Prove that for each positive integer n>=6, any square can be sub-divided in to n non-overlapping squares.

If when n = a or b, the statement is true, then it is also true when n = a+b-1
Therefore you just need to prove the cases when n = 4, 6, 8

Thanks Mr Modeler. I got it!!!

PS: I'm not sure if this kind of problem is really combinatorics.
If not, I'd like to know which topic this problem should be in.

Thanks very much


number theory
relatively prime
number theory unsolved

let p is prime and m,n is positive integer so $m^{p(n-1)}=0%Error. "modn" is a bad command.
$.Prove that $gcd(m^{n-1},n)>1$

What system is that congruence in?  I would at first think $p$, but then there are some difficulties...is it $n$ then?

it is a problem which has solution but i don't understand it :(

Well if the congruence is mod p, then $p|m$, which gives no information

If it's mod n, then we have $(m, n) > 1$, because if $(m, n)=1$, then $m^{p(n-1)}$ would be relatively prime to $n$ as well, a contradiction.  That implies what we wanted.  Is that what you wanted to prove?

The correct statement is ${m^{p(n-1)}=1(mod n)}$ then $gcd(n,m^{ n-1}-1)>1$. Now, a proof?  :)

if there is a prime q s.t. q|n and $m^{n-1}\equiv 1\ \ (\mod q\ )$ we have finished.

we pose $\alpha=max\ x: \ p^{x}|n-1$

if $m^{n-1}\neq 1\ \ (q)$ $\forall\ q$ prime s.t. q|n then $p|ord_{n}\ m$ because $m^{p(n-1)}\equiv 1\ (q)$,
but if $p^{k}|ord_{q}\ m$ and $k\leq\alpha$ then $m^{n-1}\equiv 1\ (q)$ so
$p^{\alpha+1}|ord_{q}\ m|q-1$ $\forall\ q$.

then $q\equiv 1\ (p^{\alpha+1})$ for alla q prime s.t. q|n...
then $n\equiv 1\ (p^{\alpha+1})\ \Rightarrow\ p^{\alpha+1}|n-1$

this is an absourdity then the proof is complete


Σε μια διαλεξη ο καθηγητης για να λυσει την διφορικη $ xy'\minus{}y\equal{}x^3$ την εφερε πρωτα στην γενικη μορφη $ y'\minus{}\frac{1}{x}y\equal{}x^2$ και μετα εκανε τα γνωστα για να την λυσει... Αυτο που δεν καταλαβαινω ειναι γιατι επιτρεπεται να διαιρουμε με το  x χωρις να ξερουμε το π.ο της y?

[quote="mailo"]Σε μια διαλεξη ο καθηγητης για να λυσει την διφορικη $ xy' \minus{} y \equal{} x^3$ την εφερε πρωτα στην γενικη μορφη $ y' \minus{} \frac {1}{x}y \equal{} x^2$ και μετα εκανε τα γνωστα για να την λυσει... Αυτο που δεν καταλαβαινω ειναι γιατι επιτρεπεται να διαιρουμε με το  x χωρις να ξερουμε το π.ο της y?[/quote]

Δεν εχω φτασει ακομα στο 3ο εξαμηνο οπου εχουμε τις διαφορικες, αλλα αυτη ειναι αρκετα απλη και λυνεται με γνωσεις λυκειου. Για x διαφορο του 0 μπορουμε αφοβα να διαιρεσουμε με το $ x^2$, και αμεσα προκειπτει $ y \equal{} \frac {x^3}{2} \plus{} cx$, για $ x \equal{} 0$ στη διαφορικη εχουμε το προφανες $ y(0) \equal{} 0$ (διοτι η $ y$ ειναι συνεχης στο 0).

Ψιλοασχετο: επηδη δουλευουμε σε δυο διαφορετικα διαστηματα, αν οι συνθηκες ηταν τετοιες ωστε να μη θελουμε υποχρεωτικα η $ y \equal{} y(x)$ να παραγωγιζεται στο $ 0$, τοτε νομιζω οτι θα μπορουσε η συναρτηση να ηταν δικλαδη με διαφορετικο $ c$ εκατεροθεν του 0

[quote="mailo"]Σε μια διαλεξη ο καθηγητης για να λυσει την διφορικη $ xy' \minus{} y \equal{} x^3$ την εφερε πρωτα στην γενικη μορφη $ y' \minus{} \frac {1}{x}y \equal{} x^2$ και μετα εκανε τα γνωστα για να την λυσει... Αυτο που δεν καταλαβαινω ειναι γιατι επιτρεπεται να διαιρουμε με το  x χωρις να ξερουμε το π.ο της y?[/quote]

Κανονικά πρέπει να πει από την αρχή το πεδίο ορισμού. Συνηθίζεται όμως, όπως και με τις συναρτήσεις στο σχολείο, να θεωρούμε ότι εργαζόμαστε στο ευρύτερο δυνατό σύνολο όπου έχει νόημα το y.

Αν θέλουμε να ακριβολογήσουμε θα λέγαμε:

Εξετάζουμε τη εξίσωση για $ x\ne 0$, δηλαδή την εξετάζουμε χωριστά στα
$ ( \minus{} \infty , 0)$ και $ (0, \plus{} \infty )$.  Σε κάθε ένα από αυτά τα δύο διαστήματα έχουμε (επιτρέψτε μου τις λεπτομέρειες):

$ y' \minus{} \frac {1}{x}y \equal{} x^2$ με ολοκληρωτικό παράγοντα $ e^{ \minus{} \int \frac {1}{x}dx} \equal{} \frac {1}{x}$. [color=white].[/color]  Άρα $ \frac {1}{x}y' \minus{} \frac {1}{x^2}y \equal{} x$ ,
ή  $ (\frac {1}{x}y)$ ' = x και τελικά $ y \equal{} \frac {x^3}{2} \plus{} cx$.

Εννοείται ότι  η σταθερά εξαρτάται από το διάστημα. Δηλαδή η λύση είναι

y  = $ \frac {x^3}{2}$ + $ c_1x$ στο $ ( \minus{} \infty , 0)$   και

y  = $ \frac {x^3}{2}$ + $ c_2x$ στο  $ (0, \plus{} \infty )$.

Βλέπουμε τώρα ότι η λύση αυτή επεκτείνεται και στο x = 0. Επίσης,
αν υποθέσουμε ότι η f είναι συνεχής (που αν το πρόβλημα προήλθε από φυσική εφαρμογή, τότε συνήθως αυτό έπεται από την κατάσταση που μελετάμε), είναι $ c_1 \equal{} c_2$. Δηλαδή  y  = $ \frac {x^3}{2}$ + cx στο [b]R[/b]

Τ.Π.

Ευχαριστω πολυ για τις απαντησεις, μου λυσατε την απορια !  :)


The number $d_{1}d_{2}\ldots d_{9}$ has nine (not necessarily distinct) decimal digits. The number $e_{1}e_{2}\ldots e_{9}$ is such that each of the nine 9-digit numbers formed by replacing just one of the digits $d_{i}$ of $d_{1}d_{2}\ldots d_{9}$ by the corresponding digit $e_{i}$ $(1\leq i \leq 9)$ is divisible by $7$. The number $f_{1}f_{2}\ldots f_{9}$ is related to $e_{1}e_{2}\ldots e_{9}$ in the same way: that is, each of the nine numbers formed by replacing one of the $e_{i}$ by the corresponding $f_{i}$ is divisible by $7$. Show that for each $i$, $d_{i}-f_{i}$ is divisible by $7$.

I didn't understand the question clearly, can you explain it one more time? thansk  :)

[quote="paladin8"]I think you mean $f_{1}f_{2}\ldots f_{9}$.[/quote]

what do you mean?

Its tough to explain it clearer. Just look at it again and you should get it. Or maybe specify which part is difficult to understand.

I think you had $f_{1}f_{29}$ before by mistake, which is what he's talking about.

just wanted to revive this thread because i think that this is a good problem.

It is a good problem, but if nobody could do it then, why would we be able to do it now?  Do you have a solution?

yes, i have a solution. i found it during government (i probably should have been paying attention to the teacher)


How can I check n≡ 3 or 4 mod7  has no integer solution from n=x3+y3 ? :?

A cube is always $0$ or $ \pm 1 \bmod 7$.  Do you see why that is true?  Do you see how that fact implies that your equation has no solutions?

This topic should probably be moved from the Olympiad section to the Intermediate forum.

Also it is good to remember that a cube is always 0 or $\pm 1$ modulo 9.

[quote="mikewd"]I don't see why a cube is either -1,0 or 1 mod 7 or 9.[/quote]

Just  try it: For example  (all calculations mod 7) we have cubes of  $1,2,3,4, ...$

as  $0,1,8,27,64,125,216$  which are same as  ( if you use mod 7)  $0,1,1,-1,1,-1,-1$
Similary you can try mod 9 too  and the cubes are $0,1,-1,0,1,0,1,-1 $


Let $ A\in{M_n(Z)}$. Prove that for every$ k\in{Z}, k\neq{\minus{}1,0,1}$,  the matrix $ B\equal{}I\plus{}kA$ has $ detB\neq{0}$

The determinant is $ 1$ mod $ k$, so it's not zero.


If $ i^{i^{i^{i^...}}} $ approaches a limit, find it.

How about $ (((i^i)^i)^i)^i... $ ?

The second one oscillates between $i^i$, $i^{-1}$, $i^{-i}$, and $i^1$.

The first one... well, i^i can take on a lot of different values, but the question is do they lead to the same result?  I don't know.. let someone else figure it out.

Since I will be using $e^{\pi i}=-1$ quite often, I will refer to it as E. Simplify $i^i$ to $e^{i\ln i}$. Take the square root of both sides of E to determine $\ln{i}=\frac{\pi i}{2}$. So $i^i=e^{\frac{-\pi}{2}}$, a real number! Give a positive real number r, $r^i=e^{i\ln{r}}=(-1)^{\frac{\ln{r}}{\pi}}$. Since this function oscillates between real and imaginary, it has no limit. I would be interested to know what numbers it oscillates between though...

bubala: what do you mean $i^i$ can take on many values??? It has a unique value, namely $e^{\frac{-\pi}{2}}$.

You're right--$i^i$ does indeed equal $e^{\frac{-\pi}{2}}$.  But it also equals $e^{\frac{3\pi}{2}}$.  And it also equals $e^{\frac{-5\pi}{2}}$, etc.

 Because using your method, $ln i = \frac{-3\pi}{2}$ and more.

The way to prove $e^{i\pi}+1=0$ is using DeMoive's THm: $e^{ix}=cosx+isinx$.

That's why $e^{\frac{i\pi}{2}}=i$.  And so does $e^{\frac{-3i\pi}{2}}$ and more.

For non-integer values of $a$, the function $f(z) = z^a$ takes on multiple values.  We are very used to this when $a$ is, say, a rational number -- every number (other than 0) has 2 square roots (one-half powers).  When $a$ isn't rational, though, the function actually ends up taking on infinitely many values, making a question like this rather tricky.

Also, note that alternating between imaginary and real or complex and real does not preclude a limit -- for example, the sequence 1, i/2, 1/4, i/8, 1/16, ... has the limit 0, although it doesn't approach along a straight line.

[quote="seamusoboyle"]I dont think you can use logs when dealing with imaginrys...[/quote]

You can, but it also becomes an infinite-valued function.  If we define $\log(w) = z$ if $e^z = w$, then it is clear that $\log(w) = z + 2\pi i n$ for $n \in \mathbb{Z}$.  Also, you can show that $\log(w) = \log |w| + i\arg w$, where the $\log$ on the right side is the "normal" real logarithm of a positive real number while $\arg$ is the argument (angle) function, which we know to be infinite-valued (since we can always just add $2\pi$ to it).


Four whole numbers, when added three at a time, give the sums 180, 197, 208 and 222.  What is the largest of the four numbers?
\[ \text{(A)} 77 \ \ \ \text{(B)} 83\ \ \ \text{(C)} 89\ \ \ \text{(D)} 95 \]
(E) cannot be determined from the given information

Nice problem  :P
Here's a small hint:

[hide]Write out your equations (notice how order doesn't matter, a, b, c or d could be the largest number) and see what you can do with these equations.  Should you try messy substitution?  Or should you try to get another equation with all 4 variables in it using the 4 you already have...?

[/hide]

[hide]The largest number is 89. First we list out the four numbers, A, B, C, and D. Assume that:
A+B+C =180
A+B+D = 197
A+C+D = 208
C+D+B =222
then A+B+D - (A+B+C)= 17
                              D-C =17
        A+C+D - (A+B+C) = 28
                              D-B = 28
        C+D+B - (A+B+C+) =24
                               D- A =42

Through a process of elimination (replacing D with each of the values) we find that 89 is suitable; when D is 89, C is 72 (D-17), B is 61 (D-28) and A is 47 (D-42). When adding any three of the numbers, its sum is one of the ones listed. If anyone knows of a shorter way, please do tell me. [/hide]


The graph of $ y\equal{}\log x$

$\textbf{(A)}\text{Cuts the }y\text{-axis} \qquad\\
\textbf{(B)}\ \text{Cuts all lines perpendicular to the }x\text{-axis} \qquad\\
\textbf{(C)}\ \text{Cuts the }x\text{-axis} \qquad\\
\textbf{(D)}\ \text{Cuts neither axis} \qquad\\
\textbf{(E)}\ \text{Cuts all circles whose center is at the origin}$

[hide]So, because of the wording, the answer has to be either A, C, or D.  $ x\ge 0$ for logarithms, so A is incorrect.  Taking the point $ (1,0)$, we see that C is correct.[/hide]


Homer selected an even positive integer less than 11, and Jethro selected an odd positive integer less than 20. What is the probability that the number selected by Jethro is greater than the number selected by Homer? Express your answer as a common fraction.

Homer's possibilities: $ \frac{5}{11}$

Jethro's possibilities: $ \frac{1}{2}$

Multiply to get $ \frac{5}{22}$ ??

There are $ 5$ possibilities for Homer. ($ 2,4,6,8,10$)
There are $ 10$ possibilities for Jethro. ($ 1,3,5,7,9,11,13,15,17,19$)
You don't multiply, but I'm not sure what to do next, but the answer is $ 7/10$.

I guess you have to use casework (ew). I don't really see any other way.


function
algebra
polynomial
complex analysis

A continuous function on [0,1] can be uniformly approximated by polynomials
(Weierstrass). Is it true that every continuous function on the unit disc can be approximated
uniformly by polynomials?
Any help is appreciated.

[quote="Kalle"]Yes and no. If you mean polynomials in two real variables $ x,y$ (or $ z$ and $ z^*$ for complex valued functions) then yes; this follows by Stone-Weierstrass.

No if you mean polynomials in one complex variable, because the uniform limit of polynomials is analytic (and under much weaker types of convergence you still get analytic functions[/quote]
Thanks for your help. The question says that polynomials in variable z. Does that mean that polynomials in one complex variable.


AMC USA(J)MO USAMO trigonometry AIME counting distinguishability

Well now I counted that there are 38 9th grader USAMO qualifiers, on the Teacher's Manual they say they are only going to take around 30 9th graders, so how will the selection process work this year?

Thanks

So what kind of USAMO scores do we 9th grader need?

Well they'd probably take you if you figured out how to get a negative score... can we agree on non-zero as a suitable adjective?

Oh, shoot. Considering I might get a 0... oh man... bad.

Yes it is...

 :?

So, now I'm like stuck. Everyone says it's too late to begin preparing, and I didn't expect to make it. But now I really want to see what MOP is like... sigh. Looks like I need some professional help.

Must write flattering statements on the USAMO write-up. Maybe they'll give you some credit for trying, like the USAMTS.

I am preparing to write a sob story of my life and how I have overcome adversity to get to where I am today and that I have a bet with a huge bookie that I made with my last $5000 that I would make MOP. If I win that bet, I will get $100,000 from him and that money will be sent towards feeding my 9 starving younger brothers and sisters.

How are you guys going to try to convince them?

lol same strategy yifman :).  Well, actually, a positive score is not too hard.  Just write every theorem you know and one of them is bound to be used somewhere :D.  Or actually solve the problem ;).  Oh yeah, and some 9th graders qualified through top in state.  No offense to them...  :blush:

Well I'm aiming for a 10 on USAMO, hopefully solve 1 question(last year's #1 was kind of easy) and get like 3 points by writing some "good ideas". I must redeem myself on USAMO!

10 is a good goal.  I think you can do better though because last year's #4 was also easy.  But you never know with these USAMO problems :(.

If only you got a 3 for writing good ideas. You have to connect some dots on USAMO to get 3 points. :-) Rumor has it that somebody was awarded 3 points for writing "law of cosines" on the solution to #1 last year. Lucky individual.

Fermatprime thinks he got 1 or 2 for that.  But pertaining more to the topic, do you think they'll take indices somehow?  Like 8*AIME + 7*USAMO?

Well, the problem pretty much reduced itself to law of cosines. :D

Will your AMC and AIME score count for 9th grade MOP selection?

It's really not hard to get a positive score on USAMO. Honestly, what did you do for 4 1/2 hours each day if you couldn't put together one coherent idea? 1, 2, 4, 5 are never THAT bad that basic logic alone can't get you a point on one of them.

As for the person who said "the overwhelming majority of [people taking USAMO for the first time] getting zeroes" ... I would certainly hope not. Like I've been saying, getting ONE point is not that hard. Have a little faith in yourselves.

[quote="darktreb"]It's really not hard to get a positive score on USAMO. Honestly, what did you do for 4 1/2 hours each day [/quote]

I have no idea what they do for 9 hours, but I know that the most common score for the people taking USAMO for the first time is a zero. Last year, about a fifth of all the people taking USAMO made 1 or less, and half made 3 or less. Here's the score distribution:
http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2004-ua/04usamo-results.html

[quote="texas137"]Using the index would make sense. I don't see any way they could select from among people who all made the same USAMO score, and the overwelming majority of people taking the USAMO for the first time make zero. Of the 38 ninth graders who qualified for USAMO, I'm guessing that only a very small number of them will have a score of 1 or better. Of those, a couple might do so well they qualify for main MOP, which takes them out of the pool for 9th grade MOP. And there will probably be at least a couple who have conflicts and can't come. So that would leave pretty close to 30. They might take all of them.[/quote]

Hmm although each person will cost a lot of money, so they'll try their best to kick as many people out as possible.

Ah, then I'm at a major disadvantage because my index is 209.5. Let's say someone has an index of like 230. Then I'd already need to score a 3 just to beat them if they get a 0, and they won't get a zero.  :(

[quote="beta"] Hmm although each person will cost a lot of money, so they'll try their best to kick as many people out as possible.[/quote]

I don't think so. I think they will want to have as many as they have money for. The whole idea is to train people how to do this early in their careers. The assumption is that the people participating in 9th grade MOP do not already know to do this type of math and will need to be taught. They might take a hard line about people who want to participate but have a conflict with part of the program, but I doubt if they will try to exclude 4 or 5 ninth graders who are basically indistinguishable from a much larger number of ninth graders they are taking.

[quote="MithsApprentice"]A 0 is not a bad score...[/quote]

Why not? How can u get lower than 0?
How do you calculate the scores in USAMO? U can get points deducted?

[quote="white_horse_king88"]Wait, is this only for 9th graders? Not for 8ths?[/quote]

Yes, I am 99.999% sure.  Although it's not fair; I think I did better than many 9th graders last year  :mad:

lol, no, you can not get points deducted. A zero is a perfectly respectable score considering a large portion of the scores are zeros. Anyway, just try your best and if you think about a problem long enough you should be able to piece together some ideas that will earn at least a partial score.....man, why can't I be a FRESHMEN *cries again*..... :P

[quote="Chinaboy"]For all of those who went last year: What was Freshman MOP like last year? What did you guys do there?[/quote]

The same thing as everyone else: Math. Loads and loads and loads and loads of it. You have lower-level courses and problems, but that's about it.


[quote="Jaswinder"]acceleration of block A is 2g/3 downwards and that of block B is g/3 upwards

plz reply and tell whether it is correct or not.............. :)  :wink:[/quote]
$ a_A \equal{} \frac {20}{3}, \\
a_B \equal{} \frac {10}{3}, \; considering \; \; g \equal{} 10 m/s^2$

[quote="anurags92"]Find accelerations.....

Sorry couldn't post in same topic......[/quote]
anurag, please post Physics questions in the Physics forum, search for it in the index. You have two choices, Introductory Physics and Advanced Physics. As for this question, it must go to the Introductory Physics sub-forum.

oh sorry
 :oops:
i am really sorry if i sounded rude :(

plz forgive me all mathlinkers

i am begging for ur pardon plzzzzzzzzzzz

sorryyyyyyyyyy

http://www.artofproblemsolving.com/Forum/index.php?f=332

Nice problems as you wanted. Well, I thought, reading your posts you are trying to sound a know-it-all. But since you apologise so much, I'd better change my opinion. Sorry!

One last advice anurags92,

Obviously you are smart as u qualified for NSEA Sr. in 10th itself.

You did not get these sums even though they are not hard because they were new to you.

But if u read HC verma closely u will observe that some new concepts which are not given in the theory are

introduced in the solved examples.

This concept of twice tension and two times acceleration are introduced in the solved examples.

P.S. Come on Jaswinder and Rajiv, i am not blaming any of u, but we are Indians guys, shd we not be united

and not fight over petty things.

[quote="anurags92"]thanks befuddlers...........

i am facing problems in questions where constraint relations are used..........

i solved all other questions in hcv... but a bit difficulty in visualizing the problems where there is a combination of two or three pulleys..... I will take some care ..........

As for posting sch elementary and almost same type question here.........
I wanted to see the methodology....[/quote]
When you've finished with your problems and feel confident, you may try to solve this as for good exercise:
The bigger mass weighs $ M \equal{} 12kg$, while the smaller mass weighs $ m \equal{} 5kg$.
Find individual accelerations.

And for higher level problems for practice, they are posted in advanced Physics section. If you can't find it, click on the smiley at the end of my signature just after the word "Einstein" and you'll be there.

Regards,
-Rajiv Patki.

Jai Hind!

[quote="Jaswinder"]i am getting the acceleration of block of mass 12kg as g/9 upwards and that of other block as g/9 downwards.............

plz reply and tell whether it is correct or not............[/quote]
Acceleration of $ 5kg$ block will be twice as much as that of $ 12kg$ block.
$ T_1 \plus{} T_2$ pulls $ M$ upwards while only $ T_2$ pulls $ m$ upwards. If $ M$ goes down by $ a$, you have the entire rope with $ T_1$ move with $ a$ as also rope with $ T_2$ move with $ a$. However, the second pulley is being accelerated by $ a$ ( :oops: I forgot to mention puleys are weightless, frictionless.), hence, $ a_m \equal{} 2a_M$. (Hope you get my notations :D )
Solving, I get:
$ T_1 \equal{} (70 \minus{} 22a)N$
and, $ T_2 \equal{} (10a \plus{} 50)N$.
Further work should be simple.

[quote="Jaswinder"]but i think constraint relations will gives us the accelerations to be the same..........

plz reply???? :([/quote]
If the pulley is accelerated and the rope too, the accelerations would surely add up. If you are on your cycle, both of you placed in a bus, and you accelerate your bike@$ a$, as well as the bus too accelerates@$ a'$, won't the observer on ground notice your acceleration as $ a \plus{} a'$?

And last time you requested good problems, so [url=http://www.artofproblemsolving.com/Forum/index.php?f=332]click here[/url].

[quote="Jaswinder"]but i think that since the mass m accelerates with A and that of M with a

then we have

2a=a+A

isn't that correct plz reply

i am a bit confused??????????? :?:  :([/quote]
In fact, that put me in a tizzy too. Consider the smaller pulley as your reference. Now, you are sitting on it observing $ m$. It appears to accelerate with $ a$ towards you. You are attached to another rope. You too are accelerating with $ a$.
Maybe you (perhaps me instead) got the relations wrong (either of us did).
I'll check again. but I'm confident acceleration of $ m$ is $ 2a$.

but i think i did that very carefully

i don't think so that i made wrong relations

but still i will recheck and u too

n plz reply and tell us the answer

just check from the source u got the problem

[quote="Rajiv"]The question cmes from a large question bank my friend gave me. i dont  have the answers. I think posting it in Another forum could help. Yes, I'll do that.[/quote]
I have posted the question at another, more active forum.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=250783]View Topic and Replies[/url].

yes hey rajiv

got it

infact both of us were wrong

i had been erring in the directions :P

rest my constraint relations were perfect

but in physics

directions are must.............. :)

[quote="Jaswinder"]yes hey rajiv

got it

infact both of us were wrong[/quote]
Yes. I have finally settled on a value for $ a$, and $ a_m\equal{}3a_M$. See the post the link to which I've reported previously.

[quote="Rajiv"]Acceleration of $ 5kg$ block will be twice as much as that of $ 12kg$ block.
[/quote]
HOW cud've u thoght that????

and 1 more think----this forum is the INDIAN forum where we can discuss questions among ourselves :roll:

[quote="rituraj007"][quote="Rajiv"]Acceleration of $ 5kg$ block will be twice as much as that of $ 12kg$ block.
[/quote]
HOW cud've u thoght that????
[/quote]
Well! I explained why I was thinking like that...but now I've corrected my views!


Let $ P(x)$ be a polynomial with real coefficients and $ x_0$ be any real. The line $ T(x) \equal{} mx \plus{} b$ is tangent to $ P(x)$ at $ x \equal{} x_0$. Determine $ m$ and $ b$. Do not use any calculus.

[hide="Hint for starting"]$ T(x)$ is tangent to $ P(x)$ at $ x_0$ if and only if $ (x \minus{} x_0)^2 \mid P(x) \minus{} T(x)$.[/hide]

[hide="Super Hint"]It's enough to consider $ x_0\equal{}0$. (Otherwise just define the polynomial $ Q(x)\equal{}P(x\plus{}x_0)$)[/hide]


[hide]the answer is 15.  Since the 2.73 ends with a 3, you must have at least 3 pennies. I used guess and check and when you use each coin 3 times it works.[/hide]

[quote="MCrawford"]A man has $\$$2.73 in pennies, nickels, dimes, quarters and half dollars.  If he has an equal number of coins of each kind, then the total number of coins he has is
\[ \text{(A)}\ 3 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 15  \][/quote]

[hide="Non Guess and Check Way"]$1p+5n+10d+25q+50h=273$

where $p=n=d=q=h$. set them equal to $x$

$x+5x+10x+25x+50x=273$

$91x=273$

$x=3$

There are three coins of each kind so there are $15$ in total.[/hide]

[quote="MCrawford"]A man has $\$$2.73 in pennies, nickels, dimes, quarters and half dollars.  If he has an equal number of coins of each kind, then the total number of coins he has is
\[ \text{(A)}\ 3 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 15  \][/quote]
can you explain about pennies,nickels,... . because it really don't know how much are these? :blush:

[quote="jensen"]can you explain about pennies,nickels,... . because it really don't know how much are these? :blush:[/quote]

1 penny = 1 cent.
1 nickel = 5 cents.
1 dime = 10 cents.
1 quarter = 25 cents.
1 half-dollar = 50 cents.
1 dollar = 100 cents.


logarithms
LaTeX
trigonometry
algebra proposed
algebra

proof that

$\arcsin x=-i\ln\left(x-\sqrt{x^2-1}\right)+\frac{\pi}{2}$

(note that $\arcsin x=-i\ln\left(x+\sqrt{x^2-1}\right)+\frac{\pi}{2}$  :!=: arc sin x  since there is a verticaal difference, it is still a solution to sin a = x though)

I will rewrite it using $\LaTeX$.

Prove that $\arcsin x=-i\ln\left(x-\sqrt{x^2-1}\right)+\frac{\pi}{2}$.

P.S. It is post 4001  :P

[quote="ZetaX"]Use $cos(x)=sin(x+\frac{\pi}{2})$ and $cos(x)=\frac{e^{ix}+e^{-ix}}{2}$ and solve for x[/quote]

you do not necesarily have to convert the sine to cosine, but it is correct.
You can also solve it to obtain
$\arcsin x=-i\ln\left(ix-\sqrt{1-x^2}\right)$
$\arcsin x=-i\ln\left(x-\sqrt{x^2-1}\right)+\frac{\pi}{2}$


GAME 1)
[b]Rule[/b]:
Suppose there is a 1xn (n>2) rectangular region which consists of n unit squares.
We start by putting three coins on the last three squares on the right.
Each player then takes turns to move any one of the coins to its left hand side.
The coins cannot overlap.
The first one who cannot move a coin loses.

[b]Remark[/b]:
 [ ] [ ] [ ] [ ] [ ] [ ] [ ] [C] [C] [C]      ->       [ ] [C] [ ] [ ] [ ] [ ] [ ] [C] [ ] [C]
This is a legal move.

[b]For example[/b]:
When n = 4: [ ] [C] [C] [C]
If the first player move the right most coin to the left, i.e.  [C] [C] [C] [ ], then the first player win.

When n = 5:  [ ] [ ] [C] [C] [C]
If the first player move the left most coin to the left, either  [ ] [C] [ ] [C] [C] or  [C] [ ] [ ] [C] [C].
Then no matter how the second player move any one of the two coins on the right. The first player then move the remaining coin and win the game.

[b]Question[/b]: Does any one of the players has a winning strategy for all possible values of n? If so, what is it?

Personally I agree that one of the players has a winning strategy, by the Zermelo's Theorem, this game cannot end in a draw, result follows.
But then I cannot think of a winning strategy for all n. Any comment ?

=====

GAME 2)
[b]Rule[/b]:
This is the misère tic-tac-toe, i.e. the same as ordinary tic-tac-toe but the first player who completes a row, column, or diagonal with that player's marks loses.

[b]Question[/b]: Want to show each of the two players has a non-losing strategy.

A non-losing strategy for the first player is simple.
Just put the mark at the centre first.
Then no matter how the second player put his or her mark, the first player just need to put the mark opposite to that of the second player.

However, I cannot think of a non-losing strategy for the second player, or equivalently, to show the first player has no winning strategy.

=====

GAME 3)
[b]Rule[/b]:
This is the misère SIM, i.e. the same as ordinary SIM but the first player who completes a triangle with that player's color wins.

[b]Question[/b]: Want to show one of the two players has a winning strategy and find the winning strategy.

I am not sure if my argument works.
As in the proof that the ordinary SIM cannot end in a draw, so the misère SIM cannot end in a draw too.
By the Zermelo's Theorem, one of the two players has a winning strategy.

Now, as in the ordinary SIM, the second player has a winning strategy, that is to prevent himself or herself to complete a triangle earlier than the first player.
For the misère SIM, the first player can then adopt the same idea to prevent the second player to complete a triangle earlier than himself or herself.

Thank you very much!

the coin game sounds suspiciously like nim, since it's impartial, but i can't find a quick isomorphism. (by sprague-grundy, we know there must be one)

the misere sim is trivial. here's the strategy:

1) first player plays a line: blue
2) second player plays anywhere, doesn't matter
3) first player plays a line adjacent to his last move
4) second player must block the triangle
5) first player plays a line connecting the only vertex of blue degree 2 & a vertex with degree 0. he now has a star of n=3.
6) second player must block one of the two threatened triangles.
7) first player takes the other one and wins.

I've calculated Sim game using seven vertices on my computer, and found out that the second player has the winning strategy. May be I will make Sim in Flash.

Good Luck!
Serge


Show that $\left(\begin{array}{c}2p-1\\p-1\end{array}\right)\equiv 1(mod p^2)$ if and only if $p$ is prime and odd number.

if $p$ is prime then $2{2p-1 \choose p-1}={2p \choose p}=\sum_{i=0}^{p}{p \choose i}^2 \equiv 2$ $(\mod p^2)$,so ${2p-1 \choose p-1} \equiv 1$ $(\mod p^2)$.
The proof of the other side is more difficult, but now I don't have time for it.


You can find a very nice solution of this one in the Old and New inequality book.


Given positive integers $m$, $a$, $b$, $(a,b)=1$. $A$ is a non-empty subset of the set of all positive integers, so that for every positive integer $n$ there is $an \in A$ and $bn \in A$. For all $A$ that satisfy the above condition, find the minimum of the value of $\left| A \cap \{ 1,2, \cdots,m \} \right|$

Let $a<b$.
We can suppose $S$ contains all numbers greater than $m$.
Then if $x>\frac mb$ the condition is already fulfilled.
Let $H$ be the set of all elements less than $\frac mb$ that are divisible by an even power of $b$ (i.e. either not divisible by $b$ or divisible by $b^2$ and not $b^3$ and so on). Then for each $x\in H$ either $ax$ or $bx$ is in $S$. Moreover we can't have $ax=by$ for $x,y$ in $H$ because then the powers of $b$ in $x,y$ would differ by $1$ so one would be odd. Hence $|S\bigcap \{1,2,\ldots,m\}|\geq |H|$. However te condition is satisfied by $S=bH\bigcup \{m+1,m+2,\ldots\}$: if $x\leq \frac mb$ then if $x \in H$ then $bx \in S$ otherwise $\frac{ax}b \in H$ so $ax$ in $S$.

Finally $|H|$ can be computed to be $[\frac mb]-[\frac m{b^2}]+[\frac m{b^3}]+\ldots$

I am not sure that I understood this problem right... If I am right
numbers in $A$ are all positive integers that are divisible with $a$ or $b$, and only those?
Then $|A\cap \{1,2,...,m\}|=[\frac{m}a]+[\frac{m}b]-[\frac{m}{lcm(ab)}]$.

Do I make mistake somewhere?


function
advanced fields
advanced fields unsolved

Let $ (E,||\cdot||)$ be a finite dimensional normed
space. Show that then there is always some closest point in the sense
of the theorem of the closet point property, but that this point nedd
no be unique.

If $ A$ is a nonempty closed subset and $ x\in E$, then there is $ a\in A$ such that $ \|x\minus{}a\|\equal{}\inf\{\|x\minus{}y\|\colon y\in A\}$. Use the continuity of the function $ y\mapsto \|x\minus{}y\|$ together with the compactness of [b]bounded[/b] closed subsets.

For non-uniqueness consider the norm $ \|(x_1,x_2)\|\equal{}\max\{|x_1|,|x_2|\}$, let $ A$ be the unit ball in that space, and let $ x\equal{}(2,0)$.

Can You please explain in more detail?? Thanks


function
floor function
logarithms
calculus
derivative
real analysis
integration

Let f:[0,1]$ \rightarrow$ R satisfies |f(x)-f(y)| $ \le$ |$ x^{1/3}\minus{}y^{1/3}$| for  all x, y $ \in [0,1]$. must f be absolute continuous?

Yes. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=309619]here[/url]. With the differences of a convex or concave monotone function bounding everything, it looks just like uniform continuity.

[hide="Now, if that is |x-y|^(1/3) ..."]Take the standard Weierstrass example of a continuous nowhere differentiable function $ f(x)\equal{}\sum_{n\equal{}0}^{\infty} a^{\minus{}n}\cos(b^n x)$ (with $ 1<a<b$, of course). We have $ |f(x)\minus{}f(y)|\le |x\minus{}y|\cdot\sum_{n\equal{}0}^m \left(\frac ba\right)^n \plus{}\sum_{n\equal{}m\plus{}1}^{\infty}2a^{\minus{}n}$ for all $ m$. Choose $ m\equal{}\lfloor \minus{}\log_b \frac{x\minus{}y}{2}\rfloor$ (or $ \minus{}1$ when this is negative), so that this estimate is as small as possible.
Then $ 2b^{\minus{}m\minus{}1}<|x\minus{}y|\le 2b^{\minus{}m}$, and we have estimated
$ |f(x)\minus{}f(y)|\le |x\minus{}y|\cdot a^{\minus{}m}\frac{b^{m\plus{}1}\minus{}a^{m\plus{}1}}{b\minus{}a}\plus{}\frac{2a^{\minus{}m}}{a\minus{}1}$
$ |f(x)\minus{}f(y)|\le 2a^{\minus{}m}\left(\frac{b\minus{}\frac{a^{m\plus{}1}}{b^m}}{b\minus{}a}\plus{}\frac1{a\minus{}1}\right)\le 2a^{\minus{}m}\left(\frac{b}{b\minus{}a}\plus{}\frac1{a\minus{}1}\right)$
As long as $ b^3>a$, that's less than $ |x\minus{}y|^{\frac13}$ for large $ m$. For small $ m$, it's at most a constant multiple worse, and we can replace $ f$ by some $ c\cdot f$ to get that constant back. Quantifying that, this estimate is at its worst for $ |x\minus{}y|\equal{}2$, on the cusp between $ m\equal{}0$ and $ m\equal{}\minus{}1$. There, it says that $ |f(x)\minus{}f(x\plus{}2)|\le 2a\left(\frac{b}{b\minus{}a}\plus{}\frac1{a\minus{}1}\right)$. Define $ g(x)\equal{}\left(\frac{2ab}{b\minus{}a}\plus{}\frac{a}{a\minus{}1}\right)^{\minus{}1}\cdot f(x)$. Now $ |g(x)\minus{}g(y)|\le a^{\minus{}m\minus{}1}< |x\minus{}y|^{\log_b a}$ when $ 2b^{\minus{}m\minus{}1}<|x\minus{}y|\le 2b^{\minus{}m}$, and the Holder condition is satisfied.

As for absolute continuity, the derivative of an absolutely continuous function is Lebesgue integrable, and must therefore be defined almost everywhere. That's certainly not true here.

This is not the only way to build a counterexample. If $ h$ is the classic Cantor devil's staircase function, $ |h(x)\minus{}h(y)|\le |x\minus{}y|^{\log_3 2}$, and this $ h$ is not absolutely continuous since all its variation happens in the measure-zero Cantor set.[/hide]

Hi jmerry,
 I have did the following. Is it correct away to prove f is absolute continuous?

$ g(x) \equal{} x^{1/3}$ then $ g(x) \equal{} (1/3)\int_{0}^{x}t^{ \minus{} 2/3} dt$, so g(x) is absolute continuous function , hence f is absolute continuous


Στο χώρο αυτό να βάλουμε :

   [b]α) [/b]Ασκησεις κλασσικής συνδυαστικής ανάλυσης και κλασσικής πιθανότητας.
   [b] β)[/b] Περιστεροφωλιά και  combinatorial Arithmetic

 Δε θα βάλουμε συνδυαστική Γεωμετρία, διότι θα μπεί στο άλλο topic.

   Διαλέξετε όμορφα θέματα , από μεσαία έως και δύσκολα.
 Θα βάλουμε [color=red]30 Ασκήσεις.[/color]
    Εδώ θα σπαζοκεφαλιάσουμε, αλλά και αυτό πρέπει κάποτε να γίνει !!!

 [b] Καλή Μελέτη!  [/b]

ΑΣΚΗΣΗ 1 -(Νίκος Ράπανος)
Σε ένα reality πρόκειται να επιλεχθούν 5 παίκτες από 15 αρχικούς υποψήφιους.
Μεταξύ αυτών των 15 υποψηφίων υπάρχουν 3 ανδρόγυνα, για τα οποία ισχύει ότι οι δύο σύζυγοι θα είναι μαζί από την αρχή μέχρι και το τέλος της διαδικασίας της επιλογής. Δηλαδή ή θα επιλεχθούν και οι δύο σύζυγοι ή κανένας από τους δύο.
Η επιλογή των <<5>> θα εξελιχθεί σε δύο φάσεις :
Στην πρώτη φάση θα επιλεχθούν 10 από τους 15 αρχικούς υποψήφιους και στη δεύτερη φάση θα επιλεχθούν 5 από τους 10 επιτυχόντες της πρώτης φάσης.
Με πόσους διαφορετικούς τρόπους μπορεί να πραγματοποιηθεί η επιλογή των <<5>> ;

ΛΥΣΗ (Στέλιος Δεσποτάκης)
$-\binom{3}{3}\binom{9}{4}=$ όλοι οι δυνατοί συνδυασμοί με τους οποίους στην πρώτη δεκάδα μπορούν να μπουν τρία ζευγάρια,

  $\binom{3}{2}\binom{4}{1}=$ όλοι οι δυνατοί συνδυασμοί με τους οποίους στην τελική πεντάδα μπορούν να μπουν δύο ζευγάρια,εφόσον στην πρώτη δεκάδα έχουν μπει τρία ζευγάρια,

  $\binom{3}{1}\binom{4}{3}=$ όλοι οι δυνατοί συνδυασμοί με τους οποίους στην τελική πεντάδα μπορεί να μπει ένα μόνο ζευγάρι,εφόσον στην πρώτη δεκάδα έχουν μπει τρία ζευγάρια.

$-\binom{3}{2}\binom{9}{6}=$ όλοι οι δυνατοί συνδυασμοί με τους οποίους στην πρώτη δεκάδα μπορούν να μπουν δύο μόνο ζευγάρια,

  $\binom{2}{2}\binom{6}{1}=$ όλοι οι δυνατοί συνδυασμοί με τους οποίους στην τελική πεντάδα μπορούν να μπουν δύο ζευγάρια,εφόσον στην πρώτη δεκάδα έχουν μπει δύο μόνο ζευγάρια,

  $\binom{2}{1}\binom{6}{3}=$ όλοι οι δυνατοί συνδυασμοί με τους οποίους στην τελική πεντάδα μπορεί να μπει ένα μόνο ζευγάρι,εφόσον στην πρώτη δεκάδα έχουν μπει δύο μόνο ζευγάρια,

  $\binom{6}{5}=$ όλοι οι δυνατοί συνδυασμοί με τους οποίους στην τελική πεντάδα δεν θα υπάρχει κανένα ζευγάρι,εφόσον στην πρώτη δεκάδα έχουν μπει δύο μόνο ζευγάρια.

$-\binom{3}{1}\binom{9}{8}=$ όλοι οι δυνατοί συνδυασμοί με τους οποίους στην πρώτη δεκάδα μπορεί να μπει ένα μόνο ζευγάρι,

  $\binom{1}{1}\binom{8}{3}=$ όλοι οι δυνατοί συνδυασμοί με τους οποίους στην τελική πεντάδα μπορεί να μπει ένα ζευγάρι,εφόσον στην πρώτη δεκάδα έχει μπει ένα μόνο ζευγάρι,

  $\binom{8}{5}=$ όλοι οι δυνατοί συνδυασμοί με τους οποίους στην τελική πεντάδα δεν θα υπάρχει κανένα ζευγάρι,εφόσον στην πρώτη δεκάδα έχει μπει ένα μόνο ζευγάρι.

Τελικό αποτέλεσμα:
$\binom{3}{3}\binom{9}{4}\left[\binom{3}{2}\binom{4}{1}+\binom{3}{1}\binom{4}{3}\right]+\ \binom{3}{2}\binom{9}{6}\left[\binom{2}{2}\binom{6}{1}+\binom{2}{1}\binom{6}{3}+\binom{6}{5}\right]+$

$+\binom{3}{1}\binom{9}{8}\left[\binom{1}{1}\binom{8}{3}+\binom{8}{5}\right]=\boxed{19152}$

Silouan μπορείς να μας εξηγήσεις πώς προκύπτει η λύση αυτή? Είναι ότι βασικά  μ διαφέυγει η θεωρία που χρησιμοποιήθηκε για τη λύση της άσκησής. :blush:  :blush:  :blush:  :blush:

ΑΣΚΗΣΗ 2-(Νίκος Ράπανος)
Σε ένα νησί βρίσκονται ιππότες και κατάσκοποι   συνολικά  100 άτομα . Οι ιππότες λένε πάντα την αλήθεια και οι κατάσκοποι  λένε είτε  αλήθεια και είτε ψέματα .
Οι ιππότες είναι περισσότεροι από τους κατασκόπους .
Στο νησί βρίσκεται ένας detective ο οποίος  θέλει να μάθει ποιοι είναι ιππότες και ποιοι είναι κατάσκοποι .Αυτός κάνει την ερώτηση σε κάποιον για το τι είναι κάποιος άλλος .(δηλαδή κατάσκοπος ή ιππότης ) .Είναι δυνατόν με 150 τέτοιες ερωτήσεις να βρει πόσοι είνια οι ιππότες και πόσοι οι κατάσκοποι ??

ΛΥΣΗ-(Γιάννης Παναγέας)
Αρχικά πρέπει να παρατηρήσει κάποιος ότι για να βρω πόσοι είναι οι ιππότες και πόσοι οι κατάσκοποι,
πρέπει κάποια στιγμή να βρω ένα ιππότη, γιατί οι κατάσκοποι μπορεί πάντα να λένε την αλήθεια!
Παίρνω τυχαία έναν τον $A_{1}$ από τους 100. Αρχίζω και ρωτάω τους υπόλοιπους έναν έναν για το τι είναι ο $A_{1}$.
Όσο το πλήθος αυτών που έχουν πει ότι ο $A_{1}$είναι ιππότης είναι μεγαλύτερο ή ίσο από αυτών που έχουν πει
ότι είναι κατάσκοπος , συνεχίζω να ρωτάω. Έτσι στο τέλος θα έχω κάνει $2x_{1}+1$ ερωτήσεις όπου $x_{1}$ θα έχουν
απαντήσει ότι είναι ιππότης και $x_{1}+1$ ότι είναι κατάσκοπος. Είναι εύκολο να δείξει κανείς ότι σε αυτή την ομάδα
των $2x_{1}+2$ (συμπεριλαμβανομένου και του $A_{1}$) $x_{1}+1$ θα είναι τουλάχιστον κατάσκοποι.(το $x_{1}$ μπορεί να είναι 0)
Έπειτα παίρνω έναν $A_{2}$ από τους $100-2(x_{1}+1)$ που απέμειναν και κάνω την ίδια διαδικασία. Κάνοντας αυτή τη διαδικασία παρατηρώ ότι παραμένει το πλήθος των ιπποτών να είναι μεγαλύτερο από αυτό των κατασκόπων(από αυτούς που μένουν). Κάποια στιγμή , μιας και οι ιππότες θα είναι περισσότεροι , θα ρωτάω για τον $A_{i}+1$ τους
$p = 100-2(x_{1}+x_{2}+...+x_{i}+i)$ που μένουν και θα είναι συνεχώς το πλήθος αυτών που έχουν πει ότι ο $A_{i}+1$ είναι ιππότης μεγαλύτερο ή ίσο από αυτών που έχουν πει ότι είναι κατάσκοπος. Έτσι όταν το πλήθος αυτών φτάσει το
$\frac{p}{2}$ τότε σίγουρα ο $A_{i}+1$ είναι ιππότης(έστω ότι έκανα τότε y ερωτήσεις, άρα $y-\frac{p}{2}$ είναι οι κατάσκοποι στην $i+1$ ομάδα).

Έτσι έχω βρει ότι ο $A_{i}+1$ είναι ιππότης και έχω κάνει            $2(x_{1}+x_{2}+...+x_{i})+i+y$ ερωτήσεις. Αλλά επίσης γνωρίζω
ότι στην πρώτη ομάδα υπάρχουν τουλάχιστον $x_{1}+1$ κατάσκοποι, στη δεύτερη $x_{2}+1$ ... στην ι-οστη $x_{i}+1$ και άρα με άλλες $i$ ερωτήσεις θα έχω κάνει $2(x_{1}+...+x_{i})+2i+y$ ερωτήσεις και θα ξέρω ότι έχω $x_{1}+....+x_{i}+i+y-\frac{p}{2}$ κατασκόπους και τον $A_{i}+1$ ιππότη. Άρα μένουν
$150-(2(x_{1}+x_{2}+...+x_{i})+2i+y)$ ερωτήσεις που θα κάνω στον $A_{i}+1$ για $100-(x_{1}+x_{2}+...+x_{i}+i+y)-1+\frac{p}{2}= 150-2(x_{1}+x_{2}+...+x_{i}+i)-y-1$ άτομα (είναι τα άτομα κατά 1 λιγότερα από τις ερωτήσεις που απομένουν). Άρα μπορώ με 150(μπορώ με 149) ερωτήσεις να βρω πόσοι είναι κατάσκοποι και πόσοι είναι ιππότες!

Άλλη μία λύση για την [color=red][size=150]Άσκηση 4[/size][/color]...

[size=150][color=red]Άσκηση 5[/color][/size]

Βάζω και εγώ μία κατά την γνώμη μου πολύ όμορφη και
πρωτότυπη εφαρμογή της περιστεροφωλιάς...

[size=150][color=red]Άσκηση 6[/color][/size]

Βάζω και μία άσκηση απλή, αλλά ενδεικτική της κομψότητας της συνδυαστικής επιχειρηματολογίας....

Με πόσους διαφορετικούς τρόπους μπορούμε να ανέβουμε ν σκαλιά εάν ανεβαίνουμε σε κάθε βήμα 1 ή 2 σκαλιά;

(Λύση στο προσαρτημένο pdf)

ΑΣΚΗΣΗ 7 (ΕΣΣΔ 1968)

Πεπερασμένο πλήθος καρτών τοποθετούνται σε δύο στίβες με περισσότερες κάρτες στην αριστερή στίβα παρά στη δεξιά. Κάθε κάρτα έχει γραμμένα πάνω της ένα ή περισσότερα διαφορετικά μεταξύ τους ονόματα, όμως διαφορετικές κάρτες μπορεί να έχουν κάποια ονόματα κοινά. Για κάθε όνομα, ορίζουμε ως 'ανακάτεμα' τη μεταφορά κάθε κάρτας που έχει γραμμένο αυτό το όνομα στην άλλη στίβα. Να δειχθεί ότι μπορούμε πάντοτε να καταλήξουμε με περισσότερες κάρτες στη δεξιά στίβα επιλέγοντας ορισμένα διαφορετικά ονόματα και εκτελώντας το 'ανακάτεμα' για κάθε όνομα.

Λύση: (Oaz Nir)

Θα χρησιμοποιήσουμε επαγωγή στο ν, όπου ν ο αριθμός των διαφορετικών ονομάτων που υπάρχουν.
Ονομάζουμε Λ την αριστερή στίβα και Ρ τη δεξιά.
Για ν=1 αρκεί ένα ανακάτεμα αφού υπάρχουν περισσότερες κάρτες με το όνομα αυτό στη Λ.
Υποθέτουμε ότι ισχύει για ν ονόματα για κάποιο ν θετικό ακέραιο. Θα δείξουμε ότι ισχύει για ν+1 ονόματα.
Ονομάζουμε α1, α2, α3, ..., αν τα πρώτα ν ονόματα και α το 'καινούριο'. Διακρίνουμε δύο περιπτώσεις:

1) Ο αριθμός των καρτών της Λ που περιέχουν μόνο το α είναι μικρότερος ή ίσος του αριθμού των καρτών της Ρ που περιέχουν μόνο το α.  (1)
Αγνοούμε το α και χρησιμοποιούμε την υπόθεσή μας εκτελώντας ανακατέματα για ορισμένα υποσύνολα των α1, α2,...,αν. Όμως τώρα, σύμφωνα με την υπόθεση και με τη σχέση (1), υπάρχουν περισσότερες κάρτες στη στίβα Ρ. Άρα δείχθηκε το ζητούμενο.

2) Ο αριθμός των καρτών της Λ που περιέχουν μόνο το α είναι μεγαλύτερος από τον αριθμό των καρτών της Ρ που περιέχουν μόνο το α. Κάνοντας τότε ένα ανακάτεμα με το α αναγόμαστε στην πρώτη περίπτωση οπότε ολοκληρώθηκε η απόδειξη.

Θα παρουσιάσω δύο απλές σχετικά ασκήσεις που στη λύση τους εφαρμόζεται η πολλαπλασιαστική αρχή:
ΑΣΚΗΣΗ 8
Με πόσους τρόπους μπορεί κανείς να τοποθετήσει σε μια στοίβα τα 52 φύλλα της τράπουλας αν θεωρήσουμε ότι κάθε φύλλο μπορεί να τοποθετηθεί με την πάνω ή κάτω πλευρά να φαίνεται (κάθε φύλλο είναι μοναδικό);

Λύση
Το πρώτο φύλλο, το πάνω στη στοίβα, μπορεί να επιλεγεί με 52 τρόπους και να τοποθετηθεί με 2 τρόπους (πάνω ή κάτω) άρα σύνολο $2*52=104$ πιθανές περιπτώσεις στην πρώτη θέση. Παρόμοια στη 2η θέση θα έχουμε $2*51=102$ πιθανές περιπτώσεις και σύμφωνα με την πολλαπλασιαστική αρχή θα έχουμε συνολικά $2^{52}*52!$ τρόποι να τοποθετήσει κανείς τα φύλλα σε στοίβα.

ΑΣΚΗΣΗ 9
Ένα ερωτηματολόγιο που μοιράζεται σε Ν οικογένειες ρωτάει εκτώς των άλλων και πόσα παιδιά έχει η κάθε οικογένεια και οι πιθανές απαντήσεις είναι 0,1,2,3,4,...,9. Πόσα είναι τα πιθανά αποτελέσματα που θα λάβει η εταιρία που κάνει την έρευνα αν τα αποτελέσματα θεωρηθούν διατεταγμένα (να παίζει ρόλο η σειρά με την οποία εμφανίζονται τα αποτελέσματα).

Λύση
Κάθε οικογένεια μπορεί να δώσει 10 πιθανές απαντήσεις άρα αφού οι οικογένειες είναι Ν τα αποτελέσματα μπορούν να συνδυαστούν με $10^{Ν}$ τρόπους.
Μια εναλλακτική λύση θα ήταν να τοποθετήσουμε όλα τα Ν αποτελέσματα στη σείρα και να αντιστοιχίσουμε κάθε αποτέλεσμα στο ψηφίο που αντιπροσωπεύει. Έτσι θα είχαμε ένα Ν-ψήφιο αριθμό ή μικρότερο αν διώξουμε τα 0 που τυχόν προηγούνται. Έτσι αντιστοιχίζουμε κάθε πιθανό αποτέλεσμα σε έναν αριθμό του δεκαδικου συστήματος από 0 μέχρι 999...9(Ν φορές) δηλαδή $10^{Ν}$ αριθμούς

Λοιπόν,

 με μια γερή έφοδο , όπως αυτή,  στα combinatorics θα μαζάψουμε αρκετές ασκήσεις.

  α) Να γράφετε τις λύσεις σας πολύ αναλυτικά , εξηγώντας και την τελευταία λεπτομέρεια. Να αποφεύγετε τελείως τα χάσματα και τα λεγόμενα αυτονόητα βήματα, διότι το αντικείμενο είναι πολύ δύσκολο.
  β) Είναι βέβαια δύσολα τα σχήματα , αλλά θα χρειαστούμε και μερικές ασκήσεις στη συνδυαστική γεωμετρία. Γράψτε εσείς τις ασκήσεις και όταν έρθει η ώρα θα στείλετε τις λύσεις και τα σχήματα χειρόγραφα , ώστε να γίνουν με λετπομέρεια. Είναι καλό να κρατάτε σε αρχείο με πολύ καλές λύσεις όλες τις ασκήσεις που βάζετε , ώστε όταν πληκτρολογηθούν να έχουν την πλήρη μορφή.

   Μπάμπης

Esto $x_{n}$ mia akoluthia pragmatikon arithmon ke $\forall \ n , \ m \in N$ isxiei
$|x_{n+m}-x_{m}-x_{n}| \leq \frac{1}{m+n}$
na dixte i akoluthia ine arihmitiki proodos.

Esto $n. \ m \in N$
$|(x_{m+1}-x_{m})-(x_{n+1}-x_{n})|=|(x_{m+1}-x_{m})+(x_{n+m+1}-x_{n+m+1})-(x_{n+1}-x_{n})| \leq |x_{m+n+1}-(x_{m+1}+x_{n})|+|x_{m+n+1}-(x_{n+1}+x_{m})| \leq \frac{2}{m+n+1}$

tora, i skepesi pu ine apo piso ine na kanume tin posotita $|(x_{m+1}-x_{m})-(x_{n+1}-x_{n})|$ oso mikri thelume(pws?)
k diladi telika $x_{m+1}-x_{m}=x_{n+1}-x_{n}$  $\forall \ n , \ m \in N$.

esto $t$ enas fisikos(opoosdipote thelume)
$|(x_{m+1}-x_{m})-(x_{n+1}-x_{n})|=|(x_{m+1}-x_{m})-(x_{t+1}-x_{t})+(x_{t+1}-x_{t})-(x_{n+1}-x_{n})| \leq |(x_{m+1}-x_{m})-(x_{t+1}-x_{t})|+|(x_{n+1}-x_{n})-(x_{t+1}-x_{t})| \leq \frac{2}{m+t+1}+\frac{2}{n+t+1}\leq \frac{4}{t}$

Exoume teliosi giati gia $t$ arketa megalo i posotita $|(x_{m+1}-x_{m})-(x_{n+1}-x_{n})|$ ginete oso mikri thelume.

sinexizontas briskume oti h akoluthia ine tis morfis  $x_{n}=(n-1)d, \ d \in R$

Μήπως έκανες λάθος sef και έστειλες την άσκηση άλλο topic
[color=red]ΑΣΚΗΣΗ 10[/color]
Επειδή βάζω ελληνικά ονόματα στα αρχεία που στέλνω παρατήρησα ότι η πρώτη λέξη σβήνεται :blush:  δε νομίζω να είναι πρόβλημα αυτό?

kpsychas endiaferusa paratirisi ala pu les na teriazi kalitera(ke giati)? apo algebriki skopia i moni boithitiki protasi pu xrisimopiume ine i trigoniki anisotita(dedomenu oti energume sto soma ton  pragmatikon arithmon). thelo na epanelthume
s ayti ti sizitisi ala teliunoun i monades!!!

as dume mia ali pu gia ti lisi tis xrisimopiume tin $|x| \geq x$.

brite tin elaxisti timi tis parastasis $A= |x_{2}-x_{1}|+|x_{3}-x_{2}|+\cdots+|x_{n}-x_{n-1}|+|x_{1}-x_{n}|$ opu
$x_{1}, \cdots , x_{n}$ ine mia anadiataxi ton arithmon $1,2, \cdots , n$

epidi i timi tis parastisis den alazi se mia kikliki anadiataxi ton oron $x_{1}, \cdots , x_{n}$(an to leo kala?) mporume
na ipothesume oti $x_{1}=1$ tote gia kapoio $k$, $1<k \leq n$  exume $x_{k}=n$
sinepos,

$(1) \ x_{k}-x_{1}=n-1=(x_{k}-x_{k-1})+(x_{k-1}-x_{k-2})+\cdots+(x_{2}-x_{1}) \leq |x_{k}-x_{k-1}|+|x_{k-1}-x_{k-2}|+\cdots+|x_{2}-x_{1}|$

$(2) \ x_{k}-x_{1}=n-1=(x_{k}-x_{k+1})+(x_{k+1}-x_{k+2})+\cdots+(x_{n}-x_{1}) \leq |x_{k}-x_{k+1}|+|x_{k+1}-x_{k+2}|+\cdots+|x_{n}-x_{1}|$

Apo (1),(2) pernume $2(x_{k}-x_{1})=2(n-1) \leq A$. distixos den mporo na sinexiso dite pote isxi isotita

Sef sygnwmi alla synexizw na pisteuw oti i askiseis pou esteiles tha eprepe na einai sto topic tis algebras. Exallou de nomizw oti xrisimopoieis texnikes syndiastikis. As pei kai kapoios allos ti gnwmi tou mipws kanw lathos.
As valw twra kai mia askisi pou fainomenika einai algebras alla einai syndiastiki.
Estw 2 akolouthies pragmatikwn arithmwn $a_{n}$ kai $b_{n}$ me $a_{0}=1$ kai $b_{0}=\sqrt{2}$ kai
$a_{n+1}=\frac{2a_{n}b_{n}}{a_{n}+b_{n}}$ kai $b_{n+1}=\frac{a_{n}+b_{n}}{2}$.
Na deixte oti den yparxei deiktis k fysikos etsi wste:
$a_{k}=\frac{\sqrt{2}}{2}$ kai $b_{k}=\sqrt{2}+1$

Ti lysi tha ti steilw prosexws

[quote="kpsychas"]
Estw 2 akolouthies pragmatikwn arithmwn $a_{n}$ kai $b_{n}$ me $a_{0}=1$ kai $b_{0}=\sqrt{2}$ kai
$a_{n+1}=\frac{2a_{n}b_{n}}{a_{n}+b_{n}}$ kai $b_{n+1}=\frac{a_{n}+b_{n}}{2}$.
Na deixte oti den yparxei deiktis k fysikos etsi wste:
$a_{k}=\frac{\sqrt{2}}{2}$ kai $b_{k}=\sqrt{2}+1$
[/quote]

Me epifilaksi mipws kai iparxei kapoio lathaki sto sillogismo mou (dioti den xrisimopoiisa pouthena sindiastiki) exoume:

Fanera $b_{n+1}\geq a_{n+1}$ (logw anisotitas Arithmitikou-Armonikou Mesou)

Episis $a_{n+1}\geq a_{n}$ (antikatastiste to $a_{n+1}$ kai tha kataliksete se kati profanes -ara i akolouthia $a_{n}$ einai auksousa).

Omoia $b_{n}\geq b_{n+1}$ (ara h akolouthia $b_{n}$ einai fthinousa).

Sinepws isxiei oti $\sqrt{2}=b_{0}\geq b_{n}\geq b_{n+1}\geq a_{n+1}\geq a_{n}\geq a_{0}=1$ gia kathe $n\in \mathbb{N}$.

Ara oloi oi oroi tis akolouthias einai fragmenoi apo tous arithmous $1$ (katw fragma) kai to $\sqrt{2}$ (panw fragma). Sinepws den iparxoun oroi twn akolouthiwn autwn pou na ikanopoioun tis parapanw sxeseis tis ekfwnisis.

Xanw kapou?
Alexandros

Swsta skeftikes alexandre alla egw ekana lathos ta noumera. Ta thymomoun sto peripou kai ekana to exis lathos oi arxikes times pou edwsa eprepe na itan oi telikes kai oi telikes arxikes. Dinw diorthomeni tin ekfwnisi kai ti lysi.
Estw 2 akolouthies pragmatikwn arithmwn $a_{n}$ kai $b_{n}$ me $a_{0}=\frac{\sqrt{2}}{2}$ kai $b_{0}=\sqrt{2}+1$
kai $a_{n+1}=\frac{2a_{n}b_{n}}{a_{n}+b_{n}}$ kai $b_{n+1}=\frac{a_{n}+b_{n}}{2}$.
Na deixte oti den yparxei deiktis k fysikos etsi wste:
$a_{k}=1$ kai $b_{k}=\sqrt{2}$

Paratiroume oti $a_{n+1}\cdot{b_{n+1}}=a_{n}\cdot{b_{n}}=...=a_{0}\cdot{b_{0}}$. Ara to ginomeno twn n-ostwn orwn twn dyo akolouthiwn paramenei analloioto. Eukola epalitheuoume oti $a_{0}\cdot{b_{0}}\neq{\sqrt{2}}$ diladi de tha symbei pote
$a_{k}=1$ kai $b_{k}=\sqrt{2}$

οι ακολουθιες $a_{n}$ kai $b_{n}$ εχουν το ιδιο οριο που λεγεται μεσος αριθμητικοαρμονικος και ισουται με $\sqrt{a_{0}b_{0}}$.....

 :)

kpsychas(de xriazete sixoresi9, ine diskolo se merikes periptosis na anagume askisis sti mia i stin ali katigoria.
ala tis ebala edo giati to pexnidi stis dio parapno periptosis ginete stin katalili  euresi diakritwn
orvn tis akoluthias ke sto pos tha tus balis na alilepidrun metaxi tus(poies parastasis theoris).
ala ena sxolio gia ti proti ama teriazi stin algebra teriazi exisu kala ke stin analisi. des poia
gnosti protasi xrisimopoihthike.

Επειδή έχει καιρό να μπει άσκηση συνδιαστικής σκέφτηκα να βάλω το επόμενο θεματάκι:
 Πόσοι διαφορετικοί τρόποι υπάρχουν να φτάσουμε στην κορυφή μιας σκάλας που έχει 10 σκαλοπάτια εαν μπορούμε να ανέβουμε 1 ή 2 σκαλοπάτια τη φορά;

einai η ασκηση 6....

βαλε αλλη αν ειναι... :wink:

Sorry παιδιά είμαι καινούργιος και δεν είδα ότι είχε ξαναμπεί  :oops:

Έστω 6 σπίτια τα οποία συνδέονται ανά δύο με έναν δρόμο ο οποίος  είτε είναι ασφαλτοστρωμένος είτε είναι χωματόδρομος .Δείξε ότι τουλάχιστον τρία σπίτια συνδέονται το ένα με το άλλο με δρόμους του  ίδιου είδους (είτε όλοι ασφάλτοστρόμενοι είτε όλοι χωματόδρομοι).


calculus geometry search USAMTS AMC USA(J)MO USAMO \/closed

I noticed that there was no place for new members to introduce themselves and get a little guidance on how to navigate around the site. I guess I'm saying because I'm pretty new here and don't really know where to start. I'm a senior in high school and not too bad in math. I've taken Calculus and from what I've seen in the high school threads, Calculus and other higher level maths aren't welcome. I'm pretty interested in programming as well and hoped that there would be an area that would involve higher level maths with programming. At this point, I'm not sure if I'd need to purchase the book because I don't plan on participating in any math-related competitions since I'm already a senior. So does anyone have an idea of where I could go to improve my math skills and programming skills.

There is a calculus forum, [url=http://www.artofproblemsolving.com/Forum/index.php?f=66]see here[/url].
There is also a computer science forum, [url=http://www.artofproblemsolving.com/Forum/index.php?f=331]see here[/url].

[quote="AIME15"]What you'll see around the site is very ADVANCED applications of lower maths.  The reason that calculus isn't welcome here is because there are many younger students (including me!) that are still in middle school, and don't understand calculus yet.  However, many people that have taken a basic calculus class at school still find problems here challenging.  For example, at school you don't do any "real" proofs.  Here is an example:

Prove that a natural number is either even or odd. Pretty easy.

Note that here, we want RIGOROUS proofs, which means you don't want to say:

Every number is divisible by 2 or not divisible by 2, thus it is either even or odd.

If you search around the site, especially in the Intermediate Topics and Pre-Olympiad forums, you will find many many challenging problems.

As for programming, I have no idea.[/quote]

Just for reference, what would be the "correct" proof for determining a natural number. That was actually the first thing that popped into my head when you mentioned that.

More advanced applications of lower maths is probably what I'm looking for at this point since I am trying to gain more problem solving skills and that doesn't always entitle higher level maths. So where could I find more problems like that and maybe a little more guidance in why it is correct? Should I probably purchase the book?

The even / odd proof is a bad example, because by definition, even means divisible by 2 and odd means not divisible by 2, so AIME15's mentioned "bad" proof is actually a completely legitimate proof.  Most of what is taught in schools as proofs is valid and correct; the problem is that it is very simple and often involves only definitions or very simple steps to solve it.

As for books, there are several AoPS books available, including The Art of Problem Solving Volumes 1 & 2 (click on the bookstore link on the right hand side of the website to see a list of all the books offered here).

Well my "proof' wasn't rigorous at all...

If you want to learn rigorous proofs, study previous USAMTS and USAMO problems.

That proof _is_ rigorous. And math is not about rigorous proofs at all, but about informative (but correct) proofs.

Well you can 'introduce' yourself by going to a regional forum. Usually there are Who are you? forums.


[quote="crystaljewel3"]:blush: $ \frac {(b \plus{} c \minus{} a)^{2}}{(b \plus{} c)^{2} \plus{} a^{2}} \plus{} \frac {(c \plus{} a \minus{} b)^{2}}{(c \plus{} a)^{2} \plus{} a^{2}} \plus{} \frac {(a \plus{} b \minus{} c)^{2}}{(a \plus{} b)^{2} \plus{} c^{2}}\geq \frac {3}{5}.$ :?:[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=28678

http://www.mathlinks.ro/viewtopic.php?p=1215179#1215179
See other ideal there


Let $ a,b,c,d$ be positive real numbers. Prove that the following inequality holds
$ \left.\frac{a^4\plus{}b^4}{(a\plus{}b)(a^2\plus{}ab\plus{}b^2)}\plus{}\frac{b^4\plus{}c^4}{(b\plus{}c)(b^2\plus{}bc\plus{}c^2)}\plus{}\frac{c^4\plus{}d^4}{(c\plus{}d)(c^2\plus{}cd\plus{}d^2)}\plus{}\frac{d^4\plus{}a^4}{(d\plus{}a)(d^2\plus{}da\plus{}a^2)}\geq\frac{a^2\plus{}b^2\plus{}c^2\plus{}d^2}{a\plus{}b\plus{}c\plus{}d}\right.$

[quote="marin.bancos"]Let $ a,b,c,d$ be positive real numbers. Prove that the following inequality holds
$ \left.\frac {a^4 \plus{} b^4}{(a \plus{} b)(a^2 \plus{} ab \plus{} b^2)} \plus{} \frac {b^4 \plus{} c^4}{(b \plus{} c)(b^2 \plus{} bc \plus{} c^2)} \plus{} \frac {c^4 \plus{} d^4}{(c \plus{} d)(c^2 \plus{} cd \plus{} d^2)} \plus{} \frac {d^4 \plus{} a^4}{(d \plus{} a)(d^2 \plus{} da \plus{} a^2)}\geq\frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2}{a \plus{} b \plus{} c \plus{} d}\right.$[/quote]

$ \frac {a^4 \plus{} b^4}{(a \plus{} b)(a^2 \plus{} ab \plus{} b^2)} \ge \frac {\frac {1}{2}(a^2 \plus{} b^2)^2}{(a \plus{} b)(a^2 \plus{} ab \plus{} b^2)}$

Thus , it remains to prove that

$ \sum_{cyc}\frac {(a^2 \plus{} b^2)^2}{(a \plus{} b)(a^2 \plus{} ab \plus{} b^2)} \ge \frac {2(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)}{a \plus{} b \plus{} c \plus{} d}$

Acording to Cauchy-Shwarz inequality we have :

$ LHS \ge \frac {4(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)^2}{A}$ where $ A \equal{} \sum_{cyc}(a \plus{} b)(a^2 \plus{} ab \plus{} b^2) \equal{} 2(a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3) \plus{} 2\sum_{cyc}ab(a \plus{} b)$

It suffices to show that $ 2(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)(a \plus{} b \plus{} c \plus{} d) \ge 2(a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3) \plus{} 2\sum_{cyc}ab(a \plus{} b)$
$ \Leftrightarrow \sum_{cyc}\left(2a^3 \plus{} 2a^2b \plus{} 2a^2c \plus{} 2a^2d \minus{} 2a^3 \minus{} 2ab(a \plus{} b)\right) \equal{} \sum_{cyc}2a^2c \ge 0$

My solution is basically the same as Abdek's, just note on the one hand that we can do the first step directly by Cauchy-Schwartz, i.e.
\[ \sum \frac{a^{4}+b^{4}}{(a+b)(a^{2}+ab+b^{2})}\geqslant\frac{(\sum a^2)^2+(\sum b^2)^2}{\sum(a+b)(a^{2}+ab+b^{2})}\mbox{,}\]
and on the other hand that the given inequality is strict because $ \sum_{cyc}a^2c>0$ as all the variables are positive  :P .

Good solution. It's the same as mine. Thanks. Indeed, the inequality is a strict one. Taking just three variables, $ a,b,c$, the inequality works with "$ \geq$".


geometry
3D geometry
prism
geometric transformation
rotation
geometry proposed

This requires some imagination and creative thinking:

Prove or disprove:

There exists a solid such that, for all positive integers $n$ with $n \geq 3$, there exists a "parallel projection" (I hope the terminology is clear) such that the image of the solid under this projection is a convex $n$-gon.

Dude, this is such a sweet problem! :) I'm not entirely sure of the following, since I've just made a mental picture of the whole thing, but I hope it's correct (and I hope I can explain it properly).

Take an equilateral triangle $S_1=A_{11}A_{12}A_{13}$ (I think acute-angled suffices). Now take a quadrilateral $S_2=A_{21}A_{22}A_{23}A_{24}$ s.t. $A_{21}=A_{11},A_{22}=A_{12}$, and $A_{23},A_{24}$ lie on $A_{12}A_{11},A_{13}A_{11}$ respectively, very close to $A_{11}$. Then consider a pentagon $S_3=A_{31}A_{32}A_{33}A_{34}A_{35}$, with $A_{31}=A_{21},A_{32}=A_{22}$, $A_{33},A_{35}$ lie on $A_{22}A_{23},A_{21}A_{24}$ respectively, very close to $A_{23},A_{24}$, and s.t. $A_{34}$ lies on $A_{23}A_{24}$. We continue this process for hexagons, heptagons, and so on. I think drawing a picture will make the process clear.

For each $S_n$, construct the orthogonal prism $\mathcal P_n$ having $S_n$ as a basis (we construct it on both sides of the plane containing $S_n$).

Now we start moving the polygons constructed above (they still lie in the same plane). As a polygon $S_n$ moves, the prism $\mathcal P_n$ will move with it. The polygons will be moved by rotation around the common edge $A_{11}A_{12}$, all in a certain direction we choose before we start. In the $n$'th step we move $S_{n+1},S_{n+2},\ldots$ , all at the same time, keeping them in the same plane $\pi$, and we leave $S_n$ behind.

In step $1$, move $\pi$ until $\mathcal P_2$ contains $S_1$. In step $2$, move $\pi$ until $\mathcal P_3$ contains both $S_1$ and $S_2$, and so on. I think (I hope, actually :)) that if "very close" in that first paragraph is taken to be close enough at each step of the construction of the polygons, so that $S_{n+1}$ approximates $S_n$ very well, then the intersection of all the prisms $\mathcal P_n$ will be a solid like the one we're looking for, projecting onto a polygon congruent to $S_n$ on the plane on which $S_n$ lies after all the movements have terminated.

Sorry if this turns out to be flawed, and this message wastes the time of anyone who might read it :).

Hm, I have a similar approach, I also constructed polygons and prisms in a similar way, and I moved some plane $\pi$ as well, but, I am not completely sure about it either. Can someone check? It's a lovely question though!


algebra
function
domain
Ring Theory
summer program
PROMYS
quadratics

[quote="Phelpedo"]Prove that a prime can be expressed as a sum of 2 squares iff it is congruent to 1 (mod 4)

Followup:
For x,y integers:

Prove that all primes congruent to 1 (mod 8) can be expressed as $x^{2}+16y^{2}$.

Prove that every prime congruent to 5 (mod 8) can be expressed as $(2x+y)^{2}+4y^{2}$.[/quote]

Um.. $2=1^{2}+1^{2}$?

Ok this is a rather famous problem and we don't care about trivial cases...how about if I said "distinct squares."

[hide="hint"]
to show that if $p=4k+1$, then there is an n for which pdivides $n^{2}+1=(n+i)(n-i)$. If p did not factor in Z(i), it would have to be a complex number.

also remmember that any square is congruent to 1 or 0 mod 4 .[/hide]

Monty's class from SIMUW... did he go through this proof in your year also, Amir?

Also, prime usually implies odd primes (nobody cares about 2 that much  :P ).

Sorry  :oops:

Err...[quote="amirhtlusa"][hide="hint"]
to show that if $p=4k+1$, then there is an n for which pdivides $n^{2}+1=(n+i)(n-i)$. If p did not factor in Z(i), it would have to be a complex number.
[/hide][/quote]

I'm not really understanding this  :blush: sorry if i sound stupid

Yeah amirthulsa's hint doesn't look like it has anything to do with the conventional proof.

errr...the hint was absolutely correct.

[hide]
there exists $a$ such that $p$ divides $a^{2}+1=(a+i)(a-i).$  We have that $\mathbb{Z}[i]$ is a unique factorization domain, so therefore, if $p$ is prime in $\mathbb{Z}[i]$ as well, then either $p$ divides $a+i$or $p$ divides $a-i$.  But both of these are clearly wrong.  Therefore, we must have that $p$ is not prime in $\mathbb{Z}[i]$, that is, $p=(m+ni)(m-ni)=m^{2}+n^{2}$ for some $m$ and $n$.[/hide]

There are other proofs of this result (I was thinking of euler's).

For this proof, I'm not sure about some of your notation. What does "ufd" stand for, and what is Z[i]? doesn't look like standard notation for a field to me.

Ok sorry I didn't know that one. What about parts b and c?

[quote="paladin8"]That proof is quite standard... UFD = Unique Factorization Domain, $\mathbb{Z}[i]$ is standard notation for the Gaussian Integers, $a+bi$ with $a, b \in \mathbb{Z}$.[/quote]
That's quite true, of course but to be fair not every one (even in this crowd) is born with the instict to know that UFD holds for Gaussian integers :)

There are some other well known and cute methods to prove this theorem. I can put some outlines  of some easy fun way, if anyone else is interested  but  of course any good book or Internet has this too.

On a side note, an interesting question, for those who like this stuff. Given a prime (of the type $4k+1$) how does (practically speaking) one find the numbers..

For example, if I give you a prime number of the form $4k+1$, say

$5882353$

can you find two numbers whose squares adds up to the number.

(You can use a calculator but not programming  to tryout  brute force - If you like programming I could have given a 50 digit prime, can you still find the numbers  - in reasonable time)

See http://www.mathlinks.ro/Forum/viewtopic.php?t=46300 .
There is also a interesting proof showing that $\sigma^{*}(p-a^{2})$ must be odd for some $a$. Here $\sigma^{*}(n)$ is the sum of divisors $d$ of $n$ such that $\frac{n}{d}$ is odd).
Tthe key is to show the identity \[\sigma^{*}(n) = 2 \sum_{1 \leq u < \sqrt{n}}(-1)^{u-1}\sigma^{*}(n-u^{2})\] for all $n$ not being a square (for the squares we have to add $(-1)^{n-1}n$ on the right hand side).

I found another non-Minkowski proof using just integers (to be exact, at PROMYS this year, week 3)...and in terms of actually generating the two squares, it also works pretty well if I'm given the square root of $-1$ mod p...
The proof is a little complicated, though...

[quote="Phelpedo"]what is Z[i]? doesn't look like standard notation for a field to me.[/quote]

[b]Definition:[/b]  Let $\mathcal{R}$ be some ring.  We say that $\mathcal{R}[t_{1}, t_{2}, ... t_{n}]$ (read: R adjoin t_1, t_2, ... t_n) is the intersection of the rings that contain $\mathcal{R}$ and $t_{1}, t_{2}, ... t_{n}$.

In fact, $\mathbb{Z}[i]$ is not a field.  What it is is, essentially, the "smallest" ring that contains the integers and $i$.

Thanks for the clear-up.

Everybody seems interested in the first part and not the second and third :roll:

because they are simple corollaries of the first.
For example,
if a prime = 1(mod 8), let $p = x^{2}+y^{2}$.
They have to be one odd and another even.
If the even one, say y = 2(mod 4), then $x^{2}+y^{2}\equiv 5 (mod 8)$, which is no good.
This forces $y \equiv 0 (mod 4)$, and the result follows.

The another one is similar.

Did any one find the two  numbers whose squares add up to $5882353$
(You may be in a slight surprise  :)

Indeed!
But would you (or anyone) like to put a solution (how did they find it out)..
(One method , I have in mind, can also be considered a proof)

Another related old problem (may be  a little chalanging)  is,

Prove any prime of the form $8k+3$ can be written as $a^{2}+2b^{2}$
(Yes, this problem, I am sure has visited this forum a few times, but it may still be fun)

[quote="Gyan"]Prove any prime of the form $8k+3$ can be written as $a^{2}+2b^{2}$
(Yes, this problem, I am sure has visited this forum a few times, but it may still be fun)[/quote]

There is a [b]massive[/b] generalization of these types of problems.  Can anyone spot it and prove it?

[hide="Huge hint"] Quadratic Reciprocity ;) [/hide]

Massive generalisation¿  :huh:

There is no such theorem like: any prime of the form $5n+1$ can be written as $a^{2}+5b^{2}$.
So what did you mean¿

Well we know that if $\mathbb{Z}[\sqrt{-d}]$ has UFD, then $\left(\frac{-d}{p}\right)=1 \iff x^{2}+dy^{2}=p$ has a solution, by work with Minkowski's theorem.  From there, it's a matter of working out $\left(\frac{-d}{p}\right)$ using quadratic reciprocity.  In the case where $-d=-2$, we know $\left(\frac{-d}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right)=(-1)^{\frac{p-1}{2}+\frac{p^{2}-1}{8}}=(-1)^{\frac{p^{2}+4p-5}{8}}=(-1)^{\frac{(p-1)(p+5)}{8}}$.  Some basic casework gives us the form of the primes which yield an even exponent there...
But in general, if we can show $\mathbb{Z}[\sqrt{-d}]$ has UFD (it turns out it does for d=2), then it's a question of breaking and using quadratic reciprocity...(of course, the whole question of generally determining UFD has to do with class numbers and quadratic forms and that's above me).

Sorry for having to point that out...something from PROMYS...

In fact, Minkowski's theorem can be used for sums of more squares and more complicated equations, but I don't know if you can use it for, say, r-gonal numbers...working on that with the PROMYS open-door set :D

Well, what I was pointing to: $\mathbb{Z}[\sqrt{-d}]$ with $d>0$ squarefree is only UFD if $d=1,2$ (so this is a very rare event). It is very near to be for $d=3$.

@Minkowski: I never use it  to show such theorems, but rather use norms.
A higher dimensional example is that of:
What primes can be written as $p=a^{3}+2b^{3}+4c^{3}-6abc$.
Here you have to consider $\mathbb{Z}[\sqrt[3]{2}]$.

[quote="Peter VDD"][quote="ZetaX"]There is no such theorem like: any prime of the form $5n+1$ can be written as $a^{2}+5b^{2}$.[/quote]

There is a theorem saying any prime of the form $6n+3$ can be written as $a^{3}-5b^{2}$ though.  :D[/quote]

Only prime, I know  the form $6n+3$ is $3$ :) , and $a=2,b=1$ does work. :roll:

(Any way the problem(s)  I asked above  (and no doubt have been asked before me too) does have solution, not to difficult,  suitable for this forum.)

[quote="Gyan"]Only prime, I know  the form $6n+3$ is $3$ :) , and $a=2,b=1$ does work. :roll:[/quote]That was kind of the joke, yes. ;)


What is the number of square units in the triangular region bounded by the lines $ y\equal{}2x$, $ y\equal{}\minus{}2x$, and $ x\equal{}5$?

Finding the intersection points of these lines gives

$ (5, 10)$, $ (5, \minus{}10)$, and $ (0,0)$

This is a triangle with a base (y-axis) of $ 20$ and a height (x-axis) of $ 5$

This gives an area of $ \frac{bh}{2} \equal{} \frac{20*5}{2} \equal{} \boxed{50}$


limit
function
calculus
calculus computations

hi :)
The limit of $ \sqrt{n\plus{}1}\minus{}\sqrt{n}$ while $ n\rightarrow \infty$ is obviously zero since we have,
$ \lim_{n\to \infty }\left (\sqrt{n\plus{}1}\minus{}\sqrt{n} \right )\equal{}\lim_{n\to \infty }\frac{1}{\sqrt{n\plus{}1}\plus{}\sqrt{n}}\equal{}0$
i found another approach,
$ \frac{1}{\sqrt{n\plus{}1}\plus{}\sqrt{n}}< \frac{1}{2\sqrt{n}}
\Rightarrow 0< \sqrt{n\plus{}1}\minus{}\sqrt{n}< \frac{1}{2\sqrt{n}}$
therefore the limit is zero.
would you please tell me how we got this approach.

[quote="DineshManohar"]You have handled with Squeeze theorem . It goes like this  if $ x_{n} \le y_{n} \le z_{n}, \forall n \in N$ if we are aware that $ x_{n} \to l$ and $ z_{n} \to l$ then the condition squeezes $ y_{n} \to l$.  :)

Hope this will clear you!![/quote]

thanks :) ,but i didn't understand how they got,
$ \frac{1}{\sqrt{n\plus{}1}\plus{}\sqrt{n}}<\frac{1}{2\sqrt{n}}\Rightarrow 0<\sqrt{n\plus{}1}\minus{}\sqrt{n}<\frac{1}{2\sqrt{n}}$

It is very clear $ 0$ is a lower bound for the sequence .Also the limit is $ 0$. You can adopt an upper bound sequence ( a best approximation ) to work out for the limits.

An usual procedure to use Squeeze theorem.

For this sequence we have   $ 0 <  \frac{1}{\sqrt{n\plus{}1}\plus{}\sqrt{n}}$.

You find an upper bound sequence so that the limit  converges to $ 0$.

$ \sqrt{n\plus{}1}\plus{}\sqrt{n} > 2\sqrt{n}$ . This is clear i suppose ( You note that we easily choose bounds by intuitively knowing the limit of the sequence ).

$ \frac{1}{\sqrt{n\plus{}1}\plus{}\sqrt{n}}<\frac{1}{2\sqrt{n}}\Rightarrow 0<\sqrt{n\plus{}1}\minus{}\sqrt{n}<\frac{1}{2\sqrt{n}}$.
Using Squeeze Theorem the limit is $ 0$  :)

[quote="MathObsessed"],but i didn't understand how they got,
$ \frac {1}{\sqrt {n \plus{} 1} \plus{} \sqrt {n}} < \frac {1}{2\sqrt {n}}\Rightarrow 0 < \sqrt {n \plus{} 1} \minus{} \sqrt {n} < \frac {1}{2\sqrt {n}}$[/quote]
In the LHS just multiply and divide by $ \sqrt{n\plus{}1}\minus{}\sqrt{n}$.

That algebraic technique, that of rationalizing the numerator, is a limited technique. You won't be able to use it everywhere. Here's a version of the proof that uses something a lot more general.

Consider the function $ f(x)\equal{}\sqrt{x}$ and note that $ f'(x)\equal{}\frac1{2\sqrt{x}}.$ If we look at this function on the interval $ [n,n\plus{}1],$ the Mean Value Theorem applies. In particular,
\[ \sqrt{n\plus{}1}\minus{}\sqrt{n}\equal{}1\cdot\frac{1}{2\sqrt{\xi}}\]
where $ \xi$ some point in $ (n,n\plus{}1).$

What do we do with that $ \xi?$ We don't want to know it exactly - if we knew it exactly, we wouldn't need it in the first place. What we do is as what the worst case is. Since $ \frac1{2\sqrt{xi}}$ is (by inspection) a decreasing function of $ \xi$ on the interval in question, we can make it as large as possible by making $ \xi$ as small as possible. Since $ \xi> n,$ that gives us $ \frac1{2\sqrt{\xi}}<\frac1{2\sqrt{n}}.$

Hence, $ 0<\sqrt{n\plus{}1}\minus{}\sqrt{n}<\frac1{2\sqrt{n}}.$

Now try this, as an exercise in this technique:

Prove that $ \lim_{n\to\infty}\left[\sqrt[3]{n\plus{}1}\,\ln(n\plus{}1)\minus{}\sqrt[3]{n}\,\ln(n)\right]\equal{}0.$


My parents always stress the importance of GPA, but I just can't get the 4.0.  I take all the AP and honors classes available.  So I was wondering how much difference would a 4.0 and 3.9 be when applying to the best colleges?

It definitely depends on your other things.
It helps, but when you're applying to the top colleges, a 4.0 GPA is nothing spectacular, because so many kids have it. But don't let that discourage you. I have a 3.83 (unweighted) and I got into some pretty good schools.

Yes, it does depend on the classes you take, and how bad the non-A grades are, and on an lot of other issues. A 3.9 grade average in tough courses is still competitive almost everywhere, but there is no reason not to go for the A every time if you can still fit other challenging activities into your schedule.

can you report a 3.995 as a 4.0? my english teacher says no but if you are asked to report to "one decimal place" then... come ON! who wants to put 3.9 when you have a 3.995? ouch.

do colleges take in consideration weighted grades, or will they just stick with how many A's a student has?

I think you tell them what classes you took, and they either weight your grade or just take it into consideration.

Either way, they will notice that a 4.0 with regular classes is not as good as a 3.8 with all AP/IB classes

Colleges make their own internal GPA calculations when evaluating applicants, with each college using its own rules. But it always looks better to BOTH take hard classes AND get good grades in all classes taken--because the most competitive applicants manage to do both of those things. Some high schools have a reputation for having easy, nonintellectual courses, and some high schools (e.g., Exeter) have a reputation for being very academically rigorous.

Can AIME be considered as a credit for college? or is it too "low"?

What do you mean?  Colleges give you credit for taking college level classes and sometimes other things like researching etc.  There's no way they would give you credit just because you took one test.

Would anyone know exactly how to figure out your weighted GPA?  Or does every school do it different?  (my school does unweighted GPA, they only weight it to decide class rank, but this doesnt happen until senior year)  Also, are all advanced classes weighted?  For example, I am in a math class that is a few years ahead of my grade, but it is not AP, will it be weighted?

I dont think weighted grades really matter, since when colleges look at your GPA, they also look at the classes you took. So even if you don't have a 4.0 with all AP/honors classes, you are still better off than a 4.0 in normal classes.

[quote="stupidkid"]Sorry, I don't mean the credit as in a class, but rather more of a credential. (i.e. makes you more competitive.)[/quote]

This has been discussed a lot on AoPS before.  It really depends on the college and who reviews your application.  If that person knows what the AIME is and appreciates what it is worth then, yes, it will give you a little bit of an edge.  If they don't know about it they will probably just take it as a math accomplishment of yours.  It will still give you an edge, just not as big of one, usually.

Darn that sucks, o well hopefull I can make it to USAMO next year, considering I just missed it by 1 problem this year.

The AIME is specifically asked about on the application forms of MIT and Caltech. A score up at the floor score level is impressive to those schools, and does help applications.

[quote="beta"]I have a D on AIME  :(  :) , that's like a... pretty low GPA.[/quote]

A D on the aime? So you mean like 10 right?

nah, a 10 on the aime would be an F (at least according to a decent grading scale), he'd have made like an 11 or 12

My goal on the AIME was not to flunk. And I achieved it. With a D-!

sorry.. can some1 explain to me diff. between weighted and unweighted GPA's?

The usual unweighted scale goes up to 4.0 (straight As). A weighted grading scale might give, for example, more grade points for an A in an AP course, so the student might get a 5 for that class, getting an overall GPA above 4.0 when all classes are averaged. There isn't any standard way to weight classes, just as there isn't any standard way to teach them or to grade them. So colleges usually have internal methods to recalculate the GPAs that appear on the transcripts of applicants.

The main thing is to take undeniably hard classes, and to do undeniably well in them. I have heard online stories of young people with 3.5 averages in college courses taken as a junior or senior in high school getting into better colleges than age mates who had 4.0 grade averages in strong suburban high schools, so not all grade averages are equal.

[quote="beta"][quote="xxreddevilzxx"]I think he means more like 2~4.

BTW, does USAMO qual help alot in the college app??[/quote]

Last time I checked 2~4/15 = 13~27%=F... the way we do it at our school, 60% = D, and 9/15 = 60%, maybe you do it differently at your school though(so 13% is not considered failing at your school?)[/quote]
i wish i could go to that scool!

On the topic of AIME grades... I did better on the AIME than I am doing in AP Physics! :lol: (11/15 > 70%)


calculus
derivative
search
binomial coefficients

(i) Expand $ (a \plus{} x)^{ \minus{} 2}$ in ascending powers of $ x$ up to and including the term in $ x^{2}$.

(ii) When $ (1 \minus{} x)(a \plus{} x)^{ \minus{} 2}$ is expanded, the coefficient of $ x^{2}$ is $ 0$. Find the value of $ a$.

[I get

$ a\equal{}\minus{}\frac{3}{2}$.]

[hide="(i)"]$ (a \plus{} x)^{ \minus{} 2} \equal{} \minus{} \frac {d}{dx}\left[\frac {1}{a \plus{} x}\right] \equal{} \minus{} \frac {d}{dx}\left[\frac {1/a}{1 \minus{} ( \minus{} x/a)}\right]$

$ \equal{} \minus{} \frac {d}{dx}\left[\frac {1}{a} \minus{} \frac {x}{a^2} \plus{} \frac {x^2}{a^3} \minus{} \frac {x^3}{a^4} \plus{} \frac {x^4}{a^5} \minus{} ...\right]$

$ \equal{} \minus{} \left[ \minus{} \frac {1}{a^2} \plus{} \frac {2x}{a^3} \minus{} \frac {3x^2}{a^4} \plus{} \frac {4x^3}{a^5} \minus{} ...\right]$

$ \equal{} \frac {1}{a^2} \minus{} \frac {2x}{a^3} \plus{} \frac {3x^2}{a^4} \minus{} \frac {4x^3}{a^5} \plus{} ...$[/hide]

[hide="(ii)"]Expanding and collecting the $ x^2$ terms gives

$ \frac{3x^2}{a^4}\plus{}\frac{2x^2}{a^3} \equal{} 0 \Rightarrow x^2(3\plus{}2a)\equal{}0 \Rightarrow a \equal{} \minus{}\frac{3}{2}$   :)[/hide]

Use of calculus is discouraged (other than second derivative concavity testing)

[hide="Without calculus"]
$ (x\plus{}a)^{\minus{}2}\equal{}\sum_{k\equal{}0}^{\infty}\binom{\minus{}2}{k}x^{\minus{}2\minus{}k}a^{k}\equal{}x^{\minus{}2}\minus{}2x^{\minus{}3}a\plus{}3x^{\minus{}4}a^2\minus{}\cdots$
[/hide]

[quote="grn_trtle"]Use of calculus is discouraged (other than second derivative concavity testing)

[hide="Without calculus"]
$ (x \plus{} a)^{ \minus{} 2} \equal{} \sum_{k \equal{} 0}^{\infty}\binom{ \minus{} 2}{k}x^{ \minus{} 2 \minus{} k}a^{k} \equal{} x^{ \minus{} 2} \minus{} 2x^{ \minus{} 3}a \plus{} 3x^{ \minus{} 4}a^2 \minus{} \cdots$
[/hide][/quote]

Just wondering, how do you calculate the value of a binomial if the top is negative? Wouldn't that be by definition:
\[ \binom{ \minus{} 2}{k} \equal{} \dfrac{ \minus{} 2!}{(k!)( \minus{} 2 \minus{} k)!}
\]

Edit: Thanks SimonM, I should have known to search wikipedia.

[url=http://en.wikipedia.org/wiki/Binomial_coefficient#Generalization_to_real_and_complex_argument]Generalized Binomial Coefficients[/url] gives us a definition


There are six clearly distinguishable frogs sitting in a row. Two are green, three are red, and one is blue. Green frogs refuse to sit next to the red frogs, for they are highly poisonous. In how many ways can the frogs be arranged?

green cannot be next to red. therefore, the blue frog has to be in between them. if we call the group of 3 red frogs R and green frogs G, the ordre has to be either RBG or GBR. there are 3!=6 ways to order the red frogs (distinguishable) and 2!=2 ways for green. so 2*6*2=24


advanced fields
advanced fields unsolved

If $C$ is a closed curve in plane then prove that there exist a square that it's vertexs are on the curve $C$.

Saman Gharib

I'm pretty sure this is still an open question :?.

Yes, it is. Marian Dinca or levi told me about this. ;)

It is

         C=C^2/C^2+C^2/C^2=(C^2/C^2)

         Then,
                            C=C^2   OR   C<C^2  OR   C>C^2

Which is  a complete answer.(Not Sure).


Sean and Zach are going to build a porch on their family's house. The house is36 feet wide across the front sean says they should build a porch that is 16 ft long  adn is centerd against the house. He wants to make it 8ft wide. zach says the should that idea to begin with, but they should enlarge the porch and make it in the shape of a trapezoid by connecting the front corner of the porch to the corner of the house.(I think you should try to draw this out) :?

ekk i forgot to write the rest of the question.

a) what is the area to the nearest square foot of sean's porch?
b) what is the total area, to the nearest square foot, if Zach's plan is followed

[hide]
A)128
B)i'm stuck because i can't remember the formula for area of a trapezoid[/hide]

Area of a Trapezoid: $A=\frac{h(b_1+b_2)}{2}$.


ratio
quadratics
geometry
trapezoid
similar triangles
Pythagorean Theorem

In the accompanying figure, circle A with radius of 9 and circle B with radius of 1 are externally tangent.  Circles C and D, with radii of [tex]r_1[/tex] and [tex]r_2[/tex], respectively, are tangent to circles A and B, and all four circles are tangent to a line.  Find the ratio [tex]\frac{r_1}{r_2}[/tex].

Another problem: In the accompanying figure, the distance between the centers of the circles is 65.  The radii of these circles are 36 and 20.  There are two segments tangent to these circles (determined by tangency points), one longer than the other.  Find the sum of these two segments.

I solved the second one. I put the answer as a spoiler.

[hide]I got 63 for the top segment and 33 for the lower one (sum is 96). The top is

sqrt(65^2-(36-20)^2).

The lower one I got by similar triangles (draw the segment between the two centers and also the radii to the tangent points). The ratio is 9/5. That means that the segment between the two centers is split into segments of length 65*9/14 and 65*5/14. Pythagorean theorem gives the answer of 33.[/hide]

zabelman wrote:Major hint (if you know it): [hide]Descartes Circle Theorem[/hide]

I don't really know what that is but here is how I did it(referring to the diagram below):

[hide]AB=9+1=10, AX=9-1=8. XB=6. [tex]AV=9-r_2, AD=9+r_2[/tex].[tex] VD^2 ={(9+r_2)} ^2 -(9-r_2)^2 =36r_2[/tex]. [tex]VD=6 \sqrt {r_2} [/tex]Using the same idea, [tex]DW= 2 \sqrt {r_2}. [/tex]

VD+DW=6, then  [tex]r_2= \frac {9}{16}. [/tex]

Same strategy we find [tex]ZC= 6 \sqrt {r_1}[/tex]. XB=6, and [tex]BY=2 \sqrt {r_1}[/tex].

Since XB+BY=ZC, [tex]r_1= \frac {9}{4}[/tex]. The ratio is 4.[/hide]

zabelman, can you post your solution using Descartes' Circle Theorem.

Descartes Circle Theorem (or at least the special case that I know) says that if there are 4 circles each tangent to the other 3, then the [b]curvatures[/b] of the circles satisfy

[tex](a+b+c+d)^2=2(a^2+b^2+c^2+d^2)[/tex]

Note: curvature is reciprocal of radius. For a line, the 'radius' is infinity so the curvature is 0. Furthermore, if there is one big circle outside the other 3, then the outside circle has negative curvature.

Now to the problem. We want 4 tangent circles with curvatures of 1, 1/9, 0 (the line), and x (unknown). Plug these 4 values into the formula above, and it is quadratic in x. Solve that, and there are 2 solutions for x: 16/9 and 4/9. So, the circles have radii of 9/16 and 9/4, and the ratio is 4.

In the accompanying figure, the isosceles trapezoid has side lengths AB = 1, BC = DA = 5 and CD = 7.  The points M on AD and N on BC determine a segment parallel with DC which divides the area of the trapezoid into two equal parts.  Find MN.

I got the answer to the trapezoid question by [hide]finding the area of ABCD to be 16, since the height is 4. Thus ABMN = CDMN = 8. Then I extended AD and BC to their intersection, which I called V. I found the area of VAB to be 1/3, using its 1:7 similarity to VCD.  Then I found the ratio of the areas of VMN and VCD, which is 25:49. Thus the ratio of MN to CD is 5:7, and thus MN = 5[/hide]


Two Metal Gliders are floating along a hoirzontal air track collide elastically.  There first glider (200g) is moving 32 cm/s while the second glider (300g) is moving in the opposite direction at 52 cm/s.  What is the velocity of each glider after impact?

I've Been using these eqaution:

V1f = V1f (m1-m2/m2+m1)

V2f = V1f (2m1/m1+m2)

And the answer in the back of the book is:

V1f = -68.8 cm/s
V2f = 15.2 cm/s

For some reason i keep getting these answers wrong.
Could sombody help me plz!!!

The problem is that your equations don't work. You can find the right ones here:

http://en.wikipedia.org/wiki/Elastic_collision

(Sorry, I don't remember them myself :).)

You should try to solve every problem from the root. Only when confident, you should use formula.
First apply, conservation of momentum, then for elastic collision apply conservation of energy.


Given a triangle $ \triangle ABC$ and a point $ P$ inside or on the boundary of $ \triangle ABC$. Construct a line through $ P$ that divides $ \triangle ABC$ into two regions of equal area.

[hide="Hidden by mod"]If $ P$ lies on the boundary of $ \triangle ABC$ this isn't hard. WLOG suppose that $ P$ lies on $ AC$. Let $ E$ denote the midpoint of $ AB$ and construct the line through $ C$ parallel to $ PE$ which meets $ AB$ at $ M$. Then $ PM$ is the desired line, since $ [APM] \equal{} [APE] \plus{} [PEM] \equal{} [APE] \plus{} [PEC] \equal{} [AEC] \equal{} \frac 12[ABC]$.

But I can't figure out how to do this when $ P$ lies inside the triangle...[/hide]


function
trigonometry
algebra
polynomial
complex numbers

A function $ f$ is defined by $ f(z) \equal{} i\bar z$, where $ i \equal{}\sqrt{\minus{}\!1}$ and $ \bar z$ is the complex conjugate of $ z$. How many values of $ z$ satisfy both $ |z| \equal{} 5$ and $ f (z) \equal{} z$?

$ \textbf{(A)}\ 0 \qquad
\textbf{(B)}\ 1 \qquad
\textbf{(C)}\ 2 \qquad
\textbf{(D)}\ 4 \qquad
\textbf{(E)}\ 8$

[hide]
Let $ z \equal{} r\text{cis}\, \theta$, where $ \text{cis}\, \theta \equal{} \cos \theta \plus{} i\sin \theta$. Then, $ z \equal{} \text{cis}\, \theta \equal{} f(z) \equal{} i\bar{z} \equal{} \text{cis}\, \left( \minus{} \theta \plus{} \frac {\pi}{2}\right)$. Thus $ \theta \equal{} \minus{} \theta \plus{} \frac {\pi}{2} \plus{} 2\pi n \Longrightarrow \theta \equal{} \frac {\pi}{2} \plus{} \pi n$, and we have $ z \equal{} \pm\left(\frac {1}{\sqrt {2}} \plus{} \frac {1}{\sqrt {2}}i\right)r$ which describes a line on the complex plane passing through the origin.

Alternatively, letting $ z \equal{} a\plus{}bi$, we have $ a\plus{}bi \equal{} z \equal{} f(z) \equal{} i\bar z \equal{} i(a\minus{}bi) \equal{} b \plus{} ai$, so $ a\equal{}b$.

$ |z| \equal{} 5$ describes a circle of radius $ 5$ on the complex plane, centered at the origin. These graphs have $ 2\ \mathbf{(C)}$ intersections.
[/hide]

\[ f(z) \equal{} z \Leftrightarrow z \equal{} i\bar z \Leftrightarrow z^2 \equal{} i|z|^2.\] If $ |z| \equal{} 5$, this becomes $ z^2 \equal{} 25i$ which has the two roots $ \pm 5 \sqrt{i}$ and no more by FTA since it is a second degree polynomial, making the answer $ \textbf{C}$.

Let $ z \equal{} a \plus{} bi$.  Then $ f(z) \equal{} i(a \minus{} bi) \equal{} b \plus{} ai \equal{} z \equal{} a \plus{} bi$.  So $ a \equal{} b$ and $ a^2 \plus{} b^2 \equal{} 25 \equal{} 2a^2$.  So $ a \equal{} b \equal{} \pm \sqrt {25/2}$ which gives two distinct $ z$'s.

[quote="AlanLiangTheAsian"]Ok, based on my understanding of this problem, the first sentence is not necessary to solve the problem? Because f(z) = z is just y = x on the complex plane with |z| = 5 is a circle centered at the origin with radius 5 on the complex plane... so if we interpret this through the graph, we see there are two intersections and thus two solutions?[/quote]

If $f(z)$ had been something else like $f(z)=zi$ then we would have gotten a different line when we did $zi=z$

Its just a coincidence that $z=\bar{z}i$ has solutions $a=b$ for the $z=a+bi$

Dont confuse this with $y=x$.. For example, if $f(x)=10x^4$, $f(x)=x$ has solutions the roots of $10x^4-x$ not all $x$..

$f(z)=z$ for complex $z$ is not a line; you need four dimensions to graph two variable complexes.

bruh.
you just solve maths equation a^2+b^2=25 and i(a-bi)=a+bi to get ai+b=a+bi and a=b then 2a^2=25 so a=+-5/sqrt2

How did you get from $ai+b = a+bi$ to $a=b?$

ai+b=bi+a implies a=b

because subtract bi and b from both sides to get ai-bi=a-b, which clearly shows that ai-bi has to equal 0 for this to be true. thus a=b

[quote=mod_x]ai+b=bi+a implies a=b[/quote]

Hmm, I did not know that.

[b]Solution:[/b] Let $z=a+bi$. Notice that to satisfy the problem condition, $f(z) = i(a-bi) = b + ai = z = a + bi$ so this directly implies $a=b$. Since we also have that $a^2 + b^2 = 25 \Rightarrow 2a^2 = 25 \Rightarrow a=b = \pm \frac{5}{\sqrt{2}}$. So there are $\boxed{2}$ possibilities.


How many square numbers(?) are there in Fibonacci sequence? and why?

I think the answer is 1 and 144, but I don't know why..help,,

(Sorry for my poor English)

The question is addressed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1475424965&t=1423]here[/url].  One of the posts has a reference, and the last post suggests some exercises that lead to a proof.

Thanks a lot! It's  really helpful for me~

but, isn't there any full solution?

It's very nice, but still too hard for me,