Let a tangential quadrangle $ ABCD$; $ E,F$ in $ BC$; $ AE\cap DF \equal{} G$; $ H,I,J,K$ are incenters of triangles $ ABE,EFG,CDF,DAG$. Prove that: $ HIJK$ is a cyclic quadrangle.

I also see that: If $ AE\cap HK \equal{} M,DF\cap JK \equal{} N$ then $ AMND$ is a tangential quadrangle, but I can prove it. Anybody can?

Two non-intersecting directed circles (with no common interior points) have only 2 (directed) common tangents. For example, if both circles are directed counter-clockwise, their directed common tangents are the common external tangents of undirected circles. If one circle is directed counter-clockwise and the other one clockwise, their directed common tangents are the common internal tangents of undirected circles. Two intersecting directed circles have common tangents only if they have the same direction.

[color=red]Theorem: [/color] Let $ \mathcal A, \mathcal B, \mathcal C, \mathcal D$ be four directed circles. Let $ a, a'$ be the directed common tangents of $ \mathcal A, \mathcal B,$ let $ b, b'$ be the directed common tangents of $ \mathcal B, \mathcal C,$ let $ c, c'$ be the directed common tangents of $ \mathcal C, \mathcal D,$ and let $ d, d'$ be the directed common tangents of $ \mathcal D, \mathcal A.$ Suppose the directed lines $ a, b, c, d$ are tangent to a single directed circle $ \mathcal I.$ Then the directed lines $ a', b', c', d'$ are also tangent to a single directed circle $ \mathcal I'.$

[hide="See the well known dual theorem."] Let $ \mathcal X, \mathcal Y, \mathcal Z, \mathcal T$ be four (undirected) circles. Let $ X, X'$ be intersections of $ \mathcal X, \mathcal Y,$ let $ Y, Y'$ be intersections of $ \mathcal Y, \mathcal Z,$ let $ Z, Z'$ be intersections of $ \mathcal Z, \mathcal T,$ and let $ T, T'$ be intersections of $ \mathcal T, \mathcal X.$ Suppose the points $ X, Y, Z, T$ are concyclic. Then the points $ X', Y', Z', T'$ are also concyclic.[/hide]

The proof is a bit tedious without reference to a figure. So, assume $ a, b, c, d$ form a convex quadrilateral $ ABCD,$ where $ A \equiv d \cap a,$ $ B \equiv a \cap b,$ $ C \equiv b \cap c,$ $ D \equiv c \cap d,$ so that $ a \equiv \overline{AB},$ $ b \equiv \overline{BC},$ $ c \equiv \overline{CD},$ $ d \equiv \overline{DA}$ and the circles $ \mathcal A, \mathcal B, \mathcal C, \mathcal D$ are all inside this quadrilateral and directed counter-clockwise, same as the directed circle $ \mathcal I.$ Denote intersections of the other directed common tangents $ A' \equiv d' \cap a',$ $ B' \equiv a' \cap b',$ $ C' \equiv b' \cap c',$ $ D' \equiv c' \cap d',$ so that $ a' \equiv \overline{B'A'},$ $ b' \equiv \overline{C'B'},$ $ c' \equiv \overline{D'C'},$ $ d \equiv \overline{A'D'}.$  Let $ \mathcal A$ be tangent to $ d, a$ at $ P_1, P_2$ and to $ d', a'$ at $ P_1', P_2'.$ Similarly, let $ \mathcal B$ be tangent to $ a, b$ at $ Q_1, Q_2$ and to $ a', b'$ at $ Q_1', Q_2'.$ Cyclically, define the tangency points $ R_1, R_2, R_1', R_2'$ of $ \mathcal C$ and $ S_1, S_2, S_1', S_2'$ of $ \mathcal D.$

Suppose that $ ABCD$ is tangential, i.e., the directed lines $ a, b, c, d$ touch a single directed circle $ \mathcal I.$ Then

$ \overline{AB} \plus{} \overline{CD} \equal{} \overline{BC} \plus{} \overline{DA}.$

Subtracting $ \overline{P_1A} \equal{} \overline{AP_2}, \overline{Q_1B} \equal{} \overline{BQ_2},$ etc., it follows that

$ \overline{P_2Q_1} \plus{} \overline{R_2S_1} \equal{} \overline{Q_2R_1} \plus{} \overline{S_2P_1}.$

These are tangent lengths between tangency points, therefore

$ \overline{Q_1'P_2'} \plus{} \overline{S_1'R_2'} \equal{} \overline{R_1'Q_2'} \plus{} \overline{P_1'S_2'}.$

Subtracting $ \overline{A'P_1'} \equal{} \overline{P_2'A'}, \overline{B'Q_1'} \equal{} \overline{Q_2'B'},$ etc., yields

$ \overline{B'A'} \plus{} \overline{D'C'} \equal{} \overline{C'B'} \plus{} \overline{A'D'},$

so that $ A'B'C'D'$ is also tangential, directed lines $ a', b', c', d'$ are tangent to a single directed circle. QED.

As to the posted problem: The circles $ \mathcal D, \mathcal A$ are of zero radius, degenerated to points $ D, A.$ Their common tangents $ d \equiv \overline{DA}, d' \equiv \overline {AD}$ coincide in position, but they still have opposite directions. The incircles $ (H), (J)$ of $ \triangle ABE, \triangle CDF$ are the circles $ \mathcal B, \mathcal C.$ Let $ b'$ be the common external tangent of the incircles $ (H), (J)$ other that $ b \equiv \overline{BC}.$ Let this tangent intersect $ \overline{EA}, \overline{DF}$ at $ M, N.$ According to the theorem, quadrilateral $ AMND$ formed by the lines $ d' \equiv \overline{AD}, a' \equiv \overline{EA}, b' \equiv \overline{NM}, c' \equiv \overline{DF}$ is tangential. Its incircle is the incircle $ (K)$ of the $ \triangle DAG.$

Consider now the general quadrilateral $ MEFN.$ (If $ G$ is inside of $ ABCD,$ $ MEFN$ is reflex.) Let $ k_m, k_e, k_f, k_n$ be the external bisectors of its angles at $ M, E, F, N.$ The incenters of $ \triangle ABE, \triangle EFG, \triangle CDF, \triangle DAG$ are intersections $ H \equiv k_e \cap k_m,$ $ I \equiv k_e \cap k_f,$ $ J \equiv k_f \cap k_n,$ $ K \equiv k_n \cap k_m.$ It follows that $ HIJK$ is cyclic. (This is well known and a simple angle chase.)


My organization wants to make the US more immigrant friendly; so, we are asking you guys/girls what you think should be done to make the US a better place for Korean nationals?

hmmm, i can think of one really good way
IMPEACH BUSH GD.


Maybe It is silly math question, But I really want to know
If I have cos(x)=1/5 or cos(x)=2/3
Is it possible to find x ?

Yes. If you want to do it on a calculator, use the $sin^{-1}$ function.

Well, in your case, it'd rather be $cos^{-1}$ ;)

When using inverse functions on a calculator, keep in mind that the inverse trig functions have limited range, so if you were looking for a really big angle that's well out of the inverse trig function's range, it's not going to give you the right angle.

[quote="invisal"]so $cos^{-1}$ is same as arccos ?[/quote]
Yes, you can also type [url=http://www.google.com/search?sourceid=navclient&ie=UTF-8&rls=HPIB,HPIB:2005-19,HPIB:en&q=arccos%281%2F3%29]This in google.[/url] (note the answer would be given in radians.


Recently I read TPOTO and I found it quite nice but the movie is a bit different. That's usual of course but I think that the unique characteristic is that both the book and the movie are... :!: ...brilliant and  marvelus. The plots are also different but they are both unique. The movie is not a copy or a copy that  a few pieces are changed or cut but an evolution of the book imo. What do you think?

I didn't read the book but I happened to see the movie (which I didn't like thaaaaaaaat much).
However, when I was in New York last year, I happened to see the show on Broadway (my professor there told me it has been playing on Broadway forever) and it was really good. When I return to France next year, I'll try to go watch it in Paris.

Haven't read the book or seen the movie, but we played the song in band and IT IS AWESOME!!!!!!!! :lol:

YAY I'M NOT THE ONLY PHANTOM OF THE OPERA GEEK! :lol:

I personally prefer the movie, but that's just me. Both are very good. The book makes you feel more for Raoul's point of view. The movie... well, not only does it make you almost hate Raoul, but it really lets you see the human side of the Phantom. Plus, with Gerry Butler as the Phantom... there's just no comparison!

What do you think of Emmy Rossum (Christine Daae) in the movie? Some people say her voice was not operatic enough for the role, while others (me included) like her voice because it conveys a gentle, innocent persona.


It would be the latter.

Always express in terms of a real number times the imaginary unit.

Yes, another way to see it is if $\sqrt{x}^2, x|x<0$

Then $\sqrt{x}^2 = \sqrt{x}\sqrt{x} = -\sqrt{x^2}$

the convension is that you cannot do that with negative numbers, i.e. $\sqrt{a}*\sqrt{b} \ne \sqrt{ab}$ $\forall a,b<0$, the reason that we run into a problem is that with square roots we always take the positive one, but with negative roots, it is not real there are 2 roots, and you cannot really say that one is larger than the other, so you have to convert them into positive numbers before you can do the radical thing

Actually, the point is this:

Every complex number (and real numbers are just the special case of complex numbers) have TWO values of the square root. In exponential form, we have $-2=2e^{i(\pi+2n\pi)}$, hence $\sqrt{-2}$ can be either $\sqrt{2}e^{i\pi/2}$ or $\sqrt{2}e^{-i\pi/2}$.

If you multiply SAME values of the root (which is understood as the proper way of squaring), you'll get $2e^{i\pi}=-2$, and if you multiply DIFFERENT values (which is possible, but then it's not squaring), you'll get $2e^{i0}=2.$

Hence, in the "identity" $\sqrt{-2}^2=\underline{\sqrt{-2}\cdot\sqrt{-2}}=\sqrt{(-2)\cdot(-2)}=2$, the two underlined roots are actually different numbers, though written in the same way, and that's similar to the "identity" $\sqrt{4}^2=\sqrt{4}\cdot\sqrt{4}=2\cdot(-2)=-4.$


integration
calculus
trigonometry
calculus computations

I know it's easy question but can you tell me how do you compute $ \int xdx\plus{}ydy\plus{}zdz$ when $ x\equal{}acost, y\equal{}a sint, z\equal{}t$ i don't know the definition of double integral of type 2 but i guess it's not just puting $ dx\equal{}acost(\minus{}sint)$ and so on and all becoming $ \int t^2/2$ doesn't it??

$ x \equal{} a\cos t$ so $ dx \equal{} \minus{}a\sin t \,dt$,
$ y \equal{} a\sin t$ so $ dy \equal{} a\cos t \,dt$,
and $ z \equal{} t$ so $ dz \equal{} dt$

Therefore $ \int_{\sigma(t)} xdx \plus{} ydy \plus{} zdz$
$ \equal{} \int_{t_1}^{t_2} (a\cos t)(\minus{}a\sin t \,dt) \plus{} (a\sin t)(a\cos t \,dt) \plus{} (t)(dt)$
$ \equal{} \int_{t_1}^{t_2} t \,dt$


let $0\leq a_{1}\leq b_{1}$ and sequence $(a_{n})$and$(b_{n})$ defined by $a_{n+1}=\frac{a_{n}+b_{n}}{2},,b_{n+1}=\sqrt{a_{n}.b_{n}}$
Find what limit of $a_{n}$ and$b_{n}$ when n---> $\inf$

That's the arithmetic-geometric mean, see some web resources for it.

Here's a reference:
[url]http://mathworld.wolfram.com/Arithmetic-GeometricMean.html[/url]


induction
number theory proposed
number theory

Prove among $k+m-1$ arbitraty integer number  $m \geq k \geq 2$ and $m$ divisible $k$ then exist $m$ numbers have sum divisible $k$.
Goodluck

Let $m=nk$. We perform induction on $n$. For $n=1$ it's famous theorem of Erdos (look for example here http://www.mathlinks.ro/Forum/viewtopic.php?highlight=combinatorial&t=19496 at Vesselin's post - it is corollary 2). Suposse we have prove it for $n$ -then let's prove it for $n+1$. Due to induction hypothesis from the set of cardinality $(n+1)k-1$ we can take $nk$ numbers with sum divisible by $k$. So also we can take $nk$ numbers with sum divisible by $k$ from the set of cardinality $(n+2)k-1$ and we are left with $2k-1$. But from this set we can take $k$ numbers with sum divisible by $k$ according to base of induction. Hence, all in all, we have chosen  $(n+1)k$ with sum divisible by $k$. QED


Hi everyone.
I want to practice to draw a circle with hand, without any help from any tools. It is very difficluts to draw a complete and smooth one. So do you guys have any hint or idea to help me to do that? :?
Thanks in advance.

[quote="tokenadult"]Try putting one of your fingers (maybe your thumb, maybe some other finger) down on the paper like a compass point, hold your pencil comfortably, and then rotate the piece of paper with your other hand. (I'm just thinking this up; I haven't tried the experiment.)

Does anyone else have a suggestion?[/quote]

I tried that and got more of an oval shape but i guess it sorta works.

Whoa. I tried tokenadult's idea and if you do it carefully, it works almost perfectly!
The only problem I see is if your workspace is fairly limited. That will result in the paper constantly bumping into other things and jiggling and messing things up. And it only works for fairly large circles due to the size of your fingers. Still, it's pretty cool.

if you hold your arm about the length of your fingers above the paper and rotate your wrist, some sort of circle is drawn... you need to be more careful with this method if you want a good circle though

[quote="shobber"]Maybe it is a kind of practise and experience and feel...My physics teacher can draw a perfect circle without any help on the blackborad. He just stares at the blackboard and then raise his hand and ....complete![/quote]

Maybe he's a robot... or a jedi! :starwars:

i refer you to paladin's comical graphic

[quote="shobber"]what is Jedi...?[/quote]
 :rotfl:  :rotfl:  :rotfl: shobber must watch Star Wars movies!!!

Replying to Shobber's original question, I think the key to drawing an accurate circle on a blackboard is to THINK about an ideal circle before starting to move the chalk. Don't look at what your hand is doing; just see in your mind a picture of a perfect circle, and what your hand will do will look better than if you think too much like "Now I have to move here, now I have to move here," etc.

To answer Shobber's second question, a jedi is a fictional outer-space "knight" that strongly influences the story action in a movie in the Star Wars series. No one is really a jedi, but some actors portray jedi in movies.

[quote="shobber"]what is Jedi...?[/quote]
haven't you ever seen star wars???
and also this topic is really random.
i think that if you practice drawing cricles you will get better at it.  Draw 50 a day and you will be good at it. ;)

[quote="shobber"]Maybe it is a kind of practise and experience and feel...My physics teacher can draw a perfect circle without any help on the blackborad. He just stares at the blackboard and then raise his hand and ....complete![/quote]

He suffers from a rare disease that we in America call "being artistic". A circle is like any form of art, you can get better with practice but you will reach a limit. Some people can naturally draw a circle, that's just how it is.

[quote="shobber"]Thanks for all your help!  :D
I shall follow your suggestions. First: Use my mind as tokenadult suggested. Second: Practice makes progress as math92 suggested.
I hope I shall be able to draw at least smooth circle after practice.[/quote]
Yes, shobber use mind! Be Jedi and use [u]force[/u] to guide drawing technique :D


Are there $n\in\mathbb{N}$ and $p\geq99$ a prime number so that $n^{10}+p^{n+1}$ is a perfect square?


Find the integer part of

$\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}$


Are someone know the results of brazilian team?

Well, after many hours with no news, we finally have confirmation of 2 silver and 4 bronze medals for Brazil. Were very happy with this result! Congrats to all!

and dont oyu know anything about Argentina? Is there any gold medal from latinamerica

ok.......congratulations for your performance........ on which position did you finish?

PD: Do you know Davi (imo 2001/2002/2003) because i lost his email.....

I still dont have new information, except that Brazil made 132 points total, ending up in 21st position, together with Canada. Argentina made 92 points, 39th position.
(see the other post in this forum with the complete country list).

P.S. Conozco a Davi, te envo un mensaje privado a respecto.

BRA 1 - Fbio Moreira -  Bronze ( 6 + 7 + 1 + 5 + 1 + 1= 21)
[b]BRA 2 - Gabriel Bujokas - Silver (3 + 6 + 1 + 7 + 5 + 7 = 29) [/b]
BRA 3 - Henry Hsu - Bronze (6 + 1 + 1 + 7 + 3 + 1 = 19)
[b]BRA 4 - Rafael Hirama - Silver (6 + 2 + 2 + 7 + 7 + 2 = 26) [/b]
BRA 5 - Rafael Marini - Bronze (6 + 1 + 1 + 1 + 0 + 7 = 16)
BRA 6 - Thiago Santos - Bronze (7 + 3 + 3 + 6 + 1 + 1 = 21)

[quote="luishhh"]Hi all!
I'am Luis Hernandez, spanish imo contestant in 01, 02 , 03.
I've got some news about the argentinian team. I met spanish team this morning and they told me yue yang (arg) got a GOLD! 32, just over the cutoff, exactly as Carlos did last year didn't he? I hope a gold for iberoamerica wll become a tradition into the future imos. Maybe one for spain in 2008, when we host the imo!!
Luis[/quote] wow, you will host the imo in 2008? I didn't knew that ... it's  great news :)

Hi Luis... Im Carlos, as far as i know Yue got a better gold than mine...... he got 34.... 751777....
 (es patetico tener que responder en ingles....... como andas?)


vector
combinatorics proposed
combinatorics

Suppose $V= \mathbb{Z}_2^n$ and for a vector $x=(x_1,..x_n)$ in $V$ and permutation $\sigma$.We have $x_{\sigma}=(x_{\sigma(1)},...,x_{\sigma(n)})$
Suppose $ n=4k+2,4k+3$ and $f:V \to V$ is injective and if $x$ and $y$ differ in more than $n/2$ places then $f(x)$ and $f(y)$ differ in more than $n/2$ places.
Prove there exist permutaion $\sigma$ and vector $v$ that $f(x)=x_{\sigma}+v$

First you should write the other two parts of the question!
1) If $D(x,y)$ be the number of differences between $x$ and $y\in V$ then the number of $z\in V$ such that $D(z,x)>{n\over 2}$ and $D(z,y)>{n\over 2}$ only depends on $D(x,y)$.
Proof is easy!Only you should change the members of vectors.
2) Suppose the number of shuch $z$'s in (1) is $l_k$.($k=D(x,y)$) Then
$l_1 =l_2 >l_3 =l_4 >l_5 =l_6>\ldots$
Proof of this one is also easy.By only transforming each $z$ into another $z$!
Now that we know these we come to prove the main problem.
Because $n=4k+2$ or $n=4k+3$ , $[{n\over 2}]=2k+1$ so $[{n\over 2}]+1=2k$.
Now suppose that $D(x,y)=k$. The number of $z$'s that are good (like the part 2) to $x$ and $y$ are $l_k$. Suppose $D(f(x),f(y))=r$. We have that $f(z)$ is also good to $f(x)$ and $f(y)$ and also the inverse. By part 2 we have $|r-k|<2$.
Now suppose that $D(x,y)=[{n\over 2}]+1$. Because it's an even number then $D(f(x),f(y))=[{n\over 2}]$ or $[{n\over 2}]+1$. But its more than $n\over 2$. So it is exactly $[{n\over 2}]+1$. So $D(f(x),f(y))=D(x,y)$
By the same proof we have if $D(x,y)=[{n\over 2}]$ then $D(x,y)=D(f(x),f(y))$.
Now suppouse $x\in V$ and $i\in \{1,\ldots,n\}$.There exist a $y\in V$ such that $D(x,y)=[{n\over2}]+1$ and and differs with $x$ in the $i$'th member.
Also there exist a $y'$ such that it only differs with $y$ in $i$'th member.
We have $D(x,y)=D(f(x),f(y))$ and $D(x,y')=D(f(x),f(y'))$.
Now suppose that $I$ is the set of indexes that $f(x)$ and $f(y)$ differ in.
Construct $I'$ like the $I$ for $f(x)$ and $f(y')$.
We want to prove that $I'\subset I$. If it is not then it is easily seen that $I$ and $I'$ differ in more than 2 members. But $D(y,y')=1$ so $D(f(y),f(y'))<3$.
So $I\subset I'$.
By changing $x$ we can make the $y$ to become all the members of $V$.
So for every $y\in V$ changing one member causes one change to $f(y)$.
Suppose that $x\in V$. And suppose that changing the $i$'th member causes the $g(i)$'th member of $f(x)$ to change. We have $g$ is a permutation because $f$ is injective.
We shall now prove that $g$ is uniqe for all $x\in V$.
Suppose that it is not.
So we have $x,y\in V$ and $i\in \{1,\ldots,n\}$ such that a change to $x$ in $i$ causes the $j$'th member of $f(x)$ to change, and a change to $y$ causes the $k$'th member.
Suppose that the changed $x$ is $x'$ and the changed $y$ is $y'$.
$D(x,y)=D(x',y')=t$
$|D(f(x'),f(y'))-t|<2$
But |D(f(x'),f(y'))-t| is even because we have two changes in $f(x)$ and $f(y)$ to obtain $f(x')$ and $f(y')$.
So $D(f(x'),f(y'))=t$
So $D(f(x'),f(y))-D(f(x),f(y))=1$ xor $D(f(x),f(y'))-D(f(x),f(y))=1$ and the other one is $-1$.
Suppose the first one is $-1$.
Then it is easily seen that $D(f(x),f(y))-D(x,y)=1$.(Because in other ways we have $|D(f(x'),f(y))-D(x',y)|>1$)
But from the $D(f(x),f(y'))-D(f(x),f(y))=1$ we get $D(x,y)=D(f(x),f(y))$.(Like the last part)
So $g$ is uniqe for all $x\in V$.
Now because that we can change every $x$ to every $y$ by changing members, we have that the problem is proved!

Of course Nima you got the best mark from this question.Well done. ;)  ;)


5'11" and 135

however i can bench press my weight
and dead lift like 225

not something many underweight people can do

[quote="The Herald of Doom"]I'm just curious.

i'm pretty puny for an 8th grader....  :([/quote]

Arnav, how many grades did you fail?!

six
point nine

going on seven
god knows

i'm pretty sure he's referring to weight and stuff (unless you're a girl, then you can read between his lines)

[quote="pianoforte"]size of what? there are many sizes you could give[/quote]

IT IS HOW BIG YOU ARE DUH

I don't think The Herald of Doom is ArnAv.

He might be an even bigger spammer than Arnav. :rotfl:

6'4'' 190lbs squat 450 lbs but I can only bench press 185lbs twice.

[quote="solafidefarms"]No, I'm pretty sure the moderators agree that HoD is Arnav.

In any case pretty soon his rating will be less than Treething's :|...[/quote]

Is this because of IP address, or just behavior and stuff?

guys guys

The Herald of Doom is not ArnAv.


summer program
PROMYS
geometry
AMC
USA(J)MO
USAMO
inequalities

I have a hard time to choose between PROMYS or AoPS Olympiad classes. Right now I solved/wrote up solutions to every PROMYS problems. PROMYS is a fun camp and I want to meet some good math people out there. However PROMYS only focus on number theory(which isn't my favorite area in math). I have heard that AoPS Olympiad Classes are excellent preparations for USAMO. My main goal this summer is to get better at proof-writing, which I am horrible at according to my USAMO score this year. Taking AoPS classes later is an option but I think I should concentrate more on school during school time since my grades this year are horrible.

Suggestions are appreciated!

PROMYS - will pound rigor into you like no other. Very good for proof writing, fun, meet cool people, play games, have an awesome time, etc. However, it does only focus on NT and usually this NT goes "beyond" in a sense what is required of the USAMO. To get the full package from PROMYS, all I can say is DO YOUR PROBLEM SETS EVERY DAY. Last year, some people just wasted their time away and probably left with a lot less than most of us. It's a lot of work, but it pays off both in proof writing and pure mathematical dedication.

AoPS Olympiad Classes - very useful for the USAMO, almost specific to it. Make sure to take Problem Solving and Geometry, though I would say Inequalities is the coolest (it didn't show up on USAMO, though :(). But really these don't take as much time as you might think, especially if you have no other plans over the summer. Even if you take them together it's 5 hours max instruction per week, whereas PROMYS is 6 weeks straight.

Overall, my opinion would be to go to PROMYS, build up the mentality to take on the USAMO and then, when you get back, take the Olympiad classes in the Fall/Winter/Spring. Oh, that's what I did, come to think of it... and I didn't do so hot on the USAMO, so maybe you should get another opinion.

Go to PROMYS.  You'll have all year to USAMO prep (and AoPS class if you choose to).  You won't have all year to be exposed in person all day to the caliber of students you'll be around at PROMYS.

If it's possible to do Olympiad Geom while at PROMYS, you could potentially do both, but if you had to pick one, I'd pick virtually any of the top math camps if you want to be immersed in interesting math and spend lots of time in person with cool people.


Let $ ABC$ be a triangle with I as its incenter.Circle $ k$ is centerd at $ I$ and lies inside triangle $ ABC$.Point $ A_1$ lies on $ k$ such that $ IA_1$ is perpendicular to $ BC$.Points $ B_1$ and $ C_1$ are defined anlogously.Prove that $ AA_1,BB_1,CC_1$ are concurrent.

Dear Mathlinkers,
this is the Kariya's theorem.
Sincerely
Jean-Louis

This is an application of isogonal theorem.
See here for the general:
http://www.mathlinks.ro/viewtopic.php?t=284298


[quote="sharkman"]A pond is enclosed by a wooden deck that is 3 feet wide. The fence surrounding the deck is 100 feet long.

(a) If the pond is square, what are its dimensions?

(b) If the pond is rectangular and the length of the pond is three times its width, what are the dimensions of the pond?[/quote]
try using equations

[quote="sharkman"]bpms,

I know you mean well but your hints are not helpful at all.

What equations?

Why not set up the equations for me?

I can take it from there.[/quote]

[hide="Equations"](a) Let the side of the square be $x$. Then the length of the fence is $4 x+24$, which is $100$. So $4x+12 = 100$.

(b) Let the dimensions be $y\times 3y$. Then the length of the fence is $8 y+24$, which is $100$. So $8 y+12 = 100$[/hide]

Telling us to set up equations for you is completely useless on your part, sure you can solve equations, but that's not the essence of the problem and you don't learn anything if we just tell you the equations to solve. You might as well do a worksheet of 100 problems such as

$4x+30=110$
$3x+7=16$
etc etc


If you write out the natural numbers  one, two, three, four, five, … ,  and stop when you get to sixty, how many times have you written the letter “f”?

A.   32	B.   33	C.   41	D.   42	E.   43

[hide] $ 4$ and $ 5$ are the only digits with an 'f'. Thus, just by considering numbers which end in either $ 4$ or $ 5$, we have $ 13$ 'f's (because $ 15$ has two 'f's). also, every number in the forties will contribute an extra 'f' and every number in the fifties will contribute two extra 'f's. Thus, $ 13\plus{}10\plus{}2(10)\equal{}\boxed{43}$. Thus, the answer is E.[/hide]


John attended a three-country multicultural festival. At the first exhibit, John paid a $3 admission fee. He then spent half the money he had left on souvenirs. Finally, he spent $3 on food and left. At each of the second and third exhibits, he again paid a $3 admission fee, then spent half of the money he had left for souvenirs, then spent $3 on food, and left. When he left the third exhibit, he had no money remaining. How many dollars did he spend at the festival?

I solved this problem by setting up an equation to solve for the original amount of money he had, but it took way too long and I made a lot of silly mistakes. Is there a better way to do this problem?

Just read the question "backwards". For example, when it says spent $3 on food, just add $3 to the amount you have. When it says used half the money, jsut multiply the amount by 2.


Suppose $a,b,c$ are digits. Find $[abbc]$ (where $[ \hskip .1in ]$ means the concatenated number as opposed to the product) if

$[abbc] = a-[bb]+[bc]^{2}$.

[hide="hint"] It hacks. [/hide]

[hide="Solution"]\begin{eqnarray*}&& 1000a+110b+c=a-11b+\overline{bc}^{2}\\ &\iff& 999a+121b+c=\overline{bc}^{2}\\ &\iff& 999a+111b=\overline{bc}^{2}-\overline{bc}=\overline{bc}(\overline{bc}-1)\\ &\iff& 3\cdot 37(9a+b)=\overline{bc}(\overline{bc}-1)\end{eqnarray*}

Hence $\overline{bc}\in\{37,38,74,75\}$. Out of those, only $37$ and $75$ give RHS which is divisible by $3$:

1. $\overline{bc}=37\implies 3(9a+3)=36\implies a=1$, hence $\overline{abbc}=1337$

2. $\overline{bc}=75\implies 3(9a+7)=150\implies a\not\in\mathbb{Z}$.

Hence the only solution is $\overline{abbc}=1337$

NOTE: Degenerate cases like $\overline{bc}=0\lor\overline{bc}=1$ (giving $\overline{abbc}=0000\lor\overline{abbc}=0001$) are dismissed. [/hide]


calculus
integration
trigonometry
calculus computations

Hello,

who can you integrate this? who can you simplify before you integrate?

[img]http://img296.imageshack.us/img296/6746/inte3th.jpg[/img]

Thanks.

$\left( \frac{ds}{dt} \right)^2 = (2 \sin t + 2 \sin 2t)^2 + (2 \cos t + 2 \cos 2t)^2 = 4 \sin^2 t + 8 \sin t \sin 2t + 4 \sin^2 2t + 4 \cos^2 t + 8 \cos t \cos 2t + 4 \cos^2 2t = 8 + 8 \left( \sin t \sin 2t + \cos t \cos 2t \right) = 8 + 8 \left( \cos (2t - t) \right) = 16 \frac{1 + \cos t}{2} = 16 \cos^2 \frac{t}{2}$

$\frac{ds}{dt} = 4 \cos \frac{t}{2}$

Then $\int ds = 8 \sin \frac{t}{2} + C$.  :)

Thank you very well now I put in my upper limit 2 pi and my lower limit 0

Then I get o for my definte integral but the solution in my book is 16 were do I miss ?

Maybe this problem is finding the arc length of Deltoid.

yes that is the problem but is the solution in my book not correct?

yes but what do you find at the finale solution in other words what is the lenght of the arc.

Thanks.

As kunny noted (and I forgot to specify), the integrand is not $4 \cos \frac{t}{2}$, but rather $4 | \cos \frac{t}{2} |$.

Hence the correct integral is $8 \sin \frac{t}{2} |_{0}^{\pi} - 8 \sin \frac{t}{2} |_{\pi}^{2\pi} = 8 \sin \frac{\pi}{2} + 8 \sin \frac{\pi}{2} = \boxed{16}$.

thank you verry well do someone know what kind of software (freeware) you can use to plot with parametric.

Just to have some idee of the function.

Greets.


If $ S \equal{} i^n \plus{} i^{\minus{}n}$, where $ i \equal{} \sqrt{\minus{}1}$ and $ n$ is an integer, then the total number of possible distinct values for $ S$ is:

$ \textbf{(A)}\ 1\qquad
\textbf{(B)}\ 2\qquad
\textbf{(C)}\ 3\qquad
\textbf{(D)}\ 4\qquad
\textbf{(E)}\ \text{more than 4}$

If n=0: $ i^0 \plus{} i^0 \equal{} 2$.
If n=1: $ i \plus{} \frac {1}{i} \equal{} i \minus{} i \equal{} 0$
If n=2: $ i^2 \plus{} \frac {1}{i^2} \equal{} \minus{} 1 \minus{} 1 \equal{} \minus{} 2$
If n=3: $ i^3 \plus{} \frac {1}{i^3}\equal{}\minus{}i\minus{}\frac{1}{i}\equal{}\minus{}i\plus{}i \equal{} 0$

Everything just repeats from n=4 and so forth. This gives a total of $ \boxed{3}$ distinct values for $ n$.

Answer: C

[hide="Solution"]$ n\equal{}0: 2$
$ n\equal{}1: 0$
$ n\equal{}2: \minus{}2$
$ n\equal{}3: 0$

$ n\equal{}4: 2$
$ n\equal{}5: 0$
$ n\equal{}6: \minus{}2$
$ n\equal{}7: 0$

and so on.

Therefore, $ \boxed{\textbf{(C)}\ 3}$.[/hide]

Edit: Aw, too late.


trigonometry
geometry
geometric transformation
rotation
analytic geometry

A man is standing in a courtyard of [b]square shape[/b].
The man then marks his original position in the courtyard, and  measures the distances between the mark and three corners of the courtyard one by one, and obtained the measurements of 30 feet, 40 feet and 50 feet respectively.

Question:
What is the length of the [b]diagonal [/b]of the courtyard?
Note: The answer should be [b]exact[/b],and no approximation.

Hint:
[hide]You may tackle the problem with any one of the following techniques:

1. Trigonometry, with the knowledge of Cosine Formula and also a basic Identity
$ sin^2 \theta \plus{} cos^2 \theta \equal{} 1$.

2. Geometry, with the application of a rotation of figure.

3. Coordinates Geometry, with the knowledge of "Equation of Circle".[/hide]

Ah I think I get it.

If you set this in coordinate geometry you can solve this.
Draw a circle of radius $ 3$,$ 4$,$ 5$ respectively on the vertexes.
Find the unique intersection, that gives you the point.
Then you can use trig to find the angles and then calculate the length of a side of the square.
The diagonal is $ \sqrt2$ times the side.

Is my approach correct?

Wait.. I think this is really simple..
Dot is where man is standing.
By the picture, you can tell that diagonal is $ \boxed{80}$
Am I right??
[asy]size(150); pair A,B,C,D,E; A= (0,10); B= (0,40); C= (-40,0); D= (40,0); E= (0,-40); draw(A--B); draw(A--C); draw(A--D); draw(A--E); draw(B--C); draw(B--D); draw(C--E); draw(D--E); label("30",(5,20)); label("40",(-20,0)); label("50",(5,-20)); dot(A);[/asy]

[quote="Fraenkel"]If it was $ 80$ then the edge with length $ 40$ would be on it's midpoint, which is a contradiction that half a diagonal is $ 30$ and $ 50$[/quote]
Oh.. Yeah.. You are right..

Never mind, then.

Fraenkel, you've got the correct procedure.
Find the length of the square first, then multiply it by root 2.
Have you tried to solve it by yourself?
I suggest you should try other methods too.  :)


trigonometry
inequalities
trig identities
Law of Cosines
AMC

A ship sails $ 10$ miles in a straight line from $ A$ to $ B$, turns through an angle between $ 45^{\circ}$ and $ 60^{\circ}$, and then sails another $ 20$ miles to $ C$. Let $ AC$ be measured in miles. Which of the following intervals contains $ AC^2$?
[asy]unitsize(2mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;

pair B=(0,0), A=(-10,0), C=20*dir(50);

draw(A--B--C);
draw(A--C,linetype("4 4"));

dot(A);
dot(B);
dot(C);
label("$10$",midpoint(A--B),S);
label("$20$",midpoint(B--C),SE);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,NE);[/asy]$ \textbf{(A)}\ [400,500] \qquad \textbf{(B)}\ [500,600] \qquad \textbf{(C)}\ [600,700] \qquad \textbf{(D)}\ [700,800]$
$ \textbf{(E)}\ [800,900]$

[hide]From the law of cosines, we have:

$ AC^2 \equal{} 100 \plus{} 400 \plus{} 2(10)(20)\cos{\theta} \equal{} 500 \plus{} 400\cos{\theta}$

Note that we're adding the cosine term instead of subtracting, because the actual angle between AB and BC is $ 180 \minus{} \theta$, so $ \cos{180 \minus{} \theta} \equal{} \minus{}\cos{\theta}$

We know that $ 45 \leq \theta \leq 60$, so:

$ \frac{\sqrt{2}}{2} \geq \cos{\theta} \geq \frac{1}{2}$

$ 200\sqrt{2} \geq 400\cos{\theta} \geq 200$

$ 500 \plus{} 200\sqrt{2} \geq 500 \plus{} 400\cos{\theta} \geq 700$

If you feel the need to confirm, you can estimate the LHS of the inequality, which is about 782. So our answer is $ \boxed{D}$

Also, you could probably just evaluate $ AC^2$ at the extremes, it comes to essentially the same thing.[/hide]


A farmer has four straight pieces of fencing: 1, 2, 3, and 4 metres in length. What is the maximum area, in metres, that he can enclose by connecting the pieces? Assume the land is flat

Please help me with this question.

[hide]
by Brahmagupta's formula

$A = \sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^{2}\theta}$

where $\theta$ is half the sum of the opposite angles and $s$ is the semiperimeter

$s = \frac{1+2+3+4}{2}= 5$

$A^{2}= (5-1)(5-2)(5-3)(5-4)-(1)(2)(3)(4)\cos^{2}\theta$
$A^{2}= 24-24\cos^{2}\theta$

$A$ is maximized when $A^{2}$ is maximized
$A^{2}$ is maximized when $\cos^{2}\theta$ is minimized

$\cos^{2}\theta$ is minimized when $\cos\theta = 0$ or when $\theta=90$

$\theta = 90$ is a valid angle choice because that would mean that the sum of the opposite angles would be $180$, which is possible in a quadrilateral.

plugging in $\theta=90$ yields

$A^{2}= 24$
$A = 2\sqrt{6}$
[/hide]


There are given a quadrilateral A1B1C1D1 inscribed in a circle with radii R1 and quadrilateral A2B2C2D2 inscribed in a circle with radii R2.
Prove that if
<A1B1C1=<A2B2C2
<B1C1D1=<B2C2D2
<C1D1A1=<C2D2A2
<D1A1B1=<D2A2B2
Then:
A1B1/A2B2=B1C1/B2C2=C1D1/C2D2=D1A1/D2A2=R1/R2.

I'm sorry ... don't loose your time - the statement is wrong.


phone numbers can only have 6 digits ranging from 000000-999999. what is the maximum number of phone numbers that can be in place at one time such that phone numbers differ in at least two places? e.g. 111111 and 111122 ..

Oops, I misunderstood the question. I think the wording could be better...

[quote="sen"]phone numbers can only have 6 digits ranging from 000000-999999. how many phone numbers are possible such that any two number plates differ in at least two places?[/quote]

The workding is hard to understand.. Maybe you should rephrase it better.


Eight U.S. coins are worth a total of 38 cents. How many of the coins are dimes?

If we have 1 dime, there's no way to make 28 cents out of 7 coins. (Try it and see!) However, if we use $ \boxed{2}$ dimes, the remaining six coins could be be 3 nickels and 3 pennies.


algebra
difference of squares
special factorizations

Factoring as a difference of squares:
$ A^{4}\minus{}B^{4}\equal{} (A^{2}\minus{}B^{2})(A^{2}\plus{}B^{2})$. Doing it again:
$ (A\minus{}B)(A\plus{}B)(A^{2}\plus{}B^{2})$.

If its more complicated like

$ x^{8}\minus{}y^{8}$ then let $ x^{4}\equal{}a$ and $ y^{4}\equal{}b$

so $ a^{2}\minus{}b^{2}\equal{}(a\plus{}b)(a\minus{}b)$ then since $ x^{4}\equal{}a$ and $ y^{4}\equal{}b$ sub back in and keep factoring

but then when you substitute them back in you have to factor the $ x^{4}\minus{}y^{4}$. so your answer to that is $ (x\minus{}y)(x\plus{}y)(x^{2}\plus{}y^{2})(x^{4}\plus{}y^{4})$


Let $ ABC$ be a triangle with the circumcircle $ w$. Let $ P$ be an interior point of the $ \triangle ABC$ and denote $ \left\| \begin{array}{ccc} X\in AP\cap w \\ Y\in BP\cap w \\ Z\in CP\cap w\end{array}\right\|$ .

Also denote 6 points such as $ \left\|\begin{array}{ccc} A_1\in BC\cap ZX,\ A_2\in BC\cap XY \\ B_1\in CA\cap XY,\ B_2\in CA\cap YZ \\ C_1\in AB\cap YZ,\ C_2\in AB\cap ZX\end{array}\right\|$ .

Prove that $ A_1B_2\cap B_1C_2\cap C_1A_2\equal{}\{P\}$.

[quote="Mateescu Constantin"]Let $ ABC$ be a triangle with the circumcircle $ w$. Let $ P$ be an interior point of the $ \triangle ABC$ and denote $ \left\| \begin{array}{ccc} X\in AP\cap w \\
Y\in BP\cap w \\
Z\in CP\cap w\end{array}\right\|$ .

Also denote 6 points such as $ \left\|\begin{array}{ccc} A_1\in BC\cap ZX,\ A_2\in BC\cap XY \\
B_1\in CA\cap XY,\ B_2\in CA\cap YZ \\
C_1\in AB\cap YZ,\ C_2\in AB\cap ZX\end{array}\right\|$ .

Prove that $ A_1B_2\cap B_1C_2\cap C_1A_2 \equal{} \{P\}$.[/quote]

I don't known if i am missing something, but i do believe that the concurrency of the lines AX,BY, CZ is irrelevant.
the six sidelines of two triangles with the same circumcircle are tangent to a same conic. The concurrency the main diagonals of the hexagon A_1A_2....C_2   follows from the brianchon theorem

[b]tiger100[/b], the question is not to show that $ A_1B_2, B_1C_2, C_1A_2$ are just concurrent, but that their concurrency point is the same as the concurrency point $ P$ of $ AX, BY, CZ.$


Is anyone going?  I will probably be competing as varsity for a change, because we have only managed to scrape together five people total (there are lots of conflicts and people are scared).

Hmm, looks like I won't be going after all, due to the aforementioned participation problems.


vector
invariant
algebra
polynomial
linear algebra
matrix
abstract algebra

this is sort of nice and classical:
Let $N$ be a set of nilpotent linear operation on a finite dimensional vector space .assume that for each pair $A,B$ in $N$ there exists a scalar $c$ such that $AB-cBA$ belongs to $N$ .then $N$ is triangularizable.

What do you mean by "N is triangularizable"?

I think that he means that there exists a basis in which all elements of $N$ have a triangular form.
I think you knew that, but were just hoping for a trivial problem.

I did not know that... how would I?

"triangularizable" means to me: able to be converted to triangular form, but that's all matrices.
So what you explain, I would simply call "triangular".

You can't say that a linear operator in triangular. You say that is triangularizable in the sense that there exists a basis with respect to which it has triangular form.
This is just linguistics. I'm not sure that the word triangularizable exists in english  :blush:  but you need a word different from triangular for operators.

Has anyone a solution for this problem? I have no idea how to find invariant spaces here. Sam-n???  :(

[quote="PeterVDD"]...able to be converted to triangular form, but that's all matrices. [/quote]
No, that's not all matrices. The key point is the existence of eigenvalues. If the characteristic polynomial of $A$ factors completely in the field of scalars of this vector space, then $A$ is similar to a triangular matrix. Hence this applies to all matrices over $\mathbb{C},$ or over any other algebraically complete field. But in posing this problem sam-n did not suppose such a field.

Put another way, $\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ is not similar to any triangular matrix within $M_2(\mathbb{R}).$

excuse me these days I have not access to internet properly anyway ,if we replace $cB$ in Jaboson theorem by any noncommutative polynomial in $A$ and $B$.even statment holds.pick a subset $N_0$ of $N$ whose members have a common kernel $K$ of minimal positive dimension.we can,and do ,also assume that $N_0$is the maximal such subset .for $A$ in $N_0$ let $A'$be the induced operator on the quotient space $V/K$.Observe that the set $ \ { \ A':A\in N_0 \ } \ $ satisfies the hypothesis of theorem on $V/K$,which has smaller dimension than $ V$ ;so by induction conclusion follows. we shall show $N=N_0$.assume this is not the case,so there is $B$ in $N$ with $BK\neq 0$.then $K$ is not invaraiant under $B$,because otherwise the restriction of $B$ to $K$ would be nilpotent,implying that the nonzero kernel of$N_0\cup \ { \ B \ } \ $is properly contained in $K$,for at least one $A_1$ in $N_0$ we have $A_1BK\neq 0$.let $p_1$ be the polynomial with $B_1=A_1\cdot B-p_1(A_1,B)A_1 \in N$and observe that $B_1k\neq 0$.thus $K$ is not invariant under $B$ ,just as in the case of $B$ above,and there is an $A_2$ in $N_0$ with $A_2\cdot BK\neq 0$ and $B_2=A_2B_1-p_2(A_2,B_1)A_2\in N$ so if we continue the procedurewe get $A_nA_{n-1}\dots A_1BK\neq 0$where $n$ is dimension of $V$ .this is a contradiction because the triangularizability of $N_0$ implies that $A_n\dots A_1=0$ ;)
(Edited)


Find the probability that a circle with center (x,y) and radius 1 on the xy plane will strictly enclose exactly three lattice points.

[quote="demji"]it doesn't enclose any right?
4 lie on the circle but none strictly within it.  Therefore the probability is 0, unless you mean something else.  But i would like to think i am reading this right.  or can x and y be non-integer numbers?[/quote]
No, $x$ and $y$ do not have to be integers.
[hide="Partial Solution/Hint"]Let the point be point $P$. Assume WLOG that the point is in the unit square with vertices $(0,0)$, $(0,1)$, $(1,0)$, and $(1,1)$. Draw circles of radius $1$ centered at the vertices of the unit square. The circle centered at $P$ will enclose exactly $3$ points if and only if $P$ is contained within exactly $3$ of the circles. Then we just need to calculate the area  of the region where this occurs and divide it by the total area of the square (which is $1$).[/hide]

For example, a unit circle with center (1/2, 1/2) would stricly enclose 4 lattice points.

[hide]

just looking at one lattice square which is 1x1, the diagonal length is $\sqrt{2}$ and since it is a unit circle when we place the center along that diagonal we strictly enclose 3 lattice points with an area defined by a square whose vertices are $\sqrt{2}-1$ from each lattice point along the diagonals.  As a result the diagonal length is $\sqrt{2}-2(\sqrt{2}-1=2-\sqrt{2}$  Thus the area of this square is: $\frac{(2-\sqrt{2})^{2}}{2}=3-2\sqrt{2}$  And so our probability that a circle strictly contains at least 3 lattice points equals $\frac{A of innersquare}{A of lattice square}=\frac{3-2\sqrt{2}}{1}=3-2\sqrt{2}\approx .171573$
[/hide]

I don't know if you had a typo, but the problem asks for strictly enclosing exactly 3 lattice points.

You need to find the area of the attached figure


search
function
number theory unsolved
number theory

Let $a$ and $b$ be two distinct natural numbers such that $ab(a+b)$ is divisible by $a^{2}+ab+b^{2}$. Prove that $|a-b|>\sqrt[3]{ab}$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=292
http://www.mathlinks.ro/Forum/viewtopic.php?t=18698

Note that using the [url=http://www.mathlinks.ro/Forum/search.php]search function[/url] with, as a keyword, "russia" and restricting your search to the NT subforum, you'd have found those quite quickly.


logarithms
calculus
calculus computations

Say $ \sum_{n\equal{}1}^\infty a_n$ converges. Is it necessarily the case that $ \sum_{n\equal{}1}^\infty a_n \log (a_n)$ converges?

This should be in the calculus computations and tutorials forum.  (But don't re-post it -- a moderator will probably move it.)  The answer is no.  We need an example where convergence rests on something log-size in the terms.  So a natural thought is something like $ a_n \equal{} \frac {1}{n (\log n)^2}$.  Then $ a_n \log a_n \equal{} \minus{} \frac {\log n \plus{} 2\log \log n}{n (\log n)^2}$, and the associated series does not converge.

Let me Wikipedia that for you:  http://en.wikipedia.org/wiki/Convergence .

A sequence $ (a_{n})$ converges to a point $ x$ if $ \forall\varepsilon>0$, $ \exists N>0$ such that $ \left|a_{n}\minus{}x\right|<\varepsilon$ $ \forall n\geq N$.  Basically, you can always get "most" of your sequence within any $ \varepsilon$ of your point of convergence.


geometry
circumcircle
trigonometry
power of a point
radical axis
perpendicular bisector
geometry proposed

On the sides $AB$, $BC$, $CA$ of the triangle $ABC$ consider the points $C'$, $A'$, $M$ such that $MA' = MC$ and $MC'=MA$.
 Find the geometric locus of the center of the circumcircle of the triangle $A'BC'$.

Let a = BC, b = CA, c = AB the sides of the triangle $\triangle ABC$, H be the orthocenter and R the circumradius. Let K be the intersection of the line MA' with the A-altitude AH and L the intersection of the line MC' with C-altitude CH. Since $\angle MC'A = \angle MAC' = \angle A$, we have $\angle C'BH = \angle C'LH = 90^\circ - \angle A$, which means that the quadrilateral C'BLH is cyclic. Likewise, since $\angle MA'C = \angle MC'A = \angle C$, we have $\angle A'BH = \angle A'KH = 90^\circ - \angle C$, which means that the quadrilateral A'BKH is also cyclic. Let (P), (Q) be the circumcircles of the quadrilaterals A'BKH and C'BLH and assume that these 2 circles are different. Then their radical axis is the B-altitude BH. The triangles $\triangle AMK, \triangle CML$ are both isosceles with MA = MK, MC = ML: For example, $\angle MAK = 90^\circ - \angle C$ and $\angle AMK = \angle MCA' + \angle MA'C = 2 \angle C$, hence,

$\angle MKA = 180^\circ - (90^\circ - \angle C) - 2 \angle C = 90^\circ - \angle C = \angle MAK$

Likewise, $\angle MCL = 90^\circ - \angle A$ and $\angle CML = \angle MAC' + \angle MC'A = 2 \angle A$, hence,

$\angle MLC = 180^\circ - (90^\circ - \angle A) - 2 \angle A = 90^\circ - \angle A = \angle MCL$

The power of the vertex C to the circumcircle (P) of the quadrilateral A'BKH is equal to

$CA' \cdot CB = 2\ CM \cos \widehat C \cdot a = 4\ CM \cdot R\ \sin \widehat A \cos \widehat C$

The power of the vertex C to the circumcircle (Q) of the quadrilateral C'BLH is equal to

$CQ \cdot CH = c\ \cot \widehat C \cdot 2\ CM \cos (90^\circ - \widehat A) =$

$= 4\ CM \cdot R\ \cot \widehat C \sin \widehat C \sin \widehat A = 4\ CM \cdot R\ \cos \widehat C \sin \widehat A$

Thus the power of the vertex C to the circles (P), (Q) is the same, which means that the vertex C lies on their radical axis. But their radical axis is the B-altitude BH, which does not pass through the vertex C (unless the angle $\angle C = 90^\circ$ is right, but then the isosceles triangle $\triangle MCA'$ would not exist). Consequently, the 2 circles must be identical. Let's call this common circumcircle (P). Since it passes through the points A', B, C', it is the circumcircle of the triangle $\triangle A'BC'$. Since it passes through 2 fixed points BH, the circumcircles of all triangles $\triangle A'BC'$ form a pencil with the base points B, H and common radical axis BH. Their centers all lie on the perpendicular bisector of their common chord BH.

in http://www.mathlinks.ro/Forum/viewtopic.php?t=59510

was proven by Darij that $A'$, $C'$, $B$ and $H$ are concyclic, where $H$ is the ortocenter of $ABC$.

so, the circuncircle of $A'C'B$ always pass trought $B, H$.

this implies that the circuncenter lies in the bisector of $BH$.

E.L


Determine the amount of times the mass of the molecule of water (H2O) is contained in the mass of the molecule of glucose (C6H12O6)
:

Atomic mass: H = 1 ; O = 16 ; C = 12

a) 1
b)5
c)10
d)15
e) 20

Mass of glucose $ C_6 H_{12} O _6$ = $ (12*6) \plus{} (12*1) \plus{} (16*6) \equal{} 180amu$

Mass of water $ H_2 O$ = $ (1*2) \plus{} 16*1 \equal{} 18amu$

So, 10 times the mass of one molecule of water is contained in one molecule of glucose.
Answer is c

Thank you very much ;)
Now I understand.


The difference between the cube of an integer and the square of that same integer is 100. What is the integer?

anyone have a quick and easy method? i could only solve it by brute force and it's an easy problem to take that long...

If you remember that all powers of 5 above 1 end in 25, then you can quickly realize that this will always leave a multiple of 100 as the difference.[/hide]

It's not really wimpy... it's really the best way to solve the problem without resorting to "drastic measures," don't you think?

When you look at this problem, you need to understand that the last two digits have to be the same.  If you were more lenient and would only go for the last digit, all the multiples of 5 would occur, nothing else.  So try 10 because it is an easy number.  Too high?  Yes, so it has to be lower, so only 5 works.  Doing this problem took all of 10 seconds for me.

[quote="archimedes1"]I would call this wimp force.

[hide]x^3 - x^2 = 100
(x^2)(x - 1) = 100
perfect square factors of 100 are 4, 10, 25.
then by just looking at the numbers it is obvious that x^2 = 25 and x = 5[/hide][/quote]
10 isn't a perfect square... :P


Let $ f: R \mapsto R$ and $ |f(x)| \leq 1$ for all $ x \in R$ and  $ f(x) \plus{} f(x\plus{} \frac{13}{42})\equal{} f(x\plus{} \frac{1}{6}) \plus{} f(x\plus{} \frac{1}{7})$. Prove that: f is periodic function

Put x + 1/6 , x + 2/6 + ... + x + 5/6 instead of x and sum the equations that you have we obtain f(x) + f(x+1+1/7) = f(x+1) + f(x+1/7). Now if we apply the similar process for this equation (Put x + 1/7, x + 2/7, ... , x + 6/7 instead of x) we have the following equation: f(x+2) - f(x+1) = f(x+1) - f(x). If we put a, instead of f(x+1) - f(x); summing gives that f(x+k) = f(x) + k.a . Here, if a=0, we arrive at f(x+1) = f(x). This shows that f is a periodic function. If a doesn't equal to 0 , turn back to the equation.                        f(x+k) = f(x) + k.a , since f is bounded, f(x+k) is bounded, too. But when k reaches infinity, the right of the equation reaches infinity, too. That is a contradiction. So a must be 0. Therefore f(x+1) = f(x) and it is periodic. :P


function inequalities calculus derivative integration limit logarithms

Let $f\colon [0,+\infty) \to [0,+\infty)$ a crescent and bijective function. Prove that the sum $\sum_{n=1}^\infty{1\over f(n)}$ converges if and only if the sum $\sum_{n=1}^\infty{f^{-1}(n)\over n^{2}}$ converges, where $f^{-1}$ is the inverse function of $f$.

What is the meaining of the word "crescent" here? Would it be "concave upwards," or convex in the Jensen sense? (That is, assuming it has a second derivative, that $f''>0?$) Or does it just mean "increasing"?

I think the word "crescent" means "(strict) increase".

Sorry, I meant "strictly increasing". I only noticed that "crescent" thing after posting.

Hi,
I think I have a  method for solving ~ the Brazilian problem~.

First we observe that since $f$ is an increasing function we get that

$\frac{1}{f(k+1)}<\int_{k}^{k+1}\frac{1}{f(x)}dx=\frac{1}{f(c_{k})}<\frac{1}{f(k)}$
Thus $\sum_{n=1}^{\infty}\frac{1}{f(n)}$ has the same nature as the integral $\int_{1}^{\infty}\frac{1}{f(x)}dx$.

A couple of observations:
$f$ is increasing thus, f is differentiable almost everywhere, so without loosing generality we assume that $f$ is differentiable everywhere, since the Riemann integral does not depend on a set of Lebesque measure zero.

Now, via the substitution $f(x)=y$, $x=f^{-1}(y)$, we get that
$A=\int_{1}^{\infty}\frac{1}{f(x)}dx=\int_{f(1)}^{\infty}\frac{1}{yf'(f^{-1}(y)}dy$.
We compare A to $\int_{1}^{\infty}\frac{f^{-1}(y)}{y^{2}}dy$ via the limit test.
Thus we calculate

${{L=\lim_{x\rightarrow \infty}\frac{\frac{1}{yf'(f^{-1})(y)}}{\frac{f^{-1}(y)}{y^{2}}}}}=\lim_{x\rightarrow \infty}\frac{y}{f'(f^{-1}(y)f^{-1}(y)}$.
Now we make the substitution $f^{-1}(y)=u$, observe that when $u$ converges to $\infty$ then $y$ also converges to $\infty$.
Thus we get that
$L=\lim_{u\rightarrow \infty}\frac{f(u)}{uf'(u)}$.
We distinquish here some cases:
1) Definitely that if $L\neq 0, \infty$ then we are done, via the limit test.(of course the limit exists)

2) $L=\lim_{x\rightarrow \infty}\frac{f(x)}{xf'(x)}=\infty$ then $\frac{f}{xf'}\geq \frac{1}{A}$ for $x>a$ , where $A<1$ is such taken, this follows via the limit definition
it follows that $\frac{f'}{f}\leq \frac{A}{x}$ from which it follows that
$\left(\ln f-\ln x^{A}\right)'<0$ thus we get that the function $x\rightarrow \ln\frac{f}{x^{A}}$ is decreasing and after simple calculations we get that
$f<Cx^{A}$  , where $C=\frac{f(a)}{a^{A}}$ is a constant and $x>a$.
thus $\frac{1}{f}>\frac{1}{Cx^{A}}$ and since $A<1$ we get that
$\int_{a}^{\infty}\frac{1}{f}dx>\int_{a}^{\infty}\frac{1}{Cx^{A}}dx=\infty$.
Now observe that the inequality $f(x)\leq Cx^{A}$ implies via $x=f^{-1}(y)$ that $y\leq C(f^{-1}(y))^{A}$ or $f^{-1}(y)>\frac{y^{1/A}}{C^{1/A}}$.
I will drop the constant, $\int_{b}^{\infty}\frac{f^{-1}(y)}{y^{2}}dy>\int_{b}^{\infty}\frac{y^{1/A}}{y^{2}}dy=\int_{b}^{\infty}\frac{1}{y^{2-1/A}}dy=\infty$ since $A<1$. thus the two integrals have the same nature, namely both divergent.
the third case when $L=0$ is handled similarly, namely we get that the two integrals are convergent, we use the same technique as in the second case.

Remark:
This problem implies the following integral test

Let $f: (0,\infty)\rightarrow (0,\infty)$ be a strictly increasing function. Then $\int_{0}^{\infty}\frac{1}{f}dx$ converges if and only if
$\int_{0}^{\infty}\frac{f^{-1}}{x^{2}}dx$ converges.

I think the following observation can simplify the argument, although I didn't actually push to completion.

Your argument studies the integral
\[\int \frac{f^{-1}(x)}{x^{2}}\, dx \, . \]
Let $y = f^{-1}(x)$.  By the substition rule, the integral changes to
\[\int \frac{y f'(y)}{f(y)^{2}}\, dy \, . \]
Integration by parts shows that this integral is equal to
\[\int \frac{dy}{f(y)}-\frac{y}{f(y)}\, . \]
Notice that the final integral is the other integral you are interested in.  So we have related the two integrals of interest.


How many positive integer divisors does 72 have?

Prime factorizing it, we get $ 2^3\times 3^2$.  We then do $ (3\plus{}1)(2\plus{}1)\equal{}\boxed {12}$


Let $a;b$ be real numbers satisfying $(a-\sqrt{a^{2}+1})(b-\sqrt{b^{2}+1})=2$
Caculate $a+b=?$

I am surely that i have seen the problem has posted before, but now i don't know where is it, exactly. If i don't find the link, i will post again solution. The problem is fairly easy

[quote="chien than"]Caculate $a+b=?$[/quote]
Obviously $a+b=b+a$ AND WE ARE DONE  :rotfl:

hey chien tian, can you post the answer? I can only deduce the fact that the product of the larger roots of the equations x^2-2ax-1=0 and x^2-2bx-1=0 as 2..., but it seems not sufficient enough to find the sum of the four roots in the two equation, which is 2(a+b)


vector
calculus
geometry
search
abstract algebra

Can anyone suggest me some books on General Relativity which explains things in more "physics" way than a more mathematical way . Like more emphasis on Conceptual explanations without advanced high tech complex mathematical proofs .

How much math have you taken?  Do you know special relativity well?  If you have taken math through vector calculus and differential equations, and have a strong grasp of special relativity  I would recommend Schutz's [i]A First Course in General Relativity[/i].  However, if you don't yet completely understand special relativity then I would recommend A.P French's [i]Special Relatvity[/i] (all you need is basic calculus).  Lastly, if you don't have the prerequisites in math then I would recommend Kip Thorne's [i]Black Holes and Time Warps[/i], which is good conceptual, but doesn't allow you to fully understand general relativity, because it doesn't include the maths.

Schutz's [i]A First Course in General Relativity[/i], as mentioned above, is a popular book and I would recommend it as well. If you're interested in black holes and cosmology, Hobson, Efstathiou, and Lasenby's [i]General Relativity: An Introduction for Physicists[/i] covers these in more detail than Schutz, while keeping the text very mathematically light. I would say the bare minimum of mathematical experience needed for these books is vector calculus, although a stronger background would definitely be useful.

On the other end of the spectrum, if you someday happen to be looking for a more mathematical text to read, perhaps once you have gone through a book geared towards physical intuition, I would recommend Wald's [i]General Relativity[/i]. I probably wouldn't recommend this as a first text even to someone with a good math background, but if you are really interested in general relativity you should definitely read it at some point.

Thanks for the info friends ,
er.......I am just a 10th grade student and I know some basic calculus ...I only know special relativity conceptually not with mathematics rigour......can you suggest me books which cover all the pre-requisites of general relativity ... Vector calculus ,Differential Geometry all should be learnt by me .
Please also mention the pre requisite of the pre requisite for general relativity .Ya I am also interested in the black holes book.....I feel it would be first better to understand it in a more conceptual way before going to mathematics part of it ...sometimes too much math hurts the brain (atleast for me ) :rotfl:.
The book by Robert Wald is greek and latin for me .....I feel like fainting  :D
[quote="zhyllilom"]
If you're interested in black holes and cosmology, Hobson, Efstathiou, and Lasenby's General Relativity: An Introduction for Physicists covers these in more detail than Schutz, while keeping the text very mathematically light. I would say the bare minimum of mathematical experience needed for these books is vector calculus, although a stronger background would definitely be useful.
[/quote]
Friend ,I have never even seen vector calculus in my life ,please dont scoff that I plan to learn general relativity . Can you suggest the pre requisites for Vector calculus etc...
Also I am very much interested in cosmology ,black holes ...It would be nice if I first read books which  treat stuffs in a more conceptual and humorous  way .....like the dead famous etc......
Ok humour is not that mandatory.....

Okay, so are you currently taking calculus?  Vector Calculus can be taken as soon as you complete single variable calculus.  Differential Geometry is another story, as its usually taught for advanced undergraduates or graduate students.  Also, a course in differential equations could help - again you can take it as soon as you finish single variable calculus.  However, don't underestimate the physics you must understand, especially special relativity, in order to grasp general relativity.  Again I would highly recommend you go through a special relativity textbook (like A.P French's [i]Special Relativity[/i]) before attempting to learn general relativity.  While you're waiting on it to learn the math you could also read Schutz's [i]Gravity From the Ground Up[/i] which only uses basic high school math (no calculus).

If you just want pop science books try searching for:

Leonard Susskind
Brian Greene
Stephen Hawking
Kip Thorne

They all have written popular books... Thorne's [i]Black Holes & Time Warps[/i] is probably the best "laymen" explanation of black holes that I've read.

Is Differential Geometry a must for the general theory of relativity ?
Anyway can you suggest me  books for the Vector Calculus etc......

The black holes and time warps book is cool my friend ,thank you very much .

Well in order to [b]completely[/b] understand general relativity you need differential geometry.  However, you can get a good understanding of it without differential geometry.  (general relativity isn't usually studied until grad school anyways).

As for vector calculus there are several textbooks you can use (search on amazon) - I would also highly recommend MIT's vector calculus lectures (for free) on iTunes!

What are your views regarding the book on GR by James.B.Hartle  ?
Also I need to know tensor algebra and some advanced classical mechanics right ?

I'm not familiar with Hartle's book - however you could look on amazon for reviews as to the difficulty level.

I would say, with your preparation, there is absolutely no rush. If you just want to toy with some ideas for fun, that's fine, but you won't get out much if you don't have the math prerequisites.

Learn the cool things that are withing grasp first (trust me, there are a LOT of those). There's not reason to rush things up, you only hurt yourself.

For example, after learning single variable really well, start learning vector calculus (multivariable) with a decent textbook, and practice your skills on a book like Griffiths Electrodynamics.  Learn at the same time Special Relativity, and you are well on your way on building an understanding of relativistic electrodynamics, sth I find very cool, and well within grasp of a motivated high school student (within 1-2 yrs of hard work, you'll get there).

thanks for the advice friend :lol:  ,I am not in a rush anyway. I want to learn all the math pre-requisite . After Vector calculus what is the math I require ?


Given 4 lines in general position, construct a quadrilateral with internal
    angle bisectors on these lines.

     Comd.  Pestich

I'm more interested in the problem source than the problem itself.

"Politbureau Entrance exam"  :D What is it? I've never heard that entering Politburo needs a math test  :blush:

It shows one more time that the political leadership liked to keep
   population in the dark about it's practices, and it will continue
   with this policy, especially towards comrades and their
   families who send in their census forms late.

   Lt. Col.   Pestich

   P.S.   What about the geometry problem at hand?


Find the remainder when $ 37^{100}$ is divided by $ 29$.

Please explain it  :)

[hide="Solution"]By [url=http://en.wikipedia.org/wiki/Euler%27s_theorem]Euler's Theorem[/url], $ 37^{28}\equiv 1 \mod 29$.

$ 37^{100} \equal{} (37^{28})^3\cdot37^{16}\equiv 37^{16}\equiv23\mod 29$

Answer: 23[/hide]

Or use Fermat's little theorem (which Euler generalizes)

How did you find that $ 37^{16}\equiv23\mod 29$?

Lol, calculator bash... :blush:

Well, it's easier than $ 37^{100}$, right?

You could also do

$ 37^{16} \equiv 2^{48} \equiv 2^{20} \equiv 32^{4} \equiv 3^4 \equiv 81 \equiv \minus{}6 \equiv 23 \pmod{29}.$

Yeah, here's what I did:
$ 37^{16}\equiv 6^{8}\equiv 7^{4}\equiv 20^{2}\equiv 400\equiv 23\pmod{29}.$

Sorry for being late. I understand this part $ 37^{28}\equiv 1\mod 29$ because I know that Fermat Little Theorem is $ a^{p\minus{}1}\equiv 1\mod p$.
But I didn't understand how you did the second part.  :(

You repeatedly have to apply the properties of modular arithmetic. (Search on the wiki if you don't know them.)

People are leaving out the steps of their modular reductions, so you can't tell what calculations they did or whether they did them by hand or by calculator.

In detail: $ 37^{100} \equiv 8^{100} \bmod {29}$, because $ 37 \equiv 8 \bmod {29}$.
Also $ 2^{28} \equiv 1 \bmod {29}$, by Fermat's Little Theorem.
So $ 8^{100} \equal{} 2^{300} \equal{} (2^{28})^{10} \cdot 2^{20} \equiv 2^{20} \bmod {29}$

$ 2^5 \equal{} 32 \equiv 3 \bmod {29}$
So $ 2^{20} \equal{} (2^5)^4 \equiv 3^4 \equiv 23 \bmod{29}$

Therefore, $ 37^{100} \equiv 23 \bmod {29}$


geometry proposed
geometry
U sviti matematyky

Points $ C_1,$ $ A_1$ and $ B_1$ are chosen at sides $ AB,$ $ BC$ and $ AC$ of triangle $ ABC$ in such a way that the straight lines $ AA_1,$ $ BB_1$ and $ CC_1$ are concurrent. Points $ C_2,$ $ A_2$ and $ B_2$ are chosen at sides $ A_1B_1,$ $ B_1C_1$ and $ A_1C_1$ of triangle $ A_1B_1C_1$ in such a way that the straight lines $ A_1A_2,$ $ B_1B_2$ and $ C_1C_2$ are concurrent. Prove that the straight lines $ AA_2,$ $ BB_2$ and $ CC_2$ are concurrent.

The [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=6579]Cevian Nests Theorem[/url], again.

Kostas Vittas.


I can't really find any topics about gaming (although I only looked in the first 4 pages), so discuss gaming here.

I have a 360, I'm camping out for a Wii, and I'm not getting within 30 feet of a PS3.

I dont like Xbox
the controlers are bulky and i dont give a :cursing: bout Halo
Any ways i would camp out for a Wii the games are so fun!
If you want a preview watch On the Spot at gamestop.com
Also 360_Fan why do u hate the PS3 so much it will have an amazing varity of features

I might get a Wii.  Some of my friends think that the Wii will do much better than others expect.  We were talking about that earlier today.

I agree. I think the Wii is going to establish a good base in every gamers home while the second choice of system will be between 360 and PS3.

Here is the link to see the demo of the all-new kick- :cursing: Legend of Zelda:Twilight Princess For Wii.
[url]http://www.gameinformer.com/NR/exeres/4093da30-d529-4dfa-8689-973d2bda4268.htm[/url]
This is a great site! GameInformer was the nations best game magazine these past 2 years.

That's just the description of the game. Just imagine doing all that with the nunchuck and remote!

its amazing how [size=200]BAD[/size] at advertising Nintendo is!
-jorian

Sorry, that was the wrong link. This goes directly to the videos;
[url]http://www.gameinformer.com/News/Media/Media.htm?CS_pid={814A5054-B8CB-4D70-BFC0-7F5E7EBA0E51}[/url]

There we go. :lol:

The X-Box 360 will probably lose its popularity in the next few days.

If you're getting a Wii then you're going to be within 30 feet of a PS3 obviously.

Also, Wii is overpriced.

PS3's on the other hand, are a much better deal for the price you're playing.

XBox360's are unnecessary, I already have some portable heating units.

Halo RULZ!!!!!!! the Wii sounds cool but the first PS3 i see im lighting a match...

i mean, what are you gonna get for PS3 (new games)? Assassin's Creed is now for 360 too (go 360!)....yes you can play metal gear solid or something but that can be for the 360 also.......and isn't the online play on the 360 better than the ps3?

[quote].......and isn't the online play on the 360 better than the ps3?[/quote]
Except online play is free on the PS3 and Wii, I think.

[quote]Also, Wii is overpriced.

PS3's on the other hand, are a much better deal for the price you're playing. [/quote]

Hmmm, 250 bucks for a Wii, it's funner to play (although I desperately hope they have some third-party support lined up), and 600 bucks for a PS3, and they just think that they rule the gaming world and can dictate that the future world of gaming is going to use Blu-ray.

I think I'll take the Wii.

Oh, I'm not trying to spark an argument on this new topic, I'm just a devout Nintendo fan and newly converted Microsoft believer.

Several of my friends are going to be camping out to buy a Wii.

On the PS3 vs. Wii debate: [url=http://www.vgcats.com/comics/?strip_id=212]Vegenomics[/url]

Sega Genisis.

Games: Toy Sory, Sonic the Hedgehog, Terminator II, PacMan, Tiny Toons: Bugs Bunny.

I got half-way in Tiny Toons, practically finished PacMan, got to level 4 in Terminator II, got to the last level in Sonic, got to Sid's room in Toy Story.

Yeah, I personally would go for the PS3 if it weren't for the price and fact that it's already sold out (except on ebay, where I saw one auction go up to $\$$99,999,999.99 (seriously, though I dunno if it was real))

So in that case, It's probably Wii for now, and PS3/next Microsoft console later.

$\$$ 600 for a PS3....when the 360 came out it was only 400!

I don't hate the PS3...  I just don't like it. :P

I also hear that it's made poorly and when you're actually playing it it looks and feels low quality.

stuff i read comparing the three new systems:

the PS3 has dissappoiting release titles
the 360 was released earlier and therefore the "technology" might be not up to par compared to the PS3
the WII looks like the design was made in 1950........

i've heard about the PS3 having technical problems. i saw a video of this guy that was trying to turn the PS3 on and this red light would start blinking, then  it would shut off.

[quote="Treething"]I'm planning to learn how to play Halo off my friend's 360 >.> . That's all I'm doing for video games.[/quote]

you can buy a 360 just for the halo series alone  :lol: ...

if u need help with halo and stuff let me know  :starwars:

Well, the Wii came out in North America today.  Unfortunately, I couldn't get one... :(

Meh, I'll get one maybe next week or something.

[quote="daermon"]I hate video games.  I just don't see why people would pay hundreds of dollars to waste their time  :D
I was addicted to video games once.  wasted a few years of my life.   :([/quote]

lol well it depends...are you out of school or something?

if you are it is a waste

but if you are not out then it can be fun to get rid of stress after a school year or something


As the figure below shows,let $ ABCD$ be a square and $ P$ be a movable point on the segment $ CD$.The projection points of $ A$ on $ PB$ and $ B$ on $ PA$ are $ F$ and $ E$ respectively.$ CF$ meets $ DE$ at $ Q$.

(1)Find the locus of $ Q$;
(2)Let $ G$ be the projection point of $ P$ on $ AB$.Show that $ QG$ is the bisector of $ \angle AQB$.
(3)Let $ H$ be the orthocentre of $ \triangle PAB$.Show that there exists a fixed point on $ QH$.

[quote="christ"]I think you mean the locus of Q . Let’s draw a circle by center O the middle of AB and radius OA .This circle clearly passes through E and F .Is this circle the intended locus and is these facts useful?[/quote]

That's OK. The locus of $ Q$, exactly.


calculus
integration
logarithms
calculus computations

Study the convergence of the following integral:

$\int_{t}^{\infty}\ln \frac{b+(x\ln x)^{p}}{a+(x\ln x)^{p}}dx$,

where $p>0$, $b>a>0$ and $t$ is a positive real number large enough such that the integrand makes sense.

We can write this as

$\int_{t}^{\infty}\ln\left(1+\frac{b}{(x\ln x)^{p}}\right)-\ln\left(1+\frac{a}{(x\ln x)^{p}}\right)\,dx$

For large $x,$ the integrand is comparable to $\frac{b-a}{(x\ln x)^{p}}.$ It converges for $p>1$ and diverges for $p\le 1.$

That is right. There is a conection between these two problems since I belive that  the limit of that product is $0$ if $p\leq 1$ and that the product  converges to some real number, which I think it cannot be calculated explicitely, for $p>1$.


Let be given n lines satisfying that no 2 lines parallel with each other, no 3 lines intersect together at one point. Prove that it is possible to write in each part of the plane bounded by the above lines a integer number x satisfying  x  <>  0, |x |<n so that the sum of the numbers at each side of an arbitrary line is equal to 0.

I think this is from the $2004$ Balkan Olympiad, and it's been posted and solved. It wasn't eactly the same, but surely they are strongly related.

[quote="Herodatviet"]$VERY EASY$[/quote]

I don't think this is a relevant answer. ;)

Pierre.


prove that the equation $ x^2\plus{}x\plus{}1\equal{}py$ has a solution $ (x,y)$ in the set of integers for infinitely many prime number $ p$.

Take $ y\equal{}1$ and any prime $ p \equiv 1 \mod 3$  :wink:


geometry
trigonometry
circumcircle
rectangle
parallelogram
geometry proposed

we have $ \angle A\ ,m_{a}\ ,t_{a}$ from a triangle draw it.

$ t_{a}$ is lenght of angle bisector.

In any triangle we have $ t_{a}\leq m_{a}$. If $ t_{a}=m_{a}$ then it's trivial; $ \triangle ABC$ is isosceles. So, suppose that $ t_{a}<m_{a}$

Let $ AD=t_{a}$ and $ AM=m_{a}$

We will show that $ XY=t_{a}\sin^{2}\frac{A}{2}$, where $ X$ is the projection of $ M$ to $ AD$
and $ Y$ is its inverse point wrt the circle $ (A,m_{a}\cos\frac{A}{2})$

Thus, $ M$ can be constructed with straightedge & compass
$ \hline$
[hide="Proof"]I'm sorry but I couldnt' find an elegant proof  :(

Suppose that we have constructed the triangle $ ABC$
W.L.O.G suppose that $ AB>AC$

Let $ \phi = \frac{A}{2}$

The ray $ AD$ intersects the circumcircle of $ \triangle ABC$ at $ O$. Then $ \angle OBC = \angle OAC = \phi$

Let $ L$ be the projection of $ O$ to the line $ AB$. Then $ MOBL$ is inscribed in the circle with diameter $ AB \Rightarrow$
$ \angle OLM = \angle OBM = \phi$

Notice here that, $ L,B$ are on the same side of $ OM$.
[hide="Indeed"]Let $ Q$ is the antidiametric point of $ O$ and $ L'$ be the point where the ray $ OL$ intersects again (if it does) the circle.
$ A$ must be on the arc $ BC$ not containing $ O$.
If we assume directed angles, we could say that $ \angle COA\geq 0$ and from $ \angle AOL'=90-\phi$ we get $ \angle COL'\geq 90-\phi\Rightarrow \angle QOL'\geq 0$ which means that $ L'$ can't be in the arc $ CQ$ and consequently $ L,L',B$ are in the same side of $ OM$[/hide]
Let $ O'$ be the projection of $ M$ to $ AD$. Then from the right triangles $ OLA,LO'A$ we get $ AO'=AO\cdot \cos^{2}\phi$

[img]9545[/img]
$ \hline$

Let $ M' \in AM$ such that $ O'M'\parallel OM$. The parallel line through $ M'$ to $ BC$ intersects $ AB,AC,AD$ at $ B',C',D'$ respectively.
Then the triangles $ ABC,AB'C'$ are similar
Particularly, for any point $ X \in \{B,C,M,D,O\}$ we have $ \boxed{AX' = AX \cos^{2}\phi}$

[img]9548[/img]
$ \hline$

Through $ M'$ we bring a perpendicular to $ AM'$ which intersects $ AD$ at $ F$.
Let $ w$ be the circle with diameter $ FM$.
Then $ M',O' \in w$, so $ AF \cdot AO' = AM \cdot AM' \Rightarrow \boxed{AF \cdot AO' = AM^{2}\cos^{2}\phi}$

We draw the rectangle $ FO'MH$. Then $ FO'\parallel=HM$
$ O'H$ is a diameter of $ w$, so $ O'M' \perp M' H$. But $ O'M' \perp B'C'\Rightarrow O'M' \perp D'M'$, so we get that $ D'M',H$ are collinear
$ D'D\parallel HM, \ D'H\parallel DM\Rightarrow D'DMH$ is parallelogram $ \Rightarrow D'D=HM=FO'$
But $ AD' = AD \cos^{2}\phi \Rightarrow DD' = AD(1-\cos^{2}\phi) \Rightarrow \boxed{DD' =AD\sin^{2}\phi}$

[img]9550[/img]
$ \hline$

Now it's finished! We need to construct two points $ F,O'\in AD$, such that
$ \begin{array}{l}AF \cdot AO' = AM^{2}\cos^{2}\phi\\ FO'=AD\sin^{2}\phi \end{array}\}$

In other words the points $ F,O'$ are inverse wrt the circle centered and $ A$ with radius $ AM\cos\phi$

[color=red][size=150]Theorem[/size]
Points $ X,Y$ are inverse wrt the circle $ (O,R)$ and $ XY=2d$. If $ X$ is out of the circle then $ OX=d+\sqrt{d^{2}+R^{2}}$[/color]

$ 2d = OX-OY = OX-\frac{R^{2}}{OX}= \frac{OX^{2}-R^{2}}{OX}\Rightarrow$

$ OX^{2}-R^{2}=2d\cdot OX \Rightarrow$

$ OX^{2}-(2d)\cdot OX-R^{2}=0\Rightarrow OX = \frac{2d+\sqrt{4d^{2}+4R^{2}}}{2}=d+\sqrt{d^{2}+R^{2}}$
$ \hline$

Applying the theorem with $ R=AM\cos\phi$ and $ d=\frac{AD\sin^{2}\phi}{2}$ we get the exact position of $ O'$, which is
$ AO' = \frac{AD\sin^{2}\phi}{2}+\sqrt{(\frac{AD\sin^{2}\phi}{2})^{2}+(AM\cos\phi)^{2}}$

(Of course, $ R$ and $ d$ are constructible with straightedge and compass)

Finally, we construct the circle $ s=(A,m_{a})$ and we bring a perpendicular through $ O'$ to $ AD$.
We'll find $ M$ on the intersection between the line and the circle.

[hide="We will have solution iff "]$ AO'\leq AM \iff d+\sqrt{d^{2}+R^{2}}\leq AM \iff$
$ \sqrt{d^{2}+R^{2}}\leq AM-d \ \ (1)\ \Longrightarrow^\star$
$ d^{2}+R^{2}\leq (AM-d)^{2}\ \ (2) \iff$
$ d^{2}+R^{2}\leq AM^{2}-2AM\cdot d+d^{2}\iff$
$ R^{2}\leq AM^{2}-2AM\cdot d \iff$
$ 2AM\cdot d\leq AM^{2}-R^{2}\iff$
$ 2AM\cdot d\leq AM^{2}(1-\cos^{2}\phi) \iff$
$ 2 d\leq AM\sin^{2}\phi \iff$

$ AD\sin^{2}\phi\leq AM\sin^{2}\phi \iff$
$ AD\leq AM$

$ (\star)$ If $ AD\leq AM$ then $ (2)$ holds

Does $ (2)$ imply $ (1)$  :?:

$ AM-d=AM-\frac{AD\sin^{2}\phi}{2}>AM-\frac{AD}{2}>AM-AD\geq 0$, so $ (2) \implies (1)$

[b]So we have a solution if and only iff $ AD\leq AM$[/b][/hide][/hide]


I don't know where I should put this, but feel free to move it. Please don't delete this...

A game of trivial pursuit. Someone posts a question(serious question), and someone else answers it.

Example:

{A wrote:
What puppet was featured on the August, 1980, cover of [i]Life[/i]?

B wrote:
Miss Piggy}

I'll start.

What year did Mexico City host the Summer Olympics?

If there is more than one date, the first would be accepted, not the latest.

Sounds like fun. :) Can we look on the Internet, or does this need to be from our own knowledge?

We need a next question.  Let's keep these easy:

What shape has the same number of lines as a limerick?

According to 1=2's rules, you're supposed to post the next problem.

What was the name of the second [i]Star Trek[/i] series originally going to be?

Classic Trek?

I actually don't know this one.

Nice try, 1=2. :)

It's been quite a while. I don't think anyone knows it. I'll just post the answer and come up with a different question, if that's all right.

[hide="Answer"][i]Star Trek: Phase II[/i][/hide]

New question: Who is the famous singer who was once married to actress Carrie Fisher (Leia from [i]Star Wars[/i])?


group theory
abstract algebra
linear algebra
matrix
modular arithmetic
superior algebra
superior algebra unsolved

Let $ G$ be a group of order $ pq$, where $ p$ and $ q$ are both primes and $ q<p$.   Prove that if $ q \nmid (p\minus{}1)$, then $ G$ is cyclic.    If $ q \mid (p\minus{}1)$, then $ G$ is nonabelian and it is the only nonabelian group of order $ pq$ up to isomorphism.

Let P and Q be Sylow p-, q-subgroup of G respectively. Clearly, P is normal in G.  Also the number n_q of Sylow q-subgroups is p or 1.  If n_q=1, then G is cyclic group of order pq. If n_q=p, then q| p-1 by Sylow theroem, in addition,  Q acts fixed point freely on P since G has no an element of order pq, that is, G is a Frobenius.

$ P\cong \mathbb{Z}/p\mathbb{Z}$ has automorphism group $ \cong (\mathbb{Z}/p\mathbb{Z})^\times\equal{}\mathbb{F}_p^\times$ cyclic with only one subgroup of order $ q$.
So there is only one non-trivial way $ Q$ could act on $ P$ by conjugation,
and $ G$ must be the semi-direct product $ P\rtimes Q$ with that action.

I don't understand you olorin.    I was told to use automorphism to prove that $ G$ is not abelian (the case when it is abelian I saw it was easy by using Sylow theorems :P ).    I haven't seen semi-direct product, so I cannot use that.

[quote="olorin"]The [url=http://en.wikipedia.org/wiki/Semi-direct_product]semi-direct product[/url] is one of the most basic and useful constructions in group theory
making any group action on another group into conjugation in some bigger group.
You better learn it quickly. It is often the easiest way to represent the group you are looking for.

Anyway here is another proof for you.
As I said if $ q|p \minus{} 1$, then $ \mathbb{F}_p^\times$ is cyclic with only one subgroup of order $ q$.
So $ GL_2(\mathbb{F}_p)$ has the subgroup $ H: \equal{} \left\{\begin{pmatrix}1 & \\
a & b\end{pmatrix}: a,b\in\mathbb{F}_p, b^q \equal{} 1\right\}$ nonabelian of order $ pq$.

And if $ G$ is any nonabelian group of order $ pq$ with $ P \equal{} \langle x\rangle \unlhd G, Q \equal{} \langle y\rangle$ some Sylow $ p,q$-subgroups of $ G$,
then $ G \equal{} PQ \equal{} \{x^my^n: m,n\in\mathbb{Z}\}$ and $ yxy^{ \minus{} 1} \equal{} x^t$ with some $ t\in\mathbb{Z}\setminus (1 \plus{} p\mathbb{Z})$,
such that $ x \equal{} y^qxy^{ \minus{} q} \equal{} x^{t^q}$ and $ t^q\equiv 1\pmod{p}$.

This gives the isomorphism $ G\to H, x^my^n\mapsto \begin{pmatrix}1 & \\
m \plus{} p\mathbb{Z} & t^n \plus{} p\mathbb{Z}\end{pmatrix}$ for all $ m,n\in\mathbb{Z}$.[/quote]

Thank you olorin.    It is not that I don't want to learn semi-direct products, it is just I cannot use that because I haven't seen it in my class yet.

Morova

[quote="morova"]It is not that I don't want to learn semi-direct products, it is just I cannot use that because I haven't seen it in my class yet.[/quote]
What we have here is a failure to communicate.

The term "semi-direct product" is completely unimportant for this problem.  What we are trying to prove is: [i]for any $ P\lhd G$, $ Q\leq G$ with $ P\cap Q\equal{}1$, $ PQ\equal{}G$, the group $ G$ is determined, up to isomorphism, by the isomorphism classes of $ P$ and $ Q$, and the action of $ Q$ on $ P$.[/i]

This is really not so hard: write any element of $ G$ as $ xy$, with $ x\in P$, $ y\in Q$, and write down the multiplication.

olorin's second demonstration is very clean but this is the right way to think about the earlier explanation.

Actually the Classification requires the proof of the existence of such a non-abelian group.
And to construct such a group, you can either use a representation by generators and relations
or take the semi-direct product $ (\mathbb{Z}/p\mathbb{Z})\rtimes (\mathbb{Z}/q\mathbb{Z})$ or find it in some larger known group.
And the semi-direct product is simply easiest to handle.

That's true.  Still, my point was just that describing a group of ordered pairs with a certain multiplication is all we're doing here, whether we call it the semi-direct product or not.  The abstraction is useful, of course, but only once someone has understood the situation more concretely.


calculus
integration
algebra
function
domain
Ring Theory
number theory

I have a question: if p>0 is a prime number(for example 5) is -p (in my example -5) also a prime number?
In other words, when you say prime number do you only refer to positive integers(as wikipedia says) or has the term been extended.
It would be logical that also negative numbers should be prime (-5 and 5 have the same divisors: -5, -1, 1 and 5).

Yes prime must be positive .
If we call $ \minus{}p$ is prime then the fundamental law of number theory not true:''Every natural number can represent unique in the form product of prime numbers'' .And 1 is  also not a prime.
If the fundamental is not true then it is quite hard to solve problem in number theory.

How would it contradict that fundamental truth?
Every natural number is represented unique as a product of natural prime numbers.
If you extend that to integers, how do you represent -60 in a product of prime numbers?

$ \minus{} 5$ is prime in $ \mathbb Z$. Along with $ \minus{} 2, \minus{} 3, \minus{} 7, ...$ .

To clarify this, we consider arbitrary integral domains (commutative rings with $ 1$ such that $ ab \equal{} 0$ implies $ a \equal{} 0$or $ b \equal{} 0$). At first some simple notions for those:
$ x|y$ (spoken: $ x$ [i]divides[/i] $ y$) iff there is a $ z$ with $ y \equal{} xz$ as usual.
A [i]unit[/i] $ u$ is an element such that $ u|1$, in other words: there is an inverse of multiplication.
A [i]prime element[/i] is a (nonzero nonunit) $ p$ such that if $ p|ab$, then $ p|a$ or $ p|b$.
A similar notion is that of an [i]irreducible element[/i]: a nonzero nonunit $ p$ is called irreducible iff $ p \equal{} ab$ implies that $ a$ or $ b$ is a unit.

Now prime elements $ p,q$ (and similar irreducible elements) are seen to be equivalent iff they just differe by a unit, thus if $ p \equal{} qu$ for some unit $ u$.

Just a small and easy to prove remark:
Lemma:
Every prime element is an irreducible one.

[u]Theorem[/u]:
The following are equivalent:
a) every nonzero element can be represented as a product of prime elements and units.
b) the representation from a) exists and is unique up to multiplication by units and reordering.
c) every nonzero element can be written as a product of irreducible elements and units, and this is unique as in b).
If this holds, the ring is called UFD (unique factorisation domain).

Looking back, all notions make sense if we look at the theorem. Applying the deifnitions on $ \mathbb Z$ we see: $ \minus{} 2, \minus{} 3, \minus{} 5, \minus{} 7,...$ are primes.

I often consider the "essentially different" (rational) primes to be $ \bar{\mathbb P} \equal{} \{ 0,2,3,5,7,..., \infty \}$ :D(yes, this really makes sense, e.g. as [i]places[/i]).

Thank you very much for the documented clarification ZetaX. Yet, I don't really understand the last sentence.[/img][/quote]

Another way of viewing primes is that of a (real, multiplicative) valuation, this means a function $ | \cdot | : K \to \mathbb R_0^\plus{}$ (here $ K\equal{}\mathbb Q$) such that:
$ |x| \equal{} 0 \iff x\equal{}0$
$ |a \cdot b| \equal{} |a| \cdot |b|$
$ |a\plus{}b| \leq |a| \plus{} |b|$

We call $ | \cdot |$ und $ | \cdot |'$ equivalent if $ |x|' \equal{} |x|^r$ for some $ r>0$.

Then up to equivalence there are the following valuations of $ \mathbb Q$:

- $ | \cdot |_\infty$, the standard modulus.
- $ | \cdot |_0$, the trivial valuation (given by $ |x|\equal{}1$ for $ x \neq 0$).
- for every positive natural prime $ p$ we define $ | \cdot |_p$ as follows: write a rational $ x$ as $ x\equal{} p^v \cdot \frac bc$ with $ p \nmid b,c$ and set $ |x|_p : \equal{} p^\minus{}v$.

The reasoning for choosing the indices $ 0$ (see it as a prime in the sense $ 0|ab$ implies $ 0|a$ or $ 0|b$, too) and $ \infty$ (similar as $ 0$) is a bit more complicated to explain, but this should already give the idea why to consider them prime.


[hide="Answer"]$a+bi+\sqrt{a^2+b^2}=2+8i$
The only $i$ is found in $bi$, so $b=8$. Therefore, $a+\sqrt{a^2+64}=2\Rightarrow (a-2)^2=a^2+64\Rightarrow a=-15$. Hence, $a^2+b^2=225+64=289\Rightarrow \boxed{E}$.[/hide]


hello, we have $ 2x^4\plus{}3x^3\plus{}x^2\plus{}x\plus{}4\equal{}(x^2\minus{}x\plus{}1)(2x^2\plus{}5x\plus{}4)$
Sonnhard.

can you please work out yhe factorization step-by-step???

Mos of it is just guess and check because it is hard to use the extreme and mode which is used in simple algebraic factorization. Like: 2x^2+3x+1.

we can do the factorization by taking x^3=-1
                                then x^3+1=0
                              or (x+1)(x^2-x+1)=0
                        as any of the factors could be 0.so,
                        2x^4+3x^3+x^2+x+4
                    =  2x^2(x^2-x+1)+5x(x^2-x+1)+4(x^2-x+1)
                    =  (x^2-x+1)(2x^2+5x+4)
is it good enough??? please advice on its process or show other likely ways of doing it.please...

The factorization would be in the form (2x^2+ax+b)(x^2+cx+d), where a,b,c,d integers. Then when you expand, you get a system of four equations and four variables. You can solve this.

Well, the first (but hard-to-see) step is plugging in $ \omega$ where $ \omega^3\equal{}\minus{}1$ and $ \omega\neq\minus{}1$. Observe that $ \omega$ satisfies $ \omega^2\minus{}\omega\plus{}1\equal{}0$.

Then $ 2\omega^4\equal{}\minus{}2\omega$ and $ 3\omega^3\equal{}\minus{}3$, so the expression is equivalent to $ \omega^2\minus{}\omega\plus{}1$, which is 0, so $ x^2\minus{}x\plus{}1$ is a factor.

[quote="earth"]please factorize-
     2x^4+3x^3+x^2+x+4
how can the expression be resolved into factors???[/quote]
$ 2x^4\plus{}3x^3\plus{}x^2\plus{}x\plus{}4\equal{}2x^4\plus{}2x\plus{}3(x^3\plus{}1)\plus{}(x^2\minus{}x\plus{}1)\equal{}2x(x^3\plus{}1)\plus{}3(x^3\plus{}1)\plus{}(x^2\minus{}x\plus{}1)\equal{}(2x\plus{}3)(x^3\plus{}1)\plus{}(x^2\minus{}x\plus{}1)\equal{}(2x\plus{}3)(x\plus{}1)(x^2\minus{}x\plus{}1)\plus{}(x^2\minus{}x\plus{}1)\equal{}(x^2\minus{}x\plus{}1)(2x^2\plus{}5x\plus{}4)$
 :)


Consider points A, B, C, and D in that order on a line, where AC: BD = 3:7 and BC:AD = 1:4. Determine AB:CD.

There's probably a much faster and cleaner solution, but here's one:

[hide="Solution"]First, we transform the given ratios into equations:
\begin{align}\frac {AC}{BD} = \frac {3}{7}\quad & \Longrightarrow\quad7AC = 3BD \\
\frac {BC}{AD} = \frac {1}{4}\quad & \Longrightarrow\quad4BC = AD\end{align}
Then, we simplify $ (\text{equation}\,1) - 3\cdot(\text{equation}\,2)$ to determine $ \frac {AB}{BC}$:
\begin{align*}7AC - 3(4BC) & = 3BD - 3(AD) \\
7(AB + BC) - 12BC & = 3BD - 3(AB + BD) \\
7AB - 5BC & = - 3AB \\
10AB & = 5BC \\
\frac {AB}{BC} & = \frac {1}{2}\end{align*}
Next, we simplify $ (\text{equation}\,1) + 7\cdot(\text{equation}\,2)$ to obtain $ \frac {BC}{CD}$:
\begin{align*}7AC + 7(4BC) & = 3BD + 7(AD) \\
28BC - 3(BC + CD) & = 7(AC + CD) - 7AC \\
25BC - 3CD & = 7CD \\
25BC & = 10CD \\
\frac {BC}{CD} & = \frac {2}{5}\end{align*}
Finally, $ \frac {AB}{CD} = \frac {AB}{BC}\cdot\frac {BC}{CD} = \frac {1}{2}\cdot\frac {2}{5} = \boxed{\frac {1}{5}}$.[/hide]

Let $ x \equal{} AB$, $ n \equal{} BC$, and $ y \equal{} CD$, so we want $ \frac xy$.

We already know that

$ \frac {x \plus{} n}{y \plus{} n} \equal{} \frac37$ and $ \frac n{x \plus{} n \plus{} y} \equal{} \frac14\implies4n\equal{}x\plus{}n\plus{}y\implies n \equal{} \frac {x \plus{} y}3$.

Plugging this into our first equation, we have $ \frac {x \plus{} \frac {x \plus{} y}3}{y \plus{} \frac {x \plus{} y}3} \equal{} \frac {4x \plus{} y}{x \plus{} 4y} \equal{} \frac37\implies28x \plus{} 7y \equal{} 3x \plus{} 12y\implies25x \equal{} 5y\implies\frac xy \equal{} \boxed{\frac15}$.


There are solutions to this equation since it did not mention positive integers.

[quote="seamusoboyle"]hmmm. im pretty sure he said without FLT. i could be wrong, but im not.[/quote]

sorry i didn`t see. the equations doesn`t have sols, boxedexe :wink:

What about $(2^{30},0),(0,2^{30})$?  As I said, it did not positive integers.

[quote="Andreas"]Find the integer solutions of the equation
\[x^{3}+y^{3}= 8^{30}\]
(without FLT).[/quote]

one of $x$ or $y$ must be positive, wlog $x$, then 3 cases

$y$ is negative, then $x^{3}=(2^{30})^{3}+(-y)^{3}$ has no solutions by FLT
$y$ is positive, no solutions again by flt
$y=0$ so $x=2^{30}$
then switching, we get the other solution, for a solution set of:

$(x,y)=(2^{30},0)$, $(0,2^{30})$

Nice Solution, but the problem states "without FLT."


linear algebra
matrix
group theory
superior algebra
superior algebra unsolved

Sorry, I'm new to abstract algebra. I don't know if this is the right forum. Anyway here's the problem:

Let $G : ={\left[\begin{array}{cccc}1 & n \\0 & 1\end{array}\right]}$ under matrix multiplication.

Show $f: G \mapsto Z$ where $f{\left[\begin{array}{cccc}1 & n \\0 & 1\end{array}\right]}=n$ is an isomorphism

i think they are. thanks.
[quote="boxedexe"]I have only studied group theory for a few hours, so I only know the very elementary things, more like notation.  I have never read about matrices in group sand such, so bear with me.  Tell me if I am correct.

1) Prove that $G$ forms a group with the identity $e_{G}=\left[\begin{matrix}1&0\\0&1\end{matrix}\right]$, which is very easy.
2) Prove that $f$ is a homomorphism by proving that $f\left[\begin{matrix}1&n\\0&1\end{matrix}\right]=f\left(\left[\begin{matrix}1&p\\0&1\end{matrix}\right]\left[\begin{matrix}1&q\\0&1\end{matrix}\right]\right)=f\left[\begin{matrix}1&p\\0&1\end{matrix}\right]f\left[\begin{matrix}1&q\\0&1\end{matrix}\right]$
3) Prove that $f$ is both injective and surjective.

Are these steps correct/complete?[/quote]

Assuming you want $G=\left\{\left[\begin{matrix}1&n\\ 0&1\end{matrix}\right] : n\in\mathbb{Z}\right\}$, then you can define your isomorphism in the natural way that you want to. Let $f: G\rightarrow \mathbb{Z}$ by $\left[\begin{matrix}1&n\\ 0&1\end{matrix}\right]\mapsto n$. Identity goes to identity since $f\left(\left[\begin{matrix}1&0\\ 0&1\end{matrix}\right]\right)=0$.

Let $\left[\begin{matrix}1&l\\ 0&1\end{matrix}\right]$, $\left[\begin{matrix}1&k\\ 0&1\end{matrix}\right]\in G$. Then $f\left(\left[\begin{matrix}1&l\\ 0&1\end{matrix}\right]\left[\begin{matrix}1&k\\ 0&1\end{matrix}\right]\right)= f\left(\left[\begin{matrix}1&l+k\\ 0&1\end{matrix}\right]\right)=l+k= f\left(\left[\begin{matrix}1&l\\ 0&1\end{matrix}\right]\right)+f\left(\left[\begin{matrix}1&k\\ 0&1\end{matrix}\right]\right)$

Now you need to check 1-1 and onto. Both of which should be pretty straightforward from the definition. I can do it, though, if you'd like.

[quote="boxedexe"]Oops, I have a typo. In my post, it should say $f\left[\begin{matrix}1&n\\0&1\end{matrix}\right]=f\left(\left[\begin{matrix}1&p\\0&1\end{matrix}\right]\left[\begin{matrix}1&q\\0&1\end{matrix}\right]\right)=f\left[\begin{matrix}1&p\\0&1\end{matrix}\right]+f\left[\begin{matrix}1&q\\0&1\end{matrix}\right]$[/quote]

But shouldn't $f\left[\begin{matrix}1&n\\0&1\end{matrix}\right]=f\left(\left[\begin{matrix}1&p\\0&1\end{matrix}\right]\left[\begin{matrix}1&q\\0&1\end{matrix}\right]\right)=f\left[\begin{matrix}1&p\\0&1\end{matrix}\right]f\left[\begin{matrix}1&q\\0&1\end{matrix}\right]$[/quote]

Why did f(ab) = f(a) + (b). I thought its matrix multiplication, where did the $+$ sign came? (Assuming that $a$ is the matrix with the $p$ and $b$ be the matrix with the $q$) Can you please explain, I'm quite poor at this. Thanks.

[quote="boxedexe"]No. $f\left(\left[\begin{matrix}1&p\\0&1\end{matrix}\right]\left[\begin{matrix}1&q\\0&1\end{matrix}\right]\right)=f\left[\begin{matrix}1&p+q\\0&1\end{matrix}\right]=f\left[\begin{matrix}1&p\\0&1\end{matrix}\right]+f\left[\begin{matrix}1&q\\0&1\end{matrix}\right]$[/quote]

Can you please explain how did that happen?

EDIT:
Sorry,i think i got it.


Let $a,b,c,d$ be nonegative real numbers. Prove that
\[\underset{sym}{\sum}\frac{1}{a^{3}+b^{3}}\ge \frac{243}{2(a+b+c+d)^{3}}\]
:)

@Mr.Thuan: I have solved many problems you sent to me, but i can't send them to you by mail because i have lost my USB yesterday. :(

Yes, that is correct inequality and can be solved in a similar manner proposed by Nguyen Quoc Ba Ca. Even a more general result is possible.

But I think that it's not very easy.

[quote="VIETNAM"]The similar.
Let a,b,c,d be nonegative real numbers. Prove that :
$\sum_{cyc}\frac{1}{ a^{3}+b^{3}+c^{3}}\geq \frac{24}{(a+b+c+d)^{3}}$
And it's very easy.[/quote]
No one like this? :(

That is true and NICE solution. Come on Silouan.

Thank you all . I feel happy that you liked it . :)


[quote="mr.danh"]Prove that there exists an integer $ n$ such that $ 2^n\plus{}3^n$ has exactly $ 23$ distinct prime divisors.[/quote]

who can post solution?


induction
combinatorics open
combinatorics

[quote="Iltanen"]Henry plays a game where he has to guess if a card picked from a deck is red or black:
He makes a guess and picks up a card.
If his guess was correct he gets a point.
He makes his next guess; if there are more black cards left in the deck he chooses black and vice versa and picks up a card.
If there are equally many black and red cards he always chooses black.

On the average, how many points does he get when all the 52 cards have been picked up?

I have no idea of the difficulty of this so I posted it here.[/quote]
An idea to how to solve it:

Let $ a_{r,b}$ be the average number of points he gets with $ r$ red cards and $ b$ black cards.
We know that $ a_{n,m} \equal{} a_{m,n}$ and $ a_{n,0} \equal{} n$
We can easily see that $ a_{n,n} \equal{} \frac{1}{2} \plus{} a_{n,n\minus{}1}$
And for $ m > n$ we get $ a_{m,n} \equal{} \frac{m}{m\plus{}n} (1 \plus{} a_{m\minus{}1,n}) \plus{} \frac{n}{m\plus{}n} a_{m,n\minus{}1}$

So we "just" have to calculate $ a_{26,26}$. It's almost impossible to do without a computer, so if we could derive a general formula for $ a_{n,n}$ it would help a lot.

Lets instead try to calculate $ a_{n,k}$ for $ n>k$.
$ k \equal{} 0$: $ a_{n,0} \equal{} n$ and $ a_{0,0} \equal{} 0$
$ k \equal{} 1$: $ a_{n,1} \equal{} \frac{n}{n\plus{}1}(1\plus{}a_{n\minus{}1,1})\plus{}\frac{1}{n\plus{}1}a_{n,0} \equal{} \frac{2n}{n\plus{}1} \plus{} \frac{n}{n\plus{}1}a_{n\minus{}1,1}$ and we get $ a_{n,1} \equal{} \frac{n^2 \plus{} n \plus{} 1}{n\plus{}1} \equal{} \frac{n^3\minus{}1}{n^2\minus{}1}$ by induction, and $ a_{1,1} \equal{} \frac{3}{2}$

I might complete it later, but I've gotta go :)

If i should guess i would say $ 26 < a_{26,26} < 27$


calculus
integration
abstract algebra
superior algebra
superior algebra unsolved

Hi all

I'm just working through problems in Marcus on my own time. I am stuck on problem 30 from chapter 1.

Let $ R$ be an integral domain. Show that two ideals of $ R$ are isomorphic as $ R$-modules $ iff$ they are in the same ideal class.

Now I can prove that if they are in the same ideal class they are isomorphic. But I have been unable to prove the other direction. Any help would be appreciated, but please don't use any high level machinery if possible, as it is the first chapter and none of the previous problems relied on any high-powered theorems at all.

Thanks for the help.

Let the two ideals be I and J, and let f be an isomorphism between them. Now fix an element x in I and look at its image in J. Basically, we want to identify x and its image. So, what are two easy a,b to multiply by so that ax=bf(x)?


Are there any books that are on specific subjects like algebra or combinatorics or geometry that are as easy to read as AoPS. I am trying to look for one that goes more in depth but still easy to understand.

You should do AoPS first because it gives you a good foundation on all of the subjects. Then youi can go into more specific books like the ones stated in the Resources section on this site.


Given a circle of center A and let CD be the diameter of that circle. Let CF be a chord of the circle. Show that the triangle CFD is a right triangle.

**Geogebra picture included to illustrate the above problem. (F=F1)
[geogebra]c39b98579f820f5cab1c0c97bf5536aa01419e11[/geogebra]

This is pretty standard

Line $ AF$ splits the triangle into two isoscelese triangles: $ \bigtriangleup CAF$ and $ \bigtriangleup DAF$
In each triangle there are two angles the same let them be $ x$ and $ y$.

$ x\plus{}x\plus{}y\plus{}y \equal{} 180^o$
$ 2(x\plus{}y) \equal{} 180$
$ x\plus{}y \equal{} 90$

So all such triangles are right angled triangles

[quote="kratsg"]Just to clarify, the measure of angle CDF = y, the measure of angle DCF = x, and the measure of angle CFD = x+y. And then by using the property that all inner angles of a triangle must add up to 180 degrees, you can say that for triangle CFD: x+(x+y)+y = 180.

Correct?[/quote]

yeah that's it


Define $\phi_{n}\in (l^\infty)^{*}$ by $\phi(f)=n^{-1}\sum_{j=1}^{n}f(j).$Then the sequence $\{\phi_{n}\}$
has a weak* cluster point $\phi$, and $\phi$ is an elemnet of $(l^\infty)^{*}$
that does not arise from an element of $l^{1}.$

( I think since $\parallel \phi_{n}\parallel \leq 1$, so by Alaoglu Thm, this sequence is
relatively weak* compact, so must have a cluster point, but
what is it? And why it is not from $l^{1}$?)

$\phi\notin\ell_{1}$ because it vanishes on all sequences with finite support.


Prove the Ceva Theorem using the Menelaus Theorem.

I am sure that this was already discussed in this forum.But to prove Menelaus using [b]only [/b] Ceva is much difficulter.Lets ask [b]darij[/b] to give us a link .:)

Oh well, the proof of the Ceva theorem using the Menelaos theorem is quite well-known and easy: See http://www.cut-the-knot.org/Generalization/Menelaus.shtml#CfromM .

  Darij