real analysis
function
topology
geometry
rectangle
3D geometry
real analysis unsolved

Given a Lebesgue set $F$ of $\mathbb{R}^{n}$ of finite measure (or possibly bounded?) I would like to find a closed set $H \subset F$ such that $\mu_{n}(H) > \mu_{n}(F)-\epsilon$. I know that I can find a Borel set $B\subset F$ with the same measure as $F$, so it seems likely that we can find a closed subset not much smaller than $F$. However, I can't figure out how to do it.

edit: I am actually trying to prove Lusin's theorem. I have found a sequence of continuous functions converging almost uniformly to my real valued measurable function $f$ on a bounded Lebesgue set $A$. However, to get a continuous function I would like to replace the set in the definition of almost uniform convergence with a closed set about the same size so that I have uniform convergence to a continous function which I may by Tietze's theorem extend to all of $\mathbb{R}^{n}$. Seems like I am on the right track, no?

If you define $\mu_{n}$ by infimum over countable coverings (like the Hausdorff measures), then it's easy to find an open superset $U\supset E$ with $\mu_{n}(U)<\mu_{n}(E)+\epsilon$. Just let $U$ be the union of all open balls in an appropriate covering of $E$. Carefully passing to complements, you will get a closed subset with the desired property.

Is this for a bounded set? The idea I get from your post is to take $E$ bounded and take some big closed rectangle $M$ containing $E$. Then we cover $M-E$ with an open set $G$ such that $\mu(G) < \mu(M-E)+\epsilon$. Now $M-G \subset E$ is closed and $\mu(M-G) = \mu(M)-\mu(G) > \mu(M)-\mu(M-E)-\epsilon = \mu(E)-\epsilon$

However, I don't see how to generalize this to a non-bounded set?

Divide the space into closed squares (or cubes), like a chessboard. For each square $Q_{i}$ find a closed subset $K_{i}\subset Q_{i}\cap E$ such that $\mu_{n}(K_{i})>\mu_{n}(Q_{i}\cap E)-\epsilon/2^{i}$. Observe that $\bigcup_{i=1}^{\infty}K_{i}$ is a closed set (even though it is a union of [b]infinitely [/b]many closed sets).


How many pairs (x,y) of positive integers satisfy 2x + 7y = 1000?

[hide="Answer"]Since GCD[2,7]=1, given any solution (x,y), (x-7,y+2) is also a solution.

We know that (500,0) satisfies the equation but is not positive. Starting from here, we have the solutions:
(493,2)
(486,4)
(479,6)
...

Clearly, following this procedure, y>0 will always be true. We want to see how many cases there are for which x>0 -- in other words, how many times can we subtract 7 from 500 and still have a positive number.

500/7=71.42..., so the set of positive solutions is given by (500-7n, 2n), where n is any integer from 1 to 71, inclusive. There are thus 71 solutions.[/hide]


probability
expected value
probability and stats

Have a sample a1,a2,...aT
I chose n at random from them and take the sum, call that X.
What's the expression for the variance and expected value of X?

The expectation of a sum of random variables is the sum of expectations.
$ X \equal{} \xi_1 \plus{} \dots \plus{} \xi_n$.

So $ E(X) \equal{} \sum_{k \equal{} 1}^n E(\xi_k) \equal{}  n E(\xi_1)$. (I leave you to find what $ E(\xi_1)$ is.)

Variance of $ \xi$ is just its covariance with itself. But covariance is linear in each variable, so there is a formula
for the variance of a sum, similar to the formula of the square of a sum:
\[ V(X) \equal{} \sum_{k \equal{} 1}^n V(\xi_n) \plus{} 2\sum_{1\le k < j\le n} \mathrm{cov}(\xi_i,\xi_j) \equal{} nV(\xi_1) \plus{} n(n \minus{} 1)\mathrm{cov}(\xi_1,\xi_2).\]
And again you should be smart enough to find $ V(\xi_1)$ and $ \mathrm{cov}(\xi_1,\xi_2)$ yourself.

May add my personal trick:
Here $ \xi_{1}\plus{}...\plus{}\xi_{T}\equal{}const$, so $ Cov(\xi_{1},\xi_{1}\plus{}...\plus{}\xi_{T})\equal{}0$.
Covariation is linear, so $ Cov(\xi_{1},\xi_{1})\plus{}(T\minus{}1)Cov(\xi_{1},\xi_{2})\equal{}0$.
And $ Cov(\xi_{1},\xi_{2})\equal{}\frac{\minus{}1}{T\minus{}1}Var(\xi_{1})$.

Nice trick, [b]roah[/b], I didn't know it, thank you for posting.

I don't get it.

If you write $ \xi_{1} \plus{} ... \plus{} \xi_{T} \equal{} const$, then you are treating $ \xi_{1}, ..., \xi_{T}$ as fixed, correct? Otherwise, the sum would be equal to a random variable.

In that case, $ E[ \xi_{1} ] \equal{} \xi_{1} \ne \xi_{2} \equal{} E[ \xi_{2} ]$,

and it also follows that $ Cov(\xi_{1}, \xi_{2}) \equal{} E[(\xi_{1} \minus{} \xi_{1})(\xi_{2} \minus{} \xi_{2})] \equal{} 0$

Guillaume,
the sum $ \xi_{1}\plus{}...\plus{}\xi_{T}$ is fixed, but particular values are random variables.

For example: i have two numbers 2 and 3 and if i take them in a random order i will obtain two random variables $ \xi_{1}$ - the first choosen number and $ \xi_{2}$ - the second choosen number. But for sure $ \xi_{1}\plus{}\xi_{2}\equal{}5$.

[quote="roah"]Guillaume,
the sum $ \xi_{1} \plus{} ... \plus{} \xi_{T}$ is fixed, but particular values are random variables.

For example: i have two numbers 2 and 3 and if i take them in a random order i will obtain two random variables $ \xi_{1}$ - the first choosen number and $ \xi_{2}$ - the second choosen number. But for sure $ \xi_{1} \plus{} \xi_{2} \equal{} 5$.[/quote]

Thank you - for some reason I thought that $ a_1 \equal{} \xi_1$.


Clock A loses $ 30$ seconds a day, and clock B gains $ 45$ seconds a day. If Clock A is $ 10$ minutes ahead of clock B. How many days will it take clock B to be $ 5$ minutes ahead of clock A

They will diverge 75 seconds a day.
In order to have 15 minutes change = 900 seconds change, 900/75=12 days


How many 3-digit numbers are divisible by 7?

the way i did it was that 14x7=98 so evry 100 there are 14 divisible numbers by 7 so therefore 14x9=126

but i dont know why it is wrong ????

@pink3433 This doesn't work because not every hundred has the same number of numbers divisible by 7. 100 is not divisible by 7, so you can't expect to see the same exact number of divisible numbers in a every one hundred.
Good thinking though!

[quote=deepfish007]I use this way,
trunk(1000/7) - trunk(99/7) = 142 - 14 = 128.

Looks simpler than the solution given by Alcumus.[/quote]

There was no need to bump this thread as the solution was already given


Here's an interesting question:
Set $A$ contains all the positive integers that can be values of $x$ to make $100000000000\le x\le999999999999$ true.  A subset of $A$, $B$, only contains 12-digit positive integers that have the number 2006 in them.  Examples of some numbers in set $B$ are $152006913502$, $711495820066$, $420061200679$ (to show that 2006 can appear more than once in the sequence), and $200610590365$.  If $B_n$ is the $n$th element of $B$, and $B_k=999999992006$ (obviously the largest element of $B$), compute $k$.
[hide="Simple wording"]How many twelve-digit positive integers contain the number 2006 in them?[/hide]

I can easily brute force it, but is there a more elegant way to do something like this?

[hide="If you want the answer..."][hide="Are you sure you want to peek?"][hide="Last chance to turn back!"]I think it's 819,924,800.[/hide][/hide][/hide]

Umm, $k$ definitely has to be greater than $1$...

Here's what I'm thinking:
[hide]Treat the block $2006$ like a single digit, and it's easy to see there are $9!$ ways to arrange the digits.  Now, there are $9$ possibilites for the first digit (1-9), and $10$ for all the rest (0-9), giving $9\times10^7\times9!$ possibilities...[/hide]

Hmmm, you're right.  How would you deal with that?

[quote="Adunakhor"][quote="DanK"]Umm, $k$ definitely has to be greater than $1$...

[hide="Here's what I'm thinking"]Treat the block $2006$ like a single digit, and it's easy to see there are $9!$ ways to arrange the digits.  Now, there are $9$ possibilites for the first digit (1-9), and $10$ for all the rest (0-9), giving $9\times10^7\times9!$ possibilities...[/hide][/hide][/quote]

lingomaniac88's answer is likely the probability of choosing such a number out of all $12$ digit numbers. Your method seems to double count a lot of the possibilities (what if $2006$ appears more than once in the number?), and at the same time misses the cases where $2006$ appears at the very beginning.[/quote]

Yes, I originally stated the probability as $\frac{624487547}{625000000}$.  I realized my mistake and changed it to the actual number, 819,924,800.

I used brute force, and it involved where 2006 could show up.  I looked at where 2006 could show up, and where it wouldn't.  I just eliminated various numbers for that.

What brute force are you talking about? Listing them out?

DanK, maybe apply PIE?


function
calculus
derivative
trigonometry
calculus computations

For some reason I cannot determine when this function is concave up and concave down.

f(x) = (cosx)^2 - 2*sinx

 on the interval [0,2pi]

Some help would be much appreciated!

You want the second derivative, factored.

$f(x)=\cos^2x-2\sin x.$

$f'(x)=-2\cos x\sin x -2\cos x =-2\cos x(\sin x + 1)$

$f''(x)=2\left[\sin x(\sin x+1)-\cos^2x\right]$

$=2\left[\sin^2x+\sin x-1+\sin^2x\right]$

$= 2(2\sin^2x+\sin x-1) = 2(2\sin x-1)(\sin x+1)$

That should be good enough for you to finish it.


calculus
integration
algebra
function
domain
calculus computations

$g$ is the inverse of $f$.  $f$ has domain $x\in[a,b]$.  Prove

$\int^{b}_{a}{f(x)}\,dx=b\beta-a\alpha-\int^{\beta}_{\alpha}{g(y)}\,dy$

where $\alpha=f(a)$ and $\beta=f(b)$

Hence prove that $\int^{e^{b}}_{e^{a}}{\frac{1}{x}}\,dx=\int^{e^{-a}}_{e^{-b}}{\frac{1}{x}}\,dx$

where $b>a$

*Even though the last part is obvious, it just verifies what has been shown in the first part*  :blush:

[hide="First part"]We have $y=f(x)$ and $g(f(x)) = x.$

Now, using integration by parts,

$\int_{\alpha}^{\beta}g(y)\, dy = yg(y)\Big|_{\alpha}^{\beta}-\int_{\alpha}^{\beta}yg'(y) \, dy = b\beta-a\alpha-\int_{\alpha}^{\beta}yg'(y) \, dy$

[Note that $g(\beta) = g(f(b) = b$ and $g(\alpha) = g(f(a) = a.$]

But $g(y) = x \Rightarrow g'(y)\, dy = \, dx.$ So changing the variables and limits above we get

$\int_{\alpha}^{\beta}g(y)\, dy = b\beta-a\alpha-\int_{a}^{b}f(x) \, dx$, which is what we had to show.[/hide]

Here is a [url=http://www.physicsforums.com/showthread.php?t=129324]thread on the same problem[/url].


email
LaTeX
calculus
derivative
geometry
cyclic quadrilateral

Hi guys,

here's the topic of the "[b]Oliforum contest[/b]", the competition for training that i'm arranging from some weeks  :lol: .
i'll try to sumarize all details about it, be carefully.

[b]1-[/b] The contest is made of some rounds(I think 3 max).

[b]2-[/b] Each round is made of only 3 problems, and i will give points 0 to 7: 6 for all correct solutions, 7 for [i]only[/i] elegants and correct solutions (similar to ML contest).

[b]3-[/b] It is tought for high-school student, and also for undergraduate or higher student, but you cannot use non-elementary-theory, otherwise the solution is invalid.

[b]4-[/b] All begins from italian-oliforum and the classification is unique.

[b]5-[/b] The first round starts on 12:00 of 20 september and it ends on 24:00 of 21 seprember, so saturday and sunday.
        [b]NB.[/b] All times (date and hours) are according to [i]ROME MERIDIAN (+2GTM), [/i]please attention!

[b]6-[/b] Enrollment is not necessary, it sufficies to send me solution.

[b]7-[/b] I’ll take in consideration time of arriving solutions [i]only[/i] for the first place [i]if [/i] there are more 21/21 points.

[b]8-[/b] [i]Way of sending solutions[/i].
-          You need to send solutions to both email address: pao_leo88@hotmail.it and leonettipaolo@gmail.com in a [i]unique pdf file[/i] (so, with attachment)  of a right size;
-          The pdf file must be written with [i]Latex [/i](or a way that I can understand);
-          You [i]rename the pdf file [/i]as your nickname (the same of yours in ML);
-          You need to write nothing in the email;
-          Try to be [i]clear[/i] in solutions! I’ll estimate it!

 {If [i](and only if)[/i] you are not able to create a pdf in latex model, you can send me a email in .doc format with solutions, as latex or in a way that i (and all correctors) can understand}.

Later, I'll make a list below this topic where i'll write all names of who sent me solutions in right way.

For all doubts and questions, ask me  :wink:

Thanks for the attentions
And good luck

[i]Paolo Leonetti[/i]
[url=http://olimpiadi.dm.unipi.it/oliForum/viewtopic.php?p=96490#96490]Italian version[/url]

[b]OLIFORUM CONTEST - ROUND 1[/b]

[b]Question 1[/b]
Prove that in the set $ S\equal{}\{2008,2009,. . .,4200\}$ there are $ 5^3$ elements such that any three of them are not in arithmetic progression.

[b]Question 2[/b]
Let $ ABCD$ be a cyclic quadrilateral with $ AB>CD$ and $ BC>AD$. Take points $ X$ and $ Y$ on the sides $ AB$ and $ BC$, respectively, so that $ AX\equal{}CD$ and $ AD\equal{}CY$. Let $ M$ be the midpoint of  $ XY$. Prove that $ AMC$ is a right angle.

[b]Question 3[/b]
Let $ a,b,c$ be three pairwise distinct real numbers such that $ a\plus{}b\plus{}c\equal{}6\equal{}ab\plus{}bc\plus{}ca\minus{}3$. Prove that $ 0<abc<4$.

[b]NB.[/b] [i]In problem 1, we'll not accept merely numerical solutions..[/i]

Good work  :wink:

hey boyypa wen is the last date or time for sending it!

from now you have still about 28 hours! so, it is the midnight (24:00) of tomorrow!

[u]NB in problem 3 you [i]cannot use[/i] derivatives or calculos![/u]

[i]Contestants who sent me solutions:[/i]

A Cave
sumita
Madhukar_Daftary
NickNafplio
turcas_c
mod_2
Rigel
dima_ukraine
kamyar1991
giove
bestiedda
Akashnil
fede90
String
Alex89
Broniran
julio14
silouan
quicktimeplayers
nicelbole
tomci0
eucla
mhdi
Algebert

[quote="bboypa"][quote="NickNafplio"]In problem 3, we have a,b,c not equal to zero, right?[/quote]

we have that a,b,cmust be all distinct..if they are \neq 0 may be a consequence  :wink:[/quote]

Ok, I think I found a solution, but I cannot write it in microsoft word (I dont know how to write some symbols), someone told me that I can send the solution to you by private message here, can you accept it by pm?

[quote="bboypa"]try to write in doc. as latex model (with use of dollars for your forumlas) and send it to my email, (it is the same that you're writing in a pm); sumita sent me his solution in notepad, (always as latex), it goes right too. :wink:[/quote]

Now I sent you a solution with microsoft word. Check your e-mail and tell my if my solution is right please.

If I solve the other 2 problems, I will send you the solutions in doc as latex

in problem one i prove more strong than the purpose of the question. more than 130

[quote="sumita"]i have a solutoin with more that 220.[/quote]

now, i'm too..  :wink:

[quote="bboypa"][quote="mhdi"]When will solutions be posted?
Will we receive our answers correction?[/quote]

@mhdi&all others: thanks for partecipating!
i hope the problems were good for you..
however in this days i'll post problems also on olimpic section (i'll put a link here)
and during this week i (and other correctors) try to get the final graduation for the first round :wink:
about solutions, if in the link there are no exsaustive, then i'll post the easiest and shortest ones..

still, thanks to all  :lol:
cheers
paolo[/quote]

Is there any link with the solutions?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=227401]First problem[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1260910#1260910]Second problem[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1260908#1260908]Third problem[/url]

however, when i'll post graduation, i'll attach also ufficial solution

[quote="Kamyar1991"]hi bboypa when the second round will be start?  what is the graduation??? :maybe:  totally this work is very good so i hope this is be continued and have prize....[/quote]

second round will be after two weeks from now i think..
however i'll contact you by email and i'll post dates on this post.
the graduation will be max tomorrow,
thanks of all..

[quote="Kamyar1991"]what is the graduation???[/quote]
I think he means gradation (results i.e. giving points)

sorry my english is not very good  :maybe:

here there are the results: there are only contestant whit [i]score > 7 points[/i]:

[b][size=150]I-  giove  -  (Oliforum) FULL SCORE -21 points

II-  Akashnil  - (Mathlinks) -21 points

III-  silouan  -(Mathlinks) - 21 points

IV-  tomci0  - (Mathlinks) - 21 points[/size][/b]

[b]V-(ex-aequo) - sumita -(Mathlinks : 7/6/7) -20 points[/b]

                      [b] julio14- (Oliforum: 7/7/6) - 20 points[/b]

[b]VII- Kaymar1991 - (Mathlinks: 6/7/6) - 19 points[/b]

[b]VIII-A Cave - (Yahoo: 7/7/2) - 16 points[/b]

[b]IX- (ex-aequo)- Rigel [/b]- (Oliforum: 6/6/1) -13 points

[b]mhdi -[/b] (Mathlinks: 0/6/7)- 13 points

[b]algebert [/b]- (Oliforum: 0/7/6)- 13 points

[b]XII- nicelbole [/b]- (Oliforum: 7/0/5) - 12 points

[b]XIII- turcas(linker)[/b] - (Mathlinks: 0/7/3) - 10 points

[b]XIV- dima_ukraine[/b] - (Mathlinks: 0/7/2) - 9 points

[i][u]Thanks for having partecipated to :[/u]

NickNafplio(Mathlinks)

fede90(Oliforum)

String(Oliforum)

Alex89(Oliforum)

Broniran(Mathlinks)

Eucla(Oliforum)

Bestiedda(Olifrum)

Quicktimeplayers(Oliforum)

Madhukar_daftary(Yahoo)

mod_2(Oliforum)[/i]

*[i]Honorable mention[/i] goes to [i]A Cave [/i], who sent all solutions in less of 4 hours (and invalidating half of third exercise only for a misunderstood with regulament..)

[i]NB to all others contestant i'll send an email with his personal score..![/i]

Best regard
Paolo & other correctors

The second round is for everyone or only for those who solved more than one of the three problems?

[quote="bboypa"]NickNafplio your solution to the third problem is perfect, one of the best! 7points :lol: !

however, no, everyone can partecipate to every round.. :wink:

next round will be saturday 4 october at the same hours!
cheers to all[/quote]

Ok, thanks

here there are solutions of Tomci0, a clear and elegant example of solutions  :wink:


modular arithmetic
number theory
number theory proposed

prove that

$ (\forall x.y.z \in \mathbb{Z}); x^3\plus{}y^3\plus{}z^3 \equiv 0[3]$=====> $ x\plus{}y\plus{}z \equiv 0[3]$

Consider  $ x^3 \minus{}x \plus{} y^3 \minus{}y \plus{} z^3 \minus{} z$. For any integer $ n$, $ n^3 \minus{}n \equal{}n(n\plus{}1)(n\minus{}1)$ is divible by $ 3!\equal{} 6$.
So infact, you can replace your questiob by $ 6$ and iff for the implies.

You can use $ a^3\plus{}b^3\plus{}c^3\minus{}3abc\equal{}(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)$

Or, you can simply use the fact that $ x^3\equiv x \pmod 3$, from Fermat's little theorem, so in fact if you replace $ 3$ by a prime $ p$ in both the exponent and the mod, it remains true (and becomes if and only if).


AMC USA(J)MO USAMO MATHCOUNTS LaTeX probability articles

Bob plays a game. Every round, he keeps on flipping a fair coin until he gets tails. If he plays 2008 rounds, how many times in total is he expected to flip the coin?

My Answer:
[hide]4016[/hide]

My Solution: (Don't give up too quickly!  :o ) If possible, could you rate my solution from 1 to 10? Thanks
[hide]
First, let's find out how many coins Bob is expected to flip in one tound. There is a $ \frac {1}{2}$ chance of flipping a tail on the first flip, thus flipping $ 1$ time. There is a $ \frac {1}{4}$ chance of flipping a head then a tail, flipping the coin $ 2$ times. For each one more flip, there is a half chance of getting that. To find the expected number of flips per round, we get: $ \frac {1}{2} \cdot 1 \plus{} \frac {1}{4} \cdot 2 \plus{} \frac {1}{8} \cdot 3 \plus{} \dots$ or $ \frac {1}{2} \plus{} \frac {2}{4} \plus{} \frac {3}{8} \plus{} \dots$ Let this expression equal $ x$.

To solve this expression, we can have a seperate expression: $ \frac {1}{2} \plus{} \frac {1}{4} \plus{} \frac {1}{8} \plus{} \dots$.
Solving this one is much easier, let $ y$ equal the expression. $ y \equal{} \frac {1}{2} \plus{} \frac {1}{4} \plus{} \frac {1}{8} \plus{} \dots$ ==> $ 2y \equal{} 1 \plus{} \frac {1}{2} \plus{} \frac {1}{4} \plus{} \frac {1}{8} \plus{} \dots$ ==> $ 2y \equal{} 1 \plus{} y$ ==> $ y \equal{} 1$.

We can subtract the two equations that equal x and y...:
$ x \equal{} \frac {1}{2} \plus{} \frac {2}{4} \plus{} \frac {3}{8} \plus{} \frac {4}{16} \plus{} \dots$
$ y \equal{} \frac {1}{2} \plus{} \frac {1}{4} \plus{} \frac {1}{8} \plus{} \frac {1}{16} \plus{} \dots$
...to get this. Also, I replaced y with 1:
$ x \minus{} 1 \equal{} \frac {1}{4} \plus{} \frac {2}{8} \plus{} \frac {3}{16} \plus{} \dots$
Now we multiply both sides by 2...
$ 2(x \minus{} 1) \equal{} \frac {1}{2} \plus{} \frac {2}{4} \plus{} \frac {3}{8} \plus{} \dots$
The expression on the right side of the equation above is just $ x$! :) Therefore,
$ 2(x \minus{} 1) \equal{} x$
$ 2x \minus{} 2 \equal{} x$
$ x \equal{} 2$
So Bob flips an average of $ 2$ flips per round. In $ 2008$ rounds, he flips an average of $ 2008 \cdot 2 \equal{} \boxed{4016}$ flips.
[/hide]

[hide="Click here to win a free copy of Super Smash Bros. Brawl!!!!!!!!!"](jk)There's a 1/2 expected chance of getting tails and 1/2 for heads. That means 2 times to flip in each round. 2 times 2008=4016.  :)  [/hide]

Um...well...

Another Approach:Get a T=1/2 chance=1004 times=+1004
HT=1/4=502=+502*2=2008
HHT=1/8=251=+251*3=2761
HHHT=+125.5*4=3263
HHHHT=+62.75*5=3576.75...
So I guess it would eventually be 4016...but I don't want to keep going...

And toadoncart, I think that approach is incorrect, because if the person got a tails, it would only be 1 flip.

So another way (I don't know if this is correct though):
1/2 chance of getting tails--1*1004=1004
1/2 chance of getting heads--expected to get tails--2*1004=2008??? I'm iffy on this step
So could it maybe be 3012? I dunno I still think it's 4016.

I posted my solution above. Rating my solution from 1 to 10 would be helpful, thanks.

Good solution! That's essentially what I did, just without as much rigor...

Sorry, my counting/probability sucks. :P

Bob, I really liked your solution. Good problem, too. I give you a 10.  :thumbup:

very nice solution, BOB. I give you a 10! :D

USAMO Grader:
Hmm... you made trivial progress on this. However, your solution still has many holes and flaws, for example you must ensure that your original series converges in the first place. I give this a 2.

Me:
Wow... 10! Wait, not even 10. 100! :wow:

[quote="Wickedestjr"]I do not know if this is correct or not, but it is just an expected value problem. Lets say that you have a dollar and for every flip that is heads his amount of money doubles. It turns out that the expected amount of money that you'd win is infinity, because there is a one half chance of getting two dollars, so that is $ \$$1, then there is a one fourth chance that you'd get four dollars, so that would be $ \$$1, or one fourth of 4, and if you continue this it is always going to be a one, and

1 + 1 + 1 + 1 + 1 + 1 + ... = infinity

So in conclusion the expected value is infinity, but you'd never actually get infinity, it is just a wierd, but true solution. So if you can end up flipping the coin and infinite number of times, so I think that the expected value would be infinity.

If this were on something from Mathcounts, then it is probably wrong, because I don't think that infinity has ever been an answer.

How would you rate this solution from 1 to 10?

[color=darkred][size=75]Fixed $ \text{\LaTeX}$.
-nebula42[/size][/color][/quote]
You are correct that the expected value of money obtained in this way would be infinite; however, finding the expected value for this modified problem does not tell you much about the expected value for the original problem.

For example, the sum
$ \frac12 + \frac14 + \frac18 + \dots = 1$.
Thus, you might expect that the sum
$ 2^{\frac12} + 2^{\frac14} + 2^{\frac18} + \dots$
might be equal to $ 2^1 = 2$. However, it actually does not converge (or it is 'infinite'), because
$ 2^{\frac12} + 2^{\frac14} + 2^{\frac18} + \dots > 1 + 1 + 1 + \dots$
which does not converge.

[quote="dynamo729"]I think the theoretical probability of how may times he would flip a coin would be 4016 [b][u]or[/u][/b] 4015. After 4014 rounds, The next coin could be a tails....[/quote]
[quote="Wickedestjr"]Yeah, there might be alot of solutions to this problem, but its a probability problem so you can never actually be positive about a solution.[/quote]
Note that the question does not ask if Bob will flip [i]exactly[/i] 4016 coins; instead it asks for the [i]expected value[/i] of the number of flips he needs, which is an exact value. If you haven't heard of expected value before, you might want to take a look at [url=http://www.artofproblemsolving.com/Wiki/index.php/Expected_value]this article[/url] or [url=http://en.wikipedia.org/wiki/Expected_value]this one[/url] (skipping over the calculus when necessary).

Wouldn't

 2^(1/2) + 2^(1/4) + 2^(1/8)... actually be equal to 2 rad 2? Just wondering.

No it wouldn't; it would be unbounded. Here's why.

Each of the terms is strictly larger than 1, right? (Any root of 2, no matter how big, will be greater than 1). So we can compare that sum with this one:
\[ 1 \plus{} 1 \plus{} 1 \plus{} 1 \plus{} \dots
\]
The above sum is unbounded; therefore, the one in question is too (since it is definitely "larger" than unbounded, which is still unbounded).

i rate your soluton 10! :coolspeak:

But intro c and p and aops 1 both have a lot of and your solution is similar...  :P jk..

I rate your solution a 9.9999999999...... which equals 10;  :wink:


Let $ ABCD$ bicentric quadrilateral, intersection of diagonals $ AC$ and $ BD$ is $ E$, center of incircle is $ I$, center of circumcircle is $ O$. Prove that points $ E, I, O$ are collinear.

Let $ X \equiv AD \cap BC, Y \equiv AB \cap DC$ and $ M,N,L,P$ are the tangency points of $ (I)$ with $ AB,$ $BC,$ $CD,$ $DA.$ By Newton's theorem for the tangential quadrilateral $ ABCD,$ we have $ E \equiv ML \cap NP,$ which implies that $ XY$ is the polar of $ E$ WRT $ (I)$ $ \Longrightarrow$ $ EI \perp XY.$ But since the quadrilateral $ ABCD$ is also cyclic, then $ XY$ is the polar of its diagonal intersection $ E$ WRT its circumcircle $ (O)$ $\Longrightarrow$ $EO \perp XY.$ Therefore, we deduce that $ E,I,O$ are collinear on a perpendicular to $ XY.$

Sorry, I don't understood polarity. why $ EI\perp XY$ and $ EO\perp XY$. Can you explain? thank you very much your interest.

It is a well known classical problem. For some elementary approaches, you can see :
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31693 - Proof ( the simplest I think ) of the collinearity as the proposed problem states.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=110887 - Proof of the perpenticularity mentioned by [b]Luis[/b]
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=139564 - Proof of a result of the polar theory.

Kostas Vittas.


.iam currently in INDIA doing by 12th standard ie my last year of schooling and next year i have to join a college for my undergraduate study's .i am very very deeply interested in MATHEMATICS and its my passion.i want to be a mathematician,though i haven’t even passed the REGIONAL MATHEMATICS OLYMPIAD that's conducted in my country.but the problem is i have no idea as to what i should be working so that i can become a mathematician.i have no clue on where i have to join for my under- graduation or what course i have to go for. but the thing is that i very much love the subject and i would practically do anything to achieve my goal.i jus have a worry tah whether it is too late for me to improve on my mathematical abilities as iam already sixteen.my main areas of interest include some amt of elementary calculus,elemantary sequeces and series,and i also like properties of numbers and fascinated by them. please help me and advice me on which books i should study and which fields should i develop my knowledge more on.

Please post it here:

 http://www.mathlinks.ro/index.php?f=362 [url][/url]

Hey---India Genius---I have almost the same problem as yours---I became really interested in this subject when I was 15---quite late---I was deeply fascinated by math since then---I always doubt whether I have a talent for math---
About reading stuffs I think maybe you can try "An Introduction to the Theory of Numbers "by G.H.Hardy---

@vijaymennon, hope that you are still visiting this post of yours. I also have a similar case like rather much worse. One thing I can advice (if you have real interest in it and really LOVE it ) you for sure is that it is never too late to start building up your foundation of mathematics. Although the amount of mathematics you have to learn will indeed increase and even you will find it sometimes very tedious. Whatever you study, just make sure that you understand it thoroughly and understand every theory behind it ( believe me though it seems daunting it will gradually become more and more easier as you become mathematically mature ). Just don't worry about the high-school mathematics you missed. If you ponder on it, you are bound to become depressed. Just leave where it is ans and focus on your current courses. You will become good at mathematics. Take my word for it.


Compute the numerical value of
$\left({1+\frac{2}{\sqrt[3]{9}+\sqrt[3]{3}+1}\right)}^{3}$

[hide="Hint"]
Put $\sqrt[3]{3}=x$. What can you do with that denominator? [/hide]

basically, your "algebraic urge" as my calc teacher says, is to rationalize...so once you do that, then it is trivial...


Παραθέτω το ακόλουθο πρόβλημα γεωμετρίας , το οποίο ελπίζω να το προσπαθήσετε
και να το λύσετε
. Δίνεται τρίγωνο $ABC$  και  το ύψος του $BD$ . Με διάμετρο  $BD$  γράφουμε κύκλο που τέμνει τις $AB$ , $BC$  στα  $K$  , $L$  αντίστοιχα .Έστω οι εφαπτομένες στα $K$  , $L$  στον κύκλο που τέμνονται στο  $M$ .Αποδείξτε ότι η $BM$  περνά από το μέσον της $AC$

Κανένας δεν ασχολήθηκε μ'αυτή??
Γράφω συνοπτικά τη λύση:
[b]H BM ειναι συμμετροδιάμεσος στο BKL (μια υπόδειξη:φέρουμε κάθετες από το Μ προς τις BC,KL,AB.).
Η KL είναι αντιπαράλληλη της AL,άρα η ΒΜ είναι διάμεσος στο ABC.[/b]
[b][u]Σημείωση:[/u][/b]Η άσκηση γενικεύεται για οποιαδήποτε Κ,L αρκεί το τετράπλευρο AKLC να είναι εγγράψιμο.

Ωραίος ...... ;)  . Η λύση μου είναι λίγο διαφορετική περιέχει αρμονικά τετράπλευρα αλλά πάνω στην ίδια ιδέα......

Ας δούμε και μία άλλη προσέγγιση.

Έστω Ε, το σημείο τομής των AB, DL και F, το σημείο τομής των BC, DK.

Θεωρούμε το εγγεγραμμένο τετράπλευρο BKDL, ως ειδική περίπτωση εγγεγραμμένου εξαγώνου του οποίου δύο ζεύγη κορυφών συμπίπτουν στις θέσεις των K, L.

Σύμφωνα τότε, με το θεώρημα του Pascal, τα σημεία E, M, F, είναι συνευθειακά.

Θα αποδείξουμε ότι EF//AC και ότι το M, είναι μέσον του EF.

Από την καθετότητα των ευθειών AB, DL και BC, DK, συμπεραίνουμε ότι το σημείο D, είναι ορθόκεντρο του τριγώνου BEF και επομένως έχουμε EF//AC.

Θεωρούμε ότι M', είναι το μέσον του EF και θα αποδείξουμε ότι συμπίπτει με το M.

Από το ορθογώνιο τρίγωνο KEF => M'K = M'E = M'F, (1).

Από το ορθογώνιο τρίγωνο LEF => M'L = M'E = M'F, (2).

Από (1), (2), =>  M'K = M'L => το σημείο M', είναι σημείο της μεσοκάθετης του KL και άρα συμπίπτει με το M ( ως σημείο και του EF ).

Από το τραπέζιο EACF, συμπεραίνουμε ότι η ευθεία BM, διέρχεται δια του μέσου του AC και η πρόταση έχει αποδειχθεί.

ΣΗΜΕΙΩΣΕΙΣ.

1) – Μία γενίκευση του προβλήματος είναι η εξής:

       Σημείο Κ, κινείται επί ευθείας (e) η οποία διέρχεται από την κορυφή Α, δοσμένης γωνίας <xAz. Ο κύκλος με κέντρο το Κ και ακτίνα KA, τέμνει τις πλευρές Ax, Az, στα σημεία B, C, αντιστοίχως. Ζητείται να ευρεθεί ο γεωμετρικός τόπος του σημείου Μ, τομής των εφαπτομένων του κύκλου (κ), στα σημεία B, C.

       Δια του Μ φέρνουμε κάθετη ευθεία επί την (e), η οποία τέμνει τις πλευρές Ax, Az, της δοσμένης γωνίας στα σημεία B, C, αντιστοίχως. Αποδεικνύεται ότι το Μ είναι μέσον του BC και ότι ο ζητούμενος γεωμετρικός τόπος είναι η ευθεία ΑΜ.

2) – Έχω υπόψη μου μία ενδιαφέρουσα επέκταση του προβλήματος ( συνέχεια από την εκφώνηση όπως έχει δοθεί ):

       Έστω E, F, τα σημεία τομής της πλευράς AC, από τις εφαπτόμενες στα K, L, αντιστοίχως. Εάν P, είναι το σημείο τομής των KC, LA και Q, το σημείο τομής των KF, LE, να αποδειχθεί ότι η ευθεία PQ, διέρχεται δια του μέσου Ν του AC.

     Κώστας Βήττας.

Ψιλογνωστη..... :wink: ,αλλα πολυ διδακτικη....
Η λυση μου ειναι ιδια με αυτη του Ραπανου..(μου την ειχε δειξει αυτη την ασκηση στην ΙΜΟ και την ελυσα με αναλυτικη +4   !!! σελιδες πραξεις....Χασιμο χρονου....)
Vittasko,ωραιa λυση :thumbup:
Α,Nick,αυτη δεν ειναι η ασκηση που ειχες βαλει σtον....Francεsko???....
Εχω την αμυδρη εντυπωση,(δεν ειμαι σιγουρος...)οτι η ιδια ασκηση η καποια βοηθητικη  προταση,υπαρχει στην γεωμετρια των Ισουιτων.....

ΡΑΛΛΗΣ :D

Τελικά Rakar, είναι γνωστή και η επέκταση (2) που αναφέρω; Δεν την έχω βρει πουθενά μέχρι τώρα και με ενδιαφέρει αν κάποιος έχει μία απόδειξή της.

     Έχω και εγώ μία υπόψη μου, σχετικά εύκολη, που βασίζεται στη θεωρία περί διπλού ( ή αναρμονικού ) λόγου.

     Θα την παρουσιάσω ευχαρίστως, μετά την παρέλευση εύλογου χρόνου, εκτός αν τύχει και δημοσιευτεί από κάποιον άλλον νωρίτερα.

     Κώστας Βήττας

Geia sou Ralli!!!Ti kaneis???Oxi den einai etsi ta pragmata..Einai peripou etsi!! :wink:
Oute se esena,oute ston Francesco eipa ayti tin askisi!!
Auto pou eipa kai stous dyo sas itan i  apodeiksi tou kentrikou limatos pou xrisimopoiisa stin apodeiksi kai pou stin ousia kseutilizei tin askisi se dyo grammes!!! :rotfl:  :P
Bye


Let a,b,c are positive reals.Prove that: ab.$ \large\sqrt{2(a^{2}+b^{2})}$+cb.$ \large\sqrt{2(b^{2}+c^{2})}$+ac.$ \large\sqrt{2(a^{2}+c^{2})}$ <a.a.a+b.b.b+c.c.c

[quote="duy_nguyen"]Let $ a,b,c$ are positive reals.Prove that: $ ab\sqrt{2(a^{2}+b^{2})}+cb\sqrt{2(b^{2}+c^{2})}+ac\large\sqrt{2(a^{2}+c^{2})}\leq a^{3}+b^{3}+c^{3}.$[/quote]
Maybe
 $ ab\sqrt{2(a^{2}+b^{2})}+cb\sqrt{2(b^{2}+c^{2})}+ac\large\sqrt{2(a^{2}+c^{2})}\leq a^{3}+b^{3}+c^{3}+3abc$ $ ?$ :wink:
If so, it's Anh Cuong's inequality.


The $25$ member states of the European Union set up a committee with the following rules:
1) the committee should meet daily;
2) at each meeting, at least one member should be represented;
3) at any two different meetings, a different set of member states should be represented;
4) at $n^{th}$ meeting, for every $k<n$, the set of states represented should include at least one state that was represented at the $k^{th}$ meeting.

For how many days can the committee have its meetings?

looks like we're interested in the maximal number of subsets of a set with $n$ elements (here $n=25$) s.t. each two have non-void intersection. I think this number is $2^{n-1}$. First of all, $2^{n-1}$ works, since we can take all the sets having containing a certain element. On the other hand, this number is maximal, because if we have a set, we exclude its complement, so the number of sets we don't have is at least as large as the number of sets we have, meaning that our family of chosen sets has at most $\frac{\mathcal P}2=2^{n-1}$ elements ($\mathcal P$ is the set of subsets of our set).

Am I wrong?

You're absolutely right  :cool: The answer is $2^{24}$

Yes, You Are Right.
Could I Ask A Similar Problem? (Maybe obvious, but I can't prove it)
Determine the max value of integer n such that There Exist n SubSet of Set S that none of them is subset of another.


8. What is the probabillity that, in three single draws without replacement, two red marbles and one blue marble will be drawn in that order from a bag containing six red marbles, eight yellow marbles, and seven blue marbles? Express your answer as a common fraction.

Wouldn't this be $ 2(\frac{6}{21}\cdot \frac{5}{20})\cdot \frac{7}{19}\equal{}\frac1{19}$?
[hide="Answer"]$ \frac1{38}$[/hide]

[quote="shentang"]Yeah, you don't need the X2. This is because you have 6 red marbles in the beginning then after you have to pick from 5 red marbles, then you have from 7 marbles to choose a blue. Jeez that wasn't clear.[/quote]
Oh...hehe..Okay. Thanks.


how many path components does the group $ GL(2,\mathbb{R})$ have ?  thanks

It has two; note $ \mathbb{R}_{\ne 0} \equal{} GL(2) / SO(2)$ has two connected components and $ SO(2) \equal{} S^1$ is path connected.


[quote="roadnottaken"]I noticed that eight of the solutions for test a were from students (the exceptions being the two proofs), while none of the test be solutions were.  What changed?  (maybe this is me being disappointed at only having my name on the awesomemath site once :yankchain: )[/quote]

I'm pretty sure we pre-prepared the solution file this time around. Student solutions weren't included simply because we hadn't gotten them yet.


algebra
polynomial
modular arithmetic
number theory unsolved
number theory

We have two polynomials with integer coefficients:
$W(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n$
$Q(x)=b_0+b_1x+b_2x^2+\cdots+b_nx^n$
They have following property:
$a_0\equiv{b_0}\pmod{p}$
$a_1\equiv{b_1}\pmod{p}$
$a_2\equiv{b_2}\pmod{p}$
$\quad \cdots$
$a_n\equiv{b_n}\pmod{p}$

$p$ is a prime number(I don't think it matters in this case, but let the p be a prime

Prove that polynomials: $W(x)W(x)$ and $Q(x)Q(x)$ have the same property as $W(x)$ and $Q(x)$

"have the same property" is very ambigious. What exactly do you mean?

I think he means the congruences, but that's very trivial...


calculus
integration
LaTeX
logarithms
calculus computations

One possible approach can be multiply and divide by $x$ and then integrate by parts - from this you will get a fraction that is much easier to write as partial fractions.

Carcul

[quote="benorin"][quote="vector1"]$\int\frac{5x}{(x^{3}+1)^{2}}\, dx$

I'll give my version of the answer soon.

Finally got the Latex to work![/quote]

Factor first,
\[\frac{5x}{(x^{3}+1)^{2}}=\frac{5x}{(x+1)^{2}(x^{2}-x+1)^{2}}\]
then use partial fraction decomposition
\[\frac{5x}{(x+1)^{2}(x^{2}-x+1)^{2}}=\frac{Ax+B}{x^{2}-x+1}+\frac{Cx+D}{(x^{2}-x+1)^{2}}+\frac{E}{x+1}+\frac{F}{(x+1)^{2}}\]
solving this gives $A=\frac5{9},C=0,D=\frac5{3},B=E=F=-\frac5{9}$, so the integral is
\[\int\frac{5x}{(x^{3}+1)^{2}}\, dx=\frac5{9}\int\left[ \frac{x-1}{x^{2}-x+1}+\frac{3}{(x^{2}-x+1)^{2}}-\frac{1}{x+1}-\frac{1}{(x+1)^{2}}\right] dx \]
[/quote]

Well, that expression is not much nicer, could you please simplify?

The final answer should be: $\frac{5}{18}\left[ \frac{6x^{2}}{x^{3}+1}+2\sqrt{3}\arctan{\frac{(-1+2x)}{\sqrt{3}}}+2\log{1+x}+\log{1-x+x^{2}}\right]$

[quote="Carcul"][quote="Color"]The final answer should be:(...)[/quote]

Shouldn't it be $...-2 \log\mid1+x\mid+...$?[/quote]

Yes, sorry, but I'm not sure about the abs. values.

\[\ldots-2\log(1+x)+\log(1-x+x^{2}) \]


[quote="amirhtlusa"][quote="perfect_radio"]Prove that two segments symmetrical wrt a line $d$ have the same length.

[b]Note[/b]: this problem is set in $3D$[/quote]
what is "wrt"?[/quote]

it means "with respect to"

[quote="amirhtlusa"]there would be one plane that passes through, both segments,
so woulnd't that be kind of obvious?[/quote]

no, not necesarily. it is in 3D, so all most all the time the two segments are actually not contained by a plane.

then consider this, there must be one plane that passes through the line. and ther must be at least one point that passes througt two symetrical points on the segments, that are equidistance from the line. thus have to be equidistance from the plane.


Prove that if $ a,b,c,d > 0$,then \[ a^4b\plus{}b^4c\plus{}c^4d\plus{}d^4a \ge abcd(a\plus{}b\plus{}c\plus{}d)\]

[quote="apollo"]Prove that if $ a,b,c,d > 0$,then
\[ a^4b \plus{} b^4c \plus{} c^4d \plus{} d^4a \ge abcd(a \plus{} b \plus{} c \plus{} d)
\]
[/quote]
 :lol: AM-GM:
$ a^4b\plus{}b^4c_c^4d\plus{}d^4a\equal{}\sum\frac{23a^4b\plus{}7b^4c\plus{}11c^4d\plus{}10d^4a}{51}\ge \sum \sqrt[51]{a^{102}b^{51}c^{51}d^{51}}\equal{}\sum a^2bcd$

Yes! that or:
$ \sum_{cyc}c \sum_{cyc}d \sum_{cyc}\frac{a^3}{cd}\geq \left(\sum_{cyc} a\right)^3$ by Holder


For $ 5\le n\le15$, find the sum of all the positive integers $ n$ such that $ n!\plus{}(n\plus{}1)!$ is divisible by 5.

trying $ n \equal{} 5$, we get $ 5! \plus{} 6! \equal{} 120 \plus{} 720 \equal{} 840$. We then realize that all the factorials starting with $ 5!$ end in a $ 0$ which automatically makes the number divisible by $ 5$, so we just sum up the numbers from $ 5$ to $ 15$.
$ 5 \plus{} 6 \plus{} 7 \plus{} 8 \plus{} 9 \plus{} 10 \plus{} 11 \plus{} 12 \plus{} 13 \plus{} 14 \plus{} 15 \equal{} \boxed{110}$

More algebraic way....

[hide]$(n)!+(n+1)!=n!+(n+1)n!=n!(n+2)$
Either $n!$ or $n+2$ has to be divisible by $5$.
The factorial of any number $>5$ will be divisible by five.
$5+6+7+.....+13+14+15=5+15(15-5+1)/2=20(11)/2=10(11)=110$
Thus the sum would be $\boxed{110}$[/hide]


Find real coefficient polynomial P(x) such that :
$ P(a+b+c) = 7P(a) - 5P(c) + 4P(b) $ with $ (a+b)(b+c)(c+a) = 2a^3 + b^3 - 2c^3 $

Ah, it is easier than I though.Let see...
Take [tex] a=-b=\sqrt[3]{2}c [/tex] and we have:
[tex] 6P(\frac{a}{\sqrt[3]{2}})=7P(a)+4P(-a) ,\forall a\in R [/tex]
Assume that [tex] P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_0 [/tex] where [tex] a_n \neq 0 [/tex].
And we have:
[tex] 6{a_n}(\frac{1}{\sqrt[3]{2}})^n=7a_n+(-1)^{n} \cdot 4 \cdot a_n [/tex].
Since [tex] a_n \neq 0 , n =3 [/tex].And then we have:
[tex] 6{a_{x}}(\frac{1}{\sqrt[3]{2}})^x=7{a_x}+(-1)^{x} \cdot 4 \cdot a_x;x \in \lbrace {0,1,2} \rbrace \Rightarrow a_x=0;x \in \lbrace {0,1,2}  \rbrace [/tex]
So [tex] P(x)=k \cdot x^3 ; k \neq 0[/tex]. Check again to be sure about it. :)

[quote="Anh Cuong"][tex] 6{a_n}(\frac{1}{\sqrt[3]{2}})^n=7a_n+(-1)^{n} \cdot 4 \cdot a_n [/tex].
Since [tex] a_n \neq 0 , n =3 [/tex][/quote]

I'm sorry I didn't understand this passage, can you explain me what did you do here? Thanks :P

[quote="nttu"]The solution of the jury is :
 Take $ a=3x ; b=2x ; c=x $[/quote]
Yes that is more beautiful but that is really unnatural, how can they know it, or they created the problem from that idea. ;)

Myth : Most of the french team did not solve IMO-2004/2  :(

Pierre.

[quote="Myth"]In the thread http://www.mathlinks.ro/Forum/viewtopic.php?p=136332#136332 you said that you know how to change age. So next year apply your secret method and participate in IMO. And I think French team will have at least one gold medal :)[/quote]

This would be a gold medal for him for sure. I guess even one of the higher point gold medals...  :)

[quote="Simo_the_Wolf"][quote="Anh Cuong"]We have:
[tex] 6 \cdot (\frac{1}{2})^{\frac{n}{3}}=7+(-1)^n \cdot 4 \in Z [/tex] then [tex] n=3k [/tex] , however if [tex] k\geq 2 [/tex]  then [tex] 6 \cdot (\frac{1}{2})^{\frac{n}{3}} \in [/tex] Q.
Hence $ k=1 $ and $ n=3 $. :)[/quote]

But why do you consider only a_n? Maybe there are also a_{n-1}, a_{n-2}, ecc..[/quote]
  It's easily to show the coeffient of $ x^2 , x $ and $ a_0 $ are 0:)


Let $ a,b,c >0$ and$ a\plus{}b\plus{}c\equal{}1$ .Prove that:
 $ \frac{1}{ab\plus{}2c^{2}\plus{}2c} \plus{}\frac{1}{bc\plus{}2a^{2}\plus{}2a} \plus{}\frac{1}{ca\plus{}2b^{2}\plus{}2b}  \geq  \frac{1}{ab\plus{}bc\plus{}ca}$

[quote="sukivel"]Let $ a,b,c > 0$ and$ a \plus{} b \plus{} c \equal{} 1$ .Prove that:
 $ \frac {1}{ab \plus{} 2c^{2} \plus{} 2c} \plus{} \frac {1}{bc \plus{} 2a^{2} \plus{} 2a} \plus{} \frac {1}{ca \plus{} 2b^{2} \plus{} 2b} \geq \frac {1}{ab \plus{} bc \plus{} ca}$[/quote]
$ \sum_{cyc}\frac {1}{bc \plus{} 2a^{2} \plus{} 2a}\geq\frac {1}{ab \plus{} bc \plus{} ca}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac {1}{bc \plus{} 2a^{2} \plus{} 2a}\minus{}\frac {1}{3(ab \plus{} bc \plus{} ca)}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{ab\plus{}ac\minus{}2a^2\plus{}2bc\minus{}2a^2}{bc \plus{} 2a^{2} \plus{} 2a}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{a(c\minus{}a)\minus{}a(a\minus{}b)\plus{}(a\plus{}b)(c\minus{}a)\minus{}(a\plus{}c)(a\minus{}b)}{bc \plus{} 2a^{2} \plus{} 2a}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac{(c\minus{}a)(2a\plus{}b)}{bc \plus{} 2a^{2} \plus{} 2a}\minus{}\frac{(a\minus{}b)(2a\plus{}c)}{bc \plus{} 2a^{2} \plus{} 2a}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac{(a\minus{}b)(2b\plus{}c)}{ac \plus{} 2b^{2} \plus{} 2b}\minus{}\frac{(a\minus{}b)(2a\plus{}c)}{bc \plus{} 2a^{2} \plus{} 2a}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a\minus{}b)^2(ab \plus{} 2c^{2} \plus{} 2c)(4ab\minus{}c^2\plus{}2c)\geq0,$ which true.

[quote="arqady"][quote="sukivel"]Let $ a,b,c > 0$ and$ a \plus{} b \plus{} c \equal{} 1$ .Prove that:
 $ \frac {1}{ab \plus{} 2c^{2} \plus{} 2c} \plus{} \frac {1}{bc \plus{} 2a^{2} \plus{} 2a} \plus{} \frac {1}{ca \plus{} 2b^{2} \plus{} 2b} \geq \frac {1}{ab \plus{} bc \plus{} ca}$[/quote]
$ \sum_{cyc}\frac {1}{bc \plus{} 2a^{2} \plus{} 2a}\geq\frac {1}{ab \plus{} bc \plus{} ca}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac {1}{bc \plus{} 2a^{2} \plus{} 2a} \minus{} \frac {1}{3(ab \plus{} bc \plus{} ca)}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {ab \plus{} ac \minus{} 2a^2 \plus{} 2bc \minus{} 2a^2}{bc \plus{} 2a^{2} \plus{} 2a}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {a(c \minus{} a) \minus{} a(a \minus{} b) \plus{} (a \plus{} b)(c \minus{} a) \minus{} (a \plus{} c)(a \minus{} b)}{bc \plus{} 2a^{2} \plus{} 2a}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac {(c \minus{} a)(2a \plus{} b)}{bc \plus{} 2a^{2} \plus{} 2a} \minus{} \frac {(a \minus{} b)(2a \plus{} c)}{bc \plus{} 2a^{2} \plus{} 2a}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac {(a \minus{} b)(2b \plus{} c)}{ac \plus{} 2b^{2} \plus{} 2b} \minus{} \frac {(a \minus{} b)(2a \plus{} c)}{bc \plus{} 2a^{2} \plus{} 2a}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a \minus{} b)^2(ab \plus{} 2c^{2} \plus{} 2c)(4ab \minus{} c^2 \plus{} 2c)\geq0,$ which true.[/quote]

Line 1 : I think
     $ \sum_{cyc}\frac {1}{bc \plus{} 2a^{2} \plus{} 2a}\geq\frac {1}{3(ab \plus{} bc \plus{} ca)}$

Line 2 :$ \sum_{cyc}\frac {ab \plus{} ac \minus{} 2a^2 \plus{} 2bc \minus{} 2a^2}{bc \plus{} 2a^{2} \plus{} 2a}\geq0$
Line 3 :$ \sum_{cyc}\left(\frac {(c \minus{} a)(2a \plus{} b)}{bc \plus{} 2a^{2} \plus{} 2a} \minus{} \frac {(a \minus{} b)(2a \plus{} c)}{bc \plus{} 2a^{2} \plus{} 2a}\right)\geq0$
Line 2 $ \Leftrightarrow$ Line 3   "Why ? "

[quote="Vicky and Louis Latin"]
Line 1 : I think
     $ \sum_{cyc}\frac {1}{bc \plus{} 2a^{2} \plus{} 2a}\geq\frac {1}{3(ab \plus{} bc \plus{} ca)}$[/quote]
No!
$ \frac {1}{ab \plus{} 2c^{2} \plus{} 2c} \plus{} \frac {1}{bc \plus{} 2a^{2} \plus{} 2a} \plus{} \frac {1}{ca \plus{} 2b^{2} \plus{} 2b} \geq \frac {1}{ab \plus{} bc \plus{} ca}\Leftrightarrow$
$ \Leftrightarrow\frac {1}{ab \plus{} 2c^{2} \plus{} 2c} \plus{} \frac {1}{bc \plus{} 2a^{2} \plus{} 2a} \plus{} \frac {1}{ca \plus{} 2b^{2} \plus{} 2b} \geq$
$ \geq\frac {1}{3(ab \plus{} bc \plus{} ca)}\plus{}\frac {1}{3(ab \plus{} bc \plus{} ca)}\plus{}\frac {1}{3(ab \plus{} bc \plus{} ca)}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {1}{bc \plus{} 2a^{2} \plus{} 2a}\geq\sum_{cyc}\frac {1}{3(ab \plus{} bc \plus{} ca)}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac {1}{bc \plus{} 2a^{2} \plus{} 2a} \minus{} \frac {1}{3(ab \plus{} bc \plus{} ca)}\right)\geq0.$

[quote="Vicky and Louis Latin"]
Line 2 :$ \sum_{cyc}\frac {ab \plus{} ac \minus{} 2a^2 \plus{} 2bc \minus{} 2a^2}{bc \plus{} 2a^{2} \plus{} 2a}\geq0$
Line 3 :$ \sum_{cyc}\left(\frac {(c \minus{} a)(2a \plus{} b)}{bc \plus{} 2a^{2} \plus{} 2a} \minus{} \frac {(a \minus{} b)(2a \plus{} c)}{bc \plus{} 2a^{2} \plus{} 2a}\right)\geq0$
Line 2 $ \Leftrightarrow$ Line 3   "Why ? "[/quote]
Because $ ab \plus{} ac \minus{} 2a^2 \plus{} 2bc \minus{} 2a^2\equal{}(c \minus{} a)(2a \plus{} b)\minus{}(a \minus{} b)(2a \plus{} c).$


geometry
analytic geometry
graphing lines
slope
advanced fields
advanced fields solved

In the cartezian system xoy we consider the point A(a,b) a<>0, b<>0 and two oblic lines d1 and d2. A variable line which goes through A intersects OX in M and OY in N. The paralel through M regarding d1, intersects OY in E, and the paralel through N regarding d2 intersects OX in F. Prove that EF goes through a fixed point.

source: zonal phase olympics

Simple analytic geometry should do the trick here:

Let d1 and d2 have the eqyations a1x+b1y+c1=0 and a2x+b2y+c2=0 respectively.

Let t be the variable slope of the variable line which passes through A (let's call this line d(t)). The eqn of d(t) is tx-y+b-ta=0, so M(t)(a-b/t,0) and N(t)(0,b-ta). The equation of the line which passes through M and is parallel to d1 is a1x-y+a1(b/t-a)=0, and the eqn of the line which passes through N and is parallel to d2 is a2x-y+b-ma=0 => E(t)(0,a1(b/t-a)), F(t)((ta-b)/a2,0), so the equation of the line E(t)F(t) is x+(-t/a1a2)y+(b-ta)/a2=0, so the point of coordinates (-b/a2,-a1a) is on all the lines.

Was this proposed for the 11'th grade?

yes, as well as all the other posts I made where the source is "zonal phase olympics"


Let $ m_0,m_1,m_2,...$ be a strictly increasing sequence of positive integers such that $ m_0=1$ and every $ m_i | m_{i+1}$ for all $ i\ge0$.Prove that every positive integer n can be represented uniquely in the form $ n=\sum_{i=0}^%Error. "infinitya" is a bad command.
_im_i$ where $ 0 \le a_i \le \frac{m_{i+1}}{m_i} - 1$ for all $ i\ge0$
and [b]$ m_i=0$ for all but infinitely many integers $ i$[/b]
can someone please tell me the exact meaning of the words in the bold letters in the end?
if this is some known theorem please provide the related link.

You have a first typo; the correct representation is $ n\equal{}\sum_{i\equal{}0}^{\infty} a_{i}m_{i}$. But this is an infinite summation, and we are not interested in convergence of series, or things like that - we deal with arithmetics. Therefore, with the exception of [b]finitely many integers[/b] $ i$ we must have $ a_i \equal{} 0$ (this was your second typo).

I think these are called generalized positional numeration systems; you may do searches on Google...

nope i am not sure but the book definitely only says $ m_i \equal{} 0$
but yes u were right abt the first typo


pigeonhole principle
geometry
3D geometry
combinatorics solved
combinatorics

Eight points lie in interior part of unit cub. Show that we can find two points whose distance is not greater than 1.

<i>Unknown source</i>

Namdung

It seems so obvious that it is hard to prove. It is difficult for me. Could you please give a hint?

hmm.. I think a solution may be based on PIGEONHOLE principle.. find a way to apply it!(try dividing the cube in 2^3 cubes) :D

don't know if this is a valid hint, but this is what seems to me that may work!

hope it does!

cheers! :D :D

Why are we sure that AB<=1? I think you're right, but it's not obvious, because AB isn't necessarily parallel to one of the sides of the cube.

A, B are two points of the parallelepiped of size (a, a, 1-a). So AB^2 <= a^2+a^2+(1-a)^2 = 1 + a(3a-2) < 1.

Namdung

[quote="Namdung"]The sln is very nice:

Let's consider 8 small cubs located at 8 corner with side a.(a<=1/2). When a=1/2 this cubs contains all 8 points, so we can find the smallest a with the propery that these 8 cubs contains all 8 points. If some cub contains more then 1 points then the result follows. Asumme that each cub contains exactly 1 point. By the contruction, there is at least one point (among chosen!), say A, lies on the inner bound of one small cub. We choose the another cub whose face is opposite the face with chosen point and let B be the chosen point lies in this cub, then we have AB <=1.

Sorry for my poor english. Draw the picture and you can understand easier.

Namdung[/quote]

I'm really lost  :(  Are the mini-cubes centered at the vertices of the big cube? Why is AB a parallelipiped of lengths (a, a, 1-a)? I'm totally lost  :?

I'm sorry for my late answer, I was out of city for some days. The small cub has one vertice in vertice of big cub. AB lie in parallelepiped of sixe a, a, 1-a due their choice.

Namdung


limit
logarithms
algebra proposed
algebra

The sequence $ (a_n)_{n \equal{} 1}^{\infty}$ satisfies the conditions $ a_1 > 0$ and $ a_{n \plus{} 1} \equal{} \frac {a_n}{1 \plus{} a_n^2}$ for $ n\geq 1$. Prove that

(a) For $ n>1$, $ a_n\leq \frac {1}{\sqrt {2n}}$
(b) There exists an $ n$ such that $ a_n > \frac {7}{10\sqrt {n}}$

(a) can be proved by induction. For $ n \equal{} 2,\ (a_1 \minus{} 1)^2\geq0$ leads to $ \frac {a_1}{1 \plus{} a_1^2}\leq\frac {1}{2} \equal{} \frac {1}{\sqrt {2\cdot2}}$.

Suppose $ a_n\leq\frac {1}{\sqrt {2n}}$ for some $ n > 1$.

$ a_n\leq\frac {1}{\sqrt {2n}}\ \Rightarrow\ \frac { \minus{} 1}{1 \plus{} a_n^2}\leq\frac { \minus{} 2n}{2n \plus{} 1}\ \Rightarrow\ \frac { \minus{} a_n^3}{1 \plus{} a_n^2}\leq\frac { \minus{} 1}{(2n \plus{} 1)\sqrt {2n}}$

We can write $ a_{n \plus{} 1}$ as $ a_n \minus{} \frac {a_n^3}{1 \plus{} a_n^2}$. Hence

$ a_{n \plus{} 1}\ \leq\ \frac {1}{\sqrt {2n}} \minus{} \frac {1}{(2n \plus{} 1)\sqrt {2n}}\ \equal{} \ \frac {2n}{(2n \plus{} 1)\sqrt {2n}}$

Now
$ \color{white}.\quad\quad\ \color{black}4n(n \plus{} 1) < (2n \plus{} 1)^2$

$ \Rightarrow\ \frac {4n}{(2n \plus{} 1)^2} < \frac {1}{n \plus{} 1}$

$ \Rightarrow\ \frac {(2n)^2}{(2n \plus{} 1)^2} < \frac {n}{n \plus{} 1}$

$ \Rightarrow\ \frac {2n}{2n \plus{} 1} < \sqrt {\frac {n}{n \plus{} 1}} \equal{} \sqrt {\frac {2n}{2(n \plus{} 1)}}$

Therefore $ a_{n \plus{} 1} < \frac {1}{\sqrt {2(n \plus{} 1)}}$, completing the inductive step of the proof.

I had a go at (b) and thought I had a solution, but it was wrong. :maybe:

Set $ b_n \equal{} \frac{1}{a_n}$.  Then the $ b_n$ obey the recurrence
\[ b_{n\plus{}1} \equal{} b_n \plus{} \frac{1}{b_n} \, .
\]
Squaring gives
\[ b_{n\plus{}1}^2 \equal{} b_n^2 \plus{} 2 \plus{} \frac{1}{b_n^2} \, .
\]
It follows that $ b_n^2 \ge 2n$.  That settles part (a).

From the last displayed equation, we also have
\[ \lim_n \left( b_{n\plus{}1}^2 \minus{} b_n^2 \right)
   \equal{} \lim_n \left( 2 \plus{} \frac{1}{b_n^2} \right) \equal{} 2.
\]
From the Cesaro-Stolz theorem, it follows that
\[ \lim_n \frac{b_n^2}{n} \equal{} 2.
\]
That settles part (b).

[quote="Ravi B"]
From the Cesaro-Stolz theorem, it follows that
\[ \lim_n \frac {b_n^2}{n} \equal{} 2.
\]
That settles part (b).[/quote]

How do you apply the theorem (to which sequence)? And how does it lead to the conclusion

A special case of the Cesaro-Stolz theorem says that if $ c_n$ is a sequence such that $ \lim_n c_{n \plus{} 1} \minus{} c_n \equal{} L$, then $ \lim_n \frac {c_n}{n} \equal{} L$.  We apply the theorem to $ c_n \equal{} b_n^2$ and $ L \equal{} 2$.

If you don't like Cesaro-Stolz, here is another approach.  We already showed that $ b_n^2 \ge 2n$.  Plugging that back into our recurrence for $ b_n^2$, we get
\[ b_{n \plus{} 1}^2 \le b_n^2 \plus{} 2 \plus{} \frac {1}{2n}.
\]
Now let's unwind that recurrence.  We find that $ b_n^2$ is not more than $ 2n$ plus some harmonic series plus a constant.  But the harmonic series grows only like $ \ln n$.

[quote="Ravi B"]
Now let's unwind that recurrence.  We find that $ b_n^2$ is not more than $ 2n$ plus some harmonic series plus a constant.  But the harmonic series grows only like $ \ln n$.[/quote]

Umm.... please explain this in more detail. I don't quite get it.  :(

Keep applying the recursive inequality.  We get
\begin{align*} b_n^2 & \le \left(2 + \frac {1}{2(n - 1)}\right) + b_{n - 1}^2 \\
& \le \left(2 + \frac {1}{2(n - 1)}\right) + \left(2 + \frac {1}{2(n - 2)} \right) + b_{n - 2}^2 \\
& \le \dots \\
& \le \left(2 + \frac {1}{2(n - 1)}\right) + \left(2 + \frac {1}{2(n - 2)} \right) + \dots + \left(2 + \frac {1}{2(1)} \right) + b_1^2. \end{align*}

If you like, here is another way.  For $ n \ge 50$, we get
\[ b_{n+1}^2 \le b_n^2 + 2.01.
\]
When we unwind that recurrence, we find that
\[ b_n^2 \le 2.01n + b_{50}^2.
\]
That's good enough to settle part (b).


Given $ x,y,z,a,b,c$≥0.Knowing that $ x^2\plus{}y^2\plus{}z^2$≤$ 2(xy\plus{}yz\plus{}zx)$,prove the ineq:
$ x(a\minus{}b)(a\minus{}c)\plus{}y(b\minus{}a)(b\minus{}c)\plus{}z(c\minus{}a)(c\minus{}b)$≥$ 0$

WLOG, we can assume that $ a \ge b \ge c$. Put a = c + u + v, b = c + v  c, u, v \ge 0, the given inequality equivalent to

$ xu(u \plus{} v) \minus{} yuv \plus{} z(u \plus{} v)v \ge 0 < \equal{} > xu^2 \plus{} zv^2 \plus{} (x \plus{} z \minus{} y)uv \ge 0$

But $ xu^2 \plus{} zv^2 \ge 2uv\sqrt {xz}$. So $ L.H.S \ge (x \plus{} z \plus{} 2\sqrt {xz} \minus{} y)uv \equal{} ((\sqrt {x} \plus{} \sqrt {z})^2 \minus{} y)uv \ge 0$

The last one is true due to the condition $ x^2 \plus{} y^2 \plus{} z^2 \le 2(xy \plus{} yz \plus{} zx)$ follows that $ \sqrt {x} \plus{} \sqrt {z} \ge \sqrt {y}$


function
symmetry
algebra unsolved
algebra

$f$ is a function defined on all reals in the interval $[0, 1]$ and satisfies $f(0) = 0, f(\frac{x}{3}) = \frac{f(x)}{2}, f(1 - x) = 1 - f(x)$. Find $f(\frac{18}{1991})$.

From $f(1-x)=1-f(x)\Leftrightarrow 2\cdot\frac{1}{2}-f(x)=f\left(2\cdot\frac{1}{2}-x\right)$, $y=f(x)$ is symmetry for point $(\frac{1}{2},\frac{1}{2}).$


$a$ is an integer such that $n= 2+2\sqrt{28a^{2}+1}$ is also an integer. Prove that $n$ is a perfect square.

ha, from PSS, right?? :D [hide="hint"]
solve $n$ in terms of $a$, then do some analysis.
[/hide]

isnt $n$ already given to you in terms of $a$

[quote="gotztahbeazn"]either way, i can make some progress on it, but i'm not sure how to finish it off

[hide]
we know that $28a^{2}+1 = r^{2}$ for some integer $r$

hence:

$28a^{2}= (r-1)(r+1)$ and since the LHS is even we know that $r$ is odd so lets say $r = 2b+1$ for some integer $b$

then we have:

$28a^{2}= (2b)(2b+2) \implies 7a^{2}= b(b+1)$

prior to this though, we can substitute $28a^{2}$ with $(2b)(2b+2)$ to get:

$n = 2+2 \sqrt{4b(b+1)+1}= 4(b+1)$

hence it suffices to show that $b+1$ is a perfect square

this however i wasnt able to do... after thinking about it for a bit hmm any help?

[/hide][/quote]

I'm bring this post up again because I want to know how the problem finished  :D

Can anyone finish it please?

[quote="Gyan"][quote="computer_nuke"]because I want to know how the problem finished  :D
Can anyone finish it please?[/quote]
gotztahbeazn  practically had it. (See the post)
[hide]
[quote]$28a^{2}= (2b)(2b+2) \implies 7a^{2}= b(b+1)$[/quote]
Since,   $b$ and $b+1$  are coprime  and both  can not be divisibley by $7$
we have two choices:
$b = 7k^{2}$  (and $b+1$ is a perfect square)
OR
$b$ is a perfect square (say $m^{2}$ )  and
$b+1 = 7(\dots)^{2}$
or $m^{2}+1 \equiv 0 \bmod 7$    but thats [b]not possible  [/b]
[hide](Squares  mod 7 are  (1,2,4) and never 6  (or -1))  [/hide]
So only choice we have is $b+1$ is a perfect square :)
And of course:
[quote]$n = 2+2 \sqrt{4b(b+1)+1}= 4(b+1)$

hence it suffices to show that $b+1$ is a perfect square[/quote]
By the way .. it is interesting (but not that hard) to find sequence of values  ( eg b= 63 etc  $=7 \cdot 3^{3}$  so that  b+1 is perfect square)[hide] Standard Pell equation is all you need to  be familiar with to do   all this kind of problems [/hide]

[/hide][/quote]

When you had $b = 7k^{2}$ or $b+1 = 7(\dots)^{2}$, why do you must have b = 7 times something squared? Why can't it be like b = 7k and then k times (b+1) be a perfect square? Or b+1 = 7n and then n times b be a perfect square?

[quote]When you had $b = 7k^{2}$  or$b+1 = 7(\dots)^{2}$ , why do you must have b = 7 times something squared? Why can't it be like b = 7k and then k times (b+1) be a perfect square? Or b+1 = 7n and then n times b be a perfect square?[/quote]
Because, $b$ and $b+1$ can have no common factors!. So if I ignore 7, any  prime factor which appears in $b$  has to come even number of times (because it can not appear in $b+1$  (and you know it comes even number of times in the product of $b(b+1)$ )

(This is a standard method, if product of two numbers is a perfect square, [b]and if the two numbers have no common factors[/b], then each number has to be perfect square)

[quote="brady00n"]Can you please clarify

[quote]Since,  and  are coprime and both can not be divisibley by
we have two choices:
 (and  is a perfect square)
OR
 is a perfect square (say  ) and

or  but thats not possible [/quote][/quote]

I think he just explained it to me lol  :blush:

He's saying that because b and b+1 are coprime, meaning that they obviously have no common factors (because (b+1)-b = 1), therefore when you divide out a 7 out of either b+1 or b, the rest of what's left (the rest of the prime factors) must be a perfect square since b and b+1 have no factors, then b/7 and b+1 or b and (b+1)/7 have no factors, so it must be a perfect square.


What is the positive difference between two integers whose sum is 30 and whose product is 221?

So we have these two equations:
$ x\plus{}y\equal{}30$
$ xy\equal{}221$

Let us substitute in this particular spot. Let us say that $ x\equal{}30\minus{}y$. So then we substitute that into $ xy$: $ y(30\minus{}y)\equal{}221$. Now we expand to $ \minus{}y^2\plus{}30y\equal{}221$. Let us get rid of the negative and move the 221 over: $ y^2\minus{}30y\plus{}221$. Then we factor that into $ (x\minus{}17)(x\minus{}13)$. Next we plug one x in (it doesn't matter which one.) So we get $ 17\plus{}y\equal{}30$. We can easily find out that y is 13. So the difference between 17 and 13 is $ \boxed{4}$.

Note that $ (x \minus{} y)^2 \equal{} x^2 \plus{} y^2 \minus{} 2xy$, and $ (x \plus{} y)^2 \equal{} x^2 \plus{} y^2 \plus{} 2xy$ Therefore, we can say that $ (x \minus{} y)^2 \equal{} (x \plus{} y)^2 \minus{} 4xy \equal{} 900 \minus{} 884 \equal{} 16\implies x \minus{} y \equal{} \boxed{4}$


Be

A=1kg
B=2kg
C=3kg

find acelererations for A,B and C
find T (traction)

I assume the strings are all vertical, unlike in the picture. Then, if T is the tension in the string,

$Aa_a = Ag - T$
$Ca_c = Cg - T$
$B(a_a + a_c)/2 = 2T - Bg$

(add first two, compare to third)

$(g - T/A) + (g - T/C) = 4T/B - 2g$
$4g = T(4/B + 1/A + 1/C)$
$T = \frac{4g}{(4/B + 1/A + 1/C)} \approx 12 N$

And

$a_a = g - T/A = -0.2g$
$a_c = g - T/C = 0.6g$
$a_b = -(a_a + a_c)/2 = -0.2g$


Find all continuous functions $f:R->R$ satisfying : $x+y \in Q \Rightarrow f(x)+f(y) \in Q$

For all rational r, f_r(x)=f(x)+f(r-x) is a continuous function the image of which contains only rational numbers, so it's constant. From here, by applying the result to f_{r_1},f_{r_2}, we obtain the fact that g_a(x)=f(x+a)-f(x) is constant and rational for all rational numbers a. Let f(0)=b,\ f(1)=a+b (pretty transparent notations, huh? :D). This means that f(x+1)-f(x)=f(1)-f(0)=a. On the other hand, f(x+\frac 1n)-f(x) is also constant, so a=f(x+1)-f(x)=n(f(x+\frac 1n)-f(x))\Rightarrow f(x+\frac 1n)-f(x)=\frac an$. From here we get f(x+\frac mn)-f(x)=a\frac mn, and from the density of \mathbb Q in \mathbb R we get f(x)=ax+b, with a,b\in\mathbb Q.

[TeX doesn't seem to work; I'm leaving the formulae as they are, so I can just put the dollar signs back when TeX starts working properly again]


Show that if A+B+C=180 then
$(\sqrt{3}-sinA)(\sqrt{3}-sinB)(\sqrt{3}-sinC) \ge sinAsinBsinC$

 :)  :)


Consider the equation $ A.a^x \plus{} B.b^x \equal{} A \plus{} B (1)$ where $ a,b \in R, a > 1, 0 < b < 1$.
Prove that when $ A,B$ change, the equation (1) hasn't more than 2 roots!

Please browse the forum.. The exact same question has already been answered!


induction
quadratics
algebra
number theory unsolved
number theory

1.Prove        that        (n+1)      is      either       itself        a       prime        or        a

   product      of        primes       for      every        positive         integer          n.

2.Does         the             equation         have         solution         in         nonnegative

   integers:       2006  =  $(x+y)^2$ +3x+y

[quote="Tiks"][quote="bkunchanapalli"][quote="ZetaX"]1. Seems at it shall be done by induction... just use the definition of a prime to see whether $n+1$ is one or not and continue by induction hypothesis if not.

2. $(13+31)^2+3 \cdot 13+31 = 2006$.[/quote]

     how could you find that x  and y are  13 and 31[/quote]
Here is a way to find that numbers(mybe it is same with Zetax's way)
$(x+y)^2+3x+y=2006$<=>$x^2+x(2y+3)+(y^2+y-2006)=0$this quadratic equation must have an integer solution =>
$D=(2y+3)^2-4(y^2+y-2006)=8y+4*2006+9=t^2$ but $y^2\leq 2006$=>$y\leq 44$=>
$t^2\leq 8*44+4*2006+9$=>$t\leq 91$ and$t^2\geq 4*2006+9$=> $t\geq 90$
You can easy see that $t=90$ don't give us any solution,but  when $t=91$=>$y=31$ and $x=13$ ;)[/quote]

could you clearly prove the first sum by any mathematical argument

[quote="Tiks"][quote="bkunchanapalli"]

could you clearly prove the first sum by any mathematical argument[/quote]
I am sorry,but I can't understand about what part of my solution is you question :?[/quote]

In the first sum i asked about to prove that n+1 is a prime or product of primes
and you wrote that we can do it my induction method
can you do it completely.i didnt understand how to do it,or in any other way of mathematical argument

[quote="Tiks"][quote="bkunchanapalli"]
In the first sum i asked about to prove that n+1 is a prime or product of primes
and you wrote that we can do it my induction method
can you do it completely.[/quote]
I havn't written that kind of sentense :lol:,you maybe mean Zetax's wrote :) ,but anyway I agree with him again,and if I have well understood your problem it is the basic theorem of Arifmethic,isn't it?[/quote]

can you prove that by induction or anything else .iam not finding how to do it;
and what is that basic theorem of arithmatic
zetax told to use the definition of prime but i didnt get it
can you prove it step wise

[quote="mathmanman"]Ok, well, we prove it via strong induction.
We want to show that $P_n$ is true $\forall n \geq n_0$.
We show that $P_{n_0}$ is true.
Induction Hypothesis : $P_k$ true $\forall k \in [[n_0, n]]$.
Conclusion : $P_k$ true $\forall k \in [[n_0, n+1]]$.
We have :
*The result is true for 2. ($2 \in \mathbb{P}$)
*We then assume that the result is true $\forall k \leq n$.
*Let's consider $n+1$ :
- If $n+1$ is prime, we're done.
- If $n+1$ isn't prime, then there exists a divisor $d$ of $n+1$ different from 1 and from $n+1$. So : $(n+1) = d \cdot k$, with $1 < d \leq n$, so $1 \leq k \leq n$.
From the induction hypothesis, $d$ and $k$ are products of prime numbers.
So $k \cdot d$ is a product of prime numbers. $\mathfrak{QED}$. :)[/quote]

thanks for the proof


a) Prove that the perpendicular bisectors of the sides of any regular polygon are concurrent (you can circumscribe a circle about any regular polygon).

b) Prove that the angle bisectors of the interior angles of any regular polygon meet at this [i]same[/i] point (you can inscribe a circle in any regular polygon).

:juggle:

[hide="Do not click here"]Why is this message too small? I told you not to click here.   :moose:   [/hide]

[hide="Solution for a"]

Knowing that any regular polygon can be inscribed in a circle, we see that we may divide the polygon into $n$ isocoles triangles by drawing lines from the center of the circle to the vertices.

Because each triangle is isocoles, we know that the perpendicular bisector from the base of any triangle goes through the third vertex of the triangle, which is also the center. The perpendicular bisectors are thus concurrent.

[/hide][/hide]

Can anyone prove these without having to rely on the truth of the other?  :wink:


So...anyone know what sorta things most people do for math research projects for science competitions like Siemens Westinghouse and Intel? That is, anyone have any examples of topics that people have used for these projects before? Thanks.

Fierytycoon

But what problem in or area of graph theory. Aren't these research projects supposed to delve into unknown or unexplored areas in math, and not simply go over well-known areas as textbooks do?

Fierytycoon

[quote="Fierytycoon"]Aren't these research projects supposed to delve into unknown or unexplored areas in math, and not simply go over well-known areas as textbooks do?

Fierytycoon[/quote]

Yep.

You can read a bit about this year's Intel finalists' projects here:

[url]http://www.sciserv.org/sts/62sts/finalists.asp[/url]

OK here is a new link about Intel ISEF info.

http://www.societyforscience.org/isef/document

thank you.
do you think projects regarding matters in elliptic curves cryptography could make good isef projects?

^if its new original research...you can't just regurgitate a textbook.

of course not.
i meant original research.


inequalities
geometry
LaTeX
3D geometry
pyramid
tetrahedron
trigonometry

let ABC a traingle with vertix a b c and area k    prove  that                                                                                27[a²+b²-c²]²[b²+c²-a²]²[a²+c²-b²]²  <=[4k]*6 :D  :?

$\text{\LaTeX}$ing: Let $\triangle ABC$ be a traingle with sides a, b, c, and area K. Prove that

$27 (b^2 + c^2 - a^2)^2 (c^2 + a^2 - b^2)^2 (a^2 + b^2 - c^2)^2 \le (4K)^6$

According to problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=66431]Paper Pyramid[/url] in the high school section, the volume V of a tetrahedron with all 4 faces formed by congruent triangles with the sides a, b, c is given by

$V^2 = \frac{1}{72} (b^2 + c^2 - a^2) (c^2 + a^2 - b^2) (a^2 + b^2 - c^2)$

The volume V is less than or equal to the volume $V_0$ of a regular tetrahedron with the faces of the same area K. Let e be the edge and $h = e \sqrt{\frac 2 3}$ the space altitude of the regular tetrahedron. The face area is then $K = \frac{e^2 \sqrt 3}{4}$ and volume $V_0 = \frac{hK}{3} = \frac{e^3 \sqrt 2}{12}.$

$K^6 = \frac{3^3 e^{12}}{2^{12}},\ \ \ e^{12} = \frac{2^{12} K^6}{3^3}$

$V_0^4 = \frac{e^{12}}{2^6 \cdot 3^4} = \frac{2^6 K^6}{3^7}$

Since $V \le V_0$,

$\frac{1}{2^6 \cdot 3^4} (b^2 + c^2 - a^2)^2 (c^2 + a^2 - b^2)^2 (a^2 + b^2 - c^2)^2 = V^4 \le V_0^4 = \frac{2^6 K^6}{3^7}$

${3^3 (b^2 + c^2 - a^2)^2 (c^2 + a^2 - b^2)^2 (a^2 + b^2 - c^2)^2 \le (2^2 K)^6}$

[color=darkblue][b]Yeti,[/b] the your solution is very nicely ! But it is right only for an acute triangle$ABC\ .$
In the case of an acute or obtuse triangle $ABC\ ,$ the proposed inequality is equivalently with the inequality $\left|\prod \tan A\right|\ge 3\sqrt 3\ .$
In the case of an obtuse triangle $ABC$, this inequality isn't truly ![/color][hide="Here is a proof of the above equivalence."]$27\prod (b^2+c^2-a^2)^2\le (4k)^6\Longleftrightarrow \left |3\sqrt 3\prod (2bc\cdot \cos A)\right |\le 8k^3\Longleftrightarrow$
$\left |3\sqrt 3(abc)^2\prod \cos A\right|\le 8k^3\Longleftrightarrow\left |3\sqrt 3(4kR)^2\prod\cos A\right|\le 8k^3\Longleftrightarrow$
$\left|6\sqrt 3 R^2\prod \cos A\right|\le 2R^2\sin A\sin B\sin C\Longleftrightarrow\left|\prod \tan A\right|\ge 3\sqrt 3\ .$

$1.\blacktriangleright$[u]Remark.[/u] I used the relations
$a^2=b^2+c^2-2bc\cdot \cos A$ and $k=\frac{abc}{4R}=2R^2\sin A\sin B\sin C\ .$

$2.\blacktriangleright$[u]Remark.[/u] $4k=(b^2+c^2-a^2)\tan A$ a.s.o.$\Longrightarrow$
$\prod (4k)^3=\prod (b^2+c^2-a^2)\cdot \prod \tan A\Longrightarrow$
$\prod (4k)^6=\prod (b^2+c^2-a^2)^2\cdot \left(\prod \tan A\right)^2$
and in an acute triangle $ABC$ there is the relation $\left(\prod \tan A\right)^2\ge 27\ .$[/hide]

Yes, thank you :!: It is not possible to form the desired tetrahedron from 4 congruent obtuse triangles, because $V^2$ would come out negative.


calculus
integration
vector
calculus computations

let $ \mathbf{F} \equal{} F_1\mathbf{i} \plus{} F_2\mathbf{j}$ be a $ C^1$ vector field on $ S \equal{} \mathbb{R}^2 \minus{} {(0,0)}$ such that $ \partial_1F_2 \equal{} \partial_2F_1$ on $ S$ (but $ \mathbf{F}$ may be singular at the origin.).

Let $ C_r$ be the circle of radius $ r$ about the origin, oriented counterclockwise.  Show that $ \int_{C_r}\mathbf{F}\cdot d\mathbf{x}$ is a constant, $ a$ that does not depend on $ r$.

more generally show that $ \int_{C}\mathbf{F}\cdot d\mathbf{x} \equal{} a$ for any simple closed curve $ C$, oriented ccw, that encircles the origin.

Apply Green's theorem to the region between $ C_r$ and $ C_s$ (remember that boundary has positive orientation when the interior is to the left).

For the second part, do the same fo $ C$ and $ C_r$ where $ r$ is large enough so that $ C$ does not cross $ C_r$.

[quote="mlok"]$ C$ and $ C_r$ where $ r$ is large enough so that $ C$ does not cross $ C_r$.[/quote]
Or small enough. It works either way. (We do have to assume that $ C$ avoids the origin, and hence avoids a neighborhood of the origin.)


1) Ικανή και αναγκαία συνθήκη ώστε τα πολυώνυμα $ p(x),q(x)$ βαθμών $ n,m$ αντιστοίχως να έχουν κοινή ρίζα είναι: να υπάρχουν δυο πολυώνυμα $ r(x),s(x)$ βαθμών $ n\minus{}1,m\minus{}1$ το πολύ αντιστοίχως ώστε $ p(x)s(x)\minus{}q(x)r(x)\equal{}0$ για κάθε $ x\in R$

2)Έστω πολυώνυμο $ p(x)$ με ακέραιους συντελεστές και $ p(0)\equal{}p(1)\equal{}1$ Να δείξετε ότι υπάρχει υπάρχει πολυώνυμο $ q(x)$ με  ακέραιους συντελεστές ώστε $ p(x)\equal{}x(x\minus{}1)q(x)\plus{}1$ και στη  συνέχεια αν $ x_0 \in Z$ και ισχύει $ x_n\equal{}p(x_{n\minus{}1})$ για κάθε $ n \in N$ τότε ο $ x_n$ είναι πρώτος προς τους $ x_0,x_1,...x_{n\minus{}1}$

3)Δείξτε ότι το πολυώνυμο $ p(x)\equal{}(x\minus{}a_1)^2(x\minus{}a_2)^2...(x\minus{}a_n)^2\plus{}1$ όπου όλα τα $ a_k\in Z$ δεν γράφεται στη μορφή γινομένου δυο πολυωνύμων με ακέραιους συντελεστές εκτός αν ένα απο αυτά είναι ίσο με το $ 1$ ή το $ \minus{}1$

Οι ασκήσεις αυτές ήταν θέματα πανελλαδικών εξετάσεων το 1972,1973,1974 στον πολυτεχνικό κύκλο.Νομίζω ότι είναι όμορφες και όχι πολύ εύκολες. Αυτά για να θυμούνται οι παλιότεροι και να συγκρίνουν οι νεότεροι

[quote="maniopas"]Θα δώσω μια λύση για την πρώτη άσκηση:
Για κάθε $ x \in \mathbb{R}$ είναι
$ p(x)s(x) \minus{} q(x)r(x) \equal{} 0 < \equal{} > \frac {p(x)}{r(x)} \equal{} \frac {q(x)}{s(x)} < \equal{} > x \minus{} r_1 \equal{} x \minus{} r_2 < \equal{} > r_1 \equal{} r_2$ που είναι ρίζες, γιατί $ x \minus{} r_1$ και $ x \minus{} r_2$ διαιρούν τα $ p(x), q(x)$ αντίστοιχα. Δεν έχω χρόνο για να δώσω αναλυτικά τη λύση ή να λύσω τις άλλες ασκήσεις, πάντως δε μου φάνηκαν ιδιαίτερα δύσκολες.[/quote]
Νομίζω ότι δεν είναι σωστό
Ούτε το αντίστροφο αποδεικνύει

[quote="mostel"]

Δε νομίζω πως μπορεί να γίνεται σύγκριση. Μιλάμε για άλλες εποχές. Μιλάμε για διαφορετική διδασκόμενη ύλη και αν κάτσω να ψάξω θα βρω άλλους 1002 παράγοντες που κάνουν στην ουσία ανέφικτη τη σύγκριση.

Αυτά με όλο το σεβασμό στο πρόσωπο σας, κύριε Μπόρη !  :)

Έχω ακούσει πολλές φορές τους παλιούς να λένε (ακόμη και τον πατέρα μου) πως αυτά που κάνουμε εμείς είναι "χαβαλέ" ενώ παλιά κάναν δυσκολότερα πράγματα, ότι μας δίνουν "μασημένη" τροφή και ένα σωρό. Κάθε εποχή έχει τις δικές της ανάγκες και απαιτήσεις.

Στέλιος[/quote]

Mostel, δε μου φαίνεται ότι ο r_boris ήθελε να κάνει σύγκριση [i]δυσκολίας[/i] θεμάτων μεταξύ "τότε" και "τώρα". Μία τέτοια σύγκριση είναι μάλλον αθέμιτη... ιδίως σε ένα μαθηματικό forum που δουλειά μας είναι να ξεχωρίζουμε τα μήλα από τα πορτοκάλια και να μη συγκρίνουμε ανόμοια πράγματα ;)

Εμένα μου φάνηκε πολύ περισσότερο ως σύγκριση θεματολογίας παρά ό,τιδήποτε άλλο, και μία τέτοια σύγκριση είναι σίγουρα θεμιτή. Όπως σωστά είπες, η διδακτέα ύλη είναι εντελώς διαφορετική, πράγμα που φαίνεται ξεκάθαρα στα θέματα του r_boris και που νομίζω ότι σε αυτό ήθελε να αναφερθεί. Γενικά, το να συγκρίνεις διαφορές στη θεματολογία έχει μεγάλη εκπαιδευτική αξία αφού έτσι βλέπεις το πώς εξελίσσεται η θέση των μαθηματικών στις ανάγκες του σχολείου, του παν/μίου κλπ. Αν μη τι άλλο, το να βλέπεις μέσα από μία σύγκριση θεμάτων ότι γεωμετρία και πολυώνυμα έχουν εξαφανιστεί από τις εισαγωγικές εξετάσεις τώρα, σε κάνει να σκέφτεσαι το [i]γιατί[/i] έχουν εξαφανιστεί και το γιατί έχουν αντικατασταθεί απ'ό,τι έχουν αντικατασταθεί. Όπως λες κι εσύ, διαφορετικές εποχές, διαφορετικές ανάγκες και άρα διαφορετικά θέματα... δεν είναι κακό να παρουσιάζεται μία παλαιότερη οπτική γωνία για σύγκριση :)

Heh... και όσον αφορά τη νοσταλγική αξία, είμαι σίγουρος ότι στους παλιότερους ένα θέμα του 73 θα φέρνει αρκετές αναμνήσεις ;)

Cheerio,

Παναγιώτης

Είναι αργά και δεν έχω το χρόνο για να σκεφτώ πλήρως τη λύση που μου ήρθε για το 1...
[hide]...αλλά μου φαίνεται ότι δουλεύει όπως η συνήθης παραγοντοποίηση στους ακεραίους. Δεν αρκεί να θεωρήσεις το ΕΚΠ των $ p,q$ ώστε να βρεις τα κατάλληλα $ r,s$? Μετά μου φαίνεται ότι βγαίνει αβίαστα...[/hide]

Cheerio,

Durandal 1707

[quote="mostel"][quote="r_boris"]Αυτά για να θυμούνται οι παλιότεροι και να συγκρίνουν οι νεότεροι[/quote]

Δε νομίζω πως μπορεί να γίνεται σύγκριση. Μιλάμε για άλλες εποχές. Μιλάμε για διαφορετική διδασκόμενη ύλη και αν κάτσω να ψάξω θα βρω άλλους 1002 παράγοντες που κάνουν στην ουσία ανέφικτη τη σύγκριση.

Αυτά με όλο το σεβασμό στο πρόσωπο σας, κύριε Μπόρη !  :)

Έχω ακούσει πολλές φορές τους παλιούς να λένε (ακόμη και τον πατέρα μου) πως αυτά που κάνουμε εμείς είναι "χαβαλέ" ενώ παλιά κάναν δυσκολότερα πράγματα, ότι μας δίνουν "μασημένη" τροφή και ένα σωρό. Κάθε εποχή έχει τις δικές της ανάγκες και απαιτήσεις.

Στέλιος[/quote]
Φίλε Mostel οι λόγοι τους οποίους επικαλείσαι για το ότι δεν υφίσταται σύγκριση [b]ήδη είναι μια σύγκριση[/b] Ο Durandal καλά το κατάλαβε!! περισσότερο νοσταλγικοί ήταν οι λόγοι και δεν έκρυβαν καμία "κατηγόρια" όσο για την δυσκολία των θεμάτων αναφέρω τα θέματα του 1998, ( ξέρει καλά ο Παναγιώτης ) Η σύγκριση δεν είναι αναγκαστικά απαξίωση της γενιάς σας [b]την οποία εκτιμώ ιδιαιτέρως[/b]
Παρεμπιπτόντως λύσεις θα στείλω σε λίγες μέρες

[quote="Durandal"]Πρόβλημα 1
[hide]Αν τα $ p,q$ έχουν κοινή ρίζα, τα ζητούμενα πολυώνυμα κατασκευάζονται εύκολα αν πάρουμε $ p(x)s(x) \equal{} q(x)r(x) \equal{} \textrm{lcm}[p,q](x)$ όπου $ \textrm{lcm}[p,q]$ το ΕΚΠ των $ p,q$. Αντιστρόφως, αν υπάρχουν τέτοια πολυώνυμα, θα πρέπει $ q(x)|p(x)s(x)$ και, αφού $ \deg(s) < \deg(q)$, έπεται ότι τα $ p,q$ έχουν κοινή ρίζα.
[/hide]

Πρόβλημα 2: θα πρέπει $ x_0 \neq 0,1$, αλλιώς έχουμε αντιπαράδειγμα για $ p(x) \equal{} x^2 \minus{} x \plus{} 1$ και $ x_0 \equal{} 0,1 \implies 1 \equal{} x_1 \equal{} \ldots \equal{} x_n \equal{} \ldots$
[hide]Αφού $ p(1) \equal{} p(0) \equal{} 1$, έπεται ότι $ p(x) \minus{} 1 \equal{} x(x \minus{} 1) q(x)$ και αρκεί να δείξουμε ότι το $ q$ έχει ακέραιους συντελεστές. Ωστόσο, αυτό φαίνεται άμεσα από μία ανάπτυξη καθώς οι συντελεστές του $ p(x) \minus{} 1$ είναι ακέραιοι και οι συντελεστές του $ x(x \minus{} 1) \equal{} x^2 \minus{} x$ είναι $ \pm 1$.
Tώρα, αφού $ p(x) \equal{} x(x \minus{} 1)q(x) \plus{} 1$, φαίνεται ότι για $ x\in\mathbb{Z} \minus{} \{0,1\}$, το $ x$ δεν διαιρεί το $ p(x)$ και, επαγωγικά, προκύπτει το ζητούμενο.[/hide]

Πρόβλημα 3: θα πρέπει τα $ a_i$ να είναι διαφορετικά μεταξύ τους, αλλιώς έχουμε αντιπαράδειγμα, πχ: $ x^4 \plus{} 4 \equal{} (x^2 \minus{} 2x \plus{} 2)(x^2 \plus{} 2x \plus{} 2)$.
[hide]Έστω ότι $ p(x) \equal{} q(x)r(x)$ με $ q,r\in\mathbb{Z}[x]$. Τότε, χωρίς βλάβη γενικότητας, μπορούμε να θεωρήσουμε ότι $ q,r > 0$. Περαιτέρω, αφού $ q(a_i)r(a_i) \equal{} 1$ έπεται ότι $ q(a_i) \equal{} r(a_i) \equal{} 1$ για κάθε $ i \equal{} 1,2\ldots n$. Συνεπώς, με τα $ a_i$ να είναι διαφορετικά μεταξύ τους, και εξαιρώντας την περίπτωση $ q \equal{} 1$ ή $ r \equal{} 1$, έχουμε: $ (x \minus{} a_1)\cdots (x \minus{} a_n)\big|(q(x) \minus{} 1),(r(x) \minus{} 1)$ οπότε, από σύγκριση βαθμών, προκύπτει εύκολα ότι $ q(x) \equal{} r(x) \equal{} 1 \plus{} (x \minus{} a_1)\cdots (x \minus{} a_n)$, όπερ άτοπο.[/hide]

Nice! :)

Cheerio,

Durandal 1707[/quote]
Φυσικά έχεις δίκιο Παναγιώτη θα το συμπλήρωνα σαν σχόλιο στις λύσεις

[quote="r_boris"]1) Ικανή και αναγκαία συνθήκη ώστε τα πολυώνυμα $ p(x),q(x)$ βαθμών $ n,m$ αντιστοίχως να έχουν κοινή ρίζα είναι: να υπάρχουν δυο πολυώνυμα $ r(x),s(x)$ βαθμών $ n \minus{} 1,m \minus{} 1$ το πολύ αντιστοίχως ώστε $ p(x)s(x) \minus{} q(x)r(x) \equal{} 0$ για κάθε $ x\in R$

2)Έστω πολυώνυμο $ p(x)$ με ακέραιους συντελεστές και $ p(0) \equal{} p(1) \equal{} 1$ Να δείξετε ότι υπάρχει υπάρχει πολυώνυμο $ q(x)$ με  ακέραιους συντελεστές ώστε $ p(x) \equal{} x(x \minus{} 1)q(x) \plus{} 1$ και στη  συνέχεια αν $ x_0 \in Z$ και ισχύει $ x_n \equal{} p(x_{n \minus{} 1})$ για κάθε $ n \in N$ τότε ο $ x_n$ είναι πρώτος προς τους $ x_0,x_1,...x_{n \minus{} 1}$

3)Δείξτε ότι το πολυώνυμο $ p(x) \equal{} (x \minus{} a_1)^2(x \minus{} a_2)^2...(x \minus{} a_n)^2 \plus{} 1$ όπου όλα τα $ a_k\in Z$ δεν γράφεται στη μορφή γινομένου δυο πολυωνύμων με ακέραιους συντελεστές εκτός αν ένα απο αυτά είναι ίσο με το $ 1$ ή το $ \minus{} 1$

Οι ασκήσεις αυτές ήταν θέματα πανελλαδικών εξετάσεων το 1972,1973,1974 στον πολυτεχνικό κύκλο.Νομίζω ότι είναι όμορφες και όχι πολύ εύκολες. Αυτά για να θυμούνται οι παλιότεροι και να συγκρίνουν οι νεότεροι[/quote]
Λύσεις με κάποια σχόλια γιατί δεν θυμόμουν καλά τις εκφωνήσεις στο attachment


function
superior algebra
superior algebra unsolved

Show that if f, g, h are onto functions defined from $ X \to X$, then $ f(g(x)) \equal{} f(h(x)) \implies g(x) \equal{} h(x)$.

clearly wrong (take some $ f$ which is not injective and $ g,h$ constant).

Thank you.
It was meant to be $ f(h(x)) \equal{} g(h(x)) \implies f \equal{} g$. But is this not true for all functions regardless of whether the functions are surjectives or not? Where does one use surjectivity?

One uses the surjectivity of $ h$, i.e. $ h(X)\equal{}X$.


LaTeX
geometry
geometric transformation
reflection

So I was wondering, I noticed that one of the requirements for one's paper for Siemens is that it should be in Times New Roman or Arial. Personally I think a math paper in those fonts look way too ugly and unprofessional. Is it ok if I submit it using LaTeX font? Has anyone here done that?

I'm not sure what you mean by $ \text{\LaTeX}$ font.  Times New Roman is the default font in TeXnicCenter (and, I believe, in a $ \text{\LaTeX}$ document) and it looks fine (and nearly every "professional" paper I've read has been in a default serif font of one form or another).  A math paper should avoid Arial and other sans-serif fonts, though; they look out of place, in my opinion.

Ah sorry, I meant the default LaTeX font like $ \text{Hi! This is a test.}$

That [i]is[/i] Times New Roman (as far as I know).

I believe the left flag reflects origin and not residence.

[quote]
Using $ \text{\LaTeX}$ gives nicer font edges.

However, you'll probably find that the editing process goes more smoothly if you use a realtime WYSIWYG word processor.
[/quote]

OK thanks, for the advice.

[quote]
What, Siemens allow non-US people to enter?
Unless your flags in your profile doesn't reflect where you really are.
[/quote]
Yeah the flag in my profile doesn't reflect where I really am, as I've noticed some users having a flag of Antarctica under their name, but I am Chinese

There is a provision for setting your profile so as to show two flags if you want to - note t0rajir0u's two-flag claim to being Chinese-American. Or check some post by Valentin Vornicu, or some post by fedja, for more of the same idea.


geometry
circumcircle
limit
geometric transformation
reflection
homothety
geometry proposed

$ A,B,C,D,E,F$ lie on one circle, and $ AD\cap BE\cap CF\equal{}P$.
$ X,Y,Z$ are the semetrical point of $ D,E,F$ with respect to $ BC,CA,AB$, $ H$ is the orthocenter of $ \triangle ABC$, and $ O$ is the circumcenter of $ \triangle ABC$.
Show that $ X,Y,Z,H$ lie on the same circle.

[hide="Special case of this problem"]
1)If $ P$ lie on circumcircle of $ \triangle ABC\Rightarrow$ $ X,Y,Z,H$ lie on one line.
2)If $ \lim (PO) \rightarrow \inf\Rightarrow AD\parallel BE\parallel CF$ : China TST 2006 Day 7 Problem 1[/hide]

Let $ A',B',C'$ be the circumcenter of triangle $ HBC,HCA,HAB$ respectively. Since $ B'C'\perp AH$ and $ AH\perp BC$, $ B'C'\parallel BC$ and similarly, $ C'A'\parallel CA$,$ A'B'\parallel AB$.It follows that triangle $ ABC$ and $ A'B'C'$ are homothety. Since $ X$ is reflection of $ D$ wrt $ BC$, midpoint of $ DX$ is projection of $ D$ on $ BC$,which lie Simson line of $ D$. Also, Simson line of $ D$ pass midpoint of $ DH$. Hence $ HX$ is paralllel to Simson line of $ D$. Simson line of $ D$ is perpendicular to isogonal line of $ \overline {APA'}$.It follow that $ HX$ is perpendicular to isogonal line of $ AP$. Hence line that pass through $ A,B,C$ and perpendicular to $ HX,HY,HZ$ are concurrent at isogonal conjugate point of $ P$ wrt triangle $ ABC$. Since triangle $ ABC$ and $ A'B'C'$ are homothety, line that pass through $ A',B',C'$ and perpendicular to $ HX,HY,HZ$ are concurrent. Since $ \angle BXC \equal{} \angle BDC \equal{} \angle BAC \equal{} \angle BHC$, $ B,H,X,C$ are cyclic. Hence $ A'H \equal{} A'X$. It follows that perpendicular line to $ HX$ through $ A'$ is midperpendicular of $ HX$. Then midperpendicular of $ HX,HY,HZ$ are concurrent. Hence $ H,X,Y,Z$ are cyclic.


analytic geometry
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geometry
geometric transformation
homothety
combinatorics unsolved
combinatorics

A finite set of points $X$ and an equilateral triangle $T$ are given on a plane. Suppose that every subset $X'$ of $X$ with no more than $9$ elements can be covered by two images of $T$ under translations. Prove that the whole set $X$ can be covered by two images of $T$ under translations.

can anyone post solution (or offical solution)?

[quote="filosofimenos"]I'd love to see a solution at this marvellous problem. What do you people think we could do? Induct? Set up coordinates? For me nothing works... :mad:  Is there any web-site were there exist solution of russian olympiad problems in English?[/quote]
I only have solution in Russian not English but I can't understand them.I really want a nice solution.

[quote="filosofimenos"]Put the solution I maybe can translate it.[/quote]
Please write your email,I will send me(Because I can't up file on mathlinks)

[quote="filosofimenos"]makrianixtomenos@yahoo.com[/quote]
I had send me solution all Rusian.
Solution this problem in page 400

We need two lemmas:
 Lemma 1. If two triangles T1 and T2 are homothetic with positive
 Prolonged coefficient and the triangle T2 intersects each
 line containing the side of the triangle T1, then T2 has
 T1.
 Dock and r a t e l b a TV o.Ochevidno.
 Lemma 2. For any finite set of points M and TPE -
 gon T there is a triangle T?, homothetic to T with a positive
 tion factor containing M and containing on each page -
 doy his side a spot izM.
 Prior to a r a t e l a t bc about. Consider a triangle, positive
 PRE-homothetic T and containing M. If any party fails to
 intersects with M, will reduce the homothety with center at the opposite
 Noah this side of the top and achieve that the party crossed with M and
 triangle still soderzhalM. Repeating this for each side, in
 we obtain a required result.
 Note that in the condition of the previous lemma, any positive first -
 motetichny T triangle, soderzhaschiyM, also contains the T? by Lemma
 Theorem 1.
 Teperpereydem to solve the problem. Apply Lemma 2 to T and X. In
 we obtain a triangle, we denote it for opredelennostiABC. On the sides
 BC, AC and AB ests ootvetstvenno three points xA, xB and xC from the set
 X, which may include bytsov falling.
 If ABC does not exceed the size of T, the proof obvious
 but finished. Otherwise, consider three triangles ABACA, ABBCB and
 ACBCC, positively homothetic ABC centered at its vertices and
 equal to the initial T. The subset of X
 XA = X \
 ABACA XB = X \
 ABBCB XC = X \
 ACBCC.
 Prove one more lemma:
 Lemma 3. If a triangle T?, Obtained from the T pa -
 rallelnym transfer, contains points xA and xB, then it can not
 across multiple XC. Similarly, for other pairs of points and co -
 sponding set of them.
 2002-2003 academic year, 10 Class 401
 Prior to a r a t e l a t bc about. Suppose the contrary. Since T? contains
 xA and xB, then one of these points lies on one side with a C on
 ACBC, otherwise the length of T? more ACBC.
 Then we can see that T? crosses all party lines ACBCC,
 hence contains ACBCC. Since these triangles are equal, they are joint
 fall, but then the T? can not cross XC.
 Teperprimenim Lemma 2 to the triangle T and the sets XA, XB,
 XC, we obtain the triangles TA, TB, TC, being on their sides can
 vybratto chki (xAB, xAC, xA), (xBA, xBC, xB), (xCA, xCB, xC) corre -
 respectively (some of them may coincide), lying in the corresponding
 sets XA, XB, XC.
 The points xA, xB, xC, xAB, xAC, xBA, xBC, xCA, xCB applicable condi -
 ence problem. They are contained in the triangles T1 and T2, are parallel
 lel translation operators T.
 Some two of the first three points are contained in one of the TPE -
 angles, without loss of generality, xA, xB ∈ T1. Then by Lemma 3
 triangle T1 is not crossing mnozhestvoXC and all three points xC, xCA, xCB
 contained in another triangle T2. Hence mnozhestvoXC
 contained in T2 by Lemma 1, which means that the triangles ACBCC and T2-covering
 UT entire set X.

[quote="modularmarc101"]THis is what I got from freetranslation.com . There's a couple of phrases that weren't translated though, but the majority was :) .

Two lemmas will be necessary for us:
Lemma 1. If two triangles T1 and T2 гомотетичны with поло-
жительным in factor and triangle T2 crosses everyone
The straight line containing the party of triangle T1 T2 contains
T1.
Dock and з and т е л ь with тв about. Obviously.
Lemma 2. For any final set of points M and тре-
Square T there will be triangle T, гомотетичный T with положи-
тельным in the factor, containing M and containing on каж-
дой to the party any point изM.
Up to to and з and т е л ьс т in about. We shall consider any triangle, положи-
тельно гомотетичный T and containing M. If its any party not
It is crossed with M, we shall reduce it гомотетией with the center in противополож-
ной to this party to top also we shall achieve, that the party was crossed with M and
Triangle still содержалM. Repeating it for each party,-
лучим demanded.
Let's notice, that in a condition of the previous lemma any it is positive го-
мотетичный T the triangle, содержащийM, contains as well T On лем-
ме 1.
Теперьперейдем to the decision of a problem. We shall apply a lemma 2 to T and X.-
лучим a triangle, we shall designate it for определенностиABC. On the parties
BC, AC and AB естьс оответственно three points xA, xB and xC from set
X among which can бытьсов falling.
If ABC does not surpass T in the sizes, the proof очевид-
But it is finished. Otherwise we shall consider three triangles ABACA, ABBCB and
ACBCC, it is positive гомотетичные ABC with the centers in its tops and
Equal initial T. We shall designate subsets of set X
XA = X \
ABACA XB = X \
ABBCB XC = X \
ACBCC.
Let's prove one more lemma:
Lemma 3. If any triangle T, turning out from T па-
раллельным carry, contains points xA and xB it cannot
To cross set XC. Similarly for other pairs points and со-
Set replying to it.
2002-2003, 10 CLASS 401
Up to to and з and т е л ьс т in about. We shall assume opposite. As T Contains
xA and xB one of these points lays on one party with C rather
ACBC, differently the length of party T Is more ACBC.
Then it is possible to notice, that T Crosses all straight lines of parties ACBCC,
So contains ACBCC. As these triangles are equal, they сов-
Fall, but then T Cannot cross XC.
Теперьприменим a lemma 2 to triangle T and sets XA, XB,
XC, we shall receive triangles TA, TB, TC, and on their parties it is possible
выбратьто чки {xAB, xAC, xA}, {xBA, xBC, xB}, {xCA, xCB, xC} соответ-
ственно (some of them can coincide), laying in corresponding
Sets XA, XB, XC.
To points xA, xB, xC, xAB, xAC, xBA, xBC, xCA, xCB we shall apply усло-
вие problems. They contain in triangles T1 and T2, being парал-
лельными carries T.
Any two of first three points contain in one of these тре-
Squares, without restriction of a generality xA, xB? T1. Then on a lemma 3
Triangle T1 does not cross множествоXC and all three points xC, xCA, xCB
Contain in other triangle T2. From here follows, that множествоXC
Contains in T2 on a lemma 1; means, triangles ACBCC and T2 покрыва-
ют all set X.

EDIT: Beaten[/quote]
I really want see a complete solution.

My solution had a major flaw, but I've been able to make sense of the jumbled translations above, so I'll post a clearer translation.

[b]Proof:[/b]
Let $M=ABC$ be the smallest equilateral triangle oriented the same way as $T$, that contains all of the points in $X$. There must be a point of $X$ (not necessarily all distinct) on each edge of $M$, call them $x_A,x_B,x_C$ on edges $BC$, $CA$, and $AB$ respectively. Now if $M$ is smaller than $T$ we are certainly done, so assume $M$ is larger than $T$.

Take three copies of $T$: $AB_AC_A$, $A_BBC_B$ and $A_CB_CC$, where $A,B,C$ are the vertices of $M$ and $B_A,C_A,A_B,C_B,A_C,B_C$ lie on the edges of $M$.

Also define three subsets of points $X_A = X \backslash AB_AC_A$ := points in $X$ that aren't covered by $AB_AC_A$. Similarly define $X_B$ and $X_C$.

[b]Lemma:[/b] If a translation of triangle $T$, call it $T'$, contains two points from $\{x_A,x_B,x_C\}$, say $x_A,x_B$ then it cannot contain any point from $X_C$.

[i]Proof:[/i] Since $x_A$ and $x_B$ lie on different edges of $M$, the only way for $T'$ to cover both of them is if it contains the point $C$. This means that the edge of $T'$ parallel to $AB$ lies below the line $A_CB_C$ (of triangle $CA_CB_C$) so it cannot contain any point of $X_C \qquad \square$

Now take the smallest triangles $M_A,M_B,M_C$ that contain $X_A,X_B,X_C$ respectively. Each edge of $M_A$ contains a points of $X$, one of which is $x_A$, and similarly for $M_B$ and $M_C$. In total there are at most $9$ such points, so they can be covered by two copies of $T$. By PHP one of those two copies contains at least two pints from $\{x_A,x_B,x_C\}$, say $x_A,x_B$. Therefore it follows from the lemma that no points from $X_C$ are covered by the first triangle, hence the second triangle covers all of $X_C$. Hence the second triangle and $AB_AC_A$ must cover all points in $X$.

This completes the proof $\blacksquare$


trigonometry
function
geometry
geometric transformation
reflection
rotation

What is the value of $\cos{215}$? I don't see why it's $-\frac{\sqrt2}{2}$ unless the book made a typo...

Are the graphs of $\csc{x}$ and $\sec{x}$ odd or even and why?

Just for the fun of it, I'll prove that...I'm bored
$\\\cos 225=-\cos 55=-\cos (90-45)\\=-(\cos 90\cos 45+\sin 90\sin 45)\\=-\frac{1}{\sqrt{2}}\\=-\frac{\sqrt{2}}{2}$

EDIT: The graph of $\sec x=\frac{1}{\cos x}$ because $\cos (-x)=cos x\implies even$.  The graph of $\csc x$ is odd because $\frac{1}{\sin x}$ because $\sin (-x)=-\sin x$.

Masoud Zargar

You made a typo as well. $\cos 225=-\cos 55$ should be $\cos 225=-\cos 45$. Oh, and the things you use to "prove" it calculate the value inside the proof, which, given that knowledge, would let you just get to the value from $- \cos 45$.

yup, i mixed it up with cos 125
Well, I have
$\\\cos 225=-\cos(90-45)=-\cos 45=-\frac{\sqrt{2}}{2}$

Fixed.  The idea is lines in a circle.
Thanks.  I fixed teh calculation mistakes.

You have a typo again... now you said $55$ in some other places. :roll:

I don't see teh typo.  The other 55s should be there.
Edit: That fixing is irrelevant to this.

What are you talking about? $\cos 225=-\cos 45$, not $-\cos 55$. And then $-\sin 55=-\sin (90-35)$, not $-\sin (90-45)$.

Hmmm.  I can't subtract.  Interesting to know about myself.

Oh yeah, and another note, can anyone solve the second problem?

A function is even if it is unchanged with a reflection in the $\mbox{y}$ axis, i.e $\mbox{f(x)=f(-x)}$. In the case of $\mbox{y=sec(x)}$, this is true and so it is even.

A function is odd if it is unchanged after a rotation of $\pi$ radians about the origin, i.e $\mbox{-f(x)=f(-x)}$. In the case of $\mbox{cosec(x)}$, this is correct, and hence it is an odd function.

Oh, I realize what I did wrong. Thanks! (even though I shouldn't have got it wrong).

236factorial did not post this problem he cant do it


Let $ a_1,a_2,\cdots,a_n$ be real numbers,set:$ S_k\equal{}\sum_{i\equal{}1}^na^k_i$
If we have $ S_k>0$ for $ k\equal{}1,2,\cdots,n$Can we show $ S_k>0$ for every $ k \in \mathbb{N}$

If $ n>2$, no. Imagine many $ 1$'s and one negative number just a tiny bit larger than $ 1$ in absolute value. For all small powers, it is shadowed by $ 1$'s completely but for very large powers it dominates everything.


Determine all positive integers $ n$ such that $ 3|(2^{n}\minus{}1)$ and $ \frac{2^{n}\minus{}1}{3}|(4m^{2}\plus{}1)$ for a positive integer $ m$.

It mean $ n$ is even and all prime divisor p of $ \frac{2^n\minus{}1}{3}$ is form $ p\equal{}1\mod 4$.
If $ q$ is odd prime divisor of $ n$, then $ 2^q\minus{}1|\frac{2^n\minus{}1}{3}$, but $ 2^q\minus{}1$ had prime divisor form $ 3\mod 4$.
It give, than $ n$ is power 2. If $ n\equal{}2^k$, then all prinme divisors $ \frac{2^n\minus{}1}{3}$ is form $ 1\mod 4$.


[quote="GameBot"]The sum of two numbers is 15. One number is doubled and the other is tripled. The sum of the two new numbers is 39. What is the positive difference between the original numbers?[/quote]

You know that [b]x+y=15 (1)[/b] and that [b]2x+3y=39. (2)[/b] From here, you do regular simultaneous equations. Multiply the first equation by 2 to get [b]2x+2y=30(3)[/b] and subtract (3) from (2) to get [b]y=9[/b]. Plug in y=9 into (1) to get [b]x=6[/b]. From here, you subtract 6 from 9 to get your answer, [b]3[/b].

Two numbers are x and y:

The two equations are
$ x\plus{}y\equal{}15$
$ 2x\plus{}3y\equal{}39$

Multiply the first equation by two and subtract from second equation. We get y=9. Plug it back in first equation, giving us x=6. Positive difference is 9-6=3


How would you solve the equation
$x^{x}= ax$
for $x$ (where a is a postive constant >= 1) in closed-from using the Lambert $W$ function.

I have spent more than an hour on this problem and I just can't get it to the form $C = f(x)e^{f(x)}$.

Any clever ideas? or failed attempts for that matter?

This is a double post. Euclid 3.14 has posted the same thing in the calculus-analysis section.