geometry
3D geometry
perimeter
inequalities
function
logarithms
inequalities proposed

If $ a,b,c\in \mathbb R^\plus{}$, prove that:

$ \big(\frac{a\plus{}b}{2}\big)^2\big(\frac{b\plus{}c}{2}\big)^2\big(\frac{c\plus{}a}{2}\big)^2\geq abc\big(\frac{a\plus{}b\plus{}c}{3}\big)^3$

[hide]Please avoid full expansion of the products.[/hide]

$ \frac{a\plus{}b}{2}\geq{\sqrt{ab}}$ by $ A.M.\geq{G.M.}$
similarly, $ \frac{b\plus{}c}{2}\geq{\sqrt{bc}}$  &  $ \frac{c\plus{}a}{2}\geq{\sqrt{ca}}$
and, $ \prod_{cyc}(\frac{a\plus{}b}{2})^{\frac{1}{3}}\geq{\frac{a\plus{}b\plus{}c}{2}}\geq{\frac{a\plus{}b\plus{}c}{3}}$
Now assemble the two inequalities........Done!

[quote="Sunkern_sunflora"]$ \frac {a \plus{} b}{2}\geq{\sqrt {ab}}$ by $ A.M.\geq{G.M.}$
similarly, $ \frac {b \plus{} c}{2}\geq{\sqrt {bc}}$  &  $ \frac {c \plus{} a}{2}\geq{\sqrt {ca}}$
and, $ \prod_{cyc}(\frac {a \plus{} b}{2})^{\frac {1}{3}}\geq{\frac {a \plus{} b \plus{} c}{2}}\geq{\frac {a \plus{} b \plus{} c}{3}}$
Now assemble the two inequalities........Done![/quote]
Can you detail the last ? ican understand it
My proof is used Miving variables,it quite long

[quote="Sunkern_sunflora"]$ \frac {a \plus{} b}{2}\geq{\sqrt {ab}}$ by $ A.M.\geq{G.M.}$
similarly, $ \frac {b \plus{} c}{2}\geq{\sqrt {bc}}$  &  $ \frac {c \plus{} a}{2}\geq{\sqrt {ca}}$[/quote]
I created the problem. And I think nothing can be achieved starting with these.

[quote="Sunkern_sunflora"]$ \prod_{cyc}(\frac {a \plus{} b}{2})^{\frac {1}{3}}\geq{\frac {a \plus{} b \plus{} c}{2}}\geq{\frac {a \plus{} b \plus{} c}{3}}$[/quote]
What do you mean?

My approach is to rewrite it as
$ \prod \frac {a(a \plus{} b \plus{} c) \plus{} bc}{4}\geq \prod \frac {a(a \plus{} b \plus{} c)}{3}$
But how to continue?

[quote="Sunkern_sunflora"]$ \prod_{cyc}(\frac {a \plus{} b}{2})^{\frac {1}{3}}\geq{\frac {a \plus{} b \plus{} c}{2}}\geq{\frac {a \plus{} b \plus{} c}{3}}$[/quote]actually, the opposite is true by AM-GM: $ \prod_{cyc}\left(\frac {a \plus{} b}{2}\right)^{\frac {1}{3}}\leq{\frac {\sum\frac{a\plus{}b}{2}}{3}}$

Sorry, I tried to obtain the result using volume of cube > perimeter of cube.......but it was a BOGUS approach.
Has anyone reached till some steps? How about your approach Akashnil?

[quote="Akashnil"]If $ a,b,c\in \mathbb R^ \plus{}$, prove that:

$ \big(\frac {a \plus{} b}{2}\big)^2\big(\frac {b \plus{} c}{2}\big)^2\big(\frac {c \plus{} a}{2}\big)^2\geq abc\big(\frac {a \plus{} b \plus{} c}{3}\big)^3$

[hide]Please avoid full expansion of the products.[/hide][/quote]

Assume $ a\plus{}b\plus{}c\equal{}1$ and put $ q\equal{}ab\plus{}bc\plus{}ca,r\equal{}abc$ then the inequality becomes
\[ 27\left( {\frac{{q \minus{} r}}{8}} \right)^2  \ge r
\]
By using $ r \le \frac{{q^2 }}{3}$ we will show that
\[ 27\left( {\frac{{q \minus{} \frac{{q^2 }}{3}}}{8}} \right)^2  \ge \frac{{q^2 }}{3}
\]
\[ \Leftrightarrow 81\left( {1 \minus{} \frac{q}{3}} \right)^2  \ge 64
\]
\[ \Leftrightarrow q \le \frac{1}{3}
\]

You can use Popoviciu theorem that for a convex function, we have

$ f(x)\plus{}f(y)\plus{}f(z) \plus{} 3 f\left( \frac{x\plus{}y\plus{}z}{3} \right)$
$ \ge 2 \left [ f \left(\frac{x\plus{}y}{2} \right) \plus{} f \left(\frac{y\plus{}z}{2} \right) \plus{} f \left(\frac{z\plus{}x}{2} \right) \right]$

$ \log x$ is a convex function. And hence,

$ \log a \plus{} \log b \plus{} \log c \plus{} 3 \log \left(\frac{a\plus{}b\plus{}c}{3}\right) \ge 2 \sum \log \left(\frac{a\plus{}b}{2} \right)$

From this it follows that
$ abc \left(\frac{a\plus{}b\plus{}c}{3} \right)^3 \ge \left(\frac{a\plus{}b}{2} \right)^2 \left(\frac{b\plus{}c}{2} \right)^2 \left(\frac{c\plus{}a}{2} \right)^2$

[quote="hsbhatt"]You can use Popoviciu theorem that for a convex function, we have

$ f(x) \plus{} f(y) \plus{} f(z) \plus{} 3 f\left( \frac {x \plus{} y \plus{} z}{3} \right)$
$ \ge 2 \left [ f \left(\frac {x \plus{} y}{2} \right) \plus{} f \left(\frac {y \plus{} z}{2} \right) \plus{} f \left(\frac {z \plus{} x}{2} \right) \right]$

$ \log x$ is a [b]convex[/b] function. And hence,

$ \log a \plus{} \log b \plus{} \log c \plus{} 3 \log \left(\frac {a \plus{} b \plus{} c}{3}\right) \ge 2 \sum \log \left(\frac {a \plus{} b}{2} \right)$

From this it follows that
$ abc \left(\frac {a \plus{} b \plus{} c}{3} \right)^3 \ge \left(\frac {a \plus{} b}{2} \right)^2 \left(\frac {b \plus{} c}{2} \right)^2 \left(\frac {c \plus{} a}{2} \right)^2$[/quote]

$ \log x$ is a [b]concave[/b] function. I hope popoviciu is reversed for concave functions.

Nevertheless, [b]many thanks for the two solutions[/b]. And I didn't know popoviciu before. Thanks for that too.

:lol: yeah it does. sorry for that goof

Popoviciu I encountered in Mathematics Miniatures by Titu Andreescu

I found the term "majorisation" in some inequalities like $ M\ddot{u}irheads$, can anybody explain it to me, with some examples?


Is there any literature on the rationality of log(e)?

The base-10 log, or the natural (base e) log?  If the former, it is irrational:  If $\log e = \frac{p}{q}$ $10^{\frac{p}{q}}= e$ so $e^{q}= 10^{p}$ and then $e$ is a root of the polynomial $f(x) = x^{q}-10^{p}$.  But $e$ is transcendental, a contradiction.  (Note that this extends to any rational number in place of 10.)  If the latter, $\log e = 1$ whose rationality has never been a point of much dispute.


The sum of the areas of all triangles whose vertices are also vertices of a 1x1x1 cube is m + sqrt.n + sqrt.p, where m,n, and p are integers . Find m+n+p


Hi,

I live in Nottinghamshire, England.

For reasopns which I won't bore you with, I'm looking for the (reasonably) precise "centre" point of the county - either its "centre of gravity" or the furthest point into the county...

There's a map of the county here... http://www.touristnetuk.com/EM/nottinghamshire/towns/index.htm

Any help (or the answer!!!) gratefull received.

Kind regards.

It is precisely at the midpoint of Mansfield and Newark-on-Trent.

Thank you for that!  Can you tell me how you worked it out - or would it be too tricky for a dimwit like I to understand?

MANY thanks

My guess is Sherwood forest, watch out, Robin 'ood!

[quote="MysticTerminator"]dude, posts: 0 ?[/quote]

gff = 0 posts

[quote="akalra1"]yea, posts in gff dont add to ur post count[/quote]

pls dont use chatsp3ak so much b/c its hard 2 read :P

probability1.01, how [b]did[/b] you figure that out?

dude what zhai is smart DON'T BE DISSIN THE ZHAI

it was probably triv


What's the process of subtracting, for example, .009 - .02? Normally you would align the decimals to subtract, but with .009-.02 you don't and the answer turns out to be -.011. I can't figure out why.
I don't get it. Why is it negative? And why are the the 2 and 9 added instead of subtracted?

[quote="isabella2296"]It is negative because 0.02 is greater than 0.009.[/quote]
Oh yeah.
How do I set up those subtraction problems? With other decimals problems you line up the decimals. How do I do this one?

[quote="AIME15"]You have 0.009-0.02, right?

So:

x.009
-.020

and solve.

x is just a placeholder.[/quote]
So to solve that I minus 20 from 9?

[quote="isabella2296"]Right, but don't forget that they're decimals, so it's not -11, but -0.11.[/quote]
You mean -.011?
Alright thanks guys I think I got it. I'm gonna come up with my own to practice problems and check them on the calculator. If I have anymore problems I'll be back.


For all natural numbers n, let f(n) be the number of ways to place an arbitrary number of knights (including zero) on an nxn chessboard such that no two attack each other.

Find 1/f(1) + 1/f(2) + ... 1/f(n).


I sent the wrong version of a solution when I e-mailed it last night, and I just found out that it's due today, not tomorrow (last year due dates were always on Sundays)  so can I re send it?!?!  I sent it in the mail too, which was also has the wrong one, so can I just include a note saying to disregard the mailed one.

Also, while I'm resending, can I include other solutions that I have improved and just include notes that these are the new ones?  While I'm at it, I'll ask another question:  if we use a method to solve the problem and another to verify it is true, can we just mention that we used the one to verify it was correct.  I don't really have the time to TeX up the verification since I just figured it out, but it would prove that I really did verify it completely.  Thanks.

As for resending, we strongly discourage this, but have been lenient for the first round.  For all future rounds, we [b]will not accept re-submissions[/b].

So can I send it again this time?  I rushed because I thought it was due last night, but now I have time.

Yes, by email or fax, with a clear note that this supersedes old ones.  Do note, we won't allow this again, but this is the first round we've run, so we're being a little lenient due to all the changes from last year.

Should I email all of them again or just the ones I changed?


Calculate the volume of air at $ 30^\circ C$ and $ 1.00$ atm that is needed to burn completely $ 10.0$ g of propane ($ C_3H_8$). Assume that air is $ 21.0$ percent $ O_2$ by volume.

I wrote the balanced equation:

$ C_3H_8 \plus{} 5O_2 \implies 3CO_2 \plus{} 4H_2O \nonumber$

Then, I said $ 10$ grams of $ C_3H_8$ is equal to $ 0.23$ mol $ C_3H_8$, which implies $ 5 \cdot 0.23 \equal{} 1.14$ mol $ O_2$.

Since 1 mole of a gas is 22.4 L, $ 1.14 \cdot 22.4 \equal{} 25.5$ liters $ O_2$.

Since $ O_2$ makes up $ 21.0$ percent of air, the volume of air is $ 25.5 \cdot \frac{100}{21.0} \equal{} \boxed{121.2 \text{ L}}$.

I realize that I do not use the fact that the air is at 30 degrees and 1.00 atm, but I'm not sure how to use this information.

The answer should be $ \boxed{135 \text{ L}}$.

The problem is that there are 22.4 liters per mole at STP, which is 0C at 1 atm, which is not the case here. You have to use the Ideal Gas Law, PV=nRT, (R=.08206, P in atm, V in L, n in moles, T in kelvin). Notice that at STP, this equation gives 22.4.

(Alternatively, you could "correct" your answer using Charles Law, $ 121.2L\times\frac {303K}{273K} \equal{} 135L$, which is really just a specific case of the Ideal Gas Law. This only works at constant pressure.)

Oh, right. I should have seen that. Thank you! :)


function
number theory
number theory unsolved

Define $f(n)$ is the number of divisor of $n^{2}+n+1$. Prove, there are infinite number of $n$ which satisfy $f(n)\geq f(n+1)$

2005, Korea

i don't know how i approch to solve. plz help![/code][/u][/quote]

[quote="Severius"]Let $g(n)=n^2+n+1$. Then $n^4+n^2+1=g(n^2)$, but also $n^4+n^2+1=(n^2+n+1)(n^2-n+1)=g(n)g(n-1)$, so $g(n)g(n-1)=g(n^2)$.

Now, suppose the result is false, so that there's some $N$ such that, for all $n \geq N$, we have $f(n)<f(n+1)$. Since $n^2<n^2+n+1<(n+1)^2$ is never a square, the number of divisors is even, so $f(n)+2 \leq f(n+1)$. This implies $f(n) \geq 2n-c$ for all large enough $n$ and some constant $c$.

Now, consider $f(n^2)$, the number of divisors of $g(n^2)$. Since $g(n), g(n-1)$ are clearly coprime, and the function of number of divisors is multiplicative, we have $f(n)f(n-1)=f(n^2)$. Now, a bound: the number of divisors of $k$ is at most $2\sqrt{k}$. This is obvious, because $k$ has at most $\sqrt{k}$ divisors not bigger than $\sqrt{k}$, and the larger ones are of the form $k/d$ for $d$ one of the previously noted divisors. From this it follows that, since $g(n^2)=n^4+n^2+1<(n^2+1)^2$, we have $f(n^2) \leq 2(n^2+1)$. But $f(n)f(n-1) \geq (2n-c)(2n-2-c)$ for large enough $n$, so $(2n-c)(2n-2-c) \leq 2(n^2+1)$, a contradiction, as the LHS kills the RHS for large enough $n$.[/quote]

Fantastic! I wonder how  you get this nice, so nice solution~!! so impressive~! :lol:

I'm interesting in Number Theory~~ but i couldn't find a good book about Number Theory.(text book or, problem book)

so Could you recomend me the decent Number Theory books? :blush:

I think I won't reach the idea that you posted~~

anyway, Thank you very much :P

That remember me an exercise from Engel, who based his proof on $x^2-x+1=x^2-2x+1+x-1+1=(x-1)^2+(x-1)+1$ and who studied a similar polyomial...


Just like the three mock ARMLs that Altheman made, just easier. :)
Would anybody be interested in making problems/competing?

Sure! I'd love to think up a few questions; just, don't ask me to do any for Power Rounds! :D
I don't like proofs too much...

We are only gonna make Individuals. Team events would be too difficult; where are we gonna get 45 or 60 teams?


function
ceiling function
absolute value
complex numbers

Some floor and ceiling properties like
ceil(lg(n+1))=floor(lg(n)+1)
and "Some properties of ceiling function" section under "http://en.wikipedia.org/wiki/Floor_function"

and summation properties like
"Identities" section under "http://en.wikipedia.org/wiki/Sum"
prove to me very useful for computer science study.

But I am interesting of their proof. Does any mathematics book or computer science book has provide proofs for the properties mentioned above?

Try proving them yourself...
And this is not Superior Algebra -> moved.

Certain function properties are by definition. For instance:

$ |a| = a$ if $ a > 0$
$ |a| =-a$ if $ a < 0$

Is the definition of an absolute value function. From definitions, it is usually possible to derive several useful properties.

Again, to the absolute value example, we would have $ |a| > 0$ as one property, $ \sqrt[2m]{a^{2m}}= |a|$ as another property simply by knowing that radicals are always greater than 0 (unless using complex numbers, of course) and the list goes on...

Now try and derive all of the floor and ceiling function properties based on its definition. For the wikipedia link, that would most likely be the first 1 or two properties stated.


trigonometry
AMC
AIME
function
geometry
AMC 10

[quote="confuted"][quote="dts"]On question number 23, wouldn't the 2002nd row you refer to actually be the 2001st row (that is the one that begins 1, 2001)?[/quote]
The first row is conventionally just "1"[/quote]
I thought that was referred to as row 0.

Grr I only did 12 because I accidentally thought that cos(pi) = 1, but at the last minute, I noticed that cos(pi) = -1, so that took up most of my time :(  I hope I didn't fail again.

I think there may have been a few mistakes, can I ask about them now?

this one is harder than A, I only did 16 questions!!

Yeah, it was tough. I only answered 15.

BTW, I'm just wondering if anyone else thought #12 had a mistake. We can't talk about the solution yet, but can anyone confirm this?

i personaly thought it was easier. i answered 13.

I think he PMs them to you...

I already got mine... I got a 90, exact same score as my first mock AMC

I found this one awfully hard for an AMC "10"...

or did I miss something?

Not that it matters much, but this one had one more geometry problem, defined by me as one that would require most people to draw a diagram

I got a 94.5, not the lowest score, unfortunatly :(

[quote="takuanitromars36"][/quote][quote="nr1337"][quote="JSRosen3"][quote="takuanitromars36"]Besides, both of my mock AMC A and mock AMC B scores were lower than my AMC 12B 2004 score. Maybe the AMC 12B 2004 was just easier than normal...

Anyways, that was how I qualified for the AIME this year...which I failed.[/quote]

Both of the mock tests were harder than the average AMC 12. I'm guessing Rep will lower the mock AIME qualifying score.[/quote]

I sure hope so.. mabye lower it to about 92?   :lol:[/quote][quote="takuanitromars36"]

Hopefully, it will be lower than the AIME qualifying score of the mock AMC A.[/quote]

If it is, then I qualify!  Yay!  But that's only if the qualifying score changes...

I think the fact that I answered 7 more questons than the last time makes this tests a whole lot easier.  I know I only got like 6 points higher, that was mainly becuase of stupid mistakes.

But I really really loved 14, the one with the ceiling and the wall.  That was the best question on the test.  (Except for 2 which was really easy since it was a sample problem on the AMC 10 website.)

#24 was ridiculously easy if you have a TI-89.

I thought it was the same difficulty. My score said it was easier, but that's mainly because of the AMC A.

I increased my score by 48.5 points! If I had just gotten one more question right, I would've more than doubled my old score.  :lol:

I'm handicapped with advanced calculators. How do you do that on the T-I 89?

I used the "factor" function and found that (2x-3)^2 was a factor. That's the same as (x - 1.5)^2, and you're done!

[quote="Ultimatemathgeek"]I took the test earlier than every one, so I do not know who is top 50%.
May I know who those are, so that I can ask these people questions?

Again, I bombed this test.  Life sux.  (I may turn suicidal now)[/quote]

Come on.  It is just a practice AMC.  It doesn't mean anything.  Even if it were the real thing you shouldn't be suicidal if you don't score as high as some people do.  If you want to do better put in the time and work, study, and you ll be much better.

Can you believe I got the days mixed up?!?!  I thought today was Wed. :(  :blush:

[quote="Mildorf"][quote="Ultimatemathgeek"]How come nobody posts solution for 21 thru 25 on the Mock AMC A on the official solution thread?
I need those badly, really.[/quote]

I would do that, I just need to see the questions again... speaking of which, Rep, can you please re-post them now that the contest is over, or are they posted elsewhere, in which case could someone point me to the correct thread?[/quote]
Original Problem
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14361
The solution thread
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14516

Hm, looks like I've missed a lot during my involuntary leave of absence.  Max, would you mind posting the AMC B problems again, preferably in the first post?  Thanks.


Some years ago, there were exactly four fridays and four mondays in August. On what day of the week did AUgust 20th fall that year?

Sunday, I'm the phychic guy on the TV who knows what day of the week a particular day is...for example if you say September 12, 1685 he'll answer what day of the week that was :)

Hmm, this reminds me, I should really learn the number of days that are in each month in case a calendar-type problems shows up on the AMC.

Easy, just use your knuckle: Make a fist, then look at the left most joint (bump) on your left hand. Since it is a bump, January has 31 days. All the bumps represent 31 days and all the concave places represent 30 days, except for February which is an exception (28 days; 29 days in a leap year)
You do this until you reach July, and then move on to the left most joint (bump) on your right hand and keep going from there. Why else would July and August is the only consecutive months with the same number of days? :)

Hmmmmmmmm is this question possible? August has 31 days right?

What if there were only 3 full weeks? What happens then?

You could find a pattern for August...in fact there's a pattern for every months, but I'll leave it up to you to discover the wonders of our calender :wink:


Find all positive integer solutions to
\[a^{2}+b^{2}=c^{3}\]

I think that it's posted before...use $Z[\sqrt{-1}].$


Let $ x,y,z > 0 $ and $ xyz = x+y+z+2 $ . Prove that :
 $ xy+yz+xz \geq 2(x+y+z) $

Let $x=\frac{1}{a},y=\frac{1}{b},z=\frac{1}{c}$.So the problem is equivalent with: if $ab+bc+ca+2abc=1$ then prove that $a+b+c\geq 2(ab+bc+ca)$.Here by $a=\frac{m}{n+p},b=\frac{n}{m+p},c=\frac{p}{m+n}$, by calculations reduces to a well-known one. I created this problemm 2 weeks ago and I was intending to propose it for the next year NMO. :(

Too bad you created it two weeks ago, since I posted it as my problem ( but after that found out that many others thought about it, normally)  very long time ago. There is a beautiful solution of Namdung. My solution was based on the substitution $a=\frac{x+y}{z}$ and so on. I put it in Old and New Inequalities, since I find it very nice.

Yeah , Mr Namdung gave me this problem a long time ago . It's Harazi's problem  :)

[quote="cezar lupu"]Actually, sorry, but this one is not Harazi's inequality. :P[/quote]
then who created the problem?

Siuhochung,this problem was created a long time ago.Who knowns? In fact,this is folklore. ;)

You're amazing, Cezar, really unbelievable. Some two days ago you were so proud of creating a new inequality, and now you say it's folklore. I really can't understand you. Too bad I'm not even trying too much.

Let's rewrite the inequality as $\sum (x-1)(y-1) = 3$, and let $a = x-1$, etc so $a,b,c \geq -1$.  Our given condition then becomes (after some computation) $ab+bc+ca+abc = 4$.  If one or all of $a,b,c$ are negative, then we are trivially done.  If $a,b,c \geq 0$, then AM-GM implies that $4 \geq abc+3\sqrt[3]{a^2b^2c^2}$ so $abc \leq 1$ and $ab+bc+ca \geq 3$.  If $a\geq 0$ and $b,c \leq 0$ then $0 < 4-bc = a(bc+b+c) = a((b+1)(c+1)-1)) \leq 0$, contradiction.

Excuse me, Harazi, but I found this inequality in an old G.M yesterday, so end of story. :P

Man, it's not me who is dead about names put on trivial inequalities, but this kind of attitude really lets me cold.


Correct me if I'm wrong, but there isn't an option to make the video full screen in Alcumus.  Is it possible to get such an option or has someone found a trick to viewing the videos in full screen?

The videos look pretty ugly at full screen, and not just because the instructor is funny-looking.  The writing on the board loses resolution at full screen.

I think the instructor looks pretty good. He should have gone into modeling, not teaching!


A quadrilateral ABCD, AD=BC, AC meets BD at O, the bisectors of the angles DAB and CBA meets at I. Prove that the midpoints of AB, CD, OI are colinear.

You have already been told about posting current problems in this thread:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167953[/url]

Do not do this again.


[quote="espark52"]2/3 of all mathletes like to play basketball. 3/4 of all mathletes lie to read. what is the greatest common fraction of all mathletes that could possible like neither of these activities?[/quote]
[hide="solution"]$\frac{2}{3}+\frac{3}{4}-1=\frac{5}{12}$
[/hide]
[hide="hint for those who don't get my solution"]If you have trouble following my solution pretend there are twelve people and see what you get[/hide]

[quote="humblemastermind"]That couldn't be right, because you know that at least $\frac{3}{4}$ of them like to do something. Here's what I get:
[hide="solution"] The greatest would be if everybody who liked to play basketball liked to read as well. That would leave $\boxed{\frac{1}{4}}$ left to like nothing. [/hide][/quote]
Never mind, I misread it humblemastermind is right as usual.

i got[hide]$\frac{1}{4}$ because the $\frac{2}{3}$ could be part of the $\frac{3}{4}$, leaving the $\frac{1}{4}$ that's left to be our answer[/hide]
-jorian


function
number theory proposed
number theory

Here $ \tau(n)$ denotes the positive divisors of $ n$. Show that $ \tau(n) < 2\sqrt{n}.$

Read the answer I gave you on the other problem of that kind  :wink:

Just to make it clear: Instead of complaining about my help, you could try to understand the hint or you could nicely ask for another hint.
And take one more look at the other thread now ( http://www.mathlinks.ro/viewtopic.php?t=246771 ).


function
real analysis
real analysis unsolved

Given functions $f,g:{\mathbb{R}}{\rightarrow}{\mathbb{R}}$.$f$ and $g$ are periodic and continious functions.Define $h(x)=f(xg(x)),x\in{\mathbb{R}}$.Can $h$ be periodic?

There are trivial solutions when $f$ or $g$ are constant. These are the only solutions. WLOG, assume that $f$ and $g$ each have period 1; we can scale $g$ vertically and $f$ horizontally leaving $h$ unchanged, or we can scale $g$ and $h$ horizontally. Also WLOG, assume $g$ is somewhere positive; we can use negative factors in that scaling if necessary.

Any continuous periodic function is uniformly continuous, so if $h$ is periodic it is uniformly continuous.
Suppose that $g(b)-g(a)=c$, for some $b>a>0,c>0,g(a)>0$; such values can be found if $g$ is nonconstant. We can find $b_n,a_n\in [a,b]$ such that $g(b_n)-g(a_n)\ge\frac cn$ and $b_n-a_n=\frac{b-a}{n}.$ Then $(kn+b_n)g(b_n)-(kn+a_n)g(a_n)>kn\frac cn=kc$. Choosing an integer $k$ such that $kc\ge1$, we have that $h$ takes the entire range of $f$ in the interval $[kn+a_n,kn+b_n]$, so if $f$ is nonconstant, $h$ fails to be uniformly continuous and we are done.

[quote="jmerry"] WLOG, assume that $f$ and $g$ each have period 1; we can scale $g$ vertically and $f$ horizontally leaving $h$ unchanged, or we can scale $g$ and $h$ horizontally. Also WLOG, assume $g$ is somewhere positive; we can use negative factors in that scaling if necessary.

[/quote]
Dear [b]jmerry[/b] please a little bit details.

Let the period of $f$ be $a$ and the period of $g$ be $b$. Define $F(x)=f(ax)$. If $g(x)$ is positive somewhere, define $G(x)=\frac bag(bx)$. Then $F(xG(x))=f(axG(x))=f(bxg(bx))=h(bx)$. It's not exactly what I said earlier, but it works.
If $g$ is nonpositive everywhere, use $-a$ instead of $a$: $F(x)=f(-ax)$, $G(x)=-\frac bag(bx)$


If so, $\mathbb{Q}(\sqrt[p]{3})\subseteq\mathbb{Q}(\sqrt[7]{53})$ therefore $[\mathbb{Q}(\sqrt[p]{3}): \mathbb{Q}]\leq [\mathbb{Q}(\sqrt[7]{53}): \mathbb{Q}]$, or $p\leq 7$. Now see at $p=2,3,5,7$, i think it is easy!  :wink:


Prove that among the elements of the sequence $\left( \left\lfloor n \sqrt 2 \right\rfloor + \left\lfloor n \sqrt 3 \right\rfloor \right)_{n \geq 0}$ are an infinity of even numbers and an infinity of odd numbers.

Suppose contradiction, then starting from one $N$ our numbers $a_n = [n\sqrt{2}]+[n\sqrt{3}]$ are of the same parity.
From Kronecker theorem we know that there exist infinite number of numbers $n$ such that $\displaystyle \{n\sqrt{2}\} < 3-2\sqrt{2}$.
Choose one greater than $N$.

Let  $[n\sqrt{2}]=k$ , then easy to see $[(n+1)\sqrt{2}]=k+1$, $[(n+2)\sqrt{2}]=k+2$.

Suppose $[n\sqrt{3}]=l$ , $2 | (a_{n+1}-a_n)$, hence $[(n+1)\sqrt{3}]=l+1$.

(if $[(n+1)\sqrt{3}]= l+3$ or greater :  $3=[(n+1)\sqrt{3}]-[n\sqrt{3}]<(n+1)\sqrt{3}-(n\sqrt{3}-1)=1+\sqrt{3}$ -contradiction.)

Analogously $[(n+2)\sqrt{3}]=l+2$, thus $2=[(n+2)\sqrt{3}]-[n\sqrt{3}]>((n+2)\sqrt{3}-1)-n\sqrt{3}=2\sqrt{3}-1$ - contradiction and we are done.

A much more difficult question: is it true that the density of those numbers $n$ for which the terms are even is $1/2$?


To determine your target heart rate for exercising, you subtract your age from 220 and multiply that difference by 0.75. Madeline has a target heart rate of 129. How many years old is she?

We let x be her age. So $ (220\minus{}x).75\equal{}129$. Thus $ 220\minus{}x\equal{}172$, and $ x\equal{}\boxed{48}$.


It can be proved by induction that $\sum_{1\leqslant k\leqslant n}\frac{k+1}{2^{k}}= \frac{3\cdot 2^{n+1}-n-3}{2^{n}}$.
[b]edit:[/b] I corrected two typos  :blush: .


Where could I find the 2006 AMC 12A problems?

Thank you,
lyra

Go to the links at the top of the forum and click "Contests."
Then under National and Regional Competitions, click "USA."
From there its obvious, AMC 12, 2006.

Or if you're looking for a specific problem, they've all been posted in the board recently.

Just for clarification, if you're using the mathlinks skin, it's under resources.

[quote="lyra"]Thank you [i]very[/i] much!!!

Also is it true that some of the problems on the 12A are the same on the 12B?[/quote]

No. Some problems on AMC-10A appear on AMC-12A and some problems appear on AMC-10B appear on AMC-12B. There is no corresponding problems or concepts between A and B tests (if there were clear correpsonding concepts, it will make B too easy) although the area and difficulty of the both test should be same. Over the years, it was like that but the exception did occur. For instance, 2004 AMC-12A, if I recall correctly, was far more difficult than 2004 AMC-12B.


modular arithmetic
function
algebra
binomial theorem

We have $ 2009^{2009}\equiv9^{2009}\pmod{1000}$.

[hide="Solution 1"]
Since $ \lambda(1000)\equal{}200$ ([url=http://www.artofproblemsolving.com/Wiki/index.php/Carmichael_function]Carmichael Function[/url]), $ 9^{2009}\equiv9^9\equiv9((9^2)^2)^2\equiv9(81^2)^2\equiv9(561)^2\equiv9(721)\equiv\boxed{489}\pmod{1000}$.
[/hide]
[hide="Solution 2"]
From the Binomial Theorem,

$ 9^{2009}\equiv(10\minus{}1)^{2009}\equiv\dbinom{2009}2(10)^2(\minus{}1)^{2007}\plus{}\dbinom{2009}1(10)(\minus{}1)^{2008}$

$ \plus{}\dbinom{2009}0(10)^0(\minus{}1)^{2009}\equiv\minus{}36\cdot100\plus{}9\cdot10\minus{}1\equiv\minus{}3511\equiv\boxed{489}\pmod{1000}$.
[/hide]

 :P Happy New Year!

[hide="Solution 3 :D"]You can also use Euler's totient theorem.

$ \varphi(1000)\equal{}\varphi(2^3\cdot5^3)\equal{}1000\cdot\frac{1}{2}\cdot\frac{4}{5}\equal{}400\implies 9^{400}\equiv 1\pmod {1000}$.

$ 2009\equiv 2000\plus{}9\equiv 9\pmod {400}$

And you have $ 9^9\equiv \boxed{489} \pmod{1000}$.

It works in essentially the same way, but $ \varphi(1000)$ is easier to compute than $ \lambda(1000)$.[/hide]

This thread is old, but still: regarding solution one, lambda of 1000 is 100, not 200 :D

Much love for the binomial theorem; it still seems like the most straightforward and simple solution.


Let $ f: [0;\infty]\to{\mathbb{R}}$, a continuos function and derivable on $ (0;\infty)$ so that $ f'(x)\ge{1}$ for every $ x\in{(0,\infty)}$ and $ f(0)<0$.Prove that exist and it's unique $ a\in{(0;\infty)}$ so that $ f(a)\equal{}0$.

If $ f'(x)\geq 1$ then $ f(x)\geq x \plus{} \textbf{const}$. This means that when $ x\to \plus{} \infty$ then $ f(x)\to \plus{} \infty$. And because $ f(0) < 0$ we get that exist such $ a\in (0, \plus{} \infty)$ that $ f(a) \equal{} 0$.


function
logarithms
inequalities unsolved
inequalities

If $xyz \geq abc$ then
\[f(x)+f(y)+f(z) \leq f(a)+f(b)+f(c)\]
Find that what kind of function $f(n)$.

Are you sure with the statement? What values can $x,y,z,a,b,c$ take?
Suppose $b=c$, $y=z$, then $0\cdot yz\geq 0\cdot bc$ and $f(0)+f(y)+f(y) \leq f(0)+f(b)+f(b)$, i.e. $f(y)\geq f(b)$, where $b$ and $y$ are arbitrary numbers. It follows that $f=\textrm{const}$.

I suppose $x,y,z,a,b,c$ should be positive.

For $z\ge c$ we have $1\cdot 1\cdot z\ge 1\cdot 1\cdot c$. Thus $f(1)+f(1)+f(z)\le f(1)+f(1)+f(c)$, so $f$ is a decreasing function.

Now we have $e^X\cdot e^Y\cdot 1\ge e^{X+Y}\cdot 1\cdot 1$ and $e^{X+Y}\cdot 1\cdot 1\ge e^X\cdot e^Y\cdot 1$, so $f(e^{X+Y})+f(1)+f(1)\le f(e^X)+f(e^Y)+f(1)$
and $f(e^X)+f(e^Y)+f(1)\le f(e^{X+Y})+f(1)+f(1)$, thus we have equality.
Thus the function $g(x)=f(e^x)-f(1)$ satisfies $g(X)+g(Y)=g(X+Y)$. As $g$ is a decreasing function, the only sollutions are $g(x)=\lambda x$ where $\lambda \le 0$.
This implies $f(x)= \lambda \ln x +f(1)$, and indeed all functions $f(x)= \lambda \ln x +\mu$ ($\lambda\le0$) satisfy the inequation:
For $xyz\ge abc$ we have $\lambda \ln (xyz)\le\lambda \ln (abc)$ (remember that $\lambda\le 0$) , thus $f(x)+f(y)+f(z)=\lambda \ln x+\mu+\lambda \ln y+\mu+\lambda \ln z+\mu=\lambda \ln (xyz)+3\mu\leq \lambda \ln (abc)+3\mu=\lambda \ln a+\mu+\lambda \ln b+\mu+\lambda \ln c+\mu =f(a)+f(b)+f(c)$

[quote="mathpk"]Sorry Myth :blush: . $f(n)$ is $R^+ \rightarrow R^+$ function.[/quote]
I think it should be $R^+\rightarrow R$.
Otherwise the only solution would be a constant, too.

[quote="Myth"]Just a remark.
In my opinion, $R^+$ is quite bad notation generally. I am sure, it is much better to indicate each time whether we include 0 to the set or not. For example, notation $\mathbb R^+_0$ says for me that it includes 0, meanwhile everytime I see $R^+$ i don't know lies 0 there or not.[/quote]

I like $R_{\geq 0}$, $\mathbb{Z}_{> 0}$, etc.

[b]Extremely strange Tex error (AoPS bug report)[/b]:
the subscript for $R$ displays improperly (as $>0$) when I post just the line above,
but displays correctly (as $\geq 0$) when I include the Tex code of the subscript later
in the message (" \ geq 0", as in the preceding inequality).  In other words, if you delete
everything in this message [i]after[/i] the word "etc." above, the Tex [i]preceding[/i] it will appear differently.
Try it!


I am interested in a graduate degree in applied mathematics.

I'm expected to graduate from the cooper union next year with a gpa of approximately 3.4. I did decently well (at least for some) on the putnam last year with an 11 and I'm pretty sure I got at least 1 question completely right this year.

cooper has the equivalent of grade deflation. As far as I know, the average gpa is about 2.5 and is only that high because any semester's gpa under a 1.5 as well as any 2 semesters under a 2.0 are grounds for dismissal. hopefully graduate schools will take that into account.

also, the math department at cooper is ridiculously small, with only a few classes offered every semester. I'm taking as many as I can, but it won't be anything compared to a math major at a large university.

how much preparation do I need for grad school? do they start at a lower level to allow people to catch up or do I need to hit the ground running?

how much is gpa a factor in grad. school admissions?

what schools should i expect to be able to get into? <- I realize that this question is very loaded but I'd still like to get peoples' opinions.

Thanks

I hope mb1551 don't mind me sticking in a few of my own (somewhat related) questions.

What system is used by the universities when translating international grades to a GPA score? The grades we are given in my country differ. At my present univ we are given grades in four steps (U, 3, 4, 5). Next year I am switching too another univ where they give grades in three steps (U, G, VG). I have also read a few (language) courses where you only get U (Failed) or G (Passed). More info: [url]http://en.wikipedia.org/wiki/Grade_(education)#Sweden[/url]

Thank you.

I think that having good letters of recomendation is very important in applying to graduate school. I think a GPA of 3.4 is fine, if your GPA in your classes for you major is higher make sure you point this out in you application.
Also I think finding a graduate program that fits you best is key. If you think that you may not be as prepared as if you had gone to a big university, then maybe going for your Masters at a smaller school may be for you. There a plenty of good MS programs out there that will give you the oppertunity to do research, and that will prepare you for a Ph.D program at a larger university.
Also, don't be afried to call up the graduate admissions office and ask them any questions that you may have reguarding your application.
i.e. how they would look at international grades.


Let {a_n} be a sequence of nonnegative real numbers satisfying a_{n+m)
<= a_n + a_m, for all n, m = 0, 1, 2, ... . Show that {(a_n)/n}
converges.

The proof is the same as for an existence of spectral radius for linear bounded operator ($\lim_{n\to\infty}\sqrt[n]{||A^n||}$), because the only fact it uses is $||A^{n+m}||\leq ||A^n||\cdot||A^m||$.

Let $a=\inf\frac{a_n}{n}$. Then $\frac{a_n}{n}\geq a$ for all $n\in\mathbb{N}$.
Let $\varepsilon>0$. There exists $d\in\mathbb{N}$ s.t. $\frac{a_d}{d}<a+\varepsilon$. Suppose $N>d$ and $\frac{\max\limits_{k=1..d} a_k}{N}<\varepsilon$. Consider arbitrary $n>N$. We can write $n=ds+p$, $p<d$. Then
\[a\leq\frac{a_n}{n}\leq \frac{s\cdot a_d+a_p}{n}\leq\frac{s\cdot a_d}{ds}+\varepsilon<a+2\varepsilon.\]


Sarah places four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size. She then pours half the coffee from the first cup to the second and, after stirring thoroughly, pours half the liquid in the second cup to the first. What fraction of the first cup is now cream?

Now I should get some sleep...(trying to spend a night,because it's fall break and I can.. :lol:)
P:S I posted this at 11:43 PM

[hide]
C=Amount of Coffee.
S=Amount of Cream
\begin{eqnarray*} \text{Step 1} & : & 2C + 4S\\ \text{Step 2} & : & 3C + 2S \end{eqnarray*}
So the fraction amount of cream is $ \frac {2}{5}$
[/hide]

Correct.... but this was for our MathCounts members.. but oh well..Good job..

[quote="danb80"]If you want it to be only for JMS students.  Make sure you have Jenks in the title. such as "Jenks:  Work on these"[/quote]

Okay... but I don't know if that will stop from other people coming in and answer...

[quote="hunter34"]We may after you you guys solve it.[/quote]
That is definitely fine with me, or you guys can help us by posting hints. :lol:

whats the point of solving a problem if someone else already solved it?

They just want to try them out for fun, I guess...

Spider:  The reason why the answers are hidden is so that others can try to solve it without seeing everyone else's attempts.

The point of solving a problem when someone else already has is to see if you know [u]how[/u] to solve it.  There will always be more problems like it later.


For positive integers $ a_1,a_2,...,a_n$, let $ A,G,H$ be its arithmetic, geometric, and harmonic mean, respectively.

Prove that
for $ n$ even, $ \frac{A}{H}\leq \minus{}1\plus{}(\frac{A}{G})^n$, and
for $ n$ odd, $ \frac{A}{H}\leq \minus{}\frac{n\minus{}2}{n}\plus{}\frac{2(n\minus{}1)}{n}(\frac{A}{G})^n$

Sorry, if $ n\equal{}2k$ and $ a_1\equal{}a_2\equal{}...\equal{}a_{2k}$ then $ A\equal{}G\equal{}H,$  but inequalilty $ 1 \leq
0$ is false!  :?:

Sorry... let me post again.

For positive integers $ a_1,a_2,...,a_n$, let $ A,G,H$ be its arithmetic, geometric, and harmonic mean, respectively.

Prove that
for $ n$ even, $ \frac {A}{H}\leq \minus{} 1 \plus{} 2(\frac {A}{G})^n$, and
for $ n$ odd, $ \frac {A}{H}\leq \minus{} \frac {n \minus{} 2}{n} \plus{} \frac {2(n \minus{} 1)}{n}(\frac {A}{G})^n$

Now I hope it is not wrong


We can just consider one section with the cross-like figure. This section has a total of $ 6 \times 6 \equal{} 36$ squares. Counting shows that 16 of these squares are shaded, so the answer is $ \frac{16}{36} \equal{} \boxed{\frac{4}{9}}$,

look at the 3*3 square in the upper left hand corner. 4 out of 9 of those tiles are black... so the answer is 4/9.

Try to simplify as much as possible. Of course, izzy's way work too.


Find all positive real solutions to $x^{3}+y^{3}+z^{3}= x+y+z$, $x^{2}+y^{2}+z^{2}= xyz$

[quote="sen"][hide]From AMGM, $xyz\geq27$

and perhaps power means, i dont think any solution exists.[/hide][/quote]
This is interesting because one can also prove that $xyz \leq \frac13.$ This combined with the result shown above would yield a contradiction and hence no solution. This is what I did.

By Cauchy-Schwarz, we have
${(x\cdot 1+y\cdot 1+z\cdot 1)}^{2}\leq (x^{2}+y^{2}+z^{2})(1^{2}+1^{2}+1^{2})$
$\Rightarrow{(x^{3}+y^{3}+z^{3})}^{2}\leq 3xyz$ ... (I)

But, from AM-GM, we also have
$x^{3}+y^{3}+z^{3}\geq 3xyz$ ... (II)

(I) and (II) combined gives ${(3xyz)}^{2}\leq 3xyz \Rightarrow xyz \leq \frac13.$

Could you show how you got $xyz \geq 27$  :?:

So this means there are no positive real solutions to the equations. Very cool! :)

However, I think the important question to ask here is, how do we "guess" that there might not be a solution at all to such problems? Having some kind of knowledge "beforehand" would greatly ease our efforts.

Anyone would like to comment?

Well, here's another way I think.

Let $x^{3}+y^{3}+z^{3}=x+y+z=a$ and $x^{2}+y^{2}+z^{2}=xyz=b$

We have that:

$x^{3}+y^{3}+z^{3}-3xyz=(x+y+z)[(x^{2}+y^{2}+z^{2})-3(xy+yx+yz)]=$

$a-3b=a(b-3c)=$

$(a+3)(b+1)=3-3ac$

Clearly the RHS $\leq 0$, however, the LHS is always $>0$. Therefore, we have no positive real solutions.


1. For a set $ S$ of integers, define $ S\plus{}1$ to be $ \{x\plus{}1: x \in S\}$. How many subsets of S of $ \{1,2,3, \cdots. n\}$ satisfy $ S \cup (S\plus{}1) \equal{} \{1,2,3, \cdots, n\}$? What about a generalisation?

2. Find the number of subsets of $ \{1,2,3, \cdots. n\}$ that contain no two consecutive elements of $ \{1,2,3, \cdots. n\}$.

3. Fix positive $ a,b,c$. Define a sequence of real numbers by $ x_0 \equal{} a, x_1 \equal{} b \ \mbox{and} \ x_{n\plus{}1}\equal{}cx_nx_{n\minus{}1}$. Find a formula for $ x_n$.

4. For each $ n \ge 1$, we call a sequence of $ \mbox{n}$ (s and n)s 'legal' if the parentheses match up. For example, if $ n \equal{}4$, the sequence (()()()) is legal, but ()())(() is not. Let $ l_n$ denote the number of legal arrangements for the $ 2n$ parentheses. Find a recurrence formula for $ l_n$.

Thanks

for each $ n$ the set $ S$ must

(1) contain $ 1$ and $ n\minus{}1$.
(2) not contain $ n$.
(3) not skip 2 consecutive elements of $ \{1, ..., n\}$.

let $ a_n$ be the number of such sets $ S$ for the natural number $ n$. there are two possibilities for $ S$: (1) $ n\minus{}2\in S$, (2) $ n\minus{}2\not\in S$. in case (1), there are $ a_{n\minus{}1}$ possible sets. in case (2), we must have $ n\minus{}3\in S$, and so the number of sets in this case is $ a_{n\minus{}2}$. thus we get the recursion $ a_n\equal{}a_{n\minus{}1}\plus{}a_{n\minus{}2}$. the rest is easy...

Oh I see thank you, it is just Fibonacci, what about Q 1?

[hide]
Define $ \alpha_{n}$ to be the exponent for base $ a$, $ \beta_{n}$ for $ b$, and $ \gamma_{n}$ for $ c$ then it's clear that the $ x_n = a^{\alpha_{n}}b^{\beta_{n}}c^{\gamma_{n}}$ and can be proven by induction.
So
$ x_0 = a^1 b^0 c^0$ so $ \alpha_{0} = 1$, $ \beta_{0} = 0$, $ \gamma_{0} = 0$
$ x_1 = a^0 b^1 c^0$ so $ \alpha_{1} = 0$, $ \beta_{1} = 1$, $ \gamma_{1} = 0$
are boundary constraints for $ \alpha_{n}$,$ \beta_{n}$, $ \gamma_{n}$.

$ x_{n + 1} = a^{\alpha_{n + 1}}b^{\beta_{n + 1}}c^{\gamma_{n + 1}} = cx_{n - 1}x_{n - 2} = c a^{\alpha_{n - 1}}b^{\beta_{n - 1}}c^{\gamma_{n - 1}} a^{\alpha_{n - 2}}b^{\beta_{n - 2}}c^{\gamma_{n - 2}} = a^{\alpha_{n - 1} + \alpha_{n - 2}}b^{\beta_{n - 1} + \beta_{n - 2}}c^{\gamma_{n - 1} + \gamma_{n - 2} + 1}$

We found the set of linear recurrence relations

$ \alpha_{0} = 1$
$ \alpha_{1} = 0$
$ \alpha_{n + 1} = \alpha_{n} + \alpha_{n - 1}$

$ \beta_{0} = 0$
$ \beta_{1} = 1$
$ \beta_{n + 1} = \beta_{n} + \beta_{n - 1}$

$ \gamma_{0} = 0$
$ \gamma_{1} = 0$
$ \gamma_{n + 1} = \gamma_{n} + \gamma_{n - 1} - 1$

each of this expressions can be denoted using $ f_n$ sequence where $ f_0 = 0, f_1 = 1, f_{n + 1} = f_n + f_{n - 1}$ called Fibbonaci sequence.

Is clear that $ \alpha_{n} = f_{n - 1}$ for $ n \ge 2$, $ \beta_{n} = f_{n}$ for $ n \ge 2$ and $ \gamma_{n} = f_{n - 2} - 1$ for $ n \ge 2$

So the eventual answer is $ x_n = a^{f_{n - 1}} b^{f_{n}} c^{f_{n - 2} - 1}$

Last thing to do is to prove this result by induction.

[/hide]

Let denote the number of such subsets from $ \{1,2,3,\cdots. n\}$ which not contain $ n$ $ a_n$, and which contain $ n$
$ b_n$. Simply we get $ a_n\equal{}a_{n\minus{}1}\plus{}b_{n\minus{}1}$ and $ b_n\equal{}a_{n\minus{}1}$. So $ a_n\equal{}a_{n\minus{}1}\plus{}b_{n\minus{}1}\equal{}a_{n\minus{}1}\plus{}a_{n\minus{}2}\equal{}F_n$, and $ b_n\equal{}a_{n\minus{}1}\equal{}F_{n\minus{}1}$ where $ F_n$ is a Fibonacci number, and a number we are looking for is $ c_n\equal{}a_n\plus{}b_n\equal{}F_{n\plus{}1}$


Where do I get my school CEED code? Would the website have it?

http://apps.collegeboard.com/cbsearch_code/codeSearchHighschool.jsp

I think it is called CEEB code, if that's what you were thinking of.

Oups, yeah. I was typing too quickly and lost track.


So what are AoPSers/Mathlinkers doing over the summer? (I can't think of anything to do other than math! :) )

(For the poll, put what you're doing [i]most[/i].)

[quote="Spoon"]whoa... no working? I would think that should be on there.[/quote]

Oh sorry...didn't think of that :blush:

working, math, and living on AOPS, tennis, watching tv, oh yeah, vacationing too :surf:

But Math is the number one priority.

camping, taking robotics at U of Washinton and AoPS.

local tennis camp, IS, SAT practice

sounds fun huh?

well actually the first 3 choices apply for me. i've been sleeping a lot but when i wake up i do some math and then come here to do some problems or whatnot, and then some more math followed by some reading. i make time for video games as well. i need to be preparing more for the SAT. my reading and grammer are kinda low. and then at 1 in the morning i pass out and start the day over again around 9.

[quote="furious"]...and then at 1 in the morning i pass out and start the day over again around 9.[/quote]

.... interesting :D I'm doing lots of math, AoPS, and teaching myself Pre-calc :) I'm almost done, too. I should be done in about 2 weeks hopefully.

math (AoPS as well as other stuffs) and visiting my native country for two weeks :)

4 summer college classes, 3 AOPS classes.

That about sums it up for the rest of my summer, I already went on my vacation (albeit short, it was all that would fit between my two summer sessions).

Not a bad way to go, just gotta keep up with the work, since summer college classes need to go 3 times the normal speed to get all of the stuff done. And AOPS classword wouldn't be too hard to ignore, so gotta keep focus.

I'm doing everything on there except for the math camp part. I dwell on AoPS about 3 hours a day, edging over the 2.5 hours of math I do a day.

[quote="chess64"][quote="WindSlicer"]4 summer college classes, 3 AOPS classes. [/quote]

 :o Cool! What grade are you in? And how are you taking 3 AoPS classes? Intro Counting, Intermediate Algebra, and Olympiad Geometry? :rotfl: That's kind of funny if it's true.[/quote]

I'm going into 11th.

Intro Counting/IntermediateAlgebra/AIMEA

Learning about Genetics, Biotech, Medicine + Health Care, but most importantly, learning about myself. Summer's been a blast  :lol: I hope it has been for everyone else here on the forums as well.

I taught myself next year's course, algebra.  Next would be math, then AoPS, and i had a one week vacation.  I sleep about 6 hours in the summer :o which is a lot for me

[quote="jhollenbeck"]I sleep about 6 hours in the summer :o which is a lot for me[/quote]
Does that mean you slept only 6 hours the whole summer?  ;)

Um, MOP, China, AOPS, Other math, Sleeping, Does eating count?

[quote="aznmonkey1992"]my mom gives me hmwork every day!!! it sucks!!! :mad:  :mad:  :mad:  :mad:  :mad:  :mad:[/quote]

Ah, you should be worried when your mom doesn't give you any work because then it's up to you to do stuff and improve your math and whatever else. And most people don't have enough self control - like me. And you end up doing nothing all day.  :P

Extremely sorry for the double post!

I can't find a way to delete the other one.

No, i sleep about 6 hours a day! Not the whole summer :P


AMC
AIME
geometry
analytic geometry
linear algebra
matrix

When I looked at the answer for [url=http://www.artofproblemsolving.com/Wiki/index.php/1986_AIME_Problems/Problem_15]#15 1986 AIME[/url], I figured everything out until I got to the $ \left|\frac {3}{2}ab\right|$. It says that you use the determinant product to find the area of the triangle. I searched the wiki, but couldn't find any article on finding the area of a triangle. Please help :(

Thanks :lol:

(if this is in the wrong spot, would a moderator please move it?)

It's another name for the [url=http://www.artofproblemsolving.com/Wiki/index.php/Shoelace_Theorem]Shoelace Theorem[/url].

Yes but in the problem (or the solution), it doesn't actually give you the coordinates of the points. The shoelace theorem gives you the area when you know the points. Right now, we only know the product $ ab$. So how would you find the area without knowing the points?

The Shoelace Theorem here for the points $ (a,a),(b,2b),(\minus{}a\minus{}b,\minus{}a\minus{}2b)$ can also be expressed as half the absolute value of $ \begin{vmatrix}a&a&1\\b&2b&1\\\minus{}a\minus{}b&\minus{}a\minus{}2b&1\end{vmatrix}$, and then we can use determinant properties (adding or subtracting multiples of any row or column to another row or column doesn't change the value of the determinant) to get this to $ \begin{vmatrix}a&a&1\\b&2b&1\\\minus{}a\minus{}b&\minus{}a\minus{}2b&1\end{vmatrix}\equal{}\begin{vmatrix}a&a&1\\b&2b&1\\0&0&3\end{vmatrix}\equal{}3(\minus{}1)^{3\plus{}3}\begin{vmatrix}a&a\\b&2b\end{vmatrix}\equal{}3(2ab\minus{}ab)\equal{}3ab$, and the area is $ \frac12|3ab|\equal{}\left|\frac32ab\right|$.

Sorry for being so stupid  :oops: , but I don't know how to like take the determinant for 3x3 matrices, 4x4 etc, I only know 2x2 , so can you tell me a general formula for the determinant of an n x n matrix?

Oh, sorry for not explaining. The determinant $ \begin{vmatrix}a_{11} & a_{12} & \ldots & a_{1n} \\
a_{21} & a_{22} & \ldots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & \ldots & a_{nn}\end{vmatrix}$ (where all of these are in the form $ a_{ij}$ for row $ i$ and column $ j$) is $ a_{ij}( \minus{} 1)^{i \plus{} j}\begin{vmatrix}a_{11} & \ldots & a_{1(j \minus{} 1)} & a_{1(j \plus{} 1)} & \ldots & a_{1n} \\
\vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\
a_{(i \minus{} 1)1} & \ldots & a_{(i \minus{} 1)(j \minus{} 1)} & a_{(i \minus{} 1)(j \plus{} 1)} & \ldots & a_{(i \minus{} 1)n} \\
a_{(i \plus{} 1)1} & \ldots & a_{(i \plus{} 1)(j \minus{} 1)} & a_{(i \plus{} 1)(j \plus{} 1)} & \ldots & a_{(i \plus{} 1)n} \\
\vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\
a_{n1} & \ldots & a_{n(j \minus{} 1)} & a_{n(j \plus{} 1)} & \ldots & a_{nn}\end{vmatrix}$ for any choice of $ i$ and $ j$ (so just remove row $ i$ and column $ j$ and calculate the determinant of the remaining matrix). Hence it is a good idea to use row and column operations to get as many zeroes into a single row or column as possible.

In the future, please post questions about a problem or its solution to the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=394411#p394411]original thread[/url]. That way, all the discussion is in the same place and the next person with your question won't have to post it again.


(a) Prove that for all $n \in \mathbb N^\ast$, by dividing $3^{2^{n}}: 2^{n}$ with remainder, one obtains an even quotient.
(b) Prove that the sequence $\left\{ a_{n}\right\}_{n \geq 1}$ given by \[a_{n}= \left\lfloor \frac{3^{n}}{n}\right\rfloor, \, \forall n \in \mathbb N^\ast \] has an infinity of even elements, an infinity of elements divisible by $3$ and an infinity of elements divisible by $5$.

Hey

$\frac{3{}^{2}{}^{n}-1}{2^{n}}$ is a integer number

It's better proove this, dont you think?!

because the remainder of $3{}^{2{}^{n}}: 2^{n}$ is 1

is this true?

[hide="b"]
For all $n$ in the infinite set $\{3^{k}|k\in\mathbb{Z}^{+}\}$, we find that $a_{n}= \left\lfloor\frac{3^{3^{k}}}{3^{k}}\right\rfloor = 3^{3^{k}-k}$ which is clearly divisible by 3.

For the second part, let $p>5$ be a prime in the form $4k+1$. We claim that $5|a_{p}$. We know that $3^{p-1}\equiv 1\pmod{p}\iff \frac{3^{p}-3}{p}\in\mathbb{Z}$ by Fermat's Little Theorem. Therefore, $\left\lfloor \frac{3^{p}}{p}\right\rfloor = \frac{3^{p}-3}{p}$. Notice that $3^{p}=3^{4k+1}\equiv 3\pmod{5}$, so 5 divides $3^{p}-3$. Since $5\nmid p$, we conclude that 5 divides $\frac{3^{p}-3}{p}$, and we are done. Oh yeah, and the set $\{n|n=4k+1\in\mathbb{P}\}$ is also infinite (well-known result).
[/hide]


[quote="Allan Z"]what is the minimum value of:
$\cos(\theta)+\sec(\theta)$
A.0
B.1
C.2
D.4
E. none of the above

if this is on the AMC, which question should this be?[/quote]
[hide="Without doing any calculations"]cosine can be negative so the answer is $E$[/hide]

ya i think this would be like Part A..cuz its just a thinking question really

You could also take the derivative and find critical points if answer E was not present (and a choice provided the actual answer).

[quote="Allan Z"]what is the minimum value of:
$\cos(\theta)+\sec(\theta)$
A.0
B.1
C.2
D.4
E. none of the above[/quote]

Doesn't the minimum value of $\sec{(\theta)}$ not exist?  Becuase secant can be from $-\infty$ to $-1$ and $1$ to $\infty$...

[quote="Treething"]this question's basically "what's the minimum value of $x+\frac1x$, with $-1\leq x\leq 1$"[/quote]And when $x\rightarrow0^{-}$, $x+\frac1x\rightarrow-\infty$ :wink:

[quote="Zellex"]Maybe you should change the problem to...

"What is the minimum of $\sin \theta+\cos \theta$

Yeah..something like that...dunno

Add coefficients in front of $\sin$ and $\cos$ to make it a little harder[/quote]

umm if you wanna find the min of $a\sin(\theta)+b\sin(\theta)$.. a simple cauchy would yield the answer..
I posted this because I asked many people about this question.. and all got "2".. after using AM-GM, neglecting that $\cos\theta$ can be negative

For the other proposed problem, you can also take derivative. Go Calc!

[quote="Zellex"]Maybe you should change the problem to...

"What is the minimum of $\sin \theta+\cos \theta$

Yeah..something like that...dunno

Add coefficients in front of $\sin$ and $\cos$ to make it a little harder[/quote]
[hide]$-\sqrt{1-\sin2x}$
$-1$[/hide]
Right?
EDIT: turns out I can't add.
[hide]
-$\sqrt{1+\sin2x}$
$-\sqrt{2}$[/hide]

Or you could just know this formula off the top of your head: \[{a \cos \theta+b \sin \theta = ( \sqrt{a^{2}+b^{2}}) \cdot \cos ( \theta+\arctan \frac{-b}{a}+k \pi}), \ k \in \{ 0,1 \}\] like the cool people do!  :D


Registration has begun for the 2009 [color=indigo]Purple Comet[i][size=200]![/size][/i] Math Meet [/color]which will take place the week of April 27 through May 3 at [url]http://purplecomet.org[/url]. This is a [b]FREE, on-line, international, team, mathematics contest[/b] for middle and high school students. Teams are made up of from one to six students. The students on a team can either be all from the same school and compete as a small school team or a large school team, or the students can be from different schools and compete as a mixed team. See complete rules and problems from our past contests at our web site and download a printable contest poster.

Middle School teams have 60 minutes to work on 15 problems.
High School teams have 90 minutes to work on 25 problems.
Choose the time period during the seven day contest that works best for you.

[b]NEW THIS YEAR[/b]: The contest includes a full seven days so that teams can compete on the weekend.

[b]NEW THIS YEAR[/b]: We will produce award certificates for winning team members and certificates of participation for other participants.

[b]NEW THIS YEAR[/b]: Once a team is registered, it can run one of our past contests for practice exactly like the real contest will work.

[b]NEW THIS YEAR[/b]: We are supplying our problems in [b]English, Chinese, Russian, French, Spanish, Korean, Vietnamese, Finish, Polish, German, Hungarian, Romanian, Farsi,[/b] and [b]Turkish[/b] and as many other languages for which we can find volunteers (we are hoping to get Arabic and Japanese and maybe more).

Oh, cool. That's a lot of new things this year. Looking forward to the competition!

More people should take the purple comet. Hopefully, we will be first again.

i have a question about the test,

are the problems arranged by difficulty?

easiest-hardest?

The problems in the [color=indigo][b]Purple Comet[size=150][i]![/i][/size] Math Meet[/b][/color] are roughly arranged in order of difficulty, although problem topics are also considered so that students working the problems in order will see a variety of topics. The first few problems are written so that virtually any team that considers them carefully can get them right. They are on the order of difficulty as the easy to medium problems from the AMC 10/12. The last few problems are meant to be very challenging. They are like the hardest of the AIME or even USAMO problems.

All the earlier [color=indigo][b]Purple Comet[size=150][i]![/i][/size] Math Meet[/b][/color] problems can be found on the contest web site at [url]http://purplecomet.org[/url]. This year, once a team is registered, their supervisor can have them run an old contest for practice under the same testing conditions of the actual contest.


Have you actually read all of the first few chapters in detail and understood everything? If you have, and you're still bad at MathCounts counting and probability, then I have no clue what you should do. Maybe old problems.

dude are you sure you read that book fully and done the problems

that book is amazing, i got much better from just reading and doing the problems of the first two chapters

its easy to make a mistake on cointing problems because the wording can easily causing you to make a silly mistake . if you can understand all of intro to counting then this is the only thing left to do....probably

it's not like geo where u can kinda guess what they're asking for!

read the book again, because in just the first half of "Intro to Counting and Probability" taught ME a ton, and I thought I knew it all :lol:  Ok, no I didn't, but I found ways to do a bunch of types of problems I used to really have to beat to death to get the answer to.  Do every problem!!!!!!!

[quote="indianamath"]Well, i've done the Introduction to Counting and Probability twice, and i still consistently miss an unusual large portion of the counting problems in Mathcounts.  What should I do to improve my counting?  :([/quote]
The key to doing the counting & probability problems correctly is to read and re-read the problem and intepret it and try to rephrase it in your own words. Then connect that problem with something similar to one that you have solved before. After arriving at your solution, ask yourself whether the answer makes sense for the problem at hand. With the knowledge of a few techniques that are frequently used, usually common sense and intuition will tell you whether the answer makes sense.


Let $ ABCD$ is a convex quadrilateral, let $ AB \equal{} a, BC \equal{} b, CD \equal{} c, DA \equal{} d, p \equal{} \frac {a \plus{} b \plus{} c \plus{} d}{2}$.
Let $ S$ is area of $ ABCD$. Prove that: $ \boxed{S \equal{} \sqrt {(p \minus{} a)(p \minus{} b)(p \minus{} c)(p \minus{} d) \minus{} abcd.cos^2\frac {A \plus{} C}{2}}}$

[quote="kunny"]That's [url=http://mathworld.wolfram.com/BretschneidersFormula.html]Bretshneider's formula[/url]

Here is a proof. [url]http://horibe.jp/PDFBOX/Bretscheider.pdf[/url][/quote]
[color=darkblue][b]Hi Kunny. Thank you very much. [hide="hide"]I like Japanese![/hide][/b][/color]


pigeonhole principle
geometry
rectangle
inequalities
combinatorics unsolved
combinatorics

I know this involves Pigeonhole Principle but I can't apply it.

In a square grid of 169 points, 53 are marked.  Prove that some 4 marked points form a rectangle with sides parallel to the grid lines.

Suppose in i-th row there is $ a_{i}$ points. If there is no rectangle(with paralel sides)  thеn for each two columns i,j there is exactly one row that has points in both columns(otherwise there is a rectangle).
From here we get
$ \binom{13}{2}\leq \binom{a_{1}}{2}+\binom{a_{2}}{2}...\binom{a_{13}}{2}$
Now using the  QM-AM inequality we get contradiction.


"Experience is something you don't get until just after you need it"

Considering my motto: "Surprises only show up to those not expecting them", I cannot agree to your motto.

You say that experience comes only the hard way. I think a little caution and perhaps some attention to what happenes around you, can save a lot of trouble and bring some experience.

There are many exceptions, but it is not a general truth.

/bump.......................

$\phantom{You cannot post so soon after your last post. Take a break. Do some push-ups. Solve an interesting math problem.}$


Given the locations of points (2,7), (5,10), (10,-1), and the origin, construct point (3,6) [i]without using a compass[/i]

[hide="Solution"]If we put $A(2,7), B(5,10), C(10,-1), M(3,6)$, first we note that points $B$ and $M$ satisfy ${10\over 5}={6\over 3}$, i.e. they both lie on the line $y=2x$. Then we get the idea that $M$ might be the intersection of that line and the line $AC$, because without compass all we can do is draw lines and find their intersections. So let's check the idea. The equation of $AC$ is $y-7=\frac{-1-7}{10-2}(x-2)\iff y=-x+9$. Putting in the coordinates of $M$, we see that it really belongs to that line. Hence the construction is easy: Connect $B$ to the origin and connect $A$ to $C$ - the intersection is the desired point.[/hide]


I suggest that we should be notified when the PM inbox is 100% full.

I think, supposedly, the max space for the inbox is only 100 messages, since that seems to be the case for my messages...but some people have had more than that, so I'm not sure...That would be a pretty good idea, I think, but it should be pretty close to full when you hit 100 messages.

Yes i agree, see the discussions [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=204097]here[/url]

You can't have more than 100 if you are a regular user.

I dont know, i have 416 now :maybe: :huh:

Cool, I'm not a regular user...(121 here)


ratio
trigonometry
geometry solved
geometry

Let  $P$  be an interior point of an angle  $XOY$, and let a variable circle   $C$, which passes through the points  $P$ and $O$,  intersect the sides  $OX$ and $OY$  at   $A$  and  $B$ respectively. Prove that the ratio

     $\frac{AB}{PA+PB}$ is a constant , independent of the position of circle $C$.

[u]Babis[/u]

$\frac{AB}{PA+PB}=\frac{1}{\frac{PA}{AB}+\frac{PB}{AB}}=\frac{1}{\frac{\sin\angle PBA}{\sin\angle APB}+\frac{\sin\angle PAB}{\sin\angle APB}}=\frac{\sin\angle AOB}{\sin\angle AOP +\sin\angle BOP}$ is a constant.

I have slightly edited the wording of a problem: in English, a circle "passes through" (not "comes from") some points.

Anyway, the problem is simple: By the Sine Law in triangle APB, we have $\frac{PA}{AB}=\frac{\sin\measuredangle PBA}{\sin\measuredangle APB}$. But since the points O, A, P and B lie on one circle (namely, on the circle $C$), we get < PBA = < POA, so that sin < PBA = sin < POA = sin < POX, and we get < APB = 180° - < AOB, so that sin < APB = sin < AOB = sin < XOY. Thus, $\frac{PA}{AB}=\frac{\sin\measuredangle PBA}{\sin\measuredangle APB}=\frac{\sin\measuredangle POX}{\sin\measuredangle XOY}$. Thus, it is clear that the ratio $\frac{PA}{AB}$ is constant, i. e. it does not depend on the position of the circle $C$. Similarly, the ratio $\frac{PB}{AB}$ is also constant. Thus, the sum of these two ratios, $\frac{PA}{AB}+\frac{PB}{AB}=\frac{PA+PB}{AB}$ is also constant, and consequently, its reciprocal ratio $\frac{AB}{PA+PB}$ is consant, too. $\blacksquare$

[Edit: Leepakhin was quicker - the solution is the same, of course.]

  Darij

Thanks to  both of you , for the solutions  and for corrections(Darij yes , you are right. I  did not like << comes from >>  but I ought to go out quicly , so...>>).

 Tommorow I will have the official solutions in

                   http://www.hms.gr

This competition concerned  classes 8 , 9 , 10 , 11 , 12 and is the first round. The third round is our Olympiad.
  Perhaps we' ll send all the problems (the most interesting ) for the MLinkers of all countries and ages.

 Babis


AMC
USA(J)MO
USAMO
email
AIME
AMC 12 A
AMC 12 B

Hi everyone!
I scored 112.5 on the AMC 12B and am interested in getting a regrade my score for at most 6 points. I was just wondering:
1. Has anyone successfully raised their score from a re-grade?
2. Do you think it's worth it? I think i got the write answer, as i circled it on my test, but the my scantron answers from the results sheet says something different.

to be honest, 6 points will only raise u to a 118.5.. and the chances of qualifying for USAMO aren't raised enough for the re-grade to be really worth it, and that's assuming u get it marked right.. Think about it, you'll still need a 9 for any chance, and a 10 will still be needed to be sure...

I'd put it more like feeling good about an 8.

I got many of the difficult problems, but didn't get the easy problems, including the parenthesis question, then powers of 3 question, and the 5th question. Needlessly to say, I got 108, which was worse compared to my 111 on the A test.

I'm planning to get a rescoring for my AMC 12 A next week (actual: 118.5 expected: 129 = 10.5 pts difference), unless my AMC 12 B score come out next week and is what I expected (129).

[quote="moogra"]6 pts is too little but go for it if you really want it. Maybe you'll miss out USAMO by 6 pts.[/quote]

People have missed USAMO by 1/2 a point and you'll feel really bad about it if yo umissed by a really small margin

[quote="max221"]before you ask for a regrade, i'd recommend check the official score report with your coordinator. the email includes a csv file, which indicates what the computer read off your answer sheet for each question.[/quote]
Yeah, i looked at the csv file. Apparently i answered 3/8 instead of 3/16 on the problem about lines around a cube. However, the fact that i had 3/16 circled on my test pamphlet would suggest otherwise.

do you remember changing an answer on that question? if not, it would seem as if you just made a mistake bubbling.

our school photocopies every answer sheet before sending them out se we can look at these things without bothering the MAA...that might be a good idea in the future.

OK, so i got either a 9 or 10 on the AIME, which would put me at 202.5 or 212.5
Do i protest?

azn: yes, 212.5 probably quals, 207 doesn't [i]I[/i] think.

207 probably will; protest

It was a bit harder than last year so the index should be slightly lower than last year but to be safe I'd gain teh 6 lost pts


Where does it say when the Olympiad Corner is due?  As a specific example, in the April 2006 issue I find that the "Problems" are to be received no later than November 1.  But in the Olympiad Corner (containing Japan MO etc.) I just find "send me your nice solutions and generalizations".  I probably can't read and am missing something obvious.

Btw, does anybody want 2006 issues of Crux? They come to me by subscription, but have miserably spent their life on the shelf so far. If you are in Toronto, you can probably get them from me in some way. I have the Feb-May and Nov issues, and probably more stacked away somewhere. Happy mathing!

What, nobody wants Crux? Or do you all have piles of subscriptions gathering dust on the shelves? That would be sad  :( .

I'd want them but I live in MA  :(  :wink:

I wouldnt mind having them either but i dont live in Toronto

Ya..same...

gosh! i gotta move to toronto someday...it, and waterloo are like the math capitals of canada

Is there anywya u can ship some to waterloo..lol probably not!, ya WCI used to be a powerhouse in math, they're still pretty good but i dunno if they can three-peat the euclid cuz alot of their strong mathematicians graduated or left (alex)

LOL yeah... Too bad for WCI. My new school does not have a chance to win :( Don Mills will pwn again. I hope some good ppl go to WCI in the future.

From what I heard, WCI got destroyed on COMC :(

Ya i live in waterloo and talked to a few WCI people..i heard that no one got 3b) and a lot of them screwed up. I was surprised  :o like they may not have any qualifications for the CMO (i think they will have a couple)

They will most probably one qualifier based on their scores (one guy got like 65 or more and 65 will probably be the cut-off.) Too bad for WCI. :(

k w8, u think the cutoff is 65..o man .. if i wasnt being so dumb i couldve gotten that (like 2c) i'm still mad about that i had the answer and i dunno what i was thinking and the 3b) required a little thinking which i think i couldve done)..k well i think this if the cutoffs that low then this year's people may struggle with the olympiads, cuz people who didnt get any of 4 (which is an important question) could be writing the CMO..

Well, this year's problem four wasn't a good choice, and as Peng and David have said, it is a very well known problem.  I think getting into CMO this year shows how CAREFUL you are when you are doing problems and not really how good you are at problem solving, which really sucks.
Btw, I'm really anticipating the results now, I'm getting more and more worried, I hope it'll all work out somehow for everyone.

Question 4 in Part B wasnt the best choice to put on the contest because i looked up the "tree" and i found the proofs (i think the proofs are properties of the tree..not sure) so obviously someone who knew about the stren-brocet tree (if its called taht) they probably smoked the question..however i dont here anyone saying they know about the question besides David and Peng (didnt they do it in IMO training), so i dunno..

Has anyone else already seen that question?

Crap! Just reading this number theory book i have right now....just past some properties of  Farey sequence....gosh i wish i had read it earlier...like....before Open...

I saw the Farey Sequence (but not the same questions) in a Russain book for middle school students.  :lol:
I think they should have 1 question instead of 4 parts, because this yr B.4 just takes a long time to do but is easier.

I agree, 4b in itself wasn't that difficult of a question (though I guess it really depends on the perspective of who got it and who didn't...I mean...induction practically knocks off all 3 parts).  Of course, I'm biased to this  :blush:

I don't think 4b being well known should be a problem this year (wait until COMC scores come out).  I mean, the Stern-Brocot tree just isn't something that comes to mind when I want to look up a mathematical concept.  One could also argue that if the Stern-Brocot concept is indeed known to a specific competitor on the COMC, then shouldn't he/she get credit for it because he/she is likely competent enough in Mathematics to understand such an obscure concept in the first place?  But that would be an awfully precarious argument now...

Alex, I think what was killer were the questions [i]before[/i] 4b.  Some of the Part B questions took forever this year; even though most of them were quite trivial!  I don't know about you, but spending 30+ minutes to write up 8+ parts is a real pain.

lowest_reynolds makes a good point about B4) so does alex. Once you think about it it is easier then last year's and lets face it if your good at math you probbly would have got it (or most of it) anyway.

I think this years Part B is easier year then last year's.

It's much easier but takes longer to do. :(


Prove that $\frac{1+sin \theta+icos \theta}{1+sin \theta-icos \theta}=sin \theta+icos \theta$. Hence deduce that

$(1+sin \frac{\pi}{5}+icos \frac{\pi}{5})^{5}+i(1+sin \frac{\pi}{5}-icos \frac{\pi}{5})^{5}=0$.

[hide="Solution"] $\sin \theta+i \cos \theta = \cos (\frac{\pi}{2}-\theta)+i \sin (\frac{\pi}{2}-\theta)$

Let $z = e^{i ( \frac{\pi}{2}-\theta)}$.  Then

$\frac{1+z}{1+\frac{1}{z}}= \frac{z^{2}+z}{z+1}= z$

Hence

$( \frac{1+\sin \frac{\pi}{5}+i \cos \frac{\pi}{5}}{1+\sin \frac{\pi}{5}-i \cos \frac{\pi}{5}})^{5}= \text{cis }( \frac{5\pi}{2}-\pi ) =-i$

And the result follows (I assume you meant to have a $-i \cos \frac{\pi}{5}$ in that second term). [/hide]


My book says that in locally compact metric spaces boundedness is equivalent to local-boundedness but I cannot find a proof for this. Is completeness required? If so, what is the counterexample?

I don't know what it means for a set to be locally bounded.

Sorry I meant total boundedness. I can't change the post though.


Ring Theory
abstract algebra
superior algebra
superior algebra unsolved

Let $R$ be a local, unitary and commutative ring such that all of its ideals are principal. Prove that $x, y\in R$ implies $(x)\subset(y)$ or $(y)\subset (x)$.
PS local means it has only one maximal ideal.[a typo fixed in edit.]

Do you assume, that $R$ has no zero divisors? In this case, you can proof it this way:

Since $R$ is a PID, $R$ is a UFD and the prime ideals $\neq 0$ coincide with the maximal ideals, i.e. there's exactly one prime ideal since $R$ is local. If $R$ is a field, the assertion is trivial; otherwise there $R$ has - up to associatedness - exactly one prime element, say $\pi$, and every element in $R \backslash \{0\}$ has the form $x = \pi^{n}u$ for some $n \in \mathbb{N}, u \in R^{x}$. If $y = \pi^{m}v$ is another element, we have $x | y$, ie. $(y) \subseteq (x)$ for $n \leq m$, and $(x) \subseteq (y)$ for $m \leq n$.

[quote="-oo-"]Do you assume, that $R$ has no zero divisors? [/quote]
No, I'm afraid we don't. But a nice solution for the special case. :)

Let $(a)$ be the maximal ideal in our ring, and let $x$ be an arbitrary element of $R$. If $x$ is not invertible, then $x=ax_{1}$ for some $x_{1}$. If $x_{1}$ is not invertible, then $x_{1}=ax_{2}$ for some $x_{2}$, and so on. Assume none of the $x_{n}$ is invertible. Since the sequence of ideals $(x_{n})$ is increasing and our ring is Noetherian, this sequence is stationary, i.e. $(a)(x_{n+1})=(x_{n})=(x_{n+1})$ for some $n\ge 0$. Since $(a)$ is contained in the Jacobson radical of $R$ (actually, $(a)$ [i]is[/i] the Jacobson radical of $R$), [url=http://planetmath.org/encyclopedia/NakayamasLemma.html]Nakayama's lemma[/url] says that $(x_{n+1})=0\Rightarrow x_{n+1}=0\Rightarrow x=0$.

The above shows that for all non-zero $x$ there is an $n$ and a unit $u$ of $R$ such that $x=a^{n}u$. The desired conclusion follows immediately from this.


calculus
integration
number theory unsolved
number theory

Let $ S$ be the set of positive integers which can be written in the form
$ a^2 \plus{} 5b^2$ for some co-prime integers $ a$ and $ b$. Let $ p$ be a prime congruent to
$ 3$ modulo $ 4$. Prove that if some integral multiple of $ p$ belongs to $ S$, then
$ 2p$ belongs to $ S$ as well.
[hide="what I did"]I found $ a$ and $ b$ such that $ p|a^2 \plus{} 5b^2 < 6p$ and i proved that $ a^2 \plus{} 5b^2\neq{p,4p,5p}$ but i can't prove that it can't be $ 3p$ :( .[/hide]

thx although i had searched but it was useless


http://www.mathlinks.ro/viewtopic.php?p=925374#925374

I wonder how it's possible to have 0 posts, even though he has a post outside G&F... this is a nice contribution and I wanted to rate it, but it's not counted because he has 0 posts...

[quote="Peter"]I wonder how it's possible to have 0 posts, even though he has a post outside G&F... this is a nice contribution and I wanted to rate it, but it's not counted because he has 0 posts...[/quote]Not all forums have posts counted. In particular the PEN forum was under the EBooks forum, so posts weren't counted (don't worry there will be a recalibration soon).


inequalities
geometry
circumcircle
inradius
perimeter
geometry unsolved

$a , b , c$ are the sides of triangle and$R ,r ,p$ are respectively the ray of circumcircle and inradius and semi-perimeter prove that this inequality
$\dfrac{1}{4r^2}\ (a-b)^2(b-c)^2(c-a)^2=-p^4+2(2R^2+10Rr-r^2)p^2-r(4R+r)^2$
therefore you conclude  these inequality
$1)\sqrt{3}\ p \leq\ 4R+r$

$2)16Rr-5r^2 \leq\ p^2 \leq\ 4R^2+4Rr+3r^2$

$3)p \leq\ 2R+(3\sqrt{3}\ - 4)r$

$4)2R^2+8Rr+3r^2 \leq\ p^2 \leq\ 5R^2+4Rr-r^2$

About the first one. I gave for it at least 4 solutions. Anyway it is equivalent with Hadwiger-Finsler inequality,1937
which says that: If $a, b, c$ are the lenghts of of any triangle the following inequality holds:

    $a^{2}+b^{2}+c^{2}\geq 4S\sqrt{3} +(a-b)^{2}+(b-c)^{2}+(c-a)^{2}$ ;)


I will be starting a new science fair project next year and would like to try something realted to mathematics. Please post some ideas that you think would make an excellent project.

Please do not double post..

It seems you want the replies bad but we moderators don't want forums get spammed by same topic neither.. ;)

One you posted on Getting Started Forum has been deleted.

[quote="amirhtlusa"]this is not like a very interesting project. but at our school when you are a senior, there is a requirement that you have to do something called Senior Project. and i'm going to be a senior next year. and since i'm the president of the math team, my senior project would be that i'm going to be running a Math Competition in our school for the first time... and i'll be inviting some schools over to our school...
this was just to give you an idea... ;)[/quote]

that has nothing to do with a science fair. Science Fair project is solely for a science fair, and i dont think holding a math competition is going to give much. There is a reason it is called a "science fair" project.

BTW, I would research stuff on maybe Physics-related phenomena and build up on it, with adding a little mathematical aspect.... since physics and math are very much the same... well very similar, i meant.

Physics uses a lot of math.  Anyways, here are some ideas...
1. Half-life and radiation
2. Acceleration of different objects down different declines
3. Water tension: you can do a lot with this, for instance, how much extra volume of water is formed in a glass by water tension when it's filled up as far as possible? (the little bulge at the top)
4. Light refraction indexes and stuff with light...you could pass a beam of light through a specially constructed prism so that it makes a picture...or something like that...
5. Calculate with forces...do things with electromagnetic force, or show something about the relations between nuclear forces, or something like that...

Look here:

http://www.ams.org/news-for-students/

Those were winning projects that involved little else but pure mathematics. But as you can tell, they're extremely complex, and entering such a prestigious competition is risky without the advice/help of a professional mathematician or mentor. I plan to participate, but can't find any ideas either, or a mentor for that matter. Where would you get one in the first place?

I'd appreciate some advice and hope this helps.

[quote="darkprince"]but can't find any ideas either, or a mentor for that matter. Where would you get one in the first place?[/quote]

A university?

Most of the mentors assigned to students here at the Research Science Institute are graduate students under the direction of a professor.  It's a pretty good system, although I suppose it's not so easy to go up to a professor you don't know and ask for mentorship...

t0rajir0u wrote:
although I suppose it's not so easy to go up to a professor you don't know and ask for mentorship...

Exactly! I don't know any professors personally, so I can't choose one randomly from the staff directory of a nearby university and ask!

Unless...I can ask a good math teacher at my school if she/he will help. That'd work maybe...

Thanks t0rajir0u!


[quote="nam_tran1237"]Let a, b, c be non_negative real numbers. Prove the inequality
\[ (a^2 \plus{} b^2 \plus{} c^2 )^2 \ge 4(a \minus{} b)(b \minus{} c)(c \minus{} a)(a \plus{} b \plus{} c)
\]
When does equality occur?[/quote]
The idea of hungkhtn (the entirely mixing variables method):
We denote $ f(a,b,c)\equal{}(a^2 \plus{} b^2 \plus{} c^2 )^2 \minus{} 4(a \minus{} b)(b \minus{} c)(c \minus{} a)(a \plus{} b \plus{} c)$ and consider
$ g(t)\equal{}f(a\plus{}t,b\plus{}t,c\plus{}t) \equal{} [3t^2\plus{}2t(a\plus{}b\plus{}c)\plus{}a^2\plus{}b^2\plus{}c^2]^2\minus{}4(a \minus{} b)(b \minus{} c)(c \minus{} a)(3t\plus{}a \plus{} b \plus{} c)$
for $ t \ge 0$.
We have
$ g'(t)\equal{}4(3t\plus{}a\plus{}b\plus{}c)(3t^2\plus{}2t(a\plus{}b\plus{}c)\plus{}a^2\plus{}b^2\plus{}c^2)\minus{}12(a \minus{} b)(b \minus{} c)(c \minus{} a) \\ \ge 4(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)\minus{}12(a \minus{} b)(b \minus{} c)(c \minus{} a) \\ \equal{} 4[\sum b(b\minus{}2a)^2\plus{}\sum ab^2] \\ \ge 0$
Then $ g(t) \ge g(0) \ \forall t \ge 0$ or $ f(a\plus{}t,b\plus{}t,c\plus{}t) \ge f(a,b,c) \ \forall t \ge 0$, this implies $ f(a,b,c) \ge f(a\minus{}c,b\minus{}c,0)$ with $ c\equal{}\min(a,b,c)$.
Hence if suffits to prove $ f(a,b,0) \ge 0$, which is true because
$ f(a,b,0) \equal{} (a^2\plus{}2ab\minus{}b^2)^2 \ge 0$.

assume that $ a\ge b,\ a\ge c$. If $ b\ge c$ the inequality is obvious. Assume that $ b \le c$. We have

$ \quad(a \minus{} b)(b \minus{} c)(c \minus{} a)(a \plus{} b \plus{} c) \\
\equal{} (a \minus{} b)(c \minus{} b)(a \minus{} c)(a \plus{} b \plus{} c) \\
\equal{} (a^2 \minus{} b^2 \plus{} ac \minus{} bc)(c \minus{} b)(a \minus{} c) \\
\le (a^2 \plus{} ac)\cdot c\cdot (a \minus{} c) \\
\equal{} ac(a^2 \minus{} c^2) \\
\le \frac {1}{4}\left[(2ac)^2 \plus{} (a^2 \minus{} c^2)^2\right]^2 \\
\equal{} \frac {1}{4}(a^2 \plus{} c^2)^2 \\
\le \frac {1}{4}(a^2 \plus{} b^2 \plus{} c^2)^2$

The equality occurs when $ b\equal{}0,\ a\equal{}\left(1\plus{}\sqrt{2}\right)c$.

Are you not sure about the first part or the second (or both)? The second part is just expanding the stuff inside the large square. The first is applying the AM-GM -ish (but true for all reals) $ (P\plus{}Q)^2\geq 4PQ$ to $ P\equal{}2ac,Q\equal{}a^2\minus{}c^2$.

I'm sorry but it's just a typing mistake
It should be
$ ac(a^2\minus{}c^2) \le \frac{1}{4}\left[(2ac)^2\plus{}(a^2\minus{}c^2)^2\right]\equal{} \frac{1}{4}(a^2\plus{}c^2)^2$


function
number theory unsolved
number theory

$f(x)=2{}^2{}^m{}^-{}^1-2^n+1$, show that if the function $f(x)$ is a perfect square, then $m=n$.
For all $m,n$ positive integers such that, $n<2m$.

[color=blue][u][I can't remember even the problem was like that or not, but i remember that my friend said: it is from 1st Zhayutykov International Math Olympiad's (Kazakhstan)] [/u][/color]

Davron

you are wrong. If m=m it is perfect square only if m=n=1.
This expression give perfect square when n=2m-1<2m, m=4,n=3,...


can help me
With p=1.prove that $ 2qr \plus{} q^2 \plus{} 4q^2r \geq 3r^2 \plus{} 4r$

Do you mean that $ p\equal{}a\plus{}b\plus{}c,\ q\equal{}ab\plus{}bc\plus{}ca,\ r\equal{}abc$?


On the plane are marked $ 2005$ points, no three of which are collinear. A line is drawn through any two of the points. Show that the points can be painted in two colors so that for any two points of the same color the number of the drawn lines separating them is even. (Two points are separated by a line if they lie in different open half-planes determined by the line).

[quote="Johan Gunardi"][hide="something like this should work"]For each drawn line, mark one side with + and the other side with -. Color a point red if it lies on an even number of + regions, otherwise color it blue.[/hide][/quote]

The problem with this approach is that the number of $+$ regions do not vary continuously when one travel along an edge, it may jump more than one, i may even jump an even number. Consider the following example. There are $4$ points, and one is point $P$ lying inside the triangle $ABC$. For each edge of the triangle, mark the region containing the triangle plus. Then mark the regions divided by $PA$ so that $B$ is plus, $PB$ so that $C$ is plus, and $PC$ so that $A$ is plus. Then each $A,B,C$ will all have two pluses, hence have the same color. But each pair is separated by exactly one line.

For each edge $ab$ between two points, define a cross along $ab$ to be a drawn line that separates two endpoints.

Claim $1$: The total number of crosses along all egdes of a triangle is always even.
Proof: Each line that does not pass through any vertex of the triangle crosses either $2$ edges or no edge, hence contribute an even number of crosses. Now consider the crosses of lines that pass through exactly one of the vertices. Devide the lines in to $2002$ group of three lines, such that all lines in same group pass through a point that is not a vertex. Each group contributes either one or three crosses, depending on whether the common point lies inside or outside the triangle. There are an even number of groups, hence the total number of crosses is also even in this case.

Choose a point $a$ and color red. For any $b$, color red if $ab$ has even number of crosses, blue otherwise. Claim $1$ ensures that this coloring has the desired property. In fact, we even prove that two points are of same color if and only if their edge has even number of crosses.

Also like APMO 2004
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=82830


search
function
email
number theory
complex numbers

The other day, I came across [url=http://www.seventeenorbust.com/]Seventeen or Bust[/url], a project devoted to solving the Sierpinski problem.  Seeing that this is the AoPS Forum, I thought some of you might be interested:

Basically, a Sierpinski number is a number of the form [tex]k \cdot 2^n + 1, \quad \textrm{k odd}[/tex] such that is composite for every n :ge: 1.  In 1960, Sierpinski proved that there are infinitely many such numbers, and in 1962, John Selfridge proved that k = 78557 was a Sierpinski number.  The Sierpinski problem involves finding the smallest Sierpinski number; k = 78557 is conjectured to be the smallest one, although it has not been shown.  In order to actually show this, it is necessary to show that all k < 78557 produce non-Sierpinski numbers.

Number theory methods have been used to show that most numbers < 78557 are not Sierpinski numbers.  Others have been eliminated by finding a value of n for a given k that makes [tex]k \cdot 2^n + 1[/tex] a prime number.  But certain values of k could not be eliminated; by 2002, 17 candidates less than 78557 remained, hence the name "Seventeen or Bust".  Since then, six numbers have been eliminated, leaving eleven.

Seventeen or Bust uses the spare processor time of computers to test values of k and n.  (For those of you who are familiar with the Great Internet Mersenne Prime Search or SETI@home, this is the same idea.)  It doesn't slow your computer down at all (I've been running it in the background for the past two days, and haven't seen any performance drops) and is for a good cause, at least in my biased mathematical opinion.  I strongly encourage other AoPS community members to help out by [url=http://www.seventeenorbust.com/download/]downloading the program[/url] (don't worry, it's a small file) and running it.  Every little bit counts.

I have formed an [url=http://www.seventeenorbust.com/stats/teams/team.mhtml?teamID=331]AoPS team[/url], so that we can compete for fun.   As of right now, we're ranked 86th (ot of 300+ teams) in daily production, which is pretty good, considering that the team is one day old.

SWEET!!!!!
(note the link in my sig, it's been there since before the team was formed :P)

I'd just like to add that there are also several programs out there that let you find primes on your own, in case you want to work outside of the range the program's giving you (or just want to find really huge prime numbers. Long live 4455*2^100682+1!)

Yeah, OK, I'm joining the AoPS team as soon as I get home, probably.

[quote="nr1337"]GRRR!!! The software isn't for mac!!!  Also, if the Sierpinski problem gets solved from this, is anyone going to join ZetaGrid (to try to get zeroes of the Riemann Zeta Function)?[/quote]

Pff, what's the point? We all know they have real part of 1/2 anyway. ;)

[quote="nr1337"][quote="tokenadult"][quote="Fermatprime"]Pff, what's the point? We all know they have real part of 1/2 anyway. ;)[/quote]

This is my LOL of the day. Um, if you could write up what everyone knows in a rigorous manner, I think you would have a publishable mathematics paper.[/quote]

Yes, that would be nice...  hey... I have a great idea!!!  It's called proof by examples!  :D[/quote]

Yeah! "20 billion non-trivial zeros can't be wrong!"

I think I'm gonna put that in my sig.

[quote="Miko"]er, im still new to the AoPS community, can someone tell me what the Reimann hypothesis is plz? ive taken through Calc 2, so i might or might not understand, but plz update me, osmeone! thanx[/quote]a) It should be spelled riemann.
b) [url]http://mathworld.wolfram.com/RiemannHypothesis.html[/url]

[quote="ComplexZeta"]I don't believe that the Riemann Hypothesis is true yet. The bit of information that bothers me is that [tex]\pi(x)[/tex] and Li(x) switch infinitely often. But if you look at what [tex]\pi(x)[/tex] is in terms of Li(x) and the zeros of [tex]\zeta(s)[/tex], then it doesn't look very likely, at least not to me. In number theory, sometimes small cases (or I-don't-know-how-many billion cases) don't help very much.[/quote]

(I realize that.)

Why should it be a problem that pi(x)-Li(x) crosses 0 infinitely many times?  What bothers me is that the solutions of pi(x)-Li(x)=0 don't follow any real pattern; they don't seem to be related to the zeros of [tex]\zeta(s)[/tex], or anything else for that matter. (P.S. I realize that they are actually connected with the zeta function, but you sure wouldn't think it from looking at them.)

I can't remember the equation off the top of my head, but it's something like

[tex]\displaystyle \pi(x)=\sum_{n=1}^\infty} \frac{\mu(n)}{n}\text{Li}\left(x^{1/n}\right)+\text{terms which should be less significant}.[/tex]

That means that if [tex]\pi(x)>\text{Li}(x)[/tex], then very funny things have to be happening since the rest of the Li terms should by all rights sum to something negative. At any rate, that was what I noticed when I saw that formula, and I don't see any reason now that makes my reasoning incorrect. Am I being silly?

My processor time is already going to SETI@Home. [url=http://setiathome2.ssl.berkeley.edu/fcgi-bin/fcgi?email=enlightenedcouch%40hotmail.com&cmd=user_stats_new]see here[/url]

Yeah, theone853 joined!  [size=75]And he has a FAST computer...[/size]  Now we have three members; our motley crew is growing.

AoPSers unite!

Hey, how do you log in from the program? :P

I might try this program since my laptop can't handle Prime95. It's a shame since it's a fast computer, but it gets too hot when running Prime95.

17-or-bust says that there are 11 more multipliers left to test.  Does anyone know what these numbers are?

to join the team, do you just type the team name in the 'team affiliation' box that you get to from the 'config' button or is it more complicated?

Go to preferences and go to the team affliation section.

Anyways, I do not have a fast computer. It is merely a computer where my mom doesn't let my download anything onto it, so it has lots of space and memory to use.

Wow, look at those [url=http://www.seventeenorbust.com/stats/teams/team.mhtml?teamID=331]beautiful stats[/url]!  11 people!  We're moving up in the (SB) world.

I have been trying (without too much success) to get people to put this program on their computers. Neal put it on three, and Meir put it on a bunch. I have it on my (slow) desktop and my (fast) laptop, which I should only use for this every other day or so. Let's get to #1!

Well, I got my brother to let me use his Window laptop, so now I'm in the AoPS 17-or-bust team.

edit:  I'll have to find another Windows computer to do this on.  My brother's laptop took over 20 minutes just to calculate less than 10% of a number.  How long is it supposed to take to calculate one of these?

i think you mean one block. the whole number is the bar below the block. these usually take days upon days. the blocks on my computer take about 8 minutes each

So far, the client has been running for 1 hour and 3 minutes and is 1/3 through the first block.

wow, that's hardly worth it. running on this computer a block takes about 6:30. what processor are you running it on?

It was on a Windows 98 laptop that had less than 3 gigabytes of space and a very small amount of memory.  I stopped using 17-or-bust because that computer was so slow.

Let's get some stats going!

Post your SB username (if it's different) and approximately how long it takes you to do a block.  If you know, say how far you are through your proth test (I don't think anyone's finished one yet).

By the way, you can get a banner like the one in my signature by placing [img]http://www.clanhosts.com/dev/sobsig/sig.php?yourusernamehere[/img] in your signature box, replacing 'yourusernamehere' with your SB username.

username: khon
block time: approx 13 min. with no programs running; 16 min. with Mozilla running
% done: 22.5% of 1st proth test
total work done: 58.75 G cEMs

username: nr1337
time to do a block: about 3 hours on my laptop.  I'll have to beg someone to let me use their laptop.  My mother's laptop has been having problems, so she won't let me install it, so currently I'm currently not doing anything with 17-or-bust, but I hope to start soon.

I'm ComplexZeta (surprise, surprise). On my laptop, it takes about 10 minutes to do a block. On my desktop, it takes about 40 minutes. I'm 27.9% done on my laptop and 11.2% done on my desktop.

ml2 and ml1 from two comps about 30% done on each account

mine take about 6:30 a block. i'm about 42% done on two computers.

Let's get some more people on the AOPS team.

Current stats: 151 on total production with 1.07 T. pretty amazing considering the 4 teams around us in production average age is 300 days. our age is 10 days.


induction
number theory
relatively prime
6th edition

How did you cope with this problem? I believe I found a proof, but it occured immediately to me and seemed too simple, so I believe I made a mistake.My idea was to find at least one such number and get the others by adding
zeroes at its end.I took a look at n+1 numbers 1,11,111,... and since 2 of them are cungrent mod n their difference if divisible by n.

actually, a week before this round virtually the same problem was discussed on ML... and, yeah, that's the easiest solution

Makaveli, did you solve problem 1?

No, but I think I solved the third one.Did you?

There was a slight mistake on my part, due to haste :( which made the problem a lot more trivial.
It should have been

[quote]For any integer $n$, not divisible by 10, there exist an infinity of multiples not containing the digit "0" in them.[/quote]

This looks tougher indeed. anyone has a solution?

Use my idea from above combined with the following lemmas, easily proved by induction:
1)for every natural n there is a number written only with digits 1 and 2 divisible by 2^n that has n digits
2)for every natural n there is a numbetr with n digits written only with digits from 0 to 5 divisible by 5^n
(in the induction step analyze adding one digit on the left)
Write  if you manage it

i actually know both these problems... but how to pass to the general case $2^n k$ and $5^n k$? that is (and was) my big problem !

Well to prove for k use the method with the ones and for the last n digits use the lemmas where n is the largest power of 5 or 2 dividing the number

i still don't understand :( for $2 \not| t$ and $5 \not| t$ we have a number of the form $11 \ldots 1$ divisible by $t$. (is this what you are saying ?) for $2^n$ or $5^n$ we have the two lemmas. how do we combine them? sorry to bug you

Sorry man it`s not as simple as I believed it would be. As soon as I come up with a solution I`ll post it, OK?
If you solve it first, say how .

For the $2^nk$ case (k relatively prime to 10), we can pretty much use the 11111 trick that you mentioned:

Let $x$ be a multiple of $2^n$ with exactly $m$ digits that are all non-zero (you've noted how to find such a number). Let t be any positive integer, and consider the number N formed by stringing together $t\cdot\phi(9k)$ copies of $x$ (which certainly has all non-zero digits):

$N= x\cdot(1+10^m+\cdots+10^{(t\cdot\phi(9k)-1)m})$

${}= x\cdot\frac{10^{\phi(9k)\cdot mt}-1}{9}$.

Since $9k$ is relatively prime to 10, $9k\mid 10^{\phi(9k)\cdot mt}-1$ (Fermat's Little Theorem), i.e. $xk\mid N$ and thus $2^nk\mid N$. This provides an infinite family of working integers N.

Yeah that's what I basically had in mind ...

Sorry I may be wrong, but I thought a really simple solution and I would like you to check it...
Let N be the number and let L be the number of digits of N.
Obviously N is a multiple of itself; now let's take $M=N*10^L$. Of course it is a multiple of N, and it has exactly L 0 digits. Now let's take M+N. It is a multiple of N, and it has no 0 digits, since last L digits are the same of N.
So we have a multiple. And going on this way, we should get "infinitely many" multiples of N with no "0" digits.
In other words, I take all the multiples made stringing together (thank you zabelman, because of my poor English I didn't know how to say it :) ) N.
Am I right? I don't think so, 'cause it would be a too easy problem in my opinion...
David

If $N$ contains a $0$ digit your solution doesn't work.

There was something strange :).
I didn't read the problem correctly.


LaTeX
algebra
polynomial
complex numbers

This is not the place for homework; the Olympiad forums are for Olympiad-level problems.  This should be in High School Basics.  You might want to start by factoring out the $ 5$.

[hide="solution"]
\[ 5x^4 \minus{} 45 \equal{} 5(x^4 \minus{} 9) \equal{} 5((x^2)^2 \minus{} 3^2) \equal{} 5(x^2 \plus{} 3)(x^2 \minus{} 3) \equal{} 5(x^2 \plus{} 3)(x \plus{} sqrt{3})(x \minus{} sqrt{3})
\]
[/hide]

Sorry, I don't know how to use Latex correctly.

The question asks if it's factorable, so I have to say yes

And I think most people would rather you not include square roots in factorizations

[hide]Yes its factorable.
$ 5x^4\minus{}45$
$ 5(x^4\minus{}9)$
$ 5(x^2\minus{}3)(x^2\plus{}3)$[/hide]

[quote="Cobalt"][hide="solution"]
\[ 5x^4 \minus{} 45 \equal{} 5(x^4 \minus{} 9) \equal{} 5((x^2)^2 \minus{} 3^2) \equal{} 5(x^2 \plus{} 3)(x^2 \minus{} 3) \equal{} 5(x^2 \plus{} 3)(x \plus{} \sqrt{3})(x \minus{} \sqrt{3})
\]
[/hide]

Sorry, I don't know how to use Latex correctly.[/quote]

That's not a *real* factorization. If we did that, we could keep going to $ 5(x^2 \plus{} 3)(x \plus{} \sqrt{3})(x \minus{} \sqrt[4]{3})(x\plus{}\sqrt[4]{3})\equal{}5(x^2 \plus{} 3)(x \plus{} i\sqrt[4]{3})(x \minus{} i\sqrt{3})(x \minus{} \sqrt[4]{3})(x\plus{}\sqrt[4]{3})\ldots$ to infinity. Factorizations shouldn't contain radicals or imaginary numbers unless explicitly specified.

[quote="kelstia"]Is this factorable??? 5x^4-45[/quote]
Yes. You can just do 5*(x^4-9)=5*(x^2-3)(x^2+3)=5*(x^2+3)(x-sqrt 3)(x+sqrt 3)

[quote="Temperal"]
That's not a *real* factorization. If we did that, we could keep going to $ 5(x^2 \plus{} 3)(x \plus{} \sqrt {3})(x \minus{} \sqrt [4]{3})(x \plus{} \sqrt [4]{3}) \equal{} 5(x^2 \plus{} 3)(x \plus{} i\sqrt [4]{3})(x \minus{} i\sqrt {3})(x \minus{} \sqrt [4]{3})(x \plus{} \sqrt [4]{3})\ldots$ to infinity. Factorizations shouldn't contain radicals or imaginary numbers unless explicitly specified.[/quote]

But these factorizations are wrong (note the total degree of $ x$ when expanded). A "complete" factoring for this degree-four polynomial (in complex numbers) will be just $ 5(x\plus{}i\sqrt{3})(x\minus{}i\sqrt{3})(x\plus{}\sqrt{3})(x\minus{}\sqrt{3})$, thus all the $ x$ are lowered to first degree.

Sorry, I meant to put the radicals around the xs but forgot.

Really? I always learned that factoring into radicals or imaginary numbers is wrong when not told to do so.

Unless otherwise specified, "factor" is usually taken to mean "factor with rational coefficients."  This is not to say that the factorizations with irrational or complex coefficients are incorrect; they simply go beyond the implied bounds of the problem.

[quote="Temperal"]Sorry, I meant to put the radicals around the xs but forgot.[/quote]

That isn't a polynomial factorization at all (which is the other implicit meaning of "factor").


measurements of the medians of a triangle are respectively 9,12 and 15.
 :| wich the surface of this triangle ?

[quote="e.lopes"]we have $2(b^2+c^2)-a^2 = 324$, $2(a^2+c^2)-b^2 = 576$, $2(b^2+a^2)-c^2 = 900$.

solve the system and use Heron's Formula[/quote]

   This solution follows the tracks of the best multi-page solutions on this site.

   Not so long ago Darij posted a nice summary

  http://www.mathlinks.ro/Forum/topic-59415.html

   It will be of some help with the problem at hand.

  Thank you.

  M.T.

thanks.
i find an other solution  :roll:
i want some books for solving the olympiads problems!!  :(