This is the new game i made up.

It's more like combination of the game of elimination (in this forum) and Bang! (card game).

I will take 10 ppl for the first try.

Rules:

There are 10 ppl, 1 king, 1 queen, 2 knights (one known to public and one unknown) and 6 citizens.

King tries to kill the queen and 3 citizens without losing any knights or kill all citizens and queen to win the game.

Queen must kill King firsthand to win.

Citizens have same objective as queen.

Knights can cooperate (have same goal) as the king, or kill queen, other knight, and the king.

King starts w/ 20 pts, known knight 15 pts, and others 10 pts.

Everyone will be private except King and the known knight.

If the known knight dies, unknown knight will become known.

Daily, king will receive 3 pts, known knight 2 pts, and 1 pt for rest.

A person may attack once per day, removing same number of points from both sides and removal can only be up to 2 pts, and attack 1 person only.

King can execute someone by giving up 15 pts every 3 days.

Also, king may not be attacked unless both knights are dead. Also, anyone can give points (as much as wanted, once per day) to the king, but only king can give points to knights.

I will PM who's who if all 10 ppl are signed up; and king and knight will be announced.

/in, how are you choosing the queen and king and stuff?

I want to join.

What happens if the knight wants to do the "kill queen, other knight, and the king" objective?
As long as he is alive, he can't attack the king because "king may not be attacked unless both knights are dead." I think it's impossible for that objective to be achieved if nobody, not even him, can attack the king until he's dead.

[size=59]ME ME ME![/size]

The message is too small. :rotfl:

OK. Now I have 10 people to start with.

If you are not PMed, you will know who you are on this board, or you are citizen.

Signups:
1.tinytim
2.lotsofmath
3.MustItAlwaysBeMe
4.(^_^)
5.ProtestanT
6.dynamo729
7.distracted523
8.chenhsi
9.perfect628
10.iLord

Also, ProtestanT, in case you are wondering, knights can attack king anytime if wanted. I should have made that clear.

Also, perfect628 will be the king, and (^_^) will be the known knight.

Additonal rules (i wrote them down, typed it with memory, so i am filling it up):

You may not tell anyone in AoPS who you are, but you may discuss (here or PM) who others are.

Also, random situation may occur in favor of someone different every time.

There will be quests, and first person to accomplish will get prizes.

The Game starts today (8:10 in eastern) at this moment.

1.tinytim-11
2.lotsofmath-11
3.MustItAlwaysBeMe-11
4.(^_^)-17
5.ProtestanT-11
6.dynamo729-11
7.distracted523-11
8.chenhsi-11
9.perfect628-23
10.iLord-11

Today is not king's execution day. King may not execute someone (he will be able to on June 18(,21,24,27,30 and counting).

Quest of today: none
Prize: none

Obviously, perfect628 is king, and (^_^) is knight.

[b]Attack (^_^) 2 pts[/b]

Let's get him!

...
7.distracted523-5
9.perfect628 (King)-3

..... MIABM commited suicide

7.distracted523-6
9.perfect628 (King)-6

I call that king will win. I bet 1000 rating :P

Oh darn, yeah. I think he won anyways...even if iLord stayed alive...

[b]concedes[/b]

Yeah. The game was an interesting idea, but I think we need a better way of executing it.

You can still PM me and i hope you do pm me since I really need help on revisions.....

Still, the job for mod is open, I'll take alexhhmun's idea, and my revisions are good too (since knights died so easily :P )

I will open the Revolution II thread this coming monday. I think I have enough time to mod before AMP, and if it gets too long, someone can become the mod hopefully :|

You could make 2 sets of rules, one built to extend the game, and one build to shorten it. Remember: randomising is unbiased.

My recent idea: [b]if[/b] I can get 20 ppl (need your help too) to play, we can get 2 countries w/ same composition and have a war. For a side to win, their kingdom must win (with a exception role). It should be more interesing.

Also, i made 10 events so that each country can choose their order of them in their favor.

I don't have much time right now so i'll reply later.

OK. Sorry for double posting, but I think I mentioned it.

Changes not mentioned:

If the known knight dies, it will remain that way. The unknown knight will not be revealed.

Renegade is a new position. If the kingdom falls, and renegade has least point among all other players in the same kingdom, s/he will win.

Mob Lynch: during the whole game, a Rest (those who are not known) has 1 vote to mob lynch someone. Whenever majority is met, the person will be lynched (cannot be known knight or the king). Majority does not include abstained votes.

There are 9 events (one of them happens twice, so total 10) that the player may choose in order. I have decided them, but I may change them so I am not posting yet. Every day, 1st person to post what actions to take (doesn't matter with the kingdom, same effect will go on to same kingdom regardless of the poster. The poster also may choose to not to use any of events.)

Those will be only change in the rules. You may ask me to add in some more. :|

"If the kingdom falls, and renegade has least point among all other players in the same kingdom, s/he will win."

Do you mean most points? That makes more sense; the renegade takes control of the kingdom.

I can't understand the Mob Lynch part, but I will join!

Now please sign up again if you are not the first 3 on the Revolution II: The War on this forum


I need to know how I can run a dtx file. I use TexNic Center. I have two files a dtx file and ics file. Any help welcome.

Hi,
A .dtx file is a Documented laTeX file. If you want to get a .sty file, you don't need to run the .dtx file. You need to run the Internet Communication Settings file that accompanies the .dtx file. Just compile it in the same way you would compile a normal .tex file. The ICS file will call the .dtx file, so they need to be in the same folder. It should produce a .sty file and the accompanying documentation. If you're looking for something other than the .sty file, I can't help you.
SSN.


get MSN it has microsfot photo something, encarta, money, music (not as good as AOL), adn a lot of programs.

I'm kinda planning this..
I want to get people from Mathcounts to join by instant messenger then practice for sprint round, giving maximum of 2 minutes per each question.. 3 minutes per each target question.. 45 seconds per countdown question.. So, I'll try to get AIM and then get some people and practice during this week before Friday.. I have state competition on Friday and saturday.. so..

I have some MATHCOUNTS Countdown probs, what's your sn?

sn? What do you mean? Are you talking about what kinds of stuffs I have? Um.. I don't have much, though, 2002, 2001 questions and All-Time Greatest MathCounts Questions... I hae disket but it doesn't work well, sooo...

Like what's your AIM [b]s[/b]creen [b]n[/b]ame?

I'm getting new one now...
And will you join us for practice.. I meant "us" by four(not sure, but I have a list, though) people I've already found...
If you are thankfully agreed to join on the practice, please go to MATHCOUNTS forum and then to Great Practices and post your response..

Thanks..


Graph it separately for each quadrant.

[b]Upper-right:[/b]  $x$ and $y$ are both positive, so graph $x+y \le 2$.

[b]Upper-left:[/b]  $x$ is negative, $y$ is positive, so the absolute value of $x$ (which we usually write $|x|$) is just $-x$, so graph $-x+y \le 2$.

And so forth.

[quote="espark52"]how do you graph: $|x|+|y| \leq 2$

[size=75][color=darkred]Changed to LaTeX.
-nebula42[/color][/size][/quote]

Try to plot the graph at equals, and notice the shape.

You would get a square with it's vertices at $(2,0), (0,-2), (-2,0), (0,2)$ where the answers are the inside the square and the edges, but t0rajir0u has the best way of thinking about it.

so basically its either

$x+y\leq2$ [u][color=red]and[/color][/u] (since its less than or equal to) $x+y\geq{-2}$

-jorian

Well, you've forgotten $-x+y \le 2$ and $x-y \le 2$.


Does anybody know anything about Croatian results?

Matija Basic - 17 points - BRONZE
Goran Drazic - 16 points - BRONZE
Nikola Grubisic - 12 points - HM
Rudi Mrazovic - 19 points - BRONZE
Visnja Polic - 10 points - HM
Tvrtko Tadic - 15 points - HM


Next year I will be teaching a junior chemistry olympiad class for my school. Since its a junior class I assume they will have limited knowledge of Chemistry (probably enough to write reactions and identify elements but nothing too theoretical). Can anyone suggest some sort of teaching plan? I really have no experience with this kind of stuff. Here are a few things I personally think should be taught:

-Atomic structure (and also radiation stuff)
-Concept of mole
-Periodicity
-Acid/Base
-Simple organic chemistry
-Solubility (and some common reactions while I'm at it)
-Redox
-Bonding (ionic, covalent, metallic)
-Simple gaseous properties

I think that the order you go in is pretty important.  Although I don't remember exactly, when I was in AP Chemistry, the order we went in, which I found pretty effective, was something like,
-Concept of Mole/Stoichiometry
-Periodic Table Stuff
-Atomic Structure
-Bonding
-States of Matter and Solubility
-Something I Can't Remember
-Equilibrium
-Acids/Bases
-Thermal Chemistry
-Redox/Electrochemistry
-Simple Organic

You are going to teach chemistry? How old are you? And what is the average age of your future class?

BanishedTraitor, do you feel confortable enough in Chemistry in order to teach a young class? I am asking this because there is a *big* distance between solving hard problems and teaching. In am not trying to scare you (also, I don't know you personnaly), just keep in mind that teaching is a big responsability. I can give you some more advice, but only after knowing your age and your future students ages.

It's not "teaching" in like the serious sense of the word. It's more like tutoring (if there's a difference  :huh: ). It's kind of like introducing them to concepts.

Well I'm turning 17 next year and the average age of the people would (I imagine) be around 14.

Don't forget to take your students' past history. If they've taken any level of advanced biology then you can count on some experience in the organic department - which means you can focus a little bit more on inorganic.

Chemistry is also a subject where hands on experience comes in to play. Use labs sparingly - too many labs will just turn the class into a fun period and less on learning actual chemistry.

[quote=7h3.D3m0n.117"]too many labs will just turn the class into a fun period and less on learning actual chemistry.[/quote]
Funny that you say that because for the most part, I didn't think labs were very fun.  I think the key is to do many, many problems.  The more problems you do, the better you understand.

Well, I don't think I meant "fun" - more like a "chill" period instead of an academic one. You can't possibly learn concepts through labs.

It turns out, however, that true chemistry knowledge is learned on the lab, not just by solving a bunch of problems in the paper.

BanishedTraitor, I am somewhat curious: what is the precise reason for a person of 17 years old to be assigned for "introduce concepts" to persons of 14 years old?

[quote="Carcul"]It turns out, however, that true chemistry knowledge is learned on the lab, not just by solving a bunch of problems in the paper.[/quote]
I agree, but I'm much more of a math person so I learn the material a lot better when I see the mathematical formulations behind the idea.

I guess the first thing one needs to do when he or she embarks on teaching a group of beginners a new subject is to know his or her students' educational background.  Then one should set reasonable targets as to what he or she wants his or her students to know by the end of the course.  Also, one should structure the course to formulate the best way in presenting the material.  Perhaps, one should also find a suitable textbook that is at the right level for the students.

Speaking from personal experience, I had a terrible professor for an introductory inorganic chemistry course last year.  My professor wanted to teach the stuff that he was interested in.  Hence, we spend basically the entire course on molecular geometry.  We delved into the group theory details (with the explanations, proofs omitted), and talked of its applications in things like spectroscopy etc. However, he ignored the fact that his students do not have the prerequisite mathematical knowledge (calculus I and II and general chemistry were the prerequisites) to study that stuff rigorously let alone understand it.  My professor also did not use a textbook for his course.  Instead, he chose to explain everything through his power point slides.  I did not think he was a very effective teacher.

On the other hand, I am currently taking a physical chemistry course for the biological sciences.  This professor is a physical chemist by training, yet he understands his audience.  I think he is doing a good job, and he underlines the concepts with good examples.   Also, he reinforces what he teaches by assigning readings and exercises from a good supplementary textbook "Physical Chemistry for the Chemical and Biological Sciences" by Raymond Chang.

Anyways, that's only my opinion.


search function geometry inequalities MATHCOUNTS

I desire to learn the mathematics needed for IMO.
Would someone give me some links or say what topics should be learnt and if possibly all the theorems? Also, would you recommend some textbooks and exercise books, please?

Thanks.  :P

This is probably the wrong forum to post this in. You might want to try posting this in the olympiad section or the pre-olympiad section, or just take a look around there :wink:

[quote="JBL"]This question exists, in various different forms, dozens of times on the forum.  Try using the search function or looking in the Books and Journals subforum to find past answers.[/quote]

Which terms should I search for?

hahaha

sorry, you've come to the right place :D
It's just so general a question I don't think I've heard it asked that way before...

If you go to Olympiad Section, you'll see different parts: Geometry, Inequalities, Algebra, etc. On each section, you can see the part of "Theorems and Formulas". There, you can see some useful things to use in an IMO

But I was searching for something clear and organized. Maybe there isn't.  :(

This website sells very good books. I reccomend you buy both AoPS 1 and AoPS 2. AoPS 2 if you already have mastery over basic mathcounts material.


function
modular arithmetic
number theory

Okay, I learned bases like three years ago, and um, I never realized that BASES are actually mods.  Every time I looked at a solution, and I saw mod, I cringed and scrolled down.  So, can some sum up the function of a mod and maybe provide an example?

Yeah, what am I doing on this site?  :P  (Also, I still don't understand Legendre's formula.  I even asked my teacher and she never heard of it, but IS looking it up ;))

it is a simple notation. instead of saying that both $a$ and $b$ give the same residue when divided by $n$, you say $a\equiv b \pmod n$, or the same thing $n \mid a-b$. got it now? :P

ARE YOU SERIOUS?!   :P Haha, wow... Thanks!  Bases...riiight.

SingTheSorrow you should get a book called [u]Number Theory[/u] by G.E. Andrews. It explains all these things in simple terms. Its a good book -- and very cheap too. I think you can find it in most bookstores for about 7 or 8 dollars US.

[quote="SingTheSorrow"]Wow, you rock.   :P  Haha, Valentin, you should be happy - you won't be hearing anymore stupid questions from me anymore.   :D  (That is, if the book is actually "simple"  :))[/quote] your questions did not disturb me in any way :P you can keep asking :)


Its a really cool game. View my blog for more info.

Very cool game. Played it a lot at school with my friends.

Really fun game.

Inspired me to make a mod of it: [url=http://nisnevich.com/pse/mycreations/little_fighters_25.htm]Little Fighters 2.5[/url]

Cool.Little fighter's 2.5 is simply amazing. The best part , it's a game you can go on playing. :D

It's okay but I beat the stage mode already.  Really.  I used Mario and did his summoning vampire move.  Defend-Up-Jump i think it was.

Try beating stage mode with one of the original 10 LF2 characters (Deep, John, Henry, Rudolf, Louis, Firen, Freeze, Dennis, Woody, Davis). That's what you had to do in the original game. You'll find that it is [i]much [/i]harder.

I consider Mario, LF3 Firen, LF3 Freeze, and LF3 Firzen  to be [i]way [/i]too cheap all in all, and never play as them.

I did not make LF2, nor did I make any of the extra characters. All I did was join the two together: much less work than actually making hte characters.  :D

Hmm... lots of people like LF2. Maybe we can start a Little Fighters 2.5 tournament?  :D

The old school LF1 was awesome too... But it's in Chinese.

LF2 is cool :D

Wow, this topic is really old!  :?

Didn't your mother tell you never to revive old topics?  :mad:

it hasn't lasted for over a month yet, so I don't think it's all that old.  a month is the general standard, right?

[quote="-Terminator-"]If we are going to have a LF2 tournement we need to know each others IP adresses.[/quote]\
i have no idea what my IP adress is...how can i find out?

Google for "what is my ip" or something. There's a program in Windows that does the same thing too, but I forgot which.

To find your IP click here. [url]http://www.chami.com/tips/internet/041498I.html[/url]

Please continue to visit this topic. Also don't forget to post your IP.

Uhh I don't know if it's a good idea to use third party programs to get your IP address. I think if you write "ipconfig" in WINDOWS DOS, you will get your IP. Or maybe just your static IP? I dunno.

if you have Windows, go to Start --> Run.  Type cmd, hit Enter.  Then type ipconfig.  It should say IP address somewhere, followed by a series of numbers (four, I think) later in the same line.  If you're on a Mac, I can't help you.  Macs  :mad:  :wacko:  :ewpu:  Yeah, you get the point.


A triangle has three angles, A, B, and C.  Tan A = 1, tan B = x.  What is tan C in terms of x?

Thanks!

     -- Pawel

$\tan C = \tan \left( {\pi  - A - B} \right)\mathop  = \limits_{\mathop {\tan A = 1 \Rightarrow A = 45^0 }\limits^ \uparrow  } \tan \left( {\pi  - \frac{\pi }{4} - B} \right) = $

$ = \tan \left( {\frac{{3\pi }}{4} - B} \right) = \frac{{\tan \frac{{3\pi }}{4} - \tan B}}{{1 + \tan \frac{{3\pi }}{4}\tan B}} = \frac{{ - 1 - \tan B}}{{1 - \tan B}} = $

$ = \frac{{x + 1}}{{x - 1}}$

[hide]
Well, I didn't use the easiest way, but I got an answer that seems reasonable.

First, since $tan B = x, B = arctan x$

Since the angles sum must be 180, $C = 135 - arctan x$.

We need to find $tan C$, which is

$tan (135 - arctan x)$

$\frac{tan 135 - tan(arctan x)}{1 + tan135 tan(arctan x)}$

$\frac{-1 - x}{1 - x}$

$tan C = \frac{x+1}{x-1}$
[/hide]

[quote="Bictor717"][hide]In $\triangle ABC$:\[\tan A\tan B\tan C=\tan A+\tan B+\tan C\]
\[1\cdot x\cdot\tan C=1+x+\tan C\]
\[\tan C\left(x-1\right)=1+x\]
\[\tan C=\frac{x+1}{x-1}\][/hide][/quote]

Can you post a proof of the equation you used?? Thanks!

$A+B+C=180^\circ\rightarrow C=180^\circ-\left(A+B\right)$

$\tan C=\tan \left(180^\circ-\left(A+B\right)\right)=\frac{\tan 180^\circ-\tan\left(A+B\right)}{1+\tan 180^\circ\tan\left(A+B\right)}$

$\tan 180^\circ=0\rightarrow\tan C=-\tan\left(A+B\right)=-\left(\frac{\tan A+\tan B}{1-\tan A\tan B}\right)=\frac{\tan A+\tan B}{\tan A\tan B-1}\\
\tan A\tan B-1=\frac{\tan A+\tan B}{\tan C}$

\begin{eqnarray*}\tan A\tan B\tan C&=&\left(\tan A\tan B\tan C-\tan C\right)+\tan C\\
&=&\tan C\left(\tan A\tan B-1\right)+\tan C\\
&=&\tan C\cdot\frac{\tan A+\tan B}{\tan C}+\tan C\\
\tan A\tan B\tan C&=&\tan A+\tan B+\tan C
\end{eqnarray*}


ratio
geometry
angle bisector
geometry unsolved

the triangles $ KCA$ and $ KBM$ are similar with ratio $ \frac{CK}{BK}\equal{}\frac{1}{2}$ so $ KM \equal{} 2 AK \equal{} \frac{4}{4\plus{}2} \sqrt{4 \cdot 2 \cdot \frac{11}{2} \cdot \frac{5}{2}} \equal{} \frac{2}{3} \sqrt{110}$

Are you sure your solution is correct dear gabriel?
the answer of this problem is $ 2$ $ \times$ $ \sqrt{6}$

[quote="behdad.math.math"]In $ ABC$ triangle $ AB$ $ \equal{}$ $ 4$, $ AC$ $ \equal{}$ $ 2$, $ BC$ $ \equal{}$ $ 3$.the bisector of $ < A$ meet $ BC$ at $ K$.the line that is

parallel to $ AC$ and passes $ B$, intersect the extention of $ AK$ at $ M$ point,find $ KM$.[/quote]
We see that $ \angle KMB \equal{} \angle CAK$ and $ \angle MBK \equal{} \angle KCA$. Which means that $ \triangle MBK$ is concurrent to $ \triangle MCK$ and that $ \triangle ABM$ is isoceles, which means $ BM \equal{} 4$. So $ \frac{MK}{AK} \equal{} \frac{BM}{AC} \equal{} \frac{4}{2} \equal{} 2$. So $ MK \equal{} 2AK$. By stewarts theorem we get that $ CB((AK)^2 \plus{} CK \cdot KB) \equal{} KB \cdot (AC)^2 \plus{} CK \cdot (AB)^2$.
On the other hand $ \frac{CK}{KB} \equal{} \frac{AC}{AB} \equal{} \frac{1}{2}$. So $ CK \equal{} 1$ and $ KB \equal{} 2$. Thus $ 3((AK)^2 \plus{} 1 \cdot 2) \equal{} 2 \cdot 2^2 \plus{} 1 \cdot 4^2$ $ \iff$ $ AK \equal{} \sqrt{6}$. Hence $ MK \equal{} 2AK$ gives $ \boxed{MK \equal{} 2\sqrt{6}}$

[quote="¬[ƒ(Gabriel)³²¹º]¼"][quote="¬[ƒ(Gabriel)³²¹º]¼"]the triangles $ KCA$ and $ KBM$ are similar with ratio $ \frac {CK}{BK} \equal{} \frac {1}{2}$ so $ KM \equal{} 2 AK \equal{} \frac {4}{4 \plus{} 2} \sqrt {4 \cdot 2 \cdot \frac {9}{2} \cdot \frac {3}{2}} \equal{} 2 \sqrt {6}$[/quote]

sorry, there was a stupid calcolation error  :P[/quote]
Can you explain how you calculated it?

[quote="Mathias_DK"]Can you explain how you calculated it?[/quote]
In a triangle ABC let the angle bisector from A meet BC on K, then $ AK \equal{} \frac{2}{b\plus{}c} \sqrt{bcp(p\minus{}a)}$ [and cyclic for the others].

[quote="¬[ƒ(Gabriel)³²¹º]¼"][quote="Mathias_DK"]Can you explain how you calculated it?[/quote]
In a triangle ABC let the angle bisector from A meet BC on K, then $ AK \equal{} \frac {2}{b \plus{} c} \sqrt {bcp(p \minus{} a)}$ [and cyclic for the others].[/quote]
Ok. Thanks :) (I just realised this can be proved by stewarts too :) )


I know there is well known trick for such problems, but I can't remember.

Let's solve it . I think it 'll become easily if we find its [b]rhythm[/b]

You made a small calculation mistake.
Right expression is
\[\lim_{n\to\infty}\frac{u^{2^{n+1}}+1}{u(u^2+1)(u^4+1)...(u^{2^n}+1)}=\frac{u^2-1}{u}=\sqrt{\left(u+\frac{1}{u}\right)^2-4}=\sqrt{21}.\]


[i]Notations.[/i] A circle $k$ with center $U$ and radius $\rho$ will be abbreviated as "circle $k\left(U,\rho\right)$".

For any two circles $a$ and $b$, we define a "distance" $d_{ab}$ between these two circles as follows: Let one of the two external common tangents of the circles $a$ and $b$ (it doesn't matter which one we take) touch these circles at two points $P$ and $Q$; then, we define $d_{ab}$ as $d_{ab}=PQ$. Note that this distance $d_{ab}$ is thus only defined if the circles $a$ and $b$ have external common tangents.

[b]Casey Theorem.[/b] Let five circles $C\left(O,r\right)$, $C_{1}\left(O_{1},r_{1}\right)$, $C_{2}\left(O_{2},r_{2}\right)$, $C_{3}\left(O_{3},r_{3}\right)$, $C_{4}\left(O_{4},r_{4}\right)$ be given. If the circles $C_{1}$, $C_{2}$, $C_{3}$, $C_{4}$ are all interiorly tangent to the circle $C$, and the points where they touch the circle $C$ are arranged in the same order as we have named them, then the following relation holds:

$d_{C_{1}C_{2}}\cdot d_{C_{3}C_{4}}+d_{C_{2}C_{3}}\cdot d_{C_{4}C_{1}}=d_{C_{1}C_{3}}\cdot d_{C_{2}C_{4}}$.

[i]Remark.[/i] The same result holds if the circles $C_{1}$, $C_{2}$, $C_{3}$, $C_{4}$ are all exteriorly tangent to the circle $C$.

the proof uses Ptolemy's Theorem - since it is a generalization of it :)

Proof: Let's denote T1,T2,T3 and T4 the points of tangentcy of the circles C1, C2, C3 and C4 with C and we are going to compute d_CiCj the tangent distances between the circles Ci, Cj using the radius of the circles ri,rj, and the radius r of the circle C(O,r) and the distance TiTj between the points of tangentcy with C(O,r)

Applying the cosinus theorem in the triangles: TiOTj and OiOOj we get:

cos(TiOTj)=(2r^2-TiTj^2)/2r^2 and
OiOj^2=(r-ri)^2+(r-rj)^2-2(r-ri)(r-rj)cos(TiOTj)=(r-ri)^2+ (r-rj)^2-2(r-ri)(r-rj)*(2r^2-TiTj^2)/2r^2=(r-ri)^2+(r-rj)^2-2(r-ri)(r-rj)+(r-ri)(r-rj)*TiTj^2/r^2=(ri-rj)^2+(r-ri)(r-rj)*TiTj^2/r^2.

Let AB be the common exterior tangent of the circles: C(Oi,ri) and C(Oj,rj) so d_CiCj=AB.
From ABOjOi ( which is a right angled trapezoid) we get:

d_CiCj^2=AB^2=OiOj^2-(OiA-OjB)^2=(ri-rj)^2+(r-ri)(r-rj)TiTj^2/r^2-(ri-rj)^2=(r-ri)(r-rj)*TiTj^2/r^2 =>
d_CiCj=TiTj/r * sqrt((r-ri)(r-rj)).
The relation from the theorem becomes:
T1T2/r*sqrt((r-r1)(r-r2))*T3T4/r*sqrt((r-r3)(r-r4))+T2T3/r*sqrt((r-r2)(r-r3))*T4T1/r*sqrt((r-r4)(r-r1))=
T1T3/r*sqrt((r-r1)(r-r3))*T2T4/r*sqrt((r-r2)(r-r4)) which is T1T2*T3T4+T2T3*T4T1=T1T3*T2T4 which is Ptolemy thorem.. so Casey's theorem is proved!!!! :)

cheers! :D

Is it true that if $ d_{C_{1}C_{2}}\cdot d_{C_{3}C_{4}}\plus{}d_{C_{2}C_{3}}\cdot d_{C_{4}C_{1}}\equal{}d_{C_{1}C_{3}}\cdot d_{C_{2}C_{4}}$ for some circles $ C_1(O_1,r_1),C_2(O_2,r_2),C_3(O_3,r_3),C_4(O_4,r_4)$ then there exists an circle $ C(O,r)$ tangent to all these circles  :roll:


Por Favor aquellos que saben vectores que pongan los puntos importantes del triangulo y/o del cuadrilatero en vectores.
a los puntos que me refiero son como el gravicentro, circuncentro, hortocentro, punto de nagel, punto de gergone, punto de fermat, incentro,.....
por favor al ponerlos que lo pongan con demostracion o alguna sug. para su demostracion. :lol: !!!! gracias!!!!

aqui pongo el valor vectorial del gravicentro desde cualquier origen
A(1/3)+B(1/3)+C(1/3) su demostracion la sig.
denotemos ABC un triangulo D,E puntos medios de BC y AC, respectivamente.
tenemo que D= B(1/2)+C(1/2) Y E= A(1/2)+C(1/2)
ENTONCES: G=D*r+(1-r)A =E*k+(1-k)B
por lo que r/2 = 1-k y k/2= 1-r
por lo que r= 2-2k por lo tanto k/2=1-2+2k=2k-1 por lo tanto k=2/3
y r=2/3 por lo tanto G=A(1/3)+B(1/3)+C(1/3)


Is there a proof to the fact that the sum of 2 sides of a triangle must be greater than the 3rd side and the difference of 2 sides must be smaller then the 3rd? Of there isn't then please tell me.

Much appreciated.

That's the triangle inequality. Notice that if it were violated (i.e. the sum of 2 sides is less than or equal to the 3rd) then it will just be a straight line not a triangle.

There are quite a few ways by which you prove these inequalities, but as of now, I'm just giving you one method of proving that the sum of two sides is greater than the third side....

In $\Delta ABC$, produce $BA$ to $D$ such that $AD=AC$.Join $CD$.

In $\Delta ACD$, $AC = AD$
 $\rightarrow \angle ADC = \angle ACD$

$\rightarrow \angle BCA + \angle ACD > \angle ADC$
$\rightarrow \angle BCD > \angle BDC$ ( since angle ADC = angle BDC)
$\rightarrow BD>BC$
$\rightarrow BA + AD >BC$
$\rightarrow AB + AC >BC$ (since AD = AC)


vector
superior algebra
superior algebra unsolved

If W1 and W2 are subspaces of V, which is finite-dimensional, describe A(W1+W2) in terms of A(W1) and A(W2). Describe A(W1 intersect W2) in terms of A(W1) and A(W2).

A(W) is the annihilator of W (W a subspace of vector space V). A(W)={f in dual space of V such that f(w)=0 for all w in W}.

I suggest that $ A(W_1 \plus{} W_2) \equal{} A(W_1) \cap A(W_2)$ and that $ A(W_1 \cap W_2) \equal{} A(W_1) \plus{} A(W_2)$.

An idea on how to prove this (if it is true -- try experimenting first!)

Show the dimensions are equal (this uses the fact that $ V$ is finite-dimensional) and that one side trivially includes the other.

What about

dim(W1+W2) - dim(W2) = dim ([W1+W2]/W2)= dim (W1/[W1 intersect W2]) = dim(W1)-dim(W1 intersect W2)

??

But even if that's right, that's only for the second part of the question. I can't get the first part. Would the first part just be

dim(W1+W2) is less than or equal to dim(W1)+dim(W2)? But that doesn't really say much.

Perhaps you are misunderstanding:

A possible recipe for showing $ A(W_1 \plus{} W_2) \equal{} A(W_1) \cap A(W_2)$ (again, if it is true):

1) Show $ A(W_1) \cap A(W_2) \subset A(W_1 \plus{} W_2)$ (this should be easy).
2) Show $ \dim \left( A(W_1) \cap A(W_2) \right) \equal{} \dim A(W_1 \plus{} W_2)$. (This seems true at first - can you check it?)

(EDIT : fixed a typo)

Potentially useful facts:

$ \dim A(W) \equal{} \dim V \minus{} \dim W$ for any space $ W \subset V$.
$ \dim (W_1 \plus{} W_2) \plus{} \dim (W_1 \cap W_2) \equal{} \dim W_1 \plus{} \dim W_2$

Good luck!

Dimension analysis seems like too much work for this problem.  Break it into two:

1) Show $ A(W_1) \cap A(W_2) \subset A(W_1 \oplus W_2)$.

2) Show $ A(W_1 \oplus W_2) \subset A(W_1) \cap A(W_2)$.

The second one is where you seem to be having trouble.  But if we annihilate a space, surely we have to annihilate all subspaces as well?

Yeah, that just does it directly; guess we don't need that $ V$ is finite-dimensional then.

Sorry, I put direct sum in my post, but it works just fine if the sum isn't direct.

What about the first part of the question that just asks about the annihilator of the sum in terms of each of their annihilators?

We just answered that.  Do you mean the annihilator of the intersection?  Come to think of it, you might need finite-dimensionality for that one.

I'm trying to figure out how to prove the second claim where the sum is contained in the intersection.  I understand your reasoning but how do you prove that?

Is it true that if f is in A(W1)+A(W2) then f is in A(W1) intersect A(W2)?

[quote="rawkerrxx"]Is it true that if f is in A(W1)+A(W2) then f is in A(W1) intersect A(W2)?[/quote]
You're confusing yourself.  This is an amalgamation of the two questions you're trying to ask, but isn't true itself.

To finish up the missing piece: it is easy to show that $ A(W_1) \plus{} A(W_2) \subset A(W_1\cap W_2)$.  Now, compute $ dim(A(W_1) \plus{} A(W_2)) \equal{} dim(V) \minus{} dim(A(W_1)\cap A(W_2))$ $ \equal{} dim(V) \minus{} dim(A(W_1 \plus{} W_2)) \equal{} dim(W_1 \plus{} W_2)$ $ \equal{} dim(V) \minus{} dim(W_1 \cap W_2) \equal{} dim(A(W_1 \cap W_2))$.

This thread is a mess.

I don't see how to formally prove the first inclusion, the one you said was easy.

For the second inclusion all I did was
Assume f is in A(W1) intersect A(W2).
This implies f(w1)=0 and f(w2)=0
implies f(w1)+f(w2)=0
implies f(w1+w2)=0 (since f is a homomorphism)
implies f is in A(W1+W2)

Why is this wrong?

Your proof is correct.  But you're clearly having trouble organizing your thoughts.  There are two problems in this thread, hence there are *four* inclusions to prove.  The one I just posted about involves intersecting spaces, not intersecting annihilators.  Big, big difference.

If I were you, before I wrote anything more in this thread, I would take four sheets of blank paper, and at the top of each one, I would write one of the four inclusions I need to prove.  I would then *quadruple check* for any typos, before writing down the proofs of the ones I felt I understood.  You've solved one, and another one is reasonably easy.  For the rest, you will find solutions or major hints in this thread if you read closely.

Note: exactly one of the four inclusions requires finite-dimensionality, and the use of two of the others, and it's the hardest one.


I can't stand having some lines indented and others not, and I don't want to use \noindent all the time.  How does one make it so that there is never indentation in a LaTeX document (except for when I want there to be indentation for which I could use \indent)?

Basically, I am writing out my math homework, I am using the amsthm and \begin{proof} ... \end{proof}, and the first line starts at the beginning, but after that every time I make a new line it is automatically indented, and this just looks ridiculous, especially for something that has just over a full length of line.  Is there some way to keep everything with the same (or no) indentation automatically?

For example, I keep getting things that look something like:

Proof: Let us blah blah blah

[color=white](indent)[/color]If we assume that blah blah blah, then we have that blah blah blah is true.  Then we want to show that blah blah blah....  Now we have...

[color=white](indent)[/color]Now if we assume this then we have....

QED
-------

All this indentation here look silly as I said, and that is why I want something consistent.

Also, are there any easy ways (or environments I guess) for writing up math homework nicely?

Thanks!

Yeah, same problem here.  Just set

\parindent=0in

in the header.


From Moldova Mathematical Olympiad.

For a,b,c strictly positive real numbers prove that

a^2/(a^2+2bc) + b^2/(b^2+2ca) + c^2/(c^2+2ab)  \geq 1

From the Cauchy-Schwartz inequality we have that a^2/(a^2+2bc) + b^2/(b^2+2ca) + c^2/(c^2+2ab)\geq (a+b+c) 2 /(a 2 +b 2 +c 2 +2ab+2bc+2ca)=1.

Yes, or even simpler (?), just use that 2xy  \leq x 2 +y 2 to transform each denominator into a 2 +b 2 +c 2 , and sum...

Pierre.


inequalities
vector
inequalities proposed

In  any triangle  holds the inequality \[||IO||^2+2||IH||^2\le
||OH||^2\le 2||IO||^2+2||IH||^2\; . \]

In connection with  [b] 14.14, page 124 [/b] from  [i]Geometric Inequalities[/i] by O. Bottema & Co.

[quote="Arne"]Well, that can't be too hard - use vectors, and $OH^2 = 9R^2 - (a^2 + b^2 + c^2)$ and $OI^2 = R^2 - 2Rr$? It might not be a pleasant task though...[/quote]

can anyone give me links (perhaps even with proof) of results like this ($OH^2$, $OI^$). In particular, these have been discused on ML, but i don't know how to calculate the others like $IG^2,IH^2$ (well, i have a method with barycentrics, but it's extremely messy :( - i've never actually completed my calculations; i'd like smth elegant )


Figuring a mid-term grade.

My final grades will be based on the following. How do I figure my mid-term grade based on 90% of the over all (I won't take the final until the end of the quarter)?
......................Mid-Term Avg
50% Daily Work............109.83
30% Test..................100.33
10% Class Participation...100.00
10% Final Exam............not yet% Final Exam                not yet

Note: Extra credits cause the high averages.

I am not sure how to get 100% credit for the work I have done to date with 10% not due for 4 weeks.

It seems to me that I would simply average the three averages I already have by adding all and dividing by 3. And then at the end of the quarter add the new averages and divide by 4. However, what has to be done is convert 90% into 100% somehow so the average will be even with the grading scale. I am lost, please help.

I think you just take the 50, 30, 10 out of 90 for weighing instead. For example,

$\frac{50}{90} \cdot 109.83+ \frac{30}{90} \cdot 100.33+ \frac{10}{90} \cdot 100.00 = 105.57$.

That should be your current grade.


AMC
USA(J)MO
USAMO
algebra
polynomial
algebra unsolved
equation

Prove that the product of any two roots of $x^4 + x^3 - 1 = 0$ is a root of $x^6 + x^4 + x^3 - x^2 - 1 = 0$.

Let $a$ and $b$ be two of the roots of the polynomial
$X^{4}+X^{3}-1.$ Prove that $ab$ is a root of the polynomial $X^{6}%
+X^{4}+X^{3}-X^{2}-1.$

 Let $c$ and $d$ be the other two roots of $X^{4}+X^{3}-1.$
The Viete's relations yield
\[
a+b+c+d=-1
\]
\[
ab+ac+ad+bc+bd+cd=0
\]
\[
abc+abd+acd+bcd=0
\]
\[
abcd=-1.
\]
Write these equalities in terms of $s=a+b,$ $s^{\prime}=c+d,$ $p=ab$ and
$p^{\prime}=cd$ (this is often useful) to obtain
\[
s+s^{\prime}=-1
\]
\[
p+p^{\prime}+ss^{\prime}=0
\]
\[
ps^{\prime}+p^{\prime}s=0
\]
\[
pp^{\prime}=-1.
\]
Substituting $p^{\prime}=-\frac{1}{p}$ and $s^{\prime}=-1-s$ in the second and in the third equalities yields
\[
p-\frac{1}{p}-s^{2}-s=0
\]
and
\[
p(-1-s)-\frac{s}{p}=0.
\]
It follows from the second equality that
\[
s=-\frac{p^{2}}{p^{2}+1}.
\]
Plugging this into the first equality gives
\[
p-\frac{1}{p}-\frac{p^{4}}{(p^{2}+1)^{2}}+\frac{p^{2}}{p^{2}+1}=0.
\]
A short computation shows that this is equivalent to
\[
p^{6}+p^{4}+p^{3}-p^{2}-1=0,
\]
hence $p=ab$ is a root of the polynomial
\[
X^{6}+X^{4}+X^{3}-X^{2}-1.
\]


How many ways can the letters R, R, U, U, U, and D be ordered so that the permutation starts with some permutation of R, R, U, and U, and why?

[hide]Since the R's must go in two of the first 4 "slots", there are 4C2 ways to place the R's. Likewise, the D must go in one of the last 2 slots, so there are 2C1 ways to place the D. The U's then fill in the rest. So, there are 6*2=12 ways.[/hide]

I was thinking about it and came up with the following reasoning:

[hide] Since two U's and two R's must go in the first four spots, the last two spots could be U, D or D, U (2 ways). Since the two U's and two R's in the first four spots can be arranged 6 ways, the total number of ways to arrange the 6 letters is 6*2 = 12.[/hide]

Is this reasoning correct?

[hide]There are 4!/(2!*2!) ways to arange R,R,U,U. There are two ways to arange U,D. So 6*2=12.[/hide]


for $ a,b$ positive reals, $ a \plus{} b \equal{} 1$ with both less than 1 then prove that,
$ \frac {1}{a^4}$ is greater than$ 1 \plus{} b^4$

[quote="Dimitris X"][quote="geniusbliss"]with both less than 1 [/quote]
This is always true,there is no need to say it :wink: .

He homogenise and we have to prove that:
$ (a \plus{} b)^4 \ge a^4 \plus{} a^4b^4 \Longleftrightarrow 4a^3b \plus{} 6a^2b^2 \plus{} 4ab^3 \plus{} b^4(1 \minus{} a^4) \ge 0$ which is true.[/quote]
haha i am now wondering why i even posted anything this stupid! i posted only cuz i got a nice idea to this.

The given $ \implies 1 \plus{} \frac {b}{a} \equal{} \frac {1}{a}$, so $ \frac {1}{a^4} \equal{} (1 \plus{} \frac{b}{a})^4 > (1 \plus{} b)^4$ (since $ a < 1$) $ > 1 \plus{} b^4$.


Show that
\[ \sum_{k\equal{}1}^{\infty}\frac{1}{(k^2\plus{}k)^3}\equal{}10\minus{}\pi^2\]

Write the expression as

$ \[ \sum_{k\equal{}1}^{\infty}\frac{1}{k^{3}} \minus{} \frac{1}{(k\plus{}1)^{3}} \minus{} \frac{3}{(k)^{2}} \plus{} \frac{3}{(k\plus{}1)^{2}} \plus{} \frac{6}{k(k\plus{}1)^{2}}$

After some work by telescoping we obtain

$ \minus{}2 \plus{} 6(1 \minus{} \sum_{k\equal{}1}^{\infty}\frac{1}{(k\plus{}1)^{2}})$

Which is equal to  $ 10\minus{}\pi^{2}$


[quote="toanviet"]Can you you solve for a, b, c such that:
\[ a^2 \plus{} b^2 \plus{} c^2 \equal{} 9
\]

\[ a^3 \plus{} b^3 \plus{} c^3 \equal{} 27
\]
[/quote]
any condition about $ a.b.c$  !!!!!!!!!!

[quote="toanviet"]Can you you solve for a, b, c such that:
\[ a^2 \plus{} b^2 \plus{} c^2 \equal{} 9
\]

\[ a^3 \plus{} b^3 \plus{} c^3 \equal{} 27
\]
[/quote]
ok !.this my sulotion in $ R$

 we know tha $ a<3$ .$ b<3$ . $ c<3$

and from the system we have $ a^3\plus{}b^3\plus{}c^3\equal{}3(a^2\plus{}b^2\plus{}c^2)$

so $ a^2(a\minus{}3)\plus{}b^2(b\minus{}3)\plus{}c^2(c\minus{}3) \equal{}0$  and $ a<3$ same thing for $ b$ .$ c$

so the system <=>$ a^2(a\minus{}3)\equal{}b^2(b\minus{}3)\equal{}c^2(c\minus{}3)\equal{}0$

so $ S\equal{}{(3.0.0).(0.3.0).(0.0.3) }$


geometry
combinatorics unsolved
combinatorics

A hole of area 1 is covered by 5 pieces of cloth each of which has area greater or equal to 0.5.Show that there are 2 pieces of cloth whose common area is greater or equal to 0.2.(I called common area the surface where they overlap)

$H$ is the hole, $P_{i}$ is the $i$-th piece. For any $X \subseteq H$, denote by $[X]$ its area (if it has) and by $\overline{X}=H-X$ its complementary. Note that
\[ [\overline{P_{i}} \cap \overline{P_{j}}] = [\overline{P_{i} \cup P_{j}}] = 1-[P_{i} \cup P_{j}] = 1-[P_{i}]-[P_{j}]+[P_{i} \cap P_{j}] \leq [P_{i} \cap P_{j}] \]
Hence
\[ S=\sum_{i<j} [P_{i} \cap P_{j}] \geq \frac{1}{2} \sum_{i<j} ([P_{i} \cap P_{j}]+[\overline{P_{i}} \cap \overline{P_{j}}]) = S' \]
If a point $X$ of $H$ belongs to $k$ pieces, than it is "counted"
\[ {k \choose 2}+{5-k \choose 2} \geq 4 \]
times in $S'$. So
\[ S' \geq \frac{1}{2}4[H]=2 \Rightarrow S \geq 2 \Rightarrow \max_{i<j} {[P_{i} \cap P_{j}]} \geq \frac{1}{5} \]


geometry
rectangle
parallelogram
AMC
AMC 8

Yeah, the geoboard problem was a huge time-waster if you didn't know Pick's, as well as the problem on hexagonal numbers and this one too:[img]http://www.artofproblemsolving.com/Community/MJImages/AMC8_2004_24.gif[/img]

The geoboard problem wasn't hard though, brute force worked pretty well on it.

What's the question that goes with the picture you posted?

Am I just missing something or is that one pretty quick? Find the area of the parallelogram by finding the area of the rectangle and subtracting the areas of the triangles, and then using the fact that the area of a parallelogram is base times height, anf divide by base to get height.

I don't think it was that hard.  The only reasons I got problems wrong were becuase of stupid mistakes, or not realizing things.  I was doing all these things, like finding equations of lines and stuff for that problem Efuzzy posted since iI didn't realize that the length of d was congruent to EF or HG.

[quote="RC-7th"][quote="EFuzzy"]d was perpendicular to both bases of the parrallelogram inside the rectangle and you were asked to find the length of d.

The questions I posted maybe weren't the hardest, but they were the ones I spent a lot of time on and missed.[/quote]

Do you mean d was parallel to EF and HG? If so, the answer would just be sqrt(6^2+5^2)[/quote]
No, d was the height of the parallelogram.
The geoboard wasn't that much of a time-waster if you divided it up right, but it's hard to figure out how (I didn't, in fact I just skipped that one  :( ). The hexagons were slightly annoying, but I think it was more that that problem was prone to causing stupid mistakes than anything...

[quote="Rep123max"]Was this years AMC 8 harder than most because there were much fewer perfect scores? (They list them on http://www.unl.edu/amc/)[/quote]

Thanks for reminding us to look at the AMC site for statistics for this year. Wow! One guy I coach was only a point away from being state champion. Two of the guys on my team were in the state top thirty kids. But, yes, this year's test must have been harder, or not all of the foreign tests have been graded yet, because the number of perfect scores is WAY down, and Taiwan has many fewer than last year.

I definitely think that problem 14 (the geoboard problem) was a significant time-waster for kids (most) who don't have Pick's
Theorem memorized. One guy on my team had such a hard time SEEING all those dots that I'm ordering a large-print test for him on the A day for AMC 10/12.

We also need to find a non-noisy testing room. Our room this year in our friendly local public school (which was very quiet last year) this year had kids literally bouncing basketballs on the door of our room during the test, because we were testing at a different time of the day. I'm amazed that the guys on my team were able to maintain their concentration at all with kids in the next room yelling at the top of their lungs and playing dodgeball with basketballs. I may have found a lead to a better testing site tonight--a private tutoring service's offices, if that is acceptable to AMC headquarters.

Whew! Minnesota hasn't had a 25 on the AMC 8 for as long as I have been coaching. I hope someone on my team can reach the peak next year.

According to stats on the AMC Web-site, the average score in 2004 (9.94) is just 1/3 of a point lower than the average score in 2003 (10.27).  So, for the average student, the tests were approximately the same difficulty.

Although I don't see it on the AMC Web-site, it was reported elsewhere that the "Honor Roll" cutoff is 18 in 2004, and was 20 in 2003.  For the very high student, this suggests the 2004 test may have been "harder" than the 2003 test.

Both tests show a very nice distribution of scores.  I do not personally object to the 2004 score distribution, although I do understand the disappointment :( of students who had hoped for a perfect paper.

I don't think any question was inappropriate or out of line for the AMC-8.  After having talked to a few other coaches, my conclusion is that this test seemed to have more of the difficult questions posed in an unusual manner.  I.e., not only were the hard questions hard (as they should be), but the hard questions were also harder to parse.  [Any comments from students or other coaches?]

For those who like their data in a visual form, the attached chart was produced from data on the AMC Web-Site, as of 12/21/04.  The chart shows a graph of numeric score vs percentile for each of the last two years.

I took the amc8 recently

i got 13 out of 25

the time was the huge problem

most of the problems I solved were right but

because of the time limit the last 10 problems I guessed got

them all wrong

I though it was much harder but by class surveys it was much easier

this year. too bad..

well lucky me, I have another year to take it^.^

to shinwoo: a 13 is not terrible for the first time: here's a suggestion: when you are practicing for next year, do NOT impose the time limit upon yourself
the reasons:
1. it makes you too time concious (spelling??) and could lead to you spending valuable time looking at the clock, when you take the test
2. learning HOW to solve the problems is more important than solving ther quickly (at least in the beginning)... like all music teachers say (sometimes much to our annoyance): slow down! the speed will come!

this year's test was much better for me... a) because I had gotten better at relaxing during test (especially standardized ones) and b) i knew more math than last year.

but I have to admit: 23 out of 25 is disappointing: the hexagon problem, (where it asked to add extra black and white hexagons) threw me off and I forgot the add some of them to the remaining vertices. and the last problem... I first saw a slightly skewed right triangle... but alas.

I checked the results on the site and Max was right: scores were down a bit... though the average remainer roughly the same... reason: (i could be wrong on this) but there were more questions that required multiple steps and different sorts of math (ie there would be geometry questions mixed wiht algebra.. logic and algebra)... than last year

thanks for your tips shake9991....

well.. it's amazing that we both took AMC8 and you got 23 and I got 13..

so amazing

I feel so ahsamed and jealous....

well thanks for your encouragement

Sounds like a good idea.  You can purchase copies of the AMC 8, from 1986 on, at the AMC web site: [url]http://www.unl.edu/amc[/url].
The exams come with full solutions.  Their price is a bargain.

[quote="klopter"]I intend to start a math club for my daughter and her friends.  I would like to give them old AMC 8 problems to work on collectively.

Thanks,
  Ken[/quote]

The old South Africa tests are fine for beginners, too:

[url]http://science.up.ac.za/samo/questions.html[/url]

You'll want to look at the Junior level tests.


Show that the equation x^2+y^2+z^2 = 3xyz has infinitely many solutions in positive integers.

Not finished but possibly it's the main idea :

[hide]

1. (1,1,1) is a solution

2. if (x,y,z) is a solution, (y,z,x) also

3. if (x,y,z) is a solution, (x,y,3*x*y-z) is also a solution

So we can build a sequence (xn,yn,zn) with xn -> +oo by starting from 1. and using 3. and 2. repeatedly, hence there are infinitly many solutions.[/hide]


geometry
trapezoid
circumcircle
IMO Shortlist
Charles Leytem

[hide="Solution"]Since hypotheses, we have two cyclic quadrilaterals $ AEBF$ and $ DCEF$.

$ I\in (ABK) \iff JI\cdot JK \equal{} JA\cdot JB \equal{} JE\cdot JF \equal{} KE^2 \minus{} JK^2$ $ \iff KE^2 \equal{} JI\cdot JK \plus{} JK^2 \equal{} JK\cdot IK$
$ K\in (CDJ) \iff IJ\cdot IK \equal{} IC\cdot ID \equal{} IE\cdot IF \equal{} IK^2 \minus{} KE^2$ $ \iff KE^2 \equal{} IK^2 \minus{} IJ\cdot IK \equal{} IK\cdot JK$
Hence result.[/hide]

[quote="lasha.p"]we want to prove thet BAIK is cyclic,if and only if when CJDK is cyclic... it's <=> thet to prove that JA*JB=JI*JK <=> DI*IC=KI*IJ.    FJ=z, IF=x and KI=y.. so EK=x+y. ABFE is cyclic,so JA*JB=z(2x+2y+z). <BEI=<<FAJ=<FDI  so ECFD is cyclic and DI*IC=FI*IE=x(x+2y)   so we want to prove that     x(x+2y)=y(x+z)  <=> z(2x+2y+z)=(z+x)(x+y+z)  .... z(2x+2y+z)=(z+x)(x+y+z) <=> z(x+y+z) +z(x+y)= z(x+y+z)+x(x+y+z) <=> z(x+y)=x(x+y+z) <=> zy= x(x+y) and this is  <=> x(x+2y)=y(x+z).[/quote]

[color=darkred][b]lasha.p[/b], if you repost this under your member name, I will delete this copy. Next time, please, use the [b]<Reply to topic>[/b] button when posting a solution.[/color]

[hide="Solution"]
Since $\angle{DAE}=\angle{CBF}$, $AFBE$ is cyclic, so $(AJ)(JB)=(FJ)(JE)$. Also, $\angle{JEC}=\angle{JAD}=\angle{IDA}$ since $AB||CD$, implying that $CDFE$ is cyclic. Thus, $(ID)(IC)=(IF)(IE)$.

Note that $AKBI$ is cyclic if and only if $(IJ)(JK)=(AJ)(JB)=(FJ)(JE)$ and $CDJK$ is cyclic if and only if $(IJ)(IK)=(ID)(IC)=(IF)(IE)$. Let $IF=a$, $FJ=b$, $JK=c$, and $KE=KF=FJ+JK=b+c$. Then
\[ (IJ)(JK)=(FJ)(JE)\Leftrightarrow (a+b)(c)=(b)(c+b+c)\Leftrightarrow ac=b(b+c)\]
and
\[ (IJ)(IK)=(IF)(IE)\Leftrightarrow (a+b)(a+b+c)=(a)(a+b+c+b+c)\Leftrightarrow b(b+c)=ac\]
As a result, $I$ belongs to the circumcircle of $ABK$ if and only if $K$ belongs to the circumcircle of $CDJ$. //[/hide]

Since $\angle DAE \equiv \angle CBF$, $AEBF$ is cyclic, so $AJ \cdot JB = FJ \cdot JE$. Now, $I \in \odot ABK \Leftrightarrow \underline{IJ \cdot JK} = AJ \cdot JB = \underline{FJ \cdot JE}$ $\Leftrightarrow$ $IJ(\frac{IF+IE}{2}-IJ)=(IJ-IF)(IE-IJ)$ $\Leftrightarrow$ $IJ = \frac{2IF \cdot IE}{IF + IE}$ $\Leftrightarrow$ $IF \cdot IE = IJ \cdot \frac{IE+IF}{2}$ $\Leftrightarrow$ $IE \cdot IF = IJ \cdot IK$. Since $AEBF$ is cyclic, we get $\angle FAB \equiv \angle FEB \Rightarrow \angle IDF \equiv \angle FEC$ $\Leftrightarrow$ $FECD$ cyclic $\Leftrightarrow$ $IF \cdot IE = ID \cdot IC$. So, $IJ \cdot IK = IE \cdot IF$ $\Leftrightarrow$ $IJ \cdot IK = ID \cdot IC$ $\Leftrightarrow$ $K \in \odot CDJ$.

[u]Computational solution:[/u]
first notice that $FCED$ is cyclic. in deed $\angle DFE =\angle ABC=\angle DCE$
We have to prove that (with power of a point ): \[IJ.IK=JF.JE \Leftrightarrow IE.IF=IJ.IK\]
$IJ.IK=JF.JE \Leftrightarrow IJ(\frac{FE}{2}-IF )=IE.IF \Leftrightarrow \frac{IJ.FE}{2}=IF.JE $
similarly $IE.IF=IJ.IK \Leftrightarrow  \frac{IJ.FE}{2}=IF.JE  $  so the equivalence is proved .
QED

$K \in \bigcirc CDJ \iff IJ \cdot IK=ID \cdot IC$ $\iff  IJ \cdot IK = IF \cdot IE $   (As $EFDC$ is concyclic)$ \iff (I,J;F,E)= (-1)$(As $K$ is midpoint of $FE$)$\iff IJ \cdot JK=FJ \cdot JE \iff IJ \cdot JK=AJ \cdot JB$ (As $AEBF$ is concyclic.) $\iff I \in \bigcirc AKB$.

[quote=Legend-crush][u]Inversive solution:[/u] (i think that K being the midpoint is not a necessary condition )
$ \angle DFE =\angle ABC=\angle DCE \Rightarrow FCED$ is cyclic.
Consider $\psi$ the inversion centered at J and of ratio $\sqrt{JA.JB}$. and then $\psi(A)=B$ and $\psi(E)=F$
Let $\psi(C)=C'$ and $\psi(B)=B'$, and $\psi(I)=K'$ ($K'$ is the intersection of $(AIB)$ with $JC$)
The image of $(FCED)$ must be a circle so $FC'ED'$ is cyclic.
then the radical axis of circle $(FC'ED'), \ (DC'CD')$ and $(FCED)$ are concurent. since $FE\cap DC= I$, we get $I\in D'C'$
As a consequence $JDK'C$ is cyclic.
Therefore : \[ABKI \ cyclic \Leftrightarrow K=K' \Leftrightarrow JDKC\ cyclic  \]
QED.
In other words $(JDC)$ and $(AIB)$ intersect in segment $FE\blacksquare .$[/quote]

The solution is wrong. circles (FC'ED'),  (DC'CD'), (FCED) are duplicate

Please Who can draw figure. When I draw one quadrilateral is cyclic but another is not even convex.

My solution:
From condition $FABE$ is cyclic or $JF\cdot JE=JA\cdot JB.(1).$
Also we know $180-\angle EFD=\angle EFA=180-\angle ABC=\angle BCD=180-\angle DCE\to DFCE $ is cyclic,or $DI\cdot IC=FI\cdot IE.(2).$
$ABKI$ is cyclic iff $$JI\cdot JK=JA\cdot JB=^{(1)}JF\cdot JE=(JK-FK)(JK+FK)=JK^2-FK^2\to JK\cdot IK=KF^2(3).$$
Also $CKDJ$ is cyclic iff $$JI\cdot IK=DI\cdot IC=^{(2)}FI\cdot IE=(KF-IK)(KF+IK)=KF^2-IK^2\to KF^2=JI\cdot IK+IK^2=IK\cdot JK.(4).$$
From $(3),(4)$ as desired.

[quote=anantmudgal09]*Flashback 2 years ago.*

[quote=April]Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE \equal{} \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$.

[i]Proposed by Charles Leytem, Luxembourg[/i][/quote]

Notice that $ABFE$ is a cyclic quadrilateral. Consider the following

[b]Lemma.[/b] $CDFE$ is cyclic.

(Proof) Apply the converse of Reim's Theorem to $ABEF$ and lines $\overline{BE}$ and $\overline{AF}$. This works well since $\overline{AB} \parallel \overline{CD}$. $\blacksquare$

By power of point, $$K \in (CDJ) \iff IJ \cdot IK =IC \cdot ID=IF \cdot IE \iff (FE, IJ)=-1,$$ and $$I \in (ABK) \iff JI \cdot JK=JA\cdot JB=JF\cdot JE \iff (FE, IJ)=-1.$$ Hence, both conditions are equivalent. $\blacksquare$[/quote]

What kind of Reim's theorem did you use here?

Here's a PoP bashing solution. Note the condition $\angle DAE = \angle CAF$ implies the four points $F, A, B, E$ are concyclic. Also $\angle FDC = \angle DAJ = 180 - \angle FAB = \angle FEB = \angle FEC$, hence $\omega_{FDEC}$ exists too. By PoP, we get $$ID \cdot IC = IF \cdot IE = (FK - IK)(EK+IK) = (EK-IK)(EK+IK) = EK^2 - IK^2 (\star)$$

For the first part, i.e $I \in \omega_{ABK} \Rightarrow K \in \omega_{CDJ}$. Note that $ \text{Pow}_J(\omega_{ABEF})  = JF \cdot JE =  (JK + EK)(JK - EK) = JK^2 - EK^2 $ and $\text{Pow}_J(\omega_{ABK}) = JI \cdot JK = (JK - KI) \cdot JK = JK^2 - JK \cdot KI = JK^2 - ((JI + IK) \cdot KI) = JK^2 - (JI \cdot KI + KI^2)$. But $J$ lies in the radical axis of $\omega_{ABEF}$ and $\omega_{ABK}$. Hence, the powers are equal, which gives $EK^2 = JI \cdot KI + KI^2$, or $ IJ \cdot IK= EK^2 - KI^2 \overset{\star}{=} ID \cdot IC $, which implies $J, K, D, C$ are concyclic, as desired.

For the second part (i.e the converse), assume $J, K, D, C$ are concyclic and we need to prove that $A, B, K, I$ are concyclic. Since  $C,J,  D, K$ are concylic, $IJ  \cdot IK = \text{Pow}_I(\omega_{CDJK}) = ID \cdot IC \overset{\star}{=} EK^2 - IK^2$. Also $IJ \cdot IK = (JK - IK)IK = JK \cdot IK - IK^2$. Combining these two, we get $EK^2 = JK \cdot IK (\spadesuit)$. So $JA \cdot JB = JF \cdot JE = JK^2 - EK^2 \overset{\spadesuit}{=} JK^2 - JK \cdot IK = JK(JK-IK) = JK \cdot IJ$, hence $A, B, K, I$ are concyclic, as desired.

It is the part of [b]IMOSL 2004 G8[/b].

[b]Lemma:[/b] Given a cyclic quadrilateral $ABCD$, let $M$ be the midpoint of the side $CD$. Let $BC\cap DA=F$ and $BD\cap CA=E$. Then $(ABM)\cap CD\in EF$ .
[b]Proof[/b] (by grobber): [quote=grobber]

Let $P$ be the second point of intersection between $CD$ and the circle $(ABM)$, and let $G=AB\cap CD$. A very simple computation, based on the fact that $GD\cdot GC=GA\cdot GB=GM\cdot GP$ and $M$ is the midpoint of $CD$ will show that $P$ is, in fact, the harmonic conjugate of $C,D$ art $G$, so it belongs to $EF$.
[/quote]

So in both cases $(F,E;I,J)=-1$. So we are done.

My solution: We start off with the following well known lemma.

[size=150]Lemma[/size] Let $P,R,Q,S$ be points on a line in that order. Let $T$ be the midpoint of $RS$. Suppose that $PT \cdot PQ=PR \cdot PS$. Then $(P,Q;R,S)=-1$.

[b]PROOF:[/b] Notice that $PT \cdot PQ=PR \cdot PS=(PT-TR)(PT+TS)=(PT-TR)(PT+TR)=PT^2-TR^2$ $\Rightarrow PT(PT-PQ)=TR^2 \Rightarrow TP \cdot TQ=TR^2$. This means that $P$ and $Q$ are inverses wrt $\odot (RS)$, giving that $(P,Q;R,S)=-1$. $\blacksquare$

Return to the problem at hand. Note that, from the given conditions, $ABEF$ is cyclic.
This also gives that $\measuredangle CDF=\measuredangle BAF=\measuredangle CEF \Rightarrow CEDF$ is cyclic.

Now, $ABIK$ is cyclic $\Leftrightarrow JK \cdot JI=JA \cdot JB=JE \cdot JF \Leftrightarrow (J,I;E,F)=-1$, where the last equality follows from our Lemma.

Similarly, $CDJK$ is cyclic $\Leftrightarrow IK \cdot IJ=IC \cdot ID=IE \cdot IF \Leftrightarrow (I,J;E,F)=-1$, where the last equality follows from our Lemma.

Thus, The two given statements are equivalent to each other.

How is this even a G2?
[quote=April]Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE \equal{} \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$.

[i]Proposed by Charles Leytem, Luxembourg[/i][/quote]
By looking at the angles, we find that $ABEF$ is cyclic hence so is $DECF.$ Thus,
\begin{align*}
I \in \odot(ABK) &\Leftrightarrow JI \cdot JK=JA \cdot JB= JE \cdot JF &\Leftrightarrow (J,I;F,E)=-1 \\
K \in \odot(CDJ) &\Leftrightarrow IK \cdot IJ=IC \cdot ID=IF \cdot IE &\Leftrightarrow (I,J;F,E)=-1
\end{align*}
And so we are done. $\square$

Note that $(DFEC)$ and $(AEFD)$. So, $$IJ\cdot JK=AJ\cdot JB\iff IJ\cdot JK=FJ\cdot JE=(FK)^2-JK^2\iff IK\cdot JK=JK^2$$
$$\iff IK^2-JK^2=IK^2-IK\cdot JK\iff IF\cdot IE=IK\cdot IJ\iff IJ\cdot IK=ID\cdot IC$$ as desired.

[hide=Solution]
[quote=ISL 2008 G2]Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE \equal{} \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$.
[/quote]
[b][color=#000]Solution:[/color][/b] Let $\omega$ be a circle through $F$ and center at $K$. Let $T,T'$ $\in$ $\omega$, such that $JT,JT'$ are tangent to $\omega$. Let $TT'$ $\cap$ $EF$ $=$ $I'$. Now, $\Delta TJK$ $\sim$ $\Delta I'JT$
$$JK \times JI'=JT^2=JF \times JE = JA \times JB ~ \qquad \& \qquad ~ -1=(T,T';F,E) \stackrel{T}{=} (J,I';F,E)$$
Hence, $I \in \odot (ABK)$ $\implies$ $I \equiv I'$ $\implies$ $-1=(J,I;F,E)$. If $K$ $\in$ $\odot (CDJ)$ $$\implies JI \times IK=ID \times IC=IF \times IE$$ Let perpendicular at $I$ intersect $\omega$ $=V,V'$ $$\implies JI \times IK=IF \times IE=IV^2 \implies \Delta JIV \sim \Delta VIK$$
Hence, $\angle JVK =90^{\circ}$ $\implies$ $-1= (J,I;F,E)$ $\qquad \blacksquare$
[/hide]

Note that $(ABFE)$ is cyclic. Now, here is a much more natural restatement of the problem:
[quote]Points $A,B,F,E$ lie on a circle. Let $K$ be the midpoint of $EF$ and $J=EF\cap AB$. Let $\ell$ be the line through $I$ parallel to $AB$. Let $C=\ell\cap BE, D=\ell\cap AF$. Prove $CKDJ$ cyclic. [/quote]
Note $\angle DFE=180-\angle AFE=\angle ABE = \angle DCE$, so $DCEF$ cyclic. We want to show $IK\cdot IJ=ID\cdot IC = IF\cdot IE$. Since $(EF;K\infty)=-1$, inverting at $X$ swapping $E,F$ gives us that we want to show $(FE;JI)=-1$. Since $ABFE$ cyclic, $JF\cdot JE = JA\cdot JB = JI\cdot JK$. Again, since $(EF;K\infty)=-1$, inverting at $J$ swapping $E,F$ gives $(FE;IJ)=-1$, so $(FE;JI)=-1$, and we are done.

Can somone attach a complete diagram please? Spent half an hour using Geoegebra and couldn't even get the diagram.

@above  diagram.
You may also see the [url=https://youtu.be/PoWDt3_yFLo]video[/url].

By definition $ABEF$ is cyclic. Furthermore $\angle AFE = \angle BCD = 180^{\circ} - \angle ABC$ hence $DECF$ is cyclic. Now we are ready to PoP:\[I \in (ABK) \iff JI \cdot JK = JA \cdot JB = JF \cdot JE\]\[K \in (CDJ) \iff IF \cdot IE = ID \cdot IC = IK \cdot IJ\]and in fact both of these are the same condition as $(I, J; F, E) = (J, I; F, E) = -1$ and we are done. $\blacksquare$

Note that $AEBF$ is cyclic by the given angle conditions.
Consider the negative inversion at $K$ mapping $(AEBF)$ to itself. Note that $I \in (ABK) \iff I^* = E^*F^* \cap A^*B^*$ . By butterfly theorem, $E^*F^* \cap A^*B^*$ is the reflection of $J$ over $K$, so $I \in (ABK)$ if and only if $I$ is the inverse of $J$ under a (positive) inversion at $K$ with radius $KF$. The same argument shows that $J \in (CDK)$ if and only if it is the inverse of $I$ under this inversion.

[hide=Storage]
Definitely not a G2 level problem

Note that $ABEF$ and $DFCE$ are cyclic from the given angle conditions.
Hence by power of point.
$$JI \cdot JK=JA \cdot JB=JF \cdot JE  \hspace{0.2cm}and\hspace{0.2cm} IF \cdot IE= ID \cdot IC=IJ \cdot IK$$
QED
[/hide]

First, notice that $$\angle FAE=\angle DAE=\angle CBF = 180^{\circ}-\angle EBF,$$ so $AEBF$ is cyclic. In addition, $$\angle CDF = \angle CDA=180^{\circ}-\angle BAD =180^{\circ}-\angle BAF = 180^{\circ}-\angle BEF=180^{\circ}-\angle CEF,$$ so $DFEC$ is cyclic.

Notice that $IF\cdot IE = ID\cdot IC$ by Power of a Point, and that $IJ\cdot IK = ID\cdot IC$ iff $DJKC$ is cyclic by Power of a Point and it's converse. Thus $IF\cdot IE = IJ\cdot IK$ is equivalent to $DJKC$ being cyclic. Also, $AJ\cdot JB = JF\cdot JE$ by Power of a Point, and $AKBI$ is cyclic iff $FJ\cdot JE=JI\cdot JK$ by Power of a Point and it's converse, so $AKBI$ cyclic is equivalent to $IJ\cdot JK = FJ\cdot JE$. Thus, it remains to prove the following claim:

[b][color=#f00]Claim:[/color][/b] $IF\cdot IE= IJ\cdot IK$ is equivalent to $IJ\cdot JK=FJ\cdot JE$.
[i]Proof.[/i] Use coordinates so that the lines $AB$ and $CD$ coincide with $y=0$ and $y=-1$ respectively. Thus $I$ and $J$ have y-coordinates $-1$ and $0$, respectively. Let the y-coordinates of $F$ and $E$ be equal to $a$ and $b$, respectively. Then the y-coordinate of $K$ is equal to $\frac{a+b}{2}$. Notice that the distance between two points on line $EF$ is proportional to the difference in their y-coordinates. Thus $IF\cdot IE=IJ\cdot IK$ is equivalent to $$(a+1)\cdot (b+1)=1\cdot \left(\frac{a+b}{2}+1\right)\Rightarrow ab=-\frac{a+b}{2}.$$ However, $IJ\cdot JK=FJ\cdot JE$ is equivalent to $$1\cdot \left(\frac{a+b}{2}+1\right) = -a\cdot b,$$ which reduces to the same condition, so the claim is proved. $\blacksquare$

As the claim was the last step we needed to prove in our chain of equivalences, we are done.

[hide=GeoGebra Tips]
This is not the easiest diagram to draw on GeoGebra, as mathlogician has pointed out. After making a trapezoid (which isn't hard), let $F$ be a point on segment $AD$. Even without a to-scale diagram you can see the angle condition gives $AEBF$ cyclic, so instead of using the definition given in the problem, let $E$ be the intersection of the circumcircle of $ABF$ and line $BC$ (which isn't $B$). Proceed drawing the rest of the diagram as normal.

To get $AKBI$ and $DJKC$ cyclic, input something like $\frac{IJ\cdot JK}{AJ\cdot BJ}$ (the PoP condition for $IAKB$ to be cyclic needs this to be $1$) and simply move $F$ until you get sufficiently close to $1$ such that you can't notice the difference.

The angle condition is equivalent to $AEBF$ cyclic. Now $$I \in (ABK) \iff JA \cdot JB = JK \cdot JI \iff JE \cdot JF = JK \cdot JI \iff (EF;JI)=-1.$$ $DFCE$ is cyclic by angle chasing, so $$IJ \cdot IK = ID \cdot IC \iff IJ \cdot IK = IF \cdot IE \iff (EF;JI)=-1.$$ These are clearly equivalent.

[hide=Solution][quote=April]Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE \equal{} \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$.

[i]Proposed by Charles Leytem, Luxembourg[/i][/quote]
By definition $AEBF$ is a cyclic quadrilateral and by Reim's theorem $FDEC$ is cyclic too!
If $I \in \odot(\triangle ABK)$,By P.O.P
$$JF.JE=JA.JB=JK.JI \implies (E,F;J,I)=-1$$
Again if $K \in \odot(\triangle CDJ)$,By P.O.P
$$JI.IK=DI.IC=EI.IF \implies (E,F;J,I)=-1$$
Since above conditions match,we are done $\blacksquare$

The angle condition is equivalent to $AEBF$ being cyclic, and $CDFE$ is cyclic as $$\angle FEC=\angle FAB=180-\angle CDF.$$ By PoP, \begin{align*}I\in(ABK)&\iff IJ\cdot JK=AJ\cdot JB=FJ\cdot JE\\&\iff(IJ;EF)=-1\\&\iff IJ\cdot IK=IF\cdot IE=ID\cdot IC\\&\iff K\in(CDJ)\end{align*} $\square$

Clearly $F\in \odot (ABE)$ and by Reim $F\in \odot (CDE).$ $$I\in \odot (ABK)\iff |JI|\cdot |JK|=|JA|\cdot |JB|=|JE|\cdot |JF|\iff (EFIJ)=-1\iff$$ $$\iff |IJ|\cdot |IK|=|IE|\cdot |IF|=|IC|\cdot |ID|\iff K\in \odot (CDJ)\text{ } \blacksquare$$

Note that $ABEF,CFDE$ cyclic.

Consider the statement $(*):\frac{JF-IF}{IF}=\frac{IF}{IK}.$ Note that \[JF\cdot IK-IF^2=FJ\cdot FJ-IF\cdot IJ=FJ(JE-IK)-IJ^2=JF\cdot JE+IK\cdot IF-JI\cdot JK\] which when $JI\cdot JK=JF\cdot JE$ is equal to $IK\cdot IF$ which means that $I\in(ABK)\iff (*).$

On the other hand, \[IF^2-IK\cdot FJ=IF\cdot FK-IK\cdot IJ\] which when $IE\cdot IF=IJ\cdot IK$ is equal to $-IF\cdot IK$ which means $K\in(CDJ)\iff (*).$ Thus, we are done.

the angle  condition implies  that ABEF is  cyclic   , and since AB  //  CD  ,  DECF  is also  cyclic   .
\\
\\  By  power of a  point :
$$ABIK \quad \text{cyclic } \iff  JF.JE=JA.JB =JK.JI $$

OR  $$JF.JE=JK.JI  \iff  (FE;IJ) = -1  \iff IK.IJ= IE.IF=ID.IC $$
Which  is equivalent to DECF to  be cyclic .
\\

Note that $ABEF$ is cyclic from the given angle condition. Then
by Reim's theorem, $CDEF$ must also be cyclic. From PoP,
$K$ lies on $ABI$ if and only if
\[ JI \cdot JK = JA \cdot JB = JE \cdot JF \]
which is equivalent to $(JI;EF) = -1$ (from a lemma regarding harmonic bundles).
Additionally, note from PoP that $K$ lies on $CDJ$ if and only if
\[ IE \cdot IF = IC \cdot ID = IK \cdot IJ \]
But $IE \cdot IF = KE^2 - KI^2$ from PoP, so the condition is true if and only if
$KE^2 = KI \cdot KJ$, which is true if and only if $(JI;EF) = -1$. $\blacksquare$

Note that the angle condition rewrites as points $A$, $E$, $B$ and $F$ being concyclic. By Reims, we further have $D$, $F$, $E$ and $C$ concyclic.
Then, by PoP and Mac-Laurin, both circles in the statement exist if and only if $I$, $J$, $F$ and $E$ are harmonic.
$\square$


limit
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inequalities
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Using the Squeeze Theorem to show that $ \lim_{x\to{0}}\sqrt {x^{3} + x^{2}}\sin{\frac {\pi}{x}}$, I'm having trouble visually seeing this. I know that \sin{\frac{\pi}{x}} oscillates between $ 1$ and $ - 1$ and $ x\to{0}$ and also that $ - 1\le{\sin{\frac {\pi}{x}}}\le{1}$. Is it logical to say that? If so, then $ - \sqrt {x^{3} + x^{2}}\le{\sqrt {x^{3} + x^{2}}\sin{\frac {\pi}{x}}}\le{\sqrt {x^{3} + x^{2}}}$? If I'm correct in saying that, then I'm not having trouble finding bounds for the function itself but actually seeing this graphically because it appears to me that $ \sqrt {x^{3} + x^{2}}\sin{\frac {\pi}{x}}$ simply isn't bounded between the other two functions near $ 0$...

Edit: Gah, I figured out what I was doing wrong. I was just graphing $ \sin{\frac{\pi}{x}}$ between the two instead of the actual given function. I'll leave the original question up for reference though I suppose. But I have another question:

Prove that $ \lim_{x\to{0^{+}}}\sqrt{x}e^{\sin{\frac{\pi}{x}}}=0$ using the Squeeze Theorem. I graphed the function first but I'm a little confused as to show to set up the inequality here. I considered the function $ f(x)=e^{\sin{\frac{\pi}{x}}}$ first but I could really find a bound for that function. Also, how do I go about using the squeeze theorem for only a one-sided limit?

$ e^{\minus{}1} \le e^{ \sin \frac{\pi}{x} } \le e^{1}$ (can you see why?).

The limits are solved by this way

For all x R-{0} it holds

$ |\sqrt {x^{3} + x^{2}}\sin{\frac {\pi}{x}}|\leq \sqrt {x^{3} + x^{2}}$ $ \Longleftrightarrow - \sqrt {x^{3} + x^{2} } \leq \sqrt {x^{3} + x^{2}}\sin{\frac {\pi}{x}}\leq \sqrt {x^{3} + x^{2}}$

Taking limits both sides ${{ \lim_{x\to{0}}\sqrt {x^{3} + x^{2}}} = 0 = \lim_{x\to{0}}( - \sqrt {x^{3} + x^{2}}})$

So the  Squeeze Theorem  give us $ \lim_{x\to{0}}\sqrt {x^{3} + x^{2}}\sin{\frac {\pi}{x}} = 0$
Same way with the second limit

A generalation is the following

If the functions $ f,g$ have the following properties

[1]   $ f$    is    bounded     ($ |f(x)| \leq M)$
[2] $ \lim_{x\to{x_0}}g(x) = 0$

Then  $ \lim_{x\to{x_0}}g(x)f(x) = 0$

All of the rules you know apply for one-sided limits; replacing $ \lim_{x\to a}$ with $ \lim_{x\to a^\plus{}}$ or $ \lim_{x\to a^\minus{}}$ works in all cases. The only change in the proofs is that we look only at $ x$ on one side.


Each of ten line segments is longer than 1 but shorter than 55. Prove that you can select
three line segments from these ten to form a triangle.

(i can almost see it but can't prove it)

Assume that we can't and the shortest line segment is $ a_1$ being 1 long.
Because we can't select three line segments from these ten to form a triangle, we have $ a_2\geq 1,a_3 > a_1 \plus{} a_2 \equal{} 2,a_4 > 3,a_5 > 5,a_6 > 8,a_7 > 13,a_8 > 21,a_9 > 34,$
$ a_{10} > 55$
Contradiction! :D


function
calculus
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(a)Show taht $ \frac{d}{dx}$ $ [{tan^{-1}e^{x}+tan^{-1}e^{-x}]}= 0$

(b) Hence show that ${ [\{tan^{-1}e^{x}+tan^{-1}e^{-x}}=\frac{\pi}{4}$

(c) Hence deduce that $ f(x)=tan^{-1}e^{x}-\frac{\pi}{4}$ is an odd function


quadratics
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Polynomials

Two quadratic polynomials $ f_{1},f_{2}$ satisfy $ f_{1}'(x)f_{2}'(x)\geq |f_{1}(x)|\plus{}|f_{2}(x)|\forall x\in\mathbb{R}$ . Prove that $ f_{1}\cdot f_{2}\equal{} g^{2}$ for some $ g\in\mathbb{R}[x]$.

You're right obviously. First observation: $ a1\cdot a2>0$. From the condition $ f'_{1}(x)f'_{2}(x)\ge 0$ we get that $ f'_{1}$ and $ f'_{2}$ have the same root $ x_{0}$. But if we put $ x\equal{}x_{0}$ we get that it is a root for $ f_{1}$ and $ f_{2}$ too. So
$ f_{1}(x)\equal{}a_{1}(x\minus{}x_{0})^{2}$
$ f_{2}(x)\equal{}a_{2}(x\minus{}x_{0})^{2}$
$ g(x)\equal{}\sqrt{a_{1}a_{2}}(x\minus{}x_{0})^{2}$


Find $x$ such that:$3x \equiv 5 (mod 7)$ and $2x \equiv 3 (mod 5)$and $5x \equiv 1 (mod 9)$

Bad title http://www.mathlinks.ro/Forum/viewtopic.php?t=144404 Locked!


An astronaut weighs 8.00 ×10^2 newtons on the surface of Earth. What is the weight of the astronaut 6.37 ×10^6 meters above the surface of Earth?

6.37 x10^6 m sounds much like the Earth radius. Since the gravitational acceleration is given (approximately) by

$G=\kappa\frac{M_e}{d^2}$

where $M_e$ is the mass of Earth and $d$ is the distance from the center of Earth, if you move twice as far from Earth's center, you'll get 1/4 gravitational acceleration and your weight will also be 1/4 of your weight on the surface of Earth. So I'd say 2.00 x 10^2 N.

Thank you for your tip and great information.


[quote="pigfly"]$\sum_{i=1}^3 h_i = 3;h_i \in (0;\frac{1}{2})$.[/quote]

It is a contradiction. BTW: I think there is an other fault in your sum on the right side.

Misha

Sorry,$\sum_{i=1}^3 h_i =1$.I will correct it :D


Let $a,b ,c$  be three positive reals. Prove the inequality

$\frac{a\left(b^2+c^2\right)}{\left(b+c\right)\left(a^2+bc\right)}+\frac{b\left(c^2+a^2\right)}{\left(c+a\right)\left(b^2+ca\right)}+\frac{c\left(a^2+b^2\right)}{\left(a+b\right)\left(c^2+ab\right)} \ge \frac 3 2$

[quote="zhaobin"]Let $a,b ,c$  be three positive reals. Prove the inequality

$\frac{a\left(b^2+c^2\right)}{\left(b+c\right)\left(a^2+2bc\right)}+\frac{b\left(c^2+a^2\right)}{\left(c+a\right)\left(b^2+2ca\right)}+\frac{c\left(a^2+b^2\right)}{\left(a+b\right)\left(c^2+2ab\right)} \ge \frac 3 2$[/quote]

If $a=b=c>0$ don't work  :|

Very nice one! But I think it should be

$\frac{a\left(b^2+c^2\right)}{\left(b+c\right)\left(a^2+bc\right)}+\frac{b\left(c^2+a^2\right)}{\left(c+a\right)\left(b^2+ca\right)}+\frac{c\left(a^2+b^2\right)}{\left(a+b\right)\left(c^2+ab\right)} \geq \frac 32$

(at least, this inequality is true).

  darij

Sorry,Darij is right.
And here is mine:
Just Notice $(b+c)(a^2+bc)=ba^2+ca^2+b^2c+bc^2=b(a^2+c^2)+c(a^2+b^2)$
then let $x=a\left(b^2+c^2\right)$,similar. :P
In fact is equal to Sosoar's :)


Another way to organize the forums are based on school subjects.

In U.S., it will be:

Algebra I and Geometry for Meddle Schools

Algebra II, Pre-Calc and Calc for High Schools.

There, you can promote these forums as homework helpers to attract more users.

[quote="zmli"]homework helpers[/quote]

I think I can speak on account of most of the community and say:  :what?:  :o

I think joml88 speaks well :) For now already the forum structure is stuffed. Creating extra forums is temporarily, out of the question.

This really speaks to the philosophy of the site and our target audience.  Initially, we pictured ourselves as the type of community that it would be unlikely a less-avid student would want to approach.  Our educational philosophy has expanded considerably since we started the site.  We are developing textbooks based loosely on our courses, for example.  These books will have a broader audience than the online school or the original AoPS books.  As a result, the philosophy of our message board may expand to be more inclusive.

Therefore, it is certainly possible that we would expand our community to be more inclusive of the types of students who might seek homework help.  However, before we do so, we'll have to finish some of our programming projects for message board management.  For one, we would probably want such curricular questions to have their own message boards, as these would be well below the level of most of our current forums.  However, the Forum as a whole is pretty cluttered, so we'll need to build some tools (which we are working on) that would allow users to select which Forums are visible and which aren't.  Thus, a user could elect to never have anything on the main page related to College math, or to Middle School, etc.  I'd say it's mostly for this reason that I'll agree with Valentin that adding more forums is temporarily off the table.  However, as we complete some of our Forum programming, this is something we'll revisit.

[quote="rrusczyk"]We are developing textbooks based loosely on our courses, for example.  These books will have a broader audience than the online school or the original AoPS books.[/quote]Awesome.


trigonometry
geometry
trig identities
Law of Sines

In triangle $ ABC$, $ m\angle A \equal{} 30^{\circ}$ and $ m\angle C \equal{} 105^{\circ}$.  If $ AC \equal{} 8$ units, what is the length of segment $ BC$?  Express your answer in simplest radical form.

I personally think this problem was a bit off on the difficulty, but, nonetheless...

$ m \angle A = 30 ^{\circ}$ and $ m \angle C = 105 ^{\circ}$, so $ m \angle B = 45 ^{\circ}$.
We have $ AC = 8$ and $ m \angle B = 45 ^{\circ}$. By the Law of Sines, $ \frac {8}{\sin 45 ^{\circ}}$ = $ \frac {BC}{\sin 30 ^{\circ}}$.

$ \sin 45 ^{\circ} = \frac {\sqrt{2}}{2}$ and $ \sin 30 ^{\circ} = \frac {1}{2}$, so $ 8\sqrt {2} = 2BC$. From here, we see that the length of  $ BC$ is $ \boxed{4\sqrt {2}}$ units long.

Another (way easier) way to do this is to drop down the altitude from $ \angle C$, making a 30-60-90 right triangle and a 45-90-45 right triangle. The altitude length is half of the length of $ AC$, so it is 4. The length of BC is $ \sqrt{2}$ times that value, so it is $ \boxed{4\sqrt{2}}$, our answer. (Test these with Pythagorean Theorem).

Hint : $ 30^\circ\minus{}60^\circ\minus{}90^\circ$ and $ 45^\circ\minus{}45^\circ\minus{}90^\circ$


I'm going to ask you all to submit some questions. We will use these questions to prepare for future MATHCOUNTS tests. I'm pretty sure that it will also be fun! Here are some guidelines:

1) They MUST be MATHCOUNTS level
2) Please state whether you think that question is School, Chapter, State, or National level
3) Follow the guidelines and if you are submitting a real MATHCOUNTS question or an easy AMC question, mark the year and question number. For MATHCOUNTS, include School, Chapter, State, or National. For AMC, put AMC 8 or AMC 10 (AMC 12 is undoubtedly too difficult)
4) Though you are allowed to post previous, official questions, for this post to be more effective, please try to post your own, made-up questions
5) For every question you post, you must provide an accurate solution and make sure that it is hidden
6) Others must provide a correct solution to the above question to be able to post another question by themselves

Eh, shouldn't this be under the Mathcounts marathon? Or made into mock tests? (see the Middle School Tournaments subforum)

Also, there are some noob AMC 12 problems that could go on MC.

Take, for example, AMC 12 A 2006 #1.

Sandwiches at Joe's Fast Food cost $ \$3$ each and sodas cost $ \$2$ each.  How many dollars will it cost to purchase 5 sandwiches and 8 sodas?

Well, I guess I can accept such easy AMC 12 problems. Nevertheless, the guidelines must be followed. And also, I found a sub-forum but not any mock test forum. If there was anything wrong with posting this here, a Moderator or an Administrator will inform me.

[quote="akhil0422"]I'm going to ask you all to submit some questions. We will use these questions to prepare for future MATHCOUNTS tests. I'm pretty sure that it will also be fun! Here are some guidelines:

1) They MUST be MATHCOUNTS level
2) Please state whether you think that question is School, Chapter, State, or National level
3) [b]Follow the guidelines and if you are submitting a real MATHCOUNTS question or an easy AMC question, mark the year and question number. For MATHCOUNTS, include School, Chapter, State, or National. For AMC, put AMC 8 or AMC 10 (AMC 12 is undoubtedly too difficult)[/b]
4) Though you are allowed to post previous, official questions, for this post to be more effective, please try to post your own, made-up questions
5) For every question you post, you must provide an accurate solution and make sure that it is hidden
6) Others must provide a correct solution to the above question to be able to post another question by themselves[/quote]

for the bold part of the quote, u have to make sure that u arent posting any whole tests, only partial tests of less than 25$ \%$ are allowed. Otherwise, this post is perfectly legal and sounds like a load of fun!


quadratics
number theory unsolved
number theory

If p=1(mod4) and $ p\equal{}a^{2}\plus{}b^{2}$, where a is odd, prove that a is a quadratic residue mod p

Just a note - you need Jacobi symbol to apply ZetaX's method, which is just a generalized Legendre symbol, for odd natural $ p$s (not just primes), and satisfies the same properties.

It may not be that obvious to the people who aren't aware of the existence of quadratic reciprocity for composite numbers. A brief information about Jacobi Symbol and its applications can be seen here [url]http://mathworld.wolfram.com/JacobiSymbol.html[/url]
As ZetaX have already said, this one is a direct consequence of quadratic reciprocity.
Edit: Bambaman is just faster than me :D Sorry for this accidentally double post.

A double post giving further explaination isn't one  :D

Well, I always wondered why so many books and coures don't mention the general theorem for Jacobi symbols. It's really useful (more than the one for primes only).


If you want us to tell the property it shows, then it isn't too hard. I can't tell you anything about proving these, which I know it isn't Basics level, so it's gone.

And.... 4 isn't true either  ;)

[quote="jensen"]i know  that may be you think this is very obvious but i want you to prove it (if it has a proof):
for any real numbers like a,b,c prove:
1)$\forall a$ $\exists b$ : $a \times b=1$
2)$(a+b)+c=a+(b+c)$
3)$a+b=b+a$
4)$a+0=0+a=0$
5)$\forall a$ $\exists b$ :$a+b=0$
6)$a+(-a)=0$
7)$a(b.c)=(a.b).c$
there is more but i think it is enough now. :D[/quote]You want us to prove axioms..?

yes i want you to prove these if you can because i realy want to see the proof of these. :D

You'll have to look up in a basic book on algebra how real numbers and operations are defined. There is no way to prove these very basic properties like associativity and commutativity without first having very basic definitions of real numbers.

Also, many of the properties are simply defined to be true. For example, multiplication of integers is defined in this way:
It is an operation $\cdot$ such that:
$x\cdot 1 = x$ and
$x\cdot\rho(y) = x\cdot y + x$, where $\rho(y)$ is the sequential integer following after y.
This then leads to all common properties of multiplication. But only for positive integers of course.

I don't really think this goes in GS, but I don't know where it goes, because you can't really prove axioms.... :?

They're not 'axioms', they are definitions.

The real numbers are usually defined (assuming the rational numbers) in an analysis book, and proving these axioms, order, and completeness is a standard but tedious exercise.

This development comes with one theorem, and all you ever need anywhere else is that one theorem:
There is a complete ordered field, and this field is unique up to order-preserving isomorphism.


Suppose that $ f\circ f$ is an odd/even function. Does it imply that $ f$ is also odd/even?

say $ f(x)\equal{}x^{\sqrt2}$

f(x) is definitely not even, because f(1)=1 and f(-1)=nonreal since if (-1)^x is 1, x is a multiple of 2
$ f(f(x))\equal{}(x^{\sqrt2})^{\sqrt2}\equal{}x^2$ which is even

i think the converse is true, because if $ f(x)\equal{}f(\minus{}x)\equal{}k$, then $ f(f(x))\equal{}f(f(\minus{}x))\equal{}f(k)$
and if $ f(x)\equal{}\minus{}f(\minus{}x)\equal{}k$, then $ f(f(x))\equal{}\minus{}f(f(\minus{}x))\equal{}f(k)\equal{}\minus{}f(\minus{}k)$

[quote="stevenmeow"]say $ f(x) \equal{} x^{\sqrt2}$

f(x) is definitely not even, because f(1)=1 and f(-1)=nonreal since if (-1)^x is 1, x is a multiple of 2
$ f(f(x)) \equal{} (x^{\sqrt2})^{\sqrt2} \equal{} x^2$ which is even

i think the converse is true, because if $ f(x) \equal{} f( \minus{} x) \equal{} k$, then $ f(f(x)) \equal{} f(f( \minus{} x)) \equal{} f(k)$
and if $ f(x) \equal{} \minus{} f( \minus{} x) \equal{} k$, then $ f(f(x)) \equal{} \minus{} f(f( \minus{} x)) \equal{} f(k) \equal{} \minus{} f( \minus{} k)$[/quote]
In your case, the function $ f\circ f$ is not even since the (natural) domain of $ f(x)\equal{}x^{\sqrt2}$ is $ [0,\infty)$, which is not a "symmetric set" (i.e. a set $ X$ satisfying $ x\in X\Rightarrow \minus{}x\in X$), and so is $ f\circ f$. But we only discuss oddness and evenness with a function defined on a "symmetric set".

[hide="Hint"]The answer is [hide="Are you sure you want to peek?"]"no."  An easy way to build a counterexample is to find a function with [hide="the property that"]$ f(f(x)) \equal{} 0$ for all $ x$, but $ f(x)$ is not always 0.  (There are other ways to do it, of course.)[/hide][/hide][/hide]

I'm sure this must have appeared somewhere on the forum before, but I don't know where :(


If $ a,b$ are positve reals such that $ a\plus{}b\equal{}1$ prove that $ (a\plus{}\frac{1}{a})^{2}\plus{}(b\plus{}\frac{1}{b})^{2}\ge\frac{25}{2}$

\begin{align*}\left(a + \frac 1a\right)^2 + \left(b + \frac 1b\right)^2 & \ge\frac 12\left(a + \frac 1a + b + \frac 1b\right)^2 \\
& = \frac 12\left(1 + \frac 1a + \frac 1b\right)^2 \\
& \ge\frac 12\left(1 + \frac {4}{a + b}\right)^2 = \frac {25}{2}\end{align*}

It is an inequality weak and easy to prove

$ LHS \equal{} a^2 \plus{} b^2 \plus{} \frac{1}{a^2} \plus{} \frac{1}{b^2}\plus{} 4 \equal{} 5 \minus{} 2ab \plus{} \frac{1 \minus{} 2ab}{a^2b^2}$

$ ab\le \left(\frac{a \plus{} b}{2}\right)^2 \equal{} \frac{1}{2}\Rightarrow 5 \minus{} 2ab\ge 5 \minus{} \frac{1}{2} \equal{} \frac{9}{2}$

$ \frac{1 \minus{} 2ab}{a^2b^2}\ge 16\left(1 \minus{} \frac{1}{2}\right) \equal{} 8$

$ \Rightarrow LHS \ge \frac{9}{2} \plus{} 8 \equal{} \frac{25}{2}$

Let $ a,b$ are positve reals such that $ a+2b=1.$ Prove that $$ (a+\frac{1}{b})(b+\frac{1}{a})\geq \frac{81}{8}$$
Equality holds when $a=\frac{1}{2},b=\frac{1}{4}.$
Let $ a,b$ are positve reals such that $ a+b^2=1.$ Prove that $$ (a+\frac{1}{b})(b+\frac{1}{a})\geq 2+\frac{31}{6\sqrt 3}$$
Equality holds when $a=\frac{2}{3},b=\frac{1}{\sqrt 3}.$
Let $ a,b$ are positve reals such that $ a+b^3=1.$ Prove that $$ (a+\frac{1}{b})(b+\frac{1}{a})\geq 2+\frac{3}{4\sqrt[3]{4}}+\frac{4\sqrt[3]{4}}{3}$$
Equality holds when $a=\frac{3}{4},b=\frac{1}{\sqrt[3]{4}}.$
Let $ a,b$ are positve reals such that $ a+b^4=1.$ Prove that $$ (a+\frac{1}{b})(b+\frac{1}{a})\geq 2+\frac{1}{20}\sqrt{800+\frac{3381}{\sqrt{5}}}$$
Equality holds when $a=\frac{4}{5},b=\frac{1}{\sqrt[4]{5}}.$


Prove that for any  two  random variables  we have :

  Var[Y-E(Y/X)] = E[Var(Y/X)].

That / symbol can't mean division, can it? That wouldn't make any sense.

Do you actually mean conditional expectation? That is, is this the statment?

$ \operatorname{Var}[Y\minus{}E(Y\mid X)]\equal{}E[\operatorname{Var}(Y\mid X)].$

Yes  thats  what i  mean.  sorry  for not using latex.

$ Var(X|Y) = \mathbb{E}[X^2|Y] - \mathbb{E}[X|Y]^2$, now take expectation of that, you obtain
\[ \mathbb{E}[Var(X|Y)] = \underbrace{\mathbb{E}[\mathbb{E}[X^2|Y]]}_{ = \mathbb{E}[X^2]} - \mathbb{E}[\mathbb{E}[X|Y]^2]\pm \mathbb{E}[X]^2 = \newline Var(X) - \mathbb{E}[\mathbb{E}[X|Y]^2] + \underbrace{\mathbb{E}[X]^2}_{ = \mathbb{E}[\mathbb{E}[X|Y]]^2} =\newline Var(X) - Var(\mathbb{E}[X|Y])\].

I switched $ X$ and $ Y$.


Solve:

$ \frac{3}{\sqrt{2}}\sec \theta\minus{}\sqrt{2}\csc \theta\equal{}1$

[hide]
I know that it can be easily seen that $ \theta\equal{}45^{\circ}$ is a solution but how to find by solving the equation
[/hide]

hello, simplifying your equation we get
$ -\frac{1}{2}\frac{-3\sqrt{2}\sin(x)+2\sqrt{2}\cos(x)+2\cos(x)\sin(x)}{\cos(x)\sin(x)}=0$
converting the numerator in $ \tan(x/2)$ we get
$ {6\sqrt{2}\tan(x/2)+6\sqrt{2}\tan(1/2x)^3-2\sqrt{2}+2\sqrt{2}\tan(1/2x)^4-4\tan(1/2x)+4\tan(1/2x)^3=0}$
setting $ t=\tan(1/2x)$ and factoring we get
$ (t^3+2t^2+2\sqrt{2}t^2+2t+1+\sqrt{2})(-t-1+\sqrt{2})=0$
Sonnhard.


integration
trigonometry
calculus
analytic geometry
inequalities
function
calculus computations

For natural number $ n$, prove that $ \int_{\minus{}1}^{1}e^{\minus{}n x^2}cos x \ d x < \sqrt{\frac{\pi}{n}}.$

Just ignore $ \cos x$ and extend the integration to the entire real line.

[quote="Ji Chen"]For natural number $ n$, prove that $ \int_{ \minus{} 1}^{1}e^{ \minus{} n x^2}cos x \ d x < \sqrt {\frac {\pi}{n}}.$[/quote]

note that when $ \minus{} 1 < x < 1$,$ e^{ \minus{} nx^{2}}cos x > 0$,denote $ I \equal{} \int_{ \minus{} 1}^{1}e^{ \minus{} n x^2}cos x \ d x$

It's obvious that $ I < \int_{ \minus{} 1}^{1}e^{ \minus{} n x^2} d x$.Now make an extension toward the intergral interval,it follows that

$ I < \int_{ \minus{} \infty}^{\infty}e^{ \minus{} n x^2} d x \equal{} I_{1}$.Now we strike to calculate $ I_{1}$.Weapon in the computation process is Fubini's theorem.

$ I_{1}^{2} \equal{} \int_{ \minus{} \infty}^{\infty}\int_{ \minus{} \infty}^{\infty}e^{ \minus{} n( x^2 \plus{} y^2)} d x d y$,here we make a transformation of polar coordinate,let $ x \equal{} r cos\theta,y \equal{} rsin\theta$
$ r$ goes from $ 0$ to $ \infty$ and $ \theta$ goes from $ 0$ to $ 2\pi$,therefore,
   $ I_{1}^{2} \equal{} \int_{0}^{\infty} \int_{0}^{2\pi}e^{ \minus{} nr^2}r d r d\theta\equal{} \frac {\pi}{n}$,so $ I_{1} \equal{} \sqrt {\frac {\pi}{n}}$.So the given inequality holds.

  Remark:we can use another method to tackle this problem.Property of distribution function of continuous random variable can be used here,that' is ,$ F(\infty) \equal{} 1$,maybe this method is more convenient than previous one


[quote="ehsan2004"]Given a triangle $\triangle ABC$  with sidelengths $a$, $b$ and $c$ and the area $S$.

a) Prove that there exists a triangle $\triangle A_1B_1C_1$ with the sides $\sqrt{a}$, $\sqrt{b}$ and $\sqrt{c}$.

b) If $S_1$ is the area of  $\triangle A_1B_1C_1$ , prove that $S^2_1\geq\frac{S\sqrt{3}}{4}$.[/quote]

 a) It is obvious.Because $a,b,c$ are lenghts of a triangle we have $a+b>c$ and we have to prove that $\sqrt{a}+\sqrt{b}\geq\sqrt{c}$.By squaring this last inequality and using the first one,we are done.

 About the second statement I'm thinking right now. ;)


function
real analysis
real analysis unsolved

Hello,

Actually I'm learning the Z-transform and its proprieties. The definition and proprieties can be found here :
[url]http://fourier.eng.hmc.edu/e102/lectures/Z_Transform/node14.html[/url]
Or in wiki :
[url]http://en.wikipedia.org/wiki/Z-transform[/url]

As you can see, the Initial Value Theorem says that if $ x(n)$ is causal then $ x(0) \equal{} lim_{z \to \infty} X(z)$ (with X(z) the Z-transform of x(n) )
The proof is simple and, to my opinion, incorrect. It says that

$ X[z] \equal{} \sum_{k\equal{}0}^{\infty} \frac{x(k)}{z^k} \equal{} x(0) \plus{} \frac{x(1)}{z} \plus{} \frac{x(2)}{z^2}\plus{}...  \to x(0)$ when $ z \to \plus{}\infty$

The problem is, we are not sure that $ \frac{x(k)}{z^k} \to 0$ when z goes to infinity. The only information we have on x(n) is that it is causal ( $ x(t) \equal{} 0$ when $ t<0$ ).

Can anyone make it clear for me please ? Any help will be very appreciated.

For each fixed $ k,$ $ x(k)$ is just some fixed number. Hence, term by term, each term of the series other than the $ k\equal{}0$ term tends to zero as $ z\to\infty.$

Now, as to the overall claim that $ X[z]\to x(0)$ as $ z\to\infty,$ I see two ways to look at it, which depend on how you view $ X[z].$

1. Suppose you simply consider $ X[z]$ to be a formal power series. (I should point out that "Z-transforms" and "generating functions" are essentially the same subject, used by different groups of people, with different conventions and vocabulary. But the ideas are the same.) In the vocabulary of the generating function people, the notion of considering this to be a formal series is an attractive one. In that case, the pointwise convergence of each term to zero is enough.

2. Or, suppose that you require $ X[z]$ to be a function of $ z$ in some sense - in other word, you require that the series converge for some values of $ z.$ For this to be true, $ x(k)$ is only allowed to increase as $ k\to\infty$ as some geometric rate, and there's a radius of converge $ R$ such that the series converges for $ |z|>R.$ (In fact, we have a root-test based formula: $ R\equal{}\limsup_{k\to\infty}|x(k)|^{1/k}.$ If that looks like ordinary power series to you, just try letting $ u\equal{}1/z.$) In that case, we can make the argument you want essentially by invoking uniform converges in comparison to some geometric series.

Thank you Kent,

Yes I supposed X(z) to be a function of z, in this case, if we have $ R\equal{}\limsup_{k\to\infty}|x(k)|^{1/k}$ everything is done. I just stuck at proving it.
Anyways the series idea is a very good one, and it simplifies the proof a lot. Sometimes a problem can be solved by just viewing it a different way. Thanks for showing me that.
And I succeed in proving the Final Value Theorem by myself, cooool  :)


geometry
circumcircle
inequalities
inequalities unsolved

$ \\a,b,c$ is side of traingle and $ \\R$ is radius of circumcircle then prove that
$ \\a^2\plus{}\ b^2\plus{}\ c^2 \leq \ 9R^2$

Hi ! You can use [b]Leibniz theorem[/b] for gravity point of triangle.
$ OG^2 \equal{} \frac{1}{3}(OA ^ 2 \plus{} OB^2 \plus{} OC^2) \minus{} \frac{1}{9}(a^2 \plus{} b^2 \plus{} c^2)\geq 0$

Thank you. My students just start solving geometry problems, so do you have any easy math or geometrical problems?

[quote="boroo12"]$ \\
a,b,c$ is side of traingle and $ \\
R$ is radius of circumcircle then prove that
$ \\
a^2 \plus{} \ b^2 \plus{} \ c^2 \leq \ 9R^2$[/quote]

The seguent is also true :

$ 9R^2\geq a^2\plus{}b^2\plus{}c^2\geq4[ABC]\sqrt{3}$

Actually, I think you can use a^2=(2RsinA)^2
and then we only have to prove (sinA)^2+(sinB)^2+(sinC)^2<=9/4;
then we can apply Jensen's inequality to the equation (sinX)^2 with X within 0-90degrees ,
then we have (sinA)^2+(sinB)^2+(sinC)^2<=3{sin[(A+B+C)/3]}^2=3(sin60)^2=9/4;
(if an angle A is >= 90 degrees than we have (sinA)^2+(sinB)^2+(sinC)^2
<=3{sin[(180-A+B+C)/3]}^2<=3{sin[(90+B+C)/3]}^2<=3(sin60)^2=9/4)


algebra
polynomial
superior algebra
superior algebra unsolved

$ f(x) \in \mathbb{Q}[x]$,  and  $ f(x) > 0$  for  all  $ x \in \mathbb {Q}$.  show that   there  exist    $ g(x), h(x) \in \mathbb{Q}[x]$   such  that

$ f(x) \equal{} g(x)^2 \plus{} h(x)^2$

I read $ f\equal{}g^2\plus{}h^2$. Because if the relation is $ f^2\equal{}g^2\plus{}h^2$ then an obvious solution is $ g\equal{}f,h\equal{}0$.

Read my post #2. What is the answer if $ f(x)\equal{}3x^2\plus{}1$ ?

... or even $ f(x)\equal{}3$. $ 3$ is not a sum of squares of two rationals.

You need to allow the polynomials' coefficients to be in a finite ordered extension of $ \mathbb Q$.

Factor $ f(x) \equal{} \prod (x \minus{} \alpha_i)^{v_i}$ in $ \mathbb C$. If $ \alpha_i$ is real, then $ v_i$ must be even (as otherwise, in a small neighbourhood of $ \alpha_i$, $ f$ changes sign, giving negative values at rationals). So we have $ f(x) \equal{} g(x)^2 \cdot \prod (x \minus{} \beta_i)^{v_i}$ where $ g$ collects the real roots and all $ \beta$ are nonreal. Now the complex roots come in pairs, so lets write $ \prod (x \minus{} \beta_i)^{v_i} \equal{} \prod \left((x \minus{} \beta_i)(x \minus{} \overline{\beta_i})\right)^{v_i} \equal{} \prod \left(x \minus{} \beta_i\right)^{v_i} \cdot \prod \left(x \minus{} \overline{\beta_i}\right)^{v_i}$ and call the left factor $ h$, so $ f \equal{} g^2 \cdot h \cdot \overline{h}$. But now split $ h$ into $ h \equal{} a \plus{} bi$ for real polynomials $ a,b$, thus $ f \equal{} g^2(a \plus{} bi)(a \minus{} bi) \equal{} (ga)^2 \plus{} (gb)^2$ as required.

Note that everything can be done in finite extensions of $ \mathbb Q$. In fact, this proof works in more generality for any ordered field (after adaption on some places, like replacing $ \mathbb R$ and $ \mathbb C$ be real and algebraic closures, respectively).

thank  you very  much!  I am  thinking  the  following  problem:

if $ f(x) \in \mathbb{Z}[x]$ can be expressed as a sum of squares of $ g(x),h(x) \in \mathbb{Q}[x]$, that is , $ f(x) \equal{} g(x)^2 \plus{} h(x)^2$ ,  then it can be expressed as a sum of squares of  $ g_1(x), h_1(x)$, that is , $ f(x) \equal{} g_1(x)^2 \plus{} h_1(x)^2$ ,  where $ g_1(x), h_1(x) \in \mathbb{Z}[x]$  ?????????????

please  refer   to :   http://www.mathlinks.ro/viewtopic.php?t=312346