function
inequalities
inequalities proposed

Because he has some troubles in the post,so i posted his problem.All  are from can_hang2007
 Let $ a,b,c$ be positive real numbers such that $ abc\equal{}1$. Prove that
 \[ \frac{1}{(1\plus{}a)^2}\plus{}\frac{1}{(1\plus{}b)^2}\plus{}\frac{1}{(1\plus{}c)^2}\plus{}\frac{4}{(a\plus{}b\plus{}c\plus{}1)^2} \ge 1\]

[quote="Allnames"]Because he has some troubles in the post,so i posted his problem.All  are from can_hang2007
 Let $ a,b,c$ be positive real numbers such that $ abc \equal{} 1$. Prove that
\[ \frac {1}{(1 \plus{} a)^2} \plus{} \frac {1}{(1 \plus{} b)^2} \plus{} \frac {1}{(1 \plus{} c)^2} \plus{} \frac {4}{(a \plus{} b \plus{} c \plus{} 1)^2} \ge 1
\]
[/quote]
And here is my proof.Let $ \sum a \equal{} p.\sum ab \equal{} q$
we can rewritten it
<=>$ p^4 \minus{} 4p^2q \minus{} 6p^2 \plus{} 4q^2 \plus{} 12q \plus{} 9 \equal{} f(q)\ge 0$
But $ f'(q)\le 0$ (since $ q\le \frac {p^2 \minus{} 3}{2}$
Hence $ f(q)\ge f(\frac {p^2 \minus{} 3}{2}) \equal{} 0$ we have done  :lol:

Hello Allnames! I liked your proof. Mine was becoming complex. Could u help me?
I thought :
$ \sum \frac {1}{1\plus{}a^2} \geq \sum \frac{a\minus{}1}{a^3\minus{}1}$ then taking it as $ f(a,b,c)\equal{}\sum \frac{a\minus{}1}{a^3\minus{}1}$
But it is leading to cmplicacy! :(

[quote="Sunkern_sunflora"]Hello Allnames! I liked your proof. Mine was becoming complex. Could u help me?
I thought :
$ \sum \frac {1}{1 \plus{} a^2} \geq \sum \frac {a \minus{} 1}{a^3 \minus{} 1}$ then taking it as $ f(a,b,c) \equal{} \sum \frac {a \minus{} 1}{a^3 \minus{} 1}$
But it is leading to cmplicacy! :([/quote]
I can't understand your idea,sorry :blush: ,but can you tell me about $ f(a,b,c)$
I dont think that function is not suitable

[quote="Allnames"]
 Let $ a,b,c$ be positive real numbers such that $ abc \equal{} 1$. Prove that
\[ \frac {1}{(1 \plus{} a)^2} \plus{} \frac {1}{(1 \plus{} b)^2} \plus{} \frac {1}{(1 \plus{} c)^2} \plus{} \frac {4}{(a \plus{} b \plus{} c \plus{} 1)^2} \ge 1
\]
[/quote]
$ \frac {1}{(1 \plus{} a)^2} \plus{} \frac {1}{(1 \plus{} b)^2} \plus{} \frac {1}{(1 \plus{} c)^2} \plus{} \frac {4}{(a \plus{} b \plus{} c \plus{} 1)^2} \minus{} 1\equal{}$
$ \equal{}\frac{(a^2\plus{}b^2\plus{}c^2\minus{}3)^2}{(1\plus{}a)^2(1\plus{}b)^2(1\plus{}c)^2(1\plus{}a\plus{}b\plus{}c)^2}.$ :D

[quote="arqady"][quote="Allnames"]
 Let $ a,b,c$ be positive real numbers such that $ abc \equal{} 1$. Prove that
\[ \frac {1}{(1 \plus{} a)^2} \plus{} \frac {1}{(1 \plus{} b)^2} \plus{} \frac {1}{(1 \plus{} c)^2} \plus{} \frac {4}{(a \plus{} b \plus{} c \plus{} 1)^2} \ge 1
\]
[/quote]
$ \frac {1}{(1 \plus{} a)^2} \plus{} \frac {1}{(1 \plus{} b)^2} \plus{} \frac {1}{(1 \plus{} c)^2} \plus{} \frac {4}{(a \plus{} b \plus{} c \plus{} 1)^2} \minus{} 1 \equal{}$
$ \equal{} \frac {(a^2 \plus{} b^2 \plus{} c^2 \minus{} 3)^2}{(1 \plus{} a)^2(1 \plus{} b)^2(1 \plus{} c)^2(1 \plus{} a \plus{} b \plus{} c)^2}.$ :D[/quote]
Very Nice,arqady
and the pronlem is similar
 From pvthuan
$ \frac {1}{(1 \plus{} a)^2} \plus{} \frac {1}{(1 \plus{} b)^2} \plus{} \frac {1}{(1 \plus{} c)^2} \plus{} \frac {1}{(a \plus{} b \plus{} c \plus{} 1)} \ge 1$

[quote="arqady"][quote="Allnames"]
and the pronlem is similar
 From pvthuan
$ \frac {1}{(1 \plus{} a)^2} \plus{} \frac {1}{(1 \plus{} b)^2} \plus{} \frac {1}{(1 \plus{} c)^2} \plus{} \frac {1}{(a \plus{} b \plus{} c \plus{} 1)} \ge 1$[/quote]
But it's obvious because $ \frac {1}{a \plus{} b \plus{} c \plus{} 1}\geq\frac {4}{(a \plus{} b \plus{} c \plus{} 1)^2}.$ :wink:[/quote]
Yes,arqady,My opinion is Can_hang2007 's problem is stronger than pvthuan one, also a nice ineq

[quote="Allnames"]
Yes,arqady,My opinion is Can_hang2007 's problem is stronger than pvthuan one, also a nice ineq[/quote]
I agree with you! Everything has begun here:
 http://www.artofproblemsolving.com/Forum/viewtopic.php?t=77407

[quote="Allnames"]Let $ a,b,c$ be positive real numbers such that $ abc \equal{} 1$. Prove that

$ \frac {1}{(1 \plus{} a)^2} \plus{} \frac {1}{(1 \plus{} b)^2} \plus{} \frac {1}{(1 \plus{} c)^2} \plus{} \frac {4}{(a \plus{} b \plus{} c \plus{} 1)^2} \ge 1,$

and the pronlem is similar

$ \frac {1}{(1 \plus{} a)^2} \plus{} \frac {1}{(1 \plus{} b)^2} \plus{} \frac {1}{(1 \plus{} c)^2} \plus{} \frac {1}{a \plus{} b \plus{} c \plus{} 1} \ge 1.$[/quote]$ \frac {1}{(1 \plus{} a)^2} \plus{} \frac {1}{(1 \plus{} b)^2} \plus{} \frac {1}{(1 \plus{} c)^2} \plus{} \frac {4}{(a \plus{} b \plus{} c \plus{} 1)^2} \minus{} 1$

$ \equiv\frac {1}{\left(xyz \plus{} x^3\right)^2} \plus{} \frac {1}{\left(xyz \plus{} y^3\right)^2} \plus{} \frac {1}{\left(xyz \plus{} z^3\right)^2} \plus{} \frac {4}{\left(x^3 \plus{} y^3 \plus{} z^3 \plus{} xyz\right)^2} \minus{} \frac {1}{x^2y^2z^2}$

$ \equal{} \frac {\left(x^6 \plus{} y^6 \plus{} z^6 \minus{} 3x^2y^2z^2\right)^2}{\left(yz \plus{} x^2\right)^2\left(zx \plus{} y^2\right)^2\left(xy \plus{} z^2\right)^2\left(x^3 \plus{} y^3 \plus{} z^3 \plus{} xyz\right)^2}\geq 0$

holds for [b]real[/b] numbers $ a,b,c$ such that $ abc \equal{} 1$ and $ (1 \plus{} a)(1 \plus{} b)(1 \plus{} c)(a \plus{} b \plus{} c \plus{} 1)\ne 0$.

The pronlem is similar to Question 2(i) of  IMO 2008, see : http://www.mathlinks.ro/Forum/viewtopic.php?t=215222

By the way, the following inequality

$ \frac {1}{(1 \plus{} a)^2} \plus{} \frac {1}{(1 \plus{} b)^2} \plus{} \frac {1}{(1 \plus{} c)^2} \plus{} \frac {1}{a \plus{} b \plus{} c \plus{} 1} \ge 1$

holds for [b]real[/b] numbers $ a,b,c$ such that $ abc \equal{} 1$ and $ a \plus{} b \plus{} c \plus{} 1 > 0\ne (1 \plus{} a)(1 \plus{} b)(1 \plus{} c)$.

It's nice as of old but not easy anymore because $ a \plus{} b \plus{} c \not\geq 3.$


(a) Find the maximum value of the expression $ x^2 y \minus{} y^2 x$ when $ 0 \leq x \leq 1$ and $ 0 \leq y \leq 1.$

(b) Find the maximum value of the expression
\[ x^2 y \plus{} y^2 z \plus{} z^2 x \minus{} x^2 z \minus{} y^2 x \minus{} z^2 y
\]
when $ 0 \leq x \leq 1, 0 \leq y \leq 1, 0 \leq z \leq 1$

(a) follows from (b) (there's equality in (b) when $ z\equal{}0$).

For (b), it factors into $ (x\minus{}y)(y\minus{}z)(x\minus{}z)$. It's cyclic, so we can set $ x\equal{}\max \{ x,y,z\}$. If $ y\leq z$, the expression is nonpositive, so we just need to consider $ x\geq y\geq z$. In this case,
\[ (x\minus{}y)(y\minus{}z)(x\minus{}z)\leq (x\minus{}y)yx\leq (1\minus{}y)y\leq \boxed{\frac{1}{4}}\]
where $ x\equal{}1,y\equal{}\frac{1}{2},z\equal{}0$ produces equality.

Here's my proof for the first problem(Sorry but I'm busy so I just show the first):
We want to find maximum value, so we can assume that x>y
Now applying AM-GM we have: y(x-y)<=x^2/4
Inferring from that we get: xy(x-y)<=x^3/4=<1/4(since 0=<x=<1)
The equality happens when x=1,y=1/2(Done!)


Ohhhhh vaisakhi??? i don't celebrate that b/c i am not punjabi, but regardless, happy new year to you too!  :)

Um

New Year's was like almost a month ago.

Uhh.. up to four months ago... except if you don't use the Christian calander.

Punjabi eh?
Well I am a bengali, so Poila Boisakh-er shubhechchha, shubho noboborsho etc. :)

I wonder what calendar shravas uses though? :maybe:

[quote="Ignite168"]Weirdass Indian(s?)[/quote]

Jeez you didn't have to be so offensive!

Stop these offences.

Well, my new year was 4 months ago, but, happy new year to you, bubka

[quote="Ignite168"][hide]I be Pulak (1:08:03 AM): hey sv
I be Pulak (1:08:12 AM): who's your favorite guitarrist?
spectrumvoid (1:08:15 AM): oh. Anakin from star wars
korejora (1:08:19 AM): XD
spectrumvoid (1:08:21 AM): Me.
I be Pulak (1:08:24 AM): really?
I be Pulak (1:08:27 AM): WELL YOU SUCK
thesheamuffin (1:08:30 AM): lawl
I be Pulak (1:08:31 AM): IN FACT
I be Pulak (1:08:36 AM): YOUR FAMILY SUCKS
PrimPod (1:08:49 AM):  you had it right first time, iirc
I be Pulak (1:08:49 AM): AND YOUR BLIND BROTHER
I be Pulak (1:08:53 AM): HE SUCKS AT GUITAR TOO[/hide][/quote]

Gross generalization? Sigh.

Happy new years!

[quote="Treething"][quote="Ignite168"]Weirdass Indian(s?)[/quote]

QFT [/quote]

good going  :D


[quote="AYMANE"]Find all $f$ such that:
1)$f: \mathbb{R}\tof\mathbb{R}$
2)The set $\{f(s)=0,s\in\mathbb{R}\}$ is finite.
3)$f(x^{4}+y)=x^{3}f(x)+f(f(y))$  $\forall x,y\in\mathbb{R}$[/quote]
x=0 give $f(f(y))=f(y)$. x=1 give $f(y+1)=f(y)+f(1)\Longrightarrow f(y+n)=f(y)+nf(1),n\in N$.
y=0,x=x,-x give $f(x)=-f(-x)$
$f(f(n)-n)=0 \forall n\in N \ (f(n)=nf(1))$, therefore f(1)=1.
If $f(y)\not =y$ $x=|f(y)-y|^{1/4}$ give $0=f(x)$, Therefore exist $a_{1},..,a_{k}$ suth that $f(y)=y+a_{i}, \forall y \ (a_{i}=a(y) \ one \ is \ const)$
We have $a(y+x^{4})-a(y)=x^{3}a(x)$. Therefore $a(x)=0$, it give $f(x)=x.$

Here is my solution
1)$x=0\implies f(f(y)=f(y)$
2)$y=0\implies f(x^{4})=x^{3}f(x)=f((-x)^{4})=(-x)^{3}f(-x)\implies f$ is odd
3)$x=1,y=0\implies f(0)=0$
4)Sup $\exists k\neq 0 / f(k)=0\implies f((k^{4})^{n})=0$thus the given set is infinite then $f(k)=0\implies k=0$
5)Take \[k=f(x^{4})-x^{4}\implies f(x^{4})=f(f(x^{4}))=f(x^{4}+k)=x^{3}f(x)+f(k)=f(x^{4})+f(k)\implies f(k)=0\implies k=0\implies f(x^{4})=x^{4}\] then$\forall x>0 f(x)=x$ and as $f$ is odd we have $\forall x : f(x)=x$


inequalities
vector
trigonometry
real analysis
real analysis unsolved

Let $ V$ be  a vector space with an inner product $ P$.
Let $ X,Y,U$ be vectors such that $ X,Y$ are "perpendicular " and let $ a,b$ be the "angles" defined as  : $ a\equal{}\angle (X,U) ,b\equal{}\angle (Y,U)$ .
 Prove that : $ \cos^2a\plus{}\cos^2b \leq 1$ ..  :rotfl: (related to http://www.mathlinks.ro/Forum/viewtopic.php?t=203517)
 :cool:

we can suppose the norm of $ X,Y$ are $ 1$

we have to prove that

$ (X|U)^2 \plus{} (Y|U)^2\leq \parallel{}U\parallel{}^2$

let $ F \equal{} vect(X,Y)$  with $ X,Y$ orthogonal vectors

first case $ U\in F$

$ U \equal{} cX \plus{} dY$

$ (X|U)^2 \plus{} (Y|U)^2 \equal{} c^2 \plus{} d^2 \equal{} \parallel{}U\parallel{}^2$

second case $ U \notin F$

$ U \equal{} p_{F}(U) \plus{} h$ where $ h$ is vector orthogonal to $ F$

$ (X|U)^2 \plus{} (Y|U)^2 \equal{} (X|P(U))^2 \plus{} (Y|P(U))^2 \equal{} \parallel{}p(U)\parallel{}^2 \leq \parallel{}U\parallel{}^2$

indeed Diogene it is nice.

What does mean C.B.D ?

Bravo Moubinool ! :thumbup:
An other  proof ...  Let $ P$ be the inner product .
$ P_{(X,Y)} \equal{} 0 \Longrightarrow P^2_{X,U} \equal{} P^2_{X,U \plus{} \lambda Y}$ $ \leq P_{(X,X)}P_{(U \plus{} \lambda Y,U \plus{} \lambda Y)} \equal{} P_{(X,X)}(P_{(U,U)} \plus{} 2\lambda P_{Y,U} \plus{} \lambda^2P_{(Y,Y)})$
You minimize the right term with $ \lambda \equal{} \minus{} \frac {P_{(Y,U)}}{P_{(Y,Y)}}$ , so the final inequality (C.B.D. = Created By Diogene  :D )  is :
$ P_{(X,Y)} \equal{} 0 \Longrightarrow P_{(X,X)}P^2_{(Y,U)} \plus{} P_{(Y,Y)}P^2_{(X,U)} \leq P_{(X,X)}P_{(Y,Y)} P_{(U,U)}$
Do you see how  you can obtain Moubinool inequality ?  ( http://www.mathlinks.ro/Forum/viewtopic.php?t=203517 )  :wink:
 :cool:


fellow members of the ancient aops brotherhood,
i come to seek your help in a time of great need.

please answer for me the following question which shall no doubt be a trifling for men as great as yourselves-

using only 1,3,4,6and mathematical operators +,-,*,/ and parentheses create 24.

i got (1+3)(4+6) which is 24 in base 6

is there a way to do it in base ten.

many many thanx in advance

*** Spoiler Ahead ***

I will use the ACTUAL symbols the poster presented.
>
>
.

>
.
>
>
>
>
>

6/(1 - 3/4)  Answer

= 6/(1/4)

= 6(4)

= 24

[quote="rishabh"]using only 1,3,4,6 and mathematical operators +,-,*,/ and parentheses create 24.[/quote]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=116774[/url]

[hide]$\frac{6}{1-\frac{3}{4}}$[/hide]

box of rocks you don't have to tell us about a spoiler do [bhideb] without the bs and[/bhideb] to stop hiding content


Let triangle $ABC$ be inscribed in a circle. Points $C,M$ lie on different arcs of the circle with endpoints $A$ and $B$. Chords $MK, MP$ intersect $AC,BC$ at points $H$ and $N$ respectively. $AP$ intersects $BK$ at point $I$. Show that $H,I,N$ are collinear.


function
abstract algebra
real analysis
real analysis unsolved

If a trigonometric series converge pointwisely to a L(I) function $ f$, then this series is the fourier series of $ f$.

If a sequence $ f_n$ converges to $ f$ in the $ L^1$ sense, the $ k$th Fourier coefficient of $ f_n$ converges to the $ k$th Fourier coefficient of $ f$ for each $ k$; we're integrating against something bounded.
If $ f_n$ is a truncated trigonometric series $ \sum_{j\equal{}\minus{}n}^{n}a_je^{ijx/L}$, the $ k$th Fourier coefficient of $ f_n$ is $ \begin{cases}a_k&|k|\le n\\0&|k|>n\end{cases}$. This obviously converges to $ a_k$.

I don't know if I get your answer. My post said $ S_{n}$ pointwise converges to $ f$ but in $ L^{1}$sense.
If Sn converge in $ L^{1}$ sense, it is, by definition, f's fouries series.

Well, I misread it. By definition? No, that's easy but it needs a proof (which I gave).

Pointwise convergence is a little odd here; what really makes sense is convergence pointwise a.e.

This feels like the Dominated Convergence theorem- can the partil sums be bounded by some relative of $ f$?

This property is Trigonometric series' "own" proterty. Though there're other function series share the same property, it's not so wide that can be proved by DCT.
Well by the way, what does "relative of f" mean?
And from your argument i came up with a new question: if $ f\in L^{1}$ and $ f\sim S_{n}$, is sequence $ \parallel{}S_{n}\parallel{}_{1}$ bounded?

[quote="mathiq"] if $ f\in L^{1}$ and $ f\sim S_{n}$, is sequence $ \parallel{}S_{n}\parallel{}_{1}$ bounded?[/quote]
No. This is well-known and boils down to the fact that the $ L^1$-norms of the Dirichlet kernels are unbounded as their order tends to $ \infty$. It doesn't answer your original question though :(.


[quote="mdk"]Find all functions f : N to N such that f(nm) = f(n)f(m), and such that
lim n to infinity  log(f(n))/log(n)= 1.[/quote]
there is no such fonction,
because if $ f$ exist, we have $ \lim_{n\to\plus{}\infty}\frac{f(2^{n})}{log(2^{n})}\equal{}\lim_{n\to\plus{}\infty}\frac{(f(2))^{n}}{n.log(2)}\not\equal{}1$

[quote="aviateurpilot"][quote="mdk"]Find all functions f : N to N such that f(nm) = f(n)f(m), and such that
lim n to infinity  log(f(n))/log(n)= 1.[/quote]
there is no such fonction,
because if $ f$ exist, we have $ \lim_{n\to\plus{}\infty}\frac{f(2^{n})}{log(2^{n})}\equal{}\lim_{n\to\plus{}\infty}\frac{(f(2))^{n}}{n.log(2)}\not \equal{} 1$[/quote]

Calculus error, aviateurpilot : you computed  $ \lim_{n\to\plus{}\infty}\frac{f(2^{n})}{log(2^{n})}$ instead of  $ \lim_{n\to\plus{}\infty}\frac{log(f(2^{n}))}{log(2^{n})}$

And solution exists : $ f(n) \equal{} n$
You just have to consider that $ f(n^{p}) \equal{} (f(n))^{p}$ and then $ \lim_{p\to\plus{}\infty}\frac{log(f(n^{p}))}{log(n^{p})}$ $ \equal{}\frac{\log(f(n))}{\log(n)}\equal{} 1$ and to check that this necessary condition is also sufficient.


Houses on the north side of Elm Street are numbered with consecutive even integers from $ 1,902$ to $ 2,192$. How many houses are there on the north side of Elm Street?

The house numbers from an arithmetic sequence with a common difference of 2. I.E., to count the numbers in $2,4,6,8$, we would divide by 2 to get $1,2,3,4$ (or by Mewto's method, $\frac{8-2}{2}+1$).


If $ a_1,...,a_{1988}$ are positive numbers whose arithmetic mean is $ 1988$, show that:

$ \sqrt[1988]{\displaystyle\prod_{i,j\equal{}1}^{1988} \left( 1\plus{}\frac{a_i}{a_j} \right)} \ge 2^{1988}$

and determine when equality holds.

it's obvious after using Minkowski's ineuality (or here Huygens' ineq)

Clearly $\textstyle 1+a_i/a_j\ge 2\sqrt{a_i/a_j}$. With this, we get
\[
\prod_{i,j}\left(1+\frac{a_i}{a_j}\right)\ge \prod_{i,j}2\sqrt{\frac{a_i}{a_j}} = 2^{N^2}\prod_{i,j}\sqrt{a_i}\left(\prod_{i,j}\sqrt{a_j}\right)^{-1}= 2^{N^2}.
\]
The equality holds iff $a_i=a_j$ for all $i,j$.


geometry
3D geometry
number theory proposed
number theory

Prove that infinitely many integers cannot be written as $x^2+y^2+z^9$, $(x,y,z)\in\mathbb{Z}^3$.
Related question : [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=36026]here[/url]

sorry for writing in this old topic, but you can have a look here if you are still interested in this kind of problems
http://www.comms.scitech.susx.ac.uk/fft/math/Sums_of_Squares_Cubes_and_Higher_Powers.pdf


geometry
rectangle
trigonometry
algebra
system of equations
trig identities
Law of Cosines

$ABC$ is a right triangle with $AC=5$ and $BC=12$. Point $D$ is outside of triangle $ABC$ and forms another right triangle $ADC$ with $AD=3$ and $CD=4$. What is the length of $BD$?

[hide="my answer"]$\frac{8\sqrt{85}}{5}$

Draw points E and F where triangle CED is a right triangle, triangle ADF is a right triangle, and F, D, E lie on a line. So, you have a rectangle ACEF outside triangle ACD. let $x=AF=CE$ let $y=FD$ Then, $x^2+y^2=9$ and $x^2+(5-y)^2=16$ Solving gives $y=\frac{9}{5}, x=\frac{12}{5}$ $BD^2=BE^2+DE^2$ Substituting $BE=12+x=12+\frac{12}{5}$ and $DE=5-y=5-\frac{9}{5}$ Solving gives $BD=\frac{8\sqrt{85}}{5}$[/hide]

[hide]From Law of Cosines, $BD^2=12^2+4^2-2\cdot4\cdot12\cos(BCA+ACD)$.
Since $\cos(BCA+ACD)=\cos(\pi/2+ACD)=-\sin ACD=-3/5$,
$BD=\sqrt{160+2\cdot4\cdot12\cdot(3/5)}=\sqrt{\frac{1088}{5}}=\boxed{\frac{8\sqrt{85}}{5}}$.[/hide]

[quote="mgao"][hide="solution"]2.9024127843874778748579807629094[/hide][/quote]

I assume that that answer came from a radical right.

[quote="Ignite168"][hide="Probably not the fastest or easiest way"]
Change the the 3-4-5 triangle into a rectangle forming three right triangles. Call the legs of the triangle with hypotenuse three $x$ and $y$. The legs of the triangle with hypotenuse four are $(5-x)$ and $y$. This gives you the system of equations $x^2-10x+25+y^2=16$ and $x^2+y^2=9$. Solve to get $x=\frac{9}{5}$ and $y=\frac{12}{5}$. Now just solve the new triangle with legs $\displaystyle\frac{72}{5}$ and $\displaystyle\frac{16}{5}$ to get the answer as $\boxed{\sqrt{\frac{1088}{5}}}$. Would that be simplified? If it's not, it's $\boxed{\frac{8\sqrt{85}}{5}}$[/hide][/quote]

I believe your second answer would be accepted and not your first answer.


Mrs. Goodteacher is ordering supplies for her 28 students.  The chart shows what she needs and how these items can be purchased, only in the quantities listed, through her available supplier.  A gross is 12 dozen, and a pound is 16 ounces.  What is the minimum total cost for her supplies?

$ \begin{tabular}{|c|c|c|} \hline \textbf{{I}tem} & \textbf{Quantity Needed} & \textbf{Purchase Info} \\
\hline A & Each student needs 2 pieces & 1 gross for\ \$2.38\\
\hline B & Every 2 students share 1 piece & 1 dozen for\ \$2.27\\
\hline C & Each student needs 1 ounce & 1 pound for\ \$2\\
\hline D & Every 4 students share 1 piece & \$10 each \\
\hline \end{tabular}$

It's hard to answer this because the table's all messed up.


Find the sum of each of the following infinite series. if a series does not have a sum write no sum

9-3+1-1/3+1/9 for this I got 6 3/4 is that right
1/2+1+2+4 for this I got 7 1/2

Write the first three terms of the expansion of (a-b) to the 10th
I got a to the 10th+10a to the ninth b+100a to the eigth +b squared

what are the last three terms of (a-b) to the seventh

for this I got 70a squared b to the fifth + 7ab to the sixth + b to the sventh

I would really like to know if I got these problems right before I turn them in so any help will be greatly apprciated.

Did you just add up the terms? Is it not supposed to be an infinate sequence?

Can you make your question clearer ?
It's seems to me an interessant question !


Let $ a,b,c>0$.Prove that:
$ a\plus{}b\plus{}c\plus{}\frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a}\geq \frac{6(a^2\plus{}b^2\plus{}c^2)}{a\plus{}b\plus{}c}$

[hide="Sharper inequality"]Let $ a,b,c>0$.Prove that:
$ 3(a\plus{}b\plus{}c)\plus{}2(\frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a})\geq \frac{15(a^2\plus{}b^2\plus{}c^2)}{a\plus{}b\plus{}c}$[/hide]


We have  x,y,z>0 & x+y+z=1
 Solve the inequality
\[ \sqrt {x^2  \plus{} y}  \plus{} \sqrt {y^2  \plus{} z}  \plus{} \sqrt {z^2  \plus{} x}  \le 2
\]

we have with $ a;b;c;d>0: a\plus{}b\equal{}c\plus{}d and |a\minus{}b|\leq|c\minus{}d|\Rightarrow\sqrt{a}\plus{}\sqrt{b}\geq\sqrt{c}\plus{}\sqrt{d}$
$ x\plus{}y^2\plus{}y\plus{}z^2\equal{}(x\plus{}y)^2\plus{}z\plus{}y^2$
$ z\equal{}Min{x;y;z}$
that $ x\plus{}y^2\minus{}y\minus{}z^2\geq\minus{}(1\minus{}z)^2\plus{}z\plus{}y^2\Leftrightarrow x\geq z,Right$
       $ x\plus{}y^2\minus{}y\minus{}z^2\leq(1\minus{}z)^2\minus{}z\minus{}y^2\Leftrightarrow(y\minus{}z)(1\minus{}y\minus{}z)\geq 0,Right$
$ \Rightarrow LHS\geq x\plus{}y\plus{}\sqrt{z\plus{}x^2}\plus{}\sqrt{z\plus{}y^2}\geq 1\minus{}z\plus{}\sqrt{4z\plus{}(1\minus{}z)^2}\equal{}2$


linear algebra
matrix
quadratics
algebra
polynomial
function
linear algebra theorems

a)If $ X \in M_{m,n} (C)$ prove that $ \det (XX^t ) \ge 0$
b)If $ A_1 ,A_2 ,...,A_k \in M_n (R);$ prove that $ \det (\sum\limits_{k \equal{} 1}^n {A_i^t A_i } ) \ge 0.$

Show me if posibble an elementary solution.I'm still at highschool.

What do you already know about eigenvalues and about the notion of a positive definite (or semidefinite) quadratic form?

I know the connection between eigenvalues and principal minors,what means a hermitian matrix and that a polynomial function f(x)>0 $ \forall x > 0,x < 0$ is a quadratic form when it is written as a sum of squres.

[quote="omegatheo"]a)If $ X \in M_{m,n} (C)$ prove that $ \det (XX^t ) \ge 0$[/quote]
That's false as written. Just let $ X\equal{}\begin{bmatrix}1&0\\0&i\end{bmatrix}$ so that $ XX^T\equal{}\begin{bmatrix}1&0\\0&\minus{}1\end{bmatrix}$ and has determinant $ \minus{}1.$

If you're going to do this sort of thing with complex matrices, you need to be using $ X^*\equal{}\overline{X}^T,$ the Hermitian conjugate, rather than the simple transpose.

For now, let me just assume that you're using real matrices.

The inner product $ u\cdot v$ can be written as a matrix product as $ u^Tv.$ Then $ u^Tu\equal{}\|u\|^2\ge 0$ for any $ u.$

For any real symmetric matrix $ B,$ the associated quadratic form is $ u^TBu.$ We say $ B$ is positive semidefinite if $ u^TBu\ge 0$ for all $ u.$ Positive definite would mean that $ u^TBu>0$ for all $ u\ne 0.$

A real symmetric matrix is diagonalizable. (Orthogonally diagonalizable, but we don't need that here.) It is positive semidefinite if and only if all its eigenvalues are nonnegative; positive definite iff all its eigenvalues are positive. I won't prove that here, but it's not that hard to do.

Now consider $ XX^T.$ Look at $ u^TXX^Tu\equal{}(X^Tu)^T(X^Tu)\equal{}\|X^Tu\|^2\ge 0$ for all $ u,$ so $ XX^T$ is positive semidefinite, meaning that its eigenvalues are all nonnegative. But then its determinant is the product of its eigenvalues, and the product of nonnegative numbers is nonnegative. That settles (a).

As for (b): can you show that the sum of a finite number of positive semidefinite matrices is positive semidefinite?

Another way to solve a) is to use the [url=http://en.wikipedia.org/wiki/Cauchy%E2%80%93Binet_formula]Cauchy-Binet formula[/url].  Every term works out to the determinant of a square matrix multiplied by its conjugate transpose and $ \det X^{*} \equal{} \overline{ \det X}$ for such matrices.


inclusive, if:

1) Each syllable of a number takes 0.3 second to say.

2) The word "and" is not spoken in a number.

3) No pauses are allowed in between numbers.

4) Numbers, such as 100 and 1000, have their leading "1" spoken as in the example of "one hundred."

This can probably be simplified.
[hide]
Counting from $1$ to $N$, inclusive.
Now assign a distinct variable to each digit of $N$, such as $123= 100c+10b+a$ where $a=3,b=2,c=1$.

For the units digit, $a$,
$0.3a$, unless $b=1$, if $b=1$ then $0.3 \left( a \right)$ for $a = 1,2,3$.

For the tens digit, $b$,
$0.9+2.7b+2.1b+0.6b$, simplified $0.9+5.4b$, for $a \neq 1,2,3$.

For the hundreds, $c$,
$0.9c+2.7c+2.1c+4.8c+0.9c$, simplified $11.4c$.
$\vdots$
For the $10^{n}$, $n \in \mathbb Z$, place with digit $z$,
$0.9z+2.7z+4.8z+0.9\left( 9 z \left( n-1 \right) \right)+0.9z$, simplified $1.2z+8.1nz$, for $4 \leq n \leq 33$ because past that there is more than $3$ syllables.

Then you add up parts from each digit to get the time to say it.
[/hide]

[hide="Answer"]I got 25200.9 s. Is that correct?[/hide]

Sorry, I am unable to verify this.

I will leave this for another user.


Let $ b$ be an even positive integer. Assume that there exist integer $ n > 1$ such that $ \frac {b^{n} \minus{} 1}{b \minus{} 1}$ is perfect square.
Prove that $ b$ is divisible by 8.

That can't be the only condition. Of course $ \frac{3^2\minus{}1}{3\minus{}1}$ is a square.

Thanks a lot [b]scorpius119[/b] I missed to write that $ b$ is even number.I have edited my post!
I am so sorry about this mistake!

We have
\[ \frac{b^{n} \minus{} 1}{b \minus{} 1} \equal{} b^{n \minus{} 1} \plus{} \ldots \plus{} b \plus{} 1 \equal{} k^{2}
\textrm{, \ \ }
n, k \in \mathbb{N}
\textrm{, \ }
n \neq 1
\textrm{,}
\]
from which we can get
\[ k^{2} \minus{} 1 \equal{} b \sum_{i \equal{} 0}^{n \minus{} 2}{b^{i}}
\textrm{.}
\]
Because $ \displaystyle b \equiv 0 \pmod{2}$, from the left side follows $ \displaystyle k \equiv 1 \pmod{2}$, so $ k \equal{} 2t \minus{} 1$, $ t \in \mathbb{N}$, $ t \neq 1$ (because $ b \geqslant 2$).

Now we get
\[ 4t \left( t \minus{} 1 \right) \equal{} b \sum_{i \equal{} 0}^{n \minus{} 2}{b^{i}}
\textrm{.}
\]

Notice that $ 4t \left( t \minus{} 1 \right) \equiv 0 \pmod{8}$ and
\[ \sum_{i \equal{} 0}^{n \minus{} 2}{b^{i}} \equiv 1 \pmod{2}
\textrm{,}
\]
from which directly follows $ b \equiv 0 \pmod{8}$.

Q. E. D.

[hide="Easier"]Let $ \frac{b^{n}\minus{}1}{b\minus{}1}\equal{}k^{2}$, so that $ 1\plus{}b\plus{}\cdots \plus{}b^{n\minus{}1}\equal{}k^{2}$.  The left side of this equation is odd, so $ k$ must be odd also.  Since squares are 0, 1, or 4 mod 8, we have $ k^{2}\equiv 1\pmod{8}$, so

$ b\plus{}b^{2}\plus{}\cdots \plus{}b^{n\minus{}1}\equiv 0\pmod{8}$
$ b(1\plus{}b\plus{}\cdots \plus{}b^{n\minus{}2})\equiv 0\pmod{8}$

Since $ 1\plus{}b\plus{}\cdots \plus{}b^{n\minus{}2}$ is odd, $ 8|b$.[/hide]

Easily problem but it is beautiful problem


Let $A,B,C,D\in R^{+}$ and for all real $x$,
$\sin Ax+\sin Bx=\sin Cx+\sin Dx$
Prove that $A=C$ or $A=D$

[hide]
Hint: multiply both sides by $\sin(Ax)$, and then integrate over a sufficiently large interval $[0, T]$.
[/hide]
[hide]
Assume that this is not the case. That is, $A\neq C$ and $A\neq D$. Multiply both sides of the identity by $\sin Ax$, and integrate over $[0, T]$. \[\int_{0}^{T}\left(\sin Ax+\sin Bx\right) \sin Ax dx = \int_{0}^{T}\left(\sin Cx+\sin Dx\right) \sin Ax dx.\] If $B\neq A$, the left hand side is equal to \[\left(\frac{T}{2}-\frac{\sin(2AT)}{4A}\right)+\frac12 \left(\frac{\sin((A-B)T)}{A-B}-\frac{\sin((A+B)T)}{A+B}\right) = \frac{T}{2}+O(1).\] if $B=A$, the left hand side is equal to \[2 \left(\frac{T}{2}-\frac{\sin(2AT)}{4A}\right) = T+O(1).\] But the right hand side is $O(1)$, it is equal to \[\frac12 \left(\frac{\sin((A-C)T)}{A-C}-\frac{\sin((A+C)T)}{A+C}\right)+\frac12 \left(\frac{\sin((A-D)T)}{A-D}-\frac{\sin((A+D)T)}{A+D}\right) = O(1).\] We get a contradiction.[/hide]

This problem has more elementar solution :P ,it's from All Russian Math Olimpiad,5th round!

[quote="Jumbler"]Let $a,b,c,d\in R^{+}$ and for all real $x\ ,\ \sin ax+\sin bx=\sin cx+\sin dx\ .$ Prove that $a=c$ or $a=d\ .$[/quote]
$\lim_{x\to 0}\frac{\sin ax+\sin bx}{x}=\lim_{x\to 0}\frac{\sin cx+\sin dx}{x}\Longrightarrow \boxed{\ a+b=c+d\ }\ .$

$2\sin \frac{(a+b)x}{2}\cos\frac{(a-b)x}{2}=2\sin\frac{(c+d)x}{2}\cos\frac{(c-d)x}{2}$ $\Longrightarrow$ $\cos\frac{(a-b)x}{2}=\cos\frac{(c-d)x}{2}$ $\Longrightarrow$

$\lim_{x\to 0}(a-b)^{2}\cdot \frac{1-\cos \frac{(a-b)x}{2}}{\left[\frac{(a-b)}{2}\right]^{2}}=\lim_{x\to 0}(c-d)^{2}\cdot\frac{1-\cos\frac{(c-d)x}{2}}{\left[\frac{(c-d)x}{2}\right]^{2}}\Longrightarrow$ $\boxed{\ (a-b)^{2}=(c-d)^{2}\ }\ .$

$a+b=c+d\ \ \wedge\ \ (a-b)^{2}=(c-d)^{2}\Longrightarrow$ $a=b\ \ \vee\ \ a=d\ .$

[b]Remark.[/b] I used the following remarkable limits : $\lim_{x\to 0}\frac{\sin x}{x}=1$ and $\lim_{x\to 0}\frac{1-\cos x}{x^{2}}=\frac{1}{2}\ .$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Hi,

I'm dealing with a problem asking me to prove that a group of order 21 is cyclic. I've already shown there are exactly one Sylow 7-subgroup and we are asked to assume that there is only one subgroup of order 3.

Now, clearly, we have four subgroups and they are all normal. Is there a way to use this fact to prove the group is cyclic? Thanks!

There is a wonderfully useful theorem that says that if all the Sylow subgroups are normal, then the group is a direct product of them. So in this case $ G\cong \mathbb{Z}/3\times \mathbb{Z}/7\cong \mathbb{Z}/21$

Thanks -- that is a useful theorem! :lol:

Note that you need the assumption of just one $ 3$-Sylow. Otherwise you would have another group of order $ 21$.

A group of order $ 21$ need not to be cyclic.
Theorem: if $ |G| \equal{} pq$, where $ p<q$ are primes and $ p|q\minus{}1$ then there are exactly two groups of order $ pq$ - one abelian and other nonabelian.


integration
trigonometry
logarithms
real analysis
real analysis unsolved

$ 1\blacktriangleright$ $ \int\frac{2x^{3}-1}{x^{6}+2x^{3}+2x^{2}+1}\ dx$

$ 2\blacktriangleright$ $ \int\frac{x^{2}+x}{\left(e^{x}+x+1\right)^{2}}\ dx$.

$ 3\blacktriangleright$ $ \int\frac{x^{2}}{(x^{2}-4)\sin x+4x\cos x}\ dx$.

helle, write the integrand of (1) in the following form:
$ -\frac{1}{2}\frac{\sqrt{2}\left(\sqrt{2}x+\frac{5}{2}\sqrt{2}x^{4}+\sqrt{2}\right)}{1+\left(\frac{1}{2}\sqrt{2}x^{2}+\frac{1}{2}\sqrt{2}x^{5}+\sqrt{2}x\right)^{2}}+\frac{x}{1+\frac{1}{2}x^{4}}$
Sonnhard.


Consider the following casino game: There is a set of numbers $S$, and some numbers may appear twice in the set. You get a certain amount of money when you play the game, but you must choose a number (it does not have to be in $S$). A number is chosen randomly, and you must pay the amount you were off by. What number do you pick?

Ummmm....

Is there any range on what S is?

If not, 0?


[quote="The QuattoMaster 6000"][quote="frost13"][quote="espark52"]Whic of the fractions $\frac{13}{40}$, $\frac{17}{52}$, $\frac{21}{64}$, $\frac{25}{76}$ is the smallest?

plz dont say "use calculator"[/quote]

Each fraction is almost $\frac{1}{3}$.  If you were to triple each they would each be one piece short of $\frac{1}{3}$.

The new fraction $\frac{75}{76}$ would be closest to 1 and therefore the biggest while $\frac{39}{40}$ would be the furthest from 1 whole and therefored the smallest!![/quote]
Oh!  That is a nice way in doing it.  Here is my method.
[hide] Take $\frac{13}{40}$ and multiply both the numerator and denominator by 1.3.  We get 16.9/52, which is less than 17/52.  Now multiply the numerator and denominator of $\frac{13}{40}$ by 1.6.  We get 20.8/64, which is less than 21/64.  Finally, multiply the numerator and denominator of $\frac{13}{40}$ by 1.9.  We get 24.7/76, which is less than 25/76.  Thus, $\frac{13}{40}$ is smaller than the rest of the fractions, so it is the smallest one.  [/hide][/quote]

Thats a nice way too, I just divided in my head, since I'm good at that kind of "hardcore" math.  :wink:


This is problem from school round of highest category (category A), but it's not hard:
For what positive integers $k$ has $k(k-2)\leq(k+\frac{1}{k})x\leq k^2(k+3)$  exactly $(k+1)^2$ solutions in integers (unknown is x)?

Nobody wants to solve it? Maybe it doesn't look very good, but it's quite easy... belive me :) Or you want a hint?

Wow, such a long time has passed and nobody is attempting this..
I'll give a couple of hints, though really they're not really anything that anyone trying it shouldn't already have tried...

[hide="Step 1"]Get x by itself :D[/hide]

[hide="Step 2"]Simplify each side with long division..[/hide]

[hide="Step 3"]Look at the integer parts and fiddle with the fraction parts..[/hide]

$k(k - 2) \le \left( k + \frac 1 k \right) x \le k^2(k + 3)$

implies (as $k > 0$)

$\frac{k^2(k - 2)}{k^2 + 1} \le x \le \frac{k^3(k + 3)}{k^2 + 1}$

$k - 2 - \frac{k - 2}{k^2 + 1} \le x \le k^2 + 3k - 1 - \frac{3k - 1}{k^2 + 1}$

For $k = 1$, we get

$-1 + \frac 1 2 \le x \le 3 - \frac 2 2$
$-\frac 1 2 \le x \le 2$

If $x$ is an integer, the inequality has 3 solutions 0, 1, 2, which is less than $(1 + 1)^2 = 4$.

For $k > 1$,

$0 \le \frac{k - 2}{k^2 + 1} < 1$

$0 < \frac{3k - 1}{k^2 + 1} \le 1$

with the upper left and lower right equalities both only for $k = 2$. Therefore, if $x$ is an integer,

$k - 2 \le x \le k^2 + 3k - 2$

The number of integers between $m = k - 2$ and $n = k^2 + 3k - 2$ inclusive is

$n - m + 1 =  k^2 + 3k - 2 - (k - 2) + 1 =  k^2 + 2k + 1 = (k + 1)^2$

Hence, the inequality has exactly $(k + 1)^2$ integer solutions for all $k > 1$.

I was thinking about writing solution on my own, because noone was posting it for long time, but you did it. Thanx yetti.


number theory
prime factorization
number theory unsolved

Let $n>3$ be an odd integer with prime factorization $n = {p_1}^{a_1}...{p_k}^{a_k}$.
If $m = n(1-\frac{1}{p_1})...(1-\frac{1}{p_k})$, prove that there is a prime $p$ such that $p$ divides $2^m-1$, but does not divide $m$.

Has been posted here several times, you know.

In the UK IMO TST this year one of the problems was:

Find all positive integers n such that $2^{\phi(n)} - 1$ divides $n^n$.

To solve the problem I showed that there exists a prime dividing $2^{\phi(n)} - 1$ but not dividing n (except for some small values of n).

That is of course exactly the same problem as the one you proposed just now.

I am attaching my solution for that problem (don't have time to retype it)

for instance :
[url=http://www.mathlinks.ro/Forum/topic-42271.html]www.mathlinks.ro/Forum/topic-42271.html[/url]


AMC
AMC 10
AMC 10 A
geometry
3D geometry
rhombus
rectangle

It is rather interesting to see the great discussion on AMC 12A 2008. Is it possible to post some questions from AMC10A 2008 and discuss them? It would be helpful for middle school students to learn and take AMC 10.

Is it generally harder for AMC 10A this year than the previous years?

Thanks,

Also, can someone please post questions 15 and 21 from the AMC 10A?

I thought the AMC was a lot easier this year.

Number 15: Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian.  Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian.  Han drove 70 miles more than Ian.  How many more miles did Jan drive than Ian.
[hide="Comments"]
We could simply let Ian drive 0 hours.  It is extremely easy from here.

Or you could actually solve.[/hide]
[hide="Answer"]D) 150[/hide]
Number 21: A cube with side length 1 is sliced by a plane that passes through two diagonally opposite vertices A and C and the midpoints B and D of two opposite edges not containing A or C, as shown.  What is the area of quadrilateral ABCD?

No figure, sorry.  Note the figure is a rhombus.  The area of a rhombus is given by $ \frac{1}{2}d_1d_2$ where $ d_i$ represents the length of the diagonals.

[hide="Answer"]A) $ \frac{\sqrt{6}}{2}$[/hide]

Dang, for #21 I thought it was a square and put 5/4  :(

can someone post last few problems for AMC10A 2008? Thanks.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=189172[/url]
That's the 25th problem on AMC 10A and the 22nd on AMC 12A.

24. Let $ k\equal{}2008^2\plus{}2^{2008}$. What is the units digit of $ k^2\plus{}2^k$?

[hide="Comment/Hint"]Find the units digit of $ k$ first in order to find the units digit of $ k^2$.  Also note that $ k$ is obviously divisible by $ 4$.  This should immediately give you the answer.[/hide]
[hide="Answer"]D) 6[/hide]

23. Two subsets of the set $ S\equal{}{a,b,c,d,e}$ are to be chosen so that their union is $ S$ and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter?

[hide="Hint"]There are two ways to do this problem:
1) How many ways are there to choose exactly two elements to be intersected? This should give you an idea of what to do.
2) Use casework: Only two cases to consider, as all the others overlap.[/hide]

[hide="Answer"]B) 40[/hide]

Dang... How did I get the last 3 questions right on the AMC 10A, yet miss #8 due to thinking 15$ difference between b and a instead of a and b, 21 because i thought it was a square, and 19 because i had no idea and should've left it blank.  Now i'm probably getting a 118.5, which is way too close for comfort.

i hope theres this much geometry on the 10B...

someone asked me topost the question as well:

#17 answer if anyone needs it

[hide]17 is just the combined area of 3 rectangles with width 3 and length 6 and 3 120 degree circle sectors of radius 3. It will probably be clearer if you draw it.

So 3*6*3 + 3*1/3*(pi)*(3^2) = 54 + 9pi[/hide]

the future: i hope the opposite. xD

I hope theres this much on the 12B  :wink: I fail at polynomials and number theory on the last few questions.  Easier ones are simple but.... eeeeeh

i didnt think the 10A would have that much of a difference compared to the 12A...

For example, I think numbers 3,4,5 on the 12 were like numbers 4,9,12 on the 10...

had i known i wouldve taken the 10.


The compound "Cyclopenta-1,2 diene " does not exist as a stable species.

It does not exist because there isn't room enough on a five membered ring to accomodate 3 linear atoms. Perhaps you are referring to "Cyclopenta-1,3 diene + ethyne" (actually, when one says "cyclopentadiene" we are almost always referring to cyclopenta-1,3 diene because of its use in Diels-Alder reactions.

it is actually a cyclopentane with the te lower two carbons haveing double bomd with their upward counterparts

Then a Diels-Alder reaction will occur and the product will be norbornadiene.


calculus
integration
combinatorics open
combinatorics

Find All possible (i) integral  (ii) positive integral  solutions of the equation

[img]c:\untitled.jpg[/img]

:D

You can't include datas from your harddisk that way ;)
Either attach them to your post (see the button "Add an Attachment" at the bottom of the "Post a reply" page), or upload them somewhere to the internet and link there.
But it would be much better to type down the equation instead of giving a picture...

PS: integral equations belong to number theory.


algebra
function
domain
integration
geometry
geometric transformation
reflection

Suppose $ f: \mathbb{D} \to U$ is a biconformal bijective map that extends continuously to $ \overline{\mathbb{D}}$. Then $ s \to f(e^{is})$, $ s \in [0,2\pi]$ is rectifiable.

This is supposed to follow from Brother Riesz theorem, but I am utterly lost. We are supposed to show that $ f' \in H^{1}$, right? Well I don't know where to start...

Also, is it always true that $ f$ extends continuously? This does not quite seem to be the content of Caratheodory's theorem.

I'm quite at loss too. You can extend the Riemann mapping continuously to the boundary if the domain is bounded by a closed Jordan curve. But it would be completely ridiculous to claim that every Jordan curve is rectifiable :?.

I must have forgot to jot down some condition. I have some kind of psuedo-solution written down to this problem (that I apparently don't know what it is).

It says:
"Riesz gives this if we can show
$ 0 \equal{} \int_{\Gamma} (f^{ \minus{} 1})^n dz \equal{} \int_0^{2\pi} e^{int} \, d\mu$
where $ \Gamma$ is the boundary curve of $ U$."

It doesn't enlighten me too much though. How do we integrate over something we know not to be rectifiable? Maybe this problem is completely misstated. Maybe we are supposed to show the opposite, that if it's rectifiable then $ f' \in H^1$ ? I thought this was clear though.

Alas, I'm still at loss. My mind reading skills are not good enough to figure out what was meant. Why don't you just ask your professor what the correct statement was?

Indeed, the question was that to show that if $ U$ is bounded by a jordan curve and $ s \to f(e^{is})$ is rectifiable, then $ f' \in H^1$. I was shown a solution (where the idea is essentially to use what I wrote in the post above), but I won't write it down unless someone wants me to.

I would like to edit the original title to better reflect what the question is, but as usual I am not allowed to edit it. I don't understand why we're not allowed to edit posts after a few days.


Using each of the digits exactly once, place the digits 3,4,5,8 into each box below, to make the sum as great as possible. What is the greatest possible sum? Express your answer as a common fraction.
\[ \frac{\square}{\square} \plus{} \frac{\square}{\square}\]

The largest possible sum is $ \frac{8}{3}\plus{}\frac{5}{4}\equal{}\fbox{47/12}$.


The product of two consecutive negative integers is 272. Find the larger of the two integers.

[hide="Solution"]By trial-and-error, we get $ \minus{} 16$ and $ \minus{} 17$. Since the problem asks for the [i]larger[/i] integer, our answer is $ \boxed{ \minus{} 16}$.[/hide]

Instead of doing a lot of trial and errors, you can do this:

$ \sqrt{272}$ and choose the two integers next to it. You will find that $ 16 \cdot 17\equal{}272$ Then, you make the integers negative because it ends up in the same product. So then, you find the larger in value which is $ \boxed{\minus{}16}$.


integration quadratics calculus derivative logarithms function trigonometry

My thoughts on seeing $\int_{0}^{1}\frac{dx}{1+x^{5}}$:

This is a rational function. Can we find, in closed form, the factors (linear and quadratic) of the denominator? Yes we can; I do know how to do that (lots of golden means and other uses of $\sqrt{5}$). If you can factor the denominator, then, through partial fractions, you can compute the integral in closed form.

Therefore it is possible to compute this integral in closed form. Would it be possible to do that by hand? Yes, it would. Am I personally willing to demonstrate that? Not today; maybe not ever.

It's time for your favorite computer algebra system (Maple, Mathematica, Derive, TI-92 or TI-89, ... )

[quote="Kent Merryfield"]
It's time for your favorite computer algebra system (Maple, Mathematica, Derive, TI-92 or TI-89, ... )[/quote]

I once wondered about the integral,
$\int \frac{1}{1+x^{n}}dx$
Whether it has closed form.
After using the "Integrator" for many different values I am convinced yes.
However, the size of the anti-derivative gross exponentially .

This is what my favorite computer algebra system (Mathematica 6) says:

$\int^{1}_{0}\frac{1}{1+x^{n}}dx =\frac{\frac{\Gamma ' \left( \frac{n+1}{2n}\right)}{\Gamma \left( \frac{n+1}{2n}\right)}-\frac{\Gamma ' \left( \frac{1}{2n}\right)}{\Gamma \left( \frac{1}{2n}\right)}}{2n}$

If I try an indefinite integral, it's giving it in terms of some HyperGeometric2F1 function...
$\int = x \left( \text{HyperGeometric2F1}\left(\frac{1}{n},1,1+\frac{1}{n},x^{-n}\right) \right)+C$
Looking in the help menu, HyperGeometric2F1 is series-defined, so that really doesn't help.

As for the original problem, the first formula above works; it's:

$S=-\frac{1+\sqrt{5}}{80}\left(-8 \sqrt{\frac{2}{\sqrt{5}+5}}\pi+2(\sqrt{5}-3) \text{ArcCsch}(2)-5 \sqrt{5}\ln 2+\ln 8+\ln \left( 843-377 \sqrt{5}\right)+\sqrt{5}\ln (3+\sqrt{5}) \right)$

$S \approx 0.8883135726516996 \ldots$

after using FullSimplify.

[quote="The Zuton Force"]This is what my favorite computer algebra system (Mathematica 6) says:

$S=-\frac{1+\sqrt{5}}{80}\left(-8 \sqrt{\frac{2}{\sqrt{5}+5}}\pi+2(\sqrt{5}-3) \text{ArcCsch}(2)-5 \sqrt{5}\ln 2+\ln 8+\ln \left( 843-377 \sqrt{5}\right)+\sqrt{5}\ln (3+\sqrt{5}) \right)$

$S \approx 0.8883135726516996 \ldots$

after using FullSimplify.[/quote]

holy smokes !!  :wacko:  :P

 :rotfl:

$ \sum_{k\equal{}0}^\infty \frac{(\minus{}1)^k}{2k\plus{}1}\equal{}\frac{\pi}{4} \csc\left(\frac{\pi}{2}\right)$

$ \sum_{k\equal{}0}^\infty \frac{(\minus{}1)^k}{3k\plus{}1}\equal{}\frac{\pi}{6} \csc\left(\frac{\pi}{3}\right)\plus{}\frac{\ln 2}{3}$

$ \sum_{k\equal{}0}^\infty \frac{(\minus{}1)^k}{4k\plus{}1}\equal{}\frac{\pi}{8} \csc\left(\frac{\pi}{4}\right)\plus{}\frac{\ln(1\plus{}\sqrt{2})}{2\sqrt{2}}$

$ \sum_{k\equal{}0}^\infty \frac{(\minus{}1)^k}{5k\plus{}1}\equal{}\frac{\pi}{10}\csc\left(\frac{\pi}{5}\right)\plus{}\frac{1}{\sqrt{5}}\ln\left(\frac{1\plus{}\sqrt{5}}{2}\right)\plus{}\frac{\ln 2}{5}$

$ \sum_{k\equal{}0}^\infty \frac{(\minus{}1)^k}{6k\plus{}1}\equal{}\frac{\pi}{12} \csc\left(\frac{\pi}{6}\right)\plus{}\frac{\ln(2\plus{}\sqrt{3})}{2\sqrt{3}}$

Now, the sum from $ \minus{}\infty$ to $ \infty$ is actually nice- and I posted it here a few years ago.

Using the fact that Digamma(s+1) = Digamma(s) + 1/s, we can express these series in terms of the digamma function:

10/6 + 10/16 + 10/26 + ... = Digamma(infinity) - Digamma(6/10)
10/11 +10/21 + 10/31 + ... = Digamma(infinity) - Digamma(11/10)

Hence, the sum we want is 1 - (1/6 - 1/11 + 1/16 -_... = 1 - (Digamma(11/10) - Digamma(6/10)) / 10.

Gauss's formula for the Digamma function at rational numbers can be used to calculate explicit values for these digamma expressions.
As posted earlier, the result involves sin and cos of (pi/10) and can be expressed in terms of logarithms, radicals, the gamma constant, and pi.
(Note: cos(pi/5) = (1+sqrt(5))/4).

Edit: I forgot the initial 1- in the sum (for index k = 0).

$ S \equal{} \sum_{k \equal{} 0}^\infty \frac {( \minus{} 1)^k}{5k \plus{} 1} \equal{} \frac {1}{10} \left[\psi\left(\frac {6}{10}\right) \minus{} \psi\left(\frac {1}{10}\right)\right]$

After applying the digamma theorem

$ \psi\left(\frac {p}{q}\right) \equal{} \minus{} \gamma \minus{} \ln(2q) \minus{} \frac {\pi}{2}\cot\left(\frac {p\pi}{q}\right) \plus{} \sum_{k \equal{} 1}^{q \minus{} 1} \cos\left(\frac {2\pi p k}{q}\right)\, \ln\,\sin\left(\frac {k\pi}{q}\right)$

$ S \equal{} \frac {1}{10} \left[\frac {\pi}{2}\left(\cot \frac {\pi}{10} \minus{} \cot \frac {6\pi}{10}\right) \plus{} \sum_{k \equal{} 1}^9 \left[\cos\left(\frac {2\pi 6k}{10}\right) \minus{} \cos\left(\frac {k\pi}{10}\right)\right]\,\ln\,\sin\frac {k\pi}{10} \right]$

$ \equal{} \frac {1}{10} \left[\pi \csc \frac {\pi}{5} \minus{} \frac {\sqrt {5} \plus{} 1}{2}\ln \frac {\sqrt {5} \minus{} 1}{4} \plus{} 0 \plus{} \frac {\sqrt {5} \minus{} 1}{2}\ln \frac {\sqrt {5} \plus{} 1}{4} \plus{} 0 \plus{} 0 \plus{} 0 \plus{} \frac {\sqrt {5} \minus{} 1}{2}\ln \frac {\sqrt {5} \plus{} 1}{4} \plus{} 0 \minus{} \frac {\sqrt {5} \plus{} 1}{2}\ln \frac {\sqrt {5} \minus{} 1}{4} \right]$

Hence $ S \minus{} \frac {\pi}{10} \csc\frac {\pi}{5} \equal{} \frac15\left[\frac {\sqrt {5} \minus{} 1}{2}\ln \frac {\sqrt {5} \plus{} 1}{4} \minus{} \frac {\sqrt {5} \plus{} 1}{2}\ln \frac {\sqrt {5} \minus{} 1}{4} \right]$

$ \equal{} \frac15 \left[\frac {\sqrt {5}}{2}\left[\ln \frac {\sqrt {5} \plus{} 1}{4} \minus{} \ln \frac {\sqrt {5} \minus{} 1}{4}\right] \minus{} \frac12\left[\ln \frac {\sqrt {5} \plus{} 1}{4} \plus{} \ln \frac {\sqrt {5} \minus{} 1}{4}\right]\right]$

$ \equal{} \frac15 \left[\frac {\sqrt {5}}{2}\ln\left(\frac {\sqrt {5} \plus{} 1}{2}\right)^2 \minus{} \frac12 \ln \frac14\right] \equal{} \frac {1}{\sqrt {5}} \ln\left(\frac {\sqrt {5} \plus{} 1}{2}\right) \plus{} \frac {\ln 2}{5}$

Doe John : Do you know any article where can I find digamma theorem ? I think it's can be proved by residue ,

You find a proof of Gauss' digamma theorem on [url=http://planetmath.org/?op=getobj&from=objects&id=8550]PlanetMath[/url].


Problem2
Find all the integers solutions of the equation $ x^{8}\plus{}2^{2^{x}\plus{}2}\equal{}p$, where $ p$ is a prime number,

This topic title is complety useless. Use something meaningful, describing the problem!


Let $p \geq 5$ be a prime number. Prove that the number \[ \frac
{p^2 +2 }3\] is an integer, and can be written as a sum of three
perfect squares.

[quote="rrusczyk"]We're talking primes other than 3, eh?

Feel free to post answers to the parts separately!  One part is a lot easier than the other, I think.[/quote]Oops, forgot to add $p\geq 5$ :)

Any prime greater than three is not divisible by three, so it must equal 1 or 2 (mod 3).

1 or 2(mod 3) squared both equal 1(mod 3)

1+2=3=0(mod3)

So the numerator is divisible by three, and the number is an integer

Spoiler:

[hide]Since p is a prime other than 3, it can be written as either 3x+1 or 3x-1.  I will do 3x+1 first and 3x-1 second.

1) ((3x+1)^2+2)/3 = (9x^2+6x+3)/3 = 3x^2+2x+1 (which is an integer)  This can then be written as (x+1)^2+x^2+x^2, the sum of 3 squares.

2) ((3x-1)^2+2)/3 = (9x^2-6x+3)/3 = 3x^2-2x+1 (which is an integer)  This can then be written as (x-1)^2+x^2+x^2, the sum of 3 squares.[/hide]


probability
articles
conditional probability
algebra
binomial theorem

1. Three hundred of the 500 juniors are girls, and 250 of the 400 seniors are girls. A name is drawn at random. If the name is that of a girl, find the probability that she is a senior.

2. Spinner A is divided into three equal sectors colored red, white and blue, one color per sector. Spinner B is divided into five equal sectors, one red, one white, and three blue. If each arrow is spun once what is the probability that they land in sectors of the same color?

3. Cards are drawn, one at a time with replacement, from a standard deck. What is the probability that the second spade is obtained on the sixth draw?

4.The Blue Sox win 20% of their games if they are losing going into the seventh inning, 60% of their games if they are tied going into the seventh inning. Going into the seventh inning. Going into the seventh inning, they are losing 40% of the time and winning 60% of the time. What percentage do they win?

5. For the three events A, B, C: P(A and B and C) is 0.1, P(A and B) is 0.3, P(A and C) is 0.2, P(B and C) is 0.4, and P(A) equals P(B) equals P(C). What is the value of P(A)?

6. Chris Duhon's free throw percentage is 83.1%. If the Bulls are down by one point and Chris Duhon is about to shoot two free throws. If he makes both the Bulls win if he makes one they go to overtime (50% for each team winning) and if he misses both the bulls lose. If it is known that the Bulls won the game what is the probability that the first free throw of the two was made?

You guys don't have to answer all at once these were in my book, and might be on my test, so I wanted to see the solutions, so what better place than to ask you guys? Any help is appreciated. I'm sure almost all of them are really easy and I just missed something. Thanks in advance guys.

Most of these problems seem like conditional probability, so maybe I should explain that to help?

Conditional probability is the probability of the occurrence of some event A given that some event B has occurred. This would be "the probability of A, given B," notated as $ P(A|B)$. It is also known that $ P(A|B)\equal{}\dfrac{P(A \text{ and } B)}{P(B)}$. This can be thought of in another logical way, with the typical $ \text{Probability of Success} \equal{} \dfrac{\text{Successful Outcomes}}{\text{Total Outcomes}}$. Since you know that event B must have occurred, the total outcomes would be P(B), and since A occurring would be a success (and B has already happened), the successful outcomes would be P(A and B).

Hopefully this will help, but if not, I or many other AoPSers (plenty more qualified than me) can give more problem-specific help. But for now, I'm off to my homework. :D

Good luck on your test!

Thanks this does help i already finished the first one. However now I think my problem is the order stuff like "The probability of the second spade on the sixth draw" Thanks for the luck too I'll need it lol.

EDIT: I am done with the second. I'm not really sure why it was so hard... now I feel stupid.

1, 4, and 6 are all clear examples of conditional probability. 2 is just casework. 3 is not conditional probability, and 5 just plain confuses me... (if I figure out what it's saying, I'll post it... can someone else help?)

[hide="3:"]
The important part about this problem is that cards have replacement, so draws are independent from one another. Keeping this in mind, we must find the probability of having the second spade on the sixth draw. This has two main parts:
Part 1: There is exactly one spade drawn among the first 5 cards.
Part 2: The 6th card drawn is a spade.

These probabilities can be calculated separately, and then multiplied together to give the final answer. (You can do them yourself, but I'll hide the answers in case.
[hide="Part 1"]In any draw, the probability of getting a spade is $ \frac{1}{4}$. Now we must find the probability of getting 1 spade and 4 non-spades. Using the binomial distribution $ (p\plus{}q)^5$, with $ p\equal{}\frac{1}{4}$ as "success," and $ q\equal{}\frac{3}{4}$ as "failure." The relevant term, $ 5pq^4$, simplifies to $ \frac{405}{1024}$. Alternatively, you can assume that the first card is a spade, then the other four are not spades, and find this probability. Then you multiply by 5 for all different cases of card 1,2,3,4, or 5 being the spade.[/hide][hide="Part 2"]Clearly, it's $ \frac{1}{4}$.[/hide][hide="Final Answer"]Multiply $ \frac{405}{1024}\cdot\frac{1}{4}$ to get $ \frac{405}{4096}$.[/hide][/hide]

[quote="Brut3Forc3"]1, 4, and 6 are all clear examples of conditional probability. 2 is just casework. 3 is not conditional probability, and 5 just plain confuses me... (if I figure out what it's saying, I'll post it... can someone else help?)

[hide="3:"]
The important part about this problem is that cards have replacement, so draws are independent from one another. Keeping this in mind, we must find the probability of having the second spade on the sixth draw. This has two main parts:
Part 1: There is exactly one spade drawn among the first 5 cards.
Part 2: The 6th card drawn is a spade.

These probabilities can be calculated separately, and then multiplied together to give the final answer. (You can do them yourself, but I'll hide the answers in case.
[hide="Part 1"]In any draw, the probability of getting a spade is $ \frac {1}{4}$. Now we must find the probability of getting 1 spade and 4 non-spades. Using the binomial distribution $ (p \plus{} q)^5$, with $ p \equal{} \frac {1}{4}$ as "success," and $ q \equal{} \frac {3}{4}$ as "failure." The relevant term, $ 5pq^4$, simplifies to $ \frac {405}{1024}$. Alternatively, you can assume that the first card is a spade, then the other four are not spades, and find this probability. Then you multiply by 5 for all different cases of card 1,2,3,4, or 5 being the spade.[/hide][hide="Part 2"]Clearly, it's $ \frac {1}{4}$.[/hide][hide="Final Answer"]Multiply $ \frac {405}{1024}\cdot\frac {1}{4}$ to get $ \frac {405}{4096}$.[/hide][/hide][/quote]

All are done but 5 and 6 (especially 5). Both confuse me. The only problem about 6 is the making first but not second bit. (The A in the equation I believe) And thanks Brut3Forc3 you are a life saver.

#5 looks like it can be done using the [[Principle of Inclusion Exclusion]], but there's one thing missing. Are you sure that's the entire problem?

Yeah, I figured it was a PIE problem, but it seemed to be missing something... I think you assume that P(A or B or C)=1.

Yeah you can assume that. Hopefully now it's doable (Is doable a word)?

EDIT: I also figured 6 out.

EDIT 2: I'm still lost on 5, I can't figure this PIE stuff out, it's just making me hungry. ;)

Well if we're assuming $ P(A\cup B \cup C)\equal{}1$ then:
[hide="Answer for #5?"]
\[ 1\equal{}P(A)\plus{}P(B)\plus{}P(C)\minus{}P(A\cap B)\minus{}P(A\cap C)\minus{}P(B\cap C)\plus{}P(A\cap B\cap C)\]
Since $ P(A)\equal{}P(B)\equal{}P(C)$, and considering the given, we immediately get
\[ 1\equal{}3P(A)\minus{}.9\plus{}.1 \implies 1.8\equal{}3P(A)\]
\[ \boxed{P(A)\equal{}.6}\][/hide]
I'm not too confident, but this appears to hold after looking at a venn diagram.

[quote="Pistachio"]1. Three hundred of the 500 juniors are girls, and 250 of the 400 seniors are girls. A name is drawn at random. If the name is that of a girl, find the probability that she is a senior.
[/quote]
$ 1$
[hide="#1"]
So the total amount of people is $ 900$ There are $ 250$ senior girls.
This amounts to:
$ \frac{250}{900}\equal{}\frac{5}{19}$[/hide]

[quote="moogra"][quote="Pistachio"]1. Three hundred of the 500 juniors are girls, and 250 of the 400 seniors are girls. A name is drawn at random. If the name is that of a girl, find the probability that she is a senior.
[/quote]
$ 1$
[hide="#1"]
So the total amount of people is $ 900$ There are $ 250$ senior girls.
This amounts to:
$ \frac {250}{900} \equal{} \frac {5}{19}$[/hide][/quote]
[hide="No..."]You are given that a girl's name is drawn, so the total probability is $ \frac {250}{300 \plus{} 250} \equal{} \frac {250}{550} \equal{} \frac {5}{11}$.[/hide]

Well thanks a lot guys, especially Brut3Forc3, you guys really helped me on my test today, on which I think I did alright.

[quote="Brut3Forc3"]1, 4, and 6 are all clear examples of conditional probability. 2 is just casework. 3 is not conditional probability, and 5 just plain confuses me... (if I figure out what it's saying, I'll post it... can someone else help?)

[hide="3:"]
The important part about this problem is that cards have replacement, so draws are independent from one another. Keeping this in mind, we must find the probability of having the second spade on the sixth draw. This has two main parts:
Part 1: There is exactly one spade drawn among the first 5 cards.
Part 2: The 6th card drawn is a spade.

These probabilities can be calculated separately, and then multiplied together to give the final answer. (You can do them yourself, but I'll hide the answers in case.
[hide="Part 1"]In any draw, the probability of getting a spade is $ \frac {1}{4}$. Now we must find the probability of getting 1 spade and 4 non-spades. Using the binomial distribution $ (p \plus{} q)^5$, with $ p \equal{} \frac {1}{4}$ as "success," and $ q \equal{} \frac {3}{4}$ as "failure." The relevant term, $ 5pq^4$, simplifies to $ \frac {405}{1024}$. Alternatively, you can assume that the first card is a spade, then the other four are not spades, and find this probability. Then you multiply by 5 for all different cases of card 1,2,3,4, or 5 being the spade.[/hide][hide="Part 2"]Clearly, it's $ \frac {1}{4}$.[/hide][hide="Final Answer"]Multiply $ \frac {405}{1024}\cdot\frac {1}{4}$ to get $ \frac {405}{4096}$.[/hide][/hide][/quote]
quick questin brut3, i did everything you did before looking at yours for number 3 however i don't understand why you multiply by 5, could you clarify please? Thanks in advance!

For the first method, I used a [url=http://en.wikipedia.org/wiki/Binomial_distribution]binomial distribution[/url]. (Okay, that article might be a bit confusing...) In case you don't know what that is, it's just a way of finding the probability of a certain amount of successes in some amount of trials. In such a problem, you can set $ p$=probability of success, $ q\equal{}1\minus{}p$=probability of failure, $ n$=amount of trials, and $ k$=amount of desired successes. You then expand $ (p\plus{}q)^n$ with the binomial theorem and find the relevant term, $ \binom{n}{k}p^kq^{n\minus{}k}$. For this problem, $ p\equal{}\frac{1}{4}, q\equal{}\frac{3}{4}, n\equal{}5, k\equal{}1$. The $ \binom{5}{1}$ part is the multiplication by 5.

For the second method, I used a counting argument. First, I found the probability that the first card would be a spade, and then the next 4 would not be. This probability was equal to $ \frac{1}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} \cdot \frac{3}{4}\equal{}\frac{81}{1024}$. However, it is not necessary that the first card is the spade; any of the five cards can have the spade. Each of these cases is equally likely, so you can multiply the first probability by 5 to get $ \frac{405}{1024}$, which gives the same result as the other method.

Oh, i see,
i had guessed maybe that was your thinking but wasn't quite sure. It makes sense now, thanks i had assumed that counting principles weren't that relevant and was going in circles thinking that the position of the spade didn't matter in the first 5 cards because either way you are still multiplying by the same numbers but because there are 5 different ways of doing this that adds in to the probability, thanks


group theory
superior algebra
superior algebra unsolved

Prove that if $ A$ is a set with at least three elements, then the symmetric group $ (S_A,\circ)$ is nonabelian. (question 13.10 'Mathematical Proofs' by Gary Chartrand) Sorry if this question sounds trivial but i'm really stuck and don't have anyone to turn to for help. Any help is much appreciated.

it suffices to show it for a set with 3 elements.

consider the permutations $ \sigma$ and $ \tau$ defined by

$ \sigma(1)\equal{}1,\sigma(2)\equal{}3,\sigma(3)\equal{}2$,

$ \tau(1)\equal{}2,\tau(2)\equal{}1,\tau(3)\equal{}3$

then $ \sigma \tau(1)\equal{}3$ , but $ \tau \sigma(1)\equal{}2$, so $ \sigma$ and $ \tau$ do not commute.


$(a)$ Suppose that $a \neq 0$ and $c \neq 0$, and that $ax^{3}+bx+c$ has a factor of the form $x^{2}+px+1$. Show that $a^{2}-c^{2}= ab$.

$(b)$ In this case, prove that $ax^{3}+bx+c$ and $cx^{3}+bx^{2}+a$ have a common quadratic factor.

part $a$was from iran olympiad some years ago and for first round


The probability of $ A$ winning against $ B$ is $ x$, logically the probability of $ B$ winning against $ A$ is $ 1 \minus{} x$.

The probability of $ A$ winning against $ C$ is $ y$.

If $ 1 \minus{} x > 1 \minus{} y$ So $ B$ wins agains $ C$.
If $ x \equal{} y$ is 50-50 for each.

What is the exact probability of $ B$ winning against $ C$?

It depends :)
It depends on the assumptions about the experiment to determine the winner.

One (but not unique) possible answer is:
Suppose that each player has his own strength. And the probability is proportional to this strength. So we have:
$ x\equal{}\frac{a}{a\plus{}b}$, $ y\equal{}\frac{a}{a\plus{}c}$
We need $ \frac{b}{b\plus{}c}$

[hide="answer"] $ \frac{y\minus{}xy}{x\plus{}y\minus{}2xy}$ [/hide]

Also, note that in general probabilities are NOT transitive:

Defining die A to win against die B if die A shows a higher number than B with probability > 1/2, it is possible to construct three different dice A,B,C (by assigning numbers to the faces that are different from the usual numbers 1,2,3,4,5,6) so that A wins against B, B wins against C and C wins against A.


[quote="Cezar Lupu"]Let $x, y, z$ three positive reals. Prove that \[\sqrt{xy(x+z)(y+z)}+\sqrt{yz(x+y)(x+z)}+\sqrt{zx(y+z)(x+y)}\geq xy+yz+zx+\sqrt{3(x+y+z)}.\] [/quote]
Cezar try $x=y=1,$ $z\rightarrow0^{+}.$ :wink:


[quote="khashi70"]For $a,b,c > 0$ prove that :
$3(a+\sqrt{ab}+\sqrt[3]{abc})\leq 4(a+b+c)$[/quote]
[hide="hint"]I think you should prove that $left side \ge right side$ by something like AM-GM(arithmetic mean$\ge$ geometric mean then, subtraction is possible because the difference would be $\ge0$.[/hide]

[hide="Hint"]Equality at a = 16, b = 4, c = 1.[/hide]


Find all $f:\mathbb{R}\backslash\{0\}\to \mathbb{R}$ such that for all $x,y$ that $xy \notin \{ 0,-1\}$  we have
\[ xf \left(x+\frac 1y\right)+yf(y)+\frac yx=yf\left(y+\frac 1x\right)+xf(x)+\frac xy. \]

Indeed, this one is nice, but I have a question: HOW ON EARTH SOLVE STUDENTS IN YOUR COUNTRY THESE PROBLEMS? I find them simply impossible. I have been trying to solve this one for one hour and half and I didn't get anywhere. Please, help me!

[b]Answer:[/b] $f(x)=x+\frac{a}{x}+b$ that $a,b$ are real numbers

[i]Solution:[/i] You see for real $c$ if f is solution $f+c$ is also solution. So suppose $f(1)=1$ and instead of $y$ put $\frac1y$. \[xf(x+y)+\frac1yf(\frac1y)+\frac1{xy}=\frac1yf(\frac1x+\frac1y)+xf(x)+xy\ \ (1)\] In $(1)$ change place of $x,y$: \[yf(x+y)+\frac1xf(\frac1x)+\frac1{xy}=\frac1xf(\frac1x+\frac1y)+yf(y)+xy\ \ (2)\] Multiply equation $(1)$ by y and equation $(2)$ by $x$ and subtract this two equations. \[f(\frac1x)-f(\frac1y)+xy(f(x)-f(y))=\frac1x-\frac1y+xy(x-y)\ \ (3)\] In equation $(3)$ put $y=1$ then \[f(\frac1x)+xf(x)=\frac1x+x^{2}.\] So \[xf(x)=x^{2}+\frac1x-f(\frac1x)\ \ (4)\] Now we know \[f(\frac1x)-f(\frac1y)+y(xf(x))-x(yf(y))=\frac1x-\frac1y+xy(x-y)\] Put $xf(x)$ and $yf(y)$ from equation $(4)$ \[f(\frac1x)(1-y)-f(\frac1y)(1-x)=\frac{1-y}x-\frac{1-x}y\ \ (5)\] In equation $(5)$ put $y=2$ $f(\frac1x)=\frac1x+x(f(\frac12)-\frac12)+1$
Therefore: \[f(x)=x+\frac1x(f(\frac12)-\frac12))+1\] So there are real $a,b$ that $f(x)=x+\frac{a}{x}+b$.
 :P  :P

We can shift the function by any number. So for example, if $f(x)$ is a solution, as mentioned by Omid Hatami, so if $f(x)+c$. Thus, if $f(x)$ is a solution, $g(x)=f(x)-f(1)+1$ is also a solution, and $g(1)=1$.


function
algebra
polynomial
calculus
derivative
algebra unsolved

If the function f(x)=a*(x^3)+2*(x^2)+bx-4 is <1-1> prove that it has a solution at
[0,1], a>0

Yes this shows that for every number the price of the function is different
This is a quite difficult problem. I wait your answers

$f(0)=-4$.  there's a root in between [0,1] iff $f(1)=a+b-2>0$.

using proof by contradiction: Suppose $a+b-2<0$, so $a+b<2$ (i).

since $f(x)$ is a cubic polynomial with first coefficient >0, for it to be one-to-one its derivative is non negative.  $f '(x) =3ax^2+4x+b \geq 0$.  So $16-4(3a)(b) \leq 0 \rightarrow 4/3 \leq ab$.  since $a>0$, we know that $b>0$ now.

2>a+b from eqn (i), so $1 \geq (a+b)/2\ge \sqrt{ab}$, so $1 \geq ab$.  But this is a contradiction since we established that $4/3 \leq ab$, so therefore $a+b-2>0$ and $f$ has a root on the interval [0,1].

edit
well actually its not right.  if we can show $f '(x) \neq 0$ then its right, but i am not sure how to do that yet.


let $x$and $y$ two positive real numbers and $m$is the smallest of $x,\frac{1}{y},y+\frac{1}{x}$
prove that $m \le \sqrt{2}$

Easy. Let's assume that $m>\sqrt{2}$ and all these numbers are greater than $\sqrt{2}$. Then: \[x,\frac{1}{y},\frac{1}{x}+y>\sqrt{2}.\] Of course $\frac{1}{x}<\frac{\sqrt{2}}{2}$ and also $y< \frac{\sqrt{2}}{2}$. Adding these two inequalities we get $\frac{1}{x}+y<\sqrt{2}$ - contradiction with $\frac{1}{x}+y>\sqrt{2}$. The contradiction proves that $m\geq \sqrt{2}$.


[quote="cosinerburc"]is this true
27(a^3+b^3+c^3+abc)^2>=16(a^2+b^2+c^2)^3 where a,b,c are positive reals.[/quote]

Wrong for a = 9, b = 9, c = 1.

  darij


probability
rotation
analytic geometry
geometry
symmetry
integration
calculus

The original question:

Three points are chosen randomly and independently on a circle. What is the probability that all three pairwise distances between the points are less than the radius of the circle?

Answer: 1/12

My question:

4 points are chosen randomly and independently on a circle. What is the probability that all 4 pairwise distances between the points are less than the radius of the circle?

The first question is easy to solve using geometric probability, but I don't know how to solve my problem.  Somebody help please!

[quote="Yongyi781"]Here's an even harder problem: [b]Two [/b]points are chosen randomly in a circle. What is the probability that the distance between them is less than the radius of the circle? You may want to use MathWorld for that.[/quote]

Let $ P$ be one of the points. Draw the radius through $ P$. By symmetry we need only consider the probability if $ P$ is on that one radius. Analytic geometry bash, letting the circle be centered at the origin and the radius $ P$ is on be aligned on the positive x-axis. Hence $ P$ is at $ (p,0)$. We let the radius of the circle be $ 1$ as well. Finally, let's assume $ P$ isn't the center of the circle just for convenience - it won't make a difference to our continuous probability anyways.

The locus of points satisfying the condition is the intersection of $ x^2 + y^2\le 1$ (or $ < 1$; we're talkin' continuous probability here) and $ (x - p)^2 + y^2\le 1$.

We can find the area of their intersection using integrals. I'll post the formula I used, but this is intermediate topics, so don't ask me to explain, please. $ 2(\int_{p - 1}^{p/2}p + \sqrt {1 - y^2} + \int_{p/2}^{1}\sqrt {1 - y^2})$. In fact, $ \int_{p - 1}^{p/2}p + \sqrt {1 - y^2} = \int_{p/2}^{1}\sqrt {1 - y^2}$, so we can just evaluate one and multiply it by 4 instead of 2. I'll evaluate the second, since it's much easier.

We then get $ 2[\pi - (\frac {p\sqrt {1 - \frac {p^2}{4}}}{2} + \arcsin \frac {p}{2})]$. The probability is thus for a specific $ P$ chosen as we dictated above $ 2[\pi - (\frac {p\sqrt {1 - \frac {p^2}{4}}}{2} + \arcsin \frac {p}{2})] \div \pi$. Now I suppose I have to evaluate
\[ \sum_{0\le p \le 1} (2[\pi - (\frac {p\sqrt {1 - \frac {p^2}{4}}}{2} + \arcsin \frac {p}{2})] \div \pi)
\]
.

but that looks hopelessly intimidating. Hmph.

You do mean in, right? Because if it's on, then it's trivial, and plus I would have just wasted several dozen minutes of my life.

Can somebody please solve my problem?  It's really getting on my nerves...

4 points are chosen randomly and independently on a circle. What is the probability that all 4 pairwise distances between the points are less than the radius of the circle?

[b]@ AIME15:[/b] Here's the casework skeleton for your original problem. Using this, all you have to do now is raw computation.

[b]Problem restatement:[/b] [i]Find the probability that, of four points placed on a circle, no two are separated by an acute angle of more than 60 degrees.[/i]
[hide="Casework skeleton"]Affix point 1 at angle 0. Point 2 can be anywhere in -60..60, so there is a 1/3 chance that point 2 is well-placed.
From hereon, by symmetry, let's assume [b]point 2 is placed in 0..60[/b]. Let the angle of point 2, in degrees, be t.

Point 3, with angle s, can be anywhere in t-60..60. We can tackle this part casewise:

[b]Case 1:[/b] s is in t..60
The probability that s lands in t..60, that has a probability of (60-t)/360
Point 4 must then land in s-60..60, which has a probability of (120-s)/360

[b]Case 2:[/b] s is in 0..t
The probability that s lands in 0..t is t/360
Point 4 must then land in t-60..60, which has a probability of (120-t)/360

[b]Case 3:[/b] s is in t-60..0
The probability that s lands in t-60..0 is (60-t)/360
Point 4 must then land in t-60..s+60, which has a probability of (120+s-t)/360[/hide]

[b]@ Temperal:[/b] You can't assume that $ P$ lies on one particular radius. Note that there are twice as many possible $ P$ a distance of $ 2/3$ from the origin as there are a distance of $ 1/3$ from the origin, for example, because the circumference of a circle of radius $ 2/3$ is twice that of a circle of radius $ 1/3$. The following link tells you how to [url=http://en.wikipedia.org/wiki/Polar_coordinate_system#Generalization]integrate in standard polar coordinates[/url].

[b]@ Yongyi:[/b] Assume a radius of 1, and let $ r$ be the midpoint of the segment connecting center of the circle and the first point.
[hide="Solution"]In terms of $ r$, the area of the region in which the second point may fall is $ 2 \text{arccos}(r) \minus{} 2 r \sqrt {1 \minus{} r^2}$. Thus, we need to find the integral:
\[ I \equal{} \int_{0}^{2 \pi} \int_{0}^{1/2} r \left( 2 \text{arccos}(r) \minus{} 2 r \sqrt {1 \minus{} r^2} \right) dr d \theta \equal{} \frac {\pi ^2}{4} \minus{} \frac {3 \pi \sqrt {3}}{16}
\]
We need to now scale it down appropriately. If point 2 could occur anywhere in the circle for all possible point 1's (which would result in a probability of 1), the volume would be:
\[ T \equal{} \int_{0}^{2 \pi} \int_{0}^{1/2} r \pi dr d \theta \equal{} \frac {\pi ^2}{4}
\]
The probability is thus $ \boxed{P \equal{} \frac {I}{T} \equal{} 1 \minus{} \frac {3 \sqrt {3}}{4 \pi}}$[/hide]

[quote="AIME15"][quote="mismatchtea"]Can you explain that first problem (AMC 12 #25) please......I can't seem to get 1/12 (I instead get 1/36).[/quote]

Draw our circle, and mark point A.  Then rotate the circle so A is at the top.  In 60 degree intervals, going clockwise, label points 0, 1, 2, 3 and counterclockwise: 0,-1,-2,-3.  -3=3.

Note that when we select point B, it must be between 1 and -1.  Let's say it's at -0.75.  Then point C can be anywhere from -1 to 0.25.  Pick a few more points.  We can graph this region on a coordinate grid, and will look like a hexagon centered at the origin, with area $ 3$.  Try graphing it yourself.  Anyway, the total region that is allowed is a 6x6 square, so the total area is 3/(6*6)=1/12.[/quote]

I thought the points were IN the circle and was really sweating it >_>


The mean of ten numbers is 7.2. When an 11th number is added to
the set, the mean becomes 5. What is the 11th number?

Before the 11th number, the sum of the ten numbers is $ 7.2 \times 10$ or $ 72$.  When the 11th number is added, the sum becomes $ 5 \times 11$ or $ 55$.  Thus, the added number is $ 55\minus{}72$ or $ \boxed{17}$.

He meant $ 55\minus{}72\equal{}\boxed{\minus{}17}$.


inequalities
function
trigonometry
inequalities unsolved

In acute triangle $ \triangle ABC,x,y,z$ are three inner angles,show that $ \tan x \plus{} \tan y \plus{} \tan z > \frac {27}{(\frac {1}{\sin x} \plus{} \frac {1}{\sin y} \plus{} \frac {1}{\sin z})^3}$

$ \frac{1}{\sin k}>1$ for $ k \in (0,\frac{\pi}{2})$
As $ x,y,z$ are the angles of an acute triangle, they are in $ (0,\frac{\pi}{2})$
therefore  $ \frac{1}{\sin x}\plus{}\frac{1}{\sin y}\plus{}\frac{1}{\sin z}>3$
 $ (\frac{1}{\sin x}\plus{}\frac{1}{\sin y}\plus{}\frac{1}{\sin z})^3>27$
and $ \frac{27}{(\frac{1}{\sin x}\plus{}\frac{1}{\sin y}\plus{}\frac{1}{\sin z})^3}<1$
so if we can prove that $ \tan x\plus{}\tan y\plus{}\tan z>1$, the original inequality is also true.
By dividing by three we get $ \frac{1}{3}\cdot\tan x\plus{}\frac{1}{3}\cdot\tan y\plus{}\frac{1}{3}\cdot\tan z>\frac{1}{3}$
Because $ \tan k$ is convex on $ (0,\frac{\pi}{2})$, we can use Jensen's inequality
$ \frac{1}{3}\cdot\tan x\plus{}\frac{1}{3}\cdot\tan y\plus{}\frac{1}{3}\cdot\tan z\geq\tan(\frac{x\plus{}y\plus{}z}{3})$
As $ x,y,z$ are the inner angles of a triangle, their sum is $ \pi$, so $ \tan(\frac{x\plus{}y\plus{}z}{3})\equal{}\tan\frac{\pi}{3}\equal{}\sqrt{3}\approx 1.7>\frac{1}{3}$
So we have $ \tan x\plus{}\tan y\plus{}\tan z>1>\frac{27}{(\frac{1}{\sin x}\plus{}\frac{1}{\sin y}\plus{}\frac{1}{\sin z})^3}$


geometry
angle bisector
geometry proposed

Can anyone help?
I think I've seen this problem posted before but it's nowhere to find.
I think it's enchantingly difficult.
Two circles $\odot O_1$ and $\odot O_2$ intersect at $A$ and $B$
Line $O_1O_2$ intersects $\odot O_1$ at $C$ and $D$,$D$ is on the segment $O_1O_2$.
Line $O_1O_2$ intersects $\odot O_2$ at $E$ and $F$,$E$ is on the segment $O_1O_2$.
The angle bisector of $\angle DAF$ intersects segment $DF$ at $G$
The midpoint of $\widehat {EBF}$ is $H$,ray $HG$ intersects $\odot O_2$ at $I$
The midpoint of $\widehat {CAD}$ is $J$,segment $JG$ intersects $\odot O_1$ at $K$
Prove that line $AB,KD,IF$ are concurrent
Neither $O_1$ nor $O_2$ is in the interior of the other circle

Are you sure about your problem? Or is it because I messed up with notations?

It seemed that you've made a mistake,shobber.
$K$ is on $\odot O_1$

This is the correct Figure.Seemed that you have to download it to see clearer.

Have to show that DKIF  is cyclic.  Then 3 radicals will be concurrent.

   T.Y.

   M.T.

But how?It is trivial if that cyclic is proven.

I solved it in Iranian Community,I will translate it soon.
http://www.mathlinks.ro/Forum/viewtopic.php?t=66793

I'm sorry but what's that language...I cannot understand it...


Set $b_{n}= \frac{1}{a_{n}}$.  Then $b_{n}$ obeys the recurrence \[b_{n+1}= b_{n}-\frac{1}{2006+1/b_{n}}\, .\] In particular, we get the inequality \[b_{n+1}> b_{n}-\frac{1}{2006}\, .\] Iterating that inequality shows that $b_{n}> 2-\frac{n}{2006}$, which is at least $1$ for $n \le 2006$.  In particular, we have $a_{2006}< 1$.

For $n \le 2006$, we have shown $b_{n}> 1$.  Plugging back in the original recurrence for $b_{n}$, for $n \le 2006$ we get the inequality \[b_{n+1}< b_{n}-\frac{1}{2007}\, .\] By iterating that inequality, we see that $b_{2006}< 2-\frac{2006}{2007}= \frac{2008}{2007}$.  Taking reciprocals gives $a_{2006}> \frac{2007}{2008}= 1-\frac{1}{2008}$.

[quote="shobber"] $a_{n+1}=a_{n}+\frac{1}{2006}a_{n}^{2}$ , $n \in N$, $a_{0}=\frac{1}{2}$.

Prove that $1-\frac{1}{2008}< a_{2006}< 1$.[/quote]

This is a very interesting problem.  First of all it reduces to iteration of $f(x) = x+x^{2}$, starting from $x = 1/4012$  (divide the given equation by 2006).  Next, for small $x$, $f(x)$ will be very close to $x+x^{2}+x^{3}+x^{4}+...$  and iteration of that series can be carried out explicitly; it sends $1/n$ to $1/(n-1)$.   The problem is saying that if we start from $x = 1/2n$ then this appoximation is highly accurate even for $n$ steps of the iteration.   It cannot be accurate forever, because eventually the values must become large.


Find the largest possible real number $ \lambda$ in the inequalities $ |\lambda xy \plus{} yz|\leq\frac {\surd{5}}{2}$ such that $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$ for any real numbers $ x,y,z$.

First of all,
$ y^2(2x\plus{}z)^2 \leq y^2(x^2\plus{}z^2)(2^2\plus{}1^2) \equal{} 5y^2(1\minus{}y^2) \leq \frac {5}{4}$
$ |2xy\plus{}yz|\leq \frac {\sqrt{5}}{2}$.
Therefore, taking $ \lambda \equal{} 2$ is valid.

Put $ x\equal{}\frac {\sqrt{2}}{\sqrt{5}}, y\equal{}\frac{\sqrt{2}}{2}, z\equal{}\frac {1}{\sqrt{10}}$, one has $ |\frac {\lambda}{\sqrt{5}}\plus{}\frac {1}{2\sqrt{5}}|\leq \frac {\sqrt{5}}{2} \Leftrightarrow |2\lambda\plus{}1|\leq 5\to \lambda\leq 2$.

Hence, $ \lambda_{max} \equal{} 2$


On Monday, all 435 members of the US Congress voted and Proposition  PI3714 was defeated.  Another vote was held on Tuesday.  Thirty-two Congressmen switched their votes from “No” to “Yes”.   Five Congressmen switched their votes from “Yes” to “No”.  Of the 435 votes on Tuesday, there were 7 more “Yes” votes than “No” votes.  How many more “No” votes than “Yes” votes were there on Monday?

A.25	B.29	C.34	D.35	E. 47

[hide=" click to reveal hidden content"]

On Monday, Yes $ \equal{}x$, No $ \equal{}435\minus{}x$

On Tuesday, Yes $ \equal{}x\plus{}32\minus{}5 \equal{} x\plus{}27$, No = $ 435\minus{}x\minus{}32\plus{}5 \equal{} 408 \minus{} x$

$ x\plus{}27 \equal{} 408 \minus{} x\plus{}7\equal{} 415\minus{}x$.

Therefore $ x\equal{}194, 435\minus{}x \equal{} 241$.

Therefore difference $ \equal{}47$.

[/hide]


vector
inequalities
inequalities unsolved

I'm sure it is right but I can't prove it, please help me .
 Note:  vertor AB =  (AB--->)

Let A_1, A_2, ..., A_n be n distinct points in the plane and a_1, a_2 , ... , a_n be n real numbers satisfying the condition:  a_1 + a_2 + ... +  a_n <> 0.
Let I be the point satisfying:
    (a_1)(IA_1--->) + (a_2)(IA_2--->) + ...+ (a_n)(IA_n--->) = vector 0
  For every point O in the plane:
      Prove that:
        OA_1 + OA_2 + ... + OA_n >= n* OI

I'm sorry, it is wrong with OA_1=1;OA_2=2;OA_3=3 and a_1=1;a_2=2;a_3=3.  Please don't try to prove it.

oh i think mr Thai will say about that soon
in one book about vector i have seen it,
yes mr nguyen minh ha has said about this

Can you correct this problem, I'm sure this one is wrong !!!


inequalities
number theory proposed
number theory

Find all integers $ x,y,z,t$ such that $ x\plus{}y\plus{}z\equal{}t^2$ and $ x^2\plus{}y^2\plus{}z^2\equal{}t^3$.

[quote="SUPERMAN2"]Find all integers $ x,y,z,t$ such that $ x \plus{} y \plus{} z \equal{} t^2$ and $ x^2 \plus{} y^2 \plus{} z^2 \equal{} t^3$.[/quote]

$ 3(x^2\plus{}y^2\plus{}z^2) \ge (x\plus{}y\plus{}z)^2 \Leftrightarrow 3t^3 \ge t^4 \Leftrightarrow 3 \ge t$ as $ t \ge 0$.

[b]Case 1: [/b]$ t\equal{}3 \Rightarrow$ equality for the above inequality holds $ \Leftrightarrow x\equal{}y\equal{}z\equal{}3$
[b]Case 2:[/b] $ t\equal{}2$ then $ x^2\plus{}y^2\plus{}z^2\equal{}8$, mod $ 8$ we get $ x,y,z$ even. Let $ x\equal{}2x', y\equal{}2y',z\equal{}2z'$ we get
$ \begin{cases}x'\plus{}y'\plus{}z'\equal{}2 \\ x'^2\plus{}y'^2\plus{}z'^2\equal{}2 \end{cases}$, so $ (x,y,z)\equal{}(2,2,0),(2,0,2),(0,2,2)$ are the only possible solutions for this case.
[b]Case 3:[/b] $ t\equal{}1$ , $ x^2\plus{}y^2\plus{}z^2\equal{}1$ suggest that two of $ x,y,z\equal{}0$ , so $ (x,y,z)\equal{}(0,0,1),(0,1,0),(1,0,0)$ are the solution for this case.
[b]Case 4:[/b] $ t\equal{}0 \Leftrightarrow x\equal{}y\equal{}z\equal{}0$

So $ (x,y,z,t)\equal{}(3,3,3,3),(2,2,0,2),(2,0,2,2),(0,2,2,2),(0,0,1,1),(0,1,0,1),(1,0,0,1),(0,0,0,0)$ are the only solutions.


function
complex analysis
complex analysis unsolved

$f$ an analytic injective (or meromorphic) function defined on the open unit disk $\mathbb{D}$, satisfying $f(0)=0$ and $f'(0)=1$. Can one find an absolute constant $R$ (independant of $f$) such that $f(\mathbb{D})$ constains the disk centered at the origine and with radius $R$ ? and if $f$ is not suposed injective ?

The first case is known as [url=http://eom.springer.de/K/k055660.htm]Koebe's covering theorem[/url]. In $f$ is not assumed injective, there is no such $R$. Indeed, the function $f(z)=\frac{(z+1)^{n}-1}{n}$ does not assume the value $-\frac{1}{n}$ in $\mathbb D$.

anyone with a proof ? it seems an interesting result ..


geometry
circumcircle
geometric transformation
rotation
geometry unsolved

Let $ ABCD$ be a convex quadrilateral with $ AD \equal{} BC$, such that $ AB \not \parallel CD$, $ AD \not \parallel BC$. $ M,N$ are midpoints of sides $ CD$ and $ AB$, respectively. Let $ P \equal{} MN\cap BC$, $ Q \equal{} AD\cap MN$, $ R \equal{} AD \cap BC$. Prove that triangle $ PQR$ is isosceles.
There was and point a) in the problem, but problem becomes very easy after solving it.

See [url]http://www.mathlinks.ro/viewtopic.php?t=116208[/url] for a more general problem (different notation):

Let $ ABCD$ be a convex quadrilateral. Let $ M, N$ be points on the sides $ CD, AB$, such that $ \frac{CM}{MD} \equal{} \frac{BN}{NA} \equal{} \frac{BC}{DA}$ and $ R \equiv BC \cap AD.$ Then $ MN$ is parallel to the internal bisecor of the angle $ \angle ARB \equiv \angle DRC$.

Thank Yetti. Thank for your link  http://www.mathlinks.ro/viewtopic.php?t=116208

[quote="yetti"]
Let $ ABCD$ be a convex quadrilateral. Let $ P$ and $ Q$ be points on $ BC$ and $ AD,$ respectively, such that $ \frac {BP}
{PC} = \frac {AQ}{QD} = \frac {AB}{CD}.$ If $ BA$ and $ CD$ intersect at a point $ R,$ prove that the internal bisector $ l$
of the angle $ \angle BRC$ and $ PQ$ are parallels.
[/quote]

Spiral similarity! Is  one of the keys of solving this kind of questions.$ DA \cap CB=R$.Let $ O$,be the intersection of the

circumcircles of $ \triangle RAB,\triangle RDC$,It is clear that $ O$ is the center of spiral similarity of the two

segmensts$ AD,BC$.

$ {\chi _{\alpha}}^{k}(AD)=BC$,with $ k=\frac{OA}{OB}$ as the radius and $ \alpha$,as the angle of rotation.

And according to similarity of $ \triangle OAB,\triangle ODC$,we have:

$ \frac{OA}{OD}=\frac{OB}{OC}=\frac{AB}{CD}$,but according to the hypotheses of the problem,

the last equation would let us have: $ \frac{OA}{OD}=\frac{OB}{OC}=\frac{BP}{PC}=\frac{AQ}{QD}$

and it means that $ OQ,OP$,bisect $ \angle AOD,\angle BOC$,respectively.

we should consider that in the spiral similarity $ {\chi _{\alpha}}^{k}$,the angle between the two segment which would turn

into each other by that spiral similarity,is equal to $ \alpha$,then in our problem we should have $ \angle BRC=\angle
AOD=\angle BOC$

again $ \triangle OAQ,\triangle OBP$,are similar to each other and will turn into each other by $ {\chi _{\frac{\alpha}{2}}^{k}
(AQ)=BP}$

accordingly we should have the angle between $ AB,QP$,is equal to the angle of rotation $ \frac{\alpha}{2}=\frac{\angle
BRC}{2}$

and it is proved that $ QP$,is parallel to the bisector of $ \angle BRC$

it is really enjoyable  :D

[quote="Ahiles"][color=darkred]Let $ ABCD$ be a convex quadrilateral with $ AD \equal{} BC$, such that $ R\in AD\cap BC$ . Let $ M,N$ be the midpoints of

sides $ [CD]$ and $ [AB]$ respectively and $ P \equal{} MN\cap BC$, $ Q \equal{} AD\cap MN$ . Prove that triangle $ PQR$ is isosceles.[/color][/quote]
[color=darkblue]You can use the following remarkable geometrical locus :[/color]

[quote="Virgil Nicula"][color=darkred]Let $ ABC$ be a triangle. Let $ M$ , $ N$ be two mobile points on the sidlines $ AB$ , $ AC$ respectively

so that $ BM \equal{} CN$ . Denote the middlepoint $ D$ , $ L$ of the segments $ [BC]$ , $ [MN]$ respectively.

$ 1\blacktriangleright$ The sideline $ BC$ [b][u]doesn't separate[/u][/b] $ M$ , $ N$ . Then the geometrical locus of the point $ L$ is a line $ d$

for which $ D\in d$ and the line $ d$ is parallel with the [b][u](interior) bisector[/u][/b] of the angle $ \widehat {BAC}$ .

$ 2\blacktriangleright$ The sideline $ BC$  [b][u]separate[/u][/b] $ M$ , $ N$ . Then the geometrical locus of the point $ L$  is a line $ d$

for which $ D\in d$ and the line $ d$ is parallel with the [b][u]exterior bisector[/u][/b] of the angle $ \widehat {BAC}$ .[/color][/quote]


In triangle ABC we have :$ {b^2\equal{}ac}$ and $ {a\plus{}b\plus{}c\equal{}6}$, then find the range of $ {\vec BC *\vec BA}$

:)


function
analytic geometry
graphing lines
slope
algebra unsolved
algebra

Find all functions f,g from the real numbers to the real numbers such that g is one to one and for all real x,y: $ f(g(x)\plus{}y)\equal{}g(x\plus{}f(y))$

[hide="Solution"]
Let the given equation (in the problem statement) be called the (OFE). First, let $ x \equal{} \minus{}g(y)$. Then,

\[ f(0) \equal{} g(f(\minus{}g(y))\plus{}y) \quad \quad \quad (1)\]

Since $ g$ is injective,

\[ f(\minus{}g(y_1))\plus{}y_1 \equal{} f(\minus{}g(y_2))\plus{}y_2 \quad \forall \, y_1,y_2 \in \mathbb{R}\]

Let $ h \equal{} f\circ (\minus{}g)$. Then,

\[ \frac{h(y_1)\minus{}h(y_2)}{y_1\minus{}y_2} \equal{} \minus{}1\]

implies $ h$ is a slope of line -1, so

\[ h(t) \equal{} \minus{}t\plus{}k \quad \quad \quad (2)\]

for some $ k\in \mathbb{R}$.

Letting $ y \equal{}0$ in (1),

\[ f(0) \equal{} g(f(\minus{}g(0))) \equal{} g(k) \quad \quad \quad (3)\]

Now, let $ y\equal{}k$ and $ x\equal{}\minus{}f(0)$ in (OFE).

$ \minus{}f(x\plus{}f(0)) \equal{} \minus{}g(f(x)\plus{}k)$
$ f(\minus{}f(x\plus{}f(0))) \equal{} \minus{}f(x)$
$ f(\minus{}f(0)) \equal{} \minus{}f(\minus{}f(0))$

So,

\[ f(\minus{}f(0)) \equal{} 0 \quad \quad \quad (4)\]

Now, note $ (2) \implies f\circ (\minus{}g)$ is surjective, which means $ f$ is surjective. Now, using the substitution $ u\equal{}x\plus{}g(y)$ in the (OFE), we get $ f(u) \equal{} g(f(u\minus{}g(y))\plus{}y)$, so $ g$ is surjective, and therefore a bijection. This means $ \minus{}g$ is a bijection, so that along with the fact that $ f\circ(\minus{}g)$ is a bijection imply $ f$ is a bijection (prove this yourself if you are not very comfortable with injectiveness, surjectiveness, and bijectiveness).

Now, let $ x\equal{}\minus{}f(0)$ in the (OFE). Then,

\[ f(\minus{}f(0) \plus{} g(y)) \equal{} g(y)\quad \quad \quad (5)\]

Surjectiveness of $ g$ and $ f$ let us substitute $ f(x)$ for $ g(y)$, giving, letting $ l \equal{} f(0)$,

\[ f(\minus{}f(0) \plus{} f(x)) \equal{} f(x) \implies f(x) \equal{} x\plus{}f(0)\equal{} x\plus{}l \quad \quad \quad (6)\]

Now, applying this to (2) gives, letting $ m\equal{}f(0)\minus{}k$,

\[ f(\minus{}g(t)) \equal{} \minus{}g(t)\plus{}f(0) \equal{} \minus{}t\plus{}k \implies g(t) \equal{} t\plus{}f(0)\minus{}k \equal{} t\plus{}m \quad \quad \quad (7)\]

Hence, by (6) and (7), our solution set is a subset of the set of all $ f,g$ such that,

$ f(x) \equal{} x\plus{}l$
$ g(x) \equal{} x\plus{}m$

for some $ l,m  \in \mathbb{R}$.

Plugging these into the OFE, we find that the equation always holds regardless of the value of $ l$ or $ m$. So, this is indeed the solution set of the functional equation.
[/hide]


Let $ A_{k} \in M_{n} \mathbb{(R)}$; $ \forall k \in N$
Such that :
    $ (I_{n}+\frac {A_{m}}{m})^{m}=\sum_{k=0}^{m} {\frac {I_{n}}{k!}}$
Prove that :
$ Lim_{m \to \infty} {A_{m}=I_{n}}$
Here $ I_{n}$ is identity nxn matrix .
  :roll:

It's false.
Take $ A_m\equal{}\begin{pmatrix}1&\minus{}2\pi\\2\pi&1\end{pmatrix}$.
$ e^{A_m}\equal{}eI_n$.

More exactly
take $ A_m$ s.t. $ (I_n\plus{}A_m/m)^m\equal{}\sum_{k\equal{}0}^mI_n/k!$
and $ A_m$ tends to  $ \begin{pmatrix}1&\minus{}2\pi\\2\pi&1\end{pmatrix}$.

[quote="loup blanc"]More exactly
take $ A_m$ s.t. $ (I_n + A_m/m)^m = \sum_{k = 0}^mI_n/k!$
and $ A_m$ tends to  $ \begin{pmatrix}1 & - 2\pi \\
2\pi & 1\end{pmatrix}$.[/quote]
[b] I am sorry [/b]. In fact that   $ A_{m} \in S^{ + }_{n}(R)$ ,$ Lim_{m \to \infty} {A_{m}=I_{n}}$


Any thievery going on so far¿  :D

I'm missing some comments this year...  :(

hahaha yeah...
I feel like some team is planning on the "Mathematigerin"
nowadays more n dçmore teams have mascots... lots of target

I had Mathema Tiger in my hands... in my room!! But a stupid member of my team gave it back to the germans!!!! :@ My plan was asking the germans to get the Canmoo in order to get their tiger back... but now... They'll keep the tiger in their room... I did my best, sorry

ECU1


vector
linear algebra
matrix
probability and stats

If we have a multivariable linear regression to predict a quatity say $ Q$, and we know the $ R^2$ value as well as the standard error of the coefficients can we calculate the root mean square error of the regression?  For instance we have

\[ Q\equal{}\beta_0\plus{}\beta_1\cdot x_1\plus{}\beta_2\cdot x_2\plus{}...\plus{}\beta_n\cdot x_n\]
and we know the standard error of the $ \beta_i$ and the $ R^2$ of the regresion, could we calculate the RMSE of the regression? If not, what other information would we need?

Thanks

I don't think so

The standard errors vector is
SEB=$ \sqrt {(VECDIAG(INVX))*MS(Res)}$
where VECDIAG(INX) is the vector of diagonal elements of INVX,
INVX=$ (X'X)^{ \minus{} 1}$
Now, $ R^2 \equal{} \frac {SS(Regr)}{SS(Total)}$ and SS(Res)+SS(Regr)=SS(Total) (**)
SS(Total)=Y'(I-J/n)Y
SS(Regr)=Y'(P-J/n)Y
So, by the formula (**), SS(Res)=Y'(I-P)Y
Y is the vector of y's, I is the unity matrix, J is a matrix of 1's, P=X*INVX*X'
MS(Res) is what you denoted by RMSE, which is  SS(Res) divided by a constant.

So, if you know SEB, you need VECDIAG(INVX), or even only an element in this vector.
R^2 and (**) give you an equation between SS(Res) and SS(Regr) (or equivalently, SS(Total))
But either SS(Total) or SS(Regr) give you something that involves Y's while you need X's

"If not, what other information would we need? "
You need the vector X of x's +SEB
You also have to assume X's are independent


Give $ A=\underbrace{444\ldots 444}\limits_{2m},
B=\underbrace{222\ldots 222}\limits_{m}$
Prove $ A+B+1=C^2, C\in \mathbb{Z}$

[quote="HFL"]Give $ A = \underbrace{444\ldots 444}\limits_{2m}, B = \underbrace{222\ldots 222}\limits_{m}$
Prove $ A + B + 1 = C^2, C\in \mathbb{Z}$[/quote]
Are you sure you didn't make a typo?

Last digit of $ A+B+1$ is $ 4+2+1=7$, hence it cannot possibly be a square of an integer.

I've seen a similar problem where $ B$ consisted solely of $ 4$'s...

I believe you meant that $ B\equal{}\underbrace{444\ldots444}_m$.

[hide="Solution"]$ A\equal{}\sum_{k\equal{}0}^{2m\minus{}1}4\cdot 10^k\equal{}4\cdot \frac{10^{2m}\minus{}1}{9}$

$ B\equal{}\sum_{k\equal{}0}^{m\minus{}1}4\cdot 10^k\equal{}4\cdot \frac{10^m\minus{}1}{9}$

$ A\plus{}B\plus{}1\equal{}\frac{4\cdot 10^{2m}\plus{}4\cdot 10^m\plus{}1}{9}\equal{}\left(\frac{2\cdot 10^m\plus{}1}{3}\right)^2$.

$ 2\cdot 10^m\plus{}1\equiv 2\cdot 1\plus{}1\equiv 0\bmod 3$, so this is the square of an integer $ \Box$[/hide]


1/ Given funtion $ f(x)\equal{}|x^2\plus{}px\plus{}q|$
Determine p,q so that $ \max_{x\in[\minus{}1;1]}f(x)$ is minimum.
2/ Given funtion $ g(x)\equal{}|\frac{4x\plus{}3}{x^2\plus{}1}\plus{}1\minus{}2m|$
Determine m so that $ \min_{x\in R}f(x)$ is maximum.

[hide="Solution to Q1"]
1.$ \max_{x\epsilon[ \minus{} 1,1]}f(x)\geq \frac {1}{4}(f(1) \plus{} f( \minus{} 1) \plus{} 2f(0))\geq\frac {1}{4}|(1 \plus{} p \plus{} q) \plus{} (1 \minus{} p \minus{} q) \plus{} ( \minus{} 2q)| \equal{} \frac {1}{2}$
it is easy to show $ p \equal{} 0,q \equal{} \minus{} \frac {1}{2}$,when it reaches this minimum.
[/hide]
2. :?: I don't think there is a maximum.


Does anyone know how to apply for Physics Olympiad Screening test? The website only has the one for last year.....

Thanks,

You should probably ask a teacher or someone at school.

At my school, everyone in AP Physics is registered for the screening exam (although most don't go on).

i think i'm the only one who wants to take it....
my AP physics teacher doesn't know about it....


integration
trigonometry
limit
real analysis
calculus
function
real analysis unsolved

Let $ f(x)\equal{}\int_0^x{\frac{dt}{\sqrt{\cos t\minus{}\cos x}}},0<x<\pi$.
Show that $ \lim_{x\to 0}{f(x)}\equal{}\frac{\pi}{\sqrt{2}},\lim_{x\to \pi}{f(x)}\equal{}\plus{}\infty$

First of all, we have by t=xu

\[ f(x)\equal{}\int_{0}^1  \frac{xdu}{\sqrt{\cos(xu)\minus{}\cos(x)}}\equal{}\frac{1}{\sqrt{2}} \int_{0}^1 \frac{xdu}{\sqrt{\sin(x(1\plus{}u)/2 )\sin(x(1\minus{}u)/2)}}\]

We study when $ x\rightarrow 0$, we can suppose $ 0<x<\pi/2$. Recall that $ \sin(\alpha)\geq c \alpha$ when $ c\equal{}\frac{2}{\pi}$ and $ \alpha \in [0,\pi/2]$. So

\[ \forall u \in [0,1] \qquad     \frac{x}{\sqrt{\sin(x(1\plus{}u)/2 )\sin(x(1\minus{}u)/2)}}  \leq \frac{2}{c\sqrt{1\minus{}u^2}}\]
But $ 1/\sqrt{1\minus{}u^2}$ is integrable on $ [0,1]$, so Lebesgue Theorem proves that $ \lim\limits_{x\rightarrow 0} f(x)$ equals

\[ \frac{1}{\sqrt{2}} \int_{0}^1 \lim\limits_{x\rightarrow 0}  \frac{xdu}{\sqrt{\sin(x(1\plus{}u)/2 )\sin(x(1\minus{}u)/2)}}    \equal{} \sqrt{2} \int_{0}^1 \frac{du}{\sqrt{1\minus{}u^2}}\]

We can compute the last integral  : $ \sqrt{2} \arcsin(1)\equal{}\frac{\pi}{\sqrt{2}}$.

We study when $ x\rightarrow \pi$. The easiest is probably the Fatou's Lemma because we deal with positive mesurable functions $ \int_{0}^1  \liminf\limits_{x\rightarrow \pi} g(x,u)du\leq \liminf\limits_{x\rightarrow \pi} \int_{0}^1 g(x,u)du$ with
\[ g(x,u)\equal{}   \frac{x}{\sqrt{\sin(x(1\plus{}u)/2 )\sin(x(1\minus{}u)/2)}} \geq 0\]

\[ \int_{0}^1  \frac{\pi du}{\sqrt{\sin(\pi(1\plus{}u)/2) \sin(\pi(1\minus{}u)/2 )}} \leq \sqrt{2} \liminf\limits_{x\rightarrow \pi} f(x)\]
But the first integral is $ \int_{0}^1  \frac{\pi du}{\cos(\pi u/2)}\equal{}\infty$, see $ u\equal{}1$.

[quote="hanschinaman"]Let $ f(x) \equal{} \int_0^x{\frac {dt}{\sqrt {\cos t \minus{} \cos x}}},0 < x < \pi$.
Show that $ \lim_{x\to 0}{f(x)} \equal{} \frac {\pi}{\sqrt {2}},\lim_{x\to \pi}{f(x)} \equal{} \plus{} \infty$[/quote]

For the second limit, note $ \cos t \minus{} \cos x < \cos t \plus{}1$ for $ x$ under consideration. The power series for $ \cos$ about $ \pi$ shows  $ \cos t \plus{}1 \leq (\pi\minus{}t)^2$ for $ t$ near $ \pi$. So for some $ d > 0$, $ f(x) > \int_{\pi\minus{}d}^{x}{\frac {dt}{\sqrt { (\pi\minus{}t)^2}}}$ for $ x > \pi\minus{}d$, and the latter clearly tends to $ \infty$ as $ x\rightarrow\pi^\minus{}$.

For the first limit, we can generalize: Suppose $ g$ is stricly decreasing on $ [0,1]$, and $ g(t) \equal{} a \minus{} bt^2 \plus{} h(t)$ for $ t$ near $ 0$, where $ b > 0$ and $ h(t)$ is a smooth function near $ 0$ with $ h(0) \equal{} h'(0) \equal{} h''(0) \equal{} 0$. In our problem, $ g(t) \equal{} \cos t$, where $ a \equal{} 1, b \equal{} 1/2$.

Claim: $ \int_0^x{\frac {dt}{\sqrt {g(t) \minus{} g(x)}}} \rightarrow \frac {\pi}{2\sqrt {b}}$ as $ x\rightarrow0 \plus{}$.

We need a little lemma about $ h$: There is a constant $ M$ such that $ |h(tx) \minus{} h(x)| < \equal{} Mx^3(1 \minus{} t^2)$ for all $ x, t$ in $ [0,1]$. (Proof: not hard.)

To prove the claim, make the change of variables holomorf did, getting $ x\int_0^1{\frac {dt}{\sqrt {g(tx) \minus{} g(x)}}}$. For $ x$ small, we get $ g(tx) \minus{} g(x) \equal{} bx^2(1 \minus{} t^2) \plus{} h(tx) \minus{} h(x)$. Now because of the lemma (note the $ x^3$ term), we can factor out $ x^2$, take it outside the square root, and cancel with the $ x$ upstairs. We are left with $ b(1 \minus{} t^2) \plus{} (h(tx) \minus{} h(x))/x^2$ inside the square root, which will be at least $ b(1 \minus{} t^2)/2$ for small $ x$. So we can take as our dominating function $ 1/\sqrt { b(1 \minus{} t^2)/2}$ and apply the dominated convergence theorem, noting our integrand converges to $ 1/\sqrt { b(1 \minus{} t^2)}$ pointwise on $ [0,1]$. This proves the claim.


for the first ques-is "know" mutual?(if I know someone, he must know me as well?) for the second-what is the meaning of sum free subset? ty

Yes, "know" is generally considered to be mutual.

I'm pretty sure a sum-free set is one where you cannot find three elements of the set such that two of the three sum to the third.

here is a definition which I didn't understand http://mathworld.wolfram.com/Sum-FreeSet.html

that definition says that no element of $S$ is a sum of two (not necessarily distinct!) elements of $S$. In particular, if $k$ is in $S$, $2k$ cannot be in $S$.

so they mean to split it for example to {123}{4-10}{11-13}? which isn't true cause 1+2=3?


algebra
polynomial
linear algebra
matrix
invariant

I've not done any linear algebra for ages, but I think its characteristic polynomial $K(X)$ is in $\mathbb{Q}$, and the minimum polynomial divides $K$ in $\mathbb{Q}[X]$, so I would say yes ... to be confirmed !

[quote="perfect_radio"]If $A \in \mathcal M \left( \mathbb Q \right)$, then its minimal polynomial is in $\mathbb Q \left[ X \right]$?

Sorry if the answer to this question is too obvious, but my Linear Algebra is horrible.[/quote]

I am sorry but I dont understand the notion of the minimal polynomial. perfect_radio please explain.

The minimal polynomial is the monic (leading coefficient is 1) polynomial with least degree that annihilates the matrix. That is, if $f(x)$ is the minimal polynomial of the matrix $A$, then $f(A)$ is the zero matrix.

That should be the answer... But why is this true:

[quote="julien_santini"]and the minimum polynomial divides $K$ in $\mathbb{Q}[X]$[/quote]

??? I thought of the same thing, but I can't give an argument.

the minimal polynomial is field invariant (use the canonical form to see this). So the answer is yes.

Looks like pretty advanced stuff... too bad. i was hoping for a more elementary proof.

soo... the characteristic polinomial is in Q[x]... and from Frobenius' theorem the minimal polinomial divides the characteristic polinomial...the char. is monic so with further computations i might say that the answer to your question is "yes"


Lol, i thought he would write his answer in 1337 or something.

30205 - 2*8600 = 13005

So he couldn'tve wagered 13337 with full safety. If I had confidence in my answer I would have wagered 9795 for an even 40k. But I guess wagering 1337 looks good on TV  :)