Let $ n$ be a positive integer such that :
\[ 8\sum_{k \equal{} 0}^{n} k \equal{} m^q
\]
where $ m$ is an positive integer, and $ q$ is an odd prime.

Prove that $ q\mid 2n\plus{}1$.

This means $ 4n(n\plus{}1) \equal{} m^q$ or $ (2n\plus{}1)^2 \minus{} m^q \equal{} 1$. By Catalan (I think this special case was already posted), this means $ n\equal{}1,m\equal{}2,q\equal{}3$.

Easy check $ p$ odd.
$ x\equal{}2n\plus{}1$
$ x^2\equal{}m^q\plus{}1$
then
$ x^2\equal{}(m\plus{}1)\frac{m^q\plus{}1}{m\plus{}1}$
$ \gcd(m\plus{}1,\frac{m^q\plus{}1}{m\plus{}1})\equal{}\gcd(m\plus{}1,q)$
Suppose $ \gcd(m\plus{}1,q)\equal{}1$ then
$ m\plus{}1\equal{}a^2$ and $ \frac{m^q\plus{}1}{m\plus{}1}\equal{}b^2$
Easy to check traditional.
so $ \gcd(m\plus{}1,q)\equal{}q$ so $ q|x$

how!!!  [code]Easy to check traditional. [/code]
it's the most difficult step to prove, i think!!

Let $ x=2n+1$
if we don't have $ q\mid x$ then we won't have $ q\mid m+1$.we write $ m+1=a^2$ and $ \frac{m^q+1}{m+1}=b^2$ so $ x=ab$ with $ a>0$ and $ b>0$. since $ y$ is not null, $ y$ is not a square.
Now, we will use the pell-fermat equation $ u^2-mv^2=1$(*).we have $ (a,1)$ is a minimal solution for (*) so every solution has the unique form
$ u+\sqrt{m}v=(a+\sqrt{m})^k$ for some $ k\in Z$. if we note $ \alpha = x+\sqrt{m}m^\frac{q-1}{2}$, then $ \alpha$ is a solution for (*)($ x^2-m^q=1$).
hence, there exists $ k>0$ such that $ \alpha = x+\sqrt{m}m^\frac{q-1}{2}=(a+\sqrt{m})^k$.by modulo $ m$ we have $ m\mid a^k-x$ and $ m\mid ka^k-1$. since $ m$
and $ a$ are coprime we have $ m\mid k$. since $ m$ must be even, $ k$ is even.Now, by modulo $ a$ and using the fact that $ x=ab=0(\bmod a)$ and $ m=-1(\bmod a)$
we obtain $ a\mid 1$ so $ a=1$ it's impossible because $ m$ is not null.
so we have $ q\mid x$

i mad an error
by modulo  $ m$ we have $ m\mid a^k \minus{} x$  and $ m\mid ka^k \minus{} 1$ the right expression is by modulo$ m$we have$ m\mid a^k\minus{}x$and$ m\mid ka^{k\minus{}1}$

by modulo $ m$ we have $ m\mid a^k\minus{}x$ and $ m\mid ka^{k\minus{}1}$.


Given $ e^{xt \minus{} \frac {t^2}{2}} \equal{} \sum_{n \equal{} 0}^{\infty} H_n(x)\frac {t^n}{n!}$, prove the following contexts.

(1) $ H'_n(x) \equal{} nH_{n \minus{} 1}(x)\ (n \equal{} 1,\ 2,\ \cdots)$.

(2) $ H_n(x)$ saisfies $ y'' \minus{} xy' \plus{} ny \equal{} 0$.


Let $a_n$ is a decreasing convex sequence.Morever its limit is zero.
Prove:$lim((n+1)a_(n+1) - na_n)=0$

The sequence $ \{ a_{n} \} $ is decreasing and convex. This means that
$ a_{n} \geqslant a_{n+1} $ and $ a_{n+2} - 2a_{n+1} + a_{n} \geqslant 0 $.
Also of course $ a_{n} \geqslant 0 $ ( because $ \{ a_{n} \} $ is decreasing
and $ lim_{n \rightarrow \infty } a_{n} = 0 $ ).
Therefore if we define $ b_{n} := a_{n} - a_{n+1} $ we get that
$ b_{n} \geqslant 0 $ and $ b_{n} \geqslant b_{n+1} $.
Also $ b_{n} \rightarrow 0 $ when $ n \rightarrow \infty $ because
$ 0 \leqslant b_{n} \leqslant a_{n} $.
Fix $ m\in \mathbb{N} $. Then $ \sum_{n=m}^{N} b_{n}
= a_{m} - a_{N+1} \leqslant a_{m} $ . Therefore
$ (N-m)b_{N} \leqslant \sum_{n=m}^{N} b_{n} \leqslant a_{m} $.
Now take $ \varepsilon > 0 $ and for some $ m_{0} $ we have
$ a_{m_{0}} \leqslant \varepsilon $. So for every $ N > m_{0} $ we have
$ Nb_{N} \leqslant a_{m_{0}} + m_{0}b_{N} \leqslant \varepsilon + mb_{N} $.
Now because of $ lim_{n \rightarrow \infty } b_{n} = 0 $, we have that for
$ N $ large that $ m_{0}b_{N} < \varepsilon $.
Therefore for $ N $ large we have that $ Nb_{N} < 2\varepsilon $.
Therefore $ lim_{n \rightarrow \infty} nb_{n} = 0 $, but
$ nb_{n} = n(a_{n} - a_{n+1}) $ ,so $ (n+1)a_{n+1} - na_{n} = a_{n+1} - nb_{n} $,
and because of $ lim_{n \rightarrow \infty } a_{n+1} = lim_{n \rightarrow \infty }
nb_{n} = 0 $ we obtain that $ lim_{n \rightarrow \infty }
(n+1)a_{n+1} - na_{n} = 0 $.


I'm using MikTex 2.7 and TeXnicCenter 1 Beta 7.01.  How do I jump to the corresponding error in the source code from the error reported in the build window?  Double-clicking the error in the build window does not seem to work nor does pressing the Next/Previous error/warning toolbar icons.  Do I need to configure something to get this to work?

Thanks,
Shaw

It depends on what the error is. Take the following minimal document:
[code]\documentclass{article}
\usepackage{amsmath}
\begin{document}
Text
\end{document}[/code]If you leave off the final } in \end{document} the line won't be highlighted, but if you misspell \usepackage it will. TexnicCenter seems to rely on TeX to tell it which line the error is on (so an error in spelling of \usepackage says l.2 ie line 2) but TeX doesn't always know which line the error is and also sometimes guesses wrongly.

Thanks Steve.  I  see there's quite a bit of information here and will spend some time reviewing the older threads.

Alright, now I understand . . . you have to read completely the 4 threads above in Announcement, as well as all the stickies threads above, compile a nice list of bookmarks of preferred links these thereads point too for future reference, then download MikTex first which is the LaTex compiler, then download TeXnicCenter which is the integrated programming environment for writing LaTex documents, then start with a few minimal documents which you'll know about by reading said mentioned threads, then either build the document as either a Latex=PDF document in which the Acrobat reader is used to display it, or either build it as a Latex=DVI document in which the YAP(yet another previewer) is used to display it.  You can select which type by Build/SelectOutputProfile.

Also should check very carefully all the error messages reported by TeXnicCenter and yes if you double click the error messages you'll get a nice red arrow at the left margin in the code showing where the error is although I noticed this worked only when I start with the first error message.  There is a nice description in the threads above describing how to troubleshoot errors using the "minimal working example".

Also a good idea I think to create your code in a new project template via File/NewProject which creates a blank document with all the common inititialization code then all you need to do is start adding code.  Why do I have so much trouble learning stuff?

[edit] Also, if you want to have the option of double-clicking on a spot in the YAP document and having YAP go directly to TeXnicCenter to the source code at the location corresponding to the point you clicked, then choose the View menu item in YAP, select Options, then Inverse DVI search, then select TeXnicCenter then Ok.  Now you won't have to be scrolling a lot like I've been doing.  :)

Maybe this will help others just getting started.


geometry
perpendicular bisector
angle bisector
geometry proposed

Let $AOB$ be a right triangle with right angle $\angle AOB$. $C$ and $D$ are points moving on $AO$ and $BO$ respectively such that $AC=BD$. Show that there is a fixed point $p$ through which the perpendicular bisector of $CD$ always passes.

This problem is nicely, simply and well-known.

$\blacktriangleright\ 1^{\circ}.$ [b]The analytical method.[/b] Denote the midpoint $L$ ot the segment $[CD]$
and suppose $a\ne b$. Let $m\in R$ be a real variable. Thus,
$A(a,0),\ B(0,b)\Longrightarrow C(a+m,0),\ D(0,b+m)\Longrightarrow L\left(\frac{a+m}{2},\frac{b+m}{2}\right)\Longrightarrow$
the equation of the line-bisector of the segment $[CD]$ is:
$2(a+m)x+b^2+2bm=2(b+m)y+a^2+2am\Longrightarrow$
$\left(\forall\right)m\in R,\ 2(x-y+b-a)m+2(ax-by)+(b^2-a^2)=0\Longrightarrow$
$x-y=a-b,\ ax-by=\frac{a^2-b^2}{2}\Longrightarrow$ the fixed point is $\boxed {\ F\left(\frac{a-b}{2}\ ,\ \frac{b-a}{2}\ \right)\ }\ .$
Now you can return to a synthetical interpretation !.
[color=red][u]Remark.[/u] If the line $AB$ separates the points $C$ and $D$, and in this case the line-bisector of the segment $[CD]$ passes through a fixed point. It is well-known that for an any angle $\widehat {AOB}$, the geometrical locus of the point $L$ is a parallel line to the angle-bisector of the angle $\widehat {AOB}$, if the line $AB$ don't separate the points $C$ and $D\ ($ or a perpendicular-line on the angle-bisector of the same angle, if the line $AB$ separate the points $C$ and $D).$[/color]

$\blacktriangleright\ 2^{\circ}.$ [b]The synthetical solution.[/b] I shall come back soon.


geometry
circumcircle
trigonometry
trig identities
Law of Cosines

Prove that, if a, b and c are triangle sides, and R is its circumradius, so S = a.b.c / 4R, where S is triangle area.

p.s: solve by law of cosines! :|

[hide="Why?"] This follows from $ S \equal{} \frac{1}{2} ab \sin C$ and $ \frac{c}{\sin C} \equal{} 2R$ (the law of [i]sines[/i]). [/hide]

[color=darkblue][b][u]Proof without the Cosinus' law.[/u][/b]  Denote $ D\in BC$ so that $ AD\perp BC$ and the diameter $ [AS]$ of the circumcircle.

Observe easily that $ \triangle ABD\sim\triangle ASC$ , i.e. $ \frac {AB}{AS}\equal{}\frac {AD}{AC}$ $ \implies$ $ \frac {c}{2R}\equal{}\frac {h_a}{b}$ $ \implies$ $ h_a\equal{}\frac {bc}{2R}$ $ \implies$ $ S\equal{}\frac {ah_a}{2}\equal{}\frac {abc}{4R}$ .[/color]


Let $ ABC$ be an equilateral triangle, and let $ P$ and $ Q$ be the midpoints of sides $ AB$ and $ AC$ respectively. Let $ D$ be a point on $ PQ$. Extend the lines $ CD$ and $ BD$ so that they meet $ AB$ and $ AC$ at $ E$ and $ F$ respectively. Show that the value of $ \frac{BC}{EB}\plus{}\frac{BC}{FC}$ is constant and determine its value.

Use menelaus on $ \triangle APQ$, transversal $ EDC$, then transversal $ BDF$.

[color=darkblue]I modified a bit the enunciation of the proposed problem ![/color]

[quote="knexpert"][color=darkred] Let $ ABC$ be an equilateral triangle, and let $ P$ and $ Q$ be the midpoints of sides $ AB$ and $ AC$ respectively. Let $ D$ be a [b]mobile[/b] point on $ PQ$. Extend the lines $ CD$ and $ BD$ so that they meet $ AB$ and $ AC$ at $ E$ and $ F$ respectively. [b]Ascertain the position of the point $ D$ so that the product $ EB\cdot FC$ is minimum.[/b] [/color][/quote]
[color=darkblue][b][u]Proof I.[/u][/b] Denote $ \{\begin{array}{ccc} X\in CE & , & AX\parallel BC \\
\ Y\in BF & , & AY\parallel BC\end{array}$ . Observe that $ \frac {EA}{EB} + \frac {FA}{FC} = \frac {XY}{BC}$ and $ \frac {XY}{BC} = \frac {PQ}{BC - PQ}$ .

Thus, $ \frac {EA}{EB} + \frac {FA}{FC} = 1$ , i.e. $ \frac {AB}{EB} + \frac {AC}{FC} = 3$ . In conclusion, $ \frac {1}{BE} + \frac {1}{CF} = \frac {3}{BC}$ .

[b][u]Proof II (of my student).[/u][/b] Denote $ AB = a$ . Apply the [b]Menelaus' theorem[/b] to the transversals :

$ \blacktriangleright\ \overline {AFC}$ in $ \triangle BDE\ : \ \frac {AE}{AB}\cdot\frac {FB}{FD}\cdot\frac {CD}{CE} = 1$ $ \Longrightarrow$ $ \frac {AE}{AB}\cdot \frac {BC}{DQ}\cdot\frac {BP}{BE} = 1$ $ \Longrightarrow$

$ \frac {AE}{BE} = \frac {2\cdot DQ}{a}$ $ \Longrightarrow$ $ \frac {a - BE}{BE} = \frac {2\cdot DQ}{a}$ $ \Longrightarrow$ $ \boxed {\ \frac {1}{BE} = \frac 1a + \frac {2\cdot DQ}{a^2}\ }$ .

$ \blacktriangleright\ \overline {AEB}$ in $ \triangle CDF\ : \ \frac {AF}{AC}\cdot\frac {EC}{ED}\cdot\frac {BD}{BF} = 1$ $ \Longrightarrow$ $ \frac {AF}{AC}\cdot\frac {BC}{DP}\cdot\frac {CQ}{CF} = 1$ $ \Longrightarrow$

$ \frac {AF}{CF} = \frac {2\cdot DP}{a}$ $ \Longrightarrow$ $ \frac {a - CF}{CF} = \frac {2\cdot DP}{a}$ $ \Longrightarrow$ $ \boxed {\ \frac {1}{CF} = \frac 1a + \frac {2\cdot DP}{a^2}\ }$ .

In conclusion, $ \frac {1}{BE} + \frac {1}{CF} = \frac 2a + \frac {2\cdot PQ}{a^2}$ , i.e. $ \boxed {\ \frac {1}{BE} + \frac {1}{CF} = \frac 3a\ }$ (constant).

Thus, the product $ BE\cdot CF$ is minimum $ \Longleftrightarrow$ the product $ \frac {1}{BE}\cdot\frac {1}{CF}$ is maximum $ \Longleftrightarrow$

$ \frac {1}{BE} = \frac {1}{CF}$ , i.e. $ BE = CF = \frac {2a}{3}$ , what means the point $ D$ is the middle of the segment $ [PQ]$ .[/color]

[color=darkblue][b]Remark.[/b] See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=197178]here[/url] an easy extension ![/color]


floor function
continued fraction
calculus
calculus computations

For a given irrationial a, does $\frac{n}{\sqrt{a}}-\lfloor{\frac{n}{\sqrt{a}}}\rfloor$  converge to $0$ as $n$ goes to infinity?
No yes/no answers please.

No it is not. In general for $\alpha\not\in \mathbb{Q}$ every $x\in[0,1]$ is a limit point of $\alpha n-\lfloor \alpha n\rfloor$. First we assume this lemma

[size=150]Lemma[/size]
There is a sequence of rational number $\left(\frac{p_n}{q_n}\right)$ such that
\[
0<\alpha-\frac{p_n}{q_n}<\frac{1}{q_n^2}
\].

Let $\epsilon >0$. Choose $n$ big enough such that $\frac{1}{q_n}<\epsilon$. It follows that
\[
0<a:=\alpha q_n-p_n<\frac{1}{q_n}<\epsilon
\]
Let $x\in (0,1)$ and suppose $\epsilon$ small enough such that $0<x-\epsilon<x+\epsilon<1$.
Observe that for $a<\epsilon$,
\[
\frac{x-\epsilon}{a}<\frac{x-\epsilon}{a}+2
<\frac{x-\epsilon}{a}+\frac{2\epsilon}{a}\]
So there is an integer $m$ in $\left(\frac{x-\epsilon}{a},\frac{x+\epsilon}{a}\right)$, or equivalently
\[
ma\in (x-\epsilon,x+\epsilon)
\]
Hence $m(\alpha q_n-p_n)<1$ and it follows that $\lfloor \alpha mq_n\rfloor = mp_n$ and we have that
\[
\alpha mq_n-m p_n=\alpha mq_n-\lfloor \alpha mq_n\rfloor \in (x-\epsilon,x+\epsilon)
\]
The case for $x=0$ and $x=1$ can be treared similarly.

just an observation: the sequence $q_n$ is strictly increasing and that both $p_n,q_n$ are obtained from the continued fraction expansion of $\alpha$


Please give detailed solutions for the following:

1. How many minutes elapse between crossings of the hour hand and minute hand of a properly operating clock?

2. I assign the numbers 17,19,12,13,9,23, and 7 to the variables a,b,c,d,e,f,g such that the sum of the elemnts of each of the sets below is the same and has the maximum possible value. What is that value?
{a,c,f}    {c,d,e}      {g,e,f}

[hide="hint"]A)45         B)47     C)48      D)49          E)51[/hide]
**If you looked at the choices, please do not simply tell me guess and check**

[quote="mathwizard91"]Please give detailed solutions for the following:

1. How many minutes elapse between crossings of the hour hand and minute hand of a properly operating clock?

[/quote]

The hour hand moves at a rate of 1/12 revolutions per hour.  The minute hand moves at a rate of 1 revolution per hour.  Therefore the minute hand overtakes the hour hand by a rate of 11/12 revolutions per hour.  It therefore takes 1/(11/12) hours = 12/11 hours to overtake the hour hand by a full revolution.

[quote]
2. I assign the numbers 17,19,12,13,9,23, and 7 to the variables a,b,c,d,e,f,g such that the sum of the elemnts of each of the sets below is the same and has the maximum possible value. What is that value?
{a,c,f}    {c,d,e}      {g,e,f}

[hide="hint"]A)45         B)47     C)48      D)49          E)51[/hide]
[/quote]

In this case, the best way is actually guess and check.

Omgosh!!! You were at the Grissom High School tournament right???? From the look of the problem, you are in eighth grade right? Well I know the answer for number 2 but have no idea how to do it... what place did you get and what school? It was really not fair how seventh and eighth grade basicly have the same test.

for number 2, how would you guess and check? do u actually go and check every number?

yup!
second place-algebra 1
challenger middle

u?

If this was a MathCounts problem, then there wouldn't be multiple choice.  So for #2, guess & check is not an option, unless you come up with values to plug in yourself--but then it would be more like "brute force"...

2
[hide]The sum of each set is equal to the sum of all three sets divided by three. The maximum value of all three sets would throw out the seven because the variable $b$ is not included. The sum of all three sets is also equal to $a + 2c + 2f +d +2e + g$. To achieve the maximum sum we assign the largest numbers to the variables $c, f$ and $e$ which are 23, 19, and 17. The rest of the numbers are 12, 13, and 9 giving us a total sum of $46 + 38 + 34 + 12 + 13 + 9 = 152$. This cannot be, however, because 152 is not divisible by 3. We replace the 9 with the 7 in order to give us the sum 150 which divided by 3 is equal to $50$. It can't be 50 either because we notice that no 3 of the numbers is equal to 50. Lowering that by 1 we get $49$ which seems to work.[/hide]

The only problem I see with Tarquin's answer is that it is not one of the choices.  Maybe mathwizard91 didn't give the correct solutions though *suspicious look*

I'm so sorry mathwizard91 about last night, my mom was using the computer and i forgot to log off aim, and i was going to sleep, so i had to close it out really quick for my mom and go nighty-night....

I'm in seventh grade, Pizitz Middle, 7th place.... the trophies are totally cool! Algebra I is eighth grade right?

I totally love tarquin's solution :D

tarquin is cool...
he actually spends time to post an answer with a full descriptive solution!
plus, i know him :D

[quote="tarquin"]It is.
[hide]D[/hide][/quote]
Sorry, I read too quickly...


Two cards are randomly drawn in a standard 52-card deck. What's the probability that one card is a heart and the other is a queen?

Just thought of this in my head  :D

err, I used to be good at probability. But I don't know how you want the answer...

[hide="Lets pour my puny brains out"]
hearts: 1/13 and queen: 1/4
So...
2/17?????  :huh: [/hide]

I don't know the answer but I don't think the answer is 2/17. How did you get it?

[hide] I added the hearts to queens? I don't know, I'm trying to participate in the math forums. I mean how would you combine the two probablities. AGH! I got it... I think... 17/52????[/hide]

YOu don't add probabilities in this case. Since its an and case, you should multilply, and then possibly add. 17/52 is obviously impossible.

And jsut to tell you, i don't think my brains are any "bigger" than yours.

[quote="Black_stripe*36"]My figure of speech... What I mean is that I just like math, I'm not all that experienced in it.  Well would drawing a heart be 13/52? because there are 13 hearts in a deck and 52 cards... For the queen it would be 4/52 because there are 4 queens in 52 cards...[/quote]

Yeah, but remember that after one card is drawn, there are only 51 cards left in the deck...

[quote="Black_stripe*36"]:o  *gasps* there are, aren't there...

Ok, well then now it depends on which card you draw first... Doesn't it? Let's go with the order you introduced to cards in your post, hearts then queen.

Would that be 13/52 then 4/51? Something seems wrong about this, and how I am showing the answers. This problem is quite confusing, and you don't know the answer. Hmm...  :huh:[/quote]

I don't know how to do this problem, but I have something i know:

There are at least two cases, two of these are:

1. First you draw a heart, then a queen.

2. first you draw a queen of hearts, then a queen of another type.

And these can be reversed.

And if you draw a queen of hearts then that leaves you with 3 queens left... Finally my head woke up. I think that you should maybe rephrase the question, maybe it should be like what is the probability that you draw a heart first then a queen. Or something like that...

[quote="Black_stripe*36"] I think that you should maybe rephrase the question, maybe it should be like what is the probability that you draw a heart first then a queen. Or something like that...[/quote]

the problem would then be too easy.... not really.

What's your answer now?

Food I guess... NO, I don't think I have an answer...

There are too many cases, you'd have to incorperate (I love that word for some reason) that you could draw a queen of hearts or which one to draw first into the answer somehow. Maybe you could exclude the queen of hearts from the probability of hearts? No, because it's still a heart... MAN! I can't get this, it's gonna drive me crazy now...

The problem is basically this:

* Two cards are simultaneously drawn from standard 52 playing cards. Each card has equal chance of being drawn. What is the probability that two cards drawn are one heart and one queen? *

I won't give whole solution out but here is a hint. What is the total number of ways for you to get 2 cards chosen from 52 playing cards?

[hide="That is.."]

$\binom{52}{2}$ [/hide]

That's the denominator, or total number of outcomes. Now, you have to find the numerator [color=green][size=75][he means numerator][/size][/color]. Can you do that?

I think SIlverfalcon is saying that after you find the denominator, you find the numerator.

Heres a hint for the numerator:

[hide="hint"]

  The problem says a queen AND a heart.  "AND"s" generally mean something important what is it?  Secondly, you musn't forget about the queen of hearts! [/hide]

[quote="chess64"]I fixed it for him :D[/quote]

Thanks, I edited.

Well forgive me if we've already solved this but it doesn't appear so.

[hide="my solution"]
I would split it up into three cases:

1) the first card is a non-queen heart
2) the first card is a non-heart queen
3) the first card is the queen of hearts.

For case 1)
The probability for the first card is $\frac{12}{52} = \frac {6}{26} = \frac {3}{13}$ and the probability for the second card is $\frac 4{51}$. So the total probability is $\frac {12}{663}$

For case 2)
The probability for the first card is $\frac 3{52}$ and for the second card it is $\frac{13}{51}$ so the total probability is $\frac {13}{884}$

For case 3)
The first card must be the queen of hearts so the probability for that is $\frac 1{52}$. The second card can be either a heart or a queen, so that leaves 15 cards. so that probability is $\frac {15}{51} = \frac {5}{17}$ for a total probability of $\frac {5}{884}$

So the total probability will be
$\frac {12}{663} + \frac {13}{884} + \frac {5}{884}$

$\frac{12}{663} + \frac{18}{884}$

$\frac{4}{221} + \frac{9}{442}$

$\frac{8}{442} + \frac{9}{442}$

$\frac{17}{442}$

$\boxed{\frac 1{26}}$

[/hide]

I'm not sure but I think that's right...

[hide]
Case 1: Heart not jack, Heart jack
Case 2:Heart not jack, other jack
Case 3: Heart Jack, other jack

1) (12/52)(1/51)
2)(12/52)(3/51)
3)(1/52)(3/51)

1)+2)+3)=1/52
[/hide]


induction
real analysis
real analysis unsolved

Anyone can prove ?

$\dfrac{d^n}{dx^n}(sin^4x+cos^4x)=4{}^n{}^-{}^1cos(4x+\dfrac{n\pi}{2})$

You will use induction and you have not problem. Just a little trigonometry.

a)  For n = 1 , you take

   $\dfrac d{dx}$ ($\ (sinx)^4 + \(cosx)^4) =$…. $= 2.sin2x . (-cos2x) = - sin4x = cos(4x + \pi /2)$

   b) The rest is very easy. Use the fact that  $4.4^(n-1)(-sin(4x + n\pi/2))= 4^n. cos(4x +n\pi/2 + \pi/2)$ .

  Do not forget that  $cos(x+ \pi/2) = - sinx$.

 Perhaps I have helped you.

[u]Babis[/u]


1. Joseph wrote to Superman and did not receive an answer. Assumming that one in n[u] letters is lost in the mail, find the probability that Superman received the letter. It is to be assumed that Superman would have answered the letter if he had received it.

2. ABCD is a sqaure with side length 4. Semicircle O is incribed in ABCD such that its diameter xy is a maximum. The length of xy can be expressed as p-(sqrt)q, find p+q.

3. The expression (x+y+z)^2008 + (x-y-z)^2008 is simplified by expanding it and combining like terms. How many terms are in the simplified expression?

4. ON a standard die, one of the dots is removed at random with each dot equally likely to be chosen. The die is then rolled. What is the probabality that the top face has an odd number of dots?

5. Square ABCD is inscribed in a regular hexagon PQRSTU of side length 7cm in such a way that AB is parallel to QR and DC is parallel to TU. Find the area of sqaure ABCD.

6. I scored 35, 42, and 39 in the first three 100-item test. What is the lowest possible grade that I may have gotten in any of the next three 100-item test if the average of my 6 tests exceeds 66?

7. Let A be the set of factors of 2008 and B be the product of the elements of A. If m and n are integers and B=(251^m)(4^n), find m+n.

8. How many odd integers between 1,000 and 8,000 have no repeated digits?

9. Find the real value of x in x^3+9x^2+27x+81=0.

10. Factor completely: 24x^2+54y^2-10z^2-72xy-14xz+21yz.

11. In triangle JOH, OJH is a right angle, E is the midpoint of the hypotenuse and S and P are the midpoints of OE and EH. If JS=11 and JP=13, find JE.

12. Let A and B be points, and l be a line intersecting AB at a point different from A and B, but not perpendicular tp AB. How many right triangles can be formed such that the vertices are A, B and a point on l?

13. Find the 2008th term in the sequence 1, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2,1, 2, 3, 4, 5, 4, 3, 2, 1, ......
[/u]

[hide="1"]
If you assume that Superman always replies to letters the the probability that he recieved Joseph's letter is simply $ 0$[/hide]

EDIT: I wasn't thinking, this is completely wrong :oops:

[hide="2"]
We can see that the largest semicircle that will fit in a square will be on an angle relative to the square and be symetrical around the squares diagonal.
Since the sides of the square are tangents, draw some perpendicular lines to the center of the semicircle
Notice the rightangled triangle with sidelengths $ (4 \minus{} r,4 \minus{} r,r)$

From Pythag
$ \Rightarrow 2(4 \minus{} r)^2 \equal{} r^2 \Longrightarrow r \equal{} 4(2 \minus{} \sqrt {2})$ or $ 4(2 \plus{} \sqrt {2})$
Since  $ 4(2 \plus{} \sqrt {2}) > 4$ we exclude that one

$ \Rightarrow d \equal{} 8(2 \minus{} \sqrt {2}) \equal{} 16 \minus{} \sqrt {128}$
$ \Rightarrow p \plus{} q \equal{} 144$[/hide]

One more for good measure

[hide="3"]
group the expression $ \longrightarrow (x\plus{}(y\plus{}z))^{2008} \minus{} (x\minus{}(y\plus{}z))^{2008}$

From the Binomial expansion theorem it is clear that all the even terms in this expansion cancel
Therefore we have $ 1004$ terms in the form;
$ \binom{2008}{k} (y\plus{}z)^k x^{2008\minus{}k} \longrightarrow k \in \{1,3,5,7...2007\}$

From Binom $ (x\plus{}a)^n$ has $ n\plus{}1$ terms

So in total $ (x\plus{}y\plus{}z)^{2008} \minus{} (x\minus{}y\minus{}z)^{2008}$ has $ 2\plus{}4\plus{}6\plus{}8...\plus{}2008$ terms

$ S_n \equal{} \frac{2(1004)(1005)}{2} \equal{} 1009020$
[/hide]

[hide="9"]
$ x^3\plus{}9x^2\plus{}27x\plus{}81\equal{}0$
$ \Longrightarrow (x\plus{}3)^3\plus{}54\equal{}0$
$ \Longrightarrow x\equal{}\minus{}\sqrt[3]{54}\minus{}3$
[/hide]

1. (n-1)/(2n-1)

2. 144

3. 1,010,025

9. Answer of doabli is correct! I wasn't able to spot that! Thanks!

[quote="ocha"]If you assume that Superman always replies to letters the the probability that he recieved Joseph's letter is simply $ 0$[/quote]
Hello
It can't be zero as Superman's mail can be lost too.
[hide="solution"]
Let A be the probability that Superman receive the letter and B be the probability  Joseph receive the reply.
Basically what we need is $ P(A/B')$.
$ P(A)\equal{}\frac{n\minus{}1}{n}$
$ P(B/A)\equal{}\frac{n\minus{}1}{n}$
$ \frac{P(A intersection B)}{P(A)}\equal{}\frac{n\minus{}1}{n}$
$ P(A intersection B)\equal{}\frac{n\minus{}1}{n}(P(A))$
$ P(A intersection B)\equal{}\frac{(n\minus{}1)^2}{n^2}$
now
$ P(B)\equal{}P(A).P(B/A)\plus{}P(A')P(B/A')$
$ P(B)\equal{}(\frac{n\minus{}1}{n})^2$
Now what we want is
$ P(A/B')$=$ \frac{P(A intersection B')}{P(B)}$
             =$ \frac{P(A)\minus{}P(A intersection B)}{1\minus{}P(B)}$
             =$ \frac{n\minus{}1}{2n\minus{}1}$
    [/hide]
Thank u.

Thanks kabi for the elegant solution!

Anyone else who can help me out with the other problems?

[hide="10"]
$ 24x^2\plus{}54y^2\minus{}10z^2\minus{}72xy\minus{}14xz\plus{}21yz\equal{}(2\sqrt{6}x\minus{}3\sqrt{6}y\minus{}\frac{7\sqrt{6}}{12}z)^2\minus{}\frac{289}{24}\cdot z^2$
And now you can easily factor that thing by using $ a^2\minus{}b^2\equal{}(a\minus{}b)(a\plus{}b)$[/hide]

[hide="8"]
for an odd no's unit digit we have five choices$ (1,3,5,7,9)$
take 1 in unit's place::
for thousands place there are six choices$ (2,3,4,5,6,7)$
no.of such numbers with one in unit's place=$ 6*8*7$
Same will happen for $ 3,5,7$
In case of nine
number=$ 7*8*7$
Those total number=$ 56*31$=$ 1736$
 [/hide]
Thank u


$\triangle ABC, A=45^{\circ},\ B=105^{\circ},\ M\in (AB),\ N\in (AC),\ m(\widehat {ABN})=$ $m(\widehat {CMN})=C\Longrightarrow m(\widehat {BCM})=\ ?.$

[u]It is a difficulty to find a syntetical solution for this problem. [/u]

Awh, common, it is not difficult at all. The remaining angle of the triangle $\triangle ABC$ is

$\angle C = 180^\circ - (\angle A + \angle B) = 180^\circ - (45^\circ + 105^\circ) = 30^\circ$

The triangle $\triangle NBC$ is isosceles, because $\angle BCN = \angle C = 30^\circ$,

$\angle NBC = \angle B - \angle ABN = \angle B - \angle C = 105^\circ - 30^\circ = 75^\circ$

$\angle BNC = 180^\circ - (\angle BCN + \angle NBC) = 180^\circ - (30^\circ + 75^\circ) = 75^\circ$.

Therefore, CN = CB. Since $\angle CMN = \angle C = 30^\circ$, the point M is the intersection of the line AB with the circle (O) centered at the vertex O of an equilateral triangle $\triangle CON$ on the segment CN and passing through the points C, N. That way,

$\angle CMN = \frac{\angle CON}{2} = \frac{60^\circ}{2} = 30^\circ = \angle C$

as required. Hence, CO = CN = CB. Thus the triangle $\triangle BCO$ is also isosceles. Since $\angle BCN \equiv \angle C = 30^\circ$, CB bisects the angle $\angle OCN = 60^\circ$ and the angle $\angle BCO = 30^\circ$. Thus the remaining 2 base angles of the isosceles triangle $\triangle BCO$ are equal to

$\angle COB = \angle BOC = \frac{180^\circ - \angle BCO}{2} = \frac{180^\circ - 30^\circ}{2} = 75^\circ$

Since the external angle of the triangle $\triangle ABC$ at the vertex B is equal to $180^\circ - \angle B = 180^\circ - 105^\circ = 75^\circ$, the points A, B, O are collinear. Since $M \in AB$, the points M, B, O are also collinear. The triangle $\triangle CMO$ is isosceles, because OM = OC are radii of the circle (O), with the angle $\angle COM \equiv \angle COB = 75^\circ$. Consequently, its remaining 2 base angles are equal to

$\angle OCM = \angle OMC = \frac{180^\circ - \angle COB}{2} = \frac{180^\circ - 75^\circ}{2} = 52.5^\circ$

Thus 2 angles $\angle MBC = \angle B = 105^\circ$ and $\angle BMC \equiv \angle OMC = 52.5^\circ$ of the triangle $\triangle BCM$ are known and the remaining angle $\angle BCM$ is easily obtained:

$\angle BCM = 180^\circ - (\angle MBC + \angle BMC) = 180^\circ - (105^\circ + 52.5^\circ) = 22.5^\circ$


Show that for all integer $k>1$, there are  infinitely many natural numbers $n$ such that $k \cdot 2^{2^n} + 1$ is composite.

Three problems of one day of China TST was proposed by Mr.Feng(Feng Zuming),and this problem is one of the three.
My Solution:
Let ${p_i=2^{2^{i}}}+1,(i=0,1,2,3.4)$,
$p_5=641$ $p_6|2^{2^{5}}+1,   $,$p_6\not = 641$
then we choose k has the property
$k\equiv 1 (mod p_i),i=0,...5,k \equiv -1( mod p_6)$
then for any positive integer $n=2^l*t,$,$t$ is odd
if $l<=5$ then ${2^n k=(2^{2^l})^ t}k+1\equiv -k+1 \equiv 0 (mod p_l)$
if $l>= 6$ then $(2^{2^l})^t * k+1 \equiv k+1 \equiv 0 (mod p_6)$

Hawk Tiger, it seems that you solve completely other problem. As for Peter's problem, I have no idea.

Let p is odd prime and let $ T_p \equal{} 2^rm, \ m \ \minus{} \ odd$ is period for $ 2^n\mod p$. Let $ L_p$ period for $ 2^n\mod  m$.
If $ x_n \equal{} k*2^{2^n} \plus{} 1$, then $ x_{n \plus{} L_p}\equiv x_n\mod p$.
If $ x_n \equal{} 0\mod p$ then for any s $ x_{n \plus{} sL_p} \equal{} 0\mod p$. Therefore sufficiently find one prime p and one n, suth that $ x_n \equal{} 0\mod p$.
If $ k*2^{2^{n_0}} \plus{} 1 \equal{} P$ is prime, then $ p \equal{} P,n \equal{} n_0$ work.

[quote="Fedor Petrov"][quote="Rust"]Let p is odd prime and let $ T_p \equal{} 2^rm, \ m \ \minus{} \ odd$ is period for $ 2^n\mod p$. Let $ L_p$ period for $ 2^n\mod m$.
If $ x_n \equal{} k*2^{2^n} \plus{} 1$, then $ x_{n \plus{} L_p}\equiv x_n\mod p$.[/quote]

Why? I think, it is so if $ r\le n$.[/quote]
Yes.
Let $ k\equal{}2^sk_1$, were $ k_1$ is odd. Let \[ P\equal{}k*2^{2^{n_0}}\plus{}1\] is prime. Then \[ m*2^r|k_1*2^{2^{n_0}\plus{}s}.\]
It equavalent to $ m|k_1, r\le 2^{n_0}\plus{}s.$ Therefore \[ x_{n_0\plus{}lL_p}\equal{}k_12^{s\plus{}2^{n_0\plus{}lL_p}}\plus{}1\equal{}P\plus{}k*2^{2^{n_0}}(2^{2^{n_0}(2^{lL_p}\minus{}1)}\minus{}1).\]
Because $ 2^{lL_p}\minus{}1\equal{}m*m_1$ we get \[ x_{n_0\plus{}lL_p}\equal{}x_{n_0}\plus{}(P\minus{}1)(2^{2^{n_0}mm_1}\minus{}1).\]
If k is odd, then $ r\le 2^{n_0}$ or $ 2^{n_0}m\equal{}m_2T_p$. It give $ x_{n_0\plus{}lL_p}\equal{}x_{n_0}\plus{}(P\minus{}1)(2^{T_pm_1m_2}\minus{}1)\equal{}x_{n_0}\mod P.$
Therefore my solution work for odd k.

[quote="Rust"]
If k is odd, then $ r\le 2^{n_0}$ or $ 2^{n_0}m \equal{} m_2T_p$. It give $ x_{n_0 \plus{} lL_p} \equal{} x_{n_0} \plus{} (P \minus{} 1)(2^{T_pm_1m_2} \minus{} 1) \equal{} x_{n_0}\mod P.$
Therefore my solution work for odd k.[/quote]

We need $ r\le n_0$, not $ r\le 2^{n_0}$. Because we need $ 2^r| 2^{n_0}$.

I give another solution.
Let $ k\equal{}2^sk_1, \ k_1 \ \minus{} \ odd$ and p prime divisor of $ 2^tk_1\plus{}1$, suth that $ ord_2(p\minus{}1)\le t$. Let $ T_p$ is period $ 2^n\mod p$.
Then $ p|k_12^{t\plus{}lT_p}\plus{}1$ for any integer $ l\ge 0$.
Consider $ x_n\equal{}k*2^{2^n}\plus{}1\equal{}k_12^{s\plus{}2^n}\plus{}1$. We get $ p|x_n$ if $ T_p|2^n\plus{}s\minus{}t$.
Let $ T_p\equal{}2^rm, \ m \ \minus{} \ odd$. We choose p, suth that $ r\le ord_2(p\minus{}1)\le t$. Wt need $ 2^r|s\minus{}t$.
If s is odd, then take $ t\equal{}1$ we get $ 2^r|s\minus{}t$, because $ r\le t\le 1.$ And we get p is any prime divisor $ 2k_1\plus{}1$.
If s even, we can not chose $ t\equal{}0$, because may be $ m\plus{}1$ is power of 2 (we need odd prime divisor p). But we can chose $ t\equal{}s$ for  $ s>0$ (work any prime divisor k+1) and if $ s\equal{}0$ work $ p\equal{}x_n$.
Then for $ L_p|\phi (m)$ we get $ p|x_{n_0\plus{}lL_p}$.

So, for odd $ k$ ($ s\equal{}0$) the solution is the same as earlier version? Then it looks that the same problem appears.

but I still can't catch the solution. :oops: Can you explain more about Tp is a divisor of 2^n+s--t?
(sorry about awful latex)

[quote]but I still can't catch the solution. :oops: Can you explain more about $T_p$ is a divisor of $2^n+s-t$?
(sorry about awful latex)[/quote]


You can use generating functions :)
Desired value is coefficient near $x^{3k}$ in this polynomial:
$(1+x+x^2+\ldots+x^{2k})^3=(1-\frac{x^{2k+1}}{1-x})^3=(1-x^{2k+1})^3(\frac{1}{(1-x)^3})=$
$=(1-3x^{2k+1}+3x^{4k+2}-x^{6k+3})(\binom{2}{2}+\binom{3}{2}x+\binom{4}{2}x^2+\ldots)$
We can see that this value is:
$\binom{3k+2}{2}-3\binom{k+1}{2}=3k^2+3k+1$
Is this value all right?

Or another solution, not using generating functions I've found:

First rewrite problem to folowing: Find all triples x,y,z of integers which satisfies:
a) 0 <= x <= 2k
b) 0 <= y <= 2k
c) 0 <= z <= 2k
d) x+y+z=3k
We will have 2 cases:

[b]First case:[/b] In this we choose case x>=k.
Then we can choose for this x these values of y,z:

y=0, z=(3k-x)
y=1, z=(3k-x)-1
y=2, z=(3k-x)-2
...
y=(3k-x), z=0

So for each x in this case we have 3k-x+1 solutions. We can have x=k,k+1,...2k in this case so we just sume it all:
(3k-k+1)+(3k-(k+1)+1)+(3k-(k+2)+1)+...+(3k-2k+1)=[u](k+1)+(k+2)+...+(2k+1)[/u]

[b]Second case:[/b]  In this case we choose x<k
Then chooses for y,z are:

y=k-x, z=2k
y=k-x+1, z=2k-1
y=k-x+2, z=2k-2
...
y=2k, z=k-x

So for each x in this case we have k+x+1 solutions. We can have x=0,1,2,...,k-1 so just sum it up:
(k+0+1)+(k+1+1)+(k+2+1)+...+(k+(k-1)+1)=[u](k+1)+(k+2)+...+(2k)[/u]

Summing both cases (in a tricky way, you could also sum it specially in each case)...

((k+1)+(k+2)+...+(2k+1))+((k+1)+(k+2)+...+(2k))=
(2k+1)+((k+1)+2k)+((k+2)+(2k-1))+((k+3)+(2k-2))+...+((2k)+(k+1))=
(2k+1)+k(3k+1)=
[b]3k^2+3k+1[/b]

:D

How about a solution with recurrence equation???


The qualification test, what exactly is it like? And what should you study for it?

Is it very similar to AMCs? Or should you also study SC competitions etc.?

There are two qualification tests for the South Carolina All-State Math Team (from which the SC ARML Team is chosen). The first round test is given at your home school and usually covers just the basics of algebra I, algebra II, geometry, and pre-calculus as you would learn them in the classroom. They don't require a [i]lot[/i] of ingenuity, but some of the questions you have to read carefully. For the most part though, its almost a knowledge test - sort of like you might see on the final exams of some high school honors math classes.

Normally, I would say don't sweat this round, but the scores on it have usually been pretty low (at least compared to how students do on the second round). Maybe its because people rush through the questions, or maybe its because the test isn't given under ideal conditions at some schools, or maybe its because even good contest students have some holes in their fundamentals. I don't know (shrugs). But your teacher (i.e. me) may have (has) some old tests you can look at. Or, you could use some of the problems from Part I of the [url=http://www.delta.edu/math/mmpc/previousexams.html]Michigan Mathematics Price Competition[/url]. Some of their problems are similar in nature to first round All-State Math Team problems.

The second round exam (which is given to students who score above a certain cut off on the first round exam, and which is usually given at Irmo High School a few months later) is more ARML style. The problems are given two at a time with 8 minutes to work each pair. There are between 16 and 20 questions total (i.e. 8 to 10 pairs). The questions require a little bit more ingenuity than the first round, but they aren't nearly as hard as actual ARML problems (this is because we want enough variation in the scores to see differences). The best way to prepare for this round is to do early to mid-range AMC-12 problems, NEAML problems - if you can find them.

Keep in mind that the tests may possibly be written by someone new next year, so there's no gaurantee that they will be done exactly the same as past years. But Michigan Math Prize Competition and AMC-12 problems are good preparation for nearly any high school math contest (as are the AoPS books).

who might write the questions next year?

and my favorite question from this years first round:
#2). What is the remainder when $x^{2005}-2005x+1954$?
the funny thing is that you could figure out the answer with half the question missing because the only reasonable thing to divide by is (x-1) or (x+1).

[quote="jb05"]Just wondering-what is Irmo high school? Is it kind of like NCSSM (North Carolina School of Science and Math) where all our state competitions are?[/quote]

The second round of all state is the only thing that I know of that happens at Irmo with regards to math.  Most of the competitions in SC are held at colleges.

[quote="jb05"]Just wondering-what is Irmo high school? Is it kind of like NCSSM (North Carolina School of Science and Math) where all our state competitions are?[/quote]
i think he's asking what kind of school irmo is? if that is the question, then it's just a regular high school.

[quote="furious"]who might write the questions next year?[\quote]

One of the coaches, maybe either Patrick or Dave or myself - whoever has time to do it. There are no bylaws or anything, so whoever's organizing the committee each year gets to decide how they want to do things. Next year the head of the committe will be Patrick. But basically we just e-mail each other at the beginning of the year to work out a plan and then ask for volunteers to do different things.

There's been some discussion of making the first round exam a little more problem-solving based and a little less knowledge based next year, and doing some fun things at the Irmo meeting. But nothing's set in stone yet.

By the way, I just noticed this, but one of the teams on the cover of the 1989-1994 ARML-NYSML Contests Book (the blue book) is the South Carolina  team from either 1991 or 1992!! I've had this book for years and never looked at the people on the cover - what a shame!!

But the team in white t-shirts with little logos on the left side is the South Carolina team. Unfortunately (or fortunately) I was sitting at the back of taht row so I am not pictured. :oops:

Sounds... normal. I was just wondering if it was some really wonderful science/math/tech school. I like to keep up with other states too ;) .

yea, it's normal. the only "special" school in SC is the governor's school of science/math


invariant
combinatorics unsolved
combinatorics

Prove that an automorphism of a tree, keeps a fixed vertex or a fixed edge

Just look at what the automorphism does to a maximal chain in the tree. This maximal chain turns into another maximal chain. It's easy to show that in any connected graphm (so here too), we cannot have to disjoint maximal chains, so these two have at least a vertex in common. Since if they have two vertices in common they must have the entire segment of the chain between the two vertices in common.

I'm tired; try to complete it :).

Actually, it also follows directly from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=tree+center+centre&t=22122]this discussion[/url]. The center of a tree consists either of a vertex or of two connected vertices, and it's invariant wrt the automorphism.


function
integration
calculus
calculus computations

If $ F$ is a continuous function and $ F(x)\equal{}\int^x_0F(t)dt$ . Prove that $ F(x)\equal{}0$ for every x.

Let $ M$ be the maximum of $ F$ on $ [0,1/2]$, attained at some point $ x_0\in [0,1/2]$. Since
$ M \equal{} F(x_0) \equal{} \int_0^{x_0}F(t)\,dt\le \int_0^{x_0}M\,dt\le \frac {M}{2}$,
it follows that $ M \equal{} 0$. The same argument applies to $ \minus{} F$ and yields $ F \equal{} 0$ on $ [0,1/2]$. But then $ F \equal{} 0$ on $ [1/2,1]$, and so on.


wheres the asymtote test form all i know there is an LATEX test form ><

As far as I can tell, there is no analogue of TeXer on the site for playing with Asymptote codes. You are strongly encouraged to do that on your home computer. Once you have a code that functions on your computer, you are welcome to try it in the usual Test Forum (yes, it is for Asymptote too) to see if it runs on the site as well. You do not need to make an actual post unless you want to share something with others or are looking for comments; just do the preview and see if everything is fine.


calculus
integration
algebra
polynomial
limit
calculus computations

A sequence of polynomial $f_n(x)\ (n=0,1,2,\cdots)$ satisfies $f_0(x)=2,f_1(x)=x$,

\[f_n(x)=xf_{n-1}(x)-f_{n-2}(x),\ (n=2,3,4,\cdots)\]

Let $x_n\ (n\geqq 2)$ be the maximum real root of the equation $f_n(x)=0\ (|x|\leqq 2)$

Evaluate

\[\lim_{n\to\infty} n^2 \int_{x_n}^2 f_n(x)dx\]

how do you do this? would you put your answer up kunny?

Please imagine $f_n(x)$ as something like Tchevichev's polynomial., skylarth.


When $ \minus{}2(x \plus{} \minus{}2(x \plus{} \minus{}2(x \plus{} (\minus{}2))))$ is written in the form $ ax \plus{} b$, what is the value of $ a \plus{} b$?

First start simplifying the innermost terms, then progress outwards.

$ \minus{}2(x\plus{} \minus{}2(x \plus{} \minus{}2(x \plus{} (\minus{}2))))$

$ \minus{}2(x \plus{} \minus{}2(x \minus{} 2x \plus{} 4)))$

$ \minus{}2(x \minus{} 2x \plus{} 4x \minus{} 8)$

$ \minus{}2x \plus{} 4x \minus{} 8x \minus{} 8$

$ \minus{}6x \minus{} 8$

So our answer is $ \minus{}6 \plus{} \minus{}8 \equal{} \boxed{\minus{}14}$.

[quote="isabella2296"]First start simplifying the innermost terms, then progress outwards.

$ \minus{} 2(x \plus{} \minus{} 2(x \plus{} \minus{} 2(x \plus{} ( \minus{} 2))))$

$ \minus{} 2(x \plus{} \minus{} 2(x \minus{} 2x \plus{} 4)))$

$ \minus{} 2(x \minus{} 2x \plus{} 4x \minus{} 8)$

$ \minus{} 2x \plus{} 4x \minus{} 8x \minus{} 8$ BLARGHNESS!!!!!!!! 2 times -8=-8???

$ \minus{} 6x \minus{} 8$

So our answer is $ \minus{} 6 \plus{} \minus{} 8 \equal{} \boxed{ \minus{} 14}$.[/quote]

Its actually 16-6x so 10

The question is basically asking you to find the sum of the coefficients of the expression when simplified.  Notice that if $x = 1$, then all you are left with in the resulting expression is the sum of the coefficients, $a+b$!

Therefore, all you have to do is plug in 1 for x into the expression.

$-2(1-2(1-2(1-2))) = -2(1-2(1+2)) = -2(1-6) = -2(-5) = \boxed{10}$.


Fie a,b,c trei numere complexe astfel incat
$|a|=|b|=|c|=1$, $a+b+c$ diferit de 0 si $a^{2}+b^{2}+c^{2}=0$.Aratati ca
$|a^{3}+b^{3}+c^{3}|$=1

Se noteaza cu $ s_1\equal{}z_1\plus{}z_2\plus{}z_3$, $ s_2\equal{}z_1z_2\plus{}z_2z_3\plus{}z_3z_1$, $ s_3\equal{}z_1z_2z_3$ si $ S_2\equal{}z_1^2\plus{}z_2^2\plus{}z_3^2$.Numerele $ z_k, k\equal{}{1,2,3},cu |z_k|\equal{}1$ sunt solutii reale ale ecuatiei : \[ z^3\minus{}s_1z^2\plus{}s_2z\minus{}s^3\equal{}0\] Au loc egalitatile: \[ s_1^2\equal{}S_2\plus{}2s_2\] de unde \[ s_1^2\equal{}2s_2\]
deoarece $ S_2\equal{}0$ si \[ s_2\equal{}s_3(1/z_1\plus{}1/z_2\plus{}1/z_3)\equal{}s_3(\overline{z_1}\plus{}\overline{z_2}\plus{}\overline{z_3}\equal{}s_3 \cdot \overline{s_1}\] Dupa ce se aplica modulul in ambii membri avem $ |s_1|^2\equal{}2|s_3\parallel{}\overline{s_1}|\equal{}2|s_1|$, rezulta ca $ |s_1|\equal{}2$. Trecem la calculul modulului cerut care va da 1...


Outside of equaliteral triangle but in angle BAC chosen point M such that <CMA=30 and <BMA=a. What's value of an angle ABM?

[hide]$ \angle AMC \equal{} \frac{1}{2} \angle ABC$ which means that $ M$ lies on the circle centered at $ B$ with radius $ BA\equal{}BC$ (this is easily proved since any point $ M$ lying on the circle has this property and if we take the circle centered at $ B$, then choose a point $ M$ not on the circle, we can see that the angle can not have measure $ 30^{\circ}$).

Thus, $ \angle BAM\equal{}a$ and $ \angle ABM\equal{}120\minus{}2a$[/hide]


geometry
analytic geometry
quadratics
trigonometry
geometric transformation
reflection
perpendicular bisector

Let $ ABCD$ be a convex quadrilateral such that $ AB\equal{}BC\equal{}2,\ CD\equal{}2\sqrt{3},\ DA\equal{}2\sqrt{5}.$ Let $ M,\ N$ be mid points of $ AC,\ BD$ such that $ MN\equal{}\sqrt{2}$ respectively.Find the area of the quadrilateral.

Put coordinate origin at the midpoint $ M \equal{} (0, 0)$ of $ AC$ and let the coordinates of $ A, C$ be $ A \equal{} ( \minus{} p, 0), C \equal{} ( \plus{} p, 0),$ where $ p > 0.$ Labeling the quadrilateral $ ABCD$ counter-clockwise, the coordinates of $ B, D$ are $ B \equal{} ( \minus{} q, 0), D \equal{} (u, v),$ where $ q > 0, v > 0.$ Since $ BC < DA,$ $ u > 0$ as well. Midpoint $ N$ of $ BD$ has coordinates $ N \equal{} [u/2, (v \minus{} q)/2].$ The given conditions $ AB \equal{} BC \equal{} 2,$ $ CD \equal{} 2\sqrt 3,$ $ DA \equal{} 2 \sqrt 5,$ $ MN \equal{} \sqrt 2$ yield 4 equations for 4 unknowns $ p, q, u, v$:

$ (1)\ \ p^2 \plus{} q^2 \equal{} 4$
$ (2)\ \ (u \minus{} p)^2 \plus{} v^2 \equal{} 12$
$ (3)\ \ (u \plus{} p)^2 \plus{} v^2 \equal{} 20$
$ (4)\ \ u^2 \plus{} (v \minus{} q)^2 \equal{} 8$

Subtracting (3) - (2) yields $ up \equal{} 2.$ Substituting this to (3), then subtracting (3) - (4) and using (1) to eliminate $ p^2$ from the result yields $ vq \minus{} q^2 \equal{} 2, v \equal{} q \plus{} 2/q.$ Substituting this and $ u \equal{} 2/p$ to (4) yields

$ (5)\ \ 1/p^2 \plus{} 1/q^2 \equal{} 2$

Solving 2 quadratic equations (1), (5) for 2 unknowns $ p > 0, q > 0$ leads to $ p \equal{} \sqrt {2 \pm \sqrt 2},$ $ q \equal{} \sqrt {2 \mp \sqrt 2}.$

The quadrilateral angles $ \angle A, \angle B, \angle D$ are obviously convex, we have to make sure the angle $ \angle C$ is also convex. The $ \triangle ABC$ with $ AB \equal{} BC$ is also isosceles and

$ \cos \widehat{BCM} \equal{} \frac {MB}{BC} \equal{} \frac {p}{2} \equal{} \frac {\sqrt {2 \pm \sqrt 2}}{2},$ $ \angle BCM \equal{} \frac {\pi}{8}$ or $ \frac {3\pi}{8}.$

Coordinates of the midpoint $ N$ of $ BD$ are $ N \equal{} [u/2, (v \minus{} q)/2] \equal{} (1/p, 1/q)$ and

$ CN^2 \equal{} (p \minus{} 1/p)^2 \plus{} 1/q^2 \equal{} p^2 \minus{} 2 \plus{} 2 \equal{} p^2 \equal{} MC^2$

The $ \triangle MCN$ is also isosceles and

$ \cos \widehat{CMN} \equal{} \frac {MN}{2 MC} \equal{} \frac {\sqrt 2}{2p} \equal{} \frac {\sqrt {2 \mp \sqrt 2}}{2},$ $ \angle CMN \equal{} \frac {3\pi}{8}$ or $ \frac {\pi}{8},$ $ \angle MCN \equal{} \pi \minus{} 2 \angle CMN \equal{} \frac {\pi}{4}$ or $ \frac {3\pi}{4}.$

Consequently,

$ \angle BCN \equal{} \angle BCM \plus{} \angle MCN \equal{} \frac {\pi}{8} \plus{} \frac {\pi}{4} \equal{} \frac {3\pi}{8} < \pi$ or $ \frac {3\pi}{8} \plus{} \frac {3\pi}{4} \equal{} \frac {9\pi}{8} > \pi.$

$ \angle BCN$ is convex $ \Longleftrightarrow BD$ is inside of $ ABCD \Longleftrightarrow$ $ \angle BCD$ is convex. Thus, the 1st solution for $ p, q$ (with the upper signs) yields a convex quadrilateral $ ABCD,$ while the 2nd solution for $ p, q$ (with the lower signs) yields a concave one, not acceptable.

Since $ AC \equal{} 2p, MB \equal{} q$ are base and altitude of the isosceles $ \triangle ABC,$

$ S(ABC) \equal{} pq \equal{} \sqrt {2 \plus{} \sqrt 2} \sqrt {2 \minus{} \sqrt 2} \equal{} \sqrt 2$

Since $ AC \equal{} 2p$ is the base and $ v$ the altitude length of the $ \triangle CDA,$

$ S(CDA) \equal{} pv \equal{} p \cdot \frac {q^2 \plus{} 2}{q} \equal{} \sqrt {2 \plus{} \sqrt 2} \cdot \frac {4 \minus{} \sqrt 2}{\sqrt {2 \minus{} \sqrt 2}} \equal{}$

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

$ S(ABCD) \equal{} S(ABC) \plus{} S(CDA) \equal{} 2 \plus{} 4 \sqrt 2$

[color=red]EDIT:[/color] Originally, I had the (correct) result as $ S(ABCD) \equal{} \sqrt 2 \plus{} \sqrt {22 \plus{} 12 \sqrt 2}.$

Note that $ \sqrt{22 \plus{} 12 \sqrt{2} } \equal{} 2 \plus{} 3 \sqrt{2}$.  There's probably a much shorter solution.

[quote="t0rajir0u"]There's probably a much shorter solution.[/quote]Well, yetti's solutions to geometry problems are known to be overkilling and magnificent, if not anything less :D

Reflect $ A$ into $ N$ to produce $ A'$...This produces $ \triangle A'CD$ with $ CD \equal{} 2\sqrt {2}$, $ A'D \equal{} 2$ and $ CD \equal{} 2\sqrt {3}$...which is a right triangle. Extend $ A'C$ to hit $ AB$ at $ P$. Suppose that $ BP \equal{} x$ and $ CP \equal{} y$. Note that $ \angle APA' \equal{} 90$.

Then $ x^2 \plus{} y^2 \equal{} 4$ and $ x^2 \plus{} (y \plus{} 2\sqrt {2})^2 \equal{} 20$. So $ x \equal{} y \equal{} \sqrt {2}$. Then the total area is $ \frac {1}{2}\cdot ( (2 \plus{} \sqrt {2}) \plus{} 2)\cdot (\sqrt {2} \plus{} 2\sqrt {2}) \minus{} \frac {1}{2}\cdot \sqrt {2}\cdot \sqrt {2} \minus{} \frac {1}{2}\cdot 2\cdot 2\sqrt {2} \equal{} \boxed{4\sqrt {2} \plus{} 2}$...

EDIT: computation error

The areas of the convex and concave quadrilaterals are $ S(ABCD) \equal{} 4 \sqrt 2 \pm 2.$ (Area of the convex one is not $ 3 \sqrt 2 \plus{} 5.$) You have to prove that the 1st quadrilateral is convex and the 2nd one concave !

I fixed a computation error (it does come out to what you had for the convex case). Here is how to use positioning to show that my solution is the only convex one.

If the perpendicular bisector of segment BD divides the plane into two parts, A,C are on the same side (since AB<AD and BC<CD and it is convex),. Also A, C are on opposite sides of BD.

But then C, A' are on opposite sides of that perpendicular bisector and opposite sides of BD.

Also the distance from A to BD is greater than the distance from C to BD (here we use convexity i.e. if <ABD or <CBD is obtuse, it would be concave, so those angles are acute, then angle ABD is greater than angle CBD by hinge theorems, but they are both acute, so using AB=BC and the sine formula, we get that distance statement).

So A' is outside the triangle and furthermore, A'C intersects AB at P so C is in between P and A'.

Now we proceed to do my solution (since it is the only configuration) and we get that the area is what I said it was.

You still have a typo in your solution ($ A'D \equal{} 2 \sqrt 2$ should be $ A'C \equal{} 2 \sqrt 2$). I would fix it, but I am not a moderator here.

The triangle A'CD is right for both the convex and concave quadrilaterals ABCD (nice).

[quote="Altheman"]
But then C, A' are on opposite sides of that perpendicular bisector and opposite sides of BD.
[/quote]
C, A' are on the opposite sides of BD exactly for the concave quadrilateral ABCD. They are on the same side of BD for the convex quadrilateral ABCD.

[quote="Altheman"]
So A' is outside the triangle and furthermore, A'C intersects AB at P so C is in between P and A'.
[/quote]
A' is outside of all triangles for both the convex and concave quadrilaterals ABCD, so, what is it supposed to show ?

A'C intersects the line AB at a point P in both cases. The points A', C, P follow on the line A'C in this order for both the convex and concave quadrilaterals ABCD, C is in between P and A' in both cases, so, what is it supposed to show ?

A'C intersects the segment AB and A, P, B follow on AB in this order only for the concave quadrilateral ABCD satisfying the given conditions. But it is perfectly possible to imagine a convex quadrilateral ABCD, even one with AB = BC < CD < DA, for which A, P, B also follow on AB in this order. You are suppsed to show that for the convex quadrilateral ABCD satisfying the given conditions, the points A, B, P follow on AB in this order. You did not.

I think that you really have to show that the triangle MCN is isosceles and calculate angles of the isosceles triangles ABC, MCN.

In your solution, if A, B, P follow on AB in this order (convex quadrilateral ABCD), $ AP \equal{} 2 \plus{} \sqrt 2$ and

$ S(ABCD) \equal{} S(APA') \plus{} S(AA'D) \minus{} S(BCP) \minus{} S(A'CD) \equal{}$
$ \equal{} (3 \sqrt 2 \plus{} 3) \plus{} (3 \sqrt 2) \minus{} (1) \minus{} (2 \sqrt 2)$

But if A, P, B follow on AB in this order (concave quadrilateral ABCD), $ AP \equal{} 2 \minus{} \sqrt 2$ and

$ S(ABCD) \equal{} S(APA') \plus{} S(AA'D) \plus{} S(BPC) \minus{} S(A'CD) \equal{}$
$ \equal{} (3 \sqrt 2 \minus{} 3) \plus{} (3 \sqrt 2) \plus{} (1) \minus{} (2 \sqrt 2)$

Sorry for the bad explanation...perhaps this will make it more clear

Call the x-axis BD and the y-axis the perpendicular bisector of BD

Then we can place A,C,A' in 'quadrants'

In the concave case, A,C are in the same quadrant, so C,A' are in opposite quadrants (like I, III)

But in the convex case, A',C are in adjacent quadrants (like I,II)

Then we use the fact that the distance from A' to BD is greater than the distance from C to BD...now we draw the line parallel to BD through C...this divides the plane into 6 parts...this is enough to show that A'C at AB at P where B is between A and P; and C is between P and A'

[quote="Altheman"]Then we use the fact that the distance from A' to BD is greater than the distance from C to BD...now we draw the line parallel to BD through C...this divides the plane into 6 parts...this is enough to show that A'C at AB at P where B is between A and P; ...[/quote] Really?

ha okay. I guess I give in for now. I am not sure if there is any other simple fact that I can use to prove that P is not on segment AB...thanks for taking all the time to debunk my not so rigirous proof...


[quote="daermon"]evaluate $\sqrt{\imath}$[/quote]

[hide]$(-1)^\frac14$[/hide]

true, but I was looking for a simplified answer of degree $1$  :wink:

[quote="daermon"]evaluate $\sqrt{\imath}$[/quote]

Hmm, let me have a crack at this.

[hide]Let's say $\sqrt{i}= a+bi$
Squaring both sides, we have $i = a^{2}+2abi-b^{2}$

So 2ab= 1, $a^{2}-b^{2}= 0$, a= b

Then we can say $2a^{2}= 1$, $a^{2}= \frac12$.

Thus $a = \sqrt{\frac12}$,

or $a = \frac{\sqrt{2}}{2}= b$

And that gives us our answer:

$\sqrt{i}= \frac{\sqrt{2}(i+1)}{2}$[/hide]

[quote="lotrgreengrapes7926"]Another way: [hide] $i=1\text{ cis }90^\circ$.

$i^{\frac{1}{2}}=1^{\frac{1}{2}}\text{ cis }45^\circ=1\text{ cis }45^\circ={\frac{\sqrt2+i\sqrt2}{2}}$

or

$i^{\frac{1}{2}}=1^{\frac{1}{2}}\text{ cis }225^\circ=1\text{ cis }225^\circ=-\frac{\sqrt2+i\sqrt2}{2}$.

$\boxed{\pm\frac{\sqrt2+i\sqrt2}{2}}$ [/hide][/quote]
what is cis?  does it have something to do with the complex number plane?

I know I am being hypocritical by talking about trig in the MathCounts Forum, but cisA=cosA+isinA

[quote="daermon"]oh.  Thanks  :D

how does $\imath=\text{cis }90^\circ$?[/quote]
$\cos 90+i\sin 90=0+i(1)=i$

[quote="mathgeniuse^ln(x)"]I thought trig isn't allowed in the MathCounts Forum.

And how come complex numbers is in this forum as well?[/quote]
Mathgenius is right :mad:. In the mathcounts forum no trig and especially not imaginaries.

awwwwwwww  :(

the care bears will kill you for this!!!!!! and your little dog too!!!!!

(see signature)

Okay :|

Next time try to realize that trig and complex numbers aren't allowed in the Forum.

I wouldn't be surprised if they locked this topic.


Prove if a,b,c>$ \frac{1}{4}$such that $ \sum\sqrt{a}<\equal{}3$ then
$ \sum\sqrt{a^2\plus{}3}>\equal{}a\plus{}b\plus{}c\plus{}3$

let prove    $ \sqrt{a^2\plus{}3}\minus{}(a\plus{}1) \geq f_{(\sqrt{x})}$     denote a>$ \frac{1}{4}$

$ \sqrt{a^2\plus{}3} \geq  a\minus{}\sqrt{a}\plus{}2$   with   a.$ \frac{1}{4}$
--->Done :D

$ \sqrt {a^2 \plus{} 3} \geq a \minus{} \sqrt {a} \plus{} 2$
<-> $ (4a\minus{}1)(a\minus{}1)^2 \geq o$ :rotfl:


calculus
derivative
trigonometry
function
parameterization
limit
calculus computations

Well, i've encountered 2 problems that have to do with differentiation that i need help with. One of them i don't necessarily need help with yet, i just need help with clarification of the directions and what they're asking me to do. Anyhow, here are the problems:

1) Differentiate       y = 1/(2sin(2x))

2) These equations represent the curve parametrically; find the derivative dy/dx:

x = cos^3(θ) and y = sin^3(θ)

for problem 2, what exactly am i supposed to do? and for problem 1, i got a little flustered in the middle of the problem and couldn't come up with an answer... so if anyone comes up with an answer to that, could you show how you got there also? thanks much

[quote="sh0ebox"]Well, i've encountered 2 problems that have to do with differentiation that i need help with. One of them i don't necessarily need help with yet, i just need help with clarification of the directions and what they're asking me to do. Anyhow, here are the problems:

1) Differentiate       y = 1/(2sin(2x))

[hide]$y = \frac{1}{{2\sin 2x}} = \frac{1}{2}\cos ec2x$
$\frac{{dy}}{{dx}} = - \frac{1}{2}\cos ec2x.\cot 2x.\,2$[/hide]

2) These equations represent the curve parametrically; find the derivative dy/dx:

x = cos^3(θ) and y = sin^3(θ)

When x and y are given as functions of some parameter, say $\theta$, you can find the derivative of y with respect to x, by finding the derivatives of y and x w.r.t  $\theta$ and taking their ratio.  This is called parametric differentiation.
[hide]
here, $\frac{{dy}}{{d\theta }} = 3\sin ^2 \theta \cos \theta$

$\frac{{dx}}{{d\theta }} = - 3\cos ^2 \theta \sin \theta$

$\frac{{dy}}{{dx}} = - \tan \theta$[/hide]

[quote]y = \frac{1}{{2\sin 2x}} = \frac{1}{2}\cos ec2x[/quote]

question: what is the "cos ec2x"?

woo, i figured out most of the stuff i had trouble on... although i have a couple problems that i just encountered, i'm sorry to be such a calculus illiterate person  :blush: but can someone please help me with these bad boys? it would be much appreciated!

1) suppose that the limit (as x->0) of (g(x)-g(0))/x = 1. it follows necessarily that:

a) g is not defined at x = 0
b) g is not continuous at x = 0
c) the limit of g(x) as x approaches 0 equals 1
d) g'(0) = 1
e) g'(1) = 0

EDIT: i figured out the 2nd problem, now all i have left is that one above!  :D

Definition:

$g'(0)=\lim$[something], provided that limit exists.

Fill in the blank to make this a correct defintion.


I was wondering if it's possible to acquire transcripts from previous Math Jams. I'm hoping to begin training in the summer. Enrolling in a class won't really suite me, since I don't think I'd have a lot of free time, and even if I did enroll I wouldn't really participate in the virtual classroom. So I figured I'd just get some transcripts instead.

So... Are the previous transcripts available?

Yes: http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php then click one of the green buttons that says "view".

You cannot get any of the class transcripts for free.  You have to pay for the class.

[quote="devious_"]I've seen those. I was looking for transcripts of, say, the introduction to number theory classes -- not necessarily for free. ;)[/quote]

I see what you're asking.  You can enroll in the classes and not participate during class time.  This will give you access to the transcripts and the message board (medium to hard) practice problems.  You will also get both class exams and detailed solutions as to how to solve each.  Some of our students, particularly international students, choose to take the classes this way.

[quote="devious_"]But can I buy transcripts of previous classes?[/quote]

We do not sell those, but we will soon have textbooks for each of our Introductory classes -- and after that we'll write books for others one at a time.

[quote="devious_"]Sounds great! :D

Any estimate when the books will be available?[/quote]

Hard to say at the moment.  Sometime in the Summer we will release 1 to 3 of them.

[quote="devious_"]I was wondering if it's possible to acquire transcripts from [b]previous Math Jams[/b]. [/quote]

I blame the typo for my irrevalent post.


geometry
perimeter
inequalities
triangle inequality

I live in Illinois and I was preparing for the state math competition by looking at problems from previous years. This one has me stumped:

A triangle with sides 9, 28, and 29 has one of its sides increased by m, a second by 2m, and a third by 3m. If m is an integer such that m>0, find the perimeter of the newly formed triangle if the newly formed triangle is a right triangle.

Thanks!

hello, for the sides $ 28\plus{}3m,9\plus{}m$ and $ 29\plus{}2m$ you can calculate
$ (28\plus{}3m)^2\equal{}(9\plus{}m)^2\plus{}(29\plus{}2m)^2$ and $ m\equal{}3$ fulfills this equation.
Sonnhard.

Did you arrive at those numbers through guess and check? I'm looking for a way to actually solve it.

If you don't like what Dr Sonnhard put, the only other option is casework. There are 9 cases? Then, you get the equation and try to solve it.

akhil0422 - there are $ 3! \equal{} 6$ cases.... uh actually $ 18$ cases because after assigning the $ 1,2,3$ you have to pick a side to be the hypotenuse.  but some will be pretty ridiculous - triangle inequality should rule them out.  (In the end, I ended up testing $ 8$ cases.)

here's my crack at it:

Let $ a,b,c \equal{} {1,2,3}$.  Then we have sides of $ 9\plus{}am$, $ 28\plus{}bm$, and $ 29\plus{}cm$

we could have $ 9\plus{}3m$ as the hypotenuse - only if $ a \equal{} 3$

$ (9\plus{}3m)^2 \equal{} (28\plus{}bm)^2 \plus{} (29\plus{}cm)^2$
$ 81 \plus{} 54m \plus{} 9m^2  \equal{} (b^2\plus{}c^2)m^2 \plus{} (58c\plus{}56b)m\plus{}28^2\plus{}29^2$
$ 2m^2 \equal{} (29c\plus{}28b\minus{}27)m \plus{} 772$

if $ c\equal{}1$,$ b\equal{}2$
$ 2m^2 \minus{} 58m\minus{}772 \equal{} 0$
$ m^2 \minus{} 29m \minus{}386\equal{}0$
which does not have integer solutions

if $ c\equal{}2$, $ b\equal{}1$
$ 2m^2 \minus{} 59\minus{}772 \equal{} 0$
which does not have integer solutions

so $ 9\plus{}3m$ is not our hypotenuse.

Now we try $ 28\plus{}bm$ as our hypotenuse, and we must have that that $ b>c$ so we test 3 cases:
1) $ b\equal{}3$ $ a\equal{}2$ $ c\equal{}1$ - $ (28\plus{}3m)^2 \equal{} (9\plus{}2m)^2 \plus{} (29 \plus{} m)^2$
2) $ b\equal{}3$ $ a\equal{}1$ $ c\equal{}2$ - $ (28\plus{}3m)^2 \equal{} (9\plus{}m)^2 \plus{} (29 \plus{} 2m)^2$
3) $ b\equal{}2$ $ c\equal{}1$ $ a\equal{}3$ - $ (28\plus{}2m)^2 \equal{} (9\plus{}3m)^2 \plus{} (29 \plus{} m)^2$

And finally we can try $ 29\plus{}cm$ as our hypotenuse.  Since $ 29\plus{}cm > 28\plus{}bm$, $ 1\plus{}cm > bm$.  Both sides are integers, which means $ cm \ge bm$, and this implies $ c \ge b$.  obviously equality is impossible, so we have the following cases:
1) $ c\equal{}3$, $ b\equal{}2$, $ a\equal{}1$
2) $ c\equal{}3$, $ b\equal{}1$, $ a\equal{}2$
3) $ c\equal{}2$, $ b\equal{}1$, $ a\equal{}3$

... I don't want to solve all these quadratics.  So lets try the more general case:

let $ i,j,k \equal{} {\minus{}1,\minus{}1,1}$. (note i^2=j^2=k^2=1)  Then all the equations can be represented as

$ i(9\plus{}am)^2 \plus{} j(28\plus{}bm)^2 \plus{} k(29\plus{}cm)^2 \equal{} 0$
$ (ia^2 \plus{} jb^2\plus{}kc^2)m^2 \plus{} (18ai \plus{} 56bj \plus{} 58 ck) m  \plus{} 841k \plus{} 784j\plus{}81i \equal{} 0$

Which has the determinant
$ (18ai \plus{} 56bj \plus{} 58 ck)^2 \minus{} 4(ia^2 \plus{} jb^2\plus{}kc^2)(841k \plus{} 784j\plus{}81i)$
$ \equal{} 2016abij \plus{} 6496bcjk \plus{} 2088acik\minus{}3364a^2 ik\minus{} 3364 b^2 jk \minus{}3136a^2 ij \minus{} 3136 c^2 jk \minus{}324b^2 ij \minus{} 324 c^2 ik$

which must be a perfect square... this is so big I think it probably wasn't worth the effort- besides this will probably need some casework too.  You can first note that one of $ ij$, $ jk$, and $ ik$ will be $ 1$, the rest $ \minus{}1$. (e.g. $ i\equal{}1$ implies $ jk \equal{} 1$, $ ik\equal{}\minus{}1$, $ ij\equal{}\minus{}1$).  and its even more casework from there... ugh


For $ {2x_1x_4\plus{}2x_1x_3\plus{}2x_2x_4\plus{}2x_3x_4}$ and $ {7x_1x_4\plus{}7x_2x_3}$.
Which is greater than the other if $ {x_1,x_2,x_3,x_4}$ are a set of non negative not necessarily distinct real numbers.

Do you want us to put them from greatest to least?

If you mean letting $ {x_1<x_2<x_3<x_4}$

i think $ 7x1x4\plus{} 7x2x3$ is greater

[quote="aadil"]i think $ 7x1x4 \plus{} 7x2x3$ is greater[/quote]Prove your answer.


If 14 times a natural number has a units digit of 2 when expressed in base 12, what is the units digit of the natural number when expressed in base 6?

[quote="sunehra"]EDIT: Sorry I misread the question. It's fixed now.

[hide="My Solution"]
Let the natural number in question be $ k$. Because we are talking about units digits, we can simply use modular arithmetic. We have $ 14k \equiv 2 \pmod{12}$ and dividing by 2 we have $ 7k \equiv 1 \pmod{6}$. Solving the congruence (by adding 7) we get $ \boxed{1}$.
[/hide][/quote]

what do you mean by adding 7?  all you need to do is recognize $ 7 \equiv 1 \pmod{6}$, which implies $ 7k \equiv k \pmod{6}$.

Whoops, I meant to say add 6 because $ 7k$ will be congruent to numbers $ 6k \plus{} 1$ mod $ 6$.


AMC
USA(J)MO
USAMO
AIME
email
probability
AMC 10

If you took the AMC 12 A or the AMC 12 B and AIME I or AIME II and your combined index of of AMC 12 score + 10*AIME was greater than or equal to 217.0 you will be invited to take the USAMO.

There were 240 AMC 12-takers selected by this index.

The floor value selected from that group of 12-takers was 8.  The two AIMEs were statistically equivalent, so there was no need to distinguish between the two.

Then among students in the 10th grade and below not already qualified by virtue of high index from the AMC 12 and AIME, (regardless of whether tthe student took the AMC 10 or the AMC 12) with an AIME of 8 or greater, the highest approximately 120 indices were at 212.0 or greater.  There were actually 121 students selected in this group.

Then 7 more students were selected from states not already represented.

This gives a total of 367 students selected for the USAMO.

Our office staff is now working to correlate the CEEB numbers with a school name and to pull all records including student email and address from the answer forms.  As soon as the first is finished, a list of student name, school, and state will appear on our website.  As soon as the second task is completed, students will recieve an email followed by a paper mail with the official invitation to the USAMO.  Both of these tasks will take a day or two, so please continue to be patient.  As I said before, our office staff will not answer questions about names or specific students until March 31 or even later if our tasks are not completed.  Phoning and emailing us will only delay us further.

Steve Dunbar
AMC Director

That should be 368 students selected. -- Steve Dunbar

Thanks very much for the info!

EDIT- I take credit for seeing this first, kinda ironic because I wasn't expecting it and I wasn't refreshing the page.

So does that mean I didn't make it if I got a 7 on the AIME and a 213 index and in 9th grade?

Darn

Thank you for posting this information. I know you guys are all working really hard. Thank you once again.

yes i made it!  :)

Thank you very much for post the info here, Mr. Dunbar!

Congrats to everyone who qualified! :)

YAy! Ty for getting AoPS the results ASAP  :)

I didn't qualify  :(

Wait Im confused :? , can someone explain it in plain terms for stupid people like me?  :)   Does this mean that everyone who scored above 8(floor) in the AIME below 11th grade gets to take USAMO? :roll:

EDIT: nvm

My sincerest thanks to Mr. Dunbar for posting this information.  Thanks for all your work on the AMC contests and a special thanks for taking the time to post here. :)

Thanks so much to Mr. Dunbar for posting this informat ion instead of making everyone wait even longer to find out!

Note to Mr. Dunbar: If you're reading this and everything isn't done, GET BACK TO WORK!  ;)

Seriously though, a sincere and heartfelt "thank you" to everyone who works to make this happen.

USAMO 2006, here I come!

*sigh* another year of getting owned by USAMO

See they made the final decision at latest at 2.30p.m.(Nebraska zone)
First, thank you for your hard work! :lol:
How late or How long will the AMC directors work today? Till the Nebraska time zone 4p.m.? 5p.m.? or 6p.m.? What's the probability they can work out the list today (to corellate the CEEB)?

[quote="Fermatprime"]
Note to Mr. Dunbar: If you're reading this and everything isn't done, GET BACK TO WORK!  ;)
[/quote]

That isn't very nice.

I am confused.  If I made a 9 on AIME II, but didn't get 212 index, does htat mean I didn't make USAMO?

[quote="beta"]Okay the cutoff is 217.0(I found someone who got that index and made it)[/quote]

I have a 217.0.  I was wondering because it would have been really funky if it were 217.5.

[quote="beta"][quote="frankthetank"]The website says the cutoff is 217.5, but AMCDirector said it is 217.0.  What is the actual cutoff?[/quote]

What is the actual cutoff?

Oh how come the website says 374 students today?[/quote]

Apparently they invited 7 more students that got an 8 on the AIME regardless of index score.

No, they added 57 lol.

I think the extra people are state qualifiers they forgot to add.

sad sad
i got 128.5 on AMC 10 and 8 on AIME..

[quote="me@home"]you only got 128 on AMC 10?? then got an 8??
that's crazy, I aced the AMC 10 then got a four.[/quote]

The AMC emphasizes speed a lot more than the AIME does.  On AIME there's a lot of time for checking so the emphasis is on real problem-solving skill, whereas AMC is more speed and accuracy.

so AMC is like the mathcounts/high school math bridge?

Okay, with all of the crazieness(sp?), which should one take next year if one is in middle school and did better on the AMC 12B than that AMC10A?

Smiles,
lyra

[quote="nat mc"][quote="lyra"]Okay, with all of the craz[b]i[/b]ness, which should one take next year if one is in middle school and did better on the AMC 12B than that AMC10A?

Smiles,
lyra[/quote]
I would day that you should probably again do the 10a and 12b or 12a and 10b, but if your school allows only one next year i would probably say take the 12[/quote]
Or you could see if you could take the extra one at a nearby school (fermatprime and another person did that at our school and they aren't even in the same county, but it did lead to certain complications, so be careful...).

It had to do with Sliffe awards, in our case...

[quote=pieterminate]
An excellent system:
AMC 12: 120                      AMC [b]10[/b]: 135
AIME: 9                                          AIME: 8
Status: Screwed          Status: Happy Camper[/quote]
Does that mean if I get a 132 on AMC 10 and an 8 on AIME, I won't be able to get to the next round?  Since I thought it was 212 points to get in.

Every year is different.  This was 2006 not 2015.  Anything can happen.  They usually take, approximately, 270 people for USAMO and 230 people for USAJMO so wherever the cutoff lies to get these numbers.

NOTE: Oh yeah, they didn't have USAJMO back in 2006 so . . .


Compute the smallest six-digit number with distinct digits that is a multiple of 11.

For some number if the sum of digits in even place and sum of digits in odd place differ by a multiple of $ 11$, the number is divisible by $ 11$.
$ 102345$ is the lowest $ 6$ distinct-digit number. But difference is $ 1$. If the first $ 4$ digits are unchanged last two digits must be equal to be divisible by $ 11$.  So we must change $ 3$. Smallest change is to make $ 4$. Then the last $ 2$ smallest digits are $ 65$. So the required number is $ \boxed{102465}$


analytic geometry
graphing lines
slope
quadratics
algebra
quadratic formula

Given a circle of radius $ 2$ centered at $ (3,\minus{}3)$ and the line $ y \equal{} 2x\plus{}2$, find the coordinates of all points on the circle where the tangent line is perpendicular to the given line.

PRECALC SOLUTION
[hide]
Drawing the diagram we can quickly see that there will be two points of tangency and that those two points will be collinear with the center of the circle and also parallel to the given line. Hence it has slope $ 2$ and contains the point $ (3,\minus{}3)$, and so we have:

$ y \equal{} 2x \plus{} b \implies \minus{}3 \equal{} (2)(3) \plus{} b \implies b \equal{} \minus{}9$

knowing that the equation of the circle is $ (x\minus{}3)^2 \plus{} (y\plus{}3)^2 \equal{} 4$, we substitute $ y$ to get:

$ (x\minus{}3)^2 \plus{} (2x\minus{}6)^2 \equal{} 4 \implies 5x^2 \minus{} 30x \plus{} 41 \equal{} 0$

using the quadratic formula we get:

$ x \equal{} \frac{30 \pm \sqrt{900 \minus{} 4(5)(41)}}{10} \equal{} \frac{30 \pm \sqrt{80}}{10} \equal{} 3 \pm \frac{2\sqrt{5}}{5}$

now plugging this back into the equation of the line $ y \equal{} 2x \minus{} 9$, we get:

$ y \equal{} 2\Big(3 \pm \frac{2 \sqrt{5}}{5}\Big) \minus{} 9 \equal{} \minus{}3 \pm \frac{4 \sqrt{5}}{5}$

Hence the coordinates are :

$ \Big(3 \plus{} \frac{2\sqrt{5}}{5}, \minus{}3 \plus{} \frac{4 \sqrt{5}}{5}\Big)$ and $ \Big(3 \minus{} \frac{2\sqrt{5}}{5}, \minus{}3 \minus{} \frac{4\sqrt{5}}{5}\Big)$

[/hide]

Yea, thats the nice solution. [I had posted that exact thing you did before reading...]

A little more simple is to solve $ x^2\plus{}y^2\equal{}4$ and $ y\equal{}2x$, then translate.

Cool problem, I like it!


algebra
polynomial
linear algebra
matrix

Let $A$ and $B$ be two matrices such that the products $AB$ and $BA$ are well-defined. Why do $AB$ and $BA$ have the same eigenvalues?

Because they have the same characteristic polynomial (supposing that's square matrices you're talking about there; if not, then the algebraic multiplicity of $0$ might differ).

The ideas you get by following the [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=posted&t=110427]links[/url] will lead to a solution.

Here's another argument. It's not mine; I'm relaying it from a colleague of mine. We will assume that $ A$ is an $ m\times n$ matrix and that $ B$ is an $ n\times m$ matrix. These are matrices over an arbitrary field. For the sake of the argument, assume $ m\le n.$ For any square matrix $ M,$ let $ P_{M}(x) \equal{} \det(xI \minus{} M)$ be the characteristic polynomial of $ M.$

Claim: $ x^{n \minus{} m}P_{AB}(x) \equal{} P_{BA}(x).$

Let $ U \equal{} \begin{pmatrix}0 & A \\
B & 0\end{pmatrix}.$

Let's compute the characteristic polynomial of the $ (n \plus{} m)\times(n \plus{} m)$ matrix $ U.$

$ \det(xI \minus{} U) \equal{} \det\begin{pmatrix}xI & \minus{}A \\
\minus{}B & xI\end{pmatrix}.$

Do row operations here. Add to the lower blocks $ \frac1xB$ times the upper blocks. This leaves the determinant unchanged.

$ \det\begin{pmatrix}xI & -A \\
0 & xI \minus{} \frac1xBA\end{pmatrix}$

$ \equal{} x^{m}\det\left(xI \minus{} \frac1xBA\right) \equal{} x^{m \minus{} n}\det(x^{2}I \minus{} BA) \equal{} x^{m \minus{} n}P_{BA}(x^{2})$

Now, perform the same argument on the matrix

$ V \equal{} \begin{pmatrix}0 & B \\
A & 0\end{pmatrix}$

$ P_{V}(x) \equal{} x^{n \minus{} m}P_{AB}(x^{2}).$

But $ U$ and $ V$ are similar. They can be made similar by a permutation with the structure

$ \begin{pmatrix}0 & I \\
I & 0\end{pmatrix}.$

Therefore, $ U$ and $ V$ have the same characteristic polynomial. We conclude that

$ x^{2(n \minus{} m)}P_{AB}(x^{2}) \equal{} P_{BA}(x^{2}).$

These are polynomial expressions, with $ x^{2}$ taking the role of $ x.$ In other words, they have the form $ a_{0} \plus{} a_{1}x^{2} \plus{} a_{2}x^4 \plus{} \cdots.$ Since they're the same polynomial expressions with the $ x^{2}$ in there, they'd be the same if you replaced that with $ x.$ And that's what we set out to prove.


Suppose that the polynomial $ ax^{2}+(c-b)x+(e-d)$ has two real roots, both greater than $ 1$. Prove that $ ax^{4}+bx^{3}+cx^{2}+dx+e$ has at least one real root.


function
calculus
integration
real analysis
real analysis unsolved

Suppose that $ f: [a,b] \rightarrow \mathbb{R}$ is a Riemann integrable on $ [a,b]$ and $ f(x) \geq 0$ for all $ x \in [a,b]$. Prove that $ \sqrt{f}$ is Riemann integrable on $ [a,b]$.
Relevant Theorems & Definitions
Definition - Riemann integrable - if upper integral of $ f(x)dx$= lower integral of $ f(x)dx$.
Theorem - Riemann integrable iff $ \exists A \in \mathbb{R}$ such that $ \forall \epsilon >0 \exists$ a partition $ P$ of $ [a,b]$ such that $ \forall$ marked refinements $ Q$ of $ P$, $ |S(Q, f) \minus{} A | \leq \epsilon$, where $ A\equal{}\int^a_b f(x)dx$.
Theorem - Let $ f: [a,b] \rightarrow \mathbb{R}$ be a continuous function on $ [a,b]$. Then $ f$ is Riemann integrable on $ [a,b]$.
etc.
Theorem - Riemann integrable iff for each $ \epsilon > 0 \exists$ a partition $ P$ of $ [a,b]$ such that $ U(P,f)\minus{}L(P,f) \leq \epsilon$.

I haven't gotten very far doing this problem. I need help with this problem. Thank you.

[hide]
You can use composite integral theorem : If $ f$ is Riemann integratable and $ g$ is continuous in the interval containing range of $ f$ then $ g(f)$ is Riemann Integratable.

By using the Theorem, you can look that $ g(x)\equal{}\sqrt {x}$ is Riemann integratable and $ g(x)$ is continuous on every $ x$ statisfying $ x \geq 0$, because $ f(x) \geq 0$ and $ g$ is continuous on $ x \geq 0$, then $ f(g(x))$ is Riemann integratable in the interval given.
[/hide]
I am sorry if I am false.


[quote="Victory.US"]let $ u_n$ given by: $ u_0 \in R$ and $ {u_{n \plus{} 1}} \equal{} \frac {{{u_n}^2 \minus{} 1}}{{n \plus{} 1}}$
in wich value of $ u_0$ then $ u_n$ has Limit?[/quote]

WLOG $ u_0 \ge 0$

If $ u_0\ge 2$ and $ u_{n}\ge n \plus{} 2$
then $ u_{n}^2 \ge (n \plus{} 2)^2$
then $ u_{n \plus{} 1} \ge \frac {n^2 \plus{} 4n \plus{} 3}{n \plus{} 1} \equal{} n \plus{} 3$
We have proved then by induction that $ u_n\ge n \plus{} 2$ for $ |u_o| \ge 2$

who can complete the proof? or other idea which can slove it in nice way?

mazew show, that if $ |u_0|\ge 2$, then $ u_n\to \infty$.
Let $ v_n\equal{}n\plus{}2\minus{}u_n$, then $ v_{n\plus{}1}\equal{}v_n\frac{2n\plus{}4\minus{}v_n}{n\plus{}1}>v_n.$
If $ |u_n|\le 1$, then by induction $ \minus{}\frac{1}{n\plus{}k}<u_{n\plus{}k}\le 0,k\ge 1$, therefore $ u_n\to 0$.
If $ \minus{}2<u_0<\minus{}1$, then $ 0<u_1\equal{}u_1(|u_0|)<3$ or $ 0<v_1<3$.
If $ v_n>\epsilon$, then $ v_{n\plus{}1\plus{}[log_2{2/\epsilon}]}>2$. If $ 2\le v_n\le n\plus{}1$ for $ N<n<M$, then $ v_{n\plus{}1}>v_n\frac{n\plus{}3}{n\plus{}1}\to v_M>\frac{(M\plus{}3)(M\plus{}2)}{(N\plus{}1)(N\plus{}2)}v_N>M\plus{}1.$
Therefore exist M, suth that $ u_M<1$. It give $ u_n\to 0$,


As the figure shown,the circle O with radius 2 and  if $ \angle OAB\equal{}60^{\circ} ,\angle OCB\equal{}{45^\circ }$ then  area  of $ \triangle ABC \equal{}$?

It's easy to see that $ \triangle OAB$ is equilateral and $ \triangle OCB$ is right-angled isoscele triange.
So $ [ABC] \equal{} [OCB] \plus{} [OAB] \minus{} [OCA] \equal{} \frac {2^2}{2} \plus{} \frac {2^2 sin \frac {\pi}{3}}{2} \minus{} \frac {2 \times 2cos \frac {\pi}{3}}{2} \equal{} 1 \plus{} \sqrt {3}$

[quote="ritchjp"]To stephencheng:
   Your answer is correct but aren't you making assumptions rather than proving
  your assumptions? Let's see your work proving your assumptions are correct.[/quote]There isn't much to be proved... OA=OB=OC, which immediately yields the results he stated...


geometry
calculus
derivative
advanced fields
advanced fields unsolved

Sample Problem: For which values of a(including infinity) does the curve C have singular points if its homogeneous equation is given by
x^3 + y^3 + a*[(x + y + z)^3] ?

I converted the equation from the projective plane to affine by setting z=1 :

x^3 + y^3 + a*(x+y+1)^3 + 1 = 0. Writing this in the expanded form and putting a = -1, we obtain

(x^2)*y + x*(y^2) + x^2 + 2*x*y + y^2 + x + y = 0.

The terms of lowest degree are x and y, and so the tangent line is given by x + y = 0, which is an ordinary singular point.

For the other values of a, i found that the curve is nonsingular,i.e., it has no singular points. But i don't know what to do with the infinity case? Besides, i wonder whether this method works all the time.

I'm considered to be a newbie in algebraic geometry and your help is appreciated.

When you take affine slices you lose points at infinity.  In other words, when you set $ z \equal{} 1$ you don't consider the case that there may exist singular points when $ z \equal{} 0$, so you have to check that.  The general method is just to take partial derivatives (as I recall).

I subscribe to the above response and I would like to add that you really need to pass to affine patches if you also want to determine the multiplicity of a singular point. If you just want to find the singular points, you only have to find the common zeros of the partial derivatives of the equation.


It is known that $x_i\in R$, $x_i\ge 0$, $i=1,2,..,n$ and
$\sum_{i=1}^{n}ix_i=a$
$\sum_{i=1}^{n}i^3x_i=a^2$
$\sum_{i=1}^{n}i^5x_i=a^3$.
Find all numbers $a$, for which there exist such numbers $x_1,x_2,..,x_n$.

This is problem 5, IMO 1979 (proposed by Israel).

well it was in russian olympiad. Thanks anyways

We have $(\sum ix_i)(\sum i^5 x_i)=(\sum i^3x_i)^2=a^4$.This is the form the cauchy ineq.If some $x_i$ are 0 we cut them off from the sum.Now puting the equality case in cauchy ineq we have $i=j$ if there are two numbers numbers different from 0,false so we have n-1 numbers or n equal to 0.If there n-1 equal to 0,we have $ix_i=a,i^3 x_i=a^2,i^5x_i=a^3$ so we have $a=x_i=i=1$.

My teacher said that the only solutions are.. 0,1,4,9 ???

There is a gap in mateivld's solution because he lost some cases in when equality holds in cauchy Swarthz inequality.


How do you graph equations in Asymptote without having to calculate coordinates and such? Like if I wanted to graph $ y \equal{} x^2 \minus{} 5x \plus{} 1$ or $ y \equal{} 3x \plus{} 5$.

[asy]import graph_pi;

real f(real x){return x^2-5x+1;}

graphicrules(xunit=2cm, yunit=1cm,
             xmin=-1, xmax=3, ymin=-6, ymax=f(-1));
draw(graph(f), bp+0.8*red);
grid();
cartesianaxis(arrow=None);
labeloij(UnFill);
// crop();[/asy]


:coolspeak: signiture . it says ,,art of problem solving" right?

and thaks for telinng me what colon is

 :D

[quote="BOGTRO"][quote="@nn@"][quote="buzzer11"]what do you mean by the colon?[/quote]

Well

$ (1+2+3+...............+39)$ divided by $ 13$

P.S. What's a colon? :roll:[/quote]

39+38+37+...+3+2+1+=780.

780/13=60. 60*749=7490*6=7000*6+400*6+90*6=42000+2400+540=44940.[/quote]

you know

1. you dont need to add all them numbers up theres a formula-n(n+1)/2

2. you dont need to actually find the sum. look

plug the numbers in but dont divide and multiply $ \frac{39*40}{2}$. the 2 and 40 cancel each other to get $ 20*39$ divide by 13. $ \frac{39*20}{13}$. the 39 and 13 cancel each other out.
to get $ 3*20$ so y=60. then you have 60*749=x and do that. im too lazy though

Actually to show that something divides into something you do: $ a|b$.

Does ANYONE EVER look at previous posts?
This question has already been answered. Mods, lock.

[quote="rt_08"]hey i think u could do it with the help of arithmetic progression
u c
a=1;
l=39
n=40;
s=40/2[1+39]
now solve further![/quote]
This is a year old.

it is old but somewat more conventional !
i guess!

PLEASE LOCK!!!! This post has been bumped from a year ago!!! lock it!

[quote="AIME15"]Alton don't spam to get your post count up

(sorry for adding to the spam)

But alton about half your posts are telling the mods to lock the thread.[/quote]

lol aime, u spam a whole lot.  but thats coming from me, ernie, the spammiest of them all...

[quote]lol aime, u spam a whole lot. but thats coming from me, ernie, the spammiest of them all...[/quote]

most of ur posts were spam.....

@Mods: PLEASE LOCK.
PLEASE LET THIS TOPIC DIE.
Thank you.


[hide]The length of the rail going across is $4.5\pi$. When you unravel the staircase, you see it's hieght is 10 feet and it goes across $4.5\pi$. Use the Pythagorean theorem, $\sqrt{4.5\pi^{2}+10^{2}}$, and the answer is $\approx17.3$.[/hide]

Wait... How did you do that? Why did you apply the Pythagorean Theorem? I don't think I get the problem or the solution.  :blush: Can someone explain this?

[quote="davidli"]Well, if you picture it in your head, it kinda makes a cylinder. When you open it up, it creates a rectangle going 4.5 across and 10 up.[/quote]

And the handrail is the diagonal.

[quote="espark52"]a spiral stiarcase turns 270 degrees  as it rises 10 feet. the radus of the staircase is 3 feet. wat's the number of feet in th legnth of the handrail? express ur answer as a decimal to the nearest tenth. (the handrail is always equidistant to the stairs and extends the whole way of the staircase)[/quote]

[hide]Imagine the staircase as a cylinder.  Flatten out the cylinder to get a 10 by 6pi rectangle.  The handrail only goes 3/4 of the way so multiply that by 6pi to get 9pi/2.  Now find the diagonal of the 10 by 9pi/2 rectangle, bu i dont have a calculator with me.[/hide][/hide][/list][/code]

Why does the handrail only go $\frac{3}{4}$ of the way?

because it turns $270$ degrees and $\frac{270}{360}=\frac{3}{4}$

[hide]The railing is ¾ of a circle long (not including height). The radius is 3 feet, so we get ¾ 2(pi)3=3/4*6*(pi)=4.5pi
It rises 10 feet, however, so by applying the Pythagorean theorem we get rt{(4.5pi)^2+10^2}=rt{(4.5pi)*2+100}
[/hide]


calculus
integration
trigonometry
logarithms
analytic geometry
linear algebra
matrix

$ \int_{0}^{a\sin\beta}\left\{\int_{y\cot\beta}^{\sqrt{a^2\minus{}y^2}} \ln\left(x^2\plus{}y^2\right)dx\right\}dy$

where $ 0<\beta<\frac{\pi}{2}$

change the coordinate to polar coordinate,we get:

$ \left\{\begin{matrix}
r\sin\theta \cot\beta \leq r\cos\theta\leq \sqrt{a^2\minus{}r^2\sin^2\theta}\\
0\leq r\sin\theta\leq a\sin\beta
\end{matrix}\right.$

from this ,we get :

$ \left\{\begin{matrix}
0 \leq r \leq a\\
0 \leq \theta \leq \beta
\end{matrix}\right.$

so the above integral equal: $ \int_{0}^{a}\int_{0}^{\beta}r \ln r^2drd\theta\equal{}a^2\beta\left(\ln a \minus{}\frac{1}{2} \right)$

and I want to know the graphic of the domain $ D\equal{}\left\{\left[y\cot\beta,\sqrt{a^2\minus{}y^2}\right] \times \left[0,a\sin\beta\right]\right\}$ looks like, where $ 0< \beta < \frac{\pi}{2}$ and $ 0\leq y\leq a\sin\beta$ , because i can't see it directly


group theory
abstract algebra
superior algebra
superior algebra unsolved

Can someone please show me how A_4 has no simple groups of order 6

If $ H$ is such a subgroup, then $ |A_{4} : H| \equal{} 2$ and so $ H$ is a normal subgroup. Then, if $ \alpha \in H$, $ H$ contains all the conjugates of $ \alpha$ by even permutations. Since $ 8$ of the $ 12$ elements of $ A_{4}$ are 3-cycles, $ H$ must contains a 3-cycle $ \alpha$. how many conjugates does $ \alpha$ has in $ H$? $ \alpha$ has $ 8$ conjugates in $ S_{4}$, so that the centralizer of $ \alpha$ in $ S_{4}$ has index $ 8$ and order $ 24/8 \equal{} 3$. Thus there are only 3 permutations in $ S_{4}$ that commute with $ \alpha$ and that must be $ \alpha$, $ \alpha^{\minus{}1}$ and $ (1)$. Since all this live in $ A_{4}$, the centralizer of $ \alpha$ in $ A_{4}$ has order $ 3$ and has index $ 12/3 \equal{} 4$. Then, $ \alpha$ has $ 4$ conjugates in $ A_{4}$, and all live in $ H$ dince $ H$ is normal.

Now $ H$ must contain an element $ \beta$ of order $ 2$, because if it contains a  3-cycle, it would contain its 4 conjugates and we would have more than $ 6$ elements in $ H$. Since $ \beta$ is even, $ \beta \equal{} (ab)(cd)$. We have discovered $ 6$ elements in $ H$: four 3-cycles, $ \beta$ and $ (1)$, and this is our bound. But we can show another element in $ H$!. Since $ H \triangleleft A_{4}$, $ \gamma\beta\gamma^{\minus{}1} \in H$, where $ \gamma \equal{} (acb)$, but $ \gamma\beta\gamma^{\minus{}1} \equal{} (ca)(bd) \neq \beta$. This contradiction completes the proof.

The conjugacy classes in $ A_{4}$ are {e}, {(12)(34), (13)(24), (14)(23)}, {(123), (134), (243), (142)} and {(132),(143),(234),(124)}.

Since 3+4 > 5, it follows that $ A_{4}$ can't have a subgroup of order 6: let alone a [i]simple[/i] subgroup of order 6.

Best Regards.

YOu could show that there are no simple groups of order $ 6$.

[quote="ZetaX"]... there are no simple groups of order $ 6$.[/quote]

[b]Proof.[/b] There are only two groups of order 6: $ C_{6}$ and $ \mathcal{S}_{3}$.

$ C_{6}$ is cyclic and the [i]purported[/i] result follows here from the fact that finite cyclic groups have a normal subgroup of order d, for every number d that divides its order.

On the other hand, $ \mathcal{A}_{3}$ is a subgroup of index 2 in $ \mathcal{S}_{3}$: it implies in turn that  $ \mathcal{A}_{3}$ is normal in $ \mathcal{S}_{3}$ and the proof terminates.

I suspect the word "simple" in the original post was a typo of sorts.


inequalities
algebra
functional equation
4th edition

I just wondered if there might be a typo on this question. For n=10, I got (by calculator) the LHS at about 11000, the RHS at about 7000000! Is this just my faulty arithmetic, or is there meant to be this huge disparity?

If LHS and RHS is Left Hand Side and Right Hand Side, then what's the problem? You have to show that RHS is greater than LHS $\forall n\in \mathbb{N}$ and you did not obtain any counterexample.

The question is rather whether Valentin really meant such an inequality (where the LHS is [b]a big lot[/b] smaller than the RHS) or whether there is a typo somewhere.

  darij

The inequality is correct. Please do not make any further comments about the problem :)

Ok, thanks. Sorry; it just seemed strange to me.

(It's probably not really worth the time reposting the posts in this thread, but since Darij is the other person doing most of the reposting, I can't resist :P !)

[b][u]The author of this post is: al.M.V.[/u][/b]
_____________________________________________________________

Fiachra, could you show me your solution to #2 of Round 3 please (the functional equation one: f(x+y-xy)=f(x)+f(y)-f(xy).)

I could not figure that one out.

Thanks in advance. :)

Edit: Oops! Seeing Valentin's comment, could someone move this post to a new topic like Q3/P2 of the discussion forum? Thanks

[b][u]The author of this post is: darij grinberg.[/u][/b]
_____________________________________________________________

On another topic (I can't find the thread right now) I am wondering why Valentin found my solution of #3 worth 6 points. I find it's a simple and nice synthetic solution. Is there something still much simpler?

dg

[b][u]The author of this post is: Valentin Vornicu.[/u][/b]
_____________________________________________________________

[quote="darij grinberg"]On another topic (I can't find the thread right now) I am wondering why Valentin found my solution of #3 worth 6 points. I find it's a simple and nice synthetic solution. Is there something still much simpler?

dg[/quote]

Yes :) In plus I had to use the pencil to convince myself it's ok in your case. That qualifies (under my criterions :D) for a 6. Sorry ...

[b][u]The author of this post is: darij grinberg.[/u][/b]
_____________________________________________________________

[quote="Valentin Vornicu"]Yes :) In plus I had to use the pencil to convince myself it's ok in your case. That qualifies (under my criterions :D) for a 6. Sorry ... [/quote]

Actually, where did you have to use the pencil? :? I thought my solution was complete...

dg

[b][u]The author of this post is: Valentin Vornicu.[/u][/b]
_____________________________________________________________

[quote="darij grinberg"][quote="Valentin Vornicu"]
Yes :) In plus I had to use the pencil to convince myself it's ok in your case. That qualifies (under my criterions :D) for a 6. Sorry ... [/quote]

Just out of fun, where did you have to use the pencil?

dg[/quote]
That is classified information. :D


function calculus integration parameterization

What is a state function, and why isn't work one?

Thanks in advance!

A state function is simply what is. For example, your homework is a state function, as it doesnt matter who did it, all that matters is that its done... In other words, state functions are independant of path.

As for why is work not a state function, we shall first consider what work is.
Work is defined in physics as the scalar product between a force and a distance. More specifically, if moving in some random path, Work is defined as the line integral (sum) of all the scalar products along  the path. So.... lets say u have a box at point a and u wanna move it to point b. There are millions of paths u could take to reach that point, each path takes a different amount of work. So there you have it...

I would define a state function as a function by which only the state of the parameter in question is changed; in other terms, the state function describes the what, but not the how.

A classical example is is in thermodynamics, where there is a change in the free energy of a system due to a reaction. The $ \Delta G$ is what counts; the size of the activation energy and whatnot is not important and not considered in the value of $ \Delta G$.

A state function is a function whose values between two states of a system depend only on the initial and final state, and not on the path "walked" by the system between the two positions. Examples are P, V, T, U, H, S, G, Cp, Cv.

On the other hand, the values of a process function between two states of a system depend on the path that connected those two states. Examples are W and Q (note, however, that W + Q is always a state function).

And mathematically they are exact differentials.

Mathematically there are several things we say about state and process functions. That is one of them ( I use the notation $ dS$, for example, for a state function, and $ \delta W$ for a process function).

Another thing we can write is $ \oint statefunction \equal{} 0$ and $ \oint processfunction \neq 0$, in general.