Map the region $ G \equal{} \mathbb{C}\setminus \{ \minus{} 1\leq z \leq 1\}$ onto a unit disc. Can it be one to one?
I found this: $ 1) w \equal{} \frac {1}{z \plus{} 1} \minus{} 1$ maps $ G$ onto the plane cut along $ Re \ z\leq 0$;     $ 2) \tau \equal{}e^{i\frac {\pi}{2}} \sqrt {w}$ maps onto the upper half of the plane, then 3) any mapping from the upper half onto unit disc. It seems that each step is one to one, but it supposed to be not..am i wrong?

You are wrong for exactly the same reason as I said in the other thread. G is not simply connected and so cannot be topologically mapped to the unit disc (a simply connected region). I don't know what you mean by the plane cut along $ \Re z \leq 0$, but probably this is a simply connected set, so step 1) is wrong

I see my mistake now, but anyway, it is possible to do with an analytic function(not one to one).

first do $ \frac{z\plus{}1}{z\minus{}1}$. the region you will get a a result of this is the whole plane minus the negative real axis. now apply log...what will you get? from there it should be easy.


geometry
circumcircle
angle bisector
geometry proposed

Consider a triangle $ABC$. $M$ is the midpoint of $BC$. The external angle bisector of $A$ meets $BC$ at $D$. The circumcircle of triangle $ADM$ meet $AB$ and $AC$ at $E$ and $F$, respectively. If $N$ is the midpoint of $EF$, show that $MN$ and $AD$ are parallel.

Let $A',$ be, the intersection point of the circumcircle $(O)$ of $\bigtriangleup ABC,$ from the bisector of $\angle A.$

So, because of $MA'\perp MD$ and $AA'\perp AD,$ we have that the quadrilateral $ADA'M,$ is cyclic.

That is the circumcircle $(O')$ of $\bigtriangleup ADM,$ passes through the point $A',$ taken as its diameter the segment $DA'.$

We denote as $K,$ the intersection point of $(O),$ from the segment line $AD$ and it is easy to prove that this point is the antidiametric (=diametric opposite) of $A'.$

$\bullet$ Because of $\angle EAA' = \angle A'AF,$ we have that the diameter $DA'$ of $(O'),$ passes through the midpoint $N$ of the segment $EF.$

Now, it is easy to show that the isosceles triangles $\bigtriangleup DEF$ and $\bigtriangleup KBC$ are similar,

because of $\angle EFD = \angle EAD = \angle CAK = \angle CBK.$

Hence, we conclude that $\frac{DN}{NA'}= \frac{KM}{MA'}$ $\Longrightarrow$ $MN\parallel AD$ and the proof is completed.

Kostas Vittas.


function
combinatorics unsolved
combinatorics

A town has several clubs. Given any two residents there is exactly one club that both belong to. Given any two clubs, there is exactly one resident who belongs to both. Each club has at least 3 members. At least one club has 17 members. How many residents are there?

No solution, please help.

[quote="grobber"]Should the answer be a fixed number? I took all clubs to have equal numbers k of members and all members belonging to the same number r of clubs (this is actually a conclusion). I found k=r and n=k^2-k+1, where n is the number of people (this is block-design theory), but I don't know if such arrangements can be found for arbitrarily large n's of this form. I'll post more some other time (no time right now :( ).

Conclusion: I have no idea how to solve this, although I have tried something... :D[/quote]
Do you still have no solution? :?
I guess 1985 could be an answer  :lol:  :lol:  :lol:

I think we were discussing something related to $k^2-k+1$. And I remember, it refers to only case when $k$ is a power of prime. Pierre knows it very well. Maybe he will answer us?  :P

Well, I didn't actually try to solve this after that first failure :D. I tend to give up rather quickly.

This does look like a projective plane (clubs are lines, members are points), but it still looks to me like there's not enough info supplied to give a definite answer.

[quote="grobber"]This does look like a projective plane (clubs are lines, members are points)[/quote]
It was my though too.

Also I remember nice easy problem: there are $n$ lines on plane and each points of intersection lies on at least three lines. prove that all lines intersect in the same point.

The hypothesis of the initial problem are the 'axioms' of a projective plane (at least three points on each line, any two points define a line, any two lines have exactly one common point).
In that case, as Grobber noticed, every line contains the same number $k+1$ of points, and every point belongs to the same number of lines, which is $k+1$ too.
So that there are altogether a total of $k^2+k+1$ points (and the same for the lines).
The last hypothesis leads to $k+1 = 17$, that is $k=16$. Thus the total number of people is $16^2+16+1=273.$

Note that it is known that a projective plane does exist when $k$ is the power of a prime (which is indeed the case here).

Pierre.

Myth : For the last problem, what if the lines are all parallel?  :P

Pierre.

I think I forgot one condition: not all are parallel. Is it Ok now?

Actually, as I understood grobber considered particular case. Are there other cases?

:blush:  :blush: I think I might have had a shot at solving the initial problem right away, but for some reason by "at least one club has 17 members" I understood "at least one club has [b]at least[/b] 17 members".  :blush:  :blush:

Thus, there is no other case...and we are done. As claimed.

Pierre.

But grobber didn't show how he obtained his conclusion (see first grobber's post).

The last problem :
Assume, for a contradiction, that the lines are not all concurrent. Among all the intersection points and all the lines, there exist a ponts, say $M$ and a line say $d$ such that the distance from $M$ to $d$ is positive (so that non-zero) and minimal.
Through $M$ there are at least three lines.
- If none of these lines is parallel to $d$. Then, they meet $d$ in points $A,B,C$ respectively. An wlog, we may assume that $B$ belongs to the segment $AC$. But from $B$ there must be another line than $d$ and the line $BM$. This line must meet the triangle $ACM$ in a point $P$ which is not $M$, and which does not belongs to $d$. It follows that the distance from $P$ to $d$ is strictly less than than the distance from $M$ to $d$, which contradicts the minimality of the pair $(M,d)$.

- One of the lines from $M$ is parallel to $d$, say $d'$. Then, since there are a finite number of lines there are a finite number of intersection points between $d$ and the other lines. Thus, we may consider $M$ as the left-most point on $d'$ (that is : choose $M$ such that all the other intersection points are on the same half-line with origin $M$). In that case, the two other lines through $M$ meet $d$ in $A,B$ respectively, with $A$ at the left from $B$ on $d$. The third line through $A$ must meet $d'$ (since this third line is not $d$ it is not parallel to $d'$. But, this cannot be done at the left from $M$ by our initial choice. Thus, this has to be done at  the right from $M$, so that this line has to meet the interior of the segment $MB$ which leads to the same contradiction than in the first case above, and we are done.

Pierre.

So Grobber, write your reasoning!  :P

Pierre.

[quote="pbornsztein"]The last problem :...[/quote]
You did it!  :D
So you are worthy to go to 8th grade! :lol:  :lol:  :lol:

I don' know how to prove that, guys. :) As Myth correctly understood, I considered a particular case, without being sure whether it covered all possible cases.

We have the four necessary axioms: there are at least three non-collinear points (I guess this is what "several clubs" means: more than one line), each two points lie on exactly one line, each two lines have exactly one common point, each line has at least three points. However, I don't know how to show, starting from here, that the same number of points lie on all lines and that the same number of lines pass through each point.

Oh, I believe that too, so let's wait for him to enlighten us :D.

Oh my god!  :blush: I feel like the biggest idiot around here :D.

Of course the same number of points will lie on each line, since given two lines $d_1,d_2$, we can take a point $P$ not lying on any of them, and define a bijection between $d_1,d_2$ by setting $f(A_1)=A_2$ s.t. $P,A_1,A_2$ are collinear. It's easy to see that this function is invertible, and thus bijective (it's called projectivity).

I have to put another  :blush: here.

I knew I only had to wait a bit.... :D  :P

Pierre.


function
logarithms
inequalities unsolved
inequalities

Let $a,b>0$. Prove that:
\[ \frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}} \ge \frac{2}{\sqrt{1+ab}}  \]
if either $0<a,b \le 1$ or $ab \ge 3$.

I solved it for $ab \ge 3$ only. Help!

[quote="Lovasz"]Let $a,b>0$. Prove that:
\[ \frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}} \ge \frac{2}{\sqrt{1+ab}}  \]
if neither $0<a,b \le 1$ nor $ab \ge 3$.[/quote]

This is wrong, since the function $\frac{1}{\sqrt{1+e^x}}$ is concave for $x\leq\ln 2$ and convex for $x\geq\ln 2$.

  dg

Yeah,for example:$a=\sqrt{2},b=\frac{6}{5}$.Easy to check.

Oh, my god. That's "either... or", not "neither... nor". Sorry for my careless look. :roll:

I've proved it for $ab \geq 3$ and I think it's wrong for $a \leq 1 , b \leq 1$ (for instance take a=0.1, b=0.2).


Prove that if positive numbers $a,b,c$ satisfy the inequality $5abc>a^3+b^3+c^3$, then there is a triangle with sides $a,b,c$.
Find the best constant $k$ with the following property: If $kabc>a^3+b^3+c^3$ for a positive numbers $a,b,c$, then there exists a triangle with sides $a,b,c$.

I find the best constant k is $k=5$,why is it equal to the first part,It seems impossible in a comptition :(

The second question is from other competition, but i don`t remember from where  :stink:
They are similar and that is why i put them in one topic  :)

This problem was discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=19467 .

  darij

[quote="Beat"]Do you have a solution zhaobin? If you have please post it  :) .[/quote]
My solution is similar to the tranminhhoang's in the link Darij give us :)

Yes, Belarus. That`s why it is in my mind
Thanks Darij


analytic geometry
graphing lines
slope
algebra unsolved
algebra

Hi Everyone!! i am so gald I found this site!! I have some problems I need checked because I can't figure out what I am doing wrong.

y= 3/2x+7/2 i need to write this in standard form, I got 3/2x+y=7/2?

Determine the eqaution of the line containing the point (-3,4) and have slope 1/2 I got:
y-4=1/2(x-(-3))
y-4=1/2x+3/2 (final answer)

What is the equation of the line parallel to 3x+4y=8 and containing the origin? I got:  y=-3/4x+2

A line perpendicular to 3x+4y=8 contains the point (6,-2) What is the equation?  I got 3x+4y=-12

What is the intercepts form of the equation y=-3/5x+3  I got 15x+y=-3 ( I know this is totally wrong)

2x+15y=61
10y=3x+12.5  use either substitution or addition method  to solve for x and y

5x-2y+2z=6
3x+y+5z=4
2x-2y-3z=-3  use either sub or add method solve for z  I got z=-3/5x=9/5 ( I know is wrong)

Wrong forum.  Probably belongs in Classroom Math.

I have no idea where that forum is, HELP!!

for my second problem I got

y-4=1/2(x-(-3))
y-4=1/2x+3/2
than add 4 to both sides and got
y=1/2x+7/2  Is that right now??


function
trigonometry
calculus
derivative
geometry unsolved
geometry

In a scalene$ \Delta ABC$ , $ a,b,c$ are the sides and $ A,B,C$ are the angles. What is the minimum number of distinct terms in the six-tuple $ (a,b,c,A,B,C)$

the answer is four... for an example consider a triangle of angles $ \left(\dfrac{\pi}{6},\dfrac{\pi}{3},\dfrac{\pi}{2}\right)$ and sides $ \left(\dfrac{\pi}{6},\dfrac{\pi\sqrt3}{6},\dfrac{\pi}{3}\right)$...

suppose there exists an scalene triangle with sides $ a,b,c$ and angles $ a,b,c$, respectively, consider the function $ f(x)\equal{}\dfrac{\sin x}{x}$ and its first derivative $ f'(x)\equal{}\dfrac{x\cos x\minus{}\sin x}{x^2}$.. since $ \pi\minus{}c\equal{}a\plus{}b>c$, it follows that triangle is acute-angled... then $ f'(x)\equal{}\dfrac{\cos x(x\minus{}\tan x)}{x^2}<0$, so the function is strictly decreasing, and we can't have $ f(a)\equal{}f(b)\equal{}f(c)$, and we're done...  :D


[hide=" Solution"]First , pawn to d4.
Then if king moves to h5 do Qd3. Then king goes to h4 or g4 so do Qh3.
If king moves to e4 do pawn to e4. Then king can only move to d4 so do pawn to g3.[/hide]
Nice puzzle! :)

[quote="7h3.D3m0n.117"][quote="randomdragoon"][quote="7h3.D3m0n.117"]lol first...that is one bad situation black got himself in  :rotfl: ...lots of options

[hide]Nf3 check, but you do not want black to escape up...is there an actual answer or are you just assuming there is a mate in 3?[/hide][/quote]
There is a way to force mate in 3 moves no matter what Black does. (3 moves meaning white moves, black moves, white moves, black moves, white's third and final move is checkmate) Of course, it is trivial if you have more than 3 moves, but let's assume the 50-move rule is about to kick in ;)

BTW: Nf3 is not correct.
BTW2: Like most chess puzzles, the key move is not a check.[/quote]

like most chess puzzles, they require you to sack a piece for an opponents unwell position (in this case i can't think of anything to sack...)[/quote]It's sort of hard to sacrifice anything in this position :P

[quote="randomdragoon"][/quote][quote="7h3.D3m0n.117"][quote="randomdragoon"][quote="7h3.D3m0n.117"]lol first...that is one bad situation black got himself in  :rotfl: ...lots of options

[hide]Nf3 check, but you do not want black to escape up...is there an actual answer or are you just assuming there is a mate in 3?[/hide][/quote]
There is a way to force mate in 3 moves no matter what Black does. (3 moves meaning white moves, black moves, white moves, black moves, white's third and final move is checkmate) Of course, it is trivial if you have more than 3 moves, but let's assume the 50-move rule is about to kick in ;)

BTW: Nf3 is not correct.
BTW2: Like most chess puzzles, the key move is not a check.[/quote]

like most chess puzzles, they require you to sack a piece for an opponents unwell position (in this case i can't think of anything to sack...)[/quote][quote="randomdragoon"]It's sort of hard to sacrifice anything in this position :P[/quote]

exactly so you either check...or you risk the chance of stalemate

No, not quite. Someone already has a solution posted though (hide it please!)


[b]9.1[/b] Let $ f_m: R \rightarrow R$,
    $ f_m(x)\equal{}(m^2\plus{}m\plus{}1)x^2\minus{}2(m^2\plus{}1)x\plus{}m^2\minus{}m\plus{}1.$ Where $ m \in R.$,
1) Find the fixed common piont of all this parabolas.
2) Find m such that the distance from that fixed point to Oy is minimal.
[b]9.2[/b] Find $ f(x): (0,\plus{}\infty) \rightarrow R$ such that
    $ f(x)\cdot f(y) \plus{} f(\frac{2008}{x})\cdot f(\frac{2008}{y})\equal{}2f(x\cdot y)$ and $ f(2008)\equal{}1$
  for $ \forall x \in (0,\plus{}\infty)$.
[b]9.3[/b] From the vertex A of the equitelateral triangle ABC is drowen a line that intercept [BC] in the point E. The point $ M \in (AE$ , M is external to ABC, $ m(\angle AMB) \equal{} 20 ^ o$ and $ m(\angle AMC) \equal{} 30 ^ o$. $ m(\angle MAB) \equal{} ?$
[b]9.4[/b] Let n be a positive integer different from 0. Find all $ x_1,x_2,...,x_n$ that satisfy relation :
$ \sqrt{x_1\minus{}1}\plus{}2\cdot \sqrt{x_2\minus{}4}\plus{}3\cdot \sqrt{x_3\minus{}9}\plus{}...\plus{}n\cdot\sqrt{x_n\minus{}n^2}\equal{}\frac{1}{2}(x_1\plus{}x_2\plus{}x_3\plus{}...\plus{}x_n)$

where should we post our solutions? :maybe:

[hide="9.4"]
hmmmm,I think I've seen this problem somewhere long time ago... :D

$ \sqrt{x_1\minus{}1}\plus{}2\sqrt{x_2\minus{}4}\plus{}\ldots \plus{}n\sqrt{x_n\minus{}n^2}\equal{}\frac 12(x_1\plus{}x_2\plus{}\ldots \plus{}x_n)$

$ \iff 2\sqrt{x_1\minus{}1}\plus{}4\sqrt{x_2\minus{}4}\plus{}\ldots \plus{}2n\sqrt{x_n\minus{}n^2}\equal{}x_1\plus{}x_2\plus{}\ldots \plus{}x_n$

$ \iff\left(1\minus{}\sqrt{x_1\minus{}1}\right)^2\plus{}\left(2\minus{}\sqrt{x_2\minus{}4}\right)^2\plus{}\ldots \plus{}\left(n\minus{}\sqrt{x_n\minus{}n^2}\right)^2\equal{}0$

$ \Rightarrow x_k\equal{}2k^2$ for $ k\equal{}1,2,\ldots ,n$
[/hide]

[hide="9.1"][quote="DreamTeam"][b]9.1[/b] Let $ f_m: R \rightarrow R$,
    $ f_m(x) \equal{} (m^2 \plus{} m \plus{} 1)x^2 \minus{} 2(m^2 \plus{} 1)x \plus{} m^2 \minus{} m \plus{} 1.$ Where $ m \in R.$,
1) Find the fixed common piont of all this parabolas.
[/quote]

Call fix point is $ (x_0;y_0)$ then we have:
$ y_0 \equal{} (m^2 \plus{} m \plus{} 1)x_0^2 \minus{} 2(m^2 \plus{} 1)x_0 \plus{} m^2 \minus{} m \plus{} 1$ for all $ m$
$ \Leftrightarrow (x_0^2\minus{}2x_0\plus{}1)m^2\plus{}(x_0^2\minus{}1)m\plus{}(x_0^2\minus{}2x_0\plus{}1\minus{}y_0)\equal{}0$ for all $ m$
$ \Leftrightarrow x_0^2\minus{}2x_0\plus{}1\equal{}x_0^2\minus{}1\equal{}x_0^2\minus{}2x_0\plus{}1\minus{}y_0\equal{}0\Leftrightarrow x_0\equal{}1;y_0\equal{}0$

[quote="DreamTeam"]2) Find m such that the distance from that fixed point to Oy is minimal. [/quote]

I don't understand it! Distance from fixed point to $ Oy$ is const![/hide]

[hide="9.2"][quote="DreamTeam"][b]9.2[/b] Find $ f(x): (0, \plus{} \infty) \rightarrow R$ such that
    $ f(x)\cdot f(y) \plus{} f(\frac {2008}{x})\cdot f(\frac {2008}{y}) \equal{} 2f(x\cdot y)$ and $ f(2008) \equal{} 1$
  for $ \forall x \in (0, \plus{} \infty)$.
[/quote]

-) $ x\equal{}2008;y\equal{}1$: $ 2f(2008)f(1)\equal{}2f(2008)\Leftrightarrow f(1)\equal{}1$
-) $ y\equal{}1$: $ f(x)f(1)\plus{}f(\frac{2008}{x})f(2008)\equal{}2f(x)\Leftrightarrow f(x)\equal{}f(\frac{2008}{x})$
Then our condition mean $ f(x)f(y)\equal{}f(xy)\Rightarrow f(x)f(\frac{2008}{x})\equal{}f(2008)\Leftrightarrow f(x)^2\equal{}1\Leftrightarrow f(x)\equal{}\pm 1$
And $ f(x)\equal{}f(\sqrt{x})^2\geq 0$ then $ f(x)\equal{}1$[/hide]

[quote="BaBaK Ghalebi"]where should we post our solutions? :maybe: [/quote]Check http://www.mathlinks.ro/resources.php?c=114&cid=76&year=2008


AMC
AIME
geometry
3D geometry
geometric transformation
rotation
probability

What score did you get; do you think you made AIME; what was hard?

First of all, I have to thank JBL for his [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=70211&highlight=#70211]last-minute tip[/url]. I made AIME with 18 correct and 7 unanswered (18*6+7*2.5=125.5) as a 6th grader :D  :)  :lol:  :shock:  8-)  .

AHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH I missed problem six stupid mistake i got 119.5 NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO Please tell me that the bottom line is less than 119.

A+Math, that's an impressive score for a 6th grader.  Congratulations!

I thought it was standard practice on such exams to figure out how many you needed in order to pass, no?  At Stuy, they actually gave it to everyone with their room assignments.  It's a very useful tool if you don't usually have enough time to do the entire thing through twice, or whatever.

eh i missed #2 cuz i didn't see the "2-digit number" part and counted 7... haha

o well im still pretty sure i made AIME... hopefully

I did 22 and got 3 wrong!! :cry: I now have only 121.5, which is like the worst score in the school where I took the test......

Sophomore here, 22 right, 1 wrong (the annoying sequence question, missed 16 as first term), 2 blank. Overall score is a 137 after a really depressing showing on the 12A.

Can someone post the answers or better yet, tell me where they found em?

how did you guys know your score right away?

137 here i believe... 22 right, 1 wrong, 2 blank

Can someone post the official answers?

Answers are as follows: (from the solutions/answers booklet AMC provides)
1.C
2.B
3.A
4.B
5.D
6.C
7.A
8.A
9.B
10.D
11.C
12.A
13.B
14.C
15.A
16.D
17.B
18.E
19.C
20.D
21.A
22.D
23.B
24.B
25.B
To answer a previous question, finding your score is relatively simple after you know the answers.

I got "C" for #18, did anyone else get this?

how could i forget to write down my answers?

The answer I got for 18 is E, draw a perpendicular line from F to AC and then draw a perependicular line from F to DE.  These two lines are the heights of triangle ABF and DFE and is also similar to triangle ACE.

I got a lot better score on the B than than the A.  22 right, 0 wrong I think, although I did have trouble erasing one so I don't know it that'll pick up or not.

Good job A+ Math, that's great, you'll have 4 more years, who knowns what you can do by then.

[quote="tetrahedr0n"]132. 22 right, 3 wrong. stupid mistakes on cube rotation question, the sequence question, and i misbubbled a geometry problem.[/quote]

Whoa, exactly the same as me.  I got 22 right, 3 wrong, having missed the cube rotation (was going too quickly), the circumscribed and inscribed circle question (I knew the radii were 2 and 6.5, but again, I was going too quickly and just subtracted them.  Oy.), and the triangle one with AT/DT = 3 or whatever it was (that one I just guessed wrong because I couldn't think of an easy way).

GRR... i got a 128.5.. I thought i definitely had a 140.5, but i missed both 2!!! and 4!!!.... 07 should definitely be a 2 digit number. Besides that, i didnt answer number 20 and i missed the cube probability one becuz i didnt consider rotations.

126.5   i also missed number two...it's a very frequent mathcounts question..two digit numbers... *fist shaking*  well. i only answered 20 and got one wrong. I had that cube one down but i forgot to do rotations..thankgoodness i didn't put an answer down and i was like halfway through the circumscribed/inscribed question and i that sequence question. I should have just answered one of them..but i just kind of went "eh." and went to quadruple check number 25. got it right though. i shoulda checked number two.
beans

[quote]uh i took the amc 10b at another school. do i get to be a school winner because i was the only who took it that day?[/quote]

I think that you can practically buy the awards for being the school winner with the school administer's consent.

Hi, now that my testing group has received email notification of results, it's official that we need to prepare for AIME testing, so what's the summary of the AIME preparation advice scattered among the several threads on this forum. I suppose the advice goes something like this:

1) Practice some real AIME tests under real conditions (three-hour session, no calculator).

2) Review the solutions for [b]all[/b] of the AIME problems you practice, whether you answered them correctly or not, to learn or review problem-solving techniques.

3) Think ahead about mental calculation techniques and answer-checking techniques so that you can get the right answer without a calculator.

4) Do some spot learning (or spot reviewing) of math beyond the standard curriculum.

What other suggestions do the rest of you have? Our order of old AIME tests should be coming in the mail soon. How many of you are first-time AIME takers? How many have already taken it twice?

Well that AMC Director guy hasn't posted.  I don't know if he plans to do that or if he expects us to just follow what Texas said.  But apparently then it's not official official, so many not.  lol. I love the poll. It's so accurate.

IT IS ACCURATE IF IIIIIIIIIIIIIIIIIIIIIIIIIII DON"T MAKE IT AHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH
TOKENADULTTTTTTTTTTTTTT you recieved the email of results what are  they?

Two thoughts.

1. What's with the capital letters?  We don't read it as "louder" than if you had just used regular lowercase, so capital letters are unnecessary. :D

2. Relax.  It's not official official.  And if it is, so what? No AIME this year.  Big deal.  You're only in ninth grade, you have many years ahead of you.

[quote="fubu"]Well that AMC Director guy hasn't posted.  I don't know if he plans to do that or if he expects us to just follow what Texas said.[/quote]

There hasn't been a business day for the AMC director to post since the official (emailed) score notices began going out. I know who from my testing group is going on to AIME, but I can't speak about what the cutoff is (the forwarded email I received was edited, so it didn't have the general information Texas137 received). I can't imagine why the email that he received would have anything but the correct information about the cutoff, however.

In any event, I expect we will hear from the AMC director about what the 10 B cutoff was, probably within twenty-four hours, by public post to this forum.

Good luck to everyone who still has tests coming up.

oh yeah winning the state mathcontest is more important. but fubu you are all relaxed because you made the AMC 10B. i can just "dream" of you not making it (chaos, --------------------, etc.).

MORGAN DENNIS ENTER MY BRAIN!!!!!!!!!!!!!!!!!!!!!!!!!!

[i]There will be no murdering of AMC people here.  Calm down.

- Moderator mathfanatic[/i]

in your dreams.

Haha.  Actually, official results from my school haven't come yet so..you know....  I think you're mistaken if you think i'm "all relaxed".  Oh. You naive child.

Chill Churchilljrhigh.  You're not alone.  I missed AIME too this year because of #4 (man, I wondered how many people missed that!)  Oh well, calculation error.  I was placing my bets on getting everything right to bypass the AIME mark, but messed up on one question.  (Well could have be bolder and guessed #21, I turned out to be right anyway after the test, but sigh....annoying teacher that walks down the aisle during the test every 3 minutes looking at everyone's progress!!!)

I presume Churchilljrhigh you're a freshman (me too), and there's still plenty of time to study and do well.  Missing AIME this year is nothing.  Take this year and study hard, get a good grade on AMC 10/12 next year, and work hard to do well on the AIME.  I've seen older students improve from 90's on the AMC 12 to 130's next year after a hard year of studying.  Besides, it's no use crying over spilled milk.  I'll be probably taking the AMC 12 next year (our school uses a class-to-AMC system, meaning if you are in Alg II/Trig you take AMC 10, and if you are in Precalc/Calc you take AMC 12, regardless of your grade level).  So, good luck to you next year!

PS: I have the Australian Mathematical Contest coming up later this year, so that compensates for the AMC...hoping to get a certificate of high distinction this year!


hello, expanding our second equation we get
$ x^2\plus{}2y^2\plus{}2x\plus{}1\plus{}4y\plus{}2\equal{}22$
since $ x^2\plus{}2y^2\equal{}9$ we get after simplifying
$ x\plus{}2y\equal{}5$
inserting $ x\equal{}5\minus{}2y$  in the first equation we have
$ (5\minus{}2y)^2\plus{}2y^2\equal{}9$
solving this quadratic equation in $ y$ deriving our $ x$ we get the two solutions
$ (x,y)$
$ (1,2)$ and $ (\frac{7}{3},\frac{4}{3})$.
Sonnhard.


Let $A$ denote the area of an equilateral triangle with perimeter $P$ and let $B$ denote the area of a regular hexagon with a perimeter of $2P$, where $P>0$.  Determine the ratio of $A$ to $B$.

Think about it!

The side of the equilateral triangle is $\frac{P}{3}$ and the side of the hexagon is $\frac{2P}{6}=\frac{P}{3}$.  We know that a regular hexagon is basically $6$ equilateral triangles put together, and we know that the side of each equilateral triangle that makes up the hexagon is the same length as the equilateral triangle who's perimeter is $P$.

That means, the area of the hexagon is $6$ times the area of the equilateral triangle.  That means, if the area of the equilateral triangle is $A$, then the area of the regular hexagon is $6A$.  Therefore, the ratio of the area of the equilateral triangle and the area of a regular hexagon is:
\[\frac{\cancel{A}}{6\cancel{A}}=\boxed{\frac{1}{6}}\]

[b][size=200]OR[/size][/b]...We can do it the long Algebraic way:

To fine the area of an equilateral triangle with perimeter $P$, we must find the height and the base.  The base is simply a third of the perimeter which is $\frac{P}{3}$.  Using the rule for $30-60-90$ triangles, the height is $\left(\frac{P}{3\cdot2}\right)\sqrt3$.  This makes the area of the equilateral triangle:

\[\left(\frac{1}{2}\right)\left(\frac{P}{3}\right)\left(\frac{P}{3\cdot2}\sqrt3\right)=\frac{P^2\sqrt3}{36}\]

To fine the area of a regular hexagon with perimeter $2P$. we must find the length of one side, then with that information, we can find the height of one of the equilateral triangles in the hexagon.  Thus, we may find the area of one equilateral triangle and then multiply by $6$.

The length of one side is simply $\frac{2P}{6}=\frac{P}{3}$.  The height of one of the equilateral triangles is $\left(\frac{P}{3\cdot2}\right)\sqrt3$.  This makes the area of the hexagon:

\[\cancel6\left(\frac{1}{\cancel2}\right)\left(\frac{P}{\cancel3}\right)\left(\frac{P\sqrt3}{3\cdot2}\right)=\frac{P^2\sqrt3}{6}\].

The ratio of the area of the equilateral triangle and the area of the regular hexagon is:

\[\begin{eqnarray*}\frac{\frac{P^2\sqrt3}{36}}{\frac{P^2\sqrt3}{6}}&=&\left(\frac{\cancel{P^2\sqrt3}}{\cancelto6{36}}\right)\left(\frac{\cancel6}{\cancel{P^2\sqrt3}}\right)\\&=&\boxed{\frac{1}{6}}\]


Le ABC be a triangle with A < B < C.(I) is the incicrle of ABC.
These points L,M,N on the opposite rays CB,CA,BA respectively and
satisfy the condition :
 CLI =(C-B)/2;AMI=(C-A)/2;BNI=(B-A)/2
Prove that L,M,N are collinear

Well, let, for instance, the angle bisector from A meet BC at A'. Then,

  angle A'IL = 180 - angle LA'I - angle A'LI
                 = 180 - angle AA'C - angle CLI
                 = (angle A'AC + angle ACA') - angle CLI
                 = (A/2 + C) - (C-B)/2
                 = A/2 + C/2 + B/2 = 90,

and therefore IL is perpendicular to AI. Hence, L is the meet of the perpendicular to AI through I with the line BC. Similarly for M and N.

Altogether, the points L, M, N are the intersections of the perpendiculars to AI, BI, CI through I with the lines BC, CA, AB, respectively. But there is a general theorem stating that for any point P in the plane of a triangle, the intersections of the perpendiculars to AP, BP, CP through P with the lines BC, CA, AB are collinear. Qed. (Thanks to Jean-Pierre Ehrmann for this approach.)

Ah yes, and the general theorem used above can be easily derived from Menelaos - but it can also be established more elegantly using an idea of Jacques Hadamard. See my note "Problem: The ortho-intercepts line" at http://www.cip.ifi.lmu.de/~grinberg/geometry2.html .

  Sincerely,
  Darij Grinberg


Is the following rephrasing correct?

"Compute the number of simple graphs on $n$ vertices with specified vertex $P$ which contain a unique (edge-non-duplicating) path which begins at $P$ and passes through all vertices."

One thing I am not clear on: if our path uses every bridge at most once, it forces the visits to be in order of decreasing population.  Does this mean, for instance, if it is possible to visit the islands in the order 5-3-1-2-3-4 (so no duplicated edges, but multiple visits to the same island) the setup is bad?  Or is that okay?

I think your reformulation is OK. But just one thing about the last condition: It's clear you can't pass through the same island twice, because the third condition says that each island of your path has less inhabitants that the former one. So you cannot have repeated islands in such a path :)

Did you find the problem uninteresting, or too easy? :( Please posts your solutions... I think the problem's nice... :)

[quote="Jutaro"]The [i]Olimpia[/i] country is formed by $n$ islands. The most populated one is called [i]Panacenter[/i], and every island has a different number of inhabitants. We want to build bridges between these islands, which we'll be able to travel in both directions, under the following conditions:

a) No pair of islands is joined by more than one bridge.

b) Using the bridges we can reach every island from Panacenter.

c) If we want to travel from Panacenter to every other island, in such a way that we use each bridge at most once, the number of inhabitants of the islands we visit is strictly decreasing.

Determine the number of ways we can build the bridges.[/quote]

First of all, notice that any 3 islands in Olimpia can't be connect directly or indirectly by bridges. In words of the VIII OMCC they can´t be in a cycle (truth is I had 6 in this problem, I only mentioned cycles connectd directly). If this is breaked, we would be able to travel from Panacentro to another island "revealing against c)".

After this, is easy to see: each island can olny be connect with exactly one island with more population, excep Panacentro.

Let's denote with:

    $I_{1},I_{2},...,I_{n}$ each island, where if $i<j$ means $I_{i}$ has more population
    than $I_{j}$ and vice versa.

Then:
  $I_{2}$ can be connect with only one island with more population, $I_{1}$ (Panacentro)
  $I_{3}$ can be connect with one of two islands with more population, $I_{1}$ or $I_{2}$
   in general
  $I_{k}$ can be connect with one of $k-1$ islands

We conclude there are $(n-1)!$ ways we can build the bridges.

 :)


After 36, find the next triangle number which is also a square number

The generalization of this problem is not HSB material.

[hide="Discussion"] We want to solve the Diophantine equation

$ \frac{n(n\plus{}1)}{2}\equal{} m^{2}\Leftrightarrow$
$ 4n^{2}\plus{}4n\plus{}1 \equal{} 8m^{2}\plus{}1\Leftrightarrow$
$ (2n\plus{}1)^{2}\minus{}2(2m)^{2}\equal{} 1$

This is a Pell's equation.  It has minimal solution $ (2n\plus{}1, 2m) \equal{} (3, 2)$, and the other solutions are generated in the usual way; that is, let

$ x_{k}\plus{}y_{k}\sqrt{2}\equal{} (3\plus{}2\sqrt{2})^{k}$

$ 36$ corresponds to $ k \equal{} 2$; when $ k \equal{} 3$ we have

$ x_{3}\plus{}y_{3}\sqrt{2}\equal{} 27\plus{}54\sqrt{2}\plus{}72\plus{}16\sqrt{2}\equal{} 99\plus{}70\sqrt{2}$

so $ m \equal{} 35$; then $ m^{2}\equal{}\boxed{ 1225 }$. [/hide]

Thanks, I didn't know it was going to be that hard, perhaps the book wanted a trial and error answer. Interesting formula though.


Im sorry that i dont know latex so if someone wants to convert then go for it. Im not so intrested in the answer, more important how you derived the answer. I have been studying all day, and i think im getting tierd.

xy^2+xz^2=xw^2-6 Solve for x

Thank you in advance.

move the terms with x to one side and factor and divide.

the only problem is, i dont know how to move every thing over and divide..  could some one show me step by step?
Thanks.

You don't need to download anything to make $\LaTeX$ works in this site.

[size=150]JUST PUT DOLLAR SIGNS AROUND YOUR MATH[/size]

Fantastic, thank you. Makes perfect sense. I was leaving that xw^2 on the right side.

Thank you for your help, one last question. the -6 on the right side, is $-6/1$ why does the negative sign drop when there is no reciprocal?

wut do u mean? why it literally drops? :P

Yes, i dont understand why the negative sign goes away. the 6 didnt change sides of the equation, maybe this is in the wrong forum, i am pretty new to algebra.

The minus sign doesn't drop it just gets moved to outside of the fraction.

i didnt notice it over there, i should have looked closer. thank you!


Factor $y^4+2y^3-y$ or show that it cannot be factored.

$y(y^3+2y-1)$
[hide]
Now this can't be factored further over rational coefficients because by Rational Roots, all roots must be of form $\pm 1$ and thus the solutions must be either:
$1,1,1$ or
$-1,-1,1$

and there is no way to add them such that there sum, or the $y^2$ term, is $0$.[/hide]

Factored into what?  Every polynomial can be factored over the complex numbers, if you find the roots...

He probabbly means monomials w/ integer coefficients.

$y(y^3+y^2+y^2-1)=y(y+1)(y^2+y-1)$, last part is irreducible by quadratic formula


Well, here's a marathon. It's in all the other forums, so why not bring it here?

Problem 1:

Factor: $x^2 + y^2 + z^2 - 2yz$

Bad justification.  This is for olympiad problems, not basic algebra.  Take it to Intermediate or lower.

can't we do a marathon with all olympiad problems of a particular kind?

This is one way:
$ x^2 \plus{}y^2 \plus{}z^2 \minus{} 2yz \equal{} (x \minus{} (y\minus{}z)i)(x \plus{} (y\minus{}z)i)$ but that probably doesn't count, does it?

But can we factor a sum of squares? Depends on what field..

Well actually we can even in $ R$, the only requirement is that the double-product of the two squares is also a perfect square.. $ a^4 \plus{} 4b^4 \equal{} (a^2 \plus{}2b^2  \minus{}2ab)(a^2 \plus{}2b^2 \plus{}2ab)$.

Even tho that's not the case here, he didn't specify in which field he wanted it done so I thought I'd throw this in although I'm not sure if it was what he was asking.


Given an acute triangle $ABC$, find the locus of the points of all the points inside the triangle such that

$AB-FG=\frac{MF\cdot AG+MG\cdot BF}{CM}$,

where $F$ and $G$ are the feet of the perpendiculars from $M$ to $BC$ and $AC$ respectively.

Using Ptolemy in $MGCF$ we get:

$\frac{MF\cdot AG+MG\cdot BF}{CM}=\frac{MF\cdot b+MG\cdot a}{CM}-\frac{MF\cdot CG+MG\cdot CF}{CM}=\frac{MF\cdot b+MG\cdot a}{CM}-FG$ so our condition is equivalent to:

$c \cdot CM = b \cdot MF + a \cdot MG$ or equvalently:

$c=b\sin \angle MCB+a\sin \angle MCA$ from which it can be concluded that the locus of such points is a line (as if one point $M$ satisfy this then any other on line $CM$ would satisfy this - also it is easy to see that there can't be two such lines)

all the procedure is ok... but which line?  :lol:


If you're taking the AIME at a university, should you enter just the four digit CEEB or add two 0s in front?

[quote="rrusczyk"]Just type the university name in the CEEB field.[/quote]

What if the full name gets cut off?

[quote="rrusczyk"]All students who have 68 or more points through 3 rounds have qualified for the AIME.  In order for qualified students to be registered through the USAMTS to take the AMC, you must complete the following steps:

1) Log in to [url]http://www.usamts.org[/url].  Click on the Profile tab in the My USAMTS pages

2) Enter your school's CEEB code and the name of the teacher who will administer the test.  If you are homeschooled, enter 'homeschool' for the CEEB code.  You can look up your school's code here:

[url]http://apps.collegeboard.com/cbsearch_code/codeSearchHighschool.jsp[/url]

We do not know how the AMC will compute index scores for USAMTS qualifiers.

We will send the list of qualifiers to the AMC on Friday, February 23.  If you have not completed the CEEB information, your name and score will not be submitted to the AMC.[/quote]
A question, if 70% of the people who qualify for the AIME through AMC then will the USAMTS qualifying score get lowered?

[quote="bpms"]A question, if 70% of the people who qualify for the AIME through AMC then will the USAMTS qualifying score get lowered?[/quote](a) I don't understand your question.

(b) We don't set any policy regarding AIME qualification -- the AMC does.

(c) You don't need to requote our announcement to us.

Is there a constant number of qualifiers?
FOr example, let's say everyone who gets 68 or better gets 100+ on the AMC 12.
Will the cutoff still be 68? Or will it be lowered for more people?

Just wondering, it's not going to apply to me.

I just have a feeling that the AMC's do not want the bulk of the AIME qualifiers qualify through USAMTS, because that is not an AMC competition.

I am using USAMTS as a safety net in case my brain stops working during the AMC's.


[quote="Silverfalcon"]Is it just me or that the stickies on the regular forums have been deleted due to the new forum for middle school?

Putting back is no problem. I just wondered what was happening.[/quote]I am not sure, but if you feel you have the stickies up in the Getting Started, do it.


Prove that 2^n+1 where n is a positive integer is never a multiple of 143.

11|2 n+1 => 11|2 2n-1.
But 11|2 10-1 (and this is the smallest power or in other words 10=ord11(2).
So 10|2n, or 5|n. If 2|n, then 11|2 n-1 and hence cannot divide 2 n+1, so 5|n and n is odd.

13|2 n+1 => 13|2 2n-1.
But ord13(2)=12, so 12|2n, so 6|n, so 2|n. But n was odd and we have our contradiction.


[hide]The answer is $\frac{\sqrt{x^2+y^2}}{|xy|}$.[/hide]

[hide]sqrt(1/9+1/16)=5/12.
1/9=1/3^2 and 1/16=1/4^2.
Combine fractions by 1/3^2=4^2/3^2*4^2+3^2/3^2*4^2=25/144.
But since we need to find the square root, the answer is 5/12.
As for sqrt(1/x^2+1/y^2), that equals sqrt(x^2+y^2)/xy.  The numerator can only be an positive integer if x and y are the two smaller numbers in a Pythagorian triple.[/hide]

it's actually the first time me solving, algebra question because I am really dumb. well, I received the art of problem solving text book two days ago...
and It told me how to solve algebras....

I first thought it was really hard, but, it' isn't........

this is how u solve it.
square root of (1/16 + 1/9) equals to square root of (16/144 + 9/144) and if u add them together, it's  square root of 25/144

u make the square root as 25/144 to the 1/2 power...

then the numerator would be 5 and the denomintor would be 12
so the answer is 5/12

u know what? I won't ever freak out anymore whenever I see algebras again..

Dengmi:

Can the square root of a number ever be negative? Think of how your version can be negative.


Let $ x,y,z >0$such that:$ 2\sqrt {xy}  \plus{} \sqrt {xz}  \equal{} 1$. Find minimum:
$ S \equal{} \frac{{3yz}}{x} \plus{} \frac{{4zx}}{y} \plus{} \frac{{5xy}}{z}$

Let $ a,b,c \in \left[ {\frac{1}{3},3} \right]$. Prove that:
$ \frac{a}{{a \plus{} b}} \plus{} \frac{b}{{b \plus{} c}} \plus{} \frac{c}{{c \plus{} a}} \ge \frac{7}{5}$

[quote="Gnaht"]
Let $ a,b,c \in \left[ {\frac {1}{3},3} \right]$. Prove that:
$ \frac {a}{{a \plus{} b}} \plus{} \frac {b}{{b \plus{} c}} \plus{} \frac {c}{{c \plus{} a}} \ge \frac {7}{5}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=175423


algebra
polynomial
inequalities
algebra proposed

Find all polynomials $ P(x)\in\mathbb R[x]$ such that \[ P(P(x)+x)=P(x)P(x+1)\] for all $x \in \mathbb R$.

[quote="MJ GEO"]Find all polynomial $ P(x)\in\mathbb R[X]$ such that $ P(P(x)\plus{}x)\equal{}P(x)P(x\plus{}1)$ $ \forall x$[/quote]

Let $ n\equal{}\text{degree}(P(x))$

If $ n\ge 2$, $ \text{degree}(P(x)\plus{}x)\equal{}n$ and $ \text{degree}(LHS)\equal{}n^2$ while $ \text{degree}(RHS)\equal{}2n$ and so $ n\equal{}2$

So $ n\in\{0,1,2\}$

1) $ n\equal{}0$ : $ P(x)\equal{}c$ and so $ c\equal{}c^2$ and two solutions $ P(x)\equal{}0$ and $ P(x)\equal{}1$

2) $ n\equal{}1$ : $ \text{degree}(LHS)\le 1$ while $ \text{degree}(RHS)\equal{}2$ so no solution

3) $ n\equal{}2$ : $ P(x)\equal{}ax^2\plus{}bx\plus{}c$
The equation becomes $ a(P(x)\plus{}x)^2\plus{}b(P(x)\plus{}x)\plus{}c\equal{}P(x)P(x\plus{}1)$
So $ aP(x)^2\plus{}2axP(x)\plus{}ax^2\plus{}bP(x)\plus{}bx\plus{}c\equal{}P(x)P(x\plus{}1)$
So $ aP(x)\plus{}2ax\plus{}b\plus{}1\equal{}P(x\plus{}1)$
So $ a^2x^2\plus{}abx\plus{}ac\plus{}2ax\plus{}b\plus{}1\equal{}a(x^2\plus{}2x\plus{}1)\plus{}b(x\plus{}1)\plus{}c$
So $ a(a\minus{}1)x^2\plus{}(a\minus{}1)bx\plus{}(a\minus{}1)c\minus{}(a\minus{}1)\equal{}0$

and so $ a\equal{}1$

Hence the three solutions (it's easy to check that these necessary conditions are sufficient) :

$ P(x)\equal{}0$
$ P(x)\equal{}1$
$ P(x)\equal{}x^2\plus{}bx\plus{}c$

[quote="MJ GEO"]WHAT DO YOU MEAN   LHS  OR RHS[/quote]

LHS = Left Hand Side (left part of an equality / inequality)

RHS = Right Hand Side (right part of an equality / inequality)

YES I UNDERSTANT YOU ARE SO GOOD IN ALGEBRA CAN YOU SOLVE MY OTHER PROBLEMS THANKS


A relation R is defined on the set of real no.s .
such that for each ordered pair a,b
a is less than or equal to b^3.

show that this relation is neither reflexsive, symmetric or transetive.

The empty relation is both symmetric and transitive, and it satisfies your condition ;)

^^ Im not getting that?
Ur supposed to prove its not transetive.

Ah, I misunderstood your question (though reading it again, the ambiguity is less than I thought).  To clarify:

Let $ R$ be the relation such that $ R(a, b)$ holds if and only if $ a \leq b^3$.  Prove that $ R$ is not reflexive, transitive or symmetric.

I need to learn LaTeX.
Yes , so please solution!
IM not able to find counter examples to disprove the transetive part.

Well, assume you have $ a \leq b^3$ and $ b\leq c^3$.  You want to show that it's possible to have $ a > c^3$.  So, we need $ a$ large, say $ a \equal{} 64$.  With this value of $ a$, can you force $ c$ to be not-too-large?

Non-reflexitivity: pick $ a \equal{} \minus{}2$.

Asymmetry: Pick $ a \equal{} 4$ and $ b \equal{} 5$; $ 4 \leq 125$, but $ 125 \not \leq 4$.

Non-transitivity: Pick $ a \equal{} 9$, $ b \equal{} 8$, and $ c \equal{} 2$.

[quote="JBL"]Let $ R$ be the relation such that $ R(a, b)$ holds if and only if $ a \leq b^3$.  Prove that $ R$ is not reflexive, transitive or symmetric.[/quote]
What are the definitions of those three terms?

That's the sort of question that you should go look up instead of asking here -- you are on a computer connected to the internet, after all.

Zhero, your counter-example to symmetry is flawed.

A relation $ R$ on a set $ S$ is reflexive if $ R(a, a)$ for all $ a \in S$.  It's symmetric if $ R(a, b)$ if and only if $ R(b, a)$ for all $ a, b \in S$, and it's transitive if $ R(a, b)$ and $ R(b, c)$ imply $ R(a, c)$ for all $ a, b, c \in S$.

Asymmetry:$ A \leq B^3$, and $ B \leq A^3$
$ A^3 \leq B^3A^2$, therefore $ B \leq B^3A^2$
If $ 0 < B$ then $ 1 \leq (AB)^2$

Assume $ B < 1$ then $ B^3 < B^2$
$ A < B^2$ and $ 1 < (B^2 B)^2$
$ 1 < B^6$ and $ 1 < B$, but we assumed $ B < 1$
Therefore the equation is asymmetric for all values $ 0 < B < 1$


trigonometry
geometry
circumcircle
geometry proposed

An acute triangle $ABC$ is given. Find the locus of points $M$ in the interior of $ABC$ such that $AB - FG = \frac{MF \cdot AG + MG \cdot BF}{CM}$ where $F$ and $G$ are the feet of the perpendiculars from $M$ to $BC$ and $AC$ respectively.

the locus is a line wich passes $C$.
proof:
$AB-FG=\frac{MF.AG+MG.BF}{CM}\longrightarrow \\
CM.AB-CF.FG=MF(AC-CG)+MG(BC-CF)$
and since $MGCF$ is cyclic thus by $ptolemy$ theorem we have :
$CM.AB=MF.AC+MG.BC(I)$
now let $\angle MCF=%Error. "tetha" is a bad command.
$ thus we have
$MF=CM\sin {\theta}$ and $MG=CM\sin{(A-\theta) }$
and by using $(I)$ we have :
$AB=\sin{\theta}AC+\sin {(A-\theta)}BC$
that means $\theta$ shoulde be fixed thus the locus of $M$ is a line.

I don't think its complete as it stands, you have to specifically show that only 1 value of $\theta$ works.

it's enough to show that $f(x)=\alpha \sin{x}+\beta \cos{x}$ is a one-to-one  function.
but i think it's not . :?

Now i can coomplete my solution actually the locus is altitud($CH$).
rest of proof:
$AB=\sin{\theta}AC+\sin {(A-\theta)}BC$
but $\sin{\theta}AC$ and $\sin {(A-\theta)}BC$ denote  distances bitween $(A,CM)$and$(B,CM)$ repectively.
thus we always have :
 $AB\ge\sin{\theta}AC+\sin {(A-\theta)}BC$
and the equality hold when $CM$ be a altitude.

i'm pretty sure the angles are the wrong way round: ie, it is OC... (O and H being isogonal conjugates)

[quote="payman_pm"]but $ \sin{\theta}AC $ and $ \sin {(A-\theta)}BC $ denote  distances bitween $ (A,CM) $and$ (B,CM) $ repectively.
[/quote]
I think that it is $ \sin{\theta}AC $ and $ \sin {(A-\theta)}BC $ denote  distances bitween $ (B,CM) $and$ (A,CM) $ repectively.

[i]Sir payman_pm,[/i] in my opinion, I think that MG=CMsin(C-$\theta$) and not MG=CMsin(A-$\theta$) !
[i]Sir kueh,[/i] in my opinion, I think that F,G are the feet of the perpendiculars from M to [b]BC,AC[/b] respectively and not ....to BC,AC respectively !
 In this case the locus is the diameter $d$ (of the circumcircle) for which $C\in d$.
May be I deceive.

dear Ph-An could you post a solution for this you said?

Sorry for the second post . What do you mean Pn-an with this
<<I think that F,G are the feet of the perpendiculars from M to BC,AC respectively and not ....to BC,AC respectively>>

The initial enunciation were correctly. I note $x=m(\widehat {BCM}),\ y=m(\widehat {ACM})$. From the relation $bMF+aMG=cCM$ results $\overline {\underline {\left| c=b\sin x+a\sin y,\ x+y=C\right| }}$. This sistem has a unique solution. I know that $c=b\cos A+a\cos B$, i.e. $c=b\sin (90-A)+a\sin (90-B),\ (90-A)+(90-B)=C$. In conclusion, $x=90-A,\ y=90-B$,  i.e. the point $M$ belongs to the diameter (which passes through $C$) of the circumcircle.

[quote="Ph-An"] From the relation $bMF+aMG=cCM$ results $\overline {\underline {\left| c=b\sin x+a\sin y,\ x+y=C\right| }}$. This sistem has a unique solution[/quote]
Why this system has unique solution?

Let $x = \angle FGC, y = \angle GFC$
$GCFM$ is cyclic so $FMC=x, GMC=y$.
$AB = AC \cos A + BC \cos B$
$FG=GC \cos x + FC \cos y$
$\frac{MF}{CM}= \cos x, \frac{MG}{CM} = \cos y$
The condition becomes
$AC \cos A + BC \cos B - (FG=GC \cos x + FC \cos y) = AG \cos x + BF \cos y$
$AC \cos A + BC \cos B = AC \cos x + \BC \cos y$
$AC(\cos A - \cos x) = BC(-\cos B + \cos y)$
$\frac{AC}{BC} = \frac{\cos B - \cos y}{\cos A - \cos x}$
$\frac{\sin B}{\sin A}=\frac{\cos B - \cos y}{\cos A - \cos x}$
$\sin A \cos B + \cos B \sin A = \sin A \cos y + \sin B \cos x$
$\sin (A+B)=\sin A \cos y + \sin B \cos x$
$A+B=x+y$.
Let $x=A+z, y=B-z$
$\sin A \cos y + \sin B \cos x$
$\sin(A+B)=\sin A \cos (B-z) + \sin B \cos (A+z)$
$\sin (A+B) = \sin A (\cos B \cos z + \sin B \sin z) + \sin B (\cos A \cos z - \sin A \sin z)$
$\sin(A+B)=(\sin A \cos B +\sinh \cosh) \cos z$
$\cos z = 1$
$z = 0$
$x=A, y=B$
so $GF \parallel AB$ is equivalent to the locus condition.
The circumcenter $O$ satisfies this condition. For any point $O'$ on ray $CO$ inside the triangle, $\triangle CO'G$ similar to $\triangle COG$ and $\triangle CO'F$ similar to $\triangle COF$
$\angle CG'F'= \angle CO'F' = \angle COF = A$, similarly $\angle CF'G' = B$ so $O'$ also satisfies the condition. The locus is therefore the part of ray $CO$ lying in the interior of $\triangle ABC$.


What is the name of this game going to be? Alcumus was cool, so what is AoPS gonna call this one?

Also, I was thinking that it might be a good idea if you can name your games, so it might be a bit easier to arrange tournaments or something (Nerd, create "Tourney 1, Round 2, Game 1", minicon, join it. Xevarion, create....). Yeah, I guess so...

In fact, it might be a good idea to eventually implement a tournament feature.

Feel free to forward any ideas you have for the name of the game.

Naming the individual games is something we are looking into for the future.  Just trying to eradicate bugs now, most of which are proving to be a real pain to find.

i think there should be a big ranking chart to see who has the highest rating just like in alcumus, a top 10 scoring list :D

EDIT: whoops, nevermind  :blush:  i just remembered there's a leaderboard  :rotfl:


probability
geometry
conditional probability

[quote="My textbook"]44. Professor Indiana Jones gives his class a list of 20 study questions, from which he will select 8 to be answered on the final exam. If a given student knows how to answer 14 of the questions, what is the probability that the student will be able to answer correctly the given number of questions?
(a) All 8 questions
(b) Exactly 5 questions
(c) At least 6 questions[/quote]

I get really weird answers for these, and the answers are not on the back of the book:

I didn't explain how I got them, since that would only confuse you (and probably me, too) further.

[hide="My weird (probably incorrect) answers"]Part (a)

\[ \frac{\binom{12}{6}}{\binom{20}{14}} \equal{} \frac{77}{3230} \approx 2.4 \%\]

I also obtained another answer using another way (which surprisingly gives the same result):

\[ \frac{\binom{14}{8}}{\binom{20}{8}} \equal{} \frac{77}{3230} \approx 2.4 \%\]

Part (b)

\[ \frac{\binom{12}{3}\cdot\binom{8}{5}}{\binom{20}{14}} \equal{} \frac{308}{969} \approx 31.8\%\]

I also obtained another answer using another way (I think that the above answer is probably closer, though):

\[ \frac{\binom{14}{5}}{\binom{20}{8}} \equal{} \frac{77}{4845} \approx 1.5 \%\]

Part (c)

\[ \frac{\sum_{k\equal{}6}^{8}\binom{8}{k}\binom{12}{8\minus{}k}}{\binom{20}{14}} \equal{} \frac{389}{7752} \approx 5\%\]

I also obtained another answer using another way (I think that the above answer is probably closer, though):

\[ \frac{\sum_{k\equal{}6}^{8}\binom{14}{k}}{\binom{20}{8}} \equal{} \frac{121}{1615} \approx 7.5\%\][/hide]

I know that there is a better way to do this. Does this have anything to do with the binomial formula or conditional probability? Probability is my worst area in mathematics.

Your closest answer is 31.8%, which would be at least 6.

Please ignore my last post.  My logic was wrong.

You can both edit your old posts within a certain time frame (36 hrs?) and delete your posts when they are the last post on a given thread.

Your answer to the first one is correct (I like the form $ \binom{14}{8} /\binom{20}{8}$ the best -- the number of 8-tuples for which the student could solve all of them divided by the total number of 8-tuples.)

Your first answer to part (b) is right, too.  You should be able to write down simple sentences (like the one in my preceding paragraph) to explain why these count the right things.

[quote="JBL"]Your answer to the first one is correct (I like the form $ \binom{14}{8} /\binom{20}{8}$ the best -- the number of 8-tuples for which the student could solve all of them divided by the total number of 8-tuples.)

Your first answer to part (b) is right, too.  You should be able to write down simple sentences (like the one in my preceding paragraph) to explain why these count the right things.[/quote]

Part (b)
I will count the number of ways that the student can pick exactly five problems to get correct. First, we pick 5 problems out of the 8 on the test that he can get correct, then we pick 9 problems out of the remaining 12 that will not be on the test (since he knows how to do 5+9=14 problems). So the total number of ways is $ \binom{12}{9}\cdot\binom{8}{5}$, and we divide this by the total number of 14-tuples (since he knows how to do 14 of them, or $ \binom{20}{14}$.

Part (c)
Looking back, I think that both my previous answers are incorrect. Using the same method as in part (b) (only now we account for specific cases like exactly 6, exactly 7, and exactly 8):

\[ \frac{\sum_{k\equal{}6}^{8} \binom{8}{k}\cdot \binom{12}{14\minus{}k}}{\binom{20}{14}} \approx 54.5 \%\]

and this seems to be a more reasonable answer.

Could somebody check these? Thanks!

Also, when I used this method:

[quote="JBL"]the number of 8-tuples for which the student could solve all of them divided by the total number of 8-tuples.[/quote]

why won't it work for the other parts? For example, in part (b), why can't we just pick the number of 5-tuples for which he can solve all of them divided by the total number of 8-tuples? Playing around with my answers a little, I see that this answer is off by a factor of $ 715 \equal{} \binom{13}{9}$. Why do I need to multiply my answer by this?

It will work for the other parts, you just have to be careful: you've divided a number of 5-tuples by a number of 8-tuples, and this should be a warning sign (since in this method our sample space is a set of 8-tuples).  To do question (b) this way, you need $ \frac{\binom{14}{5} \cdot \binom{6}{3}}{\binom{20}{8}}$.  And what you've done for parts (b) and (c) looks right to me.


geometry
3D geometry
Pythagorean Theorem

A circular sheet of paper with radius of 6 cm is cut into three congruent sectors. What is the height of the cone in centimeters that can be created by rolling one of the sections until the edges meet? Express your answer in simplest radical form.

John's car tires have a radius of 15.5. If his car travels 1150 revolutions per minute, how fast is he traveling in miles per hour?

How many revolutions per minute should his tires make in order to travel 65 miles per hour?

[hide="first problem"]
first, find the circumference of the base of one of the new cones. From this, you can find the radius of the new cones.
then, find the slant height (distance from vertex to a point on the base).
Finally, apply pythagorean theorem to find the height
[hide="solution"]
old circumference = $ 2 \cdot 6 \pi$
new circumference = $ \frac{2 \cdot 6 \pi}{3}\equal{}4 \pi$
new radius (denote this by r): $ 2 \pi r\equal{}4 \pi \implies r\equal{}2$
slant height= old radius = 6
new height (denote this by h) $ h^2\plus{}2^2\equal{}6^2 \implies h^2\equal{}32 \implies h\equal{}4\sqrt2$
[/hide]
[/hide]
for number 2, are the units of the 15.5 inches?

I will assume that the tires have radii of $ 15.5$ inches, as that is the most plausible possibility...it also makes the second problem more likely (you want to go faster, most of the time :wink: ).

[hide="2a"]
The circumference of a tire is $ 2\pi\cdot15.5\equal{}31\pi$ inches, so every minute, the car travels $ 1150\cdot31\pi\equal{}35650\pi$ inches.

Since there are $ 60$ minutes in an hour, the car travels $ 60\cdot35650\pi\equal{}2139000\pi$ inches an hour, and there are $ 12\cdot5280\equal{}63360$ inches per mile, we conclude that the car moves at

$ \frac{2139000\pi}{63360}\equal{}\boxed{\frac{17825\pi}{528}}$ miles per hour.
[/hide]
[hide="2b"]
Note that the speed of the car varies directly with the number of revolutions made per minute, so we have

$ \frac x{1150}\equal{}\frac{65}{\frac{17825\pi}{528}}\implies\frac{17825\pi}{528}x\equal{}74750\implies x\equal{}\boxed{\frac{68640}{31\pi}}$ revolutions.
[/hide]


There was a problem with just "9%" written on it, and the answer was 24.92
what happened? was it a glitch?

Next time something like this happens, click "Review" after the game is over, go to the glitched problem, and click "Report Error".


This is not exact wording.. but here it is:

28. How many circles with radius of 2 can be drawn with only 2 points lying on the circumference of a circle?
(use the diagram shown below)

Do you guys know the shortcut to this kind of problem?
I have the general idea.... but still.
[hide="The Solution says.."]You do it by drawing an equilateral triangles. :| I didn't quite get it.. It would be helpful if you guys can explain. Anyways, the answer is $ \boxed{18}$[/hide]

[quote="alanchou"]Build an equilateral triangle with two of the points as vertices and has a side length of 2.  The third point is the center of the circle.[/quote]
Yeah.. That's what the solution says.. Problem solved. Thanks.

[quote="alanchou"]Build an equilateral triangle with two of the points as vertices and has a side length of 2.  The third point is the center of the circle.[/quote]
Wait..
But even by doing that, I can't see 18...

[quote="alanchou"]9 total pairs, with 2 points possible for each pair to make an equilateral triangle, so 9 times 2, which is $ 18$.

I did slide over the fact that it's possible that two of these 18 points overlap.  Try proving this is not the case. (don't do it here please)

And technically, a circle's circumference has an infinite number of points[/quote]
I only see 2 pairs that are 2 units apart from each other.
And if you mean any distance, I see more than 9.
Unless were looking at different diagrams.

look closely at the diagram, your mistake right now is like having a question say "give your answer as a percent" and you give a fraction.

None of the points are 1 unit apart.  :wink:

[quote="DonkeyKong"][quote="ZhangPeijin"]I don't get it...[/quote]
By drawing a equailateral triangle with side length of 2, you have 2 points..

And do it downward and upward. They both count, I think.[/quote]
Omg...
I just noticed.
The points
are two units apart.
Not one.
THATS WHAT MESSED ME UP!
Wait..
But I still only count 7...
Hmm...

@ZhangPeijin: Here's a site you can go to and check.

[url=http://www.mnspe.org/mathcounts/oldtests/510-04solutionsS.pdf]It's #29 Sprint[/url]

Oh now I get it.
It didn't have to all be like the one with equilateral triangles.
 :lol:


How many numbers equal their reciprocals ?

[hide] We have $ x\equal{}\frac1x \implies x^2\equal{}1 \implies x\equal{}\pm1$. So, $ \boxed{2}$ numbers equal their reciprocals. [/hide]

[hide="solution?"]2: 1 and -1[/hide]

Edit: aw, beat


Given a triangle ABC and $ M \in BC,N\in AC,P\in AB$ satisfy $ \bigtriangleup ABC\sim \bigtriangleup MNP$. I is incenter of triangle MNP. Find the locus of I.

let's suppose that $ M,N,P$ are the midpoints of the triangle $ ABC$.then denote the incenter of $ ABC$ by $ I$ and the incenter of $ MNP$ by $ J$ obviously the collinearity of  $ J,I,G$ holds (where $ G$ is the centroid of $ ABC$).
also the questions says that we must have $ J \in OH$ where $ H$ is the orthocenter of $ ABC$.but we know that $ G \in OH$ and also $ J \in GI$ which implies that we must have $ I \in OH$ and so the collinearity of $ I,O,H$ which isn't always true.
am i missing sth?

Thank
Sorry, I have some mistakes. I editted problem.


FTW AMC AIME calculus algebra function domain

I'm new on this part of the forum, the part of the games, and I don't really know which game to follow, so I suggest  to start a new game. I hope that we will make a nice story. The rule is very simple:

[b]Each of you can continue the story with a sentence or more sentences[/b]

I give the start:

[i]Once upon a time, there was a king who had three sons ... [/i]

That is not a problem. It means that this story will be the second version. :D

[i]Their names were so long and so complicated that they were often called Prince 1, Prince 2 and Prince 3 ...[/i]

[i]Acunamatatadasophtilus (Prince 2) loved math problems more than his brother (Prince 1) did[/i]

[i]The third one, Nabucodonosaurusaldinosum (Prince 3), was the youngest, and he was the best problem solver in his family. The emperor of the Eastern Empire had only one daughter, and he wanted to give her daughter and the Empire to one of the princes. He chose one of the most difficult problems from the Empire, and ...[/i]  :D

He then combed the knowledge of his country, took another difficult problem (Goldbach Conjecture?), and...

by means of repeated  :wallbash: ing (but it didn't work). Then he decided to find the meaning of 42.

Then he died of surprise (and brain damage due to :wallbash: ing)

Then he encountered the ninja :ninja: :!:

The story ended quickly.

[size=150][color=green][b]THE END [/b][/color][/size]

[size=150][b][color=orange]The story that never ends (2nd version)[/color][/b][/size]

[hide="Chapter 1"]Once upon a time, there was a king who had three sons who lived in Jangaresh. Their names were so long and so complicated that they were often called Prince 1, Prince 2 and Prince 3. Although Prince Abugadamenashibahozingtalmo (Prince 1) often hated schoolwork, he loved math problems. Acunamatatadasophtilus (Prince 2) loved more math than Prince 2. The third one, Nabucodonosaurusaldinosum (Prince 3), was the youngest, and he was the best problem solver in his family. The emperor of the Eastern Empire had only one daughter, and he wanted to give her daughter and the Empire to one of the princes. He chose one of the most difficult problems from the Empire, and destroyed it. He then combed the knowledge of his country, took another difficult problem (Goldbach Conjecture?), and phialed it. Then he tried to prove the Riemann hypothesis by means of repeated wallbashing (but it didn't work). Then he decided to find the meaning of 42. Alas, the 3 princes failed. "Is there nothing you can do? I thought you three were geniuses?!?!" said the emperor. Then he died of surprise (and brain damage due to wallbashing).

The king had another son (Prince 4), with another woman. That Prince was the best problem solver from the kingdom, so he easily solved that problem, but the princess didn't love him because he was too ugly. She decided to get married with somebody else. The story continues, as the three princes didn't know that the fourth one was alive. There was a great surprise when they first met. Prince 4 told them a big secret: "You all suck at math! I am way better than you will ever be!" The three princes, who were shocked, aghast, and appalled, jumped on the fourth prince... Unfortunately, for the three princes, the fourth prince help a 1337th degree black belt in Norris-kwon-do (cookies to those who get the reference). When the three princes heard that the problem could be solved using the secret, they understood why they couldn't solve that problem.Finally, they decided to find the secret on their own. The secret was hidden in the Dark Empire, and they decided to go in a great journey ... and then solved it, became famous, and were buried alive. They drank the Potion of Life and they continued their journey when a huge monster with 42 heads appeared and charged the princes.

They killed that monster with 42 heads because it was quite small like a mouse. The land from that region was full with little monsters. But then all the little monsters ganged up on the princes and killed them all (Here is a possible END of the story). Even if the princes are dead now (in the case that the princes are dead), there are still alive other characters of the story. Chuck Norris became very mad and came down to the land of 42-headed monsters and got pwned. It never happens, but he still got pwned. But wait! Prince 3 was still alive due to the fact that the body who faced the monster was a clone! Then Prince 3 cloned himself 999 times. Prince 3 decided that this story was not good enough for him, so he chose to go with all his clones in Attack of the Clones (Episode II - Star Wars). Then Lord Sidious came with his army in this story and tried to stop the two princes, that had reborned like phoenix bird miraculously, to find the secret, because the secret was dangerous for him and eventually he found the secret, which was that Chuck Norris never gets pwned. However, that wasn't a secret, so...

The secret that Chuck Norris never gets pwned was the secret put by Lord Sidious in the box of the secret, because Lord Sidious had stolen the true secret and gone with him back in Attack of the Clones . But the prince find that there was another way to find the solution. He went to MOTHER OMIDA and asked her, but MOTHER OMIDA was too tired, and she went in Hawaii. So the prince decided to fly to Hawaii, but his plane was hijacked by al-Qaeda terrorists over the Pacific Ocean, who demanded that the plane crashed. So they jumped out the window but the plane didn't crash. And the prince survived! The prince landed next to a destroyer, whose crew thought it was being attacked. Chuck Norris came with four ninja and helped the prince to continue his journey. The destroyer was destroyed very easy, and the prince and Chuck found MOTHER OMIDA surfing. MOTHER OMIDA drowned. The prince asked Chuck Norris to threaten to roundhouse kick the world if it didn't give the prince what he wanted. The prince didn't get what he wanted because MOTHER OMIDA drowned, so Chuck Norris was dismissed. The four ninja returned back home, and the prince was ashamed, because he couldn't get what he desired. But the KING had the secret right in his room, under the bed. The secret had a terrible smell.

The prince found the secret and he was very happy ... But that was only for a moment, he became a very sad person after that because the power of the secret was too great for him, so he went in a "sejour a Paris" where he met his brothers, and he became again a happy person. Three aliens abducted the prince from the center of Paris and took him to their planet, called Greenalia ... He fall in love with an alien girl there, but that girl was a "Fiona"(from the movie SHREK) kind of girl, so he wanted to go home, because he was terrified by her. He killed Fiona and came back. He killed Fiona because she was really ugly. She was uglier than Shrek and she was very fat But, when the prince arrived home he realised that killing an alien is a bad thing and didn't want to go to jail. When he got out from the house he met two alien police officers. Who were also fat. Their obesity proved to be advantageous, though, because the prince drowned in the fat. Then the King was reborn like a phoenix , pulled out a light saber and [that kind of battle from STAR WARS] the fat people and cut their stomachs open. Lord Sidious came again with his clones from Star Wars and destroied Greenalia ... Then Lord Sidious came to Earth and the only persons who could save the planet were the King, the princes, and the princess whose husband had died two weeks earlier. The problem was that Prince 1 was too fat, so he had plastic surgery. The princess was too scared by the clones so she simply died One of the clones, an intelligent one, forced himself/herself to look like the princess. The similarity was incredible. Everybody thought that the clone was infact the princess.
[/hide]
[hide="Chapter 2"]The princess clone then took her spaceship to combat Lord Sidious. But as she approached the Death Star, where Sidious was getting ready to destroy the Earth with his and laser, the princess was zapped with the laser and died. But Sidious had to now power up the laser again, which took eternity (literally). Darth Vader, Carth Wader, Barth Xader, and Aarth Yader, the quadruplets, all got in a fight. Carth Wader won. Then, she attacked Lord Sidious and almost died. However, she was about to beheaded when Ralph Nader, her mom, popped up and chopped off Lord Sidious's head. Ralph Nader and Carth Wader went off to a little island in the middle of a lake that's in the middle of a country that is in the middle of a continent that is in the middle of the planet that is in the middle of the galaxy in the middle of the universe and lived happily ever after.

Lord Sidious didn't die because that was only a simple clone. The true Lord Sidious had been hidden in the box of the secret, but the stupid prince had liberated him when he had opened the box. Lord Sidious had took the image of a fly and he had flown to Hawaii, where he had killed MOTHER OMIDA. Then Lord Sidious took the image of a moose and went to Alaska where he started a new plan of destroying the Earth. But then, Lord Sidious, being a moose, was hunted and was shot fatally by Sarah Palin, governer of Alaska, and Todd Palin, first dude. Then sarah palin tried to swim to russia, which, unfortunately didn't work. She drowned. However, the Russians, who could see Sarah Palin from their houses, rescued her, upon which she shot them. Ralph Nader ran for President. He won, then people found out he had an affair with Palin, and a mob lynched them both a day after he was impeached. OBAMA FTW! Sarah Palin and Ralph Nader continued their career , and Lord Sidious went to American Idol. His voice was extremely bad. He won the first prize because he cheated. The jury was terrified because his voice was so bad that they lost their hearing.
Palin and Nader died. They died after they had finished their "career"...

After winning, Lord Sidious went back to Death Star to charge up the laser. When he reached the Death Star in 2012 (he was late because of his faulty GPS), he put in his American Idol trophy into the laser. Using the gold, he charged up the laser in an instant. The earth was nearing Armageddon. Chuck Norris didn't want the Earth to come to Armageddon, so he roundhouse kicked the Death Star. That year, 2012, was a bad year for Chuck Norris and Lord Sidious. The Earth changed its poles and the ADN of the humans had 12 spirals instead of 2. The day had only 16 hours instead of 24. The sky became as red as the sky from Greenalia. The team of GOOD (the princes, the King, the princess, etc) were in holiday in Hawaii. Infact they ressurected MOTHER OMIDA. A new beginning of the worlds was imminent. However, the ancient Mayans were right! The world ended. The ancient Mayans were aliens from Greenalia who saw the end of their planet, not the end of Earth. Greenalia was the first planet destroyed with the Death Star. In 2012 the Earth was full with aliens from Greenalia. The had known that prophecy. The consequences seen on Earth in 2012 were the consequences of their arrival.

Unfortunately for everyone, the Greenalians were right about Earth too. Earth blew up and either the explosion, or the shock wave, or flying debris killed all of the characters named so far (except Chuck Norris, who can't die). The characters didn't die because MOTHER OMIDA took them to Redalia, the red planet. The new beginning was imminent, as I had said before. Everyone died and the universe started all over again. Whatever that means (here is a misunderstanding between some ideas, but I put them here anyway, because of the rules) Green people walked on the oceans. Blue people walked in the forest, and pink people drowned. (remark: And AoPS people used bad grammar; I'm sorry , I'm an amateur, the point of this story for me is to exercise written English) MOTHER OMIDA had enough power to transport the characters and the clones of the prince. One of the clones, a science nerd, made a time machine. Everyone was transported back to the day the Earth was to be destroyed. But... But then they killed every single alive mayan because the mayans were wrong. After that, there was a period of peace for 1000 years. One of the princes discovered something terrible. He found why the period of peace was only for 1000 years. [/hide]
[hide="Chapter 3"]The period of peace was only 1000 years because Lord Sidious turned back to conquer the planets from all around the world. He decided to conquer the planet called AIME15fails, and destroyed it, and it is now named Sidious999993 because it is the 999993rd planet he conquered. Unfortunately, he made an arithmetic mistake, because it was only his 42nd planet. He conquered a lot of planets. Earth was the final destination. BUT it was not a coincidence that Earth was the last planet ... A shocking discovery proved that MOTHER OMIDA and Lord Sidious were twins. MOTHER OMIDA, the queen of Gaia, had the same power like his brother Lord Sidious. MOTHER OMIDA was the good one, and Lord Sidious was the bad one. It was very hard to decide which of them was the other. They had helped each other since 100000 years ago. It was impossible for Lord Sidious to destroy his sister's home. However, he was wearing adidas shoes, so "IMPOSSIBLE IS NOTHING!" He destroyed Earth, and this computer blew up.

He destroyed Earth's society, but he didn't destroy Earth completely because he wanted to begin a new order. MOTHER OMIDA, for example, survived... Many other people survived. But then the PHARAOH was Lord Sidious, he was the master of the worlds, the only king alive. And, after his assassination, the only king dead. Everybody had wanted to kill PHARAOH. PHARAOH died like a big fat creature that he was. He was very ugly, bad, and dangerous. His animalic body broke in two pieces, because he and her sister had been twins. MOTHER OMIDA ate his last pieces... MOTHER OMIDA, queen of Gaia, took his place. Then she ate the LAST piece.

MOTHER OMIDA revealed a secret. The secret was... that she didn't like the meal from her brother's body. The meal from her parents' bodies was more tasteful. So she threw his body away. However, there was a slight problem... because she used to eat people. PHARAOH MOTHER OMIDA was the worst leader of those times. The only chance was a saviour. The saviour was the son of Lord Sidious, the prince of Gaia. ...whose name was AMYFIFTEEN. AMYFIFTEEN, the daughter/son (whatever) of Lord Sidious, the prince of Gaia, tried to take MOTHER OMIDA's place. AMYFIFTEEN killed MOTHER OMIDA epicly. Then she goes to this place called America on a raft made of plastic.........somehow surviving the vacuum of space........... ...but the plastic melted in the atmosphere and she accidentally landed on an Ares-V rocket blasting off.

The rocket hit the sun, and exploded. AMYFIFTEEN was dead. But the universe lasted on... But wait! It was all an illusion! AMYFIFTEEN survived, and killed the sun, because the sun made HIM mad. Of course that AMYFIFTEEN survived, but MOTHER OMIDA survived, too. MOTHER OMIDA was invincible. She was very sly and artful. She decided to go to Earthdual, a Earth dual, for 424242 years. She was never heard of again. Ever. AMYFIFTEEN dislikes the word "earthdual" and zaps it into ash with his laser beam vision. Then AMYFIFTEEN goes to FTW and makes sure he's still at the top of the leaderboard. After that, he takes a nap. And realizes when he wakes up that 1337 people have passed him. He scrambles to regain rating and... Pwns everyone, back to the top. Business as usual. MOTHER OMIDA was pregnant, but she was a virgin in the same time.

He gets pwndizziled by this so called Mother Omida. Then, Mother Omida blasts AMYFIFTEEN with a rocket and AMYFIFTEEN dies and AIME15 never spams. And there are no illusions, this time. Then the rocket MOTHER OMIDA used to pwn AMYFIFTEEN comes back to pwn AIME15. It misses and pwns itself. Unfortunately it blows up the whole universe too except mother omida. And Chuck Norris because Chuck Norris cant be killed. MOTHER OMIDA gave birth to a new character: CAESAR. By the way, AMYFIFTEEN didn't die either. Then MOTHER OMIDA, CHUCK NORRIS, and AMYFIFTEEN decided to kill CAESAR because they hated CAESAR salad dressing.MOTHER OMIDA didn't want to kill CAESAR because he was her son Even though she ate her brother, she couldn't kill her only son. She fell in love with him. CAESAR starts world war M.

Firstly, he visited Lord Sidious' tomb from the eastern part of Gaia. There, he learned some new tricks. He learned how to be sly and artful like his mother. Then he went to Lord Insidious' tomb in the western part and unlearned everything. Finally, he went in the northern part, and then he became invincible. He was cunning, shy, artful, sly, dangerous, fearless, strong, and most of all, very jealous. Caesar didn't like MOTHER OMIDA's tricks, and he said to her: "I don't like your tricks." She replied, "I don't like your tricks either, so you can stop talking about it." Caesar got mad and pwned MOTHER OMIDA. Then MOTHER OMIDA pwned him back. Caesar then goes to candy land and upsets the...uh...inhabitants there.

Then he went to salad land and they loved him so much they trapped him there. Caesar did not like salad, however, and he...
died. AMYFIFTEEN comes back and laughs. Caesar couldn't die because he was the son of Mother Omida and he learned a lot of tricks from the tomb of Lord Sidious. Even if AMYFIFTEEN was jealous, Caesar was better than him and he was eternal. With his last forces, AMYFIFTEEN died under the shoe of Caesar. Caesar was then very proud of him, and played Monopoly all day long. AMYFIFTEEN apparently didn't die, because he is typing this post up right now! Caesar didn't like waiting for the score , so he...commited suicide. However, it didn't work. It didn't work because MOTHER OMIDA was using her magical powers to stop him. Again Caesar attempts at suicide by listening to country music and classical music for hours. Surprisingly enough, he lives on. Then...Then he decided to conquer Egypt, and he fell in love with Cleopatra, but Cleopatra was a simple cow ......ard and hid from him. She was a coward cow She was ugly, stupid, and most of all very, very, very dangerous.

She was dangerous because she could walk through any situation, like a ghost walks through walls. She simply didn't set alarm bells off in people. (From Hitchhikers' guide to the galaxy) and she had greasy fur and long pointy horns and gnarled hoofs...bah...Caesar like her anyway because he was a twisted young man. Cleopatra didn't like him because he was a twisted young man. Then Cleopatra prodded Caesar. Then Caesar accidently kisses Cleaopatra, who turns into a beautiful lady. Cleopatra tells Caesar to die because she hates him. Caesar lost, and Neo pwned Cleopatra. And then out of nowhere, an adversary known as RYU appears. RYU eats the legendary Caeser salad and gains god-like power.[/b]

Unfortunately, his god-like powers turned him EVIL!!! Cleopatra had a special feeling for snakes that were green and... they were edible... They snacked on AIME_is_hard. they were also in a liquid state of matter. But one day the liquid snakes joined together and solidified, creating a super-soldier known as SOLID SNAKE Who then sublimated into Gaseous Snake. Meanwhile, Wickedestjr has just been born in the land of "The story that never ends (2nd version)".
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[hide="Chapter 4"]The first day of school was not very good for young Wickedestjr. This was because Wickedestjr was very unpopular. Very unpopular, indeed. and because his teachers were the characters from this story. This meant, however, that he could get his revenge on them in a very strange way... A very strange way indeed.

This way was...

UNKNOWN! However, Wickedestjr got pwned by his math teacher. This made Wicked really angry. He stormed out of his math class and tripped over something. He got up and looked at what he had tripped over. It was a magical lamp. He rubbed it and a genie came out. The genie said he would give Wicked three wishes. Wicked's first wish wasn't very unpopular. This wish was granted and there weren't any strange side effects. then he wished he could have 10 more wishes... But the genie got mad at him and made him prove that Fermat's Last Theorem is false, even though its true.

So instead he proved that the genie wanted him to prove that it was true. Unfortunately, this was in an universe where it was false... The genie was out of control after this special event, and he flew away. Wicked yelled "I wish you would stay here". Then the genie flew back.
"You have ten more wishes" said the genie.
Wicked's first wish was "I wish I could have an infinite amount of wishes.
"Whatever" said the genie.
Then Wicked had an infinite amount of wishes.
"I wish that it would be impossible for me to lose my wishes".
"okay" said the genie.
Then Wicked had an infinity amount of wishes for the rest of his life. However, the rest of his life did not exist, because the genie killed him. Still, he had his wishes... Wicked wished that moldovan couldn't post in this thread ever again.

It was an unusual wish because the genie was infact moldovan. The genie, moldovan, got really angry, and flew away for a period of time, but he promised that he would return back with new forces. Wicked waited and waited, and tried to think of something to do. He decided to sit and cry, because the genie was out to kill him. But then he toughened up.
And he started getting ready.
After working so hard, he fell asleep.
But two ninjas disguised as doctors carried him away. They put him in jail, the jail run by TEH GENIE moldovan. However, he realized that he was dead. Yes, Wicked had realized that moldovan was dead. This made Wicked really sad, because moldovan had been a good genie. Then he realized that his kidnappers had killed moldovan. Wicked opened the door to his cell (which hadn't been locked ) and attacked his kidnappers.
Then Wicked ended up escaping. The jail cell was located in Japan. So he started exploring. Wicked realized he was confused and reread the thread and noticed that he had died earlier, and that he could not every be revived.

Then, he died. Wicked, you may no longer post in this thread.

Dead people cannot post.

pythag011 walked down the alley. It was a dark alley. He realized that he was immortal. Suddenly, a genie called moldovan came... And proved that pythag011 was not actually immortal. pythag011 was mad, so he became immortal and pwned the genie. The genie was immortal, too, so he summoned reinforcements. They both fought. Now pythag is blind and the genie is deaf. Since he was blind, pythag fell off a cliff. Luckily, he was immortal. Then he hit the ground. Then he heard the sound of a boulder dropping off the cliff towards him. He then went insane. The boulder missed him. Unfortunately, it made an avalanche which made a closed container of rocks around him. Pythag lost his immortality. Then he became immortal again. And, he could see again. Then he lost his immortality (again!) and was killed by the deaf genie. The genie went to a doctor, and after that he got his voice back. Then he flew to Hawaii where he spent his holiday with Caesar and Mother Omida. Wicked got on a big boat and went to Hawaii also. He met the genie again, but this time the genie was friendful after an extraordinary holiday. He gave him a big strawberry ice cream. The genie sat on his ice cream. He ate that ice cream , and the ice cream melted in his little mouth.

The genie got pwned by Chuck Norris, who appeared out of nowhere. Chuck Norris laughed. The genie was so hungry that he ate other ice-creams, and he invited Chuck to a pizza. Wicked fainted, because he had no idea how Chuck Norris had appeared. Wicked asked him and Chuck Norris revealed his magical powers to Wicked. This inspired Wicked to become a magician and then he became very rich and famous. The genie didn't like the fact that Wicked became so rich in such a short period of time, so he tried to make him poor again. Wicked won the anti-lottery and lost 10000000 dollars.

Then he won the lottery and earned 1/2 cent. After that, the genie took everything he found in Wicked's house: clothes, objects, food, glasses, and Wicked's new pair of shoes. And Wicked's other pair of shoes. On the other hand, Mother Omida and her little boy, Caesar, realised that genie's powers could help them to conquer the world again. In order to tell their wishes to the genie, they threw a grand party, where the genie was invited. However, AMYFIFTEEN stole the invitation and changed the name to AMYFIFTEEN'S party, so AMYFIFTEEN would tell wishes to the genie. The grand party was very grand. Very grand, indeed. In fact, it was so grand that it turned into a black hole and AMYFIFTEEN and the genie perished in it. Whatever, the genie didn't care about that black hole. The genie promised that he would carry out a wish for each invited character.

was one of the invited characters. AMYFIFTEEN murdered WHAT??? said the genie. What did you do? He brought back to life. and rekilled himBut then the genie became a sad genie ... and a suicidal genie. The genie went again in Hawaii, where he spent his holiday. Mother Omida and her little boy, Caesar, trew that party, so that everybody was happy. AIMEFIFTEEN went to Azbakan for 3 years, because of his attempts to destroy the genie. (it was the genie's plan) Do you mean AMYFIFTEEN?

AMYFIFTEEN broke out of Azbakan. He then pwnt the genie, Mother Omida, and Caesar ...however, he had to go to Hawaii to do so, and was hit by a flying lava chunk and brought to Honolulu, where the police brought him back to Azkaban. Azkaban collapsed, killing everyone inside and also for a great distance outside, and also caused a nuclear meltdown that irradiated the whole Earth. Yep, I edited again the post, it was AMYFIFTEEN who was a prisoner in Azkaban, whatever... The genie wasn't happy, because of the destruction of Azkaban, and he built again that prison. The genie, Mother Omida, and Caesar started a new journey in order to meet more friends. they did meet new friends. now they set off on a journey to meet new opponents that they could cheaply and easily obliterate. They met the beast, a character best known for its impressive number of heads. In fact, it had so many that nobody ever bothered to count them. but then caesar got superhuman powers and counted the number of heads the beast had. he counted to infinity. Twice. not one to be daunted by the monster, caesar asked the genie to turn into a crab... The genie turned around and walked into a crab.
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[hide="Chapter 5"]The genie didn't like the crab's life, so he turned back into a genie, and turned Caesar into a mouse. "Where is everyone?" asked the genie ......... this story is tooo long and never ending. that's a good point, and I'm proud of it, but as I said before, this story will end after two more rounds The genie became depressed and committed suicide. But then, Caesar used his superpowers and brought the genie back to life so that he could ask the genie for 1 more wish. caeser then asked the genie to make him the greatest problem solver in the world so caesar decided to go to IMO

Caesar then found himself becoming a huge nerd, and since he did not want to give in to the dark side, he asked the genie to prevent him from being a problem solver, and asked the genie to make him into a Kung-Fu master. The genie blatantly refused, and decided to curse Caesar to become an IMO gold medalist - Caesar melted the gold in his medals to fund a small army, for which he used this army to kill the genie. Then the genie discovered that Riemann hypothesis was false, so Caesar committed suicide. the genie then got bored and brought caeser back to life the genie then got bored and brought caeser back to life the genie died in paradox. Then Caesar felt remorse over his actions and committed suicide - again.

Then, james4l sat on some fluorine Fortunately, the flourine was in a compound already. A new character, called Lady Bugy, appeared from nowhere. Then Caesar and the genie tried to catch her, but they realised that this character was infact Mother Omida. She returned, but she gave in to the dark side, like her brother Lord Sidious. caeser and lady bugy repelled each other like like poles and things got worse Lady Bugy, the real Mother Omida, decided to start a minigame , with the same rules as : The story that never ends (2nd version), but with the following supplements:

1. each sentence contains 5 words

2. the minigame lasts 5 days

3. everybody can take part in this minigame

4. each sentence corresponds to 1 point

So, Mother Omida gave the start of this minigame:

Mother Omida was so hippy She made Caesar very confused. Caesar then failed the AIME. He started following his master Who was himself, making paradoxes that would only confuse the other people, who then were not listening to Caesar. Thus... they looked at him instead. the minigame became a flop The results of the minigame are:

1st place
james4l 3p
aleph0 3p

2nd place
aadil 1p
coolts 1p

While the minigame was happening, Caesar was abducted in Madagascar. and deported to Deimos but died in exile He was reincarnated in the underworld and felt very depressed at the confinements of all of the tunnels and the heat of the underground. so he ordered the nearby reconstruction firms to touch up the place. it wasnt cheap, but soon the underworld was turned into an air-conditioned, roomy room fit for a fairy princess But then the devil exiled Caesar to the normal world. He got lost on the way and got stuck halfway between worlds. he got bored and started solving calculus problems Fortunately, the calculus problems he solved were how to get unstuck. unfortunately he forgot how to do them. caeser then became a user of the art of problem solving.com and held a party for this special event, where he invited a lot of party people.

Like spiderboy95, who accidentally blew up the punch bowl. then caesar remembered he was still stuck. He solved the problems, got unstuck, and found that he accidentally went back to the underworld. Nobody at the party was happy. This time he got permanently stuck not just between the underworld and the realworld, but with time and space too. Now his brain was incapable of doing un-sticking problems. So somebody else did them for him. Unfortunately, he/she tripped over a spacetime discontinuity on the way to tell him and got stuck in a different way. Caesar turned back on Earth through a wormhole.
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[hide="Chapter 6"]Angry that Caesar turned on Earth, especially through a wormhole, Lao Tzu declared war against the country that Caesar was born in. Caesar sat on Lao Tzu. The war turned ugly and violence started spreading . All the gods came together and spread peace, but it didn't work. Then the gods changed Caesar's name to Cesar Chavez - he was now an illegal immigrant. He got really confused and then everyone called him C@esar because everyone else also got confused about whether there was an a or not. But the shift key wasn't invented so he was C2esar.

But he took the shift key to a repairman, who swiftly fixed his shift key...at the price of jamming his C key. so now he was @esar. @esar hated his new name so much that he went on a maniac murder rampage in Philly. His whole keyboard malfunctioned and he went without a name for 3652 days. Now, it's 3653 days. But that depended on the time zone. Peaple started calling him positronio,meaning one with a positronic brain. But an electron killed him. And now the electron took over the world.
But then the electron was crushed to death by an atom. (how sad Rutfl)

SHOULDN'T THIS BE IN THE FUN FACTORY?!
IT DOESNT MATTER?!

and the atom was now the ultimate king
Then the atom was crushed by the molecule and the molecule by a human.

The human decided that being king was stupid and ran off.

All the particles attacked him.

@esr came back from the dead

Mr. Stupid killed @esr.

Mr. Green became king.

Mr. Green had a son named Rotfl

after mr.green died, his son (rolling on the floor laughing) became king.
However, AMYFIFTEEN defeated Rotfl, and AMYFIFTEEN (Yoda) took over the world. BUT THEN,

AMYFIFTEEN was sued by Rotfl and had to resign being king. So Rotfl returned to throne and the world was at peace again. Until Rotfl got ka-PWN-ed - smiley limit -

KABOOM the domain of Rotfl was annihilated and tiny bits of yellow shrouded the universe for millenia. This was called the Big Bang.

AMYFIFTEEN came back to rule the universe. But he was already dead. So he was a ZOMBIE!!!!!!!! who was arrested for being a zombie lolz And died... Then all the characters played chess. Caesar moved a pawn. It was a black pawn. which was killed by a white pawn Then Caesar got really angry. and sentenced the accursed pawn to 30 years to life in prison The bishop saved the pawn, pwning Caesar. Caesar didn't like the life in that prison, so he got furious again and pwned everybody. The bishop, who was really MOTHER OMIDA, back from a few pages ago, waited until Caesar, whose whole keyboard got fixed, to eliminate the opposition and then created a stalemate, stopping the game without any hard feelings. But the chessboard doesn't like stalemates, and it gobbles up all 32 pieces into a black hole.

Mother Omida jumped in that blackhole after them. Everything evaporated into thin dust. The thin dust disappeared into thick air. The thick air was converted into a stream on a trillion photons, operation in the CERN and somehow recreating the exact shape of Caesar. But it was only an image. Caesar was in Hawaii at that time. Hawaii sank. They sank so "delicious" that Caesar turned into a fish, and went under water, playing football with the wales. Finally, with his last forces, Caesar was preparing himself for the end of the story, but ...  ...the end of the story came 23 hours and 45 minutes early and he was too late. but the end was a lie like the cake

And so was the rest of the story. Caesar wondered, was that the cake I ate at breakfast?  Then he realized that the cake he ate was a pancake, with maple syrup that was a lie. No wonder he was so hungry. He ate a strip of pork and got swine flu. He tried to get rid of it, to no avail. He then died for the eighth time. This time, he ate some lies before dying. Oh no, he blew up in a fallacy. But he was dead already... So he died for the ninth time. Caesar got tired of dying. He survived, unsurprisingly. Then he encountered the ninja. The story ended quickly.

THE END
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[b][u]Final results [/u]

 :winner_first: 60p aleph0

 :winner_second: 53p AIME15

 :winner_third: 29p james4l

[hide="Other results"]16p coolts
11p evilhamster
11p Wickedestjr
10p pythag011
9p aadil
8p mz94
6p jain_harshi
5p tcjy
4p 1=2
3p crazypianist1116
3p spiderboy95
2p splatyango
2p ak909090
1p Poincare
1p Eragon
1p Bugi
1p came
1p mathguy999
1p ernie
1p brightzhu
1p AIME_is_hard
1p joyofpi
1p sup3rcrash3r
1p aznzhou890[/b] [/hide]

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