Today, I lost my toenial (I think).  So on that one toe (the smallest toe on my left foot), there's no nail (but it's supossed to grow back).  Has this ever happenned to you?

Edit:  Actually, I found out that there's still a little piece of toenail left.  So that's good.

Uhhh...wow that was an interesting question. No.  :D

OK, one of the odder topics... but yes, last Sunday as a matter of fact.

This was very odd, but no...
How did yu lose it?

there should be a sequel to this poll "have you lost your fingernail?" :D

Never have and hopefully never will. Also, how did you lose it?

yep, i kicked a basketball with my toenail with bare feet :blush:

[quote="jhollenbeck"]yep, i kicked a basketball with my toenail with bare feet [/quote]
why would you do that?
 :rotfl:

hmm ouch must hurt like cow...
and no lol

not that i know of... and i have kicked basketballs in my bare feet before, lol.  i mustve gotten lucky.  hope it grows back soon!

I  did! Never go up concrete steps in flip-flops.... Boy was I angry at them though! Those flip-flops were history...  :bruce:  I think they're still in brasil though, or at least what's left of them... :roll:  American flip-flops don't last at all. But anyway, I ran up the stairs and hit my toe on the step, and I was stading on one foot not screaming but just making a high-pitched noise, our maid thought it was the dog so she didn't bother to worry... (in other words I didn't scream, i held the note for a long time) And so  I hopped into the house on one foot and then collapsed on the floor, and then I realized that my toe was bleeding! So I sat down on the floor infront of the fan when my cousin comes up, sees the blood and imediately turns away... (she has a problem with blood.) And then my mom comes in and says "WHAT HAPPENED?!" And then there's this long story, it hurt for about 3 days... I had to soak my foot in salt water for a long time to get rid of the pus. I took off the nail polish and my toenail was green! (this is my second toe) It took a month for it to come off, it was just hanging on a little bit, it drove me nuts! Then I finally just took it off (May I add that I am very gross and actually saved it...  :oops: )  Well anyway, it has just now grown back... It took 4 months for it to actually start to grow back, then 2 months to finish... took a long time. But losing that nail didn't hurt nearly as badly as having to have your nail cut out.... that hurt!

:rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:
i'm glad i haven't lost a nail.

yep. i tripped and my big toe landed on this small plastic toy. the toenail broke but did not completely fall off. it was messed up for years.

i slammed my finger i a car door once and my nail was "dented".  It had a small dent in it. :D

ahm, yes.  it was a long walk during a math (!) camp around Eastern. there was still snow (and not too less) and it was a very exausting route. at the end I lost one of my toenails (the doc said my shoes were wrong  :( ). it took a month or so until it go of, but it didn't hurt much. and if I can trust the doc it will all be ok soon  :)

i lost a fingernail. note to self: never challenge a guy that's twice your size to pick-up-basketball... especially if you are inside a gym filled with exercise equipment. i ended up breaking the end of my finger in the process, which i found out AFTER i had played a game of 3-on-3 for the school tournament that afternoon. i had to go to the ER and everything. it hurt like... bad. altho i think the local anesthesia was way worse than anything. like liquid fire going through my hand. and then they ripped it off. yuck. funniest part was? doctor that took off my fingernail... her name was DR. PAYNE...  :rotfl:

LOL I had a math teacher named Mr.Payne...he wan't a pain though.

No, but I've lost a fingernail. Lion bit it, and a tiny bit of flesh off. It grew back, and you can't really notice the lost flesh  unless you focus on the finger.

yep, stuck my hand through a cage at a zoo to feed a lion bread, I was pulled away just in time so the injuries weren't worse

I had to have my toenail cut off because it was ingrown and inflamed... But there were no pain killers involved. She just lifted up the toenail *screams* then cut it off almost all the way to the middle of my nail, *screamed operatically* Yes, didn't sound to bad either! What really hurt was when my mom and I were moving furniture, we were moving a queen sized mattress into another room and she lost her grip and it fell on my toe spliting the nail almost half way, that hurt too. The thing that hurt the worse when it comes to fingers and toes though was when we were moving this 27" tv from the 80's, it weighed about 80 pounds! (now I could be exagerating, but only  little bit because I don't know the exact weight) We were moving it back into the entertainment center and she fell back and I was left with the tv and then it slammed my hand into the wall, I was there for like 3 minutes going "ah ah ah ah ah ah ah ah ah ah ou ou ou ouah ouah o uah ah o ouch! this is hurting right now.!" My mom thought I was panting or something, I make very weird noises... And finally she realized that it was on my hand after 4 minutes, when she moved the tv the pressure was getting tighter! Now I'm doing my scales "aaAAHHHhhhaaa" Yeah, I had an ice pack on that for a while. Then I jammed my toe into the chair that is bolted to the ground, stupid chair... I was limping for 2 days, I figured out that I had spraned a tendon, I had a purple triangle on my foot. Do I have bad luck or what?  :|

um, windslicer? what zoo has a lion within reaching distance?  :? sounds like a big liability to me (i.e. not in america, i dont think).

Um, it was about 12 years ago , and I had climbed under the things that are supposed to prevent you from getting to the cages. I stuck my hand through the cage and voila.

Nowadays they like to put moats between the person and the lion.

And about the liability, the only reason why my parents didn't sue is because the people paid for all of the bills at the hospital and the fingernail grew back (not what the doctors had predicted). I'm sure after that incident, they made changes to the setup.


function
abstract algebra
category theory
superior algebra
superior algebra unsolved

How can I show that the fraction function preserves exactness.

What is the "fraction function"¿ Tensoring with $ \mathbb Q$ for abelian groups¿
Don't expect everyone to know some notation of that type...

I suspect you are talking about the localization functor: If $ 0\to A\to B\to C\to 0$ (*) is a short exact sequence of $ R$-modules, then, given a multiplicative subset $ S$ of $ R$, tensoring (*)  with $ S^{\minus{}1}R$ over $ R$ yields an exact sequence of $ S^{\minus{}1}R$-modules.

This should be calculated once in a life using the natural isomorphism of $ S^{\minus{}1}R$-modules $ S^{\minus{}1}A\simeq S^{\minus{}1}R\otimes_R A$, for every $ R$-module $ A$.

Yes, Third addition,  I am talking about what you are also saying. I want to see that

$ ImS^{\minus{}1} \phi \equal{} S^{\minus{}1} Im\phi$ and $ KerS^{\minus{}1} \phi \equal{} \equal{} S^{\minus{}1} Ker\phi.$

How can I see that? Please give me some suggestions since my notes are too compact and nothing is proved in details, and by the way this is not a homework problem I am just trying to understand homological algebra, can you suggest me a basic book in homological algebra I have found many but they are too advanced for me and my course. thansks.

[quote="peteryellow"]Yes, Third addition,  I am talking about what you are also saying. I want to see that

$ ImS^{ \minus{} 1} \phi \equal{} S^{ \minus{} 1} Im\phi$ and $ KerS^{ \minus{} 1} \phi \equal{} \equal{} S^{ \minus{} 1} Ker\phi.$

How can I see that? Please give me some suggestions since my notes are too compact and nothing is proved in details,[/quote]... straight from the definition of a localization; just resolve the definition, use the properties of the original mappings and code things in terms of localizations again.[quote]and by the way this is not a homework problem I am just trying to understand homological algebra, can you suggest me a basic book in homological algebra I have found many but they are too advanced for me and my course. thansks.[/quote]You might want to have a look at the corresponding chapter(s) in Lang, Algebra.


In Byzantine Basketball, there are two scoring events (making a free throw and making a field goal).  If the free throw gives  [b]a[/b]  points and the field goal gives [b]b[/b] points, and [b]a[/b] and [b]b[/b] are relatively prime, what is the largest score that cannot be attained in the game in terms of [b]a[/b] and [b]b[/b]?

I think I have a solution, but I am not sure.  Can anyone verify this?
[hide](a-1)(b-1)-1[/hide][/b]

Yeah, this is the Chicken McNugget Theorem. Google around for it (and for the Coin Problem). Your answer is correct.


[quote="kunny"]Let $ a,\ b,\ c$ be positive real numbers.

Prove that $ (ab \plus{} bc \plus{} ca)\left\{\frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2}\right\}\leq \frac {3}{4} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a}$[/quote]

Ineq
$ < \equal{} > \sum \frac {ab}{(a \plus{} b)^2} \le \frac {3}{4}$
Done  :lol:

Oh, my god! :blush: That was so trivial, thank you for your reply.


Find formulas for $\sin(na)$ and $\cos(na)$ in terms of $sin(a)$ and $cos(a)$, respectively.
[hide="Enormous hint"]
Use the identity:
$(\cos(a)+i\sin(a))^{n}= \cos(na)+i\sin(na)$[/hide]


function
calculus
derivative
integration
calculus computations

The question is:
Find all continuous functions satisfying:
$ f(x + y) = f(x) + f(y) + xy(x + y) ; x, y \in \mathbb R.$

[quote="What I did is :"]
Let $ y=x$.
Then,
$ f(2x) = 2f(x) + x^2(2x)$
$ 2f'(2x)= 2f'(x) + 3x^2$
$ 2f''(2x)= f''(x) + 6x$
$ 4f'''(2x)= f'''(x) + 6$

Therefore:
$ 4f'''(0)= f'''(0) + 6$
$ f'''(0) = 2$
Now integrating from $ 0$ to $ x$, I got $ f(x) = \frac {x^3}{3} + cx^2$
The correct answer is ${ \color{red}{f(x) = \frac {x^3}{3} + cx}}$ where $ \color{red}c = f'(0)$
[b]Where did I possibly go wrong???[/b] :huh:

[/quote]

[quote="Rajiv"]The question is:
Find all continuous functions satisfying:
$ f(x \plus{} y) \equal{} f(x) \plus{} f(y) \plus{} xy(x \plus{} y) ; x, y \in \mathbb R.$

[quote="What I did is :"]

$ f(2x) \equal{} 2f(x) \plus{} x^2(2x)$
$ 2f'(2x) \equal{} 2f'(x) \plus{} 3x^2$
[/quote][/quote]

You took the derivative incorrectly between these two steps. But you also seem to have made another mistake. The problem said that f was continuous, not necessarily differentiable. How do you justify that you´ve found all possible solutions.

[quote="Rajiv"]
Now integrating from $ 0$ to $ x$, I got $ f(x) = \frac {x^3}{3} + cx^2$
The correct answer is ${ \color{red}{f(x) = \frac {x^3}{3} + cx}}$ where $ \color{red}c = f'(0)$
[b]Where did I possibly go wrong???[/b] :huh:

[/quote][/quote]
 i don't know? how did u use integral?
I have a simple solution:
let $ g(x)=3f(x)-x^3$
 We  have: $ g(x+y)=g(x)+g(y)$
By equation Cauchy's, thus: $ g(x)=3Cx$,
Therefor, $ f(x)=\frac{x^3}{3}+Cx$


geometry
incenter
Putnam
geometry proposed

Let the incentre of an equilateral triangle ABC be I,inradii be r. Within ⊙I take any point P, and let the distance from P to BC, CA, AB be $d_1,d_2,d_3$.

Prove: $\sqrt{d_1},\sqrt{d_2},\sqrt{d_3}$ can form a triangle with area $\frac{\sqrt{3}}{4}\sqrt{r^2-PI^2}$.

This is a minor variation of the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=2101]Putnam 2003 problem B5[/url] (or the other way around). At this thread, 3.14 different solutions have been posted.


$c \geq b \geq a \geq 0 $
Show that
$(a+3b)(b+4c)(c+2a) \geq 60abc $

first we have $16(b+4c)(c+2a) \geq 15(b+3c)(c+3a)$
this is because$16(bc+4c^2+2ab+8ac) \geq 15(bc+9ac+3c^2+3ab)$
equal to $bc+19c^2 \geq 7ac+13ab$ obvious.
the rest is easy with AM >= GM

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=5436 .

  darij

first of all we expand the left side.
then by Chebyshev inequality,  we have
$(a+3b)(b+4c)(c+2a) \&= 25abc + 3b^2 c +4ac^2+12bc^2+2a^2b+6ab^2+8a^2c\#
                                 \&\geq 25abc+ 3abc+4abc+12abc+2abc+6abc+8abc
                                 \&=60abc$

I don't see any way how to apply chebyshev's inequality.I guess you have made use of the fact that [tex]C \geq \ b \geq \ a [/tex]

[quote="utopia94"]sebescu you're right :D[/quote]
then can you elaborate? How did you use chebyshev?

[quote="sebescu"]hey zhaobin
 how did you think of:
[tex]\displaystyle 16(b+4c)(c+2a) \geq 15(b+3c)(c+3a)[/tex]
and xeroxia how did u get:
[tex]\displaystyle a + 3b \geq 4\sqrt[4]{ab^3} = 4a^{\frac 14}b^{\frac 34}[/tex]
Have u applied some well known inequality???[/quote]
I think it is only $AM \geq GM$,
rewrite $3b=b+b+b$
i think there is something commom between xeroxia's solution and mine.


group theory
superior algebra
superior algebra unsolved

This is another question relating to Cayley Graphs.  Let's say that each of the $6$ elements in the symmetric group $S_{3}$ is a vertex in a graph.  Then we have the set $S=\left\{\left(1,2\right),\left(2,3\right)\right\}$.  I know that if I have $v_{1}$ and $v_{2}$ in $S_{3}$, then they are connected by an edge if $v_{2}=sv_{1}$ where $s\in S$.  My question is:  How do you multiply elements in a symmetric group?  For example, I know that

\[ \binom{1\,2\,3}{3\,1\,2}\binom{1\,2\,3}{2\,1\,3}=\binom{1\,2\,3}{3\,2\,1}. \]
But how can I perform the same type of multiplication with the elements in $S$?

Thank you.

In fact, the multiplication is a way to simplify the notation $s\circ{} s' = ss'$.

Well I understand that multiplication here is actually the composition of functions.  Would someone please take an element from $S_{3}$ and multiply it by an element of $S$ and show me how you came to the result?  Thanks a lot.


algebra
polynomial
calculus
derivative
geometry
geometric transformation
rotation

DeMoivre's Theorem and Euler's Formula seem pretty similar to me. Are there any specific advantages/disadvantages of using one or the other, especially in competition math?

Thanks.

Er..they are the same thing I believe. If you are refering to the notation e^i :theta: vs. the notation cos( :theta: ) + i*sin( :theta: ), then use whichever one is appropriate. Though a sum of complex numbers often requires use of the former since the sum can be expressed as a geometric series.

i posted this a while ago

[hide]taylor polynomials at school

summation((i*x)^n/n!,0,infinity)

also you can take the derivative of the expression a=(cos(x)+i*sin(x))

-sin(x) +i*cos(x) dx = i *(cos(x)+i*(sin(x))) dx = i*a dx

Int(da\a)=Int(i*dx). so ln(a)=i*x a=e^(i*x)[/hide]

I'm not wondering about the proof at all for Euler's Formula. It is actually quite short, simple, elegant, and easy to understand as long as you know how to use Maclaurin Polynomials (which I do).

Even though rlee isn't interested in a proof, here's a very elegant proof that doesn't use Taylor series:

When we differentiate eix, we get i*eix. Since multiplication by i is the same as rotation of :pi:/2, the tangent vector of eix is perepndicular to the curve. (Ack. I'm using differential geometry terminology, but I don't know what else to call it.) It is pretty easy (i.e. one of my past homework problems) to show that this means it's a circle centered at the origin. Since e0=1, the circle traced out by eix is the same as x2+y2=1. We can also see that it is unit-speed (meaning that the modulus of the derivative is 1) since it's just the modulus of the function multiplied by i. Thus eix=cos(x)+i*sin(x).


Nine men - Brown, White, Adams, Miller, Green, Hunter, Knight, Jones and Smith - play several positions on a baseball team.

Determine the position played by each from the information inclusive of the statements given below.

(1) Smith and Brown won $10.00 playing poker with the pitcher.
(2) Hunter was taller than Knight and shorter than White, but each of these weighed more than the first baseman.
(3) The third baseman lived across the corridor from Jones in the same apartment house.
 (4) Miller and the outfielders play bridge in their spare time.
(5) White, Miller, Brown, the right fielder and the center fielder were bachelors. The rest were married.
 (6) Of Adams and Knight, one played an outfield position.
 (7) The right fielder was shorter than the center fielder.
 (8) The third baseman was brother to the pitcher's wife.
(9) Green was taller than the infielders and the battery - except for Jones, Smith and Adams.
(10) The second baseman beat Jones, Brown, Hunter and the catcher at a game of cards.
(11) The third baseman, the shortstop, and Hunter made $150 speculating in U.S. Steel.

(12) The second baseman was engaged to Miller's sister.

(13) Adams lives in the same house as his own sister but dislikes the catcher.

(14) Adams, Brown and the shortstop lost $225 each speculating in copper.

(15) The catcher had three daughters; the third baseman had two sons; and Green was being sued for divorce.

[hide]P: Jones
C: Smith
1B: Brown
2B:
3B: Adams
SS:
LF: Green
CF: Hunter
RF: Knight
[/hide]

I can't figure out the last two, but am I right so far?

[hide]Brown - first baseman
White - second baseman
Adams - third baseman
Miller - shortstop
Green - left fielder
Hunter - center fielder
Knight - right fielder
Jones - pitcher
Smith - catcher[/hide]

How did you get shortstop and second base?

I'm sure you can find it out yourself, just take another look at the statements!

[hide]... especially (12)[/hide]

Oh yeah :rotfl:  :wallbash_red:  :wallbash_red:

I feel pretty stupid now...


jjx1 i know you will read this and remind me of TOS.

 Me when you will mute me.... :stretcher:

unessecary comments like this and overreacting is the reason why you get muted at times.

[quote="justinahmann"]The Nats are held at Dolphin and Swann Resort in Orlando for those who didn't know and might end up in Washington D.C. confused.

Oh no!!!  Now jjx1 will mute me forever because I used location!!! I better watch out! :read:

 :D[/quote]

Aren't you a freshman in high school?  So why in the world does it matter to you where nats is being held this year?

I intended it for other people if you reread the topic.  :)


Solve in positive integers the equation:

$10^a+2^b-3^c=1997$


In triangle $ABC$, we draw three semi-circles with diameter $AB$,$BC$ and $CA$ on the outside of $ABC$. A big circle with center $O$ is tangent to all three semi-circles. It touches the semi-circle with diameter $BC$ at $K$. Let $L$ be the midpoint of $BC$. Prove that $L$,$K$ and $O$ are on the same line.

It touches the with diameter $BC$ at $K$.

What does it mean?[/quote]

Well it is well known that if cirlcles $k$ and $l$ with centres $O_1$ and $O_2$ tangent each other in point $M$ then $M,O_1,O_2$ are collinear, so there you go $L$ and $O$ are centers,  and $K$ is touching point....


$ x,y,z$ are integers.
$ x^2\minus{}y^2\equal{}2xyz$

[hide]Let $ (x, y)\equal{}d$, $ x\equal{}ad$, $ y\equal{}bd$.  Then
\[ a^2\minus{}b^2\equal{}2abz\]
with $ (a, b)\equal{}1$.

But $ b^2 \equiv 0\mod{a}$, so $ a\equal{} \pm1$.  Similarly, $ b\equal{} \pm1$.  It then follows $ x\equal{}\pm y, z\equal{}0$.[/hide]

[hide="partial solution"]
Solving for $ x$, we have
\[ x\equal{}\frac{2yx \pm \sqrt{4z^2y^2\plus{}4y^2}}{2}\]
In order for $ x$ to be an integer, $ 4z^2y^2\plus{}4y^2\equal{}n^2$, where $ n$ is an integer. Now, I don't know how to prove this, but I'm pretty sure that this is true iff $ z\equal{}0$. From there, it's pretty straightforward, and I'm about to fall asleep (it's 2:08 am), so goodnight. :)
[/hide]

Good night. :D
[hide="continuing"]$ 4z^2y^2\plus{}4y^2\equal{}4y^2(z^2\plus{}1)$, which can only be a square if $ z\equal{}0$.[/hide]


We consider the equality:

$A(x) = B(x)C(x)$ where $A(x),B(x),C(x)$ are polynomials with integer coefficient , suppose that $B(x) = D(x)$ (mod $2$ ) ,
Consider our polynomial in $F_{2}$
Can we conclude that $A(x)$ is divisible by $D(x)$  ?

SOMEONE HELP ME TO ANSWER THIS QUESTION .


find x:

angle GDF = 24 degrees
arc GF = 31
arc GC = x.
GC = FE
there ya go
thanks :)

oo whoops here is geogebra diagram:  :blush:  [geogebra]1e84200fd4f5bafd7716540b9f3cd244f3223eda[/geogebra]

Where angle P =24 degrees
Arc BC = 31 degrees
and Arc BD= x
:)

Is D that unlabeled point on the circumference?

And, (I'm guessing this is true because of your original problem), is BD=CE?


modular arithmetic
number theory proposed
number theory

Let $n$ be a positive integer. $S$ is the set of nonnegative integers $a$ such that $1<a<n$ and $a^{a-1}-1$ is divisible by $n$. Prove that if $S=\{ n-1 \}$ then $n=2p$ where $p$ is a prime number.

[i]Mihai Cipu and Nicolae Ciprian Bonciocat[/i]

Should I repeat it again for you?  :|

Don't name topics like "nice", "very nice" and so on.
Read http://www.mathlinks.ro/viewtopic.php?p=140098

I can only prove that $n=2p_1...p_k$, $p_i$ are pairwise distinct prime numbers. How to prove $k=1$? I don't see something really easy?  :?

It's from a Romanian TST in 2002. I didn't solve it in the contest, but afterwards, I have solved it at least three times, and I always forget my solution and have to work it out again :). It's getting pretty annoying..

As Myth has already mentioned, $n$ must be even and square-free. To see that it's even, we notice that if $n$ were odd we would have $(n-1)^{n-2}\equiv 1\pmod n$, so $n-1\not\in S$, and if $n$ could be written as $2^{a_1}p_2^{a_2}\ldots p_k^{a_k}$ with at least one $a_i>1$, then $1<a=2p_2\ldots p_k+1<n-1$ would also belong to $S$.

We now know that $n=2p_2\ldots p_k$. Assume now that $k\ge 3$. By the Chinese Remainder Theorem, we can find $a\equiv 1\pmod 2,\equiv 1\pmod {p_2},\ldots,\equiv -1\pmod {p_k}$. This $a$ cannot be $1$, and it cannot be $n-1$, and since $a$ is odd, $a-1$ is even, so $a^{a-1}$ is $1\pmod{p_i},\ \forall i$.

Is it Ok? If it is, I don't think I've ever found this solution before. They were all more complicated :).

My fault that I tried to find explicit form of $a$ for the square free case :(

Later solved by grobber here http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=84654  :lol:


I just had the chance to upload my photos from Winter Camp. You can find them (along with photos from various other math gatherings) [url=http://photos.yahoo.com/z_yufei]here[/url] (Adrian's photos are also included in the album).

Again, I encourage all campers who brought cameras to also post their pictures. Same thing applies to all other camps.

Finally, here're some selected photos with commentaries:

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=511[/img]
Jenn and the premier of Moosehat. And many more appearances to follow.

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=512[/img]
Peng and his "random lemma" (more commonly known as the four-coins theorem)

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=513[/img]
Return of old master Jacob -- bestowing upon us the now out-of-fashion techniques of "wam-jam"

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=515[/img]
Two lovable figures of the camp -- mascot Canmoo, and his Rubik's cube.

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=514[/img]
Canmoo posing with his not-so-distant relative Moosehat.

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=517[/img]
Guess what they're playing ...

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=516[/img]
And finally, a group photo.

Nice pics!  Good to see a lot of familiar faces!  I think there might be 1 or 2 of you that I don't know, but I'm surprised how many of you were there last year (almost everyone?  Who isn't there other than me and Oleg?)

Anyways, good luck this contest season!  No Fermat this year for a lot of you eh?  It must feel weird.

A couple of notes:

- Seeing those pictures reminded me that I still have 2 very very large Canadian that we took IMO 2005.  I'll bring them to seminar or something for you guys.  REMIND ME!  I DON'T WANT THOSE FLAGS!
- The game with the finger-poking... did I teach that to you guys?  Or did one of you teach it to me..?  That's a game that I've played for a long time.
- Moosehat is a nice touch.  It looks like you guys had a lot of fun!

~ Elyot

[quote="Elyot"]
- Seeing those pictures reminded me that I still have [color=red]2 very very large Canadian[/color] that we took IMO 2005.  I'll bring them to seminar or something for you guys.  REMIND ME!  I DON'T WANT THOSE FLAGS!
[/quote]

You mean 2 very very large Canadian flag, right?

No, he means that he still has two large canadians from IMO.

 :rotfl:

Wow.  I missed a lot of words in those sentences.  I must have been really really tired when I posted that... or something.  Geez.  Yes, I meant flags.  I DID NOT take 2 very very large Canadians from the IMO...

lol... :rotfl:
Careful there, or some important people who have been viewing this forum may start worrying about your intentions. Well at least nobody is missing... :!:

BTW, for that group photo...where is Richard? :huh:

[quote="YY"]BTW, for that group photo...where is Richard? :huh:[/quote]

Good question. I have no idea. He was probably still bowling.

My Winter camp photos are finally ready! I have only mafia ones, since I remembered that I have a camera only on the last night  :D.

Only three attachments per post? O humiliation! Ok, here are more ...


binomial coefficients
algebra unsolved
algebra

[quote="BadrGhamdi"]Hello everyone ..
i wanted to ask about how how are Fibonacci Numbers are related to Binomial Coefficients ??
i was solving a problem about Fibonacci Numbers .. and I got this :
\[ {F_{n \plus{} 2}} \equal{} {F_{n \plus{} 1}} \plus{} {F_n} \equal{} {F_{n \minus{} 2}} \plus{} 2{F_{n \minus{} 1}} \plus{} {F_n} \equal{} {F_{n \minus{} 4}} \plus{} 3{F_{n \minus{} 3}} \plus{} 3{F_{n \minus{} 2}} \plus{} {F_{n \minus{} 1}}\]
it's similar to binomial coefficients.
so ,, how are they related ?? can you explain with details??
Thanks in Advance  :) :)[/quote]
Do you want a proof for $ F_n \equal{} \sum_{i \equal{} 0}^m \binom{m}{i} F_{n \minus{} m \minus{} i}$ ?
It is very easy if you note that $ F_n \equal{} \frac {\alpha^n \minus{} \beta^n}{\sqrt {5}}$ where $ \alpha$ is the positive root of $ x^2 \minus{} x \minus{} 1 \equal{} 0$ and $ \beta$ the negative.
Then $ x^n \equal{} \sum_{i \equal{} 0}^m \binom{m}{i} x^{n\minus{}m\minus{}i} \iff$
$ x^m \equal{} \sum_{i\equal{}0}^m \binom{m}{i} \left ( \frac{1}{x} \right)^i \equal{} (1\plus{}x^{\minus{}1})^m$ which is true when $ x \equal{} 1\plus{}x^{\minus{}1} \iff x^2\minus{}x\minus{}1\equal{}0$, so it holds for both $ \alpha$ and $ \beta$ and hence:
$ \frac{\alpha^n \minus{} \beta^n}{\sqrt{5}} \equal{} \sum_{i \equal{} 0}^m \binom{m}{i} \frac{\alpha^{n\minus{}m\minus{}i}\minus{}\beta^{n\minus{}m\minus{}i}}{\sqrt{5}} \iff$
$ F_n \equal{} \sum_{i \equal{} 0}^m \binom{m}{i} F_{n \minus{} m \minus{} i}$

prove that:
$ F_{n\plus{}m}\equal{}\sum_{i \equal{} 0}^{m}\binom{m}{i}F_{n\minus{}i}$
by induction.
$ m\equal{}1: F_{n\plus{}1}\equal{}\binom{1}{0}F_n\plus{}\binom{1}{1}F_{n\minus{}1}\equal{}F_n\plus{}F_{n\minus{}1}$ - by definition.
$ m\to m\plus{}1:$
\[ F_{n\plus{}m}\equal{}\sum_{i \equal{} 0}^{m}\binom{m}{i}F_{n\minus{}i}\\
F_{n\plus{}m\minus{}1}\equal{}F_{(n\minus{}1)\plus{}m}\equal{}\sum_{i \equal{} 0}^{m}\binom{m}{i}F_{(n\minus{}1)\minus{}i}\equal{}\sum_{i \equal{} 0}^{m}\binom{m}{i}F_{n\minus{}i\minus{}1}\\
F_{n\plus{}m\plus{}1}\equal{}F_{n\plus{}m}\plus{}F_{n\plus{}m\minus{}1}\equal{}\sum_{i \equal{} 0}^{m}\binom{m}{i}F_{n\minus{}i}\plus{}\sum_{k \equal{} 0}^{m}\binom{m}{k}F_{n\minus{}k\minus{}1}\equal{}\\
\binom{m}{0}F_n\plus{}\sum_{i \equal{} 1}^{m}\binom{m}{i}F_{n\minus{}i}\plus{}\sum_{k \equal{} 0}^{m\minus{}1}\binom{m}{k}F_{n\minus{}k\minus{}1}\plus{}\binom{m}{m}F_{n\minus{}m\minus{}1}\equal{}\\
(let\, k\equal{}i\minus{}1)\equal{}F_n\plus{}\sum_{i \equal{} 1}^{m}\binom{m}{i}F_{n\minus{}i}\plus{}\sum_{i \equal{} 1}^{m}\binom{m}{i\minus{}1}F_{n\minus{}i}\plus{}F_{n\minus{}m\minus{}1}\equal{}\\
F_n\plus{}\sum_{i \equal{} 1}^{m}\left(\binom{m}{i}\plus{}\binom{m}{i\minus{}1}\right)F_{n\minus{}i}\plus{}F_{n\minus{}m\minus{}1}\equal{}\sum_{i \equal{} 0}^{m\plus{}1}\binom{m\plus{}1}{i}F_{n\minus{}i}\]
Also, there is beautiful formula:
\[ F_{n\plus{}1}\equal{}\sum_{k\equal{}0}^n\binom{n\minus{}k}{k}\]


[b]Problem 6.[/b] Point $O$ is given in the plane. Find all sets $S$ with more than one point in $S$, such that for any $A\in S$, $A\neq O$, the circle with diameter $OA$ is in $S$.

[i]Nikolai Nikolov, Slavomir Dinev[/i]


function
number theory proposed
number theory

Suppose that the numbers $a,b,c$ and

$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ , $\frac{a}{c}+\frac{c}{b}+\frac{b}{a}$

are integers. Prove that $|a|=|b|=|c|$.

Consider the function:f(x)=(x-a/b)(x-b/c)(x-c/a)=x^3-Ax^2+Bx-1
But f(a/b)=f(b/c)=f(c/a)=o so with the thorem of rational roots we have that |a|/|b|=|b|/|c|=|c|/|a|=1 so |a|=|b|=|c|
(A=a/b+b/c+c/a and B=a/c+b/a+c/b)


algebra
polynomial
modular arithmetic
number theory
superior algebra
superior algebra unsolved

Prove that a prime $ p$ is of the form $ m^2+mn+41n^2$ ($ m,n\in \mathcal l Z$) iff $ (\frac{-163}{p})=1$.

Whatever cnyd writes there (formulas without text are often impossible to read; and this one just looks wrong)...

Note that $ m^2 \plus{} mn \plus{} 41n^2$ is the norm of $ m \plus{} n \cdot \frac {1 \plus{} \sqrt { \minus{} 163}}{2}$, so you ask for the values the norm can take. Now by some algebraic number theory (especially, that $ \mathbb Q[\sqrt { \minus{} 163}]$ has class number $ 1$) exactly the splitting primes work, which are the mentioned ones; more generally, a positive integer $ r$ can be written in that form iff each of the prime factors $ p$ with $ \left(\frac { \minus{} 163}{p}\right) \equal{} \minus{} 1$ occurs an even number of times in the prime factorisation.


geometry
incenter
trigonometry
geometry unsolved

let (O) be semicircle with diameter AB ,K,L be points on AB and M,N,C be points on (O) such KLMN be square and area of ABC and KLMN are equal .Prove that intersection of KN(or LM) and AM(or BN) be incenter of ABC

Without any calculations I can't solve it too... :blush:


Prove that 11^10-1  can be divided with 100!

1)Prove that (2^2)^1959-1 can be divided with 3.
2)Prove that (2^3)^n+1 can be divided with 3^(n+1).

well... for the second one try induction.. it always works .. I think..:D

[quote="Scrambled"][quote="Lagrangia"]Well perhaps this will help:

try to prove that generally we have: n^2 | [(n+1)^n-1]  :)[/quote]
how does this help? we have 100! not 100[/quote]

He means 100 instead of 100! (could be he is excited  ? :D) .... and logically we know that $100!>11^{10}-1$  :P  so it is impossible also .


Prove for all natural numbers greater than 1:

$ \displaystyle \sum_{j \equal{} 1}^n \frac {1}{\sqrt {j}} > \sqrt {n}$

(1971 Bailey Prize Examination, #8)

[hide="Although I will never understand the practice of posting problems in local forums..."] Note that $ \sqrt{n\plus{}1} \minus{} \sqrt{n} \equal{} \frac{1}{ \sqrt{n\plus{}1} \plus{} \sqrt{n} } < \frac{1}{ \sqrt{n\plus{}1} }$. [/hide]

Although I will never understand the practice of posting in local forums of states which are located on the opposite coast of the state in which one resides...

No just kidding.  We're just trying to get more people from SC on here, and use this as a place for our ARML team to practice and discuss problems.

Is there a way of actually evaluating that sum?

Doubtful.  The finite sum is a [url=http://mathworld.wolfram.com/HarmonicNumber.html]general harmonic number[/url].  The infinite sum can be approximated by an integral (which diverges); see [url=http://en.wikipedia.org/wiki/Euler-Maclaurin_formula]Euler-Maclaurin formula[/url].


limit
inequalities
calculus
calculus computations

Hint: You want to make $|a_nb_n-MN|$ small, given that you know that $|a_n-M|$ and $|b_n-N|$ are small. But $(a_nb_n-MN)$ doesn't factor. So we emply a "put and take":

$a_nb_n-MN=a_nb_n-a_nN+a_nN-MN=a_n(b_n-N)+(a_n-M)N$

There are still a number of steps to go, but that's one piece of it.

[quote="Gibbenergy"]My math teacher asks the prove of proving the product of limit:
$\lim_{n \to \infty}a_n=L$ & $lim_{n \to \infty} b_n=M$ then
$\lim_{n \to \infty} a_nb_n= LM$ by $\epsilon$ and $N$ definition. I have no idea how to approach.[/quote]

  Thanks Kent Merry field for your hint. I have done it according to your hint but I am not sure it is correct or not. Please give me your comment:
  $\lim_{n\to\infty}a_n=L$
$\forall \epsilon >0$ $\exists N_1$such that $n> N_1$ then
  $|a_n-L| < \epsilon$
$\forall \epsilon >0$ $\exists N_2$ such that $n > N_2$ then
  $|b_n-M| < \epsilon$

Now choses $N=max(N_1,N_2)$ then we have:

$|a_nb_n-ML|=|a_nb_n-b_nL+b_nL-ML|$$=|b_n(a_n-L)+L(b_n-M)| \leq |b_n(a_n-L)|+|L(b_n-M)|$ $\leq |b_n|\epsilon + L\epsilon$ $\leq \epsilon^2+(M+L)\epsilon$

From here, I don't know which $N$ I should choose to satisfy this inequality $N=max(N_1,N_2)$ or $N=\epsilon^2+(M+L)\epsilon$. Can anyone help me?

Here's how I would organize it, although some of the choices are a matter of taste:

Given $\epsilon >0:$

$\exists\,N_1$ such that if $n>N_1,$ then $|a_n-L|<\frac{\epsilon}{2(|M|+1)}.$

$\exists\,N_2$ such that if $n>N_2,$ then $|b_n-M|<\max\left(\frac{\epsilon}{2(|L|+1)},1\right).$

Let $N=\max(N_1,N_n).$ Suppse that $n>N.$

We first note that since $|b_n-M|<1,$ we have $|b_n|<|M|+1.$

$|a_nb_n-LM|=|a_nb_n-Lb_n+Lb_n-LM|$

$\le |a_nb_n-Lb_n|+|Lb_n-LM|$

$\le |b_n||a_n-L|+|L||b_n-M|$

$\le (|M|+1)|a_n-L|+|L||b_n-M|$

$<(|M|+1)\frac{\epsilon}{2(|M|+1)}+|L|\frac{\epsilon}{2(|L|+1)}<\frac{\epsilon}2 +\frac{\epsilon}2=\epsilon.$


search
number theory unsolved
number theory

Find all positive integer pair $(a,n)$ such that $\large\frac{(a+1)^{n}-a^{n}}{n}$ is an integer.

$(a+1)^{n}-a^{n}$ is odd, therefore n is odd. If (a,n) solution, and $n=p_{1}^{k_{1}}...p_{m}^{k_{m}}$, then $(a,n)=1=(a+1,n)$. Therefore $3\not |n$. If $p|(a+1)^{n}-a^{n}$, then exist minimal  1<T=T(p)\le p-1, suth that $p|(a+1)^{T}-a^{T}$ and $T|n$. If p is minimal prime divisor of n, it give T(p)=1. But $a+1-a=1$, therefore solution is only n=1, a -any integer.

Suppose $n>1$ let p be the minimal prime factor of n. Since p does not divide both a and a+1 it does not divide neither of them. Now let q be the minimal number such that $(a+1)^{q}-a^{q}$ is divisible by p. Obviously q<p. But now since p divides$(a+1)^{n}-a^{n}$ we have q|n which is a contradiction since q<p and $q\neq 1$


All it says in chapter 5 about binary is that

At the opposite extreme from al lthese digits is the lowly base 2, or binary. Here the only two digits are 0 and 1, and counting looks like 1, 10, 11, 100, 101, 110, ...

How do you multiply a number by 2 in base 2?

How do you add two base 2 numbers without converiting to decimals? I'm not understanding these conversions much yet.

haha i guess i sort of answered a bunch on aim. do you still have any questions?

basicly binary is fairly easy to understand once you understand this:

in decimal we have ones, tens hundreds and so on. ones = $10^{0}$, tens = $10^{1}$, hundreds = $10^{2}$ and so on.

In binary instead of base 10, we have base two, so the place values works like this, $2^{0}, 2^{1}, 2^{2}...$

so what do you do in base 10 when you add numbers which overflow the current place value youre adding? you carry the extra to the next place value! So 29 + 34, you add 9 and 4, but this is greater than 10, so you put 3 in the ones and carry one to the tens and then add 2 + 3 + 1 which gives us our answer of 63.

Binary is much the same, only there are two digits which makes things much easier. If you add two 1's together, you put in zero and carry the one.

011 + 101: Add the first place value, 1+1 so we put in zero and carry the one. Add the second place value, 1 + 0, but we carried the one over so now its 1 +1 so put in a zero and carry the one. Over to the next place value, we have 1 + 0 again, but again we carried a one over so we have 1 + 1, and again we put in a zero and carry the one. There is nothing else to add to this carried one so our answer is 1000.
1000 = 8 in binary, 011 = 3 and 101 = 5, which verifies our answer.

Multiplication works just the same as it does in base 10, only your only values are 1 and zero, so youre always either multiplying 1 and 1, 0 and 0 or 1 and 0, which isnt very hard to do :D


limit
trigonometry
inequalities
integration
calculus
calculus computations

1) $ \lim_{n\to\plus{}\infty}\sin\left({\pi\sqrt{4n^{2}\plus{}\sqrt{n}}}\right)$.

2)$ \lim_{n\to\plus{}\infty}n\sin(2\pi*e*n!)$.

-1/since $ n$ tends to $ \infty$ it is the same for $ \pi\sqrt{4n^{2}\plus{}\sqrt{n}}$ so this limit does not exist due to the periodicity of $ \sin(n)$.

1) [hide="Solution"]
Notice that $ \sin\alpha \equal{}\sin\left(2n\pi\minus{}\alpha\right)$ for $ n\in\mathbb{Z}_{\plus{}}$, hence we have:
$ \lim_{n}\sin\pi\left(2n\minus{}\sqrt{4n^{2}\plus{}\sqrt{n}}\right) \equal{}\lim_{n}\sin\pi\frac{4n^{2}\minus{}4n^{2}\minus{}\sqrt{n}}{2n\plus{}\sqrt{4n^{2}\plus{}\sqrt{n}}}\equal{}\sin 0 \equal{} 0$[/hide]

2) [hide="Solution"]
Notice that $ n!e \equal{} m\plus{}\frac{1}{n\plus{}1}\plus{}\frac{1}{(n\plus{}1)(n\plus{}2)}\plus{}...$ where $ m$ is an iteger, hence:

$ \sin 2\pi en! \equal{}\sin 2\pi n!\xi_{n}$ where $ \xi_{n}\equal{}\frac{1}{(n\plus{}1)!}\plus{}\frac{1}{(n\plus{}2)!}\plus{}...$, and notice:

$ (n\plus{}k)! \equal{} n!(n\plus{}1)(n\plus{}2)\cdot\ldots\cdot (n\plus{}k) < n!(n\plus{}1)^{k}$, hence:

$ 2\pi n!\frac{1}{(n\plus{}1)!}\equal{}\frac{2\pi}{n\plus{}1}< 2\pi n!\xi_{n}\equal{} 2\pi n!\sum_{k \equal{} 1}^{\infty}\frac{1}{(n\plus{}k)!}< 2\pi\sum_{k \equal{} 1}^{\infty}\frac{1}{(n\plus{}1)^{k}}\equal{}\frac{2\pi}{n}$

For $ n > 4$ that's all in $ \left(0;\frac{\pi}{2}\right)$, so we can write this inequalitys in:

$ \sin\frac{2\pi}{n\plus{}1}<\sin 2\pi n!\xi_{n}<\sin\frac{2\pi}{n}$

$ n\sin\frac{2\pi}{n\plus{}1}< n\sin 2\pi n!\xi_{n}< n\sin\frac{2\pi}{n}$, hence:

$ \lim_{n}\frac{\sin\frac{2\pi}{n\plus{}1}}{\frac{2\pi}{n\plus{}1}}\frac{2\pi}{n\plus{}1}n <\lim_{n}n\sin 2\pi n!\xi_{n}<\lim_{n}2\pi\frac{\sin\frac{2\pi}{n}}{\frac{2\pi}{n}}$

$ 2\pi <\lim_{n}n\sin 2\pi n!\xi_{n}< 2\pi$, hence our limit equals $ 2\pi$[/hide]

i talk about $ n$ is real and not relatif :!:

Ok, very good!!
Another limit:
find $ \lim_{n\to\plus{}\infty}x_{n}$, $ n\in\mathbb{N}$,
where $ x_{n}$ is the solution of the equation $ \sum_{k\equal{}1}^{n}\frac{1}{\sqrt{nx\plus{}k}}\equal{}\sqrt{n}$.

I think it's correct:[hide="Solution"]
Rewriting this:
$ \frac{1}{n}\sum_{k \equal{} 1}^{n}\frac{1}{\sqrt{x\plus{}\frac{k}{n}}}\equal{} 1$

And now let consider $ f(x) \equal{}\frac{1}{n}\sum_{k \equal{} 1}^{n}\frac{1}{\sqrt{x\plus{}\frac{k}{n}}}$ for $ x >\minus{}\frac{k}{n}$, so $ x >\minus{}\frac{1}{n}$.

Notice that $ \lim_{x\to (\minus{}\frac{1}{n})^{\plus{}}}f(x) \equal{}\infty$ and $ \lim_{x\to\infty}f(x) \equal{} 0$ and $ f'(x) < 0$ for all $ x$. So $ f$ is decreasin', so $ g(x) \equal{} f(x)\minus{}1$ has only one root ($ x_{n}$) and $ \minus{}\frac{1}{n}< x_{n}<\infty$, hence

$ \lim_{n}\frac{1}{n}\sum_{k \equal{} 1}^{\infty}\frac{1}{\sqrt{x\plus{}\frac{k}{n}}}\equal{}\int_{0}^{1}\frac{dy}{\sqrt{y\plus{}x}}\equal{}\left[2\sqrt{y\plus{}x}\right]^{1}_{0}\equal{} 2\sqrt{x\plus{}1}\minus{}2\sqrt{x}$
And now we have $ 2\sqrt{x\plus{}1}\minus{}2\sqrt{x}\equal{} 1$, hence $ \lim_{n}x_{n}\equal{}\frac{9}{16}$[/hide]


Let $ a,b\ge 0$ such that $ a\ge \sqrt{b}\left(\sqrt{\frac{a}{4}}\plus{}\sqrt{b}\right)$ . Prove that :

\[ \sqrt[4]{7}\ge \frac{a\plus{}b}{\sqrt[4]{a^4\plus{}b^4}}\]

[quote="Pain rinnegan"]Let $ a,b\ge 0$ such that $ a\ge \sqrt {b}\left(\sqrt {\frac {a}{4}} \plus{} \sqrt {b}\right)$ . Prove that :
\[ \sqrt [4]{7}\ge \frac {a \plus{} b}{\sqrt [4]{a^4 \plus{} b^4}}\]
[/quote]
$ a\ge \sqrt {b}\left(\sqrt {\frac {a}{4}} \plus{} \sqrt {b}\right) \iff$
$ a\minus{}b \ge \sqrt {\frac {ab}{4}} \Rightarrow$
$ a^2\plus{}b^2\minus{}2ab \ge \frac{ab}{4} \iff$
$ (a\plus{}b)^2 \ge \frac{17}{4}ab \iff$
$ t \equal{} \frac{(a\plus{}b)^2}{ab} \ge \frac{17}{4}$.

$ \sqrt [4]{7}\ge \frac {a \plus{} b}{\sqrt [4]{a^4 \plus{} b^4}} \iff$
$ 7(a^4\plus{}b^4) \ge (a\plus{}b)^4$
But $ a^4\plus{}b^4 \equal{} (a^2\plus{}b^2)^2\minus{}2a^2b^2\equal{}((a\plus{}b)^2\minus{}2ab)^2\minus{}2a^2b^2\equal{}(a\plus{}b)^4\minus{}4ab(a\plus{}b)^2\plus{}2a^2b^2$
So it's equivalent to:
$ 6(a\plus{}b)^4 \minus{} 28(a\plus{}b)^2ab \plus{} 14a^2b^2 \ge 0 \iff$
$ 3t^2 \minus{} 14t \plus{} 7 \ge 0$.

And it is easy to check if $ 3t^2\minus{}14t\plus{}7 \ge 0$ when $ t \ge \frac{17}{4}$
(There is never equality btw.)


ΠΡΟΤΑΣΗ  1. Δίδεται τρίγωνο ΑΒΓ και έστω Δ τυχόν σημείο επί της πλευράς ΒΓ. Προσδιορίστε σημείο Μ επί της ΑΔ, τέτοιο ώστε η διχοτόμος της γωνίας ΒΜΓ, να είναι παράλληλη προς τη διχοτόμο της γωνίας Α του τριγώνου.

Καλώς ήρθατε ... Απλά να ζητήσω κάτι .Αν έχετε την καλοσυνη μη βάζετε τόσες μαζεμένες ασκήσεις .Για κάποιον που δεν έχει ίντερνετ για κάποιο χρονικό διάστημα είναι δύσκολο να ξαναπροσαρμοστεί και να παρακολουθήσει όλες τις ασκήσεις .
Και να περιμένουμε ίσως να λυθούν οι 'ηδη υπάρχουσες και μετά να βάζουμε άλλες ....Δεν ξέρω αν έχετε αντίρρηση .

Για  την άσκηση η θέση του $M$ είναι τέτοια ώστε
$\angle{ABM}=\angle{ACM}$  ή αλλιώς το $M$ είναι το σημείο τομής δύο ίσων περιγγεγραμμένων κύκλων που περνούν ο μεν ένας από τα $A,C$  ο δε άλλος από τα $A,B$

Συμφωνώ και συγνώμη για το πλήθος των προτάσεων σε μικρό χρόνο. Ήθελα να ξέρω αν έχουν ενδιαφέρον τέτοια θέματα και βεβαίως δεν έχω κάποια αντίρρηση.

     Πολύ σωστά για τη λύση αυτού του προβλήματος χρειαζόμαστε τη λύση της πρότασης (3).


If $a,b,c>0$ prove or disprove the following inequality:

        $16(a^2 b^2+b^2 c^2+c^2 a^2)+11abc(a+b+c)\geq 27(a^3 b+b^3 c+c^3 a)$ ;)

I see you have a problem with finding counter-examples in inequalities. It is very simple. Here for example, take $a=b$ and $c$ close to 0. Conclusion:false. Not to speak that good sense would suggest that the reverse is probably true, even if I'm sick of finding counter-examples.  Edit: obviously, the reverse is also false; take $c$ close to 0 again.


calculus
derivative
function
LaTeX
calculus computations

Let f(x) be a function on the reals that is continuous everywhere and is differentiable to all orders. Let g(x)=f(x^2). Let n be a positive integer. Without using Taylor/ power series, prove that:

the (2n)th derivative of g(0) = (2n)!/n! * the nth derivative of f(0)

and

the (2n+1)th derivative of g(0)=0.

(Sorry, I dont have latex)

it's not necessary to have latex to be typing in latex.just enclose whatever u write in latex within dollar sign

For example, $ g(x) \equal{} f(x^{2})$. Scroll your mouse over this latex expression to see how it was written. Try it: just put what you want to write between dolar signs (the symbol above the "4" in your keyboard).

Another example: to write $ g(0) \equal{}\frac{(2n)!}{n!}$ type the following:

dollar sign g(0) = \frac{(2n)!}{n!} dollar sign

[quote]Without using Taylor/ power series, prove that: [/quote]
Why would I want to impose that handicap on myself?


RULES:a person posts a wish, the following one says "wish granted" and (s)he imagines the worst scenario.For example:
person1:"I wish there was peace on earth!"
person2:"Wish granted.Now the butterflies are the dominant species."  :lol:

So,here I start:

"I wish I had more maths textbooks!"

Wish granted.  Now all the math textbooks on Earth are in Arabic.

I wish there were no hippies on Earth.

Wish granted. They are all replaced by hippos.

I wish the hippies came back.

(By the way, this alex chap is taking popular games from other forums and putting them here. Smart of him, I should say!)

All the hippies come back. But the hippos stay(and i can't tell which is worse!)

I wish that I was a mouse(this should be easy..)*squeak*

Wish granted. You became a mouse. mickey mouse.

I wish I was at IMO

WISH GRANTED! But you are a trash can there.

I wish I was myself.

Wish granted! You are a mouse, so you are a mouse.

I wish I wasn't ever sick.

You die, preventing you from being sick again.

I wish nobody would screw this wish up.

Wish granted. Noone even cares about screwing [b]your wish[/b] up.

I wish i was bald.  Like DPatrick.  Seriously.

wish granted.  you are a turtle.

i wish i was arnav!!!

Do you really?  Okay...wish granted...

I wish I couldn't speak English.

wish granted, you can't speak english so i can't understand your wish, so your wish isn't granted, so you can speak english, so i can understand your wish, so your wish is granted. . . .

i wish for a sandwich.

The sandwich is actually a penguin. It eats you.

I wish you would capitalize your sentences wobster.

WISH GRANTED.  THE FIRST LETTER OF EVERY SENTENCE IS NOW CAPITALIZED.

I WISH I WERE STILL ALIVE IN MAFIA GAME 20!

WG. Your purpose in life in nonexistent, and you die.
IW that I could come back from the dead with Alexander Hamilton, and kill Shansome for oppressing my Kid Icarus love.

WG- You do come back to life but only as a spirit who cant be noticed or change the course of time in any way, shape, or form and is forgotten forever. You also get a bow to "kill me with" but does nothing.
IW for you to stop hating me

WG but everyone else still does
IW I had more AoPs friends!

WG everyone in the world loves you and you die from lack of oxygen from getting mobbed
IW for something

WG. the something is a nuke. it falls on your house and you die.

IWF a million empty shampoo bottles

You get a watermelon..which is giant and squishs you
IWF real death

[quote=bkn6]Wish granted, but then you don't die.

IWF a watermelon
------------
I just realized that this is probably the original corrupt a wish game. It was made in 2006... 1 year after I was born :o[/quote]

1yearbeforeme

WG but you are reborn
IWF another fun factory

WG- everything is a fun factory including every air molecule and everyone suffocates and gets crushed..

WG- you ae now locked in a closet.
IW you wouldn't try to end this good game


Four persons A, B, C and D of an island had just finished  a chariot  race in which they participated. It is known that the first, second, third and fourth places were assumed by A, B, C and D ( albeit, not necessarily in this order)

A visitor to the island  wants  to know the result of the  race between the four  persons by asking each the outcome of the race.

Each of the four people give three statements as response to the visitor's query.

Among these four people there is one truther (who always tells the truth), a liar (who always lies), an alternator (who tells the truth and lies alternatingly), and a pretender.

The Pretender doesn't like the truth patterns of the other three, and will never follow the order of them. Accordingly,  the Pretender  will make  at least one false statement, and at least one true statement, but not in an alternating way.

A's Statements

1. B is the Liar.
2. C won the race.
3. I came in second.

B's Statements:

1. C is the Truther..
2.I came in last place.
3.C came in second.

C's Statements:

1. D is the Alternator.
2. A won the race.
3. B came in last place.

D's Statements:

1. B is the Pretender.
2. C came in second place.
3. B came in third place.

Determine, each person's type and in what order they finished the race, from the above statements.

i got myself confused but if alternator can say t,t,f:
[hide]
1 a liar
2 c alternator
3 b pretender
4 d truther[/hide]


geometry
IMO Shortlist
incenter
Polars
radical axis
Iran Lemma
geometry solved

The incircle $ \Omega$ of the acute-angled triangle $ ABC$ is tangent to its side $ BC$ at a point $ K$. Let $ AD$ be an altitude of triangle $ ABC$, and let $ M$ be the midpoint of the segment $ AD$. If $ N$ is the common point of the circle $ \Omega$ and the line $ KM$ (distinct from $ K$), then prove that the incircle $ \Omega$ and the circumcircle of triangle $ BCN$ are tangent to each other at the point $ N$.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

I don't like this solution, but I couldn't find a better one this late at night (or this early in the morning; it's 4:15 AM here :)).

Let $S=KA\cap \Omega$, and let $T$ be the antipode of $K$ on $\Omega$. Let $X,Y$ be the touch points between $\Omega$ and $CA,AB$ respectively.

The line $AD$ is parallel to $KT$ and is cut into two equal parts by $KS,KN,KD$, so $(KT,KN;KS,KD)=-1$. This means that the quadrilateral $KTSN$ is harmonic, so the tangents to $\Omega$ through $K,S$ meet on $NT$. On the other hand, the tangents to $\Omega$ through the points $X,Y$ meet on $KS$, so $KXSY$ is also harmonic, meaning that the tangents to $\Omega$ through $K,S$ meet on $XY$.

From these it follows that $BC,XY,TN$ are concurrent. If $P=XY\cap BC$, it's well-known that $(B,C;K,P)=-1$, and since $\angle KNP=\angle KNT=\frac{\pi}2$, it means that $N$ lies on an Apollonius circle, so $NK$ is the bisector of $\angle BNC$.

From here the conclusion follows, because if $B'=NB\cap \Omega,\ C'=NC\cap \Omega$, we get $B'C'\|BC$, so there's a homothety of center $N$ which maps $\Omega$ to the circumcircle of $BNC$.

Grobber, I like your solution! Just to clarify a few points which took me some time to understand:

[quote="grobber"]and since $\angle KNP=\angle KNT=\frac{\pi}2$,[/quote]

This is because the segment KT is a diameter of $\Omega$.

[quote="grobber"]From here the conclusion follows, because if $B'=NB\cap \Omega,\ C'=NC\cap \Omega$, we get $B'C'\|BC$,[/quote]

Is this trivial? The only explanation I have is to use the intersecting secant and tangent theorem, which yields $BB^{\prime} \cdot BN = BK^2$ and $CC^{\prime} \cdot CN = CK^2$, from what we conclude $\frac{BB^{\prime} \cdot BN}{CC^{\prime} \cdot CN} = \frac{BK^2}{CK^2}$, but since the line NK bisects the angle BNC, we have $\frac{BK}{CK} = \frac{BN}{CN}$, so that we get $\frac{BB^{\prime} \cdot BN}{CC^{\prime} \cdot CN} = \frac{BN^2}{CN^2}$, and thus $\frac{BB^{\prime}}{CC^{\prime}} = \frac{BN}{CN}$, what immediately implies B'C' || BC.

As for another solution of the problem, see http://www.mathlinks.ro/Forum/viewtopic.php?t=14741 .  :D

  Darij

I like very mutch Grobber's solution too.

Another way to see that B'C'//BC, once we know that <BNK = <CNK is the following: <B'C'K = <B'NK = <C'NK = <C'KC.

An alternate solution using some computation and inversion :
Let us prove that for a circle through $B<C$ tangent to the incircle of triangle $ABC$ at point $N'$, $K, M, N'$ are collinear.
Now, it is enough if we prove that angles $BKM$ and $BKN'$ are equal.
After a few computations, we get $\tan {BKM} =  \frac { cos(\frac{B}{2}) cos(\frac{C}{2})}{sin(\frac{B-C}{2})}$
Now apply an inversion with center $K$ and radius $BK$.
Thereagain, after a few computations, we get $tan(BKN') = tan(BKM) = \frac { cos(\frac{B}{2}) cos(\frac{C}{2})}{sin(\frac{B-C}{2})}$.
So we are done.

Dear Mathlinkers,
this problem was already posted, but where?
In order to have a complete synthetic proof, we can observe that NK goes through the A-excenter...
Sincerely
Jean-Louis

Dear jayme,
I too on trying for a synthetic solution, tried by the same collinearity.
That is the midpoint of the altitude , tangency point of the incircle with the corresponding side and the corresponding excenter are collinear.
This is true by considering the diametrically opposite point of  the tangency point of the excircle with $BC$ and drawing a parallel through it to $BC$ and using homothety and semiprojection result.

But I was not able to finish the problem using this synthetic idea.
So, could you please tell your complete solution?

[u]I observed now[/u] this [b]old and nice problem[/b]. I''ll search its synthetical proof and return soon. [u]Yes[/u] [b]Jayme[/b],

your remark is very interesting.  Indeed, if denote the point $L$ where the $A$-exincircle touches the  side $[BC]$ ,

then $M\in I_aK\cap IL$ because $\{\begin{array}{c}
\frac {KD}{KL}=\frac {s-a}{a}=\frac {h_a}{2r_a}=\frac {MD}{LI_a}\\\\
\frac {LD}{LK}=\frac sa=\frac {h_a}{2r}=\frac {MD}{IK}\end{array}$.

Let $N'$ be the tangency point of the circle $\omega$ passing through $B,C$ with $\Omega.$ Let $U$ denote the antipode of $K$ WRT $ \Omega$ and $V$ the tangency point of the A-excircle $(I_a)$ with $BC.$ According to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=332584]this topic[/url], $N'U,N'K$ bisects $\angle AN'V$ internally and externally. Let $N''$ be the image of $N'$ under the homothety with center $A$ that takes $\Omega$ and $(I_a)$ into each other. Then $UN' \parallel VN''$ $\Longrightarrow$ $N'K \perp N''V$ $\Longrightarrow$ $\triangle N'VN''$ is isosceles with apex $N',$ which implies that $I_a \in N'K,$ due to $I_aV=I_aN''=r_a.$ But since $M \in KI_a,$ we deduce that  $N \equiv N'.$

Here is a somewhat longer solution...

[hide="Solution"]Denote $I$ the incenter of $\triangle ABC$, $K'$ the reflection of $K$ through $I$, and $L$ the point where the $A$-excircle touches segment $BC$. It's well known that $A$, $K'$, and $L$ are collinear. Furthermore, $BK=LC$.

Let the tangent to $\Omega$ at $N$ intersect line $BC$ at $P$. It suffices to show that
\begin{align*}
PN^2 &= PB\cdot PC \\
\Leftrightarrow PK^2 &=  PB\cdot PC \\
\Leftrightarrow PK^2 &= (PK-BK)(PK+KC) \\
\Leftrightarrow \frac{1}{PK} &= \frac{KC-BK}{KC\cdot BK} \\
\Leftrightarrow \frac{1}{PK} &= \frac{KL}{KC\cdot BK}.
\end{align*}

Because $\triangle PIK \sim \triangle MKD$, $PK=\frac{MD\cdot IK}{DK}$. Thus we want to show
\begin{align*}
\frac{DK}{MD\cdot IK} &= \frac{KL}{BK\cdot KC} \\
\Leftrightarrow \left( \frac{BK\cdot KC}{IK^2} \right) \left( \frac{IK}{KL} \right) \left( \frac{DK}{MD} \right) &= 1.
\end{align*}
Because $A$, $K'$, and $L$ are collinear, $M$, $I$, and $L$ are also collinear. Thus $\triangle MDL \sim \triangle IKL$, so $\frac{IK}{KL} = \frac{MD}{DL}$. Thus it suffices to show
\[
\left( \frac{BK\cdot KC}{IK^2} \right) \left( \frac{DK}{DL} \right) = 1.
\]
Let $A$ be the area of $\triangle ABC$, $h_a=AD$, and $s$ be the semiperimeter of $\triangle ABC$. Now we have
\begin{align*}
\frac{DK}{DL} &= \frac{AD-KK'}{AD} \\
&= \frac{h_a - 2r}{h_a} \\
&= 1 - \frac{r}{h_a/2} \\
&= 1-\frac{A/s}{A/a} \\
&= 1-\frac{a}{s} \\
&= \frac{s-a}{s}.
\end{align*}
On the other hand, we also have
\begin{align*}
\frac{BK\cdot KC}{IK^2} &= \frac{(s-b)(s-c)}{(A/s)^2}\\
&= \frac{(s-b)(s-c)}{\left( \frac{s(s-a)(s-b)(s-c)}{s^2} \right) } \\
&= \frac{s}{s-a}.
\end{align*}
We thus have
\begin{align*}
\left( \frac{BK\cdot KC}{IK^2} \right) \left( \frac{DK}{DL} \right) &= \frac{s}{s-a} \cdot \frac{s-a}{s} \\
&= 1,
\end{align*}
as desired.
[/hide]

Let $I$ be the incenter and $I_A$ be the $A$-excenter of triangle $\triangle{ABC}$. Let $\omega$ be the incircle, $\Gamma$ be the $A$-excircle and let $\Omega$ be the circle with diameter $AD$. The homothety with center $A$ sending $\omega$ to $\Gamma$ takes $K$ to $K'$ where the tangent at $K'$ to $\Gamma$ is parallel to $BC$, perpencular to $AD$ and hence parallel to the tangent to $\Omega$ at $A$. Hence the negative homothety taking $\Omega$ to $\Gamma$ takes $A$ to $K'$ and therefore has center on $AK'$. The center of this negative homothety also lies on the common tangent $BC$ to $\Omega$ and $\Gamma$ and therefore the center of the negative homothety is $K$. This implies that $M$, $I_A$ and $K$ are collinear. Now let the circle $\gamma$ with diameter $II_A$ intersect $\omega$ at $X$ and $Y$. Note that this implies that $XI_A$ and $YI_A$ are tangent to $\omega$ and hence that $XY$ is the polar of $I_A$ with respect to $\omega$. Now let $XY$ intersect $BC$ at $T$. Let $N'$ denote the point on $\omega$ such that $TN'$ is tangent to $\omega$ and $N' \neq K$. Now note that since $T$ lies on the polar of $I_A$ with respect to $\omega$, $I_A$ lies on the polar of $T$ with respect to $\omega$, which is $N'K$. Hence $I_A$, $N'$, $K$ and $M$ are collinear and thus $N=N'$. Since $BI_A$ and $CI_A$ bisect the exterior angles of the triangle at $B$ and $C$, respectively, $\gamma$ passes through $B$ and $C$. Hence $BCYX$ is cyclic and, by power of a point with respect to $\gamma$ and $\omega$, $TN^2 = TX \cdot TY = TB \cdot TC$. This implies that the circumcircle of $\triangle{BCN}$ is tangent to $\omega$ at $N$, as desired.

WLOG, let $AB<AC$ (the case $AB=AC$ is trivial.) Let $a=BC$, $b=CA$, $c=AB$. Let $I$ be the the center of $\Omega$, let $r$ be its radius, let $\Omega$ be tangent to $AB$ and $AC$ at $X$ and $Y$, respectively, and let $Z$ be the point where $XY$ meets $BC$. Let $\omega_A$ be the $A$-excircle of $ABC$, let $I_A$ be its center, $r_A$ be its radius, and let $J$ be its tangency point with $BC$, We first claim that $M$, $K$, and $J$ are collinear.

Let $J'$ be its antipode in $\omega_A$. A homothety centered at $A$ taking $\Omega$ to $\omega_A$ sends $K$ to $J'$. Thus, a homothety centered at $K$ mapping $\Omega$ to $\omega_A$ must send $A$ to $J'$ as well. Line $AD$ is sent to a line parallel to it passing through $J'$, i.e., line $JJ'$. Since $J,D$ both line on $BC$, the homothety must map $D$ to $J$. Thus, the homothety must map the midpoint of $AD$ to the midpoint of $JJ'$, so $M$, $K$, and $I_A$ must be collinear.

Now let $P$ be the midpoint of $ZK$. Because $(Z,B,K,C)$ is harmonic, we must have $PK^2 = PB \cdot PC$ (this can also be verified by computing the lengths directly, given $ZB$, which can be found through Menelaus.) Consequently, if the tangent to $\Omega$ from $P$ distinct from $PB$ is tangent to $\Omega$ at $N'$, we have $PN'^2 = PK^2 = PB \cdot PC$, whence $PN'$ is tangent to the circumcircle of $\triangle N'BC$, i.e., $N'$ is the tangency point of the circle through $B,C$ tangent to $\Omega$. We wish to show that $N'$, $K$, and $I_A$ are collinear.

We claim that $\triangle PIK \sim \triangle I_a KJ$. The result would then follow, for we would have $\angle I_a KJ = \angle ZIK = 90^{\circ} - \angle N'KI = \angle N'KD$.  Since both triangles are right, it suffices to prove $PK/KI = I_aJ / KJ \iff PK \cdot KJ = r r_A$. Now, $rr_A = \frac{K^2}{s(s-a)} = (s-b)(s-c)$, where $K$ is the area and $s$ is the semiperimeter of $\triangle ABC$. Also, $KJ = a - BK - CJ = a - 2(s-b) = b-c$, and
\[ \frac{ZB}{ZB + a} = \frac{KB}{KC} = \frac{s-b}{s-c} \implies ZK = ZB + s-b = \frac{(s-b)(a+b-c)}{b-c}. \]
Thus,
\[ PK \cdot KJ = \frac{ZK}{2} \cdot KJ = \frac{(s-b)(s-c)}{b-c} \cdot (b-c) = (s-b)(s-c) = r r_A, \]
so we are done.

Let $I_A$ be the $A$-excenter with tangency points $X_A$, $X_B$, and $X_C$ to $BC$, $CA$ and $AB$, respectively.  Define $P$ to be the midpoint of $KI_A$.  Let $r$ be the radius of the incircle and $R$ the radius of the $A$-excircle.

It is well-known that $M$, $K$ and $I_A$ are collinear.  We claim that $NBPC$ is cyclic; it suffices to prove that $2BK \cdot KC = 2KP \cdot KN = KN \cdot KI_A$.  On the other hand, by Power of a Point we have that \[ I_AK \left( I_AK + KN \right) = II_A^2 -r^2 \implies KN \cdot KI_A = II_A^2 - r^2 - I_AK^2 \] Now we need only simplify the right-hand side using the Pythagorean Theorem; it is $\left( (r+R)^2 + KX_A^2 \right) - r^2 - \left( R^2 + KX_A^2 \right) = 2Rr$.  So it suffices to prove $Rr = (s-b)(s-c)$, which is not hard.

Now, since $P$ is the midpoint of minor arc $\widehat{BC}$ of $(NBC)$ (via $BK=CX_A$), while the incircle is tangent to segment $BC$ at $K$, the conclusion follows readily.

[asy]/* DRAGON 0.0.9.6
Homemade Script by v_Enhance. */

import olympiad; import cse5; size(11cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); real xmin=-4.18; real xmax=3.94; real ymin=-2.38; real ymax=2.34;

/* Initialize Objects */
pair A = (-1.5, 2.0);
pair B = (-2.0, 0.0);
pair C = (0.5, 0.0);
pair I = incenter(A,B,C);
pair D = foot(A,B,C);
pair M = midpoint(A--D);
pair K = foot(I,B,C);
pair N = (2)*(foot(I,M,K))-K;
pair I_A = (2)*(IntersectionPoint(circumcircle(A,B,C),Line(A,I,lisf),1))-I;
pair X_B = foot(I_A,A,B);
pair X_C = foot(I_A,A,C);
path CircleNBC = circumcircle(N,B,C);
pair X_A = foot(I_A,B,C);
path NI_A = N--I_A;
pair P = IntersectionPoint(CircleNBC,NI_A,1);

/* Draw objects */
draw(A--B, rgb(0.0,0.6,0.6));
draw(B--C, rgb(0.0,0.6,0.6));
draw(C--A, rgb(0.0,0.6,0.6));
draw(incircle(A,B,C), rgb(0.0,0.8,0.0) + linewidth(1.2));
draw(A--D, dotted);
draw(CircleNBC, rgb(0.2,0.8,0.0));
draw(CirclebyPoint(I_A,X_A), rgb(0.0,0.8,0.0) + linewidth(1.2) + dashed);
draw(B--X_B, rgb(0.6,0.6,0.6) + linetype("4 4"));
draw(C--X_C, rgb(0.6,0.6,0.6) + linetype("4 4"));
draw(NI_A, rgb(0.0,0.0,0.8));

/* Place dots on each point */
dot(A);
dot(B);
dot(C);
dot(I);
dot(D);
dot(M);
dot(K);
dot(N);
dot(I_A);
dot(X_B);
dot(X_C);
dot(X_A);
dot(P);

/* Label points */
label("$A$", A, lsf * dir(135));
label("$B$", B, lsf * dir(135));
label("$C$", C, lsf * dir(45));
label("$I$", I, lsf * dir(45));
label("$D$", D, lsf * dir(135));
label("$M$", M, lsf * dir(45));
label("$K$", K, lsf * dir(45));
label("$N$", N, lsf * dir(115));
label("$I_A$", I_A, lsf * dir(45));
label("$X_B$", X_B, lsf * dir(170));
label("$X_C$", X_C, lsf * dir(45));
label("$X_A$", X_A, lsf * dir(45));
label("$P$", P, lsf * dir(45));

/* Clip the image */
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]

heres a surprisingly clean bash

Let $X$ be the midpoint of $I_AK$. It's well known that $I_A, X, K, M, N$ are collinear.
[b]Claim:[/b] $XB^2=XK\cdot XN$.
[b]Proof:[/b] By the Pythagorean Theorem we obtain
\begin{align*}
XB^2&=\frac{a^2}{4}+\left(\frac{A}{b+c-a}\right)^2 \\
&=\left(\frac{c-b}{2}\right)^2+4A^2\left(\frac{1}{b+c-a}\right)\left(\frac{1}{a+b+c}\right)+\left(\frac{A}{b+c-a}\right)^2 \\
&=\left(\frac{c-b}{2}\right)^2+\left(\frac{A}{b+c-a}\right)\left(\frac{A}{b+c-a}+\frac{4A}{a+b+c}\right) \\
&=\left(\frac{c-b}{2}\right)^2+\left(\frac{A}{b+c-a}+\frac{2A}{a+b+c}\right)^2-\left(\frac{2A}{a+b+c}\right)^2 \\
&=XI^2-r^2 \\
&=XK\cdot XN.
\end{align*}
Now the rest is simple. We have $\frac{XB}{XN}=\frac{XK}{XB}$ which implies $\triangle XBK\sim \triangle XNB$ and similarly $\triangle XCK\sim \triangle XNC$ and since $\angle XBK=\angle XCK=\angle XNC$ we have that $XBNC$ is cyclic. Then there is a homothety centered at $N$ which sends $(XBNC)$ to $\Omega$, thus $\Omega$ and $(BCN)$ are tangent at $N$. $\blacksquare$

[quote=math31415926535]@above what is shooting lemma[/quote]

a super complicated phrase which has a very simple definition, of course!

What
Let $X,Y,Z$ be the midpoints of $BC,KN$ and $KI_A$ respectively where $I_A$ is the $A-$excenter and let $I$ be the incenter of $\triangle ABC$ and let $D'$ be the $A-$excircle touchpoint
Notice that $Y \in \odot(BIC)$ because $\angle IYK = \angle IYI_A = 90^{\circ}$ (Using the well known fact that $N-K-I_A$)
Now by PoP $$KY \cdot KI_A = (2 \cdot KY) \cdot \frac{KI_A}{2} = KN \cdot KZ \implies Z \in \odot(BCN)$$
$ID' \perp BC \implies ZM \perp BC$ which means $Z$ is the midpoint of the minor arc $\overarc{BC}$ in $\odot(BCN)$, now, the shooting lemma finishes $\blacksquare$

This is the shortest solution on this thread and requires absoultely $0$ computation. Also, 199th post!

Let $J$ be the $A-$ excenter, and $P$ be the pole of $KN$ with respect to $\Omega$. Also let $Q$ be the midpoint of $KN$.
By homothety $J-K-M$ are collinear, and so $J-K-Q-N$ too. Now,
$$
PN^2=PK^2=PQ\cdot PI=PB\cdot PC,
$$
since $B,C,Q\in (\overline{IJ})$, meaning that line $PN$ is tangent to $\Omega$ and $(BCN)$ at $N$, which proves the statement.

[hide=Storage]
Let $I_A$ be the $A$-excenter and let $G$ be the $A$-excenter contact point. Then, $M, I, G$ are collinear and $M, K, I_A$ are collinear. Let $L$ be the intersection of $KI_A$ and the perpendicular bisector of $BC$. We claim that $BLCN$ is cyclic.

Let $r, r_A$ be the radius of the incenter and excenter respectively, and let $s$ denote the semiperimeter. Note that since $I_AK \cdot I_AN = I_AI^2 - r^2$, it follows that $KN = \frac{I_AI^2 - r^2 - I_AK^2}{I_AK}$. As such, it remains to show that $KB \cdot KC = KN \cdot KL$. Note that $2 \cdot KL = KI_A$. As such,
\begin{align*}
KN \cdot KL &= \frac{I_AI^2 - r^2 - I_AK^2}{I_AK} \cdot KL = \frac{I_AI^2 - r^2 - I_AK^2}{2} = \\
&= \frac{((r + r_A)^2 + GK^2) - r^2 - (GK^2 + r_A^2)}{2} = \frac{2rr_A}{2} \\
&= rr_A = r^2\frac{s}{s - a} \end{align*}. Then, since $KB \cdot KC = (s - b)(s - c)$, equality follows from area formulas.

Then, since $L$ is the midpoint of $\widehat{BC}$, the result follows by the shooting lemma.
[/hide]

I think this was one of the problems that past me fakesolved, idk tho.
This is basicaly intersections spam (prefect for my loved projective geometry :D), let $I$ incenter, $I_A$ the A-excenter, $Z$ the point of tangency of the A-excircle with $BC$, $KI \cap \Omega=K'$, $AK \cap \Omega=G$, $AI \cap BC=L$, $NB \cap \Omega=B_1$, $NC \cap \Omega=C_1$, $\Omega$ tangent to $AB,AC$ at $F,E$ respectivily, $EF \cap BC=X$, now by trivial homothety we can see that the triples $A,K',Z$ ; $M, I, Z$ ; $M, K, I_A$ are colinear, and now by doing projections.
$$-1=(A, L; I, I_A) \overset{\infty_{\perp BC}}{=} (D, L; K, Z) \overset{A}{=} (M, I; AK \cap MI, Z) \overset{K}{=} (N, K'; G, K)$$
Now since $-1=(E, F; G, K)$ we have that $NK', EF, BC$ are concurrent at $X$ so now $-1=(X, K; B, C) \overset{N}{=} (K', K; B_1, C_1)$ and this means that $B_1C_1 \parallel BC$ which gives that $(BNC)$ is tangent to $\Omega$, thus we are done :D

[hide=Diagram]
[img width=40]https://media.discordapp.net/attachments/986049049834713139/1059623205226164334/Screen_Shot_2023-01-02_at_17.04.13.png?width=1116&height=1025[/img]
[/hide]

Let $I$ be the incenter and $I_A$ be the $A$-excenter. Let $T$ be the point on $BC$ such that $NT$ is tangent to $\Omega$. It suffices to show that $TM^2=TB\cdot TC$. Note that $\triangle IKT\cong \triangle INT$, so $MK\perp IT$. Let $P$ be $MK\cap IT$, then $TM^2=TP\cdot TI$. It thus remains to show that $B$, $I$, $P$, $C$ are concyclic.

We already have that $B$, $I$, $C$, $I_A$ are concyclic, so it suffices to show that $B$, $I$, $P$, $I_A$ are concyclic, equivalent to showing that $\angle IPI_A=90^\circ$. Since $MK\perp IP$, it suffices to show that $M,K,I_A$ are collinear, which is true.

Let $I_A$ be the A-excenter. By a well-known lemma, $M$ lies on $I_AK$ so $N$ is in fact the second intersection of $I_AK$ and $\Omega$. With $P$ and $Q$ as the midpoints of $KI_A$ and $KN$ respectively, it is evident that $B$, $C$, $I$, $Q$, $N$ are concyclic (with diameter $II_A$), so

\[KP \cdot KN = KQ \cdot KI_A = KB \cdot KC\]

hence $P$, $B$, $N$, $C$ concyclic. However $P$ lies on the the perpendicular bisector of $BC$, so the circle tangent to $(PBNC)$ at $N$ and $BC$ at $K$ is exactly $\Omega$.

Let $E$ be the point on $BC$ such that $EN$ is tangent to $ \Omega$ with center $I$. It is well-known that $MK$ passes through the $A$-excenter $I_{A}$. let $\{ F, G \} = \Omega \cap (BIC) $. $\angle {IFI_{A}} = \angle{IGI_{A}} = 90^{\circ}$ and $I_{A}F = I_{A}G$ by symmetry, so $I_{A}F, I_{A}G$ are tangent to $\Omega$. $I_{A}$ lies on the polar of $E$ wrt $\Omega$ hence by La Hire, $E$ lies on the polar of $I_{A}$ wrt $\Omega$, which is $FG$. $EN^2 = EF \cdot EG = EB \cdot EC$, thus $EN$ is also tangent to $(BNC)$. $\square$

It is well known that $KM$ passes through the $A-$ excenter $I_A$. Now, we have

$$\measuredangle I_ANF = \measuredangle KNF = \measuredangle KEF = \measuredangle I_ABF$$

so $BFNI_A$ is cyclic. Similarly, $CENI_a$ is cyclic. Now let $NB$ and $NC$ intersect the incircle again at $P$ and $Q$ resp. We have

$$\angle PNF - \angle QNE = \angle BI_AF - \angle CI_AE = \angle BI_AA - \angle CI_AA = -\angle (EF, BC)$$

so $PQ\parallel BC$, implying the desired tangency.

mommmmmmmmmmm get the camera i solved a g7 (also yes this was from egmo lol)

Let $I_A$ be the $A$-excenter of $\triangle ABC$ - it is well known that $M$, $K$ and $I_A$ are collinear. Let $P$ be the midpoint of $\overline{NK}$.

We claim that $BPCI_A$ is cyclic. Proof: it is well knwon that $BICI_A$ is cyclic with diameter $II_A$. Since $P$ is the midpoint of $\overline{NK}$, $\overline{PI} \perp \overline{NKI_A}$, so $\angle IPI_A = 90^{\circ}$, hence $P$ lies on the circumcircle of $IBI_AC$. This completes the desired claim.

Now, by power of a point, it follows that $PK \cdot KI_A = BK \cdot KC$. Let $Y$ be the midpoint of $\overline{KI_A}$. We can rewrite the power of a point equation as
\[ NK \cdot KY = BK \cdot KC,\]
so $NCYB$ is cyclic.

Letting $X$ be the touch point of the $A$-excircle with $\overline{BC}$, it is well known that a homothety of scale $\frac{1}{2}$ sends $I_AX$ to the perpendicular bisector of $\overline{BC}$; hence, $Y$ is the intersection between the perpendicular bisector of $\overline{BC}$ and the circumcircle of $\triangle BNC$. So, the tangent line to $(BNC)$ passing through $Y$ is parallel to $\overline{BC}$. So, there is a homothety about $N$ that sends $(BNC)$ to the incircle of $\triangle ABC$, which proves that the two circles are tangent at $N$.

Let $I_A$ be the $A$-excenter $\triangle{ABC}$, and Let $X$ be the contact point of the $A$-excircle on $BC$. From this, we can say that $M,I,X$ are collinear, and $M,K, I_A$ are collinear.

Let $Z$ be the intersection of the perpendicular bisector of $BC$ and $KI_A$. As $I_AX\perp BC$ and that the perpendicular bisector of $BC$ is the midpoint of $KX$, $Z$ must be the midpoint of $I_AK$.

[i]Claim:[/i]

Quadrilateral $BNCZ$ is cyclic.

[i]Proof:[/i]

Let $r_a$ be the radius of the excircle, and $r$ be the radius of the incircle. By PoP, we can say that:
\[I_AI^2-r^2=I_AK\cdot I_AN,\]
\[(r_a+r)^2+KX^2-r^2=I_AK^2+I_AK\cdot KN,\]
\[r_a^2+r^2+2r_ar+KX^2-r^2=r_a^2+KX^2+2ZK\cdot KN,\]
\[ZK\cdot KN=r_ar.\]

We know that $ICI_AB$ is cyclic with center as the midpoint of arc $BC$. If we reflect $I_AX$ across the diameter of this circle parallel to $AD$, it becomes $I'K$, where $I'K=I_AX$. If use PoP on $K$, we get:
\[r_ar=BK\cdot CK.\]
This means that:
\[ZK\cdot KN=BK\cdot CK,\]
which implies that $BNCZ$ is cyclic.

Thus, by the Circles Inscribed in Segments Lemma, $(BNC)$ is tangent to the incircle at $N$.$\blacksquare$

cute af
[hide = solution]
Let $D'$ be the excircle touch point and $E, F$ the points where the incircle touches $CA, AB$. Let $I$ be the incentre. Define $EF \cap BC$ as $D''$. It is well known that $(D''K;BC) = -1$ (gergonne point exists) and that $I_AD'$ pass through $N$. Let $M_0$ be the midpoint of $NK$. Observe that $\angle IM_0I_A = 90^{\circ}$ and thus $M_0 \in \odot(IBI_AI_C)$. Inverting wrt the incircle, it follows that the $K$ midline of $KEF$ and $IM_0$ concur. Dilating from $K$ with factor two, this implies that $D''N \perp NI_A$. Suppose the tangent to $N$ at the incircle meets $BC$ at $X$. As $XK = XN$ and $\angle D''NK = 90^{\circ}$, it follows that $TN^2 = TK^2 = TD''^2 = TB \cdot TC$ because $(D''K;BC) = -1$. This implies that $TN$ is tangent to the circumcircle of $BNC$. It promptly follows that the incircle and the circumcircle of $BNC$ are tangent to each other (at $N$).
[/hide]
[hide = some random moments of solving]
[img]https://i.imgur.com/hHKsgvs.png[/img]
[/hide]

Let the tangent at $N$ to the incircle intersect $BC$ at $P$. It remains to prove that $PN$ is tangent to $(BCN)$. We proceed with the following well known result:

[color=#f00][b]Lemma:[/b][/color] Points $N$, $M$, $K$, and $I_a$ are collinear.

[hide=Follows from the Midpoint of Altitude Configuration.]
We first denote the $A$-excenter of $ABC$ as $I_A$, the contact point of the excircle with $BC$ as $J$, and the antipode of $J$ as $L$.

We know $DKJ$ collinear, and by homothety at $A$, we have $AKL$ collinear. As \[AD \parallel JL,\] there exists a homothety at $K$ sending $AD$ to $JL$. Then the midpoints of $AD$ and $JL$ are mapped to each other, so $M$, $K$, and $I_A$ are collinear. $\Box$ [/hide] $\Box$

Define $Q = PI \cap NK$. From kite $NPKI$, we see $\angle PQN = 90$, so \[PN^2 = PQ \cdot PI.\]

Because $\angle PQN = IQI_A = 90$, we know $Q$ lies on the circle with diameter $II_A$, or $(IBI_AC)$. Using the power of $P$, we find
\[PB \cdot PC = PQ \cdot PI = PN^2,\]

and thus $PN$ is tangent to $(BNC)$ at $N$. $\blacksquare$

[asy]
size(270); defaultpen(linewidth(0.4)+fontsize(10));

pair A, B, C, D, I, K, M, IA;
A = dir(130);
B = dir(210);
C = dir(330);
D = foot(A, B, C);
I = incenter(A, B, C);
K = foot(I, B, C);
M = .5A + .5D;
IA = extension(B, rotate(90, B)*I, C, rotate(90, C)*I);

pair N = IP(2M-K--M, incircle(A, B, C));
pair P = extension(B, C, N, rotate(90, N)*I);
pair Q = extension(P, I, N, K);

draw(A--B--C--cycle);
draw(incircle(A, B, C));
draw(circumcircle(B, N, C), blue);
draw(circumcircle(I, B, C), heavygreen);
draw(I--P--B--P--N--IA--cycle^^N--I--K--D--A);
draw(C--P--I--P--N, magenta);

label("$A$", A, dir(90));
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$I$", I, dir(90));
label("$K$", K, SW);
label("$N$", N, 2dir(135));
label("$I_A$", IA, S);
label("$P$", P, W);
label("$Q$", Q, dir(90));

dot(A);
dot(B);
dot(C);
dot(D);
dot(I);
dot(K);
dot(N);
dot(IA);
dot(P);
dot(Q);
[/asy]

Let the incenter be $I$. Let $(BIC)$ Intersect $(KM)$ at $Q$ and $I_A$, such that $I_A$ is outside $ABC$. Let $(BCN)$ Intersect $KM$ again at $P$. It suffices to show that $P$ is on the perpendicular bisector of $BC$ due to a homothety.

It is well known (from the "Midpoints of Altitudes Lemma") that $I_A$ must be the $A$-excenter. Then from Incenter/Excenter Lemma, $II_A$ is a diameter of $(BIC)$. It follows that
\[ 90 = \angle IQI_A = \angle IQK \]
but since $KN$ is a chord of the incircle, this implies $Q$ is the midpoint of $KN$. Now taking the power of a point from $K$, we see that
\[ KN \cdot KP = KB \cdot KC = KQ \cdot KI_A \]
But this implies $PI_A = PK$. However, since the $A$-excircle is tangent to $BC$ at the reflection of $K$ across the midpoint of $BC$, then dropping altitudes shows that $P$ lies on the perpendicular bisector of $BC$, as desired. $\blacksquare$


There are 22 black and 3 blue balls in a bag. Ahmet chooses an integer $ n$ in between 1 and 25. Betül draws $ n$ balls from the bag one by one such that no ball is put back to the bag after it is drawn. If exactly 2 of the $ n$ balls are blue and the second blue ball is drawn at $ n^{th}$ order, Ahmet wins, otherwise Betül wins. To increase  the possibility to win,  Ahmet must choose

$\textbf{(A)}\ 2  \qquad\textbf{(B)}\ 11  \qquad\textbf{(C)}\ 12  \qquad\textbf{(D)}\ 13  \qquad\textbf{(E)}\ 23$

[hide="Solution"]$ P\left( n \equal{} 2\right) \equal{} \frac {3 \cdot 2 } {25 \cdot 24}$

If $ n \geq 3$,
Once case is $ \frac {3}{25} \cdot \frac {22}{24} \cdot \frac {21}{23} \cdots  \frac {25\minus{}n} {27\minus{}n}\cdot \frac {2}{26\minus{}n}$. $ 3$ can be at one of first $ n\minus{}1$ places.

So $ P\left( n \geq 3\right) \equal{} (n\minus{}1) \cdot \frac {3 \cdot 2 \cdot (25\minus{}n) }{25 \cdot 24 \cdot 23 }$. It is maximum at $ n \minus{} 1 \equal{} 25 \minus{} n$. So $ n\equal{}13$.

Because $ P\left( n \equal{} 13\right) > P\left( n \equal{} 2\right)$, the correct answer is $ \boxed D$.[/hide]


$ 2b \plus{} b \equal{} 21$
$ 3b\equal{} 21$
$ b\equal{}7$
$ r \equal{} 14$

So the probability is just

$ \frac{7*14}{21*20} \equal{} \boxed{\frac{7}{30}}$

I just got this question, and the answer says $ \boxed{\frac{7}{15}}$.  Is that because the order does not matter, so you would multiply $ \frac{7}{30}$ by $ 2$?

@jxl28: Yes. (Note the words "without replacement")

[quote="modularmarc101"]$ 2b \plus{} b \equal{} 21$
$ 3b \equal{} 21$
$ b \equal{} 7$
$ r \equal{} 14$

So the probability is just

$ \frac {7*14}{21*20} \equal{} \boxed{\frac {7}{30}}$[/quote]

Where do you get 20?

There are $ 21$ possible choices for the first draw. Since there is no replacement, we must next select from $ 21\minus{}1\equal{}20$ choices. After that, you multiply the probabilities, then multiply by $ 2$ (since the order does not matter).

[quote="GameBot"]A bag contains 21 marbles; there are only red marbles and blue marbles in the bag. There are twice as many red marbles as there are blue marbles. If exactly two marbles were to be selected at random, without replacement, what is the probability that one would be red and one would be blue? Express your answer as a common fraction.[/quote]
$ r \equal{} 2b$
$ 2b \plus{} b \equal{} 21$
$ 3b \equal{} 21$
$ b \equal{} 7$
$ \frac{7}{21} \times \frac{14}{20} \times 2! \equal{} \frac{1}{3} \times \frac{7}{10} \times 2! \equal{} \frac{7}{15}$
Or:
$ \frac{b}{21} \times \frac{2b}{20} \times 2! \equal{} \frac{b^2}{105} \equal{} \frac{49}{105} \equal{} \frac{7}{15}$.
Done.


If you have a regular 52 card deck and you pick 5 cards, what is the probability 3 will be kings?

A. 1/881
B. 1/221
C. 2/1105
D. 1/243
E. 1/360

[quote="G-UNIT"][quote="Andreas"][hide]It is $\frac{\binom{4}{3} \cdot \binom{49}{2}}{\binom{52}{5}}$.

So it is $\frac{2}{1105}$.[/hide][/quote]

  Should it be 48 choose 2, because you want 3 kings [b] exactly [/b], the possibility of picking another king is invalid...[/quote]

That's what I did, but the answer is not the same as any of the solutions. Check it.

sorry, you are all right. my mistake for the bad wording.


[quote="lightrhee"][quote="Elyot"]So how many of you guys play (other than me, Richard, and David?)

We should do a big game sometime... any ideas?[/quote]
I'd like to :) , even though I suck at it(haven't played for years, forgot almost everything :( ).[/quote]

I used to play, but I'm not very good either.  Same as Dave, I haven't played for countless years  :blush:

Believe it or not, I used to play Starcraft too. But that was a looooooong time ago and I wasn't good at it. Also, I never had BattleNet (you can guess why  ;) )

[quote="TheDreamer"]Oh my god, now I'm actually getting excited.  So many mathies play.  We should do a big game sometime!

We all say we suck, but I'm not sure what sucking means to different people.  How about everyone say their battle-net wins to see.

I have about 100 :(, which I'm sure is crap.  Elyot has about 1000.[/quote]

My main account has about 900 wins on BWAR, and 100 on SC... my other accounts probably have a hundred or so.  Yep... close to 1000.  My win/loss ratio is almost exactly 2:1.

How good are the other guys?  It seems to me that they are way more studious than me and hence should have even fewer games.

Just got another 2:2 win.

TP vs TP.  I made tanks... ally made goons and zeals... rushed the P, killed a few probes but didn't win... got dropped vults... exped quite fast... dropped 2 vults and a tank on the T... killed lots of SCV... then T drops crap on my ally's ledge so I save him... then P hits him hard and I save his butt again... then mine the crap outta P and T so he can't expand soon.  Then T drops my ally again, this time with way more stuff... he surrenders his exp but I hit P with all I have, taking out his nat for my ally.  Then defend against a huge drop from T, which was unsuccessful but killed a few of my factories... then tank push T and own him with tank, goliath, vulture, and science vessel.  gg.  30 mins.  I have rep.

Ownage.  I have some very cool reps too.  My crazy ultralisk ownage in ZvP is one of my favs... it includes a great fake drop and some kickass anti-archon microing with lings and ultras.

I need practice though... I'm sucking from lack of games.

I play(ed), and I used to be decent.  That was three years ago though.  Unfortunately, I lost my legitimate CD-Key, so I would need one to play on battle.net

[quote="billzhao"]Believe it or not, I used to play Starcraft too.[/quote]
 :o

[quote="Elyot"]We should do a game sometime.  Richard and I are always looking for people to play with (he's ok, but like most of you, he's outta practice)

Apparently we get comps in Banff, so I'm gonna bring SC with me :D[/quote]

Nice. We could have 2vs4...and we'll still pwn you guys :rotfl:

Hi Richard!! Finally joined MathLinks/AoPS? :D

[quote="Richard Peng"]
Nice. We could have 2vs4...and we'll still pwn you guys :rotfl:[/quote]
I know. I'm so bad. :blush:
_______________________________________________________
EDITED: Elyot, you should make a poll that asks "What do you usually play with? 1)Terran 2)Zerg 3)Protoss 4)Random 5)I don't play starcraft".

I think we should play all vs me (which will surely be a loss for me!), then keep moving people to my team until it's fairly matched!

[quote="Elyot"]I think we should play all vs me[/quote]
Then our best chance of winning would be:
[hide="click to see strategy"] worker rush! :rotfl: [/hide]

Did I mention that we'll be playing on Polaris Prime?

;)  Try to worker rush me now...

[quote="11th dimension"][quote="Elyot"][quote="11th dimension"]lol, I only played SC like 10 times in my life......
I can barely reach a stalemate with Medium level computer   :(
So, If I wanna join, can some one give me a quick tutorial there?[/quote]

Medium computers?  I didn't know they came in different difficulties...[/quote]

I think they do, dunno.[/quote]

The campaign AI scripts do... but there is only one script for melee, and it's the same one that battle.net comps use.


I want to prove that for every composite number $ n$, there exists another integer $ m$ such that $ \varphi{(m)} \equal{} n$. Can I have hints on how to start off?

That's a very bold statement. Why do you think this statement "should" be true? I'll pose a different question, that I think you should explore first:

[b]Question:[/b] [i]Find a positive integer $ m$ that solves the equation $ \varphi (m) \equal{} 14$, or prove that one does not exist.[/i]

What if the equation was changed to $ k\varphi{(m)} \equal{} n$ for $ k \in \mathbb{Z}$?

You'd note that $ \varphi(2)\equal{}1$ and therefore you always have the solution $ (k,m)\equal{}(n,2)$.

[quote="t0rajir0u"]This is obviously false if $ n$ is odd.  Furthermore, note that $ n \equiv 2 \bmod 4$ if and only if $ m \equal{} 4$, an odd prime power, or twice an odd prime power, since otherwise $ m$ is the product of two relatively prime integers both of which have an even totient.  That gets you the answer to TZF's question.[/quote]
I don't get your argument... first why is 4 an odd prime power (shouldn't it be even prime power since 2²=4)?  :huh:

That's a list of cases.  $ m \equal{} 4, m \equal{} p^k, m \equal{} 2p^k$ for $ p$ an odd prime are the only possibilities, and only if $ p \equiv 3 \bmod 4$.  In any other case, $ m$ has two relatively prime factors with even totients, so the product of their totients is divisible by $ 4$.  It follows that the only way to get $ \varphi(m) \equal{} 14$ is for $ m$ to be divisible by $ 7$ - but then $ \varphi(m)$ has to be divisible by $ 6$.

[quote="grn_trtle"]You'd note that $ \varphi(2) \equal{} 1$ and therefore you always have the solution $ (k,m) \equal{} (n,2)$.[/quote]

Sorry again, I forgot to mention that $ m \not\equal{} 2$. (This lemma is part of a proof I'm working on.)

You're probably better off posting the problem then.  Otherwise, it's not at all clear in what direction this statement should be salvaged.

Ok so I am supposed to prove that $ 2^n \minus{} 1$ is composite for all composite $ n$. I started off by assuming that there exists an integer $ m$ such that it divides $ 2^n \minus{} 1$ for composite $ n$. Then we have

$ 2^n \equiv 1 \mod{m}$

Because $ n$ is composite it can be rewritten as product of two integers and I wrote it so that one of the integers was $ \varphi{(m)}$ and the other was $ k$. And that's where my original question comes from.

In this case you can write down what $ m$ is.  If $ d | n$ then you can take $ m \equal{} 2^d \minus{} 1$.

This does not in fact imply that $ \varphi(m) | n$; why?

Thanks, that helped a lot. Btw, how would I have needed to think about the problem in order to stumble upon $ m$?

And as to your question, I'm don't know why.

Consider how to factor $ x^{a} \minus{} 1$ -- if $ n$ is composite then $ 2^n \minus{} 1 \equal{} (2^b)^a \minus{} 1$ for some $ a, b$.

[quote="sunehra"]Btw, how would I have needed to think about the problem in order to stumble upon $ m$? [/quote]
If you're familiar with how to factor expressions of the form $ a^n \minus{} b^n$, that's it.  I mean, there's more going on here, but it's a little involved.

[quote="sunehra"]And as to your question, I'm don't know why.[/quote]
Take, for example, $ m \equal{} 7$.  Then $ 2^3 \equiv 1 \bmod 7$ but $ \varphi(7)$ doesn't divide $ 3$.

The problem is that $ \varphi(m)$ isn't the [i]minimal[/i] exponent such that $ 2^n \equiv 1 \bmod m$; really it's an upper bound instead.

Just out of curiosity, how would we be able to prove it purely by modular arithmetic?

$ 2^d \equiv 1 \bmod (2^d \minus{} 1) \Leftrightarrow 2^{dk} \equiv 1 \bmod (2^d \minus{} 1)$ for any positive integer $ k$.  As in any problem of this type, the easiest way to show that a number is composite is to factor it explicitly.


superior algebra
superior algebra unsolved

If I want to show that $ f\sim g$ why is it enough to show that f' =g', where f' and g' are differentiate of f and g.

Under what relation $ \sim $?  I'm not clear on why this is a number theory question, either.


Suppose the $f$ is a function that assigns to each real number $x$ a real value $f(x)$ and suppose the equation
\[ f(x_1 + x_2 + x_3 + x_4 + x_5) = f(x_1) + f(x_2) + f(x_3) + f(x_4) + f(x_5) - 8 \]
holds for all real numbers $x_1, x_2, x_3, x_4, x_5$.  What is $f(0)$?
\[ \text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2 \]

[quote="drunner2007"]I think i saw that on a Harvard-MIT tournament[/quote]
is that for their students??

[quote="joml88"]No, it's for high school students.[/quote]
Then what does it used for? To select students?

[quote="joml88"]It's a contest hosted by Harvard and MIT.  It has the same intent of other contests like ARML and mandelbrot etc.[/quote]
sounds interesting...where can i find informations and problems of it?


trigonometry
logarithms
irrational number
calculus
calculus computations


Let $ x_1 ,x_2 ,...,x_n$be real numbers.
Prove that there exist a real number $ y$such that the sum of $ \{ x_1  \minus{} y\} ,\{ x_2  \minus{} y\} ,...,\{ x_n  \minus{} y\}$ is at most $ \frac{{n \minus{} 1}}{2}$.

Can $ y$ always be chosen to ne one of $ x_1 ,x_{2,} ...,x_n$?

we may assume WLOG that $ 0\leq x_1 < x_2 < \cdots < x_n < 1$. let $ E(y) \equal{} \sum_{i \equal{} 1}^{n}\{x_i \minus{} y\}$. we'll prove that for some $ i$, $ E(x_i)\leq \frac {n \minus{} 1}{2}$, answering your question in the affirmative. assume, for a contradiction, that all of the values, $ E(x_i)$ are greater than $ \frac {n \minus{} 1}{2}$. let $ S \equal{} \sum_{i \equal{} 1}^n x_i$. note that for all $ i$,
\[ E(x_i) \equal{} \sum_{j \equal{} 1}^{i \minus{} 1} (1 \plus{} x_j \minus{} x_i) \plus{} \sum_{j \equal{} i \plus{} 1}^{n}(x_j \minus{} x_i) \equal{}
\]

\[ i \minus{} 1 \plus{} S \minus{} x_i \minus{} (n \minus{} 1)x_i \equal{} i \minus{} 1 \plus{} S \minus{} nx_i,
\]
so we get $ \frac {n(n \minus{} 1)}{2} < \sum_{i \equal{} 1}^n E(x_i) \equal{} \sum_{i \equal{} 1}^{n}(i \minus{} 1) \plus{} nS \minus{} nS \equal{} \frac {n(n \minus{} 1)}{2}$, a contradiction.


evaluate
           \sum C(n-2k,k)(-4/27)^k  for 0 \leq k \leq n/2
excuse me for this one :)  :D

I do not understand : Is it C(n-k,k) or is it n  \geq  3k, since in C(n,p) you must have n  \geq  p? Or are you considering that C(n-2k,k) = 0 for n < 3k?

Pierre.

of course you're right , but we want to tell a close form or explicit form for this summation ( so when n<3k ,equals 0)


Let $ABCD$ be a quadrilateral inscribed in a circle.$AC$ $\cap$ $BD$=$I$.Prove that
$\frac{AB}{CD}+\frac{CD}{AB}+\frac{BC}{AD}+\frac{AD}{BC}$ $\leq$ $\frac{IA}{IC}+\frac{IC}{IA}+\frac{ID}{IB}+\frac{IB}{ID}$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=62530

  dg


Let {a_n} be an unbounded sequence . [tex] \lim_{n->\inf}(a_{n+1}-a_n/2)=0 [/tex] . Prove that : [tex] \lim_{n->inf}a_n =0 [/tex]

Hmm... He said $a_n$ is an [b]unbounded[/b] sequence  :?
I think he meant "infinite sequence".

It's really is well-known that if $|\lambda|<1$ and $a_{n+1}-\lambda\cdot a_n\to 0$, then $a_n\to 0$.

nttu,
you can write
[code]\lim_{n \rightarrow \infty}[/code]
to get
\[\lim_{n \rightarrow \infty}\]

Thank you Jhaussmann5 !
 For this problem : I have one solution (just use definition) . I want somebody will post it first .

More shorter way:
[code]\lim_{n \to \infty}[/code]
to get $\lim_{n \to \infty}$.


A juice company sells its product in either a 48-ounce size or a 32-ounce size. It charges $ \$3.90$ for the 48-ounce size. how much should it charge for the smaller size if it wants the price per ounce to be $ 25\%$ more than the price per ounce of the larger size?

Find the average price for 48-ounces.

3.9/48 = 0.08125

since the smaller ones is 25% more...

0.08125 * 1.25 = 0.1015625

multiply by 32, to get [b]3.25 dollars[/b]

Or, if you don't want 6-digit decimals, you can set up a proportion:
$\frac{48}{3.90}=\frac{32}{x}$, where $x$ is the amount of money (proportional to 48 ounces, making it $100\%$ price)

That makes $x=2.60$. Since we need to find $25\%$ more of it, which is $100\%+25\%=125\%$, we do
$\$2.60\times1.25=\boxed{3.25}$.


[quote="alcachofo"]He scored 39 with 16 years old in 2001, and didn't participated again.
I'm curious.[/quote]
He went to college in the fall of 2001. Russian students usually graduate from high school at the age of 16-17.


This is a math site, so it would be nice to have buttons for some math symbols, like infinity, pi, etc.

Custom smileys...  Good idea.  I'll take a look into it in the next week or so and see what we can come up with.

What symbols would you want?

plus or minus sign and congruence sign for mods are the ones I can think of.

Pi, definitely. I'd also like a radical sign, possibly sigma, theta, and alpha. (for sums and angles.) Also a plus or minus sign, and how about squared and cubed symbols? Those are most commonly used. Also, how about a derivative sign? What else... how about a divided by sign, i.e. 2 dots above and below a dash. That would really help for dividing complex fractions. I seem to be rambling a little it, but so what. That's really all you'd need for most equations. I haven't taken calc yet so there may be some calc symbols that I'm not aware of. Also, how about an angle sign? ( < ?)

-Isaac

After looking around on the internet for a while, I found a free program called Equation Magic Lite that can do many of these equations. The only snag is that there is no direct Export to GIF function. In other words, to get a picture of the equation, you must first do a screenshot [b]with alt - print screen.[/b] After that, go into Paint or something and crop the equation from the screenshot. If you do that, it works fine. Here's an example:
[img]http://www.geocities.com/andy17null/test_equation.gif[/img]
If enough people want to do this, we could make a geocities site or something for pics...
btw, that picture should work. If it doesn't, then idk why not.

-Isaac

Can't see the image, but assuming the custom smileys are as easy to implement as I think (hope) they'll be, we should have some symbols available in a week or so.  Keep listing here what you want - I expect with the custom smileys we'll be able to essentially create a rudimentary typesetting language for maht stuff (i.e. you'll type :inf: to get an infinity symbol and such).

The link to the above pic is [img]http://www.geocities.com/andy17null/test_equation.gif[/img]
That should work... good lick with the custom smileys.

-Isaac

I don't really know too much about how easy/difficult this would be to implement, but it would be really nice if we could write equations in TeX and then have it compile. Even if people don't know TeX, it's easy to learn enough to do simple equations. I imagine this would be too much to hope for though.

I know what TeX is but I cannot find the program. Can you help me?

-Isaac

I don't know where you can get a good one on the web. Try the AMC website (http://www.unl.edu/amc) and follow up the problems link. They have something. But I wasn't able to install. Try to get someone to burn you a program. That's what I did.

I picked up MikTeX here: [url]http://www.miktex.org/[/url]

I'm told there are better versions of LaTeX for Windows (i.e. ones with clearer error messaging), but if you're not writing extremely complicated or long documents, this should be fine.

Textures is among the best for Mac, but I don't know if it's free.

You can get a pdflatex/pdftex as part of the MikTeX, which enables you to create pdf docs from your TeX files (this is how we produce most of our pdf files for our classes).[/url]

To answer Zeta's question: it is very hard.  There are a few projects out there attempting it (WebTeX, for one), but they haven't been terribly successful as far as I can tell.  We'll have to do something a little more rudimentary (though it will be TeX-like, assuming I can create and pack as many custom smileys as I want into the board, in that you'll have 'commands' you can type to get your custom images much like you have commands for TeX functions.  What you won't have is moderate/advanced typesetting such as fractions, exponents (though I will probably make a few lower powers if they look right), etc.)

"Pi, definitely. I'd also like a radical sign, possibly sigma, theta, and alpha. (for sums and angles.) Also a plus or minus sign, and how about squared and cubed symbols?"

You've got issues, I've got solutions.

Pi: Use the symbol font, and do lowercase p.

Radical: hmm...might need a gfx for this. though this is hard because you will have to put a radical over the entire expression, which might be difficult. For text purposes, I just use sqrt( ... ).

Sigma: Ditto, in font symbol, it is S.
theta: symbol q
alpha: symbol a
Plus or minus sign exists in ASCII, its ASCII code is 241.

Lord Venom, we already have these symbols. Click on the View more Emoticons link, and then you can get a variety of math symbols.

Besides, things look much nicer when you don't have to remember that S means sigma, etc etc.  Otherwise, I might say something like S(n) and you could misinterpret that...

On a side note, if you type "pi" with colons around it, or ^2 with colons around it, or ^3 with colons around it, etc., you'll get a surprise!


Okay. You want to post all of these in one large post instead of spamming up the forum. But the answer to this one is:

The string wraps around the 4-foot circumference four times for a total of 16. So, if you unwrap the string, it makes a right triangle with a height of 12 and a length of 16. Then, using the Pythagorean Theorem, we have 20 as the length.


function
geometry
calculus
derivative
logarithms
limit
calculus computations

We have the function $ f(x)\equal{}\frac{1}{x^{2}\plus{}1}$ on $ (0,1]$. Show that there exists a triangle with minimal area formed by the tangent to $ f(x)$ and the $ x\minus{}$ and $ y\minus{}$ axis.

P.s.: The procedure to solve the task is simple, however the area $ A(x)\equal{}\frac{(3x^2\plus{}1)^2}{4x(x^2\plus{}1)^2}$ seems to have a complicated derivative. Do you know any improvement, or is brute force with calculation of $ A^{\prime}\equal{}0$ the only option?

You do not need to [b]find[/b] the minimal triangle, just to show its existence. Now, why would you need any derivatives for that?

Intuitively the point that gives the minimum area is the inflection point.

fedja's solution would be using the calculus of variation.

[quote="tmrfea"]We have the function $ f(x) \equal{} \frac {1}{x^{2} \plus{} 1}$ on $ (0,1]$. Show that there exists a triangle with minimal area formed by the tangent to $ f(x)$ and the $ x \minus{}$ and $ y \minus{}$ axis.

P.s.: The procedure to solve the task is simple, however the area $ A(x) \equal{} \frac {(3x^2 \plus{} 1)^2}{4x(x^2 \plus{} 1)^2}$ seems to have a complicated derivative. Do you know any improvement, or is brute force with calculation of $ A^{\prime} \equal{} 0$ the only option?[/quote]

$ A'(x) \equal{} \frac {1}{4}\cdot \frac {2(3x^2 \plus{} 1)\cdot 6x\cdot x(x^2 \plus{} 1)^2 \minus{} (3x^2 \plus{} 1)^2\{(x^2 \plus{} 1)^2 \plus{} x\cdot 2(x^2 \plus{} 1)\cdot 2x \}}{x^2(x^2 \plus{} 1)^4}$

$ \equal{} \frac {1}{4}\cdot \frac {12x^2(3x^2 \plus{} 1)(x^2 \plus{} 1) \minus{} (3x^2 \plus{} 1)^2\{(x^2 \plus{} 1) \plus{} 4x^2\}}{x^2(x^2 \plus{} 1)^2}$

$ \equal{} \frac {1}{4}\cdot \frac {(3x^2 \plus{} 1)\{12x^2(x^2 \plus{} 1) \minus{} (3x^2 \plus{} 1)(5x^2 \plus{} 1)\}}{x^2(x^2 \plus{} 1)^2}$

$ \equal{} \minus{} \frac {1}{4}\cdot \frac {(3x^2 \plus{} 1)(3x^4 \minus{} 4x^2 \plus{} 1)}{x^2(x^2 \plus{} 1)^2}$

$ \equal{} \frac {1}{4}\cdot \frac {(x \plus{} 1)(1 \minus{} x)(\sqrt {3}x \plus{} 1)(3x^2 \plus{} 1)}{x^2(x^2 \plus{} 1)^2}\cdot (\sqrt {3}x \minus{} 1)$,

thus the sign of $ A'(t)$ is same as that of $ \sqrt {3}x \minus{} 1$, yielding $ x \equal{} \frac {1}{\sqrt {3}}$

by $ 0 < x\leq 1$, $ A(x)$ has a local minimum and minimal which gives a local and minimal value of $ A(x)$.

[color=blue]More easier calculation of $ A'(x)$[/color]

Let $ f(x) \equal{} \frac {(3x^2 \plus{} 1)^2}{4x(x^2 \plus{} 1)^2}$, use logarithmic differentiation,

$ \ln f(x) \equal{} 2\ln (3x^2 \plus{} 1) \minus{} \ln x \minus{} \ln 2(x^2 \plus{} 1)$

$ \therefore \frac {f'(x)}{f(x)} \equal{} \frac {12x}{3x^2 \plus{} 1} \minus{} \frac {1}{x} \minus{} \frac {2x}{x^2 \plus{} 1}$

$ \equal{} \frac {(x \plus{} 1)(1 \minus{} x)(\sqrt {3}x \plus{} 1)}{x(x^2 \plus{} 1)(3x^2 \plus{} 1)}(\sqrt {3}x \minus{} 1)$

[color=red]Peculiar Solution[/color]

Rewrite $ A(x) \equal{} \frac {1}{4}\cdot \frac {1}{x}\left(3 \minus{} \frac {2}{x^2 \plus{} 1}\right)^2$, let $ t \equal{} 3 \minus{} \frac {2}{x^2 \plus{} 1}\ (1 < t\leq 2)$

$ A(x)^2 \equal{} \frac {1}{16}\cdot \frac {t^4(3 \minus{} t)}{t \minus{} 1}: \equal{} \frac {1}{16}g(t)\ (1 < t\leq 2)$,

$ g'(t) \equal{} \frac {2t^3(2 \minus{} t)}{(t \minus{} 1)^2}\cdot (2t \minus{} 3)$, thus the sign of $ g'(t)$ is same as that of $ 2t \minus{} 3$, yielding $ t \equal{} \frac {3}{2}\Longleftrightarrow x \equal{} \frac {1}{\sqrt {3}}$ which gives a local and minimal value of $ A(x)$.

[quote="kunny"]
fedja's solution would be using the calculus of variation.[/quote]
 :what?: I just meant that a continuous function tending to $ \plus{}\infty$ near the endpoints certainly attains its minimal value somewhere on the interval...

I do know it, but beginner isn't apt to notice that, I think. :wink:

Anyway, usual differentiation gives easy sulution. See my solution. :lol:

Thank you very much, fedja and kunny.

I forgot to write:

[quote]
We have the function  on $ (0,1]$. Show that there exists a triangle with minimal area formed by the tangent to the curve and the $ x \minus{}$ and $ y \minus{}$axis.
[/quote]

And determine it...

Yes, the argument for existence is simple. Since $ \lim_{x \to 0} A(x) \equal{} \infty$, taking a closed interval $ [a,1]$, where $ a$ is enough near $ 0$, we can say that the minimum of $ A(x)$ on $ [a,1]$ is also a global minimum on $ (0,1]$.

But yes, here comes the "beautiful" part of the task on determining it... This was the first task on our exam.