Let $n\geq 3$ be an integer.Consider ineq.:
\[\sum_{i=1}^n \frac{x_i\cdot x_{i+1}}{x_{i+2}}\geq \sum_{i=1}^n x_i(*)(x_{n+1}=x_1,x_{n+2}=x_2)\]
[b]a)[/b] Find all $n$ such that $(*)$ right $\forall x_1,x_2,\cdots ,x_n\geq 0$
[b]b)[/b](St.Petersburg MO 2000)
Prove that for positive integer number $x_1\leq x_2\leq \cdots \leq x_n ,(*)$ right

If $n=3$ then the ineq is easy to prove.
For $n=4$,at first I think  It's still true.
Expand to:

$(\frac{a_2}{a_3}-\frac{a_4}{a_1})(a_1-a_3)+(\frac{a_3}{a_4}-\frac{a_1}{a_2})(a_2-a_4) \ge 0$
And let $a_1=k_1a_3,a_2=k_2a_4$,We have:

$(k_1-1)(k_2-\frac{1}{k_1})\frac{a_4}{a_3}+(k_2-1)(1-\frac{k_1}{k_2})\frac{a_4}{a_3} \ge 0$

But If it true for all $a_4,a_3$ then:

$(k_1-1)(k_2-\frac{1}{k_1}) \ge 0$
$(k_2-1)(1-\frac{k_1}{k_2}) \ge 0$

Only choose $k_1 \ge k_2 \ge 1$
Then the inequality will be false immediate.
For exam,$a_1=300,a_2=2,a_3=100,a_4=1$
Then the ineq true Iff $n=3$

For $b$,we canexpand to:

$\sum\frac{(a_{i-2}-a_i)(a_{i-1}-a_i)}{a_i} \ge 0$
Let $b_i=a_i-a_{i-1}(a_0=a_{n})$

Then we need to proof:

$\sum\frac{b_i^2}{a_i}+\sum\frac{b_{i-1}b_i}{a_i} \ge 0$
But $\frac{(b_1+b_n)b_1}{a_1}+\frac{(b_2+b_1)b_2}{a_2}\ge 0$

And we are done.
(Because $b_2,b_3,...,b_n \ge 0$ and $\sum{b_i}=0$)

This problem was already posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16367]here.[/url]


ratio
trigonometry
geometry
limit
calculus
calculus computations

(1) For $ \theta > 0$, prove that $ \frac {1}{6}\theta ^ 3 \minus{} \frac {1}{120}\theta ^ 5 < \theta \minus{} \sin \theta < \frac {1}{6}\theta ^ 3.$

(2) Given a circle $ C$ with the center the point $ O$ and radius 1. For $ 0 < \theta <\pi$, take two points $ P,\ Q$ such that $ \angle{POQ} \equal{} \theta$. Let $ R$ be the intersection point of two tangent lines at $ P,\ Q$ on the circle $ C$. Denote by $ S_1$ the area of the part bounded by the line segment $ PQ$ and the minor arc $ PQ$ and denote by $ S_2$ the area of the triangle $ {PQR}$. Find $ \lim_{\theta \rightarrow \plus{} 0} \frac {S_1}{S_2}$

for (1)

let $ f(x)\equal{}x\minus{} \sin x \plus{} \frac{1}{120}x^{5} \minus{}\frac{1}{6}x^{3}$, $ g(x)\equal{} x\minus{} \sin x \minus{}\frac{1}{6}x^{3}$.

take $ f''''(\theta)$ and $ g''(\theta)$, the rest follows :D

for (2)

consider $ \theta < \pi$.

then $ \frac{S_{1}}{S_{2}}\equal{}\frac{\theta \minus{} \sin \theta}{\sin \theta}$.

therefore the answer is $ 0$. :D


geometry
power of a point
radical axis
geometry unsolved

we have circle named C.P is a arbitrary point.PQ and PL are tangenc to circle and M and N are the midpoint of them.T is a arbitrary point on MN.TH is tangenc to circle.prove that TH=TP

It is an old problem, it follows from the variation of the radical axis theorem , MN is the radical axis of the circle C and the point P ( circle with  radius zero).

Please use more meanigful Subjects next time , this one could be ''equal tangents'' , ''tangents from the point to the circle'' or whatever, if you want them to pay attention to your problems then write meaningful subjects.


Find all 7-digit numbers of the form $\overline{13xy45z}$, which are divisible by 792.

$792 = 8 \cdot 9 \cdot 11$.
$8 \mid 45z$ $\Longrightarrow$ $z = 6$.
$19 + x + y = 27$ or $19 + x + y = 36$.
$x - y + 3 = 0$ or $x - y + 3 = 11$.
We see that $x - y + 3 = 0$ doesn't works, so $x - y = 8$ and $x + y = 8$.
So $1380456$ is the only such multiple.


abstract algebra
linear algebra
linear algebra unsolved

after two years back to advance linear algebra :)
A finite group of matrices with sublinear spectrum is abelian group.(by sublinear I meant $ \sigma (A\plus{}\lambda B)\subset \sigma (A)\plus{}\lambda \sigma (B)$).

Sam-n has a passion for very hard problems! Needless to say, this problem is really difficult. A complete, but very long proof can be found in the classical book Simultaneous triangularization. All I can say is that I hope someone will come up with something easy and short, but I doubt that.


Let $ ABC$ is a triangle and let $ a\equal{}BC, b\equal{}CA, c\equal{}AB$.

Prove that: $ a^2b(a\minus{}b)\plus{}b^2c(b\minus{}c)\plus{}c^2a(c\minus{}a) \ge 0$

[quote="thanhnam2902"]Let $ ABC$ is a triangle and let $ a \equal{} BC, b \equal{} CA, c \equal{} AB$.

Prove that: $ a^2b(a \minus{} b) \plus{} b^2c(b \minus{} c) \plus{} c^2a(c \minus{} a) \ge 0$[/quote]

See here please [url]http://www.mathlinks.ro/viewtopic.php?p=343813#343813[/url]


Prove that the following equation has infinitely many solutions in positive integers:
$3^{k}=m^{2}+n^{2}+1$

YOu can also take powers of 2 and use Lagrange's identity.

Yes $k=2^{l}$ work too. My case $k=3^{l}$ is not trivial.


Let two functions $ f(x),g(x)$ be

$ f(x)\equal{}x^4\plus{}x^3\plus{}\frac{1}{2}x^2\plus{}\frac{1}{6}x\plus{}\frac{1}{24}$

$ g(x)\equal{}x^5\plus{}x^4\plus{}\frac{1}{2}x^3\plus{}\frac{1}{6}x^2\plus{}\frac{1}{24}x\plus{}\frac{1}{120}$

$ (1)$ Show that $ f(x)>0$ holds for all real number $ x$

$ (2)$ Show that the equation $ g(x)\equal{}0$ has just one real solution $ \alpha$,and then $ \alpha$ satisfies $ \minus{}1<\alpha<0$

(1) $ 24f(x)\equal{}6(2x\plus{}1)^2x^2 \plus{} (2x\plus{}1)^2 \plus{} 2x^2 > 0$
(2) Let $ h(x) \equal{} 120x^5g\left(\frac{1}{x}\right)$.
$ h(x)$ is 5th-order polynomial and $ h(\minus{}1)>0$.
And, $ h'(x)\equal{}120x^4f\left(\frac{1}{x}\right)>0$.
Thus, $ g(x)$ has just one real solution $ \alpha$,and then $ \alpha$ satisfies $ \minus{}1<\alpha<0$.
($ \because \ g(x)\equal{}0 \iff h\left(\frac{1}{x}\right)\equal{}0$)

Thank you Kouichi!
I didn't know such a technical solution there is !:w00t:


How many square units are in the area defined by the set of points $ (x,y)$ in the first quadrant which satisfy $ 3\le x\plus{}y\le5$?

Subtracting the smaller triangle from the larger, one gets $ \frac12(5^2\minus{}3^2)\equal{}\frac12\cdot16\equal{}\boxed8$.

Or you can determine the points: 3-5 range, 5-3=2, 2*4 (because of 4 points), =8.
*joke* :lol:
small to large in points: (2,1) (1,2) (3,2) and (2,3). area=8. :idea:


function
inequalities
inequalities proposed

Hmm...if i did no mistake, the function f(x) =  \sqrt (1-x 2 ) is concave on [0,1], then Jensen and blablabla...

Pierre.


geometry
analytic geometry
LaTeX
quadratics
graphing lines
slope
algebra

I am a little confused with the two following problems, can someone help me out?

1.) One vertex of an equalitatel triangle is at the origin. The 2nd vertex lies on the graph of $y=x^2$. The 3rd vertex lies on the positive portion of the $x$ axis. What is the area of this triangle?

2.) Find the coordinates of the point $Q$ on a circle whose equation is $(x-6)^{2}+(y-5)^{2} = 25$ such that $Q$ is nearest to the point $(-2, 11)$.

1.
[hide]
find the point P on y = x^2 where PO and x axis form an angle of 60 degrees
P(x,x^2)
x/sqrt(x^2+x^4) = cos60 = 1/2
4x^2 = a^2 + a^4
3= x^2 , x = sqrt3

side length PO by distance formula is 2sqrt3

area is 3sqrt3
[/hide]

Okay, I am going to type this (your post w/ some additional words) in $\LaTeX$ format, so it is much easier to read...

[quote][b]Problem:[/b]

Find the point $P$ on $y = x^2$ where $PO$ and and the $x$ axis form an angle of 60 degrees

[b]Solution:[/b]

$P(x,x^2)$,

$\frac{x}{\sqrt{x^2+x^4}} = \cos{60} = \frac{1}{2}$

$4x^2 = a^2 + a^4$

$x^2 = 3$ so this means  $x = \sqrt{3}$

To find out the length of $PO$, we can use the distance formula to obtain $2\sqrt{3}$.

This means that the area is $3\sqrt{3}$.[/quote]

My only question is how did you get that $x = \sqrt{3}$ from  the equation $4x^2 = a^2 + a^4$? Shouldn't  $4x^2 = x^2 + x^4$? Also, what is $P(x, x^2)$ supposed to represent (is it supposed to represent the coordinates of $P$?)

My understanding of what you are trying to say is that in the equilateral $\triangle{POQ}$ the coordinates of $P$ are $(\sqrt{3},3)$, the coordinates of $O$ (the origin) are $(0,0)$, and the coordinates of $Q$ we haven't solved for. However, since we found out that $\overline{PO} = 2\sqrt{3}$, then all the sides have to equal to $2\sqrt{3}$ since it  the triangle has equal sides and equal angles. Then since the area of an equilateral triangle is $\frac{1}{4}s^2\sqrt{3}$, we find out that the area is nothing but $\frac{1}{4} \times (2\sqrt{3})^{2} \times \sqrt{3} = 3\sqrt{3}$

This seems to be a correct understanding, but if you could explain your solution a little bit more it would be helpful!

[quote="Ravi B"]Number 2 is asking for the closest point on a circle to a particular point (-2, 11).  Try drawing a line segment from the center of the circle to the point (-2, 11).  That will help you find the closest point -- do you see how?[/quote]

Unfortunately, even after this hint I fail to see how :blush:

The point where that line segment intersects the circle will be the closest point.  Every other point on the circle will be farther away.  One way to show why is with the triangle inequality.

Okay, I understand that but the problem is I still don't know how to solve the problem even if I have connected a line segment from the circle to the point (-2, 11). Can anyone provide a solution?

Well, one method is to find the equation of the line that passes through (-2, 11) and the center of the circle.  That equation will give y in terms of x.  Then use that equation to replace y in the equation of the circle.  You will get a quadratic equation in x, which you can then solve.  Did that make sense?  There are other methods, but that will work.

Here is the solution (I think):

The equation of the circle is $(x-6)^{2}+(y-5)^{2} = 25$, meaning that the center of the circle is located at $(6,5)$, and the radius of the circle is $5$. Since we want to find out the equation of the line passing through the center and the point $(-2, 11)$, we find (after using the slope & slope-intercept formula) that the equation of this line is: $y = -\frac{3}{4}x + \frac{25}{2}$.

Next, we need to substitute this equation in the equation of the circle, so:

$(x-6)^{2}+(y-5)^{2} = 25$

$(x-6)^{2}+(-\frac{3}{4}x + \frac{25}{2}-5)^{2} = 25$

$(x-6)^{2}+(-\frac{3}{4}x + \frac{15}{2})^{2} = 25$

$4(x-6)^{2}-(3x - 30)^{2} = 100$

But know, when I try to use my TI-89 to solve the equation, it gives me that $ x = \frac{2\sqrt{19}+66}{5}$ or $x = -\frac{2\sqrt{19}+66}{5}$

For some reason, I think this answer is wrong, so what should I do now :?

I believe the equation of the line is actually $y = -\frac{3}{4}x + \frac{19}{2}$.  You can always check your equation by plugging the two points back in.

Other than that mistake, your approach looks good to me.

After solving everything, I get  $x = 10$ or $x=2$. Is this correct?

I think that's right.  One of them corresponds to the closest point, and the other to the farthest point.

Here is another method, which is simpler but perhaps harder to come up with.  The center is (6, 5).  The given point is P = (-2, 11).  The distance from the center to P is 10.  The radius of the circle is 5.  Thus the closest point Q must be halfway between the center and P.  So we can find Q by the usual midpoint formula.


topology
limit
advanced fields
advanced fields unsolved

First, some definitions:
In a metric space $ X$, for a real number $ \alpha>0$, define $ \lambda_{\alpha}^{(\varepsilon)}(E)\equal{}\inf \sum r_{i}^{\alpha}$ where the infimum is taken over sequences of balls covering $ E$ with radii $ r_{i}<\varepsilon$. Then define $ m_{\alpha}^{*}(E)\equal{}\sup_{\varepsilon \rightarrow 0} \lambda_{\alpha}^{(\varepsilon)}(E)\equal{}\lim_{\varepsilon \rightarrow 0} \lambda_{\alpha}^{(\varepsilon)}(E)$. This induces a measure on $ X$ under which all Borel sets are measurable.

Now the question:
For a Borel set $ E$, if $ m_{\alpha}(E)$ is finite for some $ \alpha$, how do I show that $ m_{\beta}(E)\equal{}0$ for $ 0<\beta<\alpha$?

You probably meant $ \beta>\alpha$.

Observe that $ \lambda_{\beta}^{(\epsilon)}(E)\le \epsilon^{\beta\minus{}\alpha} \lambda_{\alpha}^{(\epsilon)}(E)$.


algebra
polynomial
function
real analysis
real analysis unsolved

Let $f \in C^{\infty}(I,R)$a non-polynomial function.Show that
a)there is a compact interval $J$ such that $f|J$ is non-polynomial and $f(x)\neq 0, \forall x\in J$.
b)there is $x\in I$ such that $\forall k , f^{(k)}(x)\neq 0$.

Who's that friend?  ;) . In fact, this problem in oral agregation training and it has a totally different solution than the usual one, based on Baire's theorem.  Enjoy it, it's really nice!


Support
geometry
Duke
college
search
ARML
3D geometry

Who in this forum is going to the Chiles Invatational?(Just out of curiousity)

[quote="now a ranger"]We would have gone, but then I think our teacher changed her mind probably.   :(

Good luck to fairview, I'm sure y'all will get 1st place in team

and you guys will finish in 1st and 2nd place in individual, and many more.

 :lol:[/quote]
Thanks for the support but we will have some competition. I heard that the vestavia geometry is coming as well.

im going.  and im gonna win.  :P

go fairview. woot.

[quote="Elemennop"]where is florida huh[/quote]
dude it's like that place that's under that redneck state

whacha call it

im gonna get like last place dude
cuz im like stupid dude

Hm I would go if I was invited.   :wink:

omfg we've got some activity in this forum shweet

[quote="bpms"][quote="Walk Around The River"]im going.  and im gonna win.  :P[/quote]
Not if I have anything to say about it.[/quote]
you're going? what division are you participating in?

[quote="Walk Around The River"][quote="bpms"][quote="Walk Around The River"]im going.  and im gonna win.  :P[/quote]
Not if I have anything to say about it.[/quote]

 :cursing:
now im going to lose.  thanks for ruining my hopes and dreams preet.  :noo:[/quote]
I was joking, I just wanted to see your reaction if I said I was going.

[quote="now a ranger"]don't worry about them too much.

They only dominated at their own tournament.

At grissom, they were much more spread out, with lots of people not placing, and only 4 or so in the top 10.

They don't have as much depth as you think.

And you people are from the ghetto.[/quote]

Try working on your own competence before talking about other people.

[quote="Walk Around The River"][quote="Pakman2012"]screw you guys. i'm takin the SAT.

what can i say?

I like hard math and I cannot lie.

You other math nerds can't deny.[/quote]

yeah, we have no algebra 1 team because of that.  there all taking the SAT.  you know, with the Duke talent search thing.  i owned that stuff.[/quote]
Oh I screwed up on that.

[quote="Altheman"]que\?

i wish my school had a better math program[/quote]

sup

[hide=":play_ball: :wallbash:  :wallbash_red: The many uses of a wall"]Huh?[/hide]

[hide=" :pilot:  :pilot:  :pilot: "]What is going on?? Huh???  [/hide]

Fairveiw isnt as good as Deerlake its number 1 in the state

what?
we butt owned on saturday
dont even talk
btw, who are you?

einstein... let's try not to bump posts that are years old...

wait wat...? is this forum starting again...?haha.......yea i agree. vestavia stinks :D:D they disqualified my test at the winter comp when i got third and i got a perfect score on ciphering but they didnt count that either so those pompous fairview kids beat us. :(

but i beasted amc 12 so its all ok

You call that beasting? When I was your age, I scored higher than you :P

wendy hou, u r unaturally smart and i am not. remember??

o and i met chenyu something or watever at chiles...haha...

OMG OMG OMG Steve Han?

ZOMG. And here, I thought I was trolling at random Florida people that I didn't actually know in real life lol.

Nah, I'm just picking on you. You actually did really well. But I'll wait to hear your AIME score before I pass judgement :lol:

Ahaha, yeah, Chenyu was in Florida recruiting or something over spring break. It's cool you got to meet him.

"Unnaturally smart" nothing. It's only the many years of having everything drilled in. When you're my age, you'll be way smarter than I am now.

Happy problem solving!

HAHA wow i dont think you want to know my aime score...:D:D

so yea...yay amc stuff is over!!


The well known archimedean property is proved using the completeness of real numbers. But, it doesnot appeal tro my intution that archimedean property is a consecquence of completeness. Because, it does hold even in the set of rationals, which is not complete (ofcourse, the set of rationals is a subset of reals, and so, the same proof of archimedean property holds here also, but, intutively, I feel there must be a simpler proof)

My problem is, is it possible to prove the archimedean property without using the copleteness property?

No, it isn't. There are non-archimedean ordered fields; the easiest to describe is the quotient field of $ K[x]$, where $ K$ is an ordered field. Since the real numbers are uniquely specified as a complete ordered field, this says that completeness is necessary to prove the archimedean property.

hello jmerry,

Could you suggest a source in the web for reading about quotient fields?


[quote="manuel"]Is somebody here interested in modern and postmodern plastic art and new tendencies of art?[/quote]
what you mean with "modern and postmodern plastic"??? :?

By Modern Art I mean the movements like surrealism, cubism, posmodernism, etc.

If interested...my [url=http://www.surrealArt.com]Surrealism[/url] gallery is almost ready...

i am interested in modern buildings they are fun to look at.

Wow, I love those Surrealism. One of my favorite paintings of all time is Metamorphosis of Narcissus by Dali. The dual use of the hand/body shape is absolutely amazing.


Find all functions $ f: \mathbb R\rightarrow \mathbb R$ satisfying $ f(f(x\minus{}y))\equal{}f(x)f(y)\minus{}f(x)\plus{}f(y)\minus{}xy$

set $ f(0)\equal{}a$
$ x\equal{}y\equal{}a \minus{}>f(a)\equal{}a^2$
$ x\equal{}a,y\equal{}0 ...$
$ x\equal{}y\equal{}f(a) ...$
$ \minus{}>a\equal{}0 \minus{}>f(0)\equal{}0$
$ \minus{}>f(f(x))\equal{}f(x) \minus{}> f(x)\equal{}x or f(x)\equal{}\minus{}x$
easy to see that $ f(x)\equal{}0<\minus{}> x\equal{}0$
$ \minus{}>f(x)\equal{}x \minus{}>done$ :)


With $ a,b,c>0$ such that $ a\plus{}b\plus{}c\equal{}1$.
Find Max of $ V\equal{}(a^2\plus{}b^2\plus{}c^2)\sqrt{abc}$


Let a,b,c be positive real numbers satisfy ab+bc+ca=1.Prove that:
      $\sum$ $\frac{1+a^{2}b^{2}}{(a+b)^2}$ $\geq$ $\frac{5}{2}$

http://www.mathlinks.ro/Forum/viewtopic.php?t=57552
http://www.mathlinks.ro/Forum/viewtopic.php?t=57683

  darij


Dear everyone,

       Would you please help me with the following trigo Q?  I know this should be a easy question,  but I stuck on it.

"The bisectors of the interior angles of triangle ABC make angles of x,y,z with the sides a,b,c respectively.  Show that asin2x + bsin2y + csin2z = 0."

       Thank you very much!

Best,
tine

can you draw a picture for me?  I get a contradiction somewhere...

Hello Zellex,

       I am sorry.  The question is not attached to any pictures.

tine

[quote="MellowMelon"]Hopefully I understood it correctly and you understand this restatement: Let $ I$ be the incenter of $ \triangle ABC$. Let $ a,b,c$ be the sides across from $ A,B,C$ respectively, and let $ x,y,z$ be the angle measures of $ \angle BIC, \angle CIA, \angle AIB$. Prove that $ a \sin (2x) \plus{} b \sin (2y) \plus{} c \sin (2z)$.[/quote]

You mean $ \equal{}0$?

That seems to be an incorrect interpretation (or simply an incorrect problem); consider an equilateral triangle.

[quote="MellowMelon"]Hopefully I understood it correctly and you understand this restatement: Let $ I$ be the incenter of $ \triangle ABC$. Let $ a,b,c$ be the sides across from $ A,B,C$ respectively, and let $ x,y,z$ be the angle measures of $ \angle BIC, \angle CIA, \angle AIB$. Prove that $ a \sin (2x) \plus{} b \sin (2y) \plus{} c \sin (2z)$.[/quote]

    If all of the measures of angles BIC, CIA and AIB are obtuse, then the answer will turn out negative, not 0. I think some part of the question is missing  :( . Sines are negative between 180 and 360.


[quote="i_like_pie"]New users are sometimes unaware that $\LaTeX$ is installed on this site and think they need to install it on their computers to be able to use it here. Maybe an email or PM about $\LaTeX$ to every new user would stop this confusion?[/quote]
Agreed. I was confused too but after a few experiments on the website, it became obvious. :)

Couldn't one of the admins arrange for new users to receive an automatic email or message upon joining?

I can tell you from my personal experience regarding forums: I never ever read that welcoming message whenever I sign up on a forum :)

Additionally, there are many other things that we would like for new people to know that many of them eventually ask about.  However, it's just not practical to bombard them with notices right when they join expecting them to learn it all right then and there.  Rather, we should just let them gain experience at their own pace and help them out just as you were helped by older (in AoPS years, at least) members when you first joined.

An idea I just had is to make a wiki page on AoPS FAQ's.  That would be helpful for new members to reference whenever they come across a common problem.

Yes, some new users can't quite figure out what the "hide" button does.

[quote="jmerry"]I find it sufficient to just answer the question whenever it's asked, with a link to the $\LaTeX$ forum when appropriate. Sending messages to all new users would be a lot of clutter.[/quote]You may wish to refer them to the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX FAQ[/url] which deals with this question.

[quote="joml88"]An idea I just had is to make a wiki page on AoPS FAQ's.  That would be helpful for new members to reference whenever they come across a common problem.[/quote]Feel free to copy anything from the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX FAQ[/url].


To qualify for participating in CGMO for the USA Team, only your USAMO score matters, right?
If so, do you know what are the cut-off scores for the last few years?
Thanks.

[quote="kyyuanmathcount"]I can't say I'm a total insider on CGMO, but here's what I know:
In 2007, due to probably late organization, they had to use 2006 USAMO scores.
In 2008, 2008 usamo scores were used and the cutoff was 7 to get to MOP and 11 to go to China.
In 2009, by some process of choosing (unknown to me), 10 girls with USAMO scores of at least 7 (I think...maybe 8?) were chosen and two declined, leaving 8 team members.[/quote]
thankyou (:

[quote="qvc196"]In year 2007, due to [b]an early organization[/b], the selection was made using 2006 usamo scores. However, most of the team members’ 2007 usamo scores were clustering in 12-14 range.
In year 2008, 2008 usamo scores were used and the cutoff was [b]12[/b].
I don't know about cut off for 2009. Some qualified cgmo members made the choice of declining such opportunity, because of conflictions with other opportunities they had, such as RSI, especially for some students who made the cgmo team in their 10th grade.[/quote]

O.O so the cutoff score should be around 12...


In 1994, a March snowstorm produced 18 inches of snow which melted down to 2.4 inches of water. How many inches of snow, expressed as a decimal to the nearest tenth, would have produced 3.0 inches of water? Assume that the relationship between amount of snow and amount of resulting water is constant.

Obviously every 7.5 inches of snow melts into 1 inch of water. So 22.5 inches of snow will melt into 3 inches of water.


two circle  $ (C)$ and $ (C')$ which the distance between their centre is $ l$ intersect  in $ M$ and $ N$.
let $ A$ a point in  $ (C)$ (A#M and A#N).
the rights $ (AM)$ and $ (AN)$ intersect $ (C')$ in $ B$ and $ C$.
a) prove that  the  radius of the circumscriptible circle of $ ABC$  is $ l$

See lemma 2($ 6^{th}$ Quote)of http://www.mathlinks.ro/viewtopic.php?t=160883


trigonometry
calculus
calculus computations

If f''(x) = 14 + cos(x) and f'(1) = 14 + sin(1) and f(1) = 9-cos(1)

find the value of f(2)

is this differetial equations?

$ f''(x) \equal{} 14 \plus{} \cos(x)$
$ f'(x) \equal{} 14x \plus{} \sin(x) \plus{} C$
$ f(x) \equal{} 7x^2 \minus{} \cos(x) \plus{} Cx \plus{} D$

$ f'(1) \equal{} 14 \plus{} \sin(1)$ thus $ C \equal{} 0$.
$ f(1) \equal{} 9 \minus{} \cos(1)$ thus $ D \equal{} 2$.

We get:
$ f(x) \equal{} 7x^2 \minus{} \cos(x) \plus{} 2$

$ f(2) \equal{} 30 \minus{} \cos(2)$.

[quote="iamhe"]
If,

$ f''(x) \equal{} 14 \plus{} \cos(x) \\
f'(1) \equal{} 14 \plus{} \sin(1) \\
f(1) \equal{} 9 \minus{} \cos(1)$

[/quote]

[hide="Solution:"]

Integrate $ f''(x)$

$ f'(x) \equal{} 14x \plus{} \sin x \plus{} c \\
f(1) \equal{} 14 \plus{} \sin 1 \plus{} c \\
\therefore c \equal{} 0$

Integrating $ f'(x)$,

$ f(x) \equal{} 7x^2 \minus{} \cos x \plus{} cx \plus{} d \\
c \equal{} 0 \\
f(1) \equal{} 7 \minus{} \cos 1 \plus{} d \equal{} 9 \minus{} cos 1 \\
\therefore d \equal{} 2$

You got $ f(x)$, now since you have $ d$, find $ f(2)$.

$ f(2) \equal{} 30 \minus{} \cos 2$[/hide]


We plan to have two teams, an A and B, this year. People from NV and Utah welcome. If you're from Reno, and know someone who might join, please relay this information to him/her.

Contact me, Angela Cheng, at angelanana@gmail.com

haha

[size=200][color=blue]GO INDIANA!!!!!!!!!!!!!!!!!!![/color][/size]

I love Nevada  :clap: and im from Nebraska by the way

[quote=Solid Snake]Nevada is awesome.  :ninja:[/quote]

yes, yes we are

[quote=MinionLA]Wait LOL there's people in Nevada???!!! And there's AoPSers in Vegas? Are you spoiled children? LOL jkjkjkjk[/quote]

woahwoahwoahwoah I only go gambling every NIGHT not day. [size = 2] just kiddin'


[hide]Given any term, there arre two possibilities for the previous term. This means there was $2^{10}$ possibilities for the first term[/hide]

[quote="Hexaditidom"]A recursive sequence is defined by:

$ a_1 = $ a positive integer
$ a_{n+1} = \frac{a_n}{n} $ if a_n is divisible by n
$ a_{n+1} = a_n + 1 $ if not.

If $ a_{11} = 99 $ , how many possible values are there for $ a_1 $ ?

BTW how do you make spaces work in Latex?[/quote]

The problem is very nice, and seamusoboyle seems to be totally correct. With respect to your other question, if you want spaces in general in latex, you can write \  (backslash and an empty space) between the simbols you want the space. Also, you can divide the expression between different dollar signs pairs

[code]$\$x,\ y$\$  or $\$x$\$, $\$y$\$.[/code]
$x,\ y$  or $x$, $y$.

Sorry for the kind of dollar signs in the code, it seems like $\LaTeX$ commands work inside the code enviroment.

You're answer is correct for this question, but I don't think you've given a sufficient proof of it.

By your reasoning, the answer would be independent of the value of $a_{11}$ - however, this is not the case.

For instance, if we take $a_{11}=90$, then the answer becomes $1023$, because the value $14374$ occurs twice, since $\begin{array}{lll} 14374 & = & ((((((((((90-1).9)-1)-1).6)-1)-1).3)-1)-1) \\
& = & ((((((((((90.10)-1).8)-1)-1)-1)-1)-1).2).1)\end{array}$

An interesting problem would be to characterise the numbers for which the set of solutions is/is not pairwise distinct. Presumably this would be a function of the index of the $a$ value which you start with.


Let a, b, c, d be real numbers with ad - bc = 1.

Show that a2 + b2 + c2 + d2 + ac + bc \geq \sqrt3.


prove that an irrational number times a rational number,is a irrational number.

not always
$0\cdot\sqrt{2}=0$, which is a ration number. If it's non zero we have by contradiction:
Let $a$ be a irrational number, and $p, q, m, n$ integer numbers.
$a\cdot\frac{p}{q}=\frac{n}{m}$
$a=\frac{nq}{mp}$ Condtradiction.

nice solution dondigo,i really enjoyed it. :D


inequalities
induction
inequalities proposed

Let $a>0,$ $b>0,$ $a^n+b^n=2,$ $n\in\mathbb N.$
Prove that $n(a+b)+\frac{n-1}{ab}\geq3n-1.$

If we wil prove that $a+b \geq 2 (ab)^{\frac{n-1}{2n}}$ (*) then we'll prove the inequality.
$\displaystyle{(a+b)^n =\sum_{k=0}^{n}C _n^k a^{n-k}b^k}$ , where ${C _n^k = \frac{n!}{(n-k)!k!}}$.
From here (take a=b=1) $\sum_{k=0}^{n}C_n^k = 2^n$.

$\displaystyle{(a+b)^n =a^n+b^n + \sum_{k=1}^{n-1}C _n^k a^{n-k}b^k = 2 + \sum_{k=1}^{n-1}C _n^k a^{n-k}b^k \geq}$
$\displaystyle{\geq 2 + (2^n-2)(ab)^{\frac{n(n-1)}{2(2^n-2)}} = 2 (1+(2^{n-1}-1)(ab)^{\frac{n(n-1)}{2(2^n-2)}}) \geq}$
$\displaystyle{\geq 2( 2^{n-1} (ab)^{\frac{n(n-1)(2^{n-1}-1)}{2(2^n-2)2^{n-1}}})=2^n(ab)^{\frac{n(n-1)(2^{n-1}-1)}{2^n(2^n-2)}}}$.
Thus
$a+b \geq 2 (ab)^{\frac{(n-1)(2^{n-1}-1)}{2^n(2^n-2)}}$.
$a^n+b^n \geq 2 \sqrt{(ab)^n}$ . Then $ab \leq 1$.
Then if
$\frac{(n-1)(2^{n-1}-1)}{2^n(2^n-2)} \leq \frac{n-1}{2n}$  we'll get (*). But the last is true (by induction for example).
(Here n does not equal 1. We can prove the ineqality for n=1 separatly.)
Thus, (*) is true. Then
$n(a+b)+\frac{n-1}{ab} \geq 2n (ab)^{\frac{n-1}{2n}} + \frac{n-1}{ab} \geq 3n-1$ (by Cauchy (i.e. AM-GM)).

 :whistling:
(Is the proof ok?)
--
edit: this proof doesn't work, there's a mistake. Thanks to arqady.

[quote="arqady"][quote="kar"]
$\displaystyle{ 2 + \sum_{k=1}^{n-1}C _n^k a^{n-k}b^k \geq}$
$\displaystyle{\geq 2 + (2^n-2)(ab)^{\frac{n(n-1)}{2(2^n-2)}}}$
[/quote]
I think, this is wrong. ;)[/quote]
Why,arqady??
I think it is a nice solution.

:) Thanks, [i]zhaobin[/i] and [i]Michael[/i], but this proof doesn't work, I see my mistake now. :oops:

hi,by hint of kar:I proved the following by induction:
$(a+b)^{n^2} \ge 2^{n^2-1}(ab)^{\frac{n^2-n}2} (a^n+b^n)$ :)

Another proof:
Let $ p \equal{} \frac {1}{n},$ $ a \equal{} (1 \minus{} s)^p$ and $ b \equal{} (1 \plus{} s)^p,$ where $ 0\leq s < 1.$ Hence,

$ (1 \minus{} s)^p \plus{} (1 \plus{} s)^p \equal{} 2\left( 1 \plus{} \frac {p(p \minus{} 1)}{2!}s^2 \plus{} \frac {p(p \minus{} 1)(p \minus{} 2)(p \minus{} 3)}{4!}s^4 \plus{} \ldots \right) \geq$
$ \geq 2\left( 1 \plus{} p(p \minus{} 1) \left( \frac {1}{2}s^2 \plus{} \frac {1}{4}s^4 \plus{} \frac {1}{6}s^6 \plus{} \ldots\right) \right)$ and

$ (1 \minus{} s^2)^{ \minus{} p} \equal{} 1 \plus{} ps^2 \plus{} \frac {p(p \plus{} 1)}{2!}s^4 \plus{} \frac {p(p \plus{} 1)(p \plus{} 2)}{3!}s^6 \plus{} \ldots \geq 1 \plus{} p \left( s^2 \plus{} \frac {1}{2}s^4 \plus{} \frac {1}{3}s^6 \plus{} \ldots\right).$
Thus, $ n(a \plus{} b) \plus{} \frac {n \minus{} 1}{ab} \equal{} n\left((1 \minus{} s)^p \plus{} (1 \plus{} s)^p \plus{} (1 \minus{} p)(1 \minus{} s^2)^{ \minus{} p}\right)\geq$
$ \geq n(3 \minus{} p) \equal{} 3n \minus{} 1.$

kar, your and zhaobin's proof is very nice too!   :lol:


What does Rolle's Theorem say and how to prove?

Let $f$ be differentiable  on the open interval $(a,b)$ and continuous on the closed interval $[a,b]$. Then if $f(a)=f(b)$, then there is at least one point $c$ in $(a,b)$ where $f'(c)=0$.

Well,here it is:

 If $f$ is constant the situation is clear.Let's assume that $f$ is not the constant function.But,$f$ is a continous function on a closed and bounded interval,hence $f$ is bounded and and can reach its bounds (Weierstrass criterion).So there are $x_{m}$ and $x_{M}$ from $[a,b]$ such that $f(x_m)=m$ and $f(x_{m})=M$ where
$M=sup f(x)$ and $m=inf f(x)$ and $m<M$( $f$ is not constant).We obtain that $x_m$ is inside of interval $[a,b]$
and via Fermat's theorem we have $f^{'}(x_m)=0$. If $x_m$ is either $a$ either $b$ we apply again Fermat's theorem and the conclusion follows. ;)


Let a,b,c,k >0,s.t.xy+yz+zx=3.prove that:
$ (x\plus{}y)^k\plus{}(y\plus{}z)^k\plus{}(z\plus{}x)^k\geq{3*2^k}$

Use AM-GM and the Carlson's inequality: $ \sqrt[3]{\frac{(x\plus{}y)(x\plus{}z)(y\plus{}z)}{8}}\geq\sqrt{\frac{xy\plus{}xz\plus{}yz}{3}}.$

[quote="arqady"]Carlson's inequality: $ \sqrt [3]{\frac {(y \plus{} z)(z \plus{} x)(x \plus{} y)}{8}}\geq\sqrt {\frac {yz \plus{} zx \plus{} xy}{3}}.$[/quote]Problem 3 of 15th Austrian-Polish Mathematics Competitions 1992,

see : http://www.mathlinks.ro/Forum/viewtopic.php?t=156


i remember i was at 781 because it's divisible by 11, and i posted stuff after that so i was at least at 782.  but it dropped to 98.

I'm not sure . . . I seem to still have a decent amount of posts?

[quote="10000th User"][quote="math92"]How much did your post count drop?
Mine dropped about 1000.[/quote]So this means you are going to spam some more to recover your 1k posts?  Hail math92, the Master of Spam!  :omighty:[/quote]

HAHAHAHA :rotfl:  :rotfl:

Mine probably dropped aobut 50.

I couldn't help but notice that bubka's went from 1000--->36 :o

Yeah, umm... when you lose 90% of your post count because G&FF was nuked, something's wrong... I lost about 200 or so.

I lost somewhere around 100 posts, which was about 20% of mine.

What happened. How did people's post count drop?

G&FF doesn't count anymore, and I'm pretty sure bubka was near 2000.

Mine dropped around 400 according to mathgeniuse^ln(x), I dunno as I don't pay attention to it.

I had 42 and dropped to 21 .That means 50% loss !! Still  i can't vote because there isn't "20 posts" anywhere

I lost a star, but got it back with 12 posts. :D

[quote="chess64"][quote="Captain Sugar"]0.64% isn't that much. you were lucky.[/quote]

I definitely wasn't being sarcastic. And how was I "lucky", anyway? This isn't a game of chance... :?[/quote]

lucky as in your conscience prevented you from spamming  :lol:

[quote="Captain Sugar"]I had justt gotten Riemann Hypothesis but i got pulled back down to Poincare Conjecture.[/quote]

Same here, but I only lost about 120 (not 100, as I had previously thought...).  Wish I was like Chess64 so I wouldn't have lost as many  :D

[quote="10000th User"][quote="math92"]How much did your post count drop?
Mine dropped about 1000.[/quote]So this means you are going to spam some more to recover your 1k posts?  Hail math92, the Master of Spam!  :omighty:[/quote]

lol

math92 made like 4 posts in the test forum, probably for posts because there was nothing expect a word of text in any of them.

[quote="nutz_for2.718281828"][quote="math92"]How much did your post count drop?
Mine dropped about 1000.[/quote]

Ditto!!!! I was in Navier, but I went to Poincare theorum! (Really, Poincare has been proved!"[/quote]
First, since I can't stand bad spelling (:P), [i]theorem[/i]
(disregard this if it was a typo)
Next, has it already become a theorem? I thought the proof needs to be reviewed to be made sure that it was rigorous.

[quote="bubka"] v.v. is not a bad man, unlike some of the moderators.[/quote]

I think "v. v." is a very nice man.  If I was him, I would have banned you immediately after that post in this thread.

[quote="fredbel6"]Did anyone lose a star then? :? [/quote]

I lost 2 WHOLE ones!

:D  The new rules are funny.
Have you noticed how closely Valentin followed up that new Tug of War?

[img]http://users.telenet.be/frederic_bel6/considerspam.jpg[/img]

[quote="ragnarok23"][/quote][quote="Xantos C. Guin"][quote="10000th User"][quote="math92"]How much did your post count drop?
Mine dropped about 1000.[/quote]So this means you are going to spam some more to recover your 1k posts?  Hail math92, the Master of Spam!  :omighty:[/quote]

bubka is definetly the master of spam. His post count went from 1000+ to 35, 97% decrease. math92's post count went from 3500 to 2500, a 29% decrease.

Mine went from 805 to 625, a 22% decrease.[/quote][quote="ragnarok23"]
Nutz for e and Bubka are the duo of spammingness.
Nutz went from 1100-ish to 100.
Bubka went from somewhere around 1700-ish? to 35.
[/quote]

I'm defending bubka because he had LESS than me! You got my preivious amount right, though. You were off by 650 on bubka's! You were right on the target with bubka's new post count, but mine dropped to 180!

P.S. I didn't make triple posts! Sure I made some doubles, but I didn't mean it! I actually couldn't help it in some cases!

I remember somebody telling me that I only posted 40 units of spam in the G&FF, though I remember posting around 75.  So, less than 100.


Prove that  there exists a function $f: \mathbf{N}\rightarrow\mathbf{N}$ with no fix points such that there exists n for which we have $f^{(n)}(x)=x$ ,$\forall x\in \mathbf{N}$.($f^{(n)}$ the iterate n times)

What about this one?[hide]$f(k)=k+1$ for k odd, and $f(k)=k-1$ for k even.
Then $n=2$ works.[/hide]

A slightly stronger form:

For any $n>1$ there is a function $f:{\mathbb N}\to{\mathbb N}$ without fixed points such that $f^{(n)}(x)=x$ for any $x\in{\mathbb N}$.

For the solution you partition the positive integers into blocks of $n$ numbers  and then use a permutation inside each block.

That was easy.Ok what about a function that for any of it's iterate has no fix points $f^{(n)}(x)\neq x$,$\forall x\in \mathbf{N}$,and for all n.

For example $f(x)=ax+b$, were $a,b\in N$.

:blush: O i am sorry for this.f must be bijective.

For example $f(x)=x-2$ if x is even and x>2; f(2)=1; $f(x)=x+2$ if x is odd.


[quote="amangupta2"]
I would Request All Of You to Please Post The Questions Discussd in INMO TC's of All other Regions Too!!
[/quote]
Thanks! I have no idea if the TC he actually kicked of in UP. Coz the centre for the TC was in Lucknow and in Class 11 I think 18/22 students or so are from my school and the school held a separate camp from 26 dec to 15 jan(Yeah, its a big show off  :rotfl: )

[quote="amangupta2"]This Post Contains Some Good Questions on Polynomials and Inequalities from Delhi INMO TC

I would Request All Of You to Please Post The Questions Discussd in INMO TC's of All other Regions Too!!
 :wink:  :oops:[/quote]

Thanks for the problems. The inequalities have been discussed in ML before I suppose, I liked the polynomials better though, some problems were very good!

[quote="supranta"]Well Aman, thanx a lot for ur contribution....I'm frm d N-E region which is obviously not an advanced region in this field.......and anyway INMOTC wud b held only frm 11th January[/quote]

Hey Supranta, NE region is not advanced in this case??? Who said so man....? We have all the resources we need , just that people here are not interested! Maybe u and I can create a difference!

[quote="supranta"]Manjil, of course we can make the difference. U can try your website for making some of the difference,that wud be very useful.[/quote]

I am trying my level best!


ratio
algebra
polynomial
limit
real analysis
real analysis unsolved

Test the convergence of
$\left\{\frac{n^{100}}{1.01^{n}}\right\}_{n=1,2,...}$

[hide="My Efforts"]It is intuitively clear that it goes to zero, but I can't get a concrete analytical proof for it.[/hide]

If you want to prove that the limit is 0, note that the ratio of successive terms is $\frac{(n+1)^{100}}{n^{100}}\cdot \frac{1}{1.01}$, and as $n \to \infty$, we know that $\frac{(n+1)^{100}}{n^{100}}\to 1$. This means that for large enough $n$, each term of the sequence is almost exactly $\frac{1}{1.01}$ times the previous term. All the terms are positive, and it's a decreasing sequence, so it has a limit. For large enough $n$, every term must be arbitrarily close to the limit $L$. So basically, we need $L = 1.01\cdot L$, so $L=0$.

[quote="liyi"]Expand $1.01^{n}$ as \[(1+0.01)^{n}= 1+\binom{n}{1}0.01+\binom{n}{2}0.01^{2}+\cdots+0.01^{n}\] So if $n$ is big enough, there is a term $\binom{n}{101}0.01^{101}$ in the expansion, which is polynomial of order $101$ with respect to $n$.[/quote]

Right, if $n>202$, we are essentially done. Xevarion's proof is good too, but this is sweeter and more elegant!

Actually L'Hospital does it fine, but I wanted something better.

Thanks!

[quote="bubka"]Right, if $n>202$, we are essentially done. [/quote]

There's nothing wrong with this argument, but just so you know, when $n=203$, the term has 230 digits and it's still going strong. The sequence won't start decreasing until $n=10051$, where it has 357 digits. It won't be less than 1 until $n=117309$. Not that these are very large numbers in the grand scheme of things. ;)

it does not have to be less than one, it doesn't even have to be decreasing, getting $n^{101}$ is good enough for me. Which occurs after 203 digits.

also you can prove it by:
$\lim_{x \to+\infty}\frac{x^{1}00}{1.01^{x}}=0$
by L'hospital's rule 101 times :D


The sequences of integers $ \{a_n\}$ and $ \{b_n\}$ are defined by $ a_1 \equal{} 2,$ $ b_1 \equal{} 1$ and $ a_{n \plus{} 1} \equal{} 2a_n^2 \minus{} 1,$ $ b_{n \plus{} 1} \equal{} (a_n \minus{} b_n)(a_n \plus{} 3b_n) \minus{} 1,\, n\ge1$. Prove that for every $ n\ge1$ numbers $ a_n$ and $ b_n$ are relatively prime.

We claim that $ b_{n\plus{}1} \equal{} 2a_n b_n$.  This implies $ a_n^2 \minus{} 3b_n^2 \equal{} 1$.  Our inductive step is straightforward:

$ (2a_n^2 \minus{} 1)^2 \minus{} 3(2a_n b_n)^2 \equal{} 4a_n^4 \minus{} 4a_n^2 \plus{} 1 \minus{} 12a_n^2 b_n^2 \equal{} 4a_n^2(a_n^2 \minus{} 1 \minus{} 3b_n^2) \plus{} 1 \equal{} 1$.

[quote="t0rajir0u"]We claim that $ b_{n \plus{} 1} \equal{} 2a_n b_n$.  This implies $ a_n^2 \minus{} 3b_n^2 \equal{} 1$.  Our inductive step is straightforward:

$ (2a_n^2 \minus{} 1)^2 \minus{} 3(2a_n b_n)^2 \equal{} 4a_n^4 \minus{} 4a_n^2 \plus{} 1 \minus{} 12a_n^2 b_n^2 \equal{} 4a_n^2(a_n^2 \minus{} 1 \minus{} 3b_n^2) \plus{} 1 \equal{} 1$.[/quote]

Great! :clap2:
This is also author's solution.

Unfortunately, no contestant managed to solve this problem during competition... :(


Prove that $\forall n \in Z, n>0, n \neq x^{2}$ we have $(\sqrt{n}+1)|sin \pi\sqrt{n}|>\frac{\pi}{2}$

Let $\sqrt{n}=m+\epsilon,\ |\epsilon|\le\frac12.$ Then $\epsilon=\frac{n-m^{2}}{\sqrt{n}+m}$. Hence $|\epsilon|\ge\frac1{\sqrt{n}+m}\ge\frac1{2\sqrt{n}+0.5}$
\[|\sin(\pi\sqrt{n})|=|\sin(\pi\epsilon)|\ge\sin(\frac{\pi}{2\sqrt{n}+0.5}) \]
Then use $\sin x\ge x-\frac{x^{3}}6$.


let XOY be a fixed angle .and A,B are in the interior of it.

find point P on OX,inwich if PA,PB intersect OY,in points M,N  ,   we will have:  PM=PN

here is a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=55502&highlight=]link[/url] to the solution (if you realy want it.)
nice problem. :D


Find all functions $f: \mathbb R \to \mathbb R$ such that for all reals $x$ and $y$,
\[f((x\minus{}y)^{2})\equal{}(f(x))^{2}\minus{}2xf(y) \plus{}y^{2}.\]

[quote="AndrewTom"]Find all functions $ f(x)$ such that, for all real $ x$ and $ y$, $ f(x)$ is real and

$ f((x \minus{} y)^{2}) \equal{} (f(x))^{2} \minus{} 2xf(y) \plus{} y^{2}$.[/quote]

Let $ P(x,y)$ be the assertion $ f((x\minus{}y)^2)\equal{}f(x)^2\minus{}2xf(y)\plus{}y^2$
Let $ f(0)\equal{}a$

$ P(0,0)$ $ \implies$ $ a\equal{}a^2$ and so $ a\in\{0,1\}$
$ P(0,x)$ $ \implies$ $ f(x^2)\equal{}a^2\plus{}x^2\equal{}x^2\plus{}a$ and so $ f(x)\equal{}x\plus{}a$ $ \forall x\ge 0$

Let then $ x\ge 0$ :
$ (x\minus{}y)^2\ge 0$ and so $ f((x\minus{}y)^2)\equal{}(x\minus{}y)^2\plus{}a$
$ x\ge 0$ and so $ f(x)\equal{}x\plus{}a$ and so $ f(x)^2\equal{}(x\plus{}a)^2$

Then $ P(x,y)$ becomes $ (x\minus{}y)^2\plus{}a\equal{}(x\plus{}a)^2\minus{}2xf(y)\plus{}y^2$ and so $ 2xf(y)\equal{}2x(y\plus{}a)$ and so $ f(x)\equal{}x\plus{}a$ $ \forall x$

And it is easy to check back that these two functions indeed are solutions. Hence the two solutions :

$ f(x)\equal{}x$ $ \forall x$
$ f(x)\equal{}x\plus{}1$ $ \forall x$

Patrick, I am confused by your third paragraph. I thought you meant to write "Let then $ x<0$" instead of "Let then $ x\ge0$" to deal with the case where $ x$ is negative and the next line follows from that. But then you have $ x\ge0$ again in the third line of this third paragraph. I can't see that you've shown it for negative values of $ x$. Help!

[quote="AndrewTom"]Patrick, I am confused by your third paragraph. I thought you meant to write "Let then $ x < 0$" instead of "Let then $ x\ge0$" to deal with the case where $ x$ is negative and the next line follows from that. But then you have $ x\ge0$ again in the third line of this third paragraph. I can't see that you've shown it for negative values of $ x$. Help![/quote]

At the end of second paragraph, I got $ f(x) \equal{} x \plus{} a$ $ \forall x\ge 0$

So, in the third paragraph, I considered $ x\ge 0$ so that $ f((x \minus{} y)^2) \equal{} (x \minus{} y)^2 \plus{} a$ (since $ (x \minus{} y)^2\ge 0)$ and $ f(x) \equal{} x \plus{} a$ (since $ x\ge 0$

Using these two values, I can rewite $ P(x,y)$ as $ (x \minus{} y)^2 \plus{} a \equal{} (x \plus{} a)^2 \minus{} 2xf(y) \plus{} y^2$ $ \forall x\ge 0$, $ \forall y$

And so $ 2x(y \plus{} a) \equal{} 2xf(y)$ (remember $ a^2 \equal{} a$)  $ \forall x\ge 0$, $ \forall y$ hence the result (choose $ x \equal{} 1$, for example) : $ f(y) \equal{} y \plus{} a$ $ \forall y$ (that I immediately rewrote as $ f(x)\equal{}x\plus{}a$ $ \forall x$)

Thanks Patrick, it's quite clear to me now.


Let be $ a\in \mathbb{N}^*$ and $ a_1,a_2,a_3,...,a_n$ integers numbers $ a_1 \plus{} a_2 \plus{} a_3 \plus{} ...\plus{}a_n \equal{} 15k\ ,\ k\in \mathbb{Z}$ . Prove that :

$ 15\ |\ a_1^5 \plus{} a_2^5 \plus{} a_3^5 \plus{} ... \plus{} a_n^5$

I remember that you posted a similar problem before, alex2008.

It's easy to see that $ x^5\equal{}x(mod 5)$ and $ x^5\equal{}x(mod 3)$ for all $ x$.

So $ a_1^5\plus{}a_2^5\plus{}...\plus{}a_n^5\equal{}a_1\plus{}a_2\plus{}...\plus{}a_n\equal{}0(mod 15)$


calculus
real analysis
real analysis unsolved

let
(1) $ y' \equal{} f(x,y)$
$ y(x_{0}) \equal{} y_{0}$
also
(2) $ z' \equal{} g(x,z)$
$ z_{(}x_{0}) \equal{} z_{0}$
and
$ f,g\in C(G)$ where $ G$ is area.
suppose
$ f(x,y)\geq g(x,y)$
suppose $ \alpha (x)$ and $ \beta (x)$ is solution of the $ 1$ and $ 2$ differential equation.
 and $ \alpha (x)$ not include Points of branching.
prove that :
for all $ x > x_{0}$ we have  $ \alpha (x) >\beta (x)$

what does " $ \alpha(x)$ not includes Points of branching " mean ?  :maybe:

ok. let us consider some example:
$ y'\equal{}3y^{\frac{2}{3}}$
direct calculus show that $ y\equal{}(x\minus{}C)^{3}$ is solution of our equation on interval $ (\minus{}oo,\plus{}oo)$
and also equation has solution
$ y\equal{}0$
and now
$ y\equal{}0$ Is a tangent to  $ y\equal{}(x\minus{}C)^{3}$
.
points $ A$ and $ B$ is point of branching.


Some players are getting old, and will retire soon.

What men player (ATP) do you consider most promising ranked 10 or worse?

See vote/

[quote="zapi2007"]It might have been more accurate had you put fewer aging players.
Eg. Lleyton Hewitt?  Way removed from his prime.[/quote]

Yea... Gonzalez doesn't really belong.

How old is Haas? He's my second choice (after Cilic)

Cilic is like the Croatia version of Del Potro


Emily sees this sign in a grocery store. She selects 25 oranges
and gives the clerk one nickel. According to the price on the
sign, how many cents does the clerk owe her?
$ \boxed{ORANGES: 0.12 cents}$

It says:

ORANGES: 12 Cents

I don't understand this problem. It doesn't say if EACH orange is 12 cents, or if a certain amount of oranges costs 12 cents.

If I am correct, there was a number and the answer was a decimal. :P

Oh, my bad! It says 0.12 cents, in which case if Emily bought 25 of them, she needed to pay 3 cents. However, she gave a nickel, so the clerk owes her $ 5 \minus{} 3 \equal{} \boxed{2}$ cents.

Wow. This really tripped me up hard. I'm guessing everyone thought it said $\$0.12$ dollars?


A cylindrical container $10$ units high and $4$ units in diameter is partially filled with water.  The cylinder is tilted so that the water level reaches $9$ units up the side of the cylinder at the highest but only 3 units up at the lowest.  Compute the volume of the water in the cylinder.

This is a challenge problem so if you find it extremely easy leave it to someone else.

[hide]
The height before tipping the cylinder is the average of the two heights after tipping it, or 6.
Answer=$\pi r^2h=\pi\cdot2^2\cdot6=24\pi \textnormal{units}^3$
[/hide]


Hello! How can I download these files? After clicking on link it writes me "Sorry but you are not authorized to view or download this Attachment"

Sorry... it seems we've got a new bug.
I have deleted two topics apparently linking to the same attachments (the original poster accidentally posted his message three times). It seems that this led to the deletion of the attachments as well. I have stored one of the two files on my computer, so here I reattach it. I hope someone else has the other file.

  darij

Hey Darij, I cannot open the solutions one because it has been corrupted or improperly decoded. :(

As darij has said, you need to change it to djvu first.

[quote="Euclid's Turtle"]I did and it doesn't work. THat is why I posted. Does it work for you? :([/quote]

Yes it did  :maybe:

What program do you use for opening regular djvu files, Euclid's Turtle?
Is that the program which returns you the "corrupted" message in this case?

BTW, why isn't the proper djvu file attached (is there a reason)? If there's a problem with djvu attachments or something like that, here's a rapidshare link to it (loosing the need for renaming) [url]http://rapidshare.com/files/134331499/imc2008solutions.djvu.html[/url]

EDIT: yes, apparently djvu extension is not allowed :)

Mhmm I just downloaded a new program, WinDjView, and it works fine now :)
All is good, thanks hsiljak :)

OK, finally I saw how exactly I did in that exam. As I thought, I had blown up the exam.

That was my last chance to be a contestant though. :)

There are a lot of weird numerical mistakes in the scores scan. Can you please upload the original image somewhere, so we can take a look? Thanks.

The results are out now on http://www.imc-math.org.uk  :)

Omid Hatami, a most congratulatory applause :) First price, regardless, is a definite achievement.

[quote="FOURRIER"]What's the prizes, is it Medals like IMO?[/quote]

No, just diplomas. Also books and other prizes (T-shirts, CD's with Mathematica etc.).
And money prizes for Grand Prize winners (about 1,000 Euros).


algebra
polynomial
factorization
sum of cubes
algebra proposed

A polynomial $ P(x)$ with integer coefficients is called good,if it can be represented as a sum of cubes of several polynomials (in variable $ x$) with integer coefficients.For example,the polynomials $ x^3 \minus{} 1$ and $ 9x^3 \minus{} 3x^2 \plus{} 3x \plus{} 7 \equal{} (x \minus{} 1)^3 \plus{} (2x)^3 \plus{} 2^3$ are good.
a)Is the polynomial $ P(x) \equal{} 3x \plus{} 3x^7$ good?
b)Is the polynomial $ P(x) \equal{} 3x \plus{} 3x^7 \plus{} 3x^{2008}$ good?
Justify your answers.

For [i]a)[/i]:
$ 3x\plus{}3x^7\equal{}\minus{}x^9\plus{}(x\plus{}x^3)^3\minus{}(x\plus{}x^2)^3\plus{}(x^2\plus{}1)^3\plus{}(x\minus{}1)^3\minus{}x^3$.
For [i]b)[/i]:
Apply the lemma [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=183534]from here, the second post[/url] to get the negative answer.

Can you say me who is the proposer? Because this exact problem can be found at http://reflections.awesomemath.org/reflections/2008_6/MR_6_2008_problems.pdf see O106 Proposed by Nairi Sedrakian, Yerevan, Armenia!

Well, I think the author is the same Nairi Sedrakian  :wink:.

simply because he was and to my knowledge, is going to be again in the jury of ZIMO 2009.

i think there is a little simpler solution  for the second one :D

we have that

$ (\sum_{i = 0}^n a_ix{}^i){}^3 = \sum_{i = 0}^na_i{}^3x{}^{3i} + 3\sum_{0\leq i < j\leq n}a_i^2a_jx^{2i + j} + a{}_ia{}_j{}^2x{}^{i + 2j} + 6\sum_{0\leq i < j < k\leq n}a{}_ia{}_j{}_a{}_k$

this implies that the sum of the coefficients of any "non-cube" term is divisible by $ 6$, since $ a_i^2a_k + a{}_ia{}_j^2$ is always even, and we're done  :D

what?

$ i\plus{}2j\neq 2i\plus{}j$. That invoked misconception, I suspect.

i mean that $ i \plus{} 2j$ is not congruent to $ 0$ iff $ 2i \plus{} j$ is not congruent to $ 0$, so $ x^{i \plus{} 2j}$ is a "non-cube", iff $ x^{2i \plus{} j}$ is a "non-cube"...

i just realized that i wrote something wrong, i wanted to say that "the sum of the coefficients of all "non-cube" terms is divisible by $ 6$"...

so, for any combination of cubes of polynomials we would have that the sum of the coefficients of "non-cube" terms will be divisible by $ 6$, which is not the case of $ 3 \plus{} 3 \plus{} 3$... is it clear now?

I'd like to solve only a here,for b is trivial.
it's easy to prove that if f and g are good then f+g and f-g are good
by expanding $(x^2+x+1)^3-(x^2+x)^3$ we know that $3x^4+3x$ is good
replacing $x$ with $-x$,then $3x^4-3x$ is good.
so $3x^7+3x=3x^7+3x^4-(3x^4-3x)=x^3(3x^4+3x)-(3x^4-3x)$ is good.

We will prove that the sum of of the coefficients near the exponent of the form $3k+1,3k+2$.
This follows from:
$$P(\omega)=b_0+b_1\omega+b_2\omega^2\equiv_{2} \sum (x+y)(x+z)(z+y)$$
But $(x+y)(y+z)(z+x)$ is even ,as by php there are two numbers with the same modulo class 2.
This kills b).
For a) it is possible ,namely:
$$(1+x+x^3)^3+(-x^2-1)^3+(-x^2-x)^3+4(-x)^3+(-x^2)^3+(-x^3)^3=(3x+3x^7)$$

For a), simply play around with cubes of polynomials of the form $\varepsilon_{3}x^3 + \varepsilon_{2}x^2 + \varepsilon_{1}x + \varepsilon_{0}$ for $\varepsilon_{i} \in \{-1, 0, 1\}$ to get a construction. For example:
\[(x^3+x+1)^3 + (-x^3)^3 + (-x^2-x-1)^3 + (-x^2)^3 + (-x^2)^3 + (x+1)^3 + x^3 + x^3 + (-1)^3 = 3x^7 + 3x\]
For b), we work over $\mathbb{Z}[x]/(x^3-1)$. The polynomial we want to express becomes $9x$, but if we assume there exist reduced quadratic polynomials such that $9x = \sum\limits_{i=1}^{k} p_{i}^3$, then after writing $p_{i}(x) = a_{i}x^2 + b_{i}x^2 + c_{i}$, the sum becomes
\[\sum\limits_{i=1}^{k} x^2(3a_{i}^2 b_{i}+3b_{i}^2c_{i}+3c_{i}^2a_{i})+x(3a_{i}^2c_{i}+3c_{i}^2b_{i}+3b_{i}^2a_{i})+a_{i}^3+b_{i}^3+c_{i}^3+6a_{i}b_{i}c_{i}\]
Here $3a_{i}^2 b_{i}+3b_{i}^2c_{i}+3c_{i}^2a_{i} \equiv 3a_{i}^2c_{i}+3c_{i}^2b_{i}+3b_{i}^2a_{i} \pmod 2$ for all $i$, which doesn't hold for the polynomial $9x$, so we reach a contradiction.


algebra
polynomial
number theory proposed
number theory

find all rational x, y, z such that all these expresions are integers: x+y+z;
xyz; 1/x+1/y+1/z

...As $x+y+z=k_{1},xy+yz+zx=k_{2},xyz=k_{3}$ are integers, and polynomial $p(X)=X^{3}-k_{1}X^{2}+k_{2}X-k_{3}$ have all rational zeros they must be integer.

it is not very clear to me, but thank you anyway.

It will become more clear for you [b]surshu[/b] if you consider the following statement:
if $\frac{p}{q}$ is a root of polinomial with integer coefficients$f(x)$ so p is a divisor of the f(0) and q is divisor of the leading coefficient of f. This statement is not difficult to prove.


Let $n=1667$. Then the first nonzero digit in the decimal expansion of $\sqrt{n^2+1}-n$ is

(a) 1
(b) 2
(c) 3
(d) 4
(e) 5

Looks like a good problem to brute force  :D

Hmm... solution a little lofty, but here goes

[hide]
Using the binomial theorem, we get the expression to be $\left( n + \binom{ \frac 12 }{1} n^{\frac{-1}{2}} + \binom{ \frac 12 }{2} n^{\frac{-3}{2}} + \cdots \right) - n$.  Since we only want the first digit, we can reduce it to $\frac{1}{2 \sqrt{1667}}$.  Since $40^2 < 1667 < 41^2$, we can say the expression equals $\frac 1n$ where $80 < n < 82$.  Considering long division, we see the first digit will be 1 for all such n, so the answer is $\boxed 1$.
[/hide]

[hide="Answer"]Brute-forcing gives $2\Rightarrow \boxed{B}$. :P[/hide]

[quote="scorpius119"][hide]The expression $\sqrt{n^2+1}-n=\frac{1}{n+\sqrt{n^2+1}}$. When $n=1667$, this is less than $\frac{1}{3334}<\frac{1}{\frac{10000}{3}}=\frac{3}{10000}=.0003$ but greater than $\frac{1}{5000}=\frac{2}{10000}=.0002$. Then the first nonzero digit must be 2 $\Rightarrow \boxed{B}$.[/hide][/quote]

And just for fun ...to add to above, (and writing the denominator again as $2n + \sqrt{n^2+1}- n$  the value can be written as:
$\cfrac{1}{3334+\cfrac{1}{3334+\cfrac{1}{3334+\cfrac{1}{\dots}}}}$

Obviously slightly less than $3/10000$ (even without a pen or paper one can estimate it to be 0.00029994 - (slightly less than this with error in 11th place of decimal!) :)


Evaluate $ (x\minus{}a\plus{}3)$ if the value of $ x$ is $ (a\plus{}7)$.

Knowing that $ x\equal{} a\plus{}7$.

$ x \minus{} a \plus{} 3 \equal{} (a \plus{} 7) \minus{} a \plus{} 3 \equal{} \boxed{10}$

Simply substitute. a+7-a+3=7+3=10.

Or just set some arbitrary value for a. The most convenient is 0 in which case we just get 7+3=10.

[hide=Sol]We are asked to find x-a+3. We know x = a+7. x-a+3 = a+7-a+3 = 10[/hide]


function
algebra
domain
real analysis
real analysis unsolved

Let $ \text A$ be the set of all functions from the open interval (0,1) to $ \text R \equal{}$ "all real numbers", and $ \text B$ be the set of all continuous functions from (0,1) to $ \text R$

[color=white](a) Show that |A|=/=|B|. done.[/color]
(b) Prove or disprove |B|=|R|.

Clearly, the map which restricts the domain of each $ f \in C^0 (0, 1)$ to the set of rational numbers in $ (0, 1)$ is an injection. So $ |\mathbb{R}| \leq |B| \leq |\mathbb{R}^{\mathbb{Q}}| \equal{} |\mathbb{R}^{\mathbb{N}}|$.

So the proof completes if we show $ |\mathbb{R}^{\mathbb{N}}| \equal{} |\mathbb{R}|$. This is an immediate consequence of a general formula of cardinal exponentiation, but here is a more direct approach:

For each $ 0 < x \leq 1$, let $ x \equal{} 0.x_1 x_2 x_3 \cdots$ be the unique non-terminating binary expansion of $ x$ and $ 0_n \equal{} 0$ so that $ 0 \equal{} 0.000\cdots$. Also, let $ \phi : \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ be a bijection.

Then define $ \varphi : [0, 1] \to [0, 1]^{\mathbb{N}}$ by $ \varphi(x) \equal{} \left( n \mapsto \sum_{k\equal{}1}^{\infty} \frac{x_{\phi(n, k)}}{2^k} \right)$. Clearly, $ \varphi$ is a surjection, giving $ |\mathbb{R}^{\mathbb{N}}| \leq |\mathbb{R}|$.


Find the locus of the vertex of $y=x^2+tx+1$ where $t$ changes through the real numbers

We can write the vertexes in terms of t:

$(-\frac{t}{2} , -\frac{t^2}{4}+1)$,

which turns out to be a parabola.

[hide]To add a bit more to fourierseries's solution,
we solve the parametric equations $x=-\frac{t}{2}$, $y=-\frac{t^2}{4}+1$.

Solving for $t$ in the first equation gives $t=-2x$, so

$y=-\frac{(-2x)^2}{4}+1$

$\therefore y=-x^2+1$, which is an equation of parabola. [/hide]


four girls - Mary, Alina, Tina, Hanna - sang song in a concert as trios with one girl sitting out each time. Hanna sang 7 songs which was more than any other girl and mary sang four song which was fewer than any other girl.How many songs did these trio sing?

[hide="Solution"]Three girls get their song counts incremented at a time.  Hanna sang seven songs, so they sang at least seven times.  This means $M + A + T + H = 3 \times 7 = 21$.  We know that $M = 4, H = 7$, so $A + T = 10$, and $4 < A, T < 10$.  This gives our only possible solution as $A = T = 5$.  Working out a total song history:

$Hanna, Alina, Tina$
$Hanna, Alina, Mary$
$Hanna, Tina, Mary$
$Hanna, Alina, Mary$
$Hanna, Tina, Mary$
$Hanna, Alina, Tina$
$Hanna, Alina, Tina$

This gives $H = 7, M = 4, A = 5, T = 5$, and we have our solution.[/hide]


geometry
angle bisector
perpendicular bisector
geometry proposed

Consider triangle $ABC$. Let $U$, $V$, $W$ be points such that $U$ is on the line $BC$, $V$ is on the line $CA$, $W$ is on the line $AB$. It is given that $AU$, $BV$ and $CW$ are concurrent at a point $P$. Also $AU$ is a median of the triangle, $BV$ is an altitude and $CW$ is the internal angle bisector of $\angle BCA$. Suppose that $P$ lies on the perpendicular bisector of at least one of the sides of the triangle $ABC$. Prove that the triangle $ABC$ is equilateral.

First i will restate the problem. Let median $AM$, angle bisector $BL$ and the altitude $CH$ intersect at point $P$. Prove that if $P$ is on one of the perpendicular bisectors of $\Delta ABC$ than $\Delta ABC$ is equilateral.
[hide]By Ceva we have $\frac{AH*BM*CL}{HB*MC*LA}=1$ or $\frac{AH}{HB}=\frac{AL}{LC}$ so we see that $HL \parallel BC$
If $CH$ is a perp bisector than $LH$ is a medium line so we are done.
If $AM$ is a perp bisector so $AB=AC=b$ then $AL=AH$ and $CL=BH$ or
$AL=\frac{b^2}{a+b}=b*cos \alpha=AH$ and $CL= \frac{a*b}{a+b} = a*cos \beta$
from this we get $\ alpha=\ beta$
I’m stuck with the last case (shud be easy) :oops:[/hide][/hide]

[hide="Solution"]
It suffices to show that $P$ is the orthocentre as then $AU$ would be a median and altitude, making $AB$ = $AC$ and $BV$ would be an internal angle bisector and altitude, making $BA$ = $BC$ and thus $\triangle ABC$ is isosceles.

[b]Case 1[/b]
$P$ lies on the perpendicular bisector of $AB$

As $P$ lies on the concurrency of a median from $A$ and an internal angle bisector from $C$, it is not outside the triangle ABC. Given that unless the triangle is isosceles, the internal angle bisector and the perpendicular bisector of the opposite side meet outside the triangle $AC$ = $BC$, $CW$ is an altitude and so $P$ is the orthocentre.

[b]Case 2[/b]
$P$ lies on the perpendicular bisector of $BC$

$AU$ is therefore a perpendicular median and thus an altitude. Therefore $P$ is the orthocentre.

[b]Case 3[/b]
$P$ lies on the perpendicular bisector of $CA$

$BV$ is a bisecting altitude and thus a median. Therefore $P$ is the centroid and so $CW$ is also a median. By the Angle Bisector Theorem, $CA$ = $CB$ and so $\triangle AWC \cong \triangle BWC$. Therefore $\angle AWC$ is a right angle, $CW$ is an altitude and $P$ is the orthocentre.
[/hide]

By the way, the FST 2003 was the first Olympiad style paper I ever sat and I scored 2/21 on it. My only marks came from showing Case 2 and Case 3 from this question.


How many integers satisfy the equation $ |x+7|+|x-7|=14$?

[quote="ZhangPeijin"][hide]14: -1,-2, -3, -4, -5, -6, -7, 1, 2, 3, 4, 5, 6, 7.[/hide]

[color=darkred][size=75]Don't forget to hide your answers.
-nebula42[/size][/color][/quote]

Thats not right...

you forgot that 0 works so there are 15.

The general way to do it is to set boundaries for the absolute values to be positive/negative.


calculus
integration
rotation
combinatorics proposed
combinatorics

In a plane, there is an infinite triangular grid consists of equilateral triangles whose lengths of the sides are equal to $ 1$, call the vertices of the triangles the lattice points, call two lattice points are adjacent if the distance between the two points is equal to $ 1;$
A jump game is played by two frogs $ A,B,$ "A jump" is called if the frogs jump from the point which it is lying on to its adjacent point, " A round jump of $ A,B$" is called if first $ A$ jumps and then $ B$ by the following rules:
Rule (1): $ A$ jumps once arbitrarily, then $ B$ jumps once in the same direction, or twice in the opposite direction;
Rule (2): when $ A,B$ sits on adjacent lattice points, they carry out Rule (1) finishing a round jump, or $ A$ jumps twice continually, keep adjacent with $ B$ every time, and $ B$ rests on previous position;
If the original positions of $ A,B$ are adjacent lattice points, determine whether for $ A$ and $ B$,such that the one can exactly land on the original position of the other after a finite round jumps.

Frogs can exchange positions only if they are separated by a distance of integral multiple of 3 and it is allowed for one frog to sit on other.

If they are adjacent originally like in the picture (wlog) using rule 1 does not help but to increase the distance between the frogs. By using the rule 2 , red frog can move around blue frog on the green hexagon wlog( we can have blue frog moving around red frog but it is still same as this case). Red frog can takes positions 1,3,5. While red frog sits at any of these positions, by rule 1 (second aspect) red frog moves to center to the hexagon and blue frog moves to diagonally opposite vertex that happens to be a even vertex. From here blue frog cannot reach using rule 2 as it is sitting in a even vertex.

If frogs were initially not adjacent, we have two cases:
1. If frogs are not separated by a distance of multiple of 3 we can bring them adjacent to each other on the straight line connecting them by rule 1. By a cross over(second aspect of rule 1) we create a difference of 3 between the frogs and hence frogs cannot exchange position. So one of the frog has to stay static and the other has to rotate on the hexagon to take the diagonally opposite position. This we showed in the previous case as not possible.
2. If frogs are separated by a distance of multiple of 3 we can bring them at a distance of 0 and then the cross over, we can make them exchange positions.


geometry
incenter
circumcircle
trigonometry
AMC
AIME
perpendicular bisector

In triangle $ ABC$, $ AB \equal{} 10$, $ BC \equal{} 14$, and $ CA \equal{} 16$. Let $ D$ be a point in the interior of $ \overline{BC}$. Let $ I_B$ and $ I_C$ denote the incenters of triangles $ ABD$ and $ ACD$, respectively. The circumcircles of triangles $ BI_BD$ and $ CI_CD$ meet at distinct points $ P$ and $ D$. The maximum possible area of $ \triangle BPC$ can be expressed in the form $ a\minus{}b\sqrt{c}$, where $ a$, $ b$, and $ c$ are positive integers and $ c$ is not divisible by the square of any prime. Find $ a\plus{}b\plus{}c$.

[hide="Solution"]
[asy]unitsize(6mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));

pair B=(0,0), C=(14,0);
pair A=intersectionpoint(Circle(B,10),Circle(C,16));
pair D=(5,0), IC=incenter(A,D,C), IB=incenter(A,D,B);
pair P=intersectionpoints(circumcircle(B,D,IB),circumcircle(C,D,IC))[1];

draw(A--B--C--cycle);
draw(A--D--P);
draw(B--IB--D--IC--C--P--cycle);
draw(incircle(A,D,C));
draw(incircle(A,D,B));
draw(circumcircle(IB,D,B));
draw(circumcircle(IC,D,C));

label("$A$",A,N);
label("$B$",B,W);
label("$C$",C,E);
label("$D$",D,dir(220));
label("$P$",P,1.5*dir(35));
label("$I_B$",IB,N);
label("$I_C$",IC,N);[/asy]Let $ ADC \equal{} 2\theta$ for some $ \theta$; this uniquely determines $ D$. Since $ DI_C$ bisects $ \angle ADC$, $ I_CDC \equal{} \theta$. Similarly, $ I_CCD \equal{} \frac {C}{2}$, $ I_BDB \equal{} \frac {\pi}{2} \minus{} \theta$, and $ I_BBD \equal{} \frac {B}{2}$. Since $ I_BDPB$ and $ I_CDPC$ are cyclic,
\[ BPC \equal{} I_BPB \plus{} I_BPD \plus{} I_CPD \plus{} I_CPC \equal{} \frac {\pi}{2} \plus{} \frac {B}{2} \plus{} \frac {C}{2} \equal{} \pi \minus{} \frac {A}{2}.
\]So $ BPC$ is constant and thus the locus of $ P$ is part of a circle which includes $ B$ and $ C$, as any $ P_1$ and $ P_2$ will have $ BP_1C \equal{} BP_2C$ and thus $ BP_1P_2C$ is cyclic. We maximize the area of $ \triangle BPC$ by maximizing the distance of $ P$ from $ \overline{BC}$, since $ BC$ is constant. This occurs when $ P$ is on the perpendicular bisector $ m$ of $ \overline{BC}$. We now consider only this case.

The center of the circle which contains the $ B$, $ C$, and the locus of $ P$ will be on $ m$; call it $ R$. Major arc $ BC$ on this circle will have a measure of $ 2\pi \minus{} A$ and the minor arc will have a measure of $ A$, so $ BRC \equal{} A$:
[asy]unitsize(6mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;

pair B=(0,0), C=(14,0);
pair A=intersectionpoint(Circle(B,10),Circle(C,16));
pair P=(7,7*sqrt(3)-14);
pair R=(7,7*sqrt(3));
pair M=midpoint(B--C);

draw(A--B--C--cycle);
draw(B--P--C--R--cycle);
draw(Circle(R,14));

label("$A$",A,N);
label("$B$",B,SW);
label("$C$",C,SE);
label("$P$",P,S);
label("$R$",R,NE);
label("$M$",M,NW);
label("$m$",(7,14),SE);
draw(rightanglemark(R,M,C,10));

drawline(midpoint(B--C),bisectorpoint(B,C));
clip((-1,-4)--(-1,14)--(15,14)--(15,-4)--cycle);[/asy]By the Law of Cosines,
\[ 14^2 \equal{} 10^2 \plus{} 16^2 \minus{} 2\cdot10\cdot16\cdot\cos A \\
\cos A \equal{} \frac {1}{2} \\
A \equal{} \frac {\pi}{3}
\]So $ BRC \equal{} \frac {\pi}{3}$ as well and, since $ RB \equal{} RC$, $ \triangle BRC$ is equilateral. Let the midpoint of $ BC$ be $ M$; $ RM \equal{} 7\sqrt3$. Because $ R$, $ M$, and $ P$ are collinear, $ MP \equal{} RP \minus{} RM \equal{} 14 \minus{} 7\sqrt3$. The area of $ \triangle BPC$ is thus $ \frac {(14)(14 \minus{} 7\sqrt3)}{2} \equal{} 98 \minus{} 49\sqrt3$, giving an answer of $ 98 \plus{} 49 \plus{} 3 \equal{} \boxed{150}$.[/hide]

Hello, good solution but it is not complete. You haven't shown that the locus of $ P$ is a circle arc. You have shown that $ P$ must lie on that arc, but you have not shown that every point on that arc, there is a $ D$ for which $ P$ is the intersection of those two circles.

The only thing you need trig for is the proof that $ A \equal{} \frac {\pi}{3}$. It's a relatively simple use. There are ways to prove that without trig (drop an altitude from $ B$ or $ C$ and construct an equilateral triangle), but you would only use them if you already knew $ A$. I was only trying to find $ \cos A$, which is enough to solve the problem, but it made it even easier to find $ A$ was $ \frac {\pi}{3}$. The trig solution is far better motivated.

[quote="DataBox"]Dammit, I decided that I was going to guess a permutation of 5 1 0 as my answer and guessed 015, I knew I should have guessed 150 =([/quote]

:o I, on the other hand, guessed 105 :o

[quote="worthawholebean"]That follows from geometric continuity - $ P$ approaches $ B$ and $ C$, respectively, as $ D$ does.[/quote]

Of course you can claim that for when you are solving the problem during the AIME test but "follows from geometric continuity" is not a proof...

Essentially, you have to take an arbitrary point, $ P$, on that arc and show that there is a point $ D$ on $ BC$ that produces that $ P$.

Yeah. The thing is that the intuitive idea that it works is enough on AIME, even if it is not sufficient on anything proof-based.

What does "interior of BC" mean? I don't think it means on line BC, so how can you be sure that $ADB = \pi - ADC$ as worthawholebean's solution says?

Pretty sure "Interior of BC" means between points B and C on segment BC.

Sorry for bump, but here is a solution which follows from two applications of LoC:

[hide=Law of Cosines, AM-GM]From Law of Cosines on $\triangle{ABC}$, $$\cos{A}=\frac{16^2+10^2-14^2}{2\cdot 10\cdot 16}=\frac{1}{2}\implies\angle{A}=60^\circ.$$
Now,  $$\angle{CI_CD}+\angle{BI_BD}=180^\circ+\angle{\frac{A}{2}}=210^\circ.$$ Since $CI_CDP$ and $BI_BDP$ are cyclic quadrilaterals, it follows that $$\angle{BPC}=\angle{CPD}+\angle{DPB}=(180^\circ-\angle{CI_CD})+(180^\circ-\angle{BI_BD})=360^\circ-210^\circ=150^\circ.$$ Next, applying Law of Cosines on $\triangle{CPB}$,
\begin{align*}
& BC^2=14^2=PC^2+PB^2+2\cdot PB\cdot PC\cdot\frac{\sqrt{3}}{2} \\
& \implies \frac{PC^2+PB^2-196}{PC\cdot PB}=-\sqrt{3} \\
& \implies \frac{PC}{PB}+\frac{PB}{PC}-\frac{196}{PC\cdot PB}=-\sqrt{3} \\
& \implies PC\cdot PB = 196\left(\frac{1}{\frac{PC}{PB}+\frac{PB}{PC}+\sqrt{3}}\right).
\end{align*}
By AM-GM, $\frac{PC}{PB}+\frac{PB}{PC}\geq{2}$, so $$PB\cdot PC\leq 196\left(\frac{1}{2+\sqrt{3}}\right)=196(2-\sqrt{3}).$$
Finally,  $$[\triangle{BPC}]=\frac12 \cdot PB\cdot PC\cdot\sin{150^\circ}=\frac14 \cdot PB\cdot PC,$$ and the maximum area would be $49(2-\sqrt{3})=98-49\sqrt{3},$ so the answer is $\boxed{150}$.
[/hide]

I do not have the slightest idea how to start this question. Could I please get a hint.

[hide=Not too bad for a #15] By a quick use of LoC, we see $\angle BAC=60$. Next, note that $\angle BPC=\angle BPD+ \angle DPC$, and because quadrilaterals $BI_bDC$ and $PDI_cC$ are cyclic quadrilaterals, we see $\angle BPD=180-\angle BI_bD$, and $\angle DPC=180-\angle DI_cC$. This motivates us to find some correlation between the two, and a quick angle chase shows $90=\angle BI_bD-\frac{\angle BAD}{2}$, and similarly $90=\angle CI_cD-\frac{\angle DAC}{2}$. Summing gives $\angle BI_bD+\angle CI_cD=180+\frac{\angle BAC}{2}=210.$ We now see $\angle BPC=180+180-210=150$. We desire to find $\frac{1}{2}BP*PCsin(150)=\frac{1}{4}BP*PC$. Knowing that there is a $BP*PC$ in LoC, we are now motivated to use it. $196=BP^2+PC^2+\sqrt{3}BC*PC \iff 196=(BP+PC)^2-2BP*PC+\sqrt{3}BP*PC \ge 4BP*PC-2BP*PC+\sqrt{3}BP*PC$, for which the last result is supported by AM-GM. Simplifying yields $BP*PC\le 392-196\sqrt{3}$. Because we need a quarter of that, the maximum possible value for the area of $\Delta BPC= 98-49\sqrt{3}$, and so $98+49+3=\boxed{150},$ and we are done.

[quote=worthawholebean]I don't see why you cant think of $ P(D): \mathbb{C}\rightarrow\mathbb{C}$ and show continuity and limits. It might not be the easiest way, but you *can* do it.[/quote]

[quote=Altheman]Well okay I guess you can try that. But keep in mind that it is not obvious which maps are continuous. For example, if you are looking at two fixed points A,B and a line intersection AB. Suppose C is on that line and you define a map f(C)=H (orthocenter of ABC). Then this map is not continuous. (When C crosses AB, H flips across AB). So sure you can say I'll use geometric continuity but a) it isn't always true, b) it is difficult what you are trying to do and c) there are well-established methods for doing this problems. I do find it a bit disappointing that the math jam solutions don't speak to this point either...[/quote]

This is a good point.
What if the locus of $P$ were a constant point instead of an arc? In that case area will be a constant and the max area will be equal to this constant. And this constant may not be equal to the max area of the isosceles triangle circumscribed by the arc.

The official solution also ignored the locus part.

I proposed a similar question but with a tricky point.
[url=https://artofproblemsolving.com/community/u9551h2850791p25271117]Similar to AIME 2009-I P15[/url]

[b][color=#f00]Claim.[/b][/color] The angle $\angle BPC$ is constant.

\begin{align*}
\measuredangle BPC &= \measuredangle BPD+\measuredangle DPC \\
&= \measuredangle BI_BD + \measuredangle DI_CC \\
&= 90^\circ + \frac 12 \measuredangle BAD + 90^\circ + \frac 12 \measuredangle DAC\\
&= \frac 12 \measuredangle BAC. \ \blacksquare
\end{align*}

This means that $P$ moves along a fixed arc $\widehat{BC}$ in a fixed circle. Obviously, the maximum area occurs when $P$ is the arc midpoint. In particular, $\angle BPC = 150^\circ$ as we can compute $\angle A =60^\circ$ by the Law of Cosines. Then, $$[BPC]_{\mathrm{max}} = 7 \cdot \frac 7{\tan(75^\circ)}= 98-49\sqrt 3.$$

It's weird how this is the excircle incircle config but it is never used

bruh "The circumcircles of triangles $ BI_BD$ and $ CI_CD$"

I was kidding because there aren't actually excircles but I see your point.

Note that \[\angle BI_BD=90^{\circ}+\tfrac12\angle BAD\] \[\angle CI_CD=90^{\circ}+\tfrac12\angle CAD.\] By cyclic quadrilaterals $BI_BDP$ and $CI_CDP$, we see that \[\angle BPD=90^{\circ}-\tfrac12\angle BAD\] \[\angle CPD=90^{\circ}-\tfrac12\angle CAD,\] and it follows that \[\angle BPC=\angle BPD+\angle CDP=180^{\circ}-\tfrac12 \angle BAC.\] But by the Law of Cosines on $\triangle ABC$, we have \[\cos(\angle BAC)=\tfrac{AB^2+AC^2-BC^2}{2(AB)(AC)}=\tfrac{10^2+16^2-14^2}{2(10)(16)}=\tfrac12 \Rightarrow \angle BAC=60^{\circ}.\] Plugging this into the above expression for $\angle BPC$, we see that $\angle BPC=150^{\circ}$. Now it is clear that $[BPC]$ is maximized when $BP=PC=x$ (AM-GM), from which we get $[BPC]=\tfrac12x^2\sin(150^{\circ})=\tfrac14x^2$. By the Law of Cosines on $\triangle BPC$, we have \[14^2=x^2+x^2-2x^2\cos(150^{\circ}) \Rightarrow x^2=196(2-\sqrt3).\] Therefore, the answer is $\tfrac14(196-2\sqrt3)=\boxed{98-49\sqrt3}$.

[quote=AwesomeYRY]Oops this solution is very similar to the solution right above.

[hide= No trig needed]
Firstly, by LoC
\[\cos(\angle BAC) = \frac{10^2+16^2-14^2}{2*10*16}=\frac{1}{2}\Longrightarrow \angle BAC = 60\]
Thus,
\[\angle BPC = \angle BPD + \angle DPC = 180-\angle BI_BD+ 180-\angle DI_CC =\]
Due to incircles, this can be rewritten as
\[=180-\left(90+\frac{\angle BAD}{2}\right)+180-\left(90+\frac{\angle DAC}{2}\right)=360 - 180 - \frac{\angle BAC}{2} = 150 \]

Now by Law of Sines,
\[\frac{PC}{\sin(\angle PBC)} = \frac{BP}{\sin(\angle BCP)}=\frac{BC}{\sin (\angle BPC)}=\frac{14}{\sin(150)} = 28\].
Thus,
\[[PBC] = \frac{PC\cdot PB \sin(\angle BPC)}{2}=\frac{28^2\sin(\angle PBC)\cdot \sin(\angle BCP) \cdot \sin(150)} {2}\]
\[\leq 49*4 \cdot \sin(15)^2 = 49*4 (\frac{\sqrt{3}-1}{2\sqrt{2}})^2 = 49(2-\sqrt{3}) = 98-49\sqrt{3}\]
[/hide][/quote]

no trig needed :(


Find all pairs (x,y) of positive integers with y|x^2 + 1 and x|y^3 + 1

This was posted before (by yours truly :D). I saw it in an e-book containing various NT questions (the author is Hojoo Lee).

I think that the only solutions are those of the form (x,y)=(n2+n+1,n2+1) and the ones derived from these by the transformations (x,y)->((x2+1)/y,y) and (x,y)->(x,(y3+1)/x), but I have no idea how to prove this.


[quote="cincodemayo5590"]Determine all pairs of non-negative integers $(m,n)$ such that  $3 \cdot 2^{m}+1=n^{2}$.[/quote]

[hide]$(n-1)(n+1)=3\cdot 2^{m}$

$m=0$ gives $n=2$.

Then the factors on the LHS are of the same parity so then they are both odd.

$n=2a+1$, $(2a)(2a+2)=3\cdot 2^{m}$, $a(a+1)=3\cdot 2^{m-2}$.

The factors on the LHS are of different parity so

Case 1: $a=k\cdot 2^{m-2}$. Then $(k)(k\cdot 2^{m-2}+1)=3$, $m=3$, $k=1$, $n=5$.

Case 2: $a+1=k\cdot 2^{m-2}$. Then $(k)(k\cdot 2^{m-2}-1)=3$, $m=4$, $k=1$, $n=7$

So the solutions are $(m,n)=\boxed{(0,2)}, \boxed{ (3,5)}, \boxed{(4,7)}$?[/hide]


limit
trigonometry
calculus
calculus computations

Prove that :

$ \lim_{x\to \frac{\pi}{2}}(\frac{\pi}{2}\minus{} x)\tan x\equal{}0$

Again, does not belong to Pre-Olympiad.

[hide="Solution"]Since this indeterminate product is equivalent to $ 0\cdot \infty$, we can rewrite it as $ \lim_{x\to \frac{\pi}{2}}\frac{\frac{\pi}{2}\minus{}x}{\frac{1}{\tan x}}$ and apply l'Hospital's Rule.

Doing so, we have $ \lim_{x\to \frac{\pi}{2}}\frac{\frac{\pi}{2}\minus{}x}{\frac{1}{\tan x}}\equal{}\lim_{x\to \frac{\pi}{2}}\frac{\minus{}1}{\frac{\minus{}\sec^2 x}{\tan ^2x}}\equal{}\lim_{x\to \frac{\pi}{2}} \sin^2 x\equal{}1$.[/hide]

its more simple....

$ \lim_{x \to \frac{\pi}{2}} \frac{(\frac{\pi}{2}\minus{}x)six}{cosx}$

Now we can aply d'hospital rules ($ \frac{0}{0}$) so:

$ \lim_{x \to \frac{\pi}{2}} \frac{(\frac{\pi}{2}\minus{}x)six}{cosx}\equal{}\lim_{x\to \frac{\pi}{2}}\frac{\minus{}sinx\plus{}(\frac{\pi}{2}\minus{}x)cosx}{\minus{}sinx}\equal{}1$

${ x-\frac{\pi}{2}=\theta \Longrightarrow \lim_{x \rightarrow \frac{\pi}{2}} \left(\frac{\pi}{2}-x\right)\tan x=\lim_{\theta \rightarrow 0} \frac{\theta }{\tan \theta}}=\frac{1}{1}=1$.


Let [tex]x,y,z>=0, x+y+z=1, a,b>0[/tex]. What is the maximal value of [tex]x^ay^b+y^az^b+z^ax^b[/tex]?
Is there a general solution for this one? Maybe somebody knows a solution for some particular case (when a,b are integers, they belong to some interval or one of them is fixed etc.).

For [tex]a+b \le 1[/tex], we have

[tex]\displaymode{
x^ay^b + y^az^b + z^ax^b \le x^{a+b} + y^{a+b} + z^{a+b} \le 3\left(\frac{x+y+z}{3}\right)^{a+b} = 3^{1-a-b}
}[/tex]

by Rearrangement and Jensen respectively, where equality holds iff [tex]x = y = z = \frac{1}{3}[/tex].  Unfortunately in the more difficult case [tex]a+b > 1[/tex] the maximizing conditions do not line up as nicely.  By Rearrangement and the [tex]p[/tex]-norm inequality, we have

[tex]\displaymode{
x^ay^b + y^az^b + z^ax^b \le x^{a+b} + y^{a+b} + z^{a+b} \le \left(x + y + z\right)^{a+b} = 1
}[/tex]

but equality does not line up and this inequality is strict.  I do not know the  exact maximum.

I think your argument can be generalized when $a+b>1$ . The bound is given by
$x^{a}y^{b}+y^{a}z^{b}+z^{a}x^{b}<=x^{a+b} + y^{a+b}+z^{a+b}<=1$
Ooops, sorry for this post, I forgot to refresh the page and I didn't notice your post.
I edited mine for this comment.
Now that I edited it, my opinion is that the answer should be something like the fact that we may take one as 0, the $x+y=1 , x^{a}y^{b}<={\frac{ax+by}{a+b}}^{a+b}$. Something like this. A canadian inequality I think was for case $a=1 , b = 3$.
Check out http://www.kalva.demon.co.uk/canada/can99.html

It was posted before the inequality
$\displaystyle (x^2+y^2+z^2)^2\geq 3(x^3y+y^3z+z^3x)$.
Thus, for $\displaystyle a=\frac{3}{2}$ and $\displaystyle b=\frac{1}{2}$ , we have
$\displaystyle x^ay^b+y^az^b+z^ax^b\leq \frac{1}{3}$.
I think the inequality holds
$\displaystyle xy^b+yz^b+zx^b\leq \frac{1}{3^b}$
 for $\displaystyle b\leq\frac{3}{2}$.

[quote="Ji Chen"][quote="Vasc"]Let $x,y,z\geq0,\ x+y+z=1$. I think the inequality $xy^{p}+yz^{p}+zx^{p}\leq3^{-p}$ holds for $0<p\leq\frac{3}{2}$.[/quote][/quote]
The inequality above is proved in Algebraic Inequalities (page 137) for $0<p \le m$, where $m=1.558...$ is a root of the equation $(1+m)^{1+m}=(3m)^{m}$.


algebra
polynomial
function
trigonometry
complex numbers
absolute value
AMC

The distinct roots of the polynomial $x^{47} + 2x^{46} + 3x^{45} + \cdots + 24x^{24} + 23x^{23} + 22x^{22} + \cdots + 2x^2 + x$ are $z_k$ for $1 \le k \le n$. Let $z_k^2$ have imaginary part $b_ki$ Find $|b_1|+|b_2|+\cdots + |b_n|$.

[hide="Answer"]Note that this problem simplifies to $x(1+x+x^2+...+x^{22}+x^{23})^2$. Therefore, it has the $24^{th}$ roots of unity except for $1$ as roots, these being $e^\frac{\pi}{12}$, $e^\frac{\pi}{6}$, etc. When we square these roots, we are left with two times the $12^{th}$ roots of unity. Taking the absolute value, we end with $2\left(4\cdot \frac{1}{2}+2\cdot 1+4\cdot \frac{\sqrt{3}}{2}\right)=\boxed{8+4\sqrt{3}}$.[/hide]

Well... I kind of remembered that $(1+x)(1+x)=1+2x+x^2$, $(1+x+x^2)(1+x+x^2)=1+2x+3x^2+2x^3+x^4$, etc., so I inferred it. *shrug* You can expand it if you want, it'll work.

[quote="chess64"]How did you factor that though?[/quote]

[hide="Roots of unity!"]
Let $x$ be a 24th root of unity (not equal to 1) and you can use the identities $x^{24}=1$ and $x^{23}+x^{22}+\cdots+1=0$ to see that the equation has the 23 roots of unity (excluding 1) as roots.

But here's an even cooler trend!

\[ (x^k+x^{k-1}+\cdots+1)^2 = x^{2k}+2x^{2k-1}+\cdots+(k+1)x^k+\cdots+2x^2+x+1.  \]

For example

\begin{eqnarray*} 1^2 &=& 1\\ (x+1)^2 &=& x^2+2x+1\\ (x^2+x+1)^2 &=& x^4+2x^3+3x^2+2x+1\\ (x^3+x^2+x+1)^2 &=& x^6+2x^5+3x^4+4x^3+3x^2+2x+1\\ &\vdots& \end{eqnarray*}

Roots of unity...aren't they lovely?

Also, when expanding out those LHS, just go down the line.  e.g. if you wanted to find the number of $x^3$ terms in $(x^3+x^2+x+1)^2$ start with $x^3$...it can pair up with 1, $x^2$ pairs up with $x$, $x$ pairs up with $x^2$, and 1 pairs up with $x^3$ for a total of 4.[/hide]

Edit: Want some [hide="generating functions?"]

Note that

\begin{eqnarray*} (1+x+\cdots+x^k)^2 &=& \left(\frac{x^{k+1}-1}{x-1}\right)^2\\ &=& (x^{2k+2}-2x^{k+1}+1)\cdot\frac 1{(x-1)^2}\\ &=& (x^{2k+2}-2x^{k+1}+1)(1+2x+3x^2+4x^3+\cdots)\\ &=& (x^{2k+2}+2x^{2k+3}+3x^{2k+4}+\cdots)-2(x^{k+1}+2x^{k+2}+3x^{k+3}+\cdots)+(1+2x+3x^2+\cdots)\\ &=& x^{2k}+2x^{2k-1}+\cdots+(k+1)x^k+\cdots+2x+1 \end{eqnarray*}[/hide]

Okay, cool. Let me try to solve it now. :D

[hide]
Let $p(x) = x(1+x+x^2+\cdots + x^{23})^2$. Let $\omega$ be a 24th root of unity, so that $\omega\left(\frac{\omega^{24}-1}{\omega-1}\right)=0$, where $\omega \ne 1$, so the roots are the 24th roots of unity not equal to 1. These are $z_k=e^{\pi j i/12}$, where $j=1,2,\ldots , 23$. Hence, $z_k^2 = e^{\pi j i/6}=\text {cis } (\pi j/6)$, which has imaginary part $i\sin (\pi j/6)$. The sum of the absolute values is \[ 2[|\sin (\pi/6)| + |\sin (\pi/3)| + |\sin (\pi/2)| + |\sin (2\pi/3)| + |\sin (5\pi/6)| + |\sin (\pi)| + \\|\sin (7\pi/6)| + |\sin (4\pi/3)| + |\sin (3\pi/2)| + |\sin (5\pi/3)| + |\sin (11\pi/6)| + |\sin (2\pi)|]-|\sin (4\pi)|, \] or \[ 2[1/2 + \sqrt{3}/2 + 1 + \sqrt{3}/2 + 1/2 + 0 + 1/2 + \sqrt{3}/2 +1 +\sqrt{3}/2 + 1/2 + 0], \] which is \[ 2[(1/2)(4)+1(2)+\sqrt{3}/2(4)]=\boxed {8+4\sqrt{3}}. \][/hide]

How did you evaluate the sum, JF197?

Yea, I noticed that factoring immediately just from working with generating functions occasionally (two identical g. functions will ALWAYS look like that).

[quote="chess64"]How did you evaluate the sum, JF197?[/quote]

We have the $12^{th}$ roots of unity doubled, so we obviously just deal with the $12^{th}$ roots of unity. Since they're expressed as $\cos \theta +i\sin \theta$, the imaginary part is obviously the $\sin$ portion. Then, since we're dealing with absolute values, we have $4\sin 30^\circ +4\sin 60^\circ +2\sin 90^\circ=4+2\sqrt{3}$, since $\sin 0^\circ =\sin 180^\circ =0$. ;)

[quote="joml88"][quote="zanttrang"][quote="chess64"]How did you factor that though?[/quote]
It's an extension of the formula $(x + 1)^2 = x^2 + 2x + 1$.  If you do $(x^2 + x + 1)^2 = x^4 + 2x^3 + 3x^2 + 2x + 1$, you'll start to see the pattern that emerges.  I'm sure there's a way to prove it, but intuitively you can see the result.

Actually, I just remembered how to prove it:

Suppose you have $(x^n + x^{n-1} + ... + x + 1)^2$.  There is one way to select a term with power $2n$, two ways to select a term with power $2n- 1$ ($x^{n-1} * x^{n}$ or $x^{n} * x^{n-1}$), and so on.  You just see how many ways you can select each power.[/quote]

Hmm, isn't that pretty much what I said ;)[/quote]
Yes and no - my proof takes a slightly more combinatorical approach than yours.  Also, I didn't read your post before making mine  :P


As my gift for you, silouan, let's prove the following inequality. Welcome all Mathlinkers,

Let $a,b,c$ be positive real numbers satisfying $a^{2}+b^{2}+c^{2}=3$. Prove that
\[\frac{a}{a^{2}+2b+3}+\frac{b}{b^{2}+2c+3}+\frac{c}{c^{2}+2a+3}\le \frac{1}{2}.\]

Thank you very much hungkhtn , your problem is very nice .
I found a nice solution and I hope you will like it .
By AM-GM in the denominator $a^{2}+1\geq 2a$ it suffices to prove that
$\frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\leq 1$ or equivalently

$\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}+\frac{a+1}{c+a+1}\geq 2$
or equivalenlty

$\frac{(b+1)^{2}}{(a+b+1)(b+1)}+\frac{(c+1)^{2}}{(b+c+1)(c+1}+\frac{(a+1)^{2}}{(c+a+1)(a+1)}\geq 2$

By Cauchy-Swartz in Engel form we have that
$\frac{(b+1)^{2}}{(a+b+1)(b+1)}+\frac{(c+1)^{2}}{(b+c+1)(c+1}+\frac{(a+1)^{2}}{(c+a+1)(a+1)}\geq\frac{(a+b+c+3)^{2}}{(ab+bc+ca+3(a+b+c)+6}$

so it suffices to prove that
$\frac{(a+b+c+3)^{2}}{(ab+bc+ca+3(a+b+c)+6}\geq 2$

or $(a+b+c)^{2}\geq 2(ab+bc+ca)+3$  or $(a+b+c)^{2}\geq 2(ab+bc+ca)+a^{2}+b^{2}+c^{2}$ or $(a+b+c)^{2}\geq (a+b+c)^{2}$ which is true

Thank you again .

Hi Silouan,

You solution is very nice and... it it totally the same as mine.

I found the last identity

\[(a+b+c+3)^{2}=2(ab+bc+ca+3(a+b+c)+6)\]

is very surprising.

[quote="hungkhtn"]
You solution is very nice and... it it totally the same as mine.
[/quote]

I am glad to solve your nice gift , thank you for this .
Also this indetity is beautiful and unexpectable. :wink:  Nice


[quote="kunny"]Let $ a,\ b,\ c\ ;\ x,\ y,\ z$ be real numbers  such that $ a\plus{}b > c,\ b\plus{}c > a,\ c\plus{}a > b,\ ;\ x^{2}\plus{}y^{2}\plus{}z^{2}\neq 0,\ a^{2}x\plus{}b^{2}y\plus{}c^{2}z \equal{} 0$.
Determine the sign of $ xy\plus{}yz\plus{}zx$.[/quote]
The answer is "$ \minus{}$" because $ \Delta<0.$ :D


How many permutations of {1,2,3,4,5,6,7,8} have multiple cycles?

I think I found a hint to this problem:
Such a permutation acts as a function, taking k to  f(k) (k=1,2,3,4,...8), where {f(1),f(2),...,f(8)}={1,2,3,...8}. A cycle is a list of elements  a[size=59]1[/size],a[size=59]2[/size],a[size=59]3[/size],...,a[size=59]k[/size] where f(a[size=59]1[/size])=a[size=59]2[/size], f(a[size=59]2[/size])=a[size=59]3[/size], ... ,f(a[size=59]k[/size])=a[size=59]1[/size]

I still can't solve it though... :blush:

don't you have to define the function first o something?

isnt the answer the number of total permutations (40320) minus the number of permutations with 1 cycle (idk but it might be like 5040)?
im not too sure what a cycle is

Here's my understanding of cycle. We have a permutation $ P$ of $ 1...n$, which we represent as a $ n$-tuple. For instance, one such is $ (1,2,3,4,5,6,8,7)$. Now, let $ f(x)$ be defined on $ 1,2,3,...,n$, so that $ f(x)$ is the $ x$th element of $ P$. For instance, for $ P\equal{}(1,2,3,4,5,6,8,7)$, $ f(2)\equal{}2$, $ f(7)\equal{}8$.

An [i]$ n$-cycle[/i] is a set $ S$ of elements of $ 1,2,3,...,n$ so that $ f(x)$, $ f(f(x))$, ..., $ f^{n\minus{}1}(x)$ are all in the $ n$-cycle (but not equal to $ x$) for any $ x$ in the $ n$-cycle, and $ f^n(x)\equal{}x$.  For instance, in the example, $ (7,8)$ is a 2-cycle: it goes f(7)=8, f(f(7))=7.

In common parlance, cycles are written in parenthesis: for instance, in the example, we'd write it (1)(2)(3)(4)(5)(6)(7,8) . Often we drop 1-cycles when writing in this manner, as they are slightly redundant, so the example would be (7,8).

In general, a cycle is any $ n$-cycle, including a 1-cycle.

Now then, how many are there?

(This discussion is coming close to flying to HSB.)

Think of permutations like this.
If you have a locker, people call it a combination lock.
Combinations mean that order does not matter, but on a lock, order does indeed matter.
So the right name should be Permutation lock.
So in the set {1,2,3,4,5,6,7,8}
a permuation could be changing the order because in a permutation order does matter.
i.e
{1,2,3,4,5,6,7,8} --->{1,2,3,4,5,6,8,7} counts as a permutation.
And unless they state that the starting set doesn't count, i think the starting set does
count as a permutation.

Think about it:
Let's have a fixed starting point (say 1)
we have 7 ways to choose the next point
then 6 ways
then 5 ways
...
then we have 1 way to choose the last point, for which we start over back to 1.
So we have 7*6*5*4*3*2*1=7!

I'm not so sure, but that's what I got.


Let $x$ and $y$ be two natural numbers and $x^2+2y$ be a perfect square.
Prove that we can write $x^2+y$ as a sum of two perfect squares.

Thus, there exist $k$ st, $2y = (k-x)(k+x)$ Since $k-x = k+x \mod 2$, $k=x \mod 2$ for it to be even. Thus $x^2 + y = \frac{k^2 + x^2}{2} = (\frac{k-x}{2})^2 + (\frac{k+x}{2})^2$


Where is a good place to learn to write your own style files for $\LaTeX$. Mostly I am looking for information on how to write style file that will take an optional argument, and then define some command and environments based on this.

You could try [url=http://www.latex-project.org/guides/clsguide.pdf]LaTeX2e for class and package writers[/url]. There's also advice at [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=writecls]Learning to write LaTeX classes and packages[/url].


[hide="Let this factor be d..."]
... and clearly neither 2 nor 3 divide $d$.
$11!=2^{8}\cdot 3^{4}\cdot 5^{2}\cdot 7 \cdot 11$.
Hence $d|(5^{2}\cdot 7 \cdot 11)$.
$d$ cannot equal $(5^{2}\cdot 7 \cdot 11)$ because that is one [i]less[/i] than a multiple of $6$, and the next largest number that divides $(5^{2}\cdot 7 \cdot 11)$ is $5 \cdot 7 \cdot 11 = 385$ which is one more than a multiple of $6$.
Hence $d=385$[/hide]

[quote="espark52"]what is the greatest factor of $11!$ that is one greater than a multiple of 6?[/quote]
Wasn't this a team round question (MathCounts)? Why is this in Intermediate?


calculus
integration
logarithms
limit
calculus computations

i asked this question in an extended exam last week,
and only 4 out of 40+ students got it completely right!   :(

solve this integral:

$\int_{0}^\infty\;e^{-x}\cdot\ln^{2}x\;\textbf{d}x$

[hide]
$answer: \;\;\;\quad \boxed{\gamma^{2}\;+\;\zeta{(2)}}$
[/hide]

"i asked this question in an extended exam last week"

what kind of class is this?


trigonometry
inequalities
geometry proposed
geometry

Let $ABC$ be a triangle
  Prove that :
 $\frac{\sin^{2}\frac{A}{2}}{\cos B+\cos C}+\frac{\sin^{2}\frac{B}{2}}{\cos A+\cos C}+\frac{\sin^{2}\frac{C}{2}}{\cos B+\cos A}\leq \sin^{2}\frac{A}{2}+\sin^{2}\frac{B}{2}+\sin^{2}\frac{C}{2}$

[color=blue][b]Problem.[/b] Let ABC be a triangle. Prove the inequality

$\sin^2\frac{A}{2}+\sin^2\frac{B}{2}+\sin^2\frac{C}{2}\geq\frac{\sin^2\frac{A}{2}}{\cos B+\cos C}+\frac{\sin^2\frac{B}{2}}{\cos C+\cos A}+\frac{\sin^2\frac{C}{2}}{\cos A+\cos B}$.[/color]

[i]Solution.[/i] Using the $\sum$ notation, the inequality in question rewrites as

$\sum\sin^2\frac{A}{2}\geq\sum\frac{\sin^2\frac{A}{2}}{\cos B+\cos C}$.

But after a well-known formula from trigonometry,

$\cos B+\cos C=2\cos\frac{B+C}{2}\cos\frac{B-C}{2}$.

By the sum of angles in triangle ABC, we have A + B + C = 180°, so that B + C = 180° - A, and thus $\cos\frac{B+C}{2}=\cos\frac{180^{\circ}-A}{2}=\cos\left(90^{\circ}-\frac{A}{2}\right)=\sin\frac{A}{2}$. Thus, we get

$\cos B+\cos C=2\sin\frac{A}{2}\cos\frac{B-C}{2}$.

Hence,

$\frac{\sin^2\frac{A}{2}}{\cos B+\cos C}=\frac{\sin^2\frac{A}{2}}{2\sin\frac{A}{2}\cos\frac{B-C}{2}}=\frac12\cdot\frac{\sin\frac{A}{2}}{\cos\frac{B-C}{2}}$.

By the Mollweide formula, $\frac{\cos\frac{B-C}{2}}{\sin\frac{A}{2}}=\frac{b+c}{a}$, o that $\frac{\sin\frac{A}{2}}{\cos\frac{B-C}{2}}=\frac{a}{b+c}$ and

$\frac{\sin^2\frac{A}{2}}{\cos B+\cos C}=\frac12\cdot\frac{\sin\frac{A}{2}}{\cos\frac{B-C}{2}}=\frac12\cdot\frac{a}{b+c}=\frac{a}{2\left(b+c\right)}$.

On the other hand, by the half-angle formulas,

$\sin^2\frac{A}{2}=\frac{\left(s-b\right)\left(s-c\right)}{bc}$,

where $s=\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC. Thus, the inequality that we have to prove,

$\sum\sin^2\frac{A}{2}\geq\sum\frac{\sin^2\frac{A}{2}}{\cos B+\cos C}$,

becomes

$\sum\frac{\left(s-b\right)\left(s-c\right)}{bc}\geq\sum\frac{a}{2\left(b+c\right)}$.

This inequality is clearly equivalent to

$\sum\frac{\left(s-b\right)\left(s-c\right)}{bc}-\sum\frac{a}{2\left(b+c\right)}\geq 0$.

So we have to prove this inequality. But, well, by performing some algebra, keeping in mind that $s=\frac{a+b+c}{2}$, we see that

$\sum\frac{\left(s-b\right)\left(s-c\right)}{bc}-\sum\frac{a}{2\left(b+c\right)}=\sum\left(\frac{\left(s-b\right)\left(s-c\right)}{bc}-\frac{a}{2\left(b+c\right)}\right)$
$=\sum\frac{1}{2abc}\left(\frac{\left(s-c\right)ab}{b+c}\left(a-b\right)-\frac{\left(s-b\right)ca}{b+c}\left(c-a\right)\right)$
$=\frac{1}{2abc}\sum\left(\frac{\left(s-c\right)ab}{b+c}\left(a-b\right)-\frac{\left(s-b\right)ca}{b+c}\left(c-a\right)\right)$.

Thus, in order to prove the inequality

$\sum\frac{\left(s-b\right)\left(s-c\right)}{bc}-\sum\frac{a}{2\left(b+c\right)}\geq 0$,

it will be enough to verify that

$\sum\left(\frac{\left(s-c\right)ab}{b+c}\left(a-b\right)-\frac{\left(s-b\right)ca}{b+c}\left(c-a\right)\right)\geq 0$.

But

$\sum\left(\frac{\left(s-c\right)ab}{b+c}\left(a-b\right)-\frac{\left(s-b\right)ca}{b+c}\left(c-a\right)\right)$
$=\sum\frac{\left(s-c\right)ab}{b+c}\left(a-b\right)-\sum\frac{\left(s-b\right)ca}{b+c}\left(c-a\right)=\sum\frac{\left(s-a\right)bc}{c+a}\left(b-c\right)-\sum\frac{\left(s-a\right)bc}{a+b}\left(b-c\right)$
$=\sum\left(\frac{\left(s-a\right)bc}{c+a}\left(b-c\right)-\frac{\left(s-a\right)bc}{a+b}\left(b-c\right)\right)$
$=\sum\left(\left(s-a\right)bc\left(b-c\right)\left(\frac{1}{c+a}-\frac{1}{a+b}\right)\right)$.

Now, if we have $b\geq c$, then $b-c\geq 0$ on the one hand, while $c\leq b$ on the other hand, so that $c+a\leq a+b$ and thus $\frac{1}{c+a}\geq\frac{1}{a+b}$ and thus $\frac{1}{c+a}-\frac{1}{a+b}\geq 0$. Together with $b-c\geq 0$ and s - a > 0, this yields $\left(s-a\right)bc\left(b-c\right)\left(\frac{1}{c+a}-\frac{1}{a+b}\right)\geq 0$. This was for the case when $b\geq c$; but we can similarly see that $\left(s-a\right)bc\left(b-c\right)\left(\frac{1}{c+a}-\frac{1}{a+b}\right)\geq 0$ also holds in the case $b\leq c$. Hence, we always have $\left(s-a\right)bc\left(b-c\right)\left(\frac{1}{c+a}-\frac{1}{a+b}\right)\geq 0$. Thus,

$\sum\frac{\left(s-b\right)\left(s-c\right)}{bc}-\sum\frac{a}{2\left(b+c\right)}=\sum\left(\left(s-a\right)bc\left(b-c\right)\left(\frac{1}{c+a}-\frac{1}{a+b}\right)\right)\geq 0$.

Thus, our inequality is proven.

Probably the problem is hard since one tends to writing out everything in terms of cos A, cos B, cos C, what does not seem to yield the solution directly. Using the halfangle and the Mollweide formulas, the problem becomes straightforward. Anyway, I guess there are different solutions.

Greetings from Bad Homburg (I am posting directly from the fourth IMO training seminar).

  Darij


What is the number of square inches in the area of a rectangle having dimensions $1\frac{1}{2}$ yards by $1\frac{1}{2}$ feet?

[hide]1 1/2 yards is 4.5 feet. 4.5 times 1.5 is 6.75 square feet, and since area is squared, that is 6.75 * 144 square inches, which comes down to 972 square inches.[/hide]

[hide]
1.5 yards is 54 inches. 1.5 feet is 18 inches. 18x54=972 sq. inches[/hide]

[hide]1.5 yards is equivalent to 4.5 feet.

So the area would be 4.5* 1.5

Which is 6.75 square feet.

6.75*144 would give you the sqaure inches

which is 972[/hide]


Given three positive constants $a, b, c$ and

$\frac{a}{x} + \frac{b}{y} + \frac{c}{z} = 1$

find the maximum value of $x+y+z$ with $x, y, z$ are three positive variables.

well you could use Cauchy and get

$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge{\frac{1}{a^2+b^2+c^2}}$

and from here i guess you could use AM-HM, but im not really sure.  :blush:

there certainly isn't a maximum, since $x$ and $y$ can be made arbitrarily large and $z$ can be made close enough to $c$ to satisfy $\frac ax +\frac by +\frac cz =1$, which can be done by $z=\frac{c}{1-\frac ax -\frac by}=\frac{cxy}{xy-ay-bx}$. For the simpler $x=y$ this becomes $z=\frac{cx}{x-a-b}$ which is positive for all $x>a+b$.

[hide="maybe minimum?"] $x+y+z=(x+y+z)\left( \frac ax +\frac by +\frac cz \right)\geq \left( \sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2$ by Cauchy.

In order to attain the minimum, $\frac{a}{x^2}=\frac{b}{y^2}=\frac{c}{z^2}=k$, which would make $k(x+y+z)=1$. So $k$ would have to equal $\frac{1}{\left( \sqrt{a}+\sqrt{b}+\sqrt{c} \right)^2}$ which gives $x=a+\sqrt{ab}+\sqrt{ac}$. Similarly $y=b+\sqrt{ab}+\sqrt{bc}$ and $z=c+\sqrt{ac}+\sqrt{bc}$.

These values of $x,y,z$ attain the minimum, so it is $\boxed{\left( \sqrt{a}+\sqrt{b}+\sqrt{c} \right)^2}$.[/hide]

Actually I think what you say is true.  May be the book is wrong, minimum instead of maximum.  This book is quite buggy. :)