FES+\angle TNM+\angle ABC\\
    &= \angle FXS+\angle TOM+\angle ABC\\
    &= (90^\circ-\angle XSF)+(90^\circ-\angle MTO)+\angle ABC\\
    &= (90^\circ-\angle BSC+\angle BSX)+(90^\circ-\angle ATB+\angle OTB)+\angle ABC\\
    &= 180^\circ-(\angle BSC+\angle ATB)+(\angle BSX+\angle OTB)+\angle ABC\\
    &= 180^\circ-(\angle TBS-\angle CDA)+(\angle TBS-90^\circ)+\angle ABC\\
    &= 90^\circ+\angle CDA+\angle ABC\\
    &= 270^\circ.
\end{align*} (Here we use the fact that $\overline{TO}\perp\overline{SX},$ which follows from Brokard's Theorem)\\\\Therefore, $\overline{MN}\perp\overline{EF},$ as desired.

This problem can be solved fairly simply using the [url=https://artofproblemsolving.com/community/q1h1882783p12816991]Gliding Principle[/url].
Let $X$ be the reflection of $P$ over $AB$.
Then $\triangle AXB \cong \triangle ABP \sim \triangle DPC$. By the Gliding Principle, since $\triangle AXB \sim \triangle DPC$, and the midpoints of $AD$, $BC$, and $XP$ are $M$, $N$, and $E$, $\triangle MNE \sim \triangle BPC$.
By the same argument, $\triangle MNF \sim \triangle APB \sim \triangle BPC$.
Thus, $\triangle MNE \sim \triangle MNF$, but they both share side $MN$, so we must have $\triangle MNE \cong \triangle MNF$.
Then $MENF$ is a kite, so $MN \perp EF$.

Swapping $C$ and $D$:
[quote]In cyclic quadrilateral $ABCD$ with $AD \cap BC=P$, show the line through the feet from $P$ to $AB,CD$ is perpendicular to the Gauss line of $ABCD$.[/quote]
If $M$ is the miquel point of $ABCD$ then it's well known (EGMO 10.9) that the feet from $M$ to $AB,BC,CD,DA$ are collinear along a line perpendicular to the Gauss line. The problem then follows by homothety at $AB \cap CD$ sending $M$ to $P$.

It's really easy using this lemma .
Lemma:$AB$ is perpendicular to $CD$ if and only if $AC^2$-$AD^2$=$BC^2$-$BD^2$

Let the midpoints of $AD, BC$ be $M, N$ respectively, and $P = AC \cap BD$. Now, define the orthocenters of $PAB$ and $PCD$ as $H_1$ and $H_2$ respectively.

Clearly, we have $PAB \sim PDC$. Thus, since $E, F$ and $H_1, H_2$ are both pairs of [i]corresponding points[/i] in these two similar triangles, we know $$\frac{PH_1}{PE} = \frac{PH_2}{PF}$$ so $H_1H_2 \parallel EF$. But $H_1H_2$ is the Steiner Line of $ABDC$, implying $$MN \perp H_1H_2 \parallel EF$$ or $MN \perp EF$. $\blacksquare$

[b]Remark:[/b] This problem should be quite easy for anyone who has read Chapter 10 of EGMO.

Let $Q$ be the intersection of $AB$ and $CD$. By the Newton-Gauss line, the midpoint of $PQ$ is collinear with $K$ and $L$. So, assuming that $KL$ is perpendicular to $EF$, then $KL$ also bisects $EF$. Note: What I've said above is strictly unnecessary but it motivates why we would want to prove that $KL$ is not only perpendicular but also bisects $EF$.

    We would like to prove that $FK = EK$. By doing so, by symmetry we would also have that $FL = EL$. This would imply that $LK$ is perpendicular to and bisects $EF$. We've significantly reduced the number of points we need to worry about.

    Let $F'$, $K'$, and $E'$ be the reflections of $P$ about $F$, $K$, and $E$ respectively. Let $X$ be the point that makes $PCF'X$ a rhombus, and let $Y$ be the point that makes $PBE'Y$ a rhombus. We want to show that $F'K' = E'K'$. Then, $E'K' = CY$ by parallelograms and $F'K' = BX$ similarly. So, we want to show that $CY = BX$, but this follows since triangles $XPB$ and $CPY$ are actually congruent by SAS. So, we're done.

A very intuitive and short solution:

Let $P'$ be a point such that $APB$ and $DP'C$ are similar. Then the [i]spiral average[/i] of these two triangles is precisely $LFK$, hence it is similar to $APB$. Similarly, $LEK$ is also similar to $APB$, hence the two triangles $LEK$ and $LFK$ are congruent. The result follows.

Solved with [b]mananaban[/b].

We use the standard complex numbers setup.

By the intersection formula, we have $$p = \frac{ac(b + d) - bd(a + c)}{ac - bd}$$ and after some algebra we can also get $$\overline{p}=\frac{a+c-b-d}{ac-bd}.$$ Then $$\frac{e-f}{m-n}=\frac{a + b - c - d + (cd - ab)\overline{p}}{a + d - b - c}=\frac{-(a + d - b - c)(ab + cd - ac - bd)}{(ac-bd)(a + d - b - c)} = \frac{ab + cd - ac - bd}{bd - ac}.$$ It now suffices to check $$\frac{ab + cd - ac - bd}{bd - ac}+\frac{\frac{1}{ab} + \frac{1}{cd} - \frac{1}{ac} - \frac{1}{bd}}{\frac{1}{bd} - \frac{1}{ac}}=0,$$ which is clear after a bit of simplification.

The idea here is that $MN$ is a very "bad" line in the sense that it's not easy to find angles involving it. Thus, we try to find a line parallel to $MN$.\\

Note that because the are midpoints, we should think of reflections. Let $P'_1$ and $P'_2$ denote the reflections of $P$ across $M$ and $N$. Let $AB$ and $CD$ meet at $K$. By isogonality lemma on $\triangle KBC$ and points $A$ and $D$, $KP'_1$ is isogonal to $KP.$ Similarly, $KP'_2$ is also isogonal to $KP$, so $K,P'_1,KP'_2$ are collinear. Furthermore, $P'_1P'_2$ is parallel to $MN$ due to a homothety at $P$ with factor 2.\\

Note that $PEKF$ is cyclic. Let line $KP'_1P'_2$ meet $EF$ at $R$. We then have $$\angle PKF=\angle PEF=\angle EKR,$$ and since $\angle PEK=90$ we have $\angle KRE=90$ as well, done.

We claim that $ME=MF$ and analogously $NE=NF$, which suffices as this proves $MENF$ is a kite.

Let $P$ and $Q$ be the midpoints of $\overline{BX}$ and $\overline{CX}$, respectively. Since
[list]
[*] $MP=QX=QF$
[*] $PE=PX=MQ$
[*] $\angle MPE=\angle MPX+2\angle EBX=\angle MQX+2\angle FCX=\angle MQF$,
[/list]
we have $PME \sim QFM$, so $ME=MF$, as desired. $\square$

[b][color=#9a00ff]Remark:[/color][/b] Once we guess $ME=MF$, we can restate the problem as follows.
[quote]In $\triangle XBC$, let $M$ be the midpoint of $\overline{BC}$. Let $E$ and $F$ be points on the circles with diameters $BX$ and $CX$, respectively, such that $\measuredangle EBX=\measuredangle XCF$ (in directed angles). Prove that $ME=MF$.[/quote]
This can be solved easily using moving points, but it also motivates the above synthetic solution by making it clear why the points $P$ and $Q$ are useful.

Set $(ABCD)$ be the unit circle as usual. Then, let $P=\overline{AC} \cap \overline{BD}$. Then,
\[p=\frac{ac(b+d)-bd(a+c)}{ac-bd}\]
Note that,
\begin{align*}
\overline{p} &= \frac{\frac{1}{ac}\left(\frac{1}{b}+\frac{1}{d} \right) - \frac{1}{bd}\left(\frac{1}{a}+\frac{1}{c} \right)}{\frac{1}{ac}-\frac{1}{bd}}\\
&= \frac{a-b-c+d}{ac-bd}
\end{align*}
Then,
\begin{align*}
e &= \frac{1}{2}\left( p+a+b-ab\overline{p} \right)\\
&= \frac{abc+acd-bcd-abd+a^2c-a^2b+ab^2-b^2d}{2(ac-bd)}
\end{align*}
Similarly,
\[f = \frac{abc+acd-bcd-abd+ac^2-c^2d+cd^2-bd^2}{2(ac-bd)}\]
Then, let $M_1,M_2$ be the midpoints of $AD$ and $BC$. This gives,
\[m_1=\frac{a+d}{2} \text{ and } m_2=\frac{b+c}{2}\]
Then, we observe that,
\begin{align*}
\frac{m_1-m_2}{e-f} &= \frac{\frac{a-b-c+d}{2}}{\frac{abc+acd-bcd-abd+a^2c-a^2b+ab^2-b^2d-(abc+acd-bcd-abd+ac^2-c^2d+cd^2-bd^2)}{2(ac-bd)}}\\
&= \frac{(a-b-c+d)(ac-bd)}{(a-b-c+d)(ac+bd-ab-cd)}\\
&= \frac{ac-bd}{ac+bd-ab-cd}
\end{align*}
We can then see that,
\begin{align*}
\overline{\left(\frac{m_1-m_2}{e-f} \right)} &= \frac{\frac{1}{ac}-\frac{1}{bd}}{\frac{1}{ac}+\frac{1}{bd}-\frac{1}{ab}-\frac{1}{cd}}\\
&= \frac{\frac{bd-ac}{abcd}}{\frac{bd+ac-cd-ab}{abcd}}\\
&= -\left(\frac{ac-bd}{ac+bd-ab-cd}\right)
&= -\left(\frac{m_1-m_2}{e-f}\right)
\end{align*}
which implies that $\frac{m_1-m_2}{e-f}\in i\mathbb{R}$ and thus $EF \perp M_1M_2$ which was the required conclusion.

What are y'all waffling about? This is trivial by vectors!
-----
We use vectors, where we set $K=\overline{AC} \cap \overline{BD}$ to be the origin. It suffices to show that \[ ((\vec A - \vec B)+(\vec D - \vec C)) \cdot (\vec E - \vec F) = 0. \] By the definitions of $E$ and $F$ in the problem, we have \[ \vec E \cdot (\vec A - \vec B) = 0  \ \ \text{and} \ \ \vec F \cdot (\vec C - \vec D) = 0. \] Summing the above two equalities we obtain \[ \vec E \cdot (\vec A - \vec B) - \vec F \cdot (\vec D - \vec C) = 0. \] Thus it suffices to show that \[ \vec E \cdot (\vec D - \vec C) - \vec F \cdot (\vec A - \vec B) = 0. \] We do this synthetically. Let $\overline{KF}$ intersect $\overline{AB}$ at $F'$ and $\overline{KE}$ intersect $\overline{CD}$ at $E'$. Then \[ \vec E \cdot (\vec {CD}) - \vec F \cdot (\vec {BA}) = - KE \cdot CD \cdot \cos(\angle KE'D) + KF \cdot AB \cdot \cos(\angle KF'A). \] The fact that the RHS is zero follows by similar triangles on $\triangle AKB$ and $\triangle DKC$, and we conclude.

Let the intersection of the diagonals be $P$, and let the midpoints be $M$ and $N$.

Reflect $P$ across $M$ and $N$ to points $Q$ and $R$, and let $X = \overline{AB} \cap \overline{CD}$. The isogonality lemma shows us that points $X$, $Q$, and $R$ lie on the isogonal to $\overline{XP}$ with respect to $\angle X$.

It is well-known that the orthocenter and circumcenter of a triangle are isogonal conjugates, meaning that since $\overline{XP}$ is a diameter of $\triangle XEF$, its isogonal is the altitude, which means that $\overline{XST} \perp \overline{EF}$.

Clearly, $\overline{MN} \parallel \overline{ST}$, so we are done. $\square$

Our solution is motivated by the first isogonality lemma. Reflect $X = AC \cap BD$ over $M$, $N$ to $P$, $Q$, respectively. If we let $Y = AB \cap CD$, we know $YPQ$ is isogonal to $YX$ in $\triangle YAD$. As $YX$ is the diameter of $(XEYF)$, its isogonal $YPQ$ is the altitude from $Y$ to $EF$. Homothety at $X$ then tells us
\[EF \perp YP \parallel MN. \quad \blacksquare\]

Haha complex go brrrr.

Denote by $M$ and $N$ the midpoints of $\overline{AD}$ and $\overline{BC}$. We set $(ABCD)$ as the unit circle. Compute,
\begin{align*}
p &= \frac{ac(b+d) - bd(a+c)}{ac - bd}
\end{align*}
Now we may also compute,
\begin{align*}
e &= \frac{1}{2}(a + b + p - ab\overline{p})\\
f &= \frac{1}{2}(c + d + p - cd\overline{p})\\
m &= \frac{1}{2}(a + d)\\
n &= \frac{1}{2}(b+c)
\end{align*}
Then we wish to show,
\begin{align*}
\frac{e - f}{m - n} \in i\mathbb{R}
\end{align*}
Scale all their coordinates by $2$ so that we wish to show,
\begin{align*}
\frac{a + b - c - d - \overline{p}(ab -  cd)}{a - b - c + d} \in i\mathbb{R}
\end{align*}
The term,
\begin{align*}
\overline{p} &= \overline{\left(\frac{ac(b+d) - bd(a+c)}{ac-bd} \right)}\\
&= \frac{1/ac(1/b + 1/d) - 1/bd(1/a+1/c)}{1/ac - 1/bd}\\
&= \frac{-a + b - c +d}{bd - ac}\\
&= \frac{a - b + c - d}{ac - bd}
\end{align*}
Then we find,
\begin{align*}
\frac{a + b - c - d - \frac{(a - b + c - d)(ab - cd)}{ac - bd}}{a - b - c + d} \in i \mathbb{R}\\
\iff \frac{(ac - bd)(a+b-c-d) - (a - b + c - d)(ab - cd)}{(ac - bd)(a - b - c + d)} \in i \mathbb{R}\\
\iff \frac{(a - d) (c - b) (a - b - c + d)}{(ac - bd)(a - b - c + d)} \in i \mathbb{R}\\
\iff \frac{(a - d)(c-b)}{(ac - bd)} \in i \mathbb{R}
\end{align*}
which is easily checked to be true. $\square$

We employ complex numbers with $(ABCD)$ the unit circle. Note that for the intersection $$x = \frac{ac(b+d)-bd(a+c)}{ac-bd}$$ we want $$\frac{f-e}{m-n} =\frac{c+d-a-b+\overline x(ab-cd)}{a+d-b-c}$$ to be imaginary. In particular, note that $\overline x = \frac{a+c-b-d}{ac-bd}$, so
\begin{align*}
\frac{f-e}{m-n} &= \frac{(c+d-a-b)(ac-bd)+(a+c-b-d)(ab-cd)}{(a+d-b-c)(ac-bd)} \\
&= \frac{(a-d)(b-c)(a+d-b-c)}{(a+d-b-c)(ac-bd)} \\
&= \frac{(a-d)(b-c)}{ac-bd}.
\end{align*}
This expression clearly equals the negative of its conjugate, as needed.


This is bit stupid question but what do these mean?

WLOG

LHS

RHS

(I'm guessing LHS and RHS mean left and right side of the equation but I want to be sure)

[quote="Silverfalcon"]

LHS

RHS

(I'm guessing LHS and RHS mean left and right side of the equation but I want to be sure)[/quote]
Then what do they stand for?? :?


http://www.artofproblemsolving.com/Wiki/index.php/1999_AIME_Problems/Problem_3

It says the discriminant simplifies to $ x^{2}-35$, wouldn't it be $ 4x^{2}-35$ though?

Okay, I made the change. For some reason, the answer came out the same though...


Λοιπόν, αφού βρήκα λίγο ελεύθερο χρόνο (αλλά όχι όσο χρειάζονται τα προβλήματα του anto :( ), είπα να βάλω κανα-δυό προβληματάκια του πενταλέπτου :)

Έστω σύνολο $ A$ και συνάρτηση $ x: A\to\mathbb{R}_ \plus{}$ με $ \alpha\mapsto x_a\geq0$. Αν $ \sum_{\alpha\in A} x_\alpha < \infty$, να δειχθεί ότι $ x_\alpha > 0$ το πολύ για αριθμήσιμο πλήθος $ \alpha\in A$.

Cheerio,

Durandal 1707

Cute.

Σατανικές συμπτώσεις? :diablo: :roll: (pun intended) Αν η φραγμένη συνάρτηση $ f : R \to \mathbb{R}$, $ R \equal{} [a_1,b_1] \times \cdots \times [a_n,b_n]$, $ n \geq 1$, είναι Riemann-ολοκληρώσιμη στο $ R$, τότε είναι συνεχής σχεδόν παντού (και αντίστροφα, και βγάζουμε και Lebesgue ολοκληρωσιμότητα).


$E$ is a normed vectorial space of dimension $n$.
$C$ is a connex subset of $E$. Is it true that $Conv(C)$, the minimal convex set that contains $C$, is the set of all the points that are barycentre of at most $n$ points of $C$.

If $C$ is the union of two sides of a triangle (endpoints included) in the plane, how can you get the third side?

Looks like [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=caratheodory&t=24464]it's been discussed before[/url].

I'm not sure about what I wrote there, though :).

Clarifying Xamog's counterexample:

Let $C$ be the union of the segments $OA$ and $OB$, where $O=(0,0),A=(1,0),B=(0,1)$. Then the set of barycenters (midpoints) between points of $C$ is the union of the square $0\le x,y\le\frac12$ and the two segments $OA$ and $OB$, which is not a convex set.

I'm sure that by "barycenter" alekk did not mean merely midpoints, but convex combinations in general.


group theory
abstract algebra
superior algebra
superior algebra open

So given a finite group G, a subgroup H that contains NG(P) and P a p-sylow subgroup demonstrate the following:

(a) H=NG(H) <--I was able to prove this
(b) [G:M] \equiv 1 mod p  where p is a prime[/code]


Because I'm receiving a "Forbidden
You don't have permission to access / on this server."

Works for me-- [url]http://www.naclo.cs.cmu.edu/[/url]  Which site/page were you trying to access?


[quote="dogseatcheese"]The function [i]f[/i] satisfies [i]f(ax)[/i] = [i]af(x) [/i]for all [i]a [/i]and [i]x[/i], and f(2) = 3. What is the value of f(19)?

Please show all of you work. :lol:

Thank you :)[/quote]

[hide]f(19)=28.5? I have an extremely dumb solution so i'll only embarrass myself if i'm right [/hide] :)

[quote="dogseatcheese"]That IS the right answer..... yet I would be happy to know any solution to this problem so......do you mind telling me? Unlike you, I have NO solution :oops:  :D[/quote]
?!?! whoa  :P

well i assumed the function to be linear, i don't know why. the equation can't have a y intercept because then the values of f(ax) and af(x) would be different since a is multipled by x in the first and the whole function is multiplied by a in the second and the y intercept would make them different. So if the form is y=mx, then the equation is y=1.5x and $1.5\times19=28.5$

[quote="JesusFreak197"]It has to be in the form $f(x)=mx$; you don't need to assume it. It can't have a constant on the end. Say it was $f(x)=mx+b$. Then, you have $f(ax)=m(ax)+b$ and $af(x)=a(mx+b)=m(ax)+ab$, which obviously does not satisfy $f(ax)=af(x)$ for all values. Therefore, it cannot have a constant on the end. If it is $f(x)=x^n$, then you have $f(ax)=a^nx^n$ and $af(x)=a(x^n)$, which is again not true for every value. Therefore, it must simply be $f(x)=mx$, and you go from there. ;)[/quote]

Wait, but can you automatically assume that the function is linear? other functions might not have constants at the end either

[quote="JesusFreak197"]I already addressed the other possibilities of $x$ to some power $n$. ;)[/quote]

oh oops. sorry about that.

Or in a more strict manner:

If you put $x=1$ in $f(ax)=af(x)$, you'll get $f(a)=af(1)$. And now here's the trick - the letter [b]a[/b] can be replaced by any other letter we choose, the function will not change. Hence let's put [b]x[/b] instead of [b]a[/b]. We'll get $f(x)=xf(1)$. So $f(x)$ is indeed linear, of the form $f(x)=mx$ where $m=f(1)$.

YAY :!: Eureka :!:   I understand. Thank you very much


number theory
least common multiple
number theory proposed

Prove that for any $ n$, there is a subset $ \{a_1,\dots,a_n\}$ of $ \mathbb N$ such that for each subset $ S$ of $ \{1,\dots,n\}$, $ \sum_{i\in S}a_i$ has the same set of prime divisors.

Let $ b_i$, $i = 1,2,\ldots,n$, be any distinct positive integers, and $ P = \operatorname{rad}\left (\operatorname{lcm}\limits_{S \in \{1,2,\ldots,n\}}\left [\sum_{i\in S}b_i\right ]\right )$. Take $ a_i = Pb_i$.

Simpler, let $ b_i$, $i = 1,2,\ldots,n$, be any distinct positive integers, and $ P = \prod_{S \in \{1,2,\ldots,n\}}\sum_{i\in S}b_i$. Take $ a_i = Pb_i$.


Prove that an infinite number of triangles each having a given interior point as centroid can be inscribed in a given circle.

[hide="Method 1 (constructive)"]The naive approach is to guess that any point on the circle can be a vertex of a triangle with its centroid at any given interior point.  It turns out this doesn't work (see Method 2), but it comes suprisingly close: the second-most naive approach is to just take any chord passing through the given internal point $G$.  Say it has endpoints $A$ and $P$ on the circle, and without loss of generality $AG < GP$.  (If $G$ happens to be the midpoint of segment $AP$, either $G$ is the center of the circle, for which the problem is trivial (all equilateral triangles work) or $G$ is not the center, in which case we happened to pick the one chord of which $G$ is the midpoint, and we can pick any other chord instead.)  Let $M$ be the point on $GP$ such that $2GM = AG$, and let $BC$ be a chord(actually "the chord," unless $M$ happens to be the center of the circle) passing through $M$ such that $BM = MC$.  Then $G$ is the centroid of triangle $ABC$, and since the original chord $AP$ was arbitrary, we can in fact repeat this infinitely many times to get infinitely many such triangles, Q.E.D.[/hide]

[hide="Method 2 (non-constructive)"]Take any point $A$ on the circle.  The set of points which are the centroid of some triangle inscribed in the circle with a vertex at $A$ is a disk whose boundary is internally tangent to the given circle at $A$, with diameter two thirds that of the diameter of the given circle.  Any interior point of the given circle is covered by infinitely many of these disks.  The fact that each triangle is counted 3 times obviously doesn't matter, and we're done.[/hide]


conics
ellipse
geometry
circumcircle
ratio
incenter
power of a point

An ellipse-shaped billiard table doesn't have holes. When a ball hits the table border in a point $P$, it follows the symmetric line in respect to the normal line to the ellipse in $P$. Prove that is a ball starts from a point $A$ of the ellipse and, after hitting the table at $B$ and $C$, returns to $A$, then it will return to $B$.

Consider a different triangle $\triangle ABC$ with circumcircle (O). Let the tangent to (O) at A meet BC at L. A circle (P) with diameter OL meets BC at the midpoint M of BC, because $\angle OML$ is right. Inversion in (O) carries (P) passing through $A \in (O)$ and through the inversion center O into the line AU, where U is the intersection of tangents to (O) at B, C so that AU is the A-symmedian of $\triangle ABC.$  Let $K \in BC$  be the foot of the A- symmedian AU. Since the line AU is also the radical axis of (O), (P), $AK \equiv AU \perp OP \equiv OD.$ It is a well known fact that the points K, L divide the segment BC harmonically, $\frac{KB}{KC}\cdot \frac{LC}{LB}=-1.$

[hide="proof"]K is on the radical axis of (O), (P), hence $KB \cdot KC = KM \cdot KL.$ Denote a = BC, b = CA, c = AB. Assuming c > b,

$KB = \frac{ac^{2}}{c^{2}+b^{2}},\ \ KC = \frac{ab^{2}}{c^{2}+b^{2}}$

$KM = KB-MB =\frac{ac^{2}}{c^{2}+b^{2}}-\frac{a}{2}= \frac{a(c^{2}-b^{2})}{2(c^{2}+b^{2})}$

$KL = \frac{KB \cdot KC}{KM}= \frac{2ac^{2}b^{2}}{(c^{2}+b^{2})(c^{2}-b^{2})}$

$LM = KM+KL = \frac{a(c^{2}+b^{2})}{2(c^{2}-b^{2})}$

$LC = LM-CM = \frac{a(c^{2}+b^{2})}{2(c^{2}-b^{2})}-\frac{a}{2}= \frac{ab^{2}}{c^{2}-b^{2}}$

$LB = LM+MB = \frac{a(c^{2}+b^{2})}{2(c^{2}-b^{2})}+\frac{a}{2}= \frac{ac^{2}}{c^{2}-b^{2}}$

For oriented segments KB, KC, LB, LC we then get $\frac{KB}{KC}\cdot \frac{LC}{LB}=-1$ as expected.[/hide]
Using a parallel projection, we can project the $\triangle ABC$ inscribed in the circle (O) into a $\triangle A'B'C'$ so that the line OD stays in place, the line $AK \perp OD$ is carried into itself, and the angle $\angle DA'K'$ becomes right. The circumcircle (O) is carried into an ellipse $\mathcal E$ circumscribed around the triangle $\triangle A'B'C'$ and with the main axes lines OD, AK. Since the cross ratio $\frac{K'B'}{K'C'}\cdot \frac{L'C'}{L'B'}= \frac{KB}{KC}\cdot \frac{LC}{LB}=-1$ is preserved and the angle $\angle DA'K'$ is right, the lines $A'K' \equiv AK$ and $A'D$ are the internal and external bisectors of $\angle C'A'B',$ their feet K', D dividing the segment B'C' harmonically. Conversely, if we project the $\triangle ABC$ into the $\triangle A'B'C'$ so that the line $A'K' \equiv AK$ bisects the angle $\angle C'A'B',$ then $A'D$ is the external bisector of this angle, which makes the angle $\angle DA'K'$ right. In either case, the tangents BU, CU of the circle (O) at B, C are carried into tangents B'U', C'U' of the ellipse $\mathcal E$ at B', C' and the angle bisector A'K' passes through their intersection U'. Thus we proved the following lemma:

[color=darkblue][b]Lemma:[/b] Let $A, B, C$ be 3 points on an ellipse. The normal $l_{a}$ to the ellipse tangent $t_{a}$ at $A$ bisects the angle $\angle CAB,$ iff it passes through the intersection $U$ of the ellipse tangents $t_{b},\ t_{c}$ at $B,\ C.$[/color]

By the problem condition, the normals $l_{b},\ l_{c}$ to ellipse tangents $t_{b},\ t_{c}$ at B, C bisect the angles $\angle ABC,\ \angle BCA,$ hence they meet at the triangle incenter I and $l_{a}\equiv AI$ is the remaining angle bisector. By the above lemma, $l_{b}\equiv BI,\ l_{c}\equiv CI$ pass through the intersections $V \equiv t_{c}\cap t_{a},$ $W \equiv t_{a}\cap t_{b}$ of the ellipse tangents $t_{a},\ t_{b},\ t_{c}$ at A, B, C. It is therefore possible to project the $\triangle ABC$ inscribed in the given ellipse with center O into a $\triangle A'B'C'$ inscribed in a circumcircle (O), so that the incenter I of $\triangle ABC$ is carried into the symmedian point I' of the $\triangle A'B'C'.$ The angle bisectors AI, BI, CI are then carried into the symmedians A'I', B'I', C'I' and the remaining symmedian A'I' passes through the intersection $U' \equiv t_{b}' \cap t_{c}'$ of the circumcircle tangents at B', C'. Consequently, the original angle bisector AI passes through the intersection $U \equiv t_{b}\cap t_{c}$ of the ellipse tangents at B, C, hence by the above lemma, it is perpendicular to the ellipse tangent $t_{a}$ at A.

Hmm... I used the well-known fact that the tangent to an ellipse in $P$ is the external bisector of $\angle FPF'$, where $F$ and $F'$ are the foci of the ellipse. From this fact it's not hard to see that $F$ and $F'$ are isogonal conjugates respect to the triangle $ABC$, and the result follows.

"The billiard sharp who any one catches,
His doom's extremely hard —
He's made to dwell —
In a dungeon cell
On a spot that's always barred.
And there he plays extravagant matches
In fitless finger-stalls
On a cloth untrue
With a twisted cue
And elliptical billiard balls!"

(from [i]The Mikado[/i].  Apologies for irrelevance -- I couldn't resist.)


function calculus integration probability geometry search algebra

Here's the function:

(x^n)*[(1-x)^(d-n)]

I'm trying to evaluate an indefinite integral with respect to x (d and n are constants).  It's been several years since I've done any calculus, so I've been looking at integration tables online, but they haven't helped me.  I tried the Wolfram Mathematica Online Integrator, but it failed (even when I had it formatted properly).  That suggests to me that this function might not be integratable...

If it helps, I am only interested in integrals on the interval 0 <= x <= 1.

This is the integral that defines the (incomplete) [url=http://en.wikipedia.org/wiki/Beta_function]beta function[/url].  If $ n, d \minus{} n$ are non-negative integers, the answer over $ [0, 1]$ is particularly simple:  it is $ \frac {n!(d \minus{} n)!}{(d \plus{} 1)!}$.  There is also a pretty neat probabilistic proof in this case.

Thanks for the quick response!  And thanks for the link; I've never heard of beta functions before.

EDIT: I re-read your response more closely and I see that it only supplies part of my answer.  Let me be more specific about what I am trying to do here.  Yes, n and d are positive integers.  I am using the function as a probability histogram over the interval [0,1], so I need to divide the function by its indefinite integral over [0,1] to ensure that the area is 1.  I then wish to (for a given d and n) find a mean and a confidence interval, which is what I would need the indefinite integral for.  Is there a nifty way to get an indefinite integral too?

PS.   The full function has this coefficient:

d!/[(d-n)!n!]

but I omitted it because it doesn't vary.  It looks like with the coefficient, the integral over [0,1] would become d!/(d+1)!, which simplifies to 1/(d+1)

$ \frac{d!}{(d\plus{}1)!} \equal{} \frac1{d\plus{}1}$  :roll:

Here is yet more background on what I'm trying to do here.  I found a formula in my stats textbook that gave error estimates for the true chance of a binomial event, given an experiment.  However, I thought it would be neat to find a formula for the EXACT confidence interval, but I couldn't find that formula anywhere (perhaps I'm just not looking hard enough).

I started with the binomial probability formula:

P(n) = d!/(d-n!)n!*p^n*(1-p)^(d-n)

, where d is the number of trials and n is the number of successes.  I've seen this formula derived and it makes sense to me.

So here was my reasoning after that: we have an experiment with d trials and n successes, and we're trying to find a probability distribution for p, which can be any number from 0 to 1.  For any given p, there we can use this formula to find the chance that that p would have resulted in this n for this d.  I would simply graph P(n) as a function of p, normalize it over the interval [0,1] (by dividing by the definite integral over [0,1]), and I would have my probability distribution for p.

I could then use an indefinite integral to establish my confidence interval.  For example, if I wanted a 90% confidence interval, I would find my left error bound (a) such that the integral from a to the peak was 0.45.  The right error bound would be found in a similar fashion.

First of all, does this reasoning make sense?  Secondly, how do I find the integral :)

[quote="Tyl"]$ \frac {d!}{(d \plus{} 1)!} \equal{} \frac1{d \plus{} 1}$  :roll:[/quote]Yeah, that.  :blush:  I've edited my post above.

[quote="hypehuman"]I could then use an indefinite integral to establish my confidence interval. [/quote]
Yes, that's pretty much what I thought you were doing.  Scroll down further in the Wikipedia article.

Okay, I assume you're hinting that I look at the "Incomplete beta function" subheading.  I don't really understand how they worked out the integral, but I'll figure that out later if I feel like it.  So basically, I'm supposed to use this formula: (a+b-1)!/[j!(a+b-1-j)!]*x^j*(1-x)^(a+b-1-j), substituting for j all integer values from a to a+b-1, and adding all the results.  Since, in my case, a=n and b=(d-n), I would have (d-1)!/[j!(d-1-j)!]*x^j*(1-x)^(d-1-j), with j taking values from n to (d-1).  And to normalize it, I would divide the whole thing by (d+1).

I would then find the value of the integral at the peak (x=n/d), and find a such that I(n/d)-I(a) = 0.45.  Trouble is... I'm not sure how I would do that except by trial and error, since that sigma is scary to me.

EDIT: hold up... they're saying that I(0) = 0 and I=(1) = 1.  Doesn't that mean that I don't have to use the normalization factor of 1/(d+1)?

[quote="hypehuman"]I don't really understand how they worked out the integral, but I'll figure that out later if I feel like it.[/quote]
Binomial theorem and term-by-term integration.

[quote="hypehuman"]I would then find the value of the integral at the peak (x=n/d), and find a such that I(n/d)-I(a) = 0.45.  Trouble is... I'm not sure how I would do that except by trial and error, since that sigma is scary to me.[/quote]
Numerically.  I'm pretty doubtful that there's a "neat" way to do this.

I did some monkeying around with a grapher, and I found that this function:

(d+1)*(dCn)*p^n*(1-p)^(d-n)

gives the appropriately normalized function every time (that is, the integral over [0,1] is 1).  If I'm doing this on a computer, should I just give up and use Euler's method? :(

Okay, I've decided to take a different approach to finding the error limits.  Instead of going 0.45 to the left and right, I think it would make more sense to cut off from the bottom.  Imagine a bell curve with a horizontal line cut through it.  Draw vertical lines where the line intersects the curve, and cut off the tails.  I think that would make the most sense because you're eliminating the least likely values.

In the computer program, all I would have to do is find the height such that it leaves an area of 90%.  I would start by setting it to half the height of the peak, finding where it intersects the probability curve, and plugging those x values into the integral equation and subtracting the two values.  I would do a binary search until I come as close to 90% as my accuracy requires.

Does that look like it would come up with a correct answer?  Even if it does, is there a more efficient way of getting it?


Given:

(1) an angle $ ABC$
(2) a point $ M$ on ray $ BA$, and a point $ P$ on ray $ BC$, so that $ |MB| \equal{} |PB|$
(3) a point $ N$ distinct from $ M$ on ray $ BA$, and a point $ Q$ distinct from $ P$ on ray $ BC$, so that $ |NB| \equal{} |QB|$
(4) a point $ E$ a the intersection of $ MQ$ and $ NP$

Prove that $ BE$ bisects angle $ ABC$.[/img]

In $ \triangle BNP$ apply Menelaus theorem $ \implies \frac {NM}{MB}\cdot\frac {BQ}{QP}\cdot\frac {PE}{EN} \equal{} 1$

$ \Longleftrightarrow \frac {PE}{EN} \equal{} \frac {MB}{BQ}$, because $ PQ \equal{} MN$ .

But $ MB \equal{} BP$ and $ BQ \equal{} BN$, so $ \frac {PE}{EN} \equal{} \frac {BP}{BN}$ which means that $ BE$ is the bisector of angle $ \angle ABC$ .

Thank you, Constantin.

I haven't yet learned Menelaus's Theorem.  Is there a solution that involves congruent triangles, CPCTC, and perhaps some isosceles-triangle theorems?  I believe there must be a solution that doesn't rely on Menelaus.

[quote="Bictor717"]You can prove it using three pairs of congruent triangles, in this order: $ \triangle NBP\cong\triangle MBQ,\ \triangle NEM\cong\triangle QEP,\ \triangle NEB\cong\triangle QEB$.[/quote]

Excellent, thanks.


A number $ B$ is obtained from a positive integer number $ A$ by permuting its decimal digits. The number $ A\minus{}B\equal{}11...1$ ($ n$ of $ 1's$). Find the smallest possible positive value of $ n$.