hi,
solve the equation
\[ x^{2n} \plus{} a_{1}x^{2n\minus{}1} \plus{} ... \plus{} a_{2n \minus{} 2}x^{2} \minus{} 2nx \plus{} 1 \equal{} 0,\]
known that the equation has real positive roots.

Maybe you wanted to say that [b]all[/b] the roots are positive reals. If so, use Viete's relations :

$ 2n\equal{}S_{2n}\equal{} \sum_{k\equal{}1}^{2n} \frac{P_{2n}}{x_k}$ , where $ x_k$ are the roots, and $ P_{2n}\equal{} x_1 x_2 \cdots x_{2n}$

and $ P_{2n}\equal{}1$

But from AM-GM we know that $ S_{2n} \geq 2n \sqrt[2n] P_{2n}$ , but in our case we have equality , therefore $ \frac{1}{x_1} \equal{} \cdots \frac{1}{x_{2n}}\equal{}1$.So all roots are equal to $ 1$.


Let a quadrangle $ ABCD$ with $ AB\cap CD \equal{} E,AD\cap BC \equal{} F$. Prove that three circles with diameter $ AC, BD, EF$ have two common points.

Dear Mathlinkers,
I think that this result is not true for any kind of quadrangle...Make a figure and toys with one point.
Sincerely
Jean-Louis

As far as I remember, if $ABCD$ was cyclic, then the statement is true.

Best regards,
sunken rock


My results for Silver:

[code]
                    -- case number --
           1  2  3  4  5  6  7  8  9 10 11 12
counfr     *  *  *  *  *  *  t  t  t  t
events     *  *  *  *  *  *  *  *  *  *
wall       *  x  *  *  t  t  *  *  *  t  t  t
[/code]

Mediocre compared to you guys but I thought it was okay.  Events was my only iterative solution... I made quick recursive solutions for the other two (slow, but I got a couple right for them at least).  I hate optimization.

[quote="Xuan768"]Hmm... I just changed the file names for Silver's countyfr and submitted for Gold's cfair and got the same results.  Does that happen often?[/quote]

Happens occasionally.

My results:

cfair       *  *  *  *  *  *  *  *  *  *
mqueue      *  *  x  x  x  *  t  x  *  x
twohead     *  *  *  *  *  t  t  t  t  t

Meh, kinda botchy. Oh well, hopefully it's still enough to make USAICO.

Well, I haven't.  Not that I was even close to getting in.

[quote="JSteinhardt"](though clearly I accepted immediately upon receiving the invitation, there are apparently some people who actually have to think about whether they want to miss a week of school to go to the highest level of pre-college computer science instruction in the world).[/quote]

There are reasons, like maybe their graduation is that week, or something.  You never know.

[quote="h_s_potter2002"][quote="JSteinhardt"][quote="PenguinMath"]This will be my first year at USAICO.  JS, I assume you're going too?  Do you know when they will publish a list of all the qualifiers?[/quote]

Yeah, I'm also going. Qualifiers may not be published for a week or so, as people have to respond yes or no to the invitation, and that doesn't necesarilly happen immediately (though clearly I accepted immediately upon receiving the invitation, there are apparently some people who actually have to think about whether they want to miss a week of school to go to the highest level of pre-college computer science instruction in the world).

However if you would like a probable list of people, I would say:

2007  Richard McCutchen
2006  Kevin Modzelewski
2007  John Pardon
2006  Richard Ho
2006  George Boxer
2007  Thomas Morgan
2007  Bohua Zhan
2007  Zarathustra Brady
2008  Jacob Steinhardt
2007  Michael Gummelt
2007  Peiyu Wang
2009  Thomas Wang

P.S. Are you Bohua, Thomas, or Michael?[/quote]

List is posted - [url]http://oldweb.uwp.edu/academic/mathematics/usaco/2006/camp06/finalists.htm[/url][/quote]

Thanks for the heads up.

Hm...so it looks like I was right for everyone except Zimmerman going instead of Thomas.

Although I must admit that I am surprised that Filip Wolski is not coming from the international group.

[quote="PenguinMath"]I'm Michael.  Congrats on being the only sophomore to qualify!  We need more underclassmen or else the team isn't looking so hot for the future.  Oh, and I'm not from Virginia!  I'm a proud Texan!  They said I was from VA on that website.  I guess there's so many good math/sci students from TJHSST that if they just assume that an arbitrary student is from there then odds are they're right.

Oh and pray for me and my attempt to switch from Java -> C++ in the next few weeks.  I was a C++ programmer back in the day, but its been a while.  I have NO familiarity with the STL so that'll be the hardest part.  Best of Luck![/quote]

Lol yeah I'm in about the same position. Tom went to camp last year knowing only JAVA and didn't do too poorly. But preparation is never a bad thing.

I agree with you about the whole lack of underclassmen. I was actually sure Kevin Li at least would have made it, if not Thomas Wang, but Kevin told me personally that he didn't make it. It seems to me to be very illogical, since he has done roughly as well as Peiyu/Jonathan, who is going to camp, but Kevin is only a Sophomore. Go figure.

Good luck to you too! I guess I'll see you in June. From what I hear USAICO is awesome in that there are pretty good classes but also some nice down time as well.

Ya I've heard good things as well.  I'm excited that there's so many internationals coming, that'll be interesting.

Dont worry, I'll make it next year. [\blatent lie]

What's the best thing you can do to prepare for the USACO?

jSteinhardt said in one of his posts something like "working through the [usaco] training pages are the best thing you can do to train for the usaco.  Took me from bronze to gold in about a year."

jSteinhardt said in one of his posts something like "working through the [usaco] training pages are the best thing you can do to prepare for the usaco.  Took me from bronze to gold in about a year."

Thanks. Well, I'm in a different situation. I'm a freshman, and I have no computer science experience whatsoever though I plan on taking AP Computer Science (not sure which level my school offers, A or AB) next year. Is this enough knowledge to help me do well in USACO? Also, where can I get the USACO training pages?

[url=http://train.usaco.org/usacogate]http://train.usaco.org/usacogate[/url]
Google stuff when you want to find things-just a click away ;)
Yeah, you do need to know how to program to use the pages (i kind of overlooked that when i registered before realizing i had no experience), so take whatever class they have at your school that you think will help.

I agree, training site is really great. It took me to gold in 10 months.

Did you have experience, r2rd? Also, what's an ICQ number?

I have no programming experience at all, so all this stuff looks like Chinese to me.

http://java.sun.com/docs/books/tutorial/java/TOC.html

That teaches JAVA, although personally C++ is better.

You can just go to amazon.com or some store and buy a book on C++ if you really want to, although I guess it's a lot easier with a teacher. I am guessing that the best person to contact is Rob Kolstad, the USACO head coach who is almost always willing and able to help out. Make sure all emails have USACO in their subject, his e-mail is kolstad@ace.DELOS.COM and his AIM is RobAtDelos if you prefer that. Explain your situation, he probably has some good recommendations.

Thanks a bunch, JS.

Wow, from what friends have told me, C++ is usually taught before Java, at least at my school.


function
algebra
polynomial
trigonometry
IMO
IMO Shortlist

[b][size=150]IMO 1975 ShortListed Problems: (Posted in Forums)[/size][/b]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2025364#p2025364]1.(FRA)[/url]There are six ports on a lake. Is it possible to organize a series of routes satisfying the following conditions ?:

(i)	Every route includes exactly three ports;
(ii)	No two routes contain the same three ports;
(iii)	The series offers exactly two routes to each tourist who desires to visit two different arbitrary ports.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367460&#p367460]2.(CZS)[/url] [b]IMO1[/b] We consider two sequences of real numbers $x_{1} \geq x_{2} \geq \ldots \geq x_{n}$ and $\ y_{1} \geq y_{2} \geq \ldots \geq y_{n}.$ Let $z_{1}, z_{2}, .\ldots, z_{n}$ be a permutation of the numbers $y_{1}, y_{2}, \ldots, y_{n}.$ Prove that $\sum \limits_{i=1}^{n} ( x_{i} -\ y_{i} )^{2} \leq \sum \limits_{i=1}^{n}$ $( x_{i} - z_{i})^{2}.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2025366#p2025366]3.(USA)[/url] Find the integer represented by $\left[ \sum_{n=1}^{10^9} n^{-2/3} \right] $. Here $[x]$ denotes the greatest integer less than or equal to $x.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2025368#p2025368]4.(SWE)[/url] Let $a_1, a_2, \ldots , a_n, \ldots  $ be a sequence of real numbers such that $0 \leq a_n \leq 1$ and $a_n - 2a_{n+1} + a_{n+2} \geq  0$ for $n = 1, 2, 3, \ldots$. Prove that
\[0 \leq (n + 1)(a_n - a_{n+1}) \leq 2 \qquad \text{ for } n = 1, 2, 3, \ldots\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2025369#p2025369]5.(SWE)[/url] Let $M$ be the set of all positive integers that do not contain the digit $9$ (base $10$). If $x_1, \ldots , x_n$ are arbitrary but distinct elements in $M$, prove that
\[\sum_{j=1}^n \frac{1}{x_j} < 80 .\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=849354&#p849354]6.(USS)[/url] [b]IMO4[/b] When $4444^{4444}$ is written in decimal notation, the sum of its digits is $ A.$ Let $B$ be the sum of the digits of $A.$ Find the sum of the digits of $ B.$ ($A$ and $B$ are written in decimal notation.)

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2025371#p2025371]7.(GDR)[/url] Prove that from $x + y = 1 \  (x, y \in \mathbb R)$ it follows that
\[x^{m+1} \sum_{j=0}^n \binom{m+j}{j} y^j + y^{n+1} \sum_{i=0}^m \binom{n+i}{i} x^i = 1 \qquad (m, n = 0, 1, 2, \ldots ).\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367456&#p367456]8.(NET)[/url] [b]IMO3[/b] In  the plane of a triangle $ABC,$ in its exterior$,$ we draw the triangles $ABR, BCP, CAQ$ so that $\angle PBC = \angle CAQ = 45\,^{\circ},$ $\angle BCP = \angle QCA = 30\,^{\circ},$ $\angle ABR = \angle RAB = 15\,^{\circ}.$

Prove that

[b]a.)[/b] $\angle QRP = 90\,^{\circ},$ and

[b]b.)[/b] $QR = RP.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2025374#p2025374]9.(NET)[/url] Let $f(x)$ be a continuous function defined on the closed interval $0 \leq x \leq 1$. Let $G(f)$ denote the graph of $f(x): G(f) = \{(x, y) \in \mathbb R^2 | 0 \leq x \leq 1, y = f(x) \}$. Let $G_a(f)$ denote the graph of the translated function $f(x - a)$ (translated over a distance $a$), defined by $G_a(f) = \{(x, y) \in \mathbb R^2 | a \leq x \leq a + 1, y = f(x - a) \}$. Is it possible to find for every $a, \  0 < a < 1$, a continuous function $f(x)$, defined on $0 \leq x \leq 1$, such that $f(0) = f(1) = 0$ and $G(f)$ and $G_a(f)$ are disjoint point sets ?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367453&#p367453]10.(GBR)[/url] [b]IMO6[/b] Determine the polynomials P of two variables so that:

[b]a.)[/b] for any real numbers $t,x,y$ we have $P(tx,ty) = t^n P(x,y)$ where $n$ is a positive integer, the same for all $t,x,y;$

[b]b.)[/b] for any real numbers $a,b,c$ we have $P(a + b,c) + P(b + c,a) + P(c + a,b) = 0;$

[b]c.)[/b] $P(1,0) =1.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367455&#p367455]11.(GBR)[/url] [b]IMO2[/b] Let $a_{1}, \ldots, a_{n}$ be an infinite sequence of strictly positive integers, so that $a_{k} < a_{k+1}$ for any $k.$ Prove that there exists an infinity of terms $ a_{m},$ which can be written like $a_m = x \cdot a_p + y \cdot a_q$ with $x,y$ strictly positive integers and $p \neq q.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2025379#p2025379]12.(GRE)[/url] Consider on the first quadrant of the trigonometric circle the arcs $AM_1 = x_1,AM_2 = x_2,AM_3 = x_3, \ldots , AM_v = x_v$ , such that $x_1 < x_2 < x_3 < \cdots < x_v$. Prove that
\[\sum_{i=0}^{v-1} \sin 2x_i - \sum_{i=0}^{v-1} \sin (x_i- x_{i+1}) < \frac{\pi}{2} + \sum_{i=0}^{v-1} \sin (x_i + x_{i+1})\]

13.(ROM) Let $A_0,A_1, \ldots , A_n$ be points in a plane such that
(i)    $A_0A_1 \leq \frac{1}{ 2} A_1A_2  \leq  \cdots  \leq  \frac{1}{2^{n-1} } A_{n-1}A_n$ and
(ii) $0 < \measuredangle A_0A_1A_2 < \measuredangle A_1A_2A_3 < \cdots < \measuredangle A_{n-2}A_{n-1}A_n < 180^\circ,$
where all these angles have the same orientation. Prove that the segments $A_kA_{k+1},A_mA_{m+1}$ do not intersect for each $k$ and $n$ such that $0 \leq k \leq m - 2 < n- 2.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2025382#p2025382]14. (YUG)[/url] Let $x_0 = 5$ and $x_{n+1} = x_n + \frac{1}{x_n} \ (n = 0, 1, 2, \ldots )$. Prove that
$45 < x_{1000} < 45. 1.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367461&#p367461]15. (USS)[/url] [b]IMO5[/b] Can there be drawn on a circle of radius $1$ a number of $1975$ distinct points, so that the distance (measured on the chord) between any two points (from the considered points) is a rational number?


[quote="color"]Find value for this sum
\[6+66+666+6666+\ldots+\underbrace{666\ldots 666}_{n\, \rm{digit number}}\]
How about the general problem?
\[aa+aa+aaa+aaaa+\ldots+\underbrace{aaa\ldots aaa}_{n\, \rm{digit number}}\]
[/quote]

$S=a+aa+aaa+aaaa+\ldots+\underbrace{aaa\ldots aaa}_{n\, \rm{digit number}})$
$S=\frac{a}{9}(9+99+999+9999+\ldots+\underbrace{999\ldots 999}_{n\, \rm{digit number}})$
$S=\frac{a}{9}(10^{1}+10^{2}+10^{3}+\ldots+10^{n}-n)$

$S=\frac{a}{9}(10\frac{10^{n}-1}{9}-n)$

--
Patrick


How many duplets of prime numbers, (p,q) are possible such that
p^2 + 7pq + q^2 is a whole square, i.e. square of some integer?

For infinitely many couples.

Just take the $ p\equal{}q$

 :)

So let's make the problem harder: describe all such $ (p,q)$.

I got that $ p \neq q$ has no solutions with some nasty casework, but there's probably a much better way to determine that.

EDIT: And this is why you don't solve problems with nasty casework, because I missed $ (3,11)$ as shown in the links below.

I thought this looked familiar:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=58233[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80152[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=19090[/url]


geometry
geometric transformation
rotation
vector
analytic geometry
complex numbers
superior algebra

Hi,

First, I would like to apologise if I have misplaced this topic - I hope superior algebra is the right forum for my question.
My question concerns quaternion rotations. I know the formulas that need to be applied to rotate a certain object around the x-,y- and z-axis, but I do not know how to do this 'practically'.
For example, one point I want to rotate is the point (1,1,1).
I want to rotate this around the point (1,2,3), 90 degrees, so pi/2 radians.
I believe the formula is:
(I qoute from a non-math site now)
*******************************
If the axis of rotation is (ax, ay, az)- must be a unit vector
and the angle is theta (radians)

    w   =   cos(theta/2)
    x   =   ax * sin(theta/2)
    y   =   ay * sin(theta/2)
    z   =   az * sin(theta/2)
*******************************
w=cos(pi/4)= sqrt(2)/2
x   =   1x * sin(pi/4) =   x*sqrt(2)/2
y   =   2y * sin(pi/4) =   2y*sqrt(2)/2   = ysqrt(2)
z   =   3z * sin(pi/4)  =   3z*sqrt(2)/2   = 3/2zsqrt(2)

But how would I apply this to my (1,1,1) point?
Do you know of any other examples?

Thanks,
RicOh

When working in the three-dimensional space $\mathbb R^{3}$, it is helpful to use the standard basis vectors $\vec\i$, $\vec\j$, and $\vec k$ - these are simply vector of unit length that point in the direction Ox, Oy, Oz, respectively. So, your point $(1,1,1)$ can be identified with the vector $\vec\i+\vec\j+\vec k$. (In general, a point $(x,y,z)$ will be represented by the vector $x\vec\i+y\vec\j+z\vec k$ --- the [i]position vector[/i] of the point.)

Quaternions are objects of the form $a+bi+cj+dk$, where $i$, $j$, and $k$ are imaginary units, and $a\dots d$ are real numbers. There are addition/subtraction and multiplication/division rules that you can [url=http://mathworld.wolfram.com/Quaternion.html]look up[/url]. The notation suggests that you can interpret vectors as quaternions with real part $a$ equal to $0$. General quaternions can be written as pairs $(a,\vec v)=a+v_{1}\vec\i+v_{2}\vec\j+v_{3}\vec k$. Multiplying two vectors $\vec v$ and $\vec w$ [b]as quaternions[/b] you get the quaternion $(-\vec v\cdot\vec w, \vec v\times\vec w)$ - this multiplication contains both the dot product and the cross product.  :o

You have already computed the rotation quaternion $q=w+xi+yj+zk$. The position of a point $p$ (such as $i+j+k$ in your example) after rotation by $q$ is given by the quaternion $p'=qpq^{-1}$. Since $q$ is a unit quaternion, its inverse $q^{-1}$ is the same as its conjugate, which is $\bar q=w-xi-yj-zk$. The quaternion $p'=qp\bar q$ has zero real part, so it indeed represents a vector in 3D.

Thank you for replying. My problem was how to 'enter' the point in the formula. By representing (1,1,1) as i+j+k, my problem is solved.
Also, I'm sorry for my inferior knowledge, I am just a Belgian hugh school student with weird interests.  :roll:

No problem. I suspect, but don't know for sure, that we now use the letters i,j,k for basis vectors because Hamilton used them for quaternions. A long tim ago, quaternions were much more en vogue than they are now.

So now, step by step
1) Coordinates (a,b,c) --> ai+bj+ck
2) Rotation --> quaternion using the formula
3) Apply the rotation to the ai+bj+ck
4)Result: di+ej+fk--> (d,e,f)

I do not fully understand why you still have to multiply the result with the conjugate, but that's not (very) important.

Of course, it is not obvious why $qp\bar q$ gives the coordinates of the rotated point. I can't give a proof off the top of my head. But it is clear that $qp$ would not work - this is in general a quaternion with nonzero real part, so it does not represent a vector.

I'll Google some more then - it's the experience that counts.  :D


function
complex analysis
complex analysis unsolved

Let $ D \equal{} \{ z \in \mathbb{C} : |z|<2  \}$. Suppose that $ f: D \to \mathbb{C}$ is a continuous function that is analytic over $ D\minus{}[0,1]$. Show that $ f$ is analytic in $ D$.

Use Morera's theorem (the version with triangles). Partition the triangle into a few, so that the segment lies on the boundary.


Find $\sum_{r=n}^{2n} n^2$

Can I say

$\sum_{r=n}^{2n} n^2$ = $\sum_{r=1}^{2n} n^2$ - $\sum_{r=1}^{n} n^2$

?

I don't get some sigma notation things sometimes. (like when you can add them or multiply them or things)

Also, you have another notational problem -- most of your n's are supposed to be r's.

[quote="blahblahblah"]Not quite.

That statement is saying that

$(2n)^2+(2n-1)^2+\ldots+n^2=((2n)^2+(2n-1)^2+\ldots+1)-(n^2+(n-1)^2+\ldots+1)$

Can you see why it's not true?[/quote]
Yes I see now,
the statement would be saying:
$(2n)^2+(2n-1)^2+\ldots+n^2=((2n)^2+(2n-1)^2+\ldots+n^2+(n-1)^2+\ldots+1)-(n^2+(n-1)^2+\ldots+1)$
in which the $n^2$ term occurs on the left hand side but not the right. so:

$\sum_{r=n}^{2n} {r^2} = \sum_{r=1}^{2n} {r^2} - \sum_{r=n}^{2n} {r^2} + n^2$

thanks also, I just realised the $n^2$ should say $r^2$ after the sigma sign.
need.more.practice.

ah I get it now. always making mistakes somehow

Thanks.


[hide="Solution"]
Floor functions are usually hard to deal with directly.  So just straight algebra wont cut it.  Since we're looking at intersection points, let's try graphing.

First, note that $x-\lfloor x\rfloor=\{ x\}$ (i.e. the fractional part of $x$).  So our original equation is equivalent to $\{x\}^2 - \{x\}+y^2=0$ or $(\{x\}-1/2)^2 + y^2 = 1/4$.  By definition though, the fractional part of $x$ is on $[0,1)$.  This means that the graph is periodical with period 1.  We also note it will consist of a bunch of tangent circles by looking at the equation.  Just concentrating on quadrant I, we want to find where $y=x/5$ "leaves" the circles.  We know that the highest point the circles achieve is at $y=.5$.  It just so happens that $(2.5,.5)$ is on both graphs so this is our last point.  So the line will intersect the first circle in the first quadrant once (exclude the origin for now), the second circle twice, and the third circle twice for a total of 5 times.  By symmetry, there are 5 intersections in quadrant 4.  Add in the origin for a total of $\boxed{11.}$

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=760[/img][/hide]

[quote="4everwise"]Compute the number of intersection points of the graphs of $(x-\lfloor x\rfloor)^2+y^2=x-\lfloor x \rfloor$ and $y=\frac15x$.[/quote]
[hide]
Graph it!
[url=http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=761]Link to Picture[/url]
We see that there are 5 intersections on each side, and then the origin.
5+5+1=[b]11[/b]
[/hide]


The sum of AMC and AIME is the No.1 of the state of my highschool. Does it help?
I know one year the winners of each state got 1000 dollars.
Any awards this year for states winners?

I'm assuming that you mean your index (AMC+10xAIME) the highest in the state, not just the sum of AMC+AIME.  I don't know about any monetary prize, but I think you automatically qualify for USAMO.

that's true, you do automatically qualify. i don't know abt cash prizes tho..

It is unlikely you are highest in Virginia though given that TJ probably had multiple indices that were close to 300.

Does TJ select people from the state or country like a magnet school or is it just an average high school with a really good math program?

I believe TJ draws from students all over VA?
I'm not sure...

It's a county magnet school as far as I know, but leave it to one of the many AoPSers from TJ to really answer that question.

TJ is one of Virginia's "governor's schools" each of which draws students from a small number of surrounding counties. I'm fairly sure that in practice most of TJ is Fairfax county. The high school I went to was another such governor's school in central Virginia.

I got a plaque from WI for having highest AMC 10 B score

same for me last year, good for college i suppose

what score did u guys need to get? to get the plaque

That varies from state to state, if you're interested, AMC stats are up at their website, you can find out the top scores needed for each state there:
[url]http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2009-12a/09-1-1012-StateStats.shtml[/url]


Through an internal point $O$ of $\Delta ABC$ one draws 3 lines, parallel to each of the sides, intersecting in the points shown on the picture.

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=289[/img]

Find the value of $\frac{|AF|}{|AB|}+\frac{|BE|}{|BC|}+\frac{|CN|}{|CA|}$.

an easy nonrigorous way is just to note that the problem statement implies that the expression is constant for any choice of O, then choose something convenient.

You can [b]make[/b] it rigorous by showing that it's invariant...

OK // AC, and MN // BC $\Rightarrow$ ONKC is a paralellogram.
Thus, $CN$ = $OK$

$\angle EOK = \angle BAC, \angle OKE = \angle ACB \Rightarrow \Delta ABC$ is similar to $\Delta OEK$

\[ \Rightarrow \dfrac{CN}{CA}=\dfrac{OK}{CA}=\dfrac{EK}{BC} \]

Because AF // OD, and OF // AD, AFOD is a parallellogram.
Therefor, AF = OD.
Because $\angle ODN = \angle BAC$, and $\angle DON = \angle ABC, \Delta ODN$ is similar to $\Delta ABC$
\[ \Rightarrow \dfrac{AF}{AB}=\dfrac{OD}{AB}=\dfrac{ON}{BC}=\dfrac{KC}{BC} \]

\[ \dfrac{AF}{AB}+\dfrac{BE}{BC}+\dfrac{CN}{CA}=\dfrac{BE+EK+KC}{BC}=1 \]

Through an internal point $O$ of $\Delta ABC$ one draws $3$ lines, parallel to each of the sides, intersecting in the points shown on the picture.
[img]https://cdn.artofproblemsolving.com/attachments/e/3/03d4d1bb61eda8b4a72ff84466d700de47c147.png[/img]
Find the value of $\frac{|AF|}{|AB|}+\frac{|BE|}{|BC|}+\frac{|CN|}{|CA|}$.


inequalities
calculus
derivative
function
inequalities proposed

PVThuan is my friend in Viet Nam (actually, he teaches English in my school). The following problem is to him with my compliment... It's also for the other if you are interested in it.

Prove that if $a,b,c,d$ are non-negative real numbers and $a^2+b^2+c^2+d^2=4$ then
\[ \frac{1}{\sqrt{a^3+bcd}}+\frac{1}{\sqrt{b^3+cda}}+\frac{1}{\sqrt{c^3+dab}}+\frac{1}{\sqrt{d^3+abc}} \ge 2\sqrt{2}  \]
And a similar one is that
\[ \frac{1}{\sqrt{a^3+abc}}+\frac{1}{\sqrt{b^3+bcd}}+\frac{1}{\sqrt{c^3+cda}}+\frac{1}{\sqrt{d^3+dab}} \ge 2\sqrt{2}  \]
I want to imply a stronger one. But first try to prove the inequalities above...

[quote="Albanian Eagle"]for the second one we can use just AM-GM
 :)[/quote]

Yes... It's my first solution... But may you post a detail solution?

$\sum_{cyc} \frac{1}{\sqrt{a^3+abc}}\geq 4 \left(\frac{1}{abcd \prod_{cyc} (a^2+bc)}\right)^{\frac{1}{8}}$
then we use the fact that $8\geq \sum a^2+bc \geq 4\prod (a^2+bc) ^{\frac{1}{4}}$

Ok... It's easy. The first one can be solved by the same method, but fairly harder.

[quote="Vasc"]Using Jensen for the first, we need to prove a very nice inequality:

$a^3+b^3+c^3+d^3+abc+bcd+cda+dab \leq 8$

for $a^2+b^2+c^2+d^2=4$.
Can someone post a solution to this nice and hard inequality?[/quote]

maybe we can use derivative

WLOG let $a\geq b\geq c\geq d$

$f(a)=a^3+a(bc+bd+cd)+(b^3+c^3+d^3+bcd)$

$f'(a)=3a^2+(bc+bd+cd)>0$ for $0\leq a,b,c,d\leq 2$

so $f(a)$ is increasing . If we fix $bc+bd+cd=k$ then $4=a^2+b^2+c^2+d^2\geq a^2+(bc+bd+cd)=a^2+k$

so $a\leq \sqrt{4-k}$ . And since $f(a)$ is increasing , the maximum is when

$a=\sqrt{4-k}$ and $bc+bd+cd=k$ $\implies b=c=d=\sqrt{k/3}$ ($\because b^2+c^2+d^2=k$)

which means we only need to check the case for $a\geq b=c=d$

and we reduce the question into proving $a^3+3ab^2+4b^3\leq 8$ with constrain $a^2+3b^2=4$

or equivalent to $(a^3+3ab^2+4b^3)^2\leq (a^2+3b^2)^3$ $\Longleftrightarrow (a-b)^2b^2(3a^2-2ab+11b^2)\geq 0$

interesting to see that the equality case at $a=b=c=d=1$ or $a=2,b=c=d=0$(or its permutation)

We have
$f'(a)=3a^2+(bc+bd+cd)$
if $b,c,d$ are fixed. Then, $a^2+b^2+c^2+d^2=4$  implies $a$ fixed, and $f(a)$ is not function. :o

[quote="Vasc"]Using Jensen for the first, we need to prove a very nice inequality:

$a^3+b^3+c^3+d^3+abc+bcd+cda+dab \leq 8$

for $a^2+b^2+c^2+d^2=4$.
Can someone post a solution to this nice and hard inequality?[/quote]
$abcd=const$ and with $f(a)=a^3+\frac{M}{a}$ while using (n-1)EV
we see that the maximum ocurrs when 3 variables are equal
so we need to show that
$a^3+4b^3+3ab^2\le 8$ when $a^2+3b^2=4$ which is equivalent to $b^3+\sqrt{4-3b^2}\le 2$ which is true by AM-gM

[quote="Vasc"]Using Jensen for the first, we need to prove a very nice inequality:

$a^3+b^3+c^3+d^3+abc+bcd+cda+dab \leq 8$

for $a^2+b^2+c^2+d^2=4$.
Can someone post a solution to this nice and hard inequality?[/quote]

Yes, VASC. This is exactly the stronger inequality which I want to mention in this topic. It was appear in chapter I in my book, one exercises. My solution is using SOS method (and truly, it's nice but not hard).

I hope someone gives the solution.

[quote="Albanian Eagle"] $abcd=const$ and with $f(a)=a^3+\frac{M}{a}$ while using (n-1)EV
we see that the maximum ocurrs when 3 variables are equal [/quote] The function is $f(u)=u^{3/2}+M.u^{-1/2}$ if we replace $a^2,b^2,c^2,d^2$ by $a,b,c,d$, respectively. But
$g(x)=f'(1/x)$ is not convex, and then the method fails.

[quote="hungkhtn"] My solution is using SOS method (and truly, it's nice but not hard).
I hope someone gives the solution.[/quote]
To use SOS method, did you square the inequality or not ?

[quote="arqady"]If I am not wrong then
$a^3+b^3+c^3+d^3+abc+bcd+cda+dab \leq 8\Leftrightarrow$
$\Leftrightarrow(a^3+b^3+c^3+abc+bcd+cda+dab)^2 \le (a^2+b^2+c^2+d^2)^3\Leftrightarrow$
$\Leftrightarrow\sum_{sym}(a-b)^2(3c^4+c^3d+3c^2d^2+cd^3+3d^4+3a^2b^2-c^2ab-d^2ab)\geq0,$
which obviously true.[/quote]

Yes, Arqady. I only remember the coefficient all $S_{a,b},S_{a,c},...,S_{c,d}$ is non-negative, so the problem is easy.

Prom this result, We get a direct corollary
\[ (a+b+c+d)^6 \ge 2^8(1+a^2)(1+b^2)(1+c^2)(1+d^2) \]
for all $abcd=1$. I wait for a nice solution for this nice one.

[quote="hungkhtn"]Of course, my inital problem is
\[ (a^3+b^3+c^3+abc+bcd+cda+dab)^2 \le (a^2+b^2+c^2+d^2)^3  \]
But the reason why I post the first problem (in square-root form) is that L: This one has a very nice solution only by AM-GM.[/quote]
We choose $a^2+b^2+c^2+d^2=4$. The inequality has the form.
\[ 4(a+b)+(c^2-cd+d^2+ab)(c+d-a-b)\leq8. \]
I have tried to use AM-GM several ways but I failed: Grouping, supposing $a\geq b\geq c\geq d$.....

Can you give a hint?

[quote="hungkhtn"][quote="pvthuan"]From the condition, we have $a+b+c+d\leq4$. WLOG, we suppose that $a+b\geq c+d$. Then $a+b\leq2$. Thus,
\[ 4(a+b)+(c^2-cd+d^2+ab)(c+d-a-b)\leq8.  \]
We are done.

Is that right?[/quote]

No, why $a+b\le 2$?[/quote]

Yes, I was wrong.  :oops:

Only by $AM-GM$ inequality, we can make a very nice solution for
\[ 64(a^3+bcd)(b^3+cda)(c^3+dab)(d^3+abc) \le (a^2+b^2+c^2+d^2)^6 \]
I hope to see this one.

I wonder if a proof of the inequality shown in the preceding post is available elsewhere. If so, please give me a reference. I would have guessed that the inequality that was required to establish the initial inequality posed by hungkhtn would have required the exponent on the sum of the squares comprising the right hand side of the preceding inequality to be 5 , not 6. Unfortunately, the result is false when the 6 is replaced by a 5.


integration
trigonometry
combinatorics unsolved
combinatorics

$ {2}^{2n}\equal{}C(0,0).C(2n,n)\plus{}C(2,1).C(2(n\minus{}1),n\minus{}1)\plus{}C(4,2).C(2(n\minus{}2),n\minus{}2)\plus{}\cdots\plus{}C(2n,n).C(0,0)$, where $ n$ is a positive integer. Thanks in advance.

Okay there are two ways to do this - one is to consider \[ \int(\sin^2\theta \plus{}cos^2\theta)^n d\theta\] and look at it two ways. Another is to prove it combinatorially as follows:
Consider the number of paths on the integer line starting from zero of length 2n. The number of such paths is $ 2^{2n}$. We count this quantity in another way.
Claim: The number of paths that 'last cross' the origin after $ 2k$ moves is $ \binom{2k}{k}\binom{2n\minus{}2k}{n\minus{}k}$.
Proving the above claim relies on the task of finding a bijection between the number of paths of length $ 2k$ that end at $ 0$ and the number of paths of length $ 2k$ that never cross zero.
 Proving the last bit requires some work so I'll put it up after I work it out again. The general idea seems clear though.


logarithms
real analysis
real analysis unsolved

Eeew! Nonlinear, I do remember a technique that goes like this, though...

$1+y'^{2}=y\cdot y''$, Let $v=y'$ so $y''=\frac{dv}{dx}= \frac{dv}{dy}\cdot\frac{dy}{dx}=v\frac{dv}{dy}$ So this transforms the D.E. to

$1+v^{2}=yv\frac{dv}{dy}$ which is separable $\frac{dy}{y}=\frac{v}{1+v^{2}}dv$, so $\ln y=\frac{1}{2}\ln(1+v^{2})+C \Rightarrow Cy=(1+v^{2})^{1/2}\Rightarrow v=\sqrt{Cy^{2}-1}$, where $C$ isn't necessarily the same constant between all those steps.

Now sub back $\frac{dy}{dx}=\sqrt{Cy^{2}-1}$ which is separable: $\frac{dy}{\sqrt{Cy^{2}-1}}=dx$, which isn't all that nice, so I'll let you finish it.


Find all the roots of the equation $ x\,(3\minus{}x)^2\equal{}1.$

Actually, the solutions are $ 4\sin^2\frac{\pi}{18}.\ 4\sin^2\frac{5\pi}{18},\ 4\sin^2\frac{7\pi}{18}.$

Yeah, Mary Barton is right.

Sonnhard, did you read the problem wrong? Because I checked, and those numbers were not equivalent to Mary's. (They were complex...).

hello, i have also checked may solutions. For example on can show that
$ 4\sin\left(\frac{\pi}{18}\right)^2$ is
$ \left(\frac{1}{2}(\minus{}1\minus{}i\sqrt{3})(\frac{1}{2}(\minus{}1\plus{}i\sqrt{3}))^{\frac{1}{3}}\plus{}\frac{1}{2}(\frac{1}{2}(\minus{}1\minus{}i\sqrt{3}))^{\frac{1}{3}}(\minus{}1\plus{}i\sqrt{3})\right)^2$
Sonnhard.

Solutions for any answers would be appreciated  :) ...me wants to know  :maybe:

Is it just me, or do those messy solutions look like the cubic equation?

When in doubt, brute-force it. Then try to prove it nicely now that you have the solutions :D

For this problem, there is no need for brute force. There is actually an elegant solution that is no more than three steps long. See if you can find it. ;)

If we substitute $ x$ for $ y^2$ then $ x(3-x)^2 = (3y-y^3)$ looks almost like triple angle identity for sine. So let $ x=4\sin^2k$. Then we have $ 4(3\sinh - 4\sin^2k)^2$ = $ 4(\sin3k)^2$= $ 1$

So we have $ \sin3k = \frac{1}{2}$ => $ k=\frac{\pi}{18}, \frac{5\pi}{18}, \frac{7\pi}{18}$.

Yes, you’ve got the right idea. :-D

[hide="Full three-step solution"][b][u]Step 1[/u]:[/b] $ \sin{3\theta}\ \equal{} \ 3\sin{\theta} \minus{} 4\sin^3{\theta}\ \equal{} \ \sin{\theta}\left(3 \minus{} 4\sin^2{\theta}\right)$

[b][u]Step 2[/u]:[/b] $ 4\sin^2{3\theta}\ \equal{} \ 4\sin^2{\theta}\left(3 \minus{} 4\sin^2{\theta}\right)^2\ \equal{} \ x\,(3 \minus{} x)^2$ where $ x \equal{} 4\sin^2{\theta}$

[b][u]Step 3[/u]:[/b] Put $ \theta\ \equal{} \ \frac {\pi}{18},\ \frac {5\pi}{18},\ \frac {7\pi}{18}$ (these are all in the range $ \left[0,\tfrac{\pi}2\right]$ so their sines are all positive and distinct; hence the squares of their sines are all distinct)[/hide]

[quote="Dr Sonnhard Graubner"]hello, i have also checked may solutions. For example on can show that
$ 4\sin\left(\frac {\pi}{18}\right)^2$ is
$ \left(\frac {1}{2}( \minus{} 1 \minus{} i\sqrt {3})(\frac {1}{2}( \minus{} 1 \plus{} i\sqrt {3}))^{\frac {1}{3}} \plus{} \frac {1}{2}(\frac {1}{2}( \minus{} 1 \minus{} i\sqrt {3}))^{\frac {1}{3}}( \minus{} 1 \plus{} i\sqrt {3})\right)^2$
Sonnhard.[/quote]

How did you get this expression?
By the way, if we can't memorize the identity $ \sin{3\theta}\ \equal{} \ 3\sin{\theta} \minus{} 4\sin^3{\theta}\ \equal{} \ \sin{\theta}\left(3 \minus{} 4\sin^2{\theta}\right)$, does it mean we can't solve this problem?  :roll:

you can derive it from the sine and cosine addition formulas (i think thats what they are called)

An easier way to derive it is expanding $ \cos{3x}\plus{}i\sin{3x}\equal{}(\cos{x}\plus{}i\sin{x})^3$ and equate the real parts.

I know how to derive the identity $ \sin{3\theta}\ \equal{} \ 3\sin{\theta} \minus{} 4\sin^3{\theta}$, but what if we couldn't recognize it in this problem (I don't like to memorize formulas!)?  :( :D


I wrote this out not too long ago.. Inspiration came up while doing English paper ( :rotfl: ). I'm not quite sure if this problem is solvable but there is an answer that works so.. hopefully, it is valid.  :D

No Calculator.

Triangle $ ABC$ has $ \angle ACB \equal{} 90^\circ$ and $ AB \equal{} \sqrt{21}$. If $ BC \equal{} 4$ and $ \angle ABC \equal{} \theta,$ then $ \tan \theta \equal{} \sin 2 \alpha \cos \alpha$ where $ 0^\circ < \alpha < 60^\circ$. Find the measure of $ \alpha$.

[quote="Silverfalcon"]I wrote this out not too long ago.. Inspiration came up while doing English paper ( :rotfl: ). I'm not quite sure if this problem is solvable but there is an answer that works so.. hopefully, it is valid.  :D

No Calculator.

Triangle $ ABC$ has $ \angle ACB \equal{} 90^\circ$ and $ AB \equal{} \sqrt {21}$. If $ BC \equal{} 4$ and $ \angle ABC \equal{} \theta,$ then $ \tan \theta \equal{} \sin 2 \alpha \cos \alpha$ where $ 0^\circ < \alpha < 60^\circ$. Find the measure of $ \alpha$.[/quote]

[hide]
$ \tan{\theta}\equal{}\frac{\sqrt{5}}{4}$ by pythag
then the rhs of equation 2 is $ 2(\sin{\alpha}\minus{}\sin^3{\alpha})$
call $ \sin{\alpha}\equal{}x$
$ x^3\minus{}x\plus{}\frac{\sqrt{5}}{8}\equal{}0$

this is a depressed cubic
unfortunately i dont know how to solve those
[/hide]

[quote="Altheman"]i'm getting 18 degrees[/quote]

That's correct.

[hide="Way I Created It"]
I was goofing around and realize how similar that $ \sin 36$ and $ \cos 18$ looked.

$ \sin 36^\circ \equal{} \frac{\sqrt{5 \minus{} \sqrt{5}}}{2 \sqrt{2}}$

$ \cos 18^\circ \equal{} \frac{\sqrt{5 \plus{} \sqrt{5}}}{2 \sqrt{2}}$

Multiplying those two simplifies to $ \frac{\sqrt{5}}{4}$.  But, working backwards, I got the cubic that was mentioned earlier and did not know how to proceed either. Knowing these numbers help but I don't know what's other way. [/hide]

Can you show how?


Hi every one. Any one here know any thing about maple?

What is the maple command for writing for example

2*X  congruent   1 mod17   i.e   2*x=1mod17
I searched the net but didn't find any thing

First of all , maybe this should have been better placed in superior algebra
but then again, there isn't really any official for maple questions( i once tried in the latex help forum but got no response)

what exactly do you want
do you actually mean you just want to write it down as text ?

then i suggest you put this between dollar signs : 2^x \equiv 1\bmod{17} :
$2^x \equiv 1\bmod{17}$

or you could type this in maple

2^x=1 mod 17

you'll have to elaborate I guess

It's nice to see someone from afghanistan here (are there others?)
http://en.wikipedia.org/wiki/Afghanistan
why is your flag different from the flag here?

$2x \equiv 1\bmod{17}$

yes thats exactly what i want. I want the command equivalent to that in maple. and then ask maple to find $x$

My county's flag has changed a bit every now and then in the past but now its permanent. The administrator of this site should change Afghanistans flag to the current one.

actually in that case I suggest you type

msolve(2*x=1,17);

or am I not understand your question correctly?


function
integration
inequalities
real analysis
real analysis unsolved

Let $f: [a,b]\rightarrow R$ be an integrable function having the property that

$0\leq f(x)\leq \int_{a}^{x}f(t)dt$ for all $x\in [a,b]$.

Prove that $f=0$.

This is indeed a [url=http://en.wikipedia.org/wiki/Gr%C3%B6nwall%27s_inequality]known result[/url].


trigonometry
algebra
polynomial
function
vector
calculus
calculus computations

Find the general solutions of
(i) $ y'' \plus{} 9y \equal{} \cos 3x$
(ii) $ y'' \minus{} 2y' \plus{} y \equal{} (x \minus{} 1)e^x$

In general how do we solve for inhomogeneous ODEs when the particular solution isn't easy to guess?

You don't ;)  But in these cases it actually isn't so hard: for the second one, you want to guess something like "polynomial times $ e^x$" where "polynomial" is of degree high enough that it isn't part of the homogeneous solution.  For the first, you want to guess trigonometric function, but it turns out that the natural guesses are in the homogeneous solution.  In this case, you do the same thing: replace your guess of "trigonometric thing" with "polynomial times trigonometric thing" where "polynomial" is of the minimal degree necessary to kick you out of the homogeneous solution.

See also Kent's recent posts here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=311377

I see - I had problems guessing for the first function but then I plugged it into WolframAlpha. Thanks for the link - looks enlightening!

Also, I wanted to ask about the solution space determined by solution vectors $ {y_1 \choose y_1'}$ and $ {y_2 \choose y_2'}$. What is the meaning of "solution space/solution vector" and why are all particular solutions of linear, second-order inhomogeneous ODEs of the form $ y_p \equal{} u(x)y_1 \plus{} v(x)y_2$ where $ y_1$ and $ y_2$ are homogeneous solutions?

hello, for (i) i have got
$ y(x)\equal{}C_1\cos(3x)\plus{}C_2\sin(3x)\plus{}\frac{1}{18}\cos(3x)\plus{}\frac{1}{6}x\sin(3x)$.
Sonnhard.

I think we can combine the $ \frac{1}{18} \cos 3x$ into the $ C_1 \cos 3x$? Otherwise I get the same answer  :)


Twelve square tiles, one inch on a side, are arranged in a rectangle,
without overlapping. What is the number of inches in the sum of all
possible distinct perimeters?

We can have

1,12
2,6
3,4

The distinct perimeters are $ 26\plus{}16\plus{}14\equal{}\boxed{56}$.


If $b^2-4ac \leq 0$, prove that $ax^2+bx+c \geq 0$ for all reals $x$.
What can you tell if $b^2-4ac \geq 0$?

i dont think that first statement is necessarily correct. if a>0 then it would be, i believe.

Arkmastermind is correct.  If the discriminant is less than 0, there are no roots to the equation, thus the graph of the polynomial will never touch the x-axis.  So, depending on a's sign, $ax^2+bx+c$ will always either be greater or less than 0.  If the discrimant is 0, the graph will touch the x-axis at one point, so the value will always be greater or less than or equal to 0.

If the discriminant is positive or zero, then we know $ax^2+bx+c$ is greater than/less than (depending on a's sign) or equal to $\frac{-b}{2a}$

Oops... I forgot the condition of $a$. :blush:


What common fraction is $ 50\%$ less than $ \frac{5}{8}$?


limit
trigonometry
calculus
calculus computations

$ \lim\limits_{x \to 0} \frac {\sin (a \plus{} 2x) \minus{} 2 \sin (a \plus{} x) \plus{} \sin a}{x^2} \equal{} \lim\limits_{x \to 0} \frac {\sin a \cos 2x \plus{} \sin 2x \cos a \minus{} 2\sin a \cos x \minus{} 2 \cos a \sin x \plus{} \sin a}{x^2} \equal{} \lim\limits_{x \to 0} \frac { 2\cos a \sin x (\cos x \minus{} 1)}{x^2} \equal{} 0$
But the true  solution: $ \minus{} \sin a$

Well, The factored form ($ 2\cos a \sin x \ldots$) doesn't seem to be correct. Here's my factoring:

$ \sin(a \plus{} 2x) \minus{} 2\sin(a \plus{} x) \plus{} \sin a \equal{} \sin(a \plus{} 2x) \minus{} \sin(a \plus{} x) \plus{} \sin a \minus{} \sin(a \plus{} x) \equal{}$
$ 2\cos\left(a \plus{} \frac {3x}{2}\right)\sin\left(\frac {x}{2}\right) \minus{} 2\cos\left(a \plus{} \frac {x}{2}\right)\sin\left(\frac {x}{2}\right) \equal{}$
$ 2\sin\left(\frac {x}{2}\right)\left(\cos\left(a \plus{} \frac {3x}{2}\right) \minus{} \cos\left(a \plus{} \frac {x}{2}\right)\right) \equal{}$
$ 2\sin\left(\frac {x}{2}\right)\left( \minus{} 2\sin\left(\frac {x}{2}\right)\sin(a \plus{} x)\right) \equal{}$
$ \minus{} 4\sin^2\left(\frac {x}{2}\right)\sin(a \plus{} x)$

which gives your limit $ \lim_{x\to 0} \frac { \minus{} \sin^2\left(\frac {x}{2}\right)\sin(a \plus{} x)}{\left(\frac {x}{2}\right)^2} \equal{} \minus{} \sin a$


calculus
induction
number theory unsolved
number theory

The sequence  $ a_1\equal{}20$,$ a_2\equal{}30$, $ a_{n\plus{}1}\equal{}3a_n\minus{}a_{n\minus{}1}$ .
 Find all n so that $ 1\plus{}5a_na_{n\plus{}1}$ is a perfect square.

wonderful
Thanhs you very much :lol:  :roll:

[quote="pco"][quote="shortlist"]The sequence  $ a_1 \equal{} 20$,$ a_2 \equal{} 30$, $ a_{n \plus{} 1} \equal{} 3a_n \minus{} a_{n \minus{} 1}$ .
 Find all n so that $ 1 \plus{} 5a_na_{n \plus{} 1}$ is a perfect square.[/quote]

I'm sure there is a smarter method, but here is one :

Direct resolution gives $ a_n \equal{} (15 \minus{} 5\sqrt 5)\left(\frac {3 \plus{} \sqrt 5}2\right)^n \plus{}$ $ (15 \plus{} 5\sqrt 5)\left(\frac {3 \minus{} \sqrt 5}2\right)^n$

And then (after some calculus) $ 1 \plus{} 5a_na_{n \plus{} 1} \equal{} 501 \plus{} u_n^2$ where $ u_n \equal{} (25 \minus{} 5\sqrt 5)\left(\frac {3 \plus{} \sqrt 5}2\right)^n \plus{} (25 \plus{} 5\sqrt 5)\left(\frac {3 \minus{} \sqrt 5}2\right)^n$

We have $ u_1 \equal{} 50$ and $ u_2 \equal{} 100 > u_1$ and $ u_{n \plus{} 2} \equal{} 3u_{n \plus{} 1} \minus{} u_n$ and so $ u_n$ is a strictly increasing sequence made of positive integers

$ u_1 \equal{} 50$ and $ 1 \plus{} 5a_1a_{2} \equal{} 501 \plus{} u_1^2 \equal{} 3001$ is not a perfect square
$ u_2 \equal{} 100$ and $ 1 \plus{} 5a_2a_{3} \equal{} 501 \plus{} u_2^2 \equal{} 10501$ is not a perfect square
$ u_3 \equal{} 250$ and $ 1 \plus{} 5a_3a_{4} \equal{} 501 \plus{} u_3^2 \equal{} 63001 \equal{} 251^2$ is a perfect square
$ \forall n > 3$ : $ u_n > u_3 \equal{} 250$ and so $ u_n^2 < 501 \plus{} u_n^2 < (u_n \plus{} 1)^2$ and $ 501 \plus{} u_n^2$ can not be a perfect square.

Hence the answer : $ \boxed{n \equal{} 3}$[/quote]
I've read your post with much interest. I'm especially interested in how you came up with the idea of letting $ 1 \plus{} 5a_na_{n \plus{} 1} \equal{} 501 \plus{} u_n^2$. That was very nice!
I've found a (in my opinion) "nicer" formula for $ a_n$: $ a_n \equal{} 10 \left ( \left(\frac {3 \plus{} \sqrt 5}2\right)^{n\minus{}1} \plus{} \left(\frac {3 \minus{} \sqrt 5}2\right)^{n\minus{}1} \right )$
One can see that $ u_n \equal{} a_{n \plus{} 1} \plus{} a_{n}$, and then $ 1 \plus{} 5a_na_{n \plus{} 1} \equal{} 501 \plus{} (a_n \plus{} a_{n \plus{} 1})^2 \iff a_{n \plus{} 1}^2 \minus{} 3a_na_{n \plus{} 1} \plus{} a_{n}^2 \equal{} \minus{} 500$.
This is very interesting as it looks like $ x^2 \minus{} 3x \plus{} 1 \equal{} 0$! And one can actually quite easily see (i did it with induction) that if $ c_{n \plus{} 2} \equal{} ac_{n \plus{} 1} \plus{} bc_n$ then $ c_{n \plus{} 2}^2 \minus{} ac_{n \plus{} 2}c_{n \plus{} 1} \minus{} bc_{n \plus{} 1}^2 \equal{} ( \minus{} b)^n (c_{2}^2 \minus{} ac_2c_1 \minus{} bc_1^2)$.
(In this case $ b \equal{} \minus{} 1$ so $ a_{n \plus{} 1}^2 \minus{} 3a_{n \plus{} 1}a_n \plus{} a_n^2$ is constant)

Can I ask what motivation you had for writing $ 1 \plus{} 5a_na_{n \plus{} 1} \equal{} 501 \plus{} u_n^2$ ? It was not obvious (for me at least), and i used some time to prove it, so I'm curious if you just played around with the numbers, or if you had it in mind from the start.

Just as a reference - this is the second problem of Balkan Mathematical Olympiad, 2002.

[quote="Mathias_DK"] ...Can I ask what motivation you had for writing $ 1 \plus{} 5a_na_{n \plus{} 1} \equal{} 501 \plus{} u_n^2$ ? It was not obvious (for me at least), and i used some time to prove it, so I'm curious if you just played around with the numbers, or if you had it in mind from the start.[/quote]

No  :blush:  , I had nothing in mind. Just computed $ 1 \plus{} a_na_{n \plus{} 1}$ and saw that this quantity was $ 1501 \plus{} 25(5 \minus{} \sqrt 5)^2r_1^{2n} \plus{} 25(5 \minus{} \sqrt 5)^2r_2^{2n}$ (where $ r_1$ and $ r_2$ are the roots of $ x^2 \minus{} 3x \plus{} 1 \equal{} 0$)

And then I saw that this expression could be written as a square + something, hence the result.

And, btw, I think your simpler formula for $ a_n$ is wrong : it gives $ a_1 \equal{} 30$ instead of $ 20$

[quote="pco"]And, btw, I think your simpler formula for $ a_n$ is wrong : it gives $ a_1 \equal{} 30$ instead of $ 20$[/quote]
Thanks for correcting me! I wrote $ n$ instead of $ n\minus{}1$ by mistake.


search
inequalities unsolved
inequalities

$ a,b,c > 0$, $ a\plus{}b\plus{}c\equal{}3$ then prove:

$ 4(a^2\plus{}b^2\plus{}c^2) \plus{}abc \ge 13$

[hide="HINT"]MV[/hide]

[quote="Erken"]APMO of some year.Use search function. :wink:[/quote]

Ok, I changed the problem.

[hide="Solution"]$ \sigma_1\equal{}\minus{}(a\plus{}b\plus{}c) \\ \sigma_2\equal{}ab\plus{}bc\plus{}ca \\ \sigma_3\equal{}\minus{}abc \\ 4(a^2\plus{}b^2\plus{}c^2) \plus{}abc \ge 13 \iff 8 \sigma_2 \plus{} \sigma_3 \le 23$

But:
$ P(t)\equal{}(t\minus{}a)(t\minus{}b)(t\minus{}c)\equal{}t^3\plus{} \sigma_1 t^2 \plus{} \sigma_2 t \plus{} \sigma_3\equal{}t^3 \minus{}3t^2 \plus{} \sigma_2 t \plus{} \sigma_3$

Let $ t\equal{}8$:
$ P(8)\equal{}8^3 \minus{} 3 \cdot 8^2\plus{}8 \sigma_2 \plus{} \sigma_3\equal{}320\plus{}8\sigma_2\plus{}\sigma_3\equal{}(8\minus{}a)(8\minus{}b)(8\minus{}c) \le (\frac{(8\minus{}a)\plus{}(8\minus{}b)\plus{}(8\minus{}c)}{3})^3\equal{}343$

So:
$ 8 \sigma_2 \plus{} \sigma_3 \le 23$[/hide]

My first post :)

[quote="FOURRIER"]$ a,b,c > 0$, $ a \plus{} b \plus{} c \equal{} 3$ then prove:

$ 4(a^2 \plus{} b^2 \plus{} c^2) \plus{} abc \ge 13$

[hide="HINT"]MV[/hide][/quote]
$ 4(a^2 \plus{} b^2 \plus{} c^2) \plus{} abc \ge 13$
$ \Longleftrightarrow 36(a^2\plus{}b^2\plus{}c^2)(a\plus{}b\plus{}c)\plus{}27abc \geq 13(a\plus{}b\plus{}c)^3$
$ \Longleftrightarrow 23(a^3\plus{}b^3\plus{}c^3) \geq 3[a^2(b\plus{}c)\plus{}b^2(c\plus{}a)\plus{}c^2(a\plus{}b)]\plus{}51abc$
It's right because :
$ 17(a^3\plus{}b^3\plus{}c^3) \geq 51abc$
and $ 6(a^3\plus{}b^3\plus{}c^3) \geq 3[a^2(b\plus{}c)\plus{}b^2(c\plus{}a)\plus{}c^2(a\plus{}b)]$

[b]lemma:[/b]for $ x,y,z>0$ and $ x\plus{}y\plus{}z\equal{}3$ we have $ x^2\plus{}y^2\plus{}z^2\plus{}xyz\geq 4$.

now according to this lemma,its sufficient to show that:

$ x^2\plus{}y^2\plus{}z^2\geq 3$

which is true since:

$ x^2\plus{}y^2\plus{}z^2\geq \frac{(x\plus{}y\plus{}z)^2}3\equal{}3$

we can write A=4[(a+b+c)²-2(ab+bc+ac)]+abc;
 we have A=36-8(ab+bc+ac)+abc

I think it is easier to prove!!


Prove that any quadrilateral can be divided into at least 8 isosceles triangles.

Is this from WOOT? (and also, do you want at least 8 or for all greater than 8)?

As a hint, trying experimenting with splitting an arbitrary triangle into 4 isosceles triangles.

I think its from woot (I don't do woot but I was given the problem by a friend). It says 8 or more.
GOT IT!!!! zomg this is the first problem at this level that I solved.
[hide]
This is sorta an algorithmic process.
We know that any quadrilateral can be split into 2 triangles. Either of these triangles can be made into 4 isoceles triangles by drawing the altitude from one of the vertices, creating 2 right triangles which can be turned into 2 isoceles triangles by drawing the median to the hypotenuse. That's for 8.

Now how do we get more than 8?

We can see that there are a few things we can do with modifying the steps we perform before and after that can change the isoceles triangle count.

If we take the original quadrilateral and draw in a line segment somewhere so that it creates another isoceles triangle, we have an isoceles triangle and a new quadrilateral left (which we can split by the previous method into 8 or more). This means we can add however many multiples of 1 that we want, and that solves the problem.
This is assuming however, that no sides are congruent. Special cases:
2 sides congruent: easy to find a way to draw quadrilateral so chopping off an isoceles triangle will yield a quadrilateral instead of triangle...find a different side to cut.
3 sides congruent: easy also, if last side is longer than the other 3, chop of isoceles triangle with congruent sides that are the same length as the 3 congruent sides. If shorter, ''''''''''''''''''' that are the same length as the non congruent side.
4 sides: It is a rhombus, very easy to make 8 with. It might be a little annoying with the fact that its already got 2 isoceles triangles. That doesn't matter though, because if its not 60-120-60-120, we can draw an isoceles triangle with congruent sides NOT congruent to the sides of the rhombus, by drawing in a segment from one of the vertices with the greater angle to one of the opposite segments so it forms an isoceles triangle with a totally new side length, and a new irregular quadrilateral. If it is a square, its really too easy because any triangle cut from segments drawn to existing vertices will be a 45-45-90 and when you split those in half you add 1 to the count. If it is the 60 120 60 120, 8 can be made obviously. 10 or more can be made by pinching off a smaller equilateral triangle (side length must be irrational when compared to rhombus side length, go play with drawings to see why) from the rhombus to create a trapezoid, and 2 equilateral triangles.
The trapezoid can now be split by the original method into 8 or more, which added to the 2 equilateral triangles makes 10 or more. For 9, we can specifically pick the pinched off equilateral so that its side length is 3/5 the rhombus' side length. If you do some drawing you will see how it can be made into 7 small equilateral triangles.

[/hide]
I know its not THAT clear because I didn't draw in those fancy figures however you do that...but this method works for all quadrilaterals so whatever :P


calculus
integration
trigonometry
calculus computations

Let $ f(x) \equal{} cosx \minus{} \int^x_0f(t)f(x \minus{} t)dt$
Find $ f(x) \plus{} f^"(x)$

I think your problem is :

Let $ f(x)\equal{}\cos x\minus{}\int_0^x (x\minus{}t)f(t)\ dt$.

Find $ f(x)\plus{}f''(x)$.

[hide="Answer"]-\cos x[/hide]


Let M be a compact metric space. Show that two point a,b $ \in$ M belong to distinct connected components of M $ \Longleftrightarrow$ exists a split      M = A $ \cup$ B, with a $ \cup$ A and b $ \cup$ B. Then, conclude that in a metric space, the connected component of a point is a intersection of open-close subsets that have one point.

Use the fact that if two connected subsets intersect then their union is connected.


Given $ a, b, c \geq\ 0$. Prove that:
$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}  \geq\ \frac{a^2}{a^2\plus{}2bc} \plus{} \frac{b^2}{b^2\plus{}2ca} \plus{} \frac{c^2}{c^2\plus{}2ab}$

[quote="nguoivn"]Given $ a, b, c \geq\ 0$. Prove that:
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \geq\ \frac {a^2}{a^2 \plus{} 2bc} \plus{} \frac {b^2}{b^2 \plus{} 2ca} \plus{} \frac {c^2}{c^2 \plus{} 2ab}$[/quote]
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \geq\sum_{cyc}\frac {a^2}{a^2 \plus{} 2bc}\Leftrightarrow\sum_{cyc}\left(\frac{a^2}{ab\plus{}ac\plus{}bc}\minus{}\frac {a^2}{a^2 \plus{} 2bc}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{a^2(a\minus{}b)(a\minus{}c)}{a^2\plus{}2bc}\geq0.$
Let $ a\geq b\geq c.$ Then $ \frac{a^2}{a^2\plus{}2bc}\geq\frac{b^2}{b^2\plus{}2ac}$ and $ a\minus{}c\geq b\minus{}c.$
Thus, $ \sum_{cyc}\frac{a^2(a\minus{}b)(a\minus{}c)}{a^2\plus{}2bc}\geq(a\minus{}b)(b\minus{}c)\left(\frac{a^2}{a^2\plus{}2bc}\minus{}\frac{b^2}{b^2\plus{}2ac}\right)\plus{}$
$ \plus{}\frac{c^2(c\minus{}a)(c\minus{}b)}{c^2\plus{}2ab}\geq0.$


Jane and John, with masses of 50 kg and 60 kg, respectively, stand on a frictionless surface 10 m apart.  John pulls on a rope that connects him to Jane, giving Jane an acceleration of 0.92 m/s² toward him.  (a) What is John's acceleration? (b) If the pulling force is applied constantly, where will Jane and John meet?

My classmate said that $ m_{John} a_{John} \equal{} m_{Jane} a_{Jane}$, but I don't really see why that's true.  Shouldn't the acceleration of John be $ (F\minus{}T)/a$ where F is the force he pulls Jane and T the tension?  And $ T \equal{} m_{Jane} a$ where $ a$ is the same for both Jane and John?
I just don't understand the concept of tension well; could anyone explain it to me?  Thanks.

[quote="mathwizarddude"]My classmate said that $ m_{John} a_{John} = m_{Jane} a_{Jane}$, but I don't really see why that's true.  Shouldn't the acceleration of John be $ (F - T)/a$ where F is the force he pulls Jane and T the tension?  And $ T = m_{Jane} a$ where $ a$ is the same for both Jane and John?
.[/quote]
u r cent pc correct :)
just tell where's the problem?? :)

[quote="mathwizarddude"]My classmate said that $ m_{John} a_{John} \equal{} m_{Jane} a_{Jane}$, but I don't really see why that's true.[/quote]

Newton's 3rd Law

[quote]Shouldn't the acceleration of John be $ (F \minus{} T)/a$ where F is the force he pulls Jane and T the tension?[/quote]

Typo; you probably mean $ (F\minus{}T)/m$ ;)

[quote]And $ T \equal{} m_{Jane} a$ where $ a$ is the same for both Jane and John?[/quote]

I think so. I don't understand it very well either though :(


AMC
AIME
modular arithmetic
geometry
3D geometry
number theory

Here's the link if the way it copied out is confusing: http://www.artofproblemsolving.com/Wiki/index.php/1989_AIME_Problems/Problem_9

The problem is:

One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that 133^5+110^5+84^5+27^5=n^{5}. Find the value of n.

The solution is:
By Fermat's Little Theorem, we know {n^{5}} is congruent to n modulo 5. Hence,
3 + 0 + 4 + 7 \equiv n\pmod{5}
4 \equiv n\pmod{5}

Continuing, we examine the equation modulo 3,
-1 + 1 + 0 + 0 \equiv n\pmod{3}
0 \equiv n\pmod{3}

Thus, n is divisible by three and leaves a remainder of four when divided by 5. It's obvious that n>133, so the only possibilities are n = 144 or n = 174. It quickly becomes apparent that 174 is much too large, so n must be 144.

A couple issues:

1. Where did modulo 3 come from? Ok, modulo 5 came from the theorem and the equation being in the 5th power, but where did 3 come from?

2. Barring that, they said the only possibilities are 144 and 174. What about 159, 189, 204, etc?

1) mod 3 is good when dealing with squares, cubes, etc.

2) $ 133^5 \plus{} 110^5 \plus{} 84^5 \plus{} 27^5 \equal{} n^5 < 133^5 * 4$ so $ n < 175.494...$

and 159 is out of the question because n must be even.

Thanks, your answer to my second question made perfect sense.

However, about the first question, does that mean mod 3 is good for all powers? Is there a theorem about this?

No, not really. Mod 4 is also really good when dealing with squares(since the residues are just 0 and 1); mod 8 too. But in this problem, I think we could've used any mods, except mod 3 was more convenient because 2 of the 4 numbers on the LHS were divisible by 3. But try using 2 other mods to see if it still works, like 4 and 7.


limit
real analysis
real analysis unsolved

Let $ \sum_{n\equal{}0}^{\infty} a_n z^n$ with radius of convergence $ r>0$ and $ a_n\geq 0$ non increasing, and $ \lim_{\infty} a_n \equal{}0$.
Clearly if $ s>r$, $ (a_n s^n)_{n \geq 0}$ cannot be bounded, but I cannot show that $ \lim_{\infty} a_n s^n \equal{} \plus{}\infty$, and I suspect that it is not always the case, but I cannot prove that either.
Any idea?

It is indeed not always the case. That monotonicity makes it tricky, but here's an example:

$ a_n \equal{} 2^{ \minus{} 2^k}$ for $ 2^{k \minus{} 1} < n\le 2^k$. The values $ a_n^{ \minus{} 1/n}$ range from $ 2$ to $ 4$ in each block, so the radius of convergence is 2 but there are infinitely many $ n$ for which $ a_n\cdot 3^n < 1$.

Thanks jmerry,
Is there some (general) conditions we could add to $ a_n$ so that the result $ \lim_{\infty} a_n s^n \equal{} \plus{} \infty$ becomes true?

I thought about decreasing instead of non increasing but I'm afraid it is not going to work either.

No, making it strictly monotone wouldn't help; just take the antiderivative of the example I just used.

There's nothing there except to essentially assume the result.


Since school is starting soon for most ppl, how about we post tips to make school easier?

 :?:

Relax, and don't do your homework last minute (commonly called procrastination).

relax, chill out, don't ever take school seriously

watch out for classes you choose. don't take any that you know has a reputation of doing unneccesary amounts of work or just bs work in general. if you're in high school, there shouldn't ever be a reason to have to do homework at home. occasionally you might not feel like doing an essay at school and decide to do it at home, but that's at your own discretion. most school hw cna be done in 10 minutes or so and you can easily do it in hte period directly before the class that it's due in. this way, you cna cycle all your homework so you're doign it all a period before it's due. and what do youget? no homework, easy A's, easy high school.

@ Elemennop - Commonly known as procrastination...

Don't get stressed out about stuff, just take it easy but never leave work unfinished until the last minute. Once you start procrastinating, it's hard to get back into a normal cycle. If you're getting too much homework (judge by the normal amount, not by [i]your[/i] idea of a normal amount), then try to figure out how to do it faster instead of just wasting all your time on schoolwork and having no time left for other more important things.

[quote="i_like_pie"]@ Elemennop - Commonly known as procrastination...
[/quote]

whatever yo uwant to call it, it's still an extremely good method for getting schoolwork done and i don't see why people act as if it's a bad thing to do homework not at home.

What Elemenop is suggesting - do all of your homework at school and in class, and never any of it at home - may well work in high school. But habits can be hard to break in the long run, and you need to remember that it won't work in college.

It's reasonably common for a college class to cover in one semester, in 45 to 60 class hours, a subject that would have taken a whole year and 160 or more class hours in high school. No one's going to give you any in-class time for finishing your homework. And we start expecting more from you than you high school teachers did - more depth, more independence. Yes, you might actually have to read the textbook yourself.

-Prioritize. If you have a lot of homework and there is no way you can get it done that night (sports, etc.) then put do the toughest homework and leave the easier ones for that following school day (period before it is due).

-Know your teachers a bit. Obviously, doing your next period's homework needs a bit of skill and a teacher that is amazing at spotting things behind his back will be a problem to your procrastinating needs. It also helps having friends in higher grades so they can warn you about teachers that may be mean.

Basically what elemmenop said, plus find out the workload before you sign up for any given class, (why take AP english in 12th grade...is it really gonna get you into Harvard??)

Here's what I'm doing: Find out what books you're going to be reading in English class this year, and then read as many as you can during the last few long days of summer vacation. This way, you won't have to spend as much time reading class books during the school year, and it will reduce the overall stress level.

[quote="aanjohri"] Do what you know will please your teachers, getting on their good side is important.[/quote]

Yeah umm anybody who's hung out with me enough knows that I've learned this lesson the very hard way...

[quote="aanjohri"]Taking really good notes and reviewing them  10 minutes every day is a really good idea.

Staying orgainized is important too. Do you hw and then play.

 Do what you know will please your teachers, getting on their good side is important.

Prepare for pop quizes.[/quote]

uh...why are u answering yr own question?

-dont cram the night b4 on test
-get to become friends w/ yr teacher. it can really be good help
-DONT BE AFRAID TO ASK FOR HELP

Just do things as early as possible, NOT quickly.

Look at your notes everyday for maybe 15 minutes. For a few days, it will start to stick in your head. Cramming the night before won't get you an A. Only as high as a B. I learned that the hard way.

[quote="Bicameral"]Cramming the night before won't get you an A. Only as high as a B.[/quote]

i don't really care what point you're trying to make, but lying isn't going to help you get it across.

Procrastination is fun!

You should take notes, save them, and study them.

Try to enjoy the class, the teacher usually notices it (Note: You may get more work for this)

Ask for help, it will never do anything wrong, except if you ask the mafia.

Most of these things I never really did but procrastinate, and still got all As.

[quote="The Zuton Force"]But how many people can do that?[/quote]
I used to do that before, and now I can do homework very quickly, so your reasoning is absolutely correct.

I never thought of it as procrastination, however, because usually that refers to the "I'll do it later" category.

[quote="mz94"]
uh...why are u answering yr own question?
[/quote]

i dunno, i had some ideas, so i posted them

-Procrastinate. Rush in everything at the last second, you'll still get a perfect on that test.
-Don't study. This stresses you out for tests and lowers your grade. Only look something up if you're sure you can't remember and that it will be on the test. Then repeat that twenty times to yourself and close the book.
-Don't get expelled.

This is only half serious, but i do endorse all three of these... well maybe not the first.

[quote="The Zuton Force"]Since most of what I have to say has already been said, here's something else to think about.

I respectfully disagree with those against procrastination.

While there's the bad kind of procrastination (I'm too lazy to do it now, I'll do it "later"), controlled procrastination is actually very useful. For example, if you know you can easily finish some really stupid assignment in 45 minutes (especially writing), start it with 40 minutes of time left. Now why would someone do that?

It develops a lot of skills that aren't developed by blowing it off on time:
1. [b]It teaches you to be more creative [/b]- you need to pull out the same creativity and ideas under time pressure
2. [b]It improves your English[/b] - by being forced to express your ideas into clear sentences under time pressure, your English communication skills rise dramatically
3. [b]It improves your fast decision making[/b] - so there are two kinds of fast decision making: intelligent and rash. When you're writing under pressure, you have to make fast, intelligent decisions. Over time, this will improve the quality of your fast decisions to make them less rash.
4. [b]It teaches you to be more focused[/b] - because of the time pressure and the fact that the assignment will be graded, you are forced to be more focused - a valuable skill
5. Finally, [b]in the long run, it saves you time[/b] because you complete your work in a compact timeframe with no distractions. This is certainly faster than doing homework while on AIM or watching TV (or AoPSing! xD)

Now if you can force yourself to do the assignment like that [b]without[/b] saving it until teh last minute, then that is, of course, much better. But how many people can do that?[/quote]

I think procrastination isn't that bad, as long as you can handle heavy work on some days. I did it last year, and got all A's.  :P

equation: let a-z be 1-26 respectively.

procrastination = 16+18+15+3+18+1+19+20+9+14+1+20+9+15+14=192. Procrastination>hard work=98>knowledge=96.

Therefore, procrastination is bigger than knowledge.

Do you see the point that I'm trying to make here?

[hide="if you don't,"]procrastination is obviously less than knowledge.[/hide]

Shouldn't you multiply those...isn't this a math forum...sometimes i'm not so sure

Ummm.... did a moderator move this thread to games and fun factory or did i actually post this here and am completely clueless?


If it is 12 o clock (noon) and we need to go to a fuction that starts at 7, but need to arrive there 10 minutes early and have to go to the library before (which is 20 minutes from your house and 10 minutes from the fuction) and you need 12 minutes at the library, what time should you leave home?

[hide]7.00pm-10minutes=6.50pm
6.50-(20minutes+10minutes)=6.20pm
6.20pm-12minutes=6.08pm.
we should leave at 6.08pm.[/hide]

[hide]6:08, cool question, im kinda noob but this is ok for me lol[/hide]

[hide]$10 minutes+ 20 minutes + 10 minutes+ 12 minutes= 52 minutes$
$7: 00 pm - 52 minutes = 6: 08$[/hide]

Please hide your answers so that people can do the problem without looking at the answer first (temptation).  It's not cool to do it for eveyone.  :)  :D


[quote="greatestmaths"]let Ψ is a function  Ψ: ]-infini ;-2[ ------> R

             such that  f(x) =  x^2/(x+1)

is this function injective? :blush:[/quote]

Yes, because $ f'(x)\equal{}\frac{x(x\plus{}2)}{(x\plus{}1)^{2}}$ shows that $ f(x)$ is strictly increasing for $ x\in]\minus{}\infty,\minus{}2[$

[hide]
\[ \frac{a^{2}}{a\plus{}1}\equal{}\frac{b^{2}}{b\plus{}1}\Rightarrow a^{2}b\plus{}a^{2}\equal{}ab^{2}\plus{}b^{2}\]
\[ \Rightarrow (a\minus{}b)(ab\plus{}a\plus{}b)\equal{}0\]
Either $ a\equal{}b$, or $ ab\plus{}a\plus{}b\equal{}0$. In the second case, this would mean $ (a\plus{}1)(b\plus{}1)\equal{}1$. Now $ a,b\leq\minus{}2$ imply $ a\plus{}1,b\plus{}1\leq\minus{}1$ so $ (a\plus{}1)(b\plus{}1)\geq 1$ with equality only when $ a\equal{}b\equal{}\minus{}2$. In this case, $ a\equal{}b$ anyway.

Therefore, $ a,b\leq\minus{}2$ and $ \frac{a^{2}}{a\plus{}1}\equal{}\frac{b^{2}}{b\plus{}1}$ imply $ a\equal{}b$, so the function is injective.
[/hide]


calculus
derivative
function
real analysis
real analysis theorems

Darboux Theorem states that the derivative of a function has the Intermediate Value Property
--------------------------------------------------------------------------
Darboux property.

Let E be an interval. It is said that the function f:E-->R has the property of Daboux on E, if for any a<b in E and any real number alpha situated between f(a) and f(b), there exists a least one \alpha from ]a,b[ so that f(x_\alpha)= \alpha.

thank you Lagrangia

[quote="Moubinool"]
Let E be an interval. It is said that the function f:E-->R has the property of Daboux on E, if for any a<b in E and any real number alpha situated between f(a) and f(b), there exists a least one \alpha from ]a,b[ so that f(x_\alpha)= \alpha.
[/quote]
f'(a) and f'(b) ... so that f'(x_\alpha)= \alpha.

Let's first proof this.

LEMMA f(x) is differential on [a,b]. f'(a)f'(b) < 0. there exists \xi \in (a,b) such that f'(\xi) = 0.

suppose f'(a)>0 and f'(b)<0. (the similar argument works for f'(a)<0 and f'(b)>0).

f(x) is differential on [a,b] implies that it is continuous. Thus it has a maximum value. f(a) is not the maximum value since f'(a)>0, f(x) is increasing at a. Simiarly, f(b) is not either since f(x) decreases at b.
Hence there is a \xi \in (a,b) s.t. f(\xi) is the maximum value. By Fermat's theorem, f(\xi)=0.

Now back to the general cases.

suppose c is between f'(a) and f'(b).
Let g(x) = f(x) - cx, g'(x) = f'(x) - c. g'(a)g'(b) < 0. we know from the lemma that there is a \xi such that g'(\xi) = 0, which is f'(\xi) = c.

Can someone tell me what Fermat's theorem say (maybe we have the same in french in a different name)

FERMAT THEOREM: suppose f(x) reaches its maximum at x_0 (x_0 is an interior point of a interval) and f'(x_0) exists, then f'(x_0) = 0.


Define $ a@b \equal{} ab \minus{} b^{2}$ and $ a\#b \equal{} a \plus{} b \minus{} ab^{2}$. What is $ \frac {6@2}{6\#2}$?

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

After just substituting values you get $-\frac{1}{2}$ Which as A


Let $a,b,c>0$.

Prove that
$\sum_{cyc}a^6 + 2 \sum_{cyc}a^5 b + \sum_{cyc}a^4 b^2 \geq \sum_{cyc} a^2 b^4 +\sum_{cyc}a^3 b^3 +\sum_{sym} a^3 b^2 c$.

Look like a part in Japan 1997's proof for ineq problem.
Try AM-GM...

In all the sums you have <<cyc>>  exept for the  6th  where you have  <<sym>>

 :huh:

If all are symmetric sums then use Muirhead


[hide="Click to reveal hidden content"]The question is asking for how many values of n is: $\frac{n!} {\frac{n(n+1)} {2}}$ which can be transformed into:

$\frac{2(n-1)!} {n+1}$ It becomes obvious that the bottom will not cancel if the result is a prime, so count the number of primes under 25.

2,3,5,7,11,13,17,19,23. That's 9 primes, but when the denominator is 2, it will cancel out with the 2 on the top.

So there are 8 values of n that it will not cancel out, therefore there are $\boxed {16}$ values of n such that it will cancel out.[/hide]

[quote="ch1n353ch3s54a1l"]Source: AMC 10B 2005

For how many positive integers $n$ less than or equal to $24$ is $n!$ evenly divisible by $1+2+...+n$?[/quote]

[hide] $\frac{n!}{1+2+...+n}=\frac{2*(n-1)!}{n+1}$.

We see that if the denominator is prime, the result is not integrer. So, we have $9$ primes wich are less or equal to $25$, but one of those primes is $2$, were the result is an integrer, because both $2's$ (one in the numerator, and one in the denominator) cancel out each other. So, we have $8$ numbers that can't be, so the result is $\boxed {16}$[/hide]


[url]http://video.on.nytimes.com/ifr_main.jsp?nsid=b26f34614:10f9606731b:-71e7&rf=bm&fr_story=d14603c1e23e6ce37920a8134a2e27b1405a4991&st=1166453285801&mp=FLV&cpf=false&fvn=9&fr=121806_094751_26f34614x10f9606731bxw71e6&rdm=175553.69351783473[/url]

HAHAHAHAHAHAHAHAHAHAHA!!!!!!! If you don't laugh, nothing is funny to you.

"widgets≠gadgets"

:D  :D HAHAHA I agree. That was really funny. "right and left are not the same thing".

Yeah, it's stupid that Windows copies good features of other operating systems. It would be a lot better if it instead added features that were left out of other operating systems because they were bad. I mean, seriously, a desktop search? Why would anyone want that?


inequalities
algebra
polynomial
inequalities unsolved

In Old and New inequalities there is a really nice solution for this one.

We have: $\displaystyle \frac {1}{2} \leq \frac {a_i}{b_i} \leq 2$ then $\displaystyle (\frac {a_i}{b_i} -\frac {1}{2})(\frac {a_i}{b_i}-2) \leq 0$.
$\displaystyle \Leftrightarrow {a_i}^2+{b_i}^2 \leq \frac {5}{2}a_ib_i ; \Rightarrow \sum_{i=1}^na_ib_i \geq \frac {4}{5}\sum_{i=1}^n{a_i}^2$;$(1)$
And: $\displaystyle \frac {{a_i}^3}{b_i} + a_ib_i \leq \frac {5}{2}{a_i}^2 \Rightarrow $$\displaystyle \sum_{i=1}^n(\frac {{a_i}^3}{b_i} + a_ib_i) \leq \frac {5}{2}\sum_{i=1}^n{a_i}^2$;$(2)$
Combining $(1)$ and $(2)$; then we get: $\displaystyle \sum_{i=1}^n \frac {{a_i}^3}{b_i} \leq \frac {17}{10}\sum_{i=1}^n{a_i}^2$.
QED

Hey Harazi, would you please send all problem in "Old and New inequalities" to me? :D
Thank you very much!

this one is a bit old...
it appeared in the Chinese Provincial Contest 1998, Part II.
in that problem, $a_i,b_j\in [1,2]$.

[quote="hxtung"]Solution was posted by Myth. But my soln is nice.
I'll post may son[/quote]
btw, what's Myth's solution?

Here is a slightly more compressed solution of the above argument:
===========================================================================
For 1/2 <= x <= 2, the polynomial [tex] $P(x) = (x+2/5)(x-1/2)(x-2)$ [/tex] is non-positive.
Hence, [tex] $P(x) = x^3 -\frac{21}{10} x^2 +\frac{2}{5} \le 0$ [/tex].

Plugging in x= a_i/b_i, and multiplying by (b_i)^2 yields
[tex]$(a_i)^3/b_i - (21/10) *(a_i)^2 +(2/5)*(b_i)^2 <= 0$.[/tex]

Adding this inequality up over all i, and using sum a_i^2 = sum b_i^2
yields the desired inequality.


Okay, I know this topic has been posted before on this forum, but I tried the pause thingy, but it still doesn't work for the videos. It would go to about 1 minute, then show the advertisement for every single video. It's getting really frustrating!!! I would be watching the video, then all of a sudden, this advertisement pops up and I have to start over from the beginning. I would first pause, wait like 45 sec, then play the video. Once again, it gets stuck at the same place. Please help!!!!!

Please provide information on your connection (e.g., dial-up, dsl, cable modem, and download speed). Please also provide operating system, browser, browser version, and flash version. If you are using Windows, TO obtain the flash version right click anywhere over the video and select About.

Unfortunately, this is very likely due to a speed issue on your end or some sort of network issue between your ISP and our server host and is out of our control. However, by providing the information, we can look into and see if there is anything we can do on our end.

Anyone else experiencing similar difficulties?

Uh, we have Time Warner Cable for internet, so cable modem? I'm not too sure about the download speed, but I tried both Firefox and IE. I normally use Firefox 3. The IE was I think version 6 or 7. My OS is Windows XP, and Adobe Flash Player.

What Flash version are you using? You can obtain the Flash version by right clicking anywhere over the video and selecting About.

This is definitely sounding like an issue somewhere between your computer and up to, but not including our hosting environment. Try doing it with all virus software, internet security software, and other running programs disabled. Also, when trying to get it all to work, hook your modem directly to your computer rather the through a router. Routers can often have trouble with video. (I have to reboot my router every couple days to keep things working well.)


The bisectors BB1 and CC1 of a triangle ABC meet at point I. The
line B1C1 intersects the circumcircle of triangle ABC at points M and N.
Prove that the circumradius of the triangle MIN is twice as long as the
circumradius of △ABC.

[hide]No. $ R=R_{IMN}$
Denote $ AI\cap MN={\J\}$. You can prove that $ AJ=IJ$[/hide]

u r missing something maybe because it is ARO 2006(final round grade 11 first day problem 4)

Sorry, I'm stupid  :blush:
Here is my solution:
Let $ I_a, I_b, I_c$ be the excenters of triangle ABC then I is the orthocenter of triangle $ I_aI_bI_c$.
We get $ R_{II_bI_c}\equal{}R_{I_aI_bI_c}\equal{}2R$ (because $ (ABC)$ is the [b]Nine-point [/b]circle of triangle $ I_aI_bI_c$)
We will show that M, N lie on $ (II_bI_c)$
We have $ \angle I_cAB\equal{}\frac{\angle B}{2}\plus{}\frac{\angle C}{2}\equal{}\angle I_cIB$ thus $ I_cAIB$ is cyclic quadrilateral.
$ \Rightarrow C_1I.C_1I_c\equal{}C_1A.C_1B\equal{}C_1M.C_1N \Rightarrow I_cMIN$ is cyclic
Similarly, $ MINI_b$ is cyclic.
$ \Rightarrow \angle II_cN\equal{}\angle IMN\equal{}\angle II_bN \Rightarrow M,N$ lies on $ (II_bI_c)$
We are done.

I proved this problem in Geometry Marathon. See here:-
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=331763&start=761


geometry
integration
calculus
calculus computations

Curve with parametric equations:
x= t+sint , y =sint   , 0_<t_<pi

show that the region bounded by the curve and the x-axis has area 2.

Let $ S$ be the area and $ C$ be the anticlockwise contour.
$ S\equal{}1/2\int_{C}(xdy\minus{}ydx)$; the integral over the $ x$-axis is zero. then $ S\equal{}1/2\int_{\pi}^{0}(tcos(t)\minus{}sin(t))dt\equal{}2$.


calculus
integration
function
geometry
trapezoid
real analysis
calculus computations

Use the table of values to estimate definite integral of f from 0 to 6.
Use three equal subintervals and the (a) left endpoints, (b)right endpoints, and (c) midpoints. If f is an increasing function, how does each estimate compare with the actual value? explain your reasoning.
Table:
x        0   1   2   3     4     5     6
f(x)   -6   0   8   18   30   50   80

What is 3 equal subintervals and a left endpoints consider in the table? also, 3 equal subintervals and right endpoint and midpoints.( it isn't midpoint only one that is when x=3, why there is plural for midpoint?)

Three equal subintervals would be $ [0,2],[2,4],$ and $ [4,6].$

Using those intervals would give you a Riemann sum with three terms, and each term would contain a factor of $ \Delta x\equal{}2.$

The left-endpoint Riemann sum is $ 2(\minus{}6\plus{}8\plus{}30)\equal{}64.$

The right-endpoint Riemann sum is $ 2(8\plus{}30\plus{}80)\equal{}236.$

The midpoint Riemann sum is $ 2(0\plus{}18\plus{}50)\equal{}136.$

The trapezoidal rule on these intervals is the simple average of the left and right Riemann sums: $ \frac{64\plus{}236}{2}\equal{}150.$

The Simpson's rule estimate based on this data is the weighted average of the midpoint and trapezoid rule estimates: $ \frac{2\cdot 136\plus{}150}{3}\equal{}140.7.$

Draw pictures! That should help you understand the second part of the question -  why the left-hand rule estimate is low and the right-hand estimate is high. Furthermore, the function (if reasonably well-behaved between the sampling points) appears to be concave upward; that helps explain the relationship of the midpoint rule to the trapezoid rule.


[quote="mthiyaga"]There are an equal number of pennies, nickels, dimes and quarters in a bag. What is the probability that the combined value of four coins randomly selected with replacement will be 41 cents? Express your answer as a common fraction.[/quote]

$\frac{4!}{4^{4}}=\boxed{\frac{3}{32}}$

To get the 41 cents, you need 1 of each coin.

[hide]You need to have 1 of each coin. The first coin doesn't matter, so the probability of that is 1. The second coin has to be different from the first, so the probability of that is 3/4. The third coin has to be different from the first 2, so the probability of that is 2/4. Finally, using the same reasoning, the probability of the 4th coin is 1/4. Multiplying all these fractions, you get:
(3/4)(2/4)(1/4) = 3/32[/hide]

[hide]The possibilities are any arrangement of a penny, a nickel, a dime, and a quarter.

So it is $\frac{4!}{4^{4}}=\frac{3}{32}$[/hide]

[quote="life=tennis+math"][quote="Treething"][quote="Fanatic"]With replacement... TT. Was wondering why I couldn't solve it.[/quote]

lol it's easy to solve without replacement either, given that there's n of each demonination

hey solve that new problem[/quote]

Here is the answer to Treething's problem: [hide=" :censored: "]
Let us say there are $n$ amount of each denomination. You need one of each. So for the first amount, you can pick any coin so the probability of picking the right coin is $\frac{4n}{4n}$. For the second one, since you have 3 types to work with, so $3n$ amount of coins. You have one less than $4n$ so the probability of picking the right coin is $\frac{3n}{4n-1}$. Same reasoning to end up with:
$(\frac{4n}{4n})$ $(\frac{3n}{4n-1})$ $(\frac{2n}{4n-2})$ $(\frac{n}{4n-3})$. This we can simplify to $\frac{24n^{4}}{256n^{4}-384n^{3}+176n^{2}-24n}$.
ANSWER: $\frac{24n^{4}}{256n^{4}-384n^{3}+176n^{2}-24n}$[/hide][/quote]

That answer will still tend to $\frac{3}{32}$ as n approaches infinity.

Well yeah, give that there's an amount. I went blindly into it thinking there was a certain number "x". At that point I realized that there was no point in just multiplying  24x(x/4x-1)(x/4x-2)(x/4x-3) so I might as well just check the answer.


Prove that the smallest number of colors required to color the finite regions of a planar graph such that no two neighboring regions are colored in the same way is 2 if and only if all the vertices of the graph have even degree.

I think the probles is wrong(but I can mistake, because I don't really know exactly what a planar graph is, so I'd like a good tutorial about planar graphs and dual graphs). Look at the image, the X vertex.

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=265[/img]

then, maybe "inside" vertices must have even degree ("inside"=vertices that are not vertices of the outside face)

No more discussion, try to solve the problem in one of these ways, I now it is true. ;)


Prove that $ 72^{2n\plus{}2} \minus{}47^{2n} \plus{}28^{2n\minus{}1}$ is divisible by $ 25$ for any $ n \in \mathbb{N}$.

[hide="Solution"]Working mod 25, we can write the expression as:

$ ( \minus{} 3)^{2n \plus{} 2} \minus{} ( \minus{} 3)^{2n} \plus{} (3)^{2n \minus{} 1}$

And making everything of the same power:

$ \minus{} 27( \minus{} 3)^{2n \minus{} 1} \plus{} 3( \minus{} 3)^{2n \minus{} 1} \minus{} 1( \minus{} 3)^{2n \minus{} 1}$

Which simplifies to give:

$ \minus{} 25( \minus{} 3)^{2n \minus{} 1}$

Which is clearly congruent to 0 mod 25.[/hide]

Thanks, KrazyFK. Apart from taking out the minuses, to get $ (25)(3^{2n\minus{}1})$, I have the same thing.


An unknown polynomial yields remainder of 2 upon division by $ x\minus{}1$, and remainder of 1 upon division by $ x\minus{}2$. What remainder is obtained if this polynomial is divided by $ (x\minus{}1)(x\minus{}2)$?

[hide="hint"]
Let$ f$ be the poynomial.
Then $ f(1)\equal{}2,f(2)\equal{}1$
$ f\equal{}(x\minus{}1)(x\minus{}2)q\plus{}ax\plus{}b$
$ f(1)\equal{}a\plus{}b\equal{}2,\ f(2)\equal{}2a\plus{}b\equal{}1$ and you have to solve the system for a and b.
[/hide]


Prove the following inequality.

\[2^n<\frac{(2n)!}{(n!)^2}<2^{2n}\ (n=2,3,\cdots)\]

The middle is 2n choose n.

We know $\sum_{r=0}^n $ ${n \choose r} =  2^n$ by Binomial theorem.

Now, ${2n \choose n}$  $< \sum_{r=0}^{2n} $ ${2n \choose r}$  $= 2^{2n}$

and ${2n \choose n} = \binom{2n-1}{n} + \binom{2n-1}{n-1} > \binom{2n-2}{n} + \binom{2n-2}{n-1} + \binom{2n-2}{n-2} > \dots$

$\dots > $ $\sum_{r=0}^n {n \choose r} = 2^n$, as desired.

[quote="blahblahblah"]Another solution to the left proceeds as follows

We want to show that $(n!)^2\cdot 2^n< (2n!)$

But $(n!)^2\cdot 2^n=n!\cdot (2\cdot 4\cdot 6\cdot\ldots 2n)<n!\cdot ((n+1)(n+2)(n+3)\ldots(2n))=(2n)!$[/quote]Why only to the left, blahblahblah? The same solution is true to the right as well. :)

I guess so, but I used

$\binom{2n}{n}<\sum^{2n}_{i=0}\binom{2n}{i}=2^{2n}$,

as singular did.


Prove that $ \forall n \in \mathbb{N} $ with $ n \geq 3 $, there always exists a permutation $ \left(a_1,a_2,...,a_n\right)$ of the list $\left(1,2,...,n\right)$ satisfying $ \forall i<j<k ,a_i+a_k \neq 2a_j $.

I'm defining a notion, hence I rewrite the problem:

[i]We will say that a finite sequence $\left(x_1, x_2, \ldots, x_n\right)$ of numbers is [b]nicely-ordered[/b] if for all integers $i$, $j$, $k$ with $1 \leq i < j < k \leq n$, we have $x_i+x_k \neq 2x_j$.

Then prove that for every natural number $n$ with $n \geq 3$, there always exists a nicely-ordered permutation of the sequence $\left(1, 2, \ldots, n\right)$.[/i]

I will prove this assertion using Cauchy induction. This means that I prove it for $n = 3$, then I show that

[b](*)[/b] if it holds for some $n$ then it also holds for $2n$,

and then I show that

[b](**)[/b] if it holds for some $n$ then it also holds for $n - 1$.

From this we can conclude that the assertion holds for all $n \geq 3$.

In fact, the proof for $n = 3$ is trivial: The permutation $\left(1, 3, 2\right)$ is nicely-ordered.

Now I assume that for some $n$, there exists a nicely-ordered permutation of the sequence $\left(1, 2, \ldots, n\right)$, and I will prove that there also exists a nicely-ordered permutation of the sequence $\left(1, 2, \ldots, 2n\right)$.

In fact, from our nicely-ordered permutation of the sequence $\left(1, 2, \ldots, n\right)$, we trivially obtain a nicely-ordered permutation of the sequence $\left(1, 3, \ldots, 2n-1\right)$ and a nicely-ordered permutation of the sequence $\left(2, 4, \ldots, 2n\right)$. Now combine these two permutations by writing them in a row, the first one on the left and the second one on the right. What you get is a permutation of $\left(1, 2, \ldots, 2n\right)$. This permutation is nicely-ordered, as can be readily shown (use the fact that if $x_i$ is odd and $x_k$ is even, then $x_i+x_k$ will be odd and hence never will equal $2x_j$). Hence you have received a nicely-ordered permutation of the sequence $\left(1, 2, \ldots, 2n\right)$.

Finally it remains to make the step from $n$ to $n - 1$. In other words, we have to show that if there exists a nicely-ordered permutation of the sequence $\left(1, 2, \ldots, n\right)$, then we also can find a nicely-ordered permutation of the sequence $\left(1, 2, \ldots, n-1\right)$. But this is obvious (just remove the element $n$).

This completes the proof.

  Darij

Nice!

Mine isn't as good. I did:
[i]please note: if $(a_k-1) + (a_i-1) \not= 2(a_j-1)$ then $a_k + a_i \not= 2(a_j)$ and vice versa.
please note: if $(2a_k) + (2a_i) \not= 2(2a_j)$ then $a_k + a_i \not= 2(a_j)$ and vice versa.[/i]

I split up in odd and even, but then I reduced all odd elements by 1. So the elements there are also even. Divide them all by 2. You now have 2 sequences 1,2,3,... again and if you keep repeating the stuff (and placing them like you said) splitting up you end with the same pattern as you did :)

(of course the -1 and *0.5 was only used for the order of the elements, afterwards you still need to check if the found pattern is ok)


algebra
polynomial
AMC
AIME
USA(J)MO
USAMO
AIME I

Let
\[P(x) = 24x^{24} + \sum_{j=1}^{23} (24 - j) (x^{24-j} + x^{24+j}).\]
Let $z_1$, $z_2$, $\ldots$, $z_r$ be the distinct zeros of $P(x)$, and let $z_k^2 = a_k + b_k i$ for $k = 1, 2, \dots, r$, where $i = \sqrt{-1}$, and $a_k$ and $b_k$ are real numbers.  Let
\[\sum_{k=1}^r |b_k| = m + n \sqrt{p},\]
where $m$, $n$, and $p$ are integers and $p$ is not divisible by the square of any prime.  Find $m + n + p$.

heh, i can see why this problem didn't make the cut.  do you have a solution...?

Yes, the Alternate AIME is the same as the AIME 2.  The booklets from AMC say "Alternate AIME" on them, so I followed suit.

Notationally, this problem looks like a monster, but if you break it down, it's not too bad.

[hide="Hint"]It helped me to write P using "..." notation instead of sigma notation, arranging the terms by order of degree.  After doing so, you may discover that P can be factored.
[/hide]

Yeah, like Ravi said, you can factor the expression....hint: after factoring out the obvious (the x) you are left with a perfect square... and from there, it becomes a "standard" problem...

KD

yeah, it is actually a pretty easy #15; i was just stupidly using an unnecessarily hard approach; til i realized what was REALLY happening. (aime's are getting SO much easier though, from what i see on this year's aimes 1 and 2, for instance).  kd, we woulda surely gotten 15's if these were the aime's they were giving back then, rather than the puny 13's and 14's we were getting.  heh ;)

maybe so for you, pete, but remember, even when I was getting less than a 10 on the AIME, I would generally have the last 5 right. it was always the first 5 that got me, or at least a few in the first 5.
Same game at nysml, I got only #1 wrong in 99 and 00. Come to think of it, same with everything. on the mandelbrot in 99 i got only 1-pointers wrong, and even the ahsme(amc for you babies) in 99, my 142 was leaving # 2 blank, and getting #1 wrong. oh my dear lord. What in gods name IS this?

but yeah, I definitely feel as a whole, the AIME's are way easier than they were in, say, the 90's, and the sheer number of students that score above a 10 should show this. It used to mean something to get a 12 or 13 on the AIME. just like it used to mean something to get a 150 on the AHSME, or AMC. But nowadays, everyone and their goat seems to be pullin it off. remember the days when a 200 index was commendable, and actually was almost certain to qualify even a senior for the USAMO?

I guess the AMC peeps didn't like the median aime score being a 0 ...

well, back to playing some serious 'hold em for me. I'm all in...

yeah, i think i remember you getting one of the first couple of questions wrong, a couple of times.  that's most probably 'cause you used to work backwards, from the last question to the first, and your computational skills were worn out and jaded by the time you got to questions 5 through 1.  oh goodness.

hehe, maybe that could explain my rediculous wrong answers for the AMC, and maybe the AIME, and hell, perhaps even the USAMO. but that wouldn't explain nysml 2 years in a row getting only #1 wrong. going backwards is good practice for heart rate control....it's almost like yoga in a twisted non-delusional manner...

[quote="penny77114"]
well, back to playing some serious 'hold em for me. I'm all in...[/quote]

I love to play hold'em also!

...flushing bus ... hold em game, KD takes everyones lunch money, hehe.

fine fine, i guess we can play without money...


1. A man leaves for work at a random time b/w 7:00AM and 8:00Am. His paper arrives at his house at a random time b/w 6:30 AM and 7:30 AM. What is the probability that he has his newspaper before he leaves for work?

I draw 2 axes, one horizontal and one vertical, and label the horizontal one 6:30, 7:00, and 7:30 for the newspaper, and the vertical one 7:00, 7:30, and 8:00. What I'm confused about is what to shade in for the region where he gets the newspaper before he leaves for work. Could you please explain this one to me?

Draw a line connecting (7:00,7:00) and (7:30,7:30). The region where he gets the newspaper first will be above this line, because then (time left for work)>(time newspaper arrives).


MATHCOUNTS
geometry
geometric transformation
reflection
\/closed

Perhaps that the content in that forum doesn't necessarily correlate to MATHCOUNTS? Perhaps it fits better under Classroom Math?

And does it really matter?

[quote="7h3.D3m0n.117"]

And does it really matter?[/quote]

I'm just used to it being in the MATHCOUNTS forum.. Oh well.

I think that the tournaments in the "MATHCOUNTS" sub-forum were too hard for even MATHCOUNTS level!

I don't really understand the appropriateness of the shift; although, the administrators and/or moderators might've looked at a bigger picture than I.

This is really an useless topic. Oh no, it's been moved, what now? Has anything changed that can really reflect on your abilities?
Useless...

[quote="ZetaX"]In general, either ask the moderators directly or in the thread. Or even better: don't ask at all. Just imagine if moderators need to justify every act of moving...[/quote]

Yeah, or if it concerns whole sub-forums, ask an admin.


Find all $x$ such that
\[ x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor=54 \]
Masoud Zargar

absolutely (trying to use long words that teh post can actually post :P)

[hide="hint"]use the identity: $x=\lfloor x \rfloor+\{x\}$ where $\{y\}$ is the fractional part of $y$[/hide]

[hide="question"]Why pick a number for which there is no real solution?[/hide]

[hide="not sure"]$\frac{54}{x}$ must be an integrer?????[/hide]

[hide="hmm"]
Put $a=\lfloor x\rfloor$ and $b=x-\lfloor x \rfloor$.

The equation simplifies to \[ (a+b)\lfloor a^3+2a^2b+ab^2\rfloor=54, \] unless I made a mistake. Now we take advantage of the fact that $\lfloor z \rfloor \in \mathbb{Z} \forall z\in \mathbb{R}$?
[/hide]

[hide="no solution"]
Let $n^4 = 54$, $x$ is about $2.7108$. This will be our starting number.

Clearly, $x$ must be bigger than this value. Note that $x=3$ overshoots by achieving a value of 81 so it is too high.

Also note that this function has the property that a higher x value will produce a higher final result.

We therefore conclude that $2.7108<x<3$

Let $k$ equal the smallest positive real number, and let $x=3-k$.
$\lfloor x \rfloor = 2$

Computing the value above results in $42-k$

This means that if $x=3-k$, it will be too low, and if $x=3$, it will be too high. There are no numbers in between, so there is no solution.
[/hide]

Okay...now solve it when it equals $88$. :)

[hide]$88^{\frac{1}{4}}=3.0628$ and $4^4=256$, so clearly the answer lies between $3.0628$ and $4$. Since the number is much closer to $3^4=81$ than to $4^4=256$, it is advantageous for us to try out numbers closer to $3$ first. Trying $\pi$ (that's just how I fly) gives $87.96$, so we know we need something just slightly above $\pi$. Note that  $\lfloor \pi\lfloor \pi\lfloor \pi\rfloor\rfloor\rfloor=28$, suggesting that $\frac{88}{28}=\frac{22}{7}$, which is just above $\pi$, will give $88$.

$\frac{22}{7}$ does give $88$ as a result. Since increasing $x$ must increase the final value, we know that $\boxed{x=\frac{22}{7}}$ is the only value.[/hide]

[quote="JBL"]$x \lfloor x \rfloor = 3$ for $x \in \bf R$.[/quote][hide="Solution"]$x=-1.5$ hence the negativeness comment.[/hide]

[quote="JBL"][quote="The Zuton Force"] Since increasing $x$ must increase the final value, we know that [hide]$\boxed{x=\frac{22}{7}}$[/hide] is the only value.[/quote]

Actually, not necessarily -- $x$ could, after all, be negative.  (It turns out there aren't any negative solutions, but there is no theoretical reason there couldn't be.)  You also neglected negative values in your original argument.

Just to make the point that what I'm saying isn't entirely stupid, solve
$x \lfloor x \rfloor = 3$ for $x \in \bf R$.

While I'm at it, here's a related question:  find all integers $n$ such that the equation $x \lfloor x \rfloor = n$ has
(a) a positive real solution
(b) a negative real solution[/quote]

Very true very true  :wallbash:

:cleaning:


BF,CE are the altitude of triangle ABC , K is a point on segment EF .D is the midpoint of segment BC .DF meet BK at point G . CE ,BF meet at point P . then prove :AG⊥ KP

Dear Mathlinkers,
1. let (1) be the circle with diameter AP ; it goes through E and F
2. DF ist angent to (1) at F  (consider the circle with diameter BC and applie Reim's theorem)
3. Let Tf be this tangent,  P' the meetpoint of PK and AG.
4. BKG is the pascal's line of the hexagone AEF Tf  PP'A...
then P' is on (1) and we are done
Sincerely
Jean-Louis


There is a 22-digit number that, when multiplied by 7, is the same number but with the 7 at the end transferred to the first digit place. What is that number?

I sure don't want to waste my time finding a 22-digit number like this. I do know it's less than $1.142857\cdot10^{22}$ (since when it's multiplied by 7, it has to start with 7). Doesn't seem to help. Still too much possibilities.
after a while, I also found the number to start with 102...... or 101.....

this problem is very perplexing and i think there is some pattern but i can't find it.  Also this topic has been posted twice.

Great work, but that's got to be tedious  :D

[quote="jsc"]If 1014492753623188405797 is the smallest answer, what's the 2nd smallest number?[/quote]

What do you mean by second  smallest ?   :huh:

[quote="jsc"]2nd smallest answer as in:If you put the numbers with these properties (except digits) in a smallest to greatest list, which number would be second?[/quote]
10144927536231884057971014492753623188405797 :D
But this number does not have 22 digits

Don't know if it has already been mentioned, but let the number be $10x+7$
so we have $7(10x+7) = 7 \cdot 10^{21} +x$ which gives x right away as $\frac{7\cdot 10^{21} -49}{69} = 101449275362318840579$

If you did not know the number of digits, you have to find out $n$ such that $10^n-7$ is divisible by $23$ (as it is always divisible by 3)  etc..

(You know that $10^{22} \equiv 1 \bmod 23$ so  next answer would be 22+22 digit,  etc  ...)
(Edited later : As expected, this has been mentioned by  gunes and others ..)

Also it is nothing but the decimal expression of $\frac 7 {69} = 0.10144927\dots$
Similar problems are asked for the expression of say  (1/7)  or (2/19) etc...(When they are expanded in decimal notation  etc..)