geometry
inequalities
number theory
prime numbers
IMO
IMO 2005

These were the results of our national team :)

BEL1	0 	1 	0 	7 	0 	0 	8 	Honor (Maarten De Boeck)
BEL2	2 	7 	0 	7 	3 	7 	26 	Silver (Guolong Li)
BEL3	0 	1 	0 	0 	0 	7 	8 	Honor (Stijn Vermeeren)
BEL4	2 	1 	1 	1 	1 	0 	6 	- (Francois Crucifix)
BEL5	3 	1 	1 	2 	0 	0 	7 	- (David Frenay)
BEL6	2 	7 	6 	2 	0 	2 	19 	Bronze (Francois Gonze)
TOTAL	74

This is a strange IMO. These are not common results for belgium... we have more points on 6 than on 1 :?

I saw this on another topic :

[quote="Stijn"][As mentioned by msn:
BELGIUM
Score P1 P2 P3 P4 P5 P6 Final Medal
BEL1  0   1   0  7   0   0   8      Honor   Maarten De Boeck
BEL2  2   7   0  7   3   7   26    Silver   Guolong Li
BEL3  0   1   0  0   0   7   8      Honor   Stijn Vermeeren (me)
BEL4  2   1   1  1   1   0   6                  Francois Crucifix
BEL5  3   1   1  2   0   0   7                  David Frenay
BEL6  2   7   6  2   0   2   19    Bronze  Francois Gonze
TOTAL 74

Medal cuts are 12-23-35 I believe.[/quote]

Hey all,

I'm back in our cosy 'Belgenlandje', with my Honorable Mention, especially honorable because I achieved it on the 6 :-).

Why our score on the problem 1 is low? Well, our team scores quite bad at both geometry problems. But the 1 certainly wasn't the easiest question on the exam (even the yury rated it 3rd-easy after 4 and 2). Look at China: even they have a 4 and a 0 (!) on the first problem, so don't blame us for scoring bad on this one :-).

Myself, I didn't have any clue on how to attack these problems. But in general, I'm much better in things like combinatorics than in geometry, and I didn't train very much on IMO-level geometry questions. That shows...
I should've been able to solve the second problem, it was very easy to do, but unfortonately, I didn't have the good point of view on it.
I had a nice strategy on the 3, a bit similar to the special price solution, but there was a big error in it, so I got 0 on that.
The 4: it was not difficult, but I didn't think of applying Fermats Little Theorem. Actually I should be a bit ashamed of that, because I wrote an 'eindwerk' (school paper) on prime numbers, which included FLT :-). But I just haven't applied FLT enough in olympiad problems to see the clue to the solution of the 4.
And then the six: to the horror of Bart (our leader) I wrote eleven pages, but I included a correct 7-point solution. In fact the problem wasn't that difficult. I was just looking immediately in the correct direction, the solution was a lot work, but I never lost track of it. In fact my solution can be simplified: I did it modulo 15 instead of modulo 5 :-). But it still is a nice solution, I'm very proud of scoring a 7 on this on :-). And I spoke to Arne (Smeets): he did the test himself and solved all problems BUT the 6, which makes me even prouder of it :-).

The rest of the team did well, and we're all happy. David didn't score as good as espected (he had bronze last year), but he doesn't seem to mind. David trained not so much this year aparently, and he just had two bad days. Francois Crucifix is a bit disappointed because he was equal with another person for the selection, only he made the team, now he's somewhat sorry for the other who didn't get selected, not scoring many points. But these are just little negative notes that are overwhelmed by our satisfaction with the result. Ok, a bit less points for Belgium than last year, but we have more points than The Netherlands, more than ('Arne's') South Africa... That Flanders beat Wallony is nice too, but doesn't matter that much.

Today on the airport in Mexico city I spoke to Ukraines Sergii Slobodianiuk. He participated (like me) for the first and only time, and he just made a perfect score :o. That was really like 'oh, wtf, that's really a different level!'. That just seems so inpossible to me :-). I'm just very happy with my mention, so I didn't come home empty handed today. The result was great, the experience probably even greater!

If you want me to tell more, just ask me things!

BTW: I was very surprised to hear FIVE people talking to me: "Hey, you're Stidgn (sic) from Mathlinks isn't it?". Wow, I'm only around here for 2 months, I've posted not even 40 messages, and people already know me! The power of MathLinks! :lol:

Another little thing: Who are you, Anh khoa?

[quote="Stijn"][quote="Anh khoa"][quote="Stijn"]

Another little thing: Who are you, Anh khoa?[/quote]

I 'm a student in Wépion , like David ,Francois Gonze , and francois Crucifix , but I didn't go to the IMO. Maybe you could remember me, I was the only asiatic ( except Guolong  :roll: ) in Wépion.  :)[/quote]

Ah, ok, I have some vague remembrance. How are your chances for being selected next year? :-)[/quote]

I don't know; I don't think I'll be selected next year but I could be selected in two years  :)

Yes, i think if you work a lot, you have a good chance to be selected in two years

Hi François, welcome to the forum and congratz again with your bronze!

Wow, I am impressed by the number of Belgians flooding Mathlinks :) I remember being one of the first Mathlinks users and the only Belgian on the forum!

Congratulations Francois, especially with Q3, well done.

I am definitely going to try to take Guolong to a gold medal. I am thinking about increasing the amount of training/coaching I have been giving him, don't know how yet, but he doesn't live that far from me... He has got two more years and I am going to do everything I can!

Glad to hear that there are some very good people coming from the French-speaking sides. On the Dutch speaking side we also have some VERY young people doing well. I hope we can get at least two silver medals next year (Guolong and Francois). We'll see, I hope I will be involved with the trainings.

[quote="Arne"]Glad to hear that there are some very good people coming from the French-speaking sides. On the Dutch speaking side we also have some VERY young people doing well. I hope we can get at least two silver medals next year (Guolong and Francois). We'll see, I hope I will be involved with the trainings.[/quote]
Why not a gold with guolong? :) He's only 9 points off... that sounds doable in one year?

Besides, who are those "very young" people? I thought except for guolong no one had another year? And from the other training stage I didn't really hear great sagas either.. (except for Stijn solving the shortlist question of course :D)

I don't really know any JWO participants well, but at least Bert (JWO first prize, also at stage) should be able to do some nice things in de next years.

Goulong himself is so confident (you may as wel say arrogant) to EXPECT a gold medal next year ("otherwise I would note have improved and that would be bad"). He thinks he is capable of everything and sets very high targets, but the point is: he just achieves these targets anyway! He was saying he wanted to score more than 50% (his classmate were laughing at him because he "failed" (-50%) at the physics olympiad), but he just DID it.
And this anekdote about Guolong: he had a solution for problem 1, and after the second test Bart already told him he would get a 7 for that. But Guolong repied: "No, can't be, the solutions is wrong." Coordination: 0. Guolong was right, apparently it was a solution many people had, but there was a problem with it only being valid for all angles of the hexagon > 90° of something like that. He noticed it, Bart not. Can you imagine!
He was also the first to point out my error in my problem 3-solution.
He tends a bit to behave like a child sometimes, but he's a math genius, who will score gold at least in two years.

Oh, BTW, Francois: Arne is good enough to be his own personal coach!

And Arne, are you back in Belgium now?

Oh yes, another anekdote about Goulong: Goulong was so much paying attention at the medal ceremony, he didn't even notice his own name was called. Francois had to hurry him to the podium!  :D

And another astonishing anekdote: Maarten got a Gold Medal! In fact: he both got a Honorable Mention AND a Gold Medal. :D At least, that was mentioned on one of the diploms! He just honorfully distroyed the Gold Medal diplom, but there should be a picture of it.

[quote="FrançoisG"]I think that Guolong was really nervous, and that's why he didn't hear his own name :roll: , but it's sure he's someone special ;) .[/quote]

Were you nervous ?  :P  :P

Well, Guolong is definitely brilliant but I think he doesn't always have the necessary technical knowledge in for example geometry and inequality problems. I don't always have the flashes of inspiration he has but I believe I have a very deep knowledge of geometry, inequalities and so on which he doesn't have. I hope I can work on that with him, because IMO is not only about brilliant ideas but also about mathematical background and knowledge. I guess that is why I didn't get gold last year and why he didn't get gold this year... But I am sure I can teach him a lot and I'll do everything I can. It will be a lot of hard work though. I gained a lot of coaching experience by training the South African IMO team for three weeks (setting and marking tests, lecturing (mainly geometry), ...) and of course I have some correspondence training with some Belgian students as well.

And there are a few very young promising mathematicians in Flanders indeed - I am talking about Bert and Alexander Lemmens, for example (they are brothers). I remember the 2004 prizegiving: Bert was in grade 7 and Alexander in grade 8 (or the other way around) and they both did very well in the competition for grades 9 and 10.

I have a few ideas about detecting mathematical talent in Belgium at an early stage and training people (and so does Stijn). We didn't start our project yet but we will see how it goes.

No, I am not back in Belgium yet, I am flying tonight.

OK, since 2005 is some kind of Einstein's year, let's relativize a few things...   :lol:

[quote] I don't know; I don't think I'll be selected next year but I could be selected in two years [/quote]

Everything can happen !  Just work hard enough...

[quote] And this anekdote about Guolong: he had a solution for problem 1, and after the second test Bart already told him he would get a 7 for that. But Guolong repied: "No, can't be, the solutions is wrong." Coordination: 0. Guolong was right, apparently it was a solution many people had, but there was a problem with it only being valid for all angles of the hexagon > 90° of something like that. He noticed it, Bart not. Can you imagine! [/quote]

Well, I have another reading of the situation.  The problem with all Guolong's proofs was that, despite some progress in that matter (I was told), he hardly ever wrote any detail.  In several cases, Bart had to work very hard to be ready to answer the coordinators' questions and convince them that Guolong's claims were not coming out of thin air (as you could easily have believed, looking at his solutions), but were substantiated by some little details that gave enough evidence that he actually knew what he was talking about.  So, that Bart missed a little detail in problem 1 is not a surprise : what's really marvellous is that he didn't miss anything else !  Besides, after the re-coordination of question 1, Guolong ended up with a 2, not a 0...

So if I have but one advice for Arne concerning Guolong's "preparation", it's the following : work hard on his style !!!!

It's well-known that Guolong hardly writes anything down :-).

But, Quick, who are you actually, you seem to be even better up to date than me.

I guess you are our deputy leader Philippe?


Yesterday Kyrgyz TST passed in which I took place ,so here are the problems :

                                                              \[\text{Kyrgyz IMO Team Selection Test 2005.}\]
                                      19'th April, 10:00-14:00

        [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=212394#p212394=]Problem 1.[/url]
 Show that for any rational  $x$  the following equality doesn't hold: $\{ x^2\}+\{x\}=1.$

        [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=212397#p212397=]Problem 2.[/url]
 If reals $a,b,c$ satisfy $a^2+b^2+c^2=3$ then prove that $|a|+|b|+|c|-abc\leq4$.

        [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=212399#212399=]Problem 3.[/url]
 A sequence $x_n,n\ge 0$ ,satisfy $x_{n+1}=x_n+\frac{1}{\sqrt{x_n}}$  .Find $\displaystyle\lim_{n\to\infty}x_n$  if  $x_0> 0$.

        [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=212393#212393=]Problem 4.[/url]
 Find all the increasing functions $f:{1,2,...,10}\longrightarrow\{1,2,...,100\}$ satisfying $x+y|xf(x)+yf(y)$ for all $1\leq x,y\leq 10$.

        [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=212400#212400=]Problem 5.[/url]
 Show that a triangle with side lengths  $a\ge b,c$  is right-angled if and only if  $(\sqrt{a+b}+\sqrt{a-b})\cdot(\sqrt{a+c}+\sqrt{a-c})=\sqrt{2}\cdot (a+b+c)$.

        [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34221=]Problem 6.[/url]
 Given a point  $M$  inside the square $ABCD$  and $N-$ midpoint of  $AB$ such that $\angle{MAD}=\angle{NMD}=\angle{MDC}=x$ .Find the value of $x$.

I suppose that all  the problems were taken from other national olympiads 2003-2004 years  :D  .

I made PDF and DVI documents.

please try to respect the guidelines I wrote in this topic http://www.mathlinks.ro/Forum/viewtopic.php?t=34216 :P :P
anyway thanks for the questions

PS it seems all the questions were from the Romanian Olympiad :D :D :D

Here are the national olympiads in PDF and DVI versions.

Thanks! What about providing the problems as HTML as well and maybe posting the single problems in the respective olympiad forums?  ;)


Here you'll find a quote from "The autobiography of Emilio Segrè" italian Nobel prize winner in physics in 1959
[quote]On a different occasion, while I was waiting to be called to an oral examination, Majorana gave me a synthetic proof for the existence of Villarceau's circles on a torus. I did not fully understand it, but memorized it on the spot. As I entered the examination room, Professor Pittarelli asked me, as was his wont, whether I had prepared a special topic. "Yes, on Villarceau's circles," I said, and I proceeded immediately to repeat Majorana's words before I forgot them. The professor was impressed and congratulated me on such an elegant proof, which was new to him.
[/quote]
The question is: does anybody know  "a synthetic proof for the existence of Villarceau's circles on a torus" (and not the obvious analytical proof that you can find in mathworld).
If you don't know what the Villarceau circles are click here below
[url=http://mathworld.wolfram.com/VillarceauCircles.html]mathworld.wolfram.com/VillarceauCircles.html[/url]

It's almost impossible! Sure Majorana was an unmeasurable genius!

[quote]Sure Majorana was an unmeasurable genius[/quote]
No doubt about it , but try to find this beautiful proof, you are not supposed to be a genius to discover it!


calculus
derivative
function
real analysis
real analysis unsolved

Assume that $ f(x)$ is twice differentiable and bounded function on $ ( \minus{} \infty, \plus{} \infty)$.
Show that there is $ x_0$ such that $ f^{''}(x_0) \equal{} 0$.

Just a logical argument. If $ f''(x) \equal{} 0$ had no roots, then $ f'(x)$ would not have a maxima or a minima, in which case $ f'(x)$ can be said to be monotonic, which cannot be the case, since $ f(x)$ is bounded. I'm sure there are a couple of loopholes here, but I'm just giving a logical approach.

If you meant to say, "in which case $ f'(x)$ can be said to be monotonic", then your argument is basically correct, otherwise it's nonsense.  Anyway, from there, it's not too hard to fill in the details:

Basically, to prove that $ f'$ is monotone, you assume otherwise, and use the mean value theorem to demonstrate a point where $ f''$ is negative, and a point where it is positive.  By the intermediate value theorem, it has a zero, contradiction.

Then there's the implication $ f'(x)$ monotone $ \implies f(x)$ unbounded.  Without loss of generality, assume that $ f'(x)$ is increasing, and $ f'(0)>0$.  Then we have $ f(x)>f'(0)x$, (the mean value theorem) so $ f$ is unbounded.  The only room for any loopholes here is when $ f'(x)\equal{}0$, but clearly this is impossible if $ f''$ is never zero.

[quote="LydianRain"]If you meant to say, "in which case $ f'(x)$ can be said to be monotonic", then your argument is basically correct, otherwise it's nonsense.  [/quote]

Oops. Yeah that was a typo. :D

[quote="LR"] The only room for any loopholes here is when $ f'(x) \equal{} 0$, but clearly this is impossible if $ f''$ is never zero.[/quote]

Actually $ f'(x) \equal{} 0$ is quite possible. Just because its monotonic does not mean it cannot be equal to zero.

[quote="Cyber-Shreyas"]Actually $ f'(x) \equal{} 0$ is quite possible. Just because its monotonic does not mean it cannot be equal to zero.[/quote]

Sorry, I meant "$ f'$ is identically equal to zero."  This is the only case when $ f'$ can be monotone while $ f$ is bounded, but it's impossible.


Hi,

I would just like to know when the new and improved Alcumus is coming out. I've been reading that you can choose what type of problems you want.

Thanks,

rts2007

We will announce any future versions in the sticky post area when we are aware of a timeframe.


the inequality is equivalent to:

$ [2(a^{2}\plus{}b^{2})\minus{}(a\plus{}b)^{2}]^{2}\plus{}2(a^{2}\plus{}b^{2})^{2}\minus{}(a^{2}\plus{}b^{2})(a\plus{}b)^{2}\geq 0$

so its sufficient to show that:

$ 2(a^{2}\plus{}b^{2})^{2}\geq (a^{2}\plus{}b^{2})(a\plus{}b)^{2}$

but $ a^{2}\plus{}b^{2}\geq 0$ so we are left to prove that:

$ 2(a^{2}\plus{}b^{2})\geq (a\plus{}b)^{2}$

$ \Rightarrow a^{2}\plus{}b^{2}\geq 2ab$

$ \Rightarrow (a\minus{}b)^{2}\geq 0$

which is true...


probability
modular arithmetic
expected value
number theory
relatively prime
algebra
binomial theorem

I'm not sure how to do this problem, the solution probably involves recursion but I can't seem to find it. Help?

Problem: I roll a fair 10-sided die (with sides numbered one through ten) until I get a number greater than or equal to my previous roll. If the expected value for the number of rolls is $ \frac{m}{n}$ where m and n are relatively prime positive integers, find the remainder when $ m\plus{}n$ is divided by 1000.

Given a previous roll of $ k$, let $ a_k$ be the expected number of rolls that it takes to get a number greater than your previous roll (given the $ 10$ sided die).

Then you always have one more roll, so add 1. Then if you get less than $ k$, you use the expected value from the recursion. \[ a_k\equal{}1\plus{}\frac{1}{10}*(a_1\plus{}a_2\plus{}...\plus{}a_{k\minus{}1})\] and you have $ a_1\equal{}1$ (of course if you had a one previously, you will get a number greater than or equal to $ 1$).

If you do some subtraction, $ a_{k\plus{}1}\minus{}a_k\equal{}\frac{1}{10}a_{k}$. Then $ a_{k\plus{}1}\equal{}\frac{11}{10}\cdot a_k$ so $ a_k\equal{}1.1^{k\minus{}1}$.

Now if we go back to the problem, we have the first roll, then we will roll any of the other with equal probability, so we need to find:
\[ 1\plus{}\frac{1}{10}\cdot (a_1\plus{}a_2\plus{}...\plus{}a_{10})\equal{}a_{11}\equal{}\frac{11^{10}}{10^{10}}\]

This fraction does not reduce, and we must find $ 11^{10}\pmod{11}$, and we can use the binomial theorem as fedja said.

I was real happy with how this problem came out.

[quote="fedja"]The main thing to understand is that the probability that you make more than $ k$ rolls is $ {10\choose k}10^{ \minus{} k}$ (indeed, it just means that the first $ k$ rolls result in a decreasing sequence of numbers and the number of different decreasing sequences of length $ k$ is the same as the number of $ k$-subsets of the set $ \{1,2,\dots,10\}$). The next step is to realize that the expectation in question is just the sum of all such probabilities. Now the binomial formula tells you that $ \frac mn \equal{} \left(\frac {11}{10}\right)^{10}$, so you need to find the remainder of $ 11^{10}$ modulo $ 1000$. Since $ (1 \plus{} 10)^{10} \equal{} 1 \plus{} 10\cdot 10 \plus{} 45\cdot 10^2 \plus{} \text{something divisible by }1000$, the answer is $ 601$.[/quote]
Can you tell me how you got $ \frac mn \equal{} \left(\frac {11}{10}\right)^{10}$?


induction
arithmetic sequence
algebra unsolved
algebra

find all sequences of [b]positive integers [/b]$a_1,a_2,a_3,...$ satisfying:
$a_{n}+a_{n+1}=a_{n+2}a_{n+3}-1996 \forall n\ge 1$

Nice one...:) We write \[ a_n+a_{n+1}=a_{n+2}a_{n+3}-1996 \] \[ a_{n-1}+a_n=a_{n+1}a_{n+2}-1996 \] By substracting them we get \[ a_{n+1}-a_{n-1}=a_{n+2}(a_{n+3}-a{n+1}) \] so \[ a_{n+3}-a_{n+1}=\frac{a_{n+1}-a_{n-1}}{a_{n+2}} \] Of course for every $n$ , $a_{n+1}-a_{n-1}$ is an integer. Now if $a_{n+1}-a_{n-1}<0$ for some $n$ then $a_{n+1}-a_{n-1}\leq -1$ and so on by induction for every $n$. This means that in this sequence exists $a_k<0$ which is contradiction. Now assume $a_3-a_1>0$ This means that for every even $n$ We need to have $a_{n+1}-a_{n-1}\geq 0$. We see that from our equation that if $a_{n+2}>1$ then $a_{n+1}-a_{n-1}>a_{n+3}-a_{n+1}$ and so on...but all those differences have to be positive integers, so can't have infinitely many terms $a_{n+2}$ ( for even $n$) such that $a_{n+2}>1$ it follows that there exists $n_0$ such that for every even $n>n_0$, $a_n=1$ In analogical way , if $a_4-a_2>0$ then there exists $n_1$ such that for every odd $n>n_1$, $a_n=1$ .This means that for every $n>max(n_0,n_1)$, $a_n=1$ which is contradiction with our first equation. Now if only one of differences $a_3-a_1$ and $a_4-a_3$ is  is also greater then $0$ we also get contradiction (if for example $a_3-a_1=0$ then $a_1=a_3=a_5=a_7=...$ let $a_1=a$ then after some finite numbers of terms we have $.......a,1,a,1,a,1....$ and we get a contradiction :) ) This means that $a_3-a_1=0$ and $a_4-a_2=0$ Let $a_1=a$, $a_2=b$. Then our sequence has form $a,b,a,b,a,b,a,b.......$ Our equation means that $a+b=ab-1996$ so $(a-1)(b-1)=1997$ so we have two possible sequences : either $a_1=2,a_2=1998,a_3=2$ or $a_1=1998, a_2=2, a_3=1998$ :) I hope there aren't many errors in this solution

Sure ...you're right... :oops:  of course we can also have $a_1=a_3=...=1$ and $a_2,a_4,a_6$ form an arithmetic sequence ( and also when $a_2=a_4=...=1$ and $a_1,a_3,a_5$ form an arithmetic sequence ...with difference equal to $1997$)....the mistake in my first solution was that after proving that $a_1=a_3=...=1$ in one case we are not able to get a contradiction...and why aren't we able to get contradicton? Cos' we get kind of solution :)


i want to download science lectures on to my computer and ipod, ive found a few teacher lectures and they are usually horrible, back ground noise, poor audio, talking half the time about handouts and lame class procedures, all of which should be on a website or another handout, so students class time isnt wasted if they are there for a lecture...the way its going it seems that quality lectures will be downloaded and then the teacher is there to offer private tutor assistance or maybe to sweep up the room...as it is now the typical class room format is still very inefficient unless you have a rare good teacher....anyway, does anyone have a list of good science lectures that can be downloaded free....i found a site that had the feynman lectures from like 40 years ago and what i heard was very poorly editied, but maybe i only heard samples....feynman was supposed to be one of the best...anyway, i know they are out there, does anyone have any good sites that they would like to recommend.........thanks

ocw.mit.edu

Check out physics 8.01, 8.02, 8.03

They have video lectures there.

thanks, i went there MIT is a great resource and the courses were all set up for downloads and free, the only problem with it is that i am looking for audio downloads...i want to be able to listen to lectures on my computer headphones or ipod headphones as i am resting or working so i dont have to stare at a screen or book while im learning....if you know where good sources are for free audio downloads id be interested otherwise thanks for the info

ok i went back there again and now i see that have small lists of video and audio only lectures, i went to the audio list, it had like 20 courses or i dont remember and i clicked on the nanotechnology course....i also downloaded the corresponding pdf file...it was all pretty low grade but mit is a great resource and they explained that they were just starting out with this program....the audio lecturer wasnt bad, it was just amature level audio, like a mic clipped to the professors tie or whatever.....thanks again for your participation...i think this is a very promising area, the way educational courses are run today is like off of 1910 paradijms at best and they are slowly beginning to utilize these new technologies, id say in another ten years these new options will have become more functionally assimilated.

i used to use limewire and i was able to get all kinds of stuff (file sharing), i didnt try math/science but i got a lot of random stuff like lou holtz halftime speech, a churchill speach, a ginsberg reading.  but whether or not you'd get math/science i'd have to say would be a shot in the dark.


Source:Sosmath.com/CBB

Solve x^[log_{2}(x+4)]=32 ,x is a real number

I might be missing a trick for this one...

I think you'll find the trick you were missing was a misplaced bracket. The original question said x^log_2(x+4)=32. Later on, the poster said the answers were 1/32 or 2 so I expect the problem should have been x^(log_2 x+4)=32 ie x^{\log_{2}x +4}=32 which by contrast is easy to solve.


geometry
logarithms
AMC
triangle inequality

The lengths of the sides of a triangle with positive area are $\log_{10} 12$, $\log_{10} 75$, and $\log_{10} n$, where $n$ is a positive integer. Find the number of possible values for $n$.

[hide]First, we need $\log_{10}12+\log_{10}n>\log_{10}75$.  Combine logs to find n>6.  Also, we need $\log_{10}12+\log_{10}75>\log_{10}n$.  Here we find that 900>n.  893 n satisfy the requirements.[/hide]

Using triangle inequality , log 10¹²+log10⁷⁵>log10^n ,  so,12*75>n , 900>n
Similarly , log10⁷⁵–log10¹²<log10^n , so 75/12<n so ,6<n<900 so no. Of values of 'n' =893

p9: We must have $\log_{10}{12} + \log_{10}{75} > \log_{10}{n}$ and $\log_{10}{12} + \log_{10}{n} > \log_{10}{75}$. The first equation gives $n < 900$ and the second gives $n > 6.25$, for a total of $899-7+1 = \boxed{893}$ possibilities.

aime p7

By triangle inequalities, $\log_{10}(12)+\log_{10}(75)>\log_{10}(n)\implies \log_{10}(900)>\log_{10}(n)\implies n<900$.

By triangle inequalities again, $\log_{10}(12n)> \log_{10}(75)\implies n>6$.

By triangle inequalities again, $\log_{10}(75n)>\log_{10}(12)$, so all $6<n<900$ work, thus answer is $900-6-1=\boxed{893}$.


Let $ a,b,c$ be non-negative numbers such that $ a\plus{}b\equal{}c\equal{}1$ and no two of which are zero. Prove that :
$ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\plus{}\frac{4(a^2\plus{}b^2\plus{}c^2\minus{}1)}{a^3\plus{}b^3\plus{}c^3}\geq 5$

[quote="vo thanh van"]Let $ a,b,c$ be non-negative numbers such that $ a \plus{} b \equal{} c \equal{} 1$ and no two of which are zero. Prove that :
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {4(a^2 \plus{} b^2 \plus{} c^2 \minus{} 1)}{a^3 \plus{} b^3 \plus{} c^3}\geq 5$[/quote]
If you mean $ a\plus{}b\plus{}c\equal{}1$
It's obviously wrong.

Can you show that give a wrong mistake to me?

take $ a\equal{}b\equal{}c\equal{}\frac 13$


projective geometry
geometry
analytic geometry
algebra
polynomial
3D geometry
tetrahedron

Who can post an element proof of the Desargues Theorem?

[b]Moderator edit:[/b]

[b]Desargues theorem.[/b] Let ABC and A'B'C' be two triangles. Let the lines BC and B'C' intersect at a point X, let the lines CA and C'A' intersect at a point Y, and let the lines AB and A'B' intersect at a point Z. Then, the lines AA', BB', CC' concur at one point if and only if the points X, Y, Z are collinear.

It is easiest to view the perspectivity of the triangles in 3-dimensions.

I think that it is also true if the segments connecting the vertics are paralel to each other. Who can prove it?

Desargues theorem is usually considered as a theorem of projective geometry, so the statement (and any projective proof) will automatically include this case.  Three parallel lines intersect "at infinity", so from the projective point of view there is no difference, because there is no concept of parallelism.

Suppose, however, that a proof is given of Desargues theorem for triangles in Euclidean plane geometry, using non-projective concepts like angle, distance, and parallelism.   Then the projective version is also true, because any configuration in the projective plane (Euclidean plane + line at infinity) can be moved away from the line at infinity while preserving all relationships between points and lines.  A theorem about points and lines is true in the original configuration if and only if it is true in the moved configuration.

It is a similar idea to moving the coordinate axes to a position convenient for the problem.

I am trying to find a proof with Menelaus but I seem to lost my way...

It's Darij posting again. ;)

Here is a proof of the Desargues theorem using Menelaos.

[color=blue][b]Desargues theorem.[/b] Let ABC and A'B'C' be two triangles. Let the lines BC and B'C' intersect at a point X, let the lines CA and C'A' intersect at a point Y, and let the lines AB and A'B' intersect at a point Z. Then, the lines AA', BB', CC' concur at one point if and only if the points X, Y, Z are collinear.[/color]

[i]Proof.[/i] We will use directed segments.

First, we will prove one direction of the Desargues theorem: We will show that if the lines AA', BB', CC' concur, then the points X, Y, Z are collinear.

In fact, let the lines AA', BB', CC' concur at the point P. Then, the Menelaos theorem, applied to the triangle BPC and the collinear points B', C', X on its sides BP, CP, BC yields

$\frac{BX}{XC}\cdot\frac{CC^{\prime}}{C^{\prime}P}\cdot\frac{PB^{\prime}}{B^{\prime}B}=-1$.

Similarly, from triangles CPA and APB we infer

$\frac{CY}{YA}\cdot\frac{AA^{\prime}}{A^{\prime}P}\cdot\frac{PC^{\prime}}{C^{\prime}C}=-1$;
$\frac{AZ}{ZB}\cdot\frac{BB^{\prime}}{B^{\prime}P}\cdot\frac{PA^{\prime}}{A^{\prime}A}=-1$.

Multiplying these three equations, we get

$\left(\frac{BX}{XC}\cdot\frac{CC^{\prime}}{C^{\prime}P}\cdot\frac{PB^{\prime}}{B^{\prime}B}\right)\cdot\left(\frac{CY}{YA}\cdot\frac{AA^{\prime}}{A^{\prime}P}\cdot\frac{PC^{\prime}}{C^{\prime}C}\right)\cdot\left(\frac{AZ}{ZB}\cdot\frac{BB^{\prime}}{B^{\prime}P}\cdot\frac{PA^{\prime}}{A^{\prime}A}\right)=\left(-1\right)^3$.

In other words,

$\left(\frac{BX}{XC}\cdot\frac{CC^{\prime}}{C^{\prime}P}\cdot\frac{B^{\prime}P}{BB^{\prime}}\right)\cdot\left(\frac{CY}{YA}\cdot\frac{AA^{\prime}}{A^{\prime}P}\cdot\frac{C^{\prime}P}{CC^{\prime}}\right)\cdot\left(\frac{AZ}{ZB}\cdot\frac{BB^{\prime}}{B^{\prime}P}\cdot\frac{A^{\prime}P}{AA^{\prime}}\right)=\left(-1\right)^3$,

what simplifies to

$\frac{BX}{XC}\cdot\frac{CY}{YA}\cdot\frac{AZ}{ZB}=-1$.

Thus, applying the Menelaos theorem to the triangle ABC and the points X, Y, Z on its sides BC, CA, AB, we see that the points X, Y, Z are collinear.

Remains to prove the converse direction; that is, to prove that if the points X, Y, Z are collinear, then the lines AA', BB', CC' concur. But this converse direction can be easily inferred from the direction we already showed. In fact, consider the two triangles AA'Y and BB'X. The point of intersection of the lines A'Y and B'X is the point C', and the point of intersection of the lines YA and XB is the point C. The point of intersection of the lines AA' and BB' will be denoted by Q. The lines AB, A'B' and YX concur at one point (in fact, they concur at the point Z - since Z was defined as the point of intersection of the lines AB and A'B' and since the points X, Y, Z are collinear). Hence, by the direction of the Desargues theorem we already showed, the points C', C and Q are collinear. Thus, the point Q lies on the line CC'. In other words, the point of intersection of the lines AA' and BB' lies on the line CC'. Hence, the lines AA', BB', CC' concur. This proves the second direction of the Desargues theorem. And thus, the proof of the Desargues theorem is complete.

  Darij

[quote="fleeting_guest"]But as explained above, the proof with Menelaus' theorem (or any other method) for the Euclidean Desargues theorem implies the projective Desargues theorem, which implies the Euclidean Desargues theorem in the case where the lines are parallel.[/quote]
Well, I knows nothing about the projective geometry, so i don't know how to imply from one to another.

[quote="shobber"]But he did not mention the paralel case...
I was able to do what Darij did in this problem, but i don't know the special case.[/quote]

Shobber, what exactly do you mean by the "parallel case"?

If you mean the case when, for instance, the lines BC and B'C' are parallel, then their point of intersection X is an infinite point, and the proof I gave in post #6 works perfectly well, assumed one uses the standard convention to write $\frac{P_1P_{\infty}}{P_{\infty}P_2}=-1$, where $P_1$ and $P_2$ are two finite points and $P_{\infty}$ is the infinite point on the line $P_1P_2$.

However, if you are talking about the case when the lines AA', BB', CC' are parallel, then indeed, the proof from post #6 doesn't work anymore. However, there are two possibilities to handle this case:

Either one says that, since the Desargues theorem is proven in the generic case (when the lines AA', BB', CC' are concurrent at a finite point), it must also hold for the special case (when the lines AA', BB', CC' are parallel, i. e. are concurrent at an infinite point). This is clear from continuity or, to remain elementary, from algebraicity (the Desargues theorem can be interpreted as a polynomial identity, and once it is proven for a generic case, it must hold for degenerate cases as well).

But one can also simply adjust the proof I gave in post #6 for the case of a finite point of concurrence to the case when the lines AA', BB', CC' are parallel:

The only thing we have to change in the whole proof is the derivation of the equality

$\frac{BX}{XC}\cdot\frac{CY}{YA}\cdot\frac{AZ}{ZB}=-1$.

In fact, this time we cannot use Menelaos in the triangles BPC, CPA, APB, since the point P is infinite. But actually, the equality in question is almost trivial: Since BB' || CC', Thales yields $\frac{XB}{XC}=\frac{BB^{\prime}}{CC^{\prime}}$, so that $\frac{BX}{XC}=-\frac{XB}{XC}=-\frac{BB^{\prime}}{CC^{\prime}}$. Similarly, $\frac{CY}{YA}=-\frac{CC^{\prime}}{AA^{\prime}}$ and $\frac{AZ}{ZB}=-\frac{AA^{\prime}}{BB^{\prime}}$. Multiplying these three equations yields

$\frac{BX}{XC}\cdot\frac{CY}{YA}\cdot\frac{AZ}{ZB}=\left(-\frac{BB^{\prime}}{CC^{\prime}}\right)\cdot\left(-\frac{CC^{\prime}}{AA^{\prime}}\right)\cdot\left(-\frac{AA^{\prime}}{BB^{\prime}}\right)=-1$,

and the proof is done.

  Darij

[quote][quote]But as explained above, the proof with Menelaus' theorem (or any other method) for the Euclidean Desargues theorem implies the projective Desargues theorem, which implies the Euclidean Desargues theorem in the case where the lines are parallel.[/quote]
Well, I know nothing about the projective geometry, so i don't know how to imply from one to another.[/quote]

You need to know almost nothing about projective geometry to understand this idea.

What is needed is: (1) the relationship between Euclidean and projective geometry (projective plane = Euclidean plane + line-at-infinity) and (2) what happens to Euclidean-parallel lines in projective geometry (they intersect, but not in the Euclidean part of the plane; the intersection is on the line at infinity).

This way you can recognize a many limit cases of theorems in Euclidean geometry as special cases of the projective theorems.... which are true because the non-limit, "generic" case was proven in Euclidean geometry.

[quote="ZetaX"]

$ \frac {BX}{XC}\cdot\frac {CY}{YA}\cdot\frac {AZ}{ZB} \equal{} \minus{} 1$.

Thus, applying the Menelaos theorem to the triangle ABC and the points X, Y, Z on its sides BC, CA, AB, we see that the points X, Y, Z are collinear.

[/quote]

umm how can one apply Menelaos to the equality above?

Dear Mathlinkers,
an elementary proof of the Desargues Theorem can also be obtain with the Pappus theorem.
Sincerely
Jean-Louis

It is better to consider 3-dimensional space instead of 2-dimensional plane.
Indeed,if we take any tetrahedron $ XABC$ with base $ ABC$ and intersect it with another plane,name it $ A'B'C'$,where $ A'\in XA,\dots$.
Then [b]Desargues[/b] theorem will be nothing more than just an obvious fact,which tell us that intersection of two planes is a line.

[quote="glorfs"]But is the proof for 3d space proof enough for 2d plane also?[/quote]
yup,we can always project 3d space to some plane and conversely we can consider some tetrahedron instead of quadrilateral. :roll:

Thanks Erken!

You just made my life a lot easier =)


Why is it that when we integrate $ \minus{} d[R]/[R] \equal{} kdt$ with respect to $ t$ on both sides we get $ ln([R]t/[R]o) \equal{} \minus{} kt$?

Why is it that the LHS contains $ R_t$ and $ R_0$ yet the RHS has no such indexes?
I'm wondering if we are assuming $ t_0 \equal{} 0$ sec

Hello
when we integrate the equation
$ \minus{}\frac {d[R]}{[R]} \equal{} kdt$
we get
$ \minus{}lnR \equal{} kt \plus{} c$
now we assume at $ t \equal{} 0 R \equal{} R_0$
thus by putting $ t \equal{} 0$ $ \minus{}lnR \equal{} c$
now if at time=t $ R \equal{} R_0$
we get
$ \minus{}lnR \equal{} kt \minus{} lnR_0$
or
$ ln(\frac {[R_0]}{[R]}) \equal{} kt$
Hope it would be somehow better.
Thank u.

thanks :) I understand the issue I was originally confused about. I just realized that I still don't understand something: if we are integrating with respect to $ t$ on both sides, then how do we go about integrating the left hand side?

Why does it appear that we have integrated with respect to $ [R]$ on the left hand side when it also looks like we have integrated with respect to $ t$ on the right side?

Hello aufha
Sorry but u r somehow wrong .
When we r doing integration on LHS we are integrating it with respect to R as there is written $ dR$ and on RHS we are integrating with respect with to t as it is written $ dt$ that side.
Is it fine now?
Thank u.

[quote="aufha"] if we are integrating with respect to $ t$ on both sides, then how do we go about integrating the left hand side?
[/quote]

Unfortunately or fortunately, integration is not like differentiation. You dont integrate "both sides" wrt the same parameter, you simply integrate over a variable.

[quote="hell_ever"]

Unfortunately, integration is not like differentiation. You dont integrate "both sides" wrt something, you simply integrate.[/quote]
Well is it really unfortunate? :rotfl:


Every city in Berland is situated on Ox axis. The government of the country decided to build new telecasting station. After many experiments Berland scientists came to a conclusion that in any city citizens displeasure is equal to product of citizens amount in it by distance between city and TV-station. Find such point on Ox axis for station so that sum of displeasures of all cities is minimal. Berland have N<15000 cities.

Consider any possible point for the station. Moving it forwards changes the sum at a rate of (number of people below it)-(number of people above it). Thus, you can assume it lies at a city by moving it in the correct direction. Then, you just need to find the first city with (number of people <= it) >= (number of people >= it).

Compute a weighted average of the positions of the cities.


A cork floats on the surface of a 4 metre deep pond of liquid of refractive index 1.414
As the sun's inclination changes from noon to sunset ,find the maximum distance from the shadow of the cork to the normal through the cork.
Assume that the normal  through the cork and the sun's path are in the same plane.

By the law of light refraction,

$n_I \sin \theta_I = n_F \sin \theta_F$

where $n_I, n_F$ are the refraction indices of the initial and final medium and $\theta_I, \theta_F$ the angles of the incident and refracted light rays from a normal to the boundary between the 2 media. In our case, $n_I \doteq 1$ for air and $n_F = 1.414 \doteq \sqrt 2$ for the liquid. The distance L of the cork shadow from the normal to the liquid surface at the depth D = 4 m is

$L = D |\tan \theta_F| = \frac{D |\sin \theta_F|}{\sqrt{1 - \sin^2 \theta_F}} = \frac{D n_F |\sin \theta_F|}{\sqrt{n_F^2 - n_F^2 \sin^2 \theta_F}} = \frac{D n_I |\sin \theta_I|}{\sqrt{n_F^2 - n_I^2 \sin^2 \theta_I}}$

Substituting $D= 4\ \text m,\ n_I = 1,\ n_F = \sqrt 2$,

$L = \frac{4 |\sin \theta_I|}{\sqrt{2 - \sin^2 \theta_I}}$

This expression is obviously maximum, when $|\sin \theta_I|$ is maximum, i.e.,  when $|\sin \theta_I| = 1,\ \sin \theta_I = \pm 1,\ \theta_I = \pm 90^\circ$, i.e., at sunrise and at sunset.

$L_{\text{max}} = \frac{4}{\sqrt{2 - 1}} = 4\ \text m$


I heard tha Naoki Sato has published some useful papers for math lectures. Does any one know where i can get them? :maybe:

I've written some notes on number theory, which you can find in the articles section: [url]http://www.artofproblemsolving.com/Resources/AoPS_R_Articles.php[/url]


geometric series
combinatorics unsolved
combinatorics

Hi,

I've been trying to solve this one for a while. Hopefully someone could help me out here.

If d(m) is the number of divisors of the positive integers m, prove:

$ \sum_{n \equal{} 1}^{\infty}\frac{z^{n}}{1\minus{}z^{n}}$ = $ \sum_{m \equal{} 1}^{\infty}d{(m)}z^{m}$

Thankyou

First note that for $ |z|<1$ all sums involved are absolute convergent and that $ d(m)\equal{}|\{(a,b)\,|\,ab\equal{}m\}|$ holds.

Using a geometric series expansion we get via direct calculation $ \sum_{n\equal{}1}^{\infty}\frac{z^{n}}{1\minus{}z^{n}}\equal{}\sum_{n\equal{}1}^{\infty}\sum_{m\equal{}1}^{\infty}z^{nm}\equal{}\sum_{k\equal{}1}^{\infty}\sum_{(n,m):\,nm\equal{}k}z^{k}\equal{}\sum_{k\equal{}1}^{\infty}d(k)z^{k}$

[quote="hjbrasch"]First note that for $ |z| < 1$ all sums involved are absolute convergent and that $ d(m) \equal{} |\{(a,b)\,|\,ab \equal{} m\}|$ holds.

Using a geometric series expansion we get via direct calculation $ \sum_{n \equal{} 1}^{\infty}\frac{z^{n}}{1\minus{}z^{n}}\equal{}\sum_{n \equal{} 1}^{\infty}\sum_{m \equal{} 1}^{\infty}z^{nm}\equal{}\sum_{k \equal{} 1}^{\infty}\sum_{(n,m):\,nm \equal{} k}z^{k}\equal{}\sum_{k \equal{} 1}^{\infty}d(k)z^{k}$[/quote]

Hi hjbrasch,

Thankyou for your reply.

I want to ask you a few questions:

1)  The sum $ \sum_{n \equal{} 1}^{\infty}\sum_{m \equal{} 1}^{\infty}z^{nm}$

gives:

$ \sum_{m \equal{} 1}^{\infty}{z^{m}}\plus{}\sum_{m \equal{} 1}^{\infty}{z^{2m}}\plus{}\sum_{m \equal{} 1}^{\infty}{z^{3m}}\plus{}...$

$ \equal{}$

$ \frac{1}{1\minus{}z}\plus{}\frac{1}{1\minus{}z^{2}}\plus{}\frac{1}{1\minus{}z^{3}}\plus{}...$

but the sum $ \sum_{k \equal{} 1}^{\infty}d(k)z^{k}$ gives:

$ d(1)z\plus{}d(2)z^{2}\plus{}d(3)z^{3}\plus{}d(4)z^{4}\plus{}...$

= $ z\plus{}2z^{2}\plus{}2z^{3}\plus{}3z^{4}\plus{}...$

Could you please explain how to get from:
$ \sum_{n \equal{} 1}^{\infty}\sum_{m \equal{} 1}^{\infty}z^{nm}$

to

$ \sum_{k \equal{} 1}^{\infty}\sum_{(n,m):\,nm \equal{} k}z^{k}$

Thankyou, I would very much appreciate it.

it is a counting argument: you sum over all $ z^{nm}$ with $ n,m\in\mathbb{N}$ and this is equivalent to sum over all $ z^{k}$ with $ k\in\mathbb{N}$ and weight it with the number of occurences in the other sum, i.e. with the number of different (ordered) product representations $ k\equal{}nm$


calculus
integration
trigonometry
calculus computations

The answer is $+\infty$ because $1-\cos x\cos y\approx \frac{1}{2}(x^{2}+y^{2})$ near $(0,0)$ and $\int_{0}\int_{0}\frac{dx\,dy}{x^{2}+y^{2}}=+\infty$.

There's also another singularity at $(\pi,\pi)$, with the same behavior.


To Mod : Can You Remove this Topic to NumBer Theory Box  :lol: Thank You :D

[quote="Harry Potter"]x;y;z are 3 positve interger numbers such that : x+y+z=20 Find Max and Min Value of $ T \equal{} x^{y^{z}}$

P/s : Min T =1 When x=1 >.<[/quote]
Dear  Harry Potter We have  Min $ T \equal{} 14^{3^3}$ and Max $ T \equal{} 3^{6^{11}}$
Note :
         $ x^{y^z}\geq 14^{3^3}$ <-> $ y^z lnx\geq 27 ln 14$
  and $ y^z ln x\leq y^{17 \minus{} y} ln3$


One of the factors of $ x^4\plus{}4$ is:
$ \textbf{(A)}\ x^2\plus{}2 \qquad\textbf{(B)}\ x\plus{}1 \qquad\textbf{(C)}\ x^2\minus{}2x\plus{}2 \qquad\textbf{(D)}\ x^2\minus{}4\\
\textbf{(E)}\ \text{none of these}$


geometry
ratio
area of a triangle
similar triangles

[quote="d.lam.86"]The area of a triangle with base 24 cm is trisected by two segments parallel to the base. What is the length of the longer of these segments?

I was told it is 8*sqrt(6) cm, but I don't get why.

Thanks.[/quote]
[hide]Call the triangle ABC with BC as the length of 24. Parallel lines calls for similar triangles so call the first line closest to the base DE with D on AB and E on AC. This line is obviously the longer one. Because the area is trisected with the other line I'll call FG, the area of triangle ADE is 2/3 of ABC. The ratio of the sides of similar triangles squared is the ratio of the areas of similar triangles. We have that the ratio of the sidelengths is $ \frac{\sqrt{6}}{3}$. Since the side length of the base is 24, the length of the longer segment, or DE should be $ 8\sqrt{6}$.[/hide]


what is the probability, in which all of the angles of a triangle be acute?

It depends on what method you are using to pick your "random" triangles. Are you determining them by picking angles that add up to 180 degrees? Or are you picking three points in the plane and then connecting them? Or some other method.

How you construct your triangles will give different answers.

Of course, if you're picking three points in the plane randomly, that requires its own method.

just a side note,
the title of this thread "konkoor 85", is refering to the Iranian University Entry Exam  ;)


[hide]Let the desired value be $ S$.
First, observe that
\[ \left\lfloor \frac {2^1}{1!} \plus{} \frac {2^2}{2!} \plus{} \frac {2^3}{3!} \plus{} \frac {2^4}{4!} \right\rfloor \le \left\lfloor \frac {2^1}{1!} \plus{} \frac {2^2}{2!} \plus{} \cdots \plus{} \frac {2^{100}}{100!} \right\rfloor \iff 6 \le S.
\]
In addition, the expression inside the floor can be written as
\[ \frac {2^1}{1!} \plus{} \frac {2^2}{2!} \plus{} \frac {2^3}{3!} \left( 1 \plus{} \sum_{n \equal{} 4}^{100} \prod_{k \equal{} 4}^{n} \frac {2}{k} \right)
\]
which is less than
\[ \frac {2^1}{1!} \plus{} \frac {2^2}{2!} \plus{} \frac {2^3}{3!} \left( 1 \plus{} \sum_{n \equal{} 4}^{\infty} \prod_{k \equal{} 4}^{n} \frac {2}{4} \right) \equal{} \frac {2^1}{1!} \plus{} \frac {2^2}{2!} \plus{} \frac {2^3}{3!} \left( 1 \plus{} \sum_{n \equal{} 1}^{\infty} \frac {1}{2^{n}} \right) \equal{} \frac {20}{3}
\]
after summing an infinite geometric series. This implies that $ S \le \left\lfloor \frac {20}{3} \right\rfloor \equal{} 6$.

Therefore, $ 6 \le S \le 6$, i.e., $ S \equal{} \boxed{6}$.[/hide]


Well
[hide]2sqr((n)(n-3)/2)=n, so 2(n)(n-3)=n^2, 2n^2-6n=n^2, n^2=6n, and since n is not zero we can cancel and get n=6 [/hide]

[quote="slamchess"]how do you do this problem?[/quote]
[hide]Set up an equation.
$2\sqrt{\frac{n(n-3)}{2}}=n$
$\sqrt{\frac{n(n-3)}{2}}=\frac{n}{2}$
$\frac{n(n-3)}{2}=\frac{n^{2}}{4}$
$\frac{n^{2}-3n}{2}=\frac{n^{2}}{4}$
$2n^{2}-6n=n^{2}$
$n^{2}=6n$
$n=6$,$n=0$. But a polygon cannot have 0 sides so $6$ is the answer.[/hide]

If what you were confused about was the number of diagonals, the number of diagonals is

[hide]$\frac{n(n-3)}{2}$[/hide]


Suppose that $E$ is a connected closed subset set of $\mathbb{R}^{2}$, that contains $(0,0)$ and $(0;1)$. Suppose that $H^{1}(E)<2$ (here $H^{1}$ is the usual Hausdorf measure). Prove that there exists $0<r<1$ such that $E$ contains only one point with modulus $r$.

I don't think it matters whether the set is connected or not (or, for that matter, whether it contains those points or not :)).

First notice that for any subset $ S$ of the plane containing at least one point on each circle of radius $ < 1$ centered at the origin (call such a set "good") we have $ H^{1}(S)\ge 1$. This follows from the fact that the Lipschitz map (with Lipschitz constant $ 1$) sending each point of the plane to its Euclidean norm maps $ S$ onto a real set which contains $ (0,1)$. The desired result now follows from the observation that any closed plane set which contains at least two points on each circle of radius $ < 1$ centered at the origin contains two disjoint good Borel sets.


function
trigonometry
algebra
polynomial
calculus
integration
geometric series

Hi I have another problem regarding sums.  Does anyone know of an easy way to evaluate the following?

Question:  Evaluate $ \displaystyle\frac{1}{1^3}\plus{}\frac{1}{3^3}\minus{}\frac{1}{5^3}\minus{}\frac{1}{7^3}\plus{}\frac{1}{9^3}\plus{}\frac{1}{11^3}\minus{}\cdots$

[hide]You can use the sinus Fourier series for the function $ f(x) \equal{} x(\pi \minus{} x)$. The final answer is $ S \equal{} \frac{3\pi^3 \sqrt{2}}{128}$[/hide]

Hi Carcul, is there a thought process that you follow when you attempt problems like these? :?:  I do not understand how one would know what functions to expand?  More importantly, how does one know that the problem can be approached by Fourier series in the first place?

Anyways, I have calculated the sine expansion of $ x(\pi\minus{}x)$.  If my work is correct, it should be $ \displaystyle x(\pi\minus{}x) \equal{} \frac{8}{\pi}\left(\frac{\sin x}{1^3}\plus{}\frac{\sin 3x}{3^3}\plus{}\frac{\sin 5x}{5^3}\plus{}\cdots\right)$.  Now, how should I proceed?  It is not quite in the form of the required series yet.

You'll need to plug in some carefully-chosen values of $ \sin x$ and sum some series.  (The obvious choices to test are things like $ x \equal{} \pi, \frac{\pi}{2}$ - where do those get you?)

Okay so $ \displaystyle x(\pi\minus{}x) \equal{} \frac{8}{\pi}\left(\frac{\sin x}{1^3}\plus{}\frac{\sin 3x}{3^3}\plus{}\frac{\sin 5x}{5^3}\plus{}\cdots\right)$ if we substitute $ x\equal{}\pi/4$, we can get the desired sum.

However, I still do not know how Carcul realized that this sum could be evaluated by way of Fourier series.  Also, I do not know how he knew to find the sine expansion of $ f(x)\equal{}x(\pi\minus{}x)$.

Can anyone please explain this to me?  Thanks.

When faced with a sum, and if the usual methods (geometric series, telescoping, Mengoli series, hypergeometric series, and so on) don't work, you can always try other methods: Fourier series is just one of them.

I'm thinking the OP wants to know how one decided on what function to expand.

Its a bit of a run with inference, from the way the terms alternate one would choose an odd function as a polynomial of degree 2 (to get the third degree term in the Fourier coefficients on integration that we require) and so on.

Thank you all for input on this problem.  I am just wondering should I familiarize or "remember" the Fourier expansions of some common/basic functions?  Considering the fact that we all have a few Laplace/Fourier transforms in our minds already.  Or do you guys think that knowing Fourier series expansions is a bit that too much?

Memorising stuff doesn't really help any in mathematics in general, but sometimes there [i]are[/i] functions more likely for instructors to use (the classic case seems to be the 'sawtooth function' in Fourier analysis)- though even then, a more analytic outlook to approaching problems is always better! :)


1. rezolvati ecuatia:
2x=[x]{x}

2. gasiti functiile f:R->R cu proprietatea:

f(x+y) = f(x) + f(y) + f({x}+{y}) , pt orice x,y din R si f(1) = 1.

problemele sunt facute de mine. sper sa nu mai fi compus cineva pana acum functia asta si sa fiu acuzat de plagiatura :lol:

[b]pb 2[/b]

Daca $x\leftarrow 0$ si $y\leftarrow 1$ $\rightarrow$ $f(1)=f(0)+f(1)+f(0)$ $\rightarrow$ $f(0)=0$
Daca $x,y \in \mathbb Z$ $\rightarrow$ $f(x+y)=f(x)+f(y)$ $\rightarrow$ $f(x)=f(1)x,\forall x\in \mathbb Z$
Daca $x\leftarrow \{ x \}$ si $y\leftarrow 0$ atunci $f(\{x\})=0,\forall x \in \mathbb R$
Mai departe avem
$f(x)=f([x]+\{x\})=f([x])+f(\{x\})+f(\{x\})$ $\rightarrow$ $f(x)=f([x]),\forall x \in \mathbb R$
Dar $f([x])=f(1)[x],\forall x \in \mathbb R$ $\rightarrow$ $f(x)=f(1)[x]$
Deci $f(x)=[x],\forall x \in \mathbb R$

pb. 1 : $x \geq [x]$, deci $[x]\left\{x\right\}\geq 2[x]$, deci $x=0$

pb. 2: putin mai simplu, dar se arata cu inductie tinandu-l pe $y=1$ fix, se arata ca $f(x)=x$ pentru orice $x$ intreg $(1)$, iar punand $y=:-[x]$, obtinem $f(x)=-f(-[x])$, dar $-f(-[x])=[x]$, conform $(1)$, deci $f(x)=[x]$.

la problema 1 nu e bine ce ai gasit tu....uite de exemplu 1/2 nu verifica

mai postez aici o problema:
demonstrati ca in orice triunghi ascutitunghic are loc inegalitatea:

9S< pi (ab+bc+ca)

$\sum ab \geq 4\sqrt{3}S>\frac{9S}{\pi}$

[quote="Zamfirmihai"]
9S< pi (ab+bc+ca)[/quote]

da, inegalitatea este foarte slaba, ca idee: $ab+bc+ca \geq 3 \sqrt[3]{a^{2}b^{2}c^{2}}$, dar $a^{2}b^{2}c^{2}> S^{3}\Longleftrightarrow 16R^{2}> S \Longleftrightarrow \sin{A}\sin{B}\sin{C}< 8$, deci este foarrrrteeee slaba.

dap, stiu ca e slaba :D  astept inca o solutie pt problema 1. nu mi se pare chiar asa grea

fainutze problemele...da cam simplutze...la cls a 7-8..mai mult ..nu..:)

totusi cred ca inegalitatile alea de la problema 1 cad atunci cand $x<0$ (la ambele solutii), sau nu inteleg eu ce se intampla.
[hide="solutia mea"]
fie $[x]=a$, $\{x\}=b$. atunci $a\in\mathbb{Z}$, $b\in[0,1)$, si sunt doua variabile independente.
ecuatia este $2(a+b)=ab$.
$a=2$ nu este solutie, asadar pot scrie ca $b=\frac{2a}{a-2}$.
din conditia $0\leq b<1$ se obtine un sistem de doua inecuatii in $a$.
din $b\geq 0$ se obtine $\frac{2a}{a-2}\geq 0$ este echivalent cu $2a(a-2)\geq 0$, adica $a\notin(0,2)$.
din $b<1$ se obtine $\frac{2a}{a-2}<1$ este echivalent cu $\frac{2a-(a-2)}{a-2}<0$, adica $(a+2)(a-2)<0$, deci $a\in(-2,2)$.
din cele doua se obtine $a\in(-2,0]$, si cum $a$ este intreg, rezulta ca $a\in\{-1,0\}$.
daca $a=0$, se obtine $b=0$, si se verifica faptul ca $x=0$ este solutie.
daca $a=-1$, se obtine $b=2/3$, adica $x=-1/3$. pentru $x=-1/3$, egalitatea din enunt se transforma in $2\cdot\left(-\frac{1}{3}\right)=(-1)\cdot\frac{2}{3}$, o relatie adevarata.
asadar solutiile sunt $x=0$ si $x=-1/3$.
[/hide]

[quote="alex_tataran"]mei..daca n-am gresit atuncea..pr 1 iese cam asha...
Presupunem ca {x}<>0 si [x]<>0.Atun cea avem inegalitatea: x>=[x].deoarece x<>0 putem inmulti cu {x} si obtinem....x*{x}>={x}*[x]....dar {x}*[x]=2x =>...reducem x'u..si obtinem...{x}>=2...ceea ce este fals...presupunerea facuta este gresita...=>x=0 solutie.:P[/quote]

usoare, usoare, dar macar stii unde ai gresit la solutia ta? tentativa de fapt

stiu unde...aia am facuto asha...la prima vedere..:p...da apoi miam dat seama...:))


Tomorrow night, at 7:30 PM ET, 4:30 PM PT, Matt McGann, Assistant Director of Admissions at MIT, and Prof. Kiran Kedlaya of the math department at MIT (and a past IMO medalist and 3-time Putnam winner) will be hosting a Math Jam here at AoPS to discuss life and admissions at MIT.

Probably what life at MIT is like for the standard student as well as what it takes to be accepted. This is probaby geared towards high schoolers rather than middle schoolers  :wink: .

As for me, I won't attend because it'll make me more depressed that I'm not going to be accepted at MIT.  :noo: .

Spam: I really want to get into MIT!!!
Will the jam go in depth about the admissions process?

What's a "Math Jam"?
What's ET and PT

[hide]I know I sound extremely stupid[/hide]

http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php

ET = Eastern Time
PT = Pacific Time

Hmm interviews.. I should start talking more so my vocal cord doesn't decompose.

[quote="AstroPhys"]Recommendation : [u][i][b]DO THE[/b][/i][/u] [size=75]optional[/size] [u][i][b]INTERVIEW![/b][/i][/u][/quote]

are you in college?

[quote="7h3.D3m0n.117"][quote="mathemonster"]u might what are your math scores like?[/quote]

Haha I'm sure they look at a lot more things other than math scores. Like GPA (speaking of which mine isn't so good right now) or class rank (speaking of which mine is barely decent).[/quote]

Leadership helps too when applying for college! Like if you can lead your high school's math team  :roll:

if your math scores are beastly and you have like a ok GPA then you might get in....

Colleges want well-rounded out people, not people who made IMO but can't do anything else...

OK, they might want someone like htat.

But you get the point.

Mr. Rusczyk has 5991 posts...almost 6K!

sorry to interupt a running conversation but i have a class at 730 and i was wondering if u can post the math jam transcript on this thread. :|

you can always look at the transcript through the usual link after the math jam is finished (also when your class is finished probably)

[quote="ushavinodh"]is the audio stream on ? i am unable to listen to the discussion.[/quote]There is no audio.

Is there a way to see the entire discussion now? could anyone give it to me? I couldn't attend the entire discussion. (It was a odd time at night in here)

Where will I find it when it's available?
I've never been on an AOPS class or anything, don't know much......

[quote="Akashnil"]Where will I find it when it's available?[/quote]

http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=243

To find it, hover your mouse over the "Math Jams" link. The Transcripts should appear below the Schedule.


Hey, i think I saw this on the site before, but does anyone know where to find past AMC contests and descriptions for the answers? Thanks

[quote="t0rajir0u"]The "Contests" link at the top might not be a bad place to start.  (Although, truth be told, it's not very complete.)[/quote]

ehh

I wish there were pictures.

There's an announcement thread in this forum with the title "New Compact Disk with all prior contests 2001-present".  That sounds like what you're looking for.

there are tests AMCs from 2001+ floating around everywhere on the Web. Buying isn't exactly necessary if your teacher has it, or you can find it online.

Or if you're looking for AIME, you can go to kalva.


linear algebra
matrix
algebra
polynomial
advanced fields
advanced fields solved

Here's another problem form that contest, Gheorghe Titeica, the one I was speaking abt in the previous topic:

How many solutions does

X^(p-1)+X^(p-2)+...+X^2+X+I2=O2 have in M2(Zp) where p is a prime?

I don't se the equation, is there any =0 somewhere grobber ?

Sorry about that. Looks like I forgot about the most important thing: the actual equation  :D I fixed it.

[quote="grobber"]... contest, Gheorghe Titeica
How many solutions does
X^(p-1)+X^(p-2)+...+X^2+X+I2=O2 have in M2(Zp) where p is a prime?[/quote]
p prime Z_p is a field
If X is a solution of the equation, tX the transpose is also a solution.
--------------------------------------------------------------------------
If p=2, X+I=0=> X=-I=I in M2(Z_2) one solution.
-------------------------------------------------------------------------
If p=3, we are in M_2(Z_3),
X+X+I=0
(X+2I)=0
X+2I is nilpotent order 2, matrix of form B(a), tB(c)
B(a)=
[0  a]
[0  0]
X=
[-2 a]
[0  -2]  with a \in Z_3 so three choices for a
Consider the transpose we get 6 solutions
----------------------------------------------------------------------
Remark:
Let X a solution of the equation
X.(-X^(p-2)-X^(p-3)-...-I)=I so X is invertible  and
 \sum ={X\in M_2(Zp); X^(p-1)+X^(p-2)+...+X+X+I=0} is contained
in GL_2(Z_p) matrix of size two with coefficients in Z_p
Card  \sum  \leq Card GL_2(Z_p) =(p-p)(p-1)

oubi|ibuo

[quote="grobber"]
[21]  [22]  [01]  [02]
[20]  [10]  [22]  [12][/quote]

For p=3
These three matrix are the case when all the coefficients
of X+X+I are divisible by 3, i forgot to consider this situation.
so 9 = 3 =p solutions.

You're right: the answer is p^2. To prove this I used the fact that in Zp x^(p-1)+..+x+1=(x-1)^(p-1).

[quote="grobber"] the answer is p^2[/quote]
But for p=2 i found 1 soluce,
p=2=4=/=1  :?

Yeah, 2 is an exception. That's not a big problem, is it?

Here a solution of Mr D.

Consider the case p \geq 3 prime
In Z_p[X]
X^(p-1)+...+X+X+1=(X-1)^(p-1)  since X^p -1=(X-1)^p
the minimal polynomial is X-1 or (X-1), and the characteristic polynomial
is (X-1)=X-2X+1
X is a solution of the equation if and only if trace(X)=2, det(X)=1
X is of the form
[(1 +a)  b]
[c    (1-a)]

det(X)=1 gives a+bc=0 in the field Z_p
the number of solution of the last equation is
1+(p-1)+(p-1)+(p-1)=p

Another solution of Mr C.

The case p \geq 3 prime
X-I is nilpotent in dimension 2, look for nilpotent matrix is equivalent
that the trace=det =0.So X is of the form
[a c]
[d -a]

determinant gives a+bc = 0, the number of this equation is p

This is really nice problem for grade 12th.
oubi

grobber, can you post your soluce if it is different  than these two.

why the number of solution af a²+bc=0 in the field Z/PZ is 1+(p-1)+(p-1)+(p-1)²    :?


function
algorithm
induction
algebra
polynomial
combinatorics proposed
combinatorics

This is not my creation, it is a combinatorial restatement of a problem (rather well-known fact) from a quite different field:

Let $\mathbb{N}_{0}$ denote the set of nonnegative integers.

For a positive integer $n$, we define an [i]$n$-Catalan landscape[/i] as a function $f: \left\{0;\;1;\;2;\;...;\;2n\right\}\to\mathbb{N}_{0}$ such that $f\left(0\right)=f\left(2n\right)=0$ and such that $\left|f\left(i\right)-f\left(i+1\right)\right|=1$ for any integer $i$ with $0\leq i<2n$. The [i]index[/i] of this $n$-Catalan landscape will mean the product of the terms $f\left(i\right)+1$ over all integers $i$ (with $0\leq i<2n$) satisfying $f\left(i\right)-f\left(i+1\right)=1$.

Prove that, for a given integer $n$, the sum of the indices of all $n$-Catalan landscapes equals $1\cdot 3\cdot 5\cdot ...\cdot\left(2n-1\right)=\frac{\left(2n\right)!}{n!\cdot 2^{n}}$.

Try to give a combinatorial proof. I have one, and I find it pretty nontrivial.

  Darij

I think this works
[hide]
To each perfect matching on $\{0, 1, \dots 2n-1\}$ we associate a catalan landscape as follows:

If $(i,j)$ is a pair in the matching with $i<j$, $f(i+1)=f(i)+1$, $f(j+1)=f(j)-1$.

It's easy to check that $f(2n)=0$, and the statement that $f(k) \geq 0$ follows from how each pair's increase comes before it's decrease.  Now the question becomes: How many matchings map to each landscape?

Given the landscape, we can divide the numbers into two sets of size $n$: The "last numbers" (those appearing last in their pairs, recognizable since $f$ decreases from $i$ to $i+1$), and the "first numbers" (those appearing first in their pair, i.e. those where $f$ increases).  We examine the first numbers in decreasing order.

Suppose $j$ is a first number.  It must be matched with a last number which is larger than $j$, and which hasn't been matched yet.  Of the numbers larger than $j$, there must be $f(j)+1$ more last numbers than first numbers, since $f$ has to decrease from $f(j)+1$ to 0 over the remaining numbers.  This means there are $f(j)+1$ unmatched numbers which we can choose to match $j$ with.

This means for any landscape that the number of matchings which map to the landscape is the index of the landscape.  Therefore the sum of the indices over all landscapes is the number of matchings, which is $\frac{(2n)!}{n! 2^{n}}$[/hide]

Nice! I had something very similar in mind, which, however, would take me much longer to write up. Roughly speaking, I constructed an n-Catalan landscape by a recursive algorithm, starting with the 0-Catalan landscape $f\left(0\right)=0$, and in each step, extending a k-Catalan landscape to a $\left(k+1\right)$-Catalan landscape by inserting a "go up" edge at the position 0 (this means at the very beginning of the landscape) and a "go down" edge at some arbitrary other position, ranging from 0 (this means directly after the new "go up" edge) to 2k (this means at the very end of the landscape). There are 2k + 1 different choices at this step. I showed that the number of different ways we can obtain a given n-Catalan landscape by this algorithm is equal to the index of the landscape. This is proven by induction, considering every possible last step of the algorithm.

So far my proof idea. Here is the original problem which I solved using the above:

Let $\left(P_{n}\right)_{n\geq 0}$ be a sequence of polynomials defined by $P_{0}\left(x\right)=1$ and $P_{n+1}\left(x\right)=-2xP_{n}\left(x\right)+P_{n}^{\prime}\left(x\right)$. (This is actually a modification of the Hermite polynomials.) Prove that $P_{n}\left(0\right)=0$ for odd n, and that $P_{n}\left(0\right)=\left(-1\right)^{n/2}\frac{n!}{\left(n/2\right)!}$ for even n.

This is problem 16 in Chapter 6 of Titu Andreescu, Dorin Andrica, [i]360 Problems for Mathematical Contests[/i], Zalau: Gil, 2003.

  Darij


Who can prove it with mathematical induction:
(1+x)(1+x^2)(1+x^4)...(1+x^2n)=1-x^2^(n+1)/(1-x)

x^n means the nth power of x. n is a natural number and x doesn't equal 1.
thx a lot :D

I think the products in the LHS should approach $(1+x^{2^n})$ and the RHS should have a + on the numerator.
And if that's the case, multiplying both sides by $(x-1)$ gives us
$1+x^{2^{(n+1)}}$ on the LHS, which is the same as the RHS.

edit: ah crap didn't see the induction requirement.


What is the sum of the integers from -30 to 50 inclusive any easy ways to do this or any standard for these types of problems????? plz help

You start by adding the first and last positive numbers together, 50 + 1 = 51, then the second and second to last, 2 + 49 = 51 and you start to notice that you are going to have 50/2 or 25 such pairs of 51.

So the sum of the numbers from 1 to 50 is 25 x 51 = 1275

By the way, this method can be summarized in the formula $ \frac{n(n\plus{}1)}{2}$ where n is the highest number you are adding, starting with one.

[color=red]Can you come up with a way to add numbers that do not start with one?
[/color]

To complete your problem, do the same as above with the numbers from 1 to 30 and subtract from 1275 to get your answer.

[hide="Answer"]810[/hide]


10 marbles (0 to 9). 3 bowls (order does not count). 3 marbles in a bowl. How many combinations? A bowl combination made of 3 successive marbles is not valid.

Valid combination example: 035 247 681 (same as 274 035 168)
Invalid combination example: 035 867 249 (7, 6 and 8 are successive).

Be gentle, I have finished the school a long time ago.

Thank you.

We can't have 3 successively-numbered marbles in one bowl, or 2 successively-numbered marbles? Would 124 357 680 be valid?

[quote="solafidefarms"]We can't have 3 successively-numbered marbles in one bowl, or 2 successively-numbered marbles?[/quote] 3

[quote="solafidefarms"]Would 124 357 680 be valid?[/quote] Yes, it is valid.


trigonometry
geometry
circumcircle
trig identities
Law of Sines
geometry solved

Let a quagrangle $ABCD$ with $\angle B = \angle D$. We draw $BM\perp AC, DP \perp AC , BN \perp AD , DQ \perp BC$. Prove that  $MN = PQ$

[u] Babis[/u]

[color=blue][b]Problem.[/b] Let ABCD be a quadrilateral with < ABC = < ADC. Let M, P, N, Q be the feet of the perpendiculars from the points B, D, B, D to the lines AC, AC, AD and BC, respectively. Prove that MN = PQ.[/color]

Here is the first [i]solution[/i] that came into my mind (probably not the shortest possible one):

The [b]extended law of sines[/b] states: If ABC is a triangle and d is the diameter of its circumcircle, then $BC=d\cdot\sin\measuredangle CAB$.

Now, in our problem, we have < AMB = 90° and < ANB = 90°, and thus the points M and N lie on the circle with diameter AB. Hence, the circumcircle of triangle AMN has diameter AB. Hence, by the extended law of sines in triangle AMN, we have $MN=AB\cdot\sin\measuredangle MAN$. Since < MAN = 180° - < DAC or < MAN = < DAC, we have sin < MAN = sin < DAC, and thus $MN=AB\cdot\sin\measuredangle DAC$. Similarly, $PQ=CD\cdot\sin\measuredangle ACB$.

Thus, in order to prove MN = PQ, it is enough to show that $AB\cdot\sin\measuredangle DAC=CD\cdot\sin\measuredangle ACB$. This is equivalent to $\frac{AB}{\sin\measuredangle ACB}=\frac{CD}{\sin\measuredangle DAC}$. But by the sine law in triangle ABC, we have $\frac{AB}{\sin\measuredangle ACB}=\frac{AC}{\sin\measuredangle ABC}$, and by the sine law in triangle ADC, we have $\frac{AC}{\sin\measuredangle ADC}=\frac{CD}{\sin\measuredangle DAC}$. Hence,

$\frac{AB}{\sin\measuredangle ACB}=\frac{AC}{\sin\measuredangle ABC}=\frac{AC}{\sin\measuredangle ADC}=\frac{CD}{\sin\measuredangle DAC}$,

and the proof is complete.

  Darij

In my opinion, the [i]level[/i] of this proposed problem is ** from ******.
I think the my solution is similarly with the [u]Darij's solution[/u].

The quadrilaterals $BNDQ,\ ABMN,\ CDPQ$ are cyclically.

Therefore, $PQ=CD\sin \widehat {ACB},\ MN=AB\sin \widehat {CAD}$. But

$\widehat {ABC}\equiv \widehat {ADC},\ \frac{AC}{\sin \widehat {ABC}}=\frac{AB}{\sin \widehat {ACB}},\ \frac{AC}{\sin \widehat {ADC}}=\frac{CD}{\sin \widehat {CAD}}\Longrightarrow$

$\frac{AB}{\sin \widehat {ACB}}=\frac{CD}{\sin \widehat {CAD}}\Longrightarrow AB\sin \widehat {CAD}=CD\sin \widehat {ACB}\Longrightarrow MN=PQ$.

I wanted to use trig. too, but I left it as the last attemt. Now I see , that trig/try is necessary. Any way, thanks to both of you.

[u]Babis[/u]


ok, so i was skimming through old topics (REALLY old topics) out of boredom, and i came upon these questions that MC came up with, that no one ever (to my knowledge) answered. So here goes. Discuss at your pleasure.

Is 2 an optimal number of eyes?
Why have no new body types evolved on Earth for hundreds of millions of years?
If life developed on another planet, would it have dna?
Is it worth the cost of sending a human to Mars anytime soon?
How long will it be before computers have the power of a human brain?
How do I learn how to solve every IMO problem in an hour and a half (er, I still can't)?

[quote="bleumoose"]ok, so i was skimming through old topics (REALLY old topics) out of boredom, and i came upon these questions that MC came up with, that no one ever (to my knowledge) answered. So here goes. Discuss at your pleasure.
[/quote]
Ok.
[quote]
Is 2 an optimal number of eyes? [/quote]Yes. Any more and there will be computation difficulties.[quote]
Why have no new body types evolved on Earth for hundreds of millions of years? [/quote]Because that would be a really big mutation, and not cost effective.[quote]
If life developed on another planet, would it have dna? [/quote]No.[quote]
Is it worth the cost of sending a human to Mars anytime soon? [/quote]No.[quote]
How long will it be before computers have the power of a human brain? [/quote]10-100 years.[quote]
How do I learn how to solve every IMO problem in an hour and a half (er, I still can't)?[/quote]
Sacrifice a goat.

It would probably have something like DNA, and seeing as DNA happened here, it could happen elsewhere.

why are there no computation difficulties with two eyes?

define "big mutation" and "cost effective" in terms of "new body types." im not sure i follow you on that one.

what if you knew that if life developed on another planet, it would start with amino acids? does that change your answer any?

CAN we send a human to mars any time soon?

so will the world become like the film A.I.?

what if i don't have a goat? will a rabbit work? how about a younger sibling?

For the brain power, it depends on what you mean. In terms of calculations performed per second, computers have long beaten human brains. Perhaps you mean when will AI be at the level of human intelligence?

[quote="bleumoose"]why are there no computation difficulties with two eyes?
[/quote]
There still are, but it is easier to interpolate 2 data sets than 3.
[quote="bleumoose"]
define "big mutation" and "cost effective" in terms of "new body types." im not sure i follow you on that one.
[/quote]
Its not worth it. Any small steps towards one are a waste of energy, what we have works fine. In any case, going from 2 to 3 legs is rediculous
[quote="bleumoose"]
what if you knew that if life developed on another planet, it would start with amino acids? does that change your answer any?
[/quote]
Yes
[quote="bleumoose"]
CAN we send a human to mars any time soon?
[/quote]
Yes.
[quote="bleumoose"]
so will the world become like the film A.I.?
[/quote]
Probably.
[quote="bleumoose"]
what if i don't have a goat? will a rabbit work? how about a younger sibling?[/quote]
Anything with intelligence, that means no siblings.

Then how about flies? They have way more than 3 eyes.

[quote="PenguinIntegral"][quote="G-UNIT"][quote]How long will it be before computers have the power of a human brain?
[/quote]

  I've forgotten the name of the fellow, but he proposed a theory that declared no computer will every reach the power, or intelligence for that matter, of a human brain.  He talked about how the human brain was such a sophisticated tool that it could never be outmatched, yada, yada, yada....  The computer might be faster than us, but it will never outmatch our wits!!!!![/quote]
Find out how we make our brain, then do that electronicaly emulated. Done.[/quote]

  Yes, I'd like to see you figure out how the human mind works, let alone how it is created.  Then somehow muster this thechnology, if you'd call it, on some electronic databse.

  ( Visualizes penguinintegral plugging an disembodied brain into his laptop)

[quote="white_horse_king88"]I believe that in a study, it was found that an optimum number of limbs was 4 - do you agree?[/quote]

Why not two for holding(objects at two ends) and three for standing(better balance)?

[quote="peter"][quote="white_horse_king88"]I believe that in a study, it was found that an optimum number of limbs was 4 - do you agree?[/quote]
 three for standing(better balance)?[/quote]

How would you move?

Ever wonder why (most) robots have 4 or more legs?  Two legs is about the least stable configuration of all (well, I suppose 1 leg would be even worse...)


A taxicab offers the rates below. If three people share a
12.5-mile ride, how many dollars is the total cost of the ride?

First person: $ \$3$ for first mile; $ \$0.20$ for each additional $ \frac{1}{10}$ mile.
Additional people: $ \$2$ for each additional person

So the first person pays 3 dollars for the first mile, and then 20 cents for each 0.1 mile. There are $ 11.5 \div .1 \equal{} 115$ tenths of a mile in 11.5 miles, so the total cost of one person is $ 3 \plus{} (0.2 \times 115) \equal{} 26$ dollars. There are two other people, so we add two more dollars for each of them, giving us a total of $ \boxed{30}$.


logarithms
LaTeX
number theory unsolved
number theory

is there rational no. x s.t. e^(transcendetal no.)=x?

[color=red]Edit by Megus: moved from Algebra[/color]

[url=http://en.wikipedia.org/wiki/Trascendental_number]Wikipedia[/url] says that $\ln(a)$ is transcendental for every positive rational $a\neq 1$. I don't know how it was proved, though.
[url=http://mathworld.wolfram.com/TranscendentalNumber.html]Mathworld[/url] only says that $\ln (2)$ is transcendental, proved by Hardy and Wright (1979, p. 162).

Could sb please write this in latex form because i cant understand the problem? :huh:

no,but you can understand is there rational x such that ln(x)=transcendetal no.

[quote]Could sb please write this in latex form because i cant understand the problem?[/quote]

I think the problem is:
Does there exist a [url=http://www.artofproblemsolving.com/Wiki/index.php/Transcendental_number][b]transcendental number[/b][/url] $y$ and a rational number $x$ such that

$e^{y}=x$

The theorem of Lindemann-Weierstrass gives that $e^{a}$ is transcedental for every algebraic number $a$. Thus when $e^{b}$ is algebraic, then $b$ is transcedental.

[quote]Mathworld only says that \ln (2) is transcendental, proved by Hardy and Wright (1979, p. 162).[/quote]
Well, surely not first time proved by them ;)

zetax will you please post ideas of the proof?


Two kids are playing cards. There are 32  cards in the deck,

and every card has its pair. At the beginning every player gets 5 cards,

and then they take one card at a time from the rest of the deck. When

the deck is gone they take a card from each other, alternating. If you

have a pair (or two pairs), you must put it (them) on the table in front of

you. Prove that at the end of the game each player has the same number

of pairs.

[hide="If we change..."]
... the rules slightly, by letting each player start off with 16 cards rather than taking some from a pile, this does not change anything. (The point at which the player puts a card on the table seems to be unimportant, so, after taking the cards from the pile the players each have 16 cards, minus those in the pairs they have put down).

It is clear that after this point, each player has put down the same number of pairs- out of 16 cards one player has, for every pair he has there is another pair that he has neither card of, which means the other player has a pair. Now, every time the players swap cards, either they swap each card of one pair, in which case no-one gets a new pair, or the swap cards from different pairs, in which case both players gain one pair. Either way, the both still have the same umber of pairs, continuing to the end of the game.[/hide]


The locus of all points P such that |AP| = [i]k[/i] |BP| for points A, B and the positive scalar [i]k[/i].

When [i]k[/i] = 1 the locus is a straight line.

Show that, in general, this locus is a circle (when [i]k[/i] does not = 1) using a circular inversion.

Show that the image under inversion of this locus is again a circular locus which is now defined with respect to images A' & B' of A & B ... i.e. Show the image locus is defined as the set of all points P' such that |A'P'| = [i]k[/i]'|B'P'|.

Let us perform an inversion with centre $ A$ and radius $ r$. Then, for a pt $ P$ which belongs to the locus of the required points, it will be inverted to pt $ P'$, while pt $ B$ will be inverted to pt $ B'$. Then by the property of inversion, $ \triangle{APB}$ is similar to $ \triangle AB'P'$ so $ \frac{AB'}{AP'}\equal{}\frac{B'P'}{BP}\Rightarrow \frac{AB'}{B'P'}\equal{}\frac{AP}{PB}\equal{}k$.
Therefore, regardless of the location of pt $ P$ as long as it belongs to the locus, then $ \frac{AB'}{B'P'}\equal{}k$. Because $ AB'$ is always constant, we must have $ B'P'$ also constant hence all points $ P'$ are equidistant from point $ B'$ and therefore lies on a circle with centre $ B'$. Then, the points $ P$ belonging to the locus also lie on a circle because they are the inverted images of points $ P'$ with respect to the circle with centre $ A$, radius $ r$.
The result follows.