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[quote="FOURRIER"]Let $ ABC$ be a triangle, and let $ a\equal{}BC$, $ b\equal{}CA$, $ c\equal{}AB$ be its sidelengths.
Assume that $ a\plus{}b\plus{}c\equal{}1$.
Prove that $ a^2\plus{}b^2\plus{}c^2\plus{}4abc<\frac12$.[/quote]

For you!
We must to prove that
$ (a \plus{} b \plus{} c)(a^{2} \plus{} b^{2} \plus{} c^{2}) \plus{} 3abc\le \frac {1}{2}(a \plus{} b \plus{} c)^{3}$
Now, applying PID, we have the result
P/s or you can directly expanding
$ (a \plus{} b \plus{} c)(a^{2} \plus{} b^{2} \plus{} c^{2}) \plus{} 3abc \minus{} \frac {1}{2}(a \plus{} b \plus{} c)^{3} \equal{} \frac {1}{2}(\sum a^{2}(a \minus{} b \minus{} c))$

[quote="FOURRIER"]U used 3abc instead of 4abc

Wat is PID please[/quote]
Sorry, what's a big mistake!
You can look in Hojoo Lee manuscript. Search in Inequalities Theorem box

Can anyone tell me wat is LHS and RHS?
and thanks

LHS denotes the left hand side and RHS denotes the right hand side.

[quote="bilot4"][quote="Little Gauss"]$RHS-LHS=\frac{1}{2}(b+c-a)(c+a-b)(a+b-c)$.[/quote]

Could you explain this? From where did you derive this equality?
If you did it from $1/2-a^{2}-b^{2}-c^{2}-4abc$ I would like please to explain it :oops:[/quote]
It came form
$1/2-a^{2}-b^{2}-c^{2}-4abc=\frac{1}{2}(a+b+c)^{3}-(a+b+c)(a^{2}+b^{2}+c^{2})-4abc$ :D

Oh nice :wink: now i think I've understood :)

we know that $ abc\equal{}4pRr$ and $ a^2\plus{}b^2\plus{}c^2\equal{}2(p^2\minus{}r^2\minus{}4Rr)$ where $ p,r,R$ denote the semiperimeter,inradius and circumradius of the triangle respectively,and also we have $ 2p\equal{}1$ now our inequality is equivalent to:

$ 2(p^2\minus{}r^2\minus{}4Rr)\plus{}16pRr<\frac{1}{2}$

$ \Rightarrow 1\minus{}4r^2\minus{}16Rr\plus{}16Rr<1$

$ \Rightarrow 0<4r^2$

which is obviously true...

[quote="BaBaK Ghalebi"]we know that $ abc \equal{} 4pRr$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2(p^2 \minus{} r^2 \minus{} 4Rr)$  [/quote]

Give a proof of this please.

[quote="BaBaK Ghalebi"]@FOURRIER:[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=151066[/url][/quote]

I didn find the Lemas I asked about  :play_ball:

the $ 14^{th}$ post of the link...
anyway here it goes:
let $ H$ be a point on $ BC$ such that $ AH\perp BC$ and let $ AH\equal{}h_a$,then denote as $ S$ the area of the triangle,then we know that:

$ S\equal{}\frac{h_a.a}{2}$

now in triangle $ \triangle AHC$ we know that

$ \sin C\equal{}\frac{h_a}{b}$

thus $ h_a\equal{}b.\sin C$

so we have:

$ S\equal{}\frac{h_a.a}{2}\equal{}\frac{ab.\sin C}{2}$

also we know that

$ \frac{a}{\sin A}\equal{}\frac{b}{\sin B}\equal{}\frac{c}{\sin C}\equal{}2R$

therefore we have:

$ \boxed{S\equal{}\frac{abc}{4R} (I)}$

also let the incircle touch $ BC,AC,AB$ at $ D,E,F$ respectively then we have:

$ S_{IBC}\plus{}S_{ICA}\plus{}S_{IAB}\equal{}S$

$ \Rightarrow \frac{r.a}{2}\plus{}\frac{r.b}{2}\plus{}\frac{r.c}{2}\equal{}S$

$ \Rightarrow r(\frac{a\plus{}b\plus{}c}{2})\equal{}S$

$ \Rightarrow \boxed{S\equal{}pr (II)}$

now from $ (I),(II)$ we get that:

$ abc\equal{}4Rrp$

----------------------------------------------

also we know that:

$ S\equal{}\sqrt{p(p\minus{}a)(p\minus{}b)(p\minus{}c)}$

also we had $ S\equal{}pr$ therefore we have:

$ pr^2\equal{}(p\minus{}a)(p\minus{}b)(p\minus{}c)$

$ \Rightarrow p^3\minus{}p^2(a\plus{}b\plus{}c)\plus{}p(ab\plus{}bc\plus{}ca)\minus{}abc\equal{}pr^2$

now we had $ abc\equal{}4Rrp$ and $ a\plus{}b\plus{}c\equal{}2p$ so we get that:

$ ab\plus{}bc\plus{}ca\equal{}p^2\plus{}r^2\plus{}4Rr$

now we have:

$ (a\plus{}b\plus{}c)^2\equal{}a^2\plus{}b^2\plus{}c^2\plus{}2(ab\plus{}bc\plus{}ca)$

$ \Rightarrow 4p^2\equal{}a^2\plus{}b^2\plus{}c^2\plus{}2(p^2\plus{}r^2\plus{}4Rr)$

$ \Rightarrow a^2\plus{}b^2\plus{}c^2\equal{}2(p^2\minus{}r^2\minus{}4Rr)$

as wanted...

$ a,b,c$ are three roots of the equation :

$ x^3 \minus{} 2px^2 \plus{} (p^2 \plus{} r^2 \plus{} 4Rr)x \minus{} 4prR \equal{} 0$

Where :

$ p$ is a half of perimeter, $ R$ is radius of circumscribed circle , $ r$ is radius of inscribed circle .

Proof :

We have $ a \equal{} 2R\sin A \equal{} 4R\sin\frac {A}{2}\cos\frac {A}{2}\;(\bigstar _1)$

and :

$ p \minus{} a \equal{} r\frac {\cos\frac {A}{2}}{\sin\frac {A}{2}}\;(\bigstar_ 2)$

From $ (\bigstar _1)$ and $ (\bigstar_ 2)$ we have :

$ \frac {ar}{4R(p \minus{} a)} \plus{} \frac {a(p \minus{} a)}{4Rr} \equal{} \sin\frac {A}{2}^2 \plus{} \cos\frac {A}{2}^2 \equal{} 1$

Then :

$ a^3 \minus{} 2pa^2 \plus{} (p^2 \plus{} r^2 \plus{} 4Rr)a \minus{} 4prR \equal{} 0$

Simirlary:

$ b^3 \minus{} 2pb^2 \plus{} (p^2 \plus{} r^2 \plus{} 4Rr)b \minus{} 4prR \equal{} 0$

and

$ c^3 \minus{} 2pc^2 \plus{} (p^2 \plus{} r^2 \plus{} 4Rr)c \minus{} 4prR \equal{} 0$

Hence we're done . Use Vieta jumping , we get :

$ ab \plus{} bc \plus{} ca \equal{} p^2 \plus{} r^2 \plus{} 4Rr,a^2 \plus{} b^2 \plus{} c^2 \equal{} 2(p^2 \minus{} r^2 \minus{} 4Rr)$

[quote="quangpbc"]$ a,b,c$ are three roots of the equation :

$ x^3 \minus{} 2px^2 \plus{} (p^2 \plus{} r^2 \plus{} 4Rr)x \minus{} 4prR \equal{} 0$

Where :

$ p$ is a half of perimeter, $ R$ is radius of circumscribed circle , $ r$ is radius of inscribed circle .

Proof :

We have $ a \equal{} 2R\sin A \equal{} 4R\sin\frac {A}{2}\cos\frac {A}{2}\;(\star 1)$

and :

$ p \minus{} a \equal{} r\frac {\cos\frac {A}{2}}{\sin\frac {A}{2}}\;(\star 2)$

From $ (\star 1)$ and $ (\star 2)$ we have :

$ \frac {ar}{4R(p \minus{} a)} \plus{} \frac {a(p \minus{} a)}{4Rr} \equal{} \sin\frac {A}{2}^2 \plus{} \cos\frac {A}{2}^2 \equal{} 1$

Then :

$ a^3 \minus{} 2pa^2 \plus{} (p^2 \plus{} r^2 \plus{} 4Rr)a \minus{} 4prR \equal{} 0$

Simirlary:

$ b^3 \minus{} 2pb^2 \plus{} (p^2 \plus{} r^2 \plus{} 4Rr)b \minus{} 4prR \equal{} 0$

and

$ c^3 \minus{} 2pc^2 \plus{} (p^2 \plus{} r^2 \plus{} 4Rr)c \minus{} 4prR \equal{} 0$

Hence we're done . Use Vieta jumping , we get :

$ ab \plus{} bc \plus{} ca \equal{} p^2 \plus{} r^2 \plus{} 4Rr,a^2 \plus{} b^2 \plus{} c^2 \equal{} 2(p^2 \minus{} r^2 \minus{} 4Rr)$[/quote]
thats the same argument which was used in the link I gave...
([url]http://www.mathlinks.ro/Forum/viewtopic.php?t=151066[/url])
from post $ 14$ till the end

[quote="BaBaK Ghalebi"]the $ 14^{th}$ post of the link...
anyway here it goes:
let $ H$ be a point on $ BC$ such that $ AH\perp BC$ and let $ AH \equal{} h_a$,then denote as $ S$ the area of the triangle,then we know that:

$ S \equal{} \frac {h_a.a}{2}$

now in triangle $ \triangle AHC$ we know that

$ \sin C \equal{} \frac {h_a}{b}$..............
[/quote]

Now, I see it :w00t: ,
The use of geometry trivializes some inequalities

Thank you veeeery much for helping me. :bye:


Prove or disprove:there exist infinitely many $ n$ such that $ s(x) \plus{} x \equal{} 10^{n}$ has no solution over natural number where $ s(x)$ denotes the sum of digits of $ x$

I think;$ x\equal{}10^{n}\minus{}9n\plus{}4$  satisfies this condition

$ n \equal{} 7$: $ 10^7 \minus{} 9 \cdot 7 \plus{} 4 \equal{}  9999941$; the sum of the digits of 9999941 is 50. But 50 + 9999941 is not a power of 10.

Indeed we can prove the much stronger statement:for every $ b\in N \plus{}$ there are infinitely many $ n$ such that for every $ x < 10^n,x \plus{} s(x)\neq 10^n \plus{} b$

See Lemma 2 of [url=http://www.mathlinks.ro/viewtopic.php?p=1743162#1743162]16.23[/url]

By the way,I created this problem when dealing with 16.23 :D


Determine the number of $ n$-term sequences of positive integers in which  for each
$ i\equal{}1,2,...,n$,the number of term greater than $ i$ is at most $ n\minus{}i$

gwen01, I'm not sure what you're doing, exactly, but check the case $ n \equal{} 2$ ;)

The objects described are known as "parking functions."  It happens that they do enumerate very nicely: [url=http://www.research.att.com/~njas/sequences/A000272]A000272[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url] is the relevant sequence (though it has a weird extra leading 1 for some reason).


If $ a,b,c>0$, $ a\plus{}b<2$, $ b\plus{}c<2$ and $ c\plus{}a<2$, then

$ \prod(a\sqrt{\frac{b}{a^2\plus{}1}}\plus{}b\sqrt{\frac{a}{b^2\plus{}1}})\leq2\sqrt{2}$

The inequality is equivalent to
\[ \prod \left(\frac {a\sqrt {\frac {b}{a^{2} \plus{} 1}} \plus{} b\sqrt {\frac {a}{b^{2} \plus{} 1}}}{a \plus{} b}\right)^2\leqslant\frac {8}{\prod(a \plus{} b)^2}\mbox{.}\]
Using weighted power means and the obvious $ a^2 \plus{} 1\geqslant 2a$, the remaining inequality is
\[ \prod(a \plus{} b)\leqslant 8\mbox{,}\]
which is obvious due to the conditions.  :)


trigonometry
geometry
geometric transformation
rotation
trig identities
Law of Cosines
congruent triangles

$1.$ Take any triangle $ABC$.
$2.$ On each side of this triangle, construct an equilateral triangle such that this triangle does not overlap the triangle $ABC$.
$3.$ Let the three equilateral traingles be $RAB$, $QAC$ and $PBC$.

Prove that line segments $AP$, $BQ$ and $CR$ are equal in length

[hide="hint for another solution"]the complex plane+rotation[/hide]

[hide="hint for another solution"]find a pair of congruent triangles with 2 of those segments[/hide]

Extension: Prove that the segments are concurrent.


Let a,b,c$ \epsilon$R Prove that
$ a(a\plus{}b)(a^2\plus{}b^2) \geq0$

Where's $ c$?
And try $ a\equal{}\minus{}1,b\equal{}2$.

I guess it is $ \sum_{sym}a(a \plus{} b)(a^2 \plus{} b^2) \geq0$.

$ \sum_{cyc}a(a\plus{}b)(a^2\plus{}b^2)\geq0\Leftrightarrow\sum_{cyc}(a\plus{}b)^2(a^2\plus{}b^2)\geq0.$
Well come back, Myth! :lol:

The Mith's inequality is equivalent to the following:
$ a^4 \plus{} b^4 \plus{} c^4 \plus{} a^3b \plus{} a^3c \plus{} b^3c \plus{} b^3a \plus{} c^3a \plus{} c^3b \plus{} a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2\geq0.$
The following old inequality is true too:
\[ a^4 \plus{} b^4 \plus{} c^4 \plus{} a^3b \plus{} b^3c \plus{} c^3a \plus{} abc(a \plus{} b \plus{} c)\geq a^3c \plus{} b^3a \plus{} c^3b
\]

Yes!$ \sum a(a\plus{}b)(a^2\plus{}b^2)\geq0$

  and Prove that $ \sum a(a\plus{}b)^3 \geq 0$

[quote="tran300491"]Prove
$ a^4 \plus{} b^4 \plus{} c^4 \plus{} a^3b \plus{} b^3c \plus{} c^3a\geq0$[/quote]

$ 2(a^4 \plus{} b^4 \plus{} c^4 \plus{} a^3b \plus{} b^3c \plus{} c^3a)\geq0$

$ \equal{} \sum_{cyc} a^2(a\plus{}b)^2 \plus{} \frac 12 \sum_{cyc} (a^2\minus{}b^2)^2 \geq 0$


Eh, never played in a Mafia game, but I've been watching some

and they look cool. I want to play in one, but I dunno where or

when they appear. If anyone is making a Mafia Game can you send

me a PM?

Thanks

Check the top stickied thread of this forum.


Consider the following inequalities with  are the side lengths of a triangle.

        $ 3\left( \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \right) \ge 2\left( \frac {b}{a} \plus{} \frac {c}{b} \plus{} \frac {a}{c}\right) \plus{} 3$

and
     $ 2\left( \frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a} \right) \ge \frac{b^2}{a}\plus{}\frac{c^2}{b}\plus{}\frac{a^2}{c} \plus{}a\plus{}b\plus{}c$

see also here  :wink: (use SS)
http://www.mathlinks.ro/Forum/viewtopic.php?t=216277


I don't know if this goes in the pre-olympiad thread or not, but anyway...

Let's say you have the following recursive equation

\[ f(x)\equal{}2f(x\minus{}1)\plus{}f(x\minus{}2)\]

Now, how do I solve for $ f(x)$ without requiring another $ f(something else)$ in my equation.

Any help will be appreciated.

It's convenient to use sequence notation $ f_n$; then we are looking at the sequence $ f_n \equal{} 2f_{n\minus{}1} \plus{} f_{n\minus{}2}$.

This is called a [i]second-order homogeneous linear recurrence[/i], and its solution is well-known.  See the second part of my post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=180918501&t=164449]here[/url] for a demonstration.


geometry AMC AIME trigonometry USA(J)MO USAMO induction

In general, are the more advanced geometric principles such as inversion and projection used a lot on national olympiads?

what do you  mean by inversion and projection?

I think in general, olympiad geometry is just like AIME geometry, except you have to think more. You need to learn more tricks (extending medians, rotating triangles, algebraic manipulation with trig equations, analytic geometry, yadda yadda yadda) and how to apply them.

I mean, surely, u can use these principles, like barycentrics or something, but they're not necessary, or even preferred i think.

I mean inversive and projective geometry. I know they're used sometimes in the IMO/Putnam, but how about in national olympiads?

[color=#00FF00]There has never been a USAMO problem, or even an IMO problem I would bet, that required inversion.  It can make some of them easier, but they can all be done other ways.[/color]

Weird mathematical principles are never required

[color=#0FF000]Okay.  This is not going to be very helpful.  Inversion is a transformation in a euclidean plane plus a "point at infinity," which every line is considered to pass through in both directions at the limit.  You start with a circle.  Every point is transformed according to the following rule:

1) the point at infinity ends up in the center of the circle
2) the center of the circle ends up in the point at infinity
3) if the center of the circle is O, a point P is transformed to the point P' with properties OPP' or OP'P and OP * OP' = r^2, where r is the radius of the circle.

I recommend you go look at mathworld.wolfram.com for more information.  I'll just give the two most important properties of inversion here:  every line and every circle is transformed into another line or circle.  That is, every line becomes either a line or a circle, and every circle becomes either a line or a circle.  Also, angles are preserved in value and sense.  So, if two lines/circles met at a given angle beforehand, they will meet at the same angle after being transformed.
To compute angles of circles, use their tangent lines at the points of intersection.[/color]

how the **** would htis be useful on contest?

The point of inversion is that you're transforming circles into lines. Things with lines tend to be much easier to prove than things with lots of weird circles.

Mmm...I can't even differentiate between "inversion" and "induction" which I don't really have any clue on.

"I am such a douche bag" :)

I don't like inversion. I never manage to apply the right inversion ! :cry:

hehe
i don't even know how to use inversion simply

anyway

induction is nice
definitely
VERY useful

It takes a lot of practice or amazing intuition to be able to use inversion to solve olympiad problems.

I dont know what some of this posts are talking about, but if you read through one of the AMC's National Olympiad Books, there are definitely more advanced geometry probs than just more advanced aime probs. I personally saw several projection problems. I dont know about inversion, but im sure it would be useful on a few. Also, vector geometry, as well as directed line segments/angles were there (and in IMO problems ive seen). I even saw a problem with the radical axis theorem in the solution These are not "weird" geometry principles, just more modern techniques.

I believe using inversion also transform the vertices of cyclic quadrilaterals to collinear points. This lets you use theorems for cyclic quadrilaterals on a collinearity problem, or vice versa. I think so at least...

Yes, that is indeed the case. One proof of Ptolemy's Theorem involves inversion about a big circle centered at one of the vertices and showing that if X, Y, and Z are points on a line (in that order), then XY+YZ=XZ. It's a testimony to the power of inversion if you can turn a fairly deep and completely nonintuitive theorem like Ptolemy into something so completely obvious.

My personal opinion is that inversion is COOL! I simply love it, and I'm not even that good at applying it! :D Btw, it works miraculously on many prbs that would be really hard without it. If you combine it with proj geom it's even better.

Unfortunately I'm in the 12'th grade now, which means that I'll never have anything to do with geometry in olympiad problems ever again. I didn't know inversion in the 9'th grade, so I never got to apply it for a contest prb, but I solved lots of problems using nversion in my spare time. I only solve geom prbs for fun, but they tend to be a big waste of time since I have to learn integrals, group theory, rings and stuff like that (I mean for olympiads).

You can find cool geom prbs solved by inversion on mathlinks. Not too many, though..


A point $ P$ is taken in the interior of a right triangle $ ABC$ with angle $ C \equal{} 90$ such hat $ AP \equal{} 4$, $ BP \equal{} 2$, and $ CP \equal{} 1$.
 Point $ Q$ symmetric to $ P$ with respect to $ AC$ lies on the circumcircle of triangle $ ABC$.
 Find the angles of triangle $ ABC$.


is there a stategy for winning at tic-tac-toe. my friend showed me one but it only works if you go first and the other person goes in the center.  :?

There is no strategy that will always win tic-tac-toe. The most common result is a draw.

1 2 3
4 5 6
7 8 9

One ideal strategy (for both players) is 1,5,9,{2,4,8,6}.  Any deviation from this by the second player results in a win for the first player.

My friend and I have once played so often we knew certain moves to make that would always result in a win for yourself. The key is to force moves. They usually require the other person to go in the squares on a side but not on a corner, usually experimentally.

You can always force atleast a draw regardless of which player you are in a 3*3 game.

However in a 3*3*3 game there exists a strategy enabling the first player to always win.

tic tac toe is stupid, if your playing anyone with half a set of brains, then you know that there is 0% you'll win, 100% draw....


:!:

The letter finally came in the mail today: "CONGRATULATIONS". I may have been one of the first to receive it since I live in Maryland.

I got at most A1, A2, A3, B1, and half of B2, so many of you Juniors have a very good chance of making the team even if you don't think you got all of the questions.

Has anyone else been notified yet?

Congrats!

But I don't think I'll make it.

EDIT: do you know if the cutoff is different for different grades?

No idea, the letter didn't really say anything about the selection process.

I made it as well. I got my letter yesterday via FedEx Express..

There is definitely hope. I got all of A1 and A2, a small glitch on A3 and totally bombed A4 because I forgot to factor in the speed of sound into the time (thus making the problem kinda trivial). Then I got some of B1 and almost all of B2, save half a letter.

(I'm a junior)

I think I got everything except the tape recorder one (which was because I forgot to factor in speed of sound...)

But I made it.  Hm, they implicitly say in the letter that you can't go to MOP if you accept their invitation, anyone know anything about this?

From what I have heard, you should contact them immediately to see if they can work out a way for you to go with both.

I have not yet, and am not expecting to make it to camp, but I wish they would post a full list...

You know what, I think you should go to MOP instead of the physics camp.  That might allow some other people who wouldn't otherwise have made the physics camp (like me) to get in. :P

[quote="massacresph"]okay, i shall spend my virgin post commenting about this situation.

it is possible to attend both mop and physics training camp, aapt will not kill you for that. however, near the end of the camp, they will ask you to fill out a questionnaire about whether you will be doing mop, have other commitments, etc. etc. if you say, yes i will be going to mop, then they will consider that and you will not be a candidate for the traveling team. if you don't tell them that you will have other commitments and they pick you, they'll be pretty mad. ask people what happened in 06[/quote]

Oh wow, this makes me so relieved.  Thanks so much!

Yeah, I know about what happened in 2006.  I actually spoke to the person in question (whose name I shall not mention here, though it might be well known) yesterday because I wanted advice on what I should do. (I know him from MOP)

i was a member of the team in 2006 (and a member of the traveling team in 2007), so i can confirm what has been said.  i will also remark that these comments hold for things like RSI, too.

i urge you, however, to consider your situation carefully.  i know you were in black mop last year, so you're a front-runner for the imo team this year.  if you don't really give a crap about physics and you have absolutely no intention of going to the ipho, it is kind of disrespectful to others who may value the opportunity more, and to the coaches, who ultimately need to find people to represent the united states.  having asked around/talked to people (the math and physics communities are pretty tight), i fear that this may be a bigger issue than normal this year.  on one hand, it's great that the physics olympiad is growing and getting more attention from people of all disciplines.  but on the other hand, the physics olympiad is an olympiad, too!  not nearly as glamorous or well-publicized (or even, albeit arguably, as difficult) as the imo, but think of how those of us who genuinely prefer physics to mathematics feel!

[quote="iostream.h"]
i urge you, however, to consider your situation carefully.  i know you were in black mop last year, so you're a front-runner for the imo team this year.  if you don't really give a crap about physics and you have absolutely no intention of going to the ipho, it is kind of disrespectful to others who may value the opportunity more, and to the coaches, who ultimately need to find people to represent the united states.  having asked around/talked to people (the math and physics communities are pretty tight), i fear that this may be a bigger issue than normal this year.  on one hand, it's great that the physics olympiad is growing and getting more attention from people of all disciplines.  but on the other hand, the physics olympiad is an olympiad, too!  not nearly as glamorous or well-publicized (or even, albeit arguably, as difficult) as the imo, but think of how those of us who genuinely prefer physics to mathematics feel![/quote]

This is exactly the moral dilemma I face...if I don't really intend to go to IPhO at all, what business do I have going to a training camp?  It made me wonder why they'd offer to let someone do that.  It's not like I don't care about physics: on the contrary, I love it.  But math is just so much more important to me (the olympiad so much more because I've been a part of that process for a while)...

They eventually do post one, but I think its safe to say if someone (like me  :(  ) hasn't gotten the letter yet, they are not on the team.

On a side note, does anyone know if alternates are notified now or only after spots open up?

I'm not sure, but I would guess whenever the spots open up.  In any case, it should be soon, since they'll want to finalize the team as quickly as possible.

Also, they changed the site.  It has "team" and "traveling team" listed, but none of the links work.  Perhaps they will post the final list soon?

Edit: When I say none of the links work, I mean that where you'd expect links, there are none.

If alternates want to know, I'll say that I now know one spot is open.

Do we know who exactly will be listed on the site?  Just the people who go to the camp?

yeah, in the end, they'll just have all the people who actually attended the camp.

when is the reply deadline for those who have been selected?

I remember reading somewhere that transcripts and recommendation letters are also taken into account for selection of the team.  Is this not the case?  (or is this how alternates could be selected?)

No, i mean by what date do people selected have to let AAPT know if they are going?

confirmation/faxing over eligibility form was due this Tuesday.. April 15.

[quote="mathclass"]That was the case last year. This year, they haven't asked for any extra recs or transcripts.[/quote]

Haha, they may have decided that these things were superfluous, as everyone who qualifies for the team probably has a pretty good academic record already.

pfft not likely

the overall academic record is irrelevant to the physics team though, and this would seem to be a better reason for dropping it

also yo what happened in 2006 was [i]totally retarded[/i]

there's no official link to it, but via substituting an 8 for a 7, you get this list

[url]http://aapt.org/olympiad2008/team.cfm[/url]

gj to all

[quote="genericme"]Hey, Alex Zhai made the team.  That confirms my awesome powers of prediction.

And it looks like K81o7 decided not to go after all.[/quote]

Yeah, it conflicted with school stuff, and the IPhO conflicts with math stuff, so I figured somebody else had better go.

I imagine Zhai must be getting tired of MOP at this point, and wants to go to a different camp as well? :)  Just kidding.


Find all numbers $ k$ with distinct digits $ a,b,c,d$ such that $ a>b>c>d$ and $ abcd\minus{}dcba \equal{} k$ (positional, base 10)

This number is called base 10 Kaprekar Constant.

Denote $ \overline{abcd} = 10^3a + 10^2b + 10c + d$. Then
\begin{align*} & \overline{abcd} - \overline{dcba} \\
= & 10^3(a - d) + 10^2(b - c - 1) + 10(9 - b + c) + (10 - a + d) \\
= & \overline{(a - d)(b - c - 1)(9 - b + c)(10 - a + d)} \\
= & \overline{a'b'c'd'}\end{align*}
Therefore $ a,b,c,d$ must equal $ a - d, b - c - 1, 9 - b + c, 10 - a + d$ in some order. But note that $ a > a - d, a > b > b - c - 1$. Also, since $ a > b > c > d$ implies $ b - c < a - d - 1$, we get $ 9 - (b - c) > 10 - (a - d)$. Thus we get the following possibilities
\begin{align*} a' & = a - d = b,c \\
b' & = b - c - 1 = c,d \\
c' & = 9 - (b - c) = a,b \\
d' & = 10 - (a - d) = b,c\end{align*}
From the second and the third equality we see that $ b - c - 1 = d$ and $ 9 - (b - c) = a$ are the only possibilities. Therefore $ (a',b',c',d') = (b,d,a,c)$ or $ (c,d,a,b)$. Hence
\[ a + d = b' + c' = 8 \\
b + c = a' + d' = 10
\]
From here we see that the only possibility is $ (a,b,c,d) = (7,6,4,1)$ and thus $ k = 7641 - 1467 = \boxed{6174}$.

You can try the same method for $ 3$ digit nos. as well. You will obtain the no. $ 495$ as constant.
Moreover, you will only find unique constants for three digit and four digit nos. For other digits, you will find a kaprekar series or more than one constants.

[hide="Links for Kaprekar constant"]
http://en.wikipedia.org/wiki/Kaprekar_constant
http://en.wikipedia.org/wiki/495_(number)
http://plus.maths.org/issue38/features/nishiyama/index.html
http://mathpoint.blogspot.com/2006/12/mysterious-6174-revisited.html
http://kaprekar.sourceforge.net/[/hide]
 :)

Interesting game:

Take a 4-digit number($ \overline {abcd}$), then write 2 numbers which have digits $ a,b,c$ and $ d$ each one time, but one number has the digits ascending and the other has them descending. Subtract them and you get a new number. Do the same and continue until you get $ 6174$(can somebody prove that Kaprekar constant will always appear in this game?).

Yes....it can be proved that Kaprekar's constant will always appear in the game.
I have proved it for $ 3$-digit number, but I don't know how to prove it for the four digit.

Consider any three-digit no., say $ xyz$, which is actually $ 100x\plus{}10y\plus{}z$.
Let $ x>y>z$. So, after substraction we have \[ (100x\plus{}10y\plus{}z)\minus{}(100z\plus{}10y\plus{}x)\equal{}99(x\minus{}z)\]
In other words, we will always obtain a multiple of $ 99$.
So, let us call it $ 99x$, which can be written otherwise as $ 100(x\minus{}1)\plus{}90\plus{}(10\minus{}x)$.
Again, if we subtract according to our game, we are bound to get back another multiple of $ 99$.
So, when will the game end?? It will end iff the subtracted result is a permutation of the previous multiple of $ 99$.
[hide="Example"]Consider any multiple of $ 99$, say $ 198$. See it has only one possible permutation, that is a multiple of $ 99$, viz. $ 891$.
Thus, for any no. $ 99x\equal{}100(x\minus{}1)\plus{}90\plus{}(10\minus{}x)$, the permutation is just $ 99(11\minus{}x)$.[/hide]
Consider $ x>6$, i.e. $ (x\minus{}1)>(10\minus{}x)$ and using our condition you can find an equation $ 99(x\minus{}1)\equal{}99x$ which has clearly no solution.
So, the case left to check is $ x<6$. In this case, the equation obtained is $ 99(10\minus{}x)\equal{}99x$, which has one and only one solution $ x\equal{}5$.
So, our game will end next to the step where we obtain $ 99\times 5\equal{}594$.
Test: \[ 954\minus{}459\equal{}495\]

It will be more than easy to prove that $ 594$ is bound to apppear.
 :)


trigonometry
geometry
circumcircle
ratio
geometric transformation
homothety
inequalities

Two circles $\omega_{1}$  and   $\omega_{2}$   intersect at two points $A$ and $B$ . Construct a line $EF$ through the point $A$ such that $E$ lies on on $\omega_{1}$ and $F$ lies on $\omega_{2}$, and such that the value $EA\cdot AF$ is maximal.

$EA\cdot FE$ is power of the point $E$ for circle $\omega_2$ and that is equal
$EO_2^2-R_2^2$ where $O_2$ is center of $\omega_2$ and $R_2$ is its radius. How $R_2$
is constant, $EA\cdot FE$ is maximal when $EO_2$ is maximal. So we choose point $E$
such that it is crossing point of line $O_1O_2$ and circle $\omega_1$ where $O_1$ is
its center. So only we have to prove that $E$ is really that point, that for every
point $X$ on $\omega_1$,it stands $O_2E>O_2X$. You just construct circle with center
$O_2$ and radius $O_2E$, that circle tangent $\omega_1$ and there you go.

Am I right???

P.S. Nice problem :D

I think amir ment :$EA*FA$ be max?

But if i am wrong ,then your solution is right M@re

Oh sorry m@re I have edited it. ur solution is right.
But this problem is nicer.

Well i relly could forsee that, because that problem was too easy, i'll try this one later

 :D  :D

This last one is even easier, actually :). Just notice that all the triangles $BEF$ are similar to each other, so $EF$ is proportional to $BE$, which reaches its maximum when $E$ is the antipode of $B$ in $\omega_1$.

Hmm.grobber :
We want $(EA)*(FA)$ be maximal not $EA+FA=EF$. :?:

:D Oops! :) I don't know what I was thinking, sorry.

:cool: Hmm.By the remark of grobber ,I find something

Let $O_1,R_1$ and $O_2,R_2$ be the centers and raduises of $w_1,w_2$.
Let $EBA=\alpha_1,FBA=\alpha_2$
In the triangle$ABE$:
$EA=Sin(\alpha_1)2(R_1)$
In the triangle$ABF$:
$FA=Sin(\alpha_2)2(R_2)$
so
$EF.FA=4(R_1R_2)Sin(\alpha_1)Sin(\alpha_2)$
so we shall found the condition which
$Sin(\alpha_1)Sin(\alpha_2)$ is maximul

As grobber said
$\alpha_1+\alpha_2=$Constant $\angle O_1BO_2$.

So this is the problem now:
Let $k$ be a constant.
$\alpha_1+\alpha_2=k < 180^{\circ}$

When $Sin(\alpha_1)Sin(\alpha_2)=$Maximum?

[color=red]IF this is true,
Lemma[/color]

$Sin(\alpha_1)Sin(\alpha_2)\le Sin(\frac{\alpha_1+\alpha_2=k}{2})^2$.

then Maximum ,occurs when $\alpha_1=\alpha_2$,so we will have our line like this:

Let the $\mathcal F  $ be a point such that $\mathcal FBA=\frac{k}{2}$  then
$\mathcal FA$ is our line.

Lomos, sorry but I don't think that we should leave such nice problems to a brut force of trigonometry  :D  :D... that, and, i don't know trigonometry yet, so if someone post the solution it better be without trigonometry or with very little of it  :D  :D  :D

The following construction works for finding the line $EF$ for which $AE\cdot AF$ is maximal, and $E,F$ are on opposite sides of $A$. For the problem to be complete, we have to investigate the case when they are on the same side of $A$, but I won't do that :).

Let $C\in(BA$ be the second intersection of the circumcircle $(EBF)$ with $AB$. Since $AE\cdot AF=AB\cdot AC$, the aim is to maximize $AC$. Notice that $\angle ACE=\angle AFB,\angle ACF=\angle AEB$ are constant.

When $E,F$  move on $\omega_1,\omega_2$ s.t. they are on opposite sides of $A$ and $A,E,F$ are collinear, $k=\frac{BE}{BF}$ stays constant, and tis the ratio of the radii of $\omega_1,\omega_2$. Perform a homothety centered at $A$ wih ratio $-\frac 1k$. This will take $\omega_1$ into a circle tangent to it in $A$, of radius equal to that of $\omega_2$, which cuts $\omega_2$ in a point different from $A$, which we will call $F$ :). Let $E$ be the second intersection of $FA$ with $\omega_1$.

The point of that was to construct $E,F$ s.t. $BA$ is actually the angle bisector of $\angle EBF$. We will then find ($C$ is as above) $\angle CEA=\angle CBE=\angle CBF=\angle CFE$, which means that $CE,CF$ are, in fact, tangent to $\omega_1,\omega_2$ in $E,F$ respectively. This clearly maximizes $CA$ (remember that $\angle ACE,\angle ACF$ were constant); just draw a picture and you'll see what I mean :).

Nice solution ,grobber.
So by this solution ,I can admit that Post 10 is true ,despite the fact that i can not prove that lemma. :D

EDIT:ha-ha I proved the lemma.

Let S be the homogenety center of two circles. AS will intersects
$ \omega_1$ and $ \omega_2$ in E,F such that AE.AF be the max(?).(dont worry i will give you my solution later) ;)

Yes ,really nice amir here is my solution:

We construct circle $C$ such that $C$ be tangent at 2 circles at $A,B$ and $A,B,C$ be collinear. :lol:

It's clear that every line we construct from point C $EA.FA$ is smaller than $AC.AB$.
 :lol:  :D
to construct circle $C$ draw a line from homothecy center of $\omega_1$ and $\omega_2$ that pass through point $C$.

to :lomos_lupin here is rest of your solution if I didn't took any mistake:

$\sin\alpha_1.\sin\alpha_2\leq \frac {(\sin\alpha_1+\sin\alpha_2)^2}{4}=\frac {(2\sin(\frac {\alpha_1+\alpha_2}{2}).\cos(\frac{\alpha_1-\alpha_2}{2}))^2}{4}=\frac{(2\sin(\frac {k}{2}).\cos(\frac{\alpha_1-\alpha_2}{2}))^2}{4}$
so Inequality holds when $\alpha_1=\alpha_2$  $(\star)$ and $\cos(\frac{\alpha_1-\alpha_2}{2})=1$.

from $(\star)$ we can get that  $EF$ pass through homothecy center.
I think we are done  :)

Thanks Mathx for taking my solution to the shore.
I by myself managed to complete the solution.
The lemma i gaved is true,cause of the Karamata inequality.
Here is The link(post26)[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=305220#p305220[/url]

hey mathx ,Perfect solution.
I have just understand  your way of thinking.
You make that circle to have that AC*CB as the power of C WRT the big circle.
Nice job with the Power of the point.
Do you have any similar problem.


complex numbers
inequalities proposed
inequalities

I think that it is quie obvious that mod (sum z_i*z_i+1*z_i+2)=mod(sum z_i*z_i+1), since 1/z=conj(z) for all |z|=1. So we must find the maximum value of mod(sum z_i*z_i+1), which is trivial. I don't know exactly if I hadn't made any mistake.


The side BC of acute-angled triangleABC is diameter of a circle k.From point A 2 tangehts of this circle are drawn.Prove that the intersection points of these tangents with k and the orthocenter of triangle ABC are collinear

Classical indeed, particularly on this forum :) . See http://www.mathlinks.ro/Forum/viewtopic.php?t=5850 and http://www.mathlinks.ro/Forum/viewtopic.php?t=19707 .

  darij


First, note that the function is continuous; all maxima and minima occur when $F^{\prime}(x)=0$.

$0=F^{\prime}(x)=(sin(x)+2005)^{\prime}(cos(x)+2006)+(sin(x)+2005)(cos(x)+2006)^{\prime}=cos(x)^2+2006cos(x)-sin(x)^2-2005sin(x)$

Are you sure that $F(x)$ is not $(sin(x)+2005)(cos(x)+2005)$?

When I use my computer to find the maximum of the equation you gave, I get a maximum of about $4.025\times10^7$ occuring at $x\approx0.78565$. (Sort of close to $\frac{\pi}{4}$?)

However, if you use $F(x)=(sin(x)+2005)(cos(x)+2005)$, getting an exact answer is much easier.
In that case, $F^{\prime}=0$ when $x=\left(\frac14+k\right)\pi$ for $k=...-2,-1,0,1,2...$
$F(x)$ is periodic with period $2\pi$.
If $k$ is even, then
$F\left(\left(\frac14+k\right)\pi\right)=F(\pi/4)=(sin(\pi/4)+2005)(cos(\pi/4)+2005)=(\frac{1}{\sqrt{2}}+2005)^2=\frac{8040051}{2}+2005\sqrt2$.
If $k$ is odd, then
$F\left(\left(\frac14+k\right)\pi\right)=F(5\pi/4)=(sin(5\pi/4)+2005)(cos(5\pi/4)+2005)=(\frac{1}{\sqrt{2}}+2005)(-\frac{1}{\sqrt{2}+2005}=\frac{8040049}{2}$

So, these are the maximum and minimum of $F(x)$, respectivally.


What is the sum of all the [b]digits[/b] in the following sequence:
1, 2, 3, 4, 5, 6...97, 98, 99, 100?

I keep getting 496, but my teacher insists that my answer is wonrg. Can anyone help me?

I think your teacher is right.  I would add separately the units digits, the tens digits, and the hundreds digits.  What did you get for each of those three sums?

for the units it's 45*10
for the tens it's 45*1
and for the hundreds it's 1

add those up and you get 496...where's the flaw in my logic???

what do you mean???

there's 10, 20, 30, 40, 50, 60, 70, 80, and 90

I apologize fo rmy stupidity, but I just don't get it...

Aren't you forgetting the tens digits in the other numbers, such as 11, 21, 31, etc.?

oops...that was a pretty stupid move of me...

so it's 10*45 for the tens???

Yes.  There is a symmetry between the ones digits and tens digits, which is why their sums are equal.

1+2+3+...+9=45

For 10-19, 20-29, and so on until 90-99, the sum of the digits will be for the respective sequences, 10+45, 20+45, ... and up to 90+45.  Summing these all up, from 10-99, the sum of all the digits will be 45*19 = 855.  Including 100, the sum of the digits of numbers from 10-100 will be 856.  Adding that with 45 from above, the sum of the digits in the numbers from 1-100 is 901.


I am just curious, are any of stretching lately like Yoga, Tai-chi, Pilates etc?

Stretching is good for you (if you are careful and do it right). I don't stretch regularly but I am going to start doing it. I have some stretches that I learned in physical therapy after I injured my back and some from when I took a kung fu class. At Princeton they offer several yoga classes, but I'm not sure I'll take one next semester. I might do the kung fu class again.

Yeah, I stretch pretty often. I guess it comes with doing kung fu, but stretching has pretty cool benefits (read: mansplits).

yeah, a post game stretch does wonders to your recovery.

Yea. I stretch at the end of every run. It helps me avoid injury. I probably could do the splits during marathon season...

By the way, mansplit proofs were spawned from the same cesspit as Adam Hesterberg's factorization. (Red 1)

yeah, I also stretch at the end of every basketball game, whether it is a varsity game or a pick up game, else as soon as I get home I'll be to tired to do anything. On the other hand, a good 5-10 static stretch gives me the energy to work after the game(s).


Let $ ABC$ be an equilateral triangle. Let $ \Omega$ be its incircle (circle inscribed in the triangle) and let $ \omega$ be a circle tangent externally to $ \Omega$ as well as to sides $ AB$ and $ AC$. Determine the ratio of the radius of $ \Omega$ to the radius of $ \omega$.

[hide]Let the side length of $ ABC$ be $ m$. Then, using $ 30\minus{}60\minus{}90$ triangles, we find that the inradius is $ \frac{m\sqrt{3}}{6}$. Also, the altitude of $ ABC$ is $ \frac{m\sqrt{3}}{2}$. Therefore, the ratio of the radii is :
\[ \frac{\frac{m\sqrt{3}}{6}}{\frac{m\sqrt{3}}{18}}\equal{}\boxed{3}\]Is this right?[/hide]

[hide]

We can find the diameter (and radius) of the incircle using the formula $ A\equal{}rs$.

Subtracting the diameter from the altitude of $ ABC$ we obtain the altitude of the triangle that circumscribes $ \omega$. We can calculate the length of the sides of this triangle by knowing it's altitude and use the formula $ A\equal{}rs$ once again to calculate the radius of $ \omega$.

The ratio of the two radii is $ 3: 1$

[/hide]


I would probably say that AoPs should not have stocks as it would make things overly complicated. Please do not mock this post saying that I'm a n00b (i know i am) and I don't know 2 cents about the stock market.

Haha interesting thread there Poincare :P

Personally, I think of AoPS of more of a private enterprise, and would not attract many investors (although it is backed by government agencies).

I wouldnt like the thought of Aops becoming an enterprise... then there will be so many complications ad we may have to pay for our accounts  :P  anyway I think that aops needs more funding but stocks is taking it too far.

yeah, what if AoPs sells more than 50 percent of the company and one guy has all of it, and if that guy is just a rich business man that doesn't give a flying whoop about mathematical problem solving, he'll be CEO and art of problem solving won't be related to math any more!!! NOOO!

First: takeovers are not engineered with the idea of expanding a field. They are mostly done to diversify-unless someone else in the mathematics business was trying to eliminate competition. But I can't think of many who have the money to do so.
Second: AoPS is already thriving. There is no reason to change the way the company already is operating.
Third: Businessmen recognize the importance of education. The new owner would probably be interested more in looking for talent.
So, if someone wanted to take over AoPS, we would probably see little effect. The site and forum would remain the same, the books would continue to sell, the classes would still continue, etc., etc.
Besides, there are methods that a company can employ to stop takeovers. Richard Ruszcyk could, and would, use them if he was worried.

I personally don't care whether AoPS sells stocks, and I think it doesn't matter:

It probably wouldn't hurt much. AoPS is already a rather large enterprise, and as evidenced by the fact that new members join every day (correct me if I'm wrong), I think the company has good chances of going up.
However, stocks [i]would[/i] complicate things. It would probably be costly to offer stocks for trade. Besides, the owners probably already have enough money.
If Richard Ruszcyk decides to do so, probably waiting until mid-2009 would be best, to catch the bottom of the market.
Disclamer: I'm a noob too  :lol: so I can't say I'm saying anything remotely true (though I hope some might be).


my rating is 1402 and i usually go to tournament once or twice a month.(except during mathcounts season)

there is already a topic about this...please double check that a similar topic isn't already created

mines only 1100 i never get out much

[quote="now a ranger"]well... I plan to go to individual and team tournaments next year for HS credit. Mine's 1300 ish. Haven't played for an year. Didn't have a team to go with this year, so better luck next year.[/quote]

you need HS credit in chess????


Hi, I got stuck on this problem:
Prove that, for $ a, b, c\in\mathbb{R}^\plus{}$, e\we have the following inequality:
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\ge \frac{c\plus{}a}{c\plus{}b}\plus{}\frac{a\plus{}b}{a\plus{}c}\plus{}\frac{b\plus{}c}{b\plus{}a}$ :blush:

Can anyone help me? :?:
[hide="P/s"] I got this from the book "Old & New Inequalities" by Andresscu.[/hide]

$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\minus{}3\ge \frac{c\plus{}a}{c\plus{}b}\plus{}\frac{a\plus{}b}{a\plus{}c}\plus{}\frac{b\plus{}c}{b\plus{}a}\minus{}3$
set $ c\equal{}min(a,b,c);$
Use that $ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\minus{}3\equal{}\frac{(a\minus{}b)^2}{ab}\plus{}\frac{(a\minus{}b)(b\minus{}c)}{ac}$
So the inequality is equivalent to: $ \frac{(a\minus{}b)^2}{ab}\plus{}\frac{(a\minus{}b)(b\minus{}c)}{ac}\ge \frac{(a\minus{}b)^2}{(a\plus{}c)(b\plus{}c)}\plus{}\frac{(a\minus{}b)(b\minus{}c)}{(a\plus{}b)(a\plus{}c)}$
Which is true because c is minimal

Hello friend! your inequality is really nice, and I find a beautiful solution,
put that, $ \frac{a}{b} \equal{} x$ and $ \frac {b}{c} \equal{} y$ and $ \frac {c}{a} \equal{} z$
we have $ \sum\frac {c \plus{} a}{c \plus{} b} \equal{} \sum \frac {xy \plus{} 1}{y \plus{} 1} \equal{} \sum\left(x \plus{} \frac {1 \minus{} x}{y \plus{} 1}\right)$
then the inequality becomes,
$ \sum \frac {x \minus{} 1}{y \plus{} 1} \geq 0$ with $ xyz \equal{} 1$
$ \Leftrightarrow \sum \frac {x}{y \plus{} 1} \geq \sum \frac {1}{y \plus{} 1}$ wich is easy,
suppose that, $ a\geq b\geq c$ then with Chebychev,
$ \sum \frac {x}{y \plus{} 1} \geq \frac {1}{3}(x \plus{} y \plus{} z)\left(\sum\frac {1}{x \plus{} 1}\right)$
by AM-GM $ x \plus{} y \plus{} z\geq 3$ therefor, $ \sum \frac {x}{y \plus{} 1} \geq \sum \frac {1}{y \plus{} 1}$
and finally , we deduct that: $ \sum \frac {x \minus{} 1}{y \plus{} 1} \geq 0$ wich implies your inequality,

See also
http://www.mathlinks.ro/Forum/viewtopic.php?t=32718
http://www.mathlinks.ro/Forum/viewtopic.php?t=5263

  darij

Thanks to you all for your kind coordination

Hey i think we cant assume justok's assumption as the inequality is not symmetric .
Also i've got a nice solution using homogeneity

Hey i have a doubt regarding this
for all positive reals $ a,b,c$
$ \frac{a}{b} \le \frac{a\plus{}c}{b\plus{}c}$
so $ \sum \frac{a}{b} \le \sum \frac{a\plus{}c}{b\plus{}c}$
this reverses the inequality :huh:  :maybe:
so please help

[quote="darij grinberg"][quote="Rijul saini"]Hey i think we cant assume justok's assumption as the inequality is not symmetric .[/quote]

No, but it is cyclic, so we can "move" $ a$, $ b$, $ c$ one or two steps forward cyclically until the smallest of the elements $ a$, $ b$, $ c$ occupies the third position.

  darij[/quote]
please show me the way.......

[quote="Rijul saini"]for all positive reals $ a,b,c$
$ \frac {a}{b} \le \frac {a \plus{} c}{b \plus{} c}$[/quote]

for ALL ? check your computations...

[quote="Rijul saini"][quote="darij grinberg"][quote="Rijul saini"]Hey i think we cant assume justok's assumption as the inequality is not symmetric .[/quote]

No, but it is cyclic, so we can "move" $ a$, $ b$, $ c$ one or two steps forward cyclically until the smallest of the elements $ a$, $ b$, $ c$ occupies the third position.

  darij[/quote]
please show me the way.......[/quote]

Do you agree that we can replace $ a$, $ b$, $ c$ by $ b$, $ c$, $ a$, and the inequality will not change? And that we can also replace $ a$, $ b$, $ c$ by $ c$, $ a$, $ b$, and the inequality still will not change? Now at least one of the triples $ \left(a,b,c\right)$, $ \left(b,c,a\right)$, $ \left(c,a,b\right)$ will have its minimal entry at the end...

  darij

Inequality equivalents to $ a^4c^2\plus{}b^4a^2\plus{}c^4b^2\plus{}a^3b^3\plus{}b^3c^3\plus{}a^3c^3 \geq abc(ab^2\plus{}bc^2\plus{}ca^2\plus{}3abc)$

$ a^3b^3\plus{}b^3c^3\plus{}a^3c^3 \geq 3a^2b^2c^2$ and we take $ x^3\equal{}ab^2 , y^3\equal{}bc^2 , z^3\equal{}ca^2  \Longrightarrow$

$ a^4c^2\plus{}b^4a^2\plus{}c^4b^2\minus{}abc(ab^2\plus{}bc^2\plus{}ca^2)\equal{}x^6\plus{}y^6\plus{}z^6\minus{}x^4yz\minus{}y^4xz\minus{}z^4xy \geq0$ by Muirhead. And solution is done. :wink:


inequalities
induction
algebra solved
algebra

Which number is the greater : 2001! or 1001^2001 ?

Since k(2001 - k) \leq (2001/2)2, we have 2001! < (2001/2)2000 < 10012001.

In 1986, there was a (very simple, as usal) problem on the Flanders olympiad:

Prove that for any positive integer n > 1, n! < [(n + 1)/2]n.

[quote="Arne"]Prove that for any positive integer n > 1, n! < [(n + 1)/2]n.[/quote]

Very easy indeed, it's just AM-GM  :D
(the inegality is strict only for n > 1  :))

Flemish students do not know the AM - GM inequality (in general). I'm an exception :D
They had to look for an induction proof or for a proof using k(n - k) \leq (n/2)2.

[quote="Arne"][Flemish] students do not know the AM - GM inequality .[/quote]

I have the feeling that 99% of the use of AM-GM comes from olympiads .... So is it useful for a student ?

Well, I'd say that very few of the students taking part in the final know it. :)


induction
number theory proposed
number theory

The real sequence $\{a_n|n=0,1,2,3,...\}$ defined $a_0=1$ and
\[ a_{n+1}=\frac{1}{2}\left (a_{n}+\frac{1}{3 \cdot a_{n}} \right ).  \]
Denote
\[ A_n=\frac{3}{3 \cdot a_n^2-1}.  \]
Prove that $A_n$ is a perfect square and it has at least $n$ distinct prime divisors.

[quote="Rust"]It is obviosly: \[ A_{n+1}=\frac{12}{3(a_n+1/(3a_n))^2-4}=\frac{4}{(a_n-1/(3a_n))^2}=(\frac{6a_n}{3a_n^2-1})^2=(2a_nA_n)^2.  \][/quote]
As I understood, you meaned $a_nA_n$ integer? Why not? :D $a_n$ is not necessary integer

the solution of Rust is wrong.Here is my solution :) :
We will prove $3A_{n+1}=4A_{n}(3+A_{n})$ (for all $n \in N^{*}$)
We have $3+A_{n}=3+\frac{3}{3a_{n}^2-1}=\frac{9a_{n}^2}{3a_{n}^2-1}$
So $4A_{n}(3+A_{n})=36A_{n}\frac{a_{n}^2}{3a_{n}^2-1}=\frac{108a_{n}^2}{(3a_{n}^2-1)^2}$
     $3A_{n+1}=\frac{9}{3a_{n+1}^2-1}=\frac{9}{3(\frac{1}{2})^2(a_{n}+\frac{1}{3a_{n}})^2-1}=\frac{108a_{n}^2}{(a_{n}^2-1)^2}$
Hence: $3A_{n+1}=4A_{n}(3+A_{n})$ (for all $n \in N^{*}$) $(A_{1}=9)$
From this we easily to prove $A_{n} \equiv 0 (mod 3)$ for all $n \in N^{*}$
We will prove $\frac{A_{n}}{3}+1$ is a square number for all $n\in N^{*}$ by induction.
When $n=1$ then $\frac{A_{1}}{3}+1=4$ is a perfect square
We suppose that statement hold when $n=k(k\geq 1)$
We must prove the statement when $n=k+1$.From here:$\frac{A_{k+1}}{3}+1=\frac{4A_{k}(3+A_{k})}{9}+1=4\frac{A_{k}}{3}(1+\frac{A_{k}}{3})+1$
We have $\frac{A_{k}}{3}+1$ is a perfect square (from the conjecture) so we can put $\frac{A_{k}}{3}+1=j^2$.From here $4\frac{A_{k}}{3}(1+\frac{A_{k}}{3})+1 =4j^2(j^2-1)+1=(2j^2-1)^2$
So we complete the induction.
Come back to the problem:to prove $A_{n}$ is perfect square is obviously (By use $\frac{A_{n}}{3}+1$ is a perfect square)
Now we will prove $A_{n}$ has at least $n$ distinct prime divisors by use induction
when $n=1$ it's obviously true
We suppose it's true when $n=k ( k \geq 1)$
We will prove it's still true when $n=k+1$.we partition into two cases:
              +Case 1:If $A_{k}$ has at least $(k+1)$ distinct prime divisors then the induction is complete
              +Case 2:If $A_{k}$ has exactly $k$ distinct prime divisors.We call them are $p_{1},p_{2},\cdots,p_{k}$
It clear that $(A_{k}+3,p_{i})=1 \vee 3$.So if $A_{k+1}$ has exactly $k$ distinct prime divisors then $A_{k}+3=3^{m} (m \in N^{*}) \rightarrow A_{k}=3^{k}-3 \equiv -3 (mod 9)$
 $\rightarrow A_{k}$ is not a perfect square (Contradition)
Therefore $A_{k+1}$ has at least $k+1$ distinct prime divisors (this complete the induction)
SoThat:the problem is proved absolutely  :D  :D

It's easy to have:
$A_{n+1}=\frac{A_n}{9}\left(4A_{n-1}+6\right)^2$
:D

Define b_k = sqrt 3 / 2 * [ ( 2 + sqrt 3 ) ^ k - (2 - sqrt 3 ) ^ k ].
it is evident that sqrt [ b_k ^2 / 3 + 1 ] = 1/2 [ ( 2 + sqrt 3 ) ^ k + ( 2 - sqrt 3 ) ^ k ) ]
therefore: b_k * 2 * sqrt [ b_k^2 / 3 + 1 ] = b _ 2k ... ( *).
define B _ t = b _ (2^t).
it is evident from (*) that: B_(t+1) = B_t * 2 * sqrt ( (B_t)^2 / 3 + 1 ).
Let A'_t = (B_t)^2. it is evident that A'_1 = 9 and A'_(t+1) = A'_t * 4/3 * ( A'_t + 3) .. ( * )
by bashing the expression A_t, we get the following:
A_(t+1) = [ 6*a_n / 3*(a_n)^2 - 1 ] ^ 2
            = 4/3 * A_t ( A_t + 3 ), and A_1 = 9 is evident.
Therefore {A_t} and {A'_t} are identical --> A_t = A'_t for all t. .
By construction, A_t is a perfect square. it remains to see A_t has at least n prime divisors.
But this is trivial, since it is sufficient to prove A_t+1 / A_t has at least one prime divisor that A_t does not have.
since A_t+1/A_t = 4/3 * (A_t + 3); it follows that since (A_t, A_t + 3) = 1, A_t+3 has a prime divisor not present
in A_t, so by a trivial induction the problem is complete.

@epitomy01, is it really so hard to make a liitle effort and write your solution in LaTeX? Now nobody wants to read it because it's not clear...


inequalities
symmetry
inequalities unsolved

Prove that if a,b,c≥0,then
$ a^3\plus{}b^3\plus{}c^3\plus{}3abc$≥$ ab\sqrt{2a^2\plus{}2b^2}\plus{}bc\sqrt{2b^2\plus{}2c^2}\plus{}ca\sqrt{2c^2\plus{}2a^2}$

[quote="SUPERMAN2"]Prove that if a,b,c≥0,then
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc$≥$ ab\sqrt {2a^2 \plus{} 2b^2} \plus{} bc\sqrt {2b^2 \plus{} 2c^2} \plus{} ca\sqrt {2c^2 \plus{} 2a^2}$[/quote]
It certainly does look like schur! And we can use this fact :)
Subtract $ \sum_{cyc} a^2b$. From both sides at first, to get it even closer to schur:
$ \sum_{cyc} a^3 \plus{} abc \minus{} a^2b \minus{} a^2c \ge \sum_{cyc} ab\sqrt {2a^2 \plus{} 2b^2} \minus{} a^2b \minus{} ab^2$
You probably know how to write the $ LHS$ as sum of square. And it is not hard either to do that with the $ RHS$:
$ ab\sqrt {2a^2 \plus{} 2b^2} \minus{} a^2b \minus{} ab^2 \equal{} ab ( \sqrt {2(a^2 \plus{} b^2)} \minus{} a \minus{} b ) \equal{}$ $ \frac {ab (2(a^2 \plus{} b^2) \minus{} (a \plus{} b)^2) )}{\sqrt {2(a^2 \plus{} b^2)} \plus{} a \plus{} b} \equal{} (a \minus{} b)^2 \left ( \frac {ab}{\sqrt {2(a^2 \plus{} b^2)} \plus{} a \plus{} b} \right )$

And $ LHS \equal{} \sum_{cyc} (a \minus{} b)^2 \left ( \frac {a \plus{} b \minus{} c}{2} \right )$.

So $ LHS \minus{} RHS \equal{} \sum_{cyc} (a \minus{} b)^2 \left ( \frac {a \plus{} b \minus{} c}{2} \minus{} \frac {ab}{\sqrt {2(a^2 \plus{} b^2)} \plus{} a \plus{} b} \right )$

The square root, isn't really helpfull in this case. So use that $ \sqrt {2(a^2 \plus{} b^2)} \ge 2(a \plus{} b)$ and the fact that $ x \to \frac { \minus{} 1}{k \plus{} x}$ is increasing in $ (0; \plus{} \infty)$ for $ k > 0$. And thus we just have to prove:

$ \sum_{cyc} (a \minus{} b)^2 \left ( \frac {a \plus{} b \minus{} c}{2} \minus{} \frac {ab}{2(a \plus{} b)} \right ) \ge 0 \iff$

$ \sum_{cyc} (a \minus{} b)^2 \left ( \frac {a^2 \plus{} ab \plus{} b^2 \minus{} ac \minus{} bc}{2(a \plus{} b)} \right ) \ge 0$

Let $ S_c \equal{} \frac {a^2 \plus{} ab \plus{} b^2 \minus{} ac \minus{} bc}{2(a \plus{} b)}$ and cyclic. Assume wlog because of symmetry that $ a \ge b \ge c$. Then $ S_b, S_c \ge 0$. And since $ (a \minus{} c)^2 \ge (a \minus{} b)^2 \plus{} (b \minus{} c)^2$, we get:

$ (a \minus{} b)^2S_c \plus{} (b \minus{} c)^2S_a \plus{} (c \minus{} a)^2S_b \ge (a \minus{} b)^2(S_c \plus{} S_b) \plus{} (b \minus{} c)^2(S_a \plus{} S_b) \ge 0$

Since $ S_c \plus{} S_b \ge 0$ it suffices to prove that $ S_a \plus{} S_b \ge 0$. Which is:

$ S_a \plus{} S_b \equal{} \frac {c^2 \plus{} cb \plus{} b^2 \minus{} ba \minus{} ca}{2(b \plus{} c)} \plus{} \frac {c^2 \plus{} ca \plus{} a^2 \minus{} ba \minus{} bc}{2(a \plus{} c)} \ge 0 \iff$

$ (c^2 \plus{} cb \plus{} b^2 \minus{} ba \minus{} ca)(a \plus{} c) \plus{} (c^2 \plus{} ca \plus{} a^2 \minus{} ba \minus{} bc)(b \plus{} c) \ge 0 \iff$

$ c^2(a \plus{} b \plus{} 2c) \ge 0$

Which is obvious. :)

[quote="SUPERMAN2"]Prove that if a,b,c≥0,then
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc$≥$ ab\sqrt {2a^2 \plus{} 2b^2} \plus{} bc\sqrt {2b^2 \plus{} 2c^2} \plus{} ca\sqrt {2c^2 \plus{} 2a^2}$[/quote]

HI!!

You see Thomas Mildorf "Olympiad Inequalities" pg 29 number 39 - august 4, 2006.
Problem created by ANH-CUONG.

____________________
Panagiote Ligouras


logarithms
calculus
integration
ratio
limit
function
analytic geometry

Not sure if this works... first take the natural log to get

$\ln{L}= N\ln{2}+\sum_{k=1}^{N}\ln{k}-N\ln{N}$.

But we know $N\ln{N}-N \le \sum_{k=1}^{N}\ln{k}\le N\ln{N}-N+\ln{N}$ by integrals, so

$N(\ln{2}-1) \le \ln{L}\le N(\ln{2}-1)+\ln{N}$.

Both the upper and lower bound have limit $-\infty$, however, so $\ln{L}\rightarrow-\infty$ and $L \rightarrow 0$.

This won't be as good as paladin8's answer, but it is the way I would approach it in front of a Calculus II class:

Consider the series $\sum_{n}\frac{2^{n}n!}{n^{n}}.$

Does the series converge? Try the ratio test:

$\lim_{n\to\infty}\frac{2^{n+1}(n+1)!}{(n+1)^{n+1}}\cdot \frac{n^{n}}{2^{n}n!}$

$=\lim_{n\to\infty}\frac{2(n+1)n^{n}}{(n+1)^{n+1}}= \lim_{n\to\infty}\frac{2n^{n}}{(n+1)^{n}}$

$=\lim_{n\to\infty}\frac2{\left(1+\frac1n\right)^{n}}=\frac2e<1.$

Hence the series converges, and since the series converges, the terms tend to zero.

In fact, we know that the terms tend to zero at a rate favoribly comparable to some geometric sequence, a fact that could also be gleaned from paladin8's argument. Note in particular the terms $N(\ln2-1)$ in his argument and compare them to the $\frac2e$ in mine. (Both of us have in effect proved a crude simple version of Stirling's approximation: $n!\sim\left(\frac{n}{e}\right)^{n}$ mutilplied by something which does not grow or shrink geometrically fast.)

How would you show that

$\lim_{n \rightarrow \infty}\frac{e^{n}\cdot n!}{n^{n}}$

diverges?

Well, now we're closing in on a more detailed version of Stirling's approximation.

Go back to your argument, where you used an integral. You approxiated an integral by left-endpoint and right-endpoint Riemann sums. What if you tried using midpoint Riemann sums and/or trapezoidal rules in that estimate?

(For midpoint rules, you might try starting an integral at $\frac12$ instead of at 1.)

Ah, interesting. From midpoint approximations, we can get

\begin{eqnarray*}\ln{L}&\ge& \lim_{n \rightarrow \infty}\left(n\ln{n}-n+\int_{\frac{1}{2}}^{n+\frac{1}{2}}\ln{x}dx\right) \\ &=& \lim_{n \rightarrow \infty}\left(n\ln{n}-n+\left(n+\frac{1}{2}\right)\ln{\left(n+\frac{1}{2}\right)}-\left(n+\frac{1}{2}\right)-\frac{1}{2}\ln{\frac{1}{2}}+\frac{1}{2}\right) \\ &\approx& \lim_{n\rightarrow \infty}\frac{1}{2}\ln{n}\end{eqnarray*}

so $n! \sim \sqrt{n}\left(\frac{n}{e}\right)^{n}$ (which shows that the limit diverges as well).

[quote="maokid7"][quote="paladin8"]
But we know $N\ln{N}-N \le \sum_{k=1}^{N}\ln{k}\le N\ln{N}-N+\ln{N}$ by integrals,[/quote]

how do u show this w/ integrals? i don't follow.[/quote]

Consider the right and left hand Riemann sums of $\int_{1}^{N}\ln{x}dx$.

does the convexity of the function determine which of the left/right hand sums is larger?

For right and left hand Riemann sums, it's merely the slope (and the error estimate depends on the first derivative.) For a strictly increasing function, left-hand sums underestimate the integral and right hand sums overestimate it. (For strictly decreasing functions, reverse that.)

Concavity matters when you consider either midpoint sums or the trapezoidal rule. Both of those have error estimates that depend on the second derivative.

For a function that is concave up, midpoint sums underestimate the integral and the trapezoidal rule overestimates the sums. (The error for the trapezoidal rule is roughly twice the size of the error for midpoint sums.) For a function that is concave down, reverse the order of those inequalities.

paladin8 was applying the midpoint rule to the integral of a function which is concave down, accounting for the direction of his inequality sign. If he wanted an inequality pointing the other direction, he would need to use the trapezoidal rule (to a slightly different integral, for the sake of practicality.)


ratio
geometry
factorial
modular arithmetic
floor function

pls help. i cant answer some questions here..

23.) twenty-five different positive integers add to 2008. what is the largest value that the least of them can have?

24.) PQR is an equilateral triangle. the point U is the midpoint of PR. points T and S divide QP and QR in the ration 1:2. The point of intersection of PS, RT and QU is X. If the are of QSX is 1 square unit, what is the area in square units of triangle PQR?

25.) a two digit number n has the property that the sum of the digits of n is the same as the sum of the digits of 6n. how many such numbers are there?

28.) with how many zeroes does 2008!(factorial 2008) end?

these 4 questions really bothers me.. thx a lot in advance!

For the last one, think of how many ways you can make a 0. You can make a 0 if you have a 2 and a 5. There are a lot more 2's than 5's in 2008!, so all you have to do is figure out the number of 5's. Don't forget squares, like 25, or 625.

-rts2007

[quote="henyomzky"]pls help. i cant answer some questions here..

23.) twenty-five different positive integers add to 2008. what is the largest value that the least of them can have?
[/quote]
[hide]
If all the integers are consecutive, than the equation could be written as X-12, X-11,....,X,....X+11, X+12. Summing these, the positive and negative cancel out, leaving 25X
$ 25X\equal{}2008$
$ X\equal{}80.32$
X needs to equal an integer for this to make sense, so let $ X\equal{}80$
$ 25(80)\equal{}2000$
The sum of 68,69...80...91,92 equals 2000. You can add 8 to the last 8 numbers, to get 68 through 84 and 86 through 93. The maximum possible is 68.
[/hide]

23) 80<2008/25<81
that provides a good estimate for the integers. try  consecutive integers with an average of 80 (68-92). they add to 2000. to achieve 2008, add 8 to the largest integer. the answer is the lowest integer or 68
24) suppose the triangle has side length s and height a. then UX=a/5 and QX=4a/5. The perpendicular from S to QX has length s/3, so the area of QSX is 2as/15, which is 4/15 of the area of PQR. therefore PQR has area 15/4

[hide="Number 25"]
Using the fact that a number is equivalent to the sum of its digits $ \mod 9$, $ n\equiv 6n\pmod 9\implies n\equiv 0\pmod 9$, so we want all two digit numbers divisible by 9. $ 9\cdot 12$ has more than 2 digits, and $ 9\cdot 1$ has less than 2 digits, so $ \boxed{10}$ numbers work.
[/hide]

[hide="Number 28"]
It is well known that the largest $ k$ such that $ p^k|n!$ is $ \sum_{k=0}^{\infty}\left\lfloor{\frac{n}{p^k}\right\rfloor}$ (see if you can prove this).

In this case, we want to count the number of fives, so we have: $ 401+80+16+3=\boxed{500}$.
[/hide]

I'm a little unsure about number 25.

[hide="24"]
we look for triangles with collinear bases and the same height.

First $ \bigtriangleup QXS$ and $ \bigtriangleup RXS$ have collinear bases (with ratio $ 1: 2$) and the same hight. Therefore the Area $ \bigtriangleup RSX \equal{} 2$

By symetry $ \bigtriangleup QXT$ and $ \bigtriangleup PXT$ have areas $ 1$ and $ 2$ respectively

Now triangles $ \bigtriangleup QXR$ and $ \bigtriangleup QXT$ have areas $ 3$ and $ 1$ as wells as collinear basses with the same ratio ($ 3: 1$)

Therefore $ \bigtriangleup PXR$ and $ \bigtriangleup PXT$ have area ratio $ 3: 1$ and since $ \bigtriangleup PXT \equal{} 2 \Longrightarrow \bigtriangleup PXR \equal{} 6$

$ \therefore [PQR] \equal{} 1 \plus{} 1 \plus{} 2 \plus{} 2 \plus{} 6 \equal{} 12$[/hide]


What sort of simplification can be made in order to solve for the general solution of the following differential equations?

$ x^{2}y'+y^{2}=xyy'$

$ y'=\frac{a^{2}}{(x+y)^{2}}$

hello, write your equation in the form
$ \frac{x}{y}y'+\frac{y}{x}=y'$ and substitute $ \frac{y}{x}=u$. After this you get :
$ \frac{u-1}{u}du=\frac{dx}{x}$.
Sonnhard.


LaTeX articles parameterization integration

I'd like to ask you for a little help: I used a converter from Word to Latex, GrindEq. The problem is with the right numbered displayed equations. They are too crowded, on vertical. Too close one to the other from above. More precisely, there are more equations which form one group, and only the group is right numbered.

How can I add an extra space in vertical alignment of equations?

Thanks,
Jean

P.S. let me give an example:

% Generated by GrindEQ Word-to-LaTeX 2007
% LaTeX/AMS-LaTeX

\documentclass{article}

%%% remove comment delimiter ('%') and specify encoding parameter if required,
%%% see TeX documentation for additional info (cp1252-Western,cp1251-Cyrillic)
%\usepackage[cp1252]{inputenc}

%%% remove comment delimiter ('%') and select language if required
%\usepackage[english,spanish]{babel}

\usepackage{amssymb}
\usepackage{amsmath}
\usepackage[dvips]{graphicx}
%%% remove comment delimiter ('%') and specify parameters if required
%\usepackage[dvips]{graphics}

\begin{document}

%%% remove comment delimiter ('%') and select language if required
%\selectlanguage{spanish}

fbngnfgng dnfgfhfh rhtjhjty ghgfhg nggfnf gngfnf ngfnhg gfhf fnh hhg hg ghghgh hghghg hgmhghg

\begin{equation} \label{1.1)} \begin{array}{l} {\sum _{i=1}^{n}X_{i}^{2}  \frac{1}{n} \sum _{i=1}^{n}(X_{i} -\bar{X})^{2}  \; } \\ {\int _{g}^{t}hds +\int _{ry}^{y}jdf \; } \\ {\sum _{i=1}^{n}X_{i} Y_{i}  \sum _{i=1}^{n}X_{i} Y_{i}  \frac{x-\mu }{\sigma } \; } \\ {\sum _{i=1}^{n}X_{i} Y_{i}  \frac{1}{n} \sum _{i=1}^{n}(X_{i} -\bar{X})^{2}  X_{1} ,\ldots ,X_{n} \; } \end{array} \end{equation}

bbgfbgfgfb ggfnnh fnjhh gfhjhjn jjjhhg kjn ,kn,mn, ,mnm,n, lmnm,n, nm,m ,mnm,n ,mn,mn miujhfe ,n,m

\end{document}

Not sure why an array was used instead of align but then automated systems will never be perfect. On the other hand the inter-line spacing is easy to change for arrays using
\renewcommand{\arraystretch}{1.5}
or even
\renewcommand{\arraystretch}{2}
The parameter is a multiplying factor, so 1.5 means multiply the inter-line spacing in the array by 1.5, thus increasing the spacing by 50%.

Thank you for your answer.

1) About the renew command you said: where should I type it, in the preamble? Do I need to type a command before the array? I don't get it.

2) Is it possible to use instead the align command, or I must type the equations once again?

[quote="jeanvaljean"]1) About the renew command you said: where should I type it, in the preamble? Do I need to type a command before the array? I don't get it.[/quote]Put \renewcommand{\arraystretch}{1.5 or 2} anywhere you like before the array. You can change \arraystretch back again afterwards if you don't want it to affect other arrays.
[code]\renewcommand{\arraystretch}{2}
\begin{equation} \label{1.1)} \begin{array}{l} {\sum _{i=1}^{n}X_{i}^{2} \frac{1}{n} \sum _{i=1}^{n}(X_{i} -\bar{X})^{2} \; } \\ {\int _{g}^{t}hds +\int _{ry}^{y}jdf \; } \\ {\sum _{i=1}^{n}X_{i} Y_{i} \sum _{i=1}^{n}X_{i} Y_{i} \frac{x-\mu }{\sigma } \; } \\ {\sum _{i=1}^{n}X_{i} Y_{i} \frac{1}{n} \sum _{i=1}^{n}(X_{i} -\bar{X})^{2} X_{1} ,\ldots ,X_{n} \; } \end{array} \end{equation} [/code]
[quote="jeanvaljean"]2) Is it possible to use instead the align command, or I must type the equations once again?[/quote]The equations can stay the same, just replace the equation and array environments with align:
[code]\begin{align} {\sum _{i=1}^{n}X_{i}^{2} \frac{1}{n} \sum _{i=1}^{n}(X_{i} -\bar{X})^{2} \; } \\ {\int _{g}^{t}hds +\int _{ry}^{y}jdf \; } \\ {\sum _{i=1}^{n}X_{i} Y_{i} \sum _{i=1}^{n}X_{i} Y_{i} \frac{x-\mu }{\sigma } \; } \\ {\sum _{i=1}^{n}X_{i} Y_{i} \frac{1}{n} \sum _{i=1}^{n}(X_{i} -\bar{X})^{2} X_{1} ,\ldots ,X_{n} \; } \end{align}[/code]
This is entriely optional but the code is messy so you could tidy it up to get in your example:
[code]\begin{align}
 &\sum _{i=1}^{n}X_{i}^{2} \frac{1}{n} \sum _{i=1}^{n}(X_{i} -\bar{X})^{2} \\
 &\int _{g}^{t}hds +\int _{ry}^{y}jdf  \\
 &\sum _{i=1}^{n}X_{i} Y_{i} \sum _{i=1}^{n}X_{i} Y_{i} \frac{x-\mu }{\sigma }  \\
 &\sum _{i=1}^{n}X_{i} Y_{i} \frac{1}{n} \sum _{i=1}^{n}(X_{i} -\bar{X})^{2} X_{1} ,\ldots ,X_{n}
\end{align}[/code]
All I've done is get rid of the {... \; } for each equation which aren't necessary, put each equation on a new line and added & at the beginning of each line so they are left aligned which looks better (if you have any with = in them put the & before that instead)..

If you don't want all the equations numbered then use \notag on the end of a line where you don't want a number.

Dear Sir,

Thank you for your advices, all three have worked. I wish you all the best.

Jean


Find the number of square units in the area of the region determined by the system: $ x\plus{}y\leq6,0\leq x\leq3,0\leq y\leq5$.

Graph it and stuff to get $ 5\cdot3\minus{}\frac12\cdot2\cdot2\equal{}15\minus{}2\equal{}\boxed{13}$. Alternatively, $ \frac12\cdot6\cdot6\minus{}\frac12\cdot1\cdot1\minus{}\frac12\cdot3\cdot3\equal{}\boxed{13}$.

Here is a link to a Desmos graph:
[url]https://www.desmos.com/calculator/0ecasa2rz8[/url]


Some questions....

$1.$ Show that a power of $3$ can never be a perfect number.
$2.$ More generally, if $p$ is an odd prime, show that a power $p^{k}$ can never be a perfect number.
$3.$ Show that a number of the form $3^{i}\times 5^{j}$ can never be a perfect number.
$4.$ More generally, if $p$ is an odd prime number greater that $3$, show that the product $3^{i}\times p^{j}$ can never be a perfect number.
$5.$ Even more generally, if $p$ and $q$ are distinct odd primes, show that the number of form $q^{i}\times p^{j}$ can never be a perfect number.

EDIT1: Please move it to other place if its not suitable here.

[quote="Rust"]Let \[n=p_{1}^{k_{1}}p_{2}^{k_{2}}...p_{s}^{k_{s}}\] is odd perfect number. We have \[\sigma (n)=n\prod_{i}\frac{1-1/p_{i}^{k_{i}+1}}{1-\frac{1}{p_{i}}}<n\prod_{i}\frac{1}{1-\frac{1}{p_{i}}}\] .
Therefore (from $\sigma (n)=2n$ ) \[\prod_{i}(1-\frac{1}{p_{i}}) <\frac{1}{2}.\] It give $s\ge 3.$[/quote]

really sorry, but could not understand your proof... little too much for my level of understanding... I am software engineer by profession but have interest in number theory... but dot so conversant with it... so if its not trouble to you, can you explain it in easier terms.

[quote="Kondr"]Let
$n=p_{1}^{k_{1}}p_{2}^{k_{2}}...p_{s}^{k_{s}}$
is odd perfect number. We have the sum of ALL its divisors
$\sigma (n)=\prod_{i=1}^{s}{(1+p_{i}+...+p_{i}^{k_{i}})}= =\prod_{i=1}^{s}\frac{p_{i}^{k_{i}+1}-1}{p_{i}-1}< \prod_{i=1}^{s}\frac{p_{i}^{k_{i}+1}}{p_{i}-1}= =n\prod_{i=1}^{s}\frac{p_{i}}{p_{i}-1}$.
But since n is perfect, we have $2n=\sigma (n)$, so
$2n=\sigma (n)<n\prod_{i=1}^{s}\frac{p_{i}}{p_{i}-1}$, divided by n
$2<\prod_{i=1}^{s}\frac{p_{i}}{p_{i}-1}$
But the terms of the product can be 3/2, 5/4, 7/6, 11/10,... this sequence is decreasing with limit 1. Each term is <2, so we need more than one prime to divide n. If we took just two oft the terms, the product would be 15/8<2, so we need at least three primes to divide n.
This solves the most general problem.[/quote]

Thanks... I could understand it now.


Is there a way to increase the font size in LaTeX?
I want to make a big letter t but i dont know how.

Thanks.

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


function
limit
factorial
calculus
calculus computations

Are there any tools that can help us find the limit of a certain sequence or function beside rule of L'Hospital?
For example, find $ \lim_{n\to\infty}{\frac{\prod_{i=1}^{n}{(2n-1)}}{ \prod_{i=1}^{n}{(2n)}}\cdot\sqrt{2n+1}}$ :blush:

Yes, but there's no set of rules that lets you find the limit of an arbitrary convergent sequence, i.e., any method depends on the actual nature of the sequence you are considering.  In this case, (assuming you meant to write $ \lim_{n\to\infty}{\frac{\prod_{i=1}^{n}{(2i-1)}}{ \prod_{i=1}^{n}{(2i)}}\cdot\sqrt{2n+1}}$), we can rewrite the product in the numerator as $ \frac{(2n)!}{2^n n!}$ and the product in the denominator as $ 2^n n!$, so the entire expression is $ \lim_{n \to \infty} \frac{(2n)!}{4^n (n!)^2} \cdot \sqrt{2n + 1}$.  Now we can apply Stirling's approximation to the factorials: our limit is equal to $ \lim_{n \to \infty}
\frac{\frac{(2n)^{2n} \sqrt{4\pi n}}{e^{2n}}} {4^n \cdot \frac{n^{2n} \cdot 2\pi n}{e^{2n}}} \cdot \sqrt{2n + 1}$, and enough things cancel that you can read off the result from there.


[hide]For the triangle $ABC$, let $x$ be the length of the hypotenuse $AB$, so the length of $BC$ is $x\sin15^\circ$ and the length of $AC$ is $x\cos15^\circ$. Then you get the equation $x= 0.5x^{2}\sin15^\circ\cos15^\circ$ which you can simplify with a double angle formula
[/hide]

[quote="jsc"]sorry if this problem is tooo easy or difficult... i can't gauge it myself  :blush:

[b]A right triangle with integral side lengths, and with one angle 15 degrees, has a hypotenuse with length numerically equal to the triangle's area. find this area.[/b]

please post solutions or suggestions if you can!  :)

thanks![/quote]

I don't think the right triangle can have integral side lengths...
If the hypotenuse is $x$, then the other two sides are $x\left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)$ and $x\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)$.

Suppose the hypotenuse is an integer. Then clearly, the other two sides cannot be integers.

The problem is easy if we ignore the integer condition, so we'll ignore the 15-degree condition.

[hide="Solution"] We are looking at a right triangle $a, b, c$ with the property that $ab = 2c$.  Now, because

$a^{2}+b^{2}= c^{2}$

We have

$(a+b)^{2}= c^{2}+4c = (c+2)^{2}-4$
$(a-b)^{2}= c^{2}-4c = (c-2)^{2}-4$

But $0, 4$ are the only squares that differ by $4$.  The equations together imply $c = 0$, absurd.  Hence no such triangles exist. [/hide]
[b]Follow-up:[/b]  What happens when we ignore the right condition?  (The "hypotenuse" is then just the largest side.)

[quote="t0rajir0u"][b]Follow-up:[/b]  What happens when we ignore the right condition?  (The "hypotenuse" is then just the largest side.)[/quote]

[hide="Simple"]Let $a$ be the hypotenuse.
Three cases, depending on which two sides the $15^{\circ}$ angle is between.

CASE 1: Between $b,c$
This would imply that $a$ is not the largest side (since the angle is opposite $a$), an impossibility.

CASE 2: Between $a,b$
Using the area formula, we obtain
$a = \frac{ab\sin(15^{\circ})}{2}$
But this means $b$ is not an integer. Impossible.

CASE 3: Between $a,c$
Similar to case 2. Impossible.

Therefore, no such triangles exist.[/hide]


algebra
polynomial
search
Galois Theory
superior algebra
superior algebra unsolved

Let $ E$ be the splitting field of $ X^7\minus{}6$ over $ \mathbb{Q}$

Prove that:
$ \zeta_7\plus{}\sqrt[7]{6}$ is a primitive element of $ E$.

You have at least three possibilities:
a) Verify that the field generated by this element contains $ \zeta_7$ and $ \sqrt[7]6$ by computations.
b) Use the primitive element theorem, it gives a checkable condition if an element is primitive.
c) Show that the Galois group maps this element to as many different elements as possibly (here 42).

Hi, thanks
I've already shown that $ E\equal{}\mathbb{Q}(\zeta_7,\sqrt[7]{6})$,
but I cannot go further at the computation that those two are generated by $ \zeta_7\plus{}\sqrt[7]{6}$. Is there some trick here? :(

By the primitive element theorem, I've got the existence of those primitive elements, but how to check one given number?

There is a proof of primitive element theorem which constructs a $ \gamma$ such that $ F(\gamma) \equal{} F(\alpha,\beta)$ (at least in the infinite case, with char=0). Let $ a_1,...,a_k$ be the roots of the minimum polynomial of $ \alpha$, and $ b_1,..,b_r$ the roots of the min pol of $ \beta$, let $ c \in F \smallsetminus \{ (\alpha \minus{} a_i)/(\beta \minus{} b_j)\}$, end $ \gamma \equal{} a \plus{} cb$. In your case, try if $ c\equal{}1$ is good

Let $ p$ be a prime and $ q$ be an integer s.t. $ 2\leq{q}\leq{p\minus{}1}$..
$ E\equal{}\mathbb{Q}(\zeta_p,\sqrt[p]{q})$ is the splitting field of $ x^p\minus{}q$.
$ \zeta_p$ is an algebraic number of degree $ p\minus{}1$ over $ \mathbb{Q}$
$ \sqrt[p]{q}$ is an algebraic number of degree $ p$ over $ \mathbb{Q}$.
$ igcd(p\minus{}1,p)\equal{}1$ then $ \mathbb{Q}(\zeta_p\plus{}\sqrt[p]{q})\equal{}E$ (it is a general result).
Remark: the galois group of $ x^p\minus{}q$ is the Frobenius group of order $ p(p\minus{}1)$.
see
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1386309843&t=137762

[quote="pic"]There is a proof of primitive element theorem which constructs a $ \gamma$ such that $ F(\gamma) \equal{} F(\alpha,\beta)$ (at least in the infinite case, with char=0). Let $ a_1,...,a_k$ be the roots of the minimum polynomial of $ \alpha$, and $ b_1,..,b_r$ the roots of the min pol of $ \beta$, let $ c \in F \smallsetminus \{ (\alpha \minus{} a_i)/(\beta \minus{} b_j)\}$, end $ \gamma \equal{} a \plus{} cb$. In your case, try if $ c \equal{} 1$ is good[/quote]

that's a really useful explanation, thanks :)


[quote="boria"]can you solve $2^{m}+3^{n}=5^{l}+2$ in positive integers ?[/quote]
Consider $m<3$. $m=0$ -contradition (left -even, right -odd)
$m=1$ $3^{n}=5^{l}$ - unique solution $m=1,n=0,l=0$.
$m=2$ $3^{n}+2=5^{l}$ give l,n - are odd. From $5*(5^{a})^{2}-3*(3^{b})^{2}=2$ we get unique solution $m=2,n=l=1$.
$m\ge 3$. $3^{n}=5^{l}+2(mod 8)$ - contradition.

thanks
one more question : how $5\cdot 25^{a}-3\cdot 9^{b}=2$ implies $a=b=0$ ?


We have two different masses.They are to be dragged on a rough floor .We do this in two ways -either by connecting the two bodies by a light inextensible thread or by connecting them with a light spring .In which case will the horizontal force  applied on one body  so that the whole system moves ,be minimum?

it's rather obvious that the force required is the same.
 this is easily deduced from analysing what happens step by step.

Err, well actually, if $\mu_{s}> \mu_{k}$, then you need less with the spring.

yup you're right. i ignored that point. i s'pose it's due to continuously solving problems where we're supposed to take $\mu_{s}= \mu_{k}$

then i think it is same?

Assume ${{\mu}_{s}}={{\mu}_{s}}$
At first glance we feel that the force required is the same.
But I feel we are ignoring the fact that when we pull [b]different[/b] masses connected by a spring ,part of the work done by the force goes into the elastic potential energy of the spring while nothing of this sort happens to the spring.

Using usual symbols we have,
Work energy theorem for the first mass
$F.x = \frac{kx^{2}}{2}+{\mu}Mg$
When the second mass is about to slip
$kx ={\mu}mg$

Hence
$F ={\mu}g(m/2+M)$

In case of a string the force is larger and is equal to ${\mu}g(m+M)$

I'm afraid you've forgotten the KE in the first equation, and that could lead to a non linear differential equation.
May be using Newton's second law and coming to an harmonic oscillator equation...?

Could you write your equations please?I have assumed that the body is moved slowly so that the change in kinetic energy is negligible.

The force is the same because in order to move the 2 bodies you have to exert a force equal to the friction forces.the answer stands even if friction coeficient is bigger when it's static,because we analyse the case when the bodies tend to move.elastic and tension force just equal the friction force on the second body(not the body where you apply the force).involving energy equations is useless and deviating from original answer.

ashwinkrishnan

To get the body moving slowly you need to adjust F, al the time, in order to have the resultant null or almost null. In that case F is a function of x and you can't write, any more, its work as F.x


geometry
function
symmetry
ratio
geometry proposed

In triangle $ ABC$, $ AB\equal{}AC$, the inscribed circle $ I$ touches $ BC, CA, AB$ at points $ D,E$ and $ F$ respectively. $ P$ is a point on arc $ EF$ opposite $ D$. Line $ BP$ intersects circle $ I$ at another point $ Q$, lines $ EP$, $ EQ$ meet line $ BC$ at $ M, N$ respectively. Prove that
(1) $ P, F, B, M$ concyclic
(2)$ \frac{EM}{EN} \equal{} \frac{BD}{BP}$

(P.S. Can anyone help me with using GeoGebra, the incircle function of the plugin doesn't work with my computer.)

hmm...seems to be evident as well.
since $ QFPD$ is a harmonic quadrilateral,it follows that $ MD \equal{} DN$.Therefore, $ P,F,N$ are collinear,it follows that $ \angle FBP \equal{} \angle PMF$ due to symmetry and a really simple angle chasing.
Recalling the property of the harmonic quadrilateral,$ QFBD$, we obtain that $ \boxed{BD \equal{} BN}$.
Simple angle chasing yields that $ \triangle BPN\sim\triangle EMN$,so combining it with the boxed quality we are finishing the proof.

There is clearly a typo with the ratio equality in Part 2 of the question, but I can't figure what.

Part 1 is correct though. Connect $ DE$ and $ DR$ and $ PF$. It remains to show that $ \angle EDF \equal{} \angle MBF$. But $ \Delta DBF$ and $ \Delta CDE$ are congruent and isoceles, so $ \angle EDF \equal{} 180 \minus{} \angle FDB \minus{} \angle EDC \equal{} 180 \minus{} 2 \angle FDB \equal{} \angle DBF \equal{} \angle MBF$.

Edit: Sorry, bad error by GSP. [hide]@Erken: Your solution is wrong. Firstly $ MD$ is not equal to $ DN$, neither are $ P$, $ F$ and $ N$ collinear.. similarly $ BD$ is not equal to $ BN$ as you claim.[/hide]

according to what KenHungKK wrote:$ P$ is a point on arc $ EK$ opposite $ D$, it seems to me that he meant to write that $ P$ and $ D$ are diametrical opposite points,otherwise,the part would not be correct...

It still fails even if $ P$ and $ D$ are diametrically opposite. Refer to the attached .png. Edit: Attached .png was an erroneous GSP sketch as well.

As the question asks us to prove $ \frac{EM}{EN} \equal{} \frac{BD}{BP}$, I believe that you'd better compute $ \frac{EM}{EN} \times \frac{BP}{BD}$ but not $ \frac{EM}{EN} \times \frac{BD}{BP}$ as in your diagram

@brian: In my diagram I had computed $ \frac{EM}{EN} \minus{} \frac{BD}{BP}$, the - is a tad bit small.

Sorry for the confusion, apparently my copy of GSP is messed up. In particular my computations indicated that $ QPFD$ was not harmonic unless $ \Delta ABC$ was equilateral.

as a conclusion:
so at long last my solution isn't completely correct, unless points $ P,D$ are diametrically opposite.let's just wait till [b]KenHungKK[/b] will clarify the statement.
on the other hand if they are not opposite, the statement of the second part does not hold as well, so I'm inclined to suspect that my solution is correct and "satisfies" to the true statement...

Actually a very similar problem appeared long time ago  :)

http://www.mathlinks.ro/viewtopic.php?p=1251781&search_id=1803384435#1251781

I have an elementary solution.

[b]Part 1 (For completeness):[/b] Connect $ DE$ and $ DR$ and $ PF$. It remains to show that $ \angle EDF \equal{} \angle MBF$. But $ \Delta DBF$ and $ \Delta CDE$ are congruent and isoceles, so $ \angle EDF \equal{} 180 \minus{} \angle FDB \minus{} \angle EDC \equal{} 180 \minus{} 2 \angle FDB \equal{} \angle DBF \equal{} \angle MBF$.

[b]Part 2:[/b] Let us note 2 results: (Bear in mind $ PFBM$ is concyclic.)

1. $ \Delta EMN \backsim \Delta FQP$. Firstly $ \angle MEN \equal{} \angle QFP$. Next, $ \angle MEC \equal{} \angle AEP \equal{} \angle EQP \equal{} \angle EDP$, by Alternate Segment Theorem. Then $ \angle EMN \equal{} \angle ECD \minus{} \angle MEC \equal{} \angle EDF \minus{} \angle MEC \equal{} \angle EDF \minus{} \angle EDP \equal{} \angle PDF \equal{} \angle PQF$.

2. $ \Delta FQB \backsim \Delta BFP$. This is because $ \angle FBP \equal{} \angle FBQ$ (trivially), and $ \angle BFQ \equal{} \angle BPF$ by Alternate Segment Theorem.

By 1, $ \frac {EM}{EN} \equal{} \frac {QF}{PF}$. By 2, $ \frac {QF}{PF} \equal{} \frac {BF}{BP}$. But $ \frac {BF}{BP} \equal{} \frac {BD}{BP}$, we are done.

An (accurate) diagram is attached for reference.

[i]For the record: the question KenHungKK posted is correct. We do not require $ P$ and $ D$ to be diametrically opposite.[/i]

for the firs part even faster: since ABC is isoceles $ EF\parallel{}BC$ and then $ \angle PFA\equal{}\angle PEF\equal{}\angle PMB$ and we are done


integration
logarithms
function
real analysis
advanced fields
advanced fields unsolved

KL divergence is defined as follows:

For two prob. measures $ P$ and $ Q$ on a measure space $ (\mathbb{X}, \mathcal{X})$,

\begin{eqnarray*} D(P\|Q) & = & \int \log\left(\frac {dP}{dQ}\right)dP \, \, \, \text{if} \, \, P\ll Q \\
& = & \infty \, \, \, \text{otherwise} \end{eqnarray*}

I want know whether the level sets. i.e. $ \{P: D(P\|Q)\le k\}$ compact w.r. to the total variation distance?

All we know about KL divergence $ D(P\|Q)$ is that it is convex in the pair $ (P,Q)$ and is a lower semi continuous function of $ P$ for every fixed $ Q$.

No. Let $ Q$ be the Lebesgue measure on $ [0,1]$. Then $ P$ is just a nonnegative function with some control on $ \int \log P$. You ask for compactness in $ L^1$ metric... that's too much to ask for. For example, the sequence of functions $ P_n\equal{}n\chi_{[0,1/n]}\plus{}1$ have uniformly bounded $ \int \log P$, but does not have a convergent subsequence.

[quote="mlok"]Then $ P$ is just a nonnegative function with some control on $ \int \log P$.[/quote]
1) Actually control on $ \int P \log P$, because $ D(P\|Q) \equal{} \int \frac {dP}{dQ} \log\left(\frac {dP}{dQ}\right)dQ$

2) $ P_{n} \equal{} n\chi_{[0,1/n]} \plus{} 1$ are not valid densities right?

3) Moreover I think the result seems to be true, for example see (in the second paragraph of proof of theorem2) [url=http://www-syscom.univ-mlv.fr/~vignat/Signal/CLT.pdf]here[/url].


The equation $x+\sqrt{x-2}=4$ has
\[
\textrm{(A)} 2 \textrm{ real roots}\qquad
\textrm{(B)} 1 \textrm{ real and 1 imaginary root}\qquad
\textrm{(C)} 2 \textrm{ imaginary roots}\qquad
\textrm{(D)} \textrm{no roots}\qquad
\textrm{(E)} 1 \textrm{ real root}
\]

[hide="To JesusFreak197"]
You made a small mistake in your arithmetic.
[quote]$(\sqrt{x-2})^2=(4-x)^2 \Rightarrow x-2=x^2-8x+16 \Rightarrow x^2-7x+18=0$[/quote]When you correct that and plug the values back in, you should get one real root and one imaginary root.
[/hide]
[hide="Another way to think about it..."]
For $x\ge2$, both $x$ and $\sqt{x-2}$ are strictly increasing, so the graph $x+\sqrt{x-2}$ will pass through $f(x)=4$ only once since $\sqrt{x-2}$ is undefined for $x<2$.
[/hide]

[quote="JesusFreak197"][hide="Answer"]$x+\sqrt{x-2}=4 \Rightarrow \sqrt{x-2}=4-x$
$(\sqrt{x-2})^2=(4-x)^2 \Rightarrow x-2=x^2-8x+16 \Rightarrow x^2-9x+18=0$
$\frac{9\pm \sqrt{81-72}}{2}=\frac{9\pm 3}{2}=x^2$
$x=\sqrt{3}$, $x=\sqrt{6}$

Therefore, all roots of the equation are real, and the answer is $A$.[/hide]

[b]EDIT:[/b] Arithmetic error fixed. But... now I get two real roots; did I still go wrong somewhere?[/quote]

$\sqrt{3} < 2$ so it is not a valid answer.

Ohh, okay. So that would be the imaginary root.

When I do it I get:
$x+\sqrt{x-2}=4$
$\sqrt{x-2}=4-x$
$x-2=x^2-8x+16$
$x^2-9x+18=0$
$(x-6)(x-3)=0$
$x=6$ $x=3$
but $6$ does not work when you plug it back in

So I think it is $E)$

Squaring is an 'irreversible' operation, so sometimes you get imaginary roots which works with the transformed equation but not the original, which is the case here. Here is another example:

$ \displaystyle x+2=-5x+2 \Rightarrow x^2+4x+4=25x^2-20x+4 \Rightarrow 0=24x^2-24x $
$ x=1, x=0 $

But 1 does not fit into the original equation, so it is an imaginary root, whereas 0 is the real root.

[quote="towersfreak2006"]Just to clarify, when I earlier said:

[quote="towersfreak2006"][hide="Another way to think about it..."]
For $x\ge2$, both $x$ and $\sqt{x-2}$ are strictly increasing, so the graph $x+\sqrt{x-2}$ will pass through $f(x)=4$ only once since $\sqrt{x-2}$ is undefined for $x<2$.
[/hide][/quote]
This approach confirms the existence of only one real root (which is not a double root). But since the function has degree two, it must have another root, which turns out to be somewhere on the complex plane.

Consider this, nat mc.
How many solutions are there to $x=1$? Square both sides, $x^2=1$. Are both values that satisfy this second equation roots of the first one?
Also, if you take the negative square root of $\sqrt{x-2}$ for $x=6$, then it does actually work. But since square roots are defined to be positive, squaring accounts for the extra negative value.[/quote]
Thats why I took the 6 out, isn't it...  You can only count 3 because 6 deals with a negative square root, right...  So isn't 3 the only answer?

[quote="JesusFreak197"][hide="Answer"]$x+\sqrt{x-2}=4 \Rightarrow \sqrt{x-2}=4-x$
$(\sqrt{x-2})^2=(4-x)^2 \Rightarrow x-2=x^2-8x+16 \Rightarrow x^2-9x+18=0$
$\frac{9\pm \sqrt{81-72}}{2}=\frac{9\pm 3}{2}=x^2$
$x=\sqrt{3}$, $x=\sqrt{6}$

Therefore, the equation has one real and one imaginary root, so the answer is $B$.[/hide]

[b]EDIT:[/b] Okay, everything fixed now.[/quote]
Just a quick note. You made 2 mistakes(which haven't been pointed out). You set your quadratic equation equal to $\displaystyle\ x^2$ rather than $\displaystyle\ x$.
Another thing. Whenever you have an equation that has an imaginary root, generally speaking its conjugate is one of the other roots if the equation has rational coefficients.  I think its $\displaystyle\ x$=3 and therefore E because 6 is an extraneous root.

whoa that was weird
[quote="shobber"]$f(x)=x+\sqrt{x-2}$ is an monotone increasing function, so $3$ is the only real root.[/quote]
thank you.  Thats what I thought but I got unsure when they said my solution migt not be usable

wow Pakman2012, you better fix your post, [b]\x[/b]? you dont need backslash for x. and answer should be E because square roots are defined to be positive, the other one is called an [b]extraneous root[/b] and is not part of the answer.

[quote="kunny"]Rewrite the given equation as $\sqrt{x-2}=-x+4$, sketching two graphs $y=\sqrt{x-2}$ and $y=-x+4$,
we can find only one intersection point, yielding the answer $(E)$.

kunny[/quote]

But one intersection point does not neccesarily imply one root. (because complex root will not be shown as an intersection on the graph in the real plane)

[quote="IntrepidMath"][hide]
$\sqrt{x-2}=(4-x)$
$Squaring\ldots$
$x-2=16-8x+x^2$
$Rearranging..$
$x^2-9x+18=0$
$(x-3)^2=0$
$x=3$

The answer is A, because there is a root 3, of multipicity 2.

[/hide][/quote]
IntrepidMath??? :what?:  please TRIPLE check your answer, especially the part near the end of your solution

IntrepidMath:

The bottom part should read $x=(9\pm\sqrt{5})/2$

No, it's $\frac{9\pm \sqrt{81-72}}{2}=\frac{9\pm \sqrt{9}}{2} \Rightarrow x=3$, $x=6$, which is what the rest of us have gotten.

I have no idea what happened with that post..

Here are a couple of MAO questions that I thought would be suitable for a topic like this to test your equation abilities.

1) Solve for $x$ in the following equation: $4x^2+9=0$, where $x\in\Re$

2) How many rational roots are in the equation: $x^4-x^3-5x^2+3x+6=0$

3) Let A= the highest number of possible positive real roots of $f(x)$ and Let B= the highest number of possible negative real roots of $f(x)$, according to Descartes' Rule of Signs. Find $(A+B)^3$ for the following $f(x)$:

$f(x)=x^{2005}+1966x^{2002}-819x^{2001}-1122x^{1998}+1025x^{1991}-227x^{1980}-725x^{1966}+1$

(Note: I believe they wish for you to assume that anytime there is a 0 it wil have an addition sign in front of it...otherwise this question is very disputable)

1: No solution. Since x is imaginary and $x \in R$.

You're correct heh. The question has in slightly smaller font solve for x over the reals and its just annoying lol.

That is one root, but there's likely more than one.

[quote="shadysaysurspammed"]i dont think i'm right,but i'm getting x=2 in question 2 :huh:[/quote]

If Im not mistaken there are only $2$ roots of the equation $x+1$ and $x-2$
because after dividing it by those it becomes $x^2-3$ which gives one imaginary and one irrational root.

oh yeahhh...got my calc wronghehe :blush:

[quote="JesusFreak197"][quote="riddler"]If Im not mistaken there are only $2$ roots of the equation $x+1$ and $x-2$
because after dividing it by those it becomes $x^2-3$ which gives one imaginary and one irrational root.[/quote]

No, it gives two irrational roots...

$x^2-3=(x+\sqrt{3})(x-\sqrt{3}) \Rightarrow x=\pm \sqrt{3}$[/quote]
It gives 2 rational and 2 irrational roots
$x=-1$ and $x=2$ and $x=+-sqrt(3)$

I know. You said that $x^2-3=0$ gave one irrational and one imaginary root, which is incorrect.


not much of suggests, but a few requests.
SENCE OF CLOSURE.----when you encounter a hard problem, how long
can you keep plodding? For lower level members, It's helpful to post a full solution after, say, two to three weeks. Some people do need a sense of closure and move on.For those who want to keep the problems open, they can simply not look at the answers.
WHEN YOU WRITE SOLUTION, PLEASE LOWER DOWN YOUR LEVEL.--give a good exposition and don't poo-poo wordiness.Many lower level members are trying to learn from you. If  one continues to keep elitism in the higher math, can you blame most of people do not want to learn math?

A lot of solutions require higher mathematics, but I do think it's a good idea to avoid higher math if possible or at least give a (very vague if necessary) general idea in simpler terms.  Many people here do not know higher math (e.g., me) but still visit the Advanced forum.  As for the sense of closure, I think you can just ask for a solution (PMed if they're really reluctant).

I agree with probability1.01 -- ask people for nice, clean, lower math solutions and you will sometimes get what you need, though sometimes people might not have the time to help.  As this community grows there will be more students capable of helping with such solutions and who also have the time.

More advance math students might say something you consider elitist, but the easiest thing is to ignore that part and soak up what they have to offer.  When students criticize verbose solutions, they are sometimes making a good point, though they might not realize that the verbose vs. elegant solution is exactly what you're interested in since the elegant solution uses the "higher math" that you don't yet know.  If you explain this you are likely to get help from some of the more dedicated problem solvers who often know more than one way to write a particular proof.


Euler
function
number theory
number theory solved

Given:2003 2002 2001

Which are the last 3 digits of this number?

(its supposed to be 2003 at 2002 at 2001)

Thx Belenos!

Yes,it should be posted in the number theory section...(i havent slept for 3 days and I'm a bit confused.Also i do hallucinated a bit lately  :D )


AMC
AIME
MATHCOUNTS
USA(J)MO
USAMO
AMC 12
\/closed
noo it locked

I would like to see an AMC math jam...regularly...I need the practice...being in KY I never did get practice for it and only got a 98.5 on the AMC 12...I guess I shoulda taken the 10 but oh well...(I'm a sophomore)...will there be one?

Yup - there will be plenty of AMC Math Jams of all levels (AMC 8/10/12, AIME, USAMO).  The Mandelbrot Math Jam is roughly similar in difficulty and question style, so you'll probably want to check that out, too.

i would most like to see a mathcounts math jam on more of a national level.  the amc thing would be nice too. but keep it up. this site rocks!

ya AMC jam would be sweet. i screwed up on the amc 10... got something like a 118... shoulda guessed on 2 probs... silly me, and other 8th graders...
o yeah by the way, how much harder is the AMC 12 than the AMC 10?

I would like to see a few AIME problems on the site too.  I had an AMC 12 score of 123 and had a shot at making USAMO before I messed the AIME and only got a 4.

It's hard to make it with a 123 on AMC12. You need 11 on AIME, which means more than anything extreme carefulness. I miss quite a few AIME questions in each year due to carelessness.

So how do they choose the people for USAMO?  I know they have an index score system of AMC score plus 10*AIME score.  It seemed that if you had an index score of above like 200 there was no way you couldn't make USAMO especially if you were in like 9th grade or below. i'd really like to know an ideal score for making USAMO.

This year the way it worked was that for 11th and 12th grade students, an index (AMC12+10*AIME) score of 226 was the lowest for qualification for the USAMO. For students in 10th grade and below, a score of 8 or more on the AIME was enough to qualify for the USAMO regardless of AMC10 or AMC12 score.

ComplexZeta,

Hmm, I am a tad bit confused here. On the AMC site at http://www.unl.edu/amc/e-exams/e8-usamo/usamo.html, there is a statement saying that "To adjust for variations in contest difficulty, the number of students selected from A & B contests will be proportional to the number of students who took the (A & B) Contests." Is this actually true, or is there just one AIME cutoff score for both ppl who take any AMC/AIME A or B combination? I personally took the AIME B, and the cutoff score was, as you said, 226. Thanks.

Fierytycoon

The way they determine USAMO qualification has little resemblance to what they write on the website. The process I described was exactly the process used this year for both AMC A and B as well as AIME A and B.

"The way they determine USAMO qualification has little resemblance to what they write on the website."
 :lol:  :lol:  :lol:

Thanks for the info. Is this the way they have been doing it before this year as well? That is, I am interested whether the policy changes a lot, and so following, whether the policy may change for next year.

Fierytycoon

They did it that way in 2002 and 2003. In 2001 they did something very strange that I still don't understand, but I'm not complaining because it let me qualify. I'm not sure what they did in 2000 or before.

If you live in a small state, you don't need a score of 8 to go on.  I got a 7, but because I live in a small state that sends the best score that is gotten in it, so I got to take the USAMO, not that I'm complaining.

Thanks for the info.

Since we happen to be on the topic of USAMO qualification, I was just wondering what everyone here thought of the fact that there are two different AIME's, the A and the B. Over at the Mathcounts Upper Deck forum (which has been considerably less active due to the moderated posting), I have heard many people complaining how the AIME A is usually easier than the AIME B, and how this could be the cause of quite some unfairness, since each student can only take one of the AIME's. I mean, just 1 or 2 problems that are easier could make or break a lot of students who are right on the borderline. Personally, my school is on vacation during the week of the AIME A, so we always take the AIME B, and since I don't happen to have copies of all the recent AIME's, both A and B, I cannot compare their difficulty levels myself. What do you all think?

Fierytycoon

I don't think there's really much in it in general. I saw both contests this year, and I thought maybe the B was a bit easier, but that might just be my preference for some certain types of problems over others. I took the A every year.


Hello everyone. I am also here in Australia so I thought I might as well join the Australian Forum. It would be nice to know what school everyone goes to and what year they are in. So I might as well start.

I go to Killara High School and I am in Year 11.

good day mate.

i'm an wannabe australian too

and i miss steve irwin too.


Hi this question appeared in my country's last year's olympiad round 1 (only answer, no solution required).  However if someone knows how to do it, I would definitely appreciate a solution too.  Thanks.

For each positive integer $ n$, let $ a_{n}$ denote the number of $ n$-digit integers formed by some or all of the digits 0, 1, 2, and 3 which contain neither a block of 12 nor a block of 21.  Evaluate $ a_{9}$.

It's unclear whether leading 0s are allowed -- I'm assuming they are, but if they aren't it just changes the initial conditions.

Let $ b_n$ be the number of such strings ending in 0 or 3 and $ c_n$ be the number ending in 1 or 2.  Then $ a_n \equal{} b_n \plus{} c_n$, $ b_n \equal{} 2a_{n \minus{} 1}$ and $ c_n \equal{} 2b_{n \minus{} 1} \plus{} c_{n \minus{} 1}$, with initial values $ b_1 \equal{} c_1 \equal{} 2, a_1 \equal{} 4$.  9 is small enough that probably the best this is to just do the multiplications and additions to get $ a_9 \equal{} 102010$.