1/ Find the angles of triangle ABC if: $ cos(2A)\plus{}\sqrt{3}(cos(2B)\plus{}cos(2C)\plus{}\frac{5}{2})\equal{}0$
2/ Prove that triangle ABC is right if: $ cos\frac{A}{2}\cdot cos\frac{B}{2}\cdot cos\frac{C}{2}\minus{}sin\frac{A}{2}\cdot sin\frac{B}{2}\cdot sin\frac{C}{2}\equal{}\frac{1}{2}$

These problems are too hard for me.
Someone helps me, please! :(

[hide]Big Hint for #1:
Proving that with a,b,c are real numbers we have: a.cosA+b.cosB+c.cosC=<1/2(a^2+b^2+c^2)
[/hide]

[quote="Kyoshiro"]1/ Find the angles of triangle ABC if: $ cos(2A) \plus{} \sqrt {3}(cos(2B) \plus{} cos(2C) \plus{} \frac {5}{2}) \equal{} 0$
2/ Prove that triangle ABC is right if: $ cos\frac {A}{2}\cdot cos\frac {B}{2}\cdot cos\frac {C}{2} \minus{} sin\frac {A}{2}\cdot sin\frac {B}{2}\cdot sin\frac {C}{2} \equal{} \frac {1}{2}$[/quote]

for the second one , i think you can assume one of the A, B, C is 90. Then obviously, $ cos(\frac{90}{2})\equal{}sin(\frac{90}{2})\equal{}\frac{\sqrt{2}}{2}$
and problem becomes:

$ \frac{\sqrt{2}}{2} \cdot cos\frac {B}{2}\cdot cos\frac {C}{2} \minus{} \frac{\sqrt{2}}{2} \cdot sin\frac {B}{2}\cdot sin\frac {C}{2} \equal{} \frac {1}{2}$

Another hint for you would be, for any right triangle, $ sin(B)\equal{}cos(C)$ and you can use this identity :

$ sinB\equal{}2sin\frac{B}{2}\cdot cos\frac{B}{2}$   which  immediately follows from

$ sin2B \equal{} 2sinBcosB$

Now , it is your turn to try !  :)

[quote="Kyoshiro"]1/ Find the angles of triangle ABC if: $ cos(2A) \plus{} \sqrt {3}(cos(2B) \plus{} cos(2C) \plus{} \frac {5}{2}) \equal{} 0$
2/ Prove that triangle ABC is right if: $ cos\frac {A}{2}\cdot cos\frac {B}{2}\cdot cos\frac {C}{2} \minus{} sin\frac {A}{2}\cdot sin\frac {B}{2}\cdot sin\frac {C}{2} \equal{} \frac {1}{2}$[/quote]
2>
$ \sum(sinA) \minus{} \sum(cosA) \equal{} 1$
=> $ \sum (\sqrt {2}*sin(A \minus{} 45)) \equal{} 1$
=>$ \sum(sin(A \minus{} 45)) \equal{} \frac {1}{\sqrt2}$
=>$ 4sin(\frac {A \plus{} B \minus{} 90}{2})*4sin(\frac {B \plus{} C \minus{} 90}{2})*4sin(\frac {A \plus{} C \minus{} 90}{2}) \equal{} 0$
=>PS


Umm I am quiet familiar with the basic elementary mathematics.  and I would like to start some of the more higher level stuff (calc, rings, fields, etc)  can anyone recommend any books/sites that gives an intro to these topics?  the more contest related (PUTNAM) the better

Topics in Algebra by Herstein, Principles of Mathematical Analysis by Rudin, Real Mathematical Analysis by Pugh, Complex Analysis by Ahlfors.


Extend the rays $BC, AD$ until they intersect at a point $F$. Denote $\phi = \angle AFB$ and $\triangle_L = |\triangle ABF|,\ \triangle_M = |\triangle KLF|,\ \triangle_S = |\triangle DCF|$ the areas of the large, medium and small triangles $\triangle ABF,\ \triangle KLF,\ \triangle DCF$. The ratio of areas of the quadrilaterals $ABLK,\ ABCD$ is equal to

$\frac{|ABLK|}{|ABCD|} = \frac{\triangle_L - \triangle_M}{\triangle_L - \triangle_S} = \frac{1 - \frac{\triangle_M}{\triangle_L}}{1 - \frac{ \triangle_S}{\triangle_L}}$

The triangle areas and their ratios are

$\triangle_L = \frac 1 2 FA \cdot FB\ \sin \phi,\ \ \ \triangle_M = \frac 1 2 FK \cdot FL\ \sin \phi,\ \ \ \triangle_S = \frac 1 2 FD \cdot FC\ \sin \phi$

$\frac{\triangle_M}{\triangle_L} = \frac{FK \cdot FL}{FA \cdot FB},\ \ \ \frac{\triangle_S}{\triangle_L} = \frac{FD \cdot FC}{FA \cdot FB}$

All we have to calculate are the ratios $\frac{FK}{FA},\ \frac{FD}{FA},\ \frac{FL}{FB},\ \frac{FC}{FB}$. Using Menelaus' theorem for the triangles $\triangle AED,\ \triangle BEC$ cut by the same line $KL$, we get

$\frac{PA}{PE} \cdot \frac{QE}{QD} \cdot \frac{KD}{KA} = 1,\ \ \ \frac{PC}{PE} \cdot \frac{LB}{LC} \cdot \frac{QE}{QB} = 1$

Dividing these 2 equations leads to

$\frac{PA}{PE} \cdot \frac{QE}{QD} \cdot \frac{KD}{KA} \cdot \frac{PE}{PC} \cdot \frac{LC}{LB} \cdot \frac{QB}{QE} = \frac{PA}{PC} \cdot \frac{QB}{QD} \cdot \frac{KD}{KA} \cdot \frac{LC}{LB} = \frac{KD}{KA} \cdot \frac{LC}{LB} = 1$

where we used the fact that $P, Q$ are the midpoints of the diagonals $AC, BD$, i.e., $\frac{PA}{PC} = \frac{QB}{QD} = 1$. Thus we obtained $\frac{LB}{LC} = \frac{KD}{KA} = \frac 1 3$. Using Menelaus' theorem for the triangles $\triangle KLF, \triangle DBF$ cut by the same line $AC$, we get

$\frac{AK}{AF} \cdot \frac{CF}{CL} \cdot \frac{PL}{PK} = 1,\ \ \ \frac{AD}{AF} \cdot \frac{CF}{CB} \cdot \frac{EB}{ED} = \frac{AD}{AF} \cdot \frac{CF}{CB} \cdot \frac 1 2  = 1$

where we used the fact that the point $E$ divides the diagonal $BD$ in the ratio $\frac{ED}{EB} = \frac 1 2$. Dividing these 2 equations leads to

$\frac{AK}{AF} \cdot \frac{CF}{CL} \cdot \frac{PL}{PK} \cdot \frac{AF}{AD} \cdot \frac{CB}{CF} \cdot \frac 1 2 = \frac{AK}{AD} \cdot \frac{CB}{CL} \cdot \frac{PL}{PK} \cdot \frac 1 2 = \frac{PL}{PK} \cdot \frac 1 2 = 1$

where we again used the fact that $ \frac{KD}{KA} \cdot \frac{LC}{LB} = 1$ obtained previously. Thus we deduced $\frac{PL}{PK} = 2$. Let a line parallel to $KL$ through the point $C$ meet the line $AD \equiv AF$ at a point $M$. From the similar triangle $\triangle MCF \sim \triangle KLF$, we get $\frac{MF}{MK} = \frac{CF}{CL}$. Since $P$ is the midpoint of the diagonal $AC$ and $MC \parallel KL \equiv KP$, $KP$ is a midline of the triangle $\triangle ACM$, hence, $AK = MK$. Consequently,

$\frac{AF}{AK} = \frac{AK + MK + MF}{MK} = \frac{2MK + MF}{MK} = 2 + \frac{MF}{MK} = 2 + \frac{CF}{CL}$

Substituting these results into the equation $\frac{AK}{AF} \cdot \frac{CF}{CL} \cdot \frac{PL}{PK} = 1$ obtained previously,

$\frac{AF}{AK} \cdot \frac{CL}{CF} \cdot \frac{PK}{PL} = \frac{AF}{AK} \cdot \frac{1}{\frac{AF}{AK} - 2} \cdot \frac 1 2 = 1$

$\frac{AF}{AK} = 2 \left( \frac{AF}{AK} - 2 \right),\ \ \ \frac{AF}{AK} = 4$

Denote $u = KD$ ("u" for a unit). Since $AK = 3 KD = 3u$, we immediately obtain

$AD = AK + KD = 3u + u = 4u,\ \ FA = 4 AK = 12u$

$FD = FA - AD = 12u - 4u = 8u$

$FK = FD + KD = 8u + u = 9u$

and the desired ratios are

$\frac{FK}{FA} = \frac{9u}{12u} = \frac 3 4,\ \ \frac{FD}{FA} = \frac{8u}{12u} = \frac 2 3$

Combining the results $\frac{LB}{LC} = \frac{KD}{KA} = \frac 1 3$ and $\frac{CF}{CL} = \frac{AF}{AK} - 2 = 4 - 2 = 2$, we get the remaining 2 desired ratios. Denote $v = BL$.

$CL = 3 BL = 3v,\ \ BC = BL + CL = v + 3v = 4v,\ \ FC = 2 CL = 6v$

$FL = FC + CL = 6v + 3v = 9v,\ \ FB = FL + BL = 9v + v = 10v$

$\frac{FL}{FB} = \frac{9v}{10v} = \frac{9}{10},\ \ \ \frac{FC}{FB} = \frac{6v}{10v} = \frac 3 5$

As a result, the desired area ratio is

$\frac{|ABLK|}{|ABCD|} = \frac{1 - \frac{\triangle_M}{\triangle_L}}{1 - \frac{ \triangle_S}{\triangle_L}} = \frac{1 - \frac{FK}{FA} \cdot \frac{FL}{FB}}{1 - \frac{FD}{FA} \cdot \frac{FC}{FB}} = \frac{1 - \frac 3 4 \cdot \frac{9}{10}}{1 - \frac 2 3 \cdot \frac 3 5} = \frac{5(40 - 27)}{40(5 - 2)} = \frac{13}{24}$

This problem does not seem to fit in the Pre-Olympiad section.

Yetti


how do i make bittorrent faster?  ive looked everywhere but it usually only goes at around 20 kb/s

Port forwarding helps, as does optimizing the speed settings. I dunno, I use utorrent, but the basic concepts should apply.

for some reason utorrent lags my internet a LOT but bittorrent doesnt.

Really? It's the exact opposite for me. I haven't used BitTorrent in years, though, so I wouldn't be sure on how to get it to work for your computer.

Google always helps, though.

BitTorrent? Do you mean the [i]client[/i] BitTorrent? Dude that like totally sucks.
Use Bit Comet, my past favourite, or Bit Lord, my current client.
My bandwidth is 256 kbps only, but still I got 30 KB/s.

bitlord and bitcomet are more or less identical clients.  neither of them are particularly good.  the most robust client is azureus, but it is a huge resource hog and utorrent, which has a [i]tiny[/i] footprint and an excellent list of features, is by far the best option for the pc.  there's a good chance the op has it configured wrong.

if you are using a mac instead, the closest analogue to utorrent is probably transmission, but it is banned from a bunch of trackers and doesn't have some of the options that utorrent/azureus.  i use azureus on my mac.

The only way for Bittorrents to go faster s you have to have a lot of seeds and swarms.

It could have to do with download/upload limits to.

wth, it like stays at 20 kb/s at half the day then jumps to 100-200kb/s

It's most likely because you're getting more seeds, but maybe there's something screwy with your ISP.

If its like that you should be happy, right now my bittorrent is 1.2 bytes. :(

Try this [url=http://en.wikipedia.org/wiki/Port_forwarding]Wikipedia[/url]

If that does not make sense try
[url=http://portforward.com/routers.htm]Other[/url]

Port forwarding only applies mostly if you have a router, do you have a router?


Find the equation of the plane that passes through (1, 1, -5/4), (2, 1, -2), and (2, 3, -3).

I know the answer already because I made the problem up.  Try using the fact that the general form of the plane is Ax + By + Cz + D = 0 and solve for A, B, C, and D.  Good luck.
Also try using vectors.

[hide="solution"]

Express the equation as $ z\equal{}\frac{\minus{}A}{C}x \plus{} \frac{\minus{}B}{C}y \plus{} \frac{\minus{}D}{C}$
Let a=coefficient of x, b=coefficient of y and c= constant.

Then, we have the three equations:
a+b+c= -5/4
2a+b+c= -2
2a+3b+c= -3

Solving for a,b,c, we obtain the equation z= -6.25x - 7.5y + -11.5. Thus, we have 25x+30y+4z+46=0.

[/hide]

The general technique for finding the equation of a plane given three points $ A, B, C$ on the plane is to compute two independent vectors $ U, V$ (say, $ U \equal{} A \minus{} B, V \equal{} B \minus{} C$) in the plane; then $ N \equal{} U \times V$ is a normal vector, and the equation of the plane is, say, $ N \cdot (X \minus{} A) \equal{} 0$.

Suppose the required equation of the plane $ \pi$ is $ Ax\plus{}By\plus{}Cz\plus{}D\equal{}0$ and for any $ k\in\overline {1,3}$ we have  $ P_k(x_k,y_k,z_k)\in \pi$ ,  i.e.

$ A\cdot x_k\plus{}B\cdot y_k\plus{}C\cdot z_k\plus{}D\cdot 1\equal{}0$  . Then $ \frac {A}{\left|\begin{array}{ccc}
y_1 & z_1 & 1\\\
y_2 & z_2 & 1\\\
y_3 & z_3 & 1\end{array}\right|}\equal{}$ $ \frac {B}{\minus{}\left|\begin{array}{ccc}
z_1 & 1 & x_1\\\
z_2 & 1 & x_2\\\
z_3 & 1 & x_3\end{array}\right|}\equal{}$ $ \frac {C}{\left|\begin{array}{ccc}
1 & x_1 & y_1\\\
1 & x_2& y_2\\\
1 & x_3 & y_3\end{array}\right|}\equal{}$ $ \frac {D}{\minus{}\left|\begin{array}{ccc}
x_1 & y_1 & z_1\\\
x_2& y_2 & z_2\\\
x_3 & y_3 & z_3\end{array}\right|}\equal{}\lambda\ne 0$

(with the convention : $ \blacktriangleright$ [b]if the denominator is equally with zero, then and the numerator is equally with zero[/b] $ \blacktriangleleft$ ).

Concisely, the equation of the plane which is defined by the points $ P_k$ , $ k\in\overline {1,3}$ is : $ \left|\begin{array}{cccc}
x & y & z & 1\\\
x_1 & y_1 & z_1 & 1\\\
x_2 & y_2 & z_2 & 1\\\
x_3 & y_3 & z_3 & 1\end{array}\right|\equal{}0$ .

In the our particular case, the plane which is defined by the points $ \left\|\begin{array}{cccc}
P_1\ \left(\ 1\ ,\ 1\ ,\ \minus{}\frac 54\ \right)\\\\
P_2\ \left(\ 2\ ,\ 1\ ,\ \minus{}2\ \right)\\\\
P_3\ \left(\ 2\ ,\ 3\ ,\ \minus{}3\ \right)\end{array}\right\|$ has the equation  $ \boxed {\ 3x\plus{}2y\plus{}4z\equal{}0\ }$ .


I'm new to MATHCOUNTS (my schools doesn't have it, stupid school), so I really don't know how to register for it.  I would really appreciate any help, since this is a contest I would really like to do (more of my level).

Thanks.

http://www.mathcounts.org/Program/FAQ.html
there, it says you have to participate with your school. You may want to do as it says and next year, let your principle take a look at starting a mathcounts program. Even better, get some kids and go to him.

If your school doesn't have a mathcounts program then you should figure out who your local mathcounts organizer is. I think that their names and contacts are posted on the state mathcounts websites. For example I live in new jersey so my state website would be http://www.mathcounts-nj.org. You could also try to assemble a  school team but don't bother enetering if you can't get four people. You can compete with three but you will be penalized.

Yif man. You can compete without a school. I did. They changed the rule.

How did you compete without a school tandjsnell?

That's correct h_s_potter2002. Kinda had to. I'm homeschooled. I tried to form a team but there weren't that many homeschoolers interested.

Well, than that probably won't work since I'm not homeschooled.

I'll just e-mail my Social Studies teacher to ask about forming a team.  Thanks.

It still will work, as long as your school dosen't have a team. I didn't enter as home schooled. I made up a school name. Having home school next to your
name yields to many questions. But still I think that your best bet would be to try to get some other people (your friends for example) and form a school team. It doesn't have to be started by a school official. You will just have to pay the fees yourself and not expect a school coach. But yiff man is right. You should probaly get your principle's okay then enter with your school name.If not this is a free country so you could simply to what I did. Make up a school name and make it sound like a private school.

[quote=mathleticguyyy][quote=AwesomeKidMe]U may practice here ... It's a great website:
https://artofproblemsolving.com/community/c545817h1531516_free_talk[/quote]

Hey he posted in 2004[/quote]

Why bump though...

[quote=jlordhe]@above,
[quote=tauros]

Revive! just want to discuss mathcounts here.[/quote][/quote]

Yeah but that was still about a year ago we have tons of MATHCOUNTS discussion threads :D

I don't even know, probably the user looked up how to participate for MATHCOUNTS or something and scrolled all the way down in the search results and found this.

But we shouldn't continue this thread unless someone has a genuine question relating to the topic...
Oh sorry I didn't see the question...

I am actually not sure. I go to a school that does MATHCOUNTS, so I participate there.

[quote=mathgeek03]I'm homeschooled and think i may be able to get enough interest to have a team but can i even participate w/ a homeschool team?[/quote]

lol the first post is 5 months older than me

Yes you can participate in a homeschool team. Whovever is going to be your coach has to fill out [url = https://www.mathcounts.org/sites/default/files/u1706/1718_Homeschool_Virtual_School_Participation.pdf]this[/url] though

[quote=jlordhe]Oh, how did the "bumper" find this thread in the first place?[/quote]

maybe the person was searching up some key words

or maybe the person has the OP as a friend and was looking through their posts

Possibly. You should talk to the school math coach at the school where you take your advanced math courses.

[quote=RYang2]I take advanced math classes, and the school that I'm enrolled in doesn't do Mathcounts  :diablo: , but the school that that I have advanced math classes does do Mathcounts. Will I still be able to do Mathcounts?[/quote]

It depends on what grade you're in. If you're 6th-8th then yes if the coach lets you but otherwise probably not.

I'm older than the Original Post, but anything we post now won't help the Original Poster anymore since the last time he got on was in 2016

Also the OP is most definetly not in Middle School anymore

[quote=MathMan5000][quote=Mathlete2017][quote=Barry.B.Benson]Homeschools usually don have a team, sorry.[/quote]
An extremely famous exception is the Robitaille Homeschool (which won Mathcounts nationals twice).[/quote]

robitaille homeschool is just Luke

it doesnt actually have a team of 4 that was repped in chapter and state (at least im psure, ac_math or any1 from Texas, pls confirm)[/quote]

It still has a team (1 person :P).

Luke is an individual,  not a part of a team

Tbh Luke can probably get 1st place team by himself.

This would only be for chapter and state though, the Robitaille homeschool did not win the national team round twice. ;)

[quote=RYang2]I take advanced math classes, and the school that I'm enrolled in doesn't do Mathcounts  :diablo: , but the school that that I have advanced math classes does do Mathcounts. Will I still be able to do Mathcounts?[/quote]

I am pretty sure that the MATHCOUNTS team must be from the school in which you mainly go to for your classes. For instance, if you go to your advanced school for only a couple classes, then the other school you go to must the school that administers Mathcounts. Sorry to burst your bubble.

My school didn't have a mathcounts team either, but I convinced them to start one. See if you can do that.

[quote=a28546]Hah the OP must be an adult now[/quote]
Wait just realized original post was from 2004 :wacko:.

Oops never realized that, why is this thread still being revived, it only has 40 posts!!

Notice: there was a 13 year bump at post 10 because he/she just wanted to 'discuss mathcoutnts.'

[quote=tauros][quote=AGeneson]You just need a "sponsor" (this term may be loosely used in your case due to stupid school circumstances). Some of the websites will have information, also mathcounts can be started at your school if you talk a few teachers and a few parents into doing it (its hard but worth it). Good Luck![/quote]

Revive! just want to discuss mathcounts here.[/quote]
Like how he also quoted someone...


Hello, this is a homework problem I have been having difficulty with. I don't know if it is the arithmetic or I got the concept right.

1) Find the matrix representations of the linear transformations below with respect to the given basis.

$ T: \mathbb{C}^2 \rightarrow \mathbb{C}^2 , T(x,y) \equal{} (x \minus{} iy , 3x \plus{} 2iy)$

$ ii) [T]_{\gamma}, \gamma \equal{} { (i, 1) , (1, i) }$

Edit: so far the progress I made is the first column in the matrix is either $ (0, 5i)$ (column)  or $ {(\frac{5i}{2} , \frac{5}{2})}$

Mhmk my final answer is

$ T = ({ \begin{array}{cc}
\frac{5}{2}i & \frac{1}{2}-i \\
\frac{5}{2} & 1-\frac{i}{2}\\
\end{array} })$


linear algebra
matrix
LaTeX
linear algebra theorems

If $ A\in{M_n{\mathbb{(C)}}}$ and all $ n$ eingenvalues of $ A$ are $ 0$. Then $ A \equal{} O_n$?

Nope :D.We can only say that $ A^n = O_n$. An example of a matrix that has all its eigenvalues $ 0$ take this one :
\[ A = \left(\begin{array}{c}0\ 1\ 0\ ...\ 0\ 0 \
0\ 0\ 1\ ...\ 0 \ 0 \
....\ ...\ ....\ ..... \
0\ 0\ 0\ ...\ 0\ 1 \
0\ 0\ 0\ ...\ 0\ 0 \
\end{array} \right)
\]

Some funny $ \text{\LaTeX}$ there. Svejk meant to say
\[ A=\begin{pmatrix}0&1&0&\cdots&0&0\\ 0&0&1&\cdots&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&0&1\\ 0&0&0&\cdots&0&0\end{pmatrix}\]


What exactly is a function symbol? :huuh:  :?:

f(x) = x^2 + 2x + 1

$f(x) = x^{2}+2x+1$

Just a regular letter. :)

There is no "function symbol." It is a letter used to define the function. I'm not that good at explaining. You should ask t0rajir0u or some other smart guy.

Yes. But it's not a symbol. It's just f(x). I told you I'm not good at explaining. :ninja:

I figured that out after looking at my algebra book from last year... Thank you!

fuction symbol? its usually f(x) as in f of x


function
induction
algebra proposed
algebra

Hello

I have two questions :

[quote="nayel"]Let $f$ be a bijective function ...[/quote]
$f(x)$ is defined on what set D?. Could we chooe any set .

[quote="nayel"] $1.$ $f^{-1}$ is a function defined on $\mathbb N$.[/quote]
Do you mean $f(D)=\mathbb{N}$ or $\mathbb{N}\subset f(D)$ ?

--
Patrick

[quote="nayel"]1. $f$ takes values from $\mathbb R^{+}$

2. $f^{-1}$ takes values from $\mathbb N$ and maps them in $\mathbb N$.

sorry for my poor english.[/quote]

Since f is bijective, you mean $f$ and $f^{-1}$ are from $\mathbb{N}\rightarrow\mathbb{N}$ ?

--
Patrick

[quote="nayel"]by bijective i meant one-to-one. don't they mean the same?[/quote]

I'm sorry but I don't really understand.
If $f$ is a bijection $\mathbb{N}\rightarrow\mathbb{N}$ and $f^{-1}$ too, then, by definition, it exists $k\in\mathbb{N}$ such that $f(k)=1$ ...

I miss something....

--
Patrick

let me rephrase the problem:

Let $f$ be a one-to-one function that takes values from $\mathbb R^{+}$ and satisfies

$1.$ $f^{-1}$ is a function $\mathbb N\to \mathbb N$.

$2.$ $f(x)+f(y)=f(xy)$

Prove that $\exists k\in \mathbb N$ such that $f(k)=1$.

[quote="nayel"]if $f$ is one-to-one, $f: \mathbb R^{+}\to \mathbb R$ and $f(x)+f(y)=f(xy)$ then can you prove that $\exists k$ such that $f(k)=1$?[/quote]

[quote="in the other post, olorin"]...In fact it even exists $f: \mathbb R^{+}\to \mathbb R$ injective with $f(x)+f(y)=f(xy)$ for all $x,y\in \mathbb R^{+}$ and $f(k)\not =1$ for all $k\in \mathbb R^{+}$...[/quote]

So no.

--
Patrick

[quote="nayel"]Let $f$ be a bijective function that satisfies

$1.$ $f^{-1}$ is a function defined on $\mathbb N$.

$2.$ $f(x)+f(y)=f(xy)$

Prove that $\exists k\in \mathbb N$ such that $f(k)=1$.

compare this problem with [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=142443]this one[/url]. i'm not very good with functional equations so i'm not sure if the problem is true. but i have a solution and i'll post it later.[/quote]
If $f: N\to N$ satisfyed 2. $f(x)+f(y)=f(xy)$, then $f$ is not injective.
By induction $f(x^{n})=nf(x)$, therefore if $n=\prod_{i}p_{i}^{k_{i}}$, then $f(n)=\sum_{i}k_{i}f(p_{i})$.
Therefore $f(p^{f(q)})=f(q^{f(p)})$, but $p^{f(q)}\not =q^{f(p)}.$
Obviosly $f(1)=0$, if $0\in N$, then $f(x)+f(0)=f(x*0)\to f(x)\equiv 0.$


geometry
circumcircle
concurrency
IMO
imo 1995

I note the intersection $R\in AM\cap DN$. We will analize the all radical axisis between  the circumcircles of the (inscribed) quadrilaterals $AMPZ,\ DNPZ$ and the circumcircles with the diameters $AC,\ BD$ a.s.o.

[u]I note:[/u] $C(XY)$ - the circle with the diameter $XY;$
$\ R\in AM\cap DN,\ R_1\in AM\cap XY,\ R_2\in DN\cap XY$;
the circles $a=C(AC),\ b=C(BD),\ m=C(AP),\ n=C(DB)$;
the radical axises $d_1,\ d_2$ for the pairs of the circles $(a,n),\ (b,m)$ respectively.
--------------------------------------------------------------------------------------------------
The radical center of the circles $(a,b,m)$ is $R_1\in XY\cap AM\cap d_2\Longrightarrow R_1\in d_2$;
The radical center of the circles $(a,b,n)$ is $R_2\in XY\cap DN\cap d_1\Longrightarrow R_2\in d_1$;
The radical center of the circles $(a,m,n)$ is $R_1\in AM\cap XY\cap d_1\Longrightarrow R_1\in d_1$;
The radical center of the circles $(b,m,n)$ is $R_2\in DN\cap XY\cap d_2\Longrightarrow R_2\in d_2.$
---------------------------------------------------------------------------------------------------
Therefore, $R_1\equiv R_2\in d_1\cap d_2\cap XY\cap AM\cap DN\Longrightarrow R_1\equiv R_2\equiv R\in XY.$

we know that ${XY}\perp{AD}$ at $Z$.

let ${AM}\cap{XY} = R$.

we have $\angle{RMC} = 90$ (AC is diameter)

so, $RMCZ$ is cyclic, that implies $PR.PZ = PM.PC$.

let ${ND}\cap{XY} = S$.

analogous, we have $PS.PZ = PN.PB$.

$XY$ is the radical axes of the two circles, so, $PM.PC = PN.PB \Longrightarrow PR = PS \Longrightarrow R = S$.

I view the four circles have the same eje radical  :D

Suppose $E = AM \cap XY$ and $N' = ED \cap XY.$  We have $\angle AMC = 90^\circ.$ Let $\angle MAC = \alpha, \angle EDZ = \beta, \angle AEZ = 180^\circ - \alpha = \theta_1,$ and $\angle DEZ = 180^\circ - \beta = \theta_2.$

Now if we show that $\angle PBZ = \theta_2,$ we show that $N' = N$ and we are done. Since $Z$ lies on the radical axis $XY$ of the two given circles, $(AZ)(CZ) = (BZ)(ZD).$ It can easily be seen that $\triangle EAZ \sim \triangle CPZ$ from which we get $(EZ)/(CZ) = (AZ)/(ZP).$ Thus, $(EZ)(ZP) = (AZ)(CZ) = (BZ)(ZD),$ so $(EZ)/(BZ) = (ZD)/(ZP),$ so $\triangle BPZ \sim \triangle EDZ,$ and we are done.

Let $R = AM \cap DN$. $P$ is a point on radical axis of the circlec with diameter $AC$ and $BC$.
So we have $PC\cdot PM = BP \cdot PN $ which implies $B,C,N,M$ are concyclic.
That is $\angle NMC = \alpha = \angle NBD \Rightarrow \angle BDN = 90^{\circ}-\alpha$ and $\angle AMN=90^{\circ}+\alpha$ which also implies $A,M,N,D$ are concyclic.
Thus $RM\cdot RA = RN \cdot RD$ which means $R$ is a point on the radical axis of those circles. So $R,X,P,Y$ are concurrent.

[b]Solution using Menelaus' theorem: [/b]

Let $AM \cap  XY = T$. Equivalently we have to prove that $T,N,D$ are collinear. By Menelaus' theorem  $T,N,D$ will be collinear if and only if  $\frac{ZT}{PT}\cdot \frac{PN}{BN} \cdot \frac{BD}{DZ} = -1$

Observe  that $ \triangle AMC \sim  \triangle PZC, \implies \frac{CA}{CM}= \frac{PC}{ZC} $ and  $\triangle BND \sim \triangle PZB \implies \frac{PB}{ZB}= \frac{BD}{BN} $ . By intersecting chord theorem $ PB \cdot PN = PM\cdot PC $ and $  ZB \cdot ZD = ZA\cdot ZC $ .

 Note that    $\frac{PM}{ZA}\cdot \frac{CA}{CM} = \frac{PM\cdot PC}{ZC \cdot ZA} = \frac{PM\cdot PC}{ZB\cdot ZA} = \frac{PB\cdot PN}{ZB \cdot ZD}= \frac{BD\cdot PN}{ZD \cdot BN }$.

Since $T,M,A $ are collinear $\frac{ZT}{PT}\cdot \frac{PM}{CM} \cdot \frac{CA}{AZ} = \frac{ZT}{PT}\cdot \frac{PN}{BN} \cdot \frac{BD}{DZ} =-1$. $Q.E.D$

Let first establish that:

(1) $\frac{ZA}{ZB}=\frac{ZD}{ZC}$.

From the right triangle $AXC$ we get $ZA.ZC=ZX^2$. Similarly, from the right triangle $BXD$ it follows: $ZB.ZD = ZX^2$. Hence $ZA.ZC=ZB.ZD$, which implies (1).

Now the homothety with center $Z$ and ratio $k=\frac{ZA}{ZB}$ will send the altitude through $B$ to $PC$, to the line $AM$; the altitude through $C$ to $BP$ will go to the line $DN$, and the line $ZP$(which is the altitude through $P$ to $BC$) will be sent to itself.

This means that $AM, DN, ZP$ are concurent and its common point is the image of the orthocenter of $\Delta BCP$ by this homothety.

Let $AD \cap XY=L$
Notice that $ALOM$ and$NDLO$ are cyclic .So $\angle OMY=\angle MOY$ and $\angle YON=\angle YNO$.
So O is the circumcentre of $\triangle OMN$.and thus by angle chasing we get $AMND$ cyclic.THus the lines$AM$ and$DN$ are meeting on the radical axis.

This problem is easy but tasty....

[u][b]Notations[/b][/u]:$AM \cap DN=K,\angle{BNM}=\theta,\odot{AMC}=\Omega_1,\odot{BND}=\Omega_2$.

[u][b]Proof[/b][/u]Since $P$ lies on the radical axis $XY$ of $\Omega_1$ and $\Omega_2$,we have $BP*PN=CP*PM \Rightarrow B,C,N,M$ concyclic.So $\angle{MCB}=\angle{MNB}=\theta$.Hence $\angle{MAD}=90^{\circ}-\theta \Rightarrow M,A,D,N$ concyclic.Hence $KM*KA=KN*KD \Rightarrow K$ lies on the radical axis $XY$ as desired.

Let $Q=AM\cap XY$ and let $QD$ intersect the circle with diameter $BD$ at $D, N'$.  We wish to show $N'\equiv N$.
Note that
\[\angle AMC =\angle QMP =\angle BND =\angle BN'D =\angle AZQ =\angle QN'P=\frac{\pi}{2}.\]
Then $\angle QMP +\angle QN'P = \pi $, so $QMPN'$ is cyclic.  Now $\angle MPN' =\angle BPC$ and
\[\angle MN'P=\angle MQP =\frac{\pi}{2}-\angle QAC =\angle BCP,\]
so $\triangle MN'P\sim \triangle BPC$.  Hence,
\[\frac{PM}{PB}=\frac{PN'}{PC}\implies PM\cdot PC=PB\cdot PN'.\]
But since $P$ is on the radical axis,
\[PM\cdot PC=PB\cdot PN\implies PB\cdot PN=PB\cdot PN'\implies PN=PN'.\]
Hence $N\equiv N'$ and $AM,XY,DN$ concur as desired. $\blacksquare$

A very nice proof is also given here:
https://www.math.ust.hk/excalibur/v5_n3.pdf

This is from the book! There's no radical axes or bashing.
[hide="Thought Process"]
I like working with angles so let's get rid of the circle condition. The fact that $X$ lies on the circle with diameter $AC$ and the circle with diameter $BD$ means that $\angle AXC=\angle BXD=90^\circ$. The problem statement only uses line $XY$ to define $P,Z$, so we only have to determine some properties of line $XY$ instead of define both $X$ and $Y$. Since $P$ is arbitrary, we will use $X$ and $Z$. If we draw an accurate diagram we can tell that $XZ$ is perpendicular to $BC$. That will be a sufficient definition. We will start $XBC$ and define points $A$ and $D$ such that $\angle AXC=\angle BXD=90^\circ$. Then let $Z$ be the foot of the altitude from $X$ to $BC$. I can't think of any immediate progress so let's work backwards. We want to show that $AM,XY=XZ,DN$ are concurrent. Since $AM$ seems symmetric to $DN$, it feels like we should let $AM$ intersect $XZ$ at $Q$ and $DN$ intersect $XZ$ at $R$. If we show $Q=R$, we would be done. Similar things seem to work best by comparing lengths, and lengths are easiest with right angles. There are a lot of right angles at $Z$, so we will try to show that $QZ=ZR$. Comparing lengths usually involves similar triangles. Recall that $A,B,C,D,X,Z$ are defined at the beginning. Then put in $P$. Let us define $M$. Since it varies directly with $P$, and it lies on the circle with diameter $AC$, we know that $\angle AMC=90^\circ$. But we have not used $P$ yet. Notice that $\angle PZC=90^\circ$, and that $\angle ACM=\angle PCZ$, so $\triangle CZP\sim\triangle CMA$. Now let's find a similarity between $\triangle CMA$ and something that involves $Q$. Angles tell us that $\triangle AMC\sim\triangle AZQ$, and therefore $\triangle CZP\sim\triangle QZA$. We could have noticed this by trying to define $Q$ with spiral similarity, which is a common synthetic technique, but we will proceed anyway. We want to use similar triangles to compare lengths. We want something that defines $Q$ using only $QZ$, so we will try to find $\frac{QZ}{AZ}$. The similarity tells us that $\frac{CZ}{PZ}=\frac{QZ}{AZ}$. Similarly, $\frac{BZ}{PZ}=\frac{RZ}{DZ}$. Then, to get $\frac{QZ}{RZ}$, we divide the first equation by the second to get $\frac{CZ}{BZ}=\frac{QZ}{RZ}\cdot\frac{DZ}{AZ}$. What will suffice to show that $\frac{QZ}{RZ}=1$? We will show that $\frac{CZ}{BZ}=\frac{DZ}{AZ}$, which is equivalent to showing that $AZ\cdot CZ=BZ\cdot DZ$. Since we defined $X$ before $Z$, maybe expressing the products in terms of $XZ$ will work. Sure enough, it does, and we proceed to write the formal solution forwards instead of backwards.
[/hide]
[hide="Solution"]
Define $Q,R$ to be the intersections of $XY$ with $AM$ and $DN$, respectively. It suffices to show that $Q=R$, which is equivalent to proving $QZ=RZ$.

By equal radii, $BX=BY$ and $CX=CY$. Furthermore, $BC=BC$, so by SSS $\triangle BXC\cong\triangle BYC$. Then $\angle ZBX=\angle CBX=\angle CBY=\angle ZBY$. Recall that $BX=BY$ and $BZ=BZ$, so by SAS $\triangle BZX\cong\triangle BZY$. Then $\angle BZX=\angle BZY$, and since $Z$ lies on $XY$, $180^\circ=\angle BZX+\angle BZY=2\angle BZX$, so $\angle BZX=90^\circ$, and similarly $\angle BZX=\angle BZY=\angle CZX=\angle CZY=90^\circ$.

Since $X$ lies on the circle with diameter $AC$, $\angle AXZ+\angle CXZ=\angle AXC=90^\circ$, so $\angle XCZ=90-\angle CXZ=\angle AXZ$. Furthermore, $\angle AZX=\angle BZX=\angle CZX$, so by AA, $\triangle AXZ\sim\triangle XCZ$. This means $\frac{AZ}{XZ}=\frac{XZ}{CZ}$. Analogously, $\frac{DZ}{XZ}=\frac{XZ}{BZ}$. Dividing the previous equation by the one before that gives us $\frac{DZ}{AZ}=\frac{CZ}{BZ}$.

Since $M$ lies on the circle with diameter $AC$, so $\angle AMC=90^\circ$, and $\angle AZQ=\angle BZY=90^\circ$, which means $\angle PCZ=\angle ACM=90-\angle CAM=90-\angle QAZ=\angle AQZ$. Also, $\angle AZQ=90^\circ=\angle CZY=\angle CZP$, so by AA, $\triangle AQZ\sim\triangle PCZ$. Then $\frac{CZ}{PZ}=\frac{QZ}{AZ}$. Similarly, $\frac{BZ}{PZ}=\frac{RZ}{DZ}$. Dividing the previous equation by the one before that gives us $\frac{BZ}{CZ}=\frac{AZ}{DZ}\cdot\frac{RZ}{QZ}$.

Recall that $\frac{CZ}{BZ}=\frac{DZ}{AZ}$. We multiply this by the last equation of the previous paragraph to get $1=\frac{RZ}{QZ}$, which means $QZ=RZ$, as desired.
[/hide]
The second paragraph of the solution is well-known.

Let $G=AM\cap DN$. We have $\angle AMC=\angle CMG=\angle PMG=90^\circ$, since $\overline{AC}$ by Thale's theorem. Similarly $\angle GNP=90^\circ$, and $GNPM$ is cyclic. Since $P$ is on the radical axis of both of the circles described in the problem, the power of $P$ with respect to them is equal, meaning $PC\cdot PM=PB\cdot PN$, so $BMCN$ is cyclic. An easy angle angle chase gives
$$90^\circ-\angle CAM=\angle MCA=\angle MNB=90^\circ-\angle GNM$$
which results in $\angle DAM=\angle GNM$, meaning $AMND$ is cylic. Therefore, $DN$ is the radical axis of $(BND)$ and $AMND$ and $AM$ is the radical axis of $AMC$ and $AMND$. Since $AM$ and $DN$ intersect at $G$, $XY$ has to pass through $G$ as well.

We have $\angle MAD=\angle MAC=90^{\circ}-\angle ACM=90^{\circ}-\angle BCM=\angle BNM-90^{\circ}=\angle BNM-angle BND=\angle MND$. Thus, $ADMN$ is cyclic, so $AM$,$DN$, and $XY$ concur, as desired.

Note the right angles, implying the drawn circles. $\overline{XY}$ is a shared radical axis of all four circles. $BO$ and $CO$ are diameters, not sure how that would help though.

Random comment ig but you can prove that $OM \cdot OA = ON \cdot OD$ aka prove that $ANMD$ is cyclic.

Note that $MAZP$ and $DZPN$ are cyclic by the 180 sum condition. We know that the radical axis of $(MPZA)$ and $(AMC)$ is $\overline{AM}$. We know that the radical axis of $(DNPZ)$ and $(DNB)$ is $\overline{DN}$. We also know that the radical axis of $(MPZA)$ and $(DNPZ)$ is $\overline{PZ}$. Let $\overline{AM} \cap \overline{DN} = E$. From here we can see that $E$ has equal power to $(MPZA)$ and $(DNPZ)$ and this means that $E$ lies on $\overline{PZ}$ which implies that $E$ lies on $\overline{XY}$. This means the three lines are concurrent. $\blacksquare$

[hide= Is it OK?]We can reconstruct the problem in the following way.
Suppose that $K$ is a point (outside of the segment $XY$, but lies on the line) on $XY$ and,

$\bullet$ $AK$ intersects circle with diameter $AC$ on $M$

$\bullet$ $DK$ intersects circle with diameter $BD$ on $N$.

Now let $ MC \cap BN \equiv P$ and $KP \cap AD \equiv L$. It's enough to prove $KL \perp AD$.

Proof:
Lemma-1:  Since $K$ lies on the radical axis,$XY$, of the two circles, it is obvious that $AMND$ is cyclic  as the angles $\measuredangle NMK$ and $\measuredangle NDA$ holds equality.

Now notice,

$\measuredangle AMC =  \measuredangle BND = 90^\circ$

 $\implies KMPN cyclic \implies \measuredangle NMK= \measuredangle NPK =\measuredangle NDL$

$\implies NPLD cyclic \implies \measuredangle CLP= 90^\circ$
and we're done ![/hide]
[hide=A question:] Does the problem assure that the point $P$ lies on the [i]segment[/i] $XY$ ? If not, I thought of $3$ cases. One is proved above ( assuming $P$ inside), one is that, $P \equiv X$ or $Y$, where the proof is trivial.The last case should be putting $P$ outside the segment. Then we'll just swap $K$ with $P$ :)[/hide]

I guess I have a bit different solution using similarity, please check.
[hide = Storage]

So we begin with a claim.

Claim: ANMD is cyclic.

Proof: We clearly have that $\triangle$$PNM$ $\sim$ $\triangle$$PBC$ as P lies on the radical axes and we have vertically opposite angles, thus, it yields $\angle{PCB}=\angle{PNM}$, thus giving us, $\angle{DNM} = 90 + \angle{PCB}$, and since $\angle{AMC}=90$, we have that $\angle{MAC}=90-\angle{PCB}$, thus, we have $\angle{MAC}+\angle{DMN}=180$, which implies the concyclity of the claimed quadrilateral.

Now since we have $AMND$ is cyclic, construct the circumcircle of the mentioned quadrilateral, now we have that $AM$, $DN$ & $XY$ are the radical axes of these circles, due to existence of radical center, they will have to be concurrent. So, we're done. $\blacksquare$

[hide = Solution to IMO 1995/1][b][u]Proof:[/u][/b] Denote $G = \overline{AM} \cap \overline{DN}$. Denote by $\omega_1$ the circle with diameter $AC$ and $\omega_2$ as the circle with diameter $BD$. As $XY$ is the radical axis of $\omega_1$ and $\omega_2$, it suffices to show $Pow_{\omega_1}(G) = Pow_{\omega_3}(G)$ ($G$ lies on the radical axis $XY$). This is equivalent to showing $A, M, N, D$ is cyclic. This is easy, as if we notice the following cyclic quadrilaterals $PNDZ$ and $PNGM$, we have that $\angle NDB = \angle NPG = \angle NMG$, so that $A, M, N, D$ is cyclic, as desired. $\blacksquare$ [/hide]

Note that:
[b][color=red]Claim:[/color][/b] $MBCN$ is cyclic.
[i]Proof.[/i] From $PB \cdot PN = PX \cdot PY = PC \cdot PM$. $\blacksquare$

[b][color=red]Claim:[/color][/b] [Russia 1996/10.1] $AMND$ is cyclic.
[i]Proof.[/i] $\measuredangle DAM = \measuredangle CAM = 90^{\circ} - \measuredangle MCB = 90^{\circ} - \measuredangle MNB = 90 ^{\circ} + \measuredangle BNM = \measuredangle DNM$. $\blacksquare$

[asy]
pair A = dir(165); pair D = -conj(A); pair M = dir(100); pair N = dir(50);
pair C = extension(A, D, M, (M+((A-M)*dir(90)))); pair B = extension(A, D, N, (N+((D-N)*dir(90))));
filldraw(circumcircle(A, M, C), invisible, red); filldraw(circumcircle(B, N, D), invisible, red); draw(circumcircle(A, M, D), lightblue); draw(circumcircle(B, N, C), lightblue);
pair X = IP(circumcircle(A, M, C), circumcircle(B, N, D)); pair Y = OP(circumcircle(A, M, C), circumcircle(B, N, D));
pair T = extension(A, M, D, N); draw(T--Y, red); draw(A--T--D, grey); draw(A--D, red); draw(N--B, red); draw(M--C, red);
pair P = extension(B, N, C, M);
dot("$A$", A, dir(180)); dot("$D$", D, dir(0)); dot("$M$", M, dir(M)); dot("$N$", N, dir(N)); dot("$C$", C, dir(315)); dot("$B$", B, dir(225)); dot(X); dot(Y); dot("$T$", T, dir(T)); dot("$P$", P, dir(P));
/* -----------------------------------------------------------------+ |                 TSQX: by CJ Quines and Evan Chen                  | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ A 180 = dir 165 D 0 = -conj(A) M = dir 100 N = dir 50 C 315 = extension A D M (plus M (mult (minus A M) (dir 90))) B 225 = extension A D N (plus N (mult (minus D N) (dir 90))) circumcircle A M C / 0.1 yellow / red circumcircle B N D / 0.1 yellow / red circumcircle A M D / lightblue circumcircle B N C / lightblue X .= IP (circumcircle A M C) (circumcircle B N D) Y .= OP (circumcircle A M C) (circumcircle B N D) T = extension A M D N T--Y / red A--T--D / grey A--D / red N--B / red M--C / red P = extension B N C M */
[/asy]
Then the conclusion follows by radical axis on $(AC)$, $(BD)$, $(AMND)$.

[i]Solution 1:[/i]
[asy]
                size(12cm,0);
                pair A = (3,6);
                pair B = (0,0);
                pair C = (10,0);
                pair D = (4,0);
                pair E = (5.25,4.07);
                pair F = (1.67,3.35);
                pair X = (3.40,3.84);
                pair Y = (4.59,3.14);
                pair Z = (2.48,2.80);
                pair P = (-14.91,0);
                dot(A);
                dot(B);
                dot(C);
                dot(D);
                dot(E);
                dot(F);
                dot(X);
                dot(Y);
                dot(Z);
                dot(P);
                draw(A--B);
                draw(A--C);
                draw(B--C);
                draw(B--X);
                draw(C--X);
                draw(E--P);
                draw(Y--P);
                draw(C--P);
                draw(circle((3.74,2.31),2.31));
                draw(circle((3.74,1.69),1.69));
                draw(circle((3.24,4.80),2.14));
                label("A",A,N);
                label("B",B,SW);
                label("C",C,SE);
                label("D",D,S);
                label("E",E,NE);
                label("F",F,W);
                label("X",X,N);
                label("Y",Y,S);
                label("Z",Z,S);
                label("P",P,S);
                clip((-10,-1)--(-10,7)--(11,7)--(11,-1)--cycle);
[/asy]
        Denote the incircle of $\triangle ABC$ as $\omega_1$ and the incircle of $\triangle XBC$ as $\omega_2$. Note that the radical axis of $\omega_1$ and $\omega_2$ is $BC$. Assume $EFZY$ is cyclic and denote the circumcircle of $EFZY$ as $\omega_3$. If $EFZY$ is indeed cyclic, then the radical axis of $\omega_1$ and $\omega_3$ is $EF$ and the radical axis of $\omega_2$ and $\omega_3$ is $YZ$. Then, by the radical lemma, $EFZY$ is cyclic if and only if $BC,EF,$ and $XY$ are concurrent. \\ \\
        Extend $BC$ past $B$, $YZ$ past $Z$ and $EF$ past $F$ such that $YZ \cap BC=P_1$ and $EF \cap BC=P_2$. By Menelaus on $\triangle ABC$ and line $EF$, we have that
        $$\frac{AF}{BF} \cdot \frac{BP_2}{CP_2} \cdot \frac{CE}{AE}=1,$$
        and by Menelaus on $\triangle XBC$ and line $XY$, we have that
        $$\frac{XZ}{BZ} \cdot \frac{BP_1}{CP_1} \cdot \frac{CY}{XY}=1.$$
        Thus,
        $$\frac{AF}{BF} \cdot \frac{BP_2}{CP_2} \cdot \frac{CE}{AE}=\frac{XZ}{BZ} \cdot \frac{BP_1}{CP_1} \cdot \frac{CY}{XY}.$$
        Note that $AE=AF$, $XY=XZ$, $BF=BD=BZ$ and $CE=CD=CY$ by equal tangents. Thus, this simplifies to
        $$\frac{BP_2}{CP_2}=\frac{BP_1}{CP_1} \longrightarrow P_1=P_2. \text{ } \blacksquare$$
        (note the last equivalence must be true since $P_1$ and $P_2$ both lie on $BC$.) \\ \\
        [i]Solution 2:[/i]
        As in Solution 1, extend $BC$ past $B$, $EF$ past $F$ and $YZ$ past $Z$ such that $P_1=EF \cap BC$ and $P_2=YZ \cap BC$. $(P_1,B;D,C)$ is harmonic since $AD$,$BE$ and $CF$ are concurrent (their intersection being the Gergonne point of $\triangle ABC$). Similarly, $(P_2,B;D,C)$ is harmonic since $XD$,$BY$ and $CZ$ are concurrent (their intersection being the Gergonne point of $\triangle XBC$). Thus, $P_1=P_2$. Denote this common point as $P$. Note that
        $$\pi(P)=PD^2=PZ \cdot PY=PF \cdot PE,$$
        where $\pi(P)$ denotes the power of point $P$. Thus, $E,F,Z$ and $Y$ are concyclic. $\blacksquare$

[quote=orl]Let $ A,B,C,D$ be four distinct points on a line, in that order.  The circles with diameters $ AC$ and $ BD$ intersect at $ X$ and $ Y$. The line $ XY$ meets $ BC$ at $ Z$. Let $ P$ be a point on the line $ XY$ other than $ Z$. The line $ CP$ intersects the circle with diameter $ AC$ at $ C$ and $ M$, and the line $ BP$ intersects the circle with diameter $ BD$ at $ B$ and $ N$. Prove that the lines $ AM,DN,XY$ are concurrent.[/quote]

This was a good problem. It suffices to prove that ANMD is cyclic, where the motivation comes from concurrency using radical center being that one axis is XY, the other two need to contain AM and DN so we look at ANMD. Now denote the intersection of AM with DN as E, and note the right angles ENP and EMP. We have <AMN=<EMN=90-<NMP=90-<NBC (pop stuff)=<ADN (since BND is a right triangle), so ADMN is cyclic, as desired.

[b]Solution 1 (BMC).[/b] Let $E:=AM\cap XY$ and $F:=DM\cap XY$; we want to show $E=F$. Clearly $BCMN$ is cyclic because $PB\cdot PN=PX\cdot PY=PC\cdot PM$, and $BFNO,CMFO$ are cyclic by Thales'. Then $P$ is the radical center of $(BCMN),(BFNO),(CMFO)$ with $OP$ the radical axis of $(BFNO),(CMFO)$, meaning $(BFNO)\cap(CMFO)=\{E,F,O\}$ or $E=F$ as desired. $\square$

[b]Solution 2 (EGMO).[/b] As before $BCMN$ is cyclic, and $ADMN$ is cyclic since
\[\angle DAM+\angle DNM=\angle DAM+\angle BNM+90^\circ=\angle DAM+\angle BCM+90^\circ=180^\circ\]
by Thales' twice. So if $Q=AM\cap DN$, we have $QA\cdot QM=QD\cdot QN$ and so $Q\in XY$, as desired. $\square$

[hide=motivation]
Three lines are concurrent - one is the radical axis of $(AMC)$ and $(BND)$. Could $AM$ and $ND$ also be radical axis? Could $AMND$ be cyclic?
We know that $P$ is on the radical axis of $(AMC)$ and $(BND)$. Therefore, with equal power with respect to both circles, $PC*PM=PB*PN$. Hence, $BCNM$ is cyclic. We want $\angle{ADN}+\angle{AMN}=180^{\circ}$ for $AMND$ to be cyclic. Continue on with the solution.
[/hide]
[hide=solution]
Continuing on, we see that $\angle{CMN}=\angle{CBN}=90^{\circ}-\angle{ADN}$. Note that $\angle{AMC}=90^{\circ}$. Hence, $\angle{ADN}+\angle{AMN}=180^{\circ}$ is true. Therefore, $XY$ is the radical axis of $(AMC)$ and $(BND)$, $AM$ is the radical axis of $(AMC)$ and $(AMND)$, and $DN$ is the radical axis of $(BND)$ and $AMND$. The radical axes of any three circles are either concurrent or parallel, but we know that it is not the latter of the two. Hence, we are done. $\blacksquare{}$
[/hide]

Note that it suffices to prove that $AMND$ is cyclic. (By the Radical Center of Intersecting Circles theorem, or Theorem 2.9 from EGMO.)

By Power of a Point in the circle with diameter $\overline{AC}$, we have $$XP \cdot PY = CP \cdot PM,$$ and by Power of a Point in the circle with diameter $\overline{BD}$, we have $$XP \cdot PY = BP \cdot PN.$$

Hence, it follows that $$CP \cdot PM = BP \cdot PN,$$ so the converse of Power of a Point tells us that quadrilateral $BMNC$ is cyclic.

Now, we angle chase to get
\begin{align*}
\angle{MND} &= \angle{MNB} + \angle{BND} \\
&= \angle{MCB} + 90^\circ \\
&= (90^\circ-\angle{MAC}) + 90^\circ \\
&= 180^\circ - \angle{MAC},
\end{align*}
which implies that $AMND$ is cyclic, as desired. $\square$

Let $T = AM \cap DN$. Since $XY$ is the radical axis of $(AC)$ and $(BD)$, it suffices to show that $T$ also lies on the radical axis of $(AC)$ and $(BD)$.

[b]Claim:[/b] $AMND$ is cyclic.
[i]Proof.[/i] Since $P \in XY$, $P$ lies on the radical axis of $(AC)$ and $(BD)$. It follows that $\text{Pow}_{(AC)} P = \text{Pow}_{(BD)} P$, or $$PM \cdot PC = PN \cdot PB$$ Thus, $BMNC$ is cyclic. Further, since $AC$ is the diameter of $(AC)$ and $M$ lies on $(AC)$, we have $\angle AMC = 90^{\circ}$. By angle chasing, we have \begin{align*}
\angle MAD &= 90^{\circ} - \angle MCA \\
&= 90^{\circ} - \angle MNB \\
\end{align*}
However, we also have $\angle MND = 90^{\circ} + \angle MNB$, which leads to $\angle MND + \angle MAD = 180^{\circ}$, and the conclusion follows. $\blacksquare$

Then, by Power of a Point, we have $TM \cdot TA = TN \cdot TD$, which is equivalent to $T$ lying on the radical axis of $(AC)$ and $(BD)$, as desired. $\blacksquare$


MATHCOUNTS
AMC
AIME
geometry
AMC 8
AMC 10

Ok, so ive answered 7 of the admission qs and got 15th place tx state mathcounts, qualified for AIME, and have good recommendations, how many admission qs is a good number?

Don't worry, you'll get in. 7 admission questions done is fine. Test scores aren't the only thing they look at :)

I don't think that you need to worry. I would still suggest at least writing something down for each problem (even if you didn't solve it).

For anybody who made it, how many admission questions did you answer?

[quote="mathcrazed"]9. Just in case that message was too small =P. You're going to make it.[/quote]

Same here.

Yeah, I'm running out of time to do things, and I've only answered 8 of them really well, one of them....(*cough*5*cough*) seems way too simple, and geometry is killing me. But I'm fourth place in CO MathCounts (qualified second time), perfect score on AMC 8, qualified for AIME through AMC 10, think I have good recommendations, was careful on that dreadfully easy Math League and got another perfect. Do you think I'll get in?

sounds good 2 me.....
i dk tho i hav no personal experience

actually change my 1st post, ive answered 8 questions....
no proofs tho cuz i need work on that. thats 1 of the reasons im trying for this camp in the first place.

I don't think they reject anyone at all, actually, since last year they were overstaffed and could've taken more students.
Of course, this could've changed this year.


What is the arithmetic mean of all positive two-digit multiples
of 7?

There are $ \frac{98\minus{}14}{7}\plus{}1\equal{}13$ 2 digit multiples of 7. The average of all of them is the 7th number, which is $ 7(2\plus{}7)\equal{}63$

why is the average the seventh number?
 :maybe:  :whistling:

Because it is the middle of all 13 numbers. And it makes an arithmetic sequence which the mean = the median.

dragon96's solution is incorrect.

Yes, there are 13 positive two-digit multiples of 7.  Therefore, the average would be the median, or the 7th two-digit multiple.

However, the median is not $7(2+7)$, it's $7(2+6)$.
$7(2)$ is the 1st term, so $7(2+6)$ would be the (1+6)th term, or the 7th term.

Thus, the answer is $7(2+6) = \boxed{56}$.

How about this? The least $2$ digit multiple of $7$ is $14$ and the greatest is $98$. So using the formula \[\frac{n(a+b)}{2}\] Where $n$ is the number of terms in the sequence and $a, b$ are the first and last terms respectively, we get $\frac{13(14+98)}{2}$. So the average is $\frac{13(14+98)}{2*13}$ The $13$ in the denominator is there to find the average. The 2 $13$'s cancel and so the answer is just \[\frac{(14+98)}{2}=\boxed{56}\]


Prove that the following polynomial is irreducible in $ \mathbb Z[x,y]$:
\[ x^{200}y^5\plus{}x^{51}y^{100}\plus{}x^{106}\minus{}4x^{100}y^5\plus{}x^{100}\minus{}2y^{100}\minus{}2x^6\plus{}4y^5\minus{}2\]

we use generalized Esenestein .(the proof is the same as simple one)

$ f(x,y) \equal{} y^{100}(x^{51} \minus{} 2) \plus{} y^5 \times {(x^{100} \minus{} 2)}^2 \plus{} (x^6 \minus{} 2) \times (x^{100} \minus{} 2)$

but we know that $ x^{100} \minus{} 2$ is a prime in ring of polynomial so applying isenestain will get to the point.because
$ x^{100} \minus{} 2$ divide all of coeficients of polynomial(related to $ y$)expect$ x^{51} \minus{} 2$.
and also $ (x^{100} \minus{} 2)^2$dosent divide $ (x^6 \minus{} 2) \times (x^{100} \minus{} 2)$

Extended eisensteins wont work, since $ (x^{100} \minus{} 2)^2$ doesn't divide $ (x^6 \minus{} 2)(x^{100} \minus{} 2)$

the condition of eisenstein is that $ (x^{100}\minus{}2)^2$ shouldnt divide $ (x^6\minus{}2)(x^{100}\minus{}2)$

[hide=storage][color=#242423]$P(x,y) = y^{100}(x^{51} - 2) + y^5 \times {(x^{100} - 2)}^2 + (x^6 +1) \times (x^{100} - 2)$
FTSOC $P(x,y)=A(x,y)B(x,y)$ such that $A,B$ nonconstant
Now several claims east to check
Claim 1: for every $z \in \mathbb{Z}$ we have  $ gcd(z^{100} -2 , z^6 +1) =1$
Claim 2: for every odd $z \in \mathbb {Z} $  we have $gcd(z^{100}-2 , z^{51} -2)=1$
Claim 3: $gcd(x^{100} -2, x^{51} -2) =1$
Now fix $y$ and vary $x$. notice that for every $x \in \mathbb{Z}$ we have $ x^{100} -2$ has an odd prime divisor with odd degree. [/color]
[color=#087e8b] So by Eisenstein we have for every integer x the $ P(x,y)$ is irreducible. so one of $A , B$ becomes constant. Notice this can't happen that one of $A,B$ has $y$ terms in it and becomes constant for every integer $x$. So suppose wlog $B$ does not have $y$ term in it.  [/color]
[color=#f25f5c] so $P(x,y)=A(x,y).B(x)$ So consider $y$ terms in $P$. Their $ x$ coefficiant should be divisible by $B(x)$. But by claims we have $gcd(x^{100} -2,x^{51} -2)=1$ absurd $\blacksquare$ [/color] [/hide]


se cere sa rezolv urmatoarea "bucata" folosin ineg. lui Jensen

(x1 + x2 + ... + xn) / n <= sqrt[(x1^p + .... + xn^p) / n ]
p >= 2
f(x) = x^p

multumiri anticipate  :D

Cred cå e destul de cunoscutå. Avem $\frac{x_{1}+x_{2}+...+x_{n}}{n}\le \sqrt{\frac{x_{1}^{2}+...+x_{n}^{2}}{n}}\le \sqrt{\frac{x_{1}^{p}+...+x_{n}^{p}}{n}}$. Ultima e evidentå ($f$ e crescåtoare), iar prima rezultå aplicând Jensen functiei concave $g(x) = \sqrt{x}$ numerelor $x_{1}^{2}, ... , x_{n}^{2}$. Vezi si "Inegalitati elementare si mai putin elementare" (Editura Gil, Zalåu)  :) .

sau asa cum zice PhilAndrew...sau verificam ca $f(x)=x^{2}$ este convexa, apoi folosind inductie obtinem $f(x)=x^{p}, p \geq  2$ este convexa, dupa care scriem jensen pt n termeni, si anume :

$f(\frac{x_{1}+x_{2}+...+x_{n}}{n}) \leq  \frac{f(x_{1})+f(x_{2})+...+f(x_{n})}{n}$

Dupa care `tragem` un radical si a iesit  :oops:

Dar fara Jensen si eventual fara inductie o puteti dovedi ?

[quote="pohoatza"]$\frac{x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2}}{n}\geq \frac{(x_{1}+...+x_{n})^{2}}{n^{2}}$ $\Rightarrow n(x_{1}^{2}+...x_{n}^{2}) \geq (x_{1}+...x_{n})^{2}$, care e kiar chebishev pt n-uplele $(x_{1},...,x_{n})$ ; $(x_{1},...,x_{n})$ evident la fel ordonate.[/quote]
[color=darkblue]Pohoata a spus Cebasev, nu eu. Corect, insa pentru $p\ge 3$ cum demonstrezi ? Nu te supara ca insist. Suntem la sectiunea ... Liceu-Bacalaureat.
La bac sau admitere poate cere, de exemplu, sa se arate ca $a,b,c,d>0\Longrightarrow (a+b+c+d)^{3}\le 16(a^{3}+b^{3}+c^{3}+d^{3})\ .$

Fie $p\in N\ ,\ p\ge 2$ si $n\ge 2$ numere reale positive $a,b,c,\ldots\ .$ Folosind inegalitatea Cebasev obtinem urmatoarele $(p-1)$ inegalitati :

$\{\begin{array}{c}(a+b+\ldots )^{2}\le n(a^{2}+b^{2}+\ldots )\\\\ (a+b+\ldots )(a^{2}+b^{2}+\ldots )\le n(a^{3}+b^{3}+\ldots )\\\\ \ldots\ldots\ldots\ldots\ldots\\\\ (a+b+\ldots )(a^{p-1}+b^{p-1}+\ldots )\le n(a^{p}+b^{p}+\ldots )\end{array}$ $\Longrightarrow$ $\boxed{\ (a+b+\ldots )^{p}\le n^{p-1}(a^{p}+b^{p}+\ldots )\ }\ .$

Daca $A(x,y,\ldots )=\frac{x+y+\ldots}{n}$ este media aritmetica a $n$ numere pozitive $x,y,\ldots\ ,$ atunci inegalitatea precedenta devine :

$A^{p}(a,b,\ldots )\le A(a^{p},b^{p},\ldots )\ ,$ adica puterea mediei aritmetice este cel mult egala cu media aritmetica a puterilor ![/color]


Let $ a_{1},a_{2}, . . . ,a_{n}$ be an arbitrary permutation of the numbers $ 1,2, . . . ,n$, where n is an odd number. Prove that the
product $ (a_{1}-1)(a_{2}-2)...(a_{n}-n)$ is even.

[hide]Consider the sum of the factors of the final product. $ (a_{1}-1)+(a_{2}-2)+...+(a_{n}-n) = (a_{1}+a_{2}+...+a_{n})-(1+2+...+n) = 0$. Thus we have the sum of the n terms $ a_{i}-i$ being an even number. Because n is odd, the individual terms cannot all be odd, since the sum of an odd number of odd numbers is odd. So at least one is even, and the product is even.[/hide]

yeah, since the ak’s are a reordering of 1,2, . . . ,n. S is an odd number of summands of odd integers adding to the even integer 0. This is not absurd.


Let [n] be the set {1, 2, 3, ..., n} for any positive integer n.
Notation: f: A -> B means a function which assigns to each member of set A (the domain) exactly one member of set B (the range).

How many functions are there f: [4] -> [4] such that if x is odd, f(x) is even and if x is even, f(x) is odd?

What if we replace 4 in the above question with 5?  6?  n? (Do the even and odd cases for n seperately.)

How many [i]bijections[/i] are there f: [4] -> [4] with the same property?  (A bijection is a function such that for every element y of the range there exists [b][i]exactly[/b] one[/i] element x of the domain such that f(x) = y.)

What if 4 is replaced with 6?  2n?

How many functions f: [n] -> [m] are there such that if x is odd, f(x) is even and if x is even, f(x) is odd?  (There should be 4 cases for this one.)

Now, do all of the questions again, but instead take evens to evens and odds to odds.  (Note that for the bijections, you can use odd values of n here, too.)

A try.

JBL wrote:How many functions are there f: [4] -> [4] such that if x is odd, f(x) is even and if x is even, f(x) is odd? What if we replace 4 in the above question with 5?  6?  n?
[hide]
. f:[2n]->[2n]
- For each even number (n of them) there are n possible odd values
- For each odd number (n of them) there are n possible even values

So there are (n^n) such functions

. f:[2n+1]->[2n+1]
- For each even number (n of them) there are (n+1) possible odd value
- For each odd number (n+1 of them) there are n possible even value

So there are (n+1)^n + n^(n+1) such functions.[/hide]
JBL wrote:How many bijections are there f: [4] -> [4] with the same property?  What if 4 is replaced with 6?  2n?

[hide]

. First note there are no such f : f:[2n+1]->[2n+1] ;-)

. f:[2n]->[2n]

- For each even number (n of them) there are (n+1) possible odd value

- For each odd number (n+1 of them) there are n possible even value

f behaves as if there was two distinct bijections.

{1,3,..,2n-1} -> {2,4,..,2n}

{2,4,..,2n}   -> {1,3,..,2n-1}

So the number of such f is (n!)[/hide]

A good start -- one persistant error, though:  you kept adding instead of multiplying things together!


[quote="GameBot"]The radius of a circle is decreased by 10$ \%$. By what percent will
the area be decreased?[/quote]

Let the original radius be $ x$. We all know that the area is thusly $ \pi x^2$.

Decreasing the radius by $ 10\%$ gives a new area of $ \pi (.9x)^2$.

This is a $ 19 \%$ decrease in area.

Let $ R$ be the original radius, so the circle has area $ \pi R^{2}$.  Now the radius becomes $ .9R$, so the area drops to $ .81\pi R^{2}$.  Hence it drops by $ 19\%$.


inequalities
parameterization
number theory proposed
number theory

$(x+a)^2=(y+b)^2+(z+c)^2$
$\Rightarrow ax=by+cz$
On the other hand, by Cauchy-Schwarz Inequality,
$a^2x^2=(b^2+c^2)(y^2+z^2)\geq (by+cz)^2$ with equality holds iff $\frac{b}{y}=\frac{c}{z}$
$\therefore y=bt, z=ct$ where $t$ is a parameter
and $x=\frac{by+cz}{a}=\frac{b^2t+c^2t}{a}=at$
Therefore $(x,y,z)=(at,bt,ct)$
(Have I made any mistakes in the signs?)


I think that would be a great idea, but there isn't a North Carolina Forum.  I would like to get more people from North Carolina to join AoPS and this might help to get more people to come.  I would moderate the North Carolina forum, but I'm not sure how many people would come...   :(

I think a NC forum is a GREAT idea, especially considering the fact that NC now has (arguably) the best high school math team in the nation, and one of the top 10 MathCounts teams.  Also, I know John Berman posts in the SC forum, as well as a few other AoPSers (including Arnav).  So an NC forum would be a good idea ind33d.

Yay, thanks for the support.

I know there are several people that go to Arnav's highschool that are on AoPS but don't post, and he might get them to join.

The people I know for North Carolina:
jb05
b-flat
MysticTerminator
mgao
gt59 (my dad)
not_trig

I could send a PM to all of those people and see if they would be interested in joining.  Does anyone know anyone else from North Carolina on AoPS?

Yes, I agree it isn't many people, so I am trying to find more people who could join.  There are some people who MIGHT join at my school, but I don't know.  As for the ARML people, that would be a great idea.  I will email everybody on the list about it.  Might not be today, but I will get to it soon.

[quote="jb05"]Fine, but that's not many people. I guess you could e-mail all the ARMLers...

It's all SC's fault (or maybe just Joe's fault) for splitting SC out of carolinas.[/quote]

No, I think SC did need it's own forum, but NC probably needs one too.  The better thing to do would have been split the Carolina forum, not rename it to "South Carolina."

Goldturtle10 and turtlewarrior go to my school. Could I mod a Texas forum or something?

[quote=mysmartmouth]I thought it would be nice to have a flyer I could hand out at math club meetings about AoPS.  We could make a generic flyer, and just change the Community Name.  Any ideas or volunteers?  I'm not an expert at design...[/quote]

I could help. I'm pretty good with Apple Pages.


fast solutions for these problems please (countdown and sprint).

1) The Houses on Gauss Street are numbered consecutively from 1 to 678. How many brass digits are needed to form all the house numbers?

2) A math teacher wants to curve a set of test grades so that a student who scored  100  receive a score of 100 , but the student who scored 59 will receive a score of 70 . The teacher wishes to use a linear functional which turns an old grade, x , into a new grade f(x) . Write your formula in the form of f(x)=mx+b. Find the product of m and b . Express your answer as a fraction in lowest terms.

3) If a, b and c are positive integers, 0 < a < b < c < 10 and 1/a+1/b+1/c=1, what is the value of c?

[hide="1"] For the 9 1-digit numbers, we use 9 digits. For the 90 2-digit numbers, we use 90*2=180 digits. Fo the 579 3-digit numbers from 100 to 678, we use 3*579=1737 digits. Adding these together, we get 1737+180+9=$ \boxed{1926}$[/hide]

[hide="2"]We have two equations:
$ 100m+b=100$ and
$ 59m+b=70$. Subtracting these equations, and solving for m, we get $ m=\frac{30}{41}$. Substituting in and solving for b gives $ b=\frac{1100}{41}$. Multiplying gives us the final answer of $ \frac{33000}{1681}$ But this looks way too strange so I'm probably wrong...[/hide]

[hide="2"]
***$ x$ is the used for $ m$***

From xpmath you have 2 equations:
\begin{eqnarray*}100x+b&=&100\\ 59x+b&=&70 \end{eqnarray*}
Substitution gives that $ b=70-59x$. Solving the equation:
\begin{eqnarray*}100x+(70-59x)&=&100\\ 41x+70&=&100\\ 41x&=&30\\ x&=&\boxed{\frac{30}{41}}\end{eqnarray*}
Solving for $ b$ gives you:
\begin{eqnarray*}{b&=&70-59(\frac{30}{41}})\\ b&=&\boxed{\frac{1100}{41}}\end{eqnarray*}
So the answer is:$ \frac{1100}{41}\cdot\frac{30}{41}=\boxed{\boxed{\frac{33000}{1681}}}$

I have the same answer as xpmath, so it is right or we are both wrong.
[/hide]

oh thanks. it took me over a minute to figure out that

[quote="mihail911"][hide="3"]think of all the fractions 1-10 in decimal form. one of them has to be 1/2, because no combination of other three will give u 1. now the other two add up to 0.5....so with a little mental guess and check, u figure out the only possible combination that adds up to this is 1/3 and 1/6...so $ c=6$[/hide][/quote]
Instead of guess and check, theres this thing discovered by the egyptions (or someone else, can't remember) that says:
$ \frac{1}{n}=\frac{1}{n+1}+\frac{1}{n(n+1)}$. In this case, n=2, n+1=3, and n(n+1)=6.


trigonometry
absolute value
algebra unsolved
algebra

Nobody at my school can find the solution to this equation:
[img]http://nathan.bitmesh.com/equation.gif[/img]

Once x is found, it will be used to solve for the radius of a circle (x/8 = r). So x cannot be negative, zero, or undefined. This equation was derived by using known information, so there has to be solution. Any help would be appreciated.

First not that if $ x_0 $    is a root $ -x_0 $ is also a root. Then the number of positive roots coincides with that of negative roots. Also the number 0 is a root. Now:
\[
|x| = \frac {4} {3} |\sin x|\leq \frac {4} {3}
\]
Hence the absolute value of a root cannot exceed $ \frac {4} {3} $.
Between x= 0 and x = $ \frac {4} {3} $ there is only one root x = 0. That's all.

[quote="Nené"]
Between x= 0 and x = $ \frac {4} {3} $ there is only one root x = 0. That's all.[/quote]

Probably this should read "there is only one positive root $x_0$ and that's essentially all we can say". That's because, unfortunately, the root  is a transcendental number (this can be proved rigorously), and, most likely, not expressible using $\pi$ and $e$ either (though I doubt anyone in the world has the slightest idea of how to prove that). So, the only thing one can do is to find a numerical approximation on a calculator. Not every equation can be solved by a nice formula!
As to using calculaor to find a root, the simplest scheme to implement is to start with some number between $0$ and $\frac\pi 2$ (say, $1$)  and push the sequence of buttons $\boxed{\sin}$ $\boxed{*}$ $\boxed{4}$ $\boxed{/}$ $\boxed{3}$ $\boxed{=}$ many times until the number on the screen will stop to change. (That is if all your calculator can do is arithmetic and simple trig. functions).

O.K fedja. There are 2 trascendental solution positive and negative. In total 3 roots.

An approximation of the positive answer is $x=1.2756981$. Although I cannot prove it, I'm almost positive that this equation cannot be solved analytically


Find all pairs (x,y) of integer numbers such that:
\[x^3=y^3+2y^2+1\]


i think a1, a2, ..., an has to be increasing order at the least...

...what difference does it make if they are?

Yeah, why should they be in increasing order ?
The problem is stated correctly !

Arne wrote:Let a1, a2, ..., an be real numbers satisfying

a1 + a2 + ... + an :ge: n2

a12 + a22 + ... + an2 :le: n3 + 1

Prove that for 1 :le: k :le: n :

n - 1 :le: ak :le: n + 1.

[hide]write ak as n+bk. Then we have  :Sigma: bk:ge:0, and :Sigma: 2nbk+bk:^2::le:1. We want to show -1:le:bk:le:1. [/hide] would that be easier?

bugzpodder wrote:sorry...
if [hide]
Sum bk>=0 and Sum 2n(bk)+(bk)^2<=1[/hide] then [hide]1>=(bk^2)^2 which implies -1<=bk<=1[/hide]??

[hide]I don't get that last bit.  Are you just saying that

:Sigma: bk :ge: 0  and :Sigma: 2n*bk+bk:^2: :le: 1

implies :Sigma: bk:^2: :le: 1 which implies |bk| :le: 1

?  Because that makes sense to be, but you had some (bk^2)^2 going on, which I didn't get.[/hide]

For n :ge: 2, I believe the bound can be sharpened to
n - :sqrt:(1 - 1/n) :le: ak :le: n + :sqrt:(1 - 1/n).
Would anyone care to prove it? :)

basically

:Sigma: bk = 0
:Sigma: bk :^2:  = 1


Be aware that these three comic strips are heavier in
their amount of dialogue than usual strips (on purpose).

1.  "Chief Inspector Clues So?"
http://www.wittycomics.com/comic/45719

2.  "Progress in Aeronautics?"
http://www.wittycomics.com/comic/45720

3.  "Following Instructions"
http://www.wittycomics.com/comic/45721

Yeah.  I don't get number 1 at all.

Or number 2.

I get number 3, but it's not funny.

[quote="AIME15"]Yeah.  I don't get number 1 at all.
This is to slam people who think that stating that they see a solution
without working out a solution is acceptable.  These are problems
where the work has to worked out for the instructor.

Or number 2.
The advancements in aerospace were made early, but
people have been lousing up despite advanced technology.

I get number 3, but it's not funny.
Correction:  It's not funny to YOU; you don't
speak for all others.
[/quote]
Don't always go looking for a punchline with my strips.
I try to put thought-provoking ideas in there, often
including irony.  However, I'm not desiring for someone to
volunteer a rating on them, but just to experience them.

But if you claim to not understand a certain comic strip
(as in the past), then please bow out because it doesn't
make sense to rate one.

Does it seem that you pick a different username each time you submit a batch of comics? (like, in this batch all usernames are consistent only with their own batch)


integration
calculus
calculus computations

Calculate:
$ J\equal{}  \int_0^1 \frac{x^3}{1\plus{}\sqrt[3]{x^4\plus{}1}}  dx$

Sorry I'm in a rush but:

Substitute $ u\equal{}x^{4}\plus{}1$ so that $ du\equal{}4x^{3}\ dx$.

Then substitute $ v\equal{} u^{\frac{1}{3}}$.

This gives $ \int \frac{v^{2}}{v\plus{}1} \ dv$

and as the integrand can be written $ v\minus{}1\plus{}\frac{1}{v\plus{}1}$ it should work out OK.

Hope this is of some help.

I could write it out in full later, but by then I suspect someone else would have posted a far neater solution.

jus put $ x^4 \plus{} 1 \equal{} t^3$ and solve .

Hi vijaymenon. This is the same but, of course, you've killed two birds with one stone.


geometry
trigonometry
function
analytic geometry

Circles with centers A, B, and C each have radius $r$, where $1<r<2$. The distance between each pair of centers is 2. If $B'$ is the point of intersection of circle A and circle C which is outside circle B, and if $C'$ is the point of intersection of circle A and circle B which is outside circle C, then length $B'C'$ equals

(A) $3r-2$
(B) $r^{2}$
(C) $r+\sqrt{3(r-1)}$
(D) $1+\sqrt{3(r^{2}-1)}$
(E) none of these

Where did you find this problem? I can only get the AHSME's back to 1980. :maybe:

[quote="Farenhajt"][hide="Approach"]
Put $\alpha=\angle B'AC=\angle C'AB$ and apply Cosine Law to $\triangle B'AC'$, where trig functions of $\alpha$ can be found from $\triangle AB'C$.[/hide]

[hide="Answer"]
$D$[/hide][/quote]

Ah, nice. My solution was putting $A$, $B$, $C$ on a coordinate system with

$A\left(\frac{2}{\sqrt{3}},0\right)$, $B\left(-\frac{1}{\sqrt{3}},1\right)$, $C\left(-\frac{1}{\sqrt{3}},-1\right)$,

solving for the intersections of the circles, and using the distance formula :P. I agree with your answer.


arithmetic sequence
algebra
system of equations

In an arithmetic progression, the sum of the first 10 terms is 170 and the 8th term is 32. Find the first term and the common difference.

[hide]
Call the first term $a$ and the common difference $d$.

The sum of the first ten terms is $a + (a+d) + (a+2d) + \cdots + (a+9d) = 10a+(1+2+\cdots+9)d = 10a+45d = 170$.

The eighth term is $a+7d = 32$.

Solve this system of equations to get the answer.[/hide]

[quote="wolf"]Where did the 45 come from? And other figures that weren't in the question.[/quote]

He was using the formula for an A.P.:P :

Sum of  n terms:  s= n/2(2a+(n-1)d)
nth term: T=a+(n-1)d

a=first term
d=common difference

Could you show me how to solve the 'system of equations' please? I can't see how you did it.

simultaneous equations, no problem. Thanks for all your help.  :D


abstract algebra
category theory
Ring Theory
superior algebra
superior algebra unsolved

Let $ \phi: A\rightarrow B$ be a ring homomorphism of two unital rings.
Then any right $ A$-module $ M$ becomes a right $ B$-module by defining $ m\cdot b\equal{}m\phi(b)$ for $ m\in M, b\in B$,
and any right $ B$-module $ N$ becomes a right $ A$-module by defining $ n\cdot a\equal{}n\phi(a)$ for $ n\in N, a\in A$.
Let $ E_\phi$ denote the functor from the category of right $ A$-modules to the category of right $ B$-modules.
Let $ R_\phi$ denote the functor from the category of right $ B$-modules to the category of right $ A$-modules.

(i) Show that for any $ A$-module $ M$ and any $ B$-module $ N$, $ \mathrm{Hom}_B(E_\phi(M),N)\cong\mathrm{Hom}_A(M,R_\phi(N))$

(ii) Show that $ R_\phi$ sends projective modules into projective modules.

(iii) Show that for finitely-generated projective right $ A$-module $ M$, $ M\otimes_A B$ is finitely-generated projective right $ B$-module.

Edit: Oops nevermind looks like I messed up in comprehending the problem statement.

[quote="LordoftheQRings"]Let $ \phi: A\rightarrow B$ ... defining $ m\cdot b \equal{} m\phi(b)$ for $ m\in M, b\in B$, ...[/quote]

This does not make sense. owk

Sorry, right $ A$-module $ M$ becomes right $ B$-module $ M\otimes_A B$ by defining $ (m\otimes b)b' \equal{} m\otimes bb'$ for $ b,b'\in B$, $ m\in M$.


Please leave this for the more inexperienced proof writers - forum oldbies should be able to solve these problems on sight. They're meant more as explorations than anything else.

That said, how many proofs can you provide for each of these?

1) Prove that the three altitudes of a triangle intersect at one point. In other words, prove that if in triangle ABC, AD and BE are two altitudes and they intersect at H, then CH's extension is perpendicular to AB.

2) Prove that the three angle bisectors of a triangle intersect at one pont. In other words, prove that if in triangle ABC, AD and BE are two angle bisectors and they intersect at P, then CP bisects angle ACB.

3) Prove that the three medians of a triangle intersect at one point. In other words, prove that if in triangle ABC, AD and BE are two medians and they intersect at Q, then CQ extended bisects side AB.

4) Prove that the three perpendicular bisectors of a triangle intersect at one point. In other words, prove that if in triangle ABC, D, E and F are midpoints of sides AB, BC and CA, respectively, and OD :perp: AB, OE :perp: BC, then OF :perp: AC.

You can do first 3 if you know [hide]converse of Ceva's Theorem[/hide][/hide]

2. [hide]Points on an angle bisector are equidistant from the rays of the angle they bisect.  So, in triangle ABC, angle bisector of A is equidistant from AB and AC, and the same for the other two.  Therefore, the point equidistant from AB, AC, and BC will lie on all three lines.[/hide]

3. [hide]Suppose that the median to side a meets a at X, to b at Y, to c at Z.  Then BX=BY, CY=AY, AZ=BZ, since they are medians. Using Ceva's Theorem, the medians are concurrent iff (BX/BY)*(CY/AY)*(AZ/BZ)=1, which it does.[/hide]

1) [hide]case 1: triangle is acute

Let the triangle be ABC. Let the altitudes be AD, CF, BE. Since

, so

, so

, so

Multiply three equations to get

By Converse of Ceva's Theorem, three altitudes concur.

Case 2 holds because if O is the orthocenter of ABC, then triangle BOC is always obtuse, the altitude of BOC are OD, FB, EC. Now FB and CE so that they meet at A. We have already proved that AO is perpendicular to BC so QED.[/hide]

[quote="Osiris"]Now, now, Menelaus is kind of cheating, eh? ;)[/quote]

Does Menelaus make any of the problems easier?

Argh, I didn't mean Menelaus. I meant Ceva. Ceva's theorem makes the median one just waaaay too easy.  :D

See if you can use chasing angles to solve #1.

Different way to do the median one:

[hide]Let CQ intersect AB at F. Draw ED. Because [tex]\Delta EDQ \sim \Delta BAQ[/tex], .. Note that, where A(AQB) is area of AQB

So

Also

yields

So

Draw DF. By SAS similarity, ,  thus . BF/AF=1/2, so BF=AF, CQ bisect AB as desired.[/hide]

The solution for the 4 it's similar to the 2nd's solution that Wumbate wrote. We just need to see that the points on the perpendicular bisectors are equidistant from the extrems of the side they bisect. So the three bisectors will intersect in the point equidistant to A,B,C.

Can u introduce something about [b] "converse of Ceva's Theorem"?[/b]

Ceva's theorem says:

[b]In $\triangle ABC$, if the cevians $AD$, $BE$, and $CF$ are concurrent then
\[\frac{BD\cdot CE\cdot AF}{DC\cdot EA\cdot FB}=1.\][/b]

The converse of this happens to be true.  Namely that:

[b]If $AD$, $BE$, and $CF$ are cevians in $\triangle ABC$ and
\[\frac{BD\cdot CE\cdot AF}{DC\cdot EA\cdot FB}=1\]
then $AD$, $BE$, and $CF$ are concurrent.[/b]

I should probably also define what a cevian is.  A [b]cevian[/b] is a line which joins a vertex of a triangle to the opposite side, extended if necessary.  So for example, medians, angle bisectors, and altitudes are all special types of cevians.

[hide]   1) [b]D,E,P are the proiections of A,b,C on BC,AC,AB
the triangle APC  has a right angle in P =>$cosA=\frac{AP}{AC}\Rightarrow A=AC\cdot cosA$[b][/b] (1)
analog in the trianglesBCP=BP=BC $\cdot$ cosB  (2)
(1),(2)=>$ \frac{AP}{PB}=\frac{AC\cdot cosA}{BC\cdot cosC}$
analog=>$\frac{BD}{DC}=\frac{AB\cdot cosB}{CA\cdot cosC} and
\frac{EC}{AE}=\frac{BC\cdot cosC}{AB\cdot cosA} $
we multiply and  obtain $ \frac{AP}{PB}\cdot \frac{BD}{DC}\cdot \frac{EC}{AE}=\frac{AC\cdot cosA}{BC\cdot cosB}\cdot \frac{AB\cdot cosB}{CA\cdot cosC}\cdot \frac{BC\cdot cosC}{AB\cdot cosA}=1 $
coresponding to the reciprocal of Ceva =>AD,BE, and CM are concurrent[/hide]

is that correct? :D
 2) and3) are easier than1)  (after my opinion) :P

Thank you joml88! :D
I think I have proved it before but i just donnot know it has a name and such uses...
I think my proof may offer a different point of view...
[hide]
If $AD$,$BE$ and $CF$ are concurrent and they join at $O$ then
$\displaystyle\frac{AF}{FB}\cdot\frac{BD}{DC}\cdot\frac{CE}{AE}=\frac{\triangle{AOC}}{\triangle{BOC}}\cdot\frac{\triangle{AOB}}{\triangle{AOC}}\cdot\frac{\triangle{BOC}}{\triangle{AOB}}=1$[/hide]

And i tried to prove the converse but i don't know if i did alright...
[hide]
Assume $AD$ and $BE$ meet at $O$, and $CF$ and $AD$ meet at $P$. Then we will have:
$\displaystyle\frac{AF}{FB}\cdot\frac{BD}{DC}\cdot\frac{CE}{AE}=\frac{\triangle{APC}}{\triangle{BPC}}\cdot\frac{\triangle{AOB}}{\triangle{AOC}}\cdot\frac{\triangle{BOC}}{\triangle{AOB}}=\frac{\triangle{APC}}{\triangle{BPC}}\cdot\frac{\triangle{BOC}}{\triangle{AOC}}=1$

Then we have : $\frac{\triangle{AOC}}{\triangle{BOC}}=\frac{\triangle{APC}}{\triangle{BPC}}$
Extend $CO$ and label its intersection with $AB$ as $M$. Therefore:
\[\frac{AM}{BM}=\frac{AF}{BF}\]
Since $AM+BM=AF+BF$, we will obtain $AM=AF$ and $BF=BM$
Therefore $AC$, $BE$ and $CF$ are concurrent.[/hide]


limit
real analysis
real analysis unsolved

Prove that if a sequence $\{a_n\}^{\infty}_{n=0}$ with $a_n\geq 0$ for each $n$ satisfies $a_{m+n}\leq a_ma_n$ for every $m,n\in\mathbb{N}$, then $\lim_{n\to\infty} \sqrt[n]{a_n}$ exists.

I assume you also want them to be positive? Having $a_n=-n^n$ would wreck things.

And besides, it's clearly not true, even for positive sequences. Just take a decreasing sequence of positive reals for which that limit does not exist.

Wow, I butchered up the statement of the problem pretty badly.  Please see the corrected version.

already posted at least $3$ times ... (or a similar one with $u_n=ln(a_n)$)

Do you have a link?  I've had no luck with the search.


function
search
algebra proposed
algebra

Find all functions $ f:\mathbb{R}\to\mathbb{R}$ such that $ f(x^{3}\minus{}y)\plus{}2y(3f^{2}(x)\plus{}y^{2})\equal{}f(y\plus{}f(x))\forall x,y\in\mathbb{R}$.

See at http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1729487962&t=159278 :P


function
logarithms
algebra proposed
algebra

Let $ c > 0$ . Determine all functions function $f: \mathbb R \to \mathbb R$ which satisfy
 i) $f(x + y) = f(x)\cdot f(y)$ for all $x,y \in \mathbb R$, and
 ii) $|f(x)| \le c$ for all $x \in [ -1,1]$.

[quote="Victory.US"]Let $ c > 0$ . Determine function $ f$ satisfy
 $ i) f(x \plus{} y) \equal{} f(x).f(y), \forall x,y \in R$
 $ ii)|f(x)| \le c, \forall x \in [ \minus{} 1,1]$[/quote]

1) If $ \exists a$ such that $ f(a)\equal{}0$, then $ f((x\minus{}a)\plus{}a)\equal{}f(x\minus{}a)f(a)\equal{}0$ and we get $ f(x)\equal{}0$ $ \forall x$, which, indeed, is a solution.

2) If $ f(x)\ne 0$ $ \forall x$, let $ g(x)\equal{}\ln(|f(x)|)$. We got $ g(x\plus{}y)\equal{}g(x)\plus{}g(y)$ and $ g(x)$ is a solution of Cauchy equation.
Let then $ a\in\mathbb R$ and $ x\in[a\minus{}1,a\plus{}1]$. We get $ f(a)\equal{}f(x)f(a\minus{}x)$ and so $ |f(a)|\le c|f(x)|$ and so $ g(x)>g(a)\minus{}\ln(c)$ $ \forall x\in[a\minus{}1,a\plus{}1]$
So $ g(x)$ has a lower limit on a non empty open interval and so is continuous (classical result of Cauchy equation) and must be $ \lambda x$

So $ |f(x)|\equal{}e^{\lambda x}$ and since $ f(x)\equal{}f(\frac x2\plus{}\frac x2)\equal{}f(\frac x2)^2>0$, we get $ f(x)\equal{}e^{\lambda x}$

Plugging back in initial requirements we get that this is a solution as soon as $ |\lambda|\le \ln c$

Hence the answer :

If $ c<1$ the only solution is $ f(x)\equal{}0$
If $ c\equal{}1$ the only solutions are $ f(x)\equal{}0$ and $ f(x)\equal{}1$
If $ c>1$ the solutions are $ f(x)\equal{}0$ and $ f(x)\equal{}e^{\lambda x}$ where $ |\lambda|\le \ln c$


algebra
polynomial
superior algebra
superior algebra unsolved

Let $ p$ be a prime number such that $ p \equiv 3 \mod 4$. Let $ f(x) \in \mathbb{Z}_p[x]$ be an odd polynomial (all powers are odd). Show that the number of solutions $ (x,y)$ in $ \mathbb{Z}_p^2$ of the equation $ y^2 \equal{} f(x)$ is $ p$.

Since $ \minus{} 1$ is not a square, it follows that $ f(x)$ is a square iff $ f( \minus{} x) \equal{} \minus{} f(x)$ is not a square. (this is the only non-elementary part, ask for clarification if you need it)

Let $ S \equal{} \{x|f(x)\neq 0\}$.  Then, by the above, exactly half of the elements of $ S$ are solutions to $ f(x) \equal{} y^2$, and there are exactly two $ y$ that work here.

So, across all choices of $ x\in S$, the average number of $ y$ such that $ (x,y)$ is a solution is $ 1$, and it's clear that the average number across $ x\notin S$ is also $ 1$, so the average altogether is $ 1$, and there are $ p$ solutions.


Give an examample [b]non-atomic[/b] sigma algebra $ \mathcal{F} \subset 2^{X}$ such that
$ \mathrm{card}\mathcal{F} \equal{} 2^{\aleph_0}$
A set $ A\in \mathcal{F}$ is an atom, if there is no non-trivial subset of $ A$ in $ \mathcal{F}$
Any hints? I think that $ X$ should be uncountable...

What non-atomic sigma-algebras do you know?

Well, take $ X_0$ to be an uncountable product of copies of $ \{0,1\}$ and consider the $ \sigma$-algebra of all sets depending only on countably many coordinates. The $ \sigma$-algebra in question has cardinality continuum as long as the number of factors is $ c$ or less. Unfortunately, the underlying space $ X_0$ is too big, so the construction based on this example will work only if $ |X|\ge |X_0|$.

The question is whether we can do it as soon as $ |X| > \aleph_0$. You can improve the cardinality of the underlying space to $ c$ if you consider only "sequences" in $ X_0$ that have either countably many $ 0$'s, or countably many $ 1$'s but we'll still need to find out what happens between $ \aleph_0$ and $ c$.

It is quite clear that you cannot do anything with countable $ X$ because we can take any $ x\in X$, choose arbitrary sets $ X_y\ni x$ in our $ \sigma$-algebra such that $ y\notin X_y$ for all $ y\in X$ for which such sets exist at all, and observe that $ \bigcap_{y} X_y$ is an atom. But this proof fails as soon as $ |X| > \aleph_0$. :(


Is there a moderator for this forum?  If not, then I volunteer.

Thanks,

mysmartmouth

Hopefully this forum will see increased posting and require
moderation


pishnehad midam inja ro ekhtesas bedim be masayele hendese too levele marhale 2.
soale1:ABC yek mosallase hade zavie ast va M vasate zele BC mibashad.yek noghteye monhaser be fard N dakhele mosallas vojood darad ke zavieye ABN barabare BAM va hamchenin zavieye ACN barabare CAM bashad. sabet konid zavieye BAN barabare zavieye CAM khahad bood

goftam ye chizi benevisam ta balke in topic biad sadr ta behesh tavajjoh beshe. :D
age kasi hal ya ide ee dare ya hatta soale hendese(faghat dar hade marhale 2) lotfan inja bezare

do soale asoon va dar hadde marhale 2
1-agar dar mosallase ABC dashte bashim : $3A+2B = 180^\circ$ sabet konid : $a^{2}+bc = c^{2}$

2-ABC yek mosallase hadde zavie ast dayere k be ghotr AB khat haye AC va BC ra mojaddadan dar noghate M,N ghat mikonad momas haye vared bar dayere k dar noghate M,N dar noghte P moteghateand agar CP = MN zaviye ACB ra bedast avarid .

mosalase $ABC$ ra ba markaz ertefaei $H$ dar nazar begirid farz konid $D,E,F$ betartim paye ertefahaye nazire $A,B,C$ bashad az $D$ khati movazie $EF$ mikeshim ta $AC , AB$ ra betartib dar $P,Q$ ghat konad agar $PQ$  emtedade $BC$ ra dar $T$ ghat konad sabet konid dayere mohitie $PQT$ az vasate $BC$ migozarad.

HALLE SOALE TOXIC_FROG

soale (1) ba ghaziyeye kosinoosha hal mishe.
soale (2); chon BM ertefaye mosallase ABC ast. zavieye MBC mishe 90-C va dar natije zavieye MON mishe 180-2C va chon chaharzeliye MONP mohatiye,zavieye MPN mishe 2C. dar natije P markaze dayereye mohitie mosallase MCN mishe.pas zavieye C mishe 30 daraje

HALLE SOALE SEPEHR.

fek mikonam manzooret ine ke T mahale barkhorde emtedade EF ba BC ast.
HAL: TD.DM = QD.DP moadele hokme. az tarafi ba estefade az movazi boodane EF va PQ sabet mishe do mosallase BQD va DPC moteshabeh and. ke darim: QD.DP = BD.DC. pas TD.DM = BD.DC moadele hokme.
az tarafi az mohati boodane 4zeleehaye FEMD ( dayereye 9 noghte) va FECB darim: TD.TM = TF.TE = TB.TC  ke ba estefade az rabeteye akhir yani TD.TM =TB.TC va ye kam khar kari moadele hokm  be dast miad.( az inke M vasate BC ast ham bayad estefade konim).

kamelan doroste hala ino hal konid :wink
mossalase haddato zavie $ABC$ mafrooz ast.noghate $A_{1},A_{2}$ rooye zele $BC$ ($A_{2}$ beine $A_{1}$va$C$) va $B_{1},B_{2}$ rooye $AC$($B_{2}$beine $A$va$B_{1}$) va $C_{1},C_{2}$ rooye $AB$($C_{2}$beine$B$va$C_{1}$) tori entekhab shode ke  $AA_{1}A_{2}=AA_{2}A_{1}=BB_{1}B_{2}=BB_{2}B_{1}=CC_{2}C_{1}=CC_{1}C_{2}$
sabet konid $3$ noghte mahale talaghie $AA_{1},B_{1},CC_{1}$ be hamrahe $3$ noghteye mahale talaghie khotoote $AA_{2},BB_{2},CC_{2}$ rooye yek dayereand..

inam ye soal dige  :wink:
$O$,$T$be tartib marakeze dayere mohitie $ABC$,$AOC$ va$M$vasate$AC$mibashad.$D,E$ruye$AB,BC$mafruzand betorike zaviye$ABC=BEM=BDM$. sabet konid$BT$amud ast bar$DE$.

[quote="oveis"]HAL: ba tavajoh be tagharone shekl kafie baraye chahar noghte az 6 noghte sabet konim ke rooye yek dayere hastand.

farz konid    (BB2,CC2)=M , (BB2,AA2)=K , (BB1,AA1)=L, (CC1,BB1)=N  bashad. sabet mikonam ke in 4 noghte rooye yek dayere and.
farz konid  AA1A2=AA2A1=BB1B2=BB2B1=CC2C1=CC1C2=a daraje
pas khahim dasht   A2AA1=B2BB1=C2CC1=b daraje.
 be vozooh chaharzelie MNCB mohati ast. dar natije zavieye KMN barabare NCB va an ham barabare b+a-B khahad bood .
be tarighe moshabe ba tavajoh be mohati boodane AKLB khahim dasht KLN=b+a-B pas 4 noghteye KMLN ham dayere and. :wink:
rasti kasi roo soale avali ke too in topic gozashtam ide ee nadare :([/quote]soale agha oveis ham ba ghaziyeye sinoosa va neveshtane masahate invaro oonvare miane bar hasbe sinoose zavieha be rahati hal mishe.
albate vase rahat tar shodanesham ye borhane kholf bezanim bad nist.

in soala man ye kam roosh fek kardam vali hal nashod .vali yeki az doostam hal kard in halle oone:

HAL; farz konid emtedade DM emtedade BC ra dar P va emtedade EM emtedade BA ra dar K ghat kone .chon 4 zeleehaye DMTK va MEPT mohatiand( be rahati sabet mishe). pas TK bar BD va TP bar BC amood ast.
az tarafi ghovvate B nesbat be dayereye mohitie mosallashaye  DKT va EPT barabare chon 4 zeleeye DEPK mohatie (be rahati mishe sabet kard). hamchenin ghovvate T nesbat be in do dayere barabare.( chon rooye in do dayerast) pas BT bar khatol markazine anha amood ast. khatol markazin ham pare khate mianie mosallase DTE ast. pas BT bar DE ham amood mishe.
ensafan rahe hale in dooste man kheili ghashange. vasehamin goftam bezaram hame halesha bebaran. :wink:

do soale asab khord kon man ke hanooz natoonestam halleshon konam :

1-dar mosallase $ABC$ , $D$ paye nimsaze ras $A$ ast dayerei az $A$ migozarad va bar $BC$ dar $D$ momas ast in dayere $AC$ ra dar $E$ miborad va $BE$  dayere ra mojaddadan dar $F$ ghat mikonad neshan dahid $AF$ , $BD$ ra nesf mikonad

2-$G$ markaze seghle mosallase $ABC$ ast age $P,Q$ davaran yafte haye $B,C$ be markaze $G$ va ba zaviyehaye $120^\circ ,240^\circ$ (dar yek jahat) bashe sabet konid $APQ$ motesaviolazla ast . ---

soale 2 ro felan   ap=aq raftam badesh nemidoonam chikar mishe  :(  ?

dar mosalase $ABC(BC=AB)$;$CD$nimsaze zaviyeye $C$ hast va amude vared az $O$ (markaze dayere mohiti) bar $CD$zele$BC$ ra dar $E$ghat mikonad,az $E$movazie$CD$khati rasm mikonim ta$BA$ra dar F ghat konad.hokm:$BE=DF$

AF ra emtedad midahim ta BD ra dar M ghat konad kafist sabet konim $BM^{2}=DM^{2}$.
kamane $DN=DE \Rightarrow$ zaviyey $ABD=AE/2$ (kamaneAE),zaviye $BFM=AE/2$
pas mosalashaye ABM,BMF moteshabehand pas $BM^{2}=MF.MA ,MD^{2}=MF.MA$ pas $MD=BM$

HAL: chon EF movazie CD ast. BF.EC =BE^2  moadele hokme.
 mahale barkhorde BO ( ke ertefaye  nazire zele AC ham ast)  ba EF , AC ra     K , H va mahale barkhorde OE ba CD ra L minamim. be rahati mishe sabet kard zavieye BEK = KOE dar natije  BE^2=BK.BO =BK.OC   va chon mosallashaye BKF ba OCE moteshabeand ( rahat mishe sabet kard) darim BF.EC=BK.CO pas az 2 rabeteye akhir darim BF.EC=BE^2  :wink:


Lol interesting question. I think at some point the body will be too thick to swallow.

Eh. A snake a gulf down eggs much larger than any part of its body. My guess is in the process of swallowing it it would damamge some important organ and kill itself. But I really dont think any snake is dumb enough to even try.

[quote="salmononpi"]Eh. A snake a gulf down eggs much larger than any part of its body.[/quote]

It'll get thicker and thicker if you think about it. The cross-sectional area increases because it goes into its own stomach...and then it wraps around once in there and goes around again and again and...uh...okay I'm probably not explaining it very well. :D It may be able to eat huge eggs (and indeed some snakes can eat whole deer iirc), but eggs don't get bigger continuously as you swallow more and more of it.

I rest my claim: it all depends on the speed. For example if the speed is decreasing, you may need an infinity time to complete the process. And usually that will be the case: the more force needed, the slower the swallowing process.

At the end when it is spinning around and cant bend anymore and when the the tail wraps around the body in a circle 6 or 7 times

Well, it obviously can't stretch its jaws higher than its jaws, so it can't eat its own mouth, no matter how fast it goes (assuming it doesn't die).

A more interesting question might be: you have 100 snakes, with the first one eating the second, the second eating the third, etc. what happens?

The first snake dies because it gets overstuffed with 99 other snakes and gets fat

Erm... the last snake is also eating the first snake (it's a big circle).

[quote="MysticTerminator"][quote="stephenyamqc"][hide="What will happen?"]What will happen if a snake continually swallows its tail? :?:  :(  :wacko:  :huh: [/hide][/quote]

you get BENZENE WOOOOOOO[/quote]

What's that?  :?

Try eating your self from the feet up, and see what happens  :D

[quote="sponge008"][quote="zeb"]
A more interesting question might be: you have 100 snakes, with the first one eating the second, the second eating the third, etc. what happens?[/quote]

I guess it's like one of those contracting-circle toys, you have a bunch of stuffed snake heads and upper necks in a circle.[/quote]

ahahahah, you've seen "boom headshot" too? that clip was awsome!

  ......."lets go for a run"......."ok, but why are you holding a knife"......."cmon man, everyone knows you run faster with a knife"......... :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:


The sum of the reciprocals of the first four consecutive positive integers is greater than two. What is the least number of consecutive positive integers necessary to make the sum of the reciprocals greater than three?

Just keep adding and find that the answer is $ \boxed{11}$.

Are you sure??? Cause I don't want to do that... Is there an easier way? It looks hard... you have to find a common factor of the first 11 integers...    :(

This was a team round problem, so calculators are allowed :)

For trivia purposes, these sums of the reciprocals of consecutive positives are known as the [url=http://mathworld.wolfram.com/HarmonicNumber.html]harmonic numbers[/url].


function
calculus
integration
symmetry
Support
absolute value
real analysis

I'm not sure what section to put these problems in but can anybody help me with these:

1. Let $X_{1},X_{2}, . . . ,X_{20}$ be independent Poisson random variables with mean one.
(a) Compute Var($X_{1}$).
(b) Use the preceding and the central limit theorem to approximate P{${\sum_{i=1}^{20}}X_{i}> 15$}.

2. Suppose X and Y are independent, both distributed uniformly on [0 , 1]. Then prove that
$E (|X-Y|^\alpha ) = \frac{2}{(\alpha+1)(\alpha+2)}$ for $\alpha > 0$.

3(a) Compute the moment generating function of a random variable X whose mass function is as
follows:
$p(0) = \frac{1}{3}, p(2) = \frac{1}{6}, p(-1) = \frac{1}{2}$.
(b) Suppose Y is a random variable whose moment generating function is
$M_{Y}(t) = \frac{1}{2}exp(t)+\frac{1}{6}exp(-2*t)+\frac{1}{3}exp(5*t)$

Compute the mass function of Y .

Thanks

2) requires a double integral $E(|X-Y|^{\alpha})=\int_{0}^{1}\int_{0}^{1}|x-y|^{\alpha}\,dy\,dx$. To get rid of the absolute value, split the integral into two parts: $y\le x$ and $y>x$. In fact, these parts are equal by symmetry, so you only need to calculate $\int_{0}^{1}\int_{0}^{x}(x-y)^{\alpha}\,dy\,dx=\int_{0}^{1}\frac{x^{\alpha+1}}{\alpha+1}\,dx=\frac{1}{(\alpha+1)(\alpha+2)}$ and double it.

[quote="223rem"]I'm not sure what section to put these problems in but can anybody help me with these:

1. Let $X_{1},X_{2}, . . . ,X_{20}$ be independent Poisson random variables with mean one.
(a) Compute Var($X_{1}$).
(b) Use the preceding and the central limit theorem to approximate P{${\sum_{i=1}^{20}}X_{i}> 15$}.
[/quote]

For a Poisson distribution, the mean equals the variance, so ${\rm Var}[X_{i}] = 1$.  If we let $S = X_{1}+X_{2}+\cdots+X_{20}$, then we also have ${\rm E}[S] = 20\,{\rm E}[X_{1}] = 20$, and ${\rm Var}[S] = 20\,{\rm Var}[X_{1}] = 20$.  By the CLT, $S$ is approximately normal with mean ${\rm E}[S]$ and standard deviation $\sqrt{{\rm Var}[S]}$.  Therefore

$\Pr[S > 15] = \Pr\left[\frac{S-{\rm E}[S]}{\sqrt{{\rm Var}[S]}}> \frac{15-20}{\sqrt{20}}\right] \approx 1-\Phi(-5/\sqrt{20}) = \Phi(5/\sqrt{20})$,

where $\Phi$ is the CDF of the standard normal distribution.

[quote="223rem"]3(a) Compute the moment generating function of a random variable X whose mass function is as
follows:
$p(0) = \frac{1}{3}, p(2) = \frac{1}{6}, p(-1) = \frac{1}{2}$.
(b) Suppose Y is a random variable whose moment generating function is
$M_{Y}(t) = \frac{1}{2}exp(t)+\frac{1}{6}exp(-2*t)+\frac{1}{3}exp(5*t)$

Compute the mass function of Y .[/quote]

Use the definition of MGF:  $M_{X}(t) ={\rm E}[e^{tX}]$.  For a discrete random variable,

${\rm E}[f(X)] = \sum_{x \in S}f(x) \Pr[X = x]$,

where $S$ is the support of $X$; in this case, $S = \{-1,0,2\}$.  The result is so straightforward I need not write it here.  Part (b) should also be obvious from the result of part (a).


function calculus conics parabola inequalities

Is given a function with formula $ y\equal{}ax^{2}\plus{}b$ where $ x$ takes its value in $ [\minus{}2,2]$ and $ y$ takes value in $ [\minus{}1,7]$.Find $ a,b$.How many solution does the problem have

The intuitive answer is rigorously derived from calculus, but...

Calc version:
[hide]Extrema of the equation lie at endpoints and stationary points. Endpoints are $ x \equal{} \{ \minus{} 2, 2\}$, and $ \frac {dy}{dx} \equal{} 0 \equal{} 2ax, x \equal{} 0$ is the only stationary point. Which makes sense, since it's a quadratic.

Then, plug in $ x \equal{} \{ \minus{} 2, 0, 2\}$. You get $ y \equal{} 4a \plus{} b, b, 4a \plus{} b$. This implies that $ 4a \plus{} b, b \in [ \minus{} 1, 7]$. Thus $ b \in [ \minus{} 1,7]$, and $ \minus{} 1 \le 4a \plus{} b \le 7$ solves to $ \frac { \minus{} 1 \minus{} b}{4} \le a \le \frac {7 \minus{} b}{4}$. Tabulate a list of values to get the integer solutions.[/hide]

"intuitive" Algebraic version:
[hide]This is a parabola which is symmetrical about the origin, due to the fact that it's even. Since when $ x \equal{} 0$ the function evaluates to only b, $ \minus{} 1 \le b \le 7$. Depending on the value of b, a can take on any range of values as long as $ ax^2 \plus{} b$ doesn't go outside the boundaries. Since it's symmetrical and a parabola always increases/decreases in the same direction only extreme values need to be considered.

Let's take the top boundary, $ y \le 7$. Set $ ax^2 \plus{} b \le 7$, and plug in $ x \equal{} 2$ WOLOG. Simplifies to $ 4a \plus{} b \le 7$ and subsequently $ a \le \frac {7 \minus{} b}{4}$.

Now take the bottom boundary, $ y \ge \minus{} 1$. Set $ ax^2 \plus{} b \ge \minus{} 1$, and plug in $ x \equal{} 2$ WOLOG. Simplifies to $ 4a \plus{} b \ge \minus{} 1$ and subsequently $ a \ge \frac { \minus{} 1 \minus{} b}{4}$.

Because a has to satisfy both inequalities to let the function fall within the bounding box, we combine to get $ \frac { \minus{} 1 \minus{} b}{4} \le a \le \frac {7 \minus{} b}{4}$. Tabulate possible values of a for b to get the number of integer solutions.
[/hide]

EDIT> Should have 18 solutions for which $ a, b \in \mathbb{Z}$.


Source: British Math Olympiad, Round 1 (Wednesday, January 14, 1998)

ABP is an isosceles triangle with AB = AP and angle PAB acute.  PC is the line through P perpendicular to BP, and C is a point on the line on the same side of BP as A.  (You may assume that C is not on the line AB).  D completes the parallelogram ABCD. PC meets DA at M.  Prove that M is the midpoint of DA.

[quote="mathfanatic"]PC is on the line through P perpendicular to BP[/quote]
I don't understand this. Is this saying that PC is perpendicular to BP?

Yes.  That is, C is on the line through P perpendicular to BP, or PC is perpendicular to BP.

I shouldn't have put that "on" in, but it doesn't really change the meaning.  If I had said "BA" on the other hand...but I DIDN'T.

Well [quote="it"]You may assume that C is not on the line AB[/quote]
and I was wondering how that was possible...

are you mad at me?

No, why would I be mad at you for asking questions?

I don't know either.  I copied the problem as closely as stated as possible, since (in the past) there have been miscopies...

Since APB is isosceles, APB must be acute, and therefore C isn't on AP.  Maybe somebody asked "What if C is is on line AB?"  I don't know for sure.

k I got it. Drop the perpendicular from A to side BP. Let it meet at point X. Let AX meet BC at point Y. Now, triangle BAX is similar to triangle BMP, because angle B is equal to itself (duh) and there are right angles at X and P by defenition. Now, since BP=2BX, BC=2BY. This means AX bisects BC. Now, forget all about the isosceles triangle. We have a parallelogram, and two lines (AX and PC) running through it. By symmetry, since AX bisects BC, PC bisects AD. Done.

Dude, this is where I'm supposed to say if you're right or wrong.  But to tell you the truth, I don't know.

[hide]Just kidding.  Right-o![/hide]

lol. um-don't put spoiler unless it's obvious. or unless you want to be weird. which I guess is possible. I don't know. Now i'm confusing myself. sigh

Why not? Spoiler can help people who doesn't want to see the answer yet, and I don't think that's a weird thing to do. My two yens on Spoiler.

I think Mystic Terminator was meaning don't put spoiler in unless its obvious that its actually there :) referring to mathfanatics post, where many people will miss his sneaky little end comment. Of course that was done on purpose..

ha! but I caught it! i'm proud of myself....

actually, i check everyone's post for spoiler...because i like to put little sneaky end messages myself....

[hide]good job, you checked. you're catching on....[/hide]

Well some people get's mad at people who does that. I know Syntax got mad at me once...
*sigh*


geometry
parallelogram
The Geometry of Numbers

Prove that if a lattice parallelogram contains at most three lattice points in addition to its vertices, then those are on one of the diagonals.

I'm having some trouble understanding this.

If it refers to proving that in a lattice parallelogram with at most three lattice points in its [i]interior[/i], these three points lie on a same diagonal, then the problem is completely false (but it has no sense, i.e. it can't be refering to this - even if it had asked to prove that they lie on both diagonals, the problem is still false). Please review the statement.


In my $ \text{\LaTeX}$ I have a URL which has "%" in it. How do I "uncomment" everything that comes after the "%" because I do not intend to have that part commented out.


1.Deduce the following generalized arithmetic mean - geomet-
ric mean inequality from the original one: If q1,..., qn are positive rational
numbers such that q1 + ...+ qn = 1, then

(a1^q1)*(a2^q2)*...*(an^qn)<=a1*q1+a2*q2+...+an*qn

for any positive real numbers a1,..., an.
Remark: Note that for q1 = ...= qn = 1/n the above inequality is the
original arithmetic mean - geometric mean inequality.

2.Now show that the condition that q1,..., qn are rational in prob-
lem 2 can be dropped, i.e., the inequality in problem 2 holds for arbitrary
positive real numbers q1,..., qn such that q1 +... + qn = 1.

Please help!![/hide][/code]


What is the common fraction equivalent of $ 6\frac{1}{4} \%$?


superior algebra
superior algebra unsolved

Can someone show using Mordell's Theorem about Elliptic Curves that:
x^3=y^2+2
Has only a single solution,
x=3,y=5

You mean that it has only that one solution in positive integers- there's also the trivial solution $(3,-5)$, and infinitely many rational solutions such as $(1.29,.383)$.

Just in case one wants to see the unique-factorisation proof:
http://www.mathlinks.ro/Forum/viewtopic.php?t=15408
http://www.mathlinks.ro/Forum/viewtopic.php?t=80170
http://www.mathlinks.ro/Forum/viewtopic.php?t=84491