inequalities
rearrangement inequality
inequalities unsolved

Let $ a,b ,c$ be positive real numbers. Prove the inequality
$ \frac{{a^2  \plus{} bc}}{{b \plus{} 3c}} \plus{} \frac{{b^2  \plus{} ca}}{{a \plus{} 3c}} \plus{} \frac{{c^2  \plus{} ab}}{{a \plus{} 3b}} \ge \frac{{a \plus{} b \plus{} c}}{2}$

See it here:
http://www.toanthpt.net/forums/showthread.php?t=11190

[quote="thanhak5"]Let $ a,b ,c$ be positive real numbers. Prove the inequality
$ \frac {{a^2 \plus{} bc}}{{b \plus{} 3c}} \plus{} \frac {{b^2 \plus{} ca}}{{a \plus{} 3c}} \plus{} \frac {{c^2 \plus{} ab}}{{a \plus{} 3b}} \ge \frac {{a \plus{} b \plus{} c}}{2}$[/quote]
After expanding we need to prove that:
$ \sum_{cyc}(6a^4\plus{}15a^3b\minus{}7a^3c\minus{}6a^2b^2\minus{}8a^2bc)\geq0,$ which is obvious.

[quote="arqady"][quote="thanhak5"]Let $ a,b ,c$ be positive real numbers. Prove the inequality
$ \frac {{a^2 \plus{} bc}}{{b \plus{} 3c}} \plus{} \frac {{b^2 \plus{} ca}}{{a \plus{} 3c}} \plus{} \frac {{c^2 \plus{} ab}}{{a \plus{} 3b}} \ge \frac {{a \plus{} b \plus{} c}}{2}$[/quote]
After expanding we need to prove that:
$ \sum_{cyc}(6a^4 \plus{} 15a^3b \minus{} 7a^3c \minus{} 6a^2b^2 \minus{} 8a^2bc)\geq0,$ which is obvious.[/quote]
How about this, arqady? :)
\[ \sum \frac {a^2 \plus{} bc}{b \plus{} kc} \ge \frac {2(a \plus{} b \plus{} c)}{k \plus{} 1}
\]
for all $ a,b,c,k \ge 0$.

[quote="onlylove_math"][quote="arqady"][quote="thanhak5"]Let $ a,b ,c$ be positive real numbers. Prove the inequality
$ \frac {{a^2 \plus{} bc}}{{b \plus{} 3c}} \plus{} \frac {{b^2 \plus{} ca}}{{a \plus{} 3c}} \plus{} \frac {{c^2 \plus{} ab}}{{a \plus{} 3b}} \ge \frac {{a \plus{} b \plus{} c}}{2}$[/quote]
After expanding we need to prove that:
$ \sum_{cyc}(6a^4 \plus{} 15a^3b \minus{} 7a^3c \minus{} 6a^2b^2 \minus{} 8a^2bc)\geq0,$ which is obvious.[/quote]
Why is it obvious , arqady? Can U detail it ,plz? :lol:[/quote]
For you, onlylove_math!  :lol:
$ \sum_{cyc}(6a^4 \plus{} 15a^3b \minus{} 7a^3c \minus{} 6a^2b^2 \minus{} 8a^2bc) \equal{} 2\sum_{cyc}(b^4 \plus{} 2a^3b \minus{} 3a^2b^2) \plus{}$
$ \plus{} 3.5\sum_{cyc}(a^4 \plus{} a^3b \minus{} 2ab^3) \plus{} 0.5\sum_{cyc}(a^4 \minus{} a^3b) \plus{} 8\sum_{cyc}(a^3b \minus{} a^2bc)\geq0$ because
by AM-GM $ b^4 \plus{} 2a^3b \equal{} b^4 \plus{} a^3b \plus{} a^3b\geq3a^2b^2;$
$ \sum_{cyc}(a^4 \plus{} a^3b \minus{} 2ab^3) \equal{} \frac {1}{2}\sum_{cyc}(x^2 \minus{} y^2 \plus{} xy \minus{} xz)^2\geq0;$
$ \sum_{cyc}(a^4 \minus{} a^3b)\geq0$  and $ \sum_{cyc}(a^3b \minus{} a^2bc)\geq0$ are  true by rearrangement inequality for example.
[quote="can_hang2007"]
How about this, arqady? :)
\[ \sum \frac {a^2 \plus{} bc}{b \plus{} kc} \ge \frac {2(a \plus{} b \plus{} c)}{k \plus{} 1}
\]
for all $ a,b,c,k \ge 0$.[/quote]
It's true, but my proof is very ugly: let $ a\equal{}\min\{a,b,c\},$ $ b\equal{}a\plus{}x,$ $ c\equal{}a\plus{}y$ $ ...$
Maybe we need to prove your inequality for $ a\equal{}0,$  which is obvious too.
Is there nice proof, Can?

[quote="arqady"][/quote][quote="onlylove_math"][quote="arqady"][quote="thanhak5"]Let $ a,b ,c$ be positive real numbers. Prove the inequality
$ \frac {{a^2 + bc}}{{b + 3c}} + \frac {{b^2 + ca}}{{a + 3c}} + \frac {{c^2 + ab}}{{a + 3b}} \ge \frac {{a + b + c}}{2}$[/quote]
After expanding we need to prove that:
$ \sum_{cyc}(6a^4 + 15a^3b - 7a^3c - 6a^2b^2 - 8a^2bc)\geq0,$ which is obvious.[/quote]
Why is it obvious , arqady? Can U detail it ,plz? :lol:[/quote]
For you, onlylove_math!  :lol:
$ \sum_{cyc}(6a^4 + 15a^3b - 7a^3c - 6a^2b^2 - 8a^2bc) = 2\sum_{cyc}(b^4 + 2a^3b - 3a^2b^2) +$
$ + 3.5\sum_{cyc}(a^4 + a^3b - 2ab^3) + 0.5\sum_{cyc}(a^4 - a^3b) + 8\sum_{cyc}(a^3b - a^2bc)\geq0$ because
by AM-GM $ b^4 + 2a^3b = b^4 + a^3b + a^3b\geq3a^2b^2;$
$ \sum_{cyc}(a^4 + a^3b - 2ab^3) = \frac {1}{2}\sum_{cyc}(x^2 - y^2 + xy - xz)^2\geq0;$
$ \sum_{cyc}(a^4 - a^3b)\geq0$  and $ \sum_{cyc}(a^3b - a^2bc)\geq0$ are  true by rearrangement inequality for example.
[quote="can_hang2007"]
How about this, arqady? :)
\[ \sum \frac {a^2 + bc}{b + kc} \ge \frac {2(a + b + c)}{k + 1}
\]
for all $ a,b,c,k \ge 0$.[/quote][quote="arqady"]
It's true, but my proof is very ugly: let $ a = \min\{a,b,c\},$ $ b = a + x,$ $ c = a + y$ $ ...$
Maybe we need to prove your inequality for $ a = 0,$  which is obvious too.
Is there nice proof, Can?[/quote]
My solution used Cauchy Schwarz Inequality. :)
Here is it
By Cauchy Schwarz Inequality, we have
\[ \left( \sum \frac{a^2+bc}{b+kc}\right)\left( \sum (a^2+bc)(b+kc)\right) \ge \left( \sum a^2+\sum ab\right)^2\]
It suffices to prove that
\[ (k+1) \left( \sum a^2+\sum ab\right)^2 \ge 2\left( \sum a\right)\left( \sum (a^2+bc)(b+kc)\right)\]
To prove this inequality, it suffices to prove that the following 2 inequality holds
\[ \left( \sum a^2+\sum ab\right)^2 \ge 4\left( \sum a\right) \left( \sum a^2b\right)\] (1)
\[ \left( \sum a^2+\sum ab\right)^2 \ge 4\left( \sum a \right) \left( \sum ab^2\right)\] (2)
Then the above inequality will be proved (just adding (1) with $ k$ times (2)).
And to prove (1) and (2), it suffices to prove (1), which is easy by Vornicu Schur. :)

[quote="can_hang2007"]\[ \left( \sum a^2 \plus{} \sum ab\right)^2 \ge 4\left( \sum a \right) \left( \sum ab^2\right)
\]
 which is easy by Vornicu Schur. :)[/quote]

It is equivalent to

$ \sum a^4 \plus{}2\sum ab(a^2\minus{}b^2) \minus{} \sum a^2b^2 \geq 0$

How to write it in the form

$ \sum x(a\minus{}b)(a\minus{}c)$?

Thank you very much.

[quote="arqady"]
For you, onlylove_math!  :lol:
$ \sum_{cyc}(6a^4 \plus{} 15a^3b \minus{} 7a^3c \minus{} 6a^2b^2 \minus{} 8a^2bc) \equal{} 2\sum_{cyc}(b^4 \plus{} 2a^3b \minus{} 3a^2b^2) \plus{}$
$ \plus{} 3.5\sum_{cyc}(a^4 \plus{} a^3b \minus{} 2ab^3) \plus{} 0.5\sum_{cyc}(a^4 \minus{} a^3b) \plus{} 8\sum_{cyc}(a^3b \minus{} a^2bc)\geq0$ because
by AM-GM $ b^4 \plus{} 2a^3b \equal{} b^4 \plus{} a^3b \plus{} a^3b\geq3a^2b^2;$
$ \sum_{cyc}(a^4 \plus{} a^3b \minus{} 2ab^3) \equal{} \frac {1}{2}\sum_{cyc}(x^2 \minus{} y^2 \plus{} xy \minus{} xz)^2\geq0;$
$ \sum_{cyc}(a^4 \minus{} a^3b)\geq0$  and $ \sum_{cyc}(a^3b \minus{} a^2bc)\geq0$ are  true by rearrangement inequality [/quote]
In many your solutions, I saw that you alway use "expanding way". How do you do it? For example, like this problem. Are there some experiences ?

[quote="dulldrake"]
In many your solutions, I saw that you alway use "expanding way". How do you do it? For example, like this problem. Are there some experiences ?[/quote]
You mean, how I have obtained $ \sum_{cyc}(6a^4 \plus{} 15a^3b \minus{} 7a^3c \minus{} 6a^2b^2 \minus{} 8a^2bc)\geq0$ $ ?$
If so, I do it as:
$ \frac{{a^2 \plus{} bc}}{{b \plus{} 3c}} \plus{} \frac{{b^2 \plus{} ca}}{{c \plus{} 3a}} \plus{} \frac{{c^2 \plus{} ab}}{{a \plus{} 3b}} \ge \frac{{a \plus{} b \plus{} c}}{2}\Leftrightarrow$
$ \Leftrightarrow2\sum_{cyc}(a^2\plus{}bc)(3a^2\plus{}9ab\plus{}ac\plus{}3bc)\geq\sum_{cyc}a\sum_{cyc}\left(3a^2b\plus{}9a^2c\plus{}\frac{28}{3}abc\right)\Leftrightarrow$
$ \Leftrightarrow2\sum_{cyc}(3a^4\plus{}9a^3b\plus{}a^3c\plus{}3a^2bc\plus{}3a^2bc\plus{}9a^2bc\plus{}a^2bc\plus{}3a^2b^2)\geq$
$ \geq\sum_{cyc}(3a^3b\plus{}3a^2b^2\plus{}3a^2bc\plus{}9a^3c\plus{}9a^2b^2\plus{}9a^2bc\plus{}28a^2bc)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(6a^4\plus{}15a^3b\minus{}7a^3c\minus{}6a^2b^2\minus{}8a^2bc)\geq0.$
It takes 5 minutes.  :D  :)

I mean how you know exactly the way to express
$ \sum_{cyc}(6a^4 \plus{} 15a^3b \minus{} 7a^3c \minus{} 6a^2b^2 \minus{} 8a^2bc) \equal{} 2\sum_{cyc}(b^4 \plus{} 2a^3b \minus{} 3a^2b^2) \plus{}\plus{} 3.5\sum_{cyc}(a^4 \plus{} a^3b \minus{} 2ab^3) \plus{} 0.5\sum_{cyc}(a^4 \minus{} a^3b) \plus{} 8\sum_{cyc}(a^3b \minus{} a^2bc)$
And from this prove $ \sum \geq0$
If you don’t want to explain I won’t inquire. Anyway, I’ll try to learn something from your solution.

[quote="nothing"]If $ x,y,z$ be positive numbers, then

$ \frac {x^2 \plus{} yz}{y \plus{} kz} \plus{} \frac {y^2 \plus{} zx}{z \plus{} kx} \plus{} \frac {z^2 \plus{} xy}{x \plus{} ky} \geq \frac {2}{1 \plus{} k}(x \plus{} y \plus{} z)$

holds for all $ k\geq0$.[/quote]Without lost the general, we may assume $ x \equal{} {\rm mid} \{x,y,z\}$.

Using the arithmetic mean - geometric mean inequality,

$ (1 \plus{} k)^2x^2\sum_{cyc}{\frac {x^2 \plus{} yz}{y \plus{} kz}} \plus{} \sum_{cyc}{(x^2 \plus{} yz)(y \plus{} kz)}$

$ \geq2(1 \plus{} k)x\sum_{cyc}{(x^2 \plus{} yz)},$

Thus, it remains to prove that

$ 2(1 \plus{} k)x\sum_{cyc}{(x^2 \plus{} yz)}\geq 2(1 \plus{} k)x^2\sum_{cyc}{x} \plus{} \sum_{cyc}{(x^2 \plus{} yz)(y \plus{} kz)}.$
[hide="Proof."]$ 2(1 \plus{} k)x\sum_{cyc}{(x^2 \plus{} yz)}\minus{}2(1 \plus{} k)x^2\sum_{cyc}{x} \minus{} \sum_{cyc}{(x^2 \plus{} yz)(y \plus{} kz)}$

$ \equal{} \minus{} 2(x \minus{} y)(x \minus{} z)(y \plus{} kz) \geq 0.$[/hide]

yes,can_hang,the more general ineq is true.your proof is very nice but i find a nicer proof :) can you find it?


[quote="SplashD"][quote="Allan Z"]Let $ f(x)$ be a $ 98$ degrees polynomial and $ f(x)=\frac1x$ for $ x=1,2,3,...,99$
Find $ f(100)$[/quote]

[hide]Let $ g(x)=x\cdot f(x)-1$. $ g$ is 99 degrees and has 99 roots so $ g(x)=c(x-1)(x-2)\cdots (x-99)$. Letting $ x=0$, we see that $ c=\frac{1}{99!}$ Thus $ 100\cdot f(100)-1=g(100)=1\Rightarrow f(100)=\frac{1}{50}$[/hide][/quote]

Wait. What?

f(1) is clearly not 0 times a constant. Neither is f(2), f(3), nor f(99), for that matter, and you're talking about a 99 degree polynomial, not a 98 degree one.

I'm baked or something - misread what you were saying. Careless mistakes + overload from MOP = lose.


Prove that$ \sum_{cyclc}{\frac{a^{2}}{(p-b)(p-c)}}\leq\frac{6R}{r} $where a,b,c  are sides of a triangle.

No, $p$ is a half-perimeter, i.e. $\frac{a+b+c}{2}$.

I think this should help

Hint
[hide]If $p=\frac{a+b+c}{2}$ then the area of $\triangle ABC=rp$

\[[ABC]=\sqrt{p(p-a)(p-a)(p-b)}\]

Hence, $r=\frac{\sqrt{p(p-a)(p-b)(p-c)}}{p}$

Also $[ABC]=\frac{abc}{4R}$

Thus, $R=\frac{abc}{4[ABC]}$

$R=\frac{abc}{4(\sqrt{p(p-a)(p-a)(p-b)})}$[/hide]

After some simplification, the inequality is equivalent to:

$\sum_{sym} a^2(p-a)\leq \frac {3}{2}abc$

But since we have

$\sum_{sym} a^2b-\sum_{sym} a^3-3abc\leq 0$ by Schur,

we get

$\sum_{sym} a^2(p-a) = \frac {1}{2}\left(\sum_{sym} a^2b - \sum_{sym} a^3\right) \leq \frac {1}{2}\cdot (3 abc)=\frac {3}{2}abc$

as desired.


If a central angle is very small, there is little difference in length between an arc and the inscribed chord.  Approximate each of the following lengths by finding the necessary arc, the arc is said to [b]subtend[/b] the angle.

The mast of Brent Simon's boat is 32 feet hight.  If it subtends an angle of 2 degrees and 10 minutes, how far away is it?

I have no clue on how to work this trig word problem.  Will someone please help?

so the question is: what is the radius R of a circle if an  angle of 2°10' gives an arc of 32 feet.

[hide="hint"]
360°  => 2*Pi*R

1    °  => 2*Pi*R/360

a    °  => 2*Pi*R/360*a
 you get it?
[/hide]

The mast is bent in the shape of an arc?

No, it's straight, but it approximates a curve because the angle is so small.

At least that's what I think.


As specified in a title. I have a pair (given by some function). Can we get the $ x$ coordinate for that pair?

If your pair is named z, you can type z.x or xpart(z) for its x-coordinate.


[code]Greetings,

When asked to assemble a team of mathematically talented high students to compete in the 2009 National ARML Competition, you guys were the first on my mind. I know that you have grown mathematically and enjoy the company of highly capable young people. Certainly, the ARML Competition brings together the best math students in the country and is an experience that one never forgets.

I have attached your invitation and the application (in Word) for the 2009 Team. Please feel free to download the application and share with another mathematically talented student at your school.

Please let me know as soon as possible if you or a friend would like to join us. My email address is [color=red]pottergene  @  msn.com[/color]. ) ([color=red]take out spaces[/color])

Best regards,
Gene Potter
Secretary-Treasurer
MO-ARML
***-***-**** ([color=red]phone number blanked out[/color])[/code]
PM me your email if you would like this email forwarded. MO ARML has sent two teams and a few alternates, we go to Iowa by bus on Thurs Mar 28th, stay two nights, and leave after the competition on Sat Mar 30th. "The total cost of the trip is 200 dollars per student". It's fun and you meet many interesting people :) .

http://math.missouristate.edu/MissouriARML.htm


Let $ ABCDEF$ be a hexagon with parallel opposite sides, and let $ FC\cap{AB} \equal{}
X_1$, $ FC\cap{ED} \equal{} X_2$, $ AD\cap{EF} \equal{} Y_1$, $ AD\cap{BC} \equal{} Y_2$, $ BE\cap{CD} \equal{} Z_1$, $ BE\cap{AF} \equal{} Z_2$.
Prove that if $ X_1$, $ Y_1$, $ Z_1$ are collinear then $ X_2$, $ Y_2$,$ Z_2$ are also collinear and in this case
the lines $ X_1Y_1Z_1$ and $ X_2Y_2Z_2$ are parallel.

[img]http://i227.photobucket.com/albums/dd72/TPFC/hexagon.png[/img]
Call $ I\equal{}AD\cap CF$ we have
$ \frac{IX_1}{IX_2}\equal{}\frac{IA}{ID}\equal{}\frac{IF}{IC}\equal{}\frac{IY_1}{IY_2}$
Then $ X_1Y_1//X_2Y_2$
Similarly we have solution!

what program do you use for making those figures(sorry for the 'outoftopyc' question :D )


Assume we have 95 boxes and 19 balls distributed in these boxes in an arbitrary manner. We take 6 new balls at a time and place them in 6 of the boxes, one ball in each of the six. Can we, by repeating this process a suitable number of times, achieve a situation in which each of the 95 boxes contains an equal number of balls?

yes. let say you want to add one ball to box 1. then do the following: 1-6, 7-12, 13-18, 19-24, 25-30, 31-36, 37-42, 43-48, 49-54, 55-60, 61-66, 67-72, 73-78, 79-84, 85-90, 90-95&1. So every other box gains 1 ball except for box 1 which gained 2 balls, so relatively, the deficit of box 1 has been cut by 1. So you can cut the deficit of all the boxes with relation with the box with the most balls to 1.


Gauss
linear algebra
matrix
algebra
system of equations

I just did it which took about 20 minutes, but somewhere i made a mistake and got a wrong answer :wallbash_red:

[quote="lotrgreengrapes7926"][quote="jan4iceskate"][hide=" answer"]I got.. a=-2
         b=3
         c=6.[/hide]
idk if it's right i'm pretty sure it is.[/quote]
That's correct. Please post your solution. If you're not sure about an answer, just plug the values back into the original question.[/quote]

Yes. Two ways could be replacing letters, or just Gauss.

[quote="José"][quote="daermon"]what do you mean by gauss?  :huh:[/quote]

It's a method to solve systems with 3 unknowns and 3 equations.

Another could be the determinants method.[/quote]

You mean using matrices?

I thought it was a 3 by 4 augmented matrix? I learned it this year.

$ax+by+cz=d$
$a_{1}x+b_{1}y+c_{1}z=d_{1}$
$a_{2}x+b_{2}y+c_{2}z=d_{2}$

Then,

$\delta$=
$a-b-c$
$a_{1}-b_{1}-c_{1}$
$a_{2}-b_{2}-c_{2}$
$a-b-c$
$a_{1}-b_{2}-c_{2}$,

a $5*3$ matrix. If anyone has a link of how this method work, post it, because I can't explain it now because I have to go. If not, I'll explain it when I return.

You'll find sth here: http://en.wikipedia.org/wiki/Determinants#Determinants_of_2-by-2_matrices

but I can't find a link for a $3*3$ matrix

EDIT: http://alumnus.caltech.edu/~chamness/equation/equation.html

[quote="SplashD"]Don't you just use cramer for a 2 by 2? actually now that i though about it, isn't there a 3 by 3 cramer too?[/quote]

Yes! I found it: http://mathworld.wolfram.com/CramersRule.html

Please remember that this is a middle school forum and a lot of people that use it do not understand matrices and determinants yet. :)

thats in AoPS vol. 2, Right?  I'm still on chapter 6 of volume 1  :(

that is in standard alg 1 right? That's the course I took when I learned about it.

[quote="mtms5467"][hide="answer"]$(a,b,c)=(-2,3,6)$[/hide][/quote]

Please, remember you're in a middle-school forum, so, next time you post, remember to post your solution too, so little users can see how to solve it.

Can someone explain it without using matrices?

Express $a$ in the 1º equation in terms of $b$ and $c$, by passing the other terms to the other side. Then, replace this in the 2º and 3º one. Then, you have a system with 2 unknowns and two equations.

that makes sense.  I'll try using that  :lol:

You could also do elimination - eliminate one of the variables using 2 equations, then eliminate the same variable using 2 different equations, and then you have a system of equations that you know how to solve. After that, you just substitute to find the third variable.

Basically use any method of your choice to get two equations with two variables. Then once you have these two variables, plug it back in to find the third variable, and you have the solutions.


Find the limit of this sequence of number:
        \[ u_n=\prod_{i=1}^n (1+\frac1{2^i})\]

It is not difficult to prove that the sequence converges, but are you sure that the limit can be computed?

Pierre.

I'm not sure. I've tried to compute this limit, but fail.

Have you tried to make a new sequence
bi=ai*(1-1/2 i ) and find the limit of that then return to the original sequence?

[quote="darkmaster"]Find the limit of this sequence of number: \[u_{n}=\prod_{i=1}^{n}\left(1+\frac1{2^{i}}\right)\] [/quote]
Maybe it should be $u_{n}=\prod_{i=1}^{n}\left(1+\frac1{2^{2^{i}}}\right)$. ;)

For this case obviosly, that \[\lim_{n\to \infty }u_{n}=\frac{1}{1-2^{-2^{1}}}=\frac{4}{3}.\]


email
induction
real analysis
real analysis unsolved

[quote="hexa"]1^2 + 2^2 + 3^2 +... + n^2 = [n(n+1)(2n+1)]/6

if need the proof, email me

sysohn8394@naver.com[/quote]

yes I need more proof

[quote="polskimisiek"]Well, mavbe try simply induction? :wink:[/quote]

I have another question related to this:

say that I simplify the equation above to f(n) = $ (2n^3 \plus{} 3n^2 \plus{} n / 6 )$ and $ g(n) \equal{} n^3$

I would like to find the value of n0 so that f(n) < c * g(n) for every integer n >= n0 and there exists a real constant c > 0 and an integer constant n0 >= 1

so,

 $ (2n^3 \plus{} 3n^2 \plus{} n / 6 ) <\equal{} n^3$
 $ (2n^3 \plus{} 3n^2 \plus{} n / 6 ) <\equal{} n^3/3 <\equal{} c * n^3$

\[ \frac{n(n\plus{}1)(2n\plus{}1)}{6}\leq cn^3\]
\[ c\geq \frac{1}{6}\left( 1\plus{}\frac{1}{n}\right) \left( 2\plus{}\frac{1}{n}\right)\]
Maximum value of the left side is $ 1$. So, if $ c\equal{}1$ it will be true for all $ n\in N$

[quote="Yuriy Solovyov"]
\[ \frac {n(n \plus{} 1)(2n \plus{} 1)}{6}\leq cn^3
\]

\[ c\geq \frac {1}{6}\left( 1 \plus{} \frac {1}{n}\right) \left( 2 \plus{} \frac {1}{n}\right)
\]
Maximum value of the left side is $ 1$. So, if $ c \equal{} 1$ it will be true for all $ n\in N$[/quote]

what if I want to find what $ n_o$ is here?? I am interested in the value of c in terms of $ n_o$


The positives numbers $ x,y,z$ satisfy the equality : $ 4abc\equal{}a\plus{}b\plus{}c\plus{}1$ . Prove that :

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

If $ a\plus{}b\plus{}c\plus{}1\le 4abc$ prove that :

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

[b]Here is my solution to the first one[/b]:
By [b]AM-GM[/b], we get: $ 4abc \equal{} a \plus{} b \plus{} c \plus{} 1 \ge 4 \sqrt [4] {abc}$, hence $ abc\ge 1$
Also, $ a \plus{} b \plus{} c \equal{} 4abc \minus{} 1\ge 3abc$, $ (*)$. Now it is so obvious that $ x^2 \plus{} y^2\ge 2xy$ (also by [b]AM-GM[/b]).
$ \Longrightarrow LHS\ge \frac {2bc}{a} \plus{} \frac {2ca}{b} \plus{} \frac {2ab}{c} \equal{} \frac {2}{abc}((ab)^2 \plus{} (bc)^2 \plus{} (ca)^2)\ge \frac {2}{3abc}(ab \plus{} bc \plus{} ca)^2$.
Therefore, we just need to prove that $ ab \plus{} bc \plus{} ca\ge 3abc$. $ (**)$
Indeed, since $ (ab \plus{} bc \plus{} ca)^2\ge 3abc(a \plus{} b \plus{} c)\ge 3abc.3abc \equal{} 9(abc)^2$ (due to $ (*)$), which implies $ (**)$. Then our proof is completed :lol:


Find  all nonzero  integers  $a, b, c, d$   with   $a>b>c>d$  that  satisfy  $ab+cd=34$ and $ac-bd=19.$

[quote="silouan"]There are also the negative solutions. $a=-1$   , $b=-4$  ,  $c=-5$  and  $d=-6$.[/quote]
If so, then  $ac-bd=-19$ ;)

tkhtn, in fact arqady's solution is mainly correct. It is just a bit messy.

We have $(a^2 + d^2)(b^2 + c^2)=(ac - bd)^2 + (ab + cd)^2 = 19^2 + 34^2 = 1517$

But $1517 = 37*41$, with 37 and 41 prime numbers.
Since $a$,$b$,$c$, and $d$ are non zero, $a^2 + d^2>1$ and $b^2+c^2>1$, so one of them is equal to 37 and the other to 41.
We can easily see that $37=6^2+1^2$ and $41=5^2+4^2$ are the only way to write $37$ and $41$ as sums of two squares, so the numbers $a$,$b$,$c$, and $d$ are $1$,$4$,$5$, and $6$ with some sign in some order.
Since $ab+cd=34$ we get (by testing a few cases), that {|a|,|b|}={1,4} and {|c|,|d|}={5,6}.
Then from $ac-bd=19$ we get |a|=1, |b|=4, |c|=5 and |d|=6, and then, by testing cases:
$a=1, b=4, c=-5, d=-6, or a=-1, b=-4, c=5, d=6.$

Yes, thank you.This problem 's not quiet hardly.

Since a>b>c>d,we must have a=6,b=5,c=4 and lastly d=1.


Let $a,b,c\in\mathbb{R}^+$ and $abc=1$

Prove that:

$3+a^2b+b^2c+c^2a\geq a+b+c+ab+bc+ca$

You could basically subsitute $a=x/y$, etc to reduce it to Schur.

Or, apply Schur to the numbers $a^\frac{2}{3}b^\frac{1}{3}, b^\frac{2}{3}c^\frac{1}{3}, c^\frac{2}{3}a^\frac{1}{3}$ and use $abc=1$ and you'll get the result.

[edit] This was actually the problem that inspired the following thread:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16855

Try using the result to find another proof of the reciprocal inequality  ;) .

I have a proof using the method of Schur.

There are only two important cases (e.g, all others can be handled by rearrangement)

$c\geq 1\geq b\geq a$ and $c\geq b\geq 1\geq a$

Then, its equivalent to

$cb(1-b)(a-1)+ab(1-a)(c-1)+ac(1-c)(b-1)\geq 0$, which is done via some case work.

And, of course, $a=\frac {x}{y}$ and so on shows that it IS Schur.


analytic geometry
graphing lines
slope
function
calculus
calculus computations

For the next two questions, express the slope of the secant line of each function in terms of x and h. Be sure to simplify your answer.

(a) f(x) = -3x + 2

(b) f(x) = (1/x)

HINT GIVEN:

The slope of the secant line having two the points
(x, f(x)) and (x + h, f(x + h)) on the graph of the function may be given as f(x + h) - f(x)/h


Let $ a_n,\ a_{n \plus{} 1}$ be the roots of the quadratic equation : $ x^2 \plus{} 3nx \plus{} C_n \equal{} 0$.
For $ a_1 \equal{} 1$, find $ \sum_{n \equal{} 1}^{2p} C_n$.

presumably the answer should be expressed in terms of $ a_1$?

Sorry, I forgot to write $ a_1\equal{}1$.


geometry
3D geometry
AMC
USA(J)MO
USAMO
number theory
\/closed

I would like to request a math jam on selected problems in elementary number theory. Discuss what kind of difficulty you all would want. My vote is for national olympiad level (not IMO).

Start with this problem from a SF Bay Area competition:

Find the second smallest positive integer whose cube ends in 888.

It's not national olympiad level, but it's still a fun one.

I agree with Mindspa.  A Math Jam with USAMO level questions, on elementery number theory or otherwise would be great.

[quote]Find the second smallest positive integer whose cube ends in 888. [/quote]

Or the first smallest, for that matter, as that is #9 on the 1988 AIME.

Fierytycoon

We'll mix in some Olympiad level problems.  Probably the next one we schedule (after the two already scheduled).  I'll see if we can find a nice number theory question to include.

yes, a math jam with olympiad level problems would be greatly appreciated.


Hi, everyone I want to create a forum specially for the Indian Olympiad enthusiasts. So I am asking for some suggestions. If anyone is interested than you can contact me at info@manjil.co.cc

i suppose there is a forum [url]http://joigroup.freezoka.com/[/url]

but sadly that forum has died :(
why not make that active???

[quote]i suppose there is a forum http://joigroup.freezoka.com/

but sadly that forum has died Sad
why not make that active???[/quote]

well yes...but that is not for INMO alone....I meant to make a purely INMO forum!!!!

cool do it but dont include people like me who are spamming all the time (this one was also)


modular arithmetic
combinatorics proposed
combinatorics

To every vertex of a regular pentagon a real number is assigned. We may perform the following operation repeatedly: we choose two adjacent vertices of the pentagon and replace each of the two numbers assigned to these vertices by their arithmetic mean. Is it always possible to obtain the position in which all five numbers are zeroes, given that in the initial position the sum of all five numbers is equal to zero?

Probably not. It would be enough to prove it for rationals, since then we can use representations in a Hamel basis.

But for a triangle, instead of a pentagon, the configuration $2,-1,-1$ leads uniquely to $1/2,1/2,-1$, then $1/2,-1/4,-1/4$, which is the homothetic by $1/4$ of the initial one, so the process never ends. One would need to find such a "repeating" model for the pentagon, in order to infirm the thesis.

If we consider the pentagon with $\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7},-(\sqrt{2}+\sqrt{3}+\sqrt{5}+\sqrt{7})$. Supose that after $n$ steps each vertex have a $0$. Then $0=a_1\sqrt{2}+a_2\sqrt{3}+a_3\sqrt{5}+a_4\sqrt{7}-a_5(\sqrt{2}+\sqrt{3}+\sqrt{5}+\sqrt{7})$ where $a_1+a_2+a_3+a_4+a_5=1$. In particular, each $a_i=1/2^{(b_i)}$ for a natural integer $b_i$. Then for some $k$ big enough,  $2^ka^i$ is an integer for every i. Let $c_i=2^k(a_i-a_5)$ for $i=1,2,3,4$. Then $0=c_1\sqrt{2}+c_2\sqrt{3}+c_3\sqrt{5}+c_4\sqrt{7}$ for integers $c_i$. Then it is obvious that $c_i$ must be $0$. (just square a couple of times) But then $a_i-a_5=0$, so $a_i=a_5$ so $1=a_1+a_2+a_3+a_4+a_5=5a_5$. So $a_5=1/2^{(b_5)}=1/5$ and that is a contradiction.

Let $S$ be the set of rational numbers that can be expressed as $\displaystyle \frac{a}{2^b} - \frac{1}{5}$ with $a$ an integer and $b$ a non-negative integer.  Then $\displaystyle \frac{-1}{5} \in S$, $\displaystyle \frac{4}{5} \in S$, $\displaystyle 0 \notin S$.  Also note that $\displaystyle x, y \in S \Rightarrow \frac{x + y}{2} \in S$, which means if all the verticies of the pentagon start out with an element of $S$, they will forever have an element of $S$.  So if we start by assigning four verticies the value $\displaystyle \frac{-1}{5}$ and one $\displaystyle \frac{4}{5}$, the value zero will never appear.

let us define $\frac{a}{b} \pmod{5}$ as $ab^{-1}$
then we place five numbers $1,1,1,1,-4$
then after any number of operations,the five numbers will be all congruent to $1$ mod $5$,contradiction.

If we start with any pentagon that has 3 or 5 non-zero numbers in it, you'll never get 5 zeroes.
The reason for this is, that the only way to add zeroes replaces 2 non-zero numbers, $x$ and $-x$, with 2 zeroes.
Therefore, the parity of the amount of zeroes never changes, and as it started even, it can never become 5, so it's impossible in those situations.
Note: This means that even if you could choose any 2 numbers of the 5, ignoring adjacency, you'd get the same result.


limit
Euler
logarithms
number theory unsolved
number theory

Sorry, maybe it is well known, but I am interested in

$\lim_{n \to \infty}\sum_{i=1}^{n}\frac{1}{p_{i}}$

where $p_{i}$ are prime numbers.

In fact, it diverges.  Euler showed that

$\sum_{i=1}^{n}\frac{1}{p_{i}}\sim \ln \ln n$


By Euler's pentagonal number theorem,
\begin{eqnarray*} \prod^{\infty}_{n = 1} \left( 1 - \frac{1}{2^n} \right) & = & \sum_{k \in \mathbb{Z}} ( - 1 )^k \cdot \left( \frac{1}{2} \right)^{\frac{k \cdot ( 3 k + 1 )}{2}}\\ \end{eqnarray*}
Is this the form you are looking for ? Is there a nice closed form you know of ?

I don,t remember where I see this problem but I think the problem means numerical value.

That is what my computer gave me  :P
But it's not rational: every rational number has a periodic representation in any base, especially in base $2$. But by the series given by {x}, it can't be periodic (the distance of the numbers of type $\frac{k \cdot (3k+1)}{2}$ grows forever).


What is $ 43^{50}\bmod 101$?

[hide="hint"]Use Fermat's Little Theorem.[/hide]

[quote="erdogankerem123"][hide] We see that $ 43^{100}\equiv 1\bmod 101$, so since $ 43^{50}\equal{}\sqrt{43^{100}}$, we see that $ \boxed{43^{50}\equiv 1\bmod 101}$[/hide][/quote]

How do you know it is not $ \minus{}1\bmod 101$?

144 is perfect square, so 43 is quadratic residue mod 101,
$ 43\equiv a^{2}\pmod{101}$
$ 43^{50}\equiv a^{100}\equiv 1\pmod{101}$.[/b]

[quote="Icy"]Um...could someone elaborate on the post before mine? I'm not sure I get it...

Where does the 1 mod 101 come from at the end, after taking $ (a^{2})^{50}$?[/quote]

we know $ 43\equal{}\equal{}144\bmod 101\equal{}\equal{}12^{2}\bmod 101$

So $ 43^{5}0\equal{}\equal{}12^{1}00\bmod 101\equal{}\equal{}1\bmod 101$.

[quote="Icy"]Um...could someone elaborate on the post before mine? I'm not sure I get it...

Where does the 1 mod 101 come from at the end, after taking $ (a^{2})^{50}$?[/quote]

It's just Fermat's Little Theorem...

i think it's because $ 43\equiv 144\pmod{101}$ this is also $ \equiv\minus{}1\pmod{101}$ you are raising $ 43$ to the $ 50th$ power to you must also raise $ \minus{}1^{50}$ which obviously $ \equiv1\pmod{101}$

Ah, I get it. I've never seen the theorem before, but after looking on the AOPSwiki thing, it makes sense. Thanks.


The units digit of every power of $ 6$ is $ \boxed{6}$.

the answer is 6 because the

for every # times 6 the units will be 6

clear example

6*6=36*6=216 as you can see the # always end in 6 so this # will end in 6 as well


Eve and Oddie play the following game. One of them flips 2001 coins and Eve wins if the number of heads is even, and Oddie wins if the number of heads is odd. What is the probability that Eve will win the game?

[hide]By symmetry, $ \frac{1}{2}$.[/hide]

Think about it like this.

The probability you get 0 heads is the same as 0 tails, i.e. 2001 heads.

1 head and 2000 heads.

Every case where the number of heads is even has a corresponding equally likely case where the number of heads is odd.

[quote="bigman"]How can we use symmetry if we 2001 flips and not 2000. Explain[/quote]

2000 would just be... annoying; we can't use symmetry for evens in this case because $ 0$ heads (or, equivalently, $ 0$ tails) is a case.

Oh. I think I got it. Thanks. Just out of curiosity, what would 2000 be?

Also if we had 2009 flips, the probability would still be 1/2 right?

As long as the number of flips is odd, it's $ \frac12$. As Temperal said, $ 2000$ has a lot of ugly casework, and I don't think there's a nice solution.


integration
function
real analysis
calculus
real analysis unsolved

:lol:
Let $ a,b \in R$ and $ f(x)$ is continuous function.Find all $ h(x)$ and $ g(x)$ such that :
$ 1) \int_{a}^{b}g(x)dx=\int_{a}^{b}f(x)dx$
$ 2)\int_{a}^{b}f(g(x))dx=\int_{a}^{b}f(h(x))dx$

I don't know what you expect as an answer. Pretty much any $ g$ satisfies (1) after rescaling. And then, pretty much any function $ h$ can be modified to satisfy (2).

I know a root.
a=0 ,$ b=\pi/2$, $ h(x)=cosx,g(x)=sinx$.Maybe this is not probl which have not solution.

You didn't ask that question- you asked something that didn't really make sense.

As near as I can tell, the question you mean is this one:
Find (some or all) pairs of functions $ g,h$ defined on an interval $ [a,b]$, such that for any reasonably nice function $ f$, $ \int_{a}^{b}f(g(x))\,dx=\int_{a}^{b}f(h(x))\,dx$

It is a fairly large class of answers; they come from measure-preserving transformations of $ [a,b]$. For any reasonable $ g$ and any measure-preserving $ \phi$ mapping $ [a,b]$ (possibly excluding a measure zero set) to itself, $ h=g\circ \phi$ works.
That's the Lebesgue version; if we're using Riemann integrals, $ \phi$ can't be quite as wild.

Yes, i am wrong.it must be as you said.But Lebesgue version,what is it?

Look up Lebesgue measure and the Lebesgue integral. It's an essential part of analysis.


Ahh keychain problems.

There are 6! ways, but we can flip, rotate, and reflect. Thus our answer is $ \frac{6!}{6 \times 2} \equal{} 60$.

Note that if it's a table, you can only rotate, so we would have 6!/6=5!=120.

Okay.  So we have $ \binom{15}{11}\equal{}1365$ to pick the lineup, and $ \binom{11}{2}\equal{}55$ ways to pick the captains from the lineup.  Multiplying yields $ 1365 \times 55\equal{}\boxed{75075}$.

Make sure you understand why we multiplied. These events depend on each other: 14ABA is different from 14BAB(where numbers stand for captains and the letters for the line up.) For independent events, we add.

Alternatively, there are $ \binom{15}{2}\equal{}105$ ways to pick captains and then $ \binom{13}{9}\equal{}\frac{13\cdot12\cdot11\cdot10}{24}\equal{}{13\cdot11\cdot5}\equal{}65\cdot11\equal{}650\plus{}65\equal{}715$ so $ 105\cdot715\equal{}75075$


function
limit
algebra
polynomial
calculus
calculus computations

For example, $ 0^0, \frac {0}{0}, 1^\infty$. Some of these forms look like they can be determined by why are they said to be undefined? Wouldn't $ 1^\infty \equal{} 1$? Or $ 0^0 \equal{} 1$? Also, why is $ 0(\infty) \not\equal{} 0$?

"Indeterminate form" has an extremely specific meaning in calculus.  What it means is this:  taking the example of $ \frac {0}{0}$, knowing that two functions $ f(x), g(x)$ satisfy $ \lim_{x \to r} f(x) \equal{} \lim_{x \to r} g(x) \equal{} 0$ gives you no information about the value of $ \lim_{x \to r} \frac {f(x)}{g(x)}$; one must instead apply l'Hospital's rule.

Showing that these forms are, in fact, indeterminate in the above sense is a matter of providing two examples that converge to different limits, and is an exercise I'll leave to you if you want.  For example, $ 1^{\infty}$ is indeterminate because, as $ x$ approaches $ \infty$, $ f(x) \equal{} 1 \plus{} \frac {r}{x}$ converges to $ 1$ for any fixed real $ r$ and $ g(x) \equal{} x$ converges to $ \infty$ but

$ \lim_{x \to \infty} f(x)^{g(x)} \equal{} e^r$

which can take on any positive value.

Note that this is specific to calculus.  In other contexts, it's perfectly reasonable to define, for example, $ 0^0 \equal{} 1$, especially if you're working with polynomials or with combinatorial objects.

Ok thanks, I'm glad I finally got that cleared up. I see why you say that it's reasonable to say that $ 0^0 \equal{} 1$ when referring to polynomials. For example: $ f(x) \equal{} ax^0$. Obviously that function is a constant function, and it makes sense to have $ f(0) \equal{} a$ and that will only be true if $ 0^0 \equal{} 1$.

Yep.  The combinatorial justification is some variant of the following argument:  for non-negative integers $ a, b$, the quantity $ a^b$ describes the number of functions from a set of size $ b$ to a set of size $ a$.  (This is also true of cardinals.)  The number of functions from the empty set to the empty set is $ 1$:  it's the identity function.  This is related to the justification for the [url=http://en.wikipedia.org/wiki/Empty_product]empty product[/url].

[quote="sunehra"]For example, $ 0^0, \frac {0}{0}, 1^\infty$. Some of these forms look like they can be determined by why are they said to be undefined? Wouldn't $ 1^\infty \equal{} 1$? Or $ 0^0 \equal{} 1$? Also, why is $ 0(\infty) \not \equal{} 0$?[/quote]
Consider this:
$ \lim_{a \to 1} a^{\infty}.$

Now,
$ \lim_{a \to 1^ \plus{} } a^ {\infty} \equal{} \infty$

But $ \lim_{a \to 1^ \minus{} } a^ {\infty} \equal{} 0$

Hence, $ \lim_{a \to 1} a^{\infty}$ has no definite value, being discontinuous.

However if a problem involves $ \lim _{a \to \infty} 1 ^a$, it is not indeterminate and is equal to $ \infty$.

[quote="Rajiv"]if a problem involves $ \lim _{a \to \infty} 1 ^a$, it is not indeterminate and is equal to $ \infty$.[/quote]
That's not the issue at stake; the issue at stake is, roughly, computing $ \lim_{b \to 1} \lim_{a \to \infty} b^a$, which doesn't exist.


geometry
exterior angle
algebra
linear equation

Having a tough go at this question.

Anyone care to take a whack at it?

An external angle of a triangle is equal to the sum of the two non-adiacent angles of the triangle.
In this case: $3x + 50 = 5x + 2x$.

Andreas...Thank you!

But...that was as far as I got.

I guess Im confused by X ...since 2x and 5x are in the triangle
and the exterior angle is 3x + 50

I'm finding it difficult to see what X can be since it would be the same all around.

2x + 5x equalling 3x + 50 (since 2 is less than 3 finding X that would equal the amount + 50 is what is confusing)

Or am I overthinking it?

:)

Like in the example below.

I know that the answer is x=6

5x + 6x = 4x + 42

5 x 6 = 30
6 x 6 = 36
---------------
            66

4 x 6 + 42 = 66 (which is equal to the sum of the two non-adiacent angles)

[quote="Tyler Ann"]I'm not looking for the answer...I'm looking for a similiar example with an explaination...

please, someone..save me from the madness :)[/quote]

What explanation do you need, since it's a straight line, you can do $180-(4x+42)+11x=180$

[quote="Tyler Ann"]I'm finding it difficult to see what X can be since it would be the same all around.

2x + 5x equalling 3x + 50 (since 2 is less than 3 finding X that would equal the amount + 50 is what is confusing)[/quote]

What? I don't really understand the problem.

$2x+5x=3x+50$ is a linear equation!
$7x = 3x + 50$
$7x - 3x = 3x - 3x + 50$
$4x = 50$
$x = 12.5^\circ$

Yes, I really don't think this should be in this forum ;)

Yes, I think you are overthinking it.  Solve the equation you made for $x$.  Then plug your value for $x$ into each of the three expressions for the angle measures.

perhaps you should go ask your teacher, also this uses a simple geometry

the angles in a triangle added together equal 180 degrees AND if 2 angles are supplimentary, they add up to 180 degrees


[quote="Quevvy"]In a sequence of 6 positive integers, each term after the first is determined by multiplying the preceding term by 2. The mean of the 6 terms is $94.5$. What is the median of the 6 terms?[/quote]
[hide]So make the first term x. So all the numbers can me represented as x,2x,4x,8x,16x,and 32x. Since the mean is 94.5 the equation would be $\frac{63x}{6}=94.5$. Solving for x we get that x equals 9. Now we want the median of the set. So the median is the mean of 72 and 36 or $\boxed{54}$.[/hide]


[quote="redcomet46"]http://www.addictinggames.com/supuzzle.html?r=user_posted_link[/quote]     its impossible

This puzzle doesn't even know it has the word "electricity" spelled wrong in two
places -- at the top and at the bottom.  For a puzzle that has to do with utilities, that's pathetic, in my opinion.

Other than that, I think it's "fun" and educational.

It has a solution, but not without "cheating".

[quote="box of rocks"]Padmavathi,

Ihatepie ALREADY posted the correct solution.

What WAS it about THAT poster's reply that you were not
satisfied with that you had to go and seek an answer from another
user, from whom you don't know either?[/quote]

He never proved it or anything... He just said that he couldn't do it.

No, "Ihatepie" gave the correct solution, grammar notwithstanding.

And giving the "solution" can mean giving the answer to the puzzle in this case.

He didn't state "he couldn't do it."

He stated that "its [should be it's] impossible."  That is the correct solution.

Note how your reply put words in that he never stated.

Anyway, his solution means it is impossible regardless of who attempts to do it.

Now, additional typing and/or graphics in a posting showing how it is so

would back up his reply, but would be extra.

The how-to answer must be explicitly asked for, because the word "solution" here can be just a "yes" or "no" or "possible" or "it's impossible."

box of rocks - maybe padmavathi wanted the solution as to why it's impossible, which ihatepie did not supply. ihatepie gave the correct answer but had no solution

at padmavathi - it's graph theory, and had something to do with Euler's theorem. or something like that.

box of rocks is a troll and should be ignored.  This is a question concerning planar graphs, i.e. graphs which can be drawn in a plane so that no two edges intersect.  [url=http://www.personal.kent.edu/~rmuhamma/GraphTheory/MyGraphTheory/planarity.htm]This website[/url] has a nice proof that the graph in question (called $K_{3, 3}$) is not planar.  (It follows corollary 2.)  As sapphyre says, it is based on Euler's formula.

Here's another nice explanation.  :)

http://mathforum.org/dr.math/faq/faq.3utilities.html

A one-liner to the problem:

Jordan Curve Theorem

That is one line.  It is not a proof.  Further, the proof which relies on the Jordan Curve Theorem is at the link AiW gave.

Err...box of rocks, that was just plain weird...
rest of you guys-thanx! I hadn't learnt this before and I'm getting a chance to learn it now so thank u all for the info! I thought it would have an annoyingly simple answer which I wouldn't have thought of  :D
And what about the figure I've attached?(Sorry for the shoddy drawing...)
I guess this is the same reason why you can't draw it without lifting your pencil or drawing on one line more than once, right?

No, JBL is a user who doesn't consistently recognize trolls when they're around.
This is a phrase stated by a user when he won't face up to another user; it's
essentially empty and has the stater walking away.  I back up what I say, and I don't need five stars to cowardly and inaccurately call another user a "troll" either.

sapphyre5571, no, that user *may* have meant that, but you users have to be specific,  There's no assuming solution versus answer.  That's why I'm catching errors more often in the statements of your puzzles.

Padmavathi, the correct word should be "er."  You used "err" which means to make a mistake.  So ironically you made a weird post regarding me.


geometry
geometric transformation
homothety
trigonometry
circumcircle
inradius
quadratics

Let X, Y, Z be the midpoints of the sides a = BC, b = CA, c = AB. Let K, L, M be the feet of the bisectors of the angles $\angle A, \angle B, \angle C.$ Let T, U, V be the tangency points of the incircle with the sides a, b, c and $T_a, U_b, V_c$ the tangency points of the excircles $(E_a), (E_b), (E_c)$ with thae sides a, b, c, respectively. WLOG, assume that b < c, hence $\angle B < \angle C.$ Invert the incircle (I), the 9-point circle (N), and the excircle $(E_a)$ in the circle (X) with radius $r_X = XT = XT_a = s - b - \frac a 2 = \frac{c - b}{2}.$ Both the incircle (I) and the excircle $(E_a)$ are perpendicular to the inversion circle, so that they are carried into themselves. The 9-point circle (N) passes through the inversion center X (the midpoint of a = BC), so that it is carried into a common internal tangent n of the incircle (I) and excircle $(E_a)$ different from the common internal tangent $BC \equiv TT_a.$ Since the foot K of the angle bisector AK is the internal homothety center of the incircle (I) and the excircle $(E_a)$, this other common internal tangent n passes through the point K and it forms the angle $\widehat{(a, n)} = 2 (90^\circ - \angle AKC) = \angle C - \angle B$ with the triangle side a = BC. The Feuerbach point F is carrried into the tangency point F' of the line n with the incircle (I) and the points F, F' are collinear with the inversion center X. By the basic properties of inversion, $XF = \frac{r_X^2}{XF'} = \frac{(c - b)^2}{4 XF'}.$ We can calculate XF' by the cosine theorem from the triangle $\triangle KXF'$ with the angle $\angle XKF' = \angle C - \angle B$ and the sides

$KX = BK - BX = \frac{ca}{b + c} - \frac a 2 = \frac{a(c - b)}{2(c + b)}$

$KF' = KT = BT - BK = s - b - \frac{ca}{b + c} = \frac{(s - a)(c - b)}{b + c}$

$XF'^2 = \frac 1 4 \left(\frac{c - b}{b + c}\right)^2 \left[a^2 + (b + c - a)^2 - 2a(b + c - a) \cos(C - B)\right]$

Substituting $\cos(C - B) = \cos A + 2 \cos B \cos C$ and using the cosine theorem formulas $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ etc.,

$\cos(C - B) = \frac{b^2 + c^2 - a^2}{2bc} + \frac{(c^2 + a^2 - b^2)(a^2 + b^2 - c^2)}{2a^2bc} =$

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

$XF'^2 = \frac 1 4 \left(\frac{c - b}{b + c}\right)^2 \left[a^2 + (b + c - a)^2 - (b + c - a)\ \frac{a^2(b^2 + c^2) - (c^2 - b^2)^2}{abc}\right]$

After some tedious calculation, this can be factored as

$XF'^2 = \frac{(c - b)^2 [abc - (b + c - a)(c + a - b)(a + b - c)]}{4abc}$

Let $R, r, s, \triangle$ be the triangle circumradius, inradius, semiperimeter and area. Substituting $\frac{abc}{4} = R \triangle$ and

$\frac 1 4 (b + c - a)(c + a - b)(a + b - c) = 2(s - a)(s - b)(s - c) = \frac{2 \triangle^2}{s} = 2r \triangle$

$XF'^2 = (c - b)^2\ \frac{(R - 2r) \triangle}{4R \triangle} = \left(\frac{c - b}{2}\right)^2 \frac{R - 2r}{R}$

$XF' = \frac{c - b}{2} \sqrt{\frac{R - 2r}{R}}$

$XF = \frac{(c - b)^2}{4 XF'} = \frac{c - b}{2} \sqrt{\frac{R}{R - 2r}}$

and by cyclic exchange,

$YF = \frac{a - c}{2} \sqrt{\frac{R}{R - 2r}},\ \ ZF = \frac{b - a}{2} \sqrt{\frac{R}{R - 2r}}$

It is clear that $XF + YF + ZF = 0$, with the understanding that one or two segments come out negative, depending on the triangle sides a, b, c.

[quote="epitomy01"]Let F be the point of tangency of the incircle with the nine-point circle. Prove that FX=FY+FZ, where FX>FY and FX>FZ, with {X,Y,Z} being the midpoints of the segments of the triangle ABC in some order.[/quote]

I have given a synthetic proof of this fact (not by myself actually) at http://www.mathlinks.ro/Forum/viewtopic.php?t=24959 .

[quote="epitomy01"]Is this true for any point on the incircle?[/quote]

Nope. The locus of points P in the plane such that one of the segments PX, PY, PZ is the sum of the two others is a quartic curve and doesn't contain the incircle as a quadratic factor (in general).

  darij

[quote="epitomy01"]Let F be the point of tangency of the incircle with the nine-point circle. Prove that FX=FY+FZ, where FX>FY and FX>FZ, with {X,Y,Z} being the midpoints of the segments of the triangle ABC in some order.
Is this true for any point on the incircle?[/quote]
This problem has a very short solution with complex numbers ;) .

$epitomy01$ at laste can I know your name?
I think I know you. :|


algebra
polynomial
calculus
integration
number theory proposed
number theory

Given are polynomial $f(x)$ with non-negative integral coefficients and positive integer $a.$ The sequence $\{a_{n}\}$ is defined by $a_{1}=a,$ $a_{n+1}=f(a_{n}).$ It is known that the set of primes dividing at least one of the terms of this sequence is finite.
Prove that $f(x)=cx^{k}$ for some non-negative integral $c$ and $k.$

[i]Proposed by F. Petrov[/i]

[hide="For those of you who liked this problem."]
Check [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=62259]this thread[/url] out.[/hide]

[quote="mathmanman"]Given are polynomial $f(x)$ with non-negative integral coefficients and positive integer $a.$[/quote]

Hmmm...

Pierre.

Solution. We shall use the fact that eventually all terms of the sequence $a_n$ become sufficiently large unless $f(X)=X$ in which case we are done anyways. We consider the two cases separately, when $f(0) \not= 0$ and when $f(0)=0$.

In the former case, now, consider an $m$ such that for all $n \ge m$, we have $a_n > c$ where $c$ is a constant we shall mention later. Suppose that the set of all primes dividing at least one term of this sequence is $\{p_1, \dots , p_k\}$ for some $k \ge 1$. We write each term $a_l$ as the product $\prod p_i^{v_{p_i}(a_l)}$. To each $l$ assign the index $j$ such that the largest power in the product for $a_l$ is $p_j^{v_{p_j}(a_l)}$. It is clear, by the pigeonhole principle, that among the terms $\{a_n, \dots, a_{n+k}\}$ some two terms have the same index assigned to them. Therefore, we have a prime $p_i$ and an exponent $e_i$ such that $p_i^{e_i} \mid a_s$ and $p_i^{e_i} \mid a_{r+s}$ for some $m<n \le s$ and $r \le k$. Now, consider the fact that $$a_{r+s}=f^{r+s}(a)=f^r(f^s(a))=f^r(a_s) \equiv f^r(0) \,(\bmod \, a_s)$$ yielding that $p_i^{e_i} \mid f^r(0)$ and since $r \le k$, the latter is bounded above by some constant, namely, the largest term $M$ in the sequence $\{f^t(0)\}_{1 \le t \le k}$. However, $p_i^{e_i} \ge \min (\sqrt[k]{a_s},\sqrt[k]{a_{r+s}})$. We can safely choose $c=M^k$ and be done with this case.

For the latter, we write $f(X)=X^bg(X)$ for some integer $b \ge 1$ and see that $f^{j+1}(X)=\left(f^j(X)\right)^b\cdot g(f^(X))$. Let $\mathcal{P}$ denote the set of all primes dividing at least one term of the sequence $(a_n)_{n \ge 1}$. From here on, $p \in \mathcal{P}$ shall be assumed. Notice that $$f^{j+1}(X)=f(f^j(X)) \equiv f(0)=0 \, (\bmod \, f^j(X))$$ for all $j \ge 0$. Choose $N$ sufficiently large and consider the term $a_{N+1}$ such that all primes $p \in \mathcal{P}$ divide $a_{N+1}=f^N(a)$; such an $N>0$ exists for obvious reasons. Let us consider $j>N$ and see that if $p \nmid g(0)$ then $v_p(g(f^j(0)))=0$ and if $p \mid g(0)$ then $v_p(f^{j+1}(0))>v_p(f^j(0))$ and so, for $j$ adequately large, $v_p(f^j(0))>v_p(g(0))$ and hence $v_p(g(f^j(0)))=v_p(g(0))$ is bounded above. Since all prime factors if $g$ come from $\mathcal{P}$ we see that $g(f^j(0))$ is bounded above. This fails to hold as $j \rightarrow \infty$ unless $g$ is the constant polynomial.

We conclude that the only such polynomials are $f(X)=cX^k$ for some integers $k,c \ge 0$ which clearly satisfy our conditions.


The lengths of two opposite sides of a square are decreased by $ 40\%$ while the lengths of the other two sides are increased by $ 50\%$ to form a rectangle. By what percent does the square's area decrease?

suppose each side was 10.
now they are 6 and 15

area went 100-->90

10% decrease
ANSWER:10

$ A\equal{}x^{2}$ for the square.
$ A\equal{}.6x(1.5x)\equal{}.9x^{2}$ for the rectangle.

So the area decreases by $ 10\%$


3sin^2 a+2sin^2b=1
and 3sin2a-2 sin 2b=0 where a ,b belongs to (pi/2,pi)
then a +b lies in which quadrant.

hello, do you mean the following problem
$ 3\sin(a)^2\plus{}2\sin(b)^2\equal{}1$
$ 3\sin(2a)\minus{}2\sin(2b)\equal{}0$
and $ a,b \in \left(\frac{\pi}{2},\pi\right)$
Sonhard.

hello, i have found $ a\plus{}b\equal{}2\left(\arctan(3\plus{}2\sqrt{2})\plus{}\arctan(\sqrt{5\plus{}2\sqrt{6}})\right)$.
Sonnhard.

[quote="vijaymenon"]i think the question is $ 3sin^2 a \plus{} 2sin^2 b \equal{} 1$
$ 3 sin 2b \minus{} 2 sin 2b \equal{} 0$[/quote]
yes question is as given by you


algebra
polynomial
function
linear algebra
matrix

[b]Solution.[/b] Let $f,g\in\mathbb{Z}[X]$. Suppose that $\deg(f)=m$ and $\deg(g)=n$ with the zeros $\alpha_{1},\alpha_{2},\hdots, \alpha_{m}$ and $\beta_{1},\beta_{2},\hdots,\beta_{n}$, respectively.  Let the function $\rho\in\mathbb{Z}[X]$ be defined as
\[\rho(f,g)=a_{m}^{n}b_{n}^{m}\prod_{k=1}^{m}\prod_{\ell=1}^{n}(\alpha_{k}-\beta_{\ell}), \]
where $a_{m},b_{n}$ are the leading coefficients of $f,g$, respectively.
Then $\rho$ has degree $mn$ with zeros $\alpha_{1},\alpha_{2},\hdots, \alpha_{m},\beta_{1},\beta_{2},\hdots,\beta_{n}$.$\Box$

[b]Remark.[/b] $\rho$ is called the resultant of $f$ and $g$.  This resultant can be calculated using the the determinant of the Sylvester Matrix $S\in\textbf{M}_{mn}(\mathbb{Z})\ (\text{i.e.}\det S)$[Edit: $S\in\textbf{M}_{m+n}(\mathbb{Z})$] constructed using the coefficients of $f,g$.

[quote="The Zuton Force"]Prove that the sum of any two algebraic numbers (a number for which there exists a polynomial with integer coefficients of which it is a root) is necessarily algebraic. Fun stuff.[/quote]
Here is an elegant way to do it, but it is advanced stuff.

Let $\mathbb{Q}$ be the rational field.
Let $\mathbb{C}$ be the real field.

Then, $\mathbb{Q}<\mathbb{C}$

If $\alpha , \beta \in \mathbb{C}$ are algebraic over $\mathbb{Q}$ it means that the simple extensions,

$\mathbb{Q}\leq \mathbb{Q}(\alpha, \beta )$

Since $\alpha , \beta \in \mathbb{Q}(\alpha, \beta)$ trivially.
And since it is an extension field it follows that,
$(\alpha+\beta) \in \mathbb{Q}(\alpha , \beta)$ for it is a closed.

But simple extension over algebraic elements are algebraic extensions.
Thus, $\alpha+\beta$ is algebraic over $\mathbb{Q}$.

And hence an algebraic number.

[quote="sylow_theory"]But simple extension over algebraic elements are algebraic extensions.[/quote]

This is basically the content of the problem.

[quote="t0rajir0u"]
This is basically the content of the problem.[/quote]

That is easy to show. Anyway, much easier to show than direct definition approach.

Simple Extensions (over algebraic elements) implies Finite Field Extensions implies Algebraic Extensions.

So its easy.


Find all triples of real numbers $ \left(a,b,c\right)$ satisfying the equations
$ a^5\equal{}5b^3\minus{}4c$;
$ b^5\equal{}5c^3\minus{}4a$;
$ c^5\equal{}5a^3\minus{}4b$.

We obtain with maple (theory of Gröbner basis):
$ a(a^{2}\minus{}1)(a^{2}\minus{}4)p(a)\equal{}0$ where $ p$ is a polynomial of degree 120. $ p$ hasn't any real root (use the Sturm method). If $ a\equal{}0,1,\minus{}1,2$ or $ \minus{}2$ then $ a\equal{}b\equal{}c.$

I don't know Sturm method, can you explain it?

If you eliminate $ b,c$ you obtain (in the general case) a polynomial of degree $ 5*5*5\equal{}125$. Here Maple gives the result:
 $ a (a\minus{}1) (a\minus{}2) (a\plus{}2) (a\plus{}1) (a^{120}\plus{}5 a^{118}\plus{}989020121 a^{96}\plus{}21 a^{116}\plus{}87381 a^{104}\plus{}21845 a^{106}\plus{}85 a^{114}\plus{}341 a^{112}\plus{}1365 a^{110}\plus{}5461 a^{108}\plus{}44460036304525 a^{82}\plus{}15189459021224563628564665 a^{50}\plus{}\cdots).$
   Good luck for the calculation with the hand!
    For the sturm method you can see Wikipedia ("Sturm's theorem"); it's like the calculation of $ gcd(p,p').$

Can you explain, why in schedule polynomial you have certainty that you have x*(x^2-4)*(x^2-1) and why you known that a=b=c

The deadline for this problem was 8 October 2007. I am moving it back. It can be discussed now.

Shame on Kisiu111 for posting this problem before the deadline. Of course he didn't mention the source "59th Polish MO, 1st Round, problem 1"; it was inserted by myself (and I also edited the problem statement).

[quote="kisiu111"]Find all triples of real numbers $ \left(a,b,c\right)$ satisfying the equations
$ a^5 \equal{} 5b^3 \minus{} 4c$;
$ b^5 \equal{} 5c^3 \minus{} 4a$;
$ c^5 \equal{} 5a^3 \minus{} 4b$.[/quote]
Let $ a \equal{} \max\{a,b,c\}$
Case 1:$ c\geq b$ then
$ a^5 \plus{} 4c\geq c^5 \plus{} 4b$ so $ b\geq a$
So $ a \equal{} b \equal{} c$
Case 2:$ b\geq c$ then
$ b^5 \plus{} 4a\geq c^5 \plus{} 4b$ so $ c\geq a$ so $ a \equal{} b \equal{} c$
Solve the equation :$ a^5 \minus{} 5b^3 \plus{} 4c \equal{} 0$  we has $ a\in\{0,1, \minus{} 1,\sqrt {2},\minus{}\sqrt { 2}\}$

Hi TTsphn,
pretty solution, but you must replace $ \pm\sqrt{2}$ by $ \pm2$.


Find all $n \in N$ such that $\mu(n)+\mu(n+1)+\mu(n+2)=3$, given that $\mu(n)$ is the mobius function (http://mathworld.wolfram.com/MoebiusFunction.html)

There is a nice expression for them?  I'm curious.  The first few such $n$ are 33, 85, 93, 141, 201, 213, 217, 301, 393, 445, 633, 697, 869, 921.

The [url=http://www.research.att.com/~njas/sequences/A063838]OEIS[/url] gives more terms, but nothing to suggest a general description of them.


geometry
rectangle
analytic geometry
absolute value
geometry unsolved

A lattice point in the Carthesian plane is a point with both coordinates integers. A [i]rectangle minor[/i] (respectively a [i]square minor[/i]) is a set of lattice points formed by [i]all[/i] the lattice points lying inside or on the boundaries of a [i]rectangle[/i] (respectively [i]square[/i]) with vertices lattice points. We assign to each lattice point a real number, such that the sum of all the numbers in any square minor is less than 1 in absolute value.

Prove that the sum of all the numbers in any rectangle minor is less than 4 in absolute value

I think it is hard. Noonw solved it in the contest ?Me too :(


[quote="i_like_pie"]$ \frac {r^2 - r - 1 - d}{x^2} = \,?$

[hide]
\begin{align*}\frac {r^2 - r - 1 - d}{x^2} & = n \\
 \\
r^2 - r - 1 - d & = nx^2 \\
 \\
r^2 - r - 1 & = d + nx^2 \\
 \\
4r^2 - 4r - 4 & = 4d + 4nx^2 \\
 \\
4r^2 - 4r + 1 & = 5 + 4d + 4nx^2 \\
 \\
(2r - 1)^2 & = 5 + 4d + 4nx^2 \\
 \\
2r & = 1\pm\sqrt {5 + 4d + 4nx^2} \\
 \\
r & = \frac {1\pm\sqrt {5 + 4d + 4nx^2}}{2}\end{align*}
[/hide][/quote]

I think he's asking for the roots of r, and when asking for roots of a polynomial, I think it implies the polynomial is equivalent to 0.

[hide]
I don't see why the $ x^2$ is in the root...

$ \frac {r^2 - r - 1 - d}{x^2} = 0$
$ \frac {r^2}{x^2} - \frac {r}{x^2} - \frac {1 + d}{x^2} = 0$
$ r = \frac {\frac {1}{x^2} \pm \sqrt {\frac {1 + 1(4)(1 + d)}{x^4}}}{\frac {2}{x^2}}$
$ r = \frac {\frac {1 \pm \sqrt {5 + 4d}}{x^2}}{\frac {2}{x^2}}$
$ r = \frac {1 \pm \sqrt {5 + 4d}}{2}$

An easier way is just factoring out the common monomial $ \frac {1}{x^2}$ in the beginning, and that'll show why the $ x^2$ should not belong in the root.
$ \frac {r^2 - r - 1 - d}{x^2} = 0$
$ \frac {1}{x^2}(r^2 - r - 1 - d) = 0$
$ \frac {1}{x^2}(r^2 - r + ( - 1 - d)) = 0$
$ (1)r^2 + ( - 1)r + ( - 1 - d) = 0$
$ r = \frac {1 \pm \sqrt {1 - 4( - 1 - d)}}{2}$
$ r = \frac {1 \pm \sqrt {5 + 4d}}{2}$
[/hide]

If the exact equation is equal to $ 0$, then $ x$ can be anything except for $ 0$, and the numerator must be $ 0$. Therefore, $ x$ plays no role whatsoever in the roots of the equation, so that expression is incorrect.

Disclaimer: I'm not a qualified mathematician so don't think everything (or anything, for that matter) I said is indisputably correct.

[quote="ProtestanT"][quote="guile"][quote]
I think he's asking for the roots of r, and when asking for roots of a polynomial, I think it implies the polynomial is equivalent to 0.
[/quote]

Yeah. that's equal to zero. sorry. There must be some flaw somewhere in your solution. This is from the mathematics teacher so I am quite sure that it has been reviewed. I have the same solution, I was wondering where did the $ x^2$ came from.[/quote]
Was the $ x^2$ positioned exactly as you said it was?
Also, were there any extra information that we should know?[/quote]

yes, it was

[quote="guile"][/quote][quote="ProtestanT"][quote="guile"][quote]
I think he's asking for the roots of r, and when asking for roots of a polynomial, I think it implies the polynomial is equivalent to 0.
[/quote]

Yeah. that's equal to zero. sorry. There must be some flaw somewhere in your solution. This is from the mathematics teacher so I am quite sure that it has been reviewed. I have the same solution, I was wondering where did the $ x^2$ came from.[/quote]
Was the $ x^2$ positioned exactly as you said it was?
Also, were there any extra information that we should know?[/quote][quote="guile"]

yes, it was[/quote]
Has the maths teacher given a solution to which you are reffering?

Both i_l_p and Prot correctly solved the problem given their assumptions, for anyone who is interested.  Also, please don't make "quote towers" -- the assumption will always be that you are responding to the post directly before yours, so quoting it in full is pointless and obnoxious to the rest of us.  (This doesn't apply if you're breaking the preceding post into several pieces and responding to them point-by-point, but that's obviously not the case here.)

[quote="guile"]in an article i read, the equation is $ \frac {r^2 \minus{} r \minus{} 1 \minus{} d}{x^2}$. The root is $ r \equal{} \frac {1 \plus{} \sqrt {5 \plus{} \frac {4d}{x^2}}}{2}$. No idea how did this happen. help please.[/quote]

Lets suppose the article is mathematically OK and what it said is similar to what you post, the something you posted is not what appears on the article, or the article has a typo.

As in $ \frac {r^2 \minus{} r \minus{} 1 \minus{} d}{x^2} \equal{} 0$ the $ x^2$ has no sense, I guess this is what you didn't copy right, or this is the typo in the article.

I guees the root is right then. Working backwards we can see wich the initial equation must be:
$ r \equal{} \frac {1 \plus{} \sqrt {5 \plus{} \frac {4d}{x^2}}}{2}$
$ 2r \minus{} 1 \equal{} \sqrt {5 \plus{} \frac {4d}{x^2}}$
$ 4r^2 \minus{} 4r \plus{} 1 \equal{} 5 \plus{} \frac {4d}{x^2}$
$ 4r^2 \minus{} 4r \minus{} 4 \minus{} \frac {4d}{x^2} \equal{} 0$
$ r^2 \minus{} r \minus{} 1 \minus{} \frac {d}{x^2} \equal{} 0$

Wich is almost identical to what you post, so either yours or the article's typo is that $ x^2$ is only below $ d$.

 :)


function
calculus
integration
calculus computations

how do i calculate one primitive of the function
$f(t)=sin(t^2)+\frac{cos(t^2)}{2t^2}$

The answer is $-\frac{\cos(t^2)}{2t}+C$. The basic idea is to try to get one term and see whether the other term will arise too. And to get $\sin(t^2)$, you have to have $-\cos(t^2)$ somewhere. Pure $-\cos(t^2)$ gives you an extra factor $2t$, so it is natural to divide by $2t$ and see what else will come from the product rule. Interestingly enough, neither term in the sum has an elementary antiderivative, so we are very lucky here to have one for their combination.

BTW, this belongs to Calculus Computations and Tutorials, so I'm moving it there  :)

actually the solution was to try to bring all stuff to a common denominator
$4t^2$ and then to see that the fraction i have now is actually a derivate of the form
$\frac{h}{g}=\frac{h'g-hg'}{g^2}$ so i figure that $g=2t$ and that $h=-cos(t^2)$
and this should be the solution


geometry
incenter
geometric transformation
reflection
circumcircle
projective geometry
geometry proposed

[size=117][color=darkred]Let $ABC$ be a triangle. Denote the incenter $I$, the $A$- exincenter $I_{a}$ and the reflection $A'$ of the point $A$ w.r.t. the circumcenter.

Define the intersections : $E_{b}\in IA\cap BA'\ ,\ E_{c}\in IA\cap CA'\ ;\ F_{b}\in IA'\cap AC\ ,\ F_{c}\in IA'\cap AB$ $\mathrm{\ ;\ }$ $F'_{b}\in I_{a}A'\cap AC\ ,\ F'_{c}\in I_{a}A'\cap AB$ .

Prove that :[/color] $\{\begin{array}{c}1.\blacktriangleright\mathrm{\ The\ lines\ }BC\ ,\ E_{b}F_{b}\ ,\ E_{c}F_{c}\mathrm{\ are\ concurrently.}\\\\ 2.\blacktriangleright\mathrm{\ The\ lines\ }BC\ ,\ E_{b}F'_{b}\ ,\ E_{c}F'_{c}\mathrm{\ are\ concurrently.}\ \end{array}$[/size]

For the solution of this nice problem, we will use tools from the double ratio’s theory, as follows:
( In my drawing AB = 4.3, AC = 6.1, BC = 6.7 ).

PROOF OF PART 1. - We denote as K, the intersection point of the segment lines BFb, CEb and as L, the intersection point of AB, CA'.

We consider the pencil of lines A'.AIEbEc, which is intersected from the sideline AB of ABC.
So, we have that (A, I, Eb, Ec) = (A, Fc, B, L), (1).

But, (A, I, Eb, Ec) = (C.AIEbEc), (2) and (A, Fc, B, L) = (Fb.AFcBL), (3).

From (1), (2), (3), => (C.AIEbEc) = (Fb.AFcBL), (4).

From (4) and because of the pencils of lines C.AIEbEc, Fb.AFcBL, have the sideline AC of ABC, as their common line, we conclude that the points I, ( as the intersection point of CI, FbFc ) and K, (as the intersection point of CEb, FbB ) and L, ( as the intersection point of CEc, FbL ), are collinear.

Because of the collinearity of the points I, K, L, by Desarques’s theorem, we conclude that the triangles BFbFc and CEbEc, are perspective. Hence, the segment lines BC, EbFb, EcFc, are concurrent at a point so be it D and the proof, of part (1) of the problem, is completed.

PROOF OF PART 2. – By the same way as before, we denote as M, the intersection point of the segment lines BEc, CF'c  and as N, the intersection point of AC, BA'.

We consider the pencil of lines A'.AIaEcEb, which is intersected from the sideline AC of ABC.
So, we have that (A, Ia, Ec, Eb) = (A, F'b, C, N), (1).

But, (A, Ia, Ec, Eb) = (B.AIaEcEb), (2) and (A, F'b, C, N) = (F'c.AF'bCN),3).

From (1), (2), (3), => (B.AIaEcEb) = (F'c.AF'bCN), (4).

From (4) and because of the pencils of lines B.AIaEcEb, F'c.AF'bCN, have the sideline AB of ABC, as their common line, we conclude that the points Ia, ( as the intersection point of BIa, F'cF'b ) and M, (as the intersection point of BEc, F'cC ) and N, ( as the intersection point of BEb, F'cN ), are collinear.

Because of the collinearity of the points Ia, M, N, by Desarques’s  theorem, we conclude that the triangles BEbEc and CF'bF'c, are perspective. Hence, the segment lines BC, EbF'b, EcF'c, are concurrent at a point so be it D' and the proof, of part (2) of the problem, is completed.

NOTE. - I think that the points D, D', are the tangent points of the sideline BC of ABC, to the incircle and to A-excircle of ABC respectively, but I still have not any proof for this extension result. So, I am waiting for your remarks and advices.

Kostas Vittas.

PS. – Thank you mr Virgil Nicula for this nice problem and I am interested to know your approach about it ( It is beautiful when we solve a problem, but it is also beautiful to know another solution from somebody else ).

[color=darkblue]Yes, you infer very well. I present later the my solution. I see you are "very strong" in geometry.

I am glady you like this problem. I consider it is a "top" for the all own problems.[/color]

Thank you for your answer, Virgil. I think that a more general formulation of your problem, is as follows:

Through vertex A of a given quadrilateral ABCD, we draw an arbitrary line La, which intersects the sidelines CD, BC of ABCD, at points Eb, Ec, respectively. Through vertex C, we draw also an arbitrary line Lc, which intersects the sidelines AB, AD, at points Fb, Fc, respectively. Prove that the intersection point of the segment lines EbFb, EcFc, lies on the segment line BD.

For the proof of this general problem, we can see my previous message in this topic, considering as I, the intersection point of the lines La, Lc.

Kostas Vittas.

Mr Nicula what is your proof to the problem ?
P.S. Mr Kostas beautiful proof  :)

[size=117][color=darkred][b]Lemma.[/b] For $\triangle ABC$ denote the incenter $I$, the $A$- exincenter $I_{a}\ ,$ the reflection $A'$ of the point $A$ w.r.t. the circumcenter and
the second intersection $M$ of the line $\overline{AII_{a}}$ with the circumcircle. The incircle and the $A$- exincircle touch the side $[BC]$
in the points $D$ and $D'$ respectively. Then the intersections $S\in IA'\cap DM$ and $S'\in I_{a}A'\cap D'M$ belong to the circumcircle.

[b]Proof.[/b] SOON here !

[b]Enunciation.[/b] For $\triangle ABC$ denote the incenter $I$, the $A$- exincenter and the reflection $A'$ of the point $A$ w.r.t. the circumcenter.

Define the intersections : $\{\begin{array}{c}E_{b}\in IA\cap BA'\ ,\ E_{c}\in IA\cap CA'\mathrm{\ .\ }\\\\F_{b}\in IA'\cap AC\ ,\ F_{c}\in IA'\cap AB\mathrm{\ .\ }\\\\F'_{b}\in I_{a}A'\cap AC\ ,\ F'_{c}\in I_{a}A'\cap AB\ .\end{array}$ Prove that : $\{\begin{array}{c}1.\blacktriangleright\mathrm{The\ lines\ }BC\ ,\ E_{b}F_{b}\ ,\ E_{c}F_{c}\mathrm{\ are\ concurrently\ .}\\\\ 2.\blacktriangleright\mathrm{The\ lines\ }BC\ ,\ E_{b}F'_{b}\ ,\ E_{c}F'_{c}\mathrm{\ are\ concurrently\ .}\end{array}$[/color]

[color=darkblue][b]Proof.[/b] I"ll use the notations from the above lemma. Apply the [u]Pascal's theorem[/u] in the following two pairs of hexagons :

$\{\begin{array}{c}ABCA'SM\Longrightarrow D\in E_{c}F_{c}\\\\ ACBA'SM\Longrightarrow D\in E_{b}F_{b}\end{array}\Longrightarrow \boxed{\ D\in E_{b}F_{b}\cap E_{c}F_{c}\ }\mathrm{\ \ \ ;\ \ \ }\{\begin{array}{c}ABCA'S'M\Longrightarrow D'\in E_{c}F'_{c}\\\\ ACBA'S'M\Longrightarrow D'\in E_{b}F'_{b}\end{array}\Longrightarrow \boxed{\ D\in E_{b}F'_{b}\cap E_{c}F'_{c}\ }\ .$
[/color][/size]

[quote="vittasko"]Through vertex A of a given quadrilateral ABCD, we draw an arbitrary line La, which intersects the sidelines CD, BC of ABCD, at points Eb, Ec, respectively. Through vertex C, we draw also an arbitrary line Lc, which intersects the sidelines AB, AD, at points Fb, Fc, respectively. Prove that the intersection point of the segment lines EbFb, EcFc, lies on the segment line BD.
[/quote]
It's follow from Papp's theorem for rows $A, E_{b}, E_{c}$ and $C, F_{c}, F_{b}$.

Many thanks fagot. Very nice ( I think the best )  solution of Virgil’s problem (one row's solution !).

Kostas Vittas.

[color=darkred][b][u]Lemma[/u].[/b] For $\triangle ABC$ denote the incenter $I$, the $A$- exincenter $I_{a}\ ,$ the reflection $A'$ of the point $A$ w.r.t. the circumcenter and
the second intersection $M$ of the line $\overline{AII_{a}}$ with the circumcircle. The incircle and the $A$- exincircle touch the side $[BC]$
in the points $D$ and $D'$ respectively. Then the intersections $S\in IA'\cap DM$ and $S'\in I_{a}A'\cap D'M$ belong to the circumcircle.[/color]

[color=darkblue][b][u]Proof[/u].[/b] Denote $p_{w}(X)$- the power of the point $X$ w.r.t. the circumcircle $w=C(O,R)$ of the triangle $ABC\ .$ I recall the well-known relations : $-p_{w}(I)=IA\cdot IM=R^{2}-OI^{2}=2Rr$ and $p_{w}(I_{a})=I_{a}A\cdot I_{a}M=OI^{2}_{a}-R^{2}=2Rr_{a}\ .$

$\blacktriangleright$ $\widehat{DIM}\equiv\widehat{IAA'}$ and $\frac{ID}{IM}=\frac{AI}{AA'}$, i.e. $IA\cdot IM=2Rr$ $\Longrightarrow$ $\triangle IDM\sim\triangle AIA'$ $\Longrightarrow$ $\widehat{DMI}\equiv\widehat{IA'A}$ $\Longrightarrow$ $S\in w\ .$

$\blacktriangleright$ $\widehat{D'I_{a}M}\equiv\widehat{I_{a}AA'}$ and $\frac{I_{a}D'}{AI_{a}}=\frac{I_{a}M}{AA'}$, i.e. $I_{a}M\cdot I_{a}A=2Rr_{a}$ $\Longrightarrow$ $\triangle I_{a}D'M\equiv\triangle AI_{a}A'$ $\Longrightarrow$ $\widehat{D'MI_{a}}\equiv\widehat{I_{a}A'A}$ $\Longrightarrow$

$\widehat{AMS'}\equiv\widehat{AA'S'}$ $\Longrightarrow$ $S'\in w\ .$ Observe the "identity" between the two above proofs : $I\rightarrow I_{a}\ ;\ r\rightarrow r_{a}\ ;\ D\rightarrow D'\ ;\ \ S\rightarrow S'\ .$ [/color]

[color=darkred][b][u]PP[/u].[/b] For $\triangle ABC$ denote the incenter $I$, the $A$- exincenter and the reflection $A'$ of the point $A$ w.r.t. the circumcenter.

Define the intersections : $\left\{\begin{array}{c}E_{b}\in IA\cap BA'\ ,\ E_{c}\in IA\cap CA'\mathrm{\ .\ }\\\\F_{b}\in IA'\cap AC\ ,\ F_{c}\in IA'\cap AB\mathrm{\ .\ }\\\\F'_{b}\in I_{a}A'\cap AC\ ,\ F'_{c}\in I_{a}A'\cap AB\ .\end{array}\right\|$ . Prove that : $\left\{\begin{array}{c}1.\blacktriangleright\mathrm{The\ lines\ }BC\ ,\ E_{b}F_{b}\ ,\ E_{c}F_{c}\mathrm{\ are\ concurrently\ .}\\\\ 2.\blacktriangleright\mathrm{The\ lines\ }BC\ ,\ E_{b}F'_{b}\ ,\ E_{c}F'_{c}\mathrm{\ are\ concurrently\ .}\end{array}\right\|$.[/color]

[color=darkblue][b]Proof.[/b] I"ll use the notations from the above lemma. Apply the [u]Pascal's theorem[/u] in the following two pairs of hexagons :

$\left\{\begin{array}{c}ABCA'SM\Longrightarrow D\in E_{c}F_{c}\\\\ ACBA'SM\Longrightarrow D\in E_{b}F_{b}\end{array}\Longrightarrow \boxed{\ D\in E_{b}F_{b}\cap E_{c}F_{c}\ }\mathrm{\ \ \ ;\ \ \ }\{\begin{array}{c}ABCA'S'M\Longrightarrow D'\in E_{c}F'_{c}\\\\ ACBA'S'M\Longrightarrow D'\in E_{b}F'_{b}\end{array}\right\|\Longrightarrow \boxed{\ D\in E_{b}F'_{b}\cap E_{c}F'_{c}\ }\ .$
[/color]


group theory
abstract algebra
quadratics
Galois Theory
superior algebra
superior algebra unsolved

Find the splitting field, Galois group and  all intermediate fields and subgroups
of
1) f(X)=x^4-x^2 -2
2) f(X)=x^10  -1

Are there only the roots of unity in 2?

1) $ C(2)XC(2)$
  The only intermediate fields are the splitting fields of $ x^2\plus{}1$ and $ x^2\minus{}2$
2)$ (Z/5Z)^*$ which is $ C(4)$
 The only int. field is $ Q(r\plus{}r^4)$

[quote="Abel"]1) $ C(2)XC(2)$
  The only intermediate fields are the splitting fields of $ x^2 \plus{} 1$ and $ x^2 \minus{} 2$
[/quote]

What with the splitting field of $ x^2 \plus{} 2$ ?

P.

Yes but how do you get there I don't suppose you can it right away?

[quote="Abel"]You're right. $ C(2)XC(2)$ has 3 proper subgroups.[/quote]

I would give the following proof. (Perhaps there is a simpler one.)
The discriminants of $ \mathbb{Q} (\sqrt { \minus{} 1})$ and $ \mathbb{Q}(\sqrt {2})$ have only $ 2$ as prime factor, thus the discriminant of their composite (a.k.a product) $ K$ has the same property, thus the discriminant of every intermediate field between $ \mathbb{Q}$ and $ K$ has the same property. But an intermediate field is quadratic, i.e. of the form $ \mathbb{Q} (\sqrt {d})$ with $ d$ a squarefree integer $ \not \equal{} 1$. In general, for $ d$ a squarefree integer $ \not \equal{} 1$, the discriminant of the quadratic field $ \mathbb{Q} (\sqrt {d})$ is $ 4d$ or $ d$ (depending on the remainder of $ d$ modulo $ 4$), thus our $ d$ must be one of the numbers -1, 2, -2.

Friendly,
P.


calculus
integration
inequalities
trigonometry
function
real analysis
calculus computations

can anyone explain for me why :

$ |\int_{0}^{n}e^{ \minus{} nx}\frac {sinx}{x}dx|\leq \frac {1}{n}\int_{0}^{n}e^{ \minus{} nx}dx$  ?

many thanks

$ \frac {|\sin x|}{x} \leq \frac {1}{n}$ since $ |\sin x| \leq 1$. But not for all $ x$.

I still don't understand... :blush:
if you have $ |sinx|\leq 1$ then what ?
$ |\frac {sinx}{x}|\leq 1/x$ how can you prove that on $ [0,n]$ :      $ |sinx/x|\leq 1/n$

thanks

The original statement is clearly false as stated.

Let's look at this intuitively: for large $ n,$ the function $ e^{\minus{}nx}$ decays so quickly that nearly everything that matters takes place in a very small neighborhood of zero. In such a neighborhood, $ \frac{\sin x}{x}$ might as well be $ 1.$

Hence $ \int_0^ne^{\minus{}nx}\,\frac{\sin x}{x}\,dx$ is approximately $ \int_0^ne^{\minus{}nx}\,dx$

You can make that an inequality, since $ \frac{\sin x}{x}<1$ for all $ x>0.$ But you can't do any better than that, and there's certainly no extra $ \frac1n$ out front.

Had the exponential function been $ e^{\minus{}x/n}$ rather than $ e^{\minus{}nx},$ this would have been an entirely different problem.

By MVT, there exists a $ t \in (0,n)$ such that $ I \equal{} \int_{0}^{n} e^{ \minus{} nx} \frac {\sin x}{x} \ dx \equal{} e^{ \minus{} nt} \int_{0}^{n} \frac {\sin x}{x} \ dx \leq \frac {\pi}{2} e^{ \minus{} nt}$ which is another bound.

thanks Prof. Kent

That is a statement in a Real analysis book, I tried to prove it but it turned out by your post it is false


Find the value of $ \frac1{100}\plus{}\frac2{100}\plus{}\frac3{100}\plus{}\cdots\plus{}\frac{100}{100}$. Express your answer as a decimal.


[hide]The probability of the first person being a man is $\frac{6}{10}$. The probability of the second person being a man is $\frac{5}{9}$. The probability of the third person being a woman is $\frac{4}{8}$. The probability of the fourth person being a woman is $\frac{3}{7}$. The order can vary, but the final probability (achieved by multiplying the four fractions) will always be $\boxed{\frac{1}{14}}$.[/hide]

with column notation...
[hide]

$\binom{10}{4}=210$ is our denom.
$\binom42=6$
$\binom62=15$
so our answer is
$\frac{6*15}{210}=\frac{3}{7}$

[/hide]

[quote="max_tm"]So I [b]totally[/b] suck at probability but since boxedexe says the above solution is wrong could this possibly be right?

$\frac{(C(6,2)\times(C(4,2))}{C(10,4)}= \frac{3}{7}$

PS: I would use column notation but it's 7 in the morning and I don't want to go figuring out how.[/quote]

yes thats right.

1) in this problem, order does NOT matter so thats why you should use combanatorics

2) {6\choose 2}

[quote="Ointer"]i've found this in my book. but i'm not sure how to solve it.

a committe of 4 people is selected at random without replacement from a group of 6 men and 4 women. Find the probability that the committe consists of 2 men and 2 women.[/quote]

[hide]

You're picking 4 people from a group of 10.

Hence, there are $\binom{10}{4}$ choices.

But, you need 2 men and 2 women.

There are $\binom{6}{2}$ ways to pick 2 men from 6.

There are $\binom{4}{2}$ ways to pick 2 women from 4.

The answer is then $\frac{\binom{6}{2}\cdot \binom{4}{2}}{\binom{10}{4}}$.

[/hide]

[quote="i_like_pie"][quote="boxedexe"]I disagree with your solution.  Analyze your first sentence carefully.[/quote]
I see that I went wrong somewhere, but can you explain exactly where? For some reason, I can't find the mistake in my reasoning.[/quote]

the mistake is that you used permutations instead of combinations.  The order does not matter in this case.

No, combinations.

$\frac{\binom{5}{2}\cdot\binom{3}{1}\cdot\binom{2}{1}}{\binom{10}{4}}$

[quote="Ointer"]what about...

a box contains 5 red balls, 3 blue balls, and 2 white balls. if 4 balls are selected with replacement, find the probability of getting 2 red balls, one blue ball, and one white ball.

it should be permutaions ...right?[/quote]

[hide]im really bad at probability so i probably did it a really dumb method.

he can choose the balls 12 different orders (because there are are four places where the blue ball can go and three places where the whit ball can go and only one way t place the 2 red balls.  the prob. that we choose a red ball is $frac{1}{2}$, a blue ball is $\frac{3}{10}$ and a white ball is $\frac{2}{10}$

multiply them to get $(\frac{1}{2})^{2}\frac{3}{10}\frac{2}{10}=\frac{6}{400}=\frac{3}{200}$.  Now multiply by 12 to get $\frac{9}{50}$.[/hide]


search
combinatorics proposed
combinatorics

The $ 27$ element sets $ A_{1}, A_{2},...,A_{35}$ have the property that every three of them have exactly one element in common. Show that all the $ 35$ sets $ A_{i}$ have an element in common.

If you don't know whether it has been posted before and think it might have been, the appropriate thing to do is to [i]search[/i] to see if it has appeared before.  I searched for the word "Iran" in the Combinatorics forum and it took me at most a minute or two to find [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=140614]this thread[/url].

OK Thank you for advices and also many thanks for this link.


If you get more wrong, they will be transferred to "passed"...and I have no idea what's up with the "past due," so for that second part, ask an administrator.

[quote="cmpgm88@yahoo.com"]I had previously "mastered" some topics on Alcumus, however recently, after a few problems, they are listed as "passed" and even "past due".

What happened??????[/quote]
You have answered some questions in topics incorrectly that the system previously considered passed or mastered by you. By answering incorrectly, those topics went down to "working on" or "passed" respectively. It should only say past due for those topics that are now considered as "not ready" or "working on."

[quote="cmpgm88@yahoo.com"]I had previously "mastered" some topics on Alcumus, however recently, after a few problems, they are listed as "passed" and even "past due".

What happened??????[/quote]

You are in my C&P class right? That means you had to finish those topics before hand.

Yes, Dojo, that is correct (I remember you for doing EVERYTHING in $ \text{\LaTeX}$!!!!!!)

So I just need to do more problems in order to acheive a certain precentage?  (BTW, what is that Magic percentage for a topic to be passed or even  mastered?)

There is no magic percentage. It's all based on how other user's have performed on similar problems and how you have done on related topics.


how would you solve a system of equations such as:
y=6x-11
-2x-3y=-7

and would someone show me how to write that in LaTeX form? thanks!

for one thing, i do not have LaTeX.

but since the value of y is given (6x-11), i would ues the substitution method.

in the second equation, you would substitute all Y's with "6x-11", thus reaching the equation of:
-2x-3(6x-11)= -7

solve for X:

-20x+33=-7
-20x=-40
x=2

now plug in 2 in the first equation:
y=6(2)-11

so the answer, in (x,y) form, would be (2,1)

To write in Latex all you have to do is put dollor signs around the stuff you write.
Example y=6x-11
[code]$y=6x-11$[/code]
gives $ y\equal{}6x\minus{}11$

[quote="ernie"]for one thing, i do not have LaTeX.

but since the value of y is given (6x-11), i would ues the substitution method.

in the second equation, you would substitute all Y's with "6x-11", thus reaching the equation of:
$ \minus{}2x\minus{}3(6x\minus{}11)\equal{} \minus{}7$

solve for X:

$ \minus{}20x\plus{}33\equal{}\minus{}7$
$ \minus{}20x\equal{}\minus{}40$
$ x\equal{}2$

now plug in 2 in the first equation:
$ y\equal{}6(2)\minus{}11$

so the answer, in (x,y) form, would be (2,1)[/quote]

latexified for you :)

There are also various other methods of solving simultaneous equations (adding/subtracting equations to cancel variables, graphing); but yes, substitution would be most appropriate here.

Also, here's the Latex Guide:
http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:About

I wrote a lengthy explanation in response to another post concerning systems of equations that explains the two methods in great detail. Here's a link - it's post #28:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=196739&start=20

$ \text{\LaTeX}$ is a language so you can make $ \sqrt$ signs and fractions, like $ \frac{1}{2}$. To learn $ \text{\LaTeX}$, see the tutorials on this site.

sooooooooooo, it's just so u can ake math symbols!!!!!!!!!!! :huh:  that's the purpose!!!!!?????? :o

so for latex, u just put dollar signs around the equation?

$ y \equal{} 6x \minus{} 11$

nice
but how do u make square root signs?

[quote="vallon22"]so for latex, u just put dollar signs around the equation?

$ y = 6x - 11$

nice
but how do u make square root signs?[/quote]
[b]the square root sign[/b]
$ \sqrt$
first do the standard dollar signs
then put a backward slash '\' and type 'sqrt' after it

<dollarsigngoeshere>\sqrt<dollarsigngoeshere>

Please see the LaTeX link from the menu on the left or go directly [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]Here[/url]


quadratics
vector
geometry
geometric transformation
reflection
parallelogram
perpendicular bisector

Let $ M,N$ be two points inside circle $ C$ with center $ O$ such that $ O$ is the midpoint of $ MN$. Let $ S$ be a point on this circle. Let $ E,F$ be the intersections of lines $ SM,SN$ with this circle. Tangents at $ E,F$ to this circle intersect at $ I$. Prove that the perpendicular bisector of segment $ MN$ passes through midpoint of $ SI$.

The problem can be easily solved with complex numbers.
Let the cirlce $ C$ be the unit circle and $ S\equal{}1$. Denote $ M$ by $ x$ (then $ N$ is $ \minus{}x$), $ E$ by $ p$, $ F$ by $ q$ and $ I$ by $ z$. It is well known that points $ a,b,c$ are colinear iff $ \overline{a}b\plus{}\overline{b}c\plus{}\overline{c}a\equal{} a\overline{b}\plus{}b\overline{c}\plus{}c\overline{a}$. The points $ 1,x,p$ are collinear and $ \left|p\right|\equal{}1$, so the condition $ p\plus{}\overline{p}x\plus{}\overline{x}\equal{} \overline{p}\equal{}p\overline{x}\plus{}x$ gives $ p^{2}\left(1\minus{}\overline{x}\right)\plus{}p\left(\overline{x}\minus{}x\right)\plus{}x\minus{}1$. This quadratic equation has two roots, corresponding to two points of intersection of the line passing through $ 1$ and $ x$ with the circle: $ 1$ and $ \frac{1\minus{}x}{\overline{x}\minus{}1}$. The points $ S$ and $ E$ are different, so $ p\equal{}\frac{1\minus{}x}{\overline{x}\minus{}1}$. Analogously, $ q\equal{}\frac{1\plus{}x}{\minus{}\overline{x}\minus{}1}$. It is well known that inversion with respect to the circle $ C$ transforms the point $ I$ to the midpoint of $ EF$. Hence $ \left(\overline{z}\right)^{\minus{}1}\equal{} \frac{1}{2}\left(p\plus{}q\right)\equal{}\frac{1}{2}\left(\frac{1\minus{}x}{\overline{x}\minus{}1}\plus{}\frac{1\plus{}x}{\minus{}\overline{x}\minus{}1}\right)\equal{} \frac{x\overline{x}\minus{}1}{1\minus{}\overline{x}^{2}}$ and $ z\equal{}\frac{1\minus{}x^{2}}{x\overline{x}\minus{}1}$. The midpoint of $ SI$, say $ R$, is $ r: \equal{}\frac{z\plus{}1}{2}\equal{} \frac{x\left(\overline{x}\minus{}x\right)}{2\left(x\overline{x}\minus{}1\right)}$. We want to prove that the perpendicular bisector of $ MN$ passes through $ R$, i.e. $ OR\perp OM$. It is equivalent to the fact that the vectors $ r$ and $ x$ are perpendicular, which means that $ \frac{r}{x}$ is purely imaginary. But $ \overline{\left(\frac{r}{x}\right)}\equal{} \overline{\left(\frac{\overline{x}\minus{}x}{2\left(x\overline{x}\minus{}1\right)}\right)}\equal{} \minus{}\frac{\overline{x}\minus{}x}{2\left(x\overline{x}\minus{}1\right)}\equal{}\frac{r}{x}$, so $ \frac{r}{x}$ is purely imaginary, which completes the proof.

[quote="Rzeszut"]The problem can be easily solved with complex numbers.[/quote]

Thanks a lot, but actually the solution I have also uses complex numbers.

Do you have one not using it?  :)

$ SO$ interects $ (O)$ at $ P$, and the line passes through $ S$ and parallel to $ MN$ intersects $ (O)$ at $ Q$
Cause $ S(MNOQ)\equal{}\minus{}1$, the quadrilateral $ EPFQ$ is harmonic. So $ I,P,Q$ are collinear.
Let $ G$ be the midpoint of $ SI$, we have $ OG \parallel IP, PQ \perp QS$ and $ QS \parallel MN$. So $ OG \perp MN$, as desired

[quote="nguyentrang"] [color=darkred]Let $ M,N$ be two points inside circle $ w \equal{} C(O,R)$ such that $ O$ is the midpoint of $ MN$. Let $ S$ be a point on this circle.

Let $ E,F$ be the intersections of lines $ SM,SN$ with this circle. Tangents at $ E,F$ to this circle intersect at $ I$.

Prove that the perpendicular bisector of segment $ MN$ passes through midpoint of $ SI$.[/color] [/quote]
[quote][color=darkred][b][u]Lemma.[/u][/b] Let $ w \equal{} C(O)$ be the circumcircle of $ \triangle ABC$ and let $ E\in (AB)$ , $ F\in (AC)$ so that $ O\in EF$ ,

$ OE \equal{} OF$ . Denote the reflection $ T$ of the point $ A$ w.r.t. the point $ O$ . Then the lines $ EF$ , $ TT$ , $ BC$ are concurrently.[/color]

[hide="Proof of the lemma."][color=darkblue]Denote $ P_1\in TT\cap BC$ and $ P_2 \equal{} EF\cap BC$ . Observe that : $ \frac {P_1B}{P_1C} \equal{} \left(\frac {TB}{TC}\right)^2$ ; the quadrilateral $ AETF$ is

parallelogram, i.e. $ AE \equal{} TF$ , $ AF \equal{} TE$ ; the right triangles $ BET$ , $ CFT$ are similarly, i.e. $ BET\sim CFT$ $ \Longleftrightarrow$

$ \frac {BE}{CF} \equal{} \frac {ET}{FT} \equal{} \frac {TB}{TC}$ . Apply the [b]Menelaus' theorem[/b] to the transversal $ \overline {EFP_2}$ in $ \triangle ABC$ : $ \frac {P_2B}{P_2C}\cdot\frac {FC}{FA}\cdot\frac {EA}{EB} \equal{} 1$

$ \implies$ $ \frac {P_2B}{P_2C}\cdot\frac {CF}{BE}\cdot\frac {FT}{ET} \equal{} 1$ $ \implies$ $ \frac {P_2B}{P_2C} \equal{} \left(\frac {TB}{TC}\right)^2$ . In conclusion, $ \frac {P_1B}{P_1C} \equal{} \frac {P_2B}{P_2C} \equal{} \left(\frac {TB}{TC}\right)^2$ ,

i.e. $ P_1\equiv P_2$ what means the lines $ EF$ , $ TT$ , $ BC$ are concurrently.[/color][/hide][/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote the diameter $ [ST]$ of the circle $ w$ , the midpoint $ U$ of  $ [SI]$  and $ P\in EF\cap TT$ .

From the upper lemma obtain $ \boxed {\ P\in MN\ }$ . Thus, $ OU\parallel IT\ (*)$ and $ OI\perp EP$ $ \implies$

$ IE^2 \minus{} IP^2 \equal{} R^2 \minus{} OP^2$ $ \implies$ $ (IO^2 \minus{} R^2) \minus{} IP^2 \equal{} TO^2 \minus{} (R^2 \plus{} TP^2)$ $ \implies$

$ IO^2 \minus{} IP^2 \equal{} TO^2 \minus{} TP^2$ $ \implies$ $ IT\perp OP$ $ \implies$ $ IT\perp MN$ $ \stackrel {(*)}{\ \implies\ }$ $ OU\perp MN$ .[/color]

[quote="Virgil Nicula"]
[quote][color=darkred][b][u]Lemma.[/u][/b] Let $ w \equal{} C(O)$ be the circumcircle of $ \triangle ABC$ and let $ E\in (AB)$ , $ F\in (AC)$ so that $ O\in EF$ ,

$ OE \equal{} OF$ . Denote the reflection $ T$ of the point $ A$ w.r.t. the point $ O$ . Then the lines $ EF$ , $ TT$ , $ BC$ are concurrently.[/color]
[/quote][/quote]

An [color=blue]alternative solution to this lemma[/color] is the following: If $ P \equal{} EF\cap BC$ and $ PX, PY$ are the tangents to the circle (with $ Y$ in the arc $ BC$ where A is) and if $ X'$ is the antidiametrically opposite of $ X$ then $ X'.YCXB$ is harmonic and since $ X'Y//EF$ we have $ OE' \equal{} OF'$, where $ E' \equal{} X'B\cap EF$, $ F' \equal{} X'C\cap EF$. So, if $ OE < OE'$ then $ A$ is interior point of the triangle $ X'BC$ which is contradiction since $ CBAX'$ is cyclic. Similarly it can not be the case $ OE > OE'$ and so $ OE \equal{} OE'$ which means that $ X' \equal{} A$.

Now [u]returing to the initial exercise [/u](with the initial notation) we get that the points $ Y,X,I$ are colinear since the points $ E,F,P \equal{} EF\cap MN$ are($ PX,PY$ are again the tangents). If $ T$ is the mid point of $ SI$ then $ TO//PX$ so $ TO//PY$ and since $ PY \perp MN$ we take $ TO\perp MN$, which is the required result.

[color=red]Second Solution[/color]
Finally, I will give another synthetic-elementary solution:
Let $ B$ be the reflection of $ S$ with respect to $ O$. We draw the line $ RBQ\perp IB$ with $ R\in SF,Q\in SE$. Let T be the mid point of $ SI$. Then since $ TO//IB$ we have to show that $ BR \equal{} BQ \Leftrightarrow \angle BRI \equal{} \angle BQI \Leftrightarrow \angle BXI \equal{} \angle BYI \Leftrightarrow \angle FBX \equal{} \angle EBY$ (*) where $ IX\perp SF, X\in SF$ kai $ IY\perp SE, Y\in SE$ because of the cyclic quadrilaterals $ RBIX, QBIY$. We will give an elementary proof of (*). Indeed, consider points $ W,Z$ such that $ EW// \equal{} BZ \perp SB$ and $ EW \equal{} BZ \equal{} IE \equal{} IF$ then $ EBZW$ is parallelogram so $ WZ \perp SE$ and if $ WZ \cap SE \equal{} G$ then $ SGBZ$ is cyclic so $ \angle ZSB \equal{} \angle ZGB \equal{} \angle GBE \equal{} \angle EBY$ (since $ EW$ is the reflection of $ EI$ across $ EB$). Similarly, $ \angle FBX \equal{} \angle ZSB$ so $ \angle FBX \equal{} \angle EBY$, and this is the required result.

For more on (*) look at: http://www.mathlinks.ro/viewtopic.php?t=195897


Έστω $ f(x)$ με πεδίο ορισμού και τιμών το $ R$ τέτοια ώστε για κάθε [b]ε[/b]$ >0$ υπάρχει $ z > 0$ ώστε $ f(x \plus{} z) \equal{} f(x)$ για κάθε $ x$. Ισχύει ότι $ f(x) \equal{} c$ σταθερό ?

[quote="r_boris"][quote="wallace"]Έστω $ f(x)$ με πεδίο ορισμού και τιμών το $ R$ τέτοια ώστε για κάθε [b]ε[/b]$ > 0$ υπάρχει $ z > 0$ ώστε $ f(x \plus{} z) \equal{} f(x)$ για κάθε $ x$. Ισχύει ότι $ f(x) \equal{} c$ σταθερό ?[/quote]

Το ε τι ρόλο παίζει??[/quote]

Μάλλον εννοεί ότι "υπάρχει $ z<\epsilon$ ώστε..." (hehe, μάλιστα χρειάστηκε το σχόλιο αυτό για να δω ότι δεν υπάρχει γραμμένη η συνθήκη $ z<\epsilon$ :D).

Σε κάθε περίπτωση:[hide]Dirichlet[/hide]
Cheerio,

Durandal 1707

That's the spirit Durandal..... :wink: .....Dirichlet Function.....Elementary and easy αλλά δν πρέπει να την ξεχνάμε....


Help!

Why does the equations?:

$ 11x\plus{}5y$    and   $ 3x\plus{}7y$    are divisible by 62 for the same values for $ x$ and $ y$???

[quote="pascal12"]They aren't?

$ 11(1) \plus{} 5(1) \equal{} 16$
$ 3(2) \plus{} 7(2) \equal{} 20$[/quote]That's not considered a counterexample. Read the question again.

Useful fact: solve the linear diophantine equation to get the general solution. There are particular values of x and y that produces 11x+5y a multiple of 62. Use these same values of x and y to see if 3x+2y is a multiple of 62...


[quote="GameBot"]A plane is at an altitude of 20,000 feet and begins descending at a constant rate of 2,000 feet per minute. What is its altitude, in feet, after 45 seconds?[/quote]

45 seconds is 3/4 of a minute, so the plane descends $ 2000 \times \frac {3}{4} \equal{} 1500$ feet, and thus is at an altitude of $ 20000 \minus{} 1500 \equal{} \boxed{18500}$.

EDIT: Yeah, typo. Thanks.  :)


If I pass the exam for credit test, does that score count in GPA? If I don't pass it, do I need to take it next school year?
Any body took Pre-cal?What's the most important contest in this Pre-cal textbook(Advanced Mathematics A Precalculus Approach, by Ryan, Doubet, Fabricant and Rockhill)?

GPA credit depends entirely on your school

Try asking them. They are the ones who would know, not us.

I'm taking Precalc Honors, but I have no idea what a "credit test" is. Is it to get into Precalc?
Perhaps Other Problem Solving Topics would be a better place for this.

I think GPAs for most conventional schools only apply for classes taken. I don't know if it's possible that schools use a separate section to say that you got an A or something on an exam. And what if it's possible you pass and get a C? Oh crap, that'd be bad, so I doubt it. If you don't pass, it's almost implied that you have to take the class to learn the material.

I was unable to take a credit test to bypass precalc, so i just took it online. So if you don't pass just take it through BYU distance learning, its much easier that sitting through a boring class all year.


function
vector
limit
Functional Analysis
advanced fields
advanced fields unsolved

Let $ K\subset\mathbb{R}^n$ be a a compact set and $ C(K)$ the set of all continuous real-valued
functions on $ K$. $ C(K)$ is given a  vector space structure by the definitions
$ (f\plus{}g)(x): \equal{}f(x)\plus{}g(x)$, $ (\lambda f)(x): \equal{}\lambda f(x)$. Equip $ C(K)$ with the norm
\[ \parallel{} f \parallel{} \equal{} \max \left\{  |f(x)|: x\in K   \right\}.
\]
Show that $ C(K)$ is a Banach space with this norm.

if $ \parallel{} f_n \minus{} f_m\parallel{} \rightarrow 0$ then $ \forall _{x} |f_n(x)\minus{}f_m(x)| \rightarrow 0$ hence (because $ f_n(x)$ is Cauchy Sequence in $ \mathbb{R} ^n$ for each $ x$)  $ \exists _f:  \lim f_n(x) \equal{} f(x)$. Then it remains to prove that $ f \in C(K)$ and $ \parallel{}f_n \minus{}f\parallel{} \rightarrow 0$


Gauss
algebra
polynomial
topology
factorial
Ring Theory
advanced fields

I need  proofs of 2 theorems :
-D'Alembert-Gauss: each non constant polynomial has at least one root in $\mathbb{C}$
-The decompisition of a polynomial into product of irreductible polynomials is unique

The first is the fundamental theorem of algebra.

Assume there is no root of the polynomial $p(x)=x^n+...+a_1 x+a_0$ of degree $n$.

First proof:
Then $g(z)=\frac{1}{p(z)}$ is analytical and is bounded for $|z| \to \infty$. Let $C$ be such a bound.
By taking some $c$ such that $g(z) \leq C$ for all $|z| >c$ we see that $g(z)$ is bounded in the whole plane.
Thus, by Liouville's theorem, $g(z)$ is constant and so also $p(z)$.

Second proof:
Define $g_r(t) = \frac{p(e^{2 \pi i t} r)}{p(r)} \cdot \left| \frac{p(r)}{p(e^{2 \pi i t} r)} \right|$.
Then $g_r(t)$ is a loop in the unit circle.
Take $R$ to be that big that $|z^n| > |p(z)-z^n|$ for all $|z| \geq R$.
Now it's easy to see that there is a homotopy of $g_R(t)_{t \in [0,1]}$ to the loop circling around the unite circle $n$ times.
Now $g_r(t)$ gives also a homotopy when varying $r$, thus $g_R$ is homotopic to the constant loop $g_0\equiv 1$, which is only possible for $n=0$.

The second one holds for any field (more generaly every factorial ring) $K$:
The ring $K[x]$ is factorial, thus offers unique decomposition into prime/irreducible elements.

For the field case, this is easy to show be arguing about $K[x]$ being euclidean with respect to the degree of the polynomials (euclidean $\implies$ $PID$ $\implies$ factorial).


I'm using texshop on a macbook.  I cannot get bibtex to work.  I have my .bst file, but I can't figure out how to create a .bib file.  Any help would be great.  Thanks.

You are going to have to give more detail than "[i]I cannot get bibtex to work[/i]". It's not clear if it's a problem with texshop or the latex distribution or your files or what. If it's a problem with using bibtex you may find the bibtex tutorial [url=ftp://cam.ctan.org/tex-archive/info/bibtex/tamethebeast/ttb_en.pdf]Tame the BeaST[/url] useful.

Your right.  Sorry for the vagueness.  So I have the bibtex style I want to use.  I put \bibliographystyle{aer} in my document.  This is my .bst file.  I think Latex knows where to find this file because when I type %!TEX root =  aer.bst and compile the correct file comes up.  When I tell my bibliography where to start I put \bibliography{@@@} this is my .bib file.  I think this is where my problem is.  In texshop I enter my bibliographys and save it as a .bib file.  But I don't think Latex can find this file.  The above test doesn't work here.  I hope this helps make my problem clearer.  Thanks for the help.

I wonder whether Q5 at the [url=http://www.tug.org/mactex/faq/]MacTeX Distribution FAQ[/url] answers your question.


quadratics
geometry
algebra
quadratic formula

[quote="diophantient"][quote="Walk Around The River"]Factor $56x^{2}-31x+36$ into the form $(Ax+b)(Cx+d)$.

Find $a+b+c+d$

obviously i can add a+b+c+d, so really just looking for how to factor it, since you cant use the basic method that you learn in algebra 1.[/quote]
[hide="uhhh"]quadratic formula[/hide][/quote]

Can you expand on that?  Ive spent a year doing only geometry, so im a little behind on my algebraic thoughts.
As in, what do i do with the quadratic roots?

[quote="diophantient"][quote="7h3.D3m0n.117"]The Quadratic Formula is used to find roots, not for factoring.[/quote]
uh, dude, I know what the quadratic formula is, I took algebra 1...
if the roots are a and b the factorization of the quadratic is (x-a)(x-b)...
[hide="solution"]By quadratic formula, the roots are $\frac{31 \pm i \sqrt{7103}}{112}$
So the factorization is $(x-r_{1})(x-r_{2})$.
Multiplying by 112, we see that our answer is 224-(the sum of the roots).
Our answer is $224-\frac{31}{56}=\frac{12513}{56}$[/hide]
This problem ought to specify that gcf(A,b)=1, and gcf(C,d)=1, and that they all are integers by the way, otherwise there are infinitely many answers.[/quote]

It probably did state that, i just paraphrased.  And by the way, thats not the right answer.  The answer is an interger.  A small integer.

[quote="perfect628"]Well, it is [i]an[/i] answer-There is more than one way to assign $a,b,c,$ and $d$. For example, $(ax+b)(cx+d)=({a}^{2}x+ab)(\frac{c}{a}x+\frac{d}{a})$, so two possible answers are $a+b+c+d$ and ${a}^{2}+ab+\frac{c+d}{a}$ are two acceptable answers.[/quote]

they all have to be integers. sorry if i didnt say this.

im so confused.  does anyone have a solution that gives an integer?

hey eli, do you think there was a typo in that question and they meant it to be $56x^{2}+31x-36$?
Then it would factor into $(7x-4)(8x+9)$ and would be 20, which I put.

[quote="Walk Around The River"][quote="perfect628"]The equation has no real roots, so it can't be factored over integers.[/quote]

Well, the answer was an integer.

So i dont know, i get that there has to be an i, but is it possible that they cancel out or something?[/quote]

Then all of $a,b,c$ and $d$ aren't integers. It sounds like this typo was what happened.