Mole 2 is currently on hiatus, so I'm going to start Mole 3 applications early (it's been discussed that I will moderate). Mods, if you deem it too early for this, feel free to lock the thread, and bring it back when it is more appropriate.

I want to start the game on September 1 or about a week after Mole 2 ends, whichever is later. But that's tentative, of course.

Applications for the third installment of The Mole on AoPS begin now! If you are applying, please do not make it public. Applications should go to my (CatalystOfNostalgia) PM box. Main threads for the previous games can be found here, for an idea of how the game works:

Season 1: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=186949[/url]
Season 2: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=202287[/url]

Also, for an idea of my moderation style/skills, see the Big Brother thread:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=197275[/url]

To apply, simply copy/paste the application below into a PM to me, [b]with subject heading “Mole 3 Application” (or something similar)[/b] and fill in the answers (so that the questions and answers are in the PM).

-Username:
-Gender (so I can correctly refer to you has ‘he’ or ‘she’):
-Date of Birth (MM/DD/YY):
-If it is 12 PM in New York City (EST), what time is it where you live?
-Do you have AIM? If so, what is your screenname?
-Did you participate in either of the first two mole games? Briefly describe and assess your strategies and performances in both games.
-How often did you go inactive in the previous games, and for how long?
-How likely is it that you will need to be replaced?
-How often are you willing to check the game thread?
-Do you want to be The Mole? Why?

Would you rather be…
-a cat or a dog?
-too hot or too cold?
-dead or suffering endlessly?
-the betrayer or the betrayee (person betrayed)?
-stupid and popular or smart and an outcast?
-the cheater or the loser?
-The Mole or the sole holder of his/her identity?

Another bump. I believe the number of applicants is up to 18.

Question: How many applications will you take. Like the real game where they have 12 people total. Or is it some other number so the game last a little longer?

[quote="zapi2007"]Just out of curiosity, has anyone actually seen the Mole on tv?

It's actually pretty cool, because they're actually playing for money, but their ratings are so bad because people like to watch brainless stuff  :([/quote]

yes I watch it, I love it! The finally is next week at 9:00 Centeral. Bet you Craig is the mole  :D . I was rootin for Mark all the way.

Hmmm let's not let this degenerate into spam. But I'll take any applications that come, and I'll probably pick 12 to play. Unfortunately that means that I'll have to cut people. But if there is a very large number of promising applicants, I might expand it. Then again, my creativity for challenges only goes so far...I have 4-5 more to plan.

Craig is The Mole, but I wouldn't be surprised if I'm wrong. I think the beauty of this game is that people get eliminated approximately in order of how suspicious they are (backwards). Obviously there are exceptions. But in all three (non-celebrity) seasons, the three by far most suspicious people ended up in the final three. Beauty, I tell you.

Yeah, I liked Mark, too
He looks a little more humble once they burned the journals
I wonder if he has stopped taking too many notes...

I have the feeling Craig might be the mole, even though Mark did make them lose a lot of money, most in that last task so he could help himself

If I were Mark (and not The Mole, of course), I would have been almost 100% sure that Craig is The Mole, because I knew (or was told...) Clay thought it was Nicole, and promptly got executed. Then I would be aware that Nicole couldn't possibly beat me on the quiz seeing that she took no notes for the whole game and had to crack at some point in which everyone knew who The Mole is and it came down to who was the most observant.

I would have taken the money. But that's just me.

Mark definetly has the best chance of winning since he took the information rather than the cash. Whether or not the information helped him, no one knows except him and the host.

At first I thought that Nicole was the mole.
Then I realized that she was doing too good of a job being stupid and idiotic.

And by the final four, my views shifted to Craig.

It would be so twisted if Mark is the Mole...

I don't think they would make Mark The Mole, as he was the oldest man on the show and they already had the oldest man on the show be The Mole (Bill, season 2). I thought they were going to give it to a black guy, so my thought at the beginning of the show was Clay (and also Liz, because she's an old woman...). Yay for PCness. I doubted Craig at the beginning because that seemed like way too obvious of a choice, but then I realized that it just had to be him, by process of elimination. Clay didn't really do anything.

I did notice a clue, actually, on the night Alex was executed, they showed some code on one of those big barrels. They zoomed in on it for like 5 seconds. I don't remember the code and I don't know what it meant though...

Also I believe the Fibonacci numbers in the final 5 task was a clue. Again I don't know what it's in reference to. Did anyone notice that there was a big N on Mark's camera in the child's play task?

it covered like the whole camera and was in sharpie...

YES!!! MARK WON...Im typing this at the commercial break will learn the Mole in like 2 minutse.

I had Craig pegged since after ep2, when my main suspect, Liz, gone. Craig was always my #2. I picked up on several of their clues: the never had results shown, and the noises. Yay. Anyway, some random predictions of mine, from pre-season, that are being revealed FOR THE FIRST TIME.

Order of execution:
Clay
Mark
Ali
Bobby
Nicole
Paul
Victoria
Alex
Kristen
Craig
Marcie (winner-i know...)
Liz (Mole)

Oh that's it...huh.

Also, I had F5 order perfect, after Alex went. Go me.

Lol Mark escaped near execution 6 times I think. One was by a mere second.

I haven't received an app in a while, so I think it makes sense to close them and start sooner rather than later. The application deadline is now set at [b]Sunday, August 17 at 8:00 PM EST[/b]. If you are still considering applying, now is the time to decide.

I have a rough idea of who the twelve players to play will be. After the deadline, those twelve people will be informed that they have been selected and I will await confirmation from them. I also have a rough idea of who The Mole will be, but that will not be officially decided until all of the players are confirmed.

[quote="jhredsox93"]alright, can we get a pm confirmation though so we don't have to insanely check this forum to see if we're in?[/quote]

Yes, the players are meant to be kept confidential anyway until the game starts (which won't be for a few days).

[quote="erabel"]Have you sent them yet?[/quote]
[quote="CatalystOfNostalgia"]The application deadline is now set at [b]Sunday, August 17 at 8:00 PM EST[/b]. [/quote]
I don't think so...  :wink:

All PM's have been sent (sorry if I sent you two, it's hard to keep track sometimes...). If you didn't get one, I really do regret it...there were 21 applications in total. You are now on the replacement list.

Oh, and still don't reveal your standing in the game.

Awaiting three confirmations.

(edit, on August 18 at 7:14 PM) I now have all but one confirmation. To get things moving, the final person (you know who you are) now has 24 hours to get in that confirmation. If I don't get it, I'm going to have to give the final spot to someone else.


Given $ a \plus{} 5b \plus{} 10c \plus{} 25d \plus{} 50e \plus{} 100f \equal{} 100$ has no non-negative integral solutions, find the minimum value of $ a \plus{} b \plus{} c \plus{} d \plus{} e \plus{} f\in\mathbb{N}$.

[hide]

The obvious answer is f=1 and the other variables equal 0, giving us an answer of $ \boxed{1}$. Unless you are asking that a, b, c, d, e, f all must be > 0?

Oh, and do you have solutions for your infinitely nested radicals problems?
[/hide]

I'm pretty sure he's asking for a solution where at least one of the six variables isn't a positive integer or 0, so that particular explanation would be a contradiction.
[hide]However, you still get $ \boxed{1}$ as the answer using $ e \equal{} 3$ and $ d \equal{} \minus{} 2$.[/hide]

[quote="mathwizarddude"]$ a,b,c,d,e,f$ must be non-negative integers. (also they must not satisfy the first eqn with RHS=100 since it is stated it has no non-negative integral solutions)[/quote]How is it possible to solve the problem if the variables are non-negative but the equation has NO non-negative integral solutions? :huh: This is like "here is a box of RED apples and fill the package with apples that are NOT RED, but ... you are not given GREEN apples!" :ninja:

[quote="MathWhiz2009"]The variables [b]all[/b] have to be negative.[/quote]As highlighted with [b]bold[/b] above, that doesn't make sense either because then the sum on the LHS would be negative.  Perhaps [b]some[/b] would be the right word...

Sorry maybe my wording was confusing, but here's the clarification:

find the minimum value of $ x\in\mathbb{N}$ such that when x is written in the form $ a\plus{}b\plus{}c\plus{}d\plus{}e\plus{}f$, the six non-negative integers must not satisfy $ a\plus{}5b\plus{}10c\plus{}25d\plus{}50e\plus{}100f \equal{} 100$.

[quote="mathwizarddude"]Sorry maybe my wording was confusing, but here's the clarification:

find the minimum value of $ x\in\mathbb{N}$ such that when x is written in the form $ a \plus{} b \plus{} c \plus{} d \plus{} e \plus{} f$, the six non-negative integers must not satisfy $ a \plus{} 5b \plus{} 10c \plus{} 25d \plus{} 50e \plus{} 100f \equal{} 100$.[/quote]Oh OK, then $ x\equal{}1$ because $ a\equal{}1,b\equal{}c\equal{}d\equal{}e\equal{}f\equal{}0$ are non-negative integers that do not satisfy $ a\plus{}5b\plus{}10c\plus{}25d\plus{}50e\plus{}100f\equal{}100$.

$ x$ cannot be $ 1$ since we can set $ f\equal{}1$ and others $ 0$.

See [url]http://mathforum.org/library/drmath/view/56796.html[/url], the best way to do it is to make an algorithm that exchanges coins to gradually increase the impossible values of x by 4, as he did.

The answer is [hide]77[/hide].


Find the locus of the points whose sum of squares of the distances for two fixed points is constant and equal to $ k^2$.

I'm not sure if my translate is clear, so let $ A$ and $ B$ be the two fixed points and the distances to the point $ P$ for $ A$ and $ B$ ($ P$ is a point of the locus) are $ d_1$ and $ d_2$. Hence, $ d_1^2\plus{}d_2^2\equal{}k^2$.

[quote="LoboSolitario"]Find the locus of the points whose sum of squares of the distances for two fixed points is constant and equal to $ k^2$.

I'm not sure if my translate is clear, so let $ A$ and $ B$ be the two fixed points and the distances to the point $ P$ for $ A$ and $ B$ ($ P$ is a point of the locus) are $ d_1$ and $ d_2$. Hence, $ d_1^2 \plus{} d_2^2 \equal{} k^2$.[/quote]

A circle?

[hide="Ugly coordinates"]Set $ A \equal{} ( \minus{} 1,0),B \equal{} (1,0)$ and the sum of the squared distances from $ P \equal{} (x,y)$ is $ (x \minus{} 1)^2 \plus{} y^2 \plus{} (x \plus{} 1)^2 \plus{} y^2 \equal{} 2(x^2 \plus{} y^2 \plus{} 1) \equal{} \mathrm{const.}$ so $ x^2 \plus{} y^2 \equal{} \mathrm{const.}$ which is the equation for a circle.[/hide]

[hide="Elementary"]Choose $ D$ on $ AB$ with $ PD\perp AB$ and denote the midopint of $ \overline{AB}$ by $ M$. Pythagoras' theorem yields
$ |AP|^2 \plus{} |BP|^2 \equal{} |AD|^2 \plus{} |BD|^2 \plus{} 2|DP|^2 \equal{} 2\left(\left(\frac {AB}{2}\right)^2 \plus{} |MD|^2 \plus{} |DP|^2\right) \equal{} \frac {|AB|^2}{2} \plus{} 2|MP|^2 \equal{} \mathrm{const.}$
so the locus is a circle with midpoint $ M.$
(The obvious identity $ x^2 \plus{} y^2 \equal{} 2\left(\left(\frac {x \plus{} y}{2}\right)^2 \plus{} \left(\frac {x \minus{} y}{2}\right)^2\right)$ was used.)[/hide]


geometry incenter parallelogram rhombus inradius trigonometry LaTeX

[color=darkred]Let $\triangle ABC$ with the incenter $I\ .$ Denote the points $\{\begin{array}{c}D\in BC\ ,\ ID\perp IC\\\\ E\in BC\ ,\ IE\perp IB\end{array}$ $\mathrm{\ and\ }$ $\{\begin{array}{c}F\in AC\ ,\ EF\parallel AB\\\\ G\in AB\ ,\ DG\parallel AC\end{array}\ .$ Prove that $\{\begin{array}{cc}1.\blacktriangleright & \boxed{\ I\in FG\ }\ .\\\\ 2.\blacktriangleright & \frac{DE}{p-a}=\frac{BD}{p-b}=\frac{CE}{p-c}=\frac{a}{p}\ .\end{array}$

[b]Remark.[/b] Denote the intersection $S\in AI\cap BC\ ,$ then exist and the following relations : $\{\begin{array}{c}\frac{SD}{b}=\frac{SE}{c}=\frac{DE}{b+c}=\frac{a(p-a)}{p(b+c)}\mathrm{\ \ ;\ \ }\frac{ID}{IE}=\frac{\sqrt{b(p-b)}}{\sqrt{c(p-c)}}\ .\\\\ \frac{DC}{b}=\frac{BE}{c}=\frac{a}{p}\mathrm{\ \ ;\ \ }\frac{GB}{GA}=\frac{p-b}{b}\mathrm{\ ,\ }\frac{FC}{FA}=\frac{p-c}{c}\ .\end{array}$

The problem from the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=115733 is the particular case when $A=90^{\circ}\ .$[/color]

$\angle IDC = 90^\circ-\frac{\angle C}{2},\ \angle GDC = 180^\circ-\angle C,\ \Longrightarrow\ DI$ bisects $\angle GDC \Longrightarrow\ DG$ is tangent to the incircle (I).
$\angle IEB = 90^\circ-\frac{\angle B}{2},\ \angle FEB = 180^\circ-\angle B,\ \Longrightarrow\ EI$ bisects $\angle FEB \Longrightarrow\ EF$ is tangent to the incircle (I).
Let DG, EF meet at A'. Parallelogram AFA'G has incircle (I), it is a rhombus, the diagonal FG passes through its incenter I.

Let $r, s,\ \triangle$ be the inradius, semiperimeter and area of $\triangle ABC.$ $\triangle FEC \sim \triangle ABC$ are centrally similar with similarity center C, $\triangle GBD \sim \triangle ABC$ are centrally similar with similarity center B,

$\frac{CE}{a}= \frac{h_{c}-2r}{h_{c}}= \frac{\frac{2\triangle}{c}-\frac{2\triangle}{s}}{\frac{2\triangle}{c}}= \frac{s-c}{s}$

$\frac{BD}{a}= \frac{h_{b}-2r}{h_{b}}= \frac{\frac{2\triangle}{b}-\frac{2\triangle}{s}}{\frac{2\triangle}{b}}= \frac{s-b}{s}$

$DE = a-(BD+CE) = \frac{as-a(s-c+s-b)}{s}= \frac{a(s-a)}{s}$

In addition, $\triangle A'ED \sim \triangle ABC$ are centrally similar with similarity center K, foot of the bisector of $\angle A.$ (I) is the A'-excircle of $\triangle A'ED,$ B-excircle of $\triangle GBD,$ C-excircle of $\triangle FEC,$ hence again,

$\frac{ED}{a}= \frac{r}{r_{a}}= \frac{s-a}{s}$

$\frac{BD}{a}= \frac{r}{r_{b}}= \frac{s-b}{s}$

$\frac{EC}{a}= \frac{r}{r_{c}}= \frac{s-c}{s}$

where $r_{a},\ r_{b},\ r_{c}$ are exradii of $\triangle ABC$ opposite to A, B, C. Then

$\frac{DE}{s-a}= \frac{BD}{s-b}= \frac{CE}{s-c}= \frac{a}{s}$

[color=darkred][b][u]Similarly.[/u][/b] Let $\triangle ABC$ with the $A$- exincenter $I_{a}\ .$ Denote the points $\{\begin{array}{c}D\in BC\ ,\ I_{a}D\perp I_{a}C\\\\ E\in BC\ ,\ I_{a}E\perp I_{a}B\end{array}$ $\mathrm{\ and\ }$ $\{\begin{array}{c}F\in AC\ ,\ EF\parallel AB\\\\ G\in AB\ ,\ DG\parallel AC\end{array}\ .$ Prove that $\boxed{\ I_{a}\in FG\ }\ .$[/color]

[color=darkblue][b]Another proof for the first problem.[/b] Observe $m(\widehat{BID})=m(\widehat{CIE})=\frac{A}{2}$ and $m(\widehat{DIE})=\frac{B+C}{2}\ .$
Precisely, if $S\in AI\cap BC\ ,$ then $m(\widehat{DIS})=\frac{B}{2}$ , $m(\widehat{EIS})=\frac{C}{2}\ .$ Therefore,

$\{\begin{array}{c}\frac{GB}{GA}=\frac{DB}{DC}=\frac{IB}{IC}\cdot\sin\frac{A}{2}\\\\ \frac{FC}{FA}=\frac{EC}{EB}=\frac{IC}{IB}\cdot \sin\frac{A}{2}\end{array}$ $\Longrightarrow$ $b\cdot \frac{GB}{GA}+c\cdot\frac{FC}{FA}=$ $\sin\frac{A}{2}\cdot(b\cdot\frac{IB}{IC}+c\cdot\frac{IC}{IB})=$ $\sin \frac{A}{2}\cdot (b\cdot\frac{\sin\frac{C}{2}}{\sin\frac{B}{2}}+c\cdot\frac{\sin\frac{B}{2}}{\sin\frac{C}{2}})=$

$2R\sin \frac{A}{2}\cdot(\sin B\cdot \frac{\sin\frac{C}{2}}{\sin\frac{B}{2}}+\sin C\cdot\frac{\sin \frac{B}{2}}{\sin\frac{C}{2}})=$ $4R\sin\frac{A}{2}\cdot(\cos\frac{B}{2}\sin\frac{C}{2}+\cos\frac{C}{2}\sin\frac{B}{2})=$ $4R\sin\frac{A}{2}\sin\frac{B+C}{2}=$

$4R\sin\frac{A}{2}\cos\frac{A}{2}=$ $2R\sin A=a$ $\Longrightarrow$ $b\cdot\frac{GB}{GA}+c\cdot\frac{FC}{FA}=a$ $\implies$ $\boxed{\ I\in FG\ }\ .$

$\{\begin{array}{ccccc}\triangle AIB\sim\triangle IDB & \Longrightarrow & \frac{IA}{ID}=\frac{IB}{DB}=\frac{c}{IB}& \Longrightarrow & \{\begin{array}{ccc}DB=\frac{IB^{2}}{c}& \Longrightarrow & DB=\frac{a(p-b)}{p}\\\\ ID=\frac{IA\cdot IB}{c}& \Longrightarrow & ID=\frac{1}{p}\cdot\sqrt{ab(p-a)(p-b)}\end{array}\\\\ \triangle AIC\sim\triangle IEC & \Longrightarrow & \frac{IA}{IE}=\frac{IC}{EC}=\frac{b}{IC}& \Longrightarrow & \{\begin{array}{ccc}EC=\frac{IC^{2}}{b}& \Longrightarrow & EC=\frac{a(p-c)}{p}\\\\ IE=\frac{IA\cdot IB}{b}& \Longrightarrow & IE=\frac{1}{p}\cdot\sqrt{ac(p-a)(p-c)}\end{array}\end{array}$ $\mathrm{\ \ \ }\Longrightarrow$

$\{\begin{array}{c}\boxed{\ \frac{DE}{p-a}=\frac{BD}{p-b}=\frac{EC}{p-c}=\frac{a}{p}=\frac{DC}{b}=\frac{BE}{c}\ }\\\\ \frac{ID}{\sqrt{b(p-b)}}=\frac{IE}{\sqrt{c(p-c)}}=\frac{1}{p}\cdot\sqrt{a(p-a)}\end{array}$ $\Longrightarrow$ $\{\begin{array}{c}SD=SB-DB=\frac{ac}{b+c}-\frac{a(p-b)}{p}\\\\ SE=SC-EC=\frac{ab}{b+c}-\frac{a(p-c)}{p}\end{array}\Longrightarrow$ $\{\begin{array}{c}SD=\frac{ab(p-a)}{p(b+c)}\\\\ SE=\frac{ac(p-a)}{p(b+c)}\end{array}$ $\Longrightarrow$ $\boxed{\ \frac{SD}{b}=\frac{SE}{c}=\frac{a(p-a)}{p(b+c)}\ }\ .$

[b]Remark.[/b] $\{\begin{array}{c}\frac{GB}{GA}=\frac{DB}{DC}=\frac{p-b}{b}\\\\ \frac{FC}{FA}=\frac{EC}{EB}=\frac{p-c}{c}\end{array}$ $\Longrightarrow$ $b\cdot\frac{GB}{GA}+c\cdot\frac{FC}{FA}=b\cdot\frac{p-b}{b}+c\cdot\frac{p-c}{c}=a$ $\Longrightarrow$ $b\cdot\frac{GB}{GA}+c\cdot\frac{FC}{FA}=a$ $\Longrightarrow$ $I\in FG\ .$[/color]

[hide="Occasion."]
[color=red][size=134]Such, you had the ocassion to practice and in ... LaTeX ![/size][/color][/hide]


Yeah, I was stumped by that once at the San Diego MC also.

$236!$

AKA, (product of 47 5's) $*$ (a lot of 2's) $*$ other junk!

Therefore, there are 47 zeroes!

[quote="nat mc"]...

well your still wrong

Answer:[236/5]+[236/25]+[236/125]=57[/quote]

You round down after each result in brackets.
There's has been considerable discussion on my name...

What is $\frac{236!}{(236-9)!}$?
Simple.

Uh... No.

$\frac{236!}{227!} = \frac{236\cdot235\cdot...\cdot227\cdot226\cdot225\cdot...2\cdot1}{227\cdot226\cdot225\cdot...2\cdot1}$

Now it should look familiar. use a calculator.

Are you sure that symbol belongs in the MOEMS forum  :huh:
Actually I know what it means  :roll:

yeah i looked at it wrong, its just 236*...*228... doesn't seem very significant though...

[quote="math92"][quote="236factorial"][quote="math92"]what's does it mean[/quote]

It's like the symbol $\sum$, except it has something to do with multiplication.

If you want a more detailed explanation, visit
[url]http://mathworld.wolfram.com/Product.html[/url][/quote]
okay, i don't get it.  I'll just learn it in High School[/quote]
Yes you will. In algebra II I think you will learn the $\sum$ definition.

[quote="4everwise"]Wait, how is the answer 57? Nothing can create a "zero" except a 5 and an even number multiplied, so since there are 47 5's, shouldn't the answer be 47 zeroes?[/quote]
I thought that before too, but 25 has two fives and 125 has three fives!!!
So since 236 has 9 25's, we add those zeros those extra 5's produce and we get 56. What about that extra 5 in 125????  That creates yet another zero! So, the answer is 57.

Also remember that 625 has 4 fives multiplied in it, an 3125 has 5 fives, etc. Take that into consideration when solving these types of problems with numbers that are higher.

this is a very important skill. It comes up again and again in MOEMS, Mathcounts, and many other competitions. So you should know it  :)

$\paragraph{Problem.} \text {How many 0's are at the end of } 236!?$

$\lfloor 236/5 \rfloor + \lfloor 236/25 \rfloor + \lfloor 236/125 \rfloor =57.$ Therefore the answer is $\boxed {57}$.

[quote="4everwise"]Wait, how is the answer 57? Nothing can create a "zero" except a 5 and an even number multiplied, so since there are 47 5's, shouldn't the answer be 47 zeroes?[/quote]
\There are more than 47 5s.  25 has 2 5s.  125 has 3 5s

hi guys, im now in 10th class , and even then I never saw the symbols like  Sigma etc :s

yes, after 10th class, there are 2 more years.

A bit off topic, but what country are you from?

[quote="math92"]if you hold your mouse over the flag it will say what country it is :)[/quote]

Thanks!  I didn't know that.   :blush:


That trianle does not exist, the sum of the angles is not $180^{\circ}$

:rotfl:  :blush:
I read it as a positive $6$ making it $192$.

Haha...all x works, but not all x create acute or does it.

[quote="asdff"]One angle of an acute triangle is x+15, the other is 2x-6 and third is 171-3x.  How many different integer values can x be?[/quote]
[hide]
Note that $x+15+2x-6+171-3x=180$.
So $0<x+15<90 \rightarrow x<75$
$0<2x-6<90 \rightarrow 3<x<48$
$0<171-3x<90 \rightarrow 27<x<57$
So $27<x<48$, so there are $20$ values of x that work.[/hide]


AMC
AMC 12
geometry
3D geometry
octahedron

does anyone know an AMC combinatoric question?? i think combinatorics might be a little hard for the AMC but i was just wondering

a combinatoric question at that level would be helpful too

so thanks ahead of time!

Source: AMC12 2001

A spider has one sock and one shoe for each of its eight legs. In how many different orders can the spider put on its socks and shoes, assuming that, on each leg, the sock must be put on before the shoe?

Combinatorics is not too hard for the AMC.  Off the top of my head I can think of:

Compute the number of distinct ways to color the faces of an octahedron with 8 colors (each color used exactly once).

There are plenty of other problems...but I can't remember them and I don't feel like looking ones up.


geometry
circumcircle
trigonometry
trig identities
Law of Sines
geometry proposed

Given a point $ P$ on a circumcircle of a traingle $ ABC$. Draw the perpendiculars $ PM,\ PN$ from $ P$ to $ AB,\ AC$ respectively. Construcut the point $ P$ such that the length of $ MN$ is maximal.

[color=darkblue]Quadrilateral $ PMAN$ is inscribed in a circle with diameter $ AP$. Then

$ MN\equal{}AP\sin A \le 2R \sin A\equal{}BC$.

The equality holds if $ AP$ is diameter of the circumcircle of $ ABC$.[/color]

Since quadrilateral $ AMNP$ is cyclic with diameter $ AP$ we get using the law of sines, $ MN/\sin A\equal{}AP$ i.e. $ MN\equal{}AP\sin A$. Hence $ MN$ is maximal when $ P$ is diametrically opposite to $ A$.

EDIT: Oops, beaten. :oops:

In the given triangle $ \bigtriangleup ABC,$ we suppose that $ BC > AC > AB.$

As [b][size=100]Ahiles[/size][/b] and [b][size=100]nayel[/size][/b] said, the point $ A'$ as the [b][size=100]antidiametric[/size][/b] ( = [b][size=100]diametric opposite[/size][/b] ) point of $ A,$ has the property as the problem states and I will post here later more details.

Kostas vittas.


Hey guys, Im doing a research paper on Cubic equations and their solutions. I can't find any example of solving a cubic equation by depression, so can someone post a cubic equation (thats not factorable) and solve it by depression. If this doesn't belong here, please move it. Thank you.  :)

I presume you want something like [url=http://www.trans4mind.com/personal_development/mathematics/polynomials/cardanoMethodExamples.htm]this[/url]?

Yes. Exactly like that. I was hoping you could give an example and explain it, so I can use a similar example as an example in my paper. =o

P.S: I really dont understand that. It's really messy because its not in LateX.  :(


Let be $ x,y,z>0$ such that $ \frac{1}{x}\plus{}\frac{2}{y}\plus{}\frac{3}{z}\equal{}1$ . Show that :

$ xyz\ge 162$

$ a \equal{} \frac{y}{2}$, $ b \equal{} \frac{z}{3}$, substitute, HM-GM.

$ G_M\equal{}\sqrt[3]{x\cdot \frac{y}{2}\cdot \frac{z}{3}}$ and $ H_M\equal{}\frac{3}{x\plus{}\frac{y}{2} \plus{}\frac{z}{3}}\equal{}3$
 $ G_M \geq H_M$
$ \sqrt[3]{x\cdot \frac{y}{2}\cdot \frac{z}{3}}\geq \frac{3}{x\plus{}\frac{y}{2} \plus{}\frac{z}{3}}\Rightarrow xyz\geq 162$

[quote="enndb0x"]$ H_M \equal{} \frac {3}{x \plus{} \frac {y}{2} \plus{} \frac {z}{3}} \equal{} 3$[/quote]

Can you tell me from where did you got that ?

I think you made this wrong : $ \frac{1}{x}\plus{}\frac{2}{y}\plus{}\frac{3}{z}\equal{}1\Leftrightarrow \frac{1}{\frac{1}{x}\plus{}\frac{2}{y}\plus{}\frac{3}{z}}\equal{}1 , BUT \frac{1}{\frac{1}{x}\plus{}\frac{2}{y}\plus{}\frac{3}{z}}\neq x\plus{}\frac{y}{2}\plus{}\frac{z}{3}$

You can see that $ x\equal{}3,y\equal{}6,z\equal{}9$ satisfy my condition but they don't verify what you said .

[hide]Let $ a \equal{} x, b \equal{} \frac{y}{2}, c \equal{} \frac{z}{3}$

Therefore $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c} \equal{} 1$

By AM-GM

$ 1 \equal{} \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c} \ge 3 \sqrt[3]{\frac{1}{abc}} \Rightarrow abc \ge 27 \Rightarrow xyz \ge 27 \times 6 \equal{} 162$[/hide]


Let $a,$ $b$ and $c$ are positive numbers such that $abc=1.$ Prove that
$a^{2}+b^{2}+c^{2}+12(ab+ac+bc)\geq13(a+b+c).$

Arqady, it should be an old one since Vasile had proposed a stronger inequality for $abc=1$
\[a^{2}+b^{2}+c^{2}+15(ab+bc+ca)\geq 16(a+b+c)\]
which is solved by mixing variables.

What about the following ugly one: Let $a,b,c,d$ be positive reals such that $abcd=1$. Find the greatest $k$ for which the inequality is valid
\[a^{3}+b^{3}+c^{3}+d^{3}+k(abc+bcd+cda+dab)\geq (1+k)(a^{2}+b^{2}+c^{2}+d^{2}).\]

[quote="pvthuan"]Arqady, it should be an old one since Vasile had proposed a stronger inequality for $abc=1$
\[a^{2}+b^{2}+c^{2}+15(ab+bc+ca)\geq 16(a+b+c) \]
which is solved by mixing variables.
[/quote]
Thank you, pvthuan. It's also my inequality, stronger than Vasc's one. :D  See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80197.
I have forgotten about it!
It is impossible to prove an inequality and simultaneously to see Roland Garros. :D
I am sorry.


limit
logarithms
real analysis
real analysis unsolved

Let $p>0$ and let $a$ and $b$ be nonnegative real numbers such that $a<b$.

Find

$\lim_{n\rightarrow \infty}\frac{(2\ln 2)^{p}+a}{(2\ln2)^{p}+b}\cdot \frac{(3\ln 3)^{p}+a}{(3\ln 3)^{p}+b}\cdots \frac{(n\ln n)^{p}+a}{(n\ln n)^{p}+b}$

What's the connection between this topic and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=137559]this other topic?[/url]


Determine the integer numbers $ a,b,c$ which are the side lenghts of a right-angled triangle satisfying :

$ \frac{1}{a}\plus{}\frac{2}{b\plus{}c}\equal{}\frac{6}{a\plus{}b\plus{}c}$

[quote="alex2008"]Determine the integer numbers $ a,b,c$ which are the side lenghts of a right-angled triangle satisfying :

$ \frac {1}{a} \plus{} \frac {2}{b \plus{} c} \equal{} \frac {6}{a \plus{} b \plus{} c}$[/quote]

$ (b\plus{}c)(a\plus{}b\plus{}c)\plus{}2a(a\plus{}b\plus{}c)\equal{}6a(b\plus{}c)$
$ \Leftrightarrow (2a\plus{}b\plus{}c)(a\plus{}b\plus{}c)\equal{}6a(b\plus{}c)$
$ \Leftrightarrow (\minus{}2a\plus{}b\plus{}c)(\minus{}a\plus{}b\plus{}c)\equal{}0$
$ \Leftrightarrow \minus{}2a\plus{}b\plus{}c\equal{}0$

WLOG, $ b>a>c$, then $ a^2\equal{}b^2\minus{}c^2$

$ (2a)^2\equal{}(b\plus{}c)^2 \Leftrightarrow 3b^2\minus{}2bc\minus{}5c^2\equal{}0 \Leftrightarrow (b\plus{}c)(3b\minus{}5c)\equal{}0$
$ \Leftrightarrow 3b\equal{}5c$

So $ (b,c)\equal{}(5t,3t)$, where $ t$ is positive integer.

$ (a,b,c)\equal{}(4t,5t,3t), (4t,3t,5t)$, where $ t$ is any positive integer, are the solutions after checking.


I'm sorry if this question has been asked before, but does anyone know the exact process for the 2006 trail from the American Physics Competition to the Physics Olympiad to the IPhO. I looked at the AAPT site, yet couldnt get exact information about the process. My school has never done this competition before, and I wish to generate enough interest for my school to participate. Thank you for your help.

so basically,
you sign up for the olympiad qualifier
it's a multiple choice test + a free response test.
top 200 get to take the semifinal test
it's a bunch of free responses, harder than the qualifier by a bunch

then the top 24 scorers get to go to the camp, where you take tests and do labs for a week, and then they take the top 5 to go to ipho.

pretty much it.


[quote="Interval"]In Jackson County, Wyoming, license plates are made with two
letters (A through Z) followed by three digits (0 through 9). The plates
are made according to the following restrictions:

• the first letter must be J or W, and the second letter can be any of
the 26 letters in the alphabet
• no digit can be repeated

How many different license plates can be made with these restrictions?[/quote]
[hide="hint"]Consider each digit seperately, and consider the previous usages.[/hide]

[hide="hint"]
there will be 10 choices for the first digit, and 9 choices for the second digit.
[/hide]

[hide="same hint, explained more"]How many choices are there for the first letter?  How about the second?  How many choices for the first digit?  The second?[/hide]

[quote="kstan013"][hide]Choices for first letter-2
Choices for 2nd letter-26
Choices for first digit-10
Choices for 2nd digit-9
Multiply them together to get 260*18=4680[/hide][/quote]there are three digits

[hide="My Solution"]
$2\times 26\times 10\times 9\times 8=520\times 72=37440$
[/hide]

37440 Is the correct answer.  kstan013 forgot about the 5th character in the license plate.

[quote="Interval"]In Jackson County, Wyoming, license plates are made with two
letters (A through Z) followed by three digits (0 through 9). The plates
are made according to the following restrictions:

• the first letter must be J or W, and the second letter can be any of
the 26 letters in the alphabet
• no digit can be repeated

How many different license plates can be made with these restrictions?[/quote]
[hide]
there are $(2)(26)(10)(9)(8)$ different liscence plates.
the first 2 is for J or W, then 26 for the diff letters, then 10(9)8 for 10 possibilities for first digit, 9 for second, 8 for third. this includes the different arrangements. so the answer is $37440$
[/hide]


calculus
integration
inequalities
real analysis
real analysis unsolved

Let $ \textstyle m \equal{} \int_{0}^{1} f$ and $ g \equal{} f \minus{} m$ so that $ \textstyle \int_{0}^{1} g \equal{} 0$. Plugging $ f \equal{} g \plus{} m$, we find that it suffices to show that

$ \int_{0}^{1} g(x)^2 \, dx \minus{} 12 \int_{0}^{1} mx g(x) \, dx \plus{} 3m^2 \geq 0$.

Note that $ \textstyle \int_{0}^{1} (x \minus{} \frac{1}{2})^2 \, dx \equal{} \frac{1}{12}$, then it is equivalent to show that

$ \int_{0}^{1} g(x)^2 \, dx \minus{} 2 \int_{0}^{1} 6m(x\minus{}\tfrac{1}{2}) g(x) \, dx \plus{} \int_{0}^{1} 36m^2 (x\minus{}\tfrac{1}{2})^2 \, dx \geq 0$.

But since the left side of this inequality is equal to $ \textstyle \int_{0}^{1} \left[ g(x) \minus{} 6m(x \minus{} \tfrac{1}{2}) \right]^2 \, dx$, the proof is completed.


geometry
power of a point
radical axis
perpendicular bisector
geometry unsolved

Draw the perpendicular line $ d$ of $ E_aH_a$ and the line $ d'$ be the perp line to $ E_aH_a$ at $ H_a$. Then $ d'\equiv BC$ is just the sideline of triangle. Let $ X_a$ be the orthogonal projection of $ I$ on $ d'$. Draw the circle $ (I;IX_a)$, it's just the incircle. Let $ K$ be the symmetric point of $ I$ w.r.t the line $ d$. Draw the circle $ (E_a;IX_a)$. Construct the circle through $ I,K$ and tangent to $ (E_a)$. Its center is just the nine-point center of triangle, call $ N$. The symmetric point of $ E_a$ w.r.t $ N$ is just the midpoint $ M_a$ of side $ BC$.
Continue with [url=http://www.mathlinks.ro/viewtopic.php?t=204687]Ha I Ma[/url] or [url=http://www.mathlinks.ro/viewtopic.php?t=287425]EaMaI[/url].

Construct the circle through  I, K and tangent to (Ea)

how ?

 :(

It's perhaps a basic construction:
[b]Basic construction:[/b]
Construct a circle through two given points $ A,B$ and tangent to a given circle $ (O_1)$.

One solution is this. Draw any circle $ (O)$ through the given points $ A$, $ B$ and intersecting the given circle. The radical axes of the circles $ (O)$, $ (O_1)$, and the unknown circle $ (O_2)$ meet at some point $ P$. The tangent from $ P$ to the circle $ (O_1)$ serves as the radical axis of $ (O_1)$ and the unknown circle $ (O_2)$. Therefore, the point of tangency $ T$ belongs to the two circles, $ (O_2)$ in particular. Now construct the circle $ (O_2)$ through three points $ A,B,P$. That's wanted circle. (The center $ O_2$ is just the intersection of $ O_1T$ and the perp bisector of $ AB$).

could you correspond points $ O, O_{1}$ and $ O_{2}$ to the points of the problem construction ?

A=I and B=K  ?

 :(

Yes, A=I and B=K, dear friend.
It seems that you are not concentrate on understanding my constructions  :maybe:
Ok, I will post a step-by-step construction of point $ M_a$  :)
1) Draw the perpendicular bisector $ d$ of $ E_aH_a$
2) Draw the perpendicular line $ d'$ to $ E_aH_a$ at $ H_a$
3) Let $ K$ be the symmetric point of $ I$ w.r.t the line $ d$
4) Let $ X_a$ be the orthogonal projection of $ I$ on $ d'$. Draw the circle $ (E_a;IX_a)$, abbreviate by $ (E_a)$
5) Draw some circle through $ I,K$ and meets $ (E_a)$ at $ Q,R$.
6) Let $ P$ be the intersection of $ QR$ and $ IK$. Draw the tangent from $ P$ to $ (E_a)$. Let $ T$ be the tangency point.
7) The line $ E_aT$ meets the line $ d'$ at $ M_a$.
Then, $ M_a$ is just the midpoint of $ BC$. Continue with  [url=http://www.mathlinks.ro/viewtopic.php?t=204687]Ha I Ma[/url]


Find all positive integers n such that $(n+3)^n = \sum^{n+2}_{k=3} k^n$.

[i]Edited by Myth[/i]


$bigstar$ ai có thêm thông tin về những thành viên trong đội tuyển thì share cho bà con với :D

Chúc mừng các bạn nhé, đặc biệt là hai bạn Phạm Duy Tùng và Nguyễn Xuân Chương. Tùng là một thành viên trong lớp Toán, tiếng Anh mà trước đây tôi dạy thì phải, còn Chương cũng là học sinh trong lớp tập huấn SMO năm 2005. Hơi tiếc là lần này thiếu N.X Thọ nhỉ.

Một lần nữa đặc chúc mừng Tùng và Chương nhé.

Kết quả vậy là chính thức rồi hả Thuận? Bao giờ các bạn ấy bắt đầu học ấy nhỉ?

[quote="N.T.TUAN"]1-Đỗ Xuân Bách (KHTN-HN, lớp 12)
2-Phạm Duy Tùng (KHTN-HN, lớp 11)
3-Phạm Thành Thái (Hải Dương, lớp 12)
4-Nguyễn Xuân Chương (Vĩnh Phúc, lớp 12)
5-Đặng Ngọc Thanh( Quảng Bình, lớp 12)
6-Lê Ngọc Sơn ( Bắc Giang, lớp 11)

[/quote]
Chúc mừng Bắc Giang và Quảng Bình :)

[quote="N.T.TUAN"]1-Đỗ Xuân Bách (KHTN-HN, lớp 12)
2-Phạm Duy Tùng (KHTN-HN, lớp 11)
3-Phạm Thành Thái (Hải Dương, lớp 12)
4-Nguyễn Xuân Chương (Vĩnh Phúc, lớp 12)
5-Đặng Ngọc Thanh( Quảng Bình, lớp 12)
6-Lê Ngọc Sơn ( Bắc Giang, lớp 11)

Như vậy có 2 nơi lần đầu tham gia: Bắc Giang, Quảng Bình? Còn bạn năm ngoái thi IMO năm nay trượt mất rồi.[/quote]
Sư phạm ko có ai :(

Năm trước cũng không có, nói chung dạo này bên SPHN không quan tâm đến Toán Olympic nữa thì phải? Thi QG đang cải cách nữa :D

[quote="nmtruong1986"]Sư Phạm nắm bắt thông tin nhay, và cũng linh hoạt trong chiến lược sư phạm.
Hôm trước em có đọc báo thấy Thầy Trần Phương cựu học sinh Sư Phạm mở lớp luyện thi Đại học từ lớp 6. Có khi đấy lại là một hướng đi mới chăng? Thay vì luyện thi học sinh giỏi ta luyện học sinh lớp 6 đi thi đại học. :D Hình như cókhá nhiều người muốn gửi gắm con em mình cho lớp luyện thi Đại học ấy. Thật hay![/quote]
Hay á? :D Chưa thể nói trước điều gì. Bên ddth có một topic về cái này rồi đấy, vui ra phết :P

[quote="pvthuan"]Thực ra nói SP không quan tâm đến toán Olympiad nữa thì không phải, ông Tuấn à. Theo tôi, cái chính là hai lứa vừa rồi thiếu vài em thực sự có tư chất thôi. Tất nhiên, bây giờ thì học sinh năng động và có nhiều mối quan tâm khác. Đi du học ngay từ cấp 3 là một xu hướng ngày càng phổ biến. Đích đến nhiều nhất là Singapore, Anh.[/quote]
Nói hơi đúng, nhưng tôi tên là Tuân, không phải Tuấn. Thày Thuận toàn nhầm tên em thôi  :D

Em lai tuong ten thay pv... là Thuân cơ. Nói chung tại tên các bác khó đọc khó nhớ nên mới có nhầm nhọt kiểu này. :D

[quote="N.T.TUAN"]
Hay á? :D Chưa thể nói trước điều gì. Bên ddth có một topic về cái này rồi đấy, vui ra phết :P[/quote]
Day nay http://diendantoanhoc.net/forum/index.php?showtopic=32147&pid=160557&st=0&#entry160557 hoac [...da bi xoa...] :D


Let $a,b,c>0$ the sides of a right triangle. Find all real $x$ for which $a^x>b^x+c^x$, with $a$ is the longest side.

Function $f(x)=\displaystyle {(\frac{b}{a})^x+(\frac{c}{a})^x-1}$ is decreasing.
Can this help??
 :?

I'm sorry, a word was missing. :) Now it should be easy.

In English, "right triangle," not "rectangular triangle."

I know, it is their translation. :)

Ah well, edited. ;)

The equality is obvious at $x=2$
Now using $a$ is the longest side,
For $x>2$
$a^x = a^2 a^{x-2} =(b^2+c^2)(a^{x-2})>b^x+c^x$


calculus
integration
logarithms
trigonometry
Euler
function
calculus computations

(i)Compute

$\int^2_1\frac{\ln 2x-1}{x}dx$

(ii) Find $\int\frac{\sin^nx}{\sin^nx+\cos^nx}dx$

For the second one , if given interval $\left[0,\frac{\pi}{2}\right]$ then the question is trivial . What if no interval given  :?

[quote="Fermat -Euler"]$\int^2_1\frac{\ln 2x-1}{x}dx$
$=\int^2_1\frac{\ln 2x}{x}dx-\int^2_1\frac{1}{x}dx$
$=(ln 4)^2-(ln 2)^2-(ln 2)$
$=3(ln 2)^2-(ln 2)$[/quote]

Sorry Fermat-Euler , I should have make it more clear , it is

$\ln(2x-1)$

[quote="Beginner"]This is not from me, for no 1 My tutor told me that it seems we have some function that have to approximate  numerically. then he told me about $\sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{k^2}$. as a Librarian i seek for information in library and i got some information about Dilogarithm function. that is $L_i (x) = \int_{1}^{x} \frac{\ln t}{(1-t)} dt$. this related to your problem.[/quote]

Ok ,thanks . At least I have some sources to seek for  :D

The second integral I think is tough.
First try to solve the problem related to yours which I have posted.


According to Lagrange four squares problem every natural number is sum of four squares. All right. Every even natural number can be expressed sum of 8 odd squares? (of course natural number $\geq8$)

Not even close. A sum of 8 odd squares must be divisible by 8. For these numbers, it is true using the three-squares theorem.

For any positive integer $k$, $8k-5$ is not the product of a power of 4 and a number that is 7 (mod 8), so the three-squares theorem guarantees that $8k-5$ is a sum of three squares. Checking (mod 8) reveals that they are all odd, and having $8k$ as a sum of 8 odd squares follows (just add 5 1's).

hmm... maybe there's a proof without relying on such a difficult theorem as the three-squares theorem?


Suppose $ A$ and $ B$ are two angles such that $ \sin A\plus{}\sin B\equal{}1$ and $ \cos A\plus{}\cos B\equal{}0$. Find the value of $ 12\cos2A\plus{}4\cos 2B$.

[hide="Hint"]
\[ \cos A \plus{} \cos B \equal{} 0 \iff A \pm B \equal{}\pi \plus{} 2 k \pi \ (k \in \mathbb{Z})\]
[/hide]


calculus
integration
trigonometry
calculus computations

I have trouble solving an integral like this, is there anyone can help? Thanks alot!

try integration by parts.
the general form is ${\int e^{\alpha x} \sin \beta x dx = \frac{e^{\alpha x}}{\alpha^2+\beta^2}(\alpha\sin \beta x - \beta\cos \alpha x)}$.

oh, yea,you're right. I got it .Thanks alot. :)


Chinese Remainder Theorem could get you the general formula for this one. [hide="BTW"]I think this might belong to the NUmber Theorey forum[/hide]

Go Chinese remainder Theorem. BTW, i seen some very nontrivial proofs using Chinese Remainder Theorem in the past, on some very hard Olympiad problems, yet hardly any training books even slightly [i]hint[/i] at its applicability. Did I miss any good books?

You don't have to use the chinese remainder theorem to solve this, as a sledgehammer... Just by hands.
Anyway, I think this problem would be more appropriate in the number theory section  :wink:  (in the intermediate level?), so I move it...

Pierre.

[quote="tranthanhnam"]--------------------------------------------------------------------------------
find all natural number have $5$ digits such that :
$n$ divisible by$2$ ; $n+1$ divisible by $3$;$n+2$ divisible by $4$;$n+3$divisible by $5$[/quote]
It give (n-2) divisible to $3*4*5=60$, therefore all solutions given by $n=60k+2$.


This is what i can tell so far:

[hide]There is no rational solution $x$.
There is an irrational $x$ between 3 and 4 (it seems like it will be closer to 4) which works.[/hide]

This better [b]not[/b] be one of those  :cursing: -ing general cubics!

So far I've bounded it $\frac{15}{4} < x < 4$.  Yeah, I think it might be a general cubic too  :what?:

Umm. I used an equation solver and it gave me three hideous roots. Are you sure you typed the question correctly?

could someone please explain what a general cubic is.  :?

Basically, a cubic that can't be factored into quadratic and/or linear factors with rational coefficients.  So the solution's going to be a hideous radical or a nested radical.

I'm going to hazard a guess that the original question was $(x^2 + 4)(x^2 - 4) = -1$, in which case the answer is (Edited; sorry) $\sqrt[4]{15}$, which is a bit more straightforward.

[quote="daveed"]it would've been, then, in the form

$(a+4)(a-4)=-1$... where $a=x^2$

and then multiplying out would've given $a=\sqrt{3}$[/quote]

But then don't you get $a^2-16=-1 \Longleftrightarrow a = \sqrt{15} \Longleftrightarrow x=\pm \sqrt[4]{15}?$

[quote="daveed"]touché

long day...[/quote]

haha amen to that

but if it is not what daveed suggested, then there is no elementary solution to the problem... unless you're some insane mathematician. :o

Well, if you count [url=http://mathforum.org/dr.math/faq/faq.cubic.equations.html]the general solution[/url] to a cubic equation as 'elementary'... :|


How many digits are in 2^40?

hmm, must be a formula for it.

http://mathworld.wolfram.com/Digit.html

So you get $ 40\log2$ which becomes $ \boxed{13}$ digits.

Don't you mean $ 1\plus{}\lfloor 40 \log 2\rfloor\equal{}1\plus{}12\equal{}13$?

To explain,
$ number\equal{} 10^{digits\minus{}1}\plus{}stuff$
$ \left \lfloor \log(number) \right \rfloor \plus{}1\equal{}digits$

Also, for this particular problem, there are a couple other things you should know:

$ 2^{10} \equal{} 1024 \Rightarrow 2^{20} \equal{} 1,048,576$ (expand $ 1024^2 \equal{} (1000 \plus{} 24)^2$)
Then, we have $ 2^{20}\approx1,000,000 \equal{} 10^6$, so $ 2^{40}\approx 10^{12}$ and hence has 13 digits (you can be more formal here and so actual bounding, but it should be apparent).


induction
Functional Analysis
real analysis
real analysis unsolved

$E$ is a finite dimensional Banach space and $a_n$ is a sequence going to 0. Prove that we can choose signs $k_i$ such that the sum of $a_i k_i$ converges.

Since nobody answered in two weeks, I'll post a sketch of what I did.

1) All finite dimensional Banach spaces are isomorphic to the usual $\mathbb R^n$.

2) It suffices to prove that there is a constant $C_n$ such that for all sequences $a_j\in\mathbb R^n$ with $|a_j|\le 1$, we can find signs $\delta_j$ such that the partial sums of the series $\sum_j \delta_j a_j$ are bounded by $C_n$.

3) This can be proved by induction on $n$. The case $n=1$ is fairly simple. We can take $C_1=1$. Suppose that we have the result in dimension $n-1$. Choose a big number $S$. Suppose that $v=\sum_{j=1}^\ell \delta_j a_j$ is less than $S$ in norm but the next sum is greater than $S$ regardless of the choice of the sign $\delta_{\ell+1}$. Then write $a_j=b_j+c_j$ where $b_j$ are orthogonal to $v$ and $c_j$ are parallel to $v$. Choose the signs $\delta_j$, $j\ge \ell+1$ so that the sums $\sum_{j=\ell+1}^m \delta_j b_j$ are bounded by $C_{n-1}$. If the sums $\sum_{j=\ell +1}^m\delta_j c_j$ are bounded by $S$, then all sums $\sum_{j=1}^m \delta_j a_j$ are bounded by $2S+C_{n-1}$. If, for some $m$, the sum $\sum_{j=\ell+1}^m\delta_j c_j$ exceeds $S$, take the least such $m$ and, if necessary, replacing all $\delta_j$ ($j=\ell+1,\dots,m$) by $-\delta_j$, observe that the sum $\sum_{j=1}^m \delta_j a_j$ has the component parallel to $v$ of length at most $2$ and the component orthogonal to $v$ of length at most $C_{n-1}$. Thus, this sum is not greater than $C_{n-1}+2$, which is much less than $S$ if $S$ is chosen sufficiently large. So, we can start looking for the first sum greater than $S$ again and so on.

This approach results in an exponential estimate for $C_n$. I still wonder whether it is possible to get the power growth rate here.


linear algebra
matrix
limit
induction
geometric series
linear algebra theorems

F is a square matrix.Show that if norm(F)<1 then inv(I-F)=sum(F^k) for all k=0,1,2,... and norm(inv(I-F))<=1/(1- norm(F)

You didn't specify what norm - there are many possibilities, and they all work.

You do need the following facts about whatever norm you are using:

(1) $ \|AB\|\le \|A\|\,\|B\|$ for any square matrices $ A$ and $ B.$

(2) Completeness: if $ (A_n)$ is a Cauchy sequence in this norm, then there exists a matrix $ A$ such that $ \lim A_n \equal{} A.$

In practice, we'll show completeness by showing that if $ \|A \minus{} B\| < \epsilon,$ then the the corresponding entries in $ A$ and in $ B$ cannot differ by more than $ \epsilon.$

With those preliminaries out of the way:

The series $ \sum_{k \equal{} 0}^{\infty}F^k$ is a Cauchy series (that is, its partial sums form a Cauchy sequence.)

By property (1) and induction, $ \|F^k\|\le \|F\|^k.$ That means that the series $ \sum_{k \equal{} 0}^{\infty}F^k$ is termwise dominated by the convergent real series $ \sum_{k \equal{} 0}^{\infty}\|F\|^k \equal{} \frac {1}{1 \minus{} \|F\|}.$ That's how we know this is a Cauchy series and that's how we know that
\[ \left\|\sum_{k \equal{} 0}^{\infty}F^k\right\|\le\frac {1}{1 \minus{} \|F\|}
\]
Now, as for the sum itself -- multiply a partial sum by $ (1 \minus{} F):$

$ (1 \minus{} F)\sum_{k \equal{} 0}^nF^k \equal{} I \minus{} F^{n \plus{} 1}.$ (The sum telescopes.)

Take the limit as $ n\to\infty.$ We know that $ \|F^{n \plus{} 1}\|\le\|F\|^{n \plus{} 1}\to0$ so $ F^{n \plus{} 1}\to0.$ Hence

$ (1 \minus{} F)\sum_{k \equal{} 0}^{\infty}F^k \equal{} I,$ which we rewrite as $ \sum_{k \equal{} 0}^{\infty}F^k \equal{} (1 \minus{} F)^{ \minus{} 1}.$

The geometric series is one of the most powerful themes in mathematics, and has been so for a very long time.


Let $f : \mathbb{C}\to \mathbb{C}$ be a function satisfying

$f(0) = 0$
$| f(z)-f(w) | = |z-w|$

For all $z \in \mathbb{C}$ and for $w = 0, 1, i$.  Prove that

$f(z) = f(1) z \vee f(1) \bar{z}$

Where $|f(1)| = 1$.

[hide="Hint"] Consider the relation of the triangle $0, 1, i$ in the complex plane to the triangle $f(0), f(1), f(i)$. [/hide]

Looks suspiciousally like an IMC problem...


Prove that every perfect number is of the type:

P=2^(p-1)*(2^p-1) when q=2^p-1 is a Mersenne Prime.

Enjoy and good night.  ;)

BTW if any one proves that ( icluding  for an odd number)  he will be famous ... very famous indeed....:)
(It's not hard to prove for even numbers though...)

To elaborate -- it's not known if any odd perfect numbers exist.  It is known that all even numbers match the form $(2^p - 1)(2^{p-1})$ when $2^p - 1$ is prime, as keta asks you to show.

Proof that all even numbers match the form $(2^p - 1)(2^{p-1})$ when $2^p - 1$ is prime is fairly well known ( From Euler's time). Here  is the outline of an proof, if any one is interested .  (One can also google  to see other references)

[hide="proof"]If the number is even, write it as  $2^n  q $ where  $q$ is odd, and $n>0$
Then $\sigma  (2^n q) =  \sigma (2^n) \sigma (q)$
but since it is a perfect number, $\sigma$ should be equal to twice the number , so

$ 2^{n+1} q = \sigma (2^n) \sigma (q) $
or $2^{n+1} q = (2^{n+1} -1) \sigma(q) $
     $ = (2^{n+1} - 1) \sigma (q)  - q  + q $
or $(2^{n+1}-1)(\sigma(q) -q) =  q $
So $(2^{n+1} - 1)$ must be a divisor of $q$ , so let  $q = k(2^{n+1} -1 )$
So we have  $\sigma(q) - q = k$
But above can only be true if   q is prime (and  thus  k = 1)   becasue factors or q are (at least)  1 and q .
and  more if q is not prime. [/hide]


analytic geometry
graphing lines
slope
geometry
perpendicular bisector

The graph of the line $ x\plus{}y\equal{}b$ is a perpendicular bisector of the line segment from $ (1,3)$ to $ (5,7)$. What is the value of b?

It will have to pass through (3,5), so x+y=3+5=8=b

[quote="hutyputy66"]Um, why would it have to pass (3,5)?[/quote]

The line has to pass through $(3,5)$ because it is the perpendicular [b]bisector[/b] of the line segment from $(1,3)$ to $(5,7)$ and the midpoint of $(1,3)$ and $(5,7)$ is $(3,5)$.

[quote="AwesomeToad"]The line that passes through $(1,3)$ and $(5,7)$ has slope $1$ and equation $y=x+2$.

So substituting $x=3$ gives $y=5$.[/quote]

I'm not sure how this is relevant...... (Since, even though the point $(3,5)$ is on the line segment, you did not prove why it must also be on the perpendicular bisector of the line segment)

[quote="\$LaTeX\$"]I'm not sure how this is relevant...... (Since, even though the point (3,5) is on the line segment, you did not prove why it must also be on the perpendicular bisector of the line segment)[/quote]

Sorry, I thought s/he meant the original line.

Since the perpendicular bisector bisects the line from $(1,3)$ through $(5,7)$, it goes through the midpoint, or $(3,5)$.

And then you get $x+y=3+5=8$.


Three distinct digits $a$, $b$, and $c$ are used to form six distinct three-digit numbers. What is the greatest positive integer that is a divisor of the sum of those six numbers, regardless of the original numbers chosen.

[i]Do not just present three different types of numbers and show that there is only one common factor :P [/i]

the sum of them would be 222a+222b+222c, so the greatest positive integer would be [b]222[/b].

[hide="full solution"]
call the numbers a,b,c.
that means the six combinations are:
100a+10b+c
100a+10c+b
100b+10a+c
100b+10c+a
100c+10b+a
100c+10a+b
add them all up to get 222a+222b+222c
so 222[/hide]


analytic geometry
calculus
integration
MATHCOUNTS

[quote="#H34N1"]What is the sum of the coordinates of the midpoints of the line segment whose endpoints are (3,9) and (19,21)?

Hint: Look at the title :D[/quote]

umm i think i know what to do

[hide]3+4+....+19
9+10+...+21

12(30/2)
+
16(22/2)

12(15)
+16(11)
=356[/hide]

so are you asking for all the integral points of x and y added?

I really don't understand what to do. But the midpoint of the two coordinats is at least $(11, 5)$.
But can you explain what to do for me. Since I'm 15 I haven't learnt about this in school yet... and my english isn't great.

No, I don't think 11,5 is right either:
[hide] To find the midpoint a segment with endpoints $(x,y)$ and $(a,b)$, it is $(\frac{a+x}{2}, \frac{b+y}{2})$.  Plugging the values of x, y, a, and b in, we get $(\frac{3+19}{2},\frac{9+21}{2})$, which is $(11,15)$.  When you add the coordinates, you get $11+15=26$, which is our answer.  [/hide]

The midpoint formula is \[\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\] [hide]so plugging it in, you get the midpoint is (11, 15) adding up the ordinate (y) and the abscissa (x), you get 26.[/hide]
The QuattoMaster 6000 beat me to it...

[hide](11, 15) is the midpoint so the sum is 26[/hide]

Aha. I didn't understand that you was supposed to add the two midpoint coordinats. :blush:

[quote="myspacesucks"]There's an easier way, if you take the average of the coordinates (3,9) and (19,21) you get 20 and 6.  You add those together and you get the right answer in 7 seconds flat.  The formula above, submitted by Quatto, just gives you exactly what I showed, the average of the two and you add them together.  Just expressing it in "idiot terminology."[/quote]
How do you take the "average" of coordinates?

He means take the average of 3 and 9 (from coordinate (3,9))as well as taking the average of 19 and 21 (from coordinate (19,21)).  These are going to be 6 and 20 respectively.  When you add them, you get 26, which is the answer.

Huh. Good one myspace! I'll remember that for the countdown rounds in MathCounts!

edit: 500th post!

[quote="#H34N1"]What is the sum of the coordinates of the midpoints of the line segment whose endpoints are (3,9) and (19,21)?

Hint: Look at the title :D[/quote]

[hide]
$(\frac{22}{2},\frac{30}{2})$

$11+15=\boxed{26}$[/hide]


The 2009 USAMO Day I and Day II papers are on the AMC website.
We will not post any official solutions on the AMC website until after the grading which takes place May 8-9-10.
The 12 USAMO Winners, Honorable Mentions and other data will not be announced until May 11.  MOSP invitations will be shortly after that date.

Steve Dunbar
MAA Director, American Mathematics Competitions

What time on May 11 will the scores be released?

Just a question:
The USAMO teachers manual seems to say results are released on the 13th (bottom of the 3rd page), which day is it?

11th is HM and winners, then MOP invitations start from Wednesday the 13th.

I am anxiously waiting.  :(


geometry
trapezoid
circumcircle
trigonometry
geometric transformation
reflection
angle bisector

ABC is triangle with point D on side AB and F on side AC. <CDB=70 <CDF=40 and <DCB=30 <DCA=20. Find angle <DBF=?

Any idea? i worked on this problem about an hour but i couldn't get an answer.

It is not the way I like a solution, but it still works...
Counting the angles, we shall get that $ \triangle ABC$ is isosceles, with $ 50 - 80 - 50^\circ$.
Let's take $ E$ as symmetrical of $ B$ about $ AC$ and $ {P}=CD \cap BC$ ( we shall find from angle counting $ PE = AC$).
The circle $ \Gamma (E, EA)$ will pass through $ C, P$ and $ \Gamma\cap AB = {A, Q }$, with $ \angle AEQ = 20^\circ$ ( 1 ), hence $ \triangle PEQ$ is equilateral and $ BCEQ$ isosceles trapezoid, therefore cyclic, and let's call $ F'$ its circumcenter, with $ \angle F'BE = 10^\circ$.
We claim that $ F'=F$, so $ \angle DBF=30^\circ$. From $ \angle QEB=\angle QCB=60^\circ$, so $ CD$ is the angle bisector of $ \angle BCQ$, passing through R, the middle of $ BQ$ and $ \angle BF'Q=120^\circ$, so $ F'$ and $ R$ are symmetrical about $ AB$ and $ \angle BDC=\angle RDA=\angle ADF'=70^\circ$, consequently $ \angle CDF'=40^\circ$ hence, by the uniqueness of construction $ F'=F$.

Best regards,
sunken rock

[quote="Ulanbek_Kyzylorda KTL"][color=darkred]Let $ ABC$ be a triangle and $ D\in (AB)$ , $ F\in (AC)$ be two points so that $ m(\angle BDC) = 70^{\circ}$ ,

$ m(\angle CDF) = 40^{\circ}$ , $ m(\angle BCD) = 30^{\circ}$ and ${ m(\angle ACD) = 20^\circ}$ . Find $ m(\angle DBF)$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] Prove easily that $ \frac {BD}{BC} = \frac {FD}{FC} = \frac {1}{2\cos 20^{\circ}}$ . Hence the $ B$-bisector in $ \triangle CBD$ and the $ F$-bisector in $ \triangle CFD$ are concurrently in

$ K\in (CD)$ . Denote $ L\in CD$ so that $ D\in (CL)$ and $ \frac {LD}{LC} = \frac {KD}{KC}$ . Observe that the points $ B$ , $ F$ belong to the circle with diameter $ [KL]$ .

Denote $ x = m(\angle DBF)$ . Since $ \angle LBF\equiv\angle LKF$ and $ m(\angle LBF) = 50^{\circ} + x$ , $ m(\angle LKF) = 80^{\circ}$  obtain $ 50^{\circ} + x = 80^{\circ}$ , i.e. $ x = 30^{\circ}$ .[/color]

[quote="Virgil Nicula"][color=darkred][b][u]Generalization[/u].[/b] Let $ ABC$ be a triangle and $ D\in (AB)$ , $ F\in (AC)$ be two points so that $ BC\cdot FD = BD\cdot CF$,

$ m(\angle ADF) = \alpha$ , $ m(\angle BCD) = \beta$ and $ m(\angle ACD) = \gamma$ . Prove that $ m(\angle ABF) = \frac {\alpha + \gamma - \beta}{2}$ . [/color] [/quote]

Consider the right triangle $ PQR$ with $ \angle P \equal{} 50^\circ, \angle R \equal{} 40^\circ$ .Let $ O$ be the midpoint of $ PR$ . Let The circumcircle of $ ORQ$ meet the perpendicular bisector of $ OQ$ at $ X$.

$ \angle XOQ \equal{} \angle XQO \equal{} 20^\circ$ . Let $ Y$ be the reflection of $ X$ about the perpendicular bisector $ l$ of $ PR$. Observe that $ \angle XOP \equal{} \angle POQ \minus{} \angle XOQ \equal{} 60^\circ$ (Since $ O$ is the circumcenter of $ PQR$, $ \angle POQ \equal{} 80^\circ$

Hence $ OXY$ is equilateral. Since $ XQ \equal{} XO$ , $ X$ lies on the bisector of $ \angle R$. Since $ PY$ and $ RX$ are reflections of each other about $ l$, $ \angle RPY \equal{} 20^\circ$

Since $ \angle XQR \equal{} 60^\circ$ and $ XY$ is parallel to $ PR$, $ \angle QXY \equal{} 80^\circ$ and since $ XQ \equal{} XY, \angle XQY \equal{} 50^\circ \rightarrow \angle RQY \equal{} 10^\circ$

Let $ QY$ intersect $ PR$ at $ M$. We easily find that $ \angle QMP \equal{} 50^\circ, \angle PYO \equal{} 40^\circ$ and $ \angle OQM \equal{} 30^\circ$

The triangle considered in the problem is triangle $ PQM$ and $ Y$ and $ O$ are the points $ D$ and $ F$

So the required angle is $ \angle OQM \equal{} 30^\circ$

[/img][url=http://img691.imageshack.us/i/imagezqf.png/][img]http://img691.imageshack.us/img691/2098/imagezqf.png[/img][/url]

[b]O.K.[/b] [u]Nice proof[/u] ! And my extension ?!

Virgil Nicula's generalization can be solved by exactly the same way as he solved the original problem. :D

[quote="Virgil Nicula"][color=darkred][b][u]Generalization[/u].[/b] Let $ ABC$ be a triangle and $ D\in (AB)$ , $ F\in (AC)$ be two points so that $ BC\cdot FD \equal{} BD\cdot CF$,

$ m(\angle ADF) \equal{} \alpha$ , $ m(\angle BCD) \equal{} \beta$ and $ m(\angle ACD) \equal{} \gamma$ . Prove that $ m(\angle ABF) \equal{} \frac {\alpha \plus{} \gamma \minus{} \beta}{2}$ . [/color] [/quote]

[hide]Let the internal and external angle bisectors of $ \angle DBC$ intersect $ CD$ at $ K$ and $ L$ respectively. We have $ KD/KC\equal{}BD/BC\equal{}FC/FC$ and $ LD/LC\equal{}BD/BC\equal{}FC/FC$, so $ FK$ and $ FL$ are the internal and external angle bisector of $ \angle DFC$ respectively. So $ \angle KBL\equal{}\angle KFL\equal{}90^{\circ}$ which implies that $ KFLB$ is concylic. Let $ \angle CDB\equal{}\theta$. We have $ \angle DBC\equal{}180^{\circ}\minus{}\theta\minus{}\beta$, so $ \angle LBD\equal{}\frac{\theta\plus{}\beta}2$ and hence $ \angle LBF\equal{}\frac{\theta\plus{}\beta}2\plus{}x$. Also note that $ \angle FDK\equal{}180^{\circ}\minus{}\alpha\minus{}\theta$ and $ \angle DFK\equal{}\frac{\angle DFC}2\equal{}\frac{180^{\circ}\minus{}\angle FDC\minus{}\angle FCD}2\equal{}\frac{\alpha\plus{}\theta\minus{}\gamma}2$. So $ \angle LKF\equal{}\angle DKF\equal{}180^{\circ}\minus{}(180^{\circ}\minus{}\alpha\minus{}\theta)\minus{}\frac{\alpha\plus{}\theta\minus{}\gamma}2\equal{}\frac{\alpha\plus{}\theta\plus{}\gamma}2$. But $ KFLB$ is cyclic, so $ \angle LBF\equal{}\angle LKF$, which gives $ \frac{\theta\plus{}\beta}2\plus{}x\equal{}\frac{\alpha\plus{}\theta\plus{}\gamma}2$, and hence $ x\equal{}\frac{\alpha\minus{}\beta\plus{}\gamma}2$. Q.E.D.[/hide]

[u]Thanks[/u], [b]Gunardi[/b] ! Your proof is similarly with mine. And now the upper problem is a simple particular case of this extension.

virgul nicula, why F and B are bisectors in your first solution (not generalization) [u](Hence the B -bisector in   CBD and the F-bisector FDC) [/u]

For [b]Ulanbek_Kyzylorda/KTL[/b] :

$ \left\|\begin{array}{ccc} \frac {BD}{BC} \equal{} \frac {\sin \widehat {BCD}}{\sin\widehat {BDC}} \equal{} \frac {\sin 30^{\circ}}{\sin 70^{\circ}} \equal{} \frac {\frac 12}{\cos 20^{\circ}} \equal{} \frac {1}{2\cos 20^{\circ}} \\
 \\
\frac {FD}{FC} \equal{} \frac {\sin\widehat {FCD}}{\sin\widehat {FDC}} \equal{} \frac {\sin 20^{\circ}}{\sin 40^{\circ}} \equal{} \frac {\sin 20^{\circ}}{2\sin 20^{\circ}\cos 20^{\circ}} \equal{} \frac {1}{2\cos 20^{\circ}}\end{array}\right\|\ \implies\ \frac {BD}{BC}$ $ \equal{} \frac {FD}{FC}\equal{}\frac {1}{2\cos 20^{\circ}}$ .

Let $ X\in (CD)$ for which $ \frac {XD}{XC} \equal{} \frac {1}{2\cos 20^{\circ}}\ \implies\ \left\|\begin{array}{ccc} \frac {BD}{BC} \equal{} \frac {XD}{XC} & \implies & \mathrm{X\ belongs\ to\ B \minus{} bisector\ in\ triangle\ DBC} \\
 \\
\frac {FD}{FC} \equal{} \frac {XD}{XC} & \implies & \mathrm{X\ belongs\ to\ F \minus{} bisector\ in\ triangle\ DFC}\end{array}\right\|$ .

a.s.o.


Let $ n\ge 10000$ be an integer such that its last four digits are same each other(for example $ 12222$, $ 103333$).

If $ n$ is a perfect square, show that $ n$ is dividible by $ 10000$.

PS:This is a problem of univ of tokyo but it originally had next inductuonal question:

[hide="question"]
Let $ m\ge 100$ be a perfect square and its ten's digit is $ a$ and one's digit is $ b$.
If $ a\plus{}b$ is even,show that $ b\equal{}0$ or $ b\equal{}4$.
[/hide]

ok so obviously the square cannot end in 2222, 3333, 7777, or 8888 because of quadratic residue stuff in base 10
we can also rule out 1111, 5555, 6666, and 9999 because of quadratic residues in mod 100
here i will list all the quadratic residues for mod 100:
[hide]
0
1
4
9
16
25
36
49
64
81
21
44
69
96
56
89
24
61
41
84
29
76
[/hide]
EDIT: ok yeah so for -4444 divide the square by 4 and you should get a -11 square but these dont exist
woot

Mods destroy the first problem.

The possibilities to consider are the last four digits being:
0000
1111
2222
3333
4444
5555
6666
7777
8888
9999

Taking mod 4, 1111, 5555, 9999, 2222, and 6666 are ruled out because 2 and 3 are not quadratic residues mod 4.

This leaves:
0000
3333
4444
7777
8888

Taking mod 5 this time, this rules out 3333, 7777, and 8888 because 2 and 3 are not quadratic residues mod 5.

So this leaves:
0000
4444

And finally, 4444 doesn't work because 12 is not a quadratic residue mod 16.

Therefore, the only possibility is for it to end in 0000, and $ 10000 \mid n$, as desired.


A bunch of young boys and girls were standing near a lamppost. Horace pointed to his little sister, Emily and said "My mom is twice as tall as Emily. If my mom were standing where Emily is, my mom's shadow would be twice as long as Emily's shadow.

Horace apparently thinks that shadow length is directly proportional to height. Investigate, using diagrams and geometrical ideas, whether this is true.

More specifically, consider the case of a lamppost that is 20 feet tall and a person who is standing 12 feet away from the lamppost. The length 'L' of that person's shadow will be a function of his or her height 'H'.

Find an _expression for 'L' as a function of 'H' In particular find out whether 'L' is directly proprtional to 'H'?

Counterexample:  let the lamppost have height $L$.  Suppose Mom is of height $L$.  Then her shadow has infinite length.

If you have the same height as the lamppost, your shadow doesn't end.  However, if your height was half that of the lamppost, your shadow does end.

This isn't a complete solution to the problem, but it provides a counterexample to Horace's claim.

I don't know what to write for the expression though...

Draw a diagram, then think similar triangles.


I noticed a pattern of the probabilities of this question:

There are $x$ coins on a table.  The contestant's goal is to get all coins turned to heads in $x$ turns.  The contestant flips all the coins on his first turn.  The ones that are heads are set aside and remain heads the rest of the game and he flips the remaining, which were tails.  He does this $x$ times.  What is the probability that he gets all coins to be heads?

The pattern I noticed was that the probability, at least for the first few, turned out to be $\left(\frac{2^x-1}{2^x}\right)^x$

Can someone prove or disprove it?

It seems pretty hard  ;)

..after the first two times, all of the coins will be heads, right?  There's something i'm missing here...

ok, I'll try to make it clear if you don't understand it.

Lets say there are 5 coins.  You flip all of them and 2 heads come up.  set those two heads aside.  Now we flip the reaming 3.  no heads come up.  We flip the three again. 1 head comes up, set it aside with the other to heads.  Flip the reamining two. No heads come up.  Flip the two again.  Both are heads. you win.

Oh, flipping like that  :D   i was actually thinking you just flip them so if they're tails, they become heads, and if heads, become tails....

The weirdest thing is that when x is 1,2,3, and 4, this pattern works, but I don't know how to prove it

Is it the probability that he gets them all on exactly x flips?  How do they start?  Just randomly?

No it is the probability that he gets it in the time frame of x turns, he can win earlier.

What do you mean how do they start?  He flips all the coins like someone usually flips a coin, and it randomly lands on heads or tails

[quote="mathmanman"]If it works for 1, 2, 3 and 4, maybe you could try to prove it by induction?[/quote]
If you or anyone has any idea how, please show, cause I don't

I'm pretty sure your formula is correct.  Look at each individual coin.  Let's find the complement, i.e. the probability that it comes up tails every time.  This has probability $\frac 1{2^x}$.  Thus the probability of a head coming up at least once is $1-\frac 1{2^x}$. There are $x$ coins thus the desired probability is \[\left(1-\frac 1{2^x}\right)^x=\left(\frac{2^x-1}{2^x}\right)^x.\]

[quote="nat mc"]I'm not sure about that proof because after one turn and you do get a head, then there won't be $x$ coins anymore on the next turn out of the $x$ turns[/quote]
He's saying looking at each coin individually.  For a given coin, the chance of getting a head at some point is $1-\frac{1}{2^x}$

Oh, now I get it.  That was so much easier.  I was thinking partitions of $x$, and allthis other stuff, but you guys made it so much easier. Thanks


Under E2 conditions, the isomers A (trans-1-bromo-2-methylcyclohexane) and B (cis-1-bromo-2-methylcyclohexane) react at different rates to give a methylcyclohexene. Also, a different product is obtained from each isomer. Find out which isomer reacts faster and the product obtained from each one.

the cis isomer reacts fater because u need anti-periplanar arrangement for elimnation and in case of cis there are 2 such hydrogen atoms shile in trans u just have one so rest remaining same the probability factor plays a major role.
trans primarily gives $ \text{3 \minus{} Methyl Cyclohexene}$ and cis gives $ \text{1 \minus{} Methyl Cyclohexene}$

Carcul assuming i am right will there be a change in the major product of trans isomer if the temperature is too high

Your answer is correct. Regarding your question, I don't think so.


linear algebra
matrix
linear algebra unsolved

If $ \simeq$ denotes a similarity relation, then look [url=http://en.wikipedia.org/wiki/Nilpotent_matrix]here[/url], on the "Classification" paragraph, which is different from your condition.


What is the total volume in cubic feet of three boxes if each box is a cube with edge length 4 feet?

The volume of a cube is x^3 where a side is x.
Thus, the volume of 1 cube is 4*4*4=4^3=64.
BUT, it askes you for the volume of 3 cubes=64*3=192.


ABC is a triangle and F,D,E,are the feet of the altitudes from A,C,B,respectivly.

let DE and BC meet each other in point I.and let aline wich is paralell to DE,and goes through F,intersect AC in point

G.prove that:

                   the circumcircle of the triangle GFI,bisects side BC.

[quote="ashegh"]ABC is a triangle and F,D,E,are the feet of the altitudes from A,C,B,respectivly.

let DE and BC meet each other in point I.and let aline wich is paralell to DE,and goes through F,intersect AC in point G.
prove that:

                   the circumcircle of the triangle GFI,bisects side BC.[/quote]

Not hard but not posted correctly. It should be:

"Let DE and BC meet each other in point I. Let a line parallel to DE and passing through F intersect the line AC at a point G [b]and the line AB at a point H[/b]. Prove that the circumcircle of the triangle [b]GHI[/b] bisects the side BC."
_______________________________________________________________________

Denote $a = BC, b = CA, c = AB$ the triangle sides and $\alpha = \angle A, \beta = \angle B, \gamma = \angle C$ its interior angles. Let M be the midpoint of the side BC. We have to show that the quadrilateral IHMG is cyclic, i.e., that

$(?)\ \ FG \cdot FH = FI \cdot FM$

First, the quadrilateral BGCH is obviously cyclic, for example, because $\angle BCG = 180^\circ - \gamma$ and $\angle BHG = \angle BDE = 180^\circ - \gamma$. This is because the lines $HG \parallel DE$ are parallel and $\dfrac{AD}{AE} = \dfrac{CA \cos \alpha}{AB \cos \alpha} = \dfrac{b}{c}$, hence the triangles $\triangle AED \sim \triangle ABC$ are similar and the angles $\angle EDA = \angle BCA = \gamma$. Consequently,

$FG \cdot FH = FB \cdot FC = AB \cos \beta \cdot CA \cos \gamma = bc \cos \beta \cos \gamma$

Using the sine theorem for the triangle $\triangle BDI$,

$\dfrac{BI}{BD} = \dfrac{\sin (180^\circ - \gamma)}{\sin[180^\circ - (180^\circ - \gamma + \beta)]} = \dfrac{\sin \gamma}{\sin (\gamma - \beta)}$

$BI = BD\ \dfrac{\sin \gamma}{\sin (\gamma - \beta)} = BC \cos \beta\ \dfrac{\sin \gamma}{\sin (\gamma - \beta)} = \dfrac{a \cos \beta \sin \gamma}{\sin (\gamma - \beta)}$

$FI = BI - BF = BI - AB \cos \beta = \dfrac{a \cos \beta \sin \gamma}{\sin (\gamma - \beta)} - c \cos \beta$

$FM = \dfrac{BC}{2} - CF = \dfrac{BC}{2} - CA \cos \gamma = \dfrac a 2 - b \cos \gamma$

$FI \cdot FM = \left( \dfrac{a \cos \beta \sin \gamma}{\sin (\gamma - \beta)} - c \cos \beta \right) \left( \dfrac a 2 - b \cos \gamma \right) =$

$= \dfrac{a^2 \cos \beta \sin \gamma}{2 \sin (\gamma - \beta)} - \dfrac{ab \cos \beta \sin \gamma \cos \gamma}{\sin (\gamma - \beta)}\ - \dfrac{ac \cos \beta}{2} + bc \cos \beta \cos \gamma =$

$= \dfrac{a \cos \beta}{2 \sin (\gamma - \beta)} \left[a \sin \gamma - 2b \sin \gamma \cos \gamma - c \sin (\gamma - \beta) \right] + bc \cos \beta \cos \gamma$

To complete the proof, all we have to do is to show that the factor in parentheses is zero, because then

$FG \cdot FH = bc \cos \beta \cos \gamma = FI \cdot FM$

and we would be done.

$(?)\ \ a \sin \gamma - 2b \sin \gamma \cos \gamma - c \sin (\gamma - \beta) = 0$

Substituting the sine theorem formulas for the triangle $\triangle ABC$: $a \sin \gamma = c \sin \alpha$, $b \sin \gamma = c \sin \beta$, we get

$(?)\ \ c \sin \alpha - 2c \sin \beta \cos \gamma - c \sin \gamma \cos \beta + c \cos \gamma \sin \beta = 0$

$(?)\ \ \sin \alpha - \sin \beta \cos \gamma - \sin \gamma \cos \beta = 0$

$(?)\ \ \sin \alpha - \sin (\beta + \gamma) = 0$

which is an obvious identity, because $\beta + \gamma = 180^\circ - \alpha$. This means that the factor in question is really equal to zero.

I just found this topic searching for something different, and would like to note that the problem was also solved at http://www.mathlinks.ro/Forum/viewtopic.php?t=29208 .

  darij


Have any of you ever read this book by John Allen Paulos?  It is very interesting and it has a lot of cool little math tricks that can be applied to economics.  Like it talks about the trick in which there are three different cards, one black on both sides, one red on both sides, and one black on one side and red on the other.  The cards are put into a hat and one card is picked, but only one side of it is looked at.  Then the people playing the game guess which color the other side is.  It turns out there is 2/3 chance it will be the same color as the other side.

The book gives a very rational outlook on the stock market (its not a get-rich-quick book).  For any one interested in math and economics it is a good read.

[quote="rrusczyk"][quote="themonster"]
The book gives a very rational outlook on the stock market (its not a get-rich-quick book).  For any one interested in math and economics it is a good read.[/quote]

I love the image of Paulos writing a get-rich-quick book.  Now, that's funny![/quote]

I am probably "too young" to get it, but I'll ask anyway:  Who is "Paulos?"

He is a math professor at Temple University and has written some popular math books.

What are some of his accomplishments?  I mean, he must be famous for everyone to know him.  Or am I missing something?

To my knowledge its just popular math books that have some application to real life. Recently I saw an article by him in the wall street journal.

[quote="rrusczyk"][quote="themonster"]
The book gives a very rational outlook on the stock market (its not a get-rich-quick book).  For any one interested in math and economics it is a good read.[/quote]

I love the image of Paulos writing a get-rich-quick book.  Now, that's funny![/quote]

Did any body actually use the principles he said in the book to play stocks and won?  In "A beautiful mind", Nash borrowed money from his mother to play stocks but lost them all. I don't know what to say. Interesting phenomena though….

Ahhh.. if you really want to read a nice book about the stock market and life you should check out "Fooled By randomness" By Nassim Taleb.
In my opinion it's a beautifully written book.

I started to get interested in playing the stock market when I was a teen and I read like 100 books(yeah i was that curious)..  that journey also led me to studying math and statistics because I started to get more and more curious about certain things and techniques which people use.
(I ended up being an undergraduate in economics and statistics  now doing my masters but that's another story)

What is nice to notice is that with all the sharp and skilled people out there trying to "beat the market", one of the greatest traders of all times was George Soros which really
has a philosophy about trading which i forgot its name but he really was mega succesful.

Some more nice books applying some statistics to trading is the books by Victor Niederhoffer...(Although he lost all of his money)


Find all integered-valued functions f(n) taking values in the integers
satisfying the equation f[f(n)] = n + 1.

The first step of the solution is "Applying f to both sides we obtain f(n+1)=f(n)+1.
Where does it come from, i dont understand at all. Would appreciate a simple explanations.
Thanks

You obtain
f(f(f(n)))=f(n+1)

Now using f(f(n))=n+1 and putting n:=f(n)
You get f(f(f(n)))=f(n)+1

hope u get it now


Denote by $P(a,b,c)=\sum a^3(b-c)$.
We get that $P(b,b,c)=0, P(a,c,c)=0, P(a,b,a)=0$. So $(a-b)(b-a)(c-a)|P(a,b,c)$.
Now we have that $P(a,b,c)=(a-b)(b-c)(c-a)Q(a,b,c)$. The degree of $P$ is 4 so the degree of $Q$ will be 1. Now it follows easily that $P(a,b,c)=-(a-b)(b-c)(c-a)(a+b+c)$.

[quote="Myth"][quote="Krishna Kothoor"]:) Factor this expression:
$a^3(b-c) + b^3(c-a) + c^3(a-b)$[/quote]
Also I like the following one:
\[a^3+b^3+c^3-3abc.\]
 :D[/quote]

The equation with roots $a,b,c$ is $(x-a)(x-b)(x-c)=0\Longleftrightarrow x^3-(a+b+c)x^2+(ab+bc+ca)x-abc=0$
Since $a,b,c$ satisfy this equation,we have the following equations.
\[a^3-(a+b+c)a^2+(ab+bc+ca)a-abc=0\]
\[b^3-(a+b+c)b^2+(ab+bc+ca)b-abc=0\]
\[c^3-(a+b+c)c^2+(ab+bc+ca)c-abc=0\]

Adding up these equations,we have

$a^3+b^3+c^3-(a+b+c)(a^2+b^2+c^2)+(ab+bc+ca)(a+b+c)-3abc=0$
$\Longleftrightarrow a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

You can exercise more applying with your problem at [url]http://mathlinks.ro/Forum/viewtopic.php?t=19463[/url]

kunny

Similary the roots of

x^3 - 3bc X + (b^3+c^3)
are -(b+c), -(bw+cw^2), -(bw^2,cw)   where  w is cube root of unity.

gives the factors in nice form

a^3+b^3+c^3- 3abc = (a+b+c)(a+bw+cw^2)(a+bw^2+cw)
which is (a+b+c)(a^2+b^2+c^2-ab-bc-ca)
also the second factor is
(1/2)((a-b)^2+(b-c)^2+(c-a)^2) and  if a,b,c are real, zero only if a=b=c.

Thus for  real values,  a^3+b^3+c^3 = 3abc  only if a+b+c = 0 or a=b=c


I am so bad at graph theory  :blush:

[b]Problem:[/b] Given $ 6$ people in a room show that either $ 3$ are complete friends or $ 3$ are complete strangers. (If I am your friend you are my friend, so this is a [b]un[/b]directed graph).

I tried solving this and ran into a lot of case work which made me suspect this is not the elegant solution via graph theory.

Color the undirected graph with red and blue. For vertex A, either there are three or more vertexes with red edges between A, or there are three or more vertexes with blue edges between A. Assume their edges are red. Call them B,C,D. If $ \vartriangle BCD$'s three edges are all in blue, it's finished. Or there are one or more red edges in BC, BD, CD. Assume BC is red. Then $ \vartriangle ABC$'s three edges are all in red.

I am sorry I dont understand the solution  :oops:

maybe this will help

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=156813

I tried using those two hints, it doesnt work out. The Pigeonhole argument is used in some tricky way there, I cant figure it out. I tried making the holes to be the ordered 3 elemenets sets of the vertices but that doesnt work out because that is not a valid partition. And I cant understand some of the graph terminology used because I am a newbie to graph theory. Can someone please write an elementary proof of this problem.

For such a small number, it's trivial to exhaustively list all possible graphs.

Can anybody finish this argument?

Basically, we can take all connected nodes and make them unconnected and vice versa. Lets say that we have $ k$ edges. Then that implies $ 16-k$ edges.

Can anybody show $ k\implies k-1$?

Say we have $ k$ edges, and we remove one edge. If there are two $ K_{3}$s, we are done. Else, assume that our $ k-1$ edge graph has no $ K_{3}$s.

[insert argument here]

That way it suffices to show $ k=8$.

[hide] Pick a vertex. There are five edges emanating from that vertex. By pigeonhole, at least 3 are the same color, let's say red. If any of these three have a red edge between them, it forms a red triangle. Otherwise, they are all blue and that forms a blue triangle [/hide]

Altheman, are you trying to prove the existence of two monochromatic triangles?

[hide] A monochromatic triangle contributes 3 monochromatic angles, a non-monochromataic triangle contributes 1 monochromatic angle. There are a total of $ \binom{6}{3}=20$ triangles. Suppose a given vertex has $ a$ red and $ b$ blue edges emanating. Then it has $ \binom{a}{2}+\binom{b}{2}\geq \binom{3}{2}+\binom{2}{2}=4$ monochromatic angles (the $ \geq$ comes from the convexity of $ \binom{n}{k}$) so there are a total of at least $ 24$ monochromatic angles. If there is only one monochromatic triangle, there are at least $ 19+3=22$ monochromatic triangles, contradiction.[/hide]

[quote="sylow_theory"][quote="Phelpedo"][hide] Pick a vertex. There are five edges emanating from that vertex. By pigeonhole, at least 3 are the same color, let's say red. If any of these three have a red edge between them, it forms a red triangle. Otherwise, they are all blue and that forms a blue triangle [/hide]
[/quote]

I have a question.
[hide="Your solution with my commentary"]
Create this graph. Look at the [i]maximal[/i] degree vertex. If this vertex has degree $ \leq 4$ then the proof if complete (this also bothers me).  If not then it has $ 5$ edges emanating from it (that is what you assumed). So by Pigeonhole three of those are the same color. Let's introduce some notation to make it easier to follow. Again $ v$ is this vertex, and let $ a,b,c$ be the other three edges that form a red edge. So far we know that $ va,vb,vc$ are all red. If any one of $ ab,bc,ac$ are red then we have a red triangle and the proof is complete. If not then all three are blue ($ ab,bc,ac$) and we have a blue triangle, and the proof is complete.

Look at the last line "If not then all three are blue and we have a blue triangle". That's what I dont understand, you are [i]assuming they are connected[/i]. What happens if they are not? Even casework doesnt work here.  :(

[/hide][/quote]

Red = they are complete friends
Blue = they are complete strangers

This is the classification made.  We assume that every relation between two people is one of these two: that is, [b]any two vertices are connected by an edge of one of these two colors[/b].  If the two categories you gave in the problem statement do not describe all relationships between two people, then the problem statement is false.

Thanks to Pheldo and k8107 I understand now :)

Here is a more difficult version.

[b]Problem:[/b] Given 17 people in a room, between any two the following is exactly true (only one): They have never met, They are good friend, They hate eachother. Show that there exists 3 people all of whom never met, or are good friends or hate eachother.

I am not posting the solution, because its pretty easy after what Pheldepo did.

Here is my Generalized Generalized version.  :ninja:

[b]Problem:[/b] Given 67 people in a room, between any two of them the following is exactly true: they never met, they are good friends, they hate eachother, they are relatives. Show that there exists 3 people all of whom never met or are good friends or hate eachother or are relatives.

And we can even Generalize this:

[b]Problem:[/b] Given 338 people ... blah blah blah.

In general, for an $ n$ color graph can do this for some $ a_{n}$...find a recursive relation for $ a_{n}$.

Edit: In response to JBL's post below, just to clarify, this recursive relation can give you the number requred to use a simple pigeonhole-type argument: it is by no means the strongest upper bound on that Ramsey number.  You can similarly find uppoer bounds for all Ramsey numbers in which all colors are identical (of course not necessarily the strongest), and thus find bounds for all Ramsey numbers.

These numbers that you are playing with are the Ramsey numbers $ R(3, 3, \ldots, 3)$.  The classical Ramsey numbers $ R(m, n)$ (and especially $ R(m, m)$) concern the question, "How many people do you need at a party to assure that there are $ m$ who are mutually acquainted or $ n$ who are mutual strangers?"  Thus, the first question solved shows that $ R(3, 3) \leq 6$ since in any party with 6 people either 3 are acquainted or 3 are mutual strangers.  Multi-color Ramsey numbers have not such a nice formulation in terms of parties, although they can be expressed easily as graph-coloring questions.  Only a very few Ramsey numbers are actually known.  [url=http://mathworld.wolfram.com/RamseyNumber.html]Mathworld[/url] and [url=http://www.combinatorics.org/Surveys/ds1/sur.pdf]this paper[/url] (.pdf) in the [url=http://combinatorics.org]Electronic Journal of Combinatorics[/url] provide known bounds for both classical and multi-color Ramsey numbers.  (The recursion K81o7 is suggesting you look for will be a recursion on the upper bound only.)


ratio
arithmetic sequence
geometric sequence

Consider the infinite arithmetic sequence $ A$ with first term $ 5$ and common difference $ \minus{}2$. Now define the infinite sequence $ B$ so that the $ k^{th}$ term of $ B$ is $ 2$ raised to the $ k^{th}$ term of $ A$. Find the sum of all of the terms of $ B$.

Taking a look at the first few terms of $ B$, we get:

$ 2^5\plus{}2^3\plus{}2^1\plus{}2^{\minus{}1}\plus{}2^{\minus{}3}\plus{}2^{\minus{}5}\plus{}2^{\minus{}7}\plus{}2^{\minus{}9}\plus{}2^{\minus{}11}\approx42.666$. Checking out a little more, we get $ 42\frac{2}{3}$.

You could also use the sum of an infinite geometric sequence formula, where the sum is a/(1-r) where a is the first term and r is the ratio (and where r<1), so the answer is 32/(1-1/4)=32/(3/4)=[b]128/3[/b]


The dipole is centered at the origin and that a point charge Q is a distance r away along the line of the dipole. Show that the location of the positive point charge the electric field E due to the dipole is toward the dipole and find its magnitude.  :D  :D

Don't understand yours?
wonder that how old are you lah?  :D
i thought you are so old to study this kind of Phys


I don't know if another forum already mentioned this article but we covered it at Metroplex Math Circle:

http://metroplexmathcircle.wordpress.com/2008/11/23/boston-globe-colleges-scour-china-for-top-students/


i^1=i, i^2=-1, i^3=-i, i^4=1, i^5=i, and so on. These repeat every four exponents. You can use this to solve these problems.

[hide="1."]
This can be simplified to -i+i-i+i=0. Remember that you can only add exponents in multiplication.[/hide]

[hide="2."]  2(1)(1-1)=2(0)=0[/hide]

[quote="boxedexe"][quote="7h3.D3m0n.117"][quote="Interval"]Write each expression in the standard form a + bi.

(1) i^7 + i^5 + i^3 + i

(2) 2i^4 (1 + i^2)[/quote]

[hide="1"]
$\begin{eqnarray*}i^{7+8+3+1}&=& i^{19}\\ &=&-1$
[/hide]

[hide="2"]
$\begin{eqnarray*}2i^{4}(1+i^{2}) &=& 2i^{4}+2i^{6}\\ &=& 4i^{10}\\ &=&4i$
[/hide][/quote]
You cannot just make up mathematics without any logical reasoning.  I suggest you review complex numbers (just a friendly recommendation :))[/quote]

eh, i dont think complex #'s is the problem.  just review some algebra.  i think anyone made this mistake a while back

[edit]: the person anyone, not anyone :rotfl:


abstract algebra
group theory
superior algebra
superior algebra unsolved

true or false- R is a ring of order 210,then R is commutative.

i have no clue on even how to start. whats so special abt 210? it is a product of distinct primes. could that make a difference? i am trying to investigate for smaller values eg 6. please give me a hint if you solve it.

[quote="Albanian Eagle"]It is true. Every ring whose order is squarefree is commutative.
See [url=http://www.jstor.org/view/00029890/di991509/99p0012p/0]here[/url] for example.[/quote]

Hello.
If you didn't subscribe to JSTOR, you can look at [url=http://planetmath.org/encyclopedia/CyclicRing3.html]here[/url]
The additive group of a ring is abelian, thus if the number of elements of this ring is squarefree, the additive group is cyclic. (A finite abelian group with squarefree order is always cyclic.)
Now, a ring with cyclic additive group is commutative. (Look at the link.)

Friendly,
P.


Sa se demonstreze ca oricare ar fi $x,y,z>0$ cu $x+y+z=1$ are loc
$\frac{x^2}{x^3+1}+\frac{y^2}{y^3+1}+\frac{z^2}{z^3+1}\geq \frac{9}{37-27(xy+yz+zx)}$

Din Cauchy:
$(\frac{x^2}{x^3+1}+\frac{y^2}{y^3+1}+\frac{z^2}{z^3+1}) (x^3+y^3+z^3+3)\geq (x+y+z)^2$
$\Rightarrow S\geq\frac{1}{x^3+y^3+z^3+3}$
Daca aratam ca $\frac{1}{x^3+y^3+z^3+3}\geq\frac{9}{37-27(xy+xz+yz)}$, problema e rezolvata.
Chestia e echivalenta cu $37-27(xy+xz+yz)\geq 9x^3+9y^3+9z^3+27$.
Notam $s=xy+xz+yz$, deci vreau sa demonstrez ca
$37-27s\geq 9(x+y+z)(x^2+y^2+z^2-s)+27xyz+27\Leftrightarrow$
$37-27s\geq 9(x+y+z)^2 - 18s - 9s + 27xyz+27\Leftrightarrow$
$1\geq 27xyz$, care iese usor din $\frac{x+y+z}{3}\geq\sqrt[3]{xyz}$
Initial am vrut sa demonstrez ca $\frac{x^2}{x^3+1}$ e convexa si sa aplic Jensen. Si am derivat de 2 ori, si am ajuns la o oribilitate, si am renuntat :)

Dap. si eu ma gandeam la o solutie de acest gen,nu cu jensen


Let $ a,b$ and $ c$ be positives real numbers.Prove that
\[ \sum_{cyc}\frac{a^2\plus{}2bc}{(a\plus{}2b)^2\plus{}(a\plus{}2c)^2} \leq \frac{1}{2}\]

[quote="apollo"]Let $ a,b$ and $ c$ be positives real numbers.Prove that
\[ \sum_{cyc}\frac {a^2 \plus{} 2bc}{(a \plus{} 2b)^2 \plus{} (a \plus{} 2c)^2} \leq \frac {1}{2}
\]
[/quote]

$ a^2  \plus{} b^2  \ge \frac{{\left( {a \plus{} b} \right)^2 }}{2}$

[quote="Honey_S"][quote="apollo"]Let $ a,b$ and $ c$ be positives real numbers.Prove that
\[ \sum_{cyc}\frac {a^2 \plus{} 2bc}{(a \plus{} 2b)^2 \plus{} (a \plus{} 2c)^2} \leq \frac {1}{2}
\]
[/quote]

$ a^2 \plus{} b^2 \ge \frac {{\left( {a \plus{} b} \right)^2 }}{2}$[/quote]
Nice solution  :lol:
My solution is similar.


calculus
derivative
function
limit
calculus computations

show that $ D_{1}f$, where $ f(x,y) = xy \frac{x^{2}-y^{2}}{x^{2}+y^{2}}$ if $ (x,y) \neq (0,0)$ and $ f(0,0)=0$, is discontinuous at the origin.

I don't feel like computing the derivative, but it should be continuous at the origin. The function $ f$ is homogeneous of degree $ 2$, that is, $ f(cx,cy)=c^{2}f(x,y)$ for any $ c\ge 0$. Differentiation reduces the degree of homogeneity by one. Since $ D_{1}f$ is homogeneous of degree $ 1$, we have $ D_{1}f(0,0)=\lim_{(x,y)\to(0,0)}D_{1}f(x,y)=0$. But the higher-order derivatives will be discontinuous.


Let $V$ be a convex polygon.
(a) Show that if $V$ has $3k$ vertices, then $V$ can be triangulated such that each vertex is in an odd number of triangles.
(b) Show that if the number of vertices is not divisible with 3, then $V$ can be triangulated such that exactly 2 vertices have an even number of triangles.

I apologize if this is difficult to read, if I have any time I will try to clean it up later.

a) Let $ V$ be a convex polygon with $ 3k$ vertices.  We proceed by induction on $ k$.  If $ k \equal{} 1$, then $ V$ is a triangle, and the problem is trivial.

For the inductive step, assume we can triangulate a polygon with $ 3k$ vertices ($ k\geq 1$) so that every vertex is included in an odd number of triangles.  Now suppose we have a polygon $ V$ with $ 3(k \plus{} 1)$ vertices.  Consider the polygon formed by drawing a line between the first and fifth vertices of $ V$ and ignoring the second, third, and fourth vertices.  Since this new polygon $ W$ has $ 3(k \plus{} 1) \minus{} 3 \equal{} 3k$ vertices, we can triangulate it so that every one of its vertices is included in an odd number of triangles.  Now triangulate the pentagon $ P$ whose vertices are the first five vertices of $ V$ by drawing a line between the first and third vertices and a line between the third and fifth.  Then the second and fourth vertices are each included in one triangle and the third is included in three.

For the first vertex of $ V$, notice that it belongs to an odd number of triangles in $ W$ and two triangles in $ P$.  Therefore it is included in an odd number of triangles in $ V$.  A similar argument holds for the fifth vertex.  Therefore all vertices of $ V$ are included in an odd number of triangles.

b)  Let $ V$ be a convex polygon having a number of vertices not divisible by three.  Let $ n$ represent the number of vertices of $ V$.  Then we have two cases.

Case 1:  $ n \equiv 1 \;\mathrm{mod}\; 3$.  In this case, draw a line between the first and third vertices of $ V$.  Then the polygon $ W$ created by ignoring the second vertex of $ V$ has $ n \minus{} 1$ vertices, and since $ n \minus{} 1 \equiv 0 \;\mathrm{mod}\; 3$, this polygon may be triangulated so that each of its vertices are included in an odd number of triangles by part a).  Then, since the first and third vertices of $ V$ are each included in one additional triangle (the one with the first, second, and third vertices of $ V$ as vertices), they are each included in an even number of triangles.  Since the second vertex of $ V$ is included in one triangle, exactly two vertices of $ V$ are included in an even number of triangles.

Case 2: $ n \equiv 2 \;\mathrm{mod}\; 3$.  In this case, draw a line between the first and fourth vertices of $ V$.  Then the polygon $ W$ created by ignoring the second and third vertices of $ V$ has $ n \minus{} 2$ vertices, and since $ n \minus{} 2 \equiv 0 \;\mathrm{mod}\; 3$, this polygon may be triangulated so that each of its vertices are included in an odd number of triangles by part a).  Now draw another line between the first and third vertices of $ V$.  Now, in the quadrilateral $ Q$ with vertices the first four vertices of $ V$, the first and third vertices are each included in two triangles and the second and fourth vertices are included in one triangle each.  Therefore, when considering all of $ V$, the first vertex is included in an odd number of triangles (two from $ Q$ and an odd number from $ W$) and the second vertex is included in one triangle.  The third is included in two triangles, and the fourth, which is included in an odd number of triangles from $ W$ and one from $ Q$, is in an even number of triangles.  Since every other vertex is part of $ W$ only, exactly two vertices of $ V$ are included in an even number of triangles.

Therefore, if $ V$ has a number of vertices not divisible by $ 3$, it can be triangulated so that exactly two of its vertices are included in an even number of triangles.

this is problem is one of the first problems on IMC few years ago... easy one ;)


I'd like to create a math circle like the Bay Area and the San Diego one's for my area, but I have no idea on where to start, or what I need to do to get started, or anything.  Can anyone help with this?

I'd think the first thing you'd need to do is find a responsible, well-organized, math oriented adult who'd be willing to help you out and then go from there.


[url=http://www.homestarrunner.com/sbemail3.html]Click here to see some of the greatest algebra ever.[/url]

:lol: Ah yes, always a favorite. There's also [url=http://www.homestarrunner.com/sbemail39.html]The property of ones[/url].

i havent watched homestar runner in forever almost forgot about that sight :rotfl:

i thought the transitive butt property was really funny.


counting
distinguishability
pigeonhole principle
combinatorics proposed
combinatorics

Let $\{a,b\}$ is an alphabet and $w$ is a word over this alphabet. Denote by $A(w)$ the number of occurances of $a$ in $w$ and by $B(w)$ the number of occurances of $b$ in $w$. Let $L$ be a language consisting of those words $w$, for which $(A(w),B(w))=1$. Is it true that $L$ is a [i]regular[/i] language? (I don't know what [i]regular[/i] language is)

[i](5 points)[/i]

A language is regular if it can be recognized by a finite-state machine.  That is, all the words in the language lead to "accept" states, and all the words not in the language lead to "reject" states.

Aaha.
I understand now. Unfortunately, it didn't come to my mind then I was reading it for first time.
Thank you!

BTW, is there equivalence between Turing machines and finite-state machines? Since I know that I can do all what I want with Turing machine.

No, finite-state machines are much weaker than Turing machines.  For example, a finite-state machine cannot even determine whether the number of a's is equal to the number of b's.  A finite-state machine is like a Turing machine without a "tape" -- it only has a constant amount of memory.

Here is a solution to the original problem.  Consider the words of the form $\texttt{a}^p$, where $p$ is prime.  All these words are inequivalent mod $L$.  (That's because $\texttt{a}^p \texttt{b}^p$ is not in $L$, but $\texttt{a}^q \texttt{b}^p$ is in $L$, where $p$ and $q$ are different primes.)  By the Myhill-Nerode theorem, an infinite set of inequivalent words implies that the language is not regular.

It seems I dodn't understand something... I thought I am somewhat familiar with that theory.
Words $a^p$ are not in $L$. Right? Anyway...  :?

Yes, the words $\texttt{a}^p$ are not in $L$, but that doesn't matter for applying the Myhill-Nerode theorem.  I can't really explain the whole theory here, but you might want to look [url=http://www.cs.odu.edu/~toida/nerzic/390teched/regular/reg-lang/non-regularity.html]here[/url] for a discussion of the Myhill-Nerode theorem.  (That link uses the term "indistinguishable" where I use "equivalent".)

The Pumping Lemma is usually taught in "Theory of Computation" courses as the main method for proving non-regularity.  But I generally find the Myhill-Nerode theorem to be more convenient.  It ends up killing problems such as this one.

But it is an atomic bomb here. The problem is obvious itself.

Maybe, but what's your solution to the problem?

But you actually wrote it above. Since all $a^p$ are not equaivalent it means that the machine should have infinitely many states (final state for each $p$). And we don't need to use any theorems. Right?

In a way, yes.  The Myhill-Nerode theorem is trivial to prove, but has lots of nontrivial consequences.  Sort of like the Pigeonhole Principle.


floor function
ceiling function
summer program
MathPath
function
modular arithmetic

[quote="Franz Joseph"]Given that $n$ positive integers that sum to $19$. What is the maximum product of the $n$ positive integers?

Is there a general way to do problems like this?[/quote]
[hide] It's simply AM-GM.
$19=a_{1}+a_{2}+...+a_{n}\ge n\sqrt[n]{a_{1}a_{2}...a_{n}}$
$\frac{19}{n}\ge\sqrt[n]{a_{1}a_{2}...a_{n}}$
$\boxed{\left(\frac{19}{n}\right)^{n}}\ge a_{1}a_{2}...a_{n}$

Generalization: If the sum is $s$...
$s=a_{1}+a_{2}+...+a_{n}\ge n\sqrt[n]{a_{1}a_{2}...a_{n}}$
$\frac{s}{n}\ge\sqrt[n]{a_{1}a_{2}...a_{n}}$
$\boxed{\left(\frac{s}{n}\right)^{n}}\ge a_{1}a_{2}...a_{n}$ [/hide]

[quote="drunner2007"]you have to remember that we're dealing with integers though. that works if the numbers are real numbers[/quote] Oh, sorry.  :blush: And you meant $\text{positive}$ reals?

I don't know if this is right....
[hide]If $19\equiv x\bmod{n}$, $0\le x<n$, then we have that the maximum product is $\left\lfloor \frac{19}{n}\right\rfloor^{n-x}\times \left\lceil \frac{19}{n}\right\rceil^{x}$.[/hide]

Threes are good.

See, that sentence was 3 words... proving that threes are even better.

I'm being really cryptic. One of the MathPath problem sets had a problem that went something like "Given n positive integers that sum to 2005, what is the greatest possible product of the n integers?"

And we proved that to maximize the product, you should use as many threes as possible. I don't remember the details of the proof.

So would the answer be $3^{5}\cdot2^{4}=3888$?

Doubtful...

This was from the Australian Mathematics Competition for the Westpac award. So, anyone who did the Intermediate and Senior Division? They should know it.

[quote="Franz Joseph"]Given that $n$ positive integers that sum to $19$. What is the maximum product of the $n$ positive integers?[/quote] Are we...
1. finding a function in terms of $n$?
Or...
2. a. finding the value of $n$ at the maximum product, along with
b. the maximum product?
[hide="2a"] I believe the maximum is $\left\lfloor\frac{19}{e}+0.5\right\rfloor=7$. With the generalization being $\left\lfloor\frac{s}{e}+0.5\right\rfloor$ [/hide]
[hide="2b"] We want the numbers as close to $e$ as possible, which is what mathnerd was saying. $3^{5}\cdot2^{2}$ gives a nice proportion.

Generalization: If the product is in the form $2^{a}\cdot3^{b}$, we can write the equations...
$2a+3b=s$
$a+b=\left\lfloor\frac{s}{e}+0.5\right\rfloor$
Solving, we get $b=s-2\left\lfloor\frac{s}{e}+0.5\right\rfloor$ and $a=3\left\lfloor\frac{s}{e}+0.5\right\rfloor-s$ [/hide]

EDIT: The generalizations don't work, but I think they're close. :D

[quote="DanK"]Um, wouldn't it have to be $3^{5}\cdot 2^{2}=972$ if they sum to 19.  And that's not even the highest number of 3's.  That would be $3^{6}\cdot 1=729$, so that's definitely not it...[/quote]

The most efficient number (not an integer) is $e$ (can't remember why).

Thus, you want to use the maximum number of 3s.  However, if $n\equiv1\pmod3$, then you should replace a 3 with 2 twos ($2\times2>3$).

[quote="b-flat"]

The most efficient number (not an integer) is $e$ (can't remember why).
[/quote]
Does that have anything to do with $(x+\frac{1}{n})^{n}$ tends to e?

[quote="now a ranger"]i got 500 lol using 5, 5, 5, and 4. What did I do wrong, i think i might have read the question completely wrong. :mad:[/quote]

Read through the other messages - the best number for this is 3 (unless it is 1 mod 3, in which case you use two 2s instead of a 3 & a 1).  Actually, $e$ is the best (try it).

We still haven't decided why, & I think why is not a question for this forum.


function
inequalities
limit
algebra unsolved
algebra

Prove that if a function obeys Jensen's inequality for two variables, then it is continuous.

Bomb

It's easy to check that for fixed $x$, the function $g(h)=\frac{f(x+h)-f(x)}h$ is increasing both on $(0,\infty)$ and on $(-\infty,0)$. This means that $\lim_{h\to 0^+}\frac{f(x+h)-f(x)}h$ exists, and so does $\lim_{h\to 0^-}\frac{f(x+h)-f(x)}h$, and it's also easy to see that they must be finite (we use the fact that $x$ is an interior point, so it's an accumulation point both from the right and from the left), and from here the conclusion follows.