Let $ \ x\equal{}\sqrt{a^2\minus{}ab\plus{}b^2},y\equal{}\sqrt{b^2\minus{}bc\plus{}c^2},z\equal{}\sqrt{c^2\minus{}ca\plus{}a^2}$
Prove that
a.$ \ x\plus{}y\plus{}z\ge\sqrt{3(a^2\plus{}b^2\plus{}c^2)}$;
b.$ \ xy\plus{}yz\plus{}zx\ge a^2\plus{}b^2\plus{}c^2$.

Ineq(b) is stronger than ineq (a).
And it's obviously trues by Am-Gm. It's also a problem in Pham VanThuan's book.
 :)

$ Nguoivn$,can you show me your solution by AM-GM?

[quote=Understandingmathematics][quote=arqady][quote="duongptnkvn"]Let $ \ x \equal{} \sqrt {a^2 \minus{} ab \plus{} b^2},y \equal{} \sqrt {b^2 \minus{} bc \plus{} c^2},z \equal{} \sqrt {c^2 \minus{} ca \plus{} a^2}$
Prove that
b.$ \ xy \plus{} yz \plus{} zx\ge a^2 \plus{} b^2 \plus{} c^2$.[/quote]
$ xy \plus{} yz \plus{} zx\equal{}\sum_{cyc}\sqrt{(a^2\minus{}ab\plus{}b^2)(a^2\minus{}ac\plus{}c^2)}\geq$
$ \geq\sum_{cyc}\left(a^2\minus{}\frac{a(b\plus{}c)}{2}\plus{}bc\right)\equal{}a^2\plus{}b^2\plus{}c^2.$[/quote]

how $\sum_{cyc}\sqrt{(a^2-ab+b^2)(a^2-ac+c^2)}\geq$  $\sum_{cyc}\left(a^2-\frac{a(b+c)}{2}+bc\right).$ ????

can you pls add some more details to it...[/quote]

By [b] Cauchy-Schwarz's inequality [/b], we have \begin{align*}\sqrt{(a^2-ab+b^2)(b^2-bc+c^2)} &= \sqrt{\left[\left(a-\frac{b}{2}\right)^2+\frac{3b^2}{4}\right]\left[ \left(a-\frac{c}{2}\right)^2+\frac{3c^2}{4}  \right]} \\&\ge \left(a-\frac{b}{2}\right)\left(a-\frac{c}{2}\right) + \frac{3}{4}bc \\&= a^2-\frac{a(b+c)}{2}+bc\end{align*}

[quote=Understandingmathematics]

how $\sum_{cyc}\sqrt{(a^2-ab+b^2)(a^2-ac+c^2)}\geq$  $\sum_{cyc}\left(a^2-\frac{a(b+c)}{2}+bc\right).$ ????

can you pls add some more details to it...[/quote]

By [b] Cauchy-Schwarz's inequality [/b], we have \begin{align*}\sqrt{(a^2-ab+b^2)(b^2-bc+c^2)} &=\frac{1}{4} \sqrt{\left[\left(a+b\right)^2+ 3\left(a-b\right)^2
 \right] \left[\left(a+c\right)^2+ 3\left(a-c\right)^2
 \right]        } \\&\ge  \frac{\left(a+b\right)\left(a+c\right) +3\left(a-b\right)\left(a-c\right) }{4}
\\& = a^2-\frac{a(b+c)}{2}+bc    \end{align*}


A recent autobiography used 2808 digits to number the pages of the book. How many numbered pages are in the book?

$ 9*1\plus{}90*2\plus{}3x\equal{}2808$
$ 3x\equal{}2619$
$ x\equal{}873$

So total number of pages is $ 873\plus{}90\plus{}9\equal{}972$


Trong số thành viên cũ của đội tuyển Việt Nam, ai sẽ trở về IMO lần này nhỉ? Chắc chắn là không thể thiếu Le Hung Viet Bao rồi.

Nguyễn Minh Trường tham gia như thế nào em? Lâu rồi không thấy tin tức gì, chú Trường dạo này sao rồi?

[quote="pvthuan"]Trong số thành viên cũ của đội tuyển Việt Nam, ai sẽ trở về IMO lần này nhỉ? Chắc chắn là không thể thiếu Le Hung Viet Bao rồi.[/quote]
Bảo bây giờ làm ở đâu vậy Thuận? :D

[quote]Nguyễn Minh Trường tham gia như thế nào em? Lâu rồi không thấy tin tức gì, chú Trường dạo này sao rồi?[/quote]
Nó vẫn vào đây post bài mà? Mặc dù chẳng có tý Toán nào trong các post đấy cả! :P

Bảo đang theo học toán Lý thuyết tại đây http://www.trin.cam.ac.uk/.

Tự dưng lại thấy có tên mình ở đây.
Tại dạo này em quên hết tiếng Anh rồi (nói thế chứ em chỉ không nói được thôi chứ đọc và viết vẫn OK), nên vào đây chỉ đọc có 2 chỗ, là box Vietnam với France. :D Mà ở 2 chỗ đấy lấy đâu ra đất để viết cái gì liên quan tới toán học cơ chứ.
Hè này em có về Việt Nam cung với Bảo. Hình như có mấy anh đội tuyển từ cách đây nhiều năm cũng về thì phải, như anh Nhật Quang, NGô Đức Duy ...

À, hoá ra chú này đang học bên Pháp à? :D Chắc chú học cùng trường anh Tuấn khi xưa? Đợt này anh Duy về, không biết anh Tuấn có về không nhỉ? :)


Help me: does $ td(\mathbb{Q}(\sqrt2, \sqrt3, e^{\sqrt2}, e^{\sqrt3})/\mathbb{Q})$ equal 2?
And is it possible, that the field $ K$ algebraically dependent over $ \mathbb{Q}$ can be expanded (by adding to it some new numbers) to an algebraically independent field over $ \mathbb{Q}$?

Yes, the degree is $ 2$ (by Lindemann-Weierstrass).
I don't understand the second question. What does "field $ K$ algebraically dependent over $ \mathbb Q$" mean¿

Well I am not able to understand the notations. Could you please explain ?? :(

Sorry, I meant if the set of numbers $ K$ algebraically dependent over $ \mathbb{Q}$ can be expanded (by adding to it some new numbers) to an algebraically independent set over $ \mathbb{Q}$

Adding more numbers doesn't change the algebraic dependence of the starting numbers. Thus no.

Thank you, then what can be $ td({\mathbb{Q}(z_1, ..., z_n, e^{z_1}, ..., e^{z_n})/\mathbb{Q})}$ equal to, if $ z_1, ..., z_n$ are algebraically dependent over $ \mathbb{Q}$?


prove that: if the $\angle A, \angle B, \angle C$ of of a triangle satisfy, $cos3\angle A+ cos3\angle B+cos3\angle c$$=1$, then one of its angles is $\angle120$.

I am assuming you meant:
\[ \cos{3A}+\cos{3B}+\cos{3C}=1 \]
Proof:
Since: $\cos{\pi}=-1$, so we can rewrite the equation:
\[ \cos{3A}+\cos{3B}+\cos{3C}+\cos{\pi}=0 \]
\[ \Longleftrightarrow 2\cos{\frac{3A+3B}{2}}\cos{\frac{3A-3B}{2}}+2\cos{\frac{\pi+3C}{2}}\cos{\frac{\pi-3C}{2}}=0 \]
\[ \Longleftrightarrow \cos{\frac{3\pi-3C}{2}}\cos{\frac{3A-3B}{2}}+\cos{\frac{\pi+3C}{2}}\cos{\frac{\pi-3C}{2}}=0 \]
\[ \Longleftrightarrow \cos{\frac{3\pi-3C}{2}}(\cos{\frac{3A-3B}{2}}-\cos{\frac{\pi+3C}{2}})=0 \]
Now, if $\cos{\frac{3\pi-3C}{2}}=0$, we will have: $C=120^o$.

If $\cos{\frac{3A-3B}{2}}=\cos{\frac{\pi+3C}{2}}$, we will have $A=120^o$ or $B=120^o$.

QED.


You are given two perfectly straight metal rods, and a cutting tool that allows you to make “perfect
cuts” along the length of the rods. You do not have any tools that can be used for measuring, but
whenever you have two rods (either the original rods or pieces of the original rods) where one is shorter
than the other, you may mark off the length of the shorter rod on the longer one and then cut the longer
rod into two pieces–one being the length of the shorter rod, the other being the difference in length
between the two rods.
(a) Suppose the two rods you start with are of lengths five and seven centimeters. Explain how you
can use your cutting tool to get a rod one centimeter long.
(b) If you start with rods of length 39 and 51 centimeters, can you get a rod one centimeter long?
Explain.
(c) For which pairs m and n, of whole numbers, can you start with rods m and n centimeters long
and end up with a rod one centimeter long? Your goal here is to somehow characterize all pairs of
numbers that work; writing down a list of pairs will only serve as an answer if there are only finitely
many possibilities. In the event there are infinitely many possibilities, your solution should provide
some sort of description of pairs that work, and a detailed explanation of why your description
includes all, and only, the pairs that work.

How to start? By not posting this in Advanced Fields.

Try various (all?) pairs of $ m$ and $ n$ up to $ 10$ and see for which pairs you can get $ 1$ cm piece.


I am having a hard time understanding how heteroskedasticity affects a regression equation.

Can anyone explain to me the effects of heteroskedasticity on the estimates of the parameters and their variances in a normal regression model.

Thanks

Can you be a bit more specific about what exactly gives you trouble? (I mean, we can retell the corresponding passages from textbooks or send you to Wikipedia and Mathworld, but you've, most likely, tried those options already :) )

OK, so far this is what I know.

Maybe  I am thinking to much into the what heteroskedasticity represents.

The effects of heteroskedasticity on the estimates of the parameters in a normal regression model is that the OLS estimators are shown to be unbiased and consistent but not efficient.

The effects of heteroskedasticity on the variances of the estimates of the parameters in a normal regression model is that it does bias the variance of the estimated parameters, so the parameter variance estimates are biased.

This is what I can understand from reading texts, etc.

I guess my confusion comes from is it really that simple or is there more to it.

I would think that heteroskedasticity effects the parameter estimates and their variances more than just inefficiency and biasedness, but I am not sure what other outlying factors there would be on the parameters and the variances of the parameters.

If you have any other ideas of outside factors of heteroskedasticity and how it would effect the parameters and the parameters and their variances, I guess I am looking for more of a discussion in that sense.

Thanks.


I am making a MathCounts/AMC 8/AMC 10 Problem Solving Handbook.  Each page will be dedicated to a single problem, and it will have  a generalized method for solving that type of problem, as well as a specific solution.  It will then have one problem for the reader to try for himself.  Anyone want to help?

I have attached a sample page, so you can see what I need submitted.
You will get credit on the cover, if you contribute a whole page.
Please submit all material to mysmartmouth@gmail.com

This will probably be fun.  :P

Why not use tex to render it?  Why MS word?

Edit:
I think we should just post everything here.  It would allow peer editing and such.

Edit2:
I finished page 2.

This is a good idea, except you probably don't want to call it a handbook, since there already is a handbook. ;)

Also, mysmartmouth: your formula for arithmetic sequences is identical to the more common $a_n = a_1+d(n-1)$ where $a_n$ is the $n$th term and $d$ is the common difference. I think you should use this one because most people already know it, since they learned it in school in that form.

[quote="chess64"]
Also, mysmartmouth: your formula for arithmetic sequences is identical to the more common $a_n = a_1+d(n-1)$ where $a_n$ is the $n$th term and $d$ is the common difference. I think you should use this one because most people already know it, since they learned it in school in that form.[/quote]

Yes, but people who already know it don't need to read it.  I am designing this handbook for beginners.  While you have a valid point, I would find it much easier (in 7th grade) to understand [b]and memorize[/b] $a+dn-d=l$ rather than $a_n = a_1+d(n-1)$.  I have noticed that notation of the form $x_y$ readily confuses middle schoolers, though it is designed to make formulas easier.

But anyway, I am straying away from the topic of this forum.

If anyone wants to $TEX$ everything, feel free!  While I can use Latex relatively well on AoPS, I do not know the commands neccessary to typeset a, [b]clean, eye-appealing[/b] document in TeXnicCenter.

Okay, w/e...

About $\text{\TeX}$ing...it really looks much better in $\text{\TeX}$. My uncles all say that they used $\text{\LaTeX}$ to type their thesis, and it's commonly used by professional mathematicians. So don't diss $\text{\LaTeX}$ :P

Since this is a handbook, it might help to display a fast solution to the problem, as well.  For 2 gallons and 1 gallon, the 2 pull down and 1 pulls up, making the result $\frac{1}{3}$ of the way from 1 to 4, or 2%.  This method can be to speed up problems like these very quickly.


Um I think Aops books are sold in some stores, but I doubt it will be in major stores like Borders unless maybe you ask them to supply the books.

[quote="007math"]i was just wondering if you can buy AoPS books from bookstores such as Borders.  Can you?[/quote]  You can place special orders through Borders, but they'll just call us and order from us.  You'll get the books faster just ordering online through us.

So will their be a time where aops can supply its books to places like Borders like normal books?

I think AoPS should start selling their books to more nation-wide stores like Barnes and Nobles and Borders, I don't think many people have heard about any of the books they offer unless they have discovered this website, or been told by a friend or coach.  Selling them to stores would also increase income, right?

Not so sure about that, since most people who buy Aops books are really interested in math and if you are really interested in math then most likely you have already found this site.

not necessarily

i only found this website because my math team teacher told me about it

i would never have found it otherwise


I  have  installed  MikTex +Texniccenter+ Asymptote+Ghostscript+gsview.
And  I  would  like   to  try  MetaPost, so  I  have   some   question:

1.  how  to  install   MetaPost   in   Miktex?
2.   how to configure TeXnicCenter to quickly compile   the  MetaPost   code?
3.  Could  you  show  me  a  MetaPost   code,  and   how  to compile   the    code?

Many  thanks!

http://www.latex-community.org/forum/viewtopic.php?f=45&t=4859

But  I  cann't   find  [color=red] emp   [/color]package

Take a look at [url]http://wiki.contextgarden.net/MetaFun[/url] and at [url]http://melusine.eu.org/lab/bmp[/url].
Have a good read.

Thank  you very much!
I will  read  them  later.

first  I  must    find  out  how   metapost  works  in [color=red]  Miktex   [/color]and [color=red] ConText    [/color]and    write  a   simplest   code  to   get  the   picture.................


Rachel and Robert run on a circular track. Rachel runs counterclockwise and completes a lap every $ 90$ seconds, and Robert runs clockwise and completes a lap every $ 80$ seconds. Both start from the start line at the same time. At some random time between $ 10$ minutes and $ 11$ minutes after they begin to run, a photographer standing inside the track takes a picture that shows one-fourth of the track, centered on the starting line. What is the probability that both Rachel and Robert are in the picture?

$ \textbf{(A)}\ \frac{1}{16}\qquad
\textbf{(B)}\ \frac18\qquad
\textbf{(C)}\ \frac{3}{16} \qquad
\textbf{(D)}\ \frac14\qquad
\textbf{(E)}\ \frac{5}{16}$

10 minutes is 600 seconds and 11 minutes is 660 seconds. The interval of the track that is in the picture, letting 0 be the starting point, is -1/8 to 1/8. Therefore, Rachel is in the picture if the time is -11.25 to 11.25 mod 90, and Robert is in the picture if the time is -10 to 10 mod 80.

Therefore, Rachel is in the picture from 618.75 seconds to 641.25 seconds, but Robert is in the picture from 630 to 650 seconds. Therefore, they are both in the picture from 630 to 641.25 seconds, which is a time interval of 11.25 seconds. $ \frac {11.25}{60} \equal{} \frac {45}{240} \equal{} \frac {3}{16}\textbf{ (C)}$.

[hide]The picture will be taken between $ 600$ seconds and $ 660$ seconds and includes the eighth of the track on either side of the starting line.

Robert is at the starting line at all multiples of $ 80$, including $ 640$.  Running an eighth of the track takes him $ 10$ seconds, so he is in the area of photography between $ 630$ and $ 650$.

 Rachel is at the start on multiples of $ 90$, including $ 630$.  An eighth of the track takes her $ 11.25$ seconds, so she is available between $ 618.75$ and $ 641.25$.

 Both are in the picture from $ 630$ to $ 641.25$, so the answer is $ \frac{11.25}{60}\equal{}\frac{3}{16}\equal{}C$.[/hide]

[quote=fmasroor]I still don't understand why my solution is wrong:
90 seconds is 1.5 minutes, so in 10 minutes she will have done 10/1.5=6+2/3 laps, so she will be 2/3 into a lap. In 11 minutes, she will be 7+1/3 laps, so she will be 1/3 laps into the next one, so she has traveled 2/3 of the entire lap in that period.
Similarly, Robert travels 3/4 of the lap in the same period of time.
(1/4)/(2/3)=3/8
(1/4)/(3/4)=1/3
3/8*1/3=1/8 B.[/quote]

Sorry, this is kind of late, but the two people don't cross the section at the same time


consider a circle and an ngon inscripted in it.
select 3 consecutive nodes of the ngon
A,B,C(in this order they appear on the circle also).
A' is the middle of the arc AB and B' is the middle of the arc BC.
does there exist a side of the ngon with length A'B'?

You can show that the arc A'B' is the arithmetic average of arcs AB and BC because arc A'B' = arc A'B + arc BB' = 1/2 (arc AB + arc BC). If there's another side (call it DE) of the ngon with the same length as A'B', then arc DE = arc A'B' by the Law of Cosines. So what you're asking is whether there exist two adjacent vertices with an arc of arc A'B' between them - and the current information gives us no indication whatsoever of whether this is true or false.

If $A_1A_2...A_n$ is a regular n-gon, and $M_{12}$ and $M_{23}$ are the midpoints of the arcs $A_1A_2$ and $A_2A_3$ on its circumcircle, then $M_{12}M_{23}=A_1A_2$. The proof is simple: If O is the center of the circumcircle of the regular n-gon $A_1A_2...A_n$, then $\measuredangle A_1OA_2=\measuredangle A_2OA_3$ (since the n-gon $A_1A_2...A_n$ is regular), $\measuredangle A_1OM_{12}=\frac{\measuredangle A_1OA_2}{2}$ (since the point $M_{12}$ is the midpoint of the arc $A_1A_2$ on the circumcircle), $\measuredangle A_2OM_{23}=\frac{\measuredangle A_2OA_3}{2}$ (similarly), so that $\measuredangle A_2OM_{23}=\frac{\measuredangle A_2OA_3}{2}=\frac{\measuredangle A_1OA_2}{2}=\measuredangle A_1OM_{12}$. Hence, $\measuredangle M_{12}OM_{23}=\measuredangle M_{12}OA_2+\measuredangle A_2OM_{23}=\measuredangle M_{12}OA_2+\measuredangle A_1OM_{12}=\measuredangle A_1OA_2$. Thus, the chords $M_{12}M_{23}$ and $A_1A_2$ of the circumcircle of the n-gon $A_1A_2...A_n$ suspend the same angle at the center O of this circumcircle; thus, they must have equal length: $M_{12}M_{23}=A_1A_2$.

This was pretty obvious, so I am sure you meant something different. What exactly did you mean?

  Darij

uhm...that was what i meant darij :)
thank you for the wonderful proof:)
the problem i was trying to solve is this one
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=points+circle&t=73954
i can get nowhere with it,i was trying to use this observation
i asked here,but not for regular ngons but with random ngons to solve
the problem


Does anyone know what is meant by "a sequence is well defined"?

I am looking at a version of the Kantorovich Theorem (which is the multidimensional Newton Convergence Theorem) and one of the conclusions is that "the sequence produced by Newton's method, starting with $ x^0$, is well defined."

I can't seem to find a definition anywhere of what a well-defined sequence is.

Thanks!

The most direct interpretation is "the procedure works", at least to the point of getting a result. In the case of Newton's method, this means everything we need to invert is invertible (no division by zero).


Find the minimum and maximum value of $a\sin\theta+b\sin\theta$.

$a \sin \theta+b \sin \theta = (a+b) \sin \theta$

$-|a+b| \leq \sin \theta \leq |a+b|$

[b]Harder:[/b]

Find the minimum and maximum values of $a \sin \theta+b \cos \theta$

do you mean $b \cdot \cos \theta$?

[hide="If so... a hint."]
Cauchy.  :wink: [/hide]

[quote="10000th User"]If you use Cauchy, it's overkill.

Try a more elementary approach:

$a\sin\theta+b\cos\theta=\sqrt{a^{2}+b^{2}}\left(\frac{a}{\sqrt{a^{2}+b^{2}}}\sin\theta+\frac{b}{\sqrt{a^{2}+b^{2}}}\cos\theta\right)$

Let $\sin\alpha=\frac{a}{\sqrt{a^{2}+b^{2}}}$[/quote]

Or, even more overkill is [hide="a cannon"] \[a \cos \theta+b \sin \theta = \sqrt{ a^{2}+b^{2}}\cos \left( \theta-\arctan \left( \frac{-b}{a}\right)+k \pi \right), \ k \in \{ 0,1 \}\] [/hide]


A fair coin (one for which both the probability of heads and the probability of tails are 0.5) is tossed 60 times. The probability that less than 1/3 of the tosses are heads is ____?

[hide="What I Did"]
binomcdf(60,0.5,19)=0.003, but the answer key says 0.0043
[/hide]

A college basketball player makes 80% of his free throws. Over the course of the season he will attempt 100 free throws. Assuming free throw attempts are independent, what is the probability that he makes at least 90 of these attempts?

[hide="What I Did"]1 - binomcdf(100,0.8,89)=0.0057, but the answer key says 0.0062[/hide]

Thanks in advance!

My cumulative binomial tables only go up to $ n \equal{} 30$, so I used Normal approximations, getting $ 0.0034$ for the first question and $ 0.0067$ for the second. Sorry I can't be of much help.

Just found some more tables. Using $ n\equal{}100, p\equal{}0.2$ and looking at misses rather than goals as the probabilities in my table only go up to $ 0.5$, I also get $ 0.0057$ for your second question.

For the first one,
I would use
$ \frac {\binom {60}{0} \plus{} \binom {60}{2} \plus{} \binom {60}{3} \plus{}...\plus{}\binom {60}{19}}{2^60}$
Since I need to add each of the probablities 0 heads, 1 heads etc. seperatedly. Is there an easier solution?


N is the point of the side AC of triangle ABC such that AN = 2NC and M is the point of the side AB such that MN is perpendicular to AB. Knowing that AC = 12 cm and the centroid G of triangle ABC belongs to the segment MN, determine the length of BG.

Portuguese question:

seja N é o ponto do lado AC do triângulo ABC tal que AN=2NC e M o ponto do lado AB tal que MN é perpendicular a AB .
sabendo que AC é igual a 12cm e que o baricentro G do triângulo ABC pertence ao segmento MN,determine o comprimento de BG.

Let $ K \equal{} BG\cap AC$.So $ K$ is the midpoint of $ AC$.In order to find the lenght of $ BG$ it suffices to find the lenght of $ BK$.
From Stewar't theorem on $ \triangle ABC$ and $ BK$ we have that:
$ AC(BK^2 \plus{} AK \cdot KC) \equal{} BC^2 \cdot AK \plus{} AB^2KC \Longleftrightarrow 2(BK^2 \plus{} 36) \equal{} AB^2 \plus{} BC^2$ $ \boxed {1}$.

Let $ F \equal{} AG\cap BC$.
Then $ \frac {AG}{AF} \equal{} \frac {AN}{AC} \equal{} \frac {2}{3}$.
So the lines $ NM$ and $ BC$ are parellel.
So $ \angle ABC \equal{} 90$.
So $ AB^2 \plus{} BC^2 \equal{} AC^2 \equal{} 144$.
So from $ \boxed{1}$ we have that $ 2(BK^2 \plus{} 36) \equal{} 144 \Longleftrightarrow BK \equal{} 6$.
So $ BG \equal{} \frac {2}{3}BK \equal{} 4$.

Done.

Ok, but how do you know that $ \angle ABC \equal{} 90$ ?

I wrote that:
Let $ F \equal{} AG\cap BC$.
Then $ \frac {AG}{AF} \equal{} \frac {AN}{AC} \equal{} \frac {2}{3}$.
So the lines $ NM$ and $ BC$ are parellel.

We know that $ NG$ which is actually $ NM$ is perpendicular $ AB$.

We also know that  a line which is perpendicular to one of parallel line it is also peprendicular to the other :wink:


limit
trigonometry
ratio
analytic geometry
parameterization
function

No!
A classical one. I'll introduce you to a nice way .Let y=1*x,then  the result will be clear.
Check y=f(x)*x,where lim f(x)=1 as x goes to 0.
For instance f(x)=e^x ,(1-x),... .The latter will do the trick for us.

Thank you for your reply, "Captain".

The idea is very nice but when I tried y=x(1-x) I got \[
\mathop {\lim }\limits_{x\, \to \,0} \,\,f\left( {x\,,\,\,x\left( {1 - x} \right)} \right) = 0 = f\left( {0,0} \right)
\]   Therefore, if I am not wrong, it seems to me that the problem is still "open" !

P.S.: I did think about some "intrinsic relationship" between x and y to find "problematic sequences" but... I must confess it did not last long till I started trying my (impossible?) mission of proving the continuity at this point!

I just tried to give you a hint ,and I thought it would help.
Now consider this one \[ y=(1-x)x^\frac{3}{2} \]. This will lead us to the desired result.

Very nice solutions indeed.     :)

Thank you both, Captain and Jimmy.

Dear Captain,

Although the "r" missing in Jimmy´s  calculations "collapses" his solution (perhaps it could be fixed with some $ \alpha $ in $ r^{\,\alpha } \cos \theta  $ , "analogously" to your solution ?!) ,  I do not understand why the existence or not of uniform convergence "breaks" his reasoning (it certainly "disturbs" it but... does it break?) . Let me explain you why (and please reply about it):

IF his trigonometric ratio were right, let us say that for some pre-fixed $
\theta _1  \ne \theta _2  $ we got two sequences \[
\left( {r\cos \theta _{\,{\rm{i}}} \,\,,\,\,r\,sen\,\theta _{\,{\rm{i}}} } \right)\mathop  \to \limits_{r\,\, \to \,\,0} \left( {0,0} \right)\,\,\,\,\,\,\left( {{\rm{i}} = 1,2} \right)
\]   Although the limits themselves $ \left( {f\left( {r\cos \theta _{\,{\rm{i}}} \,\,,\,\,r\,sen\,\theta _{\,{\rm{i}}} } \right)} \right)_{\,r\,\, \to \,\,0}
\, \, \, \, $  could NOT BE $ \, \, \, \, \,
 - \frac{{\cos ^2 \theta _{\,{\rm{i}}} \,}}{{sen\,\theta _{\,{\rm{i}}} }} $  (yes, because of non-uniform convergence!) , couldn´t we argue that :

1) If one of them (at least) does not exist , then the discontinuity of f at the origin  follows ($
{\theta _{\,{\rm{i}}} }
$ is constant!)

2) If both of them exist, then they are different and the same conclusion applies?

Well... It seems to me that my initial question is much more interesting than I could suspect!      :)

Thank you both, Captain and Jimmy. I am looking forward to "hearing" from you!

Regards,
fskilnik

[quote="The Captain"]Dear Jimmy.You're not suggesting the right thing,I'm afraid.
First of all the statement you have mentioned, Does Not hold in general.
(If you wanna see why ,you can refer to articles about "Uniform Convergence".)
Another point is that $x^2y=r^3\cos^2\theta\sin\theta$  !! (A typo).[/quote]

Thanks for pointing that out, I'll look up uniform convergence.

Just one question, when I fixed the typo, the limit tended to 0, so is the polar substitution method not always correct? And if so how do you know?

Thanks.

I start with Jimmy's question.You see, $\theta$ is not a constant; it's a variable depending on $x $ and  $ y$.
And you want the limit not just as $r$ goes to zero, but on small disks around the pole,because you are
taking a two-dimensional limit.
You said you got $0$.The reason is that a limit of $L $as$ x $and $y $approach zero means that $f(x,y) $is near $L$ whenever $x $and $y$ are BOTH near zero.  Well,$x$ and$ y$are both near zero exactly when $r$ is near zero, but $\theta$ can have any value.
Now fskilnik question.
If the limit on the right exists for every $\theta$, that is not enough to say that the limit on the left exists.
If the limit on the right exists for every theta and is independent of$\theta$ (i.e. it's always the same limit), then that is still not enough to say that the limit on the left exists.  But if the limit on the right exists uniformly for all $\theta$, then it is the same as the limit on the left.  That is, the $\delta$ value for a given $\epsilon $can't
depend on $\theta$.

Dear Captain,

Thank you for your response. I agree with most of it, but I am afraid you missed my point. Please let me explain it a bit more and please answer it whenever you have time.

First of all, let me discuss Jimmy´s question and your argument!

Jimmy: you may use polar coordinates "whenever" you want because all sequences
$  \left( {x,y} \right) \in R^2 $ are related to some sequence(s) $ \left( {r,\theta } \right) \in \,\,R_ +   \times \,\,\left[ {0,2\pi } \right[  $ and reciprocally.

In other words, you are able to converge to the origin using rectangular coordinates or polar ones and topologically speaking there is no "preferences". We use one or the other (or others) according to the calculations involved!

The "problem" with your solution is that you got 0 (that´s right) and $ f\left( {0,0} \right) = 0 $  (by definition!) , therefore you could NOT prove discontinuity. The Captain proved it (discontinuity) , because the limit involved in his suggestion did not exist and SHOULD be zero  to make continuity POSSIBLE. (This will be clearer in my comments to the Captain.)

I thought you got the discontinuity correctly proved, because I did not check your calculations till The Captain mentioned the "r" missing but, as I will explain now, if that ratio (you got) was right, you would have done it.

Captain:

Your phrase

[quote]  $ \theta $ is not a constant ... but on small disks around the pole,because you are taking a two-dimensional limit.
[/quote] is perfect, of course. That means that we could not prove CONTINUITY with Jimmy´s argument because we should "change both polar parameters" , for sure!

But... I understand Jimmy tried to prove DISCONTINUITY, proving (using polar coordinates) that with 2 FIXED directions ($ \theta _{{\rm{ i}}} $´s  fixed) , approaching the origin by two different straight lines ($ r \to 0_ +  $) , he could get two different limits (or proving at least one does not exist) the conclusion would follow.

By the way, I should have imagined that his solution could not be right because I had already tried all lines passing through the origin...   :blush:  (Jimmy, please try $
y = mx $ with m fixed and afterwards, deal with the vertical line - y axis - separately!)

Well, I guess that´s all.

Thank you both for all these matters!

Regards,
fskilnik.

Thank you for your explanations.  My math teacher just told us that if the limit as $r\to0$ cnverged to a constant, then that was the limit, and I never really questioned it, obviously I should have. :)  And just to clarify, I was right to say that a limit depending on $\theta$ did not converge (which interestingly my math teacher never mentioned), right?

Thanks again.

Hello again, Jimmy!

The answer to your question is "right!" (I mean "yes") , in the sense that "if the limit exists, it is necessarily unique".  (Limits don´t "converge/diverge", they exist  or not!)

This problem is closely related to the following

Theorem: A function $ \, \, f: R^{\,k}  \to R^{\,m} \,\,\,\,\,\left( {for\,\,us: \,\,\,k = 2\,,\,\,m = 1} \right) \, \, $ is continuous at a certain given point $ \,
a \in R^{\,k} \, \, $  if and only if $
\mathop {\lim }\limits_{n\,\, \to \,\,\infty } f\left( {x_n } \right)\,\, \to f\left( a \right)
 \, \, $ ,  for all sequences $ \,
\left\langle {x\,_n } \right\rangle _{n\,\, \ge \,\,1} \, \, \, $ in $\, \, R^{\,k} \, \, \, \, $ such that $\, \,
\mathop {\lim }\limits_{n\,\, \to \,\,\infty } x\,_n \,\, = \,\,a  \, \,
$

Anyway, The Captain´s arguments are very important/instructive. You will learn a lot studying the "uniform converge" theme he suggested! (Before that, start with "pointwise-convergence" topics that you can find in good analysis books!)

All the best,
fskilnik.


can you post the results of China NMO 2009 please

Is this year's problems harder than last year's?

according to the last year cut-offs, a gold medal was awarded for >81 points if I am not mistaken, so it is logical to assume that this year problems are slightly easier.

[quote="Fang-jh"]There are eleven perfect scores (126 points).
Cut-offs
Gold medal: 93 points. (46 participants)
Silver medal: 51 points.(75 participants)
Bronze medal: 6 points. (all participants except for gold and silver medals. 78 participants)[/quote]

If the perfect score is 126, does that mean that you're given 18 problems at your NMO ? :shock:

[quote="Inequalities Master"][quote="Erken"]No, each problem is worth 21 points[/quote]

Strange, what's so special about this number ?  Or that number is chosen instead of 7, because there are many participants who can high scores?[/quote]

I believe each problem on CMO is worth as three problems in subsequent quizzes in the selection for IMO team.

[quote="Aravind Srinivas L"]Just 6 points are needed for Bronze medal? Too bad, it must be 26 or something maybe a typo.
And yeah, ChNMO 2009 is definitely easier than 2008.[/quote]

not a typo. essentially all participants got at least a Bronze

I really can't believe that this is the National Olympiad of China. 3rd question is just a simple calculation, 5th one is known (Russia) and I got a strong idea about 4th one is known too. It's the real National Olympiad, right?

well ya, so no wonder there are more than 10 full scorers...

Probably next year's 'll be tougher...


probability
conditional probability
probability and stats

Hello everyone!

I have this kind of problem:
[quote]there is a markov chain with n states and the probabilities of reaching each of them, in other words: I have the transitive matrix of this chain. Let T be the time after which the first state [which is out starting state] reaches the last one for the first time. How can I compute the expected value of T?[/quote]

this sounds like a standard excercise from the 'stochastic process' course to me, but I cant remember what should one do in this kind of problem. Similar or equivalent state of problem is using the term 'random walk':
[quote]we have a stochastic process with n states. for each state, there is a certain probability for going 2 steps back, 1 step back, for leaving where it is and for 1 and 2 steps further. we start at the first state. what is the expected value of the time that the process will reach the last state for the first time?
[/quote]
I would be really grateful, if someone could help me with this one.

I haven't computed the exact probabilites, cause i have problems with dealing with conditional probability, but i'm 100% positive, that from each state the process can jump only 1 or 2 states back or further, or stay where it is. i cant tell whether this information is helpful, I hope the solution is pretty trivial.

second thing I'd like to ask you will help me compute the missing data:

[quote]
we have a red ball and a black ball. we pick one at random. the probability that it is the red one equals p. If there were K red balls and L black balls, what will be the probability that taking randomly 2 balls, we would get one red ball and one black ball?[/quote]
I'm aware that this is really elementary, but I never got into that bayes's formulas, which i need to use here, right?

thanks for your time,
rahl.

For the first question, you just consider the expectations $ T_i$ of the times required to reach the $ n$-th state from the $ i$-th state. You have $ T_n\equal{}0$ and $ T_i\equal{}1\plus{}\sum_j p_{ij}T_j$ for $ i\ne n$. It remains to solve this linear system.

The second question is just a partial case of the first one.

The last question is hard to understand. Do you mean that regardless of the numbers $ K$ and $ L$, the probability to pick a red ball is $ p$, or do you mean that all balls can be picked with equal probability? Also, do we pick two balls at random simultaneously, or we take out a ball, look at it, throw it back to the bag and then pick the second one? Or do you mean something else? As written, it just doesn't seem to make much sense.

[quote]For the first question, you just consider the expectations $ T_i$ of the times required to reach the $ n$-th state from the $ i$-th state. You have $ T_n \equal{} 0$ and $ T_i \equal{} 1 \plus{} \sum_j p_{ij}T_j$ for $ i\ne n$. It remains to solve this linear system.

The second question is just a partial case of the first one.
[/quote]
big thanks! that was exactly what I was looking for.
[quote]The last question is hard to understand. As written, it just doesn't seem to make much sense.[/quote]
i know, sorry. it turned out, that the real problem was something different from what i wrote and i've already solve it, so nevermind.

thanks once more!
rahl


What is the maximum number of electrons in an atom that can have the following quantum numbers:
a) n=2, spin quantum number = -1/2
b) n= 5, l = 3
c) n = 4, l = 4, magnetic quantum number = -3
d) n=4, l=1, magnetic quantum number = 1

For a principal quantum number $ n$, the azimuthal quantum number $ l$ can take on the values $ 0,...,n\minus{}1$, the magnetic quantum number $ m_l$ can take on the values $ \minus{}l,...,l$ (that is, there are $ 2l\plus{}1$ orbitals for a given $ l$), and the spin quantum number $ m_s$ can be either $ \minus{}1/2$ or $ \plus{}1/2$ (that is, only a pair of electrons can occupy a single orbital).

(Or you could remember that $ l\equal{}0$ corresponds to an s orbital, and there is 1 per energy level, $ l\equal{}1$ corresponds to the 3 p orbitals, $ l\equal{}2$ corresponds to the 5 d orbitals, $ l\equal{}3$ corresponds to the 7 f orbitals.)
[hide]A. So, if $ n\equal{}2$, then $ l$ can be either 0 or 1, and thus, there are $ (2\times0\plus{}1)\plus{}(2\times1\plus{}1) \equal{} 4$ orbitals, and we want one from each, so 4 electrons.

B. $ l\equal{}3$ implies $ 2\times3\plus{}1 \equal{} 7$ orbitals, two electrons in each orbital, so 14 electrons.

C. If $ n\equal{}4$, then $ l$ can not be 4, so no electrons.

D. If $ n\equal{}4$,$ l\equal{}1$, and $ m_l\equal{}1$, that defines one orbital uniquely, and any particular orbital can have 2 electrons.[/hide]

[b]Pianoforte[/b], I think you may want to revise your answers. The question asked the number of electrons in the atom, not on the orbitals.


A magician deposits the same numbers of rabbits (at least one) at each of five houses. To get to the first house he crosses a magic river once, and to get to any house from another, he also crosses a magic river once. Each time he crosses a magic river, the number of rabbits he has doubles. He has no rabbits left when he leaves the fifth house. What is the minimum number of rabbits he could have.

[hide]Let n=the number of rabbits to begin with and k=the number deposited at each house.

He starts with n rabbits.
After the first house he has 2n-k rabbits
After the second house he has 2(2n-k)-k rabbits
...
After the fifth house he has 2(2(2(2(2n-k)-k)-k)-k)-k rabbits

Multiplying out and setting equal to 0 we obtain 32n-31k=0
32n = 31k

The lowest integer value of n which satisfies this equation is [b]n=31[/b] (this occurs when k=32)

In fact we can generalise this result. The expression for the number of rabbits after visiting x houses is:

(2^x)n - ((2^x)-1)k

Sorry if it was hard to read but i don't know any latex...[/hide]

Let $ r$ be the number of rabbits he starts off with, and let $ h$ be the number he gives away at each house. After crossing the first river he has $ 2r$ rabbits. Then he gives away $ h$, so he has $ 2r \minus{} h$ rabbits. Then he crosses another magic river to the second house. He has $ 4r \minus{} 2h$ by the time he reaches this house. Then he gives away $ h$ rabbits to this house. Now he has $ 4r \minus{} 3h$ rabbits. After crossing the third magic river, he has $ 8r \minus{} 6h$ rabbits. Then he gives away $ h$ rabbits to the third house. He now has $ 8r \minus{} 7h$ rabbits. Then he crosses the fourth river and ends up with $ 16r \minus{} 14h$ rabbits. Then he gives $ h$ away at the fourth house and he is left with $ 16r \minus{} 15h$ rabbits. Then he crosses the last river and ends up with $ 32r \minus{} 30h$ rabbits and gives away $ h$ rabbits again and ends up with
$ 32r \minus{} 31h$ rabbits. We are told that this is equivalent to zero so;

$ 32r \minus{} 31h \equal{} 0$

$ 32r \equal{} 31h$

$ r \equal{} 31h/32$

So as long as $ r$ is a multiple of $ 31$ and $ h$ is a multiple of $ 32$, then it's a possibility. The smallest multiple of these numbers are $ 31$ and $ 32$ respectively, so the answer is $ 31$.

Am I misunderstanding the question? Why can't he start out with 1 and give away 1 each time? After the 1st river he has 2 and then deposits 1, so he has one left. Then going to the second house it doubles, so he now has 2, and so on.

He'll be left off with one rabbit at the end, and he is supposed to get rid of all the rabbits.


[hide="Solution"]Because the common difference between each terms is 1, we get $2(a_2+a_4+a_6+a_8+\ldots+a_{98})$ by adding half the number of terms to the original sequence. For example, if we add 1 to $a_5$, we get $a_6$, and etc.

Thus, our answer is $\frac{1}{2}\left(137+ \frac{98}{2}\right)=\frac{186}{2}=\boxed{93}$.[/hide]

When I was doing this problem, I tried to find the first term $a_1$ by using the formula

$\frac{98(a_1 + a_1 +97)}{2} = 137$, but I get a $-\frac{4616}{98}$

$\frac{97({-\frac{4616}{98} + 1} -\frac{4616}{98}+96)}{2} = 135.602$

This is not the answer as it is AIME and must be an integer from 0 - 999.

I'm not sure. :?

I tried that at first also, but then used my first solution after the number turned out weirdly.

[quote="breez"][hide]Ok I got it. The solution was correct after all:

$\frac{98(a_1 + a_1 +97)}{2} = 137$, thus $a_1 = -\frac{2308}{49}$

$49\frac{(-\frac{2308}{49} + 1 -\frac{4616}{98} + 96)}{2} = \boxed{93}$[/hide]

However, 4everwise, your solution is more simple and elegant.[/quote]Cool. It worked. :)

Thanks for the compliment. I thought that the Kalva solution was uncool.  ;)  :P

4everwise, cool solution!

But I have an alternate version. This is simlar to what breez did but I never found what $a_1$ was (I could but I didn't do it).

[hide="Solution"]

If $a_1$ is the first term, then $a_1+a_2+a_3 + \cdots + a_{98} = 137$ can be rewritten as:

\[ 98a_1 + 1+2+3+ \cdots + 97 = 137  \]
\[ 98a_1 + \frac{97 \cdot 98}{2} = 137  \]

Our desired value is $a_2+a_4+a_6+ \cdots + a_{98}$ so this is:

\[ 49a_1 + 1+3+5+ \cdots + 97  \]

which is $49a_1+ 49^2$. So, from the first equation, we know $49a_1 = \frac{137}{2} - \frac{97 \cdot 49}{2}$. So, the final answer is:

\[ \frac{137 - 97(49) + 2(49)^2}{2} = \fbox{93}  \] [/hide]

[quote=4everwise]Find the value of $a_2 + a_4 + a_6 + \dots + a_{98}$ if $a_1$, $a_2$, $a_3$, $\dots$ is an arithmetic progression with common difference 1, and $a_1 + a_2 + a_3 + \dots + a_{98} = 137$.[/quote]

you can write it like a_1 + a_1+1+a_1+2+...+a_1+97 = 137
so then 98a_1 + 4753=137 so then a_1 = -4616/98
then you can just use that and solve for a_2 and do and a_4 and then use arithmetic sequence formula thing


I have a similar solution. From sine law $\frac{a}{2\sin{A}}=R$, substituting we get $b+c=2\sin{A}\sqrt{bc}$
But from $AM-GM$ we have $b+c\geq 2\sqrt{bc}$, hence $\sin{A}=1$ or $A=90^{o}$ and moreover $b=c$, giving $B=C=45^{o}$

[quote="N.T.TUAN"]The circumradius $R$ of a triangle with side lengths $a, b, c$ satisﬁes $R =\frac{a\sqrt{bc}}{b+c}$. Find the angles of the triangle.[/quote]
$R =\frac{a\sqrt{bc}}{b+c}\leq\frac{a}{2}$
$2R\leq{a}$ <=>\[ A=90{o}\]   \[B=C=45{o}\]

here is my solution its completely different one

R=abc/4S where a,b,c denote the side lengths and S denote the area. equating the two relations we get
S=(bc)^1/2(b+c)/4 and let x be the angle between the side b and c and using the formula s=bc/2 sinx
(B+C)/2=(BC)^1/2 sinx so sinx =(y+1/y)/2 where y=(b/c)^1/2  sinx x cant be negative since 0<x<180 so the only place where y+1/y is less than or equal to 1 at y=1 so sinx=1 so b=c and angle between them is 90. so the angles are 45,90,45

Here is nice solution.
From the sine law we get that $\frac{a}{\sin\angle BAC}=\frac{2a\sqrt{bc}}{b+c}$, after simple factorization we get that
 $\sin\angle BAC=\frac{(b+c)\sqrt{bc}}{2bc}$, $\cos \angle BAC=\sqrt{1-\sin^2 \angle BAC}$,
 $\cos \angle BAC= \sqrt{1-\frac{(b+c)^2 bc}{4b^2c^2}}=
\sqrt{\frac{4b^2c^2-(b+c)^2 bc}{4b^2c^2}}=
\frac{\sqrt{bc(4bc-(b+c)^2)}} {2bc}=
\frac{\sqrt{bc(2bc-b^2-c^2)}} {2bc}=
\frac{\sqrt{-bc(b^2-2bc+c^2)}} {2bc}=
\frac{\sqrt{-bc(b-c)^2}} {2bc}$
$bc$ and $(b-c)^2$ are always greater or equal than $0$ and $-bc$ and $-bc(b-c) ^2$ will always be less than $0$ $\implies$ $(b-c)^2=0$ or $b=c$. Since $b=c$, $\cos\angle BAC=0$ and $\angle BAC=90^\circ$ and $\angle ACB=\angle ABC=45^\circ$ :)

for storage:

Consider triangle $ABC$ with sidelengths $AB=c$, $AC=b$ and $BC=a$.
Let $M$ be the midpoint of $BC$. Let $O$ be the circumcentre of triangle $ABC$.

Consider triangle $CMO$: Since $MO\perp BC$, we get that $$\frac{CM}{CO}=\cos{\angle MCO}\implies \cos{\angle MCO}=\frac{b+c}{2\sqrt{bc}}\geq 1,$$ by AM-GM. Since $\cos$ is always $\leq 1$, we must have $\cos{\angle MCO}=0\implies \angle MCO=0^{\circ}$. Thus, $O$ lies on $BC$ and since we have equality on AM-GM, we get that $b=c$. Since $O$ lies on $BC$ and we have $OB=OC$, we get that $M=O$ and since $BC$ is the diameter of $ABC$, we get that $\triangle ABC$ is right triangle. Since $AC=b=c=AB$, then $\triangle ABC$ is isosceles right triangle.

Answer. $\boxed{\triangle{ABC}\text{ is a triangle with the angles } 90^{\circ},45^{\circ},45^{\circ}.}$

[quote=nayel]since $\text{area}=\frac{abc}{4R}=\frac{\sqrt{bc}(b+c)}{4}=\frac{1}{2}bc\sin A\le \frac{bc}{2}$ we obtain $\frac{b+c}{2}\le \sqrt{bc}$ hence $b=c$ and $\sin A=1\Longrightarrow \angle A=90^{\circ}$ so $\angle B=\angle C=45^{\circ}$ :D[/quote]
[b]Very good.[/b]


suppose we color each point in the plane with one of three colors red, green and blue.  Prove there exists a rectangle in the plane with all four of its verticies the same color.

For a slightly harder challenge: suppose we color every point of an MxN grid one of three colors red, green, blue.  What can you tell me about the smallest possible size of this grid that guarantees that we have a rectangle with all four of its verticies the same color?

[hide]In the first part: There are infinite number of points, so there exist a 4*19 rectangle. we will prove that there is a rectangle with all 4 vertices the same color.

A row of 4, by pigeonhole principle, we know at least 2 points in the row have the same color. Every row there are at least 2 points of same color. There are 4C2 ways to arrange 2 points of same color in a row of 4. There are 3 colors. there are total 3(4C2)=18 ways of 2 same color. In the 19th row, pigeonhole principle again, we will have a repeat.

I will take a guess that's 4*19=76= smallest size of the grid.[/hide]

What is the definition of "small" for a grid?[/hide]

[quote="zscool"][quote]What is the definition of "small" for a grid[/quote]

my bad, just minimize M and N, and I believe it can be done on a grid smaller than 4x19[/quote]

I think what JBL means is, which is smaller (for example), 4x19 or 3x20? (not saying that 3x20 works)

yea i guess that was really confusing (this is why I am a bad question writer), I think everyone realizes that it cant be done if one of N or M is < 4, so how about this:

what is the smallest N such that if every point on a 4 x N grid is colored one of the three colors red, blue or green,  it is guaranteed that we can find a rectangle with all four verticies the same color.

Then I am pretty sure N is 19

for N :le: 18  Bis blue, R is red, G is green , here is the counterexample

[hide]BBRG

BRBG

BGRB

RBBG

GBRB

RGBB

RRBG

RBRG

RGBR

BRRG

GRBR

BGRR

GGBR

GBGR

GRBG

RGBG

BGGR

RBGG[/hide]

Then I am pretty sure N is 19
for N :le: 18  Bis blue, R is red, G is green , here is the counterexample
BBRG
BRBG
BGRB
RBBG
GBRB
RGBB
RRBG
RBRG
RGBR
BRRG
GRBR
BGRR
GGBR
GBGR
GRBG
RGBG
BGGR
RBGG

my bad, i guess that is the smallest value

Well, we can extend your question: for M = 5, 6, ..., 18, compute N, the smallest number for which an MxN 3-colored grid must contain a smaller monochromatic rectangle.

Has anyone considered the posibility of a diagonally placed rectangle.
As an example, a rectangle with vertecies at (0,0), (2,1), (0,5), (-2, 4)


What is the solution to $ (1 \plus{} a_1)(1 \plus{} a_2)(1 \plus{} a_3)\cdots(1 \plus{} a_{10})$ if $ a_1,a_2,\cdots,a_{10}$ are the imaginary solutions to $ x^{11} \minus{} 1 \equal{} 0$?
Thanks.

[color=darkblue]Yeah, if you even kid about the AMC you are subjected to shameful edits. [i]~Mod.[/i][/color]

im guessing roots of unity.
i have to start reviewing complex numbers. grr.

No. Try Vieta's.  Expand that and you get $ 1+a_1+a_2+\hdots+a_10+a{}_1a{}_2+\hdots$ yep =).  Notice that the only term that isn't 0 would be the 1 and the $ a{}_1a{}_2%Error. "hdotsa" is a bad command.
_10$ xD. Also..... $ a{}_1a{}_2%Error. "hdotsa" is a bad command.
{}_10a{}_11=1$ I think. The real solution is 1 by obviousness =P.  So... yep xD


calculus
integration
geometry
derivative
analytic geometry
calculus computations

[quote="kunny"]2 curves $ y \equal{} x^3 \minus{} x$ and $ y \equal{} x^2 \minus{} a$ pass through the point $ P$ and have a common tangent line at $ P$. Find the area of the region bounded by these curves.[/quote]

derivative of $ y\equal{}x^3 \minus{} x$ is $ 3x^2\minus{}1$,derivative of $ y \equal{} x^2 \minus{} a$ is $ 2x\minus{}1$,let $ 3x^2\minus{}1\equal{}2x\minus{}1$,we get $ x\equal{}0,x\equal{}\frac{2}{3}$,it's easy to check that these two curves have a common tangent line at $ (\frac{2}{3},\minus{}\frac{10}{27})$,we can get $ a\equal{}\frac{22}{27}$,through computation ,we can obtain x-coordinates of  inersection points of the two curves are $ \frac{2}{3},\frac{1}{6}\minus{}\frac{\sqrt{5}}{2},\frac{1}{6}\plus{}\frac{\sqrt{5}}{2}$

so   the area of the region bounded by these curves is $ \int_{\frac{1}{6}\minus{}\frac{\sqrt{5}}{2}}^{\frac{2}{3}}(x^3\minus{}x\minus{}x^2\plus{}\frac{22}{27})dx\plus{}
\int_{\frac{2}{3}}^{\frac{1}{6}\plus{}\frac{\sqrt{5}}{2}}(x^2\minus{}\frac{22}{27}\minus{}x^3\plus{}x)dx\equal{}\frac{13}{12}$

Regrettably, the point of tangency is wrong.

I get the point of tangency as $ P ( 1,0 )$ ,and the area is = $ \frac{4}{3}$

The answer of area is correct, but we have another point of tangency.

Ya the other point of tangency is $ ( \frac{\minus{}1}{3} , \frac{8}{27} )$ , The answer is the same when we use either of these , The fact I used the former one was that the computation was less a pain  :P

That's right. I can guess how you found the area in your head without using paper and pen. :lol:

I`ve got a=1 or $ \frac {\minus{}1}{3}$
my answer is $ \frac {4}{3}$ anyway.
 :maybe:  :maybe:


inequalities
number theory unsolved
number theory

Show that for all $n\in N$, $\{n\sqrt7\}> \frac{3\sqrt7}{14n}$.

We have $\{n\sqrt 7\}=n\sqrt 7-[n\sqrt 7]$ and $7n^{2}-[n\sqrt7]^{2}=(n\sqrt 7-[n\sqrt 7])(n\sqrt 7+[n\sqrt 7])$. Since $7n^{2}-[n\sqrt 7]^{2}$ is an integer and $[n\sqrt 7]^{2}\equiv_{7}0,1,4$ we have $7n^{2}-[n\sqrt 7]^{2}\geq 3$. Therefore $\{n\sqrt 7\}\geq \frac{3}{n\sqrt 7+[n\sqrt 7]}>\frac{3}{2n\sqrt 7}$.


Let $ n \equal{} 11...1$, where there are $ 2007$ $ 1$'s. Determine whether there exists a function $ f$ such that for all real $ x\neq 0,1$, $ f(f(...f(f(x))...)) \equal{} (1 \minus{} \frac {1}{\sqrt [n]{x}})^n$, where in the left side, there are $ n$ $ f$'s.


I need someone to explain this to me.

The formula is
A number makes the remainder r when it is divided by p.
A= p x q + r , (0 less than or equal to r less than p)
p: divisor q: quotient r: remainder

or

A divided by p = q+ (r divided by p)

And the question is: A number makes the remainder 4 when divided by 5. What's the remainder when square of the number is divided by 5.
A= 5q +4
A^2= (5q+4)=25q^2+40q+16=5(5q^2+8q+3)+1
Therefore the remainder is 1.

Please guide me through this step by step. I'm so lost And give me one more example T.T
Thank you!!!

Well, an easy way is to pick a number with remainder 4, like the number 4.

4 squared is 16, remainder 1.  :)

Generally, any number that leaves the remainder 4 when divided by 5 can be written as $A=5q+4$, where $q$ is an integer.

[hide="reason"]
$5q$ is divisible by 5 for any number $q$.
Therefore, $5q+4$ always leaves the remainder 4 for any number $q$.[/hide]

Then, $A^2=(5q+4)^2=25q^2+40q+16=5(5q^2+8q+3)+1$
Again, $5(5q^2+8q+3)$ is divisible by 5 for any $q$, so the remainder is always 1.

Actually I didn't add much explanation to the original one... but I hope this helps anyway.


Compute exactly:

$\sum_{k=1}^{\infty}\,\tan^{-1}\left(\frac{1}{2k^{2}}\right)$.

[hide="Hint"]Telescoping Series.  :D [/hide]


This is from Jonah S.'s Problem Solving Group at Mathpath.

$\frac{1^2}{6}+\frac{2^2}{6^2}+\frac{3^2}{6^3}+\frac{4^2}{6^4}+...=?$

$\frac{F_1}{4}+\frac{F_2}{4^2}+\frac{F_3}{4^3}+\frac{F_4}{4^4}+...=?$
Where $F$ stands for the terms in the Fibonacci Sequence. $F_1=1$, $F_2=1$

for some reason i got $\displaystyle\frac{52}{125}$

$S = \displaystyle\frac{1^2}{6} + \frac{2^2}{6^2}+\frac{3^2}{6^3} + ... $

$6s = 1^2 + \displaystyle\frac{2^2}{6} + \frac{3^2}{6^2} + ... $

Subtracting:

$5s = 1 + \displaystyle\frac{3}{6} + \frac{5}{6^2} + \frac{7}{6^3} ... $

Multiplying this by 6:

$30s = 6 + 3 + \displaystyle\frac{5}{6} + \frac{7}{6^2} + ... $

Subtracting by $5s$ you get

$25s = 8 + 2(\displaystyle\frac{1}{6} + \frac{1}{6^2} + ... ) \Rightarrow 25s = 8+ 2(\displaystyle\frac{6}{5})$

$s = \displaystyle\frac{52}{125}$

Wow. How did you guys learn to do hard sums like these? The only sums I can ever do are geometric or arithmetic... :D

The solution to almost any infinite sum like those can be found by trying to find a repetition of the sum within a subset of the whole sum itself.

[quote="aappaa"][quote="gotztahbeazn"] $25s = 8 + 2(\displaystyle\frac{1}{6} + \frac{1}{6^2} + ... ) \Rightarrow 25s = 8+ 2(\displaystyle\frac{6}{5})$ [/quote]

Shouldn't it be: $25s = 8 + 2(\displaystyle\frac{1}{6} + \frac{1}{6^2} + ... ) \Rightarrow 25s = 8+ 2(\boxed{\displaystyle\frac{1}{5}})$ :?:[/quote]

aappaa is right. That's a tiny mistake that made the difference. $8+2(\frac{1}{5})$ would make $\frac{42}{5}$, s=$\frac{42}{125}$


[quote="thomaskurian89"]Let x and y be positive real numbers such that $ y^3 \plus{} y \le x \minus{} x^3$. Prove that
(a) $ y < x < 1$; and
(b) $ x^2 \plus{} y^2 < 1$.[/quote]
Since $ y^3\plus{}y>0$, we have $ x\minus{}x^3>0$, and it follows that $ x<1$. Then a) is proved.
Now, let us prove b), we have that
\[ x^3\plus{}y^3\minus{}(x\minus{}y)(x^2\plus{}y^2) \equal{}x^3\plus{}y^3\minus{}(x^3\minus{}x^2y\plus{}xy^2\minus{}y^3) \equal{}x^2y\minus{}xy^2\plus{}2y^3\equal{}xy(x\minus{}y)\plus{}2y^3>0\]
Therefore
\[ (x\minus{}y)(x^2\plus{}y^2)<x^3\plus{}y^3<x\minus{}y\]
And it follows that
\[ x^2\plus{}y^2<1\]
We have done. :)


[b]In $ \text{S}_4 \text{O}_6 ^{2 \minus{} }$, calculate the oxidation number of sulfur.
[/b]

I'm getting an oxidation of 2.5 assuming that oxygen has an oxidation state of 2-.  Thus, I'm going to assume that different sulfurs have different oxidation states.  But which ones?

[b]How would each of the following laboratory mistakes affect the calculated value of the percent NaClO in the commercial bleach (too high, too low, no change). Explain. Be sure to use your calculations as evidence and re-state the error.
[list]a. In step 1, the pipet was rinsed with distilled water immediately before being used to measure the commercial bleach solution.
b. In step 2, 3 g of KI was used instead of 2.
c. In step 3, some of the iodine that formed vaporized from solution.[/list][/b]

The complete lab procedure is [url=http://www.aaps.k12.mi.us/huron.collinsa/files/analysis_of_commercial_bleach.pdf]here[/url], but I don't think that the entire background is needed.  This looks like a typical AP type error question, but I can't figure it out.  Could someone please give me quick solution outline for each part?  Thanks in advance!

The sulfurs either have +5 or 0 oxidation states. (The ones bonded to oxygen have +5, while the two others have 0 since they are only bonded to other sulfur)

a. That would cause the bleach solution to become slightly diluted, so 5mL of "bleach" would actually be part water, so you would get a lower percent.
b. Excess KI has negligible effect on the measured result.
c. I highly doubt that it would cause any significant error in a real lab since significant quantities of iodine vaporizing from aqueous solution is almost unheard of, but vaporized iodine will nevertheless lower the amount of thiosulfate required to titrate triiodide.

For the tetrathionate: http://www.madsci.org/posts/archives/may2001/988919069.Ch.r.html

For the laboratory question, check out the last page of my lab report (attached)


geometry
geometric transformation
rotation
circumcircle
homothety
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Three circles touch each other externally and all these cirlces also touch a fixed straight line. Let $A,B,C$ be the mutual points of contact of these circles. If $\omega$ denotes the Brocard angle of the triangle $ABC$, prove that $\cot{\omega}$ = 2.

At first, let $\triangle ABC$ be arbitrary. The Brocard point $Z_1$ is the concurrency point of 3 circles $(O_A), (O_B), (O_C)$. The circle $(O_A)$ is tangent to the line AB at A and passes through the opposite vertex C, the circle $(O_B)$ is tangent to the line BC at B and passes through the opposite vertex A, the circle $(O_C)$ is tangent to the line CA at C and passes through the opposite vertex B. Then $\omega = \angle Z_1BC = \angle Z_1CA = \angle Z_1AB$ is the Brocard angle of the triangle $\triangle ABC$. Similarly, the circles $(P_B), (P_C), (P_A)$ were tangent to the lines AB, BC, CA at the other vertices B, C, A, respectively, concur at the other Brocard point $Z_2$ and $\omega = \angle Z_2CB = \angle Z_1CA = \angle Z_2BA$ is the same Brocard angle of the triangle $\triangle ABC$. We consider the 1st case of $Z_1$.

[b]Lemma 1:[/b] The triangle $\triangle O_AO_BO_C$ is spirally similar to the triangle $\triangle ABC$ with the similarity center $Z_1$ and rotation angle $\phi = \omega - 90^\circ$.

The angle $\angle AZ_1B$ s spanning the arc $\widehat {AZ_1B}$ of the circle $O_B$ is equal to $180^\circ = \angle ABC$, where $\angle ABC = \angle B$ is the angle formed by the chord AB and tangent BC of this circle at the triangle vertex B. The Brocard rays $AZ_1, BZ_1$ are the radical axes of the circle pairs $(O_B), (O_A)$ and $(O_B), (O_C)$. Hence, their center lines $O_BO_A, O_BO_C$ are perpendicular to these Brocard rays and they form the angle $\angle O_B = \angle O_AO_BO_C = 180^\circ - \angle AZ_1B = 180^\circ - (180^\circ - \angle B) = \angle B$. Similarly, the remaining 2 angles $\angle O_C, \angle O_A$ of the triangle $\triangle O_AO_BO_C$ are equal to $\angle O_C = \angle C, \angle O_A = \angle A$. Thus the triangles $\triangle O_AO_BO_C \sim \triangle ABC$ are similar. The angle $\angle Z_1O_BB$ is the central angle of the Brocard angle $\angle Z_1AB = \omega$ spanning the arc $\widehat {AZ_1B}$ of the circle $(O_B)$ , hence $\angle Z_1O_BB = 2 \omega$. Likewise, $\angle Z_1O_CC = \angle Z_1O_AA = 2\omega$. Consequently, the isosceles triangles $\triangle Z_1O_BB \sim \triangle Z_1O_CC \sim \triangle Z_1O_AA$ are all similar and

$\frac{Z_1O_B}{Z_1B} = \frac{Z_1O_C}{Z_1C} = \frac{Z_1O_A}{Z_1A}$,

$\angle O_BZ_1B = \angle O_CZ_1C = \angle O_AZ_1A = \frac{180^\circ - 2\omega}{2} = 90^\circ - \omega$,

which implies that the Brocard point $Z_1$ is the spiral similarity center of the triangles $\triangle O_AO_BO_C \sim \triangle ABC$. Since the oriented angle $\measuredangle O_BZ_1B$ ic clockwise for a triangle $\triangle ABC$ labeled counter-clockwise, the rotation angle is $\phi = - \angle O_BZ_1B = \omega - 90^\circ$.

[b]Lemma 2:[/b] Let the normals to the Brocard rays $AZ_1, BZ_1, CZ_1$ at the point $Z_1$ meet the circles $(O_A), (O_B), (O_C)$ at points D, E, F, respectively, different from $Z_1$. The triangle $\triangle DEF$ is spirally similar to the triangle $\triangle ABC$ with the similarity center $Z_1$ and rotation angle $-90^\circ$. The similarity coefficient of these 2 triangles is equal to $\cot \omega$.

Since the angle $\angle AZ_1D = 90^\circ$ inscribed in the circle $(O_A)$ is right by definition, $AD$ is a diameter of this circle, its center $O_A$ is the midpoint of of this diameter and the points $A, O_A, D$ are collinear. Similarly, BE ia a diameter of the circle $(O_B)$ and the points $B, O_B, E$ are also collinear. Thus the angle $\angle BAE = 90^\circ$ is right. Since the circle $(O_A)$ is tangent to the line AB at the vertex A, the angle $\angle BAD = 90^\circ$ is also right. As a result, the points $D, O_A, A, E$ are all collinear and similarly, the points $E, O_B, B, F$ are collinear and the points $F, O_C, C, D$ are collinear.  The angle $\angle DEF \equiv \angle AEB$ is spanning the arc $\widehat {AEB}$ of the circle $(O_B)$ opposite to the arc $\overarc{AZ1B}$, hence, $\angle DEF = 180^\circ - \angle AZ_1B = 180^\circ - (180^\circ - \angle B) = \angle B$ (see the proof of lemma 1). Likewise, $\angle EFD = \angle C, \angle FDE = \angle A$ and consequently, the triangles $\triangle DEF \sim \triangle ABC$ are similar. The angles $\angle Z_1EB = \angle Z_1AB = \omega$ spanning the same arc $Z_1B$ of the circle $(O_B)$ are equal. Likewise, $\angle Z_1FC = \angle Z_1BC = \omega$ and $\angle Z_1DA = \angle Z_1CA = \omega$. As a result, the isosceles triangles $\triangle Z_1O_BE \sim \triangle Z_1O_CF \sim \triangle Z_1O_AD$ are all similar and

$\frac{Z_1O_B}{Z_1E} = \frac{Z_1O_C}{Z_1F} = \frac{Z_1O_A}{Z_1D}$,

which implies that the Brocard point $Z_1$ is the common spiral similarity center of the triangles $\triangle DEF \sim \triangle O_AO_BO_C \sim \triangle ABC$. By definition, the rotation angle of the triangle $\triangle DEF$ with respect to the triangle $\triangle ABC$ is $90^\circ$ clockwise, i.e., $-90^\circ$ counter-clockwise. The similarity coefficient of the triangles $\triangle DEF \sim \triangle ABC$ is $\frac{Z_1E}{Z_1B}$. Since the angle $\angle BZ_1E = 90^\circ$ is right by definition and since we have shown that the angle $\angle Z_1EB = \omega$ is equal to the Brocard angle, $\frac{Z_1E}{Z_1B} = \cot \omega$.

To prove the problem proposition, we will use only lemma 1. (I got carried away before realizing that the problem was simpler that it seemed.)

(to be continued)

Let $r_1, r_2$ be the radii of the 2 given externally tangent circles $(O_1), (O_2)$. Let A be the tangency point of these 2 circles and let KL be one of their common external tangents with the tangency points $K \in (O_1), L \in (O_2)$. We follow the usual construction of the circle $(O_3)$ tangent to the line KL and to the circles $(O_1), (O_2)$ at points B, C, i.e., we invert the circles $(O_1), (O_2)$ and the line KL in a circle (A) centered at their tangency point A lying on the center line $O_1O_2$. We choose the radius of the inversion circle (A) to be $r = \sqrt{r_1r_2}$. (We could also choose $r = \sqrt{2r_1r_2}$ or $r = 2 \sqrt{r_1r_2}$, but we will stick with the original choice. This is a known trick for problems related to Archimedes' arbelos.) Since the circles $(O_1), (O_2)$ both pass through the inversion center, they are carried into straight lines $o_1, o_2$ both perpendicular to the center line $O_1O_2$. Let the 2 circles intersect the center line $O_1O_2$ at points P, Q different from their tangency point A. Since $AP = 2r_1,\ AQ = 2r_2$, the points P, Q are carried into the points $P', Q' \in O_1O_2$, such that

$AP' = \frac{r^2}{AP} = \frac{r_1r_2}{2r_1} = \frac{r_2}{2}$

$AQ' = \frac{r^2}{AQ} = \frac{r_1r_2}{2r_2} = \frac{r_1}{2}$

The lines $o_1, o_2$, i.e., the inverted circles $(O_1), (O_2)$, are the normals to the center line $O_1O_2$ at the points P', Q', recpectively. Since the triangles $\triangle PKA \sim \triangle ALQ$ are similar and right, the angle $\angle KAL = 90^\circ$ is also right. The normal to the center line $O_1O_2$ at the tangency point A is the radical axis of the circles $(O_1), (O_2)$, hence, it cuts the tangent segment KL at its midpoint R and consequently, AR is the median of the right angle triangle $\triangle KAL$ equal to half of its hypotenuse $KL = 2 \sqrt{r_1r_2}$ (the external tangent distance of 2 externally tangent circles). Thus $AR = \frac{KL}{2} = \sqrt{r_1r_2} = r$. As a result, the inversion circle (A) with radius r cuts the common external tangent KL at the midpoint R (and at some other point S). Our inversion carries the line $t \equiv KL$ into a circle (T) passing through the inversion center A, through the points R, S, and tangent to the 2 parallel lines $o_1 \parallel o_2$ separated by the distance $P'Q' = AP' + AQ' = \frac{r_1 + r_2}{2}$. Thus the radius of the circle (T) is $\frac{P'Q'}{2} = \frac{r_1 + r_2}{4}$. The unknown circle $(O_3)$ tangent to the circles $(O_1), (O_2)$ at points B, C is carried into a circle $(O_3')$ also tangent to the parallel lines $o_1 \parallel o_2$ at points B', C' and to the circle (T). The tangency points B, C of the circle $(O_3)$ are the intersections of the rays AB', AC' with the circles $(O_1), (O_2)$. Thus we obtained the defined triangle $\triangle ABC$, which is all we need. Anyway, the tangency point U of the circle $(O_3)$ with the line $t \equiv KL$ is the intersection of the ray AU' with the line KL and $(O_3)$ is the circumcircle of the triangle $\triangle BCU$. Or, the center of the circle $(O_3)$ is the intersection of the perpendicular bisector of the segment BC with the ray $AO_3'$ (because a circle and its inverted image are centrally similar with the homothety center at the inversion center).

Because of the basic properties of inversion, the triangles $\triangle AC'B' \sim \triangle ABC$ are similar and consequently, they have the same Brocard angle. Thus we do not have to worry about the triangle $\triangle ABC$ at all and we can find the Brocard angle of the inverted triangle $\triangle AB'C'$ instead. Obviously, the side B'C' of this triangle is perpendicular to the parallel lines $o_1 \parallel o_2$ and from the rectangle P'Q'C'B', $B'C' = P'Q' = \frac{r_1 + r_2}{2}$. The other 2 sides of this rectangle are equal to

$P'B' = Q'C' = \frac{AR}{2} + TU' + U'O_3' = \frac{r_1 + r_2 + \sqrt{r_1r_2}}{2}$

because $TU' = U'O_3' = \frac{r_1 + r_2}{4}$ are radii of the congruent circles $(T), (O_3')$ and the circle (T) is centered on the perpendicular bisector of the segment $AR = \sqrt{r_1r_2}$ (see above). By Pythagorean theorem for the right angle triangles $\triangle AP'B', \triangle AQ'C'$ their hypotenuses AB', AC' are

$AB'^2 = AP'^2 + P'B'^2 = \frac{r_2^2 + \left(r_1 + r_2 + \sqrt{r_1r_2}\right)^2}{4}$

$AC'^2 = AQ'^2 + Q'C'^2 = \frac{r_1^2 + \left(r_1 + r_2 + \sqrt{r_1r_2}\right)^2}{4}$

Let $(O_B)$ be a circle tangent to the side B'C' of the triangle $\triangle AB'C'$ at the vertex B' an passing through the opposite vertex A. Since $B'C' \perp P'B'$, the circle $(O_B)$ is centered on the line $o_1 \equiv P'B'$, on its intersection with the perpendicular bisector of the segment AB'. Let $(O_C)$ be a circle tangent to the side AC' at the vertex C' and passing through the opposite vertex B'. The circle $(O_C)$ is centered on the perpendicular bisector of the segment B'C', which is parallel to the lines $o_1 \equiv P'B', o_2 \equiv Q'C'$. Similarly, we could define circle $(O_A)$, but we do not need it. According to lemma 1, the side B'C' of the triangle $AB'C'$ forms the angle $90^\circ - \omega$ with the side $O_BO_C$ of the triangle $\triangle O_AO_BO_C$ ($\omega$ is the Brocard angle). Therefore, the line $O_BO_C$ forms the angle $\omega$ with the perpendicular $o_1 \equiv P'B'$ to the line B'C'. Since $O_B \in P'B'$, the angle $\angle O_CO_BB' = \omega$ is the Brocard angle of the triangles $\triangle AB'C' \sim \triangle O_AO_BO_C$. Let a normal to the line $o_1 \equiv P'B' \equiv O_BB'$ through the point $O_C$ intersects this line at a point V. Then from the right angle triangle $\triangle O_BO_CV$, $\cot \omega = \frac{O_BV}{OC_V}$. We know one leg $O_CV = O_3'B' = \frac{r_1 + r_2}{4}$ of this triangle. Now we have to calculate the other leg.

Let M, N be the midpoints of the sides AB', B'C'. (Obviously, $N \equiv O_3$ is the center of the circle $(O_3')$.) The right angle triangles $\triangle O_BMB' \sim \triangle AP'B'$ are similar, having the angle at the vertex B' common. Hence,

$\frac{O_BB'}{MB'} = \frac{AB'}{P'B'}$

$O_BB' = \frac{MB' \cdot AB'}{P'B'} = \frac{AB'^2}{2P'B'} = \frac{r_2^2 + \left(r_1 + r_2 + \sqrt{r_1r_2}\right)^2}{4\left(r_1 + r_2 + \sqrt{r_1r_2}\right)}$

Likewise, the right angle triangles $\triangle C'NO_C \sim \triangle AQ'C'$ are similar, because the angle $\angle AC'O_C = 90^\circ$ is right due to the tangency of the circle $(O_C)$ to the line B'C' at the point C'. Therefore, the angles $\angle AC'N + \angle NC'O_C = 90^\circ$ add up to a right angle and $\angle NC'O_C = 90^\circ - \angle AC'N = 90^\circ - \angle C'AP' = \angle AC'P'$. Hence,

$\frac{O_CN}{NC'} = \frac{AQ'}{Q'C'}$

$O_CN = \frac{NC' \cdot AQ'}{Q'C'} = \frac{(r_1 + r_2)r_1}{4\left(r_1 + r_2 + \sqrt{r_1r_2}\right)}$

The other leg of the right angle triangle triangle $\triangle O_BO_CV$ and $\cot \omega$ are then

$O_BV = O_BB' + B'V = O_BB' + O_CN = \frac{(r_1 + r_2)r_1 + r_2^2 + \left(r_1 + r_2 + \sqrt{r_1r_2}\right)^2}{4\left(r_1 + r_2 + \sqrt{r_1r_2}\right)}$

$\cot \omega = \frac{O_BV}{O_CV} = \frac{(r_1 + r_2)r_1 + r_2^2 + \left(r_1 + r_2 + \sqrt{r_1r_2}\right)^2}{(r_1 + r_2)\left(r_1 + r_2 + \sqrt{r_1r_2}\right)} =$

(factoring out $r_1 + r_2$)

$= \frac{(r_1 + r_2)r_1 + r_2^2 + (r_1 + r_2)^2 + 2(r_1 + r_2) \sqrt{r_1r_2} + r_1r_2}{(r_1 + r_2)(r_1 + r_2 + \sqrt{r_1r_2})} =$

$= \frac{(r_1 + r_2)\left(r_1 + r_1 + r_2 + 2 \sqrt{r_1r_2}\right) + r_2(r_1 + r_2)}{(r_1 + r_2)\left(r_1 + r_2 + \sqrt{r_1r_2}\right)} =$

$= \frac{2r_1 + 2r_2 + 2 \sqrt{r_1r_2}}{r_1 + r_2 + \sqrt{r_1r_2}} = 2$

Q.E.D.

Excellent proofs!!!!1 :D  :D  :D

Can you tell me which program you used for the wonderful diagrams and where I can sownload them???

Tx. I use The Geometer's Sketchpad to make the drawings. This program is not available for a free download, but it is very nice and very cheap ($\$$40). See the last message in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34221]Geometrical problem[/url] for more info. The label "AGif UNREGISTERED" stands just for the Active GIF Creator, which is a graphics converter. Unregistered version (not purchased, just downloaded) puts this label into the GIF images. Search the web, I got many hits for "Active GIF Creator". Normally, I use Adobe Photoshop to export the Sketchpad drawings into the transparent GIF format, but I did not have it on the computer, which I used at this time.

Yetti.


I think this question may have arose here and there previously on the forum, but I am wondering what you all think of the issue in the subject line: Does preparing for olympiad-type problems (USAMO, IMO) help you solve short-answer problems (AMC, AIME)?

I definitely think solving olympiad problems helps. Olympiad probs are more complicated and require a lot more though and creativity, so AHSME and even AIME problems seem more or less trivial. There is only one harm, starting to generalize and play around with problems on a timed test like the AMC 12.

i think playing around with olympiad problems usually screws you up on multiple choice questions where you preceives the difficulty of each question to be difficult whereas they are only trivial.  olympiad and amc-12 are very different (AIME lies in the middle)
one asks for inguinuity while the other demands speed and correctness

I think these two strands of tests are completely different. The AMC/AIME kind of tests require accuracy and speed, but not much skill while USAMO/IMO kind require ingenuitive spark over a long period of time. Studying USAMO level things might help for AMC/AIME, but you might want to glance through the AMC/AIME material too, just to remind yourself "not to make the problem difficult when it's not", a la Ms. L*****-S***** :)
Some people are good with one, some people are good with the other, so first you need to figure out which type of test you are most comfortable at, and then try to adjust yourself to the other.

i dont think they are completely different. (otherwise why have the amc 12 and aime if you risk missing out on people who are very good at olympiads but bad at simpler problem solving?). i have yet to see a person who cannot do well at the amc 12 or aime because they studied for olympiads. take a look at the amc 12 perfect scorers, and you will find most members of the us imo team there. if olympiads screw you up on other contests, how can this happen?

on a personal level, my AMC12+AIME score is not even close in qualifying for USAMO (which i believe has a cut of of 217 or something last year).  but i believe i have scored as well as a good handful of people that actually being invited to take USAMO.
i am sure there is a lot of other ppl like me who cant do arithmetic.  of all the AIME question i knew how to do during the contest, i got less than half of them right (funny at the time i thought i got them all).

i can say that most ppl that hanging around here is familiar with Olympiad level questions, such as yourself, tetrahedr0n.  obviously not everybody is at the same level as somebody like Tiankai Liu for example who gets perfect on just about everything.  now tetrahdedr0n how confident are you in making the USAMO?
AIME obviously contains questions that are similar to those found on the olympiads, but it is too calculation demanding, and one small mistake would result in a waste of time on that question.  i do not think AMC+AIME is good way of choosing IMO candidates - but on another level USA has too many good ppl that it wont make too much of a difference

Do you think there are people who should be on the team who aren't making USAMO? I personally doubt it. I do believe there are people good enough to take the USAMO who aren't qualifying, but those people probably wouldn't be making the IMO team.

actually look at it the other way.  I see ppl who scores 0 on USAMO which means they shouldn't have been invited to write the USAMO.

"Do you think there are people who should be on the team who aren't making USAMO?"  what does this have to do with ANYTHING?  how would anyone know if there are better ppl out there or not?

[quote]"Do you think there are people who should be on the team who aren't making USAMO?" what does this have to do with ANYTHING? how would anyone know if there are better ppl out there or not?[/quote]

I think Simon is saying that if indeed the AMC series of tests is for the sole purpose of choosing the US IMO team, then the fact that there are some ppl who are borderline USAMO but do not make USAMO would not affect its main goal. He is saying that it is clear that the top 6 students in the US should be able to get at least a 9 on the AIME.

Do you think that there has even been a USAMO that someone who would have been top-12 didn't take due to not qualifying? For all that it means to any of us, the USAMO is a way to determine the IMO team above all else. Of course many people have lower goals and just want to qualify, and so qualifying becomes something of a benchmark. Nonetheless, the USAMO's primary goal is to pick the twelve who get to take the TST and thus select the IMO team.


Whenever I use the

[code]\[
\begin{align} \frac{2y}{3}-3&=y\\2y-9&=3y\\y&=\boxed{-9}\end{align}
\][/code]

on TexShop (for mac) and compile, I get a message that says:

[code]! Package amsmath Error: Erroneous nesting of equation structures;
(amsmath)                trying to recover with `aligned'.

See the amsmath package documentation for explanation.
Type  H <return>  for immediate help.
 ...

l.17 ...}-3&=y\\2y-9&=3y\\y&=\boxed{-9}\end{align}[/code]

Then I hit enter, and it compiles fine.

It's not that much of a nuisance, but now in my math class we've been dealing a lot more with equations, and hitting enter 20 times gets kind of annoying...anyone know how to fix it?

Thanks!

Use eqnarray? I think it has the same purpose... (and it works fine)
\begin{align} \frac{2y}{3}-3&=y\\2y-9&=3y\\y&=\boxed{-9}\end{align}

\begin{eqnarray} \frac{2y}{3}-3&=&y\\2y-9&=&3y\\y&=&\boxed{-9}\end{eqnarray}

getting rid of the numbers:

\begin{eqnarray*} \frac{2y}{3}-3&=&y\\2y-9&=&3y\\y&=&\boxed{-9}\end{eqnarray*}

Yeah, or else it wouldn't have been able to compile at all :D

And i do use \usepackage{amsmath,amsthm}

Remove \ [ and \ ]. \begin{align} like \begin{multline} and a few other constructs contain a command of switching to math. mode already.

That worked.  But how do I get rid of the (1) and (2), etc. from the right side?

Same for eqnarray (quote my post and you will see)

Hi,

[quote="Dojo"]Use eqnarray? I think it has the same purpose... (and it works fine)
[/quote]
I wouldn't recommend eqnarray, it provides inconsistent spacing, for more reasons one could have a look at [url=http://texblog.net/latex-archive/maths/eqnarray-align-environment/]eqnarray vs. align[/url] or the longer article "[url=http://www.tug.org/pracjourn/2006-4/madsen/madsen.pdf]Avoid eqnarray![/url]". The align environment should produce better results.

Stefan

You should use "gather" and "aligned" rather than "align". "Align" doesn't (read: shouldn't!) be within displayed math, but rather itself, i.e.

[code]
\begin{align*}
\frac{2y}{3}-3 &=y \\
2y-9&=3y \\
y&=\boxed{-9}
\end{align*}
[/code]

I notice you're trying to show the steps in solving an equation. Try this:

[code]
\begin{gather*}
\frac{2y}{3} - \underset{+3}{3} = \underset{+3}{y} \\
\underset{\times 3}{\frac{2y}{3}} = \underset{\times 3}{3} + \underset{\times 3}{y} \\
2y = 9 + 3y
y = -9
\end{gather*}
[/code]

Which should get you the correct formatting with operations shown:
$ \frac{2y}{3} \minus{} \underset{\plus{}3}{3} \equal{} \underset{\plus{}3}{y}$

$ \underset{\times 3}{\frac {2y}{3}} \equal{} \underset{\times 3}{3} \plus{} \underset{\times 3}{y}$

Except it looks a bit better like this:
$ \underset{\times 3}{\frac {2y}{3}} \equal{} \underset{\times 3}{3\vphantom{\frac13}} \plus{} \underset{\times 3}{y\vphantom{\frac13}}$

You're absolutely right. However I'm not particularly accustomed to using "phantoms" (was going to say "not a fan of phantoms  :D ) -- maybe I should try more of it.


What is the length of the hypotenuse of a right trangle that has an area of 40 and its longest leg has length b?

$ 40\equal{}\frac{bh}{2}\Rightarrow h\equal{}\frac{80}{b}$ so by Pythagoras,
$ b^{2}\plus{}\left(\frac{80}{b}\right)^{2}\equal{}\frac{\sqrt{b^{4}\plus{}6400}}{b}\equal{}\mathcal{H}$


Prove this equation  without the calculator :

8*cos 20 * cos 40 * cos 80 = 1

[quote="Dr Sonnhard Graubner"]hello, do you mean $ 8\odot \cos(20^{\circ})\cdot \cos(40^{\circ})\cdot \cos(80^{\circ}) = 1$
Why don't you use $ \text{\LaTeX}$?
Sonnhad.[/quote]
yes i mean this that you have written.
sorry for this.

Easy...  :D

Just use product to sum identity..

$ 2 \cos a \cos b \equal{} \cos (a \plus{} b) \plus{} \cos (a \minus{} b)$

Expand $ 2 \cos 20^{\text{o}} \cos 40^{\text{o}}$ first, then multiply by $ 4\cos 80^{\text{o}}$, then use the same identity once again..

Fast way:

[hide="My Solution"]
Let $ x \equal{} 8\cos{20^\circ}\cdot \cos{40^\circ}\cdot \cos{80^\circ}$. Notice that are the angles are doubled from the previous. It seems like if we introduce a $ \sin{20^\circ}$, everything will reduce by repeatedly applying the double-angle formula for sine.

$ \sin{20^\circ}x \equal{} 8\sin{20^\circ}\cos{20^\circ}\cdot \cos{40^\circ}\cdot \cos{80^\circ}\\
\sin{20^\circ}x \equal{} 4\sin{40^\circ}\cdot \cos{40^\circ}\cdot \cos{80^\circ}\\
\sin{20^\circ}x \equal{} 2\sin{80^\circ}\cdot \cos{80^\circ}\\
\sin{20^\circ}x \equal{} \sin{160^\circ}$

We know that $ \sin{x} \equal{} \sin{(180 \minus{} x)}$ so $ \sin{160^\circ} \equal{} \sin{20^\circ}$. Therefore, $ x \equal{} 1$. $ \Box$
[/hide]

[quote="sunehra"]Fast way:

[hide="My Solution"]
Let $ x \equal{} 8\cos{20^\circ}\cdot \cos{40^\circ}\cdot \cos{80^\circ}$. Notice that are the angles are doubled from the previous. It seems like if we introduce a $ \sin{20^\circ}$, everything will reduce by repeatedly applying the double-angle formula for sine.

$ \sin{20^\circ}x \equal{} 8\sin{20^\circ}\cos{20^\circ}\cdot \cos{40^\circ}\cdot \cos{80^\circ} \\
\sin{20^\circ}x \equal{} 4\sin{40^\circ}\cdot \cos{40^\circ}\cdot \cos{80^\circ} \\
\sin{20^\circ}x \equal{} 2\sin{80^\circ}\cdot \cos{80^\circ} \\
\sin{20^\circ}x \equal{} \sin{160^\circ}$

We know that $ \sin{x} \equal{} \sin{(180 \minus{} x)}$ so $ \sin{160^\circ} \equal{} \sin{20^\circ}$. Therefore, $ x \equal{} 1$. $ \Box$
[/hide][/quote]

WOW...  :D

A nice way to celebrate your $10,000$-th post :) catching Dr Sonnhard misspelling his own name. Wow ...

[quote="AIME15"][quote="Dr Sonnhard Graubner"]hello, do you mean $ 8\odot \cos(20^{\circ})\cdot \cos(40^{\circ})\cdot \cos(80^{\circ})=1$
Why don't you use $ \text{\LaTeX}$?
Sonnhad.[/quote]

hello, is your name Sonnhard or Sonnhad ?
Graubner.[/quote]

Hello,
Dr Sonnhard Graubner's name is Sonnhard.
Ahirbhairav.

Lemma:$4cosa.cos(60-a).cos(60+a)$=$cos3a$
thus $4cos40.cos20.-(cos80)=-1/2$
so we have $8cos20.cos40.cos80=1$
QED

Quick typesetting tip: \cos x gives $\cos x$, while cos x gives $cos x$, which is less pretty.


The diagonal splits a rectangle into how many 30-60-90 triangles?
and how can we prove it?

[hide]It is 0 because to be 30-60-90, they must be right triangles. There are four congruent right triangles and their side ratio from long leg to short leg is $2:1$, not $\sqrt{3}:1$ as required.

The smaller triangles are not right triangles because the rectangle is not a square, and if it was, they would be 45-45-90 right triangles at that.
[/hide]

[hide]
Hmm, I think it it 0 because there are no 30 or 60 degree angles.
There are also no sides that have the correct ratio so that "proves" that there are no triangles with the given statements.[/hide]

[quote="AntonioMainenti"][quote="Done"]It's not zero/[/quote]Well so far 3 out of 3 have said it's zero. Where can you find a 30-60-90 triangle in your figure?[/quote]

well..I didn't know that correct answers are derived by majority!

Anyway, what about traingles BAD and BCD?

[quote="Zyloch"]Those are 2 of the only 4 right triangles in the drawing. Refer to my post (and everyone else's) that note that the ratios between legs of the triangles are wrong :D[/quote]

ok..so is it correct to say that, in general, the diagonal of a rectangle [i][b]might [/b][/i]split it into 30-60-90 traingles? and the maximum number of such triangles is [i][b]4[/b][/i]?

Yes, it might--if the side lenghts of the rectangle are in the ratio $\sqrt{3}:1$. This not only makes 30-60-90 right triangles, it makes [i]exactly[/i] 4.

I can't really see a way to get less than 4, so I guess it's 4 or nothing heh :)

Let me put it this way: if you place the side lengths AD and BC with $4\sqrt{3}$ then there will be a few 30-60-90 triangles.

[quote="AntonioMainenti"]Let me put it this way: if you place the side lengths AD and BC with $4\sqrt{3}$ then there will be [b]a few [/b] :? 30-60-90 triangles.[/quote]


calculus
integration
limit
real analysis
real analysis unsolved

Let $a>\frac{5}{2}$ be a fixed real; find an equivalent, when ${n\to\infty}$, of $\int^{1}_{0}\frac{t^{2a-6}}{(1+t^{a})^{n}}\,dt$.

Generalization:
Let $a>0, b>-1$ be fixed reals; find an equivalent, when ${n\to\infty}$, of $\int^{1}_{0}\frac{t^{b}}{(1+t^{a})^{n}}\,dt$.

I assumed that you are looking for the limit of that integral. The limit is 0.

Observe that the substitution $t=x^{1/a}$ transforms the integral to:

$I_{n}=\frac{1}{a}\int_{0}^{1}\frac{x^{\frac{b+1}{a}-1}}{(1+x)^{n}}dx$.

We distinguish two cases:

a) $\frac{b+1}{a}\geq 1$, then in this case it follows simply that the limit is 0 via Bounded Convergency Theorem.

b) $\frac{b+1}{a}<1$.

$I_{n}=\frac{1}{a}\left(\int_{0}^{\infty}\frac{x^{\frac{b+1}{a}-1}}{(1+x)^{n}}dx-\int_{1}^{\infty}\frac{dx}{x^{1-\frac{b+1}{a}}(1+x)^{n}}\right)$.

We get that

$I_{n}=\frac{1}{a}\left(\beta(\frac{b+1}{a}, n-\frac{b+1}{a})-B_{n}\right)$,
where
$B_{n}=\int_{1}^{\infty}\frac{dx}{x^{1-\frac{b+1}{a}}(1+x)^{n}}\leq \int_{1}^{\infty}\frac{dx}{(1+x)^{n}}\rightarrow 0$.

On the other hand, a calculation shows that $\beta(\frac{b+1}{a}, n-\frac{b+1}{a})\rightarrow 0$, hence the desired limit is 0.

OK Didilaca; the limit is 0; and the equivalent is ?
(ie asymptotical equivalence of sequences).

$I_{n}$ is said equivalent to $a_{n}$ iff $\lim_{n\to\infty}\frac{I_{n}}{a_{n}}=1$; I ask for an explicit $a_{n}$.

OK. I understand now what you are looking for. We have that:

$a_{n}=\frac{1}{n^{\frac{b+1}{a}}}$, of course this sequence should be multiplied by a constant which I will drop for simplification of the calculations. Here is the proof:

First consider the case $\frac{b+1}{a}\leq 1$. Then in this case observe that

$I_{n}=\beta(\frac{b+1}{a}, n-\frac{b+1}{a})-c_{n}$, where

$c_{n}=\int_{1}^{\infty}\frac{dx}{x^{1-\frac{b+1}{a}}(1+x)^{n}}$

Now $c_{n}\leq \frac{1}{(n-1)2^{n-1}}$ hence it follows that

$n^{\frac{b+1}{a}}c_{n}\rightarrow 0$.

To show that $\beta(\frac{b+1}{a}, n-\frac{b+1}{a})\approx \frac{1}{n^{\frac{b+1}{a}}}$ it follows from Stirling`s formula, I will skip the calculations.

For the second case when $\frac{b+1}{a}>1$ then we have the same sequence with the exception that in this case

$c_{n}=\int_{1}^{\infty}\frac{x^{\frac{b+1}{a}-1}}{(1+x)^{n}}<\frac{1}{2^{n/2}}\int_{1}^{\infty}\frac{dx}{x^{n/2-\frac{b+1}{a}+1}}=\frac{1}{2^{n/2}}\frac{1}{n/2-\frac{b+1}{a}}$
and of course the technique is the same.

In conclusion the sequence $a_{n}$ is $\frac{1}{n^{\frac{b+1}{a}}}$ multiplied by a constant which can be calculated as an application of the Stirling`s formula.
I hope that there are no mistakes.

The constant is $\Gamma(\frac{b+1}{a})$.
Proof: with his notations, Didilaca has proved that if $a>0$ and $b>-1$ then
$I_{n}\sim\beta(\frac{b+1}{a},n-\frac{b+1}{a})$.
$\beta(\frac{b+1}{a},n-\frac{b+1}{a})=\frac{\Gamma(\frac{b+1}{a})\Gamma(n-\frac{b+1}{a})}{\Gamma(n)}$$\sim\frac{\Gamma(\frac{b+1}{a})}{n^\frac{b+1}{a}}$


Euler
binomial coefficients
calculus
calculus computations

Take a row of binomial coefficients, such as

1 3 3 1

Form their gradual sums across:

1
1 + 3
1 + 3 + 3
1 + 3 + 3 + 1

to get 1 4 7 8

Now, turn them around and use them as multipliers:  8 7 4 1, and divide by the leading number

Suppose, you want a better idea of (but you won't know how well it is relative
without more particulars being said) for ln(2) than what the partial sum 1 - 1/2 + 1/3 - 1/4 gives.
It is about 0.583.

Example 1)

8 7 4 1 applied to the first four terms:

[8(1) - 7(1/2) + 4(1/3) - 1(1/4)]/8 ~ 0.6979 (overage)

Compare to ln (2) ~ 0.6931

=-=-=--=--=-=-=-=-==-=-=-=-=-=-=-=-=-=-=-=-

Example 2)

8 7 4 1 applied to the 4th, 5th, 6th, and 7th terms:

1 - 1/2 + 1/3 + [-8(1/4) + 7(1/5) - 4(1/6) + 1(1/7)]/8 ~ 0.6929 (underage)

===========================================================
EDIT:  I "discovered" this independent of others, but I mean it already "existed."

What you're doing here is similar to [url=http://en.wikipedia.org/wiki/Binomial_transform#Euler_transform]Euler's series transformation[/url].


what does $ \sum$ from 0 to 0 of something mean? is that just the function thing evaluated at 0?

Yes.  The set of indices starting at $ 0$ and ending at $ 0$ is $ \{ 0 \}$.

so would the $ \sum$ from $ {i\equal{}3}$ to $ i \equal{}{3}$ of $ f(x)\equal{} f(3)$ ?

I think:
$ \sum_{k\equal{}n}^{n} f(k) \equal{} f(n)$


geometry vector integration calculus function trigonometry search

On the given circle there are the points A1,...,An, B1,...,Bn.
Prove that there exists a point P on the circle such that:
PA1 + ... + PAn = PB1 + ... + PBn

[quote="grobber"]It's the sum of lengths, I think. Anyway, one just has to integrate $\sum PA_i$ and $\sum PB_i$ as $P$ goes around the circle.[/quote]

Is it something like \[ \int_0^{2 \pi} \left( \sum d(P,A_i) - \sum d(P,B_i) \right) d \alpha = \sum \left( \int_0^{2 \pi} d(P,A_i) d \alpha - \int_0^{2 \pi} d(P,B_i) d \alpha \right) ,  \] which must be $0$ as all $2n$ integrals have the same value. But $d(P,X)$ is a continuous function, so we must have $\sum PA_i = \sum PB_i$ for some $P$.

Right?

And is there any "geometrical" solution?

My solution is similar. The problem was for 12 grade. Maybe we should ask / post this in the geometry forum ? I am not sure if this allowed [post the same problem in 2 sub-forums only to check for 2 different solutions  :oops: ]

I found an analytic solution without calculus.

WLOG we do the research on a circle with radius 1.

We identify the points by their angle:
* $a_1, \ldots, a_n$ are the angles of the points $A_i$
* $b_1, \ldots, b_n$ are the angles of the points $B_i$
* $p$ is the angle of a point $P$

Then we have
\begin{eqnarray*} d(P, A_i) &=& \sqrt{\Big(\cos^2(p)-\cos^2(a_i)\Big)^2 + \Big(\sin^2(p)-\sin^2(a_i)\Big)^2} \\ &=& \sqrt{\cos^2(p) - 2\cos(p)\cos(a_i) + \cos^2(a_i) + \sin^2(p) - 2\sin(p)\sin(a_i) + \sin^2(a_i)} \\ &=& \sqrt{2 - 2\big(\cos(p)\cos(a_i) + \sin(p)\sin(a_i)\big)} \\ &=& \sqrt{2 - 2\cos(p-a_i)} \\ &=& 2 \sqrt{\frac{1 - \cos(p-a_i)}{2}} \\ &=& 2 \sin\big(\frac{p-a_i}{2}\big) \qquad \text{known identity?} \\ &=& 2 \Big( \sin\big(\frac{p}{2}\big) \cos\big(\frac{a_i}{2}\big) - \cos\big(\frac{p}{2}\big) \sin\big(\frac{a_i}{2}\big) \Big) \end{eqnarray*}

We search a $p$ with
\begin{eqnarray*} 0 &=& \sum_{i=1}^n \Big( d(P, A_i) - d(P, B_i) \Big) \\ &=& 2 \sum_{i=1}^n \Big( \sin\big(\frac{p}{2}\big) \cos\big(\frac{a_i}{2}\big) - \cos\big(\frac{p}{2}\big) \sin\big(\frac{a_i}{2}\big) - \sin\big(\frac{p}{2}\big) \cos\big(\frac{b_i}{2}\big) + \cos\big(\frac{p}{2}\big) \sin\big(\frac{b_i}{2}\big) \Big) \\ &=& 2 \sin\big(\frac{p}{2}\big) \sum_{i=1}^n \Big( \cos\big(\frac{a_i}{2}\big) - \cos\big(\frac{b_i}{2}\big) \Big) - 2 \cos\big(\frac{p}{2}\big) \sum_{i=1}^n \Big( \sin\big(\frac{a_i}{2}\big) - \sin\big(\frac{b_i}{2}\big) \Big) \end{eqnarray*}

This yields to
  \[ \sin\big(\frac{p}{2}\big) \sum_{i=1}^n \Big( \cos\big(\frac{a_i}{2}\big) - \cos\big(\frac{b_i}{2}\big) \Big) = \cos\big(\frac{p}{2}\big) \sum_{i=1}^n \Big( \sin\big(\frac{a_i}{2}\big) - \sin\big(\frac{b_i}{2}\big) \Big)  \]

If $\sum_{i=1}^n \Big( \cos\big(\frac{a_i}{2}\big) - \cos\big(\frac{b_i}{2}\big) \Big) = 0$, we can fulfil the equation with $p = \pi$.

In every other case we can write
  \[ \tan\big(\frac{p}{2}\big) = \frac{ \sum_{i=1}^n \Big( \sin\big(\frac{a_i}{2}\big) - \sin\big(\frac{b_i}{2}\big) \Big) } { \sum_{i=1}^n \Big( \cos\big(\frac{a_i}{2}\big) - \cos\big(\frac{b_i}{2}\big) \Big) }  \]

This gives exactly one $p \in [0,2\pi)$.


Let  $x>0$ and different from  $1$ and $n>0$ a natural number. Prove that

$\frac{1+x^2+x^4+....+x^{2n}}{x+x^3+x^5+.....+x^{2n-1}}>\frac{n+1}{n}$


modular arithmetic
function
logarithms
algorithm

find the remainder when $ 19^{92}$  is divided  by $ 92$

If you want something "innovative," let's use the Chinese Remainder Theorem with Fermat's Theorem (avoiding Euler's Theorem):

\begin{align*}
19^{92} &\equiv (-1)^{92} &\pmod{4}\\
&\equiv 1 &\pmod{4}\\
\end{align*}

\begin{align*}
19^{92} &\equiv (-4)^{92} &\pmod{23}\\
&\equiv 256^{23} &\pmod{23}\\
&\equiv 3^{23} &\pmod{23}\\
&\equiv 3 &\pmod{23}\\
\end{align*}

Finally, solving $ 23k + 3 \equiv 1 \pmod{4} \Leftrightarrow 3k \equiv 6 \pmod{4} \Leftrightarrow k \equiv 2 \pmod{4}$ implies that the remainder is $ 23\cdot 2 + 3 = 49$.

[quote="b555"]waiting  for innovative  answers[/quote]

Nothing is wrong with Euler's Theorem.

For Euler's Theorem look [url=http://lmgtfy.com/?q=Euler%27s+Theorem]here[/url].

so basically fermat's theorem. And nice website BTW.

To be precise, Euler's Theorem is a generalization of Fermat's Theorem to non-primes--clearly $ \phi(p) \equal{} p \minus{} 1$ when $ p$ is prime. Notice that to apply Euler's Theorem to $ a^{\phi(n)} \pmod{n}$, we are required that $ (a,n) \equal{} 1$. If not, you may have to use the Chinese Remainder Theorem!

You might also be interested in learning about the [url=http://en.wikipedia.org/wiki/Carmichael_function]Carmichael function[/url], because $ \phi(n)$ is not always the smallest $ b$ such that $ a^b \equiv 1 \pmod{n}$, for all $ a$ coprime to $ n$. Then you might be interested in the more general (for a [i]specific[/i] number $ a$, it solves $ a^b \equal{} c$ where $ c$ need not be $ 1$) discrete logarithm function (under the group of the integers modulo $ n$ under multiplication). Interestingly, there is no efficient algorithm for computing the discrete logarithm, unlike the algorithms for the Carmichael function and for Euler's totient function.


Let 2n distinct points on a circle be given. Arrange them into disjoint pairs in an arbitrary way and join the couples by chords. Determine theprobability that no two of these chords intersect. (All the possible arrangement into pairs are supposed to have the same probability.)

[i]Mikls Schweitzer competition 1952 problem 6[/i]

The number of desired arragements satisfies famous recurrence an=a0an-1+a1an-2+...+an-1a0, a0=a1=1. Thus an=C(2n,n)/(n+1).

The number of all arragements satisfies recurrence bn=(2n-1)bn-1, b1=1 ==> bn=(2n-1)*(2n-3)*...*3*1=(2n-1)!!=(2n)!/(2nn!).

So answer is an/bn = 2n/(n+1)!.

Is that correct?


integration
real analysis
real analysis unsolved


$1)$ How many natural numbers require $3$ digits when written in base $12$, but require $4$ digits when written in base $9$?

$2)$When the number $n$ is written in base $b$ its representation is the two digit number $AB$ where $A = b-2$ and $B=2$. What is the representation of $n$ in base $(b-1)$?

[hide="Question 1"]
A number $n$ has 3 digits in base 12 if $12^{2}\le n<12^{3}$, and it has 4 digits in base 9 if $9^{3}\le n<9^{4}$.  Then, for both of these requirements to be met, since $9^{3}>12^{2}$ and $12^{3}<9^{4}$, we know that $9^{3}=729\le n<12^{3}=1728$.  Then there are $1728-729+1=1000$ numbers that require 3 digits in base 12 but 4 in base 9.[/hide]

[hide="Question 2"]
$n=(b-2)\times b+2=b^{2}-2b+2=(b-1)^{2}+1$.  Then $n$ in base $b-1$ will be the two digit number $AB$ where $A=b-1$ and $B=1$.[/hide]

[quote="samath"][hide="Question 1"]A number $n$ has 3 digits in base 12 if $12^{2}\le n<12^{3}$, and it has 4 digits in base 9 if $9^{3}\le n<9^{4}$.  Then, for both of these requirements to be met, since $9^{3}>12^{2}$ and $12^{3}<9^{4}$, we know that $9^{3}=729\le n<12^{3}=1728$.  Then there are $1728-729+1=1000$ numbers that require 3 digits in base 12 but 4 in base 9.[/hide][/quote]
Actually, you don't need to add 1 at the end because you're not counting 1728.


uh..not sure
but [hide="2"]So the different digits are 2 1 5 0, for that there are 3*3*2=18 differnt 3 digit numbers for when all digits are distinct
Now lets have the case wehre there are already 1 1, for that you can have 1 1 5, 1 1 0, or 1 1 2

FOr the 1 1 5 and 1 1 2 cases there are 3C2 different ways so 2*3C2=6
For the 1 1 0 case there are 3C2-2C2 (for when 0 is leading digit)=2

so 18+6+2=26?[/hide]
[hide="3"]uhh this is probalby wrong, but consider the case CA as one, so we have seven spots to choose from but N and D cannot be together so there are 6C2=15 ways of them not being together.  However they are premutated so there are 15*2 (ND or DN), for the rest five spots there are 5! ways, but for the CA it is also permutated so there are 2 differnt ways, so 15*2*5!*2=7200?  This is probabky wrong because I have not considered the fact for there are two different A's.
[/hide]

[quote="macondo"]1. In how many ways can four men and four women be seated around a circular table if each man must be flanked by two women?

2. In how many ways can you form a three-digit number using the digits of the number 21150?

3. Calculate the number of permutations of the word CALENDAR in which C and A are together but N and D are not.[/quote]
[hide="1"]
The order must alternate between the men and women. We take one man as a reference point, and arrange the others from his viewpoint. There will be 4 possible women first, then 3 men, three women, etc. Thus there are $4! \cdot 3!= \boxed{144}$ distinct arrangements. [/hide]
[hide="2"] Case-work:
No ones: $3!-2!=4$
One one: $4 \cdot 3 \cdot 2-3 \cdot 2=18$. ([b]EDIT:[/b] It should be $3 \cdot 3 \cdot 2-2 \cdot 2=14$ since the first number can't be a zero, and when subtracting, there are two numbers other than the one, and two spots in which to put it.)
Two ones: Three spots for the other digit, then three numbers, so $9$. But we subtract the case $011$; thus there are $8$ of this case.
In total we have $4+14+8 = \boxed{26}$ total numbers. [/hide]
[hide="3"] Treat $CA$ as an element $X$. We are arranging the letters $XLENDAR$.
We count the total number of permutations, then subtract those with N and D together:
\[\frac{7!}{2!}-\frac{6!}{2!}= 2160 \][b]EDIT[/b]: We divide by $2!$ to account for the equivalent arrangements when either of the $A$s is placed within the element $X$. But we must also account for the distinct arrangements of element $X$ as either $CA$ or $AC$, thus we multiply by two, and the answer is $\boxed{4320}$.[/hide]

[quote="Art of Owna"]cincodemayo5590, im pretty sure there are only [b]26[/b] for number 2..for your one  1 case, there are only 14 that work, not 18 (it isnt 4*3*2-3*2, it is 3*3*2-2*2)

and im pretty sure number 3 isnt right either (i dont know if mine is right either), but im pretty sure the answer is bigger because you counted the premutations to XLENDER, but remember you only said X was CA, X can be AC too, so you forgot to multiply by 2..[/quote]

Oh, right. I'll edit it shortly.  :)

art of owna has the correct answers for 2 and 3 and you have the correct answer for number 1.

The two A's are inconsequential because if you take both of them into account, it'll be 5!*2 [for the two As] * 2 [for the flip-flop] / 2! [for the repeating As].

Hmm..another way to count for number 3 is total-the ones that dont belong..which is actually then my initial method (even if its right), so its actually 7!*2-6!*4=7200

because like what cincodemayo5590 said do, XLANDER, where X is the combination CA, so now there are 7! for that but multiply that by 2 becase it can be AC or CA.  But you dont want the ones where N and D are together so now call ND Y, so XYLAER, there are 6! but X can be AC or CA and Y can be ND or DN so 6!*2*2, so 71*2-6!*4=7200

and yea i thought the As wouldnt matter because for the X part Im only choosing one of the As (i could multiply X by 2 because there are 2 As, but later I would then divide by 2 because I doubled everything)

Your second explanation was much clearer, thanks for that. I was utilizing the indirect method as well but instead of 7! * 2 - 6! * 4, I multiplied 7! * 2 - 6! * 2. I see why there are double cases for ND being together as well.

And to cincodemayo, thanks for your explanation too. I was doing it manually, 4! x 4! [for men and then women], but didn't know why I had to divide by 4 [for the repeating positions] instead of 8. It's much easier to do it from a reference point.

Can someone help me figure out where I went wrong for #2?

I basically used permutation to find all possible 3-digit numbers using those five numbers, which is (5P3)/2! (the 2! is from the repeating 1's), and that counts to 30 possible numbers.

Then I subtract the cases where the three-digit number starts with a 0. Since we used up zero (so theres four numbers left) and we still need to fill up two digits, I used (4P2)/2! = 6.

30 - 6 = 24, but the anwer is 26...

edit:
[quote="macondo"]
I was doing it manually, 4! x 4! [for men and then women], but didn't know why I had to divide by 4 [for the repeating positions] instead of 8. It's much easier to do it from a reference point.[/quote]

I believe you divide it by 4 because it is a circular permutation. This site should explain it nicely:
http://mathworld.wolfram.com/CircularPermutation.html


OK, I know I'm not really an authority on this forum, but I thought it would be really nice if we revived the physics marathon!  The problems there are really cool, and even by looking at you guys solve problems, I learn a lot.

And plus, since I picked up some E&M/Mechanics last semester, I can dabble at it too now.. but that's another story.

So anyone out there interested?

I'll soon post a problem if I see responses  :D

I support the idea. By the way, I just noticed one interesting thing: two of three moderators (Rushil, xxreddevilzxx) have been inactive for the last 2 years, so Pardesi is de facto only mod here - I haven't noticed that before...

Pardesi is now studying at IIT (I don't know which one, perhaps Madras) so he too is [i]erm[/i]... dormant.

Hey guys!  I've posted my first problem in the marathon thread... I hope it's not too challenging, and fun to solve.  Good luck!   :D

[quote="integration"]What about having you Rajiv moderate the marathon?[/quote]

Thank you. I would love to but I'll be busy with admission procedures for quite some time. Nevertheless, I have posted a problem in the thread.