So coming in a few days is the most awaited book of the year Harry Potter and the Deathly Hallows and perhaps the most awaited movie of the year Harry potter and the Order of the Phoenix. What do you feel, as a viewer/reader is more entertaining, reading a book or watching a movie based on that book?

[quote="sonny"]What do you feel, as a viewer/reader is more entertaining, reading a book or watching a movie based on that book?[/quote]
Despite the fact that your inspiration for this question was Harry Potter, I think it's a very good question.

Personally, I think it can go either way. If I had to make a general statement, I'd probably prefer to read a book. However, I have seen some movies that were as entertaining as, if not more entertaining than, the book. A prime example is "The Swiss Family Robinson." Also, I found the screen version of James Clavell's Shogun to be very interesting and engaging.

These are two different cases: in the first one, I found the book kinda boring, and the movie producers "spiced it up" a bit; in the second, the book was wonderful, and the movie was true to the book, so the movie was wonderful as well.

You have to note that movies and books have different strengths. Books allow you to follow many more characters. Although modern CGI animation technology is pretty amazing, books can still present more fanciful things that movies can (and can explain what they are much more efficiently). On the other hand, movies give you a lot more details at once, so they are generally more intense and emotional. So a movie will improve on a book when the producer/director/whoever makes the decisions takes advantage of these differences and (a) chooses a book which doesn't do too much that a movie can't do and (b) uses the advantages of the cinema format to develop and present the story in ways that the book can't.

Anyway I would be interested to hear other opinions on this. Movies are generally "easier" to enjoy than books, I think, so some people (especially younger people) might prefer movies over books for that kind of reason. Movies are also usually much less time-consuming. (This is part of why the Swiss Family Robinson movie improved over the book. Not much happens in the book, but it's a bit long.) Ok I've really probably said far more than anyone else cared to read, so I'll shut up now.

I think it's impossible to make a general statement. Some very good movies were based on rather boring books which never achieved much acclaim -- a lot of Hitchcock's movies fall into this category.  On the other hand, almost as a rule, if I like a book, I won't like the movie made after it. For example, the Lord of the Rings: I really enjoyed the book and waited very eagerly to see some of my favorite scenes acted out, only to be disappointed after seeing the movie. The same happened with the Wizard of Oz some years ago (I read the book before watching the movie) and with several other movies, including the first Harry Potter (I haven't bothered to see other HP movies after I saw the first one).

Regarding Harry Potter, I most certainly preferred the book over the movie. As Djole said, various scenes I liked were cut out of the movie and the overall quality made it uninteresting. (I now only watch the Harry Potter movies to laugh at how bad they are :P  :rotfl: )

[quote="Xevarion"][quote="sonny"]What do you feel, as a viewer/reader is more entertaining, reading a book or watching a movie based on that book?[/quote]
Despite the fact that your inspiration for this question was Harry Potter, I think it's a very good question. [/quote]

well not exactly, i have pondered this question many times as a reader of starwars, lord of the rings and harry potter, but since harry potter is just like a few days away, i just used that as an example.....

I always like books more than movies because it seems that the movies misinterpret what the people read. But then again there are rare cases like Howl's Moving Castle.

[quote="Walk Around The River"]I dont really know how to read, so i'd say movies.

Yeah, its pretty sad, i know.

But like if i try to read, i just kinda stare at the paper and think about other things and i cant get myself to focus on the book.  So my mom reads to me whenever i have to read a book.  :oops:

And the movies are cooler because you dont have to think about it; the scenery and everything is there in front of you.[/quote]

Whoa but like.  If you can't even focus on the stuff you have to read, how the heck do you stay awake during movies?  Especially if it's in a movie theater.  All quiet, and all dark, and the seats are so comfortable... mmm

it's easier to stay awake during movies cause movies are passive

That's like saying it's easier to stay awake in MOP lectures, because they're passive.

Oh wait a minute.

But yeah, I haven't read a book in forever. I prefer going to movies cuz it's actually something social to do, as opposed to locking yourself up in a room and reading a book.

Some things are really only good in movie format. Movies like Idiocracy and the Matrix can't be books and still be as great.

[quote="Xevarion"][quote="sonny"] Although modern CGI animation technology is pretty amazing, books can still present more fanciful things that movies can (and can explain what they are much more efficiently).[/quote][/quote]
Funny, a few days ago I was watching Pearl Harbor ( :o ) and "Superman Returns", and I was thinking : CGI animation isn't really that good, is it? It's not as if it looked "real". :huh:


[quote="amir2"]prove:

$\sum_{i=0}^n \binom {n}{i} (2^k-1)^i=2^{kn}$[/quote]
nice

from binomial expansion , we know that

$(x+1)^n=\sum^n_{i=0}\binom{n}{i}x^i$ and the equality above is just by substituting $x=2^k-1$ .........

Edit : add in the $\binom{n}{i}$ thing

nice shyong .
But a little typo.
$(x+1)^n=\sum_{i=0}^n \binom {n}{i}x^{i}$


calculus
integration
function
real analysis
real analysis unsolved

On an interval [a,b], if f(x) is differentiable and f '(x) is bounded, then f(x) is integrable.  Prove this directly from the definition of integrability, without using the fact that it is continuous.

Definition of integrability:  A function f is called integrable on [a,b] if it is defined and bounded on [a,b], and it satisfies--
given e>0, L(P) e close to U(P) for all P such that |P| close to 0.

L(P) is lower sum of the partition
U(P) is upper sum of the partition

This is a weird question to me. You may just repeat the standard proof if you want? Maybe in this case the argument can be simplified a little by the fact that the mean value theorem implies that $f$ is Lipschitz continuous (http://planetmath.org/encyclopedia/LipschitzFunction.html).

You can use the Riemann condition to show that continuous functions are Riemann integrable on a compact set (it follows from uniform continuity that they are bounded). Unless you want to actually partition things in which case you can choose epsilon, use uniform continuity to get delta, and choose any partition of the closed interval whose norm is less than delta.


On a circle we write [b]p pluses[/b] and [b]q minuses. [/b]
Let [b]a[/b] be equal to the number of pair of pluses standing besides each other and
[b]b [/b]is a number of pair of minuses standing besides each other.
Prove that:
[b]|a-b| = |p-q|[/b]

Take two adjacent opposite signs and cancel them out. The quantity $|p-q|$ stays constant after such an operation, and all you need to do is check a few cases ($3$, I think) to convince yourself that $|a-b|$ also stays constant, so you can keep doing this until you're left only with $+$ or only with $-$ (or with neither), in which case the conclusion obviously holds.

A different solution:

Formalize the problem as follows:

[color=blue][b]Problem.[/b] Let $n$ be a nonnegative integer, and let $x_0$, $x_1$, ..., $x_{n-1}$ (indices are taken modulo $n$; this means that $x_n=x_0$) be $n$ numbers from the set $\left\{-1;\ 1\right\}$.

Let $p$ be the number of indices $i$ with $0\leq i<n$ satisfying $x_i=1$.
Let $q$ be the number of indices $i$ with $0\leq i<n$ satisfying $x_i=-1$.
Let $a$ be the number of indices $i$ with $0\leq i<n$ satisfying $x_i=x_{i+1}=1$.
Let $b$ be the number of indices $i$ with $0\leq i<n$ satisfying $x_i=x_{i+1}=-1$.

Prove that $a - b = p - q$.[/color]

(I have removed the modulus signs since I really don't see their use.)

Now, the [i]solution[/i] can be done easily:

For every index $i$,

- either $x_i=x_{i+1}=1$, what yields $\frac{x_i+x_{i+1}}{2}=1$;
- or $x_i=x_{i+1}=-1$, what yields $\frac{x_i+x_{i+1}}{2}=-1$;
- or one of the numbers $x_i$ and $x_{i+1}$ is $1$ and the other one is $-1$; in this case, $\frac{x_i+x_{i+1}}{2}=0$.

Hence,

$\sum_{i=0}^{n-1} x_i=\sum_{i=0}^{n-1}\frac{x_i+x_{i+1}}{2}=\sum_{i=0}^{n-1}\left\{\begin{array}{ccc}1\ \ \ \ \text{if }x_i=x_{i+1}=1\\-1\ \ \ \ \text{if }x_i=x_{i+1}=-1\\0\ \ \ \ \text{otherwise}\end{array}\right.$
$=\left(\text{number of }i\text{ such that }x_i=x_{i+1}=1\right)-\left(\text{number of }i\text{ such that }x_i=x_{i+1}=-1\right)$
$=a-b$.

On the other hand,

$\sum_{i=0}^{n-1} x_i=\sum_{i=0}^{n-1}\left\{\begin{array}{cc}1\ \ \ \ \text{if }x_i=1\\-1\ \ \ \ \text{if }x_i=-1\end{array}\right.$
$=\left(\text{number of }i\text{ such that }x_i=1\right)-\left(\text{number of }i\text{ such that }x_i=-1\right)=p-q$.

Hence, $a - b = p - q$, and the problem is solved.

By the way, I recall seeing the problem somewhere before, but I can't tell right now where...

  Darij


analytic geometry
combinatorics proposed
combinatorics

Let $\mathcal P$ be the set of all points in $\mathbb R^n$ with rational coordinates. For the points $A,B \in \mathcal l{P}$, one can move from $A$ to $B$ if the distance $AB$ is $1$. Prove that every point in $\mathcal l{ P}$ can be reached from any other point in $\mathcal{P}$ by a finite sequence of moves if and only if $n \geq 5$.

This one appeared many times here, and I solved it somewhere. Search...

Pierre.

Here it is :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=lagrange&t=2635

Pierre.


Prove (or disprove) that for  all $ x\in\mathbb{Z}$, $ 4^x\plus{}x^4$ is prime if and only if x=1.

[quote="xpmath"]Prove (or disprove) that for  all $ x\in\mathbb{Z}$, $ 4^x \plus{} x^4$ is prime if and only if x=1.[/quote]

Wasn't this a past USAMO problem?
[hide]
Yeah, it's true.
If:
If x=1, then $ 4^x\plus{}x^4\equal{}1\plus{}4\equal{}5$, which is prime.

Only If:
If $ x$ is even, then it is obvious that the term is composite.

So, if $ x\equal{}2k\plus{}1$, then notice that \[ x^4\plus{}4^x\equal{}(x^2)^2\plus{}(2^x)^2\equal{}(x^2)^2\plus{}(2^x)^2\plus{}2(x^2)(2^x)\minus{}2(x^2)(2^x)\] \[ x^4\minus{}4^x\equal{}(x^2)^2\plus{}(2^x)^2\plus{}(x^2)(2^{2k\plus{}2})\minus{}(x^2)(2^{2k\plus{}2})\equal{}(x^2\plus{}2^x)^2\minus{}[(x)(2^{k\plus{}1})]^2\] \[ x^4\plus{}4^x\equal{}[x^2\plus{}2^x\minus{}x(2^{k\plus{}1})][x^2\plus{}2^x\plus{}x(2^{k\plus{}1})]\] So, for our term to be prime, \[ x^2\plus{}2^x\minus{}x(2^{k\plus{}1})\equal{}1\] We can easily see that this is only the case when $ x\equal{}1$.

Since we have proved both cases, our solution is complete.
[/hide]

Nice job :)
I didn't realize that this was an old USAMO problem;my homework asked to find the value for which that expression was prime, and then I decided to prove it was the only value that worked.


hi. i live in england and my geomtery is dredful though i am resonably competent at other bmo maths, (i can normally anwser 1-2 questions at bmo1) so i am wishing to improve my geomtery. i have tried to work through the eucilidean geomtery book published by ukmt but i think that i first need to improve my geomtery before tackling this book as it is slightly advanced for me. do you know of any online worksheets or books for sale aimed someeone like me. thank you for your time

Just out of interest, did you start at chapter 1 of PEG? Personally I feel that the first two chapters of that book are interesting to approach in a sort of philosophical how-can-I-prove-this-thing-that-seems-unbelievably-obvious way, but that to improve your geometry you are better off starting at chapter 3. Obviously practicing axiomatic geometry (the subject of the first two chapters) does improve your geometry, but in terms of olympiad preparation, chapters 3-4 are very useful. There is always "The Mathematical Olympiad Handbook" which has all the BMO papers since the 60s, which would contain lots of geometry problems and outline solutions, but I think you are right and there is a gap in the market in the pre-Olympiad level geometry section...

thank you , do you think that if i struggled with chapters 1 and 2 i will struggle more with the later chapters? ive got the olimpiad handbook so i ll look through that too. thanks once again

Well, it is possible that you will struggle with the later chapters too, but the experience will probably be slightly more useful for you than the first two. Being able to use any theorem, rather than just those which you have just proved, may make it simpler too...!

I hate to contradict the honourable Mr Yeo, but I actually believe chapters 1 and 2 are very worthwhile unless you are actually in intensive training for the IMO, which I'm guessing you aren't. Chapter 3 onwards deals with interesting topics (like triangle geometry) which you need to know to be fluent at IMO level, but for BMO1 a more sound and secure understanding of how circle theorems, similarity, etc. works is going to be more useful than knowing about the orthocentre.

Perhaps what you might be finding offputting, Mathematix, is the fact that it isn't immediately obvious how to solve the problems. Be prepared to spend an hour or more on an "exercise" trying things out, and re-reading the relevent section a few times if necessary.

To let Dom and I help actually work out what trouble you are having, it might help if you attempt Q3 from BMO1 2005/6 here, letting us know how far you manage to get?

Ilthigore i shall give that question a go over the next few days. i think that you are correct about me not spending long enough on the questions. thanks for your help


A train is formed exactly as shown by connecting regular hexagons and regular triangles alternately, beginning with a hexagon. The side length of each polygon is 1 cm. How many hexagons are in a train that has a perimeter of 426 cm?

[asy]path A = dir(0)--dir(60)--dir(120)--dir(180)--dir(240)--dir(300)--cycle;
for(int i = 0; i < 4; ++i){
draw(shift((2*i,0))*A);}
draw(dir(240)--dir(240)+(7,0));[/asy]

upper side: perimeter of 3x
lower side: perimeter of 2x+1
Total perimeter: 5x+1=426
5x=425
x=85
[b]85[/b] hexagons total.


I have no idea of the difficulty of this question, it came to mind while I was thinking about a much simpler version.

Place n:ge:3 points inside (or on) a given triangle area 1. Let M be the minimal area of a triangle formed by three of these points (degenerate triangles included). What is the maximal value of M?

For example, n=3, M=1, n=4, M=1/3, etc.

In fact, n=5 doesn't even seem too simple..

Problems like this have been studied in a field called discrepancy theory.  For example, if the outside triangle is replaced by a rectangle, then I recall results on this problem.  Exact results were hard, but asymptotic results were possible.

If I can remember some references later, I will post.

I finally remembered something more about this problem.  It is called Heilbronn's triangle problem.  You can find an interesting discussion of it here:
[url]http://mathworld.wolfram.com/HeilbronnTriangleProblem.html[/url].


[hide="Some mumbo-jumbo ideas"]Vieta, AM-GM, binomial[/hide]

[hide]
There are 6 positive roots to this polynomial.
The sum to $6$ and multiply to $1$ based on the given coefficients.

$AM\geq GM \implies \frac{6}{6}\geq \sqrt[6]{1}\implies 1 \geq 1$.

Since we have an equality case, all 6 roots must be equal, so they are all 1.

$(x-1)^{6}=x^{6}-6x^{5}+15x^{4}-20x^{3}+15x^{2}-6x+1$
$\implies (a,b,c,d)=(15,-20,15,-6)$
[/hide]


In this island Bob is visiting, there are truth tellers and lie tellers. The truth tellers always tell the truth, and the lie tellers always tell lies, and believe whatever Bob say is a lie because of defects in their ears, and makes an answer based on it.
For example, if Bob asks a truth teller what color the sky is, and it is blue, they will say blue. But if Bob asks a lie teller, they think Bob is asking what color the sky is not, and will lie and say blue. So for this question, Bob would get the same answer from both. Bob went on a trip to the forest and is on his way back from a day of hunting to his camp, when he sees a solitary inhabitant of the island running. Bob asks him why, and the denizen says that the island will blow up in a day, and that he is running to a nearby ship. Not having a lab in his pockets, he cannot examine the inhabitant's ears, and does not have enough time to return to the main city and see if people are evacuating or to get the runner's ears examined. Also, the answer the runner gave could be given by a liar or truth teller. He asks the  runner one question and sees the runner is a liar. What question does he ask?

Wait a second...
[hide]
If Bob asks the runner why he is running, then the liar will think he asked why aren't you running, which would be because of the whole island thing...[/hide]

Really?
[hide="I thought"]
that if Bob asks the inhabitant if he's a liar, a liar would hear "are you not a liar?" or "are you a truth-teller?" and the inhabitant would respond "yes," thinking he was lying about being a truth-teller yet letting Bob know he actually is a liar, because any truth-teller would have said, "No."[/hide]


Here's another Mathcounts problem from this year's 2007 state competition team round:

In a particular list of three-digit perfect squares, the first perfect square can be turned into each of the others by rearranging its digits. What is the largest number of distinct perfect squares that could be in the list?

Please don't send me any replies copied from the rarely but sometimes lousy Mathcounts solutions explanations.  Trust me, I checked the explanation for this question that required me to know [b]all[/b] of the three-digit perfect squares. :huh:

Our team actually DID list out all the 3-digit squares, and we got it right :lol:

[hide="3-digit squares"]
100 121 144 169 196
225 256 289 324 361
400 441 484 529 576
625 676 729 784 841
900 961

Most likely, the winning group has all digits different.
The squares with all digits different are:
[b]169[/b] [b]196[/b] 256 289 324 361 529 576 625 729 784 841 [b]961[/b]

So we agreed on 169,196, 961 as the largest group.

At least, that's what I THINK we did... I can't remember  :fool:
[/hide]

yeah, i think we listed all the squares too. its not that hard, is it?

[quote="jhredsox"]i was so mad at this one. i spent about 6 minutes on it only because i couldn't find any squares that had the same digits in ALL different positions. then dan's like, "jorian, that takes two seconds!" and i'm like "oh, i was doing it wrong"

yeah, i knew all of the ones that had the same digits (please don't ask) so i just wrote those down when doing this problem

jorian[/quote]

um can you PLEASE stop ranting about how you should have gotten all the problems that you got wrong correct, because it's really annoying. It's pretty useless for you to say these types of things, no one actually cares and you're not really benefiting anyone. If you want to whine about your scores, etc., fine, but don't do it here.

[hide="alternate quote solution unquote"]Well you can note that 169, 961, 196 works, pretty easiliy, because 169 is the third that you check, and then note that in order to have more than 3, we need abc and acb to be squares for some digits a, b, c, and one can check those instead. (for example, take 324, 342 isn't a square) Those ensure that you aren't missing any, and you don't have to check groups of 3, you only have to check them by ones.[/hide]

yeah, thats what my team did.  i tried to think of 3-digit square numbers, and originally i thought it was 256 and 625.  but then i found 169, 196, and 961, and i knew that it had to be the answer.

First of all, jorian, could you stop bringing up dan every time you post? It sounds as if he's not nearly as genius as you make him out to be, and I'm sick of hearing about him.

Second, I had recently considered this problem, then it turned out to be the only one I didn't do. We still got it right though.


Putnam
geometry
function
trigonometry
area of a triangle
Heron\u0027s formula
college contests

[quote="Kent Merryfield"]For $i=1,2,$ let $T_i$ be a triangle with side length $a_i,b_i,c_i,$ and area $A_i.$ Suppose that $a_1\le a_2, b_1\le b_2, c_1\le c_2,$ and that $T_2$ is an acute triangle. Does it follow that $A_1\le A_2$?[/quote]

Yes, it does. At first, in order to prevent a conflict of notations, I rewrite the problem with different notations:

[i]Let $T_1$ be an arbitrary triangle and $T_2$ be a non-obtuse triangle. For i = 1 and i = 2, let $a_i$, $b_i$, $c_i$ be the sidelengths of triangle $T_i$. Suppose $a_1 \leq a_2$, $b_1 \leq b_2$, $c_1 \leq c_2$. Prove that $S_1 \leq S_2$, where $S_i$ is the area of triangle $T_i$ for i = 1 and i = 2.[/i]

[i]Proof.[/i] Let $A_i$, $B_i$, $C_i$ be the angles of triangle $T_i$ for i = 1 and i = 2. Then, since the angles of the triangles $T_1$ and $T_2$ have the same sum (namely $A_1+B_1+C_1=180^{\circ}$ and $A_2+B_2+C_2=180^{\circ}$), there is at least one angle of triangle $T_1$ which is less or equal to the corresponding angle of triangle $T_2$. WLOG, we have $A_1 \leq A_2$.

Since the triangle $T_2$ is non-obtuse, we have $A_2 \leq 90^{\circ}$; since $A_1 \leq A_2$, this yields that $A_1 \leq 90^{\circ}$, too. Now, the function $f\left(x\right) = \sin x$ is monotonically increasing on [0°; 90°]; thus, since $A_1 \leq 90^{\circ}$ and $A_2 \leq 90^{\circ}$, from $A_1 \leq A_2$, we can conclude $\sin A_1 \leq \sin A_2$. Together with $b_1 \leq b_2$ and $c_1 \leq c_2$, this yields

$S_1 = \frac12 b_1 c_1 \sin A_1 \leq \frac12 b_2 c_2 \sin A_2 = S_2$.

$\blacksquare$

  Darij

Prove or disprove similar statement for inscribed quadrilaterals $T_1$ and $T_2$, which contain their circumcenters.

If it were that simple, we wouldn't have to require that the triangle with larger sides is acute.

Why can we use without loss of generality here?

Because we can simply rename the vertices without changing any of the conditions.

[quote=h_s_potter2002]Uh, sorry if this is a dumb question, but the solution on the AMC website says that Heron's formula can also be used.  So if that's true, how come there were so many 0's on this problem, since I think using Heron's makes this problem kind of easy.[/quote]

(Huge revive, sorry, but I was playing around with this problem today.) While it's true that you can use Heron's formula (it was my first thought too, and my idea is below), I don't think it makes it much easier. You still have to make use of the acute-ness of the ending triangle.

Using Heron's formula is essentially equivalent to looking at the function $$f(a,b,c) = 4a^2b^2 - (a^2 + b^2 - c^2)^2,$$ and trying to show that it's "monotone-ish" (the ish being because we need to assume the ending triangle is acute). Of course, $\angle C$ being acute is equivalent to $a^2 + b^2 - c^2 > 0$. Using this formula you should then note that the function is increasing locally at $c$ if the angle opposite $c$ is acute. (Similarly for $a$, $b$, by symmetry.) Which essentially solves the problem.

[hide=Solution]
The answer is yes.

What we do is to get from smaller triangle to the larger one is as follows: we increment one side by +1 at a time. If we increment a triangle from $(a, b, c)$ to $(a, b, c+1)$, and both triangles are acute, then $\cos{C}$ is equal to $\frac{a^2+b^2-c^2}{2ab}$ in the smaller triangle and $\frac{a^2+b^2-(c+1)^2}{2ab}$ in the second triangle so that the cosine decreases while still being positive. Hence, the sine increases, and by $K = \frac{ab \sin{C}}{2}$, the area will increase. Hence, it would suffice to find a sequence that takes $(a_1, b_1, c_1)$ to $(a_2, b_2, c_2)$, all of which are acute triangles. Assume WLOG $a_i \le b_i \le c_i$ for both values of $i$ (stricter constraints only constrict the triangle pairs we can make). The simplest way to do this is by incrementing the smallest unfinished segment by $1$. However, we run into a roadblock when we cannot do this to get from $(2, 6, 6)$ to $(2, 7, 7)$. The fix to this is that when there is a tie, we increment all of the tied segments at once. Notice that in the case of incrementing two at once, the isosceles triangle merely increases in height (if the unchanged segment is the base), while for all three at once, the equilateral triangle merely expands.

For example, to get from $(4,5,6)$ to $(6,8,9)$, we go $(4,5,6), (5,5,6), (6,6,6), (6,7,7), (6,8,8), (6,8,9)$.

To finish, we need to verify that such a sequence always exists. We check all of the cases. If $c_1 \ge b_2$, then we can increment the $a$ and $b$ terms in any way we would like before finishing by increment the $c$ term as this keeps $a^2+b^2 > c^2$. If $c_1 \in [a_2, b_2)$, then we increment $a$ as normal, but take $b$ to be the same as $c$, increment them together, then let $c$ take off. Finally, in the case that $c_1 < a_2$, we perform this incrementing process until we get an equilateral triangle of side length $c_1$, increment them all to $a_2$, increment the latter two to $b_2$, then finally take the last segment off to $c_2$. This completes the proof.
[/hide]

I doubt this is rigorous/complete. I would appreciate some feedback.
[rule]
The answer is [b]yes[/b].

Consider starting at the acute $T_2$ and transforming the triangle to $T_1$ by decreasing the lengths of its sides.

To preface, consider $T_2 \equiv \triangle ABC$ with side lengths $BC=a$, $CA=b$, and $AB=c$.

If we decrease the length of a side, then the angle opposite that side will always decrease. This can be shown with the Law of Cosines. Without loss of generality, suppose that we decrease $c$. Then, $$c^2=a^2+b^2-2ab\cos(\angle C).$$ As $c$ decreases, we must have that $-2ab\cos(\angle C)$ decreases, or that $2ab\cos(\angle C)$ increases. Since $\cos$ monotonically decreases from $0^{\circ}$ to $180^{\circ}$, $\angle C$ must be decreasing as well.

If we decrease the length of a side whose opposite angle is acute, then we will decrease the area of the triangle. Likewise, if we decrease the length of a side whose opposite angle is obtuse, then we will increase the area of the triangle. This is true by considering the sine-area formula. If the area of $\triangle ABC$ is $K$, then $$K=\frac{ab\sin (\angle C)}{2}.$$ Note that $\sin$ monotonically increases from $0^{\circ}$ to $90^{\circ}$ and monotonically decreases from $90^{\circ}$ to $180^{\circ}$. Keeping $a$ and $b$ constant shows that $K$ follows this same pattern.

Now, suppose that at some point in the process of decreasing $c$, one of the other two angles becomes right. Without loss of generality, let this be $\angle A$. Then, continuing to decrease $c$ until it reaches $c - \epsilon$ will make $\angle A$ obtuse. Then, decreasing $a$ until $\angle A$ becomes right would increase the area of the triangle. However, if we never made $\angle A$ obtuse, then $\triangle ABC$ would have a greater area because $$K_{\text{before } \angle A \text{ becomes obtuse}} = \frac{bc\sin(\angle A)}{2} > \frac{b(c-\epsilon)\sin(\angle A)}{2} = K_{\text{after } \angle A \text{ becomes obtuse}}.$$ In other words, the only situation in which we can increase the area of $\triangle ABC$ is overridden by a case where we never increase the area of $\triangle ABC$. Hence, our net change in area from $T_2$ to $T_1$ can never be positive, so $A_1 \leq A_2$.


[quote="sugarwergoingdown"]:idea: Scott and Deric are playing the Shade Game. Three standard dice are rolled, and the sum of the #s rolled is the # of blank squares that must then be shaded. It is Deric's turn, and his grid is shown here. What is the probability that Deric will roll the exact sum needed to shade the rest of the grid on this turn? Express yourself as a common fraction.
 :surf:[/quote]

for ur pic
is it 6 white squares on one side and 5 on another, making it $6+5-1=10$ squares? or is it like $6+5$?

It's 6 white squares on one side and 5 white squares on the other. sorry about that!  :D  :lol:

Is that a total of 10 squares unshaded or 11?

[hide="Solving for 10 squares unshaded"]There are $6^3$ possibilities for rolling three dice.  The numbers that sum up to 10 are:

136,145,244,235,226,334 (I might be missing some :ninja: )

3 of these can be rearranged in 6 ways and 3 can be arranged in 3 ways.  So, the answer is $\frac{18+9}{216}=\frac1{8}$.[/hide]

[hide]
If there are $10$ unshaded, he needs to roll a ten.
Combinations:
136
145
226
235
244
334

3 of these can be arranged in 6 ways, and 3 in 3.
There are $6^3=216$ total ways, so the probability is:
$\frac{18+9}{216}=\frac{27}{216}=\boxed{\frac 18}$[/hide]

both of you are cccccccccccooooooooooorrect! :D


Define a sequence $ \left( x_n \right)_{n\equal{}1}^{\infty}$ by setting $ x_1 \equal{} 1$ and $ x_{n\plus{}1} \equal{} \frac{x_n}{1 \plus{} \sqrt{x_n}} \; \forall n \ge 1$. Show that $ \left( x_n \right)_{n\equal{}1}^{\infty}$ converges, and determine its limit.

I know this is monotonic decreasing since $ x_{n\plus{}1} \equal{} \frac{x_n}{1 \plus{} \sqrt{x_n}} \le \frac{x_n}{1} \equal{} x_n$. Do we need to use the formal (epsilonic) definition of a limit to prove that it converges or is it sufficient to claim there is a lower bound?

If we assume there's a limit then it is easily found to be $ 0$.


Let p and q be two distinct primes
a) Prove that the greatest common divisor of p and q is 1
b) Prove by contradiction that p doesn't divide pq+1
c) Prove by contradiction or contraposition that if pq divides c^2 for some postive integer c, then both p and q divide c.

[hide="Quick Hints"]
a) Bezout's, ie, $ ap\plus{}bq\equal{}1$
b) $ pq\plus{}1\equal{}pk$, for some integer $ k$, apply a)
c) If prime $ p$ divides $ ab$, either $ p|a$ or $ p|b$.
[/hide]

a) As $ p$ and $ q$ are distinct and their divisors which are bigger than $ 1$ is only themselves, then $ (p,q)\equal{}1$
b) Assume that $ p$ divides $ pq\plus{}1$.$ p$ divides $ pq$ and $ pq\plus{}1$ so $ (pq,pq\plus{}1)\ge p>1$. But $ (a,a\plus{}1)$ is always $ 1$. Contradiction.
c) $ c^2$ must be divisible by $ p$ and $ q$, and as $ p$ and $ q$ are primes, it follows that $ p$ divides $ c$ and $ q$ divides $ c$.

[quote="10000th User"][Bezout's, ie, $ ap \plus{} bq \equal{} 1$[/quote]
Bezout's is both a far stronger statement than you need and irrelevant to the problem (since we don't know what the $ \gcd$ is here).  Instead, apply the definition of primality as Bugi has.

By the way, sorry for not doing part c) by contradiction, but it's also correct!

[quote="Bugi"]
c) $ c^2$ must be divisible by $ p$ and $ q$, and as $ p$ and $ q$ are primes, it follows that $ p$ divides $ c$ and $ q$ divides $ c$.[/quote]

You simply restated the problem. Nothing said gives insight to [i]why[/i] the result "follows."

Assuming $ pq\nmid c$, but $ pq\mid c$ implies $ c\equal{}k\sqrt{pq}\in\mathbb{Z}$. This is only satisfied when $ p\equal{}q$, yet in reality $ p\neq q$. Contradiction.

[quote="igiul"] $ c \equal{} k\sqrt {pq}\in\mathbb{Z}$. This is only satisfied when $ p \equal{} q$[/quote]
Why?  (This is equivalent to the original problem.)

Again c)
WLOG assume that $ q\nmid c$.
Case 1) $ p\nmid c$
Then $ pq\nmid c$ and as $ p$ and $ q$ are primes, it follows that $ pq\nmid c^2$. Contradiction
Case 2) $ p\mid c$
Then $ q\nmid c^2$ and as $ p$ and $ q$ are not equal, it follows that $ pq\nmid c^2$. Contradiction


2. Find all pairs of natural numbers $(a,b)$ such that $\sqrt[3]{a+\sqrt{b}}+\sqrt[3]{a-\sqrt{b}}=1$.

Can anyone check my solution? Thanks,
[hide]
Subtract one from both sides, we get:
$\sqrt[3]{a+\sqrt{b}}+\sqrt[3]{a-\sqrt{b}}-1=0$.
Since if $x+y+z=0$, then $x^3+y^3+z^3=3xyz$.
Using this, we factor our equation and simplify, we get:
$2a-1=(-3)\sqrt[3]{a^2-b}$.
Since $\sqrt[3]{a^2-b}$ has to be positive, there's no solution.
[/hide]

[quote="kimby_102"]Since $\sqrt[3]{a^2-b}$ has to be positive[/quote]
No, it doesn't. A sample solution: $a=2,b=5$.

The solution should be easy to fix- you've got the right algebra leading up to it. A note on why we aren't introducing any false "solutions" would also help.

$\sqrt[3]{a+\sqrt{b}} + \sqrt[3]{a-\sqrt{b}}=1$
Since $(a+b)^3=a^3+b^3+3ab(a+b)$ we have (cubing both sides)

$(\sqrt[3]{a+\sqrt{b}}+\sqrt[3]{a-\sqrt{b}})^3=(a+\sqrt{b})+(a-\sqrt{b})+3(\sqrt[3]{a^2-b})(\sqrt[3]{a+\sqrt{b}}+\sqrt[3]{a-\sqrt{b}})=1$
Since the last term in parentheses is just $1$ (given), we have
$2a + 3(\sqrt[3]{a^2-b})=1$

Rearranging and cubing (again :( ) yields
$(3(\sqrt[3]{a^2-b}))^3=(1-2a)^3 \longrightarrow 27(a^2-b)=(-8a^3+4a^2-2a+1)$
 i don't really feel like finishing it up...
but, no, $\sqrt[3]{a^2-b}$ is not necessarily greater than zero, as jmerry pointed out

Oops, for some reason I thought it was sqrt... :blush:

[quote="quantum leap"] $27(a^2-b)=(-8a^3+4a^2-2a+1)$
 i don't really feel like finishing it up...
[/quote]Why give up so close to the goal?  From the above $27(b-a^2)=(2a-1)^3$ we have that $3$ must divide $2a-1$. As this is odd, we can denote $\frac{2a-1}3=: 2c-1$  with $c\in\mathbb N$, i.e. $a=3c-1$. So we have $b=(2c-1)^3+(3c-1)^2=8c^3-3c^2$. And this yields all solutions: for each $c\in\mathbb N$, we get $(a,b)=(3c-1,8c^3-3c^2).$


Princeton college counting distinguishability AMC

Well, the 5! factor implies that choice matters/we are counting permutations.

It's an issue on wording.

Admittedly, they look the same to me. If this appeared on a contest, I think you could protest that the wording is ambiguous.

[quote="1/(ln x)"]In Chapter 17 of AoPS 2,

Example 17-3 states:
In how many ways may 5 people be seated in a row of twenty chairs given that no two people may sit next to one another?

The answer is
[hide]${16 \choose 5}$[/hide]

However, Problem 290 states:
Twenty chairs are set up in a row for the Princeton garlic-eating contest (Yum, Yum). Only five eaters attend the competition, but none will sit next to any other. In how many ways can the eaters be seated?

The answer is
[hide]$5!{16 \choose 5}$[/hide]

Shouldn't the answer to Problem 290 be ${16 \choose 5}$? What is the difference between these two problems?[/quote]

Yes, you should protest since the wording is weird.  Also, we can't assume that the people aren't indistinguishable.

Well the first problem assume that they aren't, and it says nothing to that effect.

[quote="bpms"]I thought that people are distinguishable unless stated otherwise.[/quote]

Wow, I just said an incorrect $\text{triple}$ negative.

But yeah, it is really messed up that problem, but we get the general idea.


Sa se determine functiile contiunue $f: R\to\mathbb R$ pentru care $F(0)=0$ si
          $F(f(x))+f(F(x))=x^2$ unde $F$ este o primitiva a lui $f$.

$F(f(x))+f(F(x))=x^2.$ Differentiate the both sides with respect to $x,$

$F'(f(x))f'(x)+f'(F(x))F'(x)=2x,$ since $F'(x)=f(x)$ we have $f(f(x))f'(x)+f'(F(x))f(x)=2x.$

[quote="Thor"]$\ f$ or $\ F\circ f$ are not necessarly differentiable.[/quote]

Yes :P


What is |pi-3.14|+|pi-(22/7)|. Express as a fraction in lowest terms

Eh, please [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=73806649&t=195667]search[/url] in the future before posting.

Hey math888 its the azn guy from ur state. Ok... 22/7 is greater than pi.. If you type that in on a calculator, (which u can't in sprint cause i recognize this from last year) u find that me statement is true!!  So lets organize this..

l pi-3.14 l + l (22/7)-pi l

kk? i put the 22/7 in first because it's bigger and now we can clear out the absolute values...

pi - 3.14 + 22/7 - pi

and the pis cancel out which leave you 22/7 - 3.14

You convert them to fractions:

22/7 - 157/50 = 1100/350 - 1099/350 = 1/350!!!!!! yay!! :)


ABC is acute-angled. The circle diameter AC meets AB again at F, and the circle diameter AB meets AC again at E. BE meets the circle diameter AC at P, and CF meets the circle diameter AB at Q. Show that AP = AQ

AB is a diameter, so $\measuredangle AQB = 90^{\circ}$. Similarly, $AC$ is a diameter, so $\measuredangle AFQ = \measuredangle AFC = 90^{\circ}$. Hence triangles $AQB$, $AFQ$ are similar, so $AQ/AB = AF/AQ$, or $AQ^2 = AF \cdot AB$. Similarly, $AP^2 = AE \cdot AC$. But $\measuredangle BEC = \measuredangle BFC = 90^{\circ}$, so $BFEC$ is cyclic. Hence $AF \cdot AB = AE \cdot AC$


limit
integration
real analysis
real analysis unsolved

[color=darkblue]Let $ a$ , $ b$ be two positive  real numbers. Prove that :[/color]

 $ \blacktriangleright\lim_{n\rightarrow \infty}\sum_{k = 1}^{n}{\frac {1}{\sqrt {n^{2} + kn + a}} = 2\left(\sqrt 2 - 1\right)}$ ...... (see [url=http://www.mathlinks.ro/viewtopic.php?t=232927][color=red][b]here[/b][/color][/url]).

$ \blacktriangleright\lim_{n\to\infty}\ n^2\cdot\sum_{k = 1}^n\left(\frac {1}{\sqrt {n^2 + kn + a}} - \frac {1}{\sqrt {n^2 + kn + b}}\right) = \frac {2 - \sqrt 2}{2}\cdot (b - a)$ .

The first one is easy. Write:

$ \frac {1}{\sqrt {n^2 \plus{} k n \plus{} a}} < \frac {1}{\sqrt {n^2 \plus{} k n}} \equal{} \frac {1}{n} \frac {1}{\sqrt {1 \plus{} \frac {k}{n}}}$

So:

$ \lim_{n\rightarrow \infty} \frac {1}{n} \sum^n_{k \equal{} 1} \frac {1}{\sqrt {1 \plus{} \frac {k}{n}}} \equal{} \int ^1_0 \frac {1}{\sqrt {1 \plus{} x}}\,dx \equal{} 2 \sqrt {1 \plus{} x}\Big|^1_0 \equal{} 2 \left(\sqrt {2} \minus{} 1\right)$

As for the second, I believe it should be $ (b \minus{} a)\frac {2 \minus{} \sqrt {2}}{2}$. And here's my solution:

$ \frac {1}{\sqrt {n^2 \plus{} k n \plus{} a}} \minus{} \frac {1}{\sqrt {n^2 \plus{} k n \plus{} b}} \equal{}$

$ \frac {\sqrt {n^2 \plus{} k n \plus{} b} \minus{} \sqrt {n^2 \plus{} k n \plus{} a}}{\sqrt {n^2 \plus{} k n \plus{} a}\sqrt {n^2 \plus{} k n \plus{} b}} \equal{}$

$ \frac {n^2 \plus{} k n \plus{} b \minus{} (n^2 \plus{} k n \plus{} a)}{\sqrt {n^2 \plus{} k n \plus{} a}\sqrt {n^2 \plus{} k n \plus{} b}\left(\sqrt {n^2 \plus{} k n \plus{} a} \plus{} \sqrt {n^2 \plus{} k n \plus{} b}\right)} \equal{}$

$ \frac {b \minus{} a}{\sqrt {n^2 \plus{} k n \plus{} a}(n^2 \plus{} k n \plus{} b) \plus{} \sqrt {n^2 \plus{} k n \plus{} b}(n^2 \plus{} k n \plus{} a)} <$

$ \frac {b \minus{} a}{\sqrt {n^2 \plus{} k n}(n^2 \plus{} k n) \plus{} \sqrt {n^2 \plus{} k n}(n^2 \plus{} k n)} \equal{}$

$ \frac {b \minus{} a}{2\sqrt {n^2 \plus{} k n}(n^2 \plus{} k n)} \equal{} \frac {b \minus{} a}{2n^3}\left(1 \plus{} \frac {k}{n}\right)^{ \minus{} \frac {3}{2}}$

So:

$ \lim_{n\rightarrow \infty} n^2 \sum^n_{k \equal{} 1} \left(\frac {1}{\sqrt {n^2 \plus{} k n \plus{} a}} \minus{} \frac {1}{\sqrt {n^2 \plus{} k n \plus{} b}} \right)\equal{}$

$ \lim_{n\rightarrow \infty} \frac {1}{n} \sum^n_{k \equal{} 1}\frac {b \minus{} a}{2}\left(1 \plus{} \frac {k}{n}\right)^{ \minus{} \frac {3}{2}} \equal{}$

$ \equal{} \frac {b \minus{} a}{2} \int^1_0 \left(1 \plus{} x\right)^{ \minus{} \frac {3}{2}}\, dx \equal{} (a \minus{} b) (1 \plus{} x)^{ \minus{} \frac {1}{2}}\Big|^1_0 \equal{} (a \minus{} b)\left(\frac {1}{\sqrt {2}} \minus{} 1\right) \equal{}$
$ \equal{} (b \minus{} a)\frac {2 \minus{} \sqrt {2}}{2}$

It is nicely and O.K. at the right hand. And at the left hand ?!

I am sorry, I don't understand what do you mean.

He means you did half the work; your solution is incomplete

But I derived the RHS from the LHS? What is there more to prove? I showed that the subtraction of two fractions involving square roots can be shown in a certain way, and then plugged that into the limit, yielding the RHS! I'm confused..  :huh:

O.K., [b]Milin[/b] ! You are right. I didn't comprehead from the first, sorry ! It is nicely ! Thank you.


I found this interesting:
Solve for integer x,y:

x^y = y^x

in fact,it can be sloved x,y are all rational numbers.

[quote="cosinerburc"][hide]denote d=gcd(x,y) and x=da  y=db where gcd(a,b)=1. Then the equation becomes d^b.a^b=d^a.b^a. Assume that a>=b;then d^(a-b).b^a=a^b.Hence b^a divides a^b although
gcd(a,b)=1.The rest is easy.
[/hide]

zhaobin ,can you explain your solution when x,y are rational[/quote]
Yes, I will post it tomorrow.OK?

Hi - The simillar question and the answer to this question  is [url=http://artofproblemsolving.com/Forum/viewtopic.php?t=25342]here[/url] too ..

 Original equation is  $x^{(1/x)} = y^{(1/y)}$ .. so...

 If you look at the curve  $x^{(1/x)}$  which has a maximum   at $x=e$  ... and get  the bright idea ..

(One of  them has to be less than $e$ , the other greater than $e$) ... and ..

( given $n >0$):
  $(1+1/n)^n $ is slightly less than  $e$  and
 $(1+1/n)^{n+1} $ is slightly greater than $e$

so
 $x=(1+1/n)^n$ and $y=(1+1/n)^{n+1}$

are solutions, these are rational if and only if $n$  is integer (Proof is not hard .. try it).

$n=1$  gives ($x=2$ and $y=4$) and  $2^4 = 4^2$  -- the only  solution for  positive integers ...

I think you must argue so:
put k = y/x and substitution in the equation gives
\[
x^{kx}  = \left( {kx} \right)^x  \Rightarrow x^{k - 1}  = k \Rightarrow x = k^{\frac{1}{{k - 1}}} ,y = k^{\frac{k}{{k - 1}}}
\]
Let the rational number 1/(k-1) = m/n con gcd(m,n) = 1. Then
\[
k - 1 = \frac{n}{m}\;;\;k = \frac{{m + n}}{m} \Rightarrow x = \left( {\frac{{m + n}}{m}} \right)^{\frac{m}{n}} \;;\;y = \left( {\frac{{m + n}}{m}} \right)^{\frac{{m + n}}{n}}
\]
Since m and n are relativity prime, x and y are rational numbers iff n= 1. Then
\[
x = \left( {\frac{{m + 1}}{m}} \right)^{m} y = \left( {\frac{{m + 1}}{m}} \right)^{m+1}
\]
with integer different from 0 and - 1.

[quote="Gyan"]

$n=1$  gives ($x=2$ and $y=4$) and  $2^4 = 4^2$  -- the only  solution for  positive integers ...[/quote]

And if $x=y$? :D

[quote="blahblahblah"]
And if $x=y$? :D[/quote]

:blush: Okay! So some times  even getting new glasses doesn't help!   :blush:  :blush:  :blush:  :blush:

(Last time I am getting transition lesnses - even if they put it without any additional cost - it makes you miss obvious things temporarily... and you don't even realize it.. :) )

Yes, you have all forgot the trivial solution y = x right?? I am looking for the other ones too though. Is (x,y) = (2,4) the only other solution?? Interesting...I couldn't solve it myself.

my solution is similar to Nené's
so I have no choice to post it :)


trigonometry
function
Euler
complex numbers
complex analysis
complex analysis unsolved

Hello !!
I start learning complex numbers and i have a simple question.
a)$ cos(i\theta)$
b)$ isin(i\theta)$ are real or complex numbers?

if i don't know how much is θ ? how can i tell if is real or complex? I can't understand
I would like to tell me how will i understand what numbers are? Which way? What should i use?
Thanks in advance!!

Perhaps use the Taylor series for cos and sin, then you can see the result when you plug in $ i\theta$ with $ \theta$ real.

thanks for the answer
so for
a)$ cos(i\theta)$  we have:

   $ cos(x) \equal{} \sum_{n \equal{} 0}^{\infty}\frac {{ \minus{} 1}^n {x^{2n}}}{(2n!)}$

we plug in $ i\theta$ and we have:

 $ cos(i\theta) \equal{} \sum_{n \equal{} 0}^{\infty}\frac {{ \minus{} 1}^n {{(i\theta)}^{2n}}}{(2n!)}$

so all numbers in the sum are real expect $ (i\theta)^{2n}$ that we should calculate: $ \theta$ is real so $ \theta^{2n}$ is also real
$ i^{2n}$ is $ \minus{} 1$. So also $ (i\theta)^{2n}$ is real numbers.
Am i right?
same way for b)...

But if i didn't know Taylor series is there another way????
Thanks again

Well, you know this:

$ e^{ix} \equal{} \cos ix \plus{} i\sin ix$

$ e^{ \minus{} ix} \equal{} \cos ix \minus{} i\sin ix$

Now, if you add these two and divide by $ 2$, you get $ \cos x \equal{} \frac {e^{ix} \plus{} e^{ \minus{} ix}}{2}$, and if you subtract them, and then divide by two, you get $ \sin x \equal{} \frac {e^{ix} \minus{} e^{ \minus{} ix}}{2i}$. Now, plug in $ x \equal{} \theta i$.

This gives $ \cos i\theta \equal{} \frac {e^{ \minus{} \theta} \plus{} e^{\theta}}{2}$ which is real, and $ \sin i\theta \equal{} \frac {e^{ \minus{} \theta} \minus{} e^{\theta}}{2i}$ which is imaginary, so $ i\sin i\theta$ is real too.

[quote="alpar"]$ i^{2n}$ is $ \minus{} 1$. So also $ (i\theta)^{2n}$ is real numbers.
Am i right?[/quote]

almost.  $ i^{2n}$ is either $ \minus{}1$ or $ 1$.  So your conclusion that $ \cos(i\theta)$ is real is correct.

[quote="hsiljak"]Well, you know this:
$ e^{ix} \equal{} \cos ix \plus{} i\sin ix$

$ e^{ \minus{} ix} \equal{} \cos ix \minus{} i\sin ix$[/quote]

The only problem is that those equations are derived using the Taylor series of $ e^z$. I think most books define the exponent function for a complex variable in terms of its Taylor series, then define the other functions using that, including the complex exponential, and thus Euler's formula.

Indeed, it is so (it is the way the functions are introduced in my analysis courses). But here is the reason why I offered this approach: In my country, first encounter with complex numbers (Euler and de Moivre formulae included) is in tenth grade, while the Taylor series are introduced three years after that, in the first year of college.


Three faucets fill a 100-gallon tub in 6 minutes.  How long, in seconds, does it take six faucets to fill a 25-gallon tub?  Assume that all faucets dispense water at the same rate.

We have 1/4 of volume, and 2 times more faucets, so the time is shortened 8 times, or the answer is 45 seconds.

Let x be the time in minutes, we have
$ x \equal{}  \frac{(25*6/100)}{2}$
$ x \equal{} .75$
.75 minutes = 45 seconds

U r so clever quattay..

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geometry
probability
linear algebra
matrix
AMC

Two cubical dice each have removable numbers $ 1$ through $ 6$. The twelve numbers on the two dice are removed, put into a bag, then drawn one at a time and randomly reattached to the faces of the cubes, one number to each face. The dice are then rolled and the numbers on the two top faces are added. What is the probability that the sum is $ 7$?

$ \textbf{(A)}\ \frac{1}{9} \qquad
\textbf{(B)}\ \frac{1}{8} \qquad
\textbf{(C)}\ \frac{1}{6} \qquad
\textbf{(D)}\ \frac{2}{11} \qquad
\textbf{(E)}\ \frac{1}{5}$

For each combination you get on the first die, there are $ 2$ possibilities for the second die in order to sum to $ 7$.  There are, of course, $ 12 \minus{} 1 \equal{} 11$ ways to do this, so the answer is just $ \frac {2}{11}$.

Or you can do it the way I did it -_-
For the denominator, you can arrange the numbers on the dice in $ 12!$ ways.  There are $ 36$ ways to get $ 6$ numbers off of each of them for these combinations, so the denominator is just $ 12! \bullet 6^{2}$.
For the numerator, you have $ 6$ possibilities that can sum to $ 7$ (just list them out).  For each of these, you can put them on the dice in $ 6*6 \equal{} 36$ ways.  Finally, you can arrange the other $ 10$ numbers in $ 10!$ ways.  Our answer is thus: $ \frac{36 \bullet 6 \bullet 10!}{12! \bullet 36} \equal{} \frac {2}{11}$.

Note that there are only 3 non-distinct pairs of positive integers which sum to 7, these are (1,6), (2,5) and (3,4). There are a total of (12)(11)/2! = 66 combinations of the twelve dice labels. Since 6 will be (1,1), (2,2), (3,3), (4,4), (5,5) and (6,6), there are 60 pairs of the set of 1,2,3,4,5,6 in the possibilities. Since there are (6)(5)/2! = 15 non-distinct pairs, and 60/15 = 4 each possible non-distinct pair is included 4 times. Therefore (4)(3)/66 = 2/11.

Instead of putting on the faces and then rolling the dice, simply roll the dice, then put on the faces. After putting the first face on the top of one die, there are 11 other faces to put on the top of the other die. 2 of these will give the desired sum of 7, so the probability is $ \frac{2}{11}$.

[hide="What is wrong with this solution?"]
\[ \begin{tabular}{c|c} First Roll & Second Roll \\ \hline 1 & 6 \\ 2 & 5 \\ 3 & 4 \\ 4 & 3 \\ 5 & 2 \\ 6 & 1 \end{tabular} \]Each of the above can occurs twice, so there are $6\cdot2=12$ desired ways. There are a total of $12\cdot11$ ways. The probability is, thus: \[ P = \dfrac {6 \cdot 2}{12 \cdot 11} = \dfrac {1}{11}, \] so I chose $\boxed{\textbf{(D)}}$. :P
[/hide]

Thanks in advance! :)

Because each roll cannot only happen twice, but 4 times in total.  The first digit 1 and the first digit 6, the second for each, or the first 1 and second 6 or the second 1 and the first 6.  So there are in fact $6\cdot4=24$ desired ways, and this is where the extra factor of 2 enters.

This is a good example of a problem that can be solved with the probabilistic method: Let $S = \{1_a, 1_b, 2_a, 2_b, \cdots , 6_a, 6_b\}$ be the set of all $12$ numbers. Since each integer from $1$ to $6$ appears twice, we have just labeled them to distinguish between these two duplicates. Now, notice that $7$ can be written as a sum of two elements of $S$ in exactly $12$ ways (four ways as $1 + 6$, four ways as $2 + 5$, four ways as $3 + 4$). E.g. \[1_a + 6_a; \; 1_a + 6_b; \; 1_b + 6_a; \; 1_b + 6_b\] Label these twelve pairs $P_1, P_2, \cdots , P_{12}$, and attach an indicator variable $X_i$ to each $P_i$, defined by \[X_i = \bigg\{\begin{matrix} 1 \quad \text{if the two elements of} \; P_i \; \text{appear on different dice} \\
0 \quad \text{otherwise} \end{matrix}\] Notice that each pair $P_i$ for which the two elements of $P_i$ appear on different dice corresponds to exactly one chance to roll a $7.$ Since there are $6^2 = 36$ possible ways to roll the dice, the probability of rolling a $7$ is given by \[\frac{\mathbb{E}\left(X_1 + X_2 + ... + X_{12}\right)}{36}.\] By Linearity of Expectation, this is equal to \[\frac{\mathbb{E}\left(X_1\right) + \mathbb{E}\left(X_2\right) + \cdots + \mathbb{E}\left(X_{12}\right)}{36} = \frac{12\mathbb{E}\left(X_1\right)}{36} = \frac{\mathbb{E}\left(X_1\right)}{3}.\] But it is easy to compute that $\mathbb{E}\left(X_1\right) = \frac{6}{11}$, since if we consider the pair $P_1$, and fix the position of one of the elements of $P_1$, there are $11$ remaining positions to place the second element of $P_1.$ Of these $11$ positions, $6$ of them place this second element on the die opposite the first element. Therefore, our desired answer is \[\frac{\left(\frac{6}{11}\right)}{3} = \boxed{\frac{2}{11} (\text{D})}\]

Is this question an example of how AMC trolls you? The "Reatatching the faces and rolling it again" didn't even matter.

Yeah...it is just there to confuse you even more. Are there any other examples that you know of that the questions are stated to confuse us?


What is the value of
\[ (3x \minus{} 2)(4x \plus{} 1) \minus{} (3x \minus{} 2)4x \plus{} 1\]when $ x \equal{} 4$?

$ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 11 \qquad \textbf{(E)}\ 12$

$ (3x\minus{}2)(4x\plus{}1)\minus{}(3x\minus{}2)4x\plus{}1\equal{}(3x\minus{}2)(4x\plus{}1\minus{}4x)\plus{}1\equal{}3x\minus{}2\plus{}1\equal{}3x\minus{}1$. Plug in $ x\equal{}4$, and you get $ 11 \boxed D.$


Let $ ABC$ be a scalene triangle inscribed in circle (O). The circle $ (O_0)$ is internally tangent to $ (O)$ at $ T$, and $ AB, AC$ at $ E, F$ respectively. Let $ AO_0$ intersects $ (O)$ at $ M$. Prove that $ M$ is the midpoint of arc $ BC$.

AO0 pass through the incentr of triangle ABC , the midpoint of EF

I'm sorry... this problem is plain silly. Just forget it.


For how many different points (x,y) are the pints (x,y),(3,6),(5,2),(0,0) the verticies of a parallelogram?

just eyeball the problem:[hide]4?[/hide]

eyeballing doesn't always work though....and i dont think it works in this case.

[hide]If you have three vertices of a parallelogram, there is only one point that can be the fourth.[/hide]

no,...........why dont you try to graph these 3 pts?

if you really think about it (well at least i think), if you make a triangle out of these 3 pts, there should be one (x,y) off of each side. correct me if i'm wrong


This is called Schweitzer's inequality and the more general inequality is Kantorovich's inequality. Try to search.

For every $x$ in $[a,b]$ we have

$\displaystyle (x-a)(x-b) \le 0$

$\displaystyle \Leftrightarrow x^2+ab \le (a+b)x $

$\displaystyle \Leftrightarrow x+\frac{ab}x \le (a+b)$

 Then
$n(a+b) \ge (x_1+x_2+\cdots+x_n)+ab (\frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n})$

$\ge 2\sqrt[2]{ab(x_1+x_2+\cdots+x_n)\left(\frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n}\right)}$

$\displaystyle \Leftrightarrow \displaystyle (x_1+x_2+...+x_n)(\frac{1}{x_1} +\frac{1}{x_2}+...+\frac{1}{x_n}) \leq \frac{(a+b)^2}{4ab} n^2 $

 :P  :P  :P

[quote="cezar lupu"]There is a more general inequality called Greub-Reinholdt.[/quote]

can you post it?

we are still weating the general case :?

The  initial  ineq  is  from  12  th ASU , 1978 .


calculus
derivative
integration
limit
calculus computations

Hi again.

I seem to be stuck and could use some pointers...I'm probably missing something obvious (or making an incorrect assumption).

I need to show the following using the limit definition of a derivative:

Suppose f and g are continuous for all x, and that g is differentiable for all x and is one-to-one.

Letting $F(x)=\int^{g(x)}_{a}f(t)dt$ prove that $F'(x)=f(g(x))g'(x)$

I'm not sure if it matters if I apply the "limit definition of a derivative" first or the "Mean Value Theorem for Integrals" first...so here's what I did:

By the MVT, $F(x)=\int^{g(x)}_{a}f(t)dt = (g(x)-a)f(c)$

Then applying the limit definition of a derivative:

$F'(x)=\displaystyle\lim_{h\rightarrow 0}\frac{(g(x+h)-a)f(c)-(g(x)-a)f(c)}{h}$

$F'(x)=\displaystyle\lim_{h\rightarrow 0}\frac{f(c)g(x+h)-f(c)g(x)}{h}$

$F'(x)=f(c)\displaystyle\lim_{h\rightarrow 0}\frac{g(x+h)-g(x)}{h}$

$F'(x)=f(c)g'(x)$

Somehow (if I'm on the right track) I need to show $c=g(x)$ to get:

$F'(x)=f(g(x))g'(x)$

Where am I messing up?

Thanks for your help.
Bailey

Your idea to use the mean value theorem is not bad at all but the way you are doing it leads nowhere. The formal mistake is that $c$ is not the same for the integral to $g(x+h)$ and the integral to $g(x)$. It is much better to start the proof with noticing that $F(x+h)-F(x)=\int_{g(x)}^{g(x+h)}f(t)\,dt$ and applying the MVT to this integral.

I think I see my error.

Recap:

I need to show the following using the limit definition of a derivative:

Suppose f and g are continuous for all x, and that g is differentiable for all x and is one-to-one.

Letting $F(x)=\int^{g(x)}_{a}f(t)dt$ prove that $F'(x)=f(g(x))g'(x)$

Proof:

By the limit definition of a derivative:

$F'(x)=\displaystyle\lim_{h\rightarrow 0}\frac{F(x+h)-F(x)}{h}$

$F'(x)=\displaystyle\lim_{h\rightarrow 0}\frac{\int^{g(x+h)}_{a}f(t)dt-\int^{g(x)}_{a}f(t)dt}{h}$

$F'(x)=\displaystyle\lim_{h\rightarrow 0}\frac{\int^{g(x+h)}_{g(x)}f(t)dt}{h}$

Applying the MVT:

$F'(x)=\displaystyle\lim_{h\rightarrow 0}\frac{[g(x+h)-g(x)]\cdot f(c)}{h}$

where $g(x)<c<g(x+h)$

$F'(x)=\displaystyle\lim_{h\rightarrow 0}f(c)\cdot\displaystyle\lim_{h\rightarrow 0}\frac{g(x+h)-g(x)}{h}$

Applying the limit definition

$F'(x)=\displaystyle\lim_{h\rightarrow 0}f(c)\cdot g'(x)$

Since  $g(x)<c<g(x+h)$,  as $h\rightarrow 0$,  $c\rightarrow g(x)$

Therefore, $F'(x)=f(g(x))g'(x)$

I think this is correct.

Thanks for your help.
Bailey

sorry if I'm reviving an old topic or anything...

Is there a proof without using integrals or mean value theorem?


ratio
complex analysis
complex analysis unsolved

Show that any $ 4$ distinct points can be carried by a linear fractional transformation to positions $ 1,\minus{}1,k,\minus{}k$, where the value of $ k$ depends on the points. How many solutions are there, and how are thay related?

Interestingly enough, this is a crucial step in my [url=http://sciserv.org/sts/]Intel[/url] paper:  it turns out that just as the cross ratio parameterizes elliptic curves in Legendre form $ y^2 \equal{} x(x \minus{} 1)(x \minus{} \lambda)$ (the mapping is six-to-one), the value of $ k$ parameterizes elliptic curves in Edwards form $ x^2 \plus{} y^2 \equal{} k x^2 y^2 \plus{} k$ (the mapping is twelve-to-one, I think; I used the value of $ \sqrt{k}$, where the mapping is twenty-four-to-one), and this fact turned out to be the basis of a neat computation that proved my main result.


Can someone explain to me the question, I don't really get it:
Find all possible values for the sum of the digits of a perfect square. :blush:

you must find $ n\in\mathbb{N}$ such that $ \exists m\in\mathbb{N}: \ m^2\equal{}\overline{a_1a_2...a_h}^{(base\ 10)}$ and $ n\equal{}\sum_{j\equal{}1}^{h}a_j$

A square mod 9 is 0,1,4, or 7 , hence the sum of it's digits is too. This is because the remainder of a number when divided by 9 stays the same if you sum the digits. Here is a proof if you don't know this:
Let the number be $ \overline{abcdefg....}$
This is equal to $ 10^m\cdot a\cdot 10^{m \minus{} 1}\cdot b\cdot 10^{m \minus{} 2} \cdot c$.....
Since every power of 10 is one more than a multiple of 9, if we take the remainder when this is divided by 9, we get a+b+c+d..... .


1) Let $ A$ and $ B$ be arbitary $ 2$x$ 2$ matrices over $ \mathbb{C}$ .For all complex numbers $ \alpha,\beta,\gamma$
Prove that :
       $ \det(\alpha I_2 \plus{} \beta AB \plus{} \gamma BA) \equal{} det(\alpha I_2 \plus{} \gamma AB \plus{} \beta BA)$
 2) Let $ A,B \in M_2{\mathbb{(C)}}$ be such that : $ (AB)^2 \equal{} A^2B^2$ .
Prove that :
               $ \det(I_2 \plus{} AB \minus{} BA) \equal{} 1$

Maybe I will try to answer problem number 2
[hide="Hint"]
Let $ AB\minus{}BA\equal{}X$, we will prove that $ tr(X)\equal{}0$ and $ tr(X^2)\equal{}0$, the first statement is obvious, so I only prove the second statement :
$ tr(AB\minus{}BA)^2\equal{}tr(ABAB\plus{}BABA\minus{}ABBA\minus{}BAAB)\equal{}tr(2ABAB\minus{}2A^2B^2)\equal{}0$
Note that if $ x$ and $ y$ are eigenvalues of $ X$, then $ x\plus{}y\equal{}0$ and $ x^2\plus{}y^2\equal{}0$ then $ x\equal{}y\equal{}0$, then $ I\plus{}X$ has only $ 1$ as eigenvalue. Then the determinant of $ I\plus{}X$ is $ 1$.
[/hide]
I am sorry if I am wrong.


geometry
geometric transformation
rotation

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? [Hint: Arrange the men first, then the women.]

The given answer is 144.

I got an answer of 4x4x3x3x2x2x1x1/2/2 = 144.

Is this the correct way of solving it?  If so, where did the two /2's come from?

Thanks.

Note that the problem says that these men and women are sitting around a circular table. This means that rotations are indistinct, so we must divide by four since each configuration can be rotated four ways.

We can arrange the woman in $ 3!$ ways, if we fix one of the woman's position.  In the remaining spots, we can arrange the men in $ 4!$ ways.  Our answer is thus $ 4!3!\equal{}\boxed{144}$.

Another way to do this is to not fix one of the woman.  You therefore have $ 8*3!*4!$ ways to arrange the women and then the men.  However, for each possibility, we are overcounting $ 8$ times.  Once again, we get an answer of $ 4!3!\equal{}\boxed{144}$.

And yet another way to do this is as you did it.  You can choose the top spot in the circle as a woman's.  You have $ 4$ possibilities for her.  The next seat, you have $ 4$ possibilities for the boys.  Then, you have $ 3$ possibilities for the woman in the next spot, etc. etc.  The only time you are overcounting is when you have a rotation of the people sitting in the circle.  You have four seats for the first woman to sit in, and the rest of the rotation follows.  Once again, we get an answer of $ \frac{4!4!}{4}\equal{}\boxed{144}$


Positive integer numbers $ a, b, c, x, y, z$ are given.
There are:
$ a^{2}\plus{}b^{2}\equal{}c^{2}$
$ x^{2}\plus{}y^{2}\equal{}z^{2}$
and
$ |x\minus{}a|\leqslant 1$
$ |y\minus{}b|\leqslant 1$

Prove that the sets $ \lbrace a,b\rbrace$ and $ \lbrace x,y\rbrace$ are equal.

You should really say where are these problems from (this one, and the one before with number of divisors and a sequence), so we can help you with some link, couse I am sure it was discused before (if it is from some contest).

Btw, this is rather boring and simple problem: numbers $ a^2 \plus{} (b \minus{} 1)^2, (a \minus{} 1)^2 \plus{} b^2$ and $ (a \minus{} 1)^2 \plus{} (b \minus{} 1)^2$ are between $ a^2 \plus{} b^2 \equal{} c^2$ and $ (\sqrt {a^2 \plus{} b^2} \minus{} 1)^2 \equal{} (c \minus{} 1)^2$. Also numbers $ a^2 \plus{} (b \plus{} 1)^2, ...$ are bigger than $ c^2$ but smaller then $ (c \plus{} 1)^2$, so they aren't perfect squares. What about $ (a \minus{} 1)^2 \plus{} (b \plus{} 1)^2$? That is $ a^2 \plus{} b^2 \plus{} 2(b \minus{} a \plus{} 1)$, and now you have some cases maybe, nevertheless, the point stays!

Bye


probability
group theory
abstract algebra
superior algebra
superior algebra unsolved

Let E be the splitting fild of x^q -2 over Q (q prime). Show that there exists a unique intermediate field.

Wrong...
Take $ q\equal{}3$ and you have $ 4$ intermediate fields...

[hide="sorry"]i am sorry   :(  if i am disturbing the continuity f this discussion but i couldn't help myself noting this and put it down .just look at the times when abel posted and ZetaX replied to it [/hide]

nothing but an exchange of the first two numbers .seemingly highly improbable

Seemingly highly dependable on what time zone you are ;)

oh gosh i forgot that :rotfl:
hence the probability reduces even   further :wallbash:

Sorry for posting a wrong problem.
But what are the 4 intermediate fields? Isn't there 1?
Any intermediate field has group C(q-1) over Q (cyclic group of cardinality q-1). since C(q) is the [i]only[/i] normal subgroup. So there exists exactly one intermediate field that is normal over Q.

Yes, $ C_{q}$ is a normal subgroup.But then, to this subgroup belongs a field.
For $ q\equal{}3$ the intermediate fields are:
Degree $ 3$: $ \mathbb{Q}[\sqrt[3]2]$, $ \mathbb{Q}[\zeta_{3}\sqrt[3]2]$, $ \mathbb{Q}[\zeta_{3}^{2}\sqrt[3]2]$
Degree $ 2$: $ \mathbb{Q}[\zeta_{3}]$
Here $ \zeta_{3}$ is a root of $ x^{2}\plus{}x\plus{}1$, thus a primitive third root of unity.

Yes, but Q(2^(1/3)) is not a normal ext over Q.
The total no. of intermediate fields is the the no. of subgroups. But the no. of normal extensions that are  intermediate fields is the no. of normal subgroups, right? (in this case, 1)

Intermediate fields $ \leftrightarrow$ subgroups.
Normal intermediate fields $ \leftrightarrow$ normal subgroups.

You asked for intermediate fields above, not normal ones.

Is the problem about showing that there is only one normal intermediate field then¿

Now I see all the errors I was making. Actually, if d|q-1, the semi-direct product C(q)C(d) is a normal group. However, C(d) is not a normal group for any such d.
So every normal intermediate field is contained in the splitting field of  (x^q -1) over Q. Thus the no. of normal intermediate fields depends only on the no. of distinct divisors of q-1.


I found a really interesting way to do this, but I would like to see how others go about it.

Prove that $F_{nm}=\sum_{i=0}^m\binom{m}{i}F_iF_n^iF_{n-1}^{m-i}$ where $F_k$ is the kth Fibonacci number and $F_1=F_2=1$. Note that we can work backwards and find $F_0=0$, so that the first term of this sequence is actually 0.

[hide="hint"]$\phi^n=F_n\phi+F_{n-1}$[/hide]

Yes, thanks. I can get the code from your post.


Consider a solid disk of mass $m$, radius $r$, and moment of inertia $\frac{1}{2}mr^2$ and let the coefficient of friction between all surfaces be $\mu$.  Calculate the force $F$ necessary to pull the disk over the edge of height $h, \ r>h$.

Is the vector F permanently attached to the point on the picture or will this vector be on "top" of the disc for the whole time? Will F keep pointing horizonatally or will it rotate with the disc?

Since the torque due to the weight of the disc is maximum at the initial position. Hence for the disc to raise to the platform the torque due to the vector F must be greater than it. Also the torque due the vector F remains constant.
I have assumed that the friction is enough to prevent slipping of the disc.

Can anyone tell how friction prevents the disc from slipping.(I know it does, but how since it is just a point contact.

An interesting problem.

Let me help you get started.

Look at the initial conditions.  The force F is distributed normal to the tangent to the circle at the top of the distance marked h (let's call this point P -- it will be useful and gives us a sense of power over the problem).  The component of F that is normal to the tangent at P is only useful for creating sufficient friction to keep the disk from spinning.  The component of F that is parallel to the tangent at P has to be (just a little bit) larger than the opposing forces.  (Since we are talking about statics the moment of inertia is not relevant, the opposing forces are just derrived from gravity and the mass of the disk m.)

Once you have determined the parallel component of F (let's call this Fp), you need to determine the component of Fp that is useful in lifting the mass of the disk m.  (The triangle you use to solve this should be strangely familar.)  Now you know that this component of Fp has to be greater than the mass of the disk m.  Back calculation will reveil the minimum force F to lift the disk.

Since these problems always assume that friction is independent of area and is equal to the normal force times the coefficient of friction, you have a potentially limiting factor with the friction at the line segment that goes through point P.  This also must be met for the disk to start to move.

Now you need to determine if the lifting can continue with the same minimum F.  How about when the disk just reaches the top (where the center of the circle (let's call it C) is directly over P)?

From doing this, unless the disk spins starting is the hardest.  However, you have to show this with math.

Have fun.  Hope this helped get you started and on your way.


complex analysis
complex analysis unsolved

f(x) - P(x) = (x-x0)(x-x1)..(x-xn)/(n+1)! (f(n+1)(skii)

How would I use that error formula on f(x) = cos x at x = [-pi/2,pi/2].

thanks

What does (skii) mean¿
And can you please state the question more clearly, at the moment I don't see what is asked for...

[quote="ZetaX"]What does (skii) mean¿
And can you please state the question mroe clearly, at the moment I don't see what is asked for...[/quote]

ok skii is that weird symbol...there is not one here on the keyboard.
basically skii = {x0,x1,x2,x3,,,,,,,,,xn}

The question is use the error formula which was stated above and compute the error involved in f(x) = cos x for the points {-pi/2, pi/2].

See I already found the interpolating poly and now i have to find the error involved.


topology
function
integration
calculus
derivative
complex analysis
linear algebra

If V is an infinite-dimensional linear space and A a linear operator V -> V, does the fact that it is bounded on the basis of V imply that it's bounded everywhere?

Incomplete information. "Bounded" only makes sense if you tell us what kind of topology you want to place on this space. A norm? And what do you mean by "basis"? Schauder basis? Orthonormal basis? Hamel basis?

Right - V is normed and A is bounded with respect to the operator norm.

There are plenty of simple counterexamples.
Let $ X$ be the space of finitely supported $ \mathbb{R}$-valued sequences with the $ \infty$-norm, and let $ A(a_1,a_2,\dots)\equal{}(a_1\plus{}a_2\plus{}\dots,0,0,\dots)$.

OK, another question:

Let V be an infinite dimensional linear space of functions with a basis $ (e^{x}, e^{\frac{x}{2}}, \dots, e^{\frac{x}{n}}, \dots)$ and supremum norm on [0, 1]. Is derivate operator D(f)=f' bounded on this space?

This is analysis rather than algebra but the answer is "No" anyway. The way to see it immediately is to note that the linear span of $ e^{x/n}$ is dense in $ C([0,1])$. Indeed, if there were a finite Borel complex-valued measure $ \mu$ satisfying $ \int_0^1 e^{x/n}\,d\mu(x)\equal{}0$ for all $ n$, then the entire function $ F(z)\equal{}\int_0^1 e^{zx}\,d\mu(x)$ would have a zero at every point $ 1/n$ and, thereby, we would have $ F(z)\equiv 0$, so $ F(k)\equal{}0$ for all integer $ k\ge 0$. But the span of $ 1,e^x,e^{2x},e^{3x},\dots$ is dense in $ C[0,1]$ by the Weierstrass theorem, so $ \mu\equal{}0$.

Can you give any example of an infinite dimensional space with supremum norm on wich derivative operator is continuous?

I asked whether derivative operator: f -> f' is continuous (=bounded) with respect to supremum norm, and the topic you mentioned is about closed subsets of C[0,1], I don't see the connection.

It is almost exactly the same story. If $ D$ is bounded in the sup-norm on a subspace $ V$, then the closure $ W$ of $ V$ in $ C([0,1])$ consists of Lipschitz functions and, moreover, the Lipschitz constant of a function $ f\in W$ is bounded by some constant times the norm of $ f$ in $ C([0,1])$, so the unit ball in $ W$ is compact (Arzela-Ascoli) and, thereby, $ W$ is finitely dimensional. You can even get an explicit lower bound for $ \|D\|_{V\to V}$ in terms of $ \text{dim\,} V$. Finding the sharp rate of growth in this estimate would be an interesting problem.


prove that for every triangle $ABC$,we have:
$\frac{3\sqrt{3}}{2cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}}+8sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}\geq 5$

Please don't post solution, because this problem is from M&Y ([b] March, 2007[/b])

toan hoc,

Why ?   :mad:  Please see here http://www.mathlinks.ro/Forum/viewtopic.php?p=790091#790091

This Problem is proposed on the M&Y of VietNam .Please lock this topic


floor function
logarithms
function
5th edition

Is there anyone who solved at least one problem? :D  Please give the solutions! ;)

Here are my solutions to round 5. I think that the problems are nice.

{\bf Problem 1:} Find all real numbers $a>1$ such that there exists an integer $k \geq 1$ such that the sequence $\{ x_n \}_{n \geq 1}$ formed with the first $k$ digits of the number $\lfloor a^n \rfloor$ is periodical.

[hide="solution1"]
{\bf Solution:}
First, we note that WLOG, $k=1$. This is because if the sequence is periodic with a bigger $k$, then the first digits of these repeating $k$ digit numbers must also be repeating. Thus, flooring $a^n$ becomes irrelevant, since $a>1$ and we are only interested in the first digit of $a^n$. Hence, the problem becomes: Find all real numbers $a>1$ such that the sequence $\{ x_n \}_{n \geq 1}$ formed with the first digit of $a^n$ is periodical.

We will show that the answer is $a=10^{p/q}$ for some $p, q \in N$. Clearly, this works since $a^{q+n}=10^pa^n$, so the first digit is periodical with period $q$. Conversely, suppose that we have an $a$ that works. We will derive a contradiction using the following lemma.

{\bf Lemma:} Given real numbers $x$, $u$, $l$, and $b$, such that $0\leq l < u \leq 1$, and $u-l < 0.5$. If for all non-negative integers $n$, we have $l \leq \{b+nx\} < u$,  then $x$ is an integer. ($\{x \}$ denotes the fractional part of the number $x$.)

{\bf Proof of Lemma:} Suppose that $x$ is not an integer. Then according to the conditions, either $0 < \{x\} < u-l$, or $1-(u-l) < \{x\} <1 $. In the first case, let $n= \lfloor {{u- \{b\} \over \{x\} }} \rfloor$, and in the second case, let $n= \lfloor { \{b\}-l \over 1-\{x\}}\rfloor$. In both cases, since $u-l<0.5$, an easy calculation shows that $\{b+(n+1)x\}$ is not in the required interval, contradiction.

Returning to the problem, note that the first digit of $a^n$ is $d$ $(1 \leq d \leq 9)$ iff $\log_{10}d \leq \{n\log_{10}a \} < \log_{10}(d+1)$. Note that for all possible $d$ , $\log_{10}(d+1)-\log_{10}d < 0.5$. Let the period of the sequence be $g$. Our periodic condition for the first digit becomes, for some sufficiently large integer $m$, $\log_{10}d \leq \{(m\log_{10}a) + t(g\log_{10}a) \} < \log_{10}(d+1)$ for all non-negative integers $t$. By the lemma, this implies that $g\log_{10}a$ is an integer, which implies that $a$ is a power of 10 with a positive rational exponent. In conclusion, the sequence is periodic iff $a=10^{p/q}$ for some $p, q \in N$.
[/hide]

{\bf Problem 2:} Prove or disprove the existence of a function $f: S \longrightarrow R$ such that for all $x \neq y \in S$ we have $|f(x)-f(y)| \geq {1 \over x^2 + y^2}$, in each of the cases a) $S=R$; b) $S=Q$.

[hide="solution2"]
{\bf Solution:} a) When $S=R$, the function does not exist. Suppose on the contrary that it does exist. Consider the interval $I = [0,1]$. For all $x \neq y \in I$, we have $|f(x)-f(y)| \geq {1 \over x^2 + y^2} \geq {1 \over 1^2 + 1^2} = {1 \over 2}$. Let $H$ be the set $\{z | \exists x \in I,z=f(x) \}$. By our condition, $f(x)$ is different for different $x$'s, so the function is bijective and $|H| = |I|$. Thus, $|H|$  is uncountable. Now, partition the real line into intervals $a_n = [n/2, (n+1)/2)$, $n \in Z$. By our condition, for every $n$, there is at most one $x$ such that $f(x) \in a_n$. This means that $|H| \leq |Z|$, which means that $|H|$ is countable, contradiction.

b) When $S= Q$, we show that the function exists. Since $Q$ is a countable set, there exists a sequence $\{ b_n \}_{n \geq 1}$ such that every number in $Q$ is in this sequence. Using this sequence, it is trivial to construct a function that works. For each $n$, we choose $f(b_n)$ to be sufficiently large so that $f(b_n) - f(b_i) \geq {1 \over b_n^2 + b_i^2}$, for every $1 \leq i \leq n-1$. Eventually, every $x \in Q$ would be assigned a corresponding value in $f(x)$, that satisfies the condition.

[/hide]

{\bf Problem 3:} A student wants to make his birthday party special this year. He wants organize it such that among any groups of 4 persons at the party there is one that is firends with exactly another person in the group. The student only wants to invite his friends. Find the largest number of his friends that he can invite to the party.

[hide="solution3"]
{\bf Solution:} We represent the situation by a graph, where two students are friends iff there is an edge connecting the vertices representing them. The host himself is represented by vertex $O$, which is connected to all other vertices. The other vertices are labelled $A_i$, where $1 \leq i \leq n$. We want the maximum value of $n$.

This maximum value of $n$ is 6. To see that $n=6$ is possible, consider the graph with edges $(A_1, A_2)$, $(A_3, A_4)$, $(A_5, A_6)$, plus the 6 edges connecting $O$ to the 6 $A_i$'s. It is easy to check that this satisfies the condition.

To show that $n \leq 6$. Let $S = \{ A_i | 1\leq i \leq n \}$. Note that every vertex in $S$ can be connected to at most one other vertex in $S$. This is because, suppose that $(A_i, A_j)$ and $(A_i, A_k)$ are edges in the graph, then in the group $OA_iA_jA_k$, every point is connected to at least 2 other points. This contradicts the condition. Therefore, $S$ can be partitioned into disjoint sets $S_i$, where $|S_i| \leq 2$ and no edge connects vertices from two different $S_i$'s. Let the number of $S_i$'s in the partition be $m$. If $m\geq 4$, then selecting a vertex from 4 different $S_i$'s, we get a group of 4 vertices where no two vertices are connected. This contradicts the condition. Therefore, $m \leq 3$, and $n= |S| \leq 2m \leq 6$. In conclusion, the maximum value of $n$ is 6.
[\hide][/hide]

though i haven't read your solutions yet but one thing you said that for p3, n=6.

But n=7 is also possible i think.

(A,B), (C,D), (E,F), And G with other sixes. is there any contradiction in this case of n=7?

oh pengshi! I should cry and cry!!!

i haven't considered the host at all! and that was the flaw :oops_sign:

But the chance to submit the solution will not come again ;)

I had a different approach to #3. Basically assign every friend (but not the birthday student) a number, indicating how many other people excluding the student each of them is in friends with. Observe every one of these numbers is less than 2. Now there are three brief cases regarding how many 0s there can be. If there are two 0s, then have at most 2 other people (so 4 in total); if one 0, then 4 more, so 5 in total; if no 0s, then 6 in total. Just work out how many 1s there are in each case, it's pretty straightforward.  :)


for all positive integer n prove that:
$e<(1+\frac{1}{\frac{2n+1}{2}})^{\frac{2n+1}{2}}(1+\frac{1}{\frac{2n+1}{2}})$
note:$e=\lim_{n\rightarrow\infty}(1+\frac{1}{n})^{n}$

[quote="Rust"]It is not true.
$(1+\frac{1}{x})^{x}<e<(1+\frac{1}{x})^{x+1}$.[/quote]Thank Rust. I edit my post.can you prove it please?

[quote="Rust"]Let $f(x,c)=x(x+1)\frac{(x+c)ln(1+\frac{1}{x})}{dx}=\frac{1}{2}-c-sum_{k=2}^{\infty}\frac{1}{k(k+1)x^{k-1}}$.
It give $f(x,1/2)<0 \ \forall x>0$ and $f(x,0)\ge 0 \ \forall x\ge 1$. We have $\lim_{x\to \infty }(1+\frac{1}{x})^{x+c}=e.$
Therefore $(1+\frac{1}{x})^{x}<(1+\frac{1}{x})^{x+0.5}\ \forall x\ge 1$.[/quote]
I don,t understand it. :blush:


I went through the solutions to problem 4 on AIME and it asks about he number of permtations. The solution says to choose 5 out of 11 and used combinations. Why did they us combination and not permutation even if the question is asking about the number of permutations? Also, why they had to worry about 5 of thm only and not the other fives? Any explainations please?

It is because the problem conditions uniquely determine a permutation for each selection (combination).


geometry
angle bisector
geometry proposed

Let a triangle $ ABC$ and its bisector $ BE$ and let $ M$ the midpoint of $ EC$. Let a point $ K$ at the side $ AB$ such that
$ BE$ passses through the midpoint $ O$ of $ MK$ . Prove that $ \angle{BOC}>90$

We will start with the following lemma:
[b]Lemma 1[/b]: If $ V$ and $ W$ are points on line segment $ BE$ such that $ BV > BW$ and the extensions of $ MV,MW$ through $ V,W$ meet line segment $ AB$ at $ Y,Z$, respectively then $ \frac {MV}{WY} > \frac {MW}{VZ}$ (Where points $ A,B,C,E,M$ are defined in the same way as in the problem).
[b]Proof[/b]: Because points $ E,V,B$ are collinear, by the Menelaus Theorem have:
$ \frac {AE}{EM}\frac {MV}{VZ}\frac {YB}{AB} = 1$
So, $ \frac {MV}{VZ} = \frac {ME}{AE}\frac {AB}{YB}$
Because points $ E,W,B$ are collinear, by the Menelaus Theorem have:
$ \frac {AE}{EM}\frac {MW}{WZ}\frac {ZB}{AB} = 1$
So, $ \frac {MW}{WZ} = \frac {ME}{AE}\frac {AB}{ZB}$
Now because $ BV > BW$ have $ BY > BZ$ therefore:
$ \frac {ME}{AE}\frac {AB}{YB} > \frac {ME}{AE}\frac {AB}{ZB}\Rightarrow \frac {MV}{VZ} > \frac {MW}{WZ}$ and the lemma is proved.
BY lemma 1 it follows that $ MV > MW\Leftrightarrow \frac {MV}{WY} > \frac {MW}{VZ}$ [b](*)[/b]

Now let us solve the problem.
1. Drop the perpendicular $ CR$ from $ C$ to line $ BE$. If $ R$ does not lie on line segment $ BE$ then because $ \angle{CBE} < 90^o$, $ R$ must lie on teh extension of $ BE$ through $ E$ hence for any point $ O$ on line segment $ BE$, $ \angle{BOC} > \angle{CRB} = 90^o$ and the result of the problem holds.
2. If $ R$ lies on line segment $ BE$, extend $ CR$ through $ R$ to meet $ BA$ at $ T$. Then, because $ BE$ i s the angle bisector of $ \angle{B}$ we must have $ \angle{CBR} = \angle{TBR}$, also $ \angle{CRB} = \angle{BRT} = 90^o$, $ BR$ is common so $ \triangle{CBR}$ is congruent to triangle $ TBR$ so $ CR = RT$.
3. $ CR = RT$, $ CM + ME$ so $ MR$ is the middle line in triangle $ CTE$ so $ TE\parallel{}MR$ and $ TE = 2MR$. Then, if we extend $ MR$ to meet $ AB$ at $ K$, get $ ML\parallel{}ET$, and because $ AM > AE$ we must have $ ML > ET$. Now, $ MR = \frac {1}{2}ET$ so $ \frac {MR}{RL} = \frac {1}{2}\frac {ML}{ET} > \frac {1}{2}$ so $ \frac {MR}{RL} > \frac {1}{2}$.
4. Because $ \frac {MO}{OK} = \frac {1}{2}$ we have $ \frac {MO}{OK} < \frac {MR}{RL}$ therefore by (*), it follows that $ BR > BO$ therefore $ \angle{BOC} > \angle{BRC} = 90^o$ and the result follows.

Let $ D$ be midpoint of $ BC$. $ M,D$ is midpoint of $ CE,CB$ so $ MD\parallel BE$. Since $ BE$ pss midpoint of $ MK$, $ BE$ pass midpoint of $ DK$. Therefore, $ BK\equal{}BD\equal{}\frac{a}{2}$ and $ \angle KDM\equal{}90$.
$ \angle BOC>90$

<=>$ DB\equal{}DC>DO$

<=>$ a>2DO\equal{}KM($since$ OK\equal{}OD\equal{}OM)$
<=>$ a^{2}>KM^2\equal{}DK^2\plus{}DM^2\equal{}\frac{a}{2}^2\plus{}\frac{a}{2}^2\minus{}2\frac{a}{2}\cdot\frac{a}{2}cosB\plus{}\frac{1}{4}ca(1\minus{}\frac{b^{2}}{(c\plus{}a)^{2}})$ $ \equal{}\frac{a^{2}}{2}(1\minus{}cosB)\plus{}\frac{1}{4}ca(1\minus{}\frac{b^{2}}{(c\plus{}a)^{2}})$
<=>$ 2a(1\plus{}cosB)>c(1\minus{}\frac{b^{2}}{(c\plus{}a)^{2}})$
<=> $ 2a(\frac{(c\plus{}a)^{2}\minus{}b^{2}}{2ca})>c(\frac{(c\plus{}a)^{2}\minus{}b^{2}}{(c\plus{}a)^{2}})$
<=>$ (c\plus{}a)^{2}>c^{2}$(true)


trigonometry
function
inequalities
number theory
relatively prime
arithmetic sequence
real analysis

I wonder if explicit solution can be found for

$ A=\min_{a>0}\max_{x\in{R}}(\sin(x)+\sin(ax))$

Numerically, $ a_{min}\approx{0.3}$, $ A\approx{1.52}$.

Any ideas?

$ \sin(x)+\sin(3x/10)$ reaches a value of about $ 1.99$ around $ x = 26.66$.  Also, for any irrational $ a$, we should be able to find values of $ x$ which make $ \sin x+\sin ax$ take values arbitrarily close to 2.

Please look at the problem once more - we are searching for minimum in $ a$ of
\[ f(a)=\max_{x\in{R}}(\sin(x)+\sin(ax)).\]
You probably mean that for an irrational (modulo $ \pi$) values of $ a$ the function $ \sin(x)+\sin(ax)$ is not periodic and will probably take values arbitrarily close to $ 2$?  If it's true then the minimum must be attained for a rational (modulo $ \pi$) value $ a$, right? This might improve the chance that it can actually be found.  Further ideas?

[b]JBL[/b] was pointing out that $ a=0.3$ (the value mentioned in your post) gives a much larger $ A$. Of course, this is because your numerical value of $ a_{\text{min}}$ is only approximately equal to $ 0.3$. Its true value is most likely $ 1/3$.

It may be helpful to consider $ \max_{x\in\mathbb R}(\sin mx+\sin nx)$, where $ m$ and $ n$ are positive integers.

Yes, that's right - I made too rough an approximation in the original post mistakenly thinking it was not important - the actual value was $ a\approx 0.33$.  And of course we should take rationals (without modulo $ \pi$) but the basic idea was right - only then it's periodic.  So the best candidate for the moment is $ a=1/3$.

$ a=1/3$ with the maximum $ \frac 8{3\sqrt 3}$ attained at the point where $ \sin (x/3)=\frac 1{\sqrt 3}$ is, indeed, the answer. To see it, let's follow mlok's suggestion and consider $ \sin nx+\sin mx$ with integer relatively prime $ 0<n<m$. Now, consider $ x$ such that $ mx\equiv \frac \pi 2\mod 2\pi$. Then $ nx\mod 2\pi$ runs over an arithmetic progression with step $ \frac{2\pi}m$. It follows that we can choose $ x$ such that $ nx\mod 2\pi$ differs from $ \frac \pi 2$ by at most $ \frac \pi m$, so $ \sin nx\ge \sin\left(\frac \pi 2-\frac \pi m\right)\ge 1-\frac{\pi^{2}}{2m^{2}}$. Since $ 3\sqrt 3>5$, we conclude that we must have $ \frac{\pi^{2}}{2m^{2}}>\frac{2}{5}$ if we want a smaller maximum. But this inequality fails for $ m=4$ already, so we are left with three pairs: $ (1,1)$, $ (1,2)$, and $ (2,3)$ to compete with $ (1,3)$. It is quite easy to check that they all are losers :).


geometry
circumcircle
inradius
inequalities
perimeter
inequalities unsolved

Prove that

s 2  \leq  \sum (m_a)(w_a) \leq 3(2R 2 +r 2 )

where m_a,m_b,m_c and w_a,w_b,w_c are medians and bisectors of a triangle of semiperimeter s ,circumradius R and inradius r.

[color=red]Do you mean the following ineqaulities manlio :
$ s^2 \leq \sum m_{a}w_{a} \leq 3(2R^2 \plus{} r^2 )$
  The left hand side can proved base on my lemma : $ m_{a}w_{a}\ge s(s \minus{} a)$ with $ s$ is the semi perimeter of triangle.
 About the 2nd problem please give me more time  :) [/color]

[color=red] And now solution for 2nd problem :) :
   We knew that$ a^2 \plus{} b^2 \plus{} c^2\le 8R^2 \plus{} 4r^2$ and because $ m_a^2\plus{}m_b^2\plus{}m_c^2\equal{}\frac{3}{4}(a^2\plus{}b^2\plus{}c^2)$ so $ m_a^2 \plus{} m_b^2 \plus{} m_c^2\le 6R^2 \plus{} 3r^2$
  In triangle it's easy to check that $ m_a\ge w_a$ and similarly.. so we get $ \sum m_aw_a \leq 3(2R^2 \plus{} r^2 )$
[/color]


If  | [x]-2x | <= 4 then how many solutions are there for x?

thank you... :)

There are an infinite amount of solutions...does it have to be an integer?

I have to say that the question is quite confusing. It is obvious that there is whole interval of solutions in set of reals: all numbers between -4.5 and 3.5 (including them) are solutions (and, in particular x=4).

I assume that the proposer wasn't asking for integer solutions because floor function then becomes useless.

By the way, the full solution set is $ x \in \{ \minus{} 5,4,[ \minus{} 4.5,3.5] \}$


Determine all pairs of positive integers (x,y) satisfying the equation $ 2^x\equal{}3^y\plus{}5$
I'm really stuck here. :(

Hello

$ 2^x\equal{}3^y\plus{}5 <\equal{}> 2(2^{x\minus{}1}\minus{}1)\equal{}3(3^{y\minus{}1}\plus{}1) <\equal{}> 2^{x\minus{}1}\equal{}3k\plus{}1$ &$ 3^{y\minus{}1}\equal{}2k\minus{}1$ with $ k \geq  1$

therefor : $ x\equal{}\frac{ln(3k\plus{}1)}{ln(2)}\plus{}1$ & $ y\equal{}\frac{ln(2k\minus{}1)}{ln(3)}\plus{}1$

now :  $ \frac{ln(3k\plus{}1)}{ln(2)}\plus{}1$& $ \frac{ln(2k\minus{}1)}{ln(3)}\plus{}1$  should be a positive integers

I think that the only solution (x,y) are  (3,1)& (5,3)when k=1 & k=5


function
complex analysis
algebra
domain
complex analysis unsolved

Let $f$ be a function analytic on the open disc $\mathbb{D}$ with center $(0,0)$ and radius $1$, whose power series development is given by $\sum {a_{k}x^{k}}$, and which has at most $m$ roots in $\mathbb{D}$. Show that $inf(|f(z)|;\ |z|=1) \leq |a_{0}|+\dots+|a_{m}|$.

Thanks, Julien, for correcting the message :).

I'm actually assuming $f$ to be holomorphic on a domain containing the closed unit disk. Suppose $|f|$ is larger than $|a_0|+\ldots+|a_n|$ on the unit circle, and let $g(z)=a_{n+1}z^{n+1}+\ldots=f(z)-\sum_{k=0}^na_kz^k$. We have $|f(z)-g(z)|<|f(z)|$ on the unit circle, so, by Rouche's Theorem, $f$ and $g$ have the same number of zeros inside the unit disk, but $g$ has $\ge n+1$ zeros (counting multiplicities), and we get our contradiction.

Hi Grobber, thanks for the answer. I got the same and of course I guess the problem meant to assume analyticity on the closed disk (even though it's mentionned $\mathbb{D}$ is the open unit disk).


function
real analysis
real analysis unsolved

Given a sequence of function $ f_n$ that converges on $ A \subseteq \mathbb{R}$ but do not converges uniformly on $ A$.
Will the series $ \sum f_n$ converges uniformly on $ A$?
I don't think the series converges uniformly but I don't have any idea to start my proof.
can anyone give me any hints?

Of course not- the series can only converge uniformly if the sequence converges uniformly to zero.

To prove that your statement is false, all you need is an example. How about $ x^n$ on $ [0,1]$ or something similar?

It is obvious by an example.But how to prove it for general cases?
jmerry.can you give me some hints to start?


Ever read it? If you have, what are your thoughts on it? Did you like it? Hate it? Why? What did you think about the author, Sophocles? I'm starting it today.

[quote="SplashD"]Is "The Oedipus Cycle" a book or something?[/quote]

It's a collection of 3 plays. Next time, instead of spamming the thread, you could do a simple Google search.

I only read Antigone. I didn't like it that much since I'm not much of a fan of the Greek style of plays. I don't buy into the catharsis idea either.

Are you referring to the Odyssey by any chance?

EDIT: No no forget it. This has nothing to do with the Odyssey. I was just confused ...

[quote="Xevarion"]I only read Antigone.[/quote]

me too. but I know the whole story of the drama.
Have you read the prometheus (well just the fragment)

Okay I have located it. Oedipus is the queen; Antigone her daughter.

It is all Greek to me anyway ...

Oedpius is a BOY who purposely killed his father so he could marry his mother and he ended up succeeding.

but in the end he became a crippled, blind, beggarhero if I recall correctly, karma comes around to bite most people but I guess not him

He didn't know it was his father. The parents knew the prophecy, so they threw Oedipus out on some mountain. A shepherd found him and he grew up and blah blah. Then he's walking along one day (to Thebes) and the king of Thebes is going the other way in his chariot and he's like "get out of the way you noob" and Oedipus is like "who's this tool?" and kills him. Then he goes to Thebes and not knowing that Jocasta (I think that's her name) is his mom, he marries her and bangs her and they have some kids.

[quote=Then he goes to Thebes and not knowing that Jocasta (I think that's her name) is his mom, he marries her and bangs her and they have some kids.[/quote]

lol. nicely said i guess.

I think it is too common a strategy in classical stories: there's a prophecy about someone's kid doing something bad, so the parents toss him out on the hillside. Then some shepherd finds the kid and raises him and he ends up fulfilling the prophecy.

Guys I'm not trying to be annoying but is this also a book? B/c I've read the same story when preparing for a mythology exam in a short 10-15 page format. But I like the story...so if it's a book I'll read it.

It's a series of three plays. They aren't terribly long; you can still read them.

[quote="ShoeFactory"][quote="Ignite168"][quote="Ars"]Not purposefully there was a prophecy about him that he would do that and so he was thrown out of the town and later he accidently threw a spear which hit his father and married the queen of the town not knowing who she was.[/quote]

No, he killed them purposely.[/quote]

You're wrong. Completely wrong.[/quote]

Umm.. he met them on the street in a chariot, he could've decided to not kill them. So it WAS purposefully.

But he didn't know it was his father; and I'm not sure he intended to kill the man, I think things just got out of hand (though I could be wrong on that part).

Okay I have got things straight.

Oedipud - King
Jacotsa - Queen
Antigone - Son

[quote="thatin"]deimos is right. Then the Queen commits suicide and Oedipud is forced to leave the city. His 2 sons become Kings, one year each. But at the end of the n year the brother 1 refuses to give the throne and brother 2 leaves the city and then attack with an army. The 2 brothers kill each other in the battle and their uncle becomes the new King, burries the bro1 and do not allow bro2 to be burried. Antigone burries him and finally she is sentenced to death. The son of her uncle commits suicide on her grave and then his mother also commits suicide.[/quote]

yeah. there is a question hidden in the drama : should we obey the universal human rules (bury a dead man) or the government's rules (don't bury him, he betrayed his country)?

[quote="Ignite168"]Where'd you get the names Jocasta and Oedipud? I thought it was Iocaste and Oedipus.[/quote]

Oedipud was probably a typo. Jocasta isn't wrong. The play was originally written in Greek, which doesn't have a 'j'. In Greek and Latin, 'i' can be used as a consonant or a vowel. Usually, in cases where it was used as a consonant, it is transliterated like 'j' instead of 'i': consider for example the words "joke" and "justice" -- both of these come from Latin words that start with 'i' ;)

[quote="deimos"][quote="bubka"]Okay I have got things straight.

Oedipud - King
Jacotsa - Queen
[b]Antigone - Son[/b][/quote]

not the son, she is the daughter (they had two more sons and one daughter, as far as I know. brothers kill themselves over the power in the city and their uncle made some problems not allowing  a funeral for one of them, and antigona buried him but the law was strict etc.)[/quote]

Yes of course. I got that right the last time. And Oedipud was a typo indeed.

[quote="G-UNIT"]Any of yall heard of the Oedipus Complex?  Alongside to that, the cycle turns into peanuts.[/quote]

In my English class, we spent quite a few days going over a paper about Freud's Oedipus Complex. Led to some interesting discussion in class  :rotfl:

Antigone is annoying. I hate her. Creon is awesome though.

i read the oedipus cycle, translation by paul roche something or other, I thought it was really good although I preferred oedipus rex to antigone and oedipus at colonus.

i really enjoyed the language used, it evokes so much more emotion and brings out the nuances of the play really well.

Oh God, I cannot believe the kind of things I am reading here. Do you guys know ANYTHING about ANYTHING? I can not even BEGIN to explain.

Is there anybody here who understands these things at all? I know the classics probably seem boring today, but truly, saying that Oedipus killed his father and married his mother ON PURPOSE etc. etc... this must be the end of the world... :o

Unfortunately , my bad english do not allow me to explain the drama.
  This work by Sophocles is one of the best in all the human history.This is  true , not because I'm Greek , but because this is the opinion of the greatest german  and french Philosophers and not only. When I was a pupil in class 11 , I was taught this tragodie in the school for a whole school year. I had unterstood only a few things ! It was just a nice story.But when I was a student in the mathematics department in the UNI of Athens , I decided to read carefully all the works not only of Sophocles , but of all Greeks in that golden century.And after Sophocles , Eyripides and Aischylos I read Schakespear( I know , I do not write correct his name!). What do you think? Can we say that after so many centuries the world has made a big step in literature?I answered :NO.And if you read the great work of W.Goete ''Foust'' , do you thing that this is different in structure or ideas from that classical drama's ?
 You have just to read  the classic writers to recognize that that these Greek creators , were really great.
  I have nothing  to defend  , this is not my aim. I 'm just trying to say that there  are in the history  some things that we can not ignore.These things ,mainly in art and sience , belong not to Greeks , Aigyptians , Arabs , indians or Cinese , but to the whole world , to all people. If we want to understand our civilization , we must read the classics and the initial-ancient  civilisations.
 ''Antigone'' is a  drama - or better : a part of a drama - which every moment can teach us important things.
  I want you to know that I would write the same things , if Sophocles was a Persian , or a Latin or an Aigyptian or something else.I'm afraid that we can not find easy a person in the history , that was a better 'writter' - in this kind of art  -than Sophocles.
  I'm glad however , because many of you  know so much about ''Antigone '' and ''Sophocles''.
  I suggest you , the Greek students as well , to study a little classic Literature and espessially to read Sophocles and Euripides works.I was a little lucky , because my wife  is a teacher of ancient Greek language , history and Latin , so I have all such books in my library.But everybody can find good translations of these books in all languages.I'm sure , you will delight in these works.And do not  forget: they were written 2500 years ago !

  [u]Babis[/u]

[quote="deimos"][quote="Cesario"]Oh God, I cannot believe the kind of things I am reading here. Do you guys know ANYTHING about ANYTHING? I can not even BEGIN to explain.

Is there anybody here who understands these things at all? I know the classics probably seem boring today, but truly, [b]saying that Oedipus killed his father and married his mother ON PURPOSE [/b]etc. etc... this must be the end of the world... :o[/quote]

who said that?[/quote]

Well, it was mentioned and actually [b]discussed [/b] in this very topic... haven't you seen it?
I'm glad that in the meantime someone tried to explain some things, but it can hardly be compensation enough. Those are things every educated person should at least know the rudiments of. In my humble opinion, of course.


You draw 5 cards out of your deck.
What is the probability of drawing a pair of seven?
(The rest of the three has to be different. For example {club 7, spade 7, heart 8, spade 2, diamond 5} would be acceptable, but not {club 7, spade 7, heart 8, spade 8, diamond 5} nor {club 7, spade 7, heart 8, spade 8, diamond 8})

ok i might get this wrong but ill give it a try
there are $ \binom{52}{5}$ total combinations of 5 cards drawn
there are 6 ways to choose the pair of 7s ($ \binom{4}{2}$)
and there are 48 ways to choose the third card (all cards except 7s) , 44 to choose the fourth (4 less to choose from b/c you chose another number) and 40 for the last.
the answer is $ \frac{6 \cdot 48 \cdot 44 \cdot 40}{\frac{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}}$
or $ \frac{2112}{10829}$
ok i probably missed this but at least i tried

Hmm I happened to get a different answer o.O

probability = (total number of ways to meet criteria) / (total number of ways)

Total number of ways to meet criteria:

$ \binom{4}{2} \equal{} 6$ ways to choose 2 7's
$ \binom{12}{3}$ ways to choose the rest 3 numbers from other 12 numbers for the other 3 cards
$ 4^3$ ways to give each of these cards a suit

$ \binom{52}{5}$ ways to choose total number of ways

So the answer (I think) is
$ \frac {\binom{4}{2} \cdot \binom{12}{3} \cdot 4^3}{\binom{52}{5}}$

Whatever that equals I don't know because I don't have my calculator :maybe: I just hope this expresstion doesn't come out to be > 1 :D

edit: $ \frac{352}{10829}$


A regular pentagon is inscribed in a circle. How many degrees are in the measure of an angle formed by radii drawn to two consecutive vertices of the pentagon?

Such radii produce central angles. Central angles of regular polygons are congruent. There are $ 360$ total degrees in the central angles. In a pentagon, there are $ 5$ central angles, giving $ \frac{360}{5}\equal{}\fbox{72}$ degrees for each central angle.