Given three distinct digits $a,b$ and $c$, it is possible, by choosing two digits at a time, to form six two-digit numbers. Determine all possible sets $\{a,b,c\}$ for which the sum of the six two-digits numbers is 484.

[hide]It's easy to find $\overline{ab}+\overline{ba}+\overline{ac}+\overline{ca}+\overline{bc}+\overline{cb}=22(a+b+c)$. Hence, $22(a+b+c)=484\iff a+b+c=22$

We can assume $a>b>c$. At least one of the digits must be $8$ or $9$, since if they were all at most $7$, the sum wouldn't exceed $21$. Therefore, two cases:

1) $a=8\implies b+c=14\implies (b,c)\in\{(7,7)\}$. That's not valid, since all the digits must be distinct.
2) $a=9\implies b+c=13\implies (b,c)\in\{(8,5),(7,6)\}$

The desired set is $\{(9,8,5),(9,7,6)\}$[/hide]

[hide="Answer"]Since we have six two-digit numbers, we use each digit four times: two times in the tens place, and two times in the units place. Therefore, our sum will come out to be $22(a+b+c)=484\Rightarrow a+b+c=22$. Since all numbers are distinct, our only possible sets are $\boxed{(9,8,5),(9,7,6)}$.[/hide]

[hide]Each digit appears twice in each digit place.  Hence, \[ 484=22(a+b+c) \implies a+b+c=22 \]  The largest digit is $9$, because the average of the digits is greater than $7$.  So this allows two possible sets: $(5, 8, 9)$ and $(6, 7, 9)$[/hide]

[hide="solution"]There are $3P2=6$ ways to choose our numbers. We can have $10a+b+10b+a+10a+c+10c+a+10b+c+10c+b=22(a+b+c)=484 \Longrightarrow a+b+c=22$, so there are $(5,9,8)$ and $(9,7,6)$ allong with its permutations so $\boxed{12}$ possible solutions in all[/hide]


i have got a nice solution, wondering if you can solve it without jensen's:
Let $a,b,c$ be positive real numbers
$\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ac}}+\frac{c}{\sqrt{c^{2}+8bc}}\ge1$

[quote="D.P.L"]how would you write greater or equal to in $LATEX$[/quote]
$\ge$

\ge

I know for a fact you can apply isolated fudging and I think variable mixing works too.

$(a+b+c)^{2}=\left( \sum_{cyc}\frac{\sqrt{a}}{\sqrt[4]{a^{2}+8bc}}\cdot \sqrt{a}\cdot \sqrt[4]{a^{2}+8bc}\right)^{2}\le LHS \cdot \left( \sum_{cyc}a\sqrt{a^{2}+8bc}\right)$$=LHS \cdot \left( \sum_{cyc}\sqrt{a}\cdot \sqrt{a}\cdot \sqrt{a^{2}+8bc}\right)$ by Cauchy- Schwarz

$LHS \cdot \left( \sum_{cyc}\sqrt{a}\cdot \sqrt{a}\cdot \sqrt{a^{2}+8bc}\right) \le LHS \cdot (a+b+c)^{\frac{1}{2}}\cdot (a^{3}+b^{3}+c^{3}+24abc)^{\frac{1}{2}}$ again by CS.

So we have $LHS \ge \frac{(a+b+c)^{\frac{3}{2}}}{(a^{3}+b^{3}+c^{3}+24abc)^{\frac{1}{2}}}\ge 1$

and $(a+b+c)^{3}\ge a^{3}+b^{3}+c^{3}+24abc$ is true for bunching

This inequality has been posted before:
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=frac&t=17451
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=frac&t=41931


modular arithmetic
number theory proposed
number theory

Let $ m, n \geq 1$ be two coprime integers and let also $ s$ an arbitrary integer. Determine the number of subsets $ A$ of  $ \{1, 2, ..., m \plus{} n \minus{} 1\}$ such that $ |A| \equal{} m$ and $ \sum_{x \in A} x \equiv s \pmod{n}$.

If $ (m,n) \equal{} 1$, then $ \frac 1n \binom{m \plus{} n \minus{} 1}{m}$.
For any chose $ (x_1,...,x_m)$ we can find $ 0\le y<n$, suth that $ \sum_i y_i \equiv s$, were $ y_i\equal{}x_i\plus{}y \mod m\plus{}n\minus{}1$.

I think this is a really tough problem and i can't solve it

I found a solution at

http://mateforum.ro/viewtopic.php?t=2048

but it was written in romanian language (or some language else, I don't know)

so I couldn't understand it

Can someone translate this solution into english?

thanks

i think i got a solution

for a integer s, denote $ A_s$ as the set of m-tuples (X1,X2,…,Xm) ($ 1\le X1<X2<...<Xm\le m\plus{}n\minus{}1$) such that $ \ X1\plus{}X2\plus{}...\plus{}Xm \equiv s \pmod n$

now we construct a bijection between $ A_s$ and $ A_(s\plus{}m)$ :

for  $ (X1,X2,...,Xm) \in A_s$,
let (Y1,Y2,…,Ym) =
(X1+1,X2+1,…,Xm+1)  if  Xm<m+n-1

(1,2,…,t,X1+t+1,X2+t+1,…,X(m-t)+t+1) if Xm=m+n-1 and m-t is the greatest positive integer t $ (1\le t \le m)$ such that X(m-t)<m-t+n-1

(1,2,…,m) if Xm=m+n-1 and there doesn't exist a positive integer t such that Xt<t+n-1

it is trivial that $ (Y1,Y2,...,Ym) \in A_(s\plus{}m)$  and the map from (X1,X2,...,Xm) to (Y1,Y2,…,Ym) is a bijection

so we deduce that $ A_s$ and $ A_(s\plus{}m)$ have same number of elements

since (m,n)=1, we get all $ A_s$ have the same number of elements and the number is $ \frac 1n \binom {m\plus{}n\minus{}1} {m}$


Sorry if this has been posted before. I ran a search and browsed through the FAQ, etc. but I couldn't find an answer.

I'm using the AoPS board style, and when I get notification replies, I get replies from both AoPS and Mathlinks. It's just a matter of convenience, but I was wondering how I can get all my notifications to come from AoPS?

It's a small ... "bug"  :?  Anyway, they are coming from the same server, only the return address is different. It will be fixed soon, along with another series of long awaited patches.

[quote="mel"]Thanks, Valentin!

Have I mentioned that this is THE best-supported forum I belong to? :P (hehe...I only belong to 2, so technically, I don't see any competition  ;) )[/quote]No, you haven't but thanks :D :)


[b]Prove that if [/b]$r_a=r_b+r_c\vee  r_b=r_c+r_a\vee r_c=r_a+r_b$ [b]then[/b] $(4R+r)^3=4s^2(4R-r)$ [b]and [/b]
$ as=r_a^2\vee bs=r_b^2\vee cs=r_c^2.$ [b]Construct a such triangle.[/b]

Would you please explain the meaning of the mark $\vee$?


im not going to return to ftw untill someone finds me and explains whats been going on since ive been gone so if anyone wants to find me they must go on a hunt... whoever finds me can have some rating if they want  idc. so as long as i stay on this forum someone give me ideas for the hunt and then i will have angie lead it while i am gone. p.s. to find me simply pm me the website that i am hiding in i will check my pms every day and when someone finds me (except for angie she knows where i am) i will come back.

hunt as in treasure hunt...... or in this case me hunt your looking for me!!!!!!!!!

Eh... I don't have the time.

Mathblitz thinks he is treasure.  This has been a public shaming of mathblitz.  Thank you.

And I'm the stupid pirate!

[quote="1=2"]So he hides on a forum hitting F5 continuously?[/quote]

Perhaps.

nvm if this is where its going i might as well say bblf

[quote="guineapiggies"]this message is too small...whatever i found u idk if this counts[/quote]

Yay sheena!

uhhh that was before i started this....... if it that wasnt then it was cause i was telling angie to help me....

[quote="mathblitz"]ugh nvm all of you dont care ill go back every now an then[/quote]

No we don't. :P


let $ ABC$ is a triangle such that : $ sinC \equal{} cos B \plus{} cos A$
prove that $ ABC$ is a  right triangle

$ \sin{C} \equal{} \cos{B} \plus{} \cos{A}$
$ \Leftrightarrow \sin{(A \plus{} B)} \equal{} \cos{B} \plus{} \cos{A}$
$ \Leftrightarrow 2\sin{\frac {A \plus{} B}{2}}\cos{\frac {A \plus{} B}{2}} \equal{} 2\cos{\frac {A \plus{} B}{2}}\cos{\frac {A \minus{} B}{2}}$
$ \Leftrightarrow \sin{\frac {A \plus{} B}{2}} \equal{} \cos{\frac {A \minus{} B}{2}}$
$ \Leftrightarrow \sin{(\frac {\pi}{2} \minus{} \frac {C}{2})} \equal{} \cos{\frac {A \minus{} B}{2}}$ as $ A \plus{} B \plus{} C \equal{} \frac {\pi}{2}$
$ \Leftrightarrow \cos{\frac {C}{2}} \equal{} \cos{\frac {A \minus{} B}{2}}$   (*)
Because $ \frac {C}{2},\frac {A \minus{} B}{2}\in ( \minus{} \frac {\pi}{2},\frac {\pi}{2})$ then (*) implies that:
$ \frac {C}{2} \equal{} \frac {A \minus{} B}{2}$ or $ \frac {C}{2} \equal{} \minus{} (\frac {A \minus{} B}{2})$, therefore:
$ A \equal{} B \plus{} C$ or $ B \equal{} A \plus{} C$, respectively; because $ A \plus{} B \plus{} C \equal{} \frac {\pi}{2}$ then $ A \equal{} \frac {\pi}{2}$ or $ B \equal{} \frac {\pi}{2}$ and the result of the problem follows.


induction
number theory proposed
number theory

let $a,b,c$ be natural numbers. if $a+b+c|a^{2}+b^{2}+c^{2}$  prove that there are infty $n$ such that:  $a+b+c|a^{n}+b^{n}+c^{n}$

Hint:

[hide]Prove that $a + b + c$ divides $a^{2^k} + b^{2^k} + c^{2^k}$ for all positive integers $k$.

Obviously you can use induction for that :)[/hide]

A non-inductive proof here...
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=47652[/url]


how do you get people to play FTW with you. i mean I just sit there for an hour and no one comes?

You have to wait patiently.  Maybe everyone is just afk.


conics
ellipse
geometry
geometric transformation
reflection
Olimpiada de matematicas

Me he encontrado este problema, que en realidad es mas facil de lo que parece.

Prueve que si dos cuerdas de una elipse se bisecan una a la otra, entonces son diametros

Cute problem ;).

Let $AC,BD$ be the two chords, and let $A',B'$ be the reflections of $A,B$ in the center of the ellipse, respectively. We have $AB\|A'B'\|CD,AB=A'B'=CD$. If we assume that the line $CD$ lies between $AB$ and $A'B'$, then at least one of the points $C,D$ lies inside (or on the sides) of $ABB'A'$, but $A,B,A',B',C,D$ are all on the ellipse, so this contradicts the strict convexity of the ellipse.

Solo una traduccion pàra esta comunidad:

"Sean $AC$ y $BD$ las cuerdas de la elipse, y sean $A'$ y $B'$ las reflexiones de $A$ y  $B$ sobre  el centro de la elipse.  tenemos que $AB$ es paralelo a $CD$ y es paralelo a $A'B'$ y ademas sus longitudes son iguales. Si la linea $CD$ esta entre $AB$ y $A'B'$ entonces al menos uno de los puntos $C$ o $D$ cae en el interior o el borde de $ABB'A'$, pero esto contradice la convexidad de la elipse" Grobber.


AMC USA(J)MO USAMO probability trigonometry

Wow. AP stat exams are soon!

I'm not in the AP stat class, but I was planning on learning it this year. So after the USAMO, does anyone think I can learn a year's worth of stat in 5 days?

The [url=http://www.collegeboard.com/student/testing/ap/statistics/topic.html]AP Statistics Syllabus[/url] includes topics that are fair game on the USAMO, so you could afford to start your AP studying while you are still studying for USAMO. To me, a one-month cram sounds less risky than a five-day cram: time to digest the material and make it second nature is always a good idea.

Busy, busy, busy.

You can definitely learn the stats material in 5 days provided that 5 days doesn't mean an hour or two a day for five days.

Yeah the material's easy, but you have to have enough free time to learn it, which may not be the case this time of year.

well, I crammed half of AP physics C in one week and did ok...but that was spring break, so I had a lot more time.  I think it's doable since you're probably already very familiar with a lot of statistics (probability etc.)

Polymorphic, did you enroll in any previous physics course before you took the AP Physics C exam?

I was thinking about self-studying stats and taking it, but I'm glad I didn't now. I have way too much other stuff to do, but taking stats would've given me the ap national distinction award as a JUNIOR, which I think only very few students each state have (or none at all in the majority of states).

Anyway, I think it's technically possible to learn any AP class that doesn't require background knowledge (like a language) in about two weeks. Of course the amount of time spent every day for those two weeks varies with each class. With your math background, learning stats in five days wouldn't be overly hard as long as you spent a few hours a day.

What is the national ap distinction and how do you get it?

Its a good thing that only like 60-50% people pass AP Statistics.

Among people who are okay at math, AP Stats is famous for its easiness. People who have taken it place it at about the difficulty of Algebra II at my school. And if you already know a good amount of probability and stuff, stats is even easier. I looked at the free response questions on the AP tests for the last few years, and I can even do parts of a few problems without ever having looked at a Statistics textbook.

 And the AP National Distinction thing is given to students who get an average of 4 or better (out of 5) on 8 or more full-year AP classes or the equivalent (like AP Physics C Mechanics and AP Physics C E&M counts as one year's worth, since each is the equivalent of a semester of college). The vast majority of recipients of this award get it at the end of senior year.

[quote="polymorphic"]nope...I never took a physics course before taking the AP...i must have barely squeaked out a 5.  It's risky business, but it's doable.[/quote]

Woah.  Nice job.

[quote="mathnovice"]polymorphic, did you take both mechanics and E&M?

I've taken an honors physics and AP Physics B class, but my teacher never covered any electricity, fluid mechanics, gravitation, and nuclear/atomic physics during the two years.

I am studying like crazy though, but the SATII in Physics is this Saturday! The B exam is next week, and the C exam is in two weeks. I am worried about the SATII this Saturday.[/quote]

Don't worry it's not that hard.  I took the CTY course, which was 3 weeks.  I did not do [b]any[/b] practice problems, and it still got me through the AP.  I think I could've crammed 2 days before the exam and still performed ok, but it'll take a lot longer if you want to actually remember the material or go deeper.

Mathnovice, I'm not sure what you're talking about. Physics B and Physics C are both taken at the same time and have been for the last few years (I think).

And if you haven't covered any electricity, I wouldn't recommend taking the [b]electricity and magnetism[/b] section of the test :).

Heck, I wouldn't recommend taking the Physics B test unless you study a lot for that. I believe e&m is still 35% of the B test (50% of the C test).

Is SAT 2 Math 2c hard or not? (otherwords is it easy to get a 800 for a person who can easily get a 800 SAT math).. I browsed through a test prep book, and I downloaded several calc programs (the triangle solver proved that the test prep writers were smoking something when they wrote some of the law of sine problems guess they never heard of ambiguity case). I noticed logic was one of the topics. So I have spent about 20 hours learning that. Is there anything I need to learn for Saturday?

There are alternate dates if there is a conflict between 2 APs. In my case, AP Physics C is pushed back a week. Thanks for the "encouragement," rlee.  :(

I am taking Math IIC and Writing, in addition to Physics, this Saturday. I don't plan on studying for the math one and I'll do a little practice for writing. How much do colleges care about SATIIs anyway?

Colleges tend to care semi-alot about SATII. About the same as SATI.

Mathnovice, I'm not trying to be mean, but I'm telling you that a substantial portion of the B test is electricity and magnetism. If you don't know any, you're not going to do well.

IIC is extremely easy in my opinion. I winged it and got an 800. You don't need to worry much about it.

I really should have gotten an 800 on the 2c...I used the Barron's book to prepare, and took about 50 practice tests, but I think that didn't really help me much...I got an 800 on the Real Sat II practice test a couple of days before, but I got a retarded 740 on the real thing.  I had practiced so much that when I took the real thing I more or less broke down when I came to a problem I hadn't prepared specifically for.  I started trying to do everything on my calculator (bad idea), and I basically ran out of time.  That sucked...and my friend, who is about the same at math as me, and didn't even worry about studying, got an 800.   So, in conclusion, don't do what I did.

Just do a practice test or two to get acquainted with the test format and question types, and on test day, relax and get an 800.  You guys are probably  a lot better than I was then, seeing as I'm a lot better than I was then, and you're still better now.  :)   And don't get the Barron's book (it's much harder than the real thing).  Also, excuse my long and poorly-written sentences (I'm tired).


What is the greatest prime factor of 2525?

hello, we have $ 2525\equal{}5^2\cdot101$, so the greatest primfactor is $ 101$.
Sonnhard.


Find $ g(x)$ such that $ g'(x)\equal{}x^3*e^{(x^4)}$.


vector
geometry
analytic geometry
articles
linear algebra
matrix

well, i know what the shoelace method is, and i can use it efficiently, but i was wondering why it worked?

The basic idea is that you can find an explicit formula for the area of any triangle given its coordinates (it's possible to bash, but vectors work nicely here), and then break any figure where shoelace applies into triangles, whose areas you know from the formula.

You can also translate the triangle such that one vertex is at 0,0 one is at x,0 and one is at a,b.  We can find an expression  for the area of the triangle, then do what math154 said.

[quote="AIME15"]then do what math154 said.[/quote]

I was referring to breaking any polygon into triangles, which math154 said.  Sorry for being unclear.

http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf   :)

That doesn't actually prove it... :wink:

oh, ok thank you, i asked this 'cause i was taking a 1995-1996 nats sprint and like 2-3 problems relied on this method.

[quote="kunny"][quote="AIME15"]You can also translate the triangle.[/quote]

Let $ O(0,\ 0),\ A(a,\ b),\ B(c,\ d)$ with $ ad \minus{} bc \neq 0$,

$ OB: dx \minus{} cy \equal{} 0$, the altitude $ h$ of $ OAB$ is $ h \equal{} \frac {|b\cdot a \minus{} c\cdot b|}{\sqrt {d^2 \plus{} c^2}}$ by the formula of the distance of a point and a line,

$ \therefore \triangle{OAB} \equal{} \frac {1}{2}\cdot \overline{OB}\cdot h \equal{} \boxed{\frac {1}{2}|ad \minus{} bc|}$[/quote]

$ h \equal{} \frac {|d\cdot a \minus{} c\cdot b|}{\sqrt {d^2 \plus{} c^2}}$

[quote="toadoncart"]when u do shoelace, do u have to do clockwise or counterclockwise? i got kinda confused in that article, because they did both. do convex polygons and concave polygons resort to the same method. thx w/ advnace!  :)[/quote]

Both work.

You'll notice it's actually just the determinant of a 2x2 matrix multiplied by 1/2. The trick is finding out why =P


trigonometry
function
integration
calculus
calculus computations

Find antiderivative

$[1]\;\;sin(cosx)$

$[2]\;\;\sin (x^3 (\cos \sqrt x ))$
thanks

[quote="Arne"]Is this from a homework or something, or did you just pick some random function and did you try to find the "antiderivative"? Even Derive can't do it.[/quote]

[2] is a homework but I`m trying find a antiderivative but I can`t

At best, you can expand each integrand in terms of a Maclaurin series and integrate term-by-term.

[b]Edit:[/b] We must expand the second integrand at a point greater than x = 0, as the function is not infinitely differentiable about x = 0.

[quote="Kent Merryfield"]Both functions have antiderivatives.

1. $\int_0^x\sin(\cos t)\,dt,$ defined for $x\in\mathbb{R}.$

2. $\int_0^x\sin(t^3\cos\sqrt{t})\,dt.$ defined for $x\ge 0.$

What? You think I'm cheating? You're asking for closed-form, elementary antiderivatives? You can ask all you want, but you aren't getting that answer. There is simply no reason to even suspect that these functions might have elementary antiderivatives.[/quote]

 I`m looking for any antiderivative

[quote="Arne"][quote="felipesa"]

[2] is a homework but I`m trying find a antiderivative but I can`t[/quote]

I don't think anybody can. The best you can get is some approximations...[/quote]

Could you explain how?
Thanks


You want to buy the same number of grapefruits and oranges to make holiday fruit baskets.  What is the least number of oranges you can buy if grapefruits are shipped in boxes of 12, and oranges are shipped in boxes of 16?

[hide]48. To find the lowest number of fruits that you can get for each one, you need to find the LCM of the two. 12= 2*2*3 and 16=2*2*2*2. The LCM is found by taking each number and using the one with the highest exponent. so you have 3 * 16[/hide]

[hide][color=indigo]12 = 3*2*2
16 = 2*2*2*2
their least common multiple is 3*2*2*2*2, which is 48.
You can buy 48 oranges[/color][/hide]

[hide]You find the lcm of 12 and 16, which is [b]48[/b][/hide]

[hide]12*4 and 16*3 both equal 48, which is the answer[/hide]


number theory
greatest common divisor
least common multiple
Putnam
algebra
function
domain

To be comfortable in doing any olympiad level problem on GCD's or LCM's, what properties should I know between the GCD or LCM of any size group of numbers?
(such as the basic one of gcd(a, b)*lcm(a, b)=ab, that one I already know)

$ \gcd(a, b) \text{lcm}(a, b) \equal{} ab$ just expresses the fact that $ \text{min}(x, y) \plus{} \text{max}(x, y) \equal{} x \plus{} y$.  This identity can be generalized to $ n$ numbers by rewriting what $ \gcd$ and $ \text{lcm}$ do in terms of prime factors, which is often a very useful strategy (see this year's Putnam A3).  For example, the $ n \equal{} 3$ generalization is

$ \gcd(a, b, c) \text{lcm}(ab, bc, ca) \equal{} \text{lcm}(a, b, c) \gcd(ab, bc, ca) \equal{} abc$.

I have never actually seen a problem that needed this identity.  But there's a first time for everything!  Other things that are good to know:

- $ \gcd(a, b)$ is the smallest positive element of the set $ \{ ax \plus{} by | x, y \in \mathbb{Z} \}$ and divides every other member of the set.  In other words, $ \mathbb{Z}$ is a [url=http://en.wikipedia.org/wiki/Principal_ideal_domain]PID[/url].  This observation generalizes to $ n$ numbers.

- Because of the above observation, the Euclidean algorithm generalizes to an $ n$-number $ \gcd$.

- $ \gcd$ and $ \text{lcm}$ satisfy various identities that are instructive to work out (try the min-max interpretation) such as $ \gcd(\gcd(a, b), c) \equal{} \gcd(a, b, c)$.

Delete and re-post [a month] later, if you're still curious for an answer?


Hello does anyone know what is $Caysey$'s theorem ?

Maybe you mean Casey's Theorem ([url=http://mathworld.wolfram.com/CaseysTheorem.html]http://mathworld.wolfram.com/CaseysTheorem.html[/url]).


inequalities limit calculus derivative logarithms function induction

let $ n$ a positive integer , Prove that :
$ (1\plus{}\frac{1}{n})^n \leq 3$

The important step here is that $ \binom{n}{k} \frac {1}{n^k} \leq \frac {1}{k!}$. This can be shown as follows.

$ \frac {n!}{k! (n \minus{} k)! n^k} \leq \frac {1}{k!}$ $ \iff \frac {n!}{(n \minus{} k)! n^k} \leq 1$ $ \iff \frac { n (n \minus{} 1) (n \minus{} 2) \ldots (n \minus{} k \plus{} 1)}{n \cdot n \cdot n \cdot \ldots \cdot n} \leq 1$

Which is clearly true.

[quote="anas"]let $ n$ a positive integer , Prove that :
$ (1 \plus{} \frac {1}{n})^n \leq 3$[/quote]
LHS is 2 when n=1.
Let $ y(n)\equal{}\left(1\plus{}\frac1n \right)^n$
$ y' \equal{} (1\plus{}1/n)^n (\ln(1\plus{}1/n) \minus{} 1/(n\plus{}1))$

Now for $ y'\equal{}0$, we have $ \ln(1\plus{}1/n)\minus{}1/(n\plus{}1)\equal{}0$ or $ (1\plus{}1/n)^{n\plus{}1} \equal{} e$
which occurs when $ n \to \infty$.  So the upper bound of $ y$ is $ \lim_{n \to \infty} (1\plus{}1/n)^n \equal{} e < 3$.

The derivative approaching zero is not a sufficient condition for convergence.
[b]Ex:[/b] [i]Consider $ u \equal{} \log x$, which does not converge; however, $ u' \equal{} 1/x \implies \lim_{x \to \infty} u' \equal{} 0$.
[/i]
Setting the derivative to zero, however, will help you fix your second flaw, which is that even if you had established a limit, it would not imply that it is an upper bound. By showing that the derivative is positive at one point and continuous, and never zero, you prove that the derivative is strictly positive (Intermediate Value Theorem), which means you would increase to the limit and it would be an upper bound. Again, however, this only works if you have established that $ y$ does converge to a limit, which you have not.

Showing that $ y(n)$ is a strictly increasing sequence doesn't require a notion of derivative and showing that it converges to a limit is much harder than showing it's bounded.

Thanks for pointing that out.  However, I don't see the connection here
[b]Ex:[/b] [i]Consider $ u \equal{} \log x$, which does not converge; however, $ u' \equal{} 1/x \implies \lim_{x \to \infty} u' \equal{} 0$.
[/i]I never said that $ \lim y'\equal{}0$ implies the convergence of $ y$ (if I'm not misunderstood, y(n) converges if its limit exists as n tends to infty).  In this case the upper bound of $ u$ is $ \lim u \equal{} \infty$.
So what is the correct proof with derivative you have in mind?  (Sorry if I'm wrong as I'm no expert in analysis.)

The argument you used to determine the limit is far from valid; you need to know, among other things, that the function is monotonic in the first place, and the whole thing is rather backwards - understanding sequence limits occurs before understanding function limits.

let $ a_n\equal{}(1\plus{}\frac{1}{n})^n$
by AM-GM : $ (1\plus{}\frac{1}{n\minus{}1})^{n\minus{}1}\equal{}<(\frac{(n\minus{}1)(1\plus{}\frac{1}{n\minus{}1})\plus{}1}{n})^n\equal{}(1\plus{}\frac{1}{n})^n$

so that  $ a_{n\minus{}1}\equal{}<a_n$ is incresing sequence.

hence,  $ a_n\equal{}<a_{6n}$
so we have to prove that : $ a_{6n}\equal{}<3$

we have $ (\frac{6n}{6n\plus{}1})^n\equal{}(1\minus{}\frac{1}{6n\plus{}1})^n>\equal{}1\minus{}\frac{n}{6n\plus{}1}\equal{}\frac{5n\plus{}1}{6n\plus{}1}>\frac{5}{6}$

and so $ a_{6n}\equal{}(1\plus{}\frac{1}{6n})^{6n}\equal{}{(\frac{6n\plus{}1}{6n})}^{6n}<(\frac{6}{5})^6\equal{}2.98...<3$

we deduce that  $ a_n\equal{}<a_{6n}<1$
 :idea:  a proof from a friend ( :ninja: mehdi :ninja: )
puting $ k\equal{}1/n$ , so it 's equivalent to $ (1\plus{}k)^{1/k}<3$

<==> $ 3^k>(1\plus{}k)$

or $ 3^k\equal{}(1\plus{}2)^k>\equal{}1\plus{}2k>1\plus{}k$
 :roll:
(both are from morocco)

[quote="TZF"]Both the proofs in the above post assume the inequality $ (1 \plus{} a)^x \geq 1 \plus{} ax \; , \; \; \forall a \geq \minus{} 1$, which was likely derived from the first two terms of the infinite Binomial expansion.

This inequality is true iff $ x \not \in (0,1)$, and equality holds iff. $ x \in \{ 0,1 \}$. See note.

The first proof is fine because it uses $ x \equal{} n$, a positive integer.

The second proof is wrong because it uses $ x \equal{} k \equal{} 1/n \in (0,1)$. For these $ k$, we have $ 1 \plus{} k < 3^k < 1 \plus{} 2k$.

[b]Note:[/b] [i]To see why it doesn't work for, say, $ x \equal{} 1/2$, try computing $ \binom{1/2}{2}$, and you'll see that it turns out to be negative! In fact, for $ x < 1$, the binomial coefficients alternate between negative and positive.[/i][/quote]
The inequality you mentioned above is also named as Bernoulli's inequality.The sign will flip over if 0<x<1.

Yes there is a solution with induction, like as: we will prove that for any given $ n$, by induction on $ k$ that $ \left(1\plus{}\frac1n\right)^k\leq 1\plus{}\frac kn\plus{}\frac{k^2}{n^2}$ for $ 1\leq k\leq n$.
For $ k\equal{}1$ the inequality is trivial.
Assume that $ k\leq n$ and the inequality holds for any $ k'\leq k$. If $ k\equal{}n$ we are done. If $ k<n$ then
$ \left(1\plus{}\frac1n\right)^{k\plus{}1}\leq\left(1\plus{}\frac kn\plus{}\frac{k^2}{n^2}\right) \left(1\plus{}\frac1n\right)\equal{}1\plus{}\frac{k\plus{}1}{n}\plus{}\frac{k^2\plus{}k}{n^2}\plus{}\frac{k^2}{n^3}\leq1\plus{}\frac{k\plus{}1}{n}\plus{}\frac{k^2\plus{}2k}{n^2}<1\plus{}\frac{k\plus{}1}{n}\plus{}\frac{(k\plus{}1)^2}{n^2}$


analytic geometry
Asymptote
rotation
real analysis
real analysis unsolved

Does r(q) = 1/(q - a) have an asymptote?
The trivial case a = 0 is ok, when y = 1 is an asymptote.
(r(q) = 1/q, y = rsin(q) = sin(q)/q approaching 1 when q approach 0)
But what about the general case?

The curve $r(\theta)=\frac1{\theta-a}$ is just the curve $r(\theta)=\frac 1\theta$ rotated by angle $a$. So it also has an asymptote, which is just the line $y=1$ rotated (around the origin) by the same angle $a$.


geometry
geometric transformation
reflection
circumcircle
geometry proposed

Let $ABC$ be a triangle with circumcentre $O$. A point $P$ is moving on the extension of $OA$ towards $A$. Let $g$ be the reflection of the line $PB$ with respect to $BA$, $h$ be the reflection of the line $PC$ with respect to $CA$ and $Q$ be the point of intersection of $g$ and $h$.
Find the locus of $Q$ as $P$ moves on the extension of $OA$ towards $A$.

Answer: Locus of poits $Q$ is a ray.
Let be $Q,Q'$ intersections of ${AB,g}{AC,h}$ ,respectively.we will prove $Q=Q'$.Easily, we check that: $\angle OBP=\angle BQO$.so, $\triangle OBP\cong \triangle OQB$.Therefore, $OB^{2}=OP.OQ....(1)$.
Similarly,
$\angle OCP=\angle CQ'O$.so, $\triangle OCP\cong \triangle OQ'C$.Therefore, $OC^{2}=OP.OQ'....(2)$.
from $(1)$ and $(2)$, $OP.OQ=OP.OQ'\Longrightarrow OQ=OQ'$.That is $Q=Q'$.The lines $BQ,CQ,PA$ are concurrent at point $Q$.Point $Q$ is on the extension of direction of $A$ of ray $[PA$ .
Lokman GÖKÇE.

[quote="scarface"]Answer: Locus of poits $Q$ is a ray. [/quote] Well, not exactly...the locus of $Q$ is only the segment $OA$.

There is also a very nice solution using the circle of Apollonius :)

sorry, If I'm mistaken.in my figure, point $Q\in [PA$, but $Q$ is not elemenet of $(AP]$.I solved the problem according to the below figure.I checked the my solution with sketchpad program.

Ok, sorry...bad wording :blush:
$P$ is on the ray $OA$ but not on the segment $OA$, ie on the extension of $OA$ beyond $A$. (Just swap $P$ and $Q$ in your figure :D)

But this doesn't change much in your solution :)

We denote as $K,$ $L,$ the orthogonal projections of $P,$ on the sidelines $AC,$ $AB,$ of the given triangle $\bigtriangleup ABC,$ respectively.

Also, we denote as $D,$ $E,$ the reflections of $P,$ with respect to $K,$ $L$ respectively, lie on the lines $(h),$ $(g)$ and as $A',$ the antidiametric ( = diametric opposite ) point of $A,$ with respect to the circumcircle $(O),$ of $\bigtriangleup ABC.$

It is easy to show that $KL\parallel BC,$ because of $\frac{AK}{AC}= \frac{AP}{AA'}= \frac{AL}{AB}$ $($ from $A'C\parallel PK$ and $A'B\parallel PL$ $).$

We have also that $DE\parallel KL$ ( because of the points $K,$ $L,$ are the midpoints of the segments $PD,$ $PE$ respectively ) and so, we conclude that $DE\parallel BC.$

Because of now, $\angle ABO = \angle BAO = \angle PAL = \angle LAE,$ we conclude that $BO\parallel AE$ and similarly $CO\parallel AD.$

We see that the triangles $\bigtriangleup OBC,$ $\bigtriangleup AED,$ have their sidelines parallel one per one and so, they are perspective ( the intersection points of the pairs of their homologous sidelines at infinity, are collinear ).

Hence, by [b][size=100]Desarques’s theorem,[/size][/b] we conclude that the segment lines $BE,$ $CD,$ $OA,$ ( connecting the pairs of their homologous vertices ), are concurrent at one point.

That is the point $Q,$ as the intersection point of the segment lines $BE\equiv (g)$ and $CD\equiv (h),$ lies on the segment line $OA.$

It is easy to show that when the point $P,$ is variable on the extension of $OA$ towards $A,$ then the locus of $Q$ is the segment $OA$ and the solution is completed.

Kostas Vittas.


Henry took five tests, and his average score was 57 points. He
scored at least 50 points on each test. There were 100 points
possible on each test. What is the highest score that Henry could
have earned on any of the five tests?

Henry's total score on all five tests is $ 5(57)\equal{}285$.

To maximize his highest score, we must minimize his four lower scores. That minimum is $ 4(50)\equal{}200$.

Highest possible high score is $ 285\minus{}200\equal{}\boxed{85}$ points.


[quote="shadysaysurspammed"]I wanted to know the "exact" procedure for stopring questions in a .pdf.What exactly to download(the goto thing or texnicenter?).

Also,I have tried it 7 times but the goto thing doesn't get installesd.It says ionstallation complete ajnd there will be a virtualprinter in you control panel but nothing's there.[/quote]

Hmmm... I can't speak for gotoPDF, but the way CutePDF works is, you type stuff in, say, MS Word.  Then, you go to Print, and where you have the option for what printer you want, you can select CutePDF, and if you select that and hit Print, suddenly you have a PDF file with the stuff you typed up (and no actual printing takes place).  Maybe gotoPDF works the same way?  Is this what you want?

If you're asking how to compile $\LaTeX$ files into PDF using TeXnic Center, you select LaTeX->PDF at the top of the screen (it says LaTeX->DVI by default), and click the Build button (or press Ctrl+F5); you can then view the result by clicking the View Output button (or press F7).

But, I'm not exactly sure what you want here.  Could you be a little more descriptive about your problem?


Can I buy that book, Thuan? Where i can?  :maybe:

Then you can request a member of your team , i think so :)

Hi,

I live in France, can I buy it by Western Union pvthuan ? I'm really interesting in your book  :)

[quote="kunny"]To pvthuan, I also would like to buy your new book, how could I obtain it, though I'm not IMO participant.

kunny[/quote]

It is best to have someone coming to Hanoi this summer take the book home for you. That would save you shipping cost.

pvthuan, thank you for your reply.

kunny

So what about in Vietnam? Can I order via ATM? I can't go to Hanoi and don't want to cost time of my relatives.

You can post this question in Vietnam forum. He'll help you  :wink:

I'll going to Ha Noi, but I'm not IMO participant. So, where I can get this book in Ha Noi? Please give me exactly the address. Thank.

I'm sorry. I only read your problems  :D
Thank you.

oh who did delete my post ?, I only want to register that wonderful books from this new market of mathlinks  :D , sorry but now I think Harazi is true there are many books is selled on mathlink market.  :)

I gave some copies to Goran (through deputy leader of Croatia), Alex Huang (through a Chinese contestant), Tittu Andreescu (through Steven Dunbar), Marcin Kuczma (through deputy leader of Poland), Ji Chen (through Chinese observer), Darij Grinberg (through Tiks and Peter Scholes). About 25 other copies have been given for free to deputy leaders at this  IMO.

Five free copies were also given to Mathlinkers (as gift from Thuan) in an informal Mathlinks meeting with Vornicu.

Those who help your friends, coaches bring the book home, please contact at mobile phone 0919 356 939 (Pham Van Thuan).

Best wishes

oh congratulation that's really best seller book :lol:

Hojoo Lee, Ki Yin Li, Zuming Feng, Titu Andreescu, Wather Janous are among hundreds to get the book during IMO 2007.

Here is a most representative chapter in the book.

Thank you very much, pvthuan.

This seems to be very interesting to me, though written in Vietnamese.

Could we obtain the English version?

kunny

Yes,could you post the english cersion????? :)

Dear friends:

At the moment, I just have Vietnamese version available for use here. I have finished the English version of the whole book but I am not allowed to put any English part here until IMO 2008 in Spain. The whole book will be available for sales in Spain during IMO 2008.

Anway, I hope you can understand it without too much difficulty.

In April, 2009, the English version of this book was published by the Intuitive Mathematics Press, Canada. In this version, we have added some new techniques and problems that are appropriate for mathematical competitions.

Does anyone have the English version of that book?


calculus
integration
limit
logarithms
trigonometry
real analysis
real analysis unsolved

Martin - the rest of us would appreciate it if you would learn $\text{\LaTeX}.$ Is this the correct statement?

$\lim_{x\to0}\frac1x\ln\left(\int_{0}^{1}\left(\sin \left(\frac{\pi t}2\right)\right)^{x}\,dt\right)=?$

yes this is the correct statement
sorry i am not using latex. i tried downloading it as described in the site but when startitng, the program asked me to write the correct path i saved the "executable files" and when i wrote c/programe files...it wasn't correct. also i tried to find where are the exucatable files but i coudn;t. sorry

Dude, there's nothing to download! Just write your mathematical text between dollar signs.
Use this little help guide if you want.


What score do you need to earn a "master" badge or a "pass" badge? Because I mastered something when I only got 1 question in that subject and I got it wrong... :)

This is NOT :spam:

[quote="iYOA"]just wondering, how long did it take to make alcumus?[/quote]
This is quite difficult to answer as there are many parts to Alcumus.

The front end part, the part where you see the questions, the solution, report, and videos, didn't take all that long, maybe a couple weeks.

Then there is a whole bunch of of other stuff. The administrative and teacher parts of Alcumus took quite some time. The server, which is used to decide what topics you have passed and mastered, as well as picking an appropriate problem took another chunk of time. Creating the videos took quite awhile. Putting in all the questions took the most amount of time. Sorry I don't have more exact figures, but I don't really recall how long it all took.

Do you guys have an actual studio, or is it just a quiet room with a bunch of lights, a white board, and a camera?  (Weird question, but what kind off camera is used to make the videos?)

Also, could you ever make it so you can download the videos?

i am just wondering what is the difference between 'passing' and 'mastering' a badge??? :what?:

@ying:
[quote="rrusczyk"]To evaluate if you have mastered a topic, Alcumus looks at topics that depend on that topic.  You have answered a great many problems that are in much harder topics than the one you're 0-for-1 on.  Alcumus adjusts and realizes that you don't need to do a bunch of problems in that topic -- you clearly must have it down if you're able to handle these harder problems that depend on this topic.
That's part of the genius of Alcumus: no drill and kill for you when you don't need it.[/quote]
So basically, you have to get a LOT of the problems in one sction right in order to "master" it. To pass it, you have to get at least 60% of the problems right. TO master it, it's about 80%, i believe.

It is not percentage-wise decided. The passed and mastered labels require both a going-through of much of the problems and is also weighted based on the apparent difficulty of a certain topic. For computing combinations, you almost definitely need 100% to master, but one of my friends currently has "mastered" a very hard topic at 50%.

to determine whether you have mastered or passed a topic, Alcumus does not use your percentage
rather, it uses how well you have done, and compare it to others

if you have a 50% on a topic in which other people have 30% on, you will have mastered/passed it

if you have a 50% on a topic in which other people have 80% on, you will not have mastered/passed it


Έστω τρίγωνο $ ABC$ και σημεία των τμημάτων $ AC$,$ BC$ αντίστοιχα, ώστε $ AD$=$ BE$. Έστω K,L τα περίκεντρα των τριγώνων $ AEC$,$ BDC$. Αν η $ KL$ τέμνει τις $ AB$,$ AC$ στα P,Q, ν.δ.ο. $ CP$=$ CQ$.

Μιας και δεν μας έκατσαν οι γεωμετρίες στην IMO (φαντάζομαι ότι στο post σου εννοείς ότι η $ KL$ τέμνει τις $ BC$ και $ AC$):

Έστω λοιπόν ότι οι περιγεγραμμένοι κύκλοι των $ AEC$ και $ BDC$ τέμνονται στο $ M$. Τότε ισχύει ότι $ KL$ κάθετο στην $ CM$ αφού η $ CM$ είναι ο ριζικός άξονας των δύο κύκλων και $ KL$ η διάκεντρος. Αρκεί λοιπόν να δείξουμε ότι η $ CM$ είναι και διχοτόμος της γωνίας $ ACB$ αφού τότε στο $ CPQ$, η $ CM$ θα είναι ύψος και διχοτόμος, άρα το τρίγωνο θα είναι ισοσκελές και άρα $ CP \equal{} CQ$.

Ένα απλό angle chasing όμως δείχνει ότι $ MAE \equal{} MCB \equal{} MDB$, άρα τα $ M,F,D,A$ είναι ομοκυκλικά, και ομοίως $ MBD \equal{} MCA \equal{} MEA$, άρα και τα $ M,F,E,B$ είναι ομοκυκλικά. Αλλά οι δύο αυτοί κύκλοι είναι ίσοι, αφού οι ίσες (ως κατακορυφήν) γωνίες $ BFE$ και $ AFD$ βαίνουν στις ίσες (από την υπόθεση) χορδές $ BE$ και $ AD$. Έτσι, αφού οι $ MCA \equal{} MBD \equal{} MBF$ και $ MCB \equal{} MAE \equal{} MAF$ είναι εγγεγραμμένες γωνίες σε ίσους κύκλους που βαίνουν σε ίσες  χορδές (δηλαδή την κοινή χορδή $ MF$) οι $ MCA$ και $ MCB$ είναι ίσες (ή παραπληρωματικές, που προφανώς δεν ισχύει σε αυτή την περίπτωση), και άρα το ζητούμενο αποδείχτηκε.

 [geogebra]828fa69f3c17452cc63df7cbc7deea1fc2684d84[/geogebra]

Σημείωση: ξέχασα στο σχήμα να σημειώσω με $ F$ την τομή των $ AE$ και $ BD$ καθώς και να δείξω την $ KL$, καθώς και τα $ P,Q$.

Ηλίας

ΥΓ

Εδώ θα ήθελα να ευχαριστήσω όλους όσους μας έκαναν μάθημα στην προετοιμασία για την ΙΜΟ, και ιδιαίτερα τους Nick, Sil και τον κύριο Ψύχα, που τους έχω ιδιαίτερη αδυναμία!

Στο σημείο αυτό θα ήθελα και εγώ να ευχαριστήσω όλους όσους έκαναν μάθηματα στην προετοιμασία που είχα την τύχη να παρακολουθήσω!!!!
Επίσης να δόσω ξανά πολλά συγχαρητήρια στα παιδιά και είμαι βέββαιος ότι σύντομα στο μέλλον θα τα ξαναπούμε :)
Δημήτρης


Two circles intersect at points A and B. Point X lies on line AB but not on segment
AB. Prove that the lengths of all the tangents drawn from point X to the circles are equal.

This is a direct application of power of a point [like tangent-secant theorem].


inequalities
algebra
polynomial
inequalities proposed

Let $ x,y,z$ be real numbers, and let
$ A \equal{} \sum x^3y \minus{} xyz\sum x$,
$ B \equal{} \sum xy^3 \minus{} xyz\sum x$,
$ C \equal{} \sum x^2(y \minus{} z)^2$.
Prove that
$ 4AB \ge C^2$.

we have
     \[ {4AB \minus{} C^2}\equal{} 4xyz(x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} yz \minus{} zx)^2,\]
threfore  $ 4AB \minus{} C^2\ge0$  holds only for $ xyz(x \plus{} y \plus{} z)\ge{0}$.

[quote="fjwxcsl"]we have
\[ {4AB \minus{} C^2} \equal{} 4xyz(x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} yz \minus{} zx)^2,
\]
threfore  $ 4AB \minus{} C^2\ge0$  holds only for $ xyz(x \plus{} y \plus{} z)\ge{0}$.[/quote]
Indeed, the inequality holds for nonnegative $ x,y,z$, or, more general, for $ xyz(x \plus{} y \plus{} z)\ge{0}$.
Note that there is a nice solution without expanding, using the Cauchy-Schwarz inequality.

[quote="can_hang2007"]For $ x,y,z \ge 0,$ we can rewrite the inequality as
$ \left[ \sum xy(x \minus{} z)^2\right] \left[ \sum xy(y \minus{} z)^2\right] \ge \left[ \sum xy(x \minus{} z)(y \minus{} z)\right]^2,$
which is obviously true by Cauchy Schwarz Inequality. :)[/quote]
Nice. :lol:
But can we write the inequality in this form without expanding?


Find a way to generate all integral solutions to
$x^{2}+2y^{2}=z^{2}$.

[quote="boxedexe"]Use pythagorean triples.  Transform your equation into $u^{2}+y^{2}=z^{2}$, where $u^{2}=x^{2}+y^{2}$.  Pythagorean triples are generated as follows:
$\\\gcd (u,y)=\gcd (u,z)=\gcd (y,z)=1\implies\exists a,b\in\mathbb{Z}\text{ such that }\\u=2ab,\ y=|a^{2}-b^{2}|,\ z=a^{2}+b^{2}$.
Substitution and Done!

Masoud Zargar[/quote]
But $x^{2}+y^{2}$ does not have to be a perfect square...

The hardest problem is to prove or disprove if there is a nontrivial solution. But this can always be done by giving one or by modular stuff.

$x=1,y=2,z=3$ is a solution. Now use http://www.mathlinks.ro/Forum/viewtopic.php?t=87739 .

They way I solved it got
[hide]
$y=2rs$ $x=r^{2}-2s^{2}$ and $z=r^{2}+2s^{2}$ when $r^{2}>2s^{2}$
and
$y=2rs$ $x=2r^{2}-s^{2}$ and $z=2r^{2}+s^{2}$ for when $2r^{2}>s^{2}$

and $\gcd(r,s)=1$
[/hide]

However, I am not sure that this generates all solutions.  :blush:

We will proceed using Algebraic Number Theory and Unique Factorization in $\mathbb{Z}[i]$.  Hence, $x^{2}+2y^{2}=(x+iy\sqrt{2})(x-iy\sqrt{2})=z^{2}$.  Thus, $\exists a,b\in\mathbb{R}$ such that $x+iy\sqrt{2}=(a+ib)^{2}=a^{2}-b^{2}+2abi\implies x=a^{2}-b^{2}$ and $y=\sqrt{2}ab$.  From these, we have that $z=a^{2}+b^{2}$.  Now, if we find $c\in\mathbb{Z}$ such that $a=\sqrt{2}c$, then we have $y= 2cb, x=2c^{2}-b^{2}, y=2c^{2}+b^{2}$.  Thus, we have that $(x,y,z)=(2u^{2}-v^{2}, 2uv, 2u^{2}+v^{2}),\ u,v\in\mathbb{Z}$. Q.E.I.

Masoud Zargar

The following question is just a matter of interest regarding comprehension.  Isn't "to be found" the exact same thing as "wanted?"

I'm not sure, but it's possible that you missinterpreted my sentence, which has indeed a double meaning:
I just asked for the latin phrase behind it and did not say that the correct english word is wanted...


function
complex analysis
complex analysis unsolved

$ A\equal{}\{z|\text{Img}\ z>0\}\ ,\ B\equal{}\{z|\ |z|<1\ ,\ \text{Img}\ z>0\}$ find an analytic function from $ A$ to $ B$.

Are you sure this is the precise question? Should it say, "Find a [b]1-1[/b] analytic function that maps A [b]onto[/b] B"?


[quote="Treething"]Sam can squish a pile of ants in ten minutes. So can Sarah. How long does it take for them to squish a pile of ants together?

If Sam's younger brother Linda (yes, his name is Linda) can squish a pile of ants in 20 minutes, how long does it take if they all work together to squish [b]two[/b] piles of ants?[/quote]

[hide="1"]
In the first question, each person can squish 1/10 of the pile per minute. That means each minute 1/5 of the pile is squished. This means it will take 5 minutes.[/hide]

[hide="2"]
In this question, 2/20, 2/20 and 1/20 of the pile is squished per minute. That means in one minute, 1/4 of the pile is gone. This means it takes 4 minutes to squish each pile so in total it will take 8 minutes.[/hide]

[hide="1"]
i think the answer is 5 minutes because they have two of the same rates and two people so 10/2 is 5.[/hide]

[quote="Treething"]Sam can squish a pile of ants in ten minutes. So can Sarah. How long does it take for them to squish a pile of ants together?

If Sam's younger brother Linda (yes, his name is Linda) can squish a pile of ants in 20 minutes, how long does it take if they all work together to squish [b]two[/b] piles of ants?[/quote]

[hide]Sam can squish $\frac{1}{10}$ of a pile of and every minute, and so can Sarah. Therefore, together, every minute, they can squish $\frac{2}{10}=\frac{1}{5}$ of a pile of ants. Therefore they need only 5 minutes together to squish a pile of ants.[/hide]

[hide]Using the same strategy as before, $\frac{1}{5}+\frac{1}{20}=\frac{4}{20}+\frac{1}{20}=\frac{5}{20}=\frac{1}{4}$ So they, combined, can squish a pile of ants in 4 minutes. To squish two piles of ants, it'll take 8 minutes.[/hide]


function
real analysis
real analysis solved

I was thinking about f(f(x))=-x equation on the reals.

 I have proved [url=http://www.mathlinks.ro/viewtopic.php?p=12655#12655][HERE][/url] that such a function exists, but I wonder if there is one which is continuous? or at least has the Intermediate Value Property ... :?

Come on, Valentin! If such a function had the Darboux property, then since it is injective it would have been strictly monotonuos. Thus, f^2 (the composite between f and f) would be strictly increasing, false.

:blush: I admit ... I didn't think about it at all, and you are obivously right ... :blush:.

You can move this to theorems and formulas section:
Could you tell me "Darboux property"?

Darboux property here

http://www.mathlinks.ro/viewtopic.php?t=4125


geometry
perpendicular bisector
power of a point
Pythagorean Theorem

If a chord is the perpendicular bisector of a radius of length 12 in a circle, what is the length of the chord ?
Thank you for anyone who explains.

[hide="Hint"]

Try Power of a Point

[/hide]

[hide="Solution"]

Let the circle's center be $ O$. Let the diameter be $ AB$, and the perpendicular bisector of $ AO$ be $ XY$. Also, let the intersection of $ AO$ and $ XY$ be $ Z$.

By HL Congruence, $ \triangle{XOZ}\cong\triangle{YOZ}$. So, $ XZ \equal{} YZ$. Now, if we apply Power of a Point to Point $ Z$, we have:

$ (BZ)(AZ) \equal{} (XZ)(ZY) \equal{} 6\times18 \equal{} XZ^2$

Hence, $ XZ \equal{} 6\sqrt {3}$. So, the length of the chord is two times that, or $ 12\sqrt {3}$

[/hide]

Edit: woops

XZ = XY ???

What?

I think it is XZ = ZY.

[hide="Another Solution:"]

Let point O be the center of the circle.
Let X and Y be the endpoints of the chord.
Let OA be the radius which is perpendicularly bisected by the chord XY.
Let the intersection be Z.

First prove triangle XOZ is congruent to triangle YOZ using HL congruence like above post. Because the triangles are congruent,
XZ = ZY.

Since the radius is bisected,
OZ = ZA = 6.

Both OX and OY are also radii of circle O, thus,
OA = OX = OY = 12.

Lastly use Pythagorean theorem to solve for XZ and multiply by 2 to get the length of the chord.[/hide]


My soccer team has 16 players.  I have to choose a starting lineup of a goalie and 10 regular players (the regular players are interchangeable).  How many different starting lineups can I choose?

First, let's pick a goalie.  There are $ 16$ ways to do this.  Then there are $ \binom{15}{10}\equal{}3003$ ways to choose the regular players.  The final answer is $ 16 \times 3003\equal{}\boxed{48048}$.

[hide] First we choose the goalie, and any of the 16 people can be the goalie. Then we choose 10 more players from the remaining 15 players, which is the same as choosing a committee. The answer is \[16\binom{15}{10}=16\binom{15}{5}=16\times\frac{15\times 14\times 13\times 12\times 11}{5\times 4\times 3\times 2\times 1}=\boxed{48,\!048}.\][/hide]

[quote=titanium7]What's soccer?[/quote]

[quote=POKEMON123]Soccer, or like football in everyother country.
I take German, so I know soccer is like Fußball, or Fussball.
I used to take Spanish, so I know soccer is like Football. Football for americans is Football Americano.[/quote]

In Spanish, I'm pretty sure that soccer is fútbol... (just guessing)

[quote=skar02]Can anybody explain why the solution is not
$16$$ \binom{15}{9}$[/quote]
There are 10 regular players that need to be selected, not 10 players total. Soccer requires 11 players.

This problem is kind of stupid, because it doesn't tell you what you can do with the goalie, so I didn't know if you could change the goalie or not.

[quote=Aetheria]It would help to make it clear in the question that anyone can be a goalie, since that's not ordinary.[/quote]

Correct. I almost got this wrong because I thought there was only one goalie. I feel like this problem should be changed to specify that any of the 16 people can be the goalie.

[hide=Post #3 by titanium7][url=aops.com/community/user/152695][b]titanium7[/b][/url] · Nov 30, 2014, 2:21 PM [url=aops.com/community/p3667368](view)[/url][color=transparent]helo[/color]-----
What's soccer?

-----
[color=#5b7083] #3[color=transparent]hellloolo[/color] [aops]x[/aops] 10 [color=transparent]hellloolo[/color]  [aops]Y[/aops] 0 [color=transparent]hellloolo[/color] [/hide]

[tip=@titanium7][img]https://avatar.artofproblemsolving.com/avatar_152695.jpg[/img]
AoPS User[/tip] Football. :)

(Because we Americans are weird and have another game named football so we call this game soccer.)

[hide=Solution]We have $16\binom{15}{10}=\boxed{48048}$ ways to choose a starting lineup.


[quote="roadnottaken"]I hope the length requirements for the essays are guidelines rather than strict requirements, because one of my essays goes about 25 words over, and I would really dislike cutting out parts of it to fit the requirements.  Do I have anything to worry about?[/quote]
I went 36 words over on the second part, and I got in, so you should be fine.

That's what I thought, but I wanted to be [b][i]sure[/i][/b].


calculus
integration
algebra
polynomial
geometry
function
calculus computations

Let $ f_n(x) \equal{} x^n(1 \minus{} x)^n,\ I_n \equal{} \int_0^1 f_n(x)\ dx\ (n \equal{} 1,2,\cdots)$.

(1) Find a polynomial $ g(x)$ such that $ f_{n \plus{} 1}'(x) \equal{} (n \plus{} 1)f_n(x)g(x)$.

(2) Find constant numbers $ A_n,\ B_n$ such that $ f_{n \plus{} 2}''(x) \equal{} A_nf_{n \plus{} 1}(x) \plus{} B_nf_n(x)$.

(3) Find $ A_nI_{n \plus{} 1} \plus{} B_nI_n$.

(4) Let $ J_n \equal{} (2n \plus{} 1)!I_n$. Express $ J_{n \plus{} 1}$ in terms of $ J_n$.

(5) Find $ I_n$.

A computation attributed to Wallis, from the generation before Newton.

First, establish the pattern above, namely that
\[ \int_0^1x^n(1 \minus{} x)^n\,dx \equal{} \frac {(n!)^2}{(2n \plus{} 1)!}.
\]
(That's much harder if calculus hasn't been invented yet, but not impossible.)

Next: be so bold as to be willing to let $ n \equal{} \frac12.$

$ \int_0^1\sqrt {x(1 \minus{} x)}\,dx \equal{} \frac {\pi}8,$ as the area of the circle is known.

Hence, $ \frac {\left(\left(\frac12\right)!\right)^2}{2!} \equal{} \frac {\pi}8.$

Rearrange this to get $ \left(\frac12\right)! \equal{} \frac {\sqrt {\pi}}2.$

This predates Euler's definition of the gamma function by a number of decades; in retrospect, it is consistent with Euler.

[quote="Kent Merryfield"]
First, establish the pattern above, namely that
\[ \int_0^1x^2(1 \minus{} x)^n\,dx \equal{} \frac {(n!)^2}{(2n \plus{} 1)!}.
\]
[/quote]
Slight typo: it should be $ x^n$ instead $ x^2.$

[color=green][OK. Edited now. - K.M.][/color]


trigonometry ARML geometry AMC AIME ratio calculus

For all of you here like myself who want to practice for NYSML but can't find any problems anywhere, I will be creating a mock contest.

I will be posting it later today.

Please PM me your answers by Tuesday if you want your name posted here and eternal glory.

Please feel free to comment on the relative difficulty compared to past NYSML contests since this is my first ever mock contest.

If everyone's OK with it, the problems will be here soon.

You have 10 minutes for each pair of problems. Good luck!

There was a slight error in the wording of problem 7. The file should be fixed now.

thanks for making it.

Are all answers integers? What form am I suppose to leave the answer in?

No, not all the answers are integers.

The rules are similar to the official NYSML rules: "... always call for an answer in simplest form. Thus, final answers like $\frac 64$, ... $2^5$, and $2\sin 30^\circ$, for example, would not be satisfactory. In cases where there is a question as to what is "most simplified", alternate answers may be accepted. For example, $\frac 32$ and $1\frac 12$ are both acceptable."

So I guess all answers must be in "simplest radical form."

[quote="JSteinhardt"][quote="joml88"][quote="Altheman"]what is nysml???[/quote]

New York State Math League

It has the same format as ARML.

http://www.nysml.org/?page_id=2[/quote]

Is this the same as NYCIML?[/quote]

I think NYCIML is the New York City Interscholastic Math League.  Obviously, that's for people in New York City.

NYSML is the New York [i]State[/i] Math League, which isn't just for New York City people.

As nice as this topic is (and NYSML is close to my heart, as I'm from New York), it has nothing to do with the AMC, so I've moved this topic here to the ARML forum.

NYCIML is between schools in the city, pretty much used (along with Math League and the AMC/AIME) to determine who goes to NYSML and ARML as part of the NYC team. it runs at schools almost every week and is very much like ARML individual, except only 6 questions and usually easier.
NYSML is very much like ARML (ARML actually was modeled after NYSML)

Yeah, you're right, I'm an idiot.

Except it's $1-49/63$, but you got the point.

There were seven people who submitted this test.

The top scorers:

pilot: 7
frankthetank: 6
Xantos C. Guin: 6
Elemenop: 6
surge: 6

The score distribution is:

8:
7: *
6: ****
5: *
4:
3: *
2:
1:
0:

The number of people who got each respective question right:

1: *****
2: *****
3: ****
4: ******
5: *******
6: ******
7: *****
8: *

Ooh, a line through the center of mass bisects the area? cool solution, need to remember that for future use

Yes, and that could have been used to solve AIME 1, #8. I did try that, but unfortuately I got the wrong answer due to a stupid mistake.

Yes, that is certainly true. All lines in the plane going through the center of mass divide a plane figure into two pieces of equal area.

Nice mock!  I just took it cause the problems looked pretty interesting.  The problems were a little too quick, but maybe thats what the contest is supposed to be like.  I probably could've done number 6.. but doing that synthetic division looked ugly, so I just moved on.  I got a 6, missing number 6 and 8.  I saw a similar problem to 8 during a Mu Alpha Theta Interschool.

Careless error on #7, and lack of knowledge on #8. The others seemed like freebies.

[quote="Philip_Leszczynski"]Yes, that is certainly true. All lines in the plane going through the center of mass divide a plane figure into two pieces of equal area.[/quote]

Actually, this is false.  The center of mass of an equilateral triangle is its center (duh).  A line constructed parallel to one of the sides of the triangle divides it into two pieces whose areas are in the ratio 4:5.

In fact, the answer to your question is wrong, by an error slightly above 0.03.  (I don't have an exact answer to the question, by the way -- I suspect that it is incredibly ugly.)

I guess you're right... that's very strange.

I'm sure I read a theorem like that somewhere in my calculus textbook....

Okay, I guess we can all disregard problem 8 then.

JBL, is the center of mass to #8 the point $\left(\frac12-\frac{2}{3\pi},\frac12\right)$?

I don't know, I haven't tried to calculate it :-).  Quite possibly not.  I'll go do some calculatin' now.

Two theorems that I just worked out with a couple of friends over lunch (one of them a former star of the Nassau NYSML/ARML team):

Theorem 1:  If plane region A is bounded by a continuous curve and is [url=http://mathworld.wolfram.com/SimplyConnected.html]simply connected[/url], then:
There exists a point P such that any line through P divides A into two regions of equal area
if and only if
A has point-symmetry around P.  (As a result, P will be the center of mass of A.)

Theorem 2:  If A is a plane region bounded by continuous curve or curves, and there is no line $\overline{PQRST}$ such that P, R and T are in A while Q and S are not in A, then:
Any line passing through the center of mass of A divides A into two pieces of equal areas
if and only if
A has point symmetry around its center of mass.

Proof of Theorem 1:  Suppose A does not have point-symmetry around P (the other direction being trivial).  Then there are points X and Y on the boundary of A such that $\overline{XPY}$ and $XP > PY$.  By continuity,* we can take two lines through P, each passing very very close to X and Y, such that the sliver on the X side has a larger area than the sliver on the Y side.  But then the two lines can't both cut A into portions of equal area, and we're done.

* This requires an epsilon-delta argument from the definition of continuity.  I'm omitting it because I'm lazy.  I think this is fairly intuitive, however, if you draw a diagram.

I'm confident that theorem 2 is true, although I haven't tried to formally prove it -- the basic gist is that you can't preserve both the areas and the center of mass without adding an extra hole somewhere.

[i]Editted to clean up a couple typos, add some more flavor text  ;) [/i]

Actually, I note that the center-of-mass calculation in the solution to number 8 is flawed because it relies on the same non-theorem.  (A good counter-example, by the way, to get a line through a center of mass which divides a region into an arbitrarily bad ratio: take a circle of radius $n$ centered at (0, 0) and a circle of radius 1 centered at $(2n^3+2n, 0)$.  The center of mass is at $(2n, 0)$, and the line $x = 2n$ divides the region into two sections whose areas are in the ratio $n^2 : 1$.  Informally: a little mass, far enough away, makes the center of mass go whereever you want it to.)

10000th user, you correctly calculated the centers of mass of the two pieces -- however, you don't just add them to get the total center of mass, you have to weight them by the mass they are counting for.  Thus, the x-coordinate of the center of mass is $-\frac2{3\pi}\cdot\frac\pi 8 + \frac12 \cdot 1 = \frac5{12}$

Thank you for the reply, JBL. I'm trying to absorb as much as I can from your replies. And, I do agree with your answer, the semicircle has a smaller mass, which I didn't take into account.

[EDIT] I have a question: what does point-symmetry mean? Do you mean: if P has point-symmetry, then if X is a point in D, then there exists another point Y in D such that P is the midpoint of segment XY, XP=PY?

Yes, that's pretty much it.  For a 2-D region point-symmetry is the same as 180-degree rotational symmetry.  In general, one says two points P and P' are symmetric about O if O is the midpoint of segment PP', and one says a figure F has point symmetry if there is a point O such that, for every pair of points symmetric about O, either both are in F or both are not in F.  (This is almost exactly what you said, but this post would look silly if I had just stopped after the first sentence :-).)

Should I provide a fuller proof of Theorem 1?

all problems have been posted in separate threads in HSM,  I just created the related post collection

[url=https://artofproblemsolving.com/community/c3741313_2006_mock_nysml]here[/url]

enjoy  / start solving

USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
ARML mocks [url=https://artofproblemsolving.com/community/c2442143_arml_mocks]here[/url], (I added it here due to the similarity)
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)


geometry
rhombus
Putnam
combinatorics proposed
combinatorics

We are given $n$ distinct points in the plane. Consider the number $\tau(n)$ of segments of length 1 joining pairs of these points. Show that $\tau(n)\leq \frac{n^{2}}3$.

Consider a graph $G$ with $n$ verteces and two verteces $A$ and $B$ are connected iff $AB=1.$ It's easily seen  that there is no 4-clique. Then by weaker version of Turan's theorem we have that the number of edges is at most $\frac{n^{2}}{2}(1-\frac{1}{4-1})= \frac{n^{2}}{3}.$ For proof of Turan's theorem see: http://planetmath.org/?op=getobj&from=objects&id=3080

[quote="silouan"][quote="bilarev"] It's easily seen  that there is no 4-clique. [/quote]

Could you explain why this holds ?Thank you[/quote]
Suppose there is a 4-clique. Let $A$, $B$, $C$ and $D$ be these points. Then triangle $ABC$ is equilateral as well as $BCD$. Since $A\neq D$, then $ABCD$ is a rhombus with angles $60$ and $120$. Then $AD$ isn't equal to 1. Contradiction.

Thank you very much , it was simple but I couldn't see it . :wallbash_red:

Now try to prove a better bound, i.e. $\tau(n) < 2 \sqrt{n^{3}}$ (Putnam, 1978).
Also, what's the best bound known to date?

[quote="perfect_radio"]Now try to prove a better bound, i.e. $ \tau(n) < 2 \sqrt {n^{3}}$ (Putnam, 1978).
Also, what's the best bound known to date?[/quote]
If I'm not mistaken Erdos (1946) proved that $ \tau(n)<n^\frac{3}{2}$.

Spencer, Szemeredi, and Trotter (1984) showed an upper bound of $ O(n^{4/3})$.  Later, Szekely (1997) found an elegant proof of their result using crossing numbers, which you can find [url=http://en.wikipedia.org/wiki/Szemer%C3%A9di-Trotter_theorem]here[/url].


There are four lines in the plane, no three concurrent, no two parallel, and no three forming an equilateral triangle. if one of them is parallel to the Euler line of the triangle formed by the other three lines, prove that a similiar statement holds for each of the other lines.

Posted before at [url]http://www.mathlinks.ro/viewtopic.php?t=113237[/url] but with no solution.

Let $ G, H$ be the centroid and orthocenter of a $ \triangle ABC$ and let the Euler line $ GH$ meet the sideline $ BC$ at a point $ D.$ Let a parallel $ b_1$ to $ GH$ through $ A$ meet $ BC$ at $ C_1.$ Let $ H_1 \in AH$ be the orthocenter of the $ \triangle HDB.$ Then $ DH_1 \parallel CA$ (both perpendicular to $ BH$). Since $ (AH_1 \equiv AH) \perp (BC \equiv BC_1)$ and $ BH_1 \perp (DH \parallel C_1A),$ $ H_1$ is also the orthocenter of the $ \triangle ABC_1.$ Let a parallel to $ BC \equiv BC_1$ through $ G$ meet the line $ DH_1$ at a point $ G_1.$ The $ \triangle AC_1C \sim \triangle DGG_1$ are (centrally) similar, having the corresponding sides parallel. Their similarity coefficient is -3 (because $ G$ is the centroid of the $ \triangle ABC$ and their corresponding sides are oppositely oriented). Therefore, $ \overline{C_1C} \equal{} \minus{}3\ \overline{GG_1} \equal{} 3\ \overline{G_1G}$ and $ G_1$ is the centroid of the $ \triangle ABC_1.$ As a result, $ (G_1H_1 \equiv DH_1) \parallel CA$ is the Euler line of the $ \triangle ABC_1.$

If $ b_2$ is any other line parallel to the Euler line $ GH$ of the $ \triangle ABC$ (different from $ GH, b_1 \equiv C_1A$), intersecting $ AB, BC$ at $ A_2, C_2,$ the $ \triangle A_2BC_2 \sim \triangle ABC_1$ are centrally similar with center $ B$ and their Euler lines $ G_2H_2 \parallel G_1H_1$ are parallel, hence $ G_2H_2 \parallel CA$ as well.

Let the Euler line $ GH$ of the $ \triangle ABC$ meet $ CA, AB$ at points $ E, F.$ Let a parallel $ a'$ to $ BC$ through $ E$ meet $ AB$ at $ B'.$ The $ \triangle AB'E \sim \triangle ABC$ are centrally similar with center $ A,$ hence, their Euler lines $ G'H' \parallel GH$ are parallel. In exactly the same way as above, we can show that the Euler lines $ G'H', G_1'H_1'$ of the $ \triangle AB'E, \triangle AFE$ meet on $ AB'$ and that $ G_1'H_1' \parallel B'E \parallel BC.$

[b]The theorem of Zeeman[/b], view:
Goormaghtigh: "SUR LE POINT DE HERVEY D’UN QUADRILATERE ET SUR LES QUADRILATERES DE ZEEMAN" (MATHESIS, 1947 p.175-178)
and
Brabant: "Démonstration  élémentaire d’un théorème de P.ZEEMAN" (MATHESIS: 1952, p.107-108)
and
Neuberg: "SUR  LA DROITE  D’EULER  D’UN  TRIANGLE" (MATHESIS, 1908, p.233-236)
and
Neuberg: :Deux théorèmes sur la droite d’Euler"(MATHESIS, 1903, p.60-61)
or
Blanchard: "Démonstration élémentaire d’un théorème de P.ZEEMAN"(MATHESIS, 1953; p.323-324).

[b]Problem:-[/b]

Suppose, $ ABC $ and $ AB'C' $ are the two triangles, where $ B' $ lies on $ AB $ and $ C' $ on $ AC $. Given that Euler line of $ ABC $(call it $ OH $) is parallel to $ B'C' $. We have to prove that Euler line of $ AB'C' $(call it $ O'H' $) is parallel to $ BC $.

[b]Solution:-[/b]

Consider the triangle, $ AOH $.
Note that, $ AO' $ and $ AH' $ are isogonal conjugates wrt $ \angle HAO $ and also note that, $ \frac {AO}{AH}=\frac {AO'}{AH'} $. Its given that $ AH'\perp OH $ and we have to prove that $ AH\perp O'H' $.
Suppose, $ K $ is the orthocenter of $ AOH $. Clearly, $ K $ lies on $ AH' $. Suppose, $ AO'\cap OK=L $. Then note that $ \Delta AHK $ and $ \Delta AOL $ are similar. So $ \frac {AK}{AL}=\frac {AH}{AO}=\frac {AH'}{AO'} $. So $ O'H'\parallel KL $. So $ O'H'\perp AH $. So done.


The distance between the two parallel plates of a capacitor is $d = 1cm$, and each plate has a surface area of $S = 500cm^{2}$. A potential difference of 2000 V is applied between the plates, which are then isolated after equilibrium.

(a) Compute the energy stored in the capacitor.
(b) A metal sheet 2mm thick, uncharged and isolated, is introduced parallely in the middle of the capacitor, so that the distance between each plate and the metal sheet is the same. Compute the new capacitance of the capacitor.

It's quite easy.But I've no time to post the solution.Can anyone else post it?

[hide="Solution"](a) The energy stored on a capacitor is given by $E = \frac{1}{2}CV^{2}$, where C is the capacitance and V the potential difference across the plates. But the capacitance is given by $C = \frac{\epsilon_{0}S}{d}$, where S is surface area of each plate and d the distance between the plates.

(b) When a metal plate is inserted between the plates of a capacitor, the resultant system behaves like two capacitors connected in series.[/hide]


geometry
geometric transformation
rotation
perimeter

[asy]size(125); pair O,A,B,C,D,E,F,G,H; O=origin; A=2*dir(120); B=4*dir(120); C=6*dir(120); D=8*dir(120); E=2*dir(60); F=4*dir(60); G=6*dir(60); H=8*dir(60); draw(circle(O,1),red); draw(circle(A,1)); draw(circle(B,1)); draw(circle(C,1)); draw(circle(D,1)); draw(circle(E,1)); draw(circle(F,1)); draw(circle(G,1)); draw(circle(H,1));[/asy]
The red circle starts turning around the other circles, always remaining tangent to at least one circle. When it gets back into its original position, how many rotations has it made?

Two questions: Which path does the circle take? So that it touches all other circles, or goes down one "branch"?

Also, when a the red circle goes between two other circles, does it "push" the other circles apart? Like this:
[asy]size(125); pair O,A,B,C,D,E,F,G,H; O=origin; A=2*dir(120); B=4*dir(120); C=6*dir(120); D=8*dir(120); E=2*dir(60); F=4*dir(60); G=6*dir(60); H=8*dir(60); draw(circle(O,1)); draw(circle(A,1)); draw(circle(B,1),red); draw(circle(C,1)); draw(circle(D,1)); draw(circle(E,1)); draw(circle(F,1)); draw(circle(G,1)); draw(circle(H,1));[/asy]

The other circles remain stationary, while the red circle turns around their combined "perimeter".

The circle starts out going either left or right (it doesn't matter), then turns about the upper-most circle, and continues going down. Then, when it gets to the "bottom" of the V, it goes back up and around the other branch.

[quote="i_like_pie"]Anybody? :maybe:[/quote]

You only posted it yesterday...

[hide="My crack at it (probably wrong)"]Every time the circle moves to be tangent to the next pair, it makes a $\frac{1}{3}$ rotation.
Also, when it gets to the "edge", it makes a $\frac{2}{3}$ rotation
When it gets to the "corner", it makes a $\frac{1}{6}$ rotation
So counting yields: $\frac{29}{6}$.[/hide]

[hide="possible answer"]$\frac72$?[/hide]

[hide="possibel argument"]on the first move it makes a 1/3 rotation. from then onwards until it gets to the edge it makes two 1/6 rotations. then at the edge it makes a 2/3 rotation. after that it makes two more 1/6 rotations. then at the corner it makes a 1/6 rotation. then it repeats the whole process. so summing we get the desired answer. [/hide]

[hide]
well if you draw a "layer" of tangent circles around the outside, you can see that from one tangent circle to the next, a circle has to travel over a $60^\circ$ arc. When counted, there are 20 of these so $60^\circ\cdot20=1200^\circ$.

Think of this as a circle rolling over the outside of another congruent circle. $\frac{1200^\circ}{360^\circ}\cdot2=\boxed{\frac{20}{3}}$ turns

ehh that was a little too confusing.. a picture would help.
[/hide]

ehh.. sorry for the confusion.. maybe i should have posted another topic.. :blush:
[hide="Extension:"]Let the red circle have a radius that is $\frac{2\sqrt{3}-3}{3}$ times the radius of the other 8 circles. The red circle starts turning around the other circles, always remaining tangent to at least one circle. When it gets back into its original position, it has made $\frac{a\sqrt{b}+c}{d}$ rotations. Find $a+b+c+d$.
[asy]size(125); pair O,A,B,C,D,E,F,G,H, K; real r; r = (2*sqrt(3)-3)/3; O=origin; A=2*dir(120); B=4*dir(120); C=6*dir(120); D=8*dir(120); E=2*dir(60); F=4*dir(60); G=6*dir(60); H=8*dir(60); draw(circle(A,1)); draw(circle(B,1)); draw(circle(C,1)); draw(circle(D,1)); draw(circle(E,1)); draw(circle(F,1)); draw(circle(G,1)); draw(circle(H,1)); K = intersectionpoints(circle(A,r+1),circle(E,r+1))[1]; draw(circle(K,r),red);[/asy]
( :maybe: I'm pretty sure that's the right format.. please correct me if it's wrong)[/hide]

I see that the perimeter of this figure can be broken up into two parts:

1. The upper right and upper left circles.
2. The rest of the circles

 This is because the red circle will not roll over itself , and we can assume that all the circles are congruent. Since all [b]except[/b] the upper right and upper left circles have the same amount of exposed circumference(visible at the pixel level), we can guess at the  viable perimeter that the red circle can cover around the figure:

1. Approx. $\frac{2}{3}$ of the circumference is usable.

2. Approx. $\frac{1}{3}$ of the circumference is usable.

So,

$Let \ x \ = \ the \ circumference \ of \ a \ circle$

$P_{figure}\ = \ 2(\frac{2}{3}x)+6(\frac{1}{3}x)$
$\newline = \ \frac{4}{3}x+\frac{6}{3}x\newline \newline = \ \frac{10}{3}x$

 Since it seems like it is asking how many rotations the red circle makes(remember, it is rolling):

$Let \ N \ = \ the \ number \ of \ rotations$

$N_{rotations}\ = \ (P_{figure}) \div (x)$
$\newline = \ \frac{\frac{10}{3}x}{x}$

 So, if we knew the circumference, we could find out the number of rotations that the red circle makes as is rolls over the perimeter of the figure.

P.S.: First post :)

[quote="i_like_pie"]All the circles are congruent.[/quote]uh..  :( sorry i think that was misleading, but i meant to propose a similar problem (like an extension) in the same topic..  :| but if you want me to post it in a separate topic, then i will do that.

but the red circle doesn't roll over the whole perimeter, it leaves some gaps.

Which was accounted for when I went through it.

[quote="undefined117"][hide]
Refer to the attached picture. There are 7 of these points, $\frac{360^\circ\cdot8-120^\circ\cdot7}{360^\circ\cdot8}=\frac{17}{24}$ of the perimeter is passed over by the small circle.

We'll assume the larger circle's radius is 1, for simplicity.

Larger circle's circumference is $2\pi$ so the distance the smaller circle rolls over is $2\pi\cdot8\cdot\frac{17}{24}=\frac{34}{3}\pi$
Smaller circle's circumference is $\frac{4\sqrt{3}-6}{3}\pi$

So $\frac{\frac{34}{3}\pi}{\frac{4\sqrt{3}-6}{3}\pi}=\frac{34\sqrt{3}+51}{3}$

Therefore $a+b+c+d=34+3+51+3=\boxed{91}$

But I still feel like I missed something here.. when a coin rolls over another stationary coin, the coin in motion goes through [b]two[/b] rotations, not one.. how can that be applied to this?
[/hide][/quote]you have to add the number of rotations that the bigger circle goes through. what you calculated here is the number of rotations that the smaller circle does when if rolls through a distance of $\frac{34\pi}{3}$, but then it is rotating around the larger circle. you can think of it as equivalent to the small circle rotating around the large circle $\frac{17}{3}$ times or the large circle rotating around the small circle $\frac{17}{2\sqrt{3}-3}$ times..
[hide="image"][asy]size(125); pair O,A,B,C,D,E,F,G,H, K; real r; int i, n=10; r = (2*sqrt(3)-3)/3; draw(circle((0,0),1)); for(i=0;i<n;i=i+1){ draw(circle((1+r)*dir(360*i/n),r)); }[/asy][asy]size(125); pair O,A,B,C,D,E,F,G,H, K; real r; int i, n=10; r = (2*sqrt(3)-3)/3; draw(circle((0,0),r)); for(i=0;i<n;i=i+1){ draw(circle((1+r)*dir(360*i/n),1)); }[/asy][/hide]
if you think of the circles like gears, both circles rotating at the same time, the large circle makes $\frac{17}{3}$ rotations while the small circle makes $\frac{17}{2\sqrt{3}-3}$ rotations.
[asy]size(125); pair O,A,B,C,D,E,F,G,H, K; real r; int i, n=10; r = (2*sqrt(3)-3)/3; draw(circle((0,0),1)); draw(circle((1+r,0),r));[/asy]
but from here, if you hold the large circle down so that it doesn't rotate, the small circle has to account for the large circle's rotations. so i think the total number of rotations the smaller circle makes around itself is $\frac{17}{2\sqrt{3}-3}+\frac{17}{3}$. so the answer is $\frac{34\sqrt{3}+68}{3}$

( :D um... i think i just made my post so long and so confusing because i like asy)

What big and small circles? They are all congruent!

minsoens was talking about his extension to the original problem when the radius of the red circle is $\frac{2\sqrt{3}-3}{3}$ that of the other black circles..

EDIT: ok, i think i get what you mean, minsoens..

so the entire "surface" (the large circle) "rotates" $\frac{17}{3}$ times, so that is also added to the smaller circle's rotation.. (attempt at visualization - i'm a visual learner :D )

and i'm guessing this works for any shape, not just a circle, right?

EDIT#2: so the answer is $34+3+68+3=\boxed{108}$


I'm using TeXnicCenter and wish to create a document with a large number of graphic images and like using the YAP previewer so that I have the nice option of double-clicking on the document to reach the code quickly.  However it seems that YAP will only accept EPS files which seems to be very large and slow down the display dramatically and I'd rather not comment-out the images so that I know what the document looks like.  How does one implement effectively the debug-build-run cycle with a LaTex document containing a large number of graphic images?  Is commenting-out the only option?  All the images I create are done in Mathematica and can be saved in other formats such as PDF,JPG,PNIG,GIF which are much smaller in size but don't seem to be accepted by YAP.  For example here are the sizes created by Mathematica for the same image:

EPS  10Mb
PDF  1.5Mb
JPG  31Kb
PNG  32Kb
GIF  17Kb

This is a typical construct I use in my LaTex document for an image:

[code]
\begin{figure}[htbp]
	\centering
	\includegraphics[scale=.6]{figure4.eps}
	\caption{Disc of convergence for equation $\ref{eqn60}$ shown in red superimposed on log multi-sheet}
	\label{fig4}
\end{figure}
[/code]

Thanks,
Shaw

You could use the draft option fo the graphicx package which gives the final layout but just shows a box where the image is to go. But, if the quality of the images are satisfactory, I would use jpg, gif or png and convert to PDF thus avoiding the size problem.

Hello Steve.  Thanks for replying.  If I convert to PDF then the Adobe file reader is used and then I loose the double-click option of YAP which allows me to quickly go to the point in the source code where I double-clicked.  Or is their an equivalent option for the Adobe reader?

[url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=vpe]Source specials for PDF output[/url]

Ok.   Looks . . . well they all look like that.  Something I'll have to more consider a bit later.  Thanks a bunch though! :)

I did try the "draft" option  with the graphic package.  Apparently options go in brackets:
\usepackage[draft]{graphicx}

Works ok and it still displays the caption of the figure so can still see what's suppose to be there.

Just a final note and perhaps this is what Steve was talking about.  I'm reading the graphicx User Guide:  You can locally set each graphics object to "draft" mode instead of all of them.  Instead of using the command above, insert "draft" into the includegraphics command that you want excluded:

[code]
\begin{figure}[htbp]
	\centering
	\includegraphics[scale=.6,draft]{figure4.eps}
	\caption{Disc of convergence for equation $\ref{eqn60}$ shown in red superimposed on log multi-sheet}
	\label{fig4}
\end{figure}
[/code]

This figure is now excluded with a box drawn in it's place.  So then I see one method of managing a large number of graphic files:

(1) First display the image to check it's working.

(2) Once it's working, use the "draft" command like the above to exclude its inclusion into the document each time the file is built.  The caption is still displayed so you have some idea what it is.  That way can still have the speed of developing a document without graphics.


Columbia
geometry
3D geometry
inequalities
national olympiad

Here is the first day of our selection tst:

1.Let $a,b,c$ be integers such that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=3$ prove that $abc$ is a perfect cube!
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=251396#251396]here[/url]

2. In a country there are $n>3$ cities. We can add a street between $2$ cities $A$ and $B$, iff there exist $2$ cities $X$ and $Y$ different from $A$ and $B$  such that there is no street between $A$ and $X$, $X$ and $Y$, and $Y$ and $B$. Find the biggest number of streets one can construct!
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=251391#p251391]here[/url]

3.Let $A_{1}A_{2}A_{3}...A_{n}$ Let $B_{1}$ and $B_{n}$ be the midpoints of $A_{1}A_{2}$ and $A_{n-1}$ the midpoint of $A_{n-1}A_{n}$. And for $i=2,3,4...,n-1$, if $S$ is the intersection of $A_{1}A_{i+1}$ and $A_{n}A_{i}$ then $B_{i}$ is the intersection of the bisector of $<A_{i}SA_{i+1}$ and $A_{i}A_{i+1}$. Prove that:
$\sum_{i=1}^{n-1}A_{1}B_{i}A_{n}=180$
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=251393#251393]here[/url]

And secound day:

1. Prove the following inequality for positive reals $a_1,a_2...,a_n$ and $b_1,b_2...,b_n$:

$(\sum a_i)(\sum b_i)\geq (\sum a_i+b_i)(\sum\frac{a_ib_i}{a_i+b_i})$
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=251897#251897[/url]

2. Same as Moldova Test, and German Test (Moldova Tst problem 1 first test)
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=24078[/url]

3. $A$ and $B$ play a game, given an integer $N$, $A$ writes down $1$ first, then every player sees the last number written and if it is $n$ then in his turn he writes $n+1$ or $2n$, but his number cannot be bigger than $N$. The player who writes $N$ wins!, For wich values of $N$ does $B$ wins?
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=251895#p251895[/url]

And here is part of the team: there are 3 students already chose:

1: Pascual restrepo Mesa (ME!!!!!!!)
2: Jose Gabriel Acevedo
3: Esteban Gonzales

The rest of the team will be choose the 10 and 11 of january!

[quote="Pascual2005"] The rest of the team will be choose the 10 and 11 of january! [/quote]

Considering when your team fellows will be chosen are you sure that there are not only three Columbian people going to IMO?!  :D

Hehe, yeah january is a bit away :)

Congrats!

ooooooooooops  I mean June! :blush:  :blush:

Now the full team must have been announced! So who are the other team members?


What is Apollonius theorem? How to prove it?

you can find that in the topic
http://www.mathlinks.ro/Forum/viewtopic.php?t=116059

Nicer is the (practically as easy to prove) general case, Stewart's Theorem, which states that
$mb^{2}+nc^{2}=a(p^{2}+mn)$

Where a,b,c are the sides of a triangle ABC (a=BC, b=CA, c=AB), and where P is a foot of a cevian from A onto BC such that AP=p, BP=m, CP=n.

Setting m=n=a/2 clearly gives $\frac{1}{4}a(2b^{2}+2c^{2}-a)=ap^{2}$ or $\frac{1}{4}(2b^{2}+2c^{2}-a)=p^{2}$, which is Apollonius' Theorem.

There are also Apollonius circles of various kinds - for those see [url=http://mathworld.wolfram.com/ApolloniusCircle.html]this[/url]


probability
geometry
trigonometry
integration
probability and stats

I have these two probability problems that I have been working on frustratingly. Please help.

P1. Planet X is a ball with center O. Three spaceships A, B and C land at random on its surface, their positions being independent and each uniformly distributed on the surface. Spaceships A and B can communicate directly by radio if angle AOB < 90 degrees. Show that the probability that they can keep in touch (with, for example, A communicating with B via C if necessary) is $ (\pi \plus{} 2)/4\pi$.

P2. Let G be the free group with two generators a and b. Start at time 0 with the unit element 1, the empty word. At each second multiply the current word on the right by one of the four elements $ a,a^{ \minus{} 1},b,b^{ \minus{} 1}$, choosing each with probability 1/4 (independently of previous choices). The choices $ a,a,b,a^{ \minus{} 1},a,b^{ \minus{} 1},a^{ \minus{} 1},a,b$ at times 1 to 9 will produce the reduced word $ aab$ of length 3 at time 9. Prove that the probability that the reduced word 1 ever occurs at a positive time is 1/3, and explain why it is intuitively clear that (almost surely):
(length of reduced word at time $ n$)/$ n\rightarrow\frac {1}{2}$.

Thanks in advance,
FTR

The first one can't possibly be right: the probability that A and B can communicate directly is already $ \frac{1}{2} > \frac{\pi \plus{} 2}{4\pi}$.

question 2:

hint for the first part: $ \frac {1}{4} \plus{} \frac {1}{4^2} \plus{} \frac {1}{4^3} \plus{} \cdots \equal{} \frac {1}{3}$

(you can work recursively, using the fact that having the reduced word $ 1$ for the first time at time $ 2n$ requires that the first and last element are inverses of each other)

hint for the second part.  just imagine that you are summing i.i.d bernoulli random variables which have $ P(X_n \equal{} \plus{} 1) \equal{} \frac {3}{4}$ and $ P(X_n \equal{} \minus{} 1) \equal{} \frac {1}{4}$, and use the slln.

the reason, of course, that they are not 'really' identically distributed is that if $ S_n\equal{}0$, you have to add to it $ X'_n$, with $ P(X'_n\equal{}\plus{}1)\equal{}1$.  But you only hit the origin finitely many times a.s., so ...

P2: What is the probability that the new character increases the length of the reduced word? What is the probability that it decreases the length of the reduced word?

P1: All right, it's wrong. What can we say about the probabilities anyway?
There are two questions worth asking- whether a particular pair $ A,B$ can communicate, or whether all three can communicate.
I assume of course that by "random on the surface" you mean uniformly in area. That is the canonical choice here.

If we choose positions for all three spaceships, consider also the antipodal choices. Switching $ A$ with its antipode flips whether $ (A,B)$ can communicate directly and also whether $ (A,C)$ can communicate directly (we ignore $ 90^\circ$ angles as they have zero probability). Over the eight choices, we get two each in four configurations: the original, and the original with exactly two of these links flipped.
Over all choices of positions, there are two possibilities here: either an odd number of links work in each configuration, or an even number work in each configuration. In the first case, all three can communicate with probability $ \frac14$ (when all three links work) and each pair can communicate with probability $ \frac12$ (when their link works). In the second case, all three can communicate with probability $ \frac34$ and each pair can communicate with probability $ \frac34$. Two links make a complete network, while no links isolate all. Let $ p$ be the probability of the first case, that one configuration has all three central angles acute. We wish to find $ p$.

WLOG, place $ A$ and $ B$ on the equator, and let the acute central angle between them (using an antipode if necessary) be $ \theta$. Choose longitude so that $ A$ has longitude zero and $ B$ has longitude $ \theta\in (0,90^\circ)$. $ C$ then has links to both $ A$ and $ B$ if its longitude is in $ (\theta\minus{}90^\circ,90^\circ)$ and to neither if it is in the opposite longitude slice $ (\theta\plus{}90^\circ,270^\circ)$. The probability of being in one of these slices is $ p(\theta)\equal{}\frac{180^\circ\minus{}\theta}{180^\circ}$. Our desired value $ p$ is the average of $ p(\theta)$ as $ A$ and $ B$ run over the sphere.

To find this average of $ p(\theta)$, fix $ A$ at the north pole and let $ B$ be distributed uniformly over the northern hemisphere. $ \theta$ here is the colatitude of $ B$, $ \phi$ in the usual spherical coordinates. The area in a band of infinitesimal width $ d\phi$ at colatitude $ \phi$ is $ 2\pi \sin\phi\cdot d\phi$, so $ \theta$ has probability density $ \sin \phi$ on $ \phi\in \left(0,\frac{\pi}{2}\right)$. The average $ p$ of $ p(\theta)$ is $ \int_0^{\frac{\pi}{2}}p(\phi)\cdot\sin\phi\,d\phi\equal{}\int_0^{\frac{\pi}{2}}\left(1\minus{}\frac{\phi}{\pi}\right)\sin\phi\,d\phi$
$ \equal{}1\minus{}\frac1{\pi}\left[\minus{}\phi\cos\phi\plus{}\sin\phi\right]_0^{\pi/2}\equal{}1\minus{}\frac1{\pi}$

We can now finally answer the original questions. The probability that $ A$ and $ B$ somehow communicate is $ \frac12p\plus{}\frac34(1\minus{}p)\equal{}\frac12\plus{}\frac1{4\pi}$. The probability that all three ships communicate is $ \frac14p\plus{}\frac34(1\minus{}p)\equal{}\frac14\plus{}\frac1{2\pi}$.
Maybe you should have been clearer about "they"?


Can anyone recommend a good online book on polynomials (classical text are out of reach to me in printed form :( ) ?

"Polynomials" by EJ Barbeau is excellent. It is a practical guide to polynomials and all their applications. A lot of problems is given (with solutions).

I am writing a small book about polynomials myself at this moment. It will be published next year. It's a shorter introduction to polynomials than the book by Barbeau (somewhat more elementary), for high school students participating in mathematical olympiads.

I'm actually asking this question [b]because[/b] I cannot buy EJ Barbeau "Polynomials" :)
(I'm looking for an[i] online [/i]link). Anyway thanks for the interest :)

Oh, sorry! I didn't notice you wrote "online". Well, the only good online text about polynomials I know is the one written by Gabriel Carroll...

Thanks !  :D
I give the link for further reference (if anyone's else interested):
[url=http://www.people.fas.harvard.edu/~gcarroll/math/index.htm]www.people.fas.harvard.edu/~gcarroll/math/index.htm[/url]

[quote="Lovasz"][quote="Tiks"][quote="Arne"]Strange. It's likely that the page will be online again soon, but anyway, here is the text:[/quote]
What the formate is this text in? :?
I can't open it :([/quote]
It's ps file (postscript)
Try opening it by using "ghostview"
http://www.cs.wisc.edu/~ghost/ :)  :)[/quote]
Sorry,but I can't open it again :(  :? ,can you sent to me that text in PDF,or Word,file? :?

[quote="perfect_radio"]You must first install GhostScript, then GhostView. Did you do that? If yes, then it is really strange indeed...[/quote]
My computer don't want downloading and installing GhostScript :( and I don't know why :? ,mybe I did anything incorect :( .

Polynomials in PDF. I just found out that GhostView has a PDF converter :D

Edit: Any requests of the PDF will be discused via PM.

[quote="perfect_radio"]Polynomials in PDF. I just found out that GhostView has a PDF converter :D[/quote]
Thank you so much [b]perfect_radio[/b] :blush: .

does anyone know why i cannot open an pdf file in IE, but i can in mozilla?


Hey!!! Just thought this might be fun....
list your favorite book, poem, or quote!!!!! Have fun!!! =)
 :D

I have favorite book series not just books. I like Harry Potter, and Sweet Valley Twins/Junior High mainly.  :)

i also like the alex rider series and the shadow children series

O yeah I like the Shadow children books too. They're by the  author of Running Out of Time, right?

enders game

Enders Shadow>Enders Game

Bean>Ender

whats bean and yes running out of time and shadow children are both by margret haddix

gasp!?
dude...bean is ender's "shadow" in the sense that bean is uber-smart and stuff too...
bean's even IN ender's game lol

but it's definitely the whole ender's series...LOTR trilogy...Golden Compass Trilogy...HP um...septilogy?  :D

oh i thought you were refering to a different book i was thinking about enders game :blush:  but ender owns all

All above said are really good  :lol:

Tom Clancy books are great  :)  - read "The Hunt for Red October" and "Patriot Games" ATLEAST, by him.
HP of course, one that was mentioned also in the thread, Phillup Pullman's "His Dark Materials" trilogy (referred to earlier as the Golden Compass.. thats i think only the first book").  Eragon (Christopher Paulini) is a good read, but I thought it was very.. derivative of other books - not much new introduced in it. Also, if you havent read it, The Da Vinci Code is a really good read (just dont take any of it to be true - very few things in the book are). Right now I cant recall a specific other book that I remember so well (my mind.. cluttered..) - I'll edit this when I can think of others.


Find all roots to:  x^(2/3) - 8x^(1/3) - 20 = 0

confused on how to do this..

[hide]like imagenius said, make y=x^(1/3)
then the equation becomes:
y^2-8y-20=0
the roots of this are y=10 and y=-2
then substitute x^(1/3) back in for y and you have:
x^(1/3)=10 and x^(1/3)=-2
solve for x and you get the roots to be:
x=1000 and x=-8[/hide]


geometry
parallelogram
perpendicular bisector
power of a point
radical axis

In a parallelogram ABCD with $ \angle A < 90$, the circle with diameter AC meets the lines CB and CD again at E and F, respectively, and the tangent to this circle at A meets BD at P. Show that P, F, and E are collinear.

Let $O$ be the center of the parallelogram and $Q$ the orthogonal projection of $A$ on the diagonal $DB.$ Since $AF \perp DC$ and $AE \perp CB,$ the quadrilaterals $ABEQ$ and $ADFQ$ are both cyclic $\Longrightarrow$ $\angle DQF \equal{} \angle DAF \equal{} \angle BAE \equal{} \angle BQE.$  Therefore, $AQ$ and $DB$ are the internal and external bisector of $\angle EQF,$  thus the perpendicular bisector of $EF$ meets $DB$ at the midpoint of the arc $EFQ$ of $\odot(EFQ)$ $\Longrightarrow$ $OEFQ$ is cyclic. $EF$ is the radical axis of $\odot(OEFQ)$ and $\odot(O),$  $DB$ is the radical axis of $\odot(OEFQ)$ and $\odot(AQO),$ the tangent of $\odot(O)$ at $A$ is the radical axis of $(O)$ and $\odot(AQO)$ $\Longrightarrow$ $AP,DB,EF$ concur at the radical center $P$ of $(O),$ $\odot(AQO),$ $\odot(OEFQ)$ $\Longrightarrow$ $P,F,E$ are collinear.


I'm not sure if a thread on this has already started, but is Kalva dead?

Kalva seems pretty dead, but...Thank God for [b]web archive[/b]:

[url]http://web.archive.org/web/20040405065644/http://www.kalva.demon.co.uk/index.html[/url]

PS: yes, there are already two topics concernig this subject ;)

[url]http://www.mathlinks.ro/viewtopic.php?t=207824[/url]
[url]http://www.mathlinks.ro/viewtopic.php?t=207963[/url]

I'll bump this so it gets higher in the seach rank when somone searches kalva. I had to look quite a few pages down.


AMC
AMC 10
AIME
function
number theory
modular arithmetic
\/closed

Is the Intro Number Theory class going to be offered again later this year? If so, when? Also, is the material covered a pre-requisite to the AMC-10 competition? (first time participant)

[quote="gadler"]Is the Intro Number Theory class going to be offered again later this year?[/quote]

Do you mean later in the year than the September 2004 to December 2004 term on Fridays shown in the [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]class enrollment page[/url]? We plan to sign up for that section of the Introduction to Number Theory course, and I think it would be useful for a first-time AMC10 participant, but not strictly necessary. (That is, I know from a local example that it is possible to qualify for AIME off the AMC10 without first taking that course, but I think the course will make the AMC10 even easier for that same person.)

[quote="tokenadult"]Do you mean later in the year than the September 2004 to December 2004 term on Fridays shown in the [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]class enrollment page[/url]? [/quote]
Yes, I mean like winter or spring of 2005.

It is very likely we will offer the course in early 2005.

[quote="sran"]Hi,
I was wondering how hard is this class and how much homework it assigns (I thought I had already posted this but apparently not; if I double posted, forgive me). I would like to take it, but I have a lot of homework and activities this year, and am unsure that I will have the time to do it. I don't really want to do this class next year, because, aside from modular arithmetic and divisibility theory, I am intermediate level; and next year is the last year I will have the ability to do independent study.

Thanks.[/quote]

I'm not sure exactly what you are asking about.  It sounds like you are looking for some of the answers I posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13230&start=15]here[/url].

The Introductory Number Theory class would definitely help you if you are interested in learning modular arithmetic and divisibility theory.  We will also be offering an Intermediate Number Theory class next year will build on the lessons from the Introductory class and cover many new topics such as the phi function, Euler's theorem, and methods for using and solving Diophantine equations.

[quote="MithsApprentice"][quote="MCrawford"]
 We will also be offering an Intermediate Number Theory class next year...[/quote]

Would this be more likely spring or fall?[/quote]

We're aiming at both (i.e. we expect to have it finished in spring and will offer it more than once in the year).


modular arithmetic
number theory
relatively prime
Congruences

Determine all positive integers $n \ge 2$ that satisfy the following condition; For all integers $a, b$ relatively prime to $n$, \[a \equiv b \; \pmod{n}\Longleftrightarrow ab \equiv 1 \; \pmod{n}.\]

Let $n$ have that property. Then $a \equiv a \mod n$ gives $a^2 \equiv 1 \mod n$ for all $a$ coprime to $n$.
Since $2^2 \not\equiv 1 \mod 5$ (and because we always replace $2$ by some residue coprime to $n$ and $\equiv 2 \mod 5$) we get $5 \nmid n$.
Especially we are forced to have $5^2 \equiv 1 \mod n$, thus $n|24$.

It's trivial that $n=3$ and $n=8$ work, thus by the chinese remainder theorem the same holds for $n=24$. This imediately implies the property for the divisors of $24$, too.

you can easily find that if $ a\leq{\sqrt {n}}$ then $ \gcd(a,n) > 1$.
It means that all primes $ \leq{\sqrt {n}}$ divides n.
but it's known that if $ \sqrt {n} > 10$ then there exist at least two distinct primes $ p$ and $ q$ such that
 $ \frac {\sqrt {n}}{2} < p < q < \sqrt{n}$ but these primes must divide $ n$ it follows that $ n$ must be one of these forms: $ 3pq,2pq$
(because $ \frac {n}{4} < pq < n$ and $ n \equal{} kpq$ so...). the rest is trivial.


Let n be an integer on the form n=3k + 2. What can be said about the number of prime factors of n that give residue 1 and 2 upon division by 3?

Does it just mean that there is an odd number of 3n+2 primes?

this was my answer, but I would like to see a nice proof. Mine was unnecessarily complicated...

If we look at the prime factors $\mod{3}$ the.
Now 3 is not a factor, since the number is not in the form $3n$.
Write the number in its prime-factorisation. The $P_i=1 \mod{3}$ will not matter, as we are just multipling by one. So cross these off.
Now all we have left is a bunch of $P_k$'s equivilent to $2 \mod{3}$
Notice that if we had one of these primes, the number would work. Also with two of these primes, it would not
($=1 \mod{3}$).
Now since our primes is are all $2 \mod{3}$, so, finish it yourself.


Fie a,b,c trei numere complexe astfel incat:
|a|=|b|=|c|=1 si Re(a/b)=Re(b/c)=Re(c/a).Aratati ca a,b,c sunt afixele unui triunghi echilateral

sa fie bine oare ce zic eu aici ?
[hide="solutie"]evident cele trei numere nu sunt nule.
$a\overline{a}=1$, si celelalte.
deci $a/b=a\overline{b}$.
cum $2\mbox{re}(z)=z+\overline{z}$, rezulta ca relatia din ipoteza este echivalenta cu
$a\overline{b}+b\overline{a}=b\overline{c}+c\overline{b}= c\overline{a}+a\overline{c}$.
scriind doar prima egalitate, si grupand convenabil, se obtine
$b(\overline{a}-\overline{c})=\overline{b}(c-a)$, adica
$\frac{a-c}{b-0}+\frac{\overline{a}-\overline{c}}{\overline{b}-\overline{0}}=0$, adica $OB\bot AC$.
analog, se obtine ca $O$ este chiar ortocentrul triunghiului $ABC$, si cum $O$ este si centrul cercului circumscris, rezulta ca $ABC$ este echilateral.[/hide]