Hi guys!

I found an interesting number sequence on an old document.
There was 24 number and the next 6 number.
The writer wrote: the first 24 number determine the next 6 by certain math.
But I can't found any relation between the firs 24 and the 6 numbers.

Here is the example:
first 24 number - 18,24,18,14,23,33,21,31,34,29,7,22,22,36,29,31,4,32,4,36,5,6,23,3
next 6 number   - 1,2,3,4,5,6

I asked a few guy, but nobody have no idea.

Ers

Thanks Afonso! What a formula!  :lol:

but seriously
I saw an another squence where the las six was 4,5,6,28,29,30

[quote="Afonso Bandeira"]$ a_{24+n}=(4-2^{2})\left(\sum_{i=1}^{24+n}a_{n}\right)+n$???[/quote]
 :D yopu say just $ a_{n+24}=n$ :D
this isn't a solution , lol

i think it's clear that he was sarcastic ...

good point, what other answer did it deserve?

can I ask this sequence more forums or other sites? is it worth it?
or is there any data mining or analyzing software for number sequences?

or what is the next step if I can't found the relation?

Thanks guys!
ers

you take $ U_{n}=\sum_{i=1}^{30}a(i)\prod_{j=1,j\neq i}^{30}\frac{x-j}{i-j}$ where:
$ a(1)=18,a(2)=24,a(3)=18,...,a(24)=3,a(25)=1,a(26)=2,a(27)=3,a(28)=4,a(29)=5,a(30)=6$
 :blush:
but if you change $ a(25),...,a(30)$ you find another sequance where the next 6 number are not $ 1,2,3,4,5,6$
so there are unique values of the next 6 number. you must give us another condition for this sequence.

Thanks 	aviateurpilot!

You are work hard! Thanks again!
Here is the another sequence, looks like totally different:

13,30,7,10,27,9,10,6,32,10,31,7,6,13,25,19,18,34,33,34,12,28,16,21
and the last six:
4,5,6,28,29,30

I think all the two 24 number sequence made same way the last six numbers, but I can't see what is joint part.


ratio geometry incenter circumcircle trigonometry

A circle is inscribed in a equilateral triangle, which is inscribed in another circle. What is the ratio the smaller circle's area to the larger circle's area?

$1:4$
In an equilateral triangle, the incentre, circumcentre and the centroid are indeed the same point. The centroid divides the median into a ratio of 2 to 1, which is also the ratio of the circumradius to the inradius.

Hey! I remember this problem!  :o

 :D I posted it a while ago.  :D

[hide]
Without loss of generality, we can assume that the side length of the triangle is $\sqrt{3}$.

Bisect one of the sides of the triangle.
Bisect one of the adjacent angles.
Both of these lines pass throught the "center" of the triangle, which is also the center of the inscribed circle.

We have 2,  30 degree angles. the sine of the one that forms a right triangle,
is $\frac{1}{2}$.
$\sin {30}=\frac{1}{2}$.
This is the radius of the inscribed circle.

The radius of the circumscribed circle is the hypotenuse of that triangle, or $\sqrt{(\frac{1}{2})^2+(\frac{\sqrt{3}}{2})}=1$

The radius of the big circle is one.

The area of the small circle is $(\frac{1}{2})^2*\pi=\frac{\pi}{4}$.

The area of the big circle is $1^2*\pi=\pi$.

The ratio of their areas is $1:4$.
[/hide]


probability
expected value
combinatorics unsolved
combinatorics

$ A_{1}$, $ A_{2}$, $ A_{3}$ B  four guys got 2009 cards.
$ A_{1}$ wrote a number $ a_{1}$， $ A_{2}$ wrote a number $ a_{2}$,$ A_{3}$wrote a number $ a_{3}$such:$ a_{1}$>=$ a_{2}$>=$ a_{3}$ and $ a_{1}$+$ a_{2}$+$ a_{3}$=2009
B divided the cards into 3 parts with the number of  $ b_{1}$,$ b_{2}$,$ b_{3}$ satisfying:$ b_{1}$>=$ b_{2}$>=$ b_{3}$ and B didn't know what other three people wrote.
if and only if $ a_{i}$<$ b_{i}$ can $ A_{i}$ got $ a_{i}$ cards.and B get the remained cards.
ask:how many cards can B get at least?

You probably mean to ask: "what should $ (b_1, b_2, b_3)$ be, so that $ B$ maximizes the expected number of cards he will get, over all possibilities of $ (a_1, a_2, a_3)$" ?
Otherwise, for just one fixed $ (a_1, a_2, a_3)$, no matter what $ B$ does, there is a chance his partition has that number of elements, so he gets zero cards.

yes,i mean that.
but,B CAN'T get nothing in all possible situations.
$ a_{i}$<$ b_{i}$ can't be hold for i=1,2,3
or add up we will get 2009<2009.......

I guess the answer is 673....
but i don't know how to prove it


integration
calculus
calculus computations

Solve $ cx'(t) \equal{} y'(t) \plus{} y(t)$, where c is a constant.

Edited: After the first reply

This problem doesn't make sense as posed.  Do you mean $ c \equal{} \frac{dy}{dx} \plus{} y$?

I meant solve $ cx'(t) \equal{} y'(t) \plus{} y(t)$

I'm still not sure if that's the question you intend.  $ y(t)$ can be chosen arbitrarily and we then can set $ x(t) \equal{} \frac{1}{c} \int_{0}^{t} \left( y'(s) \plus{} y(s) \right) \, ds \plus{} C$ for an arbitrary constant $ C$.


Show that if $P(x)$ is a polynomial with real coefficients and if $z \in \mathbb{C}$ is a root of the polynomial, then so is $\bar{z}$.

[hide="Comment"] There is a very simple way to do this that doesn't require using any properties of a polynomial other than its definition.  :)

[hide="Hint"] What do you know about the $\bar{z}$ function? [/hide] [/hide]

[hide="Attempted answer"]Take $\overline{P(x)}$. Since the coefficients are real, $\overline{P(x)}=P(\bar{x})$. But if $P(z)=0$, then $\overline{P(z)}=P(\bar{z})=0$.[/hide]

[EDIT]Editted answer


A light year is about $5.88 \times 10^{12}$ miles.  The distance from Earth to a star $6 \times 10^6$ light years away is about $3.528 \times 10^n$ miles.  What is the value of $n$?

[hide]$19$ i had alot of work but this norton antivirus thing showed up and had to resart my comuputer.[/hide]


For nonnegative integers $n$ and $k$, define $Q(n,k)$ to be the coefficient of $x^k$ in the expansion of $(1+x+x^2+x^3)^n$.

Prove that  $\displaystyle Q(n,k)=\sum_{j=0}^k{n\choose j}{n\choose {k-2j}}$.

could you check the problem again, it does not seem to work since when $j=k$, then you have $\binom{n}{-k}$

I just checked it to make sure and that is exacly what it says.

[hide="hint"]
$(1+x+x^2+x^3)^n=(1+x)^n(1+x^2)^n$
[/hide]

yea, once you see

[hide]
$(1+x+x^2+x^3)^{n} = (1+x)^n(1+x^2)^n$
[/hide]

i think its pretty easy because it becomes a matter of equating coefficients on both sides? but try like $Q(2,1)$ in which case you have:

$(1+x+x^2+x^3)^{2}$, and the coefficient of $x$ is $2$

our sum however gives:

$\binom{2}{0}\binom{2}{1} + \binom{2}{1}\binom{2}{-1}$ which is undefined

Not undefined: $a \choose b$ is zero when $b < 0$.


vector
linear algebra
linear algebra unsolved

Let $V$ be the vector space over field $\mathbb{F}$ with $\text{dim }_{\mathbb{F}}(V) =n$. And let $\{\mathbf{x_{1}}, \mathbf{x_{2}}, \cdots, \mathbf{x_{n}}\}$ be the subset of $V$. If $X$ generates $V$, prove that $X$ is linearly independent.

Try the contrapositive: if $\{x_{1},\dots,x_{n}\}$ are linearly dependent, the space they generate has dimension less than $n$ (so they can't generate all of $V$).

Why the set with $k \leq n$ elements can't generate the vector space which has dimension $n$. I don't know such the fact, even i think its obvious. How about the mathematical proof??

thanks before.

It's a little sketchy to talk about the dimension of a space before you have that lemma proven.

What definition of dimension are you using?

[quote="Beginner"]I know it is as the cardinality of the basis (base)..  :maybe:[/quote]
And why is that well-defined? That's the key point: since each space has many possible bases, that definition is meaningless unless and until you have proved that any two bases of the same space have the same cardinality.

Yes it has been proved that for any two bases they have same number of elements.

Thx. Now i've the proof.  :)


I wonder how to prove the Brun's theorem, which states that the sum of the reciprocals of the members of each twin prime pair $ B\equiv\;(\frac{1}{3}+\frac{1}{5})+(\frac{1}{5}+\frac{1}{7})+(\frac{1}{11}+\frac{1}{13})+...$converges to a definite number even if the sum contains an infinite number a terms.

Thank you for helping. :P

A good place to look would be Tenenbaum's book,
Introduction to Analytic and Probabilistic Number Theory
(pages 56,57,...)

You can find the book on http://books.google.com


Let a_i (i=1,..,n) be real numbers in the interval [0,2] such that  \sum (i=1,..,n)a_i \geq n. Prove that

   (\sum(i=1,..,n)(a_i)^4) -n \leq 11( \sum (i=1,..,n)(a_i)^2 -n)

I didn't check my computations, so I may be wrong
  Write a_i=1+x_i. The inequality becomes sum (x_i)^4+4 sum (x_i)^3<=5 sum (x_i)^2+18 sum x_i. But (x_i)^4+4 (x_i)^3<=5 (x_i)^2 and sum x_i>=0.

Fantastic, Harazi!!!
Now I will check computations (I am sure they are correct).


alguien sabe sobre los poemas de matematicas
alguna web?  :?
ciau

hay uno muy bueno en E. Gentile, capaz que en google lo encontrás......


geometry
circumcircle
geometric transformation
reflection
projective geometry
power of a point
radical axis

The centre of the circumcircle of quadrilateral $ABCD$ is $O$ and $O$ is not on any of the sides of $ABCD$. $P=AC \cap BD$. The circumecentres of $\triangle{OAB}$, $\triangle{OBC}$, $\triangle{OCD}$ and $\triangle{ODA}$ are $O_1$, $O_2$, $O_3$ and $O_4$ respectively.

Prove that $O_1O_3$, $O_2O_4$ and $OP$ are concurrent.

Invert with respect to $O$ with power $\frac{R^2}2$, where $R$ is the circumradius of $ABCD$. The line $OP$ turns into itself, while the lines $O_1O_3,O_2O_4$ turn into the circles $(OMP),(ONQ)$, where $M,N,P,Q$ are the midpoints of $AB,BC,CD,DA$ respectively. It hence suffices to prove that those two circles meet again on $OP$.

If $T=AB\cap CD,\ S=AD\cap BC$, then $(OMP)$ is the circle with diameter $OT$, so it passes through the foot $X$ of the perpendicular through $O$ to $ST$. In the same way we prove that $(ONQ)$ passes through $X$. Since, as is well-known, the perpendicular from $O$ to $ST$ is precisely $OP,\ X\in OP$ as well, and we're done.

There is another interesting problem which is quite similar to this one: http://www.mathlinks.ro/Forum/viewtopic.php?p=462779

tangents to the circle at $A,B,C,D$ intersect at $O_1',O_2',O_3',O_4'$ and $O,O_i,O_i'$ are collinear.from brianchen theorem $O_3'O_1'$ and $O_2'O_4'$ intersect at $P$ and we are done.

See here for the generalization of the problem:
http://www.mathlinks.ro/viewtopic.php?t=187214

[quote="shobber"]The centre of the circumcircle of quadrilateral $ ABCD$ is $ O$ and $ O$ is not on any of the sides of $ ABCD$. $ P \equal{} AC \cap BD$. The circumecentres of $ \triangle{OAB}$, $ \triangle{OBC}$, $ \triangle{OCD}$ and $ \triangle{ODA}$ are $ O_1$, $ O_2$, $ O_3$ and $ O_4$ respectively.

Prove that $ O_1O_3$, $ O_2O_4$ and $ OP$ are concurrent.[/quote]
Denote by $ R_2$, $ R_4$ the reflection points of $ O$ in $ O_2$, $ O_4$, respectively. We have $ R_2B$, $ R_2C$, $ R_4A$, $ R_4D$ are all tangent to $ (O)$. Call $ Q$ the intersection of $ AD$ and $ BC$. We know that $ P$ lies on the polar line of $ Q$ with respect to $ (O)$, so we conclude that $ R_2$, $ R_4$, $ P$ are collinear. It follows $ O_2O_4$ passes through the middle point of the segment $ OP$. Similarly, $ O_1O_3$ also passes through the mid-point of segment $ OP$. Hence, $ O_1O_3$, $ O_2O_4$, and $ OP$ are concurrent at the mid-point of the segment $ OP$. $ \square$

[quote="shobber"]The centre of the circumcircle of quadrilateral $ ABCD$ is $ O$ and $ O$ is not on any of the sides of $ ABCD$. $ P \equal{} AC \cap BD$. The circumecentres of $ \triangle{OAB}$, $ \triangle{OBC}$, $ \triangle{OCD}$ and $ \triangle{ODA}$ are $ O_1$, $ O_2$, $ O_3$ and $ O_4$ respectively.

Prove that $ O_1O_3$, $ O_2O_4$ and $ OP$ are concurrent.[/quote]
[hide="Solution"]
Let the circumcircles of $ \triangle OAB$ and $ \triangle OCD$ meet at $ X$ and the circumcircles of $ \triangle OBC$ and $ \triangle ODA$ meet at $ Y$.  Also, let $ O_1O_3$ meet $ O_2O_4$ at $ Z$, $ AB$ and $ DC$ meet at $ M$, and $ AD$ and $ BC$ meet at $ N$. Notice that $ O_1O_3$ and $ O_2O_4$ are the respective perpendicular bisectors of $ OX$ and $ OY$.  Thus, $ Z$ is the circumcenter of $ \triangle OXY$.

Now, invert around $ O$ with radius being the circumradius of $ ABCD$.  Then, $ X$ maps to the intersection between $ AB$ and $ DC$, which is $ M$.  Similarly, $ Y$ maps to $ N$.  Furthermore, $ OZ$ maps to itself.  Since $ OX\cdot OM \equal{} OA^2\equal{}OY\cdot ON$, we have that $ XYNM$ is cyclic.  Therefore, the perpindicular to $ MN$ through $ O$ passes through the circumcenter of $ \triangle OXY$, which is $ Z$.  It is well-known that $ OP\perp MN$, so $ OP$ passes through $ Z$ and we conclude that $ O_1O_3$, $ O_2O_4$, and $ OP$ are concurrent. [/hide]

[url=http://img13.imageshack.us/my.php?image=chinatst2006.png][img]http://img13.imageshack.us/img13/8347/chinatst2006.png[/img][/url]

[b]Solution.[/b] Let $ A'\in O_{1}O_{4}\cap OA, \ \ B'\in O_{1}O_{2}\cap OB,\ \ C'\in O_{2}O_{3}\cap OC, \ \ D'\in O_{3}O_{4}\cap OD.$ So these points are the middlepoints of segments $ OA, \ OB, \ OC$ and $ OD$ respectively.

$ \Longrightarrow OA' \equal{} OB' \equal{} OC' \equal{} OD'.\ (*)$

Using the [u]radical axis[/u] argument we get that: $ OA'\perp O_{1}O_{4}, OB'\perp O_{1}O_{2}, OC'\perp O_{2}O_{3}$ and $ OD'\perp O_{3}O_{4}. \ (**)$

From relations $ (*)$ and $ (**)$ we obtain that the quadrilateral $ O_{1}O_{2}O_{3}O_{4}$ is tangent (the center of its incircle is $ O$).

[color=blue] Newton's Theorem[/color] applied in $ O_{1}O_{2}O_{3}O_{4}$ yieds:

$ O_{1}O_{3}, O_{2}O_{4}, A'C', B'D'$ concur.

Now it's easy to see that line $ A'C'$ meets $ [OP]$ in its middlepoint and so does line $ B'D'.$ Thus lines $ O_{1}O_{3} \ , O_{2}O_{4} \ , A'C' \ , B'D'$ and $ OP$ concur in the middlepoint of $ [OP].$

[quote="b555"]where  can i find what`s newton theorem? :?:[/quote]

What he means by Newton's Theorem is Theorem 2 in [url=http://www.cip.ifi.lmu.de/~grinberg/CircumRev.pdf]Circumscribed quadrilaterals revisited[/url] (or you can find it at http://www.mathlinks.ro/Forum/viewtopic.php?t=19549 ). I don't know if it was known to Newton, but usually people mean a [url=http://www.cut-the-knot.org/Curriculum/Geometry/NewtonTheorem.shtml]different fact (also about circumscribed quadrilaterals)[/url] by "Newton's theorem".

  darij

[b]Lemma: [/b] $O_1O_3$ bisects $OP$.
[b]Proof: [/b] Let $AB$ and $CD$ meet at $E$, and $AD$ and $BC$ meet at $F$. Let $FP$ meet $OE$ at $X$. By Brokard's Theorem, $FP$ is the polar of $E$ w.r.t. $(O)$, so $EA \cdot EB = ED \cdot EC = EX \cdot EO$. Therefore, $OXAB$ and $OXDC$ are cyclic, so $X$ is the second intersection of $(O_1)$ and $(O_3)$. Then $O_1O_3$ is the perpendicular bisector of $OX$, so $O_1O_3 \parallel FX$ and $O_1O_3$ passes through the midpoint of $OX$; hence, it also bisects $OP$, as desired.
-------------
Similarly, we can get that $O_2O_4$ bisects $OP$, so $OP$, $O_1O_3$, and $O_2O_4$ are concurrent, as desired.

I see nobody posted this solution,so i will do that! :)
Just observe that the circumcenter of APD,O,the circumcenter of BPC and P make a parallelogram(angle-chasing).After that we use aforementioned lemma(we proved that) and problem is finished :)

It's easy problem.
Show that O1 P O3 O parallelogram.

[quote=GuvercinciHoca]Problem can be generalized by taking point $O$ as any point on the plane. Here is the general form which is also true:
$ABCD$ circumscribed about a circle with center $O$ and diagonals $AC,BD$ intersect at $P$.
$Q$ is an arbitrary point on the plane and $X,Y,Z,W$ are the circumcenters of triangles $ABQ,BCQ,CDQ,DAQ$ respectively. Show that $XZ,YW,OP$ are concurrent.[/quote]

Solution:
Let $(BCQ)\cap (ADQ)=R,Q$ and $(ABQ)\cap (DCQ)=S,Q$.
 It's easy to show that $XZ\cap YW=O'$ is the circumcenter of $\triangle QRS$.
Thus, we want to prove $O,P,O'$ are collinear.
Let $E,F$ be the radical centers of $(ABQ),(DCQ),(ABCD)$ and $(ADQ),(BCQ),(ABCD)$ respectively.

$E\in QS \quad , F\in QR \implies EF$ is radical axis of $(O),(O')$.
So $OO'\perp EF$. On the other hand we have $OP\perp EF$ by brocards theorem
$\implies O,P,O'$ are collinear.

[quote=sevket12][quote=GuvercinciHoca]Problem can be generalized by taking point $O$ as any point on the plane. Here is the general form which is also true:
$ABCD$ circumscribed about a circle with center $O$ and diagonals $AC,BD$ intersect at $P$.
$Q$ is an arbitrary point on the plane and $X,Y,Z,W$ are the circumcenters of triangles $ABQ,BCQ,CDQ,DAQ$ respectively. Show that $XZ,YW,OP$ are concurrent.[/quote]

I think its very easy question   :cool:
We will just [color=red]move the point[/color] $D$ on the $(ABC)$
Then we can see that $deg(Z)=deg(W)=deg(P)=1$ Let $L=XZ \cap YW$ then $deg(L)\leq 2 $
 But if we take $D=B$ $\implies$ $XZ=YW$ because of coincide lemma $deg(L)\leq 1$
We want to prove that $O-P-L$ collinear.So its enough to show that for $0+1+1+1$ cases of point $D$
Obviously for points $D$ :$D=A$ ,$D=C$ and $AD\parallel BC$ satisfies the condition.$\blacksquare$[/quote]

Why... Just why


Check out the new redesigned MATHCOUNTS website at [url]http://www.mathcounts.org[/url]!

One really cool bit of news on the website: there will be a [b]live webcast[/b] of the National Countdown Round.  It will be on May 9th at 5 PM Eastern / 2 PM Pacific.  No other details yet.

If you are still getting the old website when you go to http://www.mathcounts.org, make sure to refresh your browser (F5 key).

Where would the live webcast be on the MATHCOUNTS website?

The new mathcounts website has dropped the mathchallenge game.  It was great for self learning at a more basic level than the head to head type of competition available here or elsewhere.  Not only less stressful but it answered a question after two wrong answers, and allowed for leveling!!  Without being easy or boring. :?:

Is it available or something similar available anywhere else?  Or any chance anyone else would pick it up?  We heard one rumor that it was so popular that it just used too much bandwidth from the sponsor --

Thanks for any help or suggestions

[quote="Math Geek"]Where can I find the link to the competition information on the site?  I can't find the list of people who qualified for nats?

PS: I like the design and style of the new site, but, not to be negative, but I'm more used to the original structure of the site.[/quote]
[url=http://mathcounts.org/NETCOMMUNITY/Page.aspx?pid=296&srcid=195]Right here.[/url]

Hi xpmath

remember
mathcrazed

lolololololololololololol

:)

Hey, are you mathcrazed with a separate account.

You can't have multiple accounts or they'll all be banned.

Where's the Math Challenge game on the MATHCOUNTS website?

Hi,
I am new here but I can solve your problem.
Thanks

the new website is Ok
i think it just doesnt have some of the other great things that the other one had. it is new and interesting and i can live with it but overall i think i liked the other one more


[quote="Silverfalcon"]I recently played a soccer video from my Windows Media Player. It was on my playlist so I played it several times but now it doesn't work. So I want to delete the file as well (so I get more space on computer) but I can't find the file. Is there anyway I can find the file?

If the file was the one on the cookie (since I got it off from internet), I deleted that one already. It didn't have anything like "Save in" window popping up. It just played in my Windows Media Player.[/quote]
The file's probably in your Temporary Internet Files. Just clear them, and it should be deleted.

File search FTW. BTW, if you need space that much, google for a garbage cleaning program that cleans out temp folders, since doing it manually is a pain. (you could just browse to c/windows/temp and clear it out though, that's easy)


[quote="max_tm"]So I was solving a problem in PPP vol. 3 and was not able to fully justify my reasoning. I looked at the solution and found that what I assumed but couldn't justify was only stated as fact:

"Since $r, p, q$ are positive constants then $(pr)(rq) > p^{2}q^{2}$"

I don't know anything about inequalities (basically, all I know is the definition of AM-GM and not even really how to use it in problem solving or how to prove it.) So I was wondering if anybody could help me out as to how this inequality works, or better yet give a name if it's a commonly known inequality? Really anything to help would be great![/quote]

You're probably missing something.

p = q = r = 1 doesn't work.

Hmm, good point. I'll post the problem and the solution as printed in the text:

Problem:
Let $p, q$ and $r$ be positive constants. Prove that at least one of the following equations has a real root:

$px^{2}+2qx+r = 0$
$rx^{2}+2px+q = 0$
$qx^{2}+2rx+p = 0$

([i]Descartes 1983[/i])

[hide=" solution"]
If $px^{2}+2qx+r = 0$ has non-real roots, then $4q^{2}-4pr < 0$ or $pr > q^{2}$
If $rx^{2}+2px+q = 0$ has non-real roots, then $4p^{2}-4rq < 0$
or $rq > p^{2}$
Since r,p, and q are positive, then $(pr)(rq) > p^{2}q^{2}$ or $r^{2}> pq$
Since, $r^{2}> pq$, then the discriminant is bigger than 0 and the equation has real roots. If two of the equations have non-real roots, the third equation must have real roots.
[/hide]

It's kindof a poorly stated proof through contradiction...

EDIT: edited for typo.

Well since $pr > q^{2}$ and $rq > p^{2}$ (was that a typo?) and all quantities are non-negative, you can assume that $(pr)(rq) > p^{2}q^{2}$.

Yes, of course! I don't know why I wasn't thinking in that way. I was convinced it was going to be something fancier than it is. Thank you.


trigonometry
inequalities
geometry
trig identities
Law of Sines
Law of Cosines
cyclic quadrilateral

Alright, I will begin by posting the names of five or so formulas / theorems.  The next user should  then post the formula for each (as much from memory as possible) and then go on to post five or so more formulas / theorems / etc.

So for example if I post:
Median to hypotenuse of right triangle:

The next user would reply:
$m = \frac{c}{2}$

Alright, so the beginning.

Divisibility by $2^n$:

Euler's Formula (relating $v$ to $e$ to $f$:

Ptolemy's Theorem for Cyclic Quadrilaterals:

Law of Sines:

Law of Cosines:

Divisibility by 2^n: nC0+nC1+nC2+...+nCn= 2^n (?)

Euler's Formula (relating v to e to f): V-E+f=2

Ptolemy's Theorem for Cyclic Quadrilaterals: In a cyclic quadrilateral ABCD, such that AB=a, BC=b, CD=c, AD=d, AC=e, and BD=f, then ac+bd=ef

Law of Sines: In a triangle ABC, a/sinA=b/sinB=c/sinC=2R

Law of Cosines: In a triangle ABC, a^2=b^2+c^2-2(b)(c)cosA (similarly for other sides).

 :idea: Here are the new formula's:-

RMS-AM-GM-HM inequality:

Cauchy's inequality:

Heron's formula:

Binomial theorem( :lol: ) for negative and one for positve integers:

Brahmagupta's Formula:

This is an excellent idea by SnowStorm, I think we should keep it going.  :|

[quote="Inspired By Nature"]Divisibility by 2^n: nC0+nC1+nC2+...+nCn= 2^n (?)[/quote]

I should have been more specific.  There isn't an explicit formula for this one, just that a number is divisible by $2^n$ if the last $n$ digits are divisible by $2^n$.  Sorry on that.

As for yours:

RMS-AM-GM-HM inequality:
$\sqrt{\frac{a_1^2 + a_2^2 + \dots + a_n^2}{n}} \geq \frac{a_1 + a_2 + \dots + a_n}{n} \geq \sqrt[n]{a_1*a_2*\dots*a_n} \geq \frac{n}{\frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n}}$

Cauchy's inequality:
$(a_1^2 + a_2^2 + \dots + a_n^2)(b_1^2 + b_2^2 + \dots + b_n^2) \geq (a_1b_1 + a_2b_2 + \dots + a_nb_n)^2$

Heron's formula:
$\sqrt{s(s-a)(s-b)(s-c)}$

Binomial theorem( Very Happy ) for negative and one for positve integers:$(a+b)^n = {n\choose 0}a^nb^0 + {n\choose 1}a^{n-1}b^{1} \dots {n\choose n}a^{0}b^n$

Brahmagupta's Formula:
$\sqrt{(s-a)(s-b)(s-c)(s-d)}$
(for cyclic quad)

Sorry I don't have time to post more questions.  I will in a couple of hours when I get back.

Hmm...I should add that Heron's is a special case of Brahmagupta's - let one side of the cyclic quad have length 0, then you have a degenerate quad (a.k.a. triangle) and we know all triangles are cyclic.  :)

[quote="SnowStorm"]Area of Regular Polygon (relating radius of inscribed circle, number of sides, and side length):

Wilson's Theorem:

Fermat's Little Theorem:

Divisibility by 11 (using the digits of a number):[/quote]
Wilson's Theorem:  $(p-1)!\equiv-1(mod p)$, where $p$ is a prime number greater than $2$.

Fermat's Little Theorem: $a^p \equiv a (mod p)$, where p is a prime number.

Divisibility by 11:  a number is divisible by $11$, if the sum of fist, third, fifth, etc. digits is equal to the sum of second, fourth, etc.. I don't think this is an elegant way to put it.....

I forgot for the formula for the area of the regular polygon...
I don't think I ever used it on math contests..

Here are more theorems...

Pick's Theorem:

Stewart Theorem:

Angle Bisector Theorem:

Binomial Theorem:

I just want to inform you about http://www.mathlinks.ro/Forum/viewtopic.php?t=76610, respectively http://www.mathlinks.ro/Forum/viewtopic.php?t=76609  :)


analytic geometry
graphing lines
slope
geometry proposed
geometry

In the usual coordinate system $ xOy$ a line $ d$ intersect the circles $ C_1:$ $ (x\plus{}1)^2\plus{}y^2\equal{}1$ and $ C_2:$ $ (x\minus{}2)^2\plus{}y^2\equal{}4$ in the points $ A,B,C$ and $ D$ (in this order). It is known that $ A\left(\minus{}\frac32,\frac{\sqrt3}2\right)$ and $ \angle{BOC}\equal{}60^{\circ}$. All the $ Oy$ coordinates of these $ 4$ points are positive. Find the slope of $ d$.

[quote="freemind"]In the usual coordinate system $ xOy$ a line $ d$ intersect the circles $ C_1:$ $ (x \plus{} 1)^2 \plus{} y^2 \equal{} 1$ and $ C_2:$ $ (x \minus{} 2)^2 \plus{} y^2 \equal{} 4$ in the points $ A,B,C$ and $ D$ (in this order). It is known that $ A\left( \minus{} \frac32,\frac {\sqrt3}2\right)$ and $ \angle{BOC} \equal{} 60^{\circ}$. All the $ Oy$ coordinates of these $ 4$ points are positive. Find the slope of $ d$.[/quote]

[i]Hello Andrew, here is my solution. [/i]

Firstly one can show easilly that $ \angle{KLA} \equal{} 60^o$ and therefore $ \angle{OBC} \equal{} 60^o\Longrightarrow \angle{OCB} \equal{} 60^o$ and finally $ \angle{OFD} \equal{} 60^o$. Now the triangle $ GDF$ is isosceles and $ \angle{GFD} \equal{} 60^o$, which means that the triangle $ GFD$ is equilateral. Therefore the abscissa of point $ D$ must be equal to $ OM$ where $ M$ the midpoint of the segment $ GF$, so $ D \equal{} (3,\sqrt {3})$ and now $ d \equal{} \frac {DT}{AT} \equal{} \frac {\sqrt {3}}{9}$.

[i]I hope that my calculations are ok, but forgive me if not.[/i]

Nick Rapanos


Consider the triangle ABC.Draw some rectangles:$ ABDE,BCFG,CAHK$ and let $ P,Q$are median points of DG and FK.Consider two lines:$ Px \perp DG$ and $ Qy \perp FK$.Find the locus of $ T\equal{}Px \cap Qy$

[quote="completely"]Consider the triangle ABC.Draw some rectangles:$ ABDE,BCFG,CAHK$ and let $ P,Q$are median points of DG and FK.Consider two lines:$ Px \perp DG$ and $ Qy \perp FK$.Find the locus of $ T \equal{} Px \cap Qy$[/quote]
First, your enunciation is incorrect (or clear), please point out all fixed lines or points. Second, you have posted this problem yesterday, so your second post is not necessary.

Sorry because I'm newbie.
Triangle ABC and rectangles ABDE,CAHK are fixed. (I'm not good at English)

[quote="completely"]Triangle ABC and rectangles ABDE,CAHK are fixed[/quote]
It transfers from original problem to the following problem:
[quote="mr.danh"]Let a quadrilateral ABCD. H,K move on the palne such that ABHK is a rectangle. Prove that the intersection P of the bisector perpendiculars of CH,DK, always lie on a fixed line. [/quote]
$ PA^2\minus{}PB^2\equal{}PK^2\minus{}PH^2\equal{}PD^2\minus{}PC^2, so PA^2\plus{}PC^2\equal{}PB^2\plus{}PD^2$.
Let M,N be the midpoints of AC,BD. Then, $ 4PM^2\plus{}AC^2\equal{}4PN^2\plus{}BD^2$.
$ PM^2\minus{}PN^2\equal{}const$, hence result.


My reading teacher is making me do this phrase thing and each phrase is related to dogs and i have to fill in the answer. There are a couple that I have no clue about and was wondering if any of you guys knew the answer.

4. A type of shrub or tree: dog   _ o _ _

5. Destruction, ruination: _ _ _ n _    _ _   _ _ _ dogs

17. An inferior type of humorous verse: dog _ e _ _ _

I have more if anyone is interested Thanks in advance  :D

(guessing that 17 is verse but again i have no idea)

yes they are real words, i THINK that number four is rose

#4 is [url=http://en.wikipedia.org/wiki/Dogwood]dogwood[/url],  #17 is [url=http://en.wikipedia.org/wiki/Doggerel]doggerel[/url] and I have no idea about #5.


ARML
geometry
summer program
Mathcamp
calculus
integration

Post what team you're going on, who you are, etc.

I'm going with the Iowa team! It's great to just be able to walk to the site.  :lol:

[quote="Rep123max"]I'm assuming that I will be going again, and it will be for Texas.[/quote]

You'll be invited again, if that's what you mean. :P I'll probably be going too, same team as the Max here.

i am pretty sure that i will be going again, i wonder how the chicago team is going to do, i believe that we lost some of the good seniors from last year...i did not even do math last year, so i would not really know though :P

I will ge going again from michigan team..

everyone watch out, michigan will own this year.  :P

im going for the first time from the Chicago team
i usually wear a red Astrocamp sweatshirt :lol:

I'm hoping chicago will win...and that I can make A team...course, when [b]was[/b] the last time a freshman made it in???

I'll be there for sure Piguy, or at least I better be considering im part of your relay.  Iowa for anyone who actually cares :lol:

First arml!

West Tennesee team (Memphis)

[quote="jackb1115"]Minnesota is going down to Chicago in the unofficial kinda-annual Friday afternoon frisbee game this year.

JB[/quote]
Oh yeah, I remember, the one that Indiana always wins.  :D

im going
west tennessee (does east TN even have a team?)
first time, since louisiana didnt send one last year.

Sorta of topic...but just wondering...is it ever technically correct to say, "I'm coming?"  :?:

[quote="jackb1115"]Minnesota is going down to Chicago in the unofficial kinda-annual Friday afternoon frisbee game this year.  [/quote]
Yes, probably, our team is kind of, well, short (height-wise) this year, and we didn't even get out to play frisbee at our last practice.   :(

[EDIT]
Cool, more MN people!  I didn't know we had that many on the forums.  Maybe we can restart the Minnesota forum again (no posts since April 2005)!

Huh. I like how it's not the ILLINOIS team, it's the CHICAGO team. So of course, that leaves me (from southern part of IL) to join up with the Missouri team. Chicago hates me...

You can join the Indiana team. We have a couple people from Illinois

Did anyone hear about the mumps outbreak at Iowa University?

Well, considering that both shogia and myself are here at the Iowa University, we are certainly the most informed. So here's our report:

YOU ALL WILL PERISH OF THE MUMPS! RUN WHILE YOU STILL CAN! DON'T COME TO ARML! THIS ISN'T A PLOT TO MAKE IOWA WIN!

But yeah, a lot of people had the mumps a few weeks ago. Though I wouldn't worry about it at all. It's basically all over now.

Good luck to the Minnesota team. No one from the local math team that I coach is going this year, but we wish all the best to Minnesota Gold and Minnesota Maroon. And to all you central states people, have a good time in Iowa.

You know what's funny?  I never knew that Iowa City existed until now.

Are AOPS people meeting somewhere for Iowa?

I think the Michigan team is having dinner at Panchero's (no idea where that is located).   :(

Pancheros is not that far from Jimmy Johns. They're in the same general downtown area, as they're both right by the Old Capitol Mall. It's possible that we could make a meeting point somewhere in there. It's not really a big mall, so it wouldn't be hard to find someone in there. Look on google maps. Iowa City might be on the map.

We should wear something... math t-shirts? Or are you guys wearing your team shirts?

I'm tall, big, dark brown hair, and I'll be wearing a mathcamp 2004 shirt if someone wants to talk to me.

oooh, a mathcamper!!  <3

maybe i should crack out the mathcamp shirt again.  i'm MC'03, and quite some time has passed since those happy days...  however, i only know of one other MC'03 who'll be at Iowa City ARML, so i'll probably be sporting another summer camp shirt instead.

I will be on the Missouri team.  I will score 6/8 on the individual part.  I will leave Iowa City six hours ago.  I will also wonder how in the world we won the T-shirt competition, but be thankful that Texas didn't.

It's pretty obvious, Ravi. Nobody else had weird-looking, different-color-from-the-rest-of-the-shirt sleeves like we did.  :D

i was wondering why Chicago didnt win the T-Shirt contest. I guess no one "got it", b/c i thought that was a cool design.

im not saying that we shouldnt have won....

Chicago's t-shirts didn't win because they weren't all that great.  Yes, it evaluates to " i <3 MATH ", which is admittedly a bit clever, but, c'mon, compare to Minnesota's!  (A tilted integral sign with the words "Just du it").  I still think that it's a wonder we didn't win.  :)

Wait, Minnesota changed its t-shirts this year???  Whoa!

I expect to see at least one of them at the summer reunion.

[quote="Sly Si"]I expect to see at least one of them at the summer reunion.[/quote]
summer reunion?

[quote="Sly Si"]Wait, Minnesota changed its t-shirts this year???  Whoa![/quote]

Yes!  And we finally got our safety green and black!  Unfortunately, it was safety green on black, instead of the other way around, so it was a little warm, but still!  Seriously, the shirts are cool.


Let $P_{1},P_{2},...,P_{n}$ be distinct points of the plane ,$n\geq 2$. Prove that \[\max_{1\leq i<j\leq n}P_{i}P_{j}>\frac{\sqrt{3}}{2}(\sqrt{n}-1)\min_{1\leq i<j\leq n}P_{i}P_{j}.\]

[quote="GacKiem"]It seems to be similar to a problem of IMO shortlist. Am I right, N.T.Tuan?[/quote]
Yes, you are right. But what is solution? Sorry, because I not remember :D

I solved it a long time ago but... now I don't remember  :oops:  Just I remember that we need to build some circles. Let me think more. But I think it's combination problem, not inequality.


[quote="Inquisitor"]$2^{16^{x}}=2^{4.2^{x}}$
So: $16^{x}=2^{4x}=2^{x+2}$.
Therefore: $4x=x+2$ so $x=\frac{2}{3}$[/quote]

Umm. could you or someone else explain a bit more?

the power thing was sort of confusing.

if you replace the decimal in 4.2 x with a multiply then it makes more sense

basically make 16 into base 2 so you get $2^{4}$ and you have to take that to the $2^{x}$. You can multiply the 4 and the $2^{x}$. Since 4=$2^{2}$, their product equals $2^{x+2}$. Hope that helps

[hide="Step by step"] We know that $16 = 2^{4}$.  Substituting in,

$2^{ (2^{4})^{x}}= (2^{4})^{2^{x}}\Leftrightarrow$
$2^{ 2^{4x}}= 2^{4 \cdot 2^{x}}$

Setting the exponents equal to each other:

$2^{4x}= 4 \cdot 2^{x}= 2^{2}\cdot 2^{x}= 2^{2+x}$

Again setting the exponents equal to each other:

$4x = 2+x \Leftrightarrow x = \frac{2}{3}$ [/hide]

Wow. It seems un-complicated when I see how it's done. Nice!

I should try more of these on my own.   :lol:


I really have hard time understanding the cutting the cubes. I know there was one another post about this also (but different one) but still, I don't seem to understand this.

A 9x9x9 cube is composed of twenty-seven 3x3x3 cubes. The big cube is "tunneled" as follows. First, the six 3x3x3 cubes which make up the center of each face as well as the center 3x3x3 cube are removed as shown. Second, each of the twenty-remaining 3x3x3 cubes is diminished in the same way. That is, the center facial unit cubes as well as each center cube are removed.
The surface area of the final figure is:

I got to the part that after first tunneling, total 6 center cubes are removed because each one from each face. And addition to that, the center cube, that isn't visible is also removed. So, that leave us 20. That 20 came from 8 cubes on the corner, which have surface area of 3*3*3 and the rest of 12 which have four faces open, resulting 3*3*4 or 36. But I don't understand the second tunneling.

If you're going to ask me to use the real cube, sorry, I don't have real cube.  :(

first step: you took out 7 3x3x3 cubes from the 9x9x9 (6 center face and the very center one).

second step: you do the same thing on the remaining 20 3x3x3 cubes, but this time, you dig out 7 1x1x1 cube on each 3x3x3 cube


"Participants may use resources such as books or the internet to do mathematical research to try to solve the problems. Participants may not use 'live' online help, e.g. they may not ask for help with the problem on online forums or tutorial services."

can we ask for theorems or formulas we "think" that can help us solve the problem?? for example, lets say the question is "find all solutions to the qudratic $3x^2 + 2x + 3 = 0$", can we ask for how to derive/prove  the discriminate formula on an online forum???

Here's what our [url=http://www.usamts.org/TipsFAQ/U_FAQ.php]FAQ[/url] says:
[quote]Whether or not to ask "general" questions when you are not supposed to get help is a difficult decision. Basically it is a "slippery slope" where it can be very easy to "cross the line" and ask an inappropriate question. We recommend that you do not do it, unless it is very clear in your mind that the question you are asking is very general.[/quote]What you describe in your example would clearly be over the line.

While asking others questions about things like this is against the rules, researching them on websites (i.e. reading, not asking) and books is ok.

Besides, you learn more from reading and figuring it out yourself than you would by asking another person, which is part of the primary purpose of the USAMTS anyway -- to learn more math through research!

I ask this question because I read something on formulas for angles in 3D, but the expression is vague and in-understandable (is that a word) because the whole things in text file (.txt).  So I need more explanation if I want to understand it.


function
inequalities
trigonometry
triangle inequality
topology

Let function $ d: \mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$ determinated by $ d(x,y) \equal{}\frac{|x\minus{}y|}{\sqrt{1\plus{}x^{2}}\sqrt{1\plus{}y^{2}}}$ .
Show that$ (\mathbb{R},d)$ is an incomplete metric space.

I think under this metric the sequence $ (n)$ is Cauchy but doesn't converge ( Just find $ d(n, m)$. First fix $ n$ and let $ m\to \infty$ and then let $ n\to \infty$

It's obvious that $ d(x,y)\ge 0$ and is equal to $ 0$ iff $ x\equal{}y$ and $ d(x,y)\equal{}d(y,x)$. I tried the triangle inequality but could not succeed.

[quote="ako"]To prove third property you can use trigonometric substitution x=Tan(a),y=Tan(b), where a,b are from [-Pi/2,Pi/2][/quote]
Can you proof,please !

I tried the triangle inequality with $ x\equal{}\tan a, y\equal{}\tan b$ but could not succeed.


Let $ a,b,c$ be positive real numbers. Prove that $ (a^2\plus{}2)(b^2\plus{}2)(c^2\plus{}2)\ge 3(a\plus{}b\plus{}c)^2$

ineq $ \Leftrightarrow a^2b^2c^2\plus{}2\sum a^2b^2\plus{}\sum a^2\plus{}8\geq 6\sum ab$
we have $ a^2b^2c^2\plus{}2\geq 4\sum ab\minus{}(\sum a)^2\equal{}2\sum ab\minus{}\sum a^2$
$ \sum (a^2b^2\plus{}1)\geq 2\sum ab$
We are done :)

[quote="karis"]i
we have $ a^2b^2c^2 \plus{} 2\geq 4\sum ab \minus{} (\sum a)^2$ [/quote]
Why?

$ a^2b^2c^2\plus{}1\plus{}1\geq 3\sqrt[3]{a^2b^2c^2}\equal{}\frac{3abc}{\sqrt[3]{abc}}\geq\frac{9abc}{\sum a}\geq \minus{}(\sum a)^2\plus{}4\sum ab$

[quote="thaithuan_GC"]Let $ a,b,c$ be positive real numbers. Prove that $ (a^2 \plus{} 2)(b^2 \plus{} 2)(c^2 \plus{} 2)\ge 3(a \plus{} b \plus{} c)^2$[/quote]
We have the generation for this one (of course, the generation is stronger than your so much  :P )
[b]Problem: [zaizai-hoang] [/b]
Let $ a,b,c$ be positive real numbers. Find the best constant such that the inequality hold:
\[ (a^2 \plus{} k)(b^2 \plus{} k)(c^2 \plus{} k)\ge k^2(a^2 \plus{} b^2 \plus{} c^2) \plus{} (k \plus{} 3)(ab \plus{} bc \plus{} ca)
\]
The answer is:
\[ k \equal{} \frac {1 \plus{} \sqrt {1 \plus{} 12\sqrt {2}}}{2\sqrt {2}}
\]

can you explain how did you found that constant or give me a link if you did that already

Zaizai, with $ a = b = c = 1; k = \frac {1 + \sqrt {1 + 12\sqrt {2}}}{2\sqrt {2}}\Rightarrow LHS < RHS$
$ a = b = c = 1;k = \frac {1 + \sqrt{1+12\sqrt {2}}}{2\sqrt {2}}\Rightarrow LHS < RHS$

Are you sure,karis?! In this case we have: $ LHS\minus{}RHS \equal{}27,4695...$. It means that you are wrong  :wink:

I'm sure
With $ a\equal{}b\equal{}c\equal{}1\Rightarrow ineq\Leftrightarrow(1\plus{}k)^3\geq 3k^2\plus{}3(k\plus{}3)\Leftrightarrow k^3\geq 8\Leftrightarrow k\geq 2$
that $ \frac{1\plus{}\sqrt{1\plus{}12\sqrt{2}}}{2\sqrt{2}}\equal{}1,85<2$


I don't think this is possible because $\log{40}$ is not equal to $5$.

Well, maybe he's asking for the base b such that $\log_b{40}=5$.

Or maybe not...

I think the question was: $\log{(40x-1)}-\log{(x-1)}=3$ which is very easy to solve ;)

Sorry for the typo, IversonFan is correct.  :D

[hide]$\boxed{x=\frac{333}{320}}$.[/hide] Wow, I'm surprised no one else even attempted this problem ;)

FYI,

[hide]999/960 can be further simplified to 333/320[/hide]


function integration limit calculus algebra polynomial AoPSwiki

Let $f: [0,\infty ) \to R$ (reals) be a function which is, for each $T \ge 0$, integrable on [0,T].

(1) Prove that if $\int_{0}^\infty{f(t)dt}$ is convergent, then $\int_{0}^\infty{f(t)dt}= \mathop{\lim }_{T \to \infty }\int_{0}^{T}{(1-\frac{t}{T})f(t)dt}$.

(2) Prove that if the function tf(t) is bounded on $[0,\infty )$, and the limit $\mathop{\lim }_{T \to \infty }\int_{0}^{T}{(1-\frac{t}{T})f(t)dt}$ exists (in the finite sense), then $\int_{0}^\infty{f(t)dt}$ is convergent.

Both statements can be rephrased in terms of the antiderivative $F=\int_{0}f$:

1) Prove that if $\lim_{x\to\infty}F(x)$ exists, then $\lim_{x\to\infty}F(x)=\lim_{T\to\infty}\frac{1}{T}\int_{0}^{T}F(t)\,dt$.

2) Prove that if $F'(t)=O(1/t)$ and $\lim_{T\to\infty}\frac{1}{T}\int_{0}^{T}F(t)\,dt$ is finite, then $\lim_{x\to\infty}F(x)$ exists.

I think this reformulation makes it easier to see what is going on.

Thanks for that clarification. What do the steps look like to verify that is true?

Note the analogies, if we were to turn integrals into sums.

(1) would be Abel's Theorem - in a Cesaro summability version.

(2) would be Littlewood's Tauberian Theorem, again, in Cesaro summability version.

Knowing that, I would expect the proof of (1) to be straightforward, with an integration by parts step (that mlok has already done in his restatement). Knowing the rather fearsome reputation of Littlewood's Tauberian Theorem, I wouldn't expect (2) to be easy, although Cesaro summability is usually a little easier than Abel summability. On the other hand, it should be a worthy exercise in its own right to weaken (2) by strengthening the hypothesis on $f(t)$ from $O(1/t)$ to $o(1/t).$

Chesaro summability is, indeed, easier than the Abel-Poisson one: For fixed $\delta>0$, write $\frac 1{\delta T}\int_{T}^{T(1+\delta)}F(x)\,dx= \frac{1+\delta}{\delta}\frac 1{T(1+\delta)}\int_{0}^{T(1+\delta)}F(x)\,dx-\frac{1}{\delta}\frac 1{T}\int_{0}^{T}F(x)\,dx$ to conclude that $\frac 1{\delta T}\int_{T}^{T(1+\delta)}F(x)\,dx$ tends to the same limit as $T\to\infty$. It remains to note that the oscillation of $F$ on $[T,T(1+\delta)]$ is $O(\delta)$, so this average is almost the same as $F(T)$ when $\delta>0$ is small. Actually, Littlewood's Tauberian theorem is just a combination of this trick with the Weierstrass theorem on polynomial approximation, so its "fearsome" reputation is really undeserved. Since various classical Tauberian theorems tend to appear on the Forum time to time, maybe we should write about them on AoPSWiki :?

I tried working through your comments, but I think I understand it better in the original question posed, can anyone go through the details to my original question. Thanks.

The whole point of mlok's reformulation was to make the question simpler and more transparent, so there is no chance you'll understand it better if we rewrite everything in terms of $f$ (if you want, you can easily do it yourself: just replace $F(x)$ by $\int_{0}^{x}f(t)\,dt$ everywhere in all the above posts and see if it helps in any way). On the other hand, if you do not understand some particular place(s) in the above argument, just ask specific questions. I'll be happy to answer :).


geometry
inradius
perimeter
inequalities
rearrangement inequality
inequalities proposed

[quote="Kent Merryfield"]Let $a, b, c$ be edge lengths of a triangle. Let $s = \frac{a+b+c}2$   be the semi-perimeter and let $r$ be the radius of the inscribed circle. Prove that \[\frac1{(s-a)^{2}}+\frac1{(s-b)^{2}}+\frac1{(s-c)^{2}}\ge\frac1{r^{2}}\] [/quote]

[hide]WLOG, let $a\le b\le c$ and let $a=x+y, b=x+z, c=y+z$

$\frac{1}{r^{2}}=\frac{s}{(s-a)(s-b)(s-c)}$ by Herons

$\frac1{(s-a)^{2}}+\frac1{(s-b)^{2}}+\frac1{(s-c)^{2}}\ge \frac{s}{(s-a)(s-b)(s-c)}$

$\Leftrightarrow \frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}\ge \frac{x+y+z}{xyz}$

$\Leftrightarrow \frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x}\ge x+y+z$

By the rearrangement inequality,

$\Leftrightarrow \frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x}\ge \frac{xy}{y}+\frac{xz}{x}+\frac{yz}{z}=x+y+z$[/hide]

Kent, see here:
http://www.mathlinks.ro/Forum/viewtopic.php?p=511017

Cezar, could you please encapsulate that url so it doesn't stretch the page? It becomes hard to read on my screen.

Here's my solution:

Let $\alpha=s-a,\beta=s-b,\gamma=s-c.$

Note that $\alpha+\beta+\gamma= 3s-(a+b+c)=s.$

Our inequality becomes:

$\frac{\alpha+\beta+\gamma}{\alpha\beta\gamma}\le \frac1{\alpha^{2}}+\frac1{\beta^{2}}+\frac1{\gamma^{2}}$

$\frac1{\beta\gamma}+\frac1{\alpha\gamma}+\frac1{\alpha\beta}\le \frac1{\alpha^{2}}+\frac1{\beta^{2}}+\frac1{\gamma^{2}}$

At this point you could appeal to rearrangement, but there's also this:

$\frac1{\beta\gamma}\le\frac12\left(\frac1{\beta^{2}}+\frac1{\gamma^{2}}\right)$

$\frac1{\alpha\gamma}\le\frac12\left(\frac1{\alpha^{2}}+\frac1{\gamma^{2}}\right)$

$\frac1{\alpha\beta}\le\frac12\left(\frac1{\alpha^{2}}+\frac1{\beta^{2}}\right)$

Now add.


I have an equation that isn't too long, but just appears at the end of a line.  Latex then breaks this equation across a space.  Can I prevent this from occurring?  I found a place that said to add the lines:

\relpenalty=9999
\binoppenalty=9999

But this caused my equation to go into the margin rather than drop down and start from the beginning of the next line.

Difficult to diagnose without the text to work with. You can stop line breaking using {[i]formula is not broken inside these brackets[/i]} or \nolinebreak or put in \slash or \* where you want the break. But the simplest method is to break the formula into two parts yourself so you get $ \$$formula 1$ \$$ $ \$$formula 2$ \$$ and then there could be a break betweeen them. But, as I say, it all depends on the text you are using.

The general problem seems to be that $ \text{\LaTeX}$ is much more afraid of underfilling boxes than of overfilling them, so setting the mentioned tolerance levels doesn't help much.

As to the particular code, it is easy to create: type aaa aaa aaa ... until you almost fill the line and then add $ f=g+h+t$. If the amount of aaa's is chosen properly (that "proper" number depends on the page width setting, so there is no point for me to try to figure it out now), the formula will be broken. What rghthndsd asks is whether it is possible to play with some global settings to make $ \text{\LaTeX}$ to prefer to start it on the second line and underfill the first line automatically in all such cases without having to deal with each formula separately (but still to hypernate long words, etc.)

[quote="rghthndsd"] I found a place that said to add the lines:
\relpenalty=9999
\binoppenalty=9999
[/quote]
Shouldn't that be
\relpenalty=10000
\binoppenalty=10000
which certainly works in the example fedja gives? Values >= 10000 turn off line breaking in maths mode.


For some reason, my private messages box is lit up and says I have two unread messages, but I don't. I even went back and read every single message again. Unless some of the messages are hidden, I don't have the foggiest clue what's going on.

Sorry if this has already been mentioned. Thanks in advance.

Delete your cookies (the link is located below the forums / topics on any forum page, in the left side on the same level as the jumpbox) and see if that works.

Hm, it still shows 2 new messages. I tried deleting the cookies that have to do with AoPS-Mathlinks, and then deleted all my computer's cookies, but it's still brokeded.

[quote="Valentin Vornicu"]Then please show us the works: OS, Browser, you know :)[/quote]

Actually, I don't know.  :D  :blush:

I really don't get any of this technology stuff. Sorry.  :blush:

You are most likely using Windows XP + IE 6.0 then :D ... can't see where the problem could be originating from ... are you sure you have no unread messages?! ... what style are you using?

Yeah, I use Windows XP Home Edition and IE 6.0
AoPS skin, if that helps.

Now I know for sure that I don't have any new messages, because just in case, I deleted all of them.  :D  But, it's still there.

[quote="Valentin Vornicu"][quote="yif man12"]Yeah, I use Windows XP Home Edition and IE 6.0
AoPS skin, if that helps[/quote]Have you tried the MathLinks skin to see if that solves the problem?

Also, does this happen only when you are logged in?[/quote]

Well, considering I can only see how many private messages I have when I'm logged in...  ;)

[quote="yif man12"]Well, considering I can only see how many private messages I have when I'm logged in...  ;)[/quote]What about ML skin?

[quote="Valentin Vornicu"][quote="yif man12"]Hm, strangely it's correct on the Mathlinks skin. Weird.[/quote]Hmm, now this is becoming more of a complicated problem than I thought :) ... I really cannot image what is going on there...  :?[/quote]

Oh well, it's okay. It's just something I noticed that was a little out of place. It's really fine the way it is right now.  :)


1. On a 3*3 Tic-Tac-Toe board, Johnny wants to use the most X's possible to not make  a Tic-Tac-Toe.

example:

[u]X| | [/u]
[u]X| |X[/u]
[u] |X|X[/u]

What is the largest possible number of X's he can use?

2. Big John and Big Jane are playing Tic-Tac-Toe. They are impressing a crowd by playing it 3D. Is it possible to end in a draw(when neither get 3 in a row)? If it is, please give a example.

3. I have a 3D Tic-Tac-Toe board myself. But, the trouble is, I have 11 marbles of the same color. Those marbles are used as markers in the Tic-Tac-Toe board. Is it possible to arrange the marbles on the board such that there is no 3 in a row?

[hide="#1"]
6.
X|X|O
X|O|X
O|X|X

The O's are the blank spaces.[/hide]

[hide="3"] arrange the marbles like this:
first square   X  X   O
                   X  O   X
                   O  O   X
second square    X  X   O
                        X  O   X
                        O  O   X
third square you could put the last one lower right corner, middle, or upper right corner to get 11 on a 3d tic tac toe.[/hide]


function
algebra
polynomial
rational function
IMC
college contests

Let f be a rational function (i.e. the quotient of two real polynomials) and suppose that $f(n)$ is an integer for infinitely many integers n. Prove that f is a polynomial.

Hey, hey , hey, not so fast! DO not forget that the coefficients are real here! Anyway, it follows from your information and another problem that I already posted and which says that such a rational fraction is the quotient of two polynomials with rational coefficients.

[quote="Arne"]This is quite ridiculous - remember IMO 2002/3, the first step of the solution?[/quote]
Which problem do you refer to?IMO 2002/3?I don't think it is the one. :(

Arne, just the proof for assuming coefficients integer gave you almost no points... the art here is to prove that the coefficients have to be rational, from which you can procede in an alike way.

My Solution at exam:
[i]Lemma:[/i] Suppose that $f=\frac{f_{1}}{f_{2}},g=\frac{g_{1}}{g_{2}}$, that $deg\ f_{1}=deg\ g_{1}=m$ and $deg\ f_{2}=deg\ g_{2}=n$ then if $f=g$ at $m+n+1$ points then $f\equiv g$.
You see that $f=\frac{f_{1}}{f_{2}}$ that $deg\ f_{1}=m,\ deg\ f_{2}=n$. Consider $a_{1},a_{2},\dots ,a_{n+m+1}\in\mathbb Z$ that $f(a_{i})\in\mathbb Z$ there exist $g=\frac{g_{1}}{g_{2}}$ that $g_{1},g_{2}\in\mathbb Z[x]$ nad $\deg\ g_{1}\leq m,\ deg\ g_{2}\leq n$. By the Lemma $f\equiv g$. Now we could finish problem easily. (You see there exist $p,q\in\mathbb Q[x]$ that $g_{1}(x)=g_{2}(x)p(x)+r(x)$ and ...)

[quote="Omid Hatami"]
You see that $f=\frac{f_{1}}{f_{2}}$ that $deg\ f_{1}=m,\ deg\ f_{2}=n$. Consider $a_{1},a_{2},\dots ,a_{n+m+1}\in\mathbb Z$ that $f(a_{i})\in\mathbb Z$ there exist $g=\frac{g_{1}}{g_{2}}$ that $g_{1},g_{2}\in\mathbb Z[x]$ nad $\deg\ g_{1}\leq m,\ deg\ g_{2}\leq n$. and ...)[/quote]
I think your mean $g=f$ at $a_{1},\cdots,a_{n+m+1}$,here.
But how to prove there exist such  $g_{1},g_{2}\in\mathbb Z[x]$?
Is there other interpolation theorem similar with Lagrange interpolation theorem?

The equations $\frac{g_{1}}{g_{2}}$ leads to a system of linear eqautions with $n+m+1$ variables and $n+m+1$ equations. So it has answer in $\mathbb Q$


In a lottery, it has been observed that 72% of people who buy lottery tickets are men, 60% are under 30 years old, and 90% of all ticket buyers are either men or under 30. The probability that a ticket buyer is a woman given that the buyer is thirty or older is?

Lots of additional information.
[hide="Answer"]10%. If 90% of all buyers are men or under thirty, then 10% of all buyers are not men and are not under thirty.[/hide]

Not quite. It says "given that", not "and". But still, not all of the information is necessary.

[hide]Since 60% are under 30, 40% are over 30.

Since 90% are either men or under 30, 10% are women over 30.

Then of all buyers over 30, $\boxed{25\%}$ are women.[/hide]

Ah, misread the question. Thought it was asking how many women over thirty there were, not how many of the people over thirty are women.


integration
trigonometry
calculus
calculus computations

See. [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=168932[/url]

Please edit the problem, Pirkuliyev Rovsen. :)


Megus.... why have you put this problem in this section???

Ok, my mistake - this should go to NT section  :)

it would be solved so easy by using the ring $Z[i]$. :D  :D

[quote=ltan]1. From (2) easy to see ab < c  < ab+2. So, must c = ab+1. Then get (a - b)^2 = 1.
2. Assume a>=b. Then get solution as {a, b, c} = {n+1, n, n(n+1)+1}
3. Because n^2+1 and (n+1)^2+1 are both prime. Must n^2+1 = 2. So, n=1. END[/quote]

Can you prove $ab<c<ab+2$ from $(2)$

[quote=pluricomplex][quote="Megus"]Find all the natural numbers $a,b,c$ such that:

1) $a^2+1$ and $b^2+1$ are primes
2) $(a^2+1)(b^2+1)=(c^2+1)$[/quote]

Well assume $a\geq b$, then $c^2+1\leq (a^2+1)^2$ and so $c<a^2+1$. Now $a^2+1\mid (c^2+1)-(a^2+1)=(c-a)(c+a)$. Note $0<c-a<c+a<a^2+a+1<2(a^2+1)$ so $a^2+1= c+a$ and $c-a=1$, so $a=2$ and then $c=3$, $b=1$.[/quote]

Can you explain why $c-a=1$ and $a^2+1=c+a$. Thank you


calculus
integration
logarithms
limit
calculus computations

For which values of $a$ is the improper integral finite?

(part a)

$\int_{0}^{1/2}\frac{1}{x|\log x|^{a}}dx$

(part b)

$\int_{2}^{\infty}\frac{1}{x|\log x|^{a}}dx$

note that for sufficiently small positive $\epsilon$,

$I = \int_{\epsilon}^{1/2}\frac{1}{x|\log x|^{a}}\, dx = \int_{2}^{1/\epsilon}\frac{1}{t (\log t)^{a}}\, dt$ (where $t = 1/x$)

so we have

1) the integral of part a converges if and only if the integral of part b and they attains same value if converges.
2) using the substitution $\log t = u$, we can calculate $I$. Then it follows that $\lim_{\epsilon \to 0^{+}}I$ converges to $\frac{(\log 2)^{1-a}}{(a-1)}$ iff $a > 1$.


We have

        \[ f_{min}\equal{}\minus{}\frac{1}{6\sqrt{3}},f_{max}\equal{}\frac{1}{6\sqrt{3}}.\]

now find the min/max of $ f$ when $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$  :lol:

[quote="outback"]Find the greatest and least values of
$ f \equal{} a(b \minus{} c)^3 \plus{} b(c \minus{} a)^3 \plus{} c(a \minus{} b)^3$,
where $ a, b, c$ are nonegative real numbers such that $ a \plus{} b \plus{} c \equal{} 1$.[/quote]
Try to solve my general one: :)
If $ a,b,c$ are nonnegative real numbers such that $ a\plus{}b\plus{}c\equal{}1$. Determin the greates and least values of
\[ P\equal{}a(b\minus{}c)^n\plus{}b(c\minus{}a)^n\plus{}c(a\minus{}b)^n\]
where $ n \in N$ is a given constant.


Fruit salad can be made with any $ 3$ of these $ 5$ fruits: apples, bananas, grapes, strawberries, and pineapples. If strawberries and pineapples do not taste good together and grapes and bananas do not look appetizing together how many possible good tasting and appetizing salads are there.

The total number of them is 5C3, or 10. Now subtract the ones that are bad-

Strawberries, Pineapples, ____
Blank fulfilled in 3 ways

Same with grapes, bananas, ________

10-3-3= $ 4$

total = 10 ways with no restrictions...
there are 6 that we counted that dont work, therefore the answer is 4


algebra
polynomial
function
algebra proposed

Let $f(n)$ be the minimal $k$ for which it is possible to write $x = f_{1}^{n}+\cdots+f_{k}^{n}$ for some polynomials $f_{1},\ldots,f_{k}\in \mathbb{C}[x]$. Show that $\sqrt{n}< f(n) \leq n$ for all $n \geq 2$.

It may be pronounced "v", but I've seen the name with a "W".

An explicit example showing that $f(n)\le n$:
$x=\sum_{j=0}^{n-1}\frac{e^{-2\pi ij/n}}{n^{2}}\left(1+e^{2\pi ij/n}x\right)^{n}$
Since all constants are $n$th powers, we can bring the constant coefficients inside so that each term is actually a cube.

The lower bound is more interesting.

Yes, sorry -- of course it's "Waring"  :blush: . How about the lower bound, won't anyone try it? There is a really magnificent proof!

Vess, do you have a solution not using the Wronskian? I saw a truly magnificient solution in Prasolov's good book on polynomials which used the Wronskian.

Do you read Russian? :)

Yes, the proof in Prasolov with the Wronskian is the magnificent one :). There is another, though, not as ingenious: use the $k$-variable Mason-Stothers' theorem.

By the way, Nevanlinna theory gives the same bound with $f_{1},\ldots,f_{n}$ only required to be meromorphic! Similarly, if $f_{1},\ldots,f_{n}$ are meromorphic functions with $f_{1}^{n}+\cdots+f_{k}^{n}\equiv 1$, one must have $k > \sqrt{n}$. This is a generalization of the fact that $f^{n}+g^{n}= h^{n}$ is not solvable in entire nonconstant functions for $n>3$ :)

No, I don't read Russian (I'd really like to), but Prasolov's book has been translated in english (fortunately). As for the others approaches, I have no idea, since I've never heard of these theorems (ok, about Nevanlina theory, just in some discussions, but never studied it).

Actually, "Mason-Stothers' theorem" just means the "ABC-theorem for polynomials" :)

Ok, I've heard about this one, but I've never heard about a $k$-variable such theorem. I do know that there exists such a result for more than three polynomials, but the proof of this result is based precisely on the Wronskian, so I don't think this is a really different approach. But maybe there is an easy proof of this $k$-variables theorem...


Is it true that for $ 2 \leq p \leq q$, $ \binom{q\minus{}2}{1}\binom{p}{2} \leq \binom{p\minus{}2}{1}\binom{q}{2}$?

$ RHS \minus{} LHS \equal{} (q \minus{} p)\{(p \minus{} 2)(q \minus{} 2) \minus{} 2\}.$

If $ (p \minus{} 2)(q \minus{} 2)\geq 2$ and $ 2\leq p\leq q$, then the inequality is true, otherwise not true.

[hide]this is equivalent to

$ p \plus{} \frac {2}{p \minus{} 2} \le q \plus{} \frac {2}{q \minus{} 2}$
if $ p \equal{} q$ it holds.

if $ p < q$, $ q$ is at least greater than $ p$ by $ 1$, but $ \frac {2}{p \minus{} 2}$ cannot be greater than  $ \frac {2}{q \minus{} 2}$ by 1.

so it holds.[/hide]

[quote="kunny"]
If $ (p \minus{} 2)(q \minus{} 2)\geq 2$ and $ 2\leq p\leq q$, then the inequality is true, otherwise not true.[/quote]

In fact $ (p \minus{} 2)(q \minus{} 2)< 2$ only when $ p \equal{} q \equal{} 3$, and equality exists.


[quote="123456789"]I saw this problem somewhere, and how do you calculate it without guessing and checking? (The exponents go on forever.)

$ x^{x^{x^{...}}}\equal{}2$

Find x.[/quote]

Since $ x^{x^{x^{...}}}\equal{}2$, we have $ x^{2}\equal{}2\Rightarrow{}x\equal{}\pm\sqrt{2}$

[quote="123456789"]I saw this problem somewhere, and how do you calculate it without guessing and checking? (The exponents go on forever.)

$ x^{x^{x^{...}}}$ = 2

Find x.[/quote]

Okay, we have to give a reasonable definition to what it means to raise an exponent readately.

Here is how I would do it.
1)Let $ a$ be a positive real number.
2)Define $ s_{1}\equal{} a$.
3)Define $ s_{n\plus{}1}\equal{} a^{s_{n}}$.

So for example,
$ s_{1}\equal{} a$,$ s_{2}\equal{} a^{s_{1}}\equal{}a^{a}$, $ s_{3}\equal{} a^{s_{2}}\equal{}a^{a^{a}}$ ...

Now we define $ a^{a^{...}}$ to be $ \lim\ s_{n}$ given that this limit exists.

Now we want to find a number $ a$ so the sequence defined as $ s_{1}\equal{}a$ and $ s_{n\plus{}1}\equal{}a^{s_{n}}$ converges to $ 2$.

Now if $ \{ s_{n}\}$ converges to $ L$ then $ \{ s_{n\plus{}1}\}$ converges to $ L$ also. And since $ f(x) \equal{} a^{x}$ is a continous function for $ a>0$ it means for any sequence $ \{x_{n}\}$converging to $ x$ it means $ \{f(x_{n})\}$ converges to $ f(x)$. Thus, if $ \lim\ s_{n}\equal{} L$ it means, $ L \equal{} a^{L}$ by taking the limit of both sides. But in this specific problem we want this limit to be $ L\equal{}2$. And so $ 2 \equal{} a^{2}$. This means $ a\equal{}\pm\sqrt{2}$. And the only possible meaning if for $ a>0$ so $ a\equal{}\sqrt{2}$.

[quote="123456789"]Sorry, but I don't really understand your reasoning. Is there any easier way to explain it?

Thanks.[/quote]

Think of it this way:

$ x^{x^{x^{...}}}$ is equal to its exponent (since both are infinite exponentiations of $ x$)

So if given $ x^{x^{x^{...}}}\equal{}L$ for any $ y$, we know that this is equal to the exponent of the left hand side. So we can find the solutions with $ x^{L}\equal{}L$. I think sylow_theory was just generalizing my technique with a more rigorous stance with limits

[quote="123456789"][quote="Anaxerzia"][quote="123456789"]Sorry, but I don't really understand your reasoning. Is there any easier way to explain it?

Thanks.[/quote]

Think of it this way:

$ x^{x^{x^{...}}}$ is equal to its exponent (since both are infinite exponentiations of $ x$)

So if given $ x^{x^{x^{...}}}\equal{} L$ for any $ y$, we know that this is equal to the exponent of the left hand side. So we can find the solutions with $ x^{L}\equal{} L$. I think sylow_theory was just generalizing my technique with a more rigorous stance with limits[/quote]

Wait, what does the y mean, and why did you set the exponent to L?[/quote]

sorry, the $ y$ shouldnt have been there, and $ L$ is just the varible i used to show parallels with sylow_theory's explanation...

these types of problems have strange restrictions. like some of them do not converge at all..that's why i believe that $ \minus{}\sqrt{2}$ is excluded from the solution set.

i think the y is unnecessary. anaxerzia is just trying to show u that assuming $ x^{x^{x^{x}}...}$= some number L. in this case L is 2. now we have $ x^{x^{x^{x}}...}\equal{}2$ or $ x^{2}\equal{}2$ u see the reasoning behind this now?

[quote="mihail911"]these types of problems have strange restrictions. like some of them do not converge at all..that's why i believe that $ \minus{}\sqrt{2}$ is excluded from the solution set.

i think the y is unnecessary. anaxerzia is just trying to show u that assuming $ x^{x^{x^{x}}...}$= some number L. in this case L is 2. now we have $ x^{x^{x^{x}}...}\equal{} 2$ or $ x^{2}\equal{} 2$ u see the reasoning behind this now?[/quote]

Yes. I just had a dumb moment there. Clearly $ x^{L}\equal{}L$, because the exponent equals the entire LHS. I do hate infinite expressions though.

Thanks for the explanations though.

Good thinking, Anaxerzia. I didn't think of that. That's weird though.

[quote="Anaxerzia"]What I cant understand is what happens if you are given $ x^{4}\equal{} 4$, who's real solutions are the same as $ x^{2}\equal{} 2$, implying that $ \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}$ converges to 2 AND 4. Makes no sense....  :ninja:  :huh:[/quote]

If you look at sylow-theory's definition of the sequence, that depends on what initial value of $ a$ you set. If you happen to choose $ a_{0}\equal{} 4$ and $ a_{n\plus{}1}\equal{}\sqrt{2}^{a_{n}}$, then $ a_{n}$ will always be 4 and will converge to 4. Of course, the natural choices for $ a_{0}$ would be $ 0,1,\sqrt{2}$ (or more generally, $ 0,1,x$).

In the case $ a_{0}\equal{} 1$ (or $ a_{0}\equal{} 0$ or $ a_{0}\equal{}\sqrt{2}$) and $ a_{n\plus{}1}\equal{}\sqrt{2}^{a_{n}}$, then straightforward induction proves that $ a_{n}< 2$ for all $ n$, so $ a_{n}$ could not possibly converge to 4. It's also increasing (it can be shown that $ 0\leq A\leq 2$ implies $ \sqrt{2}^{A}\geq A$) so that $ a_{n}$ converges to some $ L$ such that $ L^{1/L}\equal{}\sqrt{2}$. The only such $ L$ are 2 and 4, and we ruled out 4. So the sequence converges to 2.

For completeness, these infinite exponents only converge for

 $ e^{-e} < x < c^{\frac{1}{e}}$

[quote="Anaxerzia"]What I cant understand is what happens if you are given $ x^{4} \equal{} 4$, who's real solutions are the same as $ x^{2} \equal{} 2$, implying that $ \sqrt {2}^{\sqrt {2}^{\sqrt {2}^{...}}}$ converges to 2 AND 4. Makes no sense....  :ninja:  :huh:[/quote]

http://www.mathlinks.ro/viewtopic.php?t=149716

I have asked the same question some time before.

 :)


calculus
integration
trigonometry
calculus computations

show :D

$ \int_0^\infty \; \frac { \cos\, x \plus{} x\;\sin\,x } { x^2 \plus{} 1} \; dx \; \equal{} \; \boxed { \frac{\pi}{e} }$

do [b]not[/b] use complex variables.

By using Fourier transforms we have

$ \int_0^{\infty}\frac{\cos x}{x^2 \plus{} 1}\,dx \equal{} \frac{\pi}{2e}$

$ \int_0^{\infty}\frac{x \sin x}{x^2 \plus{} 1}\,dx \equal{} \frac{\pi}{2e}$

Therefore,

$ \int_0^{\infty}\frac{\cos x}{x^2 \plus{} 1}\,dx \plus{} \int_0^{\infty}\frac{x \sin x}{x^2 \plus{} 1}\,dx \equal{} \int_0^{\infty}\frac{\cos x \plus{} x \sin x}{x^2 \plus{} 1}\,dx \equal{} \frac{\pi}{2e} \plus{} \frac{\pi}{2e} \equal{} \frac{\pi}{e}$

(I have not used complex variables  :D )

Referring to [url=http://www.mathlinks.ro/viewtopic.php?t=160670]here[/url], one can find

$ \int_{0}^{\infty} \frac{e^{\minus{}\alpha} \cos(\alpha x)}{1\plus{}x^2} \, dx \equal{} \frac{\pi}{2} e^{\minus{}2\alpha}$

for $ \alpha \geq 0$ only using Fubini's theorem and some knowledge about ODE. Differentiating both sides,

$ \int_{0}^{\infty} \frac{\minus{} e^{\minus{}\alpha} \cos(\alpha x) \minus{} e^{\minus{}\alpha} x \sin (\alpha x)}{1\plus{}x^2} \, dx \equal{} \minus{} \pi e^{\minus{}2 \alpha}.$

Plugging $ \alpha \equal{} 1$ gives the desired result.


A class is forming two committees.  One is the committee of normal people and the other is the committee of stupid people.  The committee of normal people only takes people with a grade above 90 and the committee of stupid people only takes people with a grade below 10.  Both committees must have exactly 3 people.  If 7 people have a grade above 90 and 20 have a grade below 10, in how many ways can the class form the committees?

kinda easy, I know

[hide="solution"] The answer is $ \binom{7}{3} \binom{20}{3} \equal{} 39900$ [/hide]

I think it is 1175...

7 C 3 = 35

20 C 3 = 1140

Add those two together

1175

More of an explanation than xpmath gave:
You only add events if they are dependent. You multiply them if they are independent. Does the two combinations seem independent or dependent?


a,b,c are positive reals such that abc = 1. Prove that

$ (a\plus{}\frac{1}{b} \minus{} 1)(b\plus{}\frac{1}{c} \minus{} 1)(c\plus{}\frac{1}{a} \minus{} 1)\leq 1.$

Hmm opening the terms and applying AM>=GM gets us somewhere I think.

hello,make the substitution
$ a\equal{}\frac{x}{y},b\equal{}\frac{y}{z},c\equal{}\frac{z}{x}$ with $ x,y,z>0$.
Sonnhard.


Hi everyone (those of the '09 batch)!

Wish you all very good luck for all your exams, the IITJEE, AIEEE, BITSAT, VIT and the Boards too (if need be :rotfl: )  :oops: .

After the exam, do let us know about your performance and expectations.

-Rajiv

all the best for iit jee -2009.
solutions to the paper come out at various sites like http://www.brilliant-tutorials.com http://www.fiitjee.com etc..

so after our exams lets post our marks over here

but the first one to release solutions is usually some "prerna classes"  :rotfl:

Neways best of luck..

and Rajiv..u forgot  PMTs( :( )  :P

[quote="tuhin_fire"]so after our exams lets post our marks over here[/quote]

Sure!

[quote="Rituraj"]
and Rajiv..u forgot PMTs

[/quote]

I'm really sorry! :oops:

probably only one here for medicals is ritu himself :rotfl:

anyways,best of luck to all of you :)

[quote="skand"]probably only one here for medicals is [b]Ritu[/b] himself :rotfl: [/quote]

Dr. Rituraj... :mad:

Good luck!! :D

but no one excpt me needs luck here :oops:

good luck every one
i hope this is our last april as a non[b]IITIAN[/b]

@ ritu:wat u r hoping

getting dual degree(IIT D and AIIMS) :rotfl:  :rotfl:

best of luck to you all as IIT is alarmingly close

too alarmingly close...but at the same tym too far from me :rotfl:

i bet dr rituraj will conduct his own autopsy  :rotfl:

best of luck rituraj & all others who r giving the exam
[hide]please note;i have no idea as 2 when is the exam ;so if i am late i am sorry[/hide]

You are right on time. The exam is on 10th...

Less than 24 hrs for rituraj to beat all records and stand AIR 1 once again  :lol:

best of luck to ritu(and everybody else)

so wht happend?
will we meet zombie rituraj or dr rituraj  on aops :D
[hide]
i thot paper must be over by now sorry if it isn't
[/hide]

Its tomorrow and its subjective. He's assured of AIR 1 in the important exam, but tomorrow, he's a semi-God  :P

[quote="Rajiv"]
Sagar Vaze comes from stone-age. He cannot operate mobile phones.
[/quote]

wtf dude ur PM was
[quote="Rajiv"]I have to go, you [b]can[/b] SMS me on....[/quote]

i thought the emphasis on [b]can[/b] meant it was strictly optional, and as u have said before i being a [b]kanjus konknastha brahmin[/b], felt no need to waste my money on a sms  :P

PEACE

i m dying...i count just 166 breaths left...then u'll see the zombie who'll bully u for not attending his funeral :D

the xam was bad
and i m not even sure of a top 100.... :(

@rituraj007
No point in trying to fool us :o
u will b in top 10 :)

In fact, after Rituraj's scores, the'll have to invent ranks better than 1, a la Zeroth Law of Thermodynamics!

as if defining infinity :huh:

top 10 is too far.... :o

yeah
i reckon 10 is too far!
top 3 is more like it!

ur wish to think :D

but dont get disappointed wen u see the result :(


Find all b, n positive integers such that

(b + 2)^n + b^n is divisible by 2^(n+1), where b is odd.

A first step:  [hide]n has to be odd.  If n is even, then 21 is the highest power of 2 that divides (b + 2)n + bn.

In cases with n odd, b :equiv: 2n - 1 (mod 2n + 1) will always work.

I suspect that is the only solution.[/hide]

[hide]

for n even, we expand the rhs to get 2n+1|2b^n+4*(f(b)) so 2n|b^n+2*(f(b)) but b^n+2*(f(b)) is odd, so n = 0, so n is odd

then (b+2)^n + b^n = (2b+2)*((b+2)n-1 - ... + bn-1) but we note that ((b+2)n-1 - ... + bn-1) is odd, since it is the sum/difference of an odd number of odd terms, and then 2n+1|(2b+2) or 2^n | (b+1) so the solutions is the set of solutions to

b :equiv: -1 mod (2^n) with n odd, and that will be all solutions.[/hide]


$ a, b, c \geq 0$ and $ bc\plus{}ca\plus{}ab>0$
Prove that
\[ \frac{1}{b^2\minus{}bc\plus{}c^2}\plus{}\frac{1}{c^2\minus{}ca\plus{}a^2}\plus{}\frac{1}{a^2\minus{}ab\plus{}b^2}    \geq  \frac{3}{ab\plus{}bc\plus{}ca}\]

[quote="chess"]$ a, b, c \geq 0$ and $ bc \plus{} ca \plus{} ab > 0$
Prove that
\[ \frac {1}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {1}{c^2 \minus{} ca \plus{} a^2} \plus{} \frac {1}{a^2 \minus{} ab \plus{} b^2} \geq \frac {3}{ab \plus{} bc \plus{} ca}
\]
[/quote]
$ \sum_{cyc}\frac {1}{a^2 \minus{} ab \plus{} b^2} \geq \frac {3}{ab \plus{} bc \plus{} ca}\Leftrightarrow\sum_{cyc}(a\minus{}b)^4ab\plus{}\sum_{sym}(a^4bc\plus{}a^3b^2c\minus{}a^2b^2c^2)\geq0.$

Vasc's solution used as follow:
\[ f(a,b,c)\ge f(0,b,c)\ge 0\]
Easy to check it out  :wink:

yes basicaly like this
wlog $ a\geq b\geq c$
$ \frac{1}{a^2\minus{}ab\plus{}b^2} \plus{}\frac{1}{b^2\minus{}bc\plus{}c^2}\plus{} \frac{1}{a^2\minus{}ac\plus{}c^2} \geq \frac{1}{a^2\minus{}ab\plus{}b^2}\plus{}\frac{1}{a^2}\plus{}\frac{1}{b^2}\geq \frac{3}{ab} \geq \frac{3}{ab\plus{}bc\plus{}ca}$


Slove this equation: $ \displaystyle \left[ \sqrt[3]{1}\right]  \plus{} \displaystyle \left[ \sqrt[3]{2}\right]  \plus{} ... \plus{}\displaystyle \left[ \sqrt[3]{x^3\minus{}1}\right]   \equal{} 400$ , with $ x\in N$.

[quote="thanhnam2902"]Slove this equation: $ \displaystyle \left[ \sqrt [3]{1}\right] \plus{} \displaystyle \left[ \sqrt [3]{2}\right] \plus{} ... \plus{} \displaystyle \left[ \sqrt [3]{x^3 \minus{} 1}\right] \equal{} 400$ , with $ x\in N$.[/quote]
$ \displaystyle \left[ \sqrt [3]{1}\right] \plus{} \displaystyle \left[ \sqrt [3]{2}\right] \plus{} ... \plus{} \displaystyle \left[ \sqrt [3]{x^3 \minus{} 1}\right] \equal{} \sum_{k \equal{} 1}^{x \minus{} 1}k\left( (k\plus{}1)^3\minus{}k^3\right)$ the rest is easy  :wink:

note that $ (k\plus{}1)^3\minus{}k^3\equal{}|\{h\in[|1,x\minus{}1|]: \ \sqrt[3]{h}\equal{}k\}|$

[quote="thanhnam2902"]Thank you very much.[/quote]
your welcom,

generaly if we take $ S_X\equal{}\sum_{k\equal{}1}^{X}[\sqrt[p]{k}]$
then $ S_{x^{p}\minus{}1}\equal{} \sum_{h\equal{}1}^{x}h\left((h\plus{}1)^p\minus{}h^p\right)$
and $ S_{X}\equal{}S_{[\sqrt[p]{X}]^p\minus{}1}\plus{}\sum_{k\equal{}[\sqrt[p]{X}]^p}^{X}[\sqrt[p]{k}]\equal{}S_{[\sqrt[p]{X}]^p\minus{}1}\plus{}X(X\minus{}[\sqrt[p]{X}]^p\plus{}1)$

$ \bullet$ $ \displaystyle \left[ \sqrt [3]{1}\right] \plus{} \displaystyle \left[ \sqrt [3]{2}\right] \plus{} ... \plus{} \displaystyle \left[ \sqrt [3]{x^3 \minus{} 1}\right] \equal{}400 \Longleftrightarrow \sum_{k \equal{} 1}^{x \minus{} 1}k\left[(k\plus{}1)^3\minus{}k^3\right] \equal{} 400$

$ \bullet$ Let $ S_k\equal{}k[(k\plus{}1)^3\minus{}k^3]\equal{}3k^2\plus{}3k\plus{}1$. We have $ \displaystyle \left[ \sqrt [3]{1}\right] \plus{} \displaystyle \left[ \sqrt [3]{2}\right] \plus{} ... \plus{} \displaystyle \left[ \sqrt [3]{x^3 \minus{} 1}\right] \equal{}400 \Longleftrightarrow \sum_{k \equal{} 1}^{x \minus{} 1}S_k \equal{} 400$

$ \bullet$ Because $ S_k>0, k\in N^*$, $ S_1\equal{}7, S_2\equal{}38, S_3\equal{}111, S_4\equal{}244$ and $ S_1\plus{}S_2\plus{}S_3\plus{}S_4\equal{}7\plus{}38\plus{}111\plus{}244\equal{}400$

So, the equation have a root is $ \boxed{x\equal{}5}$

OK!


http://www.artofproblemsolving.com/Wiki/index.php/2004_AMC_12A_Problems/Problem_15

How come you can set up a proportion (I'm talking about the first Solution)?

I never have understood proportions well.

Let the time it takes for the first meeting be $ t_1$ and $ t_2$ for the second one. Then both the LHS and RHS are equal to $ \frac {t_2}{t_1}$ (both are of the form $ \frac {kt_2}{kt_1}$, where $ k$ is the speed of Brenda or Sally).


A price of a product is $10. If this price is 15% greater than the price to produce the product, what is the cost of production?

[quote="thedeath"]A price of a product is $\$$10. If this price is 15% greater than the price to produce the product, what is the cost of production?[/quote]
[hide]$\$ 10\cdot\frac{85}{100}=\$ 8.50$[/hide]

[quote="thedeath"]A price of a product is $\$$10. If this price is 15% greater than the price to produce the product, what is the cost of production?[/quote]
[hide]
$\text{Price }= 1.15 \times\text{ Cost}.$

$\text{ Cost}= \frac{\text{Price}}{1.15}$

$=\frac{10.00}{1.15}=8.70$
[/hide]

If the cost was 15% more than the original cost, then $\$10$ is 115% of the original cost, or 1.15 times as much.

Just clarifying definitions for the people at the beginning...

x = Production Cost

1.15x = 10
Divide both sides by 1.15.
[b]x=8.70[/b]


Oh, I see. For the first two parts I had only proved 3|a,3|b=>3|a^2+b^2, with the => and not <=>. So when I saw the solutions I thought them sorta odd since proving the statements with => was obvious.

Thanks.


There are many two kinds of black and white cards. When you have 4 cards in hand, consider the following procedure $ (\bf{A})$.

$ (\bf{A})$ You choose one card from 4 cards in hand with equal probability $ \frac {1}{4}$ and replace the card with different color one.

Suppose that you have 4 cards with 2 white and  2 black first.

Answer the following questions.

(1) Find the probabilty such that  you will first have 4 cards in same color after repeating the procedure $ \bf{(A)}$ 4 times.

(2) Find the probabilty such that  you will first have 4 cards in same color after repeating the procedure $ \bf{(A)}$ $ n$ times.

the probability that all four cards are the same color stems from a fairly simple repeating tree.

(2,2)-> (3,1)-> (1/4) (4,0) + (3/4) (2,2) -> (3,1) -> ...

so for even n, the answer is 1/4

for odd n, its always 3 of one color, and 1 of the other.


Does anyone know exactly how much the 10 graders and below are favored in the selection process for USAMO?

...b/c right now, if some of these AoPS answers are right it looks like I'm sitting right on the line...

Your score must be >= Ceiling((USAMO Index - 150)/10) probably.


(1) $GI^2=r^2 + f(a,b,c)$. find f(a,b,c)
(2) If G is on the incircle,
     a)a>=b>=c. find maximun a/c
     b) find an example that the three triangle's sides are all integers.

Solution to the first one.
Well, $r^{2}$ can be easily expressed as a functino of $a,b,c$. So it remains to express $IG^{2}$ as a function of $a,b,c$. It is natural to use Leibniz:
For any point $M$ in the plane of a triangle $ABC$ (gravity center - $G$, side lengths $a,b,c$), $MG^{2}=\frac{1}{2}(MA^{2}+MB^{2}+MC^{2})-\frac{1}{9}(a^{2}+b^{2}+c^{2})$. So,
$IG^{2}=\frac{1}{3}(IA^{2}+IB^{2}+IC^{2})-\frac{1}{9}(a^{2}+b^{2}+c^{2})$.
It suffices to find $IA^{2}+IB^{2}+IC^{2}$ as a function of $a,b,c$. Well, it can be easily proved (using repeatedly the Bisector Theorem) that $IA=\frac{\sqrt{bc}\sqrt{p(p-a)}}{p}$. Then $IA^{2}+IB^{2}+IC^{2}=\frac{bcp(p-a)}{p^{2}}+\frac{acp(p-b)}{p^{2}}+\frac{ab(p-c)}{p^{2}}=(ab+bc+ca)-\frac{3abc}{p}$. So, $IG^{2}=\frac{1}{3}(IA^{2}+IB^{2}+IC^{2})-\frac{1}{9}(a^{2}+b^{2}+c^{2})=\frac{1}{3}(ab+bc+ca-\frac{3abc}{p})$-$\frac{1}{9}(a^{2}+b^{2}+c^{2})$. Then
$f(a,b,c)=GI^{2}-r^{2}=\frac{1}{3}(IA^{2}+IB^{2}+IC^{2})-\frac{1}{9}(a^{2}+b^{2}+c^{2})=\frac{1}{3}(ab+bc+$$ca-\frac{3abc}{p})-\frac{1}{9}(a^{2}+b^{2}+c^{2})-\frac{S^{2}}{p^{2}}$.
Well, probably  :idea: this expression can be significantly reduced... :?:


Έστω $a,\ ,b,\ c,\ d$ τέσσερεις διαφορετικοί πραγματικοί αριθμοί ώστε
ι)Οι $a,\ b$ είναι ρίζες της εξίσωσης $x^{2}-3cx-8d=0$ και
ιι) Οι $c,\ d$ είναι ρίζες της εξίσωσης $x^{2}-3ax-8b=0$

Nα βρεθεί το άθροισμα $a+b+c+d$

Ναι Ράλλη αυτό είναι .Εύκολη έτσι ? :)

[quote="ambros"]μας λές την λύση σου αναλυτικά σε παρακαλώ?[/quote]
ok....
vsk, exo petaksi to diafimistiko entipo pano sto opio to elisa,ala pigene kapos etsi...
apo vieta, pernume ena sistimataki 4x4
an kapios ine 0 tote oloi ine 0 ,atopo.
tora den to eyaksa poli (mpori to a+b+c+d na prokipti me kana texnasmataki..),ala ego elisa to 4x4me diadoxikes apalifes..+vrika eykola oti....(a,b,c,d)=(-8,-8-16,-16)aporiptete,(16-4riza(12),16+4riza(12),32+16riza(12),32-16riza(12))
(16+4riza(12),16-4riza(12),32-16riza(12),32+16riza(12))... pu dinun sum 96....logika 8a vgeni +me texnasmataki omos...

 :D

[quote="RAKAR"][quote="ambros"]μας λές την λύση σου αναλυτικά σε παρακαλώ?[/quote]
ok....
vsk, exo petaksi to diafimistiko entipo pano sto opio to elisa,
 :D[/quote]
πρεπει να ηταν μικρη λυση  :P

Ωραία άσκηση Σιλουανέ. Το μόνο που χρειαζόταν ήταν μία μικρή προσοχή στις πράξεις.

Από Vieta έχουμε (σύμφωνα με την υπόθεση της άσκησης) αντιστοίχως:

$a+b=3c\quad ab=-8d$
και
$c+d=3a\quad cd=-8b$

Επίσης υψώνουμε τις ισότητες στο τετράγωνο έτσι ώστε η μία να "περιέχει" την άλλη (αυτό χρησιμεύει πολύ στην αντικατάσταση και συνεπώς στην μείωση των αγνώστων).

$(a+b)^{2}=9c^{2}\implies a^{2}+b^{2}+2ab=9c^{2}\implies a^{2}+(3c-a)^{2}-16d=9c^{2}\implies a^{2}+9c^{2}-6ac+a^{2}-16(3a-c)=9c^{2}\implies 2a^{2}-6ac-48a+16c=0\implies a^{2}-3ac-24a+8c=0\ (1)$

Ακόμη έχουμε:

$(c+d)^{2}=9a^{2}\implies c^{2}+d^{2}+2cd=9a^{2}\implies c^{2}+(3a-c)^{2}-16b=9a^{2}\implies c^{2}+9a^{2}-6ac+c^{2}-16(3c-a)=9a^{2}\implies 2c^{2}-6ac-48c+16a=0\implies c^{2}-3ac-24c+8a=0\ (2)$

Αφαιρώντας τις $(1)$, $(2)$ κατά μέλη παίρνουμε:

$a^{2}-c^{2}-3ac+3ac-24a+24c+8c-8a=0\implies a^{2}-c^{2}-32a+32c=0\implies (a-c)(a+c)-32(a-c)=(a-c)(a+c-32)=0$

Από εδώ θα έχουμε είτε $a-c=0\implies a=c$ είτε $a+c-32=0\implies a+c=32\ (3)$. Για $a=c$ έχουμε: $a+b=3c\implies a+b=3a\implies b=2a$ και $c+d=3a\implies a+d=3a\implies d=2a$. Δηλαδή $ab=-8d\implies a\cdot2a=-8\cdot2a\implies a=-8$, ή $a=c=-8, b=d=-16$, το οποίο είναι άτοπο. Για $a+c=32$ έχουμε: $a+b=3c\implies 32-c+b=3c\implies 32+b=4c\ (4)$. Επίσης: $c+d=3a\implies c+d=3\cdot(32-c)\implies c+d=96-3c\implies 96-d=4c\ (5)$. Αφαιρώντας τις $(5)$,$(4)$ κατά μέλη προκύπτει: $64-b-d=0\implies b+d=64\ (6)$. Τέλος, προσθέτοντας τις $(3)$,$(6)$ κατά μέλη παίρνουμε:
$\boxed{\mathbf{a+b+c+d=96}}$

Εύχαριστώ πάρα πολύ όλους σας για την άμεση ανταπόκριση κι δημοσίευση της λύσης

Ωραία mostel .(Απλά περίμενα μια πιο tricky λύση) .
H δική μου είναι ως εξής .

$a+b=3c$ (1)    $ab=-8d$ (2)
και
$c+d=3a$  (3)  $cd=-8b$  (4).

Αν ένας είναι μηδέν τότε όλοι είναι, άτοπο .
(2).(4) => $ca=64$ (5)
Επίσης η (2) λόγω των (1) ,(3) δίνει
$a^{2}-3ac-24a+8c=0$
από όπου προκύπτει λόγω της (5) ότι
$a^{3}-24a^{2}-192a+512=0$  ή $(a+8)(a^{2}-32a+64)=0$

από όπου $a=-8$ (apor (γιατί?))  και
$a=16+8\sqrt{3}$  ή $a=16-8\sqrt{3}$  και
$c=16-8\sqrt{3}$  ή $c==16+8\sqrt{3}$

Όμως από (1),(3)  $a+b+c+d=3(a+c)=96$

BTW, Αμβρόσιε σε είδα χθες στην τηλεόραση  :lol:

Το $a=-8$ απορρίπτεται αφού στην υπόθεση αναφέρει πέρι τεσσάρων [u]διαφορετικών[/u] πραγματικών. Τώρα όσον αφορά το πιο tricky, τι το πιο tricky έχεις στη λύση σου; Απ' ότι βλέπω, βασίζεται πάνω στην ίδια λογική. Αυτά  :D

[quote="silouan"]Επειδή δεν είχα εγώ λύση tricky περίμενα από εσάς  :wink:[/quote]

Άλλο κατάλαβα εγώ, σόρρυ. Πάντως δεν ξέρω τι το πιο tricky θα μπορούσε να βρεθεί για τη λύση της συγκεκριμένης άσκησης. Τις πράξεις όπως και να το κάνουμε δεν πρόκειται να τις γλιτώσουμε (:P). Βασικά, το πιο tricky που μπορεί να μου έρθει είναι η υλοποίηση ενός προγράμματος που βρίσκει αυτόματα το άθροισμα :rotfl:

Πίστευα (και έχω κάποιες ελπίδες ακόμα ) ότι δεδομένου του ότι τα δύο πολυώνυμα έχουν ρίζες τα $a,b$ και $c,d$ θα μπορούσαμε να βρούμε ένα πολυώνυμο που να έχει ρίζες (τα a,b,c,d ίσως ) το οποίο να έχει συντελεστή του δευτέρου όρου 96 .(καθώς κατατάσουμε το πολυώνυμο κατά τις φθίνουσες δυνάμεις ) . Και έτσι να βρισκόταν κατευθείαν  :wink: .

Πολύ καλό το σκεπτικό σου Σιλουανέ. Ομολογώ πως μου φάνηκε και πρωτότυπο. Ωστόσο, στην περίπτωση που αναφέρεις θα είχαμε να κάνουμε με ένα πολυώνυμο τετάρτου βαθμού, το οποίο δεν ξέρω πόσο εύκολα μπορεί να βρεθεί ή, τέλος πάντων, να το χειριστεί κανείς. Κατά τη γνώμη μου, ο τρόπος με τον οποίο προσεγγίσαμε το πρόβλημα αποτελεί και τη "χρυσή τομή" για τη δεδομένη άσκηση. Πάντως θα χαιρόμουν ιδιαίτερα να δω μία λύση με το σκεπτικό που ανέφερες ανωτέρω.

[quote="ambros"]ευχαριστώ πολύ Σιλουανέ εσένα κι όλους όσους με τιμήσανε ως τηλεθεατές στην Εκπομπή[/quote]

Τι εκπομπή Ambros; Γιατί δεν μας ενημέρωσες;  :(

Toις μετρητοίς .......
Βέβαια έχασε η ομάδα του  :huh:
Καλά ρε Αμβρόσιε έχουμε άτομα που έχουν διακριθεί σε διαγωνισμούς της Αμερικάνικης μαθηματικής εταιρείας  :P  :maybe:

Κρίμα! Έχασα το επισόδειο αν και ομολογώ πως παρακολουθώ συχνά το συγκεκριμένο τηλεπαιχνίδι! Αν δε σε θέλει μερικές φορές... δε σε θέλει! Αν το δείξει επανάληψη, στείλτε μου ένα pm ή κάντε ένα post!

Ευχαριστώ πολύ :)

Αν και δεν βλεπω πολυ τηλεοραση θα ηθελα να το δω.Ελπιζω να κερδισες  τιποτα απτο τηλεπαιχνιδι... :wink:

παιδια κυρίως εγώ κι η ομάδα μου κερδίσαμε εμπειρίες από το τηλεπαιχνίδι ευχαριστώ  ξανά όσους με τιμήσανε ώς τηλεθεατές κι χρόνια πολλά σε όλους σας


Anyone get tickets?

Who do you think will win?

University of Nebraska
Tulane University
University of Florida
Baylor University
University of Texas
Arizona State University
University of Tennessee
Oregon State University

Texas Longhorns are defintley going to win because i said so :lol:


Let $ a\geq 0, b\geq 0$, prove that $ 2\geq \frac {(1 \plus{} a^2b)(1 \plus{} b^2)}{(a^2 \minus{} a \plus{} 1)(1 \plus{} b^3)}$


The lateral surface area of a right circular cone is $ 260\pi \text{in}^2$, and its total surface area is $ 360\pi \text{in}^2$. Find the number of cubic inches in the volume of the cone.

[hide]We can setup equations:
\[ \pi r l \equal{} 260 \pi\]
\[ \pi r l \plus{} \pi r^2\equal{}360 \pi\]Subtracting the first equation from the second, we have $ \pi r^2\equal{}100\pi$ and $ r\equal{}10$. Also, $ l\equal{}26$, so $ h\equal{}\sqrt{26^2\minus{}10^2}\equal{}24$. The volume of the cone is $ \pi r^2 h \equal{} \pi 10^2 \cdot 24 \equal{} \boxed{2400 \pi}$.[/hide]

do that e thingy not mean anything?
b/c if it just means units then yes the answer is 2400pi

The $ \in$, is given by the code \in, which I assume was intended to mean "inches"

As far as I'm informed, $ \in$ was meant to be the symbol of relation as in $ x\in A$ meaning $ x$ is an element of the set $ A$.

Perhaps I'm missing something, but the volume of a cone is $ \dfrac {1}{3} \pi r^2 h$, so the volume would be $ 800\pi \text {in.}^3$

I believe the earlier solution was given using the volume of a cylinder. Therefore, the formula, hence the solution, was incorrect.

Sorry, I used the wrong formula. It should be $ \boxed{800\pi}$.

And BTW: where did you see that earlier "e" symbol??? Am I missing something, or is my computer not displaying correctly? I'd especially like to see an elaboration of that logarithm idea...

[quote="fishythefish"]And BTW: where did you see that earlier "e" symbol???[/quote]

I believe this was editted out in the original problem. The code \in represents $ \in$ and I believe the original poster wanted it to mean "inches" instead.

Oh...I was wondering what the earlier post meant...

*apologies to The Zuton Force


Yes, I know this is kind of a stupid question...

How do you unzip or untar something in WinRAR?

Also, do not explain in complicated language, I am not a programmer, as most of the people in this forum are.

I'm not sure if this is what you want, but I suppose it is:
Open WinRAR.
File -> Open Archive. Choose whatever you want to unzip.
Click 'Extract To' and choose a destination.

If its something else you want to know..

I am using WinAce currently. I think Winzip is the best.


[quote="mailo"]Υπαρχουν ολυμπιαδες για φοιτητες? Το ΙΜΟ ειναι μονο για μαθητες?[/quote]

H Ι.Μ.Ο είναι μόνο για μαθητές. Υπάρχει όμως και ολυμπιάδα για πρωτοετείς και δευτεροετείς φοιτητές η λεγόμενη SEEMOUS. Περισσότερες πληροφορίες θα βρείς εδώ...http://www.hms.gr/modules/news/article.php?storyid=332

Και ΒΑΣΙΚΑ,ο ΙΜC για έτη 1-4 και παγκόσμιος.http://www.imc-math.org.uk/

Ωραία, ευχαριστώ :)

Τι βιβλία προτείνεται για προετοιμασία και γενικά τι πρέπει να κάνει κάποιος για να πετύχει σε τέτοιους διαγωνισμούς?

Μηπως ξερει καποιος ποια ειναι η υλη για τους διαγωνισμους αυτους? (Seemous και IMC)


The middle school is located 3 miles east and 5 miles south of the James' home. The library is 4 miles west and 2 miles north of the James' home. If all streets run only north-south or east-west, what is the shortest street distance, in miles, from the school to the library?

it may seem like we have to use similar triangles, but the problem is much simpler than that. Since there can't be any diagonal roads, the shortest distance is $ 3\plus{}5\plus{}4\plus{}2\equal{}\boxed{14}$


inequalities
trigonometry
function
real analysis
real analysis unsolved

Prove that
\[ x\sec^2 x\minus{}\tan x\leq \frac{8\pi^2x^3}{(\pi^2\minus{}4x^2)^2}\]
for $ 0\leq x < \frac{\pi}{2}$.

Consider the strictly decreasing function $ f(x)$ in $ \left( 0,\frac {\pi }{2}\right) ,$

where $ f(x) \equal{} \frac {\tan x}{x} \plus{} \frac {\pi ^{2}}{4x^{2} \minus{} \pi ^{2}}$

In the interval $ \left( 0,\frac {\pi }{2}\right) ,$ $ f^{\prime }(x) < 0$

Thus $ \frac {x\sec ^{2}x \minus{} \tan x}{x^{2}} \minus{} \frac {8\pi ^{2}x}{\left( 4x^{2} \minus{} \pi ^{2}\right) ^{2}} < 0$

Rearranging, we have $ x\sec ^{2}x \minus{} \tan x\leq \frac {8\pi ^{2}x^{3}}{\left( 4x^{2} \minus{} \pi ^{2}\right) ^{2}}$

Equality holds when $ x \equal{} 0$


geometry
area of a triangle
Heron\u0027s formula

An isosceles triangle has legs of length 29. Given that its area is 420, find all possible values for the length of the base.

When you know all the sides of a triangle, everything is fixed.
Go to[url]http://en.wikipedia.org/wiki/Heron%27s_formula[/url]
[hide]Heron's formula is $ A=\frac{\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}}{4}$
Supposing $ a$ and $ b$ are the legs, then just solve the equation.[/hide]

[hide="Just another solution"]
Let $ x$ be the half of the base and $ y$ the height. Then we have $ xy=420$
and also from pythagorean on the triangle with sides $ x,y,29$ we get $ x^{2}+y^{2}=29^{2}$

So we have to solve the system:

$ \begin{array}{l}xy=420 \\ x^{2}+y^{2}= 29^{2}\end{array}\}\Rightarrow$

$ x^{2}+y^{2}+2xy =29^{2}+2\cdot 420 \Rightarrow (x+y)^{2}= 1681$ and
$ x^{2}+y^{2}-2xy =29^{2}-2\cdot 420 \Rightarrow (x-y)^{2}= 1$

So we get $ x+y=41$ and $ |x-y|=1$
From which $ x=21, \ y=20$ or $ x=20, \ y=21$

So the possible values of the base length are $ 42$ and $ 40$[/hide]


What is the probability that if you multiply two randomly selected two digit whole numbers, the result is greater than 100 and less than 200?  Answer as common fraction.

[hide]

Since there are 90 2 digit numbers,

Wouldn't the total possible be

89+88+87....+1 because order of the numbers doesn't matter?

that equals

(89)(90)/2 or 4005

Then, you find total possible,

10*11-19 or 9 possible
11*11-18 or 8 possible
12*12-16 or 5 possible
13*13-15 or 3 possible
14*14-14 or 1 possible

So 26 possible over 4005

26/4005

[/hide]


Let $ D,E,F$ points on sides $ BC,CA,AB$ such that $ DC\plus{}CE\equal{}EA\plus{}AF\equal{}FB\plus{}BD$
Prove that $ DE\plus{}EF\plus{}FD\geq \frac{AB\plus{}BC\plus{}CA}{2}$

also posted here:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=169692[/url]


[quote="silouan"]there exist an element of $ M$ which is neighbor of exactly one of them.[/quote]
I can't make any sense of this condition. :( Can you clarify? It sounds like this problem relates to the Hamming Code though.

[quote="JBL"]It's probably supposed to read $ A$ for $ M$ in that sentence.[/quote]

Thanks fr the correction . I have edited it .

[quote="silouan"]Let $ A$ be the set of all 4-tuples of $ 0$ and $ 1$ . two such 4-tuples are called neighbors if they coincide exactly at three positions . Let $ M$ be a subset of $ A$ with the following property :
any two elements of $ M$ are not neighbors and there exist an element of $ A$ which is neighbor of exactly one of them.
Find the minimum possible cardinality of $ M$ .[/quote]

I think you are looking for the [u]maximum [/u]cardinality of $ M$ (since the minimum is 1).

It seems to be $ 5$.

The problem is the same if we switch columns or toggle $ 0$ and $ 1$ in a column.
So we could consider WLOG that the element of $ A$ is $ (0,0,0,0)$ and its neighbor in $ M$ is $ (0,0,0,1)$

Then, if $ (0,0,0,0)$ is neighbor of nothing else in $ M$, the other elements of $ M$ has exactly $ 2,1$ or $ 0$ zeroes and could only be :
Subset $ M1$ : $ (1,0,1,0),(0,1,1,0),(1,1,0,0)$ ($ (0,0,1,1)$ has been removed because neighbor of $ (0,0,0,1)$)
Subset $ M2$ : $ (0,0,1,0),(0,1,0,0),(1,0,0,0)$ ($ (0,0,0,1)$ has been removed because neighbor of $ (0,0,0,1)$)
Subset $ M3$ : $ (1,1,1,1)$

Each element of $ M1$ is neighbor of one element of $ M2$ so we can have at most $ 3$ elements of $ M1\cup M2$ in $ M$.
So M can have at most 5 elements.

And 5 is possible :
$ (0,0,0,1),(1,0,1,0),(0,1,1,0),(1,1,0,0),(1,1,1,1)$
So the answer.