geometry
3D geometry
number theory unsolved
number theory

[quote="April"]Which positive integers are missing in the sequence $ \left\{a_n\right\}$, with $ a_n \equal{} n \plus{} \left[\sqrt n\right] \plus{}\left[\sqrt [3]n\right]$ for all $ n \ge 1$? ($ \left[x\right]$ denotes the largest integer less than or equal to $ x$, i.e. $ g$ with $ g \le x < g \plus{} 1$.)[/quote]
We miss at least one number when $a_{n+1}>a_n+1$ and so in two cases :
$[\sqrt{n+1}]\ge p>[\sqrt n]$ $\iff$ $n=p^2-1$
$[\sqrt[3]{n+1}]\ge p>[\sqrt[3] n]$ $\iff$ $n=p^3-1$

If $n+1=p^2$ is a perfect square but not a perfect cube, we skipped $p^2+p-1+[\sqrt[3]{p^2}]$

If $n+1=p^3$ is a perfect cube but not a perfect square, we skipped $p^3+p-1+[\sqrt{p^3}]$

If $n+1=p^6$ is both perfect square and perfect cube, we skipped $p6+p^3+p^2-1$ and $p^6+p^3+p^2-2$

And so missing numbers are
$p^2+p-1+[\sqrt[3]{p^2}]$ $\forall p\in\mathbb N$
$p^3+p-1+[\sqrt{p^3}]$ $\forall p\in\mathbb N$
$p^6+p^3+p^2-2$ $\forall p\in\mathbb N$


linear algebra
matrix
linear algebra theorems

I read somewhere on this forum that R(U+T) belongs to (is subset of ) R(U)+R(T)
I think they are equivalent (that the range of U+T is the sum of ranges of U and T)
Is it wrong and if it is, why?

Let $ U\equal{}\left(\begin{array}{cc} 1 & 0 \\ 0 & \minus{}1 \end{array}\right)$.  The range of the transformation corresponding to $ U$ is the whole space, and the range of the identity transformation is the whole space, but what is the range of the transformation corresponding to $ I\plus{}U$?

I get it, thanks... and the logic that R(U+T) is contained in R(U) +R(T)would be that for any particular A and B, the transformation corresponding  to A+B can be obtained by applying A combined with (added to) an action of B?


The area of a rectangular flower garden is to be 32 ft$ ^2$. Given that the dimensions must be whole numbers, how many feet are in the greatest perimeter?

32 can be factored into:

1 by 32
2 by 16
4 by 8

Greatest perimeter: 1 by 32: $ 2(1)\plus{}2(32)\equal{}2\plus{}64\equal{}\boxed{66}$ feet.


[quote="Valentin Vornicu"][quote="conceive"]Will MathLinks Contest continue in some time in future?[/quote]Definitely yes. I will have to find volunteers though to help me because I will not have the time to do it all on my own.[/quote]
Really....   :D :thumbup: ...great...It will help us a lot i hope

Thank u...but the main question is...when? :maybe:


[hide="Hint"]Draw the remaining internal tangent and use symmetry and the tangent property (e.g. $PA=PC, PD=PB$ etc.) to find the pairs of equal segments.[/hide]

yes, but the case is why $|PC|=|PA|$?
so may I please another (more concrete) hint?  :oops:

That's the general property of tangents to a circle: If you draw two tangents from the same point to the same circle, the segments are equal in length. Easy to prove - connect the point and the center and find congruent triangles.

Do you have any elementary knowledge in geometry?

I have a feeling that [hide]power of a point[/hide] is going to come in handy...

:wallbash_red:  :moose:

yeah... sometimes the most obvious thing is so hard to find.
sorry for this silly question, and thx for your help

[quote="bane"][img]http://img158.imageshack.us/img158/4851/orgqk8.png[/img]

Lines $k,l,m$ are tangent to the circles. I'm supposed to show/justify that $|AB|=|PQ|$.

Any ideas? :)[/quote]

Tangent property of circles is what you use to prove it


I recently heard about this music download website allofmp3, they charge money per MB, around .1 cent a song and 1 dollar an album.
this is too good to be true, especially compared to iTunes

Anyboday got any comments?

I really dont know how long that site will continue to exist......iTunes will surely last for sometime though.... :D

it's going to last pretty long, because it's a Russian legal site.  But in the US i don't know.

[quote="Chocoboom"]why would i wawnnna pay 1 cent for a song when i can download it illegally for zero cents   :P[/quote]

Because most of us here believe in [b]LEGAL[/b] music............... :mad:

because u don't want to be that one unlucky guy who gets sued

Well if the site is old enough to be trusted,only then is it an amazing offer.Luckily I get all my favourite songs on free downloads.

i download songs free LEGALLY!!!!!!!
limewire rules all....
[url=http://www.limewire.com/english/content/home.shtml]limewire site, remember completely LEGAL[/url]

[quote="furious"][quote="slamchess"]is it true that limewire is legal??

most people i no say it is illegal...[/quote]
no. well, technically having limewire is perfectly fine but actually using it to download is illegal (i think that's how it works)[/quote]

Using it to download copyright things without permission or reinbursement like most songs is illegal, however, downloading things in the public domain or a picture would most probably be legal.

the good thing about allofmp3 is that it's based in Russia, and the copyright law in russia have loopholes that make allofmp3 legal

what i learned is that limewire is legal as long as you don't download something and say its your own and put it on the internet.


[hide]
Let $ t\equal{}x\plus{}2$

so the given equation is :
   $ (t\minus{}1)t(t\plus{}1)(t\plus{}2)\plus{}1$
=  $ (t^2\plus{}t\minus{}2)(t^2\plus{}t) \plus{}1$  [Multiplying the 1st with 4th & 2nd with 3rd respectively]
=  $ (A\minus{}2)A\plus{}1$    Taking $ A\equal{}t^2\plus{}t$
=  $ (A^2)\minus{}2A\plus{}1$
=  $ (A\minus{}1)^2$
=  $ (t^2\plus{}t\minus{}1)^2$

and then substitute $ t\equal{}x\plus{}2$ as assumed earlier .

Hope that helps ..   :)
[/hide]

i think you mean expand:

$ (x\plus{}1)(x\plus{}2)(x\plus{}3)(x\plus{}4)\plus{}1$
because they are consecutive integers.let $ 1\equal{}a$
$ (x\plus{}a)(x\plus{}(a\plus{}1))(x\plus{}(a\plus{}2))(x\plus{}(a\plus{}3))$
$ \equal{}x^4\plus{}(4a\plus{}6)x^3\plus{}(6a^2\plus{}18a\plus{}11)x^2\plus{}(4a^3\plus{}18a^2\plus{}22a\plus{}6)x\plus{}a^4\plus{}6a^3\plus{}11a^2\plus{}6a$
$ \equal{}x^4\plus{}10x^3\plus{}35x^2\plus{}50x\plus{}24$

$ \implies (x\plus{}1)(x\plus{}2)(x\plus{}3)(x\plus{}4)\plus{}1\equal{}x^4\plus{}10x^3\plus{}35x^2\plus{}50x\plus{}25$

the answer is $ \boxed{\color{red}x^4\plus{}10x^3\plus{}35x^2\plus{}50x\plus{}25}$

Well, he might have meant factor, because you still have the extra 1.

So, binomial, if you factor your answer, then I think that might be the solution he's looking for.

This problem is very similar to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=435103]#1 on the 1989 AIME[/url].


polynomial
algebra
calculus
integration
college
greatest common divisor
algebra solved

Let p be a prime number such that:

a.) p is the greatest common divisor of a and b;
b.) p^{2} is a divisor of a.

Prove that the polynomial x^{n+2} + ax^{n+1} + bx^{n} + a + b cannot be factored as a product of two polynomials with integral coefficients, whose degrees are greater than one !

Note: see also IMO1993_1

Let f(x)=x^{n+2} + ax^{n+1} + bx^{n} + a + b. Denote p|a, p|b, p|a+b and a+b is not divisible by p2.

Suppose f(x)=(bmxm+bm-1xm-1+...+b1x+b0).(ckxk+ck-1xk-1+...+c1x+c0), where bk=ck=1.
Then a+b=b0c0 ==> exactly one of b0 or c0 is divisible by p. Suppose WLOG p|b0 and c0 is not divisible by p.

Suppose we proved that p|bl, p|bl-1, ..., p|b0, l<m. Monom xl+1 have in f(x) coefficient which is divisible by p ==> bl+1c0 + blc1 + ... is divisible by p ==> bl+1 is divisible by p ==> bm is divisible by p, but bm=1.

Isn't this exactly the proof of Eisenstein? I think in a contest we don't need to prove Eisenstein all over again. Anyway, IMO 1993_1 is KIND OF like this one, only that I haven't seen how to apply Eisenstein. Instead I obtained (if I remember correctly) something of the same kind (if that polynomial was reducible then it meant that 3|5 or sth like that).

Could smb post a proof of that prb using Eisenstein?

This problem is a consequence of the Eisenstein Irreducibility Criterion:

Let f(x) = c_{0}*x^{n}+c_{1}*x^{n-1}+...+c_{n} be a polynomial with integral coefficients. If there exists a prime p such that p is not a divisor of c_{0} but p divides c_{i} for all i = 1,2,...,n. And p^{2} is not a divisor of c_{n}, then f(x) is irreducible, i.e.  cannot be expressed as the product of two polynomials with integer coefficients of lower degrees.

For further difference check the paper polynomials from the following web site: http://www.people.fas.harvard.edu/~gcarroll/math/
This paper was written for the Berkely Math Circle at:
http://mathcircle.berkeley.edu/


For all you USAMTSers out there, how do you get commended? I heard it was about generalizing the problem. My question is though, how far/little do you have to generalize it?

IW.OAV

You don't necessarily need to generalize to be commended.  When I participated, it felt almost as though I was randomly commended. There was this one problem that I did in a REALLY idiotic brute force way that got commended and several other cute, slick proofs (some with generalizations) that didn't.

I think a lot of it has to do with whether or not your solution is original or creative relative to the others your grader has seen. So, generalize as best you can, but that won't guarantee anything.

Yes, I also felt that it was somewhat random. On one problem I used complex numbers in a nice (and completely unnecessary) way because I thought it was more fun that the basic mod method, and I was not given a commended. I had my solution published on the website for a solution I thought was trivial once too. So I don't know how it works.

I've never gotten commended :( then again, my solutions are never creative enough *sigh*. I only got 84/100 on USAMTS this past year, it being my first.


How many zeros are at the end of the decimal representation of $2002!$ ?

Wow do you mean at the end of 2002! or in the entire decimal representation?

[hide]$\lfloor 2002/5 \rfloor = 400$

$\lfloor 400/5 \rfloor = 80$

$\lfloor 80/5 \rfloor = 16$

$\lfloor 16/5 \rfloor = 3$

Add them up and get 499 total factors of 5, thus 499 total factors of 10, thus 499 zeros at the end.[/hide]


I don't get modular arithmetic - could someone explain it to me.  I read the section in the Art of Problem solving a few times, but I still don't completely get it.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11944

And that links to another useful thread.

I think I get it.  Why doesn't my schools system teach modular arithmetic?  :? Oh well.  I guess I'm not the only one.

Modular arithmetic belongs in Number Theory. School systems do not regularly offer number theory courses, which is why Aops [i]does[i] offer them.

Thanks Mr. Rusczyk and Mr. Crawford!


Show that for $a,b,c>0$ we have:

$ \sum \frac{(-a+b+c)^3}{a}  \geq  \sum a^2$.

cheers! :D :D :cool:

Brute force solution !

Given ineq is equivalent to :

a^4c + a^4b + b^4a + b^4c + c^4a + c^4b + 3a^3b^2 + 3a^2b^3 + 3b^3c^2 + 3b^2c^3 + 3a^2c^3 + 3a^3c^2 >= 8a^3bc + 8ab^3c + 8abc^3

Which is clear using rearrangement (a^4b + b^4c + c^4a >= a^3bc + ab^3c + abc^3 and the same for the others).

maybe it is easier if we denote
-a+b+c=x
a-b+c=y
a+b-c=z
and the inequality becomes:
2 \sum x/(y+z) \geq 1/4\sum (x+y)  <=>
\sum x/(y+z) \geq 1/8 \sum(x+y)

i think i solved it ...
\sum x/(y+z) \geq (\sum x)/(6 \sumx)=(\sum x)/6 \geq 0 (1)(it's not very easy to proove this)
(\sum x/(y+z) )(\sum x(y+z)) \geq (\sum x) \qeg 0 (by CBS) (2)
from (1),(2) results that \sum xy \geq 0 .(3)
from (2) and (3) results that \sum x/(y+z) \geq (\sum x)/ 2\sum xy.
Now it's enough to proove that (\sum x)/ (2\sum xy) \geq 1/8 \sum(x+y)=1/4( \sum x+\sum xy) =>
2(\sum x)-(\sum xy)(\sum x)-(\sum xy)\geq 0. (*)
Delta = (3 \sum xy)=>
r1 = \sum xy
r2= -1/2 \sum xy
we have that \sum x\geq \sum xy with equality iff x=y=z=> the inequality(*) holds for every x,y,z \in R. q.e.d.
hope that is correct. :D


geometry
rectangle
function
algebra
domain
analytic geometry
topology

Without lifting my hand, I draw a curve that connects a pair of opposite corners of a rectangular piece of paper. Prove that I cannot draw a curve that connects the other pair of corners without crossing the first curve that I drew.

(The point of this is to be rigorous...)

Offhand, it looks like this requires (or is equivalent to) the [url=http://en.wikipedia.org/wiki/Jordan_curve_theorem]Jordan curve theorem[/url].  Is that okay as an assumption?  Otherwise, I think we have to reproduce the proof of the Jordan curve theorem, which is not Pre-Olympiad material.

The curves can intersect themselves...I can draw a loop on paper...And the curve is continuous.

I'll give my idea in a bit (I doubt that I can state it rigirously though).

I think that this is easier.

"The curve is continuous" gets me thinking...  Perhaps we could invoke the Intermediate Value Theorem somehow, or modify its proof to fit the problem.  However, then we'd have to cut off the possibles loops...  This shouldn't pose much of a problem, though.

That proof doesn't work immediately...like suppose we have the vertical like x=1, then like a path can cross that in infinitely many intervals.

One supposition that I wanted to add is something along the lines of making the path finite. But you obviously cannot use length or anything like that...

I was thinking of adding this to the problem: any line intersects the path at finitely many intervals.

[hide="rigorous proof?"]
We put the rectangle in the Cartesian plane and scale it such that it becomes the unit square - obviously this will not change anything.  We also remove all loops - obviously our curves will still be everywhere continuous, although not necessarily everywhere differentiable, even implicitly.  A continuous reparameterization is fairly easily using Heaviside functions.  WLOG let $ F(s) \equal{} (f_x(s),f_y(s)): a \le s \le b$ and $ G(t) \equal{} (g_x(t), g_y(t)): c \le d$ be the two curves, such that $ F(a) \equal{} (0,1), F(b) \equal{} (1,0), G(c) \equal{} (0,0), G(d) \equal{} (1,1)$.  Note that even after removing loops, we can still create a parametric function for the curve that takes one interval as its domain.

We create functions $ s(v)$ and $ t(v)$ that take the interval $ [0,1]$ to $ [a,b]$ and $ [c,d]$, respectively, with the added condition $ f_x(s(v)) \equal{} g_x(t(v))$.  Define $ F'(v) \equal{} F(s(v))$ and $ G'(v) \equal{} G(t(v))$.  Now consider the behavior of $ H(v) \equal{} F'(v) \minus{} G'(v)$ on the interval $ [0,1]$, and set $ h_x(v) \equal{} f_x(s(v)) \minus{} g_x(t(v))$ and $ h_y(v) \equal{} f_y(s(v)) \minus{} g_y(t(v))$.  Obviously $ h_x(v) \equal{} 0$ for all $ x \in [0,1]$.  But in order to show that the curves intersect somewhere, we need only find a $ v$ such that $ H(v) \equal{} (0,0)$, which means we only have to find a point such that $ h_y(v) \equal{} 0$.  But obviously $ h_y(0) \equal{} 1$ and $ h_y(1) \equal{} \minus{}1$.  Now $ h_y(v) \equal{} f_y(s(v)) \minus{} g_y(t(v))$.  But continuity of functions is obviously preserved under scalar multiplication, addition, and function composition, with appropriate domain restrictions as necessary under function composition.  Thus $ h_y(v)$ is continuous on the interval $ [0,1]$.  Hence we just use the Intermediate Value Theorem to get a value $ 0 < v_0 < 1$ such that $ h_y(v_0) \equal{} 0$, and thus the curves intersect.

Note: the reparameterizations after removing loops could use a little more exposition, but hey.  It's not that bad.
[/hide]

I don't see why you would need to treat loops or anything else in any special way.  I think the only assumption you really need is that the paths can be parameterized (indeed, this may well be inherent to the [i]definition[/i] of what a path is, but it all depends where you're starting from).  Here's a proof sketch:

Without loss of generality say the rectangle has coordinates $ (0,0), (a,0), (0,b), (a,b)$, $ a,b > 0$, then we have some parameterized curves $ \alpha,\beta: [0,1]\to [0,a]\times[0,b]$ with
\[ \alpha(t) \equal{} (\alpha_1(t),\alpha_2(t)), \beta(t) \equal{} (\beta_1(t),\beta_2(t))
\]
and
\[ \alpha(0) \equal{} (0,0),\alpha(1) \equal{} (a,b), \beta(0) \equal{} (0,b), \beta(1) \equal{} (a,0)
\]
We want to show the existence of some $ 0\leq s,t \leq 1$ such that $ \alpha(s) \equal{} \beta(t)$ assuming $ \alpha,\beta$ are continuous.

Now for any fixed $ s_0$ we have $ \beta_1(0) \minus{} \alpha_1(s_0) \equal{} \minus{} \alpha_1(s_0)\leq 0$ and $ \beta_1(1) \minus{} \alpha_1(s_0) \equal{} a \minus{} \alpha_1(s_0)\geq 0$, so by the Intermediate Value Theorem there exists a $ t_0$ such that $ \beta_1(t_0) \equal{} \alpha_1(s_0)$.  Moreover, $ t_0 \equal{} 1$ corresponds to $ s_0 \equal{} 1$ and $ t_0 \equal{} 0$ corresponds to $ s_0 \equal{} 0$, and a $ t_0$ can be chosen for each other $ 0 < s_0 < 1$ so that $ t_0$ is a continuous function of $ s_0$.  (this assertion definitely requires proof, but it's more or less straight from the continuity of $ \alpha_1$ and $ \beta_1$)

Now define $ \gamma: [0,1]\to \mathbb{R}$,  $ \gamma(s_0) \equal{} \beta_2(t_0) \minus{} \alpha_2(s_0)$ where $ t_0$ is the number for which $ \beta_1(t_0) \equal{} \alpha_1(s_0)$.  $ \gamma$ is continuous by the continuity of $ t_0$ w.r.t. $ s_0$, $ \gamma(0) \equal{} b$, $ \gamma(1) \equal{} \minus{} b$ and hence by the Intermediate Value Theorem there is an $ s_0$ so that $ \gamma(s_0) \equal{} 0$, and thus $ \alpha(s_0) \equal{} \beta(t_0)$.

I guess this is in the spirit of Teki-Teki's proof, but again I don't think dealing with the geometry of the curves is necessary beyond stating that they are both parameterizable.

[hide="Partial solution"]WLOG the first curve goes from the bottom left to the top right. We now split the problem into two cases: either there is a farthest left intersection of the first curve with the top edge and a farthest up intersection of the first curve with the left edge or there is not.

If those intersections exist, color everything to the "left" of the first curve blue; that is, define a region $ R$ bounded by the first curve (excluding loops) between the farthest up intersection with the bottom edge and the farthest left intersection with the top edge. Include the top and left sides between the top left point and the relevant intersections. Color $ R$ blue and color the rest (including the bottom and right sides) red (this includes filling in all loops, etc). Color the part of the first curve bordering $ R$ both colors (purple). We obviously start at a blue point and end at a red point, so we must change colors at some point. Every point at which colors change is a member of the curve, so we must intersect the curve somewhere.

If there is no farthest left or farthest intersection, for any given intersection $ I$ there must be an infinite number of intersections between $ I$ and the top left corner, as with a finite number there would necessarily be a farthest left or top point. [/hide]

[quote="worthawholebean"][hide="Not sure if this is rigorous enough"]WLOG the first curve goes from the bottom left to the top right. Color everything to the "left" of this curve blue; that is, color the region bounded by the first curve (excluding loops) and the top and right sides blue. Color the rest red (this includes filling in all loops, etc). We obviously start at a blue point and end at a red point, so we must change colors at some point. The only boundary between colors is at the first curve, so we must cross it at some point.[/hide][/quote]

What do left and right mean for, say, a fractal curve, or even a space-filling curve?  The statement of the problem doesn't necessarily exclude either of those possibilities, and indeed there's no reason to think the assertion shouldn't hold in those cases.  You're also implicitly invoking the fact that the curve splits the rectangle into two parts, which is itself a very nontrivial assertion that I think would require some topology (along the lines of t0rajir0u's thinking), and is probably more difficult to prove than the original statement.

Hello, I like the thinking that is going on here. I guess I'll add a condition to help define what 'drawable' by hand means.

There exists a function from $ [0,1]$ to $ [0,a] \, x \,[0,b]$ that represents the path.

Also a rectangle should be considered to be closed.

JoeBlow's solution looks correct according to [url=http://en.wikipedia.org/wiki/Curve#Definitions]Wikipedia's definition[/url] of a curve.  It does indeed appear that this result is weaker than the Jordan curve theorem.

I still feel like there's gotta be some way to fill out the rest of my solution....


Find all integers x and y such that:

$1+x^{2}y = x^{2}+2xy+2x+y$


If there is a topic on mathworld that shows you how to compute something in the problem. Can we just cite the source or do we have to prove it ourselves?


Write one of the best and most beautiful problems you have ever seen, and the solution.
One of mine favourite problems is simple:
Find all natural numbers $a,b,c$ such that the roots of  $x^2-2ax+b=0, x^2-2bx+c=0, x^2-2cx+a=0$ are all natural numbers.

$x^n + y^n = z^n$ has no solutions for $n>2$.  The solution might be a bit long :-P

:lol:  well, you have to post the solution..  :stretcher:


salam
bel akhare emtehanate marhale sevome olampiadhaye riaziye iran be payan resid. :wink:
vali akharin soale khalaghiat kheili badihi bood  :wink:
neshandan dar $Q^{n}$ :wallbash_red:

jedi asheghali ?
Az farhat khabar dari ? gol kashte ya na ?

valla man ke shenidam terekoondeh.

sad dar sad team mishe, che berese be talla :D

Farhat vaghean bacheye Ba hoshi bod

baraye team iran shodam ,bayad kheli zahmat keshid ,Iran bache haye bahosh ziyad dare , Vali be farhat mishe omidvare bod , vaghean ba estedade.

Salam,

avval inke, aya kasi hastesh paye bashe soalaye khallaghiat ro bezane inja?

dovvom inke, zaheran emrooz 3shanbe residegi be e'terazast va farda, 4shanbe natayej e'lam mishe!

daghighan hamintore. emrooz ehtemalan saate 4.
yeki az soaleye khalaghiat:
soale 2 rooze aval:
yek loole darim keghotrash besyar barik ast.(mitavanim anra khat farz konim)
agar yek moolekoole ab dar nokhte $x$ bashad bad az $t$ saniye dar noghteye $P(t,x)$ khahad bood.
agar $P(x,t)$ yek chand jomleyiye do motaghayere bashad sabet konid sorat hameye moolekoolha sabet va barabar ast. :?:

Baraye Har $(t,k,x) \in R^{3}, t > k$ darim : \[p(t,x)-p(k,x)= p(t-k,p(t-k,x))\] $\longrightarrow : degP(t,x)=1 \longrightarrow P(x,t)=x+at$ ke $a \in R$ ye adade sabete.

merci, soale khoobi bud! khob, baghie soala chi? (be khosoos un 2 ta soali ke nadidam!!?)

age 1 nafar paye bashe, soala ro tarjome kone va tu ghesmate aslie MathLinks (yani inke tu National Forums nabashe) kheyli khoobe. chon soalaye kheyli ghashangi hastan va baghie ham mitunan estefade konan!

agaram kasi paye nist laaghal begin be tore tabi'i in soala bayad kodoom ghesmat post she, shayad khodam paye budam  :maybe:

Age Vaght Kardi ,baghiyey Soal ha ro edameye Post Omid hatami ,bede.

Hey emsal to khalaghiyat ha , Mesle On soal Havapeyma va Mohsak dashtid ? (ke moshake be havapeyma nemikhore :D )

ye zarre addresse daghigh tar midadi chon Omid chand hezar ta post dare  :oops:

na, fek konam emsal soale moadele difransieli nadashtim !

salam doostan.
man fekr mikonam mahmood soalaro behetoon dade bashe age momkene khodetoon paye shin  :wink:
oon soale havapeima fekr konam ye rahe hal gheire defransieli ham dashte bashe.

Age vaght nadarid , Bargeye Soalha ro scan konid ,bad  be posteton Attach konid.

PS.Ye Softwary hast ke mitone version scan shoda ro be word tabdilesh kone , word ro ham mitonid PDF konid. Injori Etebaresh bishtare  :) .

na, lazem be scan nist (scan vaghean khoob dar nemiad!). gharar shode agha MohammadHossein tu in 1 mahi ke bikare beshine ghashang ba LaTeX inaro type kone bezare inja  :D

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=627409#627409[/url]  :roll:

check for any mistakes!


Find the pair of positive integer such that $(1+x+y)^2=1+x^3+y^3$.

There are a lot of such problems-equations in $Z$ with two or more variables.
Nevertheless each one needs its own trick or technique in solving!! :P

Now to the problem...:

Put $x+y=a$ and $xy=b$
then the given equality becomes $(1+a)^2=1+a^3-3ab$
doing calculations one obtains $a(a+2)=a(a^2-3b)$
so either $a=0$ or $a+2=a^2-3b$
for $a=0$ we have $x+y=0$ which is impossible because $x,y>0$

if $a+2=a^2-3b$ then replacing $a$ and $b$ we get $x^2+y^2-xy-x-y-2=0$
upon multiplication with $2$ and completing squares we get
$(x-y)^2+(x-1)^2+(y-1)^2=6$

obviously $(x-1)^2=1$ or $4$ and then we easily find the pair..
 :)


Let $ (O)$ is a circle with $ AB$ is diameter. A fixed line $ (d)$ perpendicular with $ AB$ at point $ I$. Let $ M$ is a change point on circle $ (O)$. Let $ BM$ meet $ (d)$ at point $ Q$, let $ AQ$ meet $ (O)$ at point $ N$, let $ MN$ meet $ AB$ at $ S$. Prove that $ S$ is a fixed point.

Dear Mathlinkers,
BM and AN intersect on (d)
S is the pole of (d)....
Sincerely
Jean-Louis

S is always the harmonic conjugate of I wrt AB.

[quote="thanhnam2902"]Thanks ¬[ƒ(Gabriel)³²¹º]¼ and yaime very much. But I want have an other solution.

Who can solve this problem by other way for student of highschool?[/quote]
If the [b][size=100]Menelaos theorem[/size][/b], is a well known result to the high school students in Vietnam, we can show that the point $ S\equiv AB\cap MN$ is a fixed point on the line segment $ AB,$ as follows:

Because of $ AM\perp QB$ and $ BN\perp QA,$ we have that the point $ H\equiv AM\cap BN,$ as the othocenter of the triangle $ \bigtriangleup QAB,$ lies on the line segment $ QI\perp AB.$

In the triangle $ \bigtriangleup QAB$ taken the line segment $ MN$ as its transversal, applying the [b][size=100]Menelaos theorem[/size][/b],

we have that $ \frac {SA}{SB}\cdot \frac {MB}{MQ}\cdot \frac {NQ}{NA} \equal{} 1$ $ ,(1)$

From the similar right triangles $ \bigtriangleup MAQ,\ \bigtriangleup NBQ,$ we have that $ \frac {MQ}{NQ} \equal{} \frac {QA}{QB}$ $ ,(2)$

From the similar right triangles $ \bigtriangleup MBA,\ \bigtriangleup IBQ,$ we have that $ \frac {MB}{IB} \equal{} \frac {AB}{QB}$ $ ,(3)$

From the similar right triangles $ \bigtriangleup NAB,\ \bigtriangleup IAQ,$ we have that $ \frac {NA}{IA} \equal{} \frac {AB}{QA}$ $ ,(4)$

From $ (3),$ $ (4)$ $ \Longrightarrow$ $ \frac {MB}{NA}\cdot \frac {IA}{IB} \equal{} \frac {QA}{QB}$ $ ,(5)$

From $ (2),$ $ (5)$ $ \Longrightarrow$ $ \frac {IA}{IB}\cdot \frac {MB}{MQ}\cdot \frac {NQ}{NA} \equal{} 1$ $ ,(6)$

From $ (1),$ $ (6)$ $ \Longrightarrow$ $ \boxed{\frac {SA}{SB} \equal{} \frac {IA}{IB}}$ $ ,(7)$

From $ (7)$ we conclude that the point $ S$ is a constant point ( the harmonic conjugation of $ I,$ with respect to $ A,\ B,$ as mentioned by [b][size=100]Gabriel[/size][/b] ) and we are done.

Kostas Vittas.

[quote="thanhnam2902"]Hi vittasko. Thank you very much. In Viet Nam, student of hight know Menelaus theorem but not for all. Because it only for IMO. I very like your solution.

I think this problem can be done by use similar triangle. Can you solve this problem by use similar triangle or by other way?

Thank you very much.[/quote]
It is easy to show the relation $ \frac {SA}{SB}\cdot \frac {MB}{MQ}\cdot \frac {NQ}{NA} \equal{} 1$ $ ,$ I mentioned above ( which the [b][size=100]Menelaos theorem[/size][/b] states ), by [b][size=100]Thales theorem[/size][/b].

So, we denote as $ K$ the intersection point of the line segment $ MN,$ from the line through the point $ A$ and parallel to $ QB.$

From $ KA\parallel MB$ $ \Longrightarrow$ $ \frac {SA}{SB} \equal{} \frac {KA}{MB}$ $ ,(1_{a})$

From $ KA\parallel MQ$ $ \Longrightarrow$ $ \frac {KA}{MQ} \equal{} \frac {NA}{NQ}$ $ \Longrightarrow$ $ \frac {KA}{MB} \equal{} \frac {MQ}{MB}\cdot \frac {NA}{NQ}$ $ ,(1_{b})$

From $ (1_{a}),$ $ (1_{b})$ $ \Longrightarrow$ $ \frac {SA}{SB}\cdot \frac {MB}{MQ}\cdot \frac {NQ}{NA} \equal{} 1$ $ ,(1)$ and we are done ( clearly, it has just been proved, the [b][size=100]Menelaos thoerem[/size][/b]  :) ).

Kostas Vittas

[quote="vittasko"]
It is easy to show the relation $ \frac {SA}{SB}\cdot \frac {MB}{MQ}\cdot \frac {NQ}{NA} \equal{} 1$ $ ,$ I mentioned above ( which the [b][size=100]Menelaos theorem[/size][/b] states ), by [b][size=100]Thales theorem[/size][/b].

So, we denote as $ K$ the intersection point of the line segment $ MN,$ from the line through the point $ A$ and parallel to $ QB.$

From $ KA\parallel MB$ $ \Longrightarrow$ $ \frac {SA}{SB} \equal{} \frac {KA}{MB}$ $ ,(1_{a})$

From $ KA\parallel MQ$ $ \Longrightarrow$ $ \frac {KA}{MQ} \equal{} \frac {NA}{NQ}$ $ \Longrightarrow$ $ \frac {KA}{MB} \equal{} \frac {MQ}{MB}\cdot \frac {NA}{NQ}$ $ ,(1_{b})$

From $ (1_{a}),$ $ (1_{b})$ $ \Longrightarrow$ $ \frac {SA}{SB}\cdot \frac {MB}{MQ}\cdot \frac {NQ}{NA} \equal{} 1$ $ ,(1)$ and we are done ( clearly, it has just been proved, the [b][size=100]Menelaos thoerem[/size][/b]  :) ).

Kostas Vittas[/quote]

Yes, I see. Thank you very much!


trigonometry
vector
analytic geometry
algebra
system of equations

Hello!

I know this must be a subject long discussed but I couldn't find detailed information. I need to do a 2D simulation of balls colliding between them where the type of collision is elastic.

Suppose we have two balls A and B with defined position and velocities as given by: xA, yA, vxA, vyA, xB, yB, vxB, vyB. Both balls have a fixed radius r.

I don't know if I'm right but I'm trying to find a way of knowing the new vx, vy for both balls in order to find x, y but I don't know how to calculate them as well as how to calculate the direction the balls would go after the collision.

I'm trying to form a system of equations considering ball B inert and ball A with its vx, vy changed after subtracting both B's vx, vy:

Kinematic Conservation:

(1/2)*mA*(vA1^2) = (1/2)*mA*(vA2^2) + (1/2)*mB*(vB2^2) where vA1 is initial velocity and vA2, vB2 are final velocities

Momentum Conservation:

mA*vA1x = mA*vA2x + mB*vB2x
mA*vA1y = mA*vA2y + mB*vB2y

Another doubt is how I can put the initial direction between the balls given by calculating its distance d = sqrt((xA-xB)^2 + (yA-yB)) and the cosine and sine given by (xA-xB)/d and (yA-yB)/d in order to make the system of equations have 4 variables and 4 equations, that is, to make it give solutions that could be found. How these angle values affect the final direction of the balls??? What kind of vector operations should be done??

Should vA1^2, vA2^2, vB2^2 be changed to (vA1x^2 + vA1y^2), (vA2x^2 + vA2y^2), (vB2x^2 + vB2y^2) in the first equation???

Please, correct me if I made any mistakes.

Thanks a lot!!!

It's hard to read what you've done since it's not in $ \text{\TeX}$ but I can give you a basic idea:  As you said, set up equations for conservation of kinetic energy (since there is no potential energy) and conservation of linear momentum.  By doing this you can obtain the final velocities of each.  Finding the direction is a bit trickier if the collision is not head on.  If I were you, I would try to stick to the case with a head on collision.  I had to solve a similar problem in mechanics but it required different reference frames so stick to the simpler case.

Hello! Thanks for your reply!

What I really want to know is how to calculate the new direction the balls would go after collision. This seems to be kind of difficult to define. What relation between the velocity components and position coordinates from each ball gives the new directions???


vector
advanced fields
advanced fields unsolved

$ R$ is a $ G$ bundle with base $ B$ and projection $ \pi$. If $ U$ is a nbh of $ b \in B$ over which $ R$ is trivializable,
and $ \sigma : U \to {\pi}^{-1}(U)$ is a smooth section of $ R$ over $ U$. Now if $ {\sigma}'$ is another smooth section of $ R$ over $ U$,
then why: $ {\sigma}'(b)={\sigma}(b) \bullet s$, for some $ s \in G$?

Maybe because we assume that $ G$ acts transitively on the fibers?

sorry, I don't think it is correct as it is: take the principal bundle to be $ R$ with base $ [0,1]$ and group is $ Z$.
It is in the context of vector bundle associated with the principal bundle....


geometry
calculus
integration
parameterization
algebra
linear equation

Hi, I'm new to this site but it seems cool so I thought I'd see if anyone can help me with a question I'm stuck on. Well, I'm not stuck on it, I got the right answer, but I used a long-winded technique and I was wondering if there's a quicker way.

"Find the area bound between the curve with parametric equations $x=3t^{2}$ and $y=2t^{3}$, and its tangent $y=2x-8$."

So I figured the line is a tangent to the curve at point P (12, 16) and it also crosses through the curve at point Q (3, -2) so we have an enclosed area.

What I did, was drop vertical lines from the points of intersection and calculated those areas under/above the curve, then subtracted the triangles they made with the tangent. But this took me ages, surely there's a more efficient method?

This is a calculus problem, and there is a whole separate forum for calculus

but I'll try to help anyway

[hide]

I would convert those parametric equations to rectangular ones

$x = 3t^{2}$

$t = \left(\frac{y}{2}\right)^{\frac{1}{3}}$

$x = \frac{3 y^{\frac{2}{3}}}{\sqrt[3]4}$

then I would take your other linear equation and put it in terms of x

$x = \frac{y}{2}+4$

then integrate between your y values of -2 and 16

$\int_{-2}^{16}\left(\frac{3 y^{\frac{2}{3}}}{\sqrt[3]4}\right)-\left(\frac{y}{2}+4\right) dy$

There's probably a more elegant way of going about it, but this will work.
[/hide]

Elliminating parameter $t$ gives the equation of the curve; $y=\frac{2}{3\sqrt{3}}x^{\frac{3}{2}}\ (x\geq 0,\ y\geq 0)$.

Thus the desired area is $\int_{0}^{12}\frac{2}{3\sqrt{3}}x^{\frac{3}{2}}dx-\frac{1}{2}(12-4)\cdot 16=\frac{192}{15}.$


Circles $O_1$ and $O_2$ interest at two points $ B$ and $ C,$ and $ BC$ is the diameter of circle $O_1.$ Construct a tangent line of circle $O_1$ at $ C$ and intersecting circle $O_2$ at another point $ A.$ We join $ AB$ to intersect circle $O_1$ at point $ E,$ then join $ CE$ and extend it to intersect circle $O_2$ at point $ F.$ Assume $ H$ is an arbitrary point on line segment $ AF.$ We join $ HE$ and extend it to intersect circle $O_1$ at point $ G,$ and then join $ BG$ and extend it to intersect the extend line of $ AC$ at point $ D.$ Prove that \[ \frac{AH}{HF} = \frac{AC}{CD}.\]

We have:
$ \frac{AH}{HF}\equal{}\frac{S_{AEH}}{S_{FEH}}\equal{}\frac{AE.\sin\widehat{{AEH}}}{FE.\sin\widehat{{FEH}}}\equal{}\frac{CE.\sin\widehat{{BEG}}}{BE.\sin\widehat{{CEG}}}\equal{}\frac{CE.BG}{BE.CG}$
And:
$ \frac{AC}{CD}\equal{}\frac{S_{ABC}}{S_{DBC}}\equal{}\frac{AB.CE}{DB.CG}$
Combine with $ BE.BA\equal{}BC^2\equal{}BG.BD\Leftrightarrow \frac{AB}{DB}\equal{}\frac{BG}{BE}$ we have solution.

This is a nice solution which a friend of mine has come up with today.
Since $AC$ is tangent to $\odot O_2$, $\angle BCA=90^\circ$, so as $\angle AFB$.
Moreover, we also have $\angle CEB = 90^\circ$ for $BC$ is a diametre of $\odot O_1$.
So, $\angle HAE =\angle BAD$.
On the other hand,
$ \angle AHE$
$=  180^\circ - \angle FAB - \angle HEA$
$= 180^\circ - \angle ECB - \angle BEG$
$= 180^\circ - \angle EGB - \angle BEG$
$= \angle ABD$
which leads to $\triangle AHE \sim \triangle ABD$.
Hence $\frac{AH}{AB}=\frac{AE}{AD}$.
Then obviously $\triangle AEF \sim \triangle ACB$,
giving that $\frac{AB}{AF}=\frac{AC}{AE}$.
Combining the two equalities above we get $\frac{AH}{AF}=\frac{AC}{AD}$,
and the result follows.

I think this is very similar to [b]Uagu[/b] 's solution
Observe that
1)$ACBF$ is a kite
2) Hence  $\Delta AHE \sim \Delta ABD$
$\Rightarrow \frac{AH}{AE}=\frac{AB}{AD} \Rightarrow AH.AD=AE.AB =AC^2 =AC.AF \Rightarrow \frac{AH}{AF}=\frac{AC}{AD}$
Done!! :lol:

[quote=tdl]We have:
$ \frac{AH}{HF}\equal{}\frac{S_{AEH}}{S_{FEH}}$.[/quote]

$A$, $F$ and $H$ aren't collinear, so how do we know that this works?

My solution:
We know $\angle BCA=90,\angle FEA=90,\angle EAF=\angle BAC,$ then $\triangle BCA\sim \triangle FEA\to \frac{EA}{EF}=\frac{CA}{CB}(1).$
Let $\angle EFA=x,\angle FEH=y,$(Also $\angle DBC=\angle GEC=\angle FEH=y.$) then we know $$\frac{AH}{HF}=\frac{tg(x)}{tg(y)}=\frac{EA}{EF}\cdot \frac{BC}{CD}=^{(1)}\frac{AC}{BC}\cdot \frac{BC}{CD}=\frac{AC}{CD}.$$ $Q.E.D$

This is easy computation.

Let $\angle HAE= \alpha, \angle AEH= \beta$. Let $R$ denote radius of circle $O_2$.
Note that $\angle BCA=\angle AEC=\angle BGC=90^{\circ}$

By direct angle chasing, we get $\angle BAC=\alpha$ and $\angle BDC=\beta$,
which gives $AC=2R \cos \alpha$ and $CD=2R \sin \alpha \cot \beta$
$$\frac{AC}{CD}=\cot \alpha \tan \beta$$

Now we compute $AH$ and $HF$.
By sine rule on $\Delta AHE$,
\begin{align*}
AH &= \sin \angle HEA \cdot \frac{HE}{\sin \angle HAE}\\
&=\sin \beta \cdot \frac{HE}{\sin \alpha}\\
 \end{align*}
By sine rule on $\Delta FHE$,
\begin{align*}
HF&= \sin \angle FEH \cdot \frac{HE}{\sin \angle HFE} \\
&=\cos \beta \cdot \frac{HE}{\cos \alpha} \\
\end{align*}

\begin{align*}
\implies \frac{AH}{HF}
&=\frac{\sin \beta \cdot \frac{HE}{\sin \alpha}}{\cos \beta \cdot \frac{HE}{\cos \alpha}}\\
&=\tan \beta \cot \alpha\\
\end{align*}
As desired.

I want to attach a diagram but don' know how to do it.  anybody help?

[quote=tdl]We have:
$ \frac{CE.\sin\widehat{{BEG}}}{BE.\sin\widehat{{CEG}}}\equal{}\frac{CE.BG}{BE.CG}$[/quote]
how we get it?

Oh I misread the problem, it meets O1 and not O2 at G

[quote=Delray]$A$, $F$ and $H$ aren't collinear, so how do we know that this works?[/quote]

[quote=orl]Assume $ H$ is an arbitrary point on line segment $ AF.$[/quote]

Here's a synthetic sol. -
  1. Note that $ACBF$ is a kite with $AF=AE$ etc...

 2.  [b][u]Claim :[/u][/b] $\triangle AHE \sim \triangle ABD$
       [b]proof - [/b]  $\angle HAE=\angle BAD$  ;  Now let $H_1$ be the second intersection point of $EH$ and $O_2$ .  Note that the map $H$ to $H_1$ is projective and $H_1$ to $H$ is projective , so $H$ to $G$ is a projective map , choose $H$  = $F , A$ and mid point of $FA$ and we get
$\angle AEH = \angle ADB$ , so this is true for all positions of $H$  , so the claim is proved !

 Now , $AH/HB$ = $AB/AD$
           $AH = AB.AE/AD = AC^2/AD$ , since $AC = AF$ we are done !!!

[b][u]NOTE :[/u][/b] To avoid diagram issues , I use projective maps ! :D  :D

[hide=Solution]
Let $P_{\infty}$ be a point at infinity on $AC.$ Applying Law of Sine and Power of a Point, we have
\begin{align*}
\frac{AH}{HF}&=\frac{AE\sin\angle AEH}{FE\sin\angle FEH}\\
&=\frac{CE\sin\angle BEG}{BE\sin\angle CEG}\\
&=\frac{CE\cdot BG}{BE\cdot CG}\\
&=(EG,BC)\\
&\overset{B}{=}(DA,CP_\infty)\\
&=\frac{AC}{CD},
\end{align*}
as desired.
[/hide]

[quote=blacksheep2003][quote=Porky623]Hmm don't these solutions assume BECG is cyclic, which clearly isn't true because BEC are on O1 and G isn't?[/quote]

It's actually typed wrong on the AoPS wiki, but it is uneditable. I spent so much time trying to figure out what I did wrong, but really it was a flaw in the problem statement here: https://artofproblemsolving.com/community/c3667_2002_china_girls_math_olympiad - could somebody please fix this?[/quote]

I think it's a typo in EGMO as well. I am not sure if it is fixed in some updated version but the version I've bought it isn't. I just found out and I'm fuming. I spent ~2 hours redrawing the diagram and be like "there's no way my compass and ruler diagrams are this far off I'm not this bad" before looking on the contests page and realizing it was a typo. Took like 5 minutes to do after I found out, what a waste of time...

Why they keep confusing people with such ugly formulation. I just can help it but correct.
[quote]
In $\triangle ABC$ with $\angle C=90^\circ$ altitude $CE$ meet $(ABC)$ at $F$. Point $H$ is on segment $AF$. $HE$ meet $(BCE)$ at $G$. $BG$ meet $AC$ at $D$.
Prove that $FD \parallel CH$.
[/quote]
The same in [url=https://artofproblemsolving.com/community/c6h1219048p6084398]Taiwan 2016 TST1 Q1-2[/url]

Dear Mathlinkers,
a very simple proof is possible...

Sincerely
Jean-Louis

[quote=AopsUser101][hide = trivial?]Let $\angle CBE = \angle EFA = a$. Since $BC \perp CA$, $\angle BAC = 90 - a$ and since $\angle BEC = 90$, $\angle EAF = 90 - a$. Let $\angle FBC = \angle GEC = \angle FEH = b$. Since $BE \perp EC$, $\angle HEA = 90 - b$. Now, note that by law of sines:
$$\frac{HE}{\sin (90 - a)} = \frac{HA}{ \sin (90 - b)} \Longleftrightarrow \frac{HA}{HE} = \frac{\cos (b)}{\cos (a)}$$
$$\frac{HE}{\sin (a)} = \frac{HF}{\sin (b) } \Longleftrightarrow \frac{HF}{HE} = \frac{\sin (b)}{\sin (a)}$$
Thus, $\frac{AH}{HF} = \frac{\tan (a)}{\tan(b)} = \frac{\frac{CA}{BC}}{\frac{CD}{BC}} = \frac{CA}{CD}$ as desired.[/hide][/quote]

instead of $\angle FBC = \angle GEC = \angle FEH = b$ do you mean $\angle GBC = \angle GEC = \angle FEH = b$

By angle chase we easily see that $AF=AC$, $BF=BC$ and that $FE=EC$

We denote with $x=\angle BCE$ and with $y=\angle FEH$.
Then we have that:
$$\frac{HF}{\sin y}=\frac{EF}{\sin 90-y+x}=\frac{CE}{\sin 90-y+x}$$

We have that $CE$ can be expressed in terms of $BC$, this means that $\frac{CE}{\sin 90-x}=BC$, but $BC$ can be expressed in terms of $CD$, $BC=\frac{\cos y}{\sin y}CD$
This implies that $HF=\frac{\cos y \cos x}{\sin 90-y+x}$.

Similarily by noticing that $CF=2CE$ we have that $AF=\frac{\cos y \cos x}{\sin x \sin y}DC$

Thus our end goal is to prove that $(AF-HF)DC=AF.HF$, since that would imply $\frac{AH}{HF} = \frac{AC}{DC}$
Plugging in all of the information we have we get that $\cos y-x = \sin x \sin y + \cos x \cos y$ which is an identity this implies $\frac{AH}{HF} = \frac{AC}{DC}$

[hide=Solution]We know
\begin{align*}
\frac{HF}{\sin\angle FEH}&=\frac{EH}{\sin\angle EFH},\\
\frac{AH}{\sin\angle HEA}&=\frac{EH}{\sin\angle EAH},\\
\frac{CD}{\sin\angle DBC}&=\frac{BC}{\sin\angle BDC},\\
\frac{AC}{\sin\angle CBA}&=\frac{BC}{\sin\angle BAC}\\
\end{align*}
by the Law of Sines. We can also find that
\begin{align*}
\angle EFH&=\angle CBA,\\
\angle FEH&=\angle DBC,\\
\angle HEA&=\angle BDC,\\
\angle EAH&=\angle BAC\\
\end{align*}
with angle chasing.

Thus, $$\frac{AH}{HF}=\frac{\sin\angle EFH\cdot\sin\angle HEA}{\sin\angle EAH\cdot\sin\angle FEH}=\frac{\sin\angle CBA\cdot\sin\angle BDC}{\sin\angle BAC\cdot\sin\angle DBC}=\frac{AC}{CD},$$ as desired.[/hide]

By the tangent-chord theorem, we have $\angle ACE=\angle CBE\coloneq 1$. Since $B\in\Gamma_2$, $\angle CFA=\angle 1$. Therefore, $\triangle ACF$ is isoceles. Moreover, it is easy to see that $\angle ECB=\angle FCB=\angle FAB\coloneq \angle 2$. Notice that $\angle BEC=90^\circ$ since $BC$ is diameter of $\Gamma_1$, and that $BA$ is diameter in $\Gamma_2$. Since $CF$ is a chord, $AB$ bisects $CF$, implying $FE=CE$. From this it also easily follows that $\angle EAC=\angle 2$. Define $\angle AEH\coloneq \angle 4$ We now observe $\angle HEF=\angle GEC=\angle GBC\coloneq \angle 3$. It follows $\angle CDB=\angle 4$. We now use the law of sines a couple of times. We have
$$\frac{HE}{\sin(\angle 1)}=\frac{FH}{\sin(\angle 3)}$$
and
$$\frac{HE}{\sin(\angle 2)}=\frac{AH}{\sin(\angle 4)}$$
in $\triangle HEF$ and $\triangle HEA$, respectively. This gives
$$\frac{AH}{FH}=\frac{\frac{HE\sin(\angle 4)}{\sin(\angle 2)}}{\frac{HE\sin(\angle 3)}{\sin(\angle 1)}}=\frac{\sin(\angle 4)\sin(\angle 1)}{\sin(\angle 3)\sin(\angle 2)}$$
In $\triangle BCD$, we have
$$\frac{CD}{\sin(\angle 3)}=\frac{BC}{\sin(\angle 4)}$$
Since $AEC$ is right-angled, we have $AC=\frac{AE}{\sin(\angle 1)}$. Combining gives
$$\frac{AC}{CD}=\frac{AE\sin(\angle 4)}{BC \sin(\angle 1)\sin(\angle 3)}$$
We thus want to show that
$$\frac{\sin(\angle 4)\sin(\angle 1)}{\sin(\angle 3)\sin(\angle 2)}=\frac{AE\sin(\angle 4)}{BC\sin(\angle 3)\sin(\angle 1)} \Longleftrightarrow \frac{\sin(\angle 1)}{\sin(\angle 2)}=\frac{AE}{BC\sin(\angle 1)} $$
But now notice that $\triangle CEB$ is right-angled, so $EC=BC\sin(\angle 1)$
Therfore,
$$\frac{\sin(\angle 1)}{\sin(\angle 2)}=\frac{AE}{EC}\Longleftrightarrow \frac{AE}{\sin(\angle 1)}=\frac{EC}{\sin(\angle 2)}$$
which is just the law of sines in $\triangle AEC$

[[Image:2002 CGMO-4.png]]
We want to show that $$\frac{AH}{AF}=\frac{AC}{AD}$$
but $AF^2=AC^2=AB \cdot AE$ so in fact it suffices to prove that $$AH \cdot AD=AB \cdot AE$$ which is true if $$ \Delta AHE \sim \Delta ABD$$
This follows readily from angle chasing: $ \angle HAE = \angle BAD$ is obvious, and
$$\angle AEH = \angle BEG = \angle BCG = \angle ADB$$
where the last equality is true because $BC \perp AD$ and $CG \perp DB$.

Dear Mathlinkers,

[url=http://jl.ayme.pagesperso-orange.fr/Docs/Highway%20to%20Geometry%204.pdf] here [/url]   p. 20

Sincerely
Jean-Louis

OMG, I was redrawing and redrawing but couldn't find the sol but reading above, the problem statement was wrong :skull: :skull:

[quote=illogical_21]
\begin{align*}
&=\frac{CE\cdot BG}{BE\cdot CG}\\
&=(EG,BC)\\
&\overset{B}{=}(DA,CP_\infty)\\
&=\frac{AC}{CD},
\end{align*}
[/quote]

What do you mean by $(EG,BC)$ and how do you get that?

The main idea of this problem is to understand how $H$ relate to the points defined from it.\\

Claim: $CGHF$ is cyclic. We have $$\angle CGH=\angle CBE=\angle CFH.$$

We will let $\theta=\angle HEF$ be our "moving variable". Note that $$CD=BC\tan\theta$$ and from $CGHF$ cyclic we have $$HF=CG\cdot\frac{EH}{CE}=BC\sin\theta\cdot\frac{EH}{CE}.$$ Thus, $$\frac{CD}{HF}=\frac{CE}{EH\cos\theta},$$ so it suffices to show that $$\frac{CE}{EH\cos\theta}=\frac{AC}{AH}.$$ Since we want ratios to share endpoints, rearrange this to $$\frac{CE}{AC}=\frac{EH\cos\theta}{AH}.$$

By Law of Sines on $\triangle AEH$, we have $$\frac{EH}{AH}=\frac{\sin\angle BAF}{\cos\theta},$$ so $$\frac{EH\cos\theta}{AH}=\sin\angle BAF.$$ Finally, since $$\sin\angle BAF=\sin\angle BAC=\frac{CE}{AC},$$ we are done.

[color=#f0f][b]remark[/b][/color]: lol, missed the fact that $\triangle AHE\sim\triangle ABD$ just works. Lesson here: if there are two lengths that are not well understood, similar triangles can be used to reduce their ratio to a different ratio that is better understood.


I am thinking of a two-digit number less than 50 . If you double my number and subtract 12 , you get the original number with the digits reversed! What is the sum of the digits of the original number ?

(A)13 (B) 14 (C) 15 (D) 16 (E) none of these

E) None of these because your number is 48 and the sum of its digits is 12 only.


Find the least positive integer with more than $ 120$ divisors, at least 12 of which are consecutive positive integers.

[hide="Solution"]
Let $ n$ be the number with the required properties

We start of by assuming that the $ 12$ consecutive numbers which will divide $ n$ are $ \{1,2,3,\ldots,12\}$

Therefore $ n\equal{}2^3 \times 3^2 \times 5 \times 7 \times 11 \times k$ where $ k \in \mathbb{N}$

We search for the smallest $ k$ such that $ n$ has $ >120$ divisors

Let $ d(n)$ denote the number of divisors of $ n$ and $ n_k$ be $ 2^3 \times 3^2 \times 5 \times 7 \times 9 \times 11 \times k$

Clearly

$ d(n_2) \equal{} 120$
$ d(n_3) \equal{} 96$
$ d(n_4) \equal{} 144$

We claim that $ n_4$ is our desired number.

It remains to show that the least common multiple of $ 12$ consecutive numbers is $ >n_4$

If $ 13$ divides our possible number then the lowest possible is clearly $ >n_4$.
If $ 13$ doesn't divide our possible number then the next "group" from which we could obtain our number would clearly be larger than our current one.

Therefore the answer is

$ \boxed{2^5 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11 \equal{} 110880}$

[/hide]

Hello, I like your proof in some aspects. The notation $ n_k$ and such is helpful. However, I don't feel that your  proof is satisfactory later on. In particular, these two sentences have to be reconsidered:

If $ 13$ divides our possible number then the lowest possible is clearly $ > n_4$.
If $ 13$ doesn't divide our possible number then the next "group" from which we could obtain our number would clearly be larger than our current one.

Furthermore, how do you know that the set of consecutive integers doesn't end at $ 14$ or something?


ratio
factorial
Pascal\u0027s Triangle
binomial coefficients

In Pascal's Triangle, each entry is the sum of the two entries above it.  The first few rows of the triangle are shown below.

\[\begin{array}{c@{\hspace{8em}}
c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{4pt}}c@{\hspace{2pt}}
c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{3pt}}c@{\hspace{6pt}}
c@{\hspace{6pt}}c@{\hspace{6pt}}c} \vspace{4pt}
\text{Row 0: } &    &    &     &     &    &    & 1 &     &     &    &    &    &  \\\vspace{4pt}
\text{Row 1: } &    &    &     &     &    & 1 &    & 1  &     &    &    &    &  \\\vspace{4pt}
\text{Row 2: } &    &    &     &     & 1 &    & 2 &     & 1  &    &    &    &  \\\vspace{4pt}
\text{Row 3: } &    &    &     &  1 &    & 3 &    & 3  &     & 1 &    &    &  \\\vspace{4pt}
\text{Row 4: } &    &    & 1  &     & 4 &    & 6 &     & 4  &    & 1 &    &  \\\vspace{4pt}
\text{Row 5: } &    & 1 &     & 5  &    &10&    &10 &     & 5 &    & 1 &  \\\vspace{4pt}
\text{Row 6: } & 1 &    & 6  &     &15&    &20&     &15 &    & 6 &    & 1
\end{array}\]

In which row of Pascal's Triangle do three consecutive entries occur that are in the ratio $3: 4: 5$?

[hide="Solution"]In Pascal's Triangle, we know that the binomial coefficients of the nth row are $\binom{n}{0}$, $\binom{n}{1}$,...,$\binom{n}{n}$. Let our row be the nth row such that the three consecutive entries are $\binom{n}{r}$, $\binom{n}{r+1}$, and $\binom{n}{r+2}$.

After expanding and dividing one entry by another (to clean up the factorials), we see that $\frac 34=\frac{r+1}{n-r}$ and $\frac45=\frac{r+2}{n-r-1}$. Solving, $n = \boxed{062}$. [/hide]

I keep getting $n=\boxed{063}$ as my answer. Isn't the first row's coefficients for the binomial expansion $(x+y)^0$? Thus, when solving for $n$, you have to add one to get the row number. Where am I going wrong?

Sorry for the revive.
I also am getting the correct answer, but I am getting fractions for the lower part of the combination. So would it really count as a part of Pascal's triangle?

Can someone post a full solution , I keep getting $\boxed{63}$ as my answer.

[quote=champion999]Was this problem sourced from a past NYSML? I think I remember doing a similar problem in an ARML practice.[/quote]

Its an AIME problem. [hide=Sol]So you have a system, $\frac{\binom{n}{k+1}}{\binom{n}{k}}=\frac{4}{3}$ and $\frac{\binom{n}{k+2}}{\binom{n}{k+1}}=\frac{5}{4}$. Note that most of the terms cancel, $\dfrac{\cfrac{n\cdot(n-1)\cdot\ldots\cdot(n-k)}{(k+1)!}}{\cfrac{n\cdot(n-1)\cdot\ldots\cdot(n-k+1)}{k!}}$, leaving $\frac{n-k}{k+1}=\frac{4}{3}$ and $\frac{n-k-1}{k+2}=\frac{5}{4}$. From the first equation we get $k=\frac{3n-4}{7}$ and solving we get $5(\frac{3n-4}{7}+2)=4(n-\frac{3n-4}{7}-1)\implies5(3n-4+14)=4(7n-(3n-4)-7)\implies n=\boxed{62}$.[/hide]

EDIT: Note that it starts at row 0, so the $n$th row has $\binom{n}{k}$, which might have caused the off-by-one error.

[quote=champion999]Was this problem sourced from a past NYSML? I think I remember doing a similar problem in an ARML practice.[/quote]

That problem was 2016 NYSML Indivs #6.

If $2\binom{n}{k} = \binom{n}{k+1}$ and $3\binom{n}{k} = \binom{n}{k+2}$, compute $n+k$.

[hide=Solution]Note that Pascal's Triangle is made up of Binomial Coefficients, so Row $n$ is $\binom{n}{0}, \binom{n}{1}, \ldots \binom{n}{n}$. Let the three consecutive entries be $\binom{n}{a}$, $\binom{n}{a+1}$, $\binom{n}{a+2}$. Then we have $\frac{\binom{n}{r}}{\binom{n}{r+1}}=\frac{3}{4}$ and $\frac{\binom{n}{r}}{\binom{n}{r+2}}=\frac{3}{5}$. Cancelling and solving the system gives $n=\boxed{62}$.[/hide]

[hide=Full solution if you need it]Since it says [b]3 consecutive numbers[/b], and the entries in Pascals Triangle are combinations, we can consider the three entries as $\binom{n}{r}, \binom{n}{r+1}$ and $\binom{n}{r+2}$.
These numbers are in a $3:4:5$ ratio. We can pick any two to write as a fraction and simplify to make a system of equations. I would pick the numbers that are in a $3:4$ ratio, and the numbers in a $4:5$ ratio, because when we write the combinations as fractions, it'll be easy to simplify and cancel.

Ok, wish me luck with the latex:
$$\frac{3}{4} = \frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r+1)!(n-r-1)!}}$$$$\frac{3}{4} = \frac{n!}{r!(n-r)!}\cdot\frac{(r+1)!(n-r-1)!}{n!}$$$$\frac{3}{4} = \frac{r+1}{n-r}$$
$\frac{(r+1)!}{r!}$ simplifies to $r+1$ since $r+1$ is $1$ more than $r$ so the factorials cancel like that. The same logic goes with $\frac{(n-r-1)!}{(n-r)!}$.

So we have our first equation, and we need a second one to solve the system of equations. Lets simplify the combinations in the ratio of $4:5$. $$\frac{4}{5} = \frac{\frac{n!}{(r+1)!(n-r-1)!}}{\frac{n!}{(r+2)!(n-r-2)!}}$$$$\frac{4}{5} = \frac{n!}{(r+1)!(n-r-1)!}\cdot\frac{(r+2)!(n-r-2)!}{n!}$$$$\frac{4}{5} = \frac{r+2}{n-r-1}$$
So now we have our two equations, so we can solve for them and get $n$. $$\frac{3}{4} = \frac{r+1}{n-r}$$$$\frac{4}{5} = \frac{r+2}{n-r-1}$$ Cross multiplying and simplifying gives $$3n=4+7r$$$$4n=14+9r$$ Now just solve the system of equations: We'll subtract one equation from the other and obtain $$n=10+2r.$$ Now substitute in one original equation, $3n-4=7r$: $$3(10+2r)-4=7r$$$$r=26$$. We have $r$ so lets use that to find $n$, which is the numberth row that the entries are on, which is what the question asks. $$\frac{3}{4} = \frac{27}{n-26}$$$$n=62$$ So the row where there are three consecutive numbers in a $3:4:5$ ratio is the $\boxed{62}$nd row. :)[/hide]


integration
calculus
Euler
calculus computations

Find:

$ \int {\frac{{\sqrt {x^2  \plus{} x \plus{} 1} }}{{x \plus{} 1}}} dx$

 :o  :o

hello, using Euler's substitution $ \sqrt{x^2\plus{}x\plus{}1}\equal{}x\plus{}t$ we get $ x\equal{}\frac{t^2\minus{}1}{1\minus{}2t}$ and from here we have $ \sqrt{x^2\plus{}x\plus{}1}\equal{}\frac{\minus{}t^2\plus{}t\minus{}1}{1\minus{}2t}$. Further we get
$ dx\equal{}\frac{\minus{}2t^2\plus{}2t\minus{}2}{(1\minus{}2t)^2}dt$ and our integral is $ 2\int\frac{(\minus{}t^2\plus{}t\minus{}1)^2}{(t^2\minus{}2t)(1\minus{}2t)^2}\,dt$. With the partial fraction decomposition we get
the following integrals $ \int\frac{1}{2}\,dt\minus{}\frac{3}{2}\int\frac{1}{(2t\minus{}1)^2}\,dt\plus{}\int\frac{1}{t\minus{}2}\,dt\minus{}\int\frac{1}{t}\,dt\plus{}\int\frac{1}{2t\minus{}1}\,dt$.
Sonnhard.

[hide="Simplification"]\[ \frac {\sqrt {x^2 \plus{} x \plus{} 1}}{x \plus{} 1} \equal{} \frac {x^2 \plus{} x \plus{} 1}{(x \plus{} 1)\sqrt {x^2 \plus{} x \plus{} 1}} \equal{} \frac {x(x \plus{} 1) \plus{} 1}{(x \plus{} 1)\sqrt {x^2 \plus{} x \plus{} 1}}
\]

\[ \equal{} \frac {x}{\sqrt {x^2 \plus{} x \plus{} 1}} \plus{} \frac {1}{(x \plus{} 1)\sqrt {x^2 \plus{} x \plus{} 1}}
\]

\[ \equal{} \frac {2x \plus{} 1}{2 \sqrt {x^2 \plus{} x \plus{} 1}} \minus{} \frac {1}{2 \sqrt {x^2 \plus{} x \plus{} 1}} \plus{} \frac {1}{(x \plus{} 1)\sqrt {x^2 \plus{} x \plus{} 1}}
\]

The first few terms you can integrate easily, the next one maybe needs a bit more manipulation[/hide]

I'm not sure this helps, but...

$ \int {\frac {{\sqrt {x^2 + x + 1} }}{{x + 1}}}\,dx = \int {\frac {\sqrt {x^3 - 1}\cdot\sqrt {x - 1}}{x^2 - 1}\,dx = \int {\frac {\sqrt {x^3 - 1}}{\sqrt {x^2 - 1} \cdot \sqrt {x + 1}}\,dx}}$

Or...

$ \int {\frac {{\sqrt {x^2 + x + 1} }}{{x + 1}}}\,dx = \int {\frac {\sqrt {x^2 + x + 1}}{\sqrt {x^2 + 2x + 1}}}\,dx = \int {\sqrt {\frac {x^2 + x + 1}{x^2 + 2x + 1}}}\,dx = \int {\sqrt {1 - \frac x{(x + 1)^2}}}\,dx$

It might be that my manipulation won't get anywhere at all. :P

Hello

Euler's substitution:
Integrals of the form $ \int f(x)\sqrt{ax^2 \plus{} bx \plus{} c}dx$are calculated with d aid of $ 3$ Euler's  substitutions:
 $ I.\sqrt {ax^2 \plus{} bx \plus{} c} \equal{} t \plus{} (or \minus{} ) x\sqrt {a} ,if a > 0$
$ II.\sqrt {ax^2 \plus{} bx \plus{} c} \equal{} tx \plus{} \sqrt {c} ,if c > 0$
$ III.\sqrt {ax^2 \plus{} bx \plus{} c} \equal{} (x \minus{} \alpha)t$ if if $ \alpha$ is a real root of $ ax^2 \plus{} bx \plus{} c$.

Thank u.

[quote="Dr Sonnhard Graubner"]hello kabi, yes, that's right, in this case the Euler substitution is very helpful.
Sonnhard.[/quote]

Absolutely right DR. :)


I'm not sure whether or not I should take the Aime Problem Series. There are two many good classes to choose from, so it's really hard to decide.  ;)

Before, I was thinking about taking Olympiad Geometry and the Aime Problem Series.

Just yesterday however, I was posting Aime problems (with solutions a while after)and I still remember every single problem and the approach vividly right now. Since I will definately have every Aime posted by the end of Febuary, I am thinking about taking these two courses instead.

Olympiad Geometry, Intermediate Algebra

What do you guys think?

Personally, I think it's more constructive to take subject courses unless you feel that, in particular, you want to saturate yourself with AIME problems.  It should also be noted that our Intermediate courses are very much at the AIME level (for the most part) and you will see many algebraic techniques in Intermediate Algebra that will help with AIME problems.  I even wrote some AIME problems into that course because they demonstrate certain techniques so well.

The Olympiad Geometry course is the hardest course AoPS currently offers.  Be sure to test yourself with the pretest and post-test before enrolling.  It will require a greater effort.

All that said, the problem series are designed to present you with problelms on many topics.  There you can patch up any weak areas you have in a variety of subjects.

Ok, cool! I think I'll go with my new plan.

Thanks.  :)


vector
superior algebra
superior algebra unsolved

The following problem is madness.

Denote the dual space of $V$ over $k$ by $V^*=Hom(V,k)$.  Consider the natural isomorphism $\eta_V: V\to V^{**}.$  Let $\alpha\in Hom(V,W).$  Define $W^*$ and $\eta_W$ analogously.

Prove:

$\alpha^{**}\circ \eta_V=\eta_W\circ\alpha.$

Any ideas?

What do you mean by "$V$ over $k$"? Vector spaces? Or to phrase that more generally, what category are we working in?

Sorry, I was a bit vauge.  $V$ is a finite dimensional vector space over a field $k$.

-jng


What is the geometric mean of 19 and 304?

The geometric mean of $ a$ and $ b$ is defined as $ x$ such that $ x$ is a common mean in a proportion involving $ a$ and $ b$: $ \frac{a}{x}\equal{}\frac{x}{b}$, or, in other words, $ x\equal{}\sqrt{ab}$, where both the positive and negative root of the product $ ab$ count for the solution. However, when no indication is given, the positive root is implied. Thus, the geometric mean of $ 19$ and $ 304$ is $ \sqrt{19\cdot304}\equal{}\fbox{76}$.


algebra
function
domain
factorial
calculus
integration
superior algebra

how can i proove that the field of fractions of
$Z[i\sqrt{3}]$ is factorial(UFD) ?
-------------------------------
how is $S^{-1}A$ defined in the context that
A is a commutative ring(im not sure if also principal or euclidean)
-------------------------------
what is the total field of fractions? how does it differ frm
the field of fractions?
-------------------------------

In the context of commutative ring $A$. Given a multiplicative set $S$, we define the ring $S^{-1}A$ to be the set of pairs $(a,s)$, where
\[(a,s)~(b,t) \text{ iff }%Error. "there" is a bad command.
u\in S, u(at-bs)=0\]

---------------------------------------------------------
We call $S^{-1}A$ total ring of fraction if $S$ taken to be all nonzerodivisor elements. If $A$ is a domain then your total ring of fraction is a field of fraction.

[quote="spx2"]how can i proove that the field of fractions of
$Z[i\sqrt{3}]$ is factorial(UFD) ?[/quote]
The field of fractions is, as the name says, a field. Fields are always factorial.
The ring $\mathbb Z[i\sqrt{3}]$ itself is not factorial, look at $(1+i\sqrt 3)(1-\sqrt 3)=2 \cdot 2$. But it's integral closure (thus $\mathbb Z \left[\frac{1+i\sqrt{3}}2\right]$) is.


Hi I am in 8th grade. I really need help to improve my math knowledge! Ask me questions to know where I stand. (Please no geometry questions, I hate that). Thanks guys!

Ok...
when i think of one, i will post it. :)

take my mock aime...haha...i suggest you look for problems, not expect problems to come to you, your question is too vague

Yeah, like Altheman said, try searching the forum or the Resources pages for more challenging problems. I'm sure that if you are an eager problem solver, you will search for problems and not expect them to come to you. I wish money will come to me and I don't have to work when I grow up :)

[quote]I wish money will come to me and I don't have to work when I grow up

 :rotfl:  :rotfl: [/quote]


Show that the expression m^p - n^p, in which m and n are integers and p is a prime,
is either prime to p or is divisible by p2

Suppose that $m=n(modp)$ then $m^{p-1}+m^{p-2}n+...n^{p-1}=p\cdot m^{p-1}(modp)$ this proves that if  $p|m-n$ then $p^2|m^p-n^p$.Suppose that m-n is not divisible by p then we have some cases:
(1)p|m => n is not divisible by p n hence $m^{p}-n^{p}$ is not divisible by p.
(2)p|n this case is similar to (1)
(3)neither of m,n is divisible by  p => $m^p=m(mod p),n^p=n(modp)$ (by Feramat's little theorem) so $m^p-n^p$ is not divisible by p.


limit
integration
algebra
polynomial
ratio
geometry
linear algebra

As Professor Gregoris Makrides (Vice-President of MASSEE - Mathematical Society of South Eastern Europe) said,
the 3rd South-Eastern European Mathematical Olympiad for (1st and 2nd year) University Students SEEMOUS 2009 will take place on 4-9 March 2009. Arrivals are on 4 March, departures are on 9 March.

1st day: Wednesday 4 March: Arrival and first Jury meeting.
2nd day: Thursday 5 March: Opening Ceremony and Jury meeting for problem selection. Some students may arrive on this day.
3rd day: Friday 6 March: Contest Day. Papers are saved, approved by the Chairman of the Jury and numbered. Names are hidden.
4th day: Saturday 7 March: Marking of papers by the leaders and deputy leaders. Jury meeting to approve results and medals.
5th day: Sunday 8 March: Awards ceremony in the morning. Excursion in the afternoon and/or departures
6th day: Monday 9 March: Departures

The website http://www.seemous.eu will be open next week. At the moment it contains information about SEEMOUS 2007 only. Information about SEEMOUS 2008 can be found here: http://www.hms.gr/seemous08/ , http://www.mathlinks.ro/Forum/viewtopic.php?t=177869

Problems 2009 (in Romanian): http://www.mateforum.ro/viewtopic.php?t=3630

Please, don't you have the results? Thank you.

Thank you very much. The romanian team has 5 gold medals. I'm sure it was on the first place.

Nonsense.
There were about 5 different Romanian teams.
To the first place got the Israeli national team, with 105 points total.
To the second place got a Romanian team, not sure which, with just a little less points, like 103 or 102.

SEEMOUS 2009

1. a) Calculate the limit \[ \lim_{n\to\infty}\frac{(2n\plus{}1)!}{(n!)^2}\int_0^1(x(1\minus{}x))^nx^kdx,\] where $ k\in \mathbb{N}$.
b) Calculate the limit \[ \lim_{n\to\infty}\frac{(2n\plus{}1)!}{(n!)^2}\int_0^1(x(1\minus{}x))^nf(x)dx,\] where $ f: [0,1]\to\mathbb{R}$ is a continuous function.

2. Let $ P$ be a real polynomial of degree 5. Assume that the graph of $ P$ has three inflexion points lying on a straight line. Calculate the ratios of the areas of the bounded regions between this line and the graph of the polynomial $ P$.

3. Let $ \mathbf{SL}_2(\mathbb{Z})\equal{}\{A\mid A$ is a $ 2\times2$ matrix with integer entries and $ \det A\equal{}1\}$.
a) Find an example of matrices $ A,B,C\in\mathbf{SL}_2(\mathbb{Z})$  such that  $ A^2\plus{}B^2\equal{}C^2$.
b) Show that there do not exist matrices $ A,B,C\in\mathbf{SL}_2(\mathbb{Z})$ such that $ A^4\plus{}B^4\equal{}C^4$.

4. Given the real numbers $ a_1,a_2,...,a_n$ and $ b_1,b_2,...,b_n$ we define the $ n\times n$ matrices $ A\equal{}(a_{ij})$ and $ B\equal{}(b_{ij})$ by
$ a_{ij}\equal{}a_i\minus{}b_j$ and $ b_{ij}\equal{}\begin{cases}1,&\text{if }a_{ij}\ge 0,\\0,&\text{if }a_{ij}<0,\end{cases}$ for all $ i,j\in\{1,2,\ldots,n\}$.
Consider $ C\equal{}(c_{ij})$ a matrix of the same order with elements 0 and 1 such that
$ \sum_{j\equal{}1}^nb_{ij}\equal{}\sum_{j\equal{}1}^nc_{ij}$, $ i\in\{1,2,\ldots,n\}$ and $ \sum_{i\equal{}1}^nb_{ij}\equal{}\sum_{i\equal{}1}^nc_{ij}$, $ j\in\{1,2,\ldots,n\}$.
Show that:
a) $ \sum_{i,j\equal{}1}^n a_{ij}(b_{ij}\minus{}c_{ij})\equal{}0$ and $ B\equal{}C$.
b) $ B$ is invertible if and only if there exist two permutations $ \sigma$ and $ \tau$ of the set $ \{1,2,...,n\}$ such that
$ b_{\tau(1)}\le a_{\sigma(1)}< b_{\tau(2)}\le a_{\sigma(2)}<\ldots\le a_{\sigma(n\minus{}1)}<b_{\tau(n)}\le a_{\sigma(n)}$.

Unfortunately, there has been a lot of cheating on the competition.  :o

[url=http://www.seemous.eu/index.php?id=7]Regulations[/url]

[quote]7.Teams represent institutions but results per country will be computed for teams of six students made up by the best six scores of students participating from each country, except if a National team is officially participating. The National teams have to be specified, in writing, by the national Mathematical Society or the Ministry of Education of the country.[/quote]

[quote]12.National teams can participate with students not representing institutions or a mixture. If a national team participates with a leader then the leader becomes a member of the jury and has one vote.[/quote]

[b]Results by Teams:[/b]
National Team of Israel 29+18+17+15+14+12=[b]105[/b]
University of Bucharest, Romania 21+20+19+19+14+10=[b]103[/b]
Technical University of Cluj-Napoca, Romania 38+25+17+11+8+2=[b]101[/b]
POLITECHNICA BUCHAREST, Romania 30+27+17+10+5=[b]89[/b]
National Team of Greece 14+14+10+9+7+1=[b]55[/b]
National Team of Romania 24+15=[b]39[/b]
University of Sofia, Bulgaria 13+8+7+7=[b]35[/b]
University of Architecture, Civil Engineering and Geodesy, Bulgaria 9+6+6+5+5+1=[b]32[/b]
University "Al. I. Cuza Iasi", Romania [b]16[/b]
Demokritus University of Thrace, Greece 11+1+1+1+0+0(+0+0)=[b]14[/b]
West University Timisoara, Romania [b]13[/b]
University of Plovdiv, Bulgaria 6+3+2+0=[b]11[/b]
National Team of Colombia 8+1=[b]9[/b]
University of Athens, Greece [b]8[/b]
Technical University of Gabrovo, Bulgaria 0+0+0+0=[b]0[/b]
Technical University "Gheorghe Asachi" Iasi, Romania [b]disq.[/b]

[b]Results by Countries:[/b]
Israel 29+18+17+15+14+12=[b]105[/b]
Greece 14+14+10+9+7+1=[b]55[/b]
Bulgaria 13+9+8+7+7+6=[b]50[/b]
Romania 24+15=[b]39[/b]
Colombia 8+1=[b]9[/b]

[quote="darij grinberg"]Problem 4a [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5758]rings a bell[/url] (and 4b is easy with the same method).

  darij[/quote]

Problem 3 was also proposed before, at [i]XI Olimpiada Iberoamericana de Matematica Universitaria 2008[/i].


calculus integration algebra polynomial limit function logarithms

Let
\[ f(t) = \sqrt {t}\int_{ - \infty}^{\infty}{\left(\frac {1}{1 + P(x)^2\right)^tdx}
}\]
where $ P(x)$ is a polynomial with at least one real root and no real roots of multiplicity higher than 1.  Compute $ \lim_{t\to \infty}f(t)$.
[hide="what I think the answer is"]
\[ \sqrt {\pi}\sum{\frac {1}{|P'(r)|}
}\]
where the sum goes over the roots of $ P$.  Actually, I think this holds for a much wider class of functions than polynomials.  It should hold for any function $ P$ that is $ C^2$ in a neighborhood of its roots, with at least one but not infinitely many real roots, and no roots of multiplicity higher than 1.  My proof is sketchy at some points, though, so I'd love to see some analysis of this or a counterexample.
[/hide]

Clearly if $ P$ has no real roots then $ \lim_{t\to \infty}f(t) = 0$, but in this case what can we say about, for instance, $ \lim_{t\to \infty}f(t)^{\frac {1}{t}}$?  How about if we allow roots of order greater than 1?

Well if P(x)=(x-x0)Q(x) where x0 is a real root.
THEN WE CAN take an estimate 1/(1+P(x)^2)<=1/P(x)^2=1/(x-x0)^2(Q(x))^2=(Ax+B)/(x-x0)^2+(A1x+B1)/(x-x1)^2+(A2x+B2)/(x-x1*)^2+.... where * stands for complex conjugate raise it to the power t, and use the binomial equation (but in this case it's a multinomial) for t real, from there you would find what you need.

This is the only way that comes across my mind perhaps someone else had a different approach to this limit.

That approach doesn't seem promising.  For one thing, you're estimating a function that is bounded from above by $ 1$ by a function that has a non-integrable discontinuity at every root of $ P$.  Since the contribution to the integral for large $ t$ comes entirely from small neighborhoods around the roots, your estimate throws out all the useful information about the integral.

I've given the problem some more thought after I originally posted, and I think the answer to the first question is indeed correct and follows easily enough for $ C^2$ functions by steepest descent.  The case with no real roots, or multiple roots, can also be done by steepest descent, but it's a bit messier.

[quote="JoeBlow"]
[hide=what I think the answer is]
\[ \sqrt {\pi}\sum{\frac {1}{|P'(r)|}
}\]
where the sum goes over the roots of $ P$. [/quote]

yes, this is answer.
[quote="JoeBlow"]
 It should hold for any function $ P$ that is $ C^2$ in a neighborhood of its roots, with at least one but not infinitely many real roots, and no roots of multiplicity higher than 1
[/quote]
I think it's enough  to say $ f \in C^{1}$

[quote="JoeBlow"]
 but in this case what can we say about, for instance $ \lim f^{1/t}$
[/quote]
limit will exist  and equal to $ \max_{x} \frac {1}{1 + P^{2}(x)}$
since asymptoticaly this itegral is equal to
$ \approx \exp (t\ln c_{1}) t^{a_{1}} + \exp(t\ln c_{2})t^{a_{2}} + ...$
where $ c_{1} = \max_{x} \frac {1}{1 + P^{2}(x)}$ and $ c_{2} < c_{1}$...
in the case if we allow roots of order greater than 1 then answer will be equal to $ 1$ .
I will explain it tomorrow


To each vertex of a regular $ n$-gon a real number is assigned in such a way that the sum of all $ n$ numbers is positive. If three consecutive vertices are assigned the numbers $ x, y, z$ respectively and $ y<0$ then the following operation is allowed: the numbers $ x, y, z$ are replaced by $ x\plus{}y, \minus{}y, z\plus{}y$ respectively. Such an operation is performed repeatedly as long as at least one of the $ n$ numbers is negative. Determine whether this procedure necessarily terminates after a finite number of steps.

See [url=http://www.mathe.tu-freiberg.de/~wegert/Papers/Relax.pdf]Elias Wegert, Christian Reiher, [i]Relaxation procedures on graphs[/i], Discr. Appl. Math. (to appear)[/url], 1.2.

  darij


If 20$ \%$ of 10$ \%$ of a number is 12, what is 10$ \%$ of 20$ \%$ of the same number?


geometry
symmetry
calculus
derivative
trigonometry
geometry proposed

An equilateral triangle ABC of side length 1 is given, and S denotes the set of all points lying in its interior or on its boundary. For any point M in S, let a(M), b(M), c(M) denote the distances from M to BC, CA, AB respectively. Define $f\left(M\right)=\sum_{\text{cyclic}} a\left(M\right)^3\left(b\left(M\right)-c\left(M\right)\right)$. Describe geometrically the set of all points M in S for which $f\left(M\right) \geq 0$. Find the extreme values of f(M) when M moves inside S, and find the points M in S for which these extreme values are attained.

Let $A', B', C'$ be the foot of normals from the point $M$ to the sides $BC, CA, AB$ of an equilateral triangles $\triangle ABC$, such that $BC = CA = AB = 1$. The sum of the distances $(aM) + b(M) + c(M) = MA' + MB' + MC'$ is constant. This is because the sum of the areas of the triangles $\triangle BCM, \triangle CAM, \triangle ABM$ with the altitudes $MA', MB', MC'$ from the common vertex $M$ to the bases $BC = CA = AB = 1$ is equal to the area of the equilateral triangle $\triangle ABC$:

$\frac{\sqrt 3}{4} = |\triangle ABC| = |\triangle BCM| + |\triangle CAM| + |\triangle ABM| =$

$= \frac 1 2 (MA' \cdot BC + MB' \cdot CA + MC' \cdot AB) = \frac 1 2 (MA' + MB' + MC')$

$a(M) + b(M) + c(M) = \frac{\sqrt 3}{2}$

Skipping the argument $M$, the expression $f(M)$ can be factored as

$f(M) = a^3(b - c) + b^3(c - a) + c^3(a - b) = c(b^3 - a^3) - ba(b^2 - a^2) - c^3(b - a) = $

$= (b - a)\left[c(b^2 + ba + a^2) - ba(b + a) - c^3\right]$

From symmetry of the expression with respect to the cyclic exchange of $a, b, c$, it follows that it must be factorable by $(c - b)$ and $(a - c)$ as well. This can be verified by a long division:

$f(M) = (b - a)(c - b)(a - c)(a + b + c) = \frac{\sqrt 3}{2} (b - a)(c - b)(a - c)$

This expression is equal to zero, iff $a = b$ or $b = c$ or $c = a$, i.e., iff the point $M$ is on one of the altitudes of the equilateral triangle $\triangle ABC$. Let $H$ be the orthocenter and $D, E, F$ be the feet of altitudes from the vertices $A, B, C$. If the point $M$ is in the triangle $\triangle AHF$, then $b \le a,\ c \le b,\ a \ge c$ and the expression $f(M) \ge 0$ is greater than or equal to zero. If the point $M$ is in the triangle $\triangle AHE$, then $b \le a,\ c \le b,\ a \ge c$ and the expression $f(M) \le 0$ is less than or equal to zero. Similarly, if the point $M$ is in the triangles $\triangle BHD, \triangle CHE$, the expression $f(M) \ge 0$ is greater than or equal to zero and if the point $M$ is in the triangles $\triangle BFF, \triangle CHD$, the expression $f(M) \le 0$ is less than or equal to zero.

To find the extrema of the expression $f(M) = f(a, b, c)$ subject to the constraint $g(M) = g(a, b, c) = a + b + c - \frac{\sqrt 3}{2} = 0$, we use a Lagrange multiplier $\lambda$, Define

$h(a, b, c) = \frac{2}{\sqrt 3} f(a, b, c) + \lambda g(a, b, c) = (b - a)(c - b)(a - c) + \lambda \left(a + b + c - \frac{\sqrt 3}{2}\right)$

For an extremum, the partial derivatives of $h(a, b, c)$ must be all zero:

$\frac{\partial h}{\partial a} = 2ab - b^2 - 2ca + c^2 + \lambda = 0$

$\frac{\partial h}{\partial b} = 2bc - c^2 - 2ab + a^2 + \lambda = 0$

$\frac{\partial h}{\partial c} = 2ca - a^2 - 2bc + b^2 + \lambda = 0$

Subtracting the 1st pair of these equations from each other, we get

$2ab - b^2 - 2ca + c^2 = 2bc - c^2 - 2ab + a^2$

$a^2 - 4ab + b^2 = 2c\left[c - (a + b)\right] = 4c^2 - c \sqrt 3$

Similarly, subtracting the other 2 pairs of these equations from each other, we get

$b^2 - 4bc + c^2 = 4a^2 - a \sqrt 3$

$c^2 - 4ca + a^2 = 4b^2 - b \sqrt 3$

For example, subtracting the 1st pair of the last 3 equation from each other yields

$b^2 - c^2 - 4a (b - c) = \sqrt 3(b - c) - 4(b^2 - c^2)$

$(b - c)(b + c - 4a) = (b - c)\left[\sqrt 3 - 4(b + c)\right]$

First, we can assume $b \neq c$. Reducing the last equation by the factor $(b - c)$ and substituting $b + c = \frac{\sqrt 3}{2} - a$, we arrive to

$\frac{\sqrt 3}{2} - 5a = \sqrt 3 - (2\sqrt 3 - 4a),\ \ 9a = \frac{3\sqrt 3}{2},\ \ a = \frac{\sqrt 3}{6}$

Similarly, we could deduce the necessity of $b = c = \frac{\sqrt 3}{6}$. This contradicts our assumption $b \neq c$, hence, we assume $b = c$. Using the constraint $a + 2b = \frac{\sqrt 3}{2}$, the 2nd equation then becomes

$-2b^2 = 4a^2 - a \sqrt 3,\ \ -\frac 3 8 + \frac{a \sqrt 3}{4} + \frac{a^2}{4} = 4a^2 - a \sqrt 3$

$3a^2 - a\sqrt 3 + \frac 1 4 = 0,\ \ \left(a\sqrt 3 - \frac 1 2\right)^2 = 0,\ \ a = \frac{1}{2\sqrt 3} = \frac{\sqrt 3}{6}$

Therefore, these 3 equations can simultaneously hold only if $a = b = c = \frac{\sqrt 3}{6}$, i.e., if the point $M$ is identical with the orthocenter $H$ (and they do). However, $f(H) = 0$ and in any neighborhood of the orthocenter, the expression $f(M)$ attains both positive and negative values. Thus the orthocenter is a saddle point. Consequently, the extrema of $f(M)$ (if any) are on the sides of the equilateral triangle $\triangle ABC$. WLOG, consider the side $AB$. Then $c = 0$ and the expression $f(M)$ with the constraint $g(M) = 0$ becomes

$f(a, b) = \frac{\sqrt 3}{2} ab(a - b),\ \ g(a, b) = a + b - \frac{\sqrt 3}{2} = 0$

Using the Lagrange multiplier $\lambda$ again, we define $h(M)$ as before and demand that its partial derivatives are zero:

$h(a, b) = \frac{2}{\sqrt 3} f(a, b) + \lambda g(a, b) = ab(a - b) + \lambda \left(a + b - \frac{\sqrt 3}{2}\right)$

$\frac{\partial h}{\partial a} = 2ab - b^2 + \lambda = 0$

$\frac{\partial h}{\partial b} = a^2 - 2ab + \lambda = 0$

Subtracting these 2 equations from each other, we get

$2ab - b^2 = a^2 - 2ab,\ \ a^2 - 4ab + b^2 = 0,\ \ a = \frac{4b \pm \sqrt{16b^2 - 4b^2}}{2} = (2 \pm \sqrt 3) b$

and using the constraint $a + b = \frac{\sqrt 3}{2}$,

$a = \left(\frac{\sqrt 3}{2} - a\right) (2 \pm \sqrt 3),\ \ a_{\pm} = \frac{\sqrt 3 \pm 1}{4},\ \ b_{\pm} = \frac{\sqrt 3 \mp 1}{4}$

For the upper signs, $a > b$, the point $M_+$ is on the segment $AF$ and $f(M_+) > 0$, i.e., it reaches its maximum. For the lower signs, $a < b$, the points $M_-$ is on the segment $BF$ and $f(M_-) < 0$, i.e., it reaches its minimum. The values of the minimum and maximum are

$f(M_+) =\frac{\sqrt 3}{2} \cdot \frac{\sqrt 3 + 1}{4} \cdot \frac{\sqrt 3 - 1}{4} \left(\frac{\sqrt 3 + 1}{4} - \frac{\sqrt 3 - 1}{4}\right) = \frac{\sqrt 3}{32}$

$f(M_-) = - f(M_+) = -\frac{\sqrt 3}{32}$

Let $A_+', B_+'$ be the feet of normals from the points $M_+$ to the sides $BC, CA$ and $A_-', B_-'$ be the feet of normals from the points $M_-$ to the sides $BC, CA$.  From the right angle triangles $\triangle BM_+A_+', \triangle BM_-A_-'$ with the angles $M_+BA_+' = \angle M_-BA_-' \equiv \angle ABC = 60^o$, the distances of the points $M_+, M_-$ from the vertex $A$ are

$AM_+ = AB - BM_+ = 1 - \frac{a_+}{\sin{60^o}} = 1 - \frac{\sqrt 3 + 1}{4} \cdot \frac{2}{\sqrt 3} = 1 - \frac{3 + \sqrt 3}{6} = \frac{3 - \sqrt 3}{6}$

$AM_- = AB - BM_- = 1 - \frac{a_-}{\sin{60^o}} = 1 - \frac{\sqrt 3 - 1}{4} \cdot \frac{2}{\sqrt 3} = 1 - \frac{3 - \sqrt 3}{2} = \frac{3 + \sqrt 3}{6}$

Similar maxima and minima of the expression $f(M)$ can be found of the remaining 2 sides $BC, CA$ of the equilateral triangle $\triangle ABC$.


calculus
integration
function
inequalities
real analysis
real analysis unsolved

Let $f(x)$ be a continuous function at $[0,1].$ If $f(x)$ isn't 0 identically, prove that there exist $s$ such that $|f(s)|>\left|\int_{0}^{s}f(x) dx\right|\ (0<s\leq 1).$

Put $I=[0,1]$.  If $f$ is continuous on $I$, so is $|f|$.  Let $t=\max_{s\in I}|f(s)|$.

Then \[|f(t)|\geq\int^{1}_{0}|f(x)|dx\geq\int^{t}_{0}|f(x)|dx\geq\left|\int^{t}_{0}f(x)dx\right|\] The first inequality is strict unless $f$ is constant.  The second inequality is strict unless $f\equiv 0$ on $[t,1]$.  So as long as $t\not=0$, we are done.

Otherwise, $G(x)=|f(x)|-\left|\int^{x}_{0}f(t)dt\right|$ is a continuous function of $x$ with $G(0)>0$, and hence we can find some $u\in I\setminus\{0\}$ such that $G(u)>0$.


inequalities
function
induction
inequalities solved

If x_1,x_2,...,x_k (k>3) are positive real numbers prove that

 x_1/(x_k+x_2) + x_2/(x_1+x_3)+..+x_k/(x_[k-1]+x_1) \geq 2

Here is what I think:

Consider WLOG sum x_k=1
and then I applied Jensen's inequality with x1,.., xk as weights to function f(x)=1/x which is convex (it's easy to prove using derivates)

and we have that:
x_1/(x_k+x_2) + x_2/(x_1+x_3)+..+x_k/(x_[k-1]+x_1) >=1/(x1x2..x1xk)

but 2(x1x2+..+x1xk)<=(x1+..+xk)^2 (just expand)

so we get what we were looking for!

I hope it's okay! :D cheers

well.. when you post during day-time (when you are in  Romania) u hurry because the dial up costs more ,3 times more, than during the night, and mistakes like these occur, sorry again! I will try to be patient from now on! :D

[quote="wpolly"]Using Jensen we could only get LHS \geq 1/2(x1x2+x2x3+.....+xnx1).
So our goal is to prove for n \geq 4 we have (x1+x2+....+xn) 2  \geq 4(x1x2+x2x3+......+xnx1).
(A classical theorem AFAIK.)[/quote]

When n=4 the inequality is equavilent to (x1-x2+x3-x4) 2  \geq 0, which is obvious.
Then it's almost trivial to complete the proof using induction.

well this inequality you call trivial was solved only by 6 people at the Romanian TST in 2002, all of which went afterward to the IMO.

I used induction  to prove it, but it took me an hour. Of course I got to this starting from another inequality, so at first I thought that it is not true .. :)

anyway the official solution is the following:

we can suppose x_1+x_2+...+x_n=1. If n=2k is even then
x_1x_2+...+x_nx_1 \leq (x_1+x_3+...+x_{2k-1})(x_2+x_4+...+x_{2k}) \leq 1/4 by AM-GM.

now if n=2k+1, k\geq 2, we can suppose WLOG that x_1\geq x_2 then because x_1x_2+x_2x_3+x_3x_4 \leq x_1(x_2+x_3)+(x_2+x_3)x_4 and we can use the inequality for the even number of numbers: x_1, x_2+x_3, x_4, ... , x_n and we are done.

There are not many offcial solutions that this neat :D


calculus
integration
trigonometry
logarithms
calculus computations

[hide="Series representation"]From $ \arctan x \equal{}\sum_{n\equal{}0}^{\plus{}\infty}(\minus{}1)^{n}\frac{x^{2n\plus{}1}}{2n\plus{}1}$ we have that $ \frac{\arctan x}{x}\equal{}\sum_{n\equal{}0}^{\plus{}\infty}(\minus{}1)^{n}\frac{x^{2n}}{2n\plus{}1}$, and so

$ I \equal{}\int^{1/2}_{0}\frac{\arctan x}{x}\,dx \equal{}\int^{1/2}_{0}\left(\sum_{n\equal{}0}^{\plus{}\infty}(\minus{}1)^{n}\frac{x^{2n}}{2n\plus{}1}\right)\,dx \equal{}\sum_{n\equal{}0}^{\plus{}\infty}(\minus{}1)^{n}\left[\frac{x^{2n\plus{}1}}{(2n\plus{}1)^{2}}\right]^{1/2}_{0}$

$ I \equal{}\sum_{n\equal{}0}^{\plus{}\infty}\frac{(\minus{}1)^{n}}{(2n\plus{}1)^{2}\cdot 2^{2n\plus{}1}}$.[/hide]

$ \int_{0}^{\frac{1}{2}}\;\;\frac{\tan^{\minus{}1}\,x}{x}\;\;\textbf dx$

$ \equal{}\frac{1}{2}\alpha\minus{}\int_{0}^{\alpha }\ln\tan\theta\ d\theta$, where $ \tan\alpha \equal{}\frac{1}{2}$.


superior algebra
superior algebra solved

$R_1$ and $R_2$ are two rings with multiplicative identity. Prove that any ideal of $R_1\oplus R_2$ has the form $I_1\oplus I_2$, where $I_i$ is an ideal of $R_i$, $i=1,2$.

Let $I$ be the ideal, and put $I_1=I\cap R_1,I_2=I\cap R_2$. Clearly, $I_1\cap I_2=(0,0)\ (*)$. It's also clear that $I_1,I_2$ are ideals of $R_1,R_2$ respectively. Now, given $x\in I$, we have $x=(1,0)\cdot x+(0,1)\cdot x$, and $(1,0)\cdot x\in I_1,(0,1)\cdot x\in I_2$, so $I=I_1+I_2$. This, together with $(*)$, implies the result.


We have integer, not negative numbers  [b]k[1] < k[2] < ... < k[m][/b]
Let [b]n = 2^k[1] + 2^k[2] + ... + 2^k[m][/b]

Delimit the number of odd coefficients of polynomial [b]P(x) = (x+1)^n[/b]

Thanks for any help

This one is from Polish Mathematical Olympiad which is still in progress - locked till 5th of December.


It suffices to prove that:
2a^3+2b^3+2c^3=>a^3+b^3+c^3+9abc
We get
a^3+b^3+c^3=>9abc
By AM-GM
a^3+b^3+c^3/3=>3abc
thus,
a^3+b^3+c^3>=9abc
Q.E.D.

[quote="varunrocks"]It suffices to prove that:

By AM-GM
a^3+b^3+c^3/3=>3abc
[/quote]

huh? since when is this true? AM-GM gives -
$ a^3\plus{}b^3\plus{}c^3 \ge 3abc$ .

[quote="akashram"]a,b,c positive reals such that a+b>=c b+c>=a c+a>=b

2a^2[b+c] + 2b^2[c+a] +2c^2[a+b]>= a^3 + b^3 + c^3 +9abc[/quote]

Just assume that $ c\equal{}min(a,b,c)$ and take $ a\equal{}c\plus{}x,b\equal{}c\plus{}y$

use Ravi transformation, it can also solve it  :)


Discuss here, Illinoisians, how you guys think the Illinois team as a whole would place, and how might the individuals place in the written round, such as who might make top 57, how certain individuals might rank, or who might get into Nats CD. Anything that's relevant. Please be reasonable in your predictions, thanks.

Some data to concern about the Illinois team, in case if you didn't know:
(This same information is eventually going to be up on the MATHCOUNTS website)

Coach: Krish Revuluri from Plum Grove JHS, Rolling Meadows

Team:

1. Patrick Liscio, Lincoln JHS, Skokie
2. Megan Chen, Wood Oaks JHS, Northbrook
3. James Tao, Plum Grove JHS, Rolling Meadows
4. David Wang, Hinsdale MS, Hinsdale

Hmm... this is ca's goal.

1. 10/10 team round
2. everyone in top 16
3. 3 people in top 8
4. 2 people in top 4
5. 1 person in top 2.

BTW, those are "AT LEAST"

[quote="gauss1181"]Hmmm, meh. Since no one's posting I guess I'd have some Illinoisians critique our Nats goal????

Hmm...I guess we should shoot for at least 10th place team, at least one countdown qualifier, and all four members making top 57. We must not get below an 8 on the team round...

That should be reasonable for now...I guess.[/quote]

Your goal should always be #1. There's no other option if you want to do your best. Your goal should be ideal - your work should be realistic (e.g. if you want to get #1, tell yourself there's no way you're going to get #1 unless you work your butt off).

I would be quite impressed with you guys if you brought home at-most-10th place (i.e. your rank is $ \le 10$) and had a CD qualifier, although I don't know much about how good this year's kids are.

I'm pretty sure we have quite the potential.

But what's putting me in doubt are the returning kids from last year's Nats competition. :(

Which there are usually a lot of. There have been many years where there's been a lot of potential in IL, but that potential has rarely been achieved.

[quote="gauss1181"]
Coach: Krish Revuluri from Plum Grove JHS, Rolling Meadows

Team:

1. Patrick Liscio, Lincoln JHS, Skokie
2. Megan Chen, Wood Oaks JHS, Northbrook
3. James Tao, Plum Grove JHS, Rolling Meadows
4. David Wang, Hinsdale MS, Hinsdale[/quote]

OMG DR. REVULURI IS COACHING YOU GUYS??? wow. i haven't seen him in like... a long time! lol he must be so happy to coach you ppls. our math team wasn't exactly so great the last couple of years... :maybe:

[quote="gauss1181"]hm did he coach your team, guineapiggies?[/quote]

well i guess... but we epic phailed chapter XD hahahaha go us!!!  :D  :D  :D

[quote="gauss1181"]Hm...I checked the top 25% chapter list from 2008 and it apparently said that only 3 ppl made it from ur school...? :maybe: Were you the only one who didn't make it?[/quote]

made it to top 25% ? cuz if so we had more than 3 make top 25% last year... if it was make state nobody from my school made state last year... as for being the only one who didn't make it... :maybe:  :maybe:  :maybe:

[quote="gauss1181"]So then were you the only one who didn't make top 25%?

I'm confused... :maybe:[/quote]

haha yeah  :D  :D

hm guys, to kill our boredom, how bout if we play two games on this forum? (actually we have three if we include the FTW tournament)

one's an illinois math marathon (For IL peeps only!)

another's an mc nats predictor game (similar to MO's [b]but only Illinoisians who aren't going to nats this year can play, and only predictions for IL and its team members can be made[/b])

So let's begin teh predictor game:

Here are the rules.

1. Post a reply in this thread and guess the exact place that you believe IL would take as a team.

2. Guess the 4 exact places that you believe the four IL team members would place.

3. After nats is over, I will go back to your predictions and see if they're right!

4. Scoring: if you get an individual person's exact place (WRITTEN ROUND) correct with 0-4 places off, you will get 5 POINTS!!!
if you are off by 4-8 places, you will get 3 points.
if you are off by 9-12 places, you will get 1 point.
And if you're off by 13 or more places, you will get 0 points!!!
So make your bettings good.
Please note that you CAN change your predictions as time goes by (until May 7), but once May 7th begins, NO MORE EDITING!

5. IF YOU THINK SOMEONE'S GONNA MAKE COUNTDOWN:

You will get 3 points if you guess his/her exact countdown position correct (that goes for champion, runner-up, semifinalist, quarterfinalist, or participant)
You will get 2 points if you are 1 position away.
You will get 1 point if you are 2 positions away.
You will get 0 points if you are 3 or more positions away, or if the person fails to make countdown.

6. FOR GUESSING THE TEAM PLACING:
You will get 3 points by guessing the team's place with an error of 0-2 places.
2 points for 3-5 places in error.
1 point for 6-8 places in error.
0 points for 9 or more places in error.

Have fun everyone!

Hmm...so nats is only six days away. Hopefully us Illinoisians have been practicing rigorously on our own! GOOD LUCK ALL!!!!!!!!!!!!!!!!!!!! :first:  :first:  :first: AND GO ILLINOIS!

I must say

good luck gauss,

hopefully ravi will win

illinois for first place.

What are you talking about?  Ravi didn't make nats.

ask "helagha"

he's my mathcounts coach's son.

isn't ravi the alternate.

No, nats teams don't have alternates, just four team members.

w00t!!! We feel that our team is all set, with a pwnage spirit cheer!!!!!!!!!!!!!!!!!!!!!

GO ILLINOIS!!!!!!!!

GOALS:

Get at least one person into countdown

Get everyone into top 40

Get perfect team round

Bleh.

what's your cheer? i hope it's not just "go IL"

I'll PM it to you privately so that outsiders will not see it!!!! :D

I doubt anyone from illinois will make top 12 or illinois will finish in the top ten.
I bet everyone will be in the top 57, and Illinois will get about 15th-ish.
Hopefully, Illinois does much better.
Our cheer was pretty lame... Please come up with a better one.

with gauss we should get 2nd place at least.


ya estan  los problemas  en  mi  post  anterior !!!!  tuvimos un examen  perfecto  por parte  de pablín  y  41  por  gato (dave vather acá) y  yo  35  todos  nosotros  entre  otros  sacamos  oro

Felicitaciones..

 MathLinkers  :first:

 y es una alegria para esta comunidad.....
 que miembros como ustedes demuestren sus capacidades en estos eventos ..

y a continuar colaborando con la comunidad!!

Carlos Bravo :)
Lima- PERU


Can I get help on the following HW question:
If X is a standard Cauchy Random Variable then 1/X is also a standard Cauchy Random Variable. How can it be proved?
Is it correct to say that: P(1/X<x) = P(X>1/x) = 1 - P(X<1/x) for x>0?
What about if x<0?

Suppose $ U$ is uniform on $ \left[\minus{}\frac{\pi}2,\frac{\pi}2\right].$ Then $ X\equal{}\tan U$ has a Cauchy distribution. What does that make $ \frac1X\,?$

Or another way of looking at things. Suppose $ X$ and $ Y$ are independent standard normal variables, so that $ (X,Y)$ is distributed on the plane with joint density $ \frac 1{2\pi}e^{\minus{}\frac{x^2\plus{}y^2}2}.$ Then $ \frac{X}{Y}$ has a Cauchy distribution. And if that argument works for $ \frac{X}{Y},$ then it should also work for $ \frac{Y}{X}.$


Find two disjont  sets ${a,b,c}$ and ${m,n,p}$ such that $a,b,c,m,n,p \in N$ ,  $a,b,c,m,n,p \geq 2002$ and $a+b+c=m+n+p$, $a^2+b^2+c^2=m^2+n^2+p^2$.

You don't need a PC for that solutions, just divide through $2002$ and you will see.

[quote="silouan"]Please post your solution, do not put only numbers.A pc could find the numbers, but the human gives you the manner :D  Thanks[/quote]
Silouan I just take $mn=bc,a=m+n,p=b+c$ :lol:

Yes but with this way you found all the solutions ?I don't think so .


Coucou je viens harceler les [i]frenchmens[/i] ...
Aurai je des [i]replies[/i] (si je dit [i]des reponses[/i], c'est correcte ? ou c'est pas beau ?) ?

De toute maniere ce forum bouge peu, autant que les Polonais en profite  :P

:rotfl:
Oui, c'est correct.
Et, la preuve, tu as (pas "des" pour le moment..) une réponse! :D

Exact.

Et voilà tu as DEUX réponses !
et une question de logique dans la veine de la grande école polonaise (Kuratowski et co.):
C'est quoi le CPE pour un de mes élèves de 6e?[hide]des Chocolats Pour les Enfants[/hide]
Le comble c'est que cet élève s'appelle Deremetz (sans blague, il y a un polonais dans la salle ?) :rotfl:

On discute du CPE avec des $6^{\mbox{eme}}$ (en cours bien entendu) ?  :huh:

:rotfl: , pas mal (par contre, je n'ai pas compris pour "Deremetz"...)

Enfin, quand je vois que dans le collège de ma soeur certains élèves faisaient grève avec des pancartes contre M. XXX (le Conseiller Principal d'Education), je me dis que ce n'est pas mieux..


[quote="lasha"]Given, that $ a,b$ and $ c$ are three real numbers such,that $ 1/(a^{2}\plus{}1)\plus{}1/(b^{2}\plus{}1)\plus{}1/(c^{2}\plus{}1) \equal{} 2$. Prove, that $ ab\plus{}bc\plus{}ca\leq 3/2$.[/quote]
Since, $ ab\plus{}ac\plus{}bc\leq |a|\cdot|b|\plus{}|a|\cdot|c|\plus{}|b|\cdot|c|$
enough to prove it for non-negative $ a,$ $ b$ and $ c.$
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103762


6. Twenty pegs are arranged in a $2*10$ rectangular grid. You have 10 small rubber bands that can stretch as far as to enclose two adjacent pegs. (Pegs that are diagnolly opposite from one another are considered adjacent, as well). In how many ways can you stretch the rubber bands over ten pairs of of pegs to enclose all twenty pegs?

7. Triangle $ABC$ is situated within an ellipse whose major and minor axes have lengths $10$ and $8$. Point $A$ is located at one focus, point $B$ is located at the endpoint of the minor axis, and point $C$ is located on the ellipse so that the other focus lies on $BC$. Determine the inradius of triangle $ABC$.

I thought I'd found a recursion for #6, but sadly, it doesn't work. I think I got #7 correct, but I'd like to see a better solution then coordinates. Thanks!

What was the recursion you thought you found for $6$? Because I think I found one that worked for all the small cases I tested, though maybe I miscounted  :maybe:

[hide="6"]
Let $A_{i}$ denote a $2 \times i$ array of pegs. For $A_{1}$ and $A_{2}$, we can write out all possible arrangements, and find that there are $1$ and $3$, respectively.

$2 \times k$ arrangements can be formed in two different ways.

The first is by adding one more row to a $2 \times (k-1)$ array. In this case, there's only one way for the two new pegs to fit in the arrangement - they must be connected to each other by a rubber band.

The second way is to take any arrangement of a $2 \times (k-2)$ array, and arrange the $2\times 2$ remaining. $A_{2}$ is equal to $3$, but we don't include the one of the three arrangements (two horizontal rubber bands) because that one is already counted in the first part.

The recursion is then \[A_{n}= A_{n-1}+2A_{n-2}\] for all $n \geq 3$.
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Fraxia has it right for #6.

For #7, it was kinda messy, but use the fact that teh sum of the distances from a point on an ellipse to the two foci = twice the major axis, and that should give you enough to work with to find the three sides, then it's simple enough to get the inradius.

Fraxia: exact same recursion: except I had $A_{2}=4$ for some reason. Maybe that's why I got it wrong. :lol:


Suppose ABC is a scalene right triangle. P is a point on the hypotenuse such that <PBA = 45*. If AP =1 and CP =2, then what is the area of ABC?

[hide]
Let c = AB, a = BC.
Since $<PBA = 45$, BP is the angle bisector of $<ABC$.
$\frac{c}{a}= \frac{1}{2}$
$a = 2c$

The area of the triangle would be,
$\frac{1}{2}ac \implies c^{2}$

$a^{2}+c^{2}= 9 \implies 5c^{2}= 9 \implies c^{2}= \frac{9}{5}$

So the area is $\frac{9}{5}$

How do you do the angle sign in latex?
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[quote="JavaMan"]
$\frac{c}{a}= \frac{1}{2}$
[/quote]

How do you know that? Is there some theorem?

an angle bisector in a triangle divides the opposite side in the ratio of the two adjacent sides.

this can be easily proven


Given a regular heptagon ABCDEFG with side 1, show that for diagonals AC and AD it is true that:

1/AC + 1/AD = 1.

I tell tou, it is. I proved it after posting it here.
Try it.
If you can´t, I will post the answer.

Aaaa sorry I've read hexagon...  :P I'll try to prove it now.
Consider heptagon AB'C'D'E'F'G symmetric to ABCDEFG through AG. C,A and D' are collinear and $ \angle GCA\equal{}\angle GD'A\equal{}\angle ACB$. Hence GCD' and BAC are similar => $ \frac {AC}{AB} \equal{} \frac {AC \plus{} AD}{AD}$. Hence $ \frac {AC \plus{} AD}{AC\cdot AD} \equal{} 1$ QED

I can't understand what you have done with AC/AB (which obviously is AC since AB = 1) to prove that
1/AC + 1/AD = 1. (AC/AB = AC, AC + [AD/AD] = AC???)

[quote="limes123"]Aaaa sorry I've read hexagon...  :P I'll try to prove it now.
Consider heptagon AB'C'D'E'F'G symmetric to ABCDEFG through AG. C,A and D' are collinear and $ \angle GCA \equal{} \angle GD'A \equal{} \angle ACB$. Hence GCD' and BAC are similar => $ \frac {AC}{AB} \equal{} \frac {AC \plus{} AD}{AD}$. Hence $ \frac {AC \plus{} AD}{AC\cdot AD} \equal{} 1$ QED[/quote]
$ \frac {AC}{AB} \equal{} \frac {AC \plus{} AD}{AD}\iff 1\equal{}\frac{1}{AB}\equal{}\frac{AC\plus{}AD}{AC\cdot AD}\equal{}\frac{1}{AC}\plus{}\frac{1}{AD}$ ;)


Hello!

Let $ ABC$ be a triangle, $ \Gamma$ is the circumcircle and $ P$, $ Q$ are two points.
$ \left\{A'\right\}\equal{}PQ\cap BC$, $ \left\{B'\right\}\equal{}PQ\cap CA$, $ \left\{C'\right\}\equal{}PQ\cap AB$,
$ \left\{A,Pa\right\}\equal{}AP\cap\Gamma$, $ \left\{B,Pb\right\}\equal{}BP\cap\Gamma$, $ \left\{C,Pc\right\}\equal{}CP\cap\Gamma$.
1.Prove that the lines $ A'Pa$, $ B'Pb$, $ C'Pc$ are concurrent (at $ Vp$) and $ Vp\in\Gamma$.
2.Similarly construct the point $ Vq$ and $ \left\{V\right\}\equal{}PVq\cap QVp$. Prove that $ V\in\Gamma$.

Hi Petry, nice problem, here is my solution

Assume $ PA\cap BC\equal{}\{A_1\}$ we have by perspective center $ A$ $ (B'C'PA')\equal{}(BCA_1A')$, assume $ P_aA'\cap (O)\equal{}A_2$ by perspective center $ P_a$ $ (BCA_1A')\equal{}(BCAA_2)\Rightarrow (B'C'PA')\equal{}(BCA_2A)\ (1)$

Similarly $ P_bB'\cap (O)\equal{}\{B_2\},P_cC'\cap (O)\equal{}\{C_2\}$ we have $ (C'A'PB')\equal{}(CAB_2B), (A'B'PC')\equal{}(ABC_2C)$

But $ (C'A'PB')\equal{}(CAB_2B)\Rightarrow (B'C'PA')\equal{}(BCB_2A)\ (2)$ From $ (1),(2)$ we have $ A_2\equiv B_2$ similar $ B_2\equiv C_2$

So they are the same point $ V_p$, now if $ Q\in$ line $ B'C'A'P$, $ QV_p\cap (O)\equal{}I$ we consider perspective center $ V_p$ we have $ (B'C'QA')\equal{}(P_bP_cIP_a)$ but by above result $ (B'C'QA')\equal{}(BCV_qA)$ so $ (P_bP_cIP_a)\equal{}(BCV_qA)$ but triangles $ P_aP_bP_c$ and $ ABC$ are perspective triangle center $ P$ this mean $ P\in IV_q$ we are done.


Hello  :)
Given 3 variables x,y,z
We want to apply rearangement to prove
$ \sum \frac {8x^2y^2}{\sqrt {(x^2 \plus{} y^2)(y^2 \plus{} z^2)}}\geq \sum \frac {8x^2y^2}{\sqrt {(x^2 \plus{} y^2)(x^2 \plus{} y^2)}}$

How we gonna prove that
$ \frac {x^2y^2}{\sqrt {x^2 \plus{} y^2}}$,$ \frac {y^2z^2}{\sqrt {y^2 \plus{} z^2}}$,$ \frac {z^2x^2}{\sqrt {z^2 \plus{} x^2}}$ and $ \frac{1}{\sqrt{y^2\plus{}z^2}}$, $ \frac{1}{\sqrt{z^2\plus{}x^2}}$, $ \frac{1}{\sqrt{x^2\plus{}y^2}}$  are similarly ordered??

Look at http://www.mathlinks.ro/viewtopic.php?t=56071
Post #2 of Peter Scholze

[quote="FOURRIER"]

How we gonna prove that
$ \frac {x^2y^2}{\sqrt {x^2 \plus{} y^2}}$,$ \frac {y^2z^2}{\sqrt {y^2 \plus{} z^2}}$,$ \frac {z^2x^2}{\sqrt {z^2 \plus{} x^2}}$ and $ \frac {1}{\sqrt {y^2 \plus{} z^2}}$, $ \frac {1}{\sqrt {z^2 \plus{} x^2}}$, $ \frac {1}{\sqrt {x^2 \plus{} y^2}}$  are similarly ordered??
[/quote]
Maybe you are mean that
$ \frac {x^2y^2}{\sqrt {x^2 \plus{} y^2}}$, $ \frac {z^2x^2}{\sqrt {z^2 \plus{} x^2}},$ $ \frac {y^2z^2}{\sqrt {y^2 \plus{} z^2}}$ and $ \frac {1}{\sqrt {y^2 \plus{} z^2}}$, $ \frac {1}{\sqrt {z^2 \plus{} x^2}}$, $ \frac {1}{\sqrt {x^2 \plus{} y^2}}$  are similarly ordered. :wink:

If so, let $ x\geq y\geq z.$ Then $ \frac {x^2y^2}{\sqrt {x^2 \plus{} y^2}}\geq\frac {z^2x^2}{\sqrt {z^2 \plus{} x^2}}\geq\frac {y^2z^2}{\sqrt {y^2 \plus{} z^2}}$ and
$ \frac {1}{\sqrt {y^2 \plus{} z^2}}\geq\frac {1}{\sqrt {z^2 \plus{} x^2}}\geq\frac {1}{\sqrt {x^2 \plus{} y^2}}.$ :wink: