geometry
3D geometry
pyramid
Mafia
videos

he is talking about http://xkcd.com/

which is pretty funny

But this is a [i]litle[/i] spammy...

xkcd is awesome, are you kidding me? Totally deserving of a round table topic.

Yes xkcd is cool and funny, but...:
I'm very eager to see (non-mathematical) serious discussions of xkcd  :wink:
The title is awful, too...

What does xkcd stand for?

plsdon'tbeobviousiwillbeembarrassedatmylackofcommonsenseandknowledge

In fact, it's just a random string of four letters that sounds good  :wink:

Nothing, it's the most unpronounceable "word" the writer could come up with.

xkcd usually has adult-oriented references that might not be appropriate for everyone who comes here. For example, let your pointer hover over today's strip. Maybe it would be better to delete this?

Wikipedia usually has adult related contents that might not be appropriate for everyone who comes here. Maybe we should better ban everyone using Wikipedia  :wink:

But still, I'm for moving this into the G&FF.

[quote="Bictor717"]xkcd usually has adult-oriented references that might not be appropriate for everyone who comes here. For example, let your pointer hover over today's strip.[/quote]

Haha take a look at my blog. I made a reference to today's comic but I hope XKCD is still "acceptable" for the AoPS community.

[quote="ZetaX"]But still, I'm for moving this into the G&FF.[/quote]

Actually, I'd prefer it if JBL deleted this entire topic (I'm going to anyway if it's moved to G&FF).

well it is possible to have serious discussion of xkcd, you know

I put this in round table for a reason

by the way, when I said xkcd, I meant today's specific comic (although the comic in general isn't bad, either)

wow

I was in the xkcd forums

and like

my EYE randomly saw MysticTerminator's s/n in the "online" list.

Then I came here.

It was freakay.

I think it was very interesting, and something I've been thinking about recently. I've always been told that Physics is simply applied mathematics, and this seems to take that further.

What do you guys think applied sociology is?

I'd say applied sociology is politics.

We should just compile all the math-related XKCD's into this thread for easy access.

It's not just the mathematical ones. When I first read through the archives, [url=http://www.xkcd.com/228/]this strip[/url] particularly caught my eye.

this one was funny too

http://www.xkcd.com/232/

and here's a play on a famous riddle :D

http://www.xkcd.com/246/

[url=http://www.youtube.com/watch?v=eBGIQ7ZuuiU]Click to reveal hidden content[/url]
[hide]Rick Astley agrees that xkcd is fun. He is a mathematician.[/hide]

http://xkcd.com/445

I love how you can read it in practically any direction and it [i]still makes sense[/i].

I love this comic. (And this comment fits rather well.)
http://xkcd.com/442/

[quote="(^_^)"][quote="ZetaX"][quote="7h3.D3m0n.117"]Actually, I'd prefer it if JBL deleted this entire topic (I'm going to anyway if it's moved to G&FF).[/quote]
Why¿ It surely has more content than any Tug Of War...[/quote]

hey...this is like the 3rd time ive seen someone pick on the tug of war....[/quote]
true, but this topic is really gettin spammy,

[quote="myyellowducky82"][/quote][quote="(^_^)"][quote="ZetaX"][quote="7h3.D3m0n.117"]Actually, I'd prefer it if JBL deleted this entire topic (I'm going to anyway if it's moved to G&FF).[/quote]
Why¿ It surely has more content than any Tug Of War...[/quote]

hey...this is like the 3rd time ive seen someone pick on the tug of war....[/quote][quote="myyellowducky82"]
true, but this topic is really gettin spammy,[/quote]

its the GFF...

let's not make a quote pyramid, and get back on topic.

any really funny ones that no one has posted yet???

http://xkcd.com/441/ was pretty funny. It's still up on my Facebook profile.

i going to try to bump this up to the top. People need to see it.

[img]http://imgs.xkcd.com/comics/too_old_for_this_shit.png[/img]
http://xkcd.com/447/

That's kinda how I feel in a completely unsentimental way  :P  Especially since I STARTED problem solving at 13...

meh

yeah, i agree, and i only really started when I was in 8th grade and even then, I didn't get anywhere at MC chapter OR state, basically i sucked  :P

Along with bumping this thread, here's one of my [url=http://xkcd.com/247/]favorites[/url].

lol

The message is too small. Please make the message longer before submitting.

Someone should make a Mafia-like game where players move around a map and they try to eliminate the opposing will:

[img]http://imgs.xkcd.com/comics/online_communities_small.png[/img]

ah, that's really funny
but what does this have to do with xkcd?

WOW XKCD:

[img]http://imgs.xkcd.com/comics/good_morning.png[/img]

[url]http://xkcd.com/448/[/url]

Wow this compltely reads my mind for the past few weeks. For the past few weeks (since summer let out), I've basically been sleeping at any random interval from 12am-6am, and waking up anywhere from 8am-3pm. I've always been saying to my self "you are living in another timezone..." the entire time

xDDD

w00t today's (7/11) XKCD had a reference to Taiwan  :P

The comic 1=2 posted reminds me of this video:
[youtube]dGfNSA3QmWg[/youtube]


abstract algebra
group theory
superior algebra
superior algebra unsolved

Call an automorphism of a group $G$ extensible if given any extension $H$ of $G$, the automorphism can also be extended to an automorphism of $H$.

Every conjugation is clearly an extensible automorphism. Can we prove that there are no other extensible automorphisms?

(This question is pertinent for generalizing some results in group theory and is also the group theoretic equivalent of the fact that every automorphism of a field can be extended to an automorphism of its algebraic extension).

Hi

I have made some progress in the problem, namely: the only extensible automorphism of an Abelian finite group is the identity automorphism, and also every extensible automorphism of a group must act as identity on its Abelianization. The actual result I have got is somewhat stronger. Check out my homepage:

[url]http://www.cmi.ac.in/~vipul/unsolvedproblems[/url]

Currently I have put up a not-very-complete file but I do intent to put in a more up to date file soon.


The [url=http://www.mathm.org]Manhattan chapter of MathCounts[/url] held its 2005 competition yesterday at Columbia University.  It was a lot of fun.  I was there as a volunteer and grader.  I was also handing out flyers for this forum.

Here are the team results (out of 66 points):
[code]1. Hunter (53.5)
   Hunter B (50) -- unofficial
2. Simon Baruch (38.25)
3. Salk School of Science (36.25)
4. Booker T. Washington [Delta program] (36)
5. NYC Lab School (33.25)
6. Friends Seminary (27.75)
7. Manhattan Academy of Technology (26.75)
[/code]
The top 3 teams (Hunter, Baruch, and Salk) will be moving on to the State competition next month.

Here are the individual results (out of 46 points).  I didn't write down the names of the students -- just their schools.  Sorry about that.
[code] 1. Hunter student (43)
 2. Hunter student (38)
 3. Hunter student (37)
 4. Hunter student (37)
 5. Salk student (37)
 6. Hunter student (36)
 7. Lab School student (35)
 8. Hunter student (34)
 9. Hunter student (33)
10. Booker T. Washington [Delta] student (32)
11. Salk student (31)
12. Baruch student (30)
13. ? (30)
[/code]
The top 4 students (all Hunter students) received trophies.

The top 12 students competed in the Countdown Round (which is unofficial in NY state).  It was an exciting event to watch.  It had a single-elimination bracket format.  The top four students in the countdown round ended up being the same as the top four in the written rounds, except in a different order, and so each received a second trophy.

At least two members of the NYC forum were there.  metsfan001 was among the top three in the written rounds and won the countdown round!  AznXtreme418 was fourth in the written rounds and placed third in the countdown round!  Congratulations!!

Just curious . . . how did they break ties (i.e., 3rd-5th)?

The tiebreakers were specified in an instruction manual from the MathCounts National office.  I think the first tiebreaker was to count the number right in Questions 21 to 30 of the Sprint Round.  There were other tiebreaker rules after that, but I don't think we needed them.

Do you know if the top x individuals not on a winning team advance (or if x=0)?

I honestly don't know the policy of either the Manhattan chapter or New York State.  The [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=167149#p167149]post by solafidefarms[/url], which quotes the MathCounts National policy, seems pretty compelling to me.It clearly states that the top 2 students not on the winning team must move on to the State competition.  By that criterion, you are in.

I would contact (or have your coach contact) the coordinators of the Manhattan chapter.  They are Jeremy Edwards (e@edmunds.net) and LaToya Wilcher (LaToya.Wilcher@guycarp.com).  Provide for their convenience the quote from the National policy.  They may be unaware of the policy.  If that doesn't work, you can try contacting the coordinator of New York State MathCounts.  I believe that is Laura Pelizzi (lpellizi@notes.cc.sunysb.edu).  If that still doesn't work, you can try contacting the MathCounts National office. (See the [url=http://www.mathcounts.org]MathCounts web site[/url] for contact info.)

Good luck!  Let us know what happens.

I think they're going to shaft me by claiming that I was on "the winning team." They did that last year, but at least there's a new coordinator this year.

go hunter!  why is it that we always dominate 8th grade and then fade back into obscurity in the higher grades?  o right, its because most of the mathletes end up going to stuy :(


geometry
circumcircle
trigonometry
cyclic quadrilateral
trig identities
Law of Cosines
geometry unsolved

[size=150]Say that the integers $a,b,c,d,p$ and $r$ form a $cyclic$ $set$ $(a,b,c,d,p,r)$ if there exists a cyclic quadrilateral with circumradius $r$ ,sides $a,b,c$ and $d$ , and diogonals $p$ and $2r$.

$(i)$  Show that if $r{<25}$ ,no cyclic set exisits.
$(ii)$ Find a cyclic set $(a,b,c,d,p,r)$ for $r=25$.[/size]

I am not so sure about the statement.  What about $(6, 8, 6, 8, 20, 10)$?

I think he means a,b,c,d,p,r are all distinct.

In that case:

Splitting the cyclic quadrilateral along the diameter of the circumcircle, which is also a diagonal of the quadrilateral, we have two right triangles of hypotenuse $2r$ and integer legs.

If $r < 25$ then there exists maximally one set of nonzero positive integers $(a, b)$ such that $a^2 + b^2 = (2r)^2$, hence $a, b, c, d$ cannot be distinct.

At $r = 25$ we have $14^2 + 48^2 = 50^2$ and $30^2 + 40^2 = 50^2$.  It remains only to calculate $p$ (there are two distinct orientations of the two right triangles, so presumably only one of them produces an integer value of $p$).  Let us check the orientation where $40, 14$ share an angle.

The non-right angles of the cyclic quadrilateral must add to $180^{\circ}$, so Law of Cosines yields $p^2 = 40^2 + 14^2 - 2 \times 40 \times 14 \cos \theta = 30^2 + 48^2 - 2 \times 30 \times 48 \cos (180^{\circ} - \theta)$, from which we find $\cos \theta = -\frac{44}{125}$, which clearly does not yield an integer value of $p$.

For the second case, $p^2 = 40^2 + 48^2 - 2 \times 40 \times 48 \cos \theta = 14^2 + 30^2 - 2 \times 14 \times 30 \cos (180^{\circ} - \theta)$, from which we find $\cos \theta = \frac{3}{5}$; now we can find $p^2 = 1600, p = 40$.

Hence our cyclic set is given by $\boxed{ (14, 48, 30, 40, 40, 25) }$.

They're still not all distinct :(

Yeah,I think so,maybe there's some error in the question.


Why the identities of the following are called the [b]Induced Formula[/b]??

$\sin(\alpha+360)=\sin\alpha$
$\sin(-\alpha)=-\sin\alpha$
$\cos(180-\alpha)=-\cos\alpha$
....................

My guess is that it is because you can induce them visually by looking at the unit circle.  For the first one, one full circle goes back to the same spot.  For the second, going clockwise instead of counterclockwise produces a mirror effect on sine that makes it negative.  The third is much like the second except it starts at 180º.

[quote="Wumbate"]My guess is that it is because you can induce them visually by looking at the unit circle. [/quote]
Thank You! :D
 i think you are right, Wumbate. I checked the Webster's dictionary and one meaning of the word [b]Induced[/b] is "[b]Logic, to establish a proposition about a class on the basis of observations on a number of particular facts[/b].
It also means [b]"to lead"[/b] which, I think, perhaps stand for the fact that if we add a $2k\pi$ to $x$ ($\sin(x+2k\pi)$), we still have $\sin{x}$......


Given $a,b,c \geq 0$. Prove the inequality:
\[\sqrt{a \over b+4c+4a}+\sqrt{b \over c+4a+4b}+\sqrt{c \over a+4b+4c}\leq 1. \]

[hide]
Applying Cauchy to increase symmetry,
\[\left( \sum_\text{cyc}\sqrt{ \frac{a}{b+4c+4a}}\right)^{2}\leq \left( \sum_\text{cyc}c+4a+4b \right) \left( \sum_\text{cyc}\frac{a}{(b+4c+4a)(c+4a+4b)}\right)\]
We now need to show that the right hand side is no greater than 1, or
\[9(a+b+c)(a(a+4b+4c)+b(b+4c+4a)+c(c+4a+4b))\]
\[\leq (a+4b+4c)(b+4c+4a)(c+4a+4b)\]
Multiplying out will be made easier with symmetric sums: let
\[s=a+b+c,\ t=ab+bc+ca,\ u=abc\]
so that
\[\begin{eqnarray*}(a+4b+4c)(b+4c+4a)(c+4a+4b)&=&(4s-3a)(4s-3b)(4s-3c)\\ &=&16s^{3}+36st-27u\\ (a+b+c)(a(a+4b+4c)+b(b+4c+4a)+c(c+4a+4b))&=&s(a^{2}+b^{2}+c^{2}+8(ab+bc+ca))\\ &=&s^{3}+6st\\\]
Our last inequality is equivalent to
\[7s^{3}-18st-27u\geq 0\]
The left hand side is $7s(s^{2}-3t)+3(st-9u)$. Since $s^{2}-3t=\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{2}\geq 0$ and $st\geq (3u^\frac{1}{3})(3u^\frac{2}{3})=9u$ by AM-GM, the above inequality holds.
[/hide]

You needn't using the p-q-r inequality. Just letting $a+b+c=1$, hence:
\[27abc+18(ab+bc+ca) \leq 7, \]
which is obvious.


calculus
derivative
function
limit
real analysis
real analysis unsolved

Put f(0,0) = 0, and
\[f(x,y) = \frac{{xy(x^{2}-y^{2})}}{{x^{2}+y^{2}}}\]
if \[(x,y) \ne (0,0) \]

Prove that

(a) f, (D_1 f), (D_2 f) are continuous in R^2   (partial derivatives)

(b) (D_12 f) and (D_21 f) exist at every point of R^2, and are continuous except at (0,0)          (partial derivatives)

(c) (D_12 f)(0,0) = 1 and (D_21 f)(0,0) = -1
(showing partial derivatives equal these values)

(1) Use $(x-y)^{2}\geq 0$, or/and $-x^{2}-y^{2}\leq x^{2}-y^{2}\leq x^{2}+y^{2}$ ,and sendwich teorem.
The (2) and (3) are more or less the same, it is obvious that $\partial_{xy}f(x,y)$ and $\partial_{yx}f(x,y)$ are continous on $\mathbb{R}^{2}\backslash{(0,0)}$ because it is a combination of continous functions, and proving that they arnt contious at $(0,0)$ is simply differenting (which is a kill)  and looking at limits on some special sets like $\lim_{(x,x)\rightarrow (0,0)}\partial_{xy}f(x,y)$ or $\lim_{(x,-x)\rightarrow (0,0)}\partial_{xy}f(x,y)$ I believe it should be easy after you differente it twice,which is probably the "hardest" thing :wink:


function
trigonometry
calculus
derivative
geometry unsolved
geometry

[u][i][b] Problem 1[/b] [/i][/u]  Let ABC be a triangle having no obtuse angle
       prove that
                    $ tan\frac {A}{2} \plus{} tan\frac {B}{2} \plus{} tan\frac {C}{2} \plus{} tan\frac {A}{2}.tan\frac {B}{2}.tan\frac {C}{2} \geq \frac {10\sqrt {3}}{9}$
[u][i][b] Problem 2[/b] [/i][/u]  Let ABC be a triangle having no obtuse angle  but each angle of which is not less than $ \frac {\pi}{4}$
    Prove that
  cotgA+ cotgB + cotgC +3cotgA.cotgB.cotgC  $ \leq 4.(2 \minus{} \sqrt {2})$
          [i]      [b]  Y An Forever[/b][/i]
  Argady , Virgil Nicula .... :blush:

[b][i]Problem 3[/i] [/b] Find the greatest value of the function
  [i][b] f(x)[/b][/i] = $ \sqrt {2}.sinx \plus{} \sqrt {15 \minus{} 10.\sqrt {2}.cosx}$
  Virgil Nicula can you help me? :maybe:

[quote="nkht-tk14"][b][i]Problem 3[/i] [/b] Find the greatest value of the function
  [i][b] f(x)[/b][/i] = $ \sqrt {2}.sinx \plus{} \sqrt {15 \minus{} 10.\sqrt {2}.cosx}$[/quote]
It is obvious that the maximum must occur when $ \sin{x}>0$ and $ \cos{x}<0$: i.e. in the interval $ \left(\frac{\pi}{2},\pi\right)$. An intuitive choice would be $ x\equal{}\frac{3\pi}{4}$, giving the maximum value as 6.

To prove this, well …

$ \mathrm{f}'(x)\equal{}\sqrt{2}\cos{x}\plus{}\frac{5\sqrt{2}\sin{x}}{\sqrt{15\minus{}10\sqrt{2}\cos{x}}}$ and $ x\equal{}\frac{3\pi}{4}$ is indeed the only value in the interval $ \left(\frac{\pi}{2},\pi\right)$ for which this derivative vanishes. And check also that $ \mathrm{f}''\left(\frac{3\pi}{4}\right)<0$. Oh well, I’m sure there’s a much simpler way to tackle this problem than this mess. :blush:


For how many positive integers $m$ does there exist at least one positive integer $n$ such that $m\cdot n \le m+n$?
(A) 4
(B) 6
(C) 9
(D) 12
(E) infinitely many

[hide]infinetly many, because n can be 1 and mn<m+n. [/hide]

[hide="okay"]
say n=1
then $m\times{n}<m+n$[/hide]

Can you prove that? I would really like to see a proof.

[quote="SplashD"]it's still infinitely many[/quote]

Nop.

[hide]Rewrite it as $(m-1)(n-1) \leq 1$. But $m,n \geq 1$, so $(m-1),(n-1)\geq 0$ and thus $(m-1)(n-1) \geq 0$, so $(m-1)(n-1)\in\{0,1\}$. If $m\neq 1$, we have exactly one solution for $n$ if $(m-1)(n-1)=0$, i.e. $n=1$. If $m=1$, $(m-1)(n-1)=0$ is always satisfied. Now, if $(m-1)(n-1)=1$, we know that $(m-1)$ and $(n-1)$ are both positive divisors of 1. Therefore, $(m,n)=(2,2)$. So we have 2 solutions if $m=2$, infinitely many solutions if $m=1$, and exactly one solution if $m>2$. Therefore, if $n\neq 1$, only $m=1,2$ gives a solution for $n$.[/hide]

Sorry, I got the inequality mixed up as the other way around

[quote="laughinghead505"]What if n can't be 1?[/quote]

[hide]See Kurt Godel's proof.  It is good.

And for the answer.

If n=1, then it is infinitely many.

(E)[/hide]

That's interesting. So that means that at least one variable (m or n) has to be 1 or 2.


[quote="yhh3002"]
I am. :D[/quote]

We need some proofs :P  :P !!!

[quote="asianman"]koreans, blah, we are smart.
koreans, blah, we study alot.
koreans, blah, we are cool.[/quote]
Thank you. No proof needed.


Can anyone give me a solution to this problem?

Prove

\[
f^{ - 1}  \circ g^{ - 1}  = (g \circ f)^{ - 1}
\]

Thanks

Hmm...take $g \circ f$ of both sides?  that gives you an identity...

The circle notation is interpretted as "f composed of g"

Exactly.  As K81o7 stated, we just get the identity on both sides after taking

$g(f(x))$ on both sides


geometry
trapezoid
geometric transformation
homothety
ratio
geometry solved

On the bases AB and CD of a trapezoid ABCD draw two squares externally to ABCD. Let O be the intersection point of the diagonals AC and BD, and let O1 and O2 be the centers of the two squares. Prove that O1;O and O2 lie on a line.

It's enough to observe that one of the squares (the smaller one) is obtained from the other one by a homothety of center O and ratio -b/B (b is the small base and B is the large one). If a square is obtained from another one by a homothety, the center of the image square is the image of the center of the initial square by the homothety, so the 2 centers and the center of homothety are collinear.

let  I,E  is intersection of O1,O2 with AB,DC => AO1/O2C=AI/EC=AB/CD=AO/OC=AI/EC
  let E' is the intersection point of IO and DC then IA/E'C=AO/CO => E=E'=> we prove :?  :D


calculus
integration
number theory unsolved
number theory

$a-b,a^2-b^2,a^3-b^3,...$ is a infinite sequence and each member of this sequence is $\in \mathbb{N}$.prove that $a$ and $b$ are integers. :)

Indeed, they are only integers everytime when $a=b$ or both are integers, where the first is ruled out by the condition that $a-b$ is natural (assumed as positive integers; please specify in future what 'natural' means for you).

Edit: corrected the typo and want to tell that below there is a simpler proof.

Now $a=b+m , m \in \mathbb Z$.
When $m=0$, the first property holds, so assume $m \neq 0$.
Using this, we get $a^2-b^2 = b^2+2bm +m^2 -b^2 = 2bm + m^2$, thus $2bm$ is integral. Now by $2(a^3-b^3) = 6b^2m + 6 bm^2 + 2 m^3$ also $6b^2 m$ is integral.
Continuing this gives that ($(k-1)!(a^k-b^k) = k! b^{k-1} m + \sum_{s=2}^k (k-1)! \binom{k}{s} b^{k-s} m^s$) that $(k+1)! b^k m$ is integral everytime, thus the reduced denominator $d$ of $b$ fulfills $d^k|(k+1)! m$ for all $k$.
Now for all odd primes $p$ we have, when $v_p(d) \geq 1$, that $k \cdot v_p(d) \geq k$ is growing faster than $v_p(m) + v_p((k+1)!) \leq v_p(m) + \sum_{s =1}^\infty \frac{k+1}{p^s} = v_p(m) + \frac{k+1}{p-1}$ (here $v_p$ denots the multiplicity of the prime factor $p$ in $d$).
But $v_p(m)+v_p((k+1)!) = v_p(m \cdot (k+1)!)$ can't get bigger than $v_p(d)$ by the conditions, thus $v_p(d)=0$.
The same happens when $v_2(d) \geq 2$, thus we have $d=1$, done, or $d=2$.
In the last case, this means that we have odd integers $g,h$ such that $g^n-h^n$ is divisible by $2^n$ for all $n$.
But that means that especially $g^{2n+1} \equiv h^{2n+1} \mod 2^{2n+1}$ which implies ($x \to x^{2n+1}$ is injective $\mod$ powers of two when restricted to odd $x$) $g \equiv h \mod 2^{2n+1}$, thus $g=h$ since their difference is divisible by any power of two.

[quote="Rust"]$It \ is \ easy \ to \ prove:$
$if \ \ a-b, \ a^2-b^2, \ a^3-b^3 \ is \ integer, \ then \ a \ and \ b \ are \ integer.$[/quote]
STOP IT!
This has no content and is wrong, too!
Next time I will delete such posts without any more note!

$I \ am \ sorry. \ For \ all \ k \ exist \ not \ integer \ b=c/m \ and \ a=m^k+c/m, \ that$
$a^i-b^i \ is \ integer \ for \ all \ i=1,...,k.$

Just a simplification of my proof above: assume a prime divides the denominator.
Then take $k$ coprime to $p(p-1)$ and you get that $g^k \equiv h^k \mod p^k \implies g \equiv h \mod p^k$.
But this will lead to $g=h$ again.

$\mathbb{N}$ is also not the same all over the world ;)
(but it's not important here, the statement is clear; but there were problems posted for that it makes a big difference)

[quote="ZetaX"]$\mathbb{N}$ is also not the same all over the world ;)
[/quote]

Exactly because of that I use the notation $\mathbb{Z}_0^{+}$ for non-negative integers and $\mathbb{Z}^{+}$ for positive integers.

[quote="amin_68"][quote="ZetaX"] thus the reduced denominator $d$ of $b$ filfills $d^k|(k+1)! m$ for all $k$.
[/quote]
I have a question,what dose "filfills" mean? ;)
of course there is no problem but next time try to use correct word.  ;)[/quote]
This is just a little typo...
It should be read : f[b]u[/b]lfills.


Let $f\in Z[X], f(X) = X^p-X+a$ where $p$ is prime and $(a,p)=1$. Prove that $f$ is irreducible over $Z[X]$.

My prove of http://www.mathlinks.ro/Forum/viewtopic.php?t=68010 should also work for that case.

Nice seeing you again after half an year.


Solve for $ x$: $ 0.\overline{36} \equal{} \frac{24}{x}$.

$ 0.\overline{36} \equal{} \frac{36}{99}$ which simplifying to $ \frac{4}{11}$. Cross-multiplying yields $ x \equal{} \frac{264}{4}$ simplifying to $ \boxed{66}$


geometry
circumcircle
geometric transformation
incenter
homothety
radical axis
power of a point

Let $ P$ be an arbitrary point on the side $ BC$ of triangle $ ABC$. Denote by $ P'$ the second intersection of $ AP$ with the its circumcircle $ \Gamma$. Let $ \Gamma_{A}^b$ be the circle tangent to $ AP,BP$ and intouches $ \Gamma$ simultanously at $ D$; Let $ \Gamma_{A}^c$ be the circle tangent to $ AP,CP$, and intouches $ \Gamma$ simultanously at $ E$, Denote by $ A'$ the tangency point of $ A$-mixtilinear incircle of triangle $ ABC$ with $ \Gamma$. Prove that the three lines $ DE,BC, A'P'$ are concurrent.

Let $ M$, $ N$ be the tangency points of $ BC$ with $ \Gamma_{A}^{b}$ and $ \Gamma_{A}^{c}$, respectively. It is rather easy to see that the points $ D$, $ E$, $ M$, $ N$ are lying on a same circle $ \rho_{1}$. On the other hand, according to problem [url]http://www.mathlinks.ro/viewtopic.php?t=213098[/url], the points $ A^{\prime}$, $ P^{\prime}$, $ M$, $ N$ are also lying on a same circle $ \rho_{2}$. In this case, the lines $ DE$, $ BC$, $ A^{\prime}P^{\prime}$ are concurrent at the radical center of the circles $ \rho_{1}$, $ \rho_{2}$ and the circumcircle of $ ABC$.

We know that $ DE$, $ BC$, $ O_{1}O_{2}$ are concurrent at the exsimilicenter of $ \Gamma_{A}^{b}$ and $ \Gamma_{A}^{c}$. We also know that this point is the radical center of the Thebault circles of cevian $ AP$ and the circumcircle of $ ABC$ and according to the above post, we know that the midpoint of the small arc $ BC$ lies on the radical axis of $ \Gamma_{A}^{b}$ and $ \Gamma_{A}^{c}$. Actually, we can also prove that this radical axis passes through the midpoint of $ IP$, where $ I$ is the incenter of $ ABC$ (see the connection of this result with Lemma 1 of [url=http://forumgeom.fau.edu/FG2003volume3/FG200325.pdf]Jean-Louis Ayme's beautiful paper[/url]?)

Let the incircle of triangle $ ABC$ touch the side $ BC$ at $ F$, I discover an interesting conclusion $ \angle MDF\equal{}\angle NEP$.

[quote="pohoatza"]Nice. I wrote the proof of your result in the new thread posted by Zhang Fangyu at [url]http://www.mathlinks.ro/viewtopic.php?t=214687[/url]. Regarding the fact that the radical axis of the Thebault circles of cevian $ AP$ passes through the midpoint of $ IP$, consider the homothecy $ h(I,2)$ and use Lemma 1 from [url]http://forumgeom.fau.edu/FG2003volume3/FG200325.pdf[/url] (actually the "polar" equivalent of Lemma 1, i.e. since $ I$ lies on the polar of $ P$ wrt. each of the Thebault circles, then $ P$ lies on the polar of $ I$ wrt. the two circles).[/quote]
Yes!dear MR.FANG, this is exactly the solution I find! :lol:

Dear Fang-jh and Mathlinkers,
a new way for a proof of this result can be found on on my website :
http://perso.orange.fr/jl.ayme   vol. 4   A new mixtilinear incircle adventure III, p. 36, 41
Sincerely
Jean-Louis


Well guys ,what will you do in this case :D ?
You have a windows XP on your PC,and[b] someone[/b] has changed all of the passwords of your acounts ,so that you can not enter the windows normally and you desparatly need the informations in this PC  ,So what would you do?

[hide="My solution"].I will try to force that [b]someone[/b] to give me the passwords but if this doesnt work,then here is a good solution:
Restart your PC,before letting the windows to start coming up (when it will show a picture of windows symble) press F8 to go to a selection menu.select the SAFE MODE and then a new acount will apear:
its name is :
Administator
It is usually intact but if this one has password too,then start crying cause i dont know what to do this time  :( [/hide]

It's simple: reinstall windows on another partition.

Well, you can mount your partition(s) in a Linux-based environment by using mount. For more informations, when running linux in text-mode, write "man mount", or google for "man mount" whenever the X Server is loaded. In order to deal with this easier, just get your hands on a copy of the Knoppix distribution. It is very nice, it boots off the cd and in about 2-3 minutes (on my 1000 mhz Athlon), you have a running operating system, with graphical interface! Then you can go to the text mode and mount your partitions, access whatever information you need, burn it to CD's and probably the best long-term solution is to reinstall windows. I think there should be some cracks available on the internet to reset all windows passwords, but I've never run into this problem so I wouldn't know. By the way, I'm thinking about using Linux in the long term, even though it is so hard to use :)

Theres a password changer that you can dowload, just serach for password recovery, and you can download a bootable CD that allows you to change the password of any account.

Wow, that's a very insecure thing.

I don't like that idea - is there security settings installed? If not, then someone may be able to hack any computer...

Of course someone can "hack" your computer when they have the computer in their hands. The windows password has nothing to do with hacking. It's just to prevent unauthorized access. And besides the password doesnt' encrypt anything.

the Safe Mode>Administrator method works very well, I doubt the PERSON was clever enough to password that one.. besides which i doubt you can. Also, for the future, and this is what i do, make one administrative account and all the others limited, for very much the reason so the other users cant chang eyour password.


A cannonball is launched at an angle of $ 48^{o}$ and at a speed of $ 166\;m/s$. Determine the:

a.) Horizontal Component
b.) Vertical Component
c.) Distance away from the cannon the ball will land.

Neglect air resistance in your answers. Assume the starting height is $ 0$ meters.

I made these numbers up, so they might not work too well :heli:

[hide="relevant formulae"]
1.$ u_{x}\equal{}u\cos\theta$
2.$ u_{y}\equal{}u\sin\theta$
3.$ R\equal{}\frac{u^{2}\sin 2\theta}{g}$[/hide]

To solve this kind of problems start by writing the parametric equations for the position and velocity along x (horizontal) and y (vertical) axes, which we can derive by applying to each axe the equation for constant accelerated motion:

$ r \equal{} r_{0}\plus{}v_{0}t\plus{}\frac{1}{2}at^{2}$

(side issue: how can we derive this expression?)

Let's apply this to the x direction: assuming $ r_{0}\equal{} x_{0}\equal{} 0$, $ v_{0}\equal{} v_{0x}$, and as there is no force applied to the particle in the horizontal direction ($ a_{x}\equal{} 0$), then $ x \equal{} v_{0}\cos\theta t$, where $ \theta$ is the angle between the initial velocity vector and the horizontal axis. For the y direction: the only force applied is the weight, and so $ a_{y}\equal{}\minus{}g$; also let $ r_{0}\equal{} y_{0}\equal{} 0$, and so $ y \equal{} v_{0}\sin\theta t\minus{}\frac{1}{2}gt^{2}$. The parametric equations for the velocity can be derived by simply differentiating the equation for the position: $ v_{x}\equal{} v_{0}\cos\theta$ and $ v_{y}\equal{} v_{0}\sin\theta\minus{}gt$.
To find the distance at which the cannon ball will land, ask yourself "What are the coordinates of that position, and what is the total amount of time required for the ball to reach there?".


What is the value of the largest mean among the following sets of whole numbers?

A: all multiples of 3 between 1 and 50,

B: all multiples of 4 between 1 and 50,

C: all multiples of 6 between 1 and 50,

D: all multiples of 7 between 1 and 50,

E: all multiples of 9 between 1 and 50.

Mean of A: $ \frac{3\plus{}48}{2}\equal{}25.5$

Mean of B: $ \frac{4\plus{}48}{2}\equal{}26$

Mean of C: $ \frac{6\plus{}48}{2}\equal{}27$

Mean of D: $ \frac{7\plus{}49}{2}\equal{}28$

Mean of E: $ \frac{9\plus{}45}{2}\equal{}27$

Answer is $ \boxed{28}$.


induction
calculus
integration
symmetry
trigonometry
logarithms
trig identities

Coil, as in solenoid?  And does $ L$ indicate self-inductance?

Could you by any chance provide his solution?

Since a solenoid is technically a coil, we have the following:
Let $ B$ be the magnetic field in a solenoid and $ n$ the number of turns per unit length and the number of flux linkages in the length $ \ell$ is
$ N\Phi_{B}\equal{}(n\ell)(BA)$
where $ A$ is cross-sectional area.  Then we know
$ L\equal{}\frac{N\Phi_{B}}{i}$ so we have

$ L\equal{}\frac{(n\ell)\cdot (BA)}{i}$

If we really want to get nitty-gritty, we can actually stick in $ B$.  Let $ R$ be the radius of the solenoid:

$ L\equal{}\frac{\pi R^{2}\mu_{0}n^{2}\ell}{2}\left(\frac{\frac{\ell}{2}\plus{}d}{\sqrt{R^{2}\plus{}\left(\frac{\ell}{2}\plus{}d\right)^{2}}}\plus{}\frac{\frac{\ell}{2}\minus{}d}{\sqrt{R^{2}\plus{}\left(\frac{\ell}{2}\minus{}d\right)^{2}}}\right)$

I think he is not thinking of a solenoid, thats too easy for Landau to solve it...

He is probably thinking of a single turn coil...

You should start by considering the magnetic energy, and use that a magnetic field of a localized current distribution can be approximated as that of a magnetic dipole, when we are sufficiently far from the the "small" wire....

I will try to post a full solution in a while, when I have some free time.... :D

Btw. JRav please answer the "oscillations" topic, I am very interested in how you guys solved that problem... :D

Sorry but it turns out that it is not so easy as it looks.
It is actually from his book "electricity and magnetism"

I felt like reviving this thread because methinks the original problem is an interesting integration exercise (but I doubt it will help the original poster).  First, by symmetry, we know that the magnetic field within the loop varies only with the radius $ r$ from its center.  Call $ \theta$ the central angle.  Then by the scalar form of the B-S law and the law of cosines,
\[ dB = \left(\frac {\mu_0}{4\pi}\right)\frac {Idl\sin\theta}{R^2} = \left(\frac {\mu_0 I}{4\pi}\right)\frac {a\sin\alpha d\theta}{a^2 + r^2 - 2ar\cos\theta}
\]
where
\[ \frac {R}{\sin\theta} = \frac {r}{\sin\alpha} \implies dB = \left(\frac {\mu_0 I}{4\pi}\right)\frac {ar\sin\theta d\theta}{(a^2 + r^2 - 2ar\cos\theta)^{3/2}}.
\]
Thus,
\[ B(r) = 2\int_0^\pi dB = \frac {\mu_0 Iar}{2\pi}\int_0^\pi \frac {\sin\theta d\theta}{(a^2 + r^2 - 2ar\cos\theta)^{3/2}} \\
= \frac {\mu_0 Iar}{2\pi}\left[ - \frac {1}{ar}(a^2 + r^2 - 2ar\cos\theta)^{ - 1/2}\right]_0^\pi \\
= \frac {\mu_0 I}{2\pi}\left(\frac {1}{a - r} - \frac {1}{a + r}\right) \\
= \frac {\mu_0 I}{\pi}\frac {r}{a^2 - r^2}.
\]
Then,
\[ \Phi_B = 2\pi\int_0^{a - b} B(r)r dr \\
= 2\mu_0 I\int_0^{a - b} \frac {r^2}{a^2 - r^2} dr \\
= 2\mu_0 I\left[\frac {a}{2}\ln\left|\frac {a + r}{a - r}\right| - r\right]_0^{a - b} \\
= \mu_0 I\left[a\ln\left(\frac {2a}{b} - 1\right) + 2(b - a)\right] \\
\& \approx \& \mu_0 Ia\left[\ln\left(\frac {2a}{b} - 1\right) - 2\right]
\]
by the relative smallness of $ b$.  Finally, you can take $ L = \Phi_B/I$ and $ I$ cancels.  It's not a very pretty answer, but does it look correct?

Edit: Oh, another thing: in fact, we can say
\[ L \approx \mu_0 a\left[\ln\left(\frac {2a}{b}\right) - 2\right]
\]

[quote="uugan_88"]I am having trouble with understanding the solution in Landau.So Could comeone derive it in a simple way.
Calculate the induction L of a coil with radius a and made of wire with cross section radius b.
It is enough to consider the case a>>b[/quote]

Do you mean toroid?or solenoid? :roll:


Alright: 3 times sqroot(x-1) = 6th root of (2x+6)

Original question in latex:

$ 3\sqrt{x \minus{} 1} \equal{} \sqrt[6]{2x \plus{} 6}$

I don't think this has a pretty answer...

[quote="SnipedYou"]Original question in latex:

$ 3\sqrt {x \minus{} 1} \equal{} \sqrt [6]{2x \plus{} 6}$

I don't think this has a pretty answer...[/quote]

Yeah, I know for a fact that it does, but I'm having trouble because my answer doesn't match the answer given.

Uh, I tried this on my TI-89, And it was computing for about a minute.

[hide]
\begin{eqnarray*} 3\sqrt {x - 1} & = & \sqrt [6]{2x + 6} \\
729(x^3 - 3x^2 + 3x - 1) & = & 2x + 6 \\
729x^3 - 2187x^2 + 2187x - 729 & = & 2x + 6 \\
729x^3 - 2187x^2 + 2185x - 735 & = & 0 \end{eqnarray*}

I refuse to go any further.
[/hide]

I just had Mathematica solve it, and it doesn't have pretty solutions.


a merchant vistited 3 fairs.  at the firs, he doubled his money and then spent 30 dollars; at the second he tripled the remaining money and then spent 54 dollars; at hte third, he quadrupled the remaining money and then spent 72 dollars; he then had 48 dollars left.  How much money did he start with?

You start out with $29. Do you know what an arrow diagram is? I find them helpful to solve the problem, because it helps you to work backwards.
You start with x amount of money.
*2
-$30

*3

-$54
*4
-$72

= $48.
With an arrow diagram, you just make like little humps, and put what you're doing above that, and then at the points at the bottom, you put what you get when you go back, until you get to x, which is what you are solving for.
$48 + $72 = $120

$120/4 = $30

$30 + $54= $84
$84/3 = $28
$28 + $30 = $58

$58/2 = $29

[hide="um..here"](before I begin, I must say that the question was kind of vague to understand...)

beginning with the first fair ( and setting his money as 'x' ) this should be the equation...

2x -30

second fair

6x - 90 - 54 = 6x - 144

third fair

24x - 648 = 48

solving the equation...

x = 29

( that's not much of earnings..I would rather work at McDonalds..... )

[/hide]

and elementsofmathkid, please hide your answer...

I tried, but when I did, I had the BBCode enabled, and since there was a $ sign, it changed everything to the LaTex stuff that my brother did something with and won't admit to.

I'd gladly change it if you can help me figure out how to change it so I can hide my message without the LaTex stuff changing my post.

Grrr.... frustrating.  :)  :lol:

hey, guys thank you for showing your work!
 :D  :D  :D  :D


geometry
ceiling function
SFFT
special factorizations

[quote="lingomaniac88"]Prove that for any odd positive integer greater than 1, there exists two positive integers $a$ and $b$ (not necessarily distinct) so that the sum and product of the two add up to it.[/quote]
[hide="Simple_Solution"]Call the number $n = 2k+1$. Then just let $a = k$ and $b = 1$. The desired property immediately follows.[/hide]

[quote="lingomaniac88"]Prove that for any odd positive integer greater than 1, there exists two positive integers $a$ and $b$ (not necessarily distinct) so that the sum and product of the two add up to it.[/quote]

[hide]$2n+1=ab+a+b$
$2(n+1)=(a+1)(b+1)$
$a=1$, $b=n$[/hide]

Edit: Oops. Same as above..

This problem uses...

[hide]
SIMON'S FAVORITE FACTORING TRICK![/hide]

Thanks neelnal for this post's link!

Anyways, I don't get how $a=1$ and $b=n$ solves the problem... :( I am confused with your proofs.

All odd integers can be expressed as $2n+1$ where $n$ is any integer. That's how they can show the proof is complete.

How did $2(n+1)=(a+1)(b+1)$ results in $a=1$ and $b=n$?

[color=darkblue]You proved that exists at least a solution, but generally, are more. For example $(0<a\le b)$ :

- for $11$ exists two and only two solutions,  $(1;5)$ and $(2;3)\ .$
- for $23$ exists three and only three solutions, $(1;11)$ , $(2;7)$ and $(3;5)\ .$

I propose you for a fixed positive odd integer $n$ to count the all solutions of the equation $a+b+ab=n\ ,$ where $\{a,b\}\subset N^{*}$ and $a\le b\ .$[/color]

If $a =1$ and $b = n$ we have $2(n+1) = (1+1)(1+n)$
which is the same as $(2)(n+1)$
Which is the desired condition. QED.

Wow, that's so simple!  :o
Thanks DiscreetFourierTransform for helping me out!

These type of things tend to be simple in retrospect. I know many geometry proofs that are so nice and elegant *after* you've worked through them  :P
Off topic tangent:
It takes a while of staring at a triangle and not getting anywhere with your proof until you realized that
D'OH!
The stupid triangle is a right triangle and you could've done the proof in like five minutes.

Anyone else ever have this feeling of missing something important?

I get this all the time  :oops:

[quote="Virgil Nicula"][color=darkblue]I propose you for a fixed positive odd integer $n$ to count the all solutions of the equation $a+b+ab=n\ ,$ where $\{a,b\}\subset N^{*}$ and $a\le b\ .$[/color][/quote]

[hide="Solution"] Since $n+1 = (a+1)(b+1)$ we want the number of ways to factor $n+1$ into a product of non-unity integers.

Of course, the number of ways to factor $n+1$ into a product of integers is simply

$\lceil \frac{d(n+1)}{2}\rceil$

Where $d(k)$ is the number of divisors of $k$.  (Each larger divisor of $k$ is paired with a smaller divisor of $k$ except in the case $k$ a square, hence the need for the ceiling.)

Then we just ignore the factoring $1, n+1$, so we have

$\lceil \frac{d(n+1)}{2}\rceil-1$[/hide]

[quote="DiscreetFourierTransform"]The stupid triangle is a right triangle and you could've done the proof in like five minutes.[/quote][hide="Only for D.......F......T........ ."][color=darkblue][b]Example.[/b] Let $ABC$ be a  be a triangle. Denote the middlepoint $M$ of the side $[BC]$ and the points $\{X,Y\}\subset (BC)$ so that $BX=XY=YC\ .$

Prove that $\widehat{XAM}\equiv\widehat{CAM}$ $\Longleftrightarrow$ $AM\perp AB\Longleftrightarrow$ $YX^{2}=YM\cdot YB$ $\Longleftrightarrow$ $MX\cdot MC=MB\cdot MY$ $\Longleftrightarrow$ $\frac{YB}{YM}=\left(\frac{XB}{XM}\right)^{2}\ .$

I wait a solution of this simple problem and I'll come back after you "ll post it.[/color][/hide]

[quote="lingomaniac88"]Prove that for any odd positive integer greater than 1, there exists two positive integers $a$ and $b$ (not necessarily distinct) so that the sum and product of the two add up to it.[/quote]
[hide]$\begin{eqnarray*}ab+a+b=n\\ ab+a+b+1=n+1\\ \text{SIMON'S FAVORITE FACTORING TRICK!!!}\\ (a+1)(b+1)=n+1$
since $n$ is odd, $n+1$ is even.  All even numbers can be written as $2z$, so letting $a=1$ and $b=z-1$, we can always find $a$ and $b$.  :D [/hide]


inequalities
function
inequalities unsolved

The generalized Jensen's inequality states that if $f$ is a convex function and

$\displaystyle \sum^{n}_{i=1} \lambda_i=1$ with $\lambda_i\in [0,1]$

then

$\displaystyle \sum^{n}_{i=1} \lambda_if(x_i) \geq f\left(\sum^{n}_{i=1} \lambda_ix_i\right)$

Now does anyone have a proof?  (I know it for the case $\displaystyle a_i=\frac {1}{n}$)

Yes [b]blahblahblah[/b] it is Jensen's generalized inequality but there is more general inequality.Here is  the Fuchs inequality .Let$ f$ is convex function .$p_{1},p_{2},...,p_{n}$ are positive. If $x_{1}\geq{x_{2}}\geq...\geq{x_{n}}$ and $y_{1}\geq{y_{2}}\geq...\geq{y_{n}}$ such that $\sum_{i=1}^{k}{p_{i}x_{i}}\geq\sum_{i=1}^{k}{p_{i}y_{i}}       (1\leqq\leq{n-1}) $and$\sum_{i=1}^{n}{p_{i}x_{i}}=\sum_{i=1}^{nk}{p_{i}y_{i}}$ then $\sum_{i=1}^{n}{p_{i}f(x_{i})}\geq\sum_{i=1}^{n}{p_{i}f(y_{i})}.
$
We can get Jensen's inequality putting$ p_{1}+p_{2}+...+p_{n}=1$ and $y_{1}=y_{2}=...=y_{n}=p_{1}x_{1}+p_{2}x_{2}+...+p_{n}x_{n}$
Can you prove this?

Fuchs inequality was discussed in thread devoted to Karamata inequality. Its proof is much the same as Karamata.


geometry
geometric transformation
rotation
reflection
rectangle
combinatorics unsolved
combinatorics

Define a "msş" to be a figure made up of five unit squares as shown below in the picture, or any of the figures obtained by applying rotations and reflections to this figure.

Determine all  rectangles that can be covered without gaps and without overlaps with hooks such that

- the rectangle is covered without gaps and without overlaps
- no part of a msş covers area outside the rectagle.

You may find the answer in "Klarner systems and tiling boxes with polyominoes" by Michael Reid. A free copy is available on his [url=http://www.math.ucf.edu/~reid/Research/Ksystem/index.html]web page[/url] and the paper is definitely worth reading if you like such questions. :)


function
LaTeX
inequalities
trigonometry
real analysis
real analysis unsolved

Let $d= \frac{b-a}{n}$

Let $g(x) = f(x+d)-f(x)$ for $x\in [a,a+(n-1)d]$.
We want to show that $g(x)$ has a zero.

$g(a) + g(a+d) + g(a+2d) + \cdots + g(a+(n-1)d) = f(b) - f(a) = 0$.

If one of $g(a+id) = 0 (i=1,2,\dots,n-1)$ then we are done.
If none of them is 0, there must exist $i,j$ such that $g(a+id)>0$ and $g(a+jd)<0$. The conclusion follows from intermediate value property.

Another message with Darboux has my solution.

Levi,here is a proof in the particular case $a=0,b=1$.So the problem becomes:

    Let $f:[0,1]\to\mathbb{R}$ a continuous function such that $f(0)=f(1)$ and $n\in\mathbb{N}$ with $n\geq2$.
    prove that there is $x_{0}\in [0,1] $ such that $f(x_{0})=f(x_{0}+\frac{1}{n})$.

  Solution.

    If $f$ is not a continuous function and there is not $c$ such that $f(c)=0$ then $f$ has constant sign on $[0,1]$
   Let us define the function $g:[0,1-\frac{1}{n}]\to\mathbb{R}$ such that $g(x)=f(x+\frac{1}{n})-f(x)$.Obviously,$g$ is continuous.Let's assume by contradiction that $g(x)>0$ for example,for any $x$.We deduce
   that $f(x+\frac{1}{n})>f(x)$, for any $x\in [0,1-\frac{1}{n}]$.For $x=0$ we have $f(\frac{1}{n})>f(0)$
   for $x=\frac{1}{n}$ we have $f(\frac{2}{n})>\frac{1}{n})$, and so on....'til for $x=\frac{n-1}{n}$ we have
   $f(1)>f(\frac{n-1}{n})$ and from here we deduce that $f(1)>f(0)$,contradiction.
    In the other case we will obtain $f(1)<f(0)$. ;)

[b]Moderator edit: edited for a minor $\LaTeX$ error[/b]

[quote="levi"][b]Let continue function [/b]$f:[a,b]\rightarrow R,\ f(a)=f(b)\Longrightarrow $
$\forall n\in N^* ,\ \exists\ x_n\in [a,b]$ [b]  such that[/b] $f(x_n)=f\left( x_n+\frac{b-a}{n}\right).$[/quote]
It's OK, very elegant and also well-known ,    [e.g.  see Paul Levy -(1932  ??) = Chord universal theorem"=a very nice result completed by Paul Halmos [???] ]  .In the next days I inform my friends  about some refences about this beautifull theorem !!

[i]Dear Flip2004[/i], You think me! I did not know that this problem was known of approx. 80 years. Please you, give me that referencies. Here is my solution for my "own" probem:

[b]For n=1, we can choose[/b] $x_1=a\in [a,b].$[b] Let[/b] $n\in N,\ n\ge 2.$ [b]Consider the function[/b]
${F(x)=f(x)-f( x-\frac{b-a}n}),\ x\in [a,b]$.[b] But [/b]
$\sum_{k=1}^{n} F ( a+k\cdot \frac{b-a}{n}  )=$ $\sum_{k=1}^{n}  [ f  ( a+k\cdot \frac{b-a}{n} )-f ( a+(k-1)\cdot\frac{b-a}{n} ) ] =f(b)-f(a)=0.$
Thus, there are $1\le p<q\le n$ such that $F ( a+p\cdot \frac{b=a}{n} ) F ( a+q\cdot \frac{b-a}{n} )\le 0$ a.s.o.

For $ -\infty<a<b<\infty $ let us denote by  $  C[a,b] $  the linear space of all functions  $ [a,b]\to {\mathbb R}  $ which are continuous on $ [a,b]   .  $    The set of all positive integers is denoted by $ {\mathbb N}^*  . $ Let us  prove the following assertion :
 [b]Theorem 1.[/b] [i]    Let  $  f ,g \in C[a,b] $  and suppose  $ g $ strict monotonic on  $ [a,b]   .  $
[b] i).[/b] For  $  h=\frac{b-a}{n}\; ,\; n\in {\mathbb N}^*    ,
$ there exists in  $ [a,b-h] $ at least  a point $ \xi    , $ such that
\[(1)\; \; \; \;   \frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f(\xi+h)-f(\xi)}{g(\xi+h)-g(\xi)}
\; \; .
 \]
[b] ii). [/b]  If  $  h\in (0,b-a)\; ,\; h\not \in \left\{\frac{b-a}{n}\; \Big|\; n\in {\mathbb N}^*  \right\}   , $ then there exists a pair of functions  $ f^* , g^* \in C[a,b] $  for which
\[ \frac{f^*(x+h)-f^*(x)}{g^*(x+h)-g(x)}\ne \frac{f^*(b)-f^*(a)}{g^*(b)-g^*(a)}\; ,\hspace*{0.5cm} \forall
x\in [a,b-h]\; .
\]
[/i] [b]Proof.[/b]  [b]i).[/b]  Let us assume that the interval  $ [a,b-h] $ does not contains such a  point  $ \xi    .  $ According to continuity of functions
 $ f , g :[a,b]\to {\mathbb R}  $ it follows that  $  D:[a,b]\to {\mathbb R}    , $    where
\[D(x)= \left(g(b)-g(a)\right)\left(f(x+h)-f(x)\right)-\left(f(b)-f(a)\right)\left(g(x+h)-g(x)\right)
\]
preserves a constant sign on  $  [a,b-h]   .  $    For instance , let us suppose   $  D(x)>0\; ,\; \forall x\in [a,b-h]   .  $    In particular, we have
\[(2)\; \; \;
 D(x_k)>0 \; \; \; \; \mbox{where}\; \; \;  x_k=a+kh \; ,\; \; \;  k\in \{
0,1,\ldots,n-1\}\; . \]
 On the other hand, by summing  the inequalities (2) one finds
\[
\begin{array}{c}
 \displaystyle 0\; <\;  \displaystyle \sum\limits_{k=0}^{n-1}D(x_k) =
 \displaystyle
\left(g(b)-g(a)\right)\sum\limits_{k=0}^{n-1}\left(f(x_{k+1})-f(x_k)\right)-\\
- \displaystyle \left(f(b)-f(a)\right)\sum\limits_{k=0}^{n-1}\left(g(x_{k+1})-g(x_k)\right)=\\
= \displaystyle \left(g(b)-g(a)\right)\left(f(b)-f(a)\right)-
\left(f(b)-f(a)\right)\left(g(b)-g(a)\right)= 0\; ,\\
\end{array}
\]
that is a contradiction.
[b]ii).[/b] A natural question is to investigate if in (1)  the number  $ h $ may be selected in an  arbitrary way in $ (0,b-a)   .  $    This  has  a negative answer.
More precisely, we show that if    $  h\ne \frac{b-a}{n}   , $    then the mean value theorem from    (1)  does not remain  true for every pair  of functions $ f,g $ having properties  specified in  hypothesis.
 Let  $ g^*(x)=x $ and observe that    $  h\ne \frac{b-a}{n}\; ,\; n\in {\mathbb N}^*    , $    implies  $  \lambda\; :\; =\sin{\frac{\pi(b-a)}{h}}\ne 0   .  $    Further define  $ f^*:[a,b]\to {\mathbb R}   $    by
\[ f^*(x)= x-a-\frac{b-a}{\lambda^2}\sin^2{\frac{\pi(x-a)}{h}}\; \;  ,\;  \;  x\in
[a,b]\; .\] This function satisfies $ f^*(b)-f^*(a)= 0\; ,\; f^*(x+h)-f^*(x)= h\ne 0\; ,   $   which means that the equation
\[
\frac{f^*(x+h)-f^*(x)}{g^*(x+h)-g^*(x)}-
\frac{f^*(b)-f^*(a)}{g^*(b)-g^*(a)}=0\; \; ,\; \; \;  x\in [a,b-h]\] has no solution.
 This proposition is an extension of [b]universal-chord theorem [/b], the case $ g(x)=x\; , \; f(a)=f(b)   , $ see   [1]-- [4] and also [5].

[b]References.[/b]
[1] R. P. Boas , [i] A primer of real functions[/i] ,   The Carus Mathematical Monographs , MAA, vol.13, 1960.
[2] H. Hopf  , [i] \"Uber die Sehnen ebener Kontinuen und die Schliefen geschlossener Wege[/i] , Comment.Math. Helvetici 9 (1937), 303-319.
[3] R. J. Levit , [i] The finite difference extension of Rolle' s theorem[/i],  Amer. Math. Monthly 70 (1963) , 26-30 .
[4] P. Levy ,  [i] Sur une g\'en\'eralisation du th\'eor\`eme de Rolle[/i] , C. R. Acad. Sci. Paris 198 (1934) , 424-425.
[5] A. Lupas , [i]O varianta discreta a teoremei de medie a lui Cauchy[/i],  Arhimede ,  Nr.7-8/2002,pag.1-2.

We kill a butterfly with atomic bomb ! A manufacturing carpet (worked with hands) is more precious than a mechanical carpet ! I am a teacher (such you) and with my students from a high school I made, make and will make a handicraft toil.


[quote="Γιώργος"]Έστω $ a,b,c,d,e,f$ πραγματικοί αριθμοί τέτοιοι ώστε: $ a \plus{} b \plus{} c \plus{} d \plus{} e \plus{} f \equal{} 10$ και $ (a \minus{} 1)^{2} \plus{} (b \minus{} 1)^{2} \plus{} (c \minus{} 1)^{2} \plus{} (d \minus{} 1)^{2} \plus{} (e \minus{} 1)^{2} \plus{} (f \minus{} 1)^{2} \equal{} 6$ Να υπολογίσετε τη μέγιστη δυνατή τιμή του $ f$. :)[/quote]
$ maxf\equal{}10/3$ με $ BCS$

τι ειναι το BCS?

Είναι η ανισότητα Buniakovsky - Cauchy - Schwarz (Μερικές φορές λέγεται απλώς Cauchy - Schwarz):

$ (a_1^2 \plus{} a_2^2 \plus{} ... \plus{} a_n^2)(b_1^2 \plus{} b_2^2 \plus{} ... \plus{} b_n^2) >\equal{} (a_1b_1 \plus{} a_2b_2 \plus{} ... \plus{} a_nb_n)^2$
για πραγματικούς αριθμούς.

Αυτή είναι η λύση που έδωσα:
$ (a \minus{} 1)^2 \plus{} (b \minus{} 1)^2 \plus{} (c \minus{} 1)^2 \plus{} (d \minus{} 1)^2 \plus{} (e \minus{} 1)^2 \plus{} (f \minus{} 1)^2 \equal{} 6 < \equal{} >$
$ (f \minus{} 1)^2 \equal{} 6 \minus{} (a \minus{} 1)^2 \minus{} (b \minus{} 1)^2 \minus{} (c \minus{} 1)^2 \minus{} (d \minus{} 1)^2 \minus{} (e \minus{} 1)^2 < \equal{} >$(από την Andreescu)
$ (f \minus{} 1)^2 < \equal{} 6 \minus{} \frac {(a \plus{} b \plus{} c \plus{} d \plus{} e)^2}{6} < \equal{} > (f \minus{} 1)^2 < \equal{} 6 \minus{} \frac {(10 \minus{} f)^2}{6} < \equal{} >$
$ 6f^2 \minus{} 12f \plus{} 6 < \equal{} 36 \minus{} 100 \plus{} 20f \minus{} f^2 < \equal{} > 7f^2 \minus{} 32f \minus{} 30 < \equal{} 0$

Και εφαρμόζοντας θεωρία τριωνύμου βρίσκουμε ότι το f πρέπει να βρίσκεται μεταξύ των ριζών της
$ g(f) \equal{} 7f^2 \minus{} 32f \minus{} 30$

Πώς βγαίνει με B. C. S. ;

Νομίζω maniopas ότι έχεις κάνει κάποιο λάθος. Βασικά εκεί που κάνεις την Andreescu κατ' αρχάς ο παρονομαστής είναι 5 και κατά δεύτερον ο αριθμητής είναι $ (a\plus{}b\plus{}c\plus{}d\plus{}e\minus{}5)^2$

Oσον αφορά τη λύση με B.C.S που ανέφερε ο κ. Μπόρης νομίζω είναι η ίδια λογική με τη δική σου, απλά εσύ εφαρμόζεις την ανισότητα με τη μορφή του Andreescu :wink:

[quote="kunny"]Let $ a \minus{} 1 \equal{} A,\ b \minus{} 1 \equal{} B,\ c \minus{} 1 \equal{} C,\ d \minus{} 1 \equal{} D,\ e \minus{} 1 \equal{} E,\ f \minus{} 1 \equal{} F,$ the problem becomes $ A \plus{} B \plus{} C \plus{} D \plus{} E \plus{} F \equal{} 4$ and $ A^2 \plus{} B^2 \plus{} C^2 \plus{} D^2 \plus{} E^2 \plus{} F^2 \equal{} 6.$

Cauchy-Schwarz gives $ (1^2 \plus{} 1^2 \plus{} 1^2 \plus{} 1^2 \plus{} 1^2)(A^2 \plus{} B^2 \plus{} C^2 \plus{} D^2 \plus{} E^2)\geq (1\cdot A \plus{} 1\cdot B \plus{} 1\cdot C \plus{} 1\cdot D \plus{} 1\cdot E)^2$

$ \Longleftrightarrow 5(6 \minus{} F^2)\geq (4 \minus{} F)^2$, yielding $ \minus{} 1\leq F\leq \frac {7}{3}.$ By $ f \equal{} F \plus{} 1$, we have $ 0\leq f\leq \frac {10}{3}.$[/quote]
Το προβλημα ομως ζηταει την μεγιστη ΔΥΝΑΤΗ τιμη του f.
Εσυ βρηκες ΜΙΑ μεγιστη.
Πρεπει να δειξεις ομως οτι η 10/3 ειναι και δυνατη τιμη της f.

[quote="Durandal"]Hello,

Μου φαίνεται ότι η άσκηση λύνεται και με ένα απλό γεωμετρικό συλλογισμό, αρκεί να παρατηρήσει κανείς τη συμμετρία του προβλήματος. Εναλλακτικά (ιδίως για όσους δεν αρέσκονται σε μία απλή επίκληση ανισοτήτων), θεωρήστε τα παρακάτω ως τη γεωμετρική ερμηνεία του τι πραγματικά συμβαίνει με τη συγκεκριμένη εφαρμογή της Cauchy-Schwarz.

Μετά από μία προφανή μετάθεση των αξόνων κατά $ 1$, παρατηρείστε ότι οι δύο περιορισμοί γράφονται:
1. $ \sum_{i \equal{} 1}^6 x_i \equal{} 4$ που είναι ένα (υπερ)επίπεδο $ P$, κάθετο στο διάνυσμα $ e \equal{} (1,\ldots,1)$ (symmetry hint ;) ).
2. $ \sum_{i \equal{} 1}^6 x_i^2 \equal{} 6$ που είναι μία (υπερ)σφαίρα $ S$ με ακτίνα $ \sqrt {6}$.

Άρα, απλώς αναζητούμε το σημείο της τομής του $ P$ με την $ S$ που είναι πιο ψηλά στον άξονα $ x_6$. Λόγω της συμμετρίας του $ P$ (και, φυσικά, της $ S$ :P) το διάνυσμα που αναζητούμε θα έχει προφανώς τη μορφή $ (p,p,p,p,p,q)$.

Τελειώσαμε... αντικαθιστώντας στις (1) και (2) βρίσκουμε $ q \equal{} \frac {7}{3}$ ως τη μόνη αποδεκτή ρίζα της δευτεροβάθμιας $ 3q^2 \minus{} 4q \minus{} 7 \equal{} 0$ που προκύπτει. Επιστρέφοντας λοιπόν στις αρχικές μας συντεταγμένες, βρίσκουμε και το ζητούμενο $ f_{\textrm max} \equal{} \frac {10}{3}$.

Και μέγιστο και δυνατό :)

Cheerio,

Durandal 1707[/quote]
πολυ ωραία εξήγηση Παναγιώτη
Απλά υπενθυμίζω ότι .όταν  ισχύει f=10/3 ισχύει το ίσον στην BCS άρα οι συντελεστές της είναι ανάλογοι δηλαδή Α/1=Β/1=C/1=D/1=E/1 με τον συμβολισμό του Kenny γιατί την ίδια λύση έκανα και εγώ


Find, with proof, the maximum number of flops needed to row reduce an n x (n+1) matrix.
A flop is defined as carrying out a standard binary operation between two real numbers (+,-,x,/), I believe

Let's go for row echelon form, not reduced row echelon form. Don't turn the pivots into 1's, and don't do any cleanup above the pivots. It should also be clear that the maximum number of flops will occur in a matrix of maximal rank; otherwise we get extra zeros that only serve to make the process shorter. Hence I will assume maximal rank.

Look at what it takes to place a pivot on the $i$th row. By our assumption of maximal rank, that will be at the $(i,i)$ position. If we get $a_{ii}=0,$ then we must do a row interchange; however, a row interchange involves some relabeling or data transfer, but no flops.

For each of the $n-i$ rows below it, we must first determine a ratio, $\frac{a_{ji}}{a_{ii}}$ That's $n-i$ flops there.

Then for $i+1\le k\le n+1$ we must perform the operation $a_{jk}-\left(\frac{a_{ji}}{a_{ii}}\right)\cdot a_{ik}.$ That operation requires two flops (one multiplication, one subtraction). That gives $2(n-i+1)$ flops to fix the row, repeated over $n-i$ rows.

So the number of flops to place a pivot at $a_{ii}$ is $(n-i)+2(n-i)(n-i+1) = (n-i)(2n-2i+3)$

The total flops needed to reach row echelon form is than

$\sum_{i=1}^{n-1}(n-i)(2n-2i+3)$

Make the substitution $m=n-i$ to get

$\sum_{m=1}^{n-1}m(2m+3)=2\sum_{m=1}^{n-1}m^{2}+3 \sum_{m=1}^{n-1}m$

$=\frac{2(n-1)n(2n-1)}6+\frac{3(n-1)n}2= \frac{(n-1)n(4n+7)}6$

For large $n,$ that is asymptotic to $\frac23n^{3}.$

If we were doing this row reduction of an augmented matrix for the purpose of solving $n$ equations in $n$ unknowns, then we must also perform a chain of back-substitution steps on the matrix in row-echelon form. That back-substitution costs us precisely $n^{2}$ additional flops.

This is not precise, but we have counted nearly the same numbers of multiplications/divisions as we have additions/subtractions.

A careful numerical scheme would add in an extensive pivoting strategy. This might entail a substantial number of comparisons and transpositions, but no additional flops.

thanks. i had applied similar ideas, but missed a few of the subtleties.


show that for any real interger number p and q, 2^2p + 2^2q will not get a perfect square number.
please help me show, thanks a lot 1st

thanks a lot 1st,
i had some parts not understand,
why suddenly [code]2^{b}-2^{a}=2[/code]?

and how to type html code to write maths code? "=.=

[quote="yrhc@eht"]thanks a lot 1st,
i had some parts not understand,
why suddenly [code]2^{b}-2^{a}=2[/code]?[/quote]
2^a+1=m=2^b-1
then 2^b-2^a=2
[quote="yrhc@eht"]and how to type html code to write maths code? "=.=[/quote]
[code]between two "$".[/code]

ok, thanks again, i will think about the answer, thanks for teaching me how to write code.


geometry
3D geometry
algorithm
search
calculus
videos

I've worked on Rubix cubes for over 50 hours in my life. Never solved one of them...

there's an algorithm where u can solve it in less than 29 moves but ppl at competitions don't have enough time to actually do it
if u can't find a solution...just search the internet...the easiest way is solving by layers

Would you mind pointing out a (or more than one) site?

I've looked in the past and it was mostly general strategies for solving (like doing a 2x2x2 cube first, etc) and didn't realli find any "algorithm" or anything like that.

I'd just like to see how anyone could actually give steps for doing it in less than 30 moves...

I learned the algorithom at a summer camp because my RA was good at it but than forgot.

if you want to become world champion and learn all the algorithms...
[url]http://www.ws.binghamton.edu/fridrich/system.html[/url]

i can't even come close to 20 seconds using my system but mine is really easy to memorize (can finish in about 1 minute and a half)

it's hard to explain online i guess but here's the general idea:
1.)get a white cross on top (with the edge pieces matching in color with the corresponding center faces)
2.)put in the white corners in their right places (and you're done with the first layer)
3.)now put in the second layer corners in the right places (done with second layer)
4.)get a yellow cross on the bottom (now they don't have to match up with corresponding center faces)
5.)switch corners so they are in the right places(orientation doesn't matter)
6.)now orient the corners
7.)and finally switch the edge pieces that weren't matched up in step #4

i'll try to find a site with all of this (i learned it by putting together moves from different sites with my friends)
any ??s go ahead and ask

[quote="SirErnest"]My calculus teacher told us that when see was in California, she saw college students lok at the cube, put it behind their back, then solve in a minute or less.[/quote]

Yea...they can be pretty amazing. I was going over a speech with this kid once...and he would do the cube in like 20 seconds, make sure it was done, randomly dissasemble it, and then do it again in like the same amount of time...over and over and over again...and all this while looking at the piece of paper I was writing on and saying the speech for me while I made corrections and made him change things...only looked at the cube when he had solved it...

Which cube? The four by four one or the three by three one? One of my counselors at CTY did the 4 by 4 one in 5 minutes flat! That was just amazing to see. Oh yeah, why did you name yourself joml89??? He did it cuz of his initials, but I don't see why you're doing it...

Don't worry about him.  He's been sour ever since he stopped beating me last year. :P  For some reason he likes being a jerk to everyone.

How can you say you don't have the time?  All you do is study for school.  I'm sure you could find time.

get the cube with triangular sides (top corners are diamonds and top centers are triangles, now those are tough

When I was in high school (and that was a while back; now I have an 11yo son) I could solve the 3x3 cube in under a minute.  (But, not behind my back!  ;) )  I could also solve the 4x4 but not quickly.  I used to grease my cube with vaseline so I could solve it more quickly.

Took me like 30 minutes to find the guys profile so I'm gonna post it [url=http://cubefreak.hp.infoseek.co.jp/]here[/url].

[quote="yif man12"][quote="joml89"]i can solve a cube in under 3 minutes.....not that that is impressive, but it is cool to work out the cube in front of ppl who can't do it. They think you are super smart. I personally don't have the time or interest to speed cube.[/quote]

You're Genius1601, aren't you?[/quote]
yeah, but for some reason that account doen't work anymore, so i decided to make fun of joe's account.....he always makes fun of mine so what's the difference. (and no i don't always study joe...like you)
as for the cube, i can only do a 3x3. this one kid i know has a 5x5 and can solve it...he can also solve the 3x3 in under 45 seconds. maybe i'll ask for another cube for christmas :D

[quote="Kent Merryfield"]Check the profile of the user named "litoxe2718281828" and go to his linked website.[/quote]

Yeah, the video of the record solve is just plain scary.

yea these ppl have nothin else better to do than to play with a piece of plastic block....which is ok i guess as long as there are a few girls around to impress  ;)

My uncle taught me solving by layers, but he'd forgotten how to do the third layer and I haven't made much progress in figuring it out myself :( How very un-impressive...

whoa...umm u got me  :?
always a way to tie math into everthing  ;)

[quote="Kent Merryfield"]How about a mathematical question? Consider the set of all motions of the cube (say, with the centers of each face fixed) to be a group. The group multiplication is simply to do one action and then the next. There's a one-to-one correspondence between the members of this group and the possible states of the cube, with the identity element of the group identified with the "solved" configuration of the cube.

In a group, the [i]order[/i] of an element $a$ is the least postive integer $k$ such that $a^k=e,$ where $e$ is the identity element of the group.

Let $F$ be the action of turning the face closest to you by 90 clockwise. Let $R$ be the action of turning the face to your right 90 clockwise. Clearly $F$ and $R$ each have order $4.$

What is the order of $FR$?[/quote]

Oh come on, thats mean :D
Is it [hide]105[/hide]?

@@
how did you calculate that, or did you actually did an "experiment"?

how about this one
U' is top face 90 degrees counter-clockwise
R is right face 90 degrees clockwise
R' is right face 90degrees counter-clockwise
R^2 right face 180 degrees

1. what is the order of R U' R' U'
2. what is the order of R^2 U'

[edit]
by the way, i counted 310 moves, or 155 sets of FR

I can solve Rubik's cubes of any size.  3x3 in just under 1 minute, never done a 4x4, and 5x5 in about 8 minutes.  I also do not like speedcubing as anybody can memorize a whole bunch of algorithms given much time.  Finding algorithms and pretty patterns is much more fun.  I would like to be able to juggle rubiks cubes while solving them, but not even the best jugglers can do that yet.

[edit]Wow!  I went to Shotaro "Macky" Makisumi's website and he is just like me (except better at everything).  My juggling is comparable to his, I like chess, rubiks cubes, philosophy, etc.  The major difference is I think he is surrounded by people that are already skilled, whereas I had to do everything by myself.

[quote="limpet"][quote="isaacchao"]@@
how did you calculate that, or did you actually did an "experiment"?
[/quote]

Consider the 'crosses' in the front and right face first, that's the face minus the corners. They repeat after 7 FR's. Then follow the path of the top left cubie of the front face (in your head!). That takes 15 FR's. Convince yourself that the whole configuration should repeat after 15x7=105 FR's.

There should be an easier way but...[/quote]

Something like the following is pretty systematic:
[url]http://games.groups.yahoo.com/group/speedsolvingrubikscube/message/13801[/url]
That DUMB riddle is pretty amusing.  :)

check this out [url=http://www.speedcubing.com/chris/]www.speedcubing.com/chris/[/url]

pretty creapy...

[quote=JasonWu]what's ur fastest time??

if u don't have one...go GET ONE!!!!  and bring it to school and start CUBING!!!![/quote]

only 1 minute :(


inequalities
inequalities solved
3-variable inequality
Symmetric inequality
AM-GM
square root inequality

For $a,b,c$ positive reals, prove that
\[\frac1{\sqrt{1+ \frac{(a+b+c)c}{ab}}} + \frac1{\sqrt{1 + \frac{(a+b+c)a}{bc}}} +\frac1{\sqrt{1 +\frac{(a+b+c)b}{ac}}}   \leq \frac32\]

It reduces to
\[\sum_{cyc} \sqrt{\frac{ab}{(a+c)(b+c)}} \le \frac32\]
which can be obtained by summing the inequalities $\sqrt{\frac{ab}{(a+c)(b+c)}} \le \frac12\left(\frac{a}{a+c}+\frac{b}{c+b}\right)$, $\sqrt{\frac{bc}{(a+b)(a+c)}} \le \frac12\left(\frac{b}{a+b}+\frac{c}{a+c}\right)$, etc.


a,b,c non negative reals

a+b+c=2

Prove that

a^2 * b^2 * c^2 <= 2

Hi;

Use the AMGM

$ \frac {a \plus{} b \plus{} c}{3} > \equal{} \sqrt [3]{abc}$

$ \frac {2}{3} > \equal{} \sqrt [3]{abc}$

$ \frac {8}{27} > \equal{} abc$

square both sides.

$ \frac {64}{729} > \equal{} a^2b^2c^2$

Since a^2 b^2 c^2 =< 64 / 729 it is also less than 2, so we are done.


Problems can be found here: [url]http://www.cs.umd.edu/Outreach/hsContest09/questions/all.ps[/url].

Problem 6: We figured from the constraints that a solution that was asymptotically equal to or faster than $ O(n)$ was desired, especially since $ n$ could reach $ 2 \times 10^9$... but a solution that was $ O(n^{1.5})$ successfully passed. Are there solutions that are linear? Sublinear?

Problem 7: We could not think up how to obtain an optimal solution. Any help?

An obvious optimization to the problem 6 solution is to pre-compute small primes and only divide by those instead of all the numbers up to $ \sqrt N$. But I probably didn't have to tell you this... :)

As for problem 7, first construct the command tree such that commanders are the inner nodes and teams are the leaves of the tree. Then calculate V-scores for every commander (inner node). Number the leaves of the tree from left to right with numbers $ 1,2,\dots N$. You can use dynamic programming to calculate the optimal solution for first $ k$ teams:

For $ k\equal{}1$, there's only one possibility.
For $ k>1$, check all the commanders of the last ($ k$-th) team that don't command any teams with number greater than $ k$. Exactly one of them must be in the optimal command team for first $ k$ teams, checking one possibility takes $ O(1)$ time so the total complexity is $ O(N)$.

Why I can't open the link? Is there any other way to get there?

Change ".ps" to ".pdf" or ".html": http://www.cs.umd.edu/Outreach/hsContest09/questions/all.html


I have no idea how to start this problem. My thoughts sort of wrap around themselves, and leave me with nothing solid to base the rest of my ideas on. If someone could help me figure out what's true and what's not, so that I could begin the problem, I'd be really grateful.

A freight car of mass M contains a mass of sand m. At t=0 a constant horizontal force F is applied in the direction of rolling and at the same time a port in the bottom of the cart is opened to let the sand flow out at a constant rate $ \frac{dm}{dt}$. Find the speed of hte freight car when a ll the sand is gone. Assume the freight car is at rest at t=0.

Would there be a change in momentum? After all, we are only concerning ourselves with the horizontal direction in this system, and the sand particles do not change their velocity as they fall, but retain their initial velocity, correct? If there is no change in momentum, then I don't know how to relate the variables, with the exception of a simple F=ma thing:

$ F\equal{}(M\plus{}m\minus{}\frac{dm}{dt}t)a$
at time $ t$ and another one at time $ t\plus{}\Delta t$
$ F\equal{}(M\plus{}m\minus{}\frac{dm}{dt}(t\plus{}\Delta t))a_2$

Since $ F$ is constant, I set the two equal to each other and end up with
$ 0\equal{}M(a_2\minus{}a)\plus{}m(a_2\minus{}a)\minus{}\Delta ma_2$ which I rewrote as
$ 0\equal{}\Delta a(M\plus{}m)\minus{}\Delta ma_2$

I don't know how to use this, or if it's even true....help please! Just want a push in the right direction, not complete answer.

Use \[ \vec{F} \equal{} \frac{d \vec{p}}{dt} \equal{} \vec{v}(t) \frac{dm}{dt} \plus{} m(t) \frac{d \vec{v}}{dt}\] You should be able to get a separable differential equation. Notice \[ m(t) \equal{} m \minus{} \frac{dm}{dt} t\]

[quote="beezees222"] $ m(t) \equal{} m \minus{} {{dm}\over{dt}}t$[/quote]
Beware of the minus sign, otherwise you'll get quite funny a result.

Also be careful with m constant and m(t) variable ...


calculus
integration
geometry
number theory
prime factorization

How many ten digit numbers can be constructed using the numbers 1,2, and 3 so that any two consecutive digits differ by exactly 1?

Consider a pair of positive integers with sum no larger than 103 and quotient smaller than $ \frac {1}{3}$. What is the largest possible quotient of any such pair?

A large square has been divided into 2009 smaller squares whose side lengths are integers. What is the smallest possible side length of the large square?

In quadrilateral $ PQRS$, $ \overline{PQ} \equal{} 2006$, $ \overline{QR} \equal{} 2008$, $ \overline{RS} \equal{} 2007$, and $ \overline{SP} \equal{} 2009$. Which vertices correspond to interior angles whose measure must be less than $ 180^{\circ}$?

In the equation $ \frac {E\cdot I \cdot G \cdot H \cdot T}{F \cdot O \cdot U \cdot R} \equal{} T \cdot W \cdot O$

the dot represents multiplication and each letter corresponds to a different digit with the same letters having the same value. How many different values can the product $ T \cdot H \cdot R \cdot E \cdot E$ have?

In $ \triangle ABC$ the measure of $ \angle B$  is $ 20^{\circ}$ and the measure of $ \angle C$ is $ 40^{\circ}$. Moreover, the length of the bisector of $ \angle A$ is equal to 2. What is the value of $ \overline{BC} \minus{} \overline{AB}$?

EDIT: Sorry, you are correct.  :blush:  Our answer is
[hide] $ \frac {25} {76}$ [/hide]
If I didnt make another mistake.

um dynamo, if thats the case then how would you say that 25/77 doesn't work?

3.
[hide]We must find the smallest possible square that has $ 2009$ as a factor. Since the prime factorization of $ 2009$ is $ 7^2\cdot41$, the least sqaure with $ 2009$ as a factor is $ 7^2\cdot41^2$ $ \equal{}$ $ 82369$. The side length of this square is $ \sqrt{82369} \equal{} \boxed {287}.$ [/hide]

strangely, 287 wasn't on the answer choices.

Its either

a) 44
b) 45
c) 46
d) 503
e) It is impossible to divide a square into 2009 such squares

Is number $ 3$ $ \textbf{B) \ 45}$?  The square root of $ 2009$ is more than $ 44$, so even if all the squares were $ 1 \times 1$ (which is the smallest), you still couldn't make a square with side length $ 44$.  It has to be $ 1$ bigger, or $ \boxed{45}$

EDIT: Phail....2009 [b]squares[/b], not rectangles.  2008 $ 1 \times 1$ and $ 1$ $ 4\times4$ could work though, right?

EDIT: Phail again.

Uhhh... 007math, I'm not sure what you said about the 2008 and the 4X4. Keep in mind that just because it seems like it works, it may not because the positioning of the squares has to make the outside a square also.

Well, if you divide the big square into 2025 squares (such that each edge of the big square is 45 units), and then combine 9 neighboring squares into one 3 by 3 square, then do that again for another set of squares, then you will have 2009 squares.

Now, 44 doesn't work, because, if we divide our big square into 2009 squares with integral side lengths, then the area of the big square must be at least 2009. However, $ 44^2$ is 1936.

We have proven that 45 works, and that 44 doesn't work, so 45 must be the minimum.

EDIT: Fixed typo. :blush:

[hide="1."]
1. So there are two ways to do this.

You can start on 1 or 3.
You can start on 2.

Lets start with the former case.

If you start on 1 (3 is the same)
You only have one number next as a choice. Then you reach two, so you have 2 choices then. Then you reach 1 or 3, so you have 1 choice again....

so you get 1*2*1*2*1*2*1*2*1=16 ways. Since it also works for 3, you do 16*2=32.

Now starting on two, it's the same thing, but backwards.
You get 2*1*2*1*2*1*2*1*2=32 ways.

32+32=64.[/hide]

[hide="2."] so x+y<=103 and y/x< 1/3. You cannot get 1/3, and the biggest number smaller than it has an infinitely long number of digits. So take the two highest numbers possible. so 4x/3<=103 so x is basically 77. The number closest to lower than 1/3 of it is 25. So you get 25/77. Now try to lowen the denominator as much as possible before it hits 1/3. You get 25/76.[/hide]

[hide="4."] Aren't all interior angles less than 180?[/hide]

[quote="Ihatepie"]Aren't all interior angles less than 180?[/quote]
No. See
 [geogebra]bf081a174b5a5163914315fc13f327f30abbe859[/geogebra]
[hide="hint"]Moreover, note that $ AB\plus{}AD$ must be greater than $ CB\plus{}CD$, and that there can be at most one angle in a quadrilateral that is greater than 180 degrees. (prove it.)
[/hide] :)

Yes. As I said in my hint, now we have to find the vertices for which the sum of the lengths of the sides opposite to it exceed the sum of the lengths of the sides adjacent to it. :)


I'm in charge of starting my school's math club and I need some ideas of what we could cover. I have to present these ideas to my math teacher for him to approve them. Most of the members are going to be at a precalulus level or under (the one's that are in classes under precalculus are advanced and able to learn more difficult topics). I was thinking that Pascal's Triangle might be interesting for them and not too difficult. I was also thinking of sequences and having the members find the recursive rule. If you have other ideas along this line or a different one, please let me know. Any of them will be appreciated. Thanks,
Hikari

I was in charge of a small group like this at our school. I would make handouts for them with a few challenging practice problems, and then I would give them topic tests from the MAO site, which were very beneficial. Also, it's a good idea to get them to derive some of the formulas/properties that you are teaching them (example: the log properties..). As for topics, I think AoPS books have lots of great ideas :).

Here's the site:
http://mualphatheta.org/National_Convention/MAO_Past_Tests.htm


How many $ 7$ digit palindromes (numbers that read the same backward as forward) can be formed using the digits $ 2$, $ 2$, $ 3$, $ 3$, $ 5$, $ 5$, $ 5$?

$ \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 48$

since the number is 7 digits, a "5" must be in the middle (4th spot) because it is the only one that appears 3 times. there are then 6 blanks, and 6 letters. there are 3! = 3*2*1 = 6 ways to arrange the rest. (this is because if you choose one spot, then the other one of the same value must go in a corresponding spot)

An odd-digit palindrome must be "mirrored" over one number, which will be placed in the middle.
The only choice is 5 to put in the middle, otherwise we would not have enough digits to make a palindrome.
So we have the elements 532235 left.  We can now establish a  1:1 ratio that every combination on the left side has 1 complement combination on the right.

The new question is, how many ways can we order 5 3 and 2 around? The answer is $ 3*2*1$, 6.


inequalities AMC USA(J)MO USAMO function calculus derivative

\[ \begin{array}{l}
 {\rm{Prove this f}}{\bf{a}}{\rm{ntstic inequ}}{\bf{a}}{\rm{lity: }} \\
 {\rm{(}}a \plus{} b \plus{} c \plus{} d \plus{} e)\left( {\frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c} \plus{} \frac{1}{d} \plus{} \frac{1}{e}} \right) \le 25 \plus{} 6\left( {\sqrt {\frac{p}{q}}  \minus{} \sqrt {\frac{q}{p}} } \right)^2  \\
 {\rm{for }}0 < p \le a,b,c,d,e \le q \\
 \end{array}
\]

all the same can u post the solution?plz?

[hide="Hint"]A convex function assumes its maximal values at the borders of an interval.[/hide]

[quote="rt_08"]i think its a wrong hint!!!!!!![/quote]

can u give a correct hint or the soln itself. i still couldn't do it.
 :D

PS:i really stink at inequalities and especially when they r from USAMO or IMO :rotfl:

[hide="Solution"] bertram's hint is absolutely correct.  The LHS is convex in every variable, hence assumes its maximum at one of the vertices of the region $ p \le a, b, c, d, e \le q$.  If two of $ a, b, c, d, e$ are equal to $ p$ and three are equal to $ q$ we have

$ (2p \plus{} 3q) \left( \frac {2}{p} \plus{} \frac {3}{q} \right) \equal{} 4 \plus{} 9 \plus{} 6 \frac {q}{p} \plus{} 6 \frac {p}{q}$

and we can verify both that this is equivalent to the original expression and that the values of the LHS at the other vertices are less than this one (or equal to when three of $ a, b, c, d, e$ are equal to $ p$ and two are equal to $ q$).

As a side note, this problem appears in Engel. [/hide]

[quote="t0rajir0u"]The LHS is convex in every variable, hence assumes its maximum at one of the vertices of the region $ p \le a, b, c, d, e \le q$.  [/quote]
How do we prove this?

Taking a second partial derivative and showing it is always nonnegative looks like the cleanest way. Using derivatives for determining convexity is one of the few legitimate uses of calculus on olympiads. The other main way, using the definition $ \lambda f(x) \plus{} (1 \minus{} \lambda) f(y) \geq f(\lambda x \plus{} (1 \minus{} \lambda) y)$ or even the equivalent $ \frac{f(x) \plus{} f(y)}{2} \geq f \left (\frac{x\plus{}y}{2} \right)$ is almost always forbiddingly difficult to verify.

[hide="Off topic exercise"]Show that $ \frac{f(x) \plus{} f(y)}{2} \geq f \left (\frac{x\plus{}y}{2} \right)$ is equivalent to $ f^{\prime \prime} \geq 0$.[/hide]


How do you make a new command not like the one depicted in the packages section for the introduction but the one like $\sqrt{x}$ is \sqrt{x}, where you enter anything for x and it comes up with a square root?  I don't want some kind of command like well, \rttw=$\sqrt{2}$, just one where the {} is there.
The intro to $\LaTeX$ only talks about the \rttw thingy.

[quote="zabelman"]Do you mean making commands with arguments? Here's an example:

\newcommand{\cuberoot}[1]{\sqrt[3]{#1}}

this would define a function with one input (signified by the [1] in the declaration), and calling the command like this:

\cuberoot{17}

would return this:

$\sqrt[3]{17}$.[/quote]
Does it work on the forum? I tried many times but I got only
[quote]
[Unparseable or potentially dangerous latex formula. Error 2 ]
[/quote]

It does not work on the forum, for various technical reasons that I'm not going to explain.


Here is a problem about equilateral triangles:

(1) ABC is an equilateral triangle.
(2) P, Q and R are three points in  the interior of ABC such that  the segments AR, BP and CQ have the same length.
(3) Q, A and R are collinear, with Q between A and R.
(4) R, B and P are collinear, with R between B and P.
(5) P, C and Q are collinear, with P between C and Q.

Show that PQR is also an equilateral triangle.

A figure is attached below!

If you have a Java enabled broswer, you may play with this problem at

http://www.mat.puc-rio.br/~hjbortol/geometry/equilateral.html

Thanks in advance, Humberto.

Let $a = AB = BC = CA$ be the side of the equilateral triangle $\triangle ABC$. Suppose you have the segments $AR, BP, CQ$ of the equal length $d$ aligned with the triangle sides $AB, BC, CA$, i.e., the points $R, P, Q$ are on these sides. Now you rotate 2 of these segments by the same angle $0 < \alpha < \frac \pi 3$, the segment $AR$ around the vertex $A$ and the segment $CQ$ around the vertex $C$, until the rotated point $Q$ falls on the rotated segment $AR$. Since the rotated segments $AR, CQ$ form the same angle with the lines $AB, CA$, which themselves form the angle $\frac \pi 3$, the rotated segments also form the angle $\frac \pi 3$ with each other. The complementary angle $\angle CQA = \frac{2\pi}{3}$ spans the arc $CA$ of a circle $(D)$ centered at the intersection $D$ of the normals to the sides $AB, CA$ erected at the points $A, C$, respectively. But now, the triangles $\triangle CQA \cong \triangle ABR$ are congruent by SAS ($AB = CA = a,\ AR = CQ = d$, $\angle RAB = \angle QCA$). This implies that the segments $AQ = BR$ are congruent and the angles $\angle CQA = \angle APB = \frac{2\pi}{3}$ are also congruent. Let $S$ be the intersection of the line $BR, CQ$. The angle $\angle QRS$ complementary to the angle $\angle ARB$ equals to $\angle QRS = \pi - \angle ARB = \frac \pi 3$, which means that the triangle $\triangle SQR$ is equilateral (it has at least 2 interior angles equal to $\frac \pi 3$). Consequently, $RS = QR$ and $BS = BR + RS = AQ + QR = d$. So, if you now rotate the last segment $BP$ by the angle $\alpha$ around the vertex $B$, the rotated point $P$ will coincide with the point $S$, the point triples $(A, Q, R),\ (B, R, P),\ (C, P, Q)$ will be all collinear, the segments $AR = BP = CQ = d$ equal and the triangle $\triangle PQR$ equilateral.

Assume now that you rotate the segments $AR, CQ$ from their original position by different angles $0 < \alpha < \gamma < \frac \pi 3$ around the vertices $A, C$, until the rotated point $Q$ falls on the rotated segment $AR$, and then you rotate the segment $BP$ around the vertex $B$, until the rotated point $P$ falls on the rotated segment $CQ$, regardless of the position of the rotated point $R$. The ray $CQ$ intersects the circle $(B)$ with radius $d$ closer to the vertex $C$ compared to the case, when the segments $AR, CQ$ were rotated by the same angle. Since in the triangle $\triangle BPC$, the sides $BC = a,\ BP = d$ remain the same and the side $CP$ decreases, the angle $\angle PBC$ also decreases. Let $S$ be the intersection of the ray $BR$ with the ray $CQ$. In the triangle $\triangle ARB$, the sides $AB = a,\ AR = d$ remain the same, but the angle $\angle PAB = \alpha$ decreases, which means that the angle $\angle QBA$ also decreases. Hence, the angle $QBC \equiv \angle SBC$ increases (because the sum $\angle QBA + \angle QBC = \frac \pi 3$ is constant). Consequently, $\angle SBC - \angle PBC > 0$, which means that the points $B, Q, P$ cannot be collinear. The only possibility for these points to be collinear is the condition $\alpha = \angle RAB = \angle QCB = \gamma$.

Another (computational   :blush: ) solution:

WLOG, assume that $AB=1$.  Let $D = BC \cap AQ,{\rm{ }}E = AC \cap BR,{\rm{ }}F = AB \cap CF,AF = x,{\rm{ }}BD = y,CE = z$ with $x,y,z \in \left] {0,1} \right[ $.
By [b]Stewart theorem [/b]we have\[AD^2  = y^2  - y + 1 \ \ \ \ \ \ \ \ \ \ [1]  \]
By [b]Menelaus theorem[/b] relative to $\triangle ABC$ we have
\[\frac{{1 - z}}{z} \cdot \frac{1}{y} \cdot \frac{{AD - AQ}}{{AD}} = 1{\rm{           }}\ \ \ \ \ \ \ \ \ [2]\]
From [1], [2] follows
\[AQ^2  = \frac{{\left( {y^2  - y + 1} \right)\left( {1 - z} \right)^2 }}{{\left( {1 - z + yz} \right)^2 }}\]
and similarly we can find $BR^2,CF^2$. From $AQ=BR=CF$ follows
\[
\frac{{\left( {y^2  - y + 1} \right)\left( {1 - z} \right)^2 }}{{\left( {1 - z + yz} \right)^2 }} = \frac{{\left( {z^2  - z + 1} \right)\left( {1 - x} \right)^2 }}{{\left( {1 - x + xz} \right)^2 }} = \frac{{\left( {x^2  - x + 1} \right)\left( {1 - y} \right)^2 }}{{\left( {1 - y + xy} \right)^2 }}\ \ \ \  \Rightarrow
\]
\[
\frac{{\left( {y + 1} \right)^2 \left( {1 - z} \right)^2 }}{{\left( {1 - z + yz} \right)^2 }} = \frac{{\left( {z + 1} \right)^2 \left( {1 - x} \right)^2 }}{{\left( {1 - x + xz} \right)^2 }} = \frac{{\left( {x + 1} \right)^2 \left( {1 - y} \right)^2 }}{{\left( {1 - y + xy} \right)^2 }} \ \ \ \ \   \Rightarrow
\]
\[
\frac{{\left( {y + 1} \right)\left( {1 - z} \right)}}{{\left( {1 - z + yz} \right)}} = \frac{{\left( {z + 1} \right)\left( {1 - x} \right)}}{{\left( {1 - x + xz} \right)}} = \frac{{\left( {x + 1} \right)\left( {1 - y} \right)}}{{\left( {1 - y + xy} \right)}} \ \ \ \ \ [3]
\]
Solving [3] with a CAS we obtain  $x=y=z  \ \ \  \Rightarrow  \angle BAQ=\angle CBR=\angle ACP \Rightarrow PQR$ is equilateral (by the SAS postulate). $\blacksquare$


im checking the telescopical way of getting the algebraic expresion for this summation.1^n +2^n +3^n + . . . + k^n =??   i get to explaining  to n=3 but i dont know how to express a generalisation. I could use as much help posible. Thanks

Ben

Not only have you already posted this question, but you repost after there have been several responses in the first post.  Bad nettiquite....
Double posting and crying urgency is a recipe for a thread that goes ignored...

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=154838[/url]
Accept the answers you've been given and expect nothing else.
EDIT:  Make that TRIPLE POST.... that's enough ignore his questions.. :furious:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=154862[/url]
[b]EDIT: Make that a QUADRUPLE POST!!! [/b]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=870784#870784[/url]

Can a mod please delete his triple ( now quadruple) posts.

As far as I'm concerned I'm ignoring this and any future questions by him...


inequalities
geometry
circumcircle
inradius
perimeter
rearrangement inequality
inequalities solved

Prove that if x+y+z=1 and x,y,z>0 then:
   sum sqrt(z+xy)>=sqrt(1+9(xy+yz+zx)).

are u shure that this is the correct pb??
that is equivalent with z+xy>=1+9(xy+yz+zx)=x+y+z+9(xy+yz+zx)
=>0>=x+y+8xy+9(yz+zx)

harazi, do you have the solution for this?

first, note that x+yz=x(x+y+z)+yz=(x+y)(x+z).
let a= y+z,b=z+x,c=x+y, we have LHS= \sum  \sqrt  ab.

Finally, I have a solution.

Take a' = y + z, b' = z + x, c' = x + y as outlined by wpolly above.

Then we must prove  \sum  \sqrt (a'b') \geq  \sqrt (9 (\sum (a'b')) - 8), which simplifies to (after squaring and substituting (a' + b' + c')^2 = 4)

\sum a'^2 + \sum (a'\sqrt(b'c')) \geq 2 \sum a'b'.

Take a^2 = a', b^2 = b', c^2 = c'.

Then we must prove

\sum a^4 + \sum a^2 bc \geq 2 \sum a^2 b^2.

WLOG, a \geq b \geq c.

Case I: a \leq b + c.

It is easy to see that a^2 + bc \geq b^2 + ca \geq c^2 + ab.
Since a^2 \geq b^2 \geq c^2, we apply rearrangement inequality to obtain

\sum a^4 + \sum a^2 bc \geq \sum a^2 b^2 + \sum a^3 b. (*)

But by Muirhead's theorem, \sum a^3 b \geq \sum a^2 b^2, so the RHS of (*) \geq 2 \sum a^2 b^2, after which the conclusion follows.

Case II: a > b + c.

Then a^2 + bc \geq c^2 + ab \geq b^2 + ca.

Since a^2 \geq b^2 \geq c^2, we apply rearrangement inequality (choose terms differently this time) to obtain

\sum a^4 + \sum a^2 bc \geq \sum a^2 b^2 + \sum a^3 b.

From here, we continue as we did above.

Here is a simpler solution.
To prove  \sum a 4 + \sum a 2 bc \geq 2 \sum a 2 b 2 , We'll only need to prove
\sum a 2 bc  \geq 2 \sum a 2 b 2 - \sum a 4 =(a+b+c)(a-b+c)(a+b-c)(-a+b+c).

i) When a,b,c do not form a triangle, we have LHS>0 \geq RHS.
ii) When a,b,c forms a triangle, denote the triangle's area, circumradius, inradius and semi-perimeter by S, R, r, p respectively.
Then we have LHS=abc \sum a=4SR*2p=8pSR, RHS=16S 2 (by Heron's formula.)
By Euler's inequality we have LHS=8pSR \geq 16pSr=16Spr=16S 2 =RHS.
Q. E. D.


[size=150]You have received a $2,000 invoice from the Acme Swamp Water Company for raw materials to use in your beer manufacturing plant. The terms are 3/10, net 40. If yor fail to take advantage of this discount. You are paying what interest rate?

1.22.7%
2.36,735%
3.37.11%
4.55.67%
5.None of the above[/size]

Thanks

I don't quite understand the terms "3/10, net 40".

Does it mean that the ratio of the money you get to what the company gets is 3:10 plus a net of $40?


Let $a$ and $b$ be positive integers such that
$\frac{43}{197}<\frac{a}{b}<\frac{17}{77}$.
Find the minimum possible value of $b$.
 :wink:

Such problems are VERY easy to find:
http://www.mathlinks.ro/Forum/viewtopic.php?t=49420
http://www.mathlinks.ro/Forum/viewtopic.php?t=101857
keywords: 77, 197, 43

  darij

The problem had been discussed before on another link and the answer was 32. If you want to find it, go to: Contests--India--National Maths Olympiad--2005--(click on #2).

For $s,t\in \mathbb{N},$ we can write $36a=17s+43t,\ 36b=77s+197t.$


vector
ratio
trigonometry
complex numbers

Let: $ z= cis \theta \Rightarrow z^{n}= cis (n \theta)$

$ arg(z^{n}) = n \theta$
$ n.arg(z)= n \theta$.

That conclusion was one of main contributions of Caspar Wessel to the understanding of complex numbers. As you know, complex numbers can be represented by vectors. How can we multiply vectors? Let's generalize what happens with the product of real numbers. Note that, when we multiply two numbers (for example, -2 and 7) the product (-14) has the same ratio to each factor as the other factor as to 1: -14/-2 = 7/1 and -14/7 = -2/1. So we can argue that vector (which are entities that not only have a lenght but also a direction given by an angle) multiplication must also have two similar properties: defining an unit vector and considering two vectores A and B with angles $ \alpha$ and $ \beta$, then 1) the lenght of the vector product should be the product of the lenghts of A and B, and 2) the angle of the vector product should differ from the angle of each vector by the same angle as the angle of each vector differs from the unit vector. If we define the unit vector as one with 0º angle, then the product angle must be $ \alpha+\beta$. Generalizing this for n equal vectors each with an angle a, then their product angle must be n*a.

im pretty sure no complex trig stuff in HSB...


I would have to say that mine was number 1.  That was just fun to play around with.  I also liked number 3, though not as much.

I don't think 1 was hard enough to be my favorite, 3 provided the most challenge of all of them, though none were really hard...(wish I remembered to type solutions up at the end... :blush: )

I liked #4 the most. It is good when you can check an answer on your calculator.

yeah, there weren't any hard problems.... I guess #3 is my favorite since it's a bit harder than other problems.

[quote="mathkid"]Number 1. I loved my proof.[/quote]

Oh come on, you can't say that and not post the proof :-).

I demand the proof!

#2 was my favorite because I didn't notice it simplified to a^3-b^3=(a-b)^3....I ended up coming up with a rather weird, but interesting proof.

http://www.debreczeni.com/usamts161.pdf (Pgs. 2&3)

I liked problem #2 the most.  Unfortunately, that's also the one I made a silly mistake on. :(  Even so, I didn't think any of the problems were particularly difficult.  I'm just careless.

I'd have to say problem 2.  1, 4, and 5 were too easy to be my favorite; on 2, I failed to notice the simplification :? , and I ended up with a massive, probably flawed proof.  However, it was fun wrestling with it.

All of this round's problems were fun, but I really liked number 3, simply because it made me think the most.  I've seen problems similar to the other 4 before.

2 was my favorite, though I was one of the people that didn't get all the solutions (only 8).  It took forever to write up though.

I had seen a problem similar to 1 before (a story problem a friend gave me before), so I knew the general idea behind attacking it.

3 took me forever ....  and I messed it up :blush:. As for why, I overlooked the most obvious thing.

4 was probably the easiest one.

I may've done 5 incorrectly, we'll see in a few days :?

The answers to 5 were 2, sqrt 7, sqrt 13

Three was definitely the best because I could actually come up with a way of solving it by substituting 8 other variables and then simplifying it in a way that made the problem 10 times easier.  the four other problems were just way too easy to be considered as favorites.


Here are a few good websites.Please post more if you know some  :D
P---> Physics  C---> Chemistry and M--->Maths   :P
1. Take a look at Peter Doyle's website:  (P)
    http://math.dartmouth.edu/~doyle.
2. There is a great deal of discussion on the matter in the xkcd fora. You may want to check out this thread
    http://fora.xkcd.com/viewtopic.php?f=7&t=16104&start=0  (P)
3. For Math Olympiad problems look up at:
    http://www.kalva.demon.co.uk/ (M)

here is a gr8 one for physics [url=http://www.physicsforums.com/]here[/url]

http://www.iarcs.org.in

For Loads of Programming related problems for those oriented towards programming.

P.S: Is this thread only for acads related sites.


function
integration
limit
real analysis
real analysis unsolved

We have a,b and f:[a,b)$\to$ R a local integrable function Reimann ,f bounded(marginit) so that

$\int_{a}^{b-0}{f(x})dx$convergent .
Prove that if f*=f on [a,b) and f*(b)=r , r $\in$ R, then f* integrable Reimann on [a,b] and:

$\int_{a}^{b}{f*(x})dx = \int_{a}^{b-0}{f(x})dx$

Please do not post multiple copies of the same problem in different forums. I have deleted the other copies so that this will be the only one.

We must prove that having f local integrable reimann on [a,b) (that means that is continue on [a,b)) f with f*(b)=r ,(whatever r) we have f* integrable reimann on [a,b].
We also know that :

\[\exists \mathop{\lim }_{v \to b}\int_{a}^{v}{f(x)dx = I}\] because it converge

Please give me some help ,because i think that many of you know very well impropre integrale and Riemann integrale


geometry
rectangle
floor function
function
AMC
AIME
3D geometry

The graph of the equation $9x+223y=2007$ is drawn on graph paper with each square representing one unit in each direction. How many of the $1$ by $1$ graph paper squares have interiors lying entirely below the graph and entirely in the first quadrant?

omg, i had fun with this one.
[hide="My Solution (not too clear, but oh well)"]i used stabbing diagonals to find the number of squares the line crossed in the first quadrant. then i found the number of squares in the the rectangle with side lengths of the intercepts and subtracted the number of squares the diagonal crossed and divided the difference by 2 (symmetry).[/hide]

[hide="my fourth miss"]
the answer is
$\sum^{8}_{k=1}((9-k)(\left\lfloor{\frac{223k}{9}}\right\rfloor-\left\lfloor{\frac{223(k-1)}{9}}\right\rfloor))=\boxed{888}$

But I can't add and got 868
[/hide]

[quote="Ignite168"][hide]From Pick's Theorem $\frac{2007}{2}=\frac{233}{2}-\frac{2}{2}+\frac{2I}{2}$. In other words, $2I=1776$ and I is $\boxed{888}$.[/hide][/quote]

Ooh, that's way better than what I had (Used floor function a bunch of times like sapphyre571)

[quote="Infinite_Epsilon"][hide]We see that the line is the diagonal of a 223 by 9 rectangle, and doesn't pass through any lattice points other that the x- and y-intercepts.  Then, there are a total 222 intersection with vertical lines and 8 with horizontal lines.  Then, overall, there are 8+222+1 squares that the line passes through in the first quadrant.

From this, there are 223*9-222-8-1=223-223-9+1 =(223-1)(9-1) =222*8 squares that the line doesn't pass through in the rectangle.  Only half of these are below the line, so the answer is 888 squares.[/hide][/quote]

yes!! that is the way i did it (except i use different terminology). i think that is the easiest way to do it. it's like a first grade problem if you do it that way.

yeah i did what sapphyre did i just plugged in numbers for the y value and found the floor of the x's

ya I did floors

when I was doing smaller numbers I noticed that sometimes - actually most of the time - the number of squares under $ax+by=ab$ was $(a-1)(b-1)$, but I found a counterexample and was like, man, this doesn't work...

Ignite's solution is really 1337

[quote="Infinite_Epsilon"][hide]We see that the line is the diagonal of a 223 by 9 rectangle, and doesn't pass through any lattice points other that the x- and y-intercepts.  Then, there are a total 222 intersection with vertical lines and 8 with horizontal lines.  Then, overall, there are 8+222+1 squares that the line passes through in the first quadrant.

From this, there are 223*9-222-8-1=223-223-9+1 =(223-1)(9-1) =222*8 squares that the line doesn't pass through in the rectangle.  Only half of these are below the line, so the answer is 888 squares.[/hide][/quote]

i like this one.  good job dude.

the squares on line are 223+9-1=231.
so it's (2006-231)/2=888 squares.

i case bashed it (i.e. counted for how many intervals such as 0->1 1->2, etc) had 8 squares, 7, etc....

and then I miscounted the one for 4...

It's really simple if you see that in an $a$ by $b$ grid where $a,b$ are relatively prime, the diagonal from $(0,0)$ to $(a,b)$ is simply $a+b-1.$ So, we have then:

$(223*9-(223+9-1))/2 = 888$

as our answer.

AND OH MY GODS

YOU KNOW WHAT

THIS WAS VERY SIMILAR TO A #14 ON A PAST YEAR'S AIME WHERE THERE WAS A RECTANGULAR PRISM INSTEAD

that's why i got this problem quickly

i hate spamming, but I love your method Karth. This is exactly the method I use. And, if anybody cares about the generalization to non-relatively-prime integers, if we have a $a\times b$ grid, and draw the diagonal of this grid, the number of squares it passes through is $a+b-\text{GCD}(a, b)$.

Now, when I worked this problem on the actual AIME, I got

(2007 - 231)/2 = 1876/2 = 938

rotflmao pwned

I did it by Pick's Theorem like Ignite168.  I thought that IF you saw this method or Karth's stuff with $a+b-1$, then this was easy.  If you didn't then this was actually quite hard for a #5 in my opinion.

But whatever, I thought #15 was ridiculously easy and got it in under 5 minutes with no errors so whatever.

[quote="Treething"]i drew it on graph paper and counted and extrapolated

and hey guys
i got 888[/quote]

is it like even possible to count??? what do you mean "extrapolated"? was there a pattern somewhere?

Hmmm well I was doing this problem for practice and I am quite inexperienced with Pick's Theorem and all this neat higher level stuff. However what I did was that we have a 9 by 223 rectangle. Now if you draw a similar rectangle you get 1 by 24.7777777777......... Now, another similar triangle gets 2 by 49.555555555 so you have 24 whole squares below the line between these two triangles. Now between 2 by 49.5555..... and 3 by 74.3333.... you have 24*2 (since the side length of the triangle is 2) Now keep doing that and you get 8(9)/2*24=868 squares. Anyone know where I messed up since my answer was 24*36 while the real answer is 24*37.

haha i was trying to do something like Karth's method, especially because I had just proved or read a proof of that $a+b-gcd(a,b)$ thing, but couldn't really remember so I did it the bashing way.

And using that method, we could make a formula for solving the general form of this problem.  It could be the solving-problem-5-on-the-2007-alternate-AIME formula...

I brute-forced it easily in ten minutes, getting a list of numbers corresponding to the number of squares that have a top edge at a certain height (8,7,6, etc).  These summed to 888.

Unfortunately, I forgot to add one of the numbers.