modular arithmetic
quadratics
number theory unsolved
number theory

find all integer solutions of x^3 + 45=y^2.

no one wants to help?
personne ne veut m'aider?
hichkasi nist komak kone?

$ x^{3} \plus{} 45 \equal{} y^{2}$
$ 2|x$ $ \implies$  $ y^{2}\equiv 5\pmod{8}$  but $ y^{2}\equiv 0,1,4\pmod{8}$ contradiction!
$ x\not\equiv 0\pmod{2}$ $ \implies$
$ x\equiv\pm 1\pmod {4}$
$ x\equiv 1\pmod{4}$ $ \implies$  $ y^{2}\equiv 2\pmod{4}$ contradiction!
$ x\equiv \minus{} 1\pmod{4}$  $ \implies$  $ x^{3} \plus{} 52 \equal{} y^{2} \plus{} 7$ $ \implies$
$ p|x^{3} \plus{} 52 ,p\equiv \minus{} 1\pmod{4}$
$ \implies$  $ y^{2}\equiv \minus{} 7\pmod{p}$  $ \implies$
$ (\frac { \minus{} 7}{p}) \equal{} (\frac { \minus{} 1}{p})(\frac {7}{p})$
we know $ (\frac { \minus{} 1}{p}) \equal{} \minus{} 1$ for $ p\equiv \minus{} 1\pmod{4}$
$ (\frac {7}{p}) \equal{} \minus{} 1$  $ \implies$  $ x^{2}\equiv 7\pmod{p}$ has any solution
$ \implies$  if we use quadratic resiprosity  $ (\frac {p}{7}) \equal{} 1$
$ \implies$ $ x^{2}\equiv p\pmod{7}$ has a solution for $ p\equiv \minus{} 1\pmod{4}$

The only thing that i can say(and it's obvious!) is that it must be solved using quadratic resiprosity(like what cnyd did).help!

Xeyli motashakkeram---Shoki---However,I didn't learned that before---I will learn it now---


Let $ ABC$ and $ A'B'C'$ be the perspective triangles with mutually perpendicular sidelines. Prove that perspectrix $ X$ of two triangle bisects the line joining the orthocenters $ H$ and $ H'$ of triangle $ ABC,A'B'C'$ respectively.

Dear Leonhard and Mathlinkers,
I wrote on an article intitled "La droite de Simiya-Ayme-Turner" which you can find on
http://perso.orange.fr/jl.ayme    vol. 3.
In this article p. 15, this are two triangle ABC and A*B*C* like you want with J the perspector and I the center of rotation...that I proved.
There is a little way in order to find your result.
I think, that there is a way much shorter.
Sincerely
Jean-Louis

See also [url]http://www.mathlinks.ro/viewtopic.php?t=160490[/url]. The proof shows that perspector of $ \triangle ABC, \triangle A'B'C'$ is the common pole $ P \in (O), (O')$ of Simson lines of of $ \triangle ABC, \triangle A'B'C'$ parallel to their perspectrix.

Dear Mathlinkers,
an article concerning another Sondat's theorem can be found on
http://perso.orange.fr/jl.ayme/  vol. 5
Sincerely
Jean-Louis


Let the point P be (4,3). Choose Q on the line y=x and R on the x axis such that the sum of lengths PQ+QR+RP is minimized. What is this minimum total length?

A. 5 $ \sqrt{2}$
B. 3+3 $ \sqrt{2}$
C. 5
D. 8
E. 10


Prove that $ \sin ^2 20^{\circ} \sin 40^{\circ} \equal{} \sin 10^{\circ} \sin 30^{\circ} \sin 60^{\circ}$.

[hide]The relevant equations I used are: \[ \sin(x)\pm\sin(y)\equal{}2\sin\left(\frac{x\pm y}{2}\right)\cos\left(\frac{x\mp y}{2}\right)\] \[ \sin(x)\sin(y)\equal{}\frac{1}{2}\Big(\cos(x\minus{}y)\minus{}\cos(x\plus{}y)\Big)\] commonly referred to as the sum to product and product to sum identities, respectively. We start with \[ \sin 10^\circ\cos 30^\circ\equal{}\sin 10^\circ\sin 60^\circ,\] since $ \cos x\equal{}\sin(90^\circ\minus{}x)$. Using the first identity, $ \sin40^\circ\minus{}\sin20^\circ\equal{}2\sin10^\circ\cos30^\circ$, so \[ \frac{1}{2}(\sin 40^\circ\minus{}\sin20^\circ)\equal{}\sin10^\circ\sin60^\circ.\] By the double angle formula, $ \sin 2x\equal{}2\sin x\cos x$, and also, we know that $ \cos 60^\circ\equal{}\frac{1}{2}$. Hence, \[ \cos20^\circ\sin20^\circ\minus{}\sin20^\circ\cos60^\circ\equal{}\sin10^\circ\sin60^\circ.\] Since $ \sin30^\circ\equal{}\frac{1}{2}$, then \[ \frac{1}{2}\sin20^\circ(\cos20^\circ\minus{}\cos60^\circ)\equal{}\sin10^\circ\sin30^\circ\sin60^\circ.\] From the product to sum identity, $ \sin40^\circ\sin20^\circ\equal{}\frac{1}{2}(\cos20^\circ\minus{}\cos60^\circ)$, so \[ \sin^220^\circ\sin40^\circ\equal{}\sin10^\circ\sin30^\circ\sin60^\circ,\] as desired.[/hide]


Knowing that $ 2^{29}$ is a number 9-digits differents all of them, determine the digit that not appear without calculate $ 2^{29}$. Why ?

Portuguese:

Sabendo que 2^29 é um número de nove algarismos todos diferentes, determine o algarismo que
não aparece sem calcular 2^29. Se for possível, justificar a resposta.

How to do ? Explain me, please.

[hide]
First take mod 9.
$ 2^{29}\equiv (2^3)^9\times 2^2\equiv ( \minus{} 1)^9\times 4\equiv 5 \pmod{9}$.

Now, note that the sum of the numbers from 0 through 9 is 45, which is equal to 0 mod 9. So we need to get rid of the $ \boxed{4}$ to make it 5 mod 9 (because the digital sum of a number is equal to the number mod 9).[/hide]

the result of (2)^29 equal to  536,870,912

therefore , 4 is missed from  above number

Thanks, but why this is the correct answer ?
I don't understand your method. Can you explain more detalhed ?
 :blush:

Here's a more detailed explanation;

$ 2^{29}$ contains $ 9$ different digits. If you add the digits from $ 0$ to $ 9$, you get $ 45$. However, one digit is omitted. If the digit $ x$ is not present, then the total sum of the digits is then $ 45\minus{}x$. When you divide the number by $ 9$, the remainder will be $ 9\minus{}x$, or $ 0$ if the missing digit is $ 9$.

$ 2^{29} \equal{} 2^{27} \times 2^2 \equal{} 8^9 \times 2^2$. When you divide $ 8^9$ by $ 9$, you get a remainder of $ 8$ because $ 8^9 \equiv (\minus{}1)^9 \equiv \minus{}1 \equiv 8$ modulo $ 9$. $ 8 \times 4 \equiv 32 \equiv 5$ modulo $ 9$, Therefore, we seek a number $ x$ such that $ 45\minus{}x \equiv 5$ modulo $ 9$. This turns out to be $ x\equal{}4$, so the missing digit is $ \boxed{4}$.


let $ a;b;c\geq 0$ and a+b+c=3 ; prove that:
$ (3a\plus{}\frac{5}{3a\plus{}2})(3b\plus{}\frac{5}{3b\plus{}2})(3c\plus{}\frac{5}{3c\plus{}2})\leq 64$

Try  $ a \equal{} 0,b \equal{} c \equal{} \frac {3}{2}$.

[quote="wxydx00"]it maybe $ (3a \plus{} \frac {5}{3a \plus{} 2})(3b \plus{} \frac {5}{3b \plus{} 2})(3c \plus{} \frac {5}{3c \plus{} 2})\geq 64$[/quote]
No. $ a\equal{}3,$ $ b\equal{}c\equal{}0.$

[quote="dima ukraine"]$ ln((3a + \frac {5}{3a + 2})(3b + \frac {5}{3b + 2})(3c + \frac {5}{3c + 2})) = ln(3a + \frac {5}{3a + 2}) + ln(3b + \frac {5}{3b + 2}) + ln(3c + \frac {5}{3c + 2})$
$ F(x) = ln(3x + \frac {5}{3x + 2})$ is convex upwards so  for any $ c\le\ b\le\ a (a,b,c)\succ(1,1,1)$ so by Karamata ineq $ F(a) + F(b) + F(c)\le \3F(1) = 3ln(4) = ln(64)$ or $ ln((3a + \frac {5}{3a + 2})(3b + \frac {5}{3b + 2})(3c + \frac {5}{3c + 2}))\le\ln(64)$[/quote]

Iam sorry; a,b,c>0 not $ a;b;c\geq 0$
this solution is true
but if have solve by cauchy or bunhiacopxki is goog

[quote="litbon"][quote="dima ukraine"] $ ln((3a \plus{} \frac {5}{3a \plus{} 2})(3b \plus{} \frac {5}{3b \plus{} 2})(3c \plus{} \frac {5}{3c \plus{} 2}))\le\ln(64)$[/quote]

Iam sorry; a,b,c>0 not $ a;b;c\geq 0$
this solution is true
but if have solve by cauchy or bunhiacopxki is goog[/quote]
Try $ a \equal{} 0.1$ and $ b \equal{} c \equal{} 1.45$ :wink:


Με μία απλή παραγώγιση π.χ. βγάζεις τοπικό ελάχιστο στο $ (0,0)$.

Cheerio,

Durandal 1707

Edit: ναι, $ \minus{}\frac{1}{3}$.

ναι!!εγω πολλαπλασιασα με 3 προσθεσα μια μοναδα και μετα AM-GM :) ..και νομιζω βγαινει -$ 1/3$


$\ddot{x}+16\pi^{2}x = 0$, $x(0) = 0$, $\dot{x}(0)= \pi$

I know the solution has to be in the form of:

$x(t) = c_{1}\cos 4\pi t+c_{2}\sin 4\pi t$
$\dot{x}(t) =-4\pi c_{1}\sin 4\pi t+4\pi c_{2}\cos 4\pi t$

$x(t) = \frac{1}{4}\sin 4\pi t$

Now $A = \sqrt{c_{1}^{2}+c_{2}^{2}}= \sqrt{\frac{1}{16}}= \frac{1}{4}$

    The phase angle, $\delta = \tan^{-1}\frac{0.25}{0}= \frac{\pi}{2}$

The period $T = \frac{2\pi}{4\pi}= \frac{1}{2}$

But why does the answer say that $T = 8$?

Thanks


1.  Write a digit in front and at the end of the number 1998, so that the new 6-digit number is divisble by 88.

2.  Write a digit in front and at the end of the number 1998, so that the new 6-digit number is divisible by 99.

  If you are unaware of common "divisibility tricks" click here: http://www2.sunysuffolk.edu/wrightj/MA22/Num/DivRule.htm

[hide="1"]Since this number also has to be divisible by 8, the last three digits (98_) has to be divisible by 8. Only 4 works. This number has to also be divisible by 11. So
_+9+8 = 1+9+4 (or have a difference by a multiple of 11)
they can't be equal, so let's work on a 11 multiple difference.
9+4+1=14, 14+11=25. 25-(9+8) = 8.
so 819984 is divisible by 88.
[/hide]

[hide="2"]The two digits must add up to 9 or 18, since it has to be divisible by 9.
For divisibility by 11,
_+9+8 = 1+9+_ (or diff = multiple of 11)
The difference between the sums is 7.
9 = 1+8, 2+7, etc.
1 and 8 are 7 apart, so a number that would work is 119988. [/hide]

[hide="1"]To be divisible by 88, a number must be divisible by both 8 and 11. To be divisible by 8, the last three digits of a number must be evenly divisible by 8. Given the number $A1998B$, $98B$ must be divisible by $8$, so $B=4$. We now have $A+9+8=A+17$ and $1+9+4=14$. Therefore, $A=8$, and we have the number $\boxed{819,984}$.[/hide]

[hide="2"]To be divisible by 99, a number must be divisible by both 9 and 11. To be divisible by 9, the sum of the digits of a number must be a multiple of 9, so the two digits added must sum to a multiple of 9 (since $1+9+9+8=27$), equalling either 9 or 18. Given the number $A1998B$, we take $A+9+8=A+17$ and $1+9+B=B+10$. Therefore, we can have either $A=B+4$ or $B=A+7$, and we also have $A+B=9$; they cannot equal 18, because both would have to be 9, which, by the other conditions, is impossible. $A=B+4 \Rightarrow 9-B=B+4 \Rightarrow 2B=5$, which is impossible. Therefore, the only solution is $B=A+7 \Rightarrow 9-A=A+7 \Rightarrow A=1$, meaning $B=8$, and we have the number $\boxed{119,988}$. [/hide]

JesusFreak197's and 236factorial's division by 11 test doesn't  cope with the page given by G-UNIT. Would anybody mind explaining this to me.

they're using the same divisibilty test specified in the link.

The link says "subtract the sum of the even digits from the sum of the odd digits"; this means that you add together every other digit, then add together the remaining digits, subtract the second sum from the first, and if the difference is any multiple of $11$ (including 0 and negative multiples), then the number is divisible by $11$. For example, if you have the number $16,589,672$:

$1+5+9+7=22$
$6+8+6+2=22$
$22-22=0$

Therefore, the number is divisible by $11$.


Okey, I'll start it off:

name's Cinjon Resnick; I'm about 5'9 ish (i think), brown hair, brown eyes, rabble rabble.
O, also include what division you're in and what class you are in school for the upcoming year (freshman, soph, etc).
I'm Mu/Senior.

Sarah Miles/ Vero Beach High MAO/ next year Mu Senior/ currently Alpha Junior

Alan Yanuck. 5'8'' green eyes, brown hair.  Junior (as of 5-17-05)Palm Harbor University High School Delegate.  I enjoy math, alternative rock, and law.  I really look forward to making this state greater at Mu than it already is.

Jimmy Doker
Lincoln High '05
Former Lincoln President, first FAMAT President.
Mu until we leave Hawaii, then I'm just a test writer.

Jonathan Hahn
Berkeley Prep, c/o 2006
Alpha->Mu
2005-2006 FAMAT Student Delegate President

Yannik Pitcan...I'm 6'0 and black.
Class of 2007, rising junior
Mu, Lake Howell High

Wow some kid's last name sounds like my first name, =o.

I know this is like 2.5 years old, but im curious to know how the person above has 0 posts, when their post is above.  :)


calculus
integration
trigonometry
function
calculus computations

Calculate $ \int_{0}^{2a}\frac{\sqrt[4]{\sin (3a\minus{}x)}}{\sqrt[4]{\sin(3a\minus{}x)}\plus{}\sqrt[4]{\sin (a\plus{}x)}}dx$
where $ 0 < a < \frac{\pi}{3}$

use the fact $ \int _0^m f(x) \; dx \equal{} \int _m^0 f(m \minus{} x) \; d(m \minus{} x)$

$ I \plus{} I \equal{} \int_{0}^{2a}\frac {\sqrt [4]{\sin (3a \minus{} x)}}{\sqrt [4]{\sin(3a \minus{} x)} \plus{} \sqrt [4]{\sin (a \plus{} x)}}\;\; dx$

$ \plus{} \int_{2a}^0\frac {\sqrt [4]{\sin (3a \minus{} (2a \minus{} x))}}{\sqrt [4]{\sin(3a \minus{} (2a \minus{} x))} \plus{} \sqrt [4]{\sin (a \plus{} (2a \minus{} x))}}\;\; d(2a \minus{} x)$

$ \Rightarrow 2 I \equal{} \int_{0}^{2a}\frac {\sqrt [4]{\sin (3a \minus{} x)}}{\sqrt [4]{\sin(3a \minus{} x)} \plus{} \sqrt [4]{\sin (a \plus{} x)}}\;\; dx$

$ \plus{} \int_{0}^{2a}\frac {\sqrt [4]{\sin (a \plus{} x)}}{\sqrt [4]{\sin(a \plus{} x)} \plus{} \sqrt [4]{\sin (3a \minus{} x)}}\;\; dx$

$ \equal{} \int_{0}^{2a} 1\;\; dx \equal{} 2a \qquad \Rightarrow \qquad I \equal{} a$

Fast and excellent. The only thing I'm wondering is what is the intention of limitations for a?

Well, the limitations assure that subintegral function will remain real for all $ x$. Note that if $ a$ were, for instance, $ \frac{3\pi}{4}$, then for most of the $ x$ function doesn't give real values...

I'm sorry but it seems today is not a good day for me for thinking.


function
induction
number theory unsolved
number theory

Find all function $f:N^2\rightarrow N$ such that :
a)$\forall x\in N , f(x,x)=x$
b)$\forall x,y\in N, f(x,y)=f(y,x)$
c)$\forall x,y \in N,(x+y)f(x,y)=yf(x,x+y)$  :?

by induction we prove $f(k,l)={kl \over gcd(k,l)}$
1) $f(1,1)=1$
2) we prove that : $f(k,l)={kl \over gcd(k,l)} \Leftarrow f(k-l,l)={(k-l)l \over gcd(k-l,l)}$
from 2nd equation:
$(l+(k-l))f(l,k-l)=(k-l)f(l,k)$
$(k-l)f(l,k)=kf(l,k-l)={kl(k-l) \over gcd(k,k-l)}={kl(k-l) \over gcd(k,l)}$
$f(l,k)={kl \over gcd(k,l)}$


Yeah this problem is hard because I don't even know what I'm supposed to do with it :? . Perhaps you can elaborate more by providing more precise details. What kind of set is $A$? What are $f$ and $g$?

Theres definitely some information missing here.

That doesn't really improve anything.. why not just let f(x) = x for all x in A, and g(x) = x for all x in A except for, say, the fourth and fifth values, which are swapped around.

Theres some (relevant) piece of information missing here.

I think, he meant $f\circ f\circ f=g\circ g\circ g$ and $f\circ f\circ ...\circ f=g\circ g\circ ...\circ g$ (2003 times).
Is it right, mashimaro?

Thank you myth  :lol:
    I miss
     $f=f_1 $ and $ f_{n+1}=f_n(f_1(x))$

Problem is easy.

Why do no one send solution?  ;)

[quote="billzhao"]What about the functions $f, g: \{1, 2, 3\} \to \{1, 2, 3\}$ with $f \equiv 3$, and $g(1) = 2$, $g(2), g(3) = 3$.[/quote]
I think he meant $f,g$ are bijections.

Well, then.

Let $h = f_3 = g_3$. Then the condition $f_{2003} = g_{2003}$ becomes $h_{667}(f) = h_{667}(g)$. Since the composition of bijective functions is bijective, $h_{667}(f) = h_{667}(g)$ implies $f=g$.

[quote="billzhao"]Let $h = f_3 = g_3$. Then the condition $f_{2003} = g_{2003}$ becomes $h_{667}(f) = h_{667}(g)$. Since the composition of bijective functions is bijective, $h_{667}(f) = h_{667}(g)$ implies $f=g$.[/quote]
Why did you do it!? :what?:
 :lol:  :lol:  :lol:

[quote="Myth"]I meant Pre-Olympiad is not a proper place for billzhao  :P  :P  :P
Do you agree with me?[/quote]

Then why are you here?  :D  :D  :D


trigonometry
inequalities proposed
inequalities

Given $ a,b,c \geq 0$ such that $ abc\equal{}1$. Find the maximum and minximum value of
 $ P\equal{}\sqrt{a\plus{}b\plus{}c}\plus{}\sqrt{(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)(a\plus{}b\minus{}c)}$

[quote="dduclam"]Given $ a,b,c \geq 0$ such that $ abc \equal{} 1$. Find the maximum and minximum value of
 $ P \equal{} \sqrt {a \plus{} b \plus{} c} \plus{} \sqrt {(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)}$[/quote]
Try $ a\equal{}2,$ $ b\equal{}1$ and $ c\equal{}\frac{1}{2}.$ :wink:

[quote="dduclam"][quote="arqady"][quote="dduclam"]Given $ a,b,c \geq 0$ such that $ abc \equal{} 1$. Find the maximum and minximum value of
 $ P \equal{} \sqrt {a \plus{} b \plus{} c} \plus{} \sqrt {(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)}$[/quote]
Try $ a \equal{} 2,$ $ b \equal{} 1$ and $ c \equal{} \frac {1}{2}.$ :wink:[/quote]
I don't understan you,arqady.:(
Note:[b]Find the maximum and minximum value of P[/b]. :)[/quote]
OK! The maximum value of $ P$ is not exists for $ a,b,c \geq 0$ such that $ abc \equal{} 1.$  :wink:

[quote="arqady"][/quote][quote="dduclam"][quote="arqady"][quote="dduclam"]Given $ a,b,c \geq 0$ such that $ abc = 1$. Find the maximum and minximum value of
 $ P = \sqrt {a + b + c} + \sqrt {(b + c - a)(c + a - b)(a + b - c)}$[/quote]
Try $ a = 2,$ $ b = 1$ and $ c = \frac {1}{2}.$ :wink:[/quote]
I don't understan you,arqady.:(
Note:[b]Find the maximum and minximum value of P[/b]. :)[/quote][quote="arqady"]
OK! The maximum value of $ P$ is not exists for $ a,b,c \geq 0$ such that $ abc = 1.$  :wink:[/quote]

For example, $ a=b\rightarrow \infty, c=1$.

Thanks arqady and not_trig :) I edited the problem:
Given $ a,b,c \geq 0$ such that $ abc \equal{} 1$. Find the minximum value of
 $ P \equal{} \sqrt {a \plus{} b \plus{} c} \plus{} \sqrt {(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)}$


calculus
integration
function
real analysis
real analysis unsolved

Here is a similar problem to yours.

Keio University entrance exam 1975

Let $f(x)$ be the continuous function at $0\leq x\leq 1$ such that $\int_{0}^{1}(1-x)f(x)dx=0.$ Prove that there exist $a\ (0<a<1)$ for which $\int_{0}^{a}f(x)dx=0.$

[quote="Cezar Lupu"]Let $f$ be a $C^{1}[0,1]$ function. Show that there is $c\in (0,1)$ such that \[\int_{0}^{1}f(x)dx=(1-c)f(c).\] [/quote]
Is the problem correctly stated?  There seem to be counterexamples such as $f(x) = 1$ or $f(x) = x$.

Here is a proof for kunny’s problem:
Consider $F: [0,1] \rightarrow \mathbb{R}, F(x)=\int_{0}^{x}f(t)dt$. $F$ is differentiable and $F'=f$ , because $f$ is continuous.
$0=\int_{0}^{1}(1-x)f(x)dx=\int_{0}^{1}(1-x)F'(x)dx= (1-x)F(x)\mid_{0}^{1}+\int_{0}^{1}F(x)dx=$ $\int_{0}^{1}F(x)dx$.
We have $\int_{0}^{1}F(x)dx=0$. If we assume that $F(x)\not = 0, \forall x\in(0,1)$, then $F(x)<0, \forall x\in(0,1)$ or $F(x)>0, \forall x\in(0,1)$. If we consider that $F(x)<0, \forall x\in(0,1)$, then $\int_{0}^{1}F(x)dx<0$, which yields a contradiction. The case $F(x)>0, \forall x\in(0,1)$ is the same.
So there $\exists a\in(0,1)$ such that $\int_{0}^{a}f(x)dx=0$.

Correct formulation of this problem from Kyiv Mechmat Competition, 1995 is

Let $f\in C([0,1]).$ Prove that there exists
$c\in(0,1)$ such that $\int_{0}^{c}f(x)dx=(1-c)f(c).$

Sorry for the wrong formulation posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=606100#606100]here[/url] :blush:

The solution is very close to the solution of the problem mentioned by Kunny and solved by Ovidius above.

Define the function $g$ by \[g(t) = \int_{0}^{t}f(x) \, dx \, .\] Integrating $g$, we have \begin{eqnarray*}\int_{0}^{1}g(t) \, dt & = & \int_{0}^{1}\int_{0}^{t}f(x) \, dx \, dt \\ & = & \int_{0}^{1}\int_{x}^{1}f(x) \, dt \, dx \\ & = & \int_{0}^{1}f(x) \int_{x}^{1}1 \, dt \, dx \\ & = & \int_{0}^{1}f(x) (1-x) \, dx \\ & = & \int_{0}^{1}f(t) (1-t) \, dt \, . \\ \end{eqnarray*} Thus, by Rolle's theorem, we have $g(c) = f(c) (1-c)$ for some $c$ strictly between $0$ and $1$.  That's what we wanted to show.

[quote="Ravi B"]Thus, by Rolle's theorem, we have $g(c) = f(c) (1-c)$ for some $c$ strictly between $0$ and $1$.  That's what we wanted to show.[/quote]

I can't get this. Could someone explain more about it? thx.

Define a function $h$ by \[h(x) = \int_{0}^{x}g(t)-f(t) (1-t) \, dt \, .\] Then $h(0) = 0$ trivially, and $h(1) = 0$ is what I showed in my last post.  By Rolle's theorem, there is a $c$ strictly between $0$ and $1$ such that $h'(c) = 0$.  But $h'(c) = g(c)-f(c) (1-c)$.

There are other ways besides Rolle's theorem that I could have concluded the proof.  For example, ovidius had another way to conclude.  There might even be a "mean value theorem for integrals" that I could cite, but I'm not sure.  Basically, if the average (i.e., integral) of a continuous function is $0$, then the function must be $0$ somewhere.

[quote="Cezar Lupu"][color=darkred]Let $f: [0,1]\rightarrow R$ be a continuous function. Show that there is $c\in (0,1)$ such that $\int_{0}^{c}f(x)\ dx=(1-c)\cdot f(c)\ .$[/color][/quote]
[color=darkblue][b]Proof.[/b] Define the function $F(x)=(1-x)\cdot \int_{0}^{x}f(t)\ dt\ .$ Observe that $F(0)=F(1)$ and $F'(x)=-\int_{0}^{x}f(t)\ dt+(1-x)\cdot f(x)\ .$
From the Rolle's theorem obtain there is $c\in (0,1)$ so that $F'(c)=0$, i.e. $\int_{0}^{c}f(t)\ dt=(1-c)\cdot f(c)\ .$

[b]Another exercises[/b] (from the my book [u]Probleme de analiza matematica. Editura ALL Educational, 2002 Bucuresti[/u]).

$\boxed{A.}\blacktriangleright$ Let $f: [a,b]\rightarrow R$ be a continuous function. Prove that there is a point $c\in (a,b)$ so that :

$A1.\ \ \int_{a}^{c}f(x)\ dx=(b-c)\cdot f(c)\ ;$

$A2.\ \ \int_{c}^{b}f(x)\ dx=(c-a)\cdot f(c)\ ;$

$A3.\ \ \int_{a}^{c}f(x)\ dx+\int_{b}^{c}f(x)\ dx=(b+a-2c)\cdot f(c)\ .$

$\boxed{B.}\blacktriangleright$ Let $f: [0,1]\rightarrow R$ be a continuous function for which $\int_{0}^{1}f(x)\ dx=0\ .$ Prove that exists $c\in (0,1)$ so that $\int_{0}^{c}f(x)\ dx+c^{2}\cdot f(c)=0\ .$

$\boxed{C.}\blacktriangleright$ Let $f,g: [-a,a]\rightarrow R$ be two functions ($g$ is odd) from $C^{1}\ ;\ a>0$ ; for any $x\in (-a,a)\ ,\ g(x)\ne 0\ .$

Prove that there is $c\in (-a,a)$ so that $\int_{-a}^{a}f(x)g(x)\ dx=2f'(c)\int_{0}^{a}xg(x)\ dx\ .$

$\boxed{D.}\blacktriangleright$ Let $f: [0,b]\rightarrow R$ be a continuous function and $0<a<b\ .$ Prove that exists a point $c\in (0,1)$ so that $\int_{a}^{b}f(x)\ dx=bf(bc)-af(ac)\ .$[hide="Hint"] $1.\ F(x)=(b-x)\cdot\int_{a}^{x}f(t)dt\ \ ;$
$2.\ F(x)=(x-a)\int_{x}^{b}f(t)dt\ \ ;$
$3.\ F(x)=(b-x)\cdot\int_{a}^{x}f(t)dt+(x-a)\cdot\int_{x}^{b}f(t)dt\ .$[/hide][/color]

[quote="Virgil Nicula"][color=darkblue][b]Proof.[/b] Define the function $F(x)=(1-x)\cdot \int_{0}^{x}f(t)\ dt\ .$ Observe that $F(0)=F(1)$ and $F'(x)=-\int_{0}^{x}f(t)\ dt+(1-x)\cdot f(x)\ .$
From the Rolle's theorem obtain there is $c\in (0,1)$ so that $F'(c)=0$, i.e. $\int_{0}^{c}f(t)\ dt=(1-c)\cdot f(c)\ .$[/color][/quote]
Excellent proof.

[quote="Virgil Nicula"]
$\boxed{D.}\blacktriangleright$ Let $f: [0,b]\rightarrow R$ be a continuous function and $0<a<b\ .$ Prove that exists a point $c\in (0,1)$ so that $\int_{a}^{b}f(x)\ dx=bf(bc)-af(ac)\ .$[/quote]

Let $g(x) = \left( b \int_{0}^{x}f(b t ) \, dt-a \int_{0}^{x}f(a t ) \, dt \right)-x \int_{a}^{b}f(t) \, dt$ .  It's $g(1) = g(0)=0$ therefore there exist $c \in (0,1)$ such that $g'(c)=0$ .

Now $g'(x)=[bf(bx)-af(ax)]-\int_{a}^{b}f(t) \, dt$ gives $\boxed{ \int_{a}^{b}f(x)\ dx=bf(bc)-af(ac) }$ .

Hope it's correct  :)

For $B$ we must solve the first order differential equation $y = x^{2}\cdot y^\prime$. One of the solutions is (I think) $h(x) = \frac1{\exp \left( \frac{x^{-3}}3 \right)}$.
Now apply Rolle's Theorem to \[\left( \int_{0}^{x}f(t) \, dt \right) \cdot \frac1{\exp \left( \frac{x^{-3}}3 \right)}\] and you get the result.
Or something like that. $C$ looks too ugly  :(  Never liked this kind of exercises anyway.


If $ n$ runs through all the positive integers, then $ f(n) \equal{} \left[n \plus{} \sqrt {\frac {n}{3}} \plus{} \frac {1}{2} \right]$ runs through all positive integers skipping the terms of the sequence $ a_n \equal{} 3 \cdot n^2 \minus{} 2 \cdot n.$

Denote increasing positive integer sequence $\left\{ {{b_n}} \right\}_{n = 1}^{ + \infty }$ runs through all positive integers skipping the terms of the sequence of $\left\{ {{a_n}} \right\}_{n = 1}^{ + \infty }$.
Let ${b_n} = k$, ${a_m} < k < {a_{m + 1}}$, then there are $m$’s $\left\{ {{a_n}} \right\}$ and $n$’s $\left\{ {{b_n}} \right\}$ in $\left\{ {1,2,...,k} \right\}$. So $k = n + m$.
From $3{m^2} - 2m < n + m < 3({m^2} + 2m + 1) - (2m + 2)$, we have
${m^2} - m < \frac{n}{3} < {m^2} + m + \frac{1}{3}$, so ${m^2} - m + \frac{1}{3} \leqslant \frac{n}{3} \leqslant {m^2} + m$, hence ${m^2} - m + \frac{1}{4} < \frac{n}{3} < {m^2} + m + \frac{1}{4}$, then we have ${\left( {m - \frac{1}{2}} \right)^2}< \frac{n}{3} < {\left( {m + \frac{1}{2}} \right)^2}$, and $m < \sqrt {\frac{n}{3}}  + \frac{1}{2} < m + 1$, so $m = \left[ {\sqrt {\frac{n}{3}}  + \frac{1}{2}} \right]$, ${b_n} = n + m = \left[ {n + \sqrt {\frac{n}{3}}  + \frac{1}{2}} \right]$.


I wonder if you could help me to identify which function is represented by this graph
If you could write an equation for something similar to this I would be very happy.

[img]http://www.angelfire.com/planet/wolfram/graf7.JPG[/img]

Thank you for any help.

I can't think of any kind of function that would look like that...
But looking at how the curve acts as x grows you could be looking for something like:
 $y = \frac 1x$

But then it would never touch the y-axis, nor would it have the "bend" of your graph..

It could also be a function of the type:

$y = x^n - x^m$

Edit: The later seems pretty improbable though..

It can't be $\frac{1}{sinx}$.
Reciprocol trig functions have vertical asypmtotes, not horizontal...

I'm thinking something similar to a bell curve, but I can't remember the equations for those...

Just a question about "why" this function : is it coming from physics? :)

[quote="10000th User"]Its pretty hard to say which type, there are infinitely many continuous functions with a similar shape.

You can try $y=\frac3{1+x^4}$ or $y=3e^{-x^2}$ or $y=\frac3{1+\ln^2\left(1+x^2\right)}$[/quote]

On a side note, I was just wondering how you came up with the second two functions?

Yes the two last of the tree do not fit very well to say least.. Thank you anyway for your reply.

it looks like normal distribution graph to me

The graph never touches the x-axis (the x-axis is the "asíntota")... and when $x=0$ , then $y=1$

Um, that crosses the x-axis all over the place...

I agree with the previous comments about it being something like a normal distribution (bell) curve which follows something of the sort $f(x)=\frac{1}{1+x^2}$ which can be generalized, but the denominator is the numerator (a contstant) plus some even power of x.

Yeah, I just looked a thing or two that I had forgotten up, and it's definitely a bell curve, so it will look something like what Hokkage says...


A circle with area 16pi is inscribed within a triangle with an area of 12. Find the perimeter of the triangle.
 :stretcher:

[quote="xpmath"]K=rs kills the problem.[/quote]
Thx! :lol:

[quote="xpmath"]K=rs kills the problem.[/quote]

Sorry, but what do $ k\equal{}rs$ stand for? I think $ r$ is radius.

[quote="mathcrazed"]$ s$ stands for the semi-perimeter... And $ k$ represents the area of the triangle.[/quote]

$ s$ is the semi-perimeter of a triangle, right?

If so, is the answer $ 6$?

k=rs

s=semiperimeter (P/2)
r=inradius (the radius of the circle inscribed in a triangle).

Using the formula,
12=4s
s=3
so P=6

[quote="xpmath"]K=rs kills the problem.[/quote]
lol. totally. but its their fault for making that so obvious . :P


geometry
circumcircle
projective geometry
geometry proposed

Let (O) is the circumcircle of the triangle ABC, AD is a diameter of (O). The tangent of (O) at D meets BC at P. The line OP cuts AB,AC at two points M,N. Prove that OM = ON.

Dear Mathlinkers,
1. let E the second point of intersection of DN with the circumcircle and Td the tangent at D.
2. According to Pascal's theorem applied to AD Td EBCA, BE goes through O
3. According to the butterfly theorem, we are done.
Sincerely
Jean-Louis

DEAR mr.danh:
nice to see you again in mathlinks

here's solution to the nice problem:
draw the parallel line to MN through B   denote X is the midpoint of BC
1. O,X,D,P  are concyclic
2. B,Y,X,D are concyclic
3. XY//CZ
done

Let $ BD, CD\cap MN$ at $ K,L$.
Acording to [b]Menelaus[/b]'s theorem: $ \frac {MA}{MB}.\frac {KB}{KD}.\frac {OD}{OA} \equal{} 1$
$ \Rightarrow MA.MB \equal{} \frac {KD}{KB}.MB^2$, similarly, $ NA.NC \equal{} \frac {LD}{LC}.NC^2$
We need to prove that $ MA.MB \equal{} NA.NC \Leftrightarrow \frac {KD}{KB}.MB^2 \equal{} \frac {LD}{LC}.NC^2$
$ \Leftrightarrow \frac {MB^2}{NC^2} \equal{} \frac {LD}{LC}.\frac {KD}{KB} \equal{} \frac {PC}{PB}$ ([b]Menelaus[/b]'s theorem again)
$ \Leftrightarrow \frac {MB^2}{NC^2} \equal{} \frac {DB^2}{DC^2} \Leftrightarrow \frac {MB}{NC} \equal{} \frac {DB}{DC}$
$ \Leftrightarrow \Delta MBD\sim \Delta NCD (*)$
Denote $ M', N'$ the second intersections of $ MD, NC$ and $ (O)$. Applying [b]Pascal'[/b]s theorem we have $ M'C\cap N'B \equal{} \{O\}$
$ \Rightarrow \angle M'DB \equal{} \angle N'DC$ and $ \angle MBD \equal{} \angle NCD \equal{} 90^o$
So $ (*)$ is true.
We are done!

Dear mr. danh and Mathlinkers,
it is possible to consider a converse of your problem... if you want please...
Sincerely
Jean-Louis

Is it the same  :P ? see my long - time ago post: http://www.mathlinks.ro/viewtopic.php?t=254679

Dear Mathlinkers,
yes, if you make an equivalence between equality and tangent.
Thank for the reference.
Sincerely
Jean-Louis


probability
expected value
probability and stats

Each paper receives a grade from the set {A, A-, B+, B, B-, C+} with equal probability, independently of the other paper. How many papers do you expect to hand in before you receive all possible grades at least once.

p = 1/6

I don't know what to do.

[/img]

How did you get the averages? Are they expected values?

[quote="JRav"]Think of it like rolling a die.  If you want to get each face 1 time, on average, how many rolls do you need?

Well your first roll guarantees you a new number.  To get a new number, you need on average 2.  To get a third unique number, it takes on average 3.  For a fourth new number, 4 on average, and so on.

It takes an average of 1+2+3+4+5=15 rolls on average to roll each number.[/quote]

this solution is obviously wrong, since (for example) the expected number of rolls to get a new coupon after you have one unique coupon is $ 6/5$ and not $ 2$ (think of it, you have a $ 5/6$ chance of getting a new coupon and only a $ 1/6$ chance of getting the old coupon again.  now compute by conditioning or using geometric series).

if you extend that argument it's easy to get the correct answer.  see [url=http://en.wikipedia.org/wiki/Coupon_collector's_problem]here[/url].

I recall someone telling me the solution to the problem is 15.  But I obviously went about it the wrong way.

Should it be $ \frac{6}{6}\plus{}\frac{6}{5}\plus{}\frac{6}{4}\plus{}\frac{6}{3}\plus{}\frac{6}{2}\plus{}\frac{6}{1}\equal{}14.7$?

yes, that is correct (again, see the link).


vector
real analysis
real analysis unsolved

Let $ T$ be a linear functional on a normed linear space.   Show that if $ Ker(T)$ is closed, then $ T$ is continuous.

if $ H\equal{}Ker(T)$ is not dense, you can find a unit vector $ x$ such that $ x \not \in H$ and $ d(x,H)>0$. Now, every unit vector $ v$ can be decomposed $ v\equal{}\lambda x \plus{} y$ with $ y \in H$. Because $ Tv\equal{}\lambda Tx$, you just have to find a bound on $ \lambda$, which is easy since $ 1\equal{}\|v\|\equal{}\|\lambda x \plus{} y\|\equal{}|\lambda|.\|x \plus{} \frac{y}{\lambda}\|$. Do you see how to conclude ?


Find positive integer $ n$ so $ \arctan{\frac{1}{3}}\plus{}\arctan{\frac{1}{4}}\plus{}\arctan{\frac{1}{5}}\plus{}\arctan{\frac{1}{n}}\equal{}\frac{\pi}{4}$

Another hint please?

Had to use a calc :(

See Resources  :arrow: USA :arrow: AIME give solution to your questions that you have posted so far.

Hi kunny. I worked through problem 4, opened the post on problem 8 and saw your path to the solutions of both, which I followed. I was pleased to see a neater solution. Thanks.

as kunny said..

already discussed at

[url]http://www.mathlinks.ro/resources.php?c=182&cid=45&year=2008&p=1075233[/url]


for 3.6.6 in the book i dont understand how to do it. i read through the answer and i still dont understand it.

[quote]Triangle[i] ABC[/i] is equilateral in the diagram on the right above. Points D,E, and F are in triangle [i]ABC[/i] such that <CAD=<ABE=<BCF. Point D lies on line CF, point E lies on line AD, and point F lies on line BE. Prove that triangle [i]DEF[/i] is equilateral.[/quote]

$ D$ on $ CF$, $ E$ on $ AD$, and $ F$ on $ BE$? That's not what your diagram has... in fact the only way I can see that happening is if $ D \equal{} E \equal{} F$. Can you clarify?

Oh, woops.

Consider $ \triangle AEB$. Since $ \angle ABE \equal{} \angle EAC$ and $ \angle BAE \plus{} \angle EAC \equal{} \angle BAC \equal{} 60$, we have $ \angle ABE \plus{} \angle BAE \equal{} 60$. Since the angles of a triangle add up to $ 180$, this means $ \angle BEA \equal{} 120$. Then $ \angle FED$, being the supplement, is $ 60$. By repeating this argument for the triangles $ \triangle CDA, \triangle BFC$, the other two angles of $ \triangle DEF$ are also $ 60$. Thus it's equilateral.


modular arithmetic
number theory unsolved
number theory

$m$ is an integer, $\tau$ is a real, $1\leq \tau\leq m$. $(a,m)=1$. Show that
$ax\equiv y\pmod m,\quad 0\leq x\leq \tau, 0<|y|<\frac{m}{\tau}$
has a solution.

We must have $\tau < m$(You should check)
 If $\tau =1$ take $x=1,y=a(mod m)$
 Now,if $1<\tau<m$.
 There are $\left([\tau]+1\right){\left(\left[\frac m {\tau}\right]+1\right)}$(which is $>m$) numbers of form $ax+y,0\leq x\leq [\tau],0\leq y\leq {\left[\frac m {\tau}\right]}$

 So we have two distinct numbers among them such that
       $ax_1+y_1\equiv ax_2+y_2(mod m)$
 Note that $|x_1-x_2|<m,|y_1-y_2|<m,gcd(a,m)=1$ so $x_1\neq x_2,y_1\neq y_2$
 Denote $x=|x_1-x_2|,y=y_1-y_2$(or $y_2-y_1$,if $x_1>x_2$).That's the two numbers.


Let f from R[X] a polynomial . If f+n has only real roots , for every    integer n , then deg(f) = 1 .

Let f be a polynomial with degree k > 1, such that f+n has only real roots for every integer n.
Wlog, we may suppose that f has leading coefficient a > 0 (if not consider -f).

- If k is even.
Then f(x) tends to +oo as x goes to +oo and to -oo. It follows that f has a lower bound, say M.
Let n be an integer such that M + n > 0.
Then f+n is positive and then has no real roots, which contradicts the hypothesis.

- k is odd.
Then f(x) tends to -oo as x tends to -oo and f(x) tends to +oo as if goes to +oo.
It follows that there exist A,B such that f is increasing on ]-oo,A] and f is increasing on [B;+oo[ and B > 0.
Let M be the lower bound of f on [A,2B]. Then M is the lower bound of f on [A,+oo[.
Let n such that M + n > 0.
Then f+n is positive on [A,+oo[ so it has no real root on this interval, and f+n is increasing on ]-oo,A] thus it has at most one real roots on ]-oo,A].
It follows that f+n has at most one real root on |R. Since k > 1, we get the desired contradiction.

Pierre.


Eight singers participate in an art festival where m songs are performed. Each song is
performed by 4 singers, and each pair of singers performs together in the same number of songs. Find
the smallest m for which this is possible.

basically this asks for the existence of a block design with block size $4$ and point set size $8$.
counting the number of blocks one can easily see that this equals $\lambda \dfrac{8\cdot 7}{4\cdot 3}$ where $\lambda$ is the number of songs that any pair of singers together are involved in. Likewise the number of songs that any particular singer is involved in is easily seen to be $\lambda \dfrac{4}{3}$, so that $\lambda=3$ gives the min possible number of blocks.
To see that this is achievable, just take the usual projective plane or order $2$ and add an addititonal point and consider all the blocks containing this addititonal point to be the lines of the projective plane, along with this point. The remaining $7$ blocks can be completed as
$\{1,2,3,4\},\{1,2,5,6\},\{1,3,5,7\},\{1,4,6,7\},\{2,3,6,7\},\{2,4,5,7\},\{3,4,5,6\}$.


calculus
integration
function
trigonometry
complex analysis

Find
\[\int_0^\infty \frac{e^{ax}-e^{-ax}}{e^{\pi x}-e^{-\pi x}}dx\]
where $-\pi<a<\pi$.

One possible approach is to consider the integral of the function $F(z)=\frac{e^{az}-e^{-az}}{e^{\pi z}-e^{-\pi z}}$ over the boundary of the strip $0<\text{Im} z<1$ with a small semicircle around $i$ deleted. If $I$ is the integral we are looking for, we'll get the identity $2I(1+\cos a)=\pi i\text{Res}_i F$ ($\pi i$ because we integrate only over a half of the circumference around a simple pole at $i$). The residue is not hard to find. The final answer is $\frac{\sin a}{2(1+\cos a)}$ (I hope I haven't made a stupid mistake somewhere :) )


geometry
circumcircle
inradius
trigonometry
geometry proposed

Let $\triangle ABC$ be a triangle and denote $O,H,R,r,r_a,r_b,r_c$ the circumcenter, the orthocenter, circumradius, inradius, exradii respectively. Show that:

$(1) \ \ \ OH = R\sqrt {1 - 8\cos A\cos B\cos C} $

$(2) \ \ \  \cos (B - C) + \cos A = \frac{{bc}}{{2R^2 }}$

$(3) \ \ \  \frac{{bc}}{{r_a }} + \frac{{ca}}{{r_b }} + \frac{{ab}}{{r_c }} + 6R = 2R\left( {\frac{{b + c}}{a} + \frac{{c + a}}{b} + \frac{{a + b}}{c}} \right)$

[color=blue][size=150][b](1)[/b][/size][/color] From Geometry Revisited by Coxeter and Geitzer

$OH^2 = 9R^2 - (a^2 + b^2 + c^2) = R^2[9 - 4 (\sin^2{\widehat A} + \sin^2{\widehat B} + \sin^2{\widehat C})]$

$\sin^2{\widehat A} + \sin^2{\widehat B} = 1 - \cos{(\widehat A + \widehat B)} \cos{(\widehat A - \widehat B)}$

$\cos{(\widehat A + \widehat B)} + \cos{(\widehat A - \widehat B)} = 2 \cos{\widehat A} \cos{\widehat B}$

Using the above easy to verify trigonometric formulas and $\cos{\widehat C} = -\cos{(\widehat A + \widehat B)}$, this becomes

$OH^2 = R^2\{1 + 4 [\cos{(\widehat A + \widehat B)} \cos{(\widehat A - \widehat B)} + \cos^2{(\widehat A + \widehat B)}]\} =$

$= R^2\{1 + 4 \cos{(\widehat A + \widehat B)}\ [\cos{(\widehat A - \widehat B)} + \cos{(\widehat A + \widehat B)}]\} =$

$= R^2 (1 - 8 \cos{\widehat C} \cos{\widehat A} \cos{\widehat B})$

$OH = R \sqrt{1 - 8 \cos{\widehat C} \cos{\widehat A} \cos{\widehat B}}$

[color=blue][size=150][b](2)[/b][/size][/color] Proof of the 2nd formula is quick:

$\cos{(\widehat B - \widehat C)} + \cos{\widehat A} = \cos{(\widehat B - \widehat C)} - \cos{(\widehat B + \widehat C)} = 2 \sin{\widehat B} \sin{\widehat C} = 2 \frac{bc}{4R^2} = \frac{bc}{2R^2}$

[color=blue][size=150][b](3)[/b][/size][/color] Let $s, \triangle$ be the triangle semiperimeter and area.

$\frac{bc}{r_a} + \frac{ca}{r_b} + \frac{ab}{r_c} + 6R = \frac{bc}{s \tan{\frac{\widehat A}{2}}} + \frac{ca}{s \tan{\frac{\widehat B}{2}}} + \frac{ab}{s \tan{\frac{\widehat C}{2}}} + 6R =$

$= \frac{1}{sr}[bc(s - a) + ca(s - b) + ab(s - c)] + 6R =$

$= \frac{2R}{abc}[2bc(s - a) + 2ca(s - b) + 2ab(s - c) + 3abc] = $

$= \frac{2R}{abc}[bc(b + c - a) + ca(c + a - b) + ab(a + b - c) + 3abc] = $

$= \frac{2R}{abc}[bc(b + c) + ca(c + a) + ab(a + b)] = $

$= 2R \left(\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c}\right)$

where we used the following formulas

$r_a = s \tan{\frac{\widehat A}{2}}, r_b = s \tan{\frac{\widehat B}{2}}, r_c = s \tan{\frac{\widehat C}{2}}$

$r = (s - a) \tan{\frac{\widehat A}{2}} = (s - b) \tan{\frac{\widehat B}{2}} = (s - c) \tan{\frac{\widehat C}{2}}$

$\triangle = sr = \frac{abc}{4R}$

BTW, how did Ramanujan do back then ?

A simpler proof of the formula [color=blue][b](3)[/b][/color]:

Let $\Delta$ be the area and $s=\frac{a+b+c}{2}$ the semiperimeter of triangle ABC. Then, it is well-known that $\Delta=r_a\left(s-a\right)$. On the other hand, $\Delta=\frac{abc}{4R}$. Thus, $\frac{abc}{4R}=r_a\left(s-a\right)$. Consequently,

$\frac{bc}{r_a}=\frac{abc}{4R}\cdot\frac{4R}{ar_a}=r_a\left(s-a\right)\cdot\frac{4R}{ar_a}=4R\cdot\frac{s-a}{a}=2R\cdot\frac{2\left(s-a\right)}{a}$.

Since we know that 2 (s - a) = b + c - a, we thus have

$\frac{bc}{r_a}=2R\cdot\frac{b+c-a}{a}=2R\cdot\left(\frac{b+c}{a}-1\right)$.

Similarly,

$\frac{ca}{r_b}=2R\cdot\left(\frac{c+a}{b}-1\right)$;
$\frac{ab}{r_c}=2R\cdot\left(\frac{a+b}{c}-1\right)$.

Thus,

$\frac{bc}{r_a}+\frac{ca}{r_b}+\frac{ab}{r_c}=2R\cdot\left(\frac{b+c}{a}-1\right)+2R\cdot\left(\frac{c+a}{b}-1\right)+2R\cdot\left(\frac{a+b}{c}-1\right)$
$=2R\left(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}-3\right)=2R\left(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\right)-6R$,

and thus

$\frac{bc}{r_a}+\frac{ca}{r_b}+\frac{ab}{r_c}+6R=2R\left(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\right)$,

qed.

  darij


i just noted that the smilie 8) with its shortcut

[code]8)[/code]

is somehow contraproductive for a mathematical forum, since it tends to pop up in formulas where an 8 stands before a closing bracket... therefore... well... maybe a change of the shortcut would make sense... :D

  dg


For a,b,c real positive numbers prove that

 (b^2-a^2)/(c+a) + (a^2-c^2)/(b+c) + (c^2-b^2)/(a+b) \geq 0

[i]unknown source[/i]

Let x = a+b, y = b+c and z = c+a. Then a,b,c > 0 and :
(b 2 - a 2 )/(c+a) + (a 2  - c 2 )(b+c) + (c 2 - b 2 )/(a+b) = x(y-z)/z + z(x-y)/y + y(z-x)/x.
Thus it is to prove that xy/z + zx/y + yz/x  \geq x+y+z.

But using AM/GM, we have (xy/z + zx/y)/2  \geq x, and similar inequalities. Summing, leads to the result.
Note that equality occurs iff a = b = c.

Pierre.


Ok, I don't recall this beeing posted before but if it was I'm sorry:
a=(m^(m+1)+n^(n+1))/(m^m+n^n) for m,n positive integers. Prove that:
a^m+a^n \geq m^m+n^n

hehe this was actually posted! :D :D
I can't find it now, but here is the hint :
if n>=m then m<=a<=n and the inequality becomes:
(n-a)(n^(n-1)+n^(n-2)a+...+a^(n-1))<=(a-m)(a^(m-1)+a^(m-2)m+...+m^(m-1)) etc...

cheers! :D :D


[quote="kohjhsd"][quote="ProblemSets"][(x^2 +1)!]/[(x^2 +1)!]=650

solve for x[/quote]

I think it is
$\frac{(x^{2}+1)!}{(x^{2}+1)!}=650$
but that's not possible. :([/quote]
Yea, they would cancel out and then it would be 1=650, which is not true.

[quote="kohjhsd"]I think the problem was meant to be
\[\frac{(x^{2}+1)!}{(x^{2}-1)!}=650. \]
This can be solved!
[hide]
$(x^{2}-1)!$ cancel, so we have
\[x^{2}(x^{2}+1)=650. \]
which is
\[x^{4}+x^{2}-650=0\\ (x^{2}+26)(x^{2}-25)=0. \]
Only possible solution is $x=\boxed{5}$.
[/hide][/quote]
[hide="I think you meant..."]
The only possible solutions are $\boxed{\pm5}$.[/hide]


eh?
 The message is too small. Please make the message longer before submitting.

spam?
The message is too small. Please make the message longer before submitting.

If you replace "submitting" with "inserting" and "message" with

:rotfl: :rotfl: :rotfl:

lol that was hilarious

pianoforteARML??  is that a copy of pianoforte?

I believe it's an impersonator....maybe perhaps another multi-account of the Missourians? :maybe:

Pas moi = Not me.

Hmm....time to go tell pianoforte who did it.

Special thanks to Google Translate. :D


This is a classic, so don't ask for any specific sources :). I bet it can be found in lots of places on the Internet.

We are given two fuses, each of which burns, when ignited at one end, in an hour. How can we measure $45$ minutes?

We can't assume the fuses are homogenous (the first half of a fuse could burn in $5$ minutes, and the second half in $55$ minutes, for example).

But are their  burning rates  (function with respect to time) identical? (Or the fuses are identical)  if  so the solution is easy.  (I don't need f_1(t) linear, just f_1(t)=f_2(t))?)

Nope, you can't assume anything like that.

Ah, this is nice. In fact, theres so few possibilities, that you really can't miss the solution:

[hide]

Light both ends of the first fuse and one end of the second. When the first fuse has been totally burnt, 30 minutes has elapsed. Now light the second end of the second fuse - when thats burnt, 45 minutes has elapsed.

[/hide]

Okay  - Can I assume, say, for example, if I light it on the both-ends it will finish in 30 minutes?

(If yes, Then the answer is obvious) ...

...If not  can i make a pedulum  consisting of  one fuse, count the number of osscilations in 60 minutes and ....

...Or   use the first fuse to calibrate how fast I can count .one two three ...
...
...If all else fails  just  extract Cs-133 atom ( Every fuse I ever lit had that atom) and count t     24 820 105 779000  vibrations of  hyperfine transition... it will be exactly 45 minutews.. actually  ([i]That is how  45 minutes is [b]defined[/b])
[/i]

Added later: Just saw Triple-M's spoiler .. it  was same what I was thinking ...


If $ a$ and $ b$ are two unequal positive numbers, then:

$ \textbf{(A)}\ \frac {2ab}{a \plus{} b} > \sqrt {ab} > \frac {a \plus{} b}{2} \qquad\textbf{(B)}\ \sqrt {ab} > \frac {2ab}{a \plus{} b} > \frac {a \plus{} b}{2}$
$ \textbf{(C)}\ \frac {2ab}{a \plus{} b} > \frac {a \plus{} b}{2} > \sqrt {ab} \qquad\textbf{(D)}\ \frac {a \plus{} b}{2} > \frac {2ab}{a \plus{} b} > \sqrt {ab}$
$ \textbf{(E)}\ \frac {a \plus{} b}{2} > \sqrt {ab} > \frac {2ab}{a \plus{} b}$


geometry
rectangle
calculus
derivative
analytic geometry
Pythagorean Theorem
calculus computations

I had a lot of trouble figuring out this problem. I keep getting to y=0 which is not ture.

A 1 km racetrack is to be built with two straight sides semi-circles at the ends. Find the dimentions of the track that encloses the maximum area .

My attempt:

Let $y$ rep. one sidelength
Let $x$ rep. the radii of the circle (2x = width of rectangle)

$1000=2y+2\pi x$
$y=500-\pi x$ and

$A=2xy+\pi x^{2}$
$A=2x(500-\pi x)+\pi x^{2}$
$A=1000x-2\pi x^{2}+\pi x^{2}$
$A=-\pi x^{2}+1000x$
$A'=-2\pi x+1000$
$x=\frac{500}{\pi}$

Plugging this in the constraint gets me a value of $0$ for $y$. Could someone point out where i went wrong and if possible correct my mistake. Thanks!

$y=0$ is in fact the correct answer: in other words, the track should simply be a circle.

Ohh  :oops:  never thought of that hehe
While we're on the topic, I'm having some trouble finding hte equations for this problem:

Painters are painting the second floor exterior wall of the building that adjoins a busy sidewalk. A corridor 2m wide and 3 m high is built to protect pedestrians. What is the length of the shortest ladder that will reach from the ground over the corridor of the wall of the building?

I draw the diagram right but cant seem to figure out the equations. I'm also pritty sure that a 45 deg. triangle will amke the shortest distance making it $5\sqrt{2}$ But the approac is what i need

This one seems to be more complicated... Im getting into huge equations:

What I tried doing is set the height from teh corridor to the top of the ladder $x$ and the length of the corridor on the ground to the start of hte ladder $y$. Then i used pythagoras to figure out the area of the whole picture n then it gets really complicated trying to solve... Is there a better way maybe? I havent completed the simlifying just because of its complexity but if its the only way I guess I'll have to go with it...

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=24520]Look here.[/url] The 45 degree angle is not the correct answer.

The "ladder over the fence" problem is dual to the "pipe around the corner" problem. There are several other places it's been posted - try searching for the words "ladder" and "fence."

the 45 degree angle works for me when i plug in $a=3$ and $b=2$ into $\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}$ i get teh same answer as I would have gotten if I went $5\sqrt{2}$ which i got by 45 deg. triangles

The solution posted on the link you provided seems to be too complicated. There has to be a relation in which you can use derivative and optimization to solve the problem.

$5\sqrt{2}\approx\eq7.071$

$(2^{2/3}+3^{2/3})^{3/2}\approx\eq7.023$

They're not the same.

If $a=b,$ then you get the 45 degree angle. If $a$ and $b$ are reasonably close to each other, then we're not too far away.

:O samll difference i didnt catch it cuz I rounded. So the solution posted by kunny on that page is the only way to do this problem?

There are several ways to do that problem. kunny used a rectangular coordinate method, using one of the intercepts as his variable. I also posted such a method, using the (Calculus III) technique of Lagrange multipliers. It's also worth using one of the angles in the problem as the variable to differentiate with respect to. That also works.

[quote="Jeremic"][url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24520&start=20[/url]

Ok I understand Kunny's solution until the part where he says $m=-\sqrt[3]{\frac{b}{a}}$

how did he get this? I know that $f'(m)=0$ when one of the brackets is equal to zero i understand how he got that but why didnt he get the $m$ in terms of the other bracket being zero???

$\frac{2a}{m^{3}}(am-b)=0$[/quote]

Since $a>0,\ b>0,\ m<0,$ we have $\frac{2a}{m^{3}}<0,\ am-b<0,$ yielding $\frac{2a}{m^{3}}(am-b)>0.$

Thanks for the answers but you posed another question  :(
How did you get $+b$ from $-b$ I tried putting that into mathematica This is waht I get...

$Simplify[a(-a^\frac{-1}{3}b^\frac{1}{3})-b]$

Output : $-a^\frac{2}{3}b^\frac{1}{3}-b$

And also at the start of your post you said htat coordinate $Q$ is at $Q\left(0,-am+b\right)$

But then when you used it in pythagorean theorem you changed it to $(am-b)^{2}$... should'nt it be $(-am+b)^{2}$

Phew! Done with the questions! Thank you so much for your help and patience. I know that most people would flip out by now because I asked one too many questions but thanks for coping with me and helping me understand. Who knows maybe one day I'll be able to help you  :P


[quote="El fisico"]You are right, I fixed it already. :)[/quote]

No . There are more solutions...
Try to factor it... Dont delete nothing :)


geometry
3D geometry
tetrahedron
pyramid
similar triangles

Consider a circular cone with vertex $ V$, and let $ ABC$ be a triangle inscribed in the base of the cone, such that $ AB$ is a diameter and $ AC \equal{} BC$. Let $ L$ be a point on $ BV$ such that the volume of the cone is 4 times the volume of the tetrahedron $ ABCL$. Find the value of $ BL/LV$.

[hide="Solution"]Let $ r$ be the radius of the cone base. Then, $ r^2$ is the area of the pyramid's base and $ \pi r^2$ is the radius of the cone's base.

Since the cone's base is $ \pi$ times as large as the pyramid's base, and it's volume is $ 4$ times that of the pyramid, the height of the cone is $ \frac {4}{\pi}$ times that of the pyramid.

By similar triangles, $ \frac {BL}{BV} \equal{} \frac {\pi}{4} \implies \boxed{\frac {BL}{LV} \equal{} \frac {\pi}{4 \minus{} \pi}}$[/hide]


algebra
polynomial
superior algebra
superior algebra unsolved

Let $ \zeta_n\equal{}e^{2\pi i/n}$.
(a) Let $ F$ be a field with exactly eight elements. Prove that the characteristic of $ F$ is 2.
(b) Prove that $ \minus{}1$ cannot be written as a sum of squares in $ \mathbb{Q}(\zeta_3\sqrt[3]{2})$.
(c) Determine the irreducible polynomial over $ \mathbb{Q}(\zeta_3)$ of $ \zeta_6$ and $ \zeta_9$.

As for (a), I can show this by computing $ \mathbb{F}_8$, but I am wondering if there is a less clunky way without explicit computation.
As for (b), I have no idea.
As for (c), I know that $ \mathbb{Q}(\zeta_3)\equal{}\mathbb{Q}(\sqrt{\minus{}3})$, but I do not see how this would simply the factors of $ x^6\minus{}1$, $ x^9\minus{}1$.

(a) is straightforward if you look at F as a group under addition.  It has order 8, so $ 8x\equal{}0$ for all x.  Letting x be 1, we have $ (1\plus{}1)^3\equal{}8\equal{}0$, so $ 1\plus{}1\equal{}0$.

(b) Note that the minimal polynomial for $ \zeta_{3}\sqrt[3]{2}$ is $ x^3\minus{}2$.  So it suffices to prove that if $ \minus{}1$ is not a sum of squares in a field $ K$, and $ \alpha$ is the root of a degree 3 polynomial that is irreducible over $ K$, then $ \minus{}1$ is not a sum of squares in $ K(\alpha)$.  The proof of this is pretty fun.

Say $ \minus{}1 \equal{} \sum{\beta_{i}^2}$, with $ \beta_{i}\in K(\alpha)$.  Then write each $ \beta_{i} \equal{} f_{i}(\alpha)$, where each $ f_{i}$ has coefficients in $ K$ and degree at most 2.  Then the polynomial $ f(x) \equal{} 1 \plus{} \sum{(f_{i}(x))^2}$ has a root at $ \alpha$, hence the minimal polynomial of $ \alpha$ divides $ f$.  But the degree of $ f$ must be 2 or 4, and since it is divisible by a degree 3 polynomial, it follows that it has a root in $ K$.  So there is some $ z\in K$ with $ 1 \plus{} \sum{(f_{i}(z))^2} \equal{} 0$, so $ \sum{(f_{i}(z))^2} \equal{} \minus{}1$, a contradiction.

(c) It would help, to start, by finding the minimal polynomials in $ \mathbb{Q}$.  These are $ x^2\minus{}x\plus{}1$ and $ x^6\plus{}x^3\plus{}1$.  From solving the first, you can find that $ \zeta_{6}\in \mathbb(\sqrt{\minus{}3})$, so its minimal polynomial in $ \mathbb{Q}(\zeta_{3})$ is just $ x\minus{}\zeta_{6}$.  But from the latter, we see that $ \zeta_{9}$ has degree 6 and so in particular $ \zeta_{9}\notin \mathbb{Q}(\zeta_{3})$, so the polynomial $ x^3\minus{}\zeta_{3}$ has no roots, hence is irreducible, with $ \zeta_{9}$ as the root, therefore it is the minimal polynomial.

(b) $ \mathbb Q(\zeta_3\sqrt[3]2)$ is isomorphic to $ \mathbb Q(\sqrt[3]2)\subset\mathbb R$, and it is quite well-known that $ \minus{}1$ is not a sum of squares in $ \mathbb R$. owk


function
analytic geometry
graphing lines
slope
calculus
derivative
calculus computations

A car's distance traveled is a function of the time passed. If $ x$ is the time passed, and $ y$ is the distance traveled, then $ y \equal{} x^3 \minus{} x$. What is the car's velocity after $ \frac {1}{2}$ a unit of time has passed? In other words, what is the slope of the line tangent to $ y \equal{} x^3 \minus{} x$ at $ x \equal{} \frac {1}{2}$? Note that no units of measurement are present, because I do not want to get too much into physical stuff.

Is this too advanced? I expected it to go into Pre-olympiad, but it was moved to the college forum. I expected this to be much easier. Hint: use limits.

Slope of the tangent line at $ x\equal{}\frac{1}{2}$ is simply $ f'(\frac{1}{2})$.

This is a basic calculus problem which is why it was moved to the calculus forum.  The high school forum is strictly non-calculus regardless of difficulty.

Note that $ v\equal{}\frac{ds}{dt}$ where s is the distance traveled and t is the time passed.

To your problem, $ y\equal{}x^3\minus{}x$ gives the distance traveled where x is the time.

so differentiating y, we get

$ y'\equal{}3x^2\minus{}1$ which gives the velocity of the object at time x.

The velocity after $ \frac{1}{2}$ unit time is

$ y'(\frac{1}{2})\equal{}\frac{3}{4}\minus{}1\equal{}\minus{}\frac{1}{4}$

This is pretty basic. This is an introductory question of application of calculus.

[quote="takiguy"]Slope of the tangent line at $ x \equal{} \frac {1}{2}$ is simply $ f'(\frac {1}{2})$.

This is a basic calculus problem which is why it was moved to the calculus forum.  The high school forum is strictly non-calculus regardless of difficulty.[/quote]
What is that tickmark? I thought of it this way: Figure out the slope of the line connecting $ (\frac{1}{2}$,$ \frac{3}{8})$ to a point $ (x$,$ y)$. Figure out the limit the slope approaches as $ (x$,$ y)$ gets closer. That is precalculus.

[quote="DiscreetFourierTransform"]The tickmark next to f(x) represents the derivative of f.
This is a function that represents the slope of f(x) at every point in the original f's domain.

It's useful in calculus, as well as several other fields.
From what I've read, it appears as if you've synthesized the necessary ideas behind the development of the derivative.

You're looking at limits as a fixed quantity goes to zero, which is the insight you'll need to finish the development.[/quote]
I never thought of it that way. When I was setting the limit that the fixed quantity goes to zero, I was thinking rise over run. That is a cubic function over a linear function, which turns out to be a rational function. Precalculus.


probability
geometry
geometric transformation
reflection

A rich grandfather decides to give his money to his 50 grandchildren if they are clever enough to solve a problem. Everybodey goes into a room and gets a collor i their face, either red or blue. They can see every color exept their own and their task is to guess their own color. No comunication of any kind is alowed after they have entered the room, but they can discuss the problem the day before. If everyone guess their own color, they get the money. Whats the probabillity for sucsess if they choose the wisest way to do this?

If the each person's color is independent, each has a 1/2 chance of guessing correctly, so the overall probability is 1/2^50.

[quote="Ad Infinitum"]This is a very interesting question and I can't wait for somebody to figure out the answer.  I have two guesses, though I'm pretty sure they're both wrong:

1. Each person just scratches their face until some of the paint comes off and then looks at the chips to get an idea of the color.

2. If it's a brightly lit room, people can look into each others' eyes and see the reflection of their own face-paint.

But since the question was "Whats the [u]probabillity for sucsess [/u]if they choose [u]the wisest way [/u]to do this?" I have a feeling that both of my answers have nothing to do with it.[/quote]
Sorry. They can not see their own color no matter what they do. And 1/2^50 is the probability for sucsess if they dont have a plan the day before

[quote="Treething"]Parity.

:|[/quote]
Yes. Thats right. And whats the answer then?

What exactly you mean by "guess".. do they guess (a) one by one or (b) all have to guess at the same time (by some secret ballot) [hide]In case of (a) .. does the next person hear previous person's guess...etc (In that case, there is "communication" among  the grand children and obviously (1/2) is the answer - first person's guess being random) .. If there is NO communication (one does not hear what the other guesses) ..and all have to guess at the same time.. you say it is red if   $(-1)^{n}$  is +1 where  $n$ is the total number  of other reds you seen (answer still (1/2) - either all are correct or all are wrong) [/hide]

[quote="Gyan"]What exactly you mean by "guess".. do they guess (a) one by one or (b) all have to guess at the same time (by some secret ballot) [hide]In case of (a) .. does the next person hear previous person's guess...etc (In that case, there is "communication" among  the grand children and obviously (1/2) is the answer - first person's guess being random) .. If there is NO communication (one does not hear what the other guesses) ..and all have to guess at the same time.. you flip a coin  choose a color, and say it is that color if    $(-1)^{n}$  is +1 and $n$ is the number of colors you see on other heads (answer still (1/2)) [/hide][/quote]
Lets say they write it down on a paper and the grandfather is the only one who can see the paper. They can not hear what the others say

[quote="Gyan"]Okay .. that case is covered above :)[/quote]
Yes. And its correct. Took a while before somebodey managed that one.

Hmm: this seems to be a somewhat incomplete presentation of Todd Ebert's hat problem.

The key feature that's missing in Cartman's presentation: not everyone has to guess. Many of the participants may refuse to guess. The criterion for "winning" is then that at least one person guesses correctly and no one guesses incorrectly.

Given those rules, there's an obvious simple probability $\frac12$ strategy: eldest grandchild guesses randomly; everyone else doesn't guess. So the best strategy has to have at least a probability $\frac12$ of working. But can you ever do better than that?

The counterintuitive answer is that you can do a lot better than that: the 50 grandchildren can arrive at a strategy that works with probability at least $\frac{31}{32},$ although one can prove that they can't do better than $\frac{50}{51}.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=ebert&t=130354]Link to a fuller discusion.[/url]


geometry
3D geometry
algebra
polynomial
sum of roots
product of roots

Given the equation $ x^3 \minus{} 4x^2 \minus{} 2x \plus{} 10 \equal{} 0$

If two is added to each root in the above equation to form a new equation, find the sum of the cubes of the roots in the new equation.

I just know this equation has a sum of roots of 4, and a product of roots of -10. How do I make it a new equation? The sum of the roots would become 10, which would make it x^3-10x^2 or something??

Please do not post the same question in multiple forums. Everybody's questions deserve to be fair so yours do not deserve to be asked twice in the forums. Be fair. If you needed attention or that you didn't get your answer in your previous post, bump it up simply by replying in that topic and ask what you didn't understand instead of initiating another new topic of the same question in another subforum. That is considered SPAM

 :spam:

Continue the discussion in here --> [url]http://www.mathlinks.ro/viewtopic.php?t=177916[/url]

Mods definitely know what they will do to this topic :)

I've posted so much today, I forgot what I posted. =(
Delete this post., =D


calculus
integration
function
limit
real analysis
derivative
inequalities

Let the continious function $f:\mathbb{R}\to\mathbb{R}$ satisfies the following condition: $\displaystyle \sup_{x\in\mathbb{R}}\int_0^1 \frac{|f(x+t)-f(x)|}{t^2}dt<+\infty$. Prove that $f \equiv const$

Is the numerator correct? My guess is that it should rather be $|f(x+t)-f(x)|$  :?

if not
there will be a const $C>0$  such that $|f(x+t)-f(x)|>Ct$,for one  $x$,there will be $[x_1,x_2]$,when $x\in [x_1,x_2]$,$|f(x+t)-f(x)|>C_1t$,$C_1$ is another const,(because $\frac{|f(x+t)-f(x)|}{t}$ is continious.take x=x_1, we will have the contradiction.
hope it is correct.sorry for my clearless :(

[quote="zhaobin"]
hope it is correct.sorry for my clearless :([/quote] To be honest, I do not understand this solution :(. It seems like the claim is that for any continuous function that is not identically constant, we can find a point $x\in\mathbb R$ and a constant $C_1>0$ such that $|f(x+t)-f(x)|\geq C_1|t|$ for all small $t$ (which, of course would be enough to solve the problem), but I do not see how this claim is justified. Just continuity of $\frac{|f(x+t)-f(x)|}{|t|}$ for $t\neq 0$ doesn't yield anything useful... Or, maybe, I completely misunderstood zhaobin's post? :?

sorry,
I will explain it better.
if not
we will have there exist a point $x_0$,such that $\lim_{t \to 0} \frac{|f(x_0+t)-f(x_0)|}{|t|} >0$,(if not we have $f'(x)=0$ for every point.),then we have $\frac{|f(x+t)-f(x)|}{|t|}>c$,c is a const.when  t< :varepsilon:  ,take $x=x_0$,we will have a contradiction.
    but there still be a problem, when $\lim_{t \to 0} \frac{|f(x_0+t)-f(x_0)|}{|t|}  $ is not exist :( ,fedja can you complete it

I came out the idea now.
we can prove the is a point $x_0$,such that $\lim_{t \to 0}\frac{|f(x_0+t)-f(x_0)|}{|t|}$ is not equal to 0,for any positive number $a$ ,and a $t_0$such that $\frac{|f(x_0+t_0)-f(x_0)|}{|t|}> c $,when $t_0<a$,and $\frac{|f(x_0+t_0)-f(x_0)|}{|t|}$ is continious.so we can know there a number $b$,such that $\frac{|f(x+t_0)-f(x)|}{|t|}> c_1 $,when $x \in [x_0,x_0+b)$,take x=x_0,we will have the contradiction.(because $t_0$ can be very small.

[quote="zhaobin"] but there still be a problem, when $\lim_{t \to 0} \frac{|f(x_0+t)-f(x_0)|}{|t|}  $ is not exist :( ,fedja can you complete it?[/quote]
Indeed, the limit may fail to exist. My approach is slightly different. If nobody finds a solution in a week, I'll post it (it is a nice problem and I'd like to give other people an opportunity to try it first).

fedja,
can you check the solution for in #7,again.
at last,we can consider the $\int$ in $[t_0,1]$
thank you in advance

There exist M>0 s.t  for any x real
$\int_{0}^{1}\frac{|f(x+t)-f(x)|}{t^2}dt\leq M$ you have

(1) $|\int_{0}^{1}\frac{f(x+t)-f(x)}{t^2}d t|\leq M$

$f$ is continous on $J(x,t)=[x,x+t]$ fundamental theorem of calculus
$f$ is differentiable almost all points of $J(x,t)$, $f' \in L^{1}$ Lebesgue

(2) $f(x+t)-f(x)=\int_{x}^{x+t}f'(y)dy$

(1) & (2) give

$|\int_{0}^{1}\frac{\int_{x}^{x+t}f'(y)dy}{t^2}dt|\leq M$
or
$|\int_{x}^{x+1}\frac{\int_{y-x}^{1}f'(y)dt}{t^2}dy|\leq M$
or
$|\int_{x}^{x+1}f'(y)\frac{-y+x+1}{y-x}dy|\leq M$

Around x we have necessary $f'(x)=0$ (since \leq M) almost every where

$f$ is constant almost everuwhere

[quote="zhaobin"]I came out the idea now.
we can prove the is a point $x_0$,such that $\frac{|f(x_0+t)-f(x_0)|}{|t|}$ is not equal to 0,for any positive number $a$ ,and a $t_0$such that $\frac{|f(x_0+t_0)-f(x_0)|}{|t|}> c $,when $t_0<a$,and $\frac{|f(x_0+t_0)-f(x_0)|}{|t|}$ is continious[/quote] Sorry, but I fail to understand this sentence. Maybe it is just the language barrier. Could you write it in some clearer way? Right now I understand only the beginning: "There exists a point $x_0$ such that" and then I get completely lost. The next 2 words I would expect in the sentence are either "for every" or "there exists" ($t$? $a$? $c$?). Try to write it formally explicitely specifying all the quantifiers. Example (quite meaningless, but grammatically correct): "There exists $x_0$ such that there exists $a$ such that, for every $t_0<a$, there exists $c>0$ such that, for every $t<t_0$, one has $\frac{|f(x_0+t)-f(x_0)|}{|t|}\geq c$". Note that by the moment one comes to the statement, all variables in that statement have some quantifier on them. This is a very formal style of writing but it allows to prevent any misunderstanding.   :) Also, when you say "continuous", please, specify on what set and with respect to what variable. I'm still at loss whether you use continuity in $x_0$ or in $t_0$ or in $t$ or in all three. Seems like you are trying to invoke the theorem saying that a strictly positive and continuous on a compact set function is bounded by some positive constant from below on that set but the details are completely unclear to me.

[quote="Moubinool"]
$f$ is continous on $J(x,t)=[x,x+t]$ fundamental theorem of calculus
$f$ is differentiable almost all points of $J(x,t)$, $f' \in L^{1}$ Lebesgue
[/quote]
Moubinool, why is $f$ differentiable a.e. ? Why cannot it be like the Weierstrass function? Am I missing something?  :?

fedja,it's very kind of you. i should give up my this bad habit :D .
if $f=const$ is not true,we can find a point $x_0$ such that $ \lim_{t \to 0}\frac{|f(x_0+t)-f(x_0)|}{|t|} =0$ is not true.
so there exist a number $c$ for any positive number $a$,we will find a $t_0$($|t_0|<a$),
such that $\frac{|f(x_0+t_0)-f(x_0)|}{|t_0|} >c$,
let $g(t)=\frac{|f(x_0+t)-f(x)|}{|t|}$,then $g(t)$ is continuous.so we can find there exist a positive number $b$,and another positive number $c_1$,such that $\frac{|f(x+t_0)-f(x)|}{|t_0|} >c_1$,when $t \in [t_0-b,t_0+b]$
we take $x=x_0$,in $\int_0^1 \frac{|f(x+t)-f(x)|}{t^2}dt$,then  consider $\int_{t_0}^{t_0+b} \frac{|f(x+t)-f(x)|}{t^2}dt$ ,let $t_0 \to 0$
if $t_0<0$,take $x=x_0+t_0$

In Walter Rudin book real complex analysis
theorem 7.18

theorem 7.20 (chapter 7 Differentiation-fundamental theorem calculus)
If  $f$ is an absolutely continous function on [a,b] then  $f$ is differentiable at
almost all pointsof [a,b], and $f' \in L^{1}$
$f(x)-f(a)=\int_{a}^{x}f'(t)dt $

:( but there is something wrong with my solution.because $b$ is restricted by  $t_0$

[quote="fedja"]Why cannot it be like the Weierstrass function? [/quote]

Ha yes kind of function wich is continous and no where derivable

[quote="Moubinool"]
If  $f$ is an absolutely continous function on [a,b] then  $f$ is differentiable at
almost all pointsof [a,b], and $f' \in L^{1}$
$f(x)-f(a)=\int_{a}^{x}f'(t)dt $[/quote]
Yes, but how do you show that $f$ is [b]absolutely[/b] continuous?  :?

I didn't really read the other solutions to see if they are correct, but here is the approach I thought of.

Let $a$ and $b$ be real numbers such that $a \le b$.  Suppose for contradiction's sake that $f(a) \ne f(b)$.  We are given that a particular integral in $t$ is bounded away from infinity.  Let's integrate (average) that integral for all $x$ from $a$ to $b$.  We can switch integrals, and use the triangle inequality to bound the integral of the absolute value by the absolute value of an integral.  The resulting integral in $x$ then "telescopes", that is, most of its content cancels out, except for the parts near $a$ and $b$.  By continuity, we can approximate those parts by $f(a)$ and $f(b)$.  We end up with an integral in $t$ of $\frac{dt}{t}$.  But that integral is infinite, so we get a contradiction.

I know this description was very vague, and I apologize for that.  I didn't have time for a clearer description.  I hope it makes some sense.

I think that it is enough that for every $ x \in \mathbb{R} $ that
$ \int_{0}^{1} \frac{|f(x+t) - f(x)|}{t^{2}} dt < + \infty $, but may be I am wrong.
Suppose that we have that $ \int_{0}^{1} \frac{|f(x+t) - f(x)|}{t^{2}} dt < + \infty $
for every $ x \in \mathbb{R} $. Take some $ \varepsilon > 0 $.
Then for every $ x \in \mathbb{R} $, because of
$ \int_{0}^{1} \frac{|f(x+t) - f(x)|}{t^{2}} dt < + \infty $, we have that there exists
$ t > 0 $, small as we want, such that $ |f(x+t) - f(x)| \leqslant \varepsilon t $.
Indeed, otherwise there exists some $ t_{0} > 0 $, such that for every $ 0 < t < t_{0} $
we have $ |f(x+t) - f(x)| > \varepsilon t $, then $ \frac{|f(x+t) - f(x)|}{t^{2}} >
\frac{\varepsilon}{t} $, so the above integral will not converge.
So we get that for every $ x \in \mathbb{R} $ there exists
$ t > 0 $, small as we want, such that $ |f(x+t) - f(x)| \leqslant \varepsilon t $.
I claim that it is enough to prove that $ f(x) $ is constant. Indeed, take some
$ x < y $, then take $ B $ to be the set of all $ z \in [x,y] $, such that
$ |f(z) - f(x)| \leqslant \varepsilon(z-x) $. It is clearly non-empty, because in contains
$ x $, and closed, because $ f $ is continuous. Therefore if we take $ z_{0} $ to be the
maximum of $ B $, then $ z_{0} = y $ because if $ z_{0} < y $, then we can take
$ 0 < t < y-z_{0} $, such that $ | f(z_{0} + t) - f(z_{0}) | \leqslant \varepsilon t $,
so $ | f(z_{0} + t) - f(x) | \leqslant \varepsilon ( z_{0} + t -x) $, therefore
$ z_{0} < z_{0} + t < y $, and $ z_{0} + t $ belongs to $ B $, and we obtain contradiction.
So we get that $ |f(y) - f(x)| \leqslant \varepsilon |y-x| $ for every $ x,y \in \mathbb{R} $.
This is true for each $ \varepsilon > 0 $, so we get that $ f(x) = f(y) $
for every $ x,y $, so $ f \equiv const $.

I hope it is correct  :) .

[quote="Leva1980"]
I hope it is correct  :) .[/quote]
Yes, it is correct and very nice. Here is another version of almost the same solution. Assume that $a<b$ and $f(a)\neq f(b)$. Since the problem doesn't change if we replace $f(x)$ by $-f(x)$, we can assume that $f(a)<f(b)$. Let $L(x)$ be the linear function coinciding with $f$ at $a$ and $b$. Let $\delta>0$ be the slope of $L$. Let $x_0\in[a,b]$ be the point where the continuous function $g(x)=f(x)-L(x)$ attains its minimum.  Since $g(a)=g(b)=0$, we may assume that $x_0<b$ (if $x_0=b$, just change it to $a$).
Now, for all $t\in(0,b-x_0)$, we have $f(x_0+t)=g(x_0+t)+L(x_0+t)\geqslant g(x_0)+L(x_0)+\delta t =f(x_0)+\delta t$. The rest is clear.

[quote="fedja"]
how do you show that $f$ is [b]absolutely[/b] continuous?  :?[/quote]

I search for hours I don't see, I don't know either if it is possible to get
absolutely continous with the statement

Does someone try in the following direction

There exist $M>0$ such that for any $x$ real we have
$\int_{0}^{1}\frac{|f(x+t)-f(x)|}{t^2}dt \leq M$

If we do the change variable $y=\frac{1}{t}$ we get

For any $x$ real $\int_{1}^{+\infty}|f(x+\frac{1}{y})-f(x)|dy \leq M$

It means for any $x$ real the integral $\int_{1}^{+\infty}|f(x+\frac{1}{y})-f(x)|dy $
converge

For any $x$ real the Cauchy criterion is satisfied at +oo for the integral
For any $\epsilon>0$ there exist $A>0$ for any $C>B>A, \int_{B}^{C}|f(x+\frac{1}{y})-f(x)|dy \leq \epsilon$

From this point can we get $f$ is constant ?


linear algebra
matrix
algebra
polynomial
abstract algebra
linear algebra unsolved

If $ AB\equal{}BA\equal{}0$ and $ rank(A^2)\equal{}rank(A)$, show $ rank(A\plus{}B)\equal{}rank(A)\plus{}rank(B)$.

Let $ n$ be the $ A$-dimension. $ rank(A^2)\equal{}rank(A)$ implies that we may assume $ A\equal{}diag(0,U)$ where $ 0$ is the nullmatrix of dimension $ k$, $ U$ is an invertible matrix of dimension $ n\minus{}k$, and $ B\equal{}\begin{pmatrix}E&F\\G&H\end{pmatrix}$ where $ E,H$ are square matrices of dimension $ k,n\minus{}k$.
$ AB\equal{}BA\equal{}0$  implies $ UH\equal{}0,UG\equal{}0,FU\equal{}0$ that is $ H\equal{}0,F\equal{}0,G\equal{}0$.
$ rank(A\plus{}B)\equal{}n\minus{}k\plus{}rank(E)\equal{}rank(A)\plus{}rank(B)$.

loup blanc, can you explain what $ A \equal{} diag(0,U)$ means and why $ rank(A^2) \equal{} rank(A) \implies A\equal{}diag(0,U)$, where $ U$ is an invertible matrix of dimension $ n\minus{}k$?

$ diag(0,U)\equal{}\begin{pmatrix}0_k&0\\0&U_{n\minus{}k}\end{pmatrix}$.
The second point is the key of the exercise:
the $ A$-characteristic polynomial is in the form $ x^kp(x)$ where $ 0\leq{k}\leq{n}$ and $ p(0)\not\equal{}0$. Then using the theorem of decomposition of kernels we obtain that $ A$ is similar to $ diag(N,V)$ where $ N^k\equal{}0$ and $ p(V)\equal{}0$ that is $ V$ is invertible. Thus $ A^k\sim{d}iag(0,V^k)$.
Here $ rank(A^2)\equal{}rank(A)$ thus $ rank(N^2)\equal{}rank(N)$ and $ N\equal{}0$.

Another solution:
Consider $ A,B$ as left multiplication maps from $ k^n$ to itself, where $ k$ is the base field. Then clearly, $ rank(A\plus{}B) \equal{} rank(A) \plus{} rank(B) \minus{} dim(ImA \cap ImB)$. So it suffices to show that $ Im A \cap Im B \equal{} \{0\}$.
Let $ Av \equal{} Bw$ lies in the intersection. Then $ A^2v \equal{} ABw \equal{} 0$. Since $ rank(A^2) \equal{} rank(A)$, we have $ ker(A^2) \equal{} ker(A)$ by rank-nullity theorem. Thus $ Av \equal{} 0$, i.e. the common element has to be 0.

Hi Soarer,
your formula $ rank(A\plus{}B) \equal{} rank(A)\plus{}rank(B)\minus{}dim(ImA\cap ImB)$ is false.
Take $ A\equal{}\begin{pmatrix}0&0\\0&1\end{pmatrix},B\equal{}\begin{pmatrix}0&1\\0&0\end{pmatrix}$. A true equivalence is: $ rank(A\plus{}B)\equal{}rank(A)\plus{}rank(B)$ iff $ Im(A)\cap Im(B)\equal{}\{0\}$ AND $ Im(A^T)\cap Im(B^T)\equal{}\{0\}$.
     Moreover in your proof you assume only $ AB\equal{}0$ ; the preceding counterexample shows also that this condition is not sufficient.

We can reason as Soarer did : now it is sufficient to prove that $ Im(A^T)\cap Im(B^T)\equal{}\{0\}$.
$ A^Tv\equal{}B^Tw$ implies $ (A^T)^2v\equal{}A^TB^Tw\equal{}0$ because $ BA\equal{}0$. Moreover $ ker((A^T)^2)\equal{}ker(A^T)$,......
Obviously it remains to prove the NSC about $ rank(A\plus{}B)\equal{}rank(A)\plus{}rank(B)$.


Let $a,b,c$ the legth of sides of a triangle.

Show that:

\[ \left | \frac{a-b}{a+b} + \frac{b-c}{b+c} + \frac{c-a}{c+a} \right | < \frac{1}{16} . \]

Carlos Bravo :)
Lima-PERU

Also posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=51448 .
Stronger version at http://www.mathlinks.ro/Forum/viewtopic.php?t=22786 .

  darij

:o

 Thanks  .......Darij for this information ....

well ...

does anybody have informations about geometric inequalities like that??
I need more theory and methods to solve this kind of ineqs!!!!  :?

Carlos Bravo :)
LIMA-PERU


linear algebra
matrix
graph theory
combinatorics open
combinatorics

Find all graphs $G$, for which it is possible to define a real number to each vertex (not all of them zero) such that $\forall v\in G$ sum of numbers assigned to all neighbors of $v$ is zero. (Each vertex is also neighbor of itself)

Do you mean, find all classes of graphs for which we can assing a real number to each vertex ..... ... . .

The adjacency matrix A(G) of G must be singular. If a vertex isn't neighbour of itself...
Otherwise A(G)+I must be singular, i.e. det[A(G)+I]=0, so -1 must be an eigenvalue of A(G).

Yes, $-1$ must be eigenvalue of adjacency matrix of graph $G$. But can we find a Graph Theoric conditon? (This is an algebraic (linear) condition)

As far as i know, no lower-bound can be calculated for a generic graph spectrum.. but i will control again my books... :P
In algebraic graph theory the existence of an algebraic condition doesn't imply that something can be said about the graph...  :wink:

There is also another beautiful problem posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=104284]here[/url]. Some algebraic condition for all graphs.


List twelve points, each with integer coordinates, that are 5 units from (-8,1). Find an equation of the circle containing these points.

I've been wondering how to solve this problem. Are you suppose to use an equation of a circle, ( x-a)^ + (y-b)^ = r^ and plug in 12 points?

[hide="solution"]We are looking for twelve points which are on the circle with radius 5 and center (-8,1).

The equation of the circle is $(x+8)^{2}+(y-1)^{2}=25$

Here are the 12 points(I say THE 12 points because there are only 12 points.)

(-3,1), (-4,4), (-5,5), (-8,6), (-9,3), (-10,2), (-13,1), (-11,-5), (-10,-6), (-8,-4), (-7,-2), (-5,-3)

Well, please plug them back into the equation to see if I'm correct, I'm too lazy to do it myself.[/hide]

let $a = (-8,1)$.  Let $b = (\pm 3,\pm 4)$ or $(\pm 4,\pm 3)$ where the signs are independent of each other.  then $b$ takes 8 values, hence $a+b$ gives 8 points.

the other four come from adding or subtracting 5 from the x- or y-coordinate.

Dude, 1=2 some of your points are wrong.

[quote="Teki-Teki"]Dude, 1=2 some of your points are wrong.[/quote]

Thanks for checking them. I don't know any way to find the points with acccuracy except by graphing.


calculus
integration
logarithms
calculus computations

Show  :huh:
$ \int_0^a\frac{\ln{x}}{x\plus{}a}dx\equal{}\ln{2}\ln{a}\minus{}\frac{\pi^2}{12}$, $ a>0$.

The substitution $ x\equal{}at$ would solve the problem.


function
floor function
logarithms
search

King Arthur has a new way of choosing which knight is the best in mathematics. He calls a meeting every once in a while for all the knights in the kingdom to come. He orders them to sit around a round table. Then, he starts at chair #1 and counts off around the table. Every n'th knight then must leave the room and go to prison, and the chairs are removed. When he reaches the x'th chair, the last one, he returns to the first one. Finally, when there is just one knight left, the knight will receive a million dollars.

Let k(x,n) be the chair number to sit in, such that for x knights, counting off n each time, the chair is the last one to be chosen. Find a formula for k(x,n) in terms of x and n.

I haven't found a formula for a general $n$, but here is my formula for $n=2$.

We see that if $x=2^z$,with z being an integer, than $k(x,2)=1$.  Let $\lfloor(\log_2 x)\rfloor=y$.  Than, after some trial and error, we find that, for all $2^y\le x<2^{y+1}$, $k(x,2)=1+2(x-2^y)$.

I don't know where to go from here though.

Actually, I moved it from getting started to here. :D

Wow. I'm scared now. I would guess that the formula for a general n would replace the 2's with n. Can you prove your formula (or at least sketch one)?

I thought it would be simple, but is it? Nooooooooooo!

I'm not quite sure how to prove the formula, all I can do is show a lot of examples.

It seems like mods might help...it is going in a circle after all.

Mods as in moderators, or as in modulo? Bad math forum joke.

Once you get to the end, you might or might not have some left over from last time. I don't know how to apply mods (moderators?)

I remember doing this as a challenge problem in 7th grade. I think it had something to do with logs, as in the log function

Yeah. That's samath's answer.

Logs? As in logarithms or as in pieces of wood? (I'm sorry about the excruciatingly bad puns.)

I think I found a solution, but it is very difficult to explain it.
I'l try... :?

[hide="Here it is"]Let's start with an example.

$n=13$ and $x=6$

It is more convenient, to re-number the knights from 0 to 5, e.g. the first knight has the label-0, the second knight has the label-1, and so on, the 6th knight has the label-5.
We will do 5 steps. In each step, one of the knights will leave the table.

At the first step, there will take part all the six numbers, again and again, until we count n knights.
That is, the 6-plet $012345$ will be repeated as many times, as the integer part of the division $n: 6$.
(Actualy, we don't care for the integer part of this division, but only for the remainder)
 So, we need to know just the value of $n\ mod\ 6$

In this example $n\ mod\ 6 = 13\ mod\ 6 = 1$.
That is, the knight who is going to leave the table first is he who is 1st in this 6-plet.
First in this 6-plet is the number $0$. We write down number $0$ and beside we write the rest 5 items of the 6-plet, exactly in the order that they are after $0$.

After that,  we repeat the 5-plet as many times as it is needed to reach the number $13$, and we continue in the same way.

In each step we must know 2 things:
The repeated m-plet (ordered)
and the entrance point (the first knight in this m-plet)

$***$ I change the names of the steps, we will start from step-6 (because we have 6-plets at first), and we finish with the step-2 (the final step is to choose between 2 knights)

So we have:

$6th\ step: \ 0123450123450 \ \ 0 \ \ 12345$
$5th\ step:	\ 1234512345123 \ \ 3 \ \ 4512$
$4th\ step:	\ 4512451245124 \ \ 4 \ \ 512$
$3th\ step:	\ 5125125125125 \ \ 5 \ \ 12$
$2nd\ step: \ 1212121212121 \ \ 1 \ \ 2$

Finally, the winner is number-2
The only thing we should know is the place of number $2$ inside the ordered set $\{0,1,2,3,4,5\}$

Here it is easy too see that $2$ has got the 3rd place in this set.
But we know that the winner is #2 because we did this whole procedure.

We don't search for a procedure but for a general formula.

[color=red]Let's think backward[/color]

We'll find the place of the winner in each m-plet for $m=2,3,4,5,6$.

In the last step (m=2) we see that $2$ is second in the repeating pair $\"12\"$. Why this is second?

Since $13\ mod\ 2 = 1$, the pair $\"12\"$ will be repeated for some times, and we'll stop at the first element of the pair.
If $2$ was first in the pair, it would get out.
So the winner must always keep the second place in the final pair (if n is odd).

$***$ It is more convenient to refer to places starting from 0. So, In the pair $\"12\"$, $1$ has the zero-place and $2$ has the place-1.
So, it's easy to say that since $13\ mod\ 2 =1 \Rightarrow $ place of 2 in this pair is 1 (if the modulo was 0, 2 should have the place-0)
So, the place of the winner at the last pair can be found directly by $n\ mod\ 2$.

Now, let's find out the place which had the number $2$ in the 3-plet (step #3).
As we can see,the 3-plet is 512. (5 has the zero-place). The winner has got the place-2.
Now, why the winner should have the place-2?

We have $13\ mod\ 3 = 1$
So, we'll repeat some times the triplet, and after that the one who will remain (the first of the triplet) will get out.

If $2$ has the place-0 in the triplet, then it will get out at this step.
If $2$ has the place-1, then it will be one place after the looser, so it will be first (place-0) in the next step, but we don't want it.
So, $2$ sould have the place-2 if we want it to have the place-1 at the next step.

I will try to explain it more

We put 13 elements in a series $ABCABCABCABCA$, and after that we put the two elements $12$ (at the prefered order). So we have:

$ABCABCABCABCA12$

The place of first "A" in this 15-plet is 0.
The last "A" is at the place 12.
So, 1 has the place 13 and 2 has the place 14.

But 2 has the place-1 in the next step. Notice that in this 15-plet it has the place 14=13+1.
So, we can find the desired place of number-2 in previous steps with this rule:

[b][color=blue]To find the place of the winner at the m-th step, add n with his place at the step m-1 and take modulo m. $(1)$[/color][/b]

Finally, notice that if we start, for example, from step 3, with the knights $5,1,2$, then the winner will be 2 again. So, if we have 3 knights (and n=13) then the winner will be the 3rd of them. $k(3,13)=2$ (you must add 1 to find the original place)

If we start from the step 5 with the knights $1,2,3,4,5$, the winner will be the 2nd of them. $k(5,13)=1$.

$***$ We add 1 to the value of k(x,n) because it counts starting from 0.

So, with this procedure we found $k(m,n)$ for all $m=2,...,x$.

So the general rule $(1)$ can be written as:

$k(2,n) = n\ mod\ 2$
$k(m,n) = (k(m-1,n)+n)\ mod\ m, \ \forall m>2$

For our example we have
$k(2,13) = 13\ mod\ 2 = 1$
$k(3,13) = (13+1)\ mod\ 3 = 2$
$k(4,13) = (13+2)\ mod\ 4 = 3$
$k(5,13) = (13+3)\ mod\ 5 = 1$
$k(6,13) = (13+1)\ mod\ 6 = 2$

so, if we want the original (not staring from 0) place of the winner for $m=2,3,4,5,6$ we must add 1 to this values and we get the right answer (you can check it in the table above)

As I said, it is difficult to explain my idea, so if someone didn't understand something, ask me again.  :) [/hide]

By the way, this problem reminds me an old problem I was trying to solve.

I'll post it and I'll give a link here.  :D


inequalities
function
algebra unsolved
algebra

[hide="Original post"]here is the problem :)

[img]http://farm3.static.flickr.com/2661/3733079343_b6f17962db_o.jpg[/img]

or click the url below
[url]http://farm3.static.flickr.com/2661/3733079343_b6f17962db_o.jpg[/url]
[/hide]
$ S$ is a set with finitely many elements. $ R(S)$ is the set of subsets of $ S$. Define $ f: R(S)\to\mathbb R$ so as to satisfy $ f(S\setminus A)\equal{}f(A)$, $ \max\{f(A),f(B)\}\ge\max f(A\cup B),\forall A,B\subset S$.

Prove that the number of different values $ f$ can take is no more than $ |S|$.

[color=red][Edited by moderator, hopefully in a correct way.][/color]

Thanks to the moderator to help me re-edit my post
But does anybody have outline to solve this?

Why is there a "max" on the right hand side of the inequality? What is $ \max f(A \cup B)$ supposed to mean here?

......satisfy $ f(S\setminus A) \equal{} f(A)$, $ \max\{f(A),f(B)\}\ge\max f(A\cup B),\forall A,B\subset S$..........

[size=150]should be:[b]......satisfy $ f(S\setminus A) \equal{} f(A)$, $ \max\{f(A),f(B)\}\ge\ f(A\cup B),\forall A,B\subset S$..........[/b]
[b]f(S\A)=(x,x belongs to S and not belongs to A)

look my Original post . I'm sure it is actually right!  :)
[/b][/size]

Let $ A' : \equal{} S \setminus A$.

Lemma: $ \max(f(A),f(B)) \ge f(A \cap B)$.
Proof: $ \max(f(A),f(B)) \equal{} \max(f(A'),f(B')) \ge f(A' \cup B') \equal{} f((A \cap B)') \equal{} f(A \cap B)$.

Let $ d$ be the maximal  value that $ f$ takes on $ S$. Notice $ f(A) \equal{} \max(f(A),f(A')) \ge f(S)$ for
every $ A$, so $ f(S)$ is minimal. So $ f(S) \equal{} d$, $ f$ takes on only one value, namely $ d$, so
we are done.

Supposing otherwise, let $ M \subset S$ be a minimum size set such that $ f(M) \equal{} d$.
We claim $ M \equal{} \{x\}$ for some $ x \in S$. Indeed, if $ x \in M$, then $ \max(f(\{x\}), f(M \setminus \{x\})) \ge d$,
an impossibility unless $ M \equal{} \{x\}$.

Then, let $ T \subset S$ be minimal, containing $ x$, satisfying $ f(T) \ne d$. $ T$ must exist since $ f(S) \ne d$. Then if $ x \in R$ and $ f(R) \ne d$,
we must have that $ T \subset R$, for $ d > \max(f(T),f(R)) \ge f(T \cap R) \equal{} d$, since $ T \cap R$ is a strict
subset of $ T$ containing $ x$. If $ x \notin R$ and $ f(R) \ne d$, then $ x \in R'$ and so $ T \subset R'$, implying
$ T \cap R \equal{} \emptyset$. Thereby, it follows that every set $ A$ satisfying $ f(A) \ne d$ must obey $ T \subset A$ or $ T \cap A \equal{} \emptyset$.

Consider the set $ S / T$, gotten from $ S$ by "gluing together" the elements of $ T$. $ f$ restricts to a
function on subsets of $ S / T$ in an obvious way, since subsets of $ S/T$ correspond uniquely to
subsets $ A \subset S$ satisfying either $ T \subset A$ or $ T \cap A \equal{} \emptyset$. Call this
restricted $ f$ by $ f'$.

$ f'$ obeys the same constraints as $ f$, and so we may apply an inductive hypothesis on $ f'$,
since $ f'$ is defined on subsets of the $ |S| \minus{} |T| \plus{} 1$ element set $ S/T$. It follows that $ f'$
takes on at most $ |S| \minus{} |T| \plus{} 1 \le |S| \minus{} 1$ distinct values, where the inequality follows from $ |T| \ge 2$.

But $ f$ takes on at most the values taken on by $ f'$, with the addition of $ d$. Therefore, $ f$ takes on at
most $ |S|$ values.

(Edited for clarity).


assume that $n\geq 3$ is prime number and $a_1<a_2<...<a_n$ are integer.prove that $a_1,...,a_n$ constitute Arithmetic Progressions  iff there exist partition of {0,1,....} into n class $A_1,A_2,...,A_n$ s.t:
$a_1+A_1=a_2+A_2=...=a_n+A_n                                (x+S={x+y|y\in S})$

yes n should be prime otherwise  u can't conclude conversely

I has solved this problem yesterday. My solution is quite different from yours, and very different from the official one ( I am not sure that is the official solution  :D ). The official's solution is long and long...If you want, I will post it, it needs time to do.

My "solution" is not a solution, since it is not correct  :).

I haven't read it, but i think it is easy for you to resolve it. I said your "solution" is different from me because it look different from mine  :D .  The thing I want to say here is the official's sollution. It very strange with odd idea. I haven't understood it yet.
Btw,  we need a prime here ?

Yes, I remember reading the official solution in a booklet and they used the fact that $n$ was a prime, and, at the end of the proof, gave a counterexample in the case when $n$ was not a prime.


My mom said "Your never using this book"
Guess what book she was talking about?
Yep, my brother's advanced algebra book.
Then she said "You can do intermediate alogebra from AoPS, but not this book."

[hide]:rotfl:  :rotfl:  :rotfl:  :rotfl: [/hide]

[hide] :rotfl:  :rotfl:   :rotfl:  [/hide]

[quote="snee"]My mom said "Your never using this book"
Guess what book she was talking about?
Yep, my brother's advanced algebra book.
Then she said "You can do intermediate alogebra from AoPS, but not this book."
[/quote]

Your mom sounds wise to me.

I dont think so- She lied about knowing everything xD


Sanırım  forumu takip eden var fakat sadece okumakla yetinildiğinden aktif görünmüyor gibi...
Özellikle e azından popüler matematik sorularının tartışılmasının faydalı olacağını düşünüyorum. şimdiye kadar yazdığım forumların hiçbirinde cevap alamadığım soruları soracağım.
İlk soru:

...................................................

  Simdi elinize 10x10 tane kare olusturacak sekilde duzgunce dizilmis noktalar alin. Her noktayi kendisine yatay ve dikey komsu noktalara dogru parcalariyla birlestirin (cikan sekil 9x9 = 81 tane kucuk kare olacak, ayni kareli defterdeki gibi).

1) Ust kenarda yer alan noktalar kumesine A diyelim. Sag alt kosedeki nokta da B olsun. Simdi bu ag uzerinde soyle bir rastgele hareket yapiyoruz: Eger A veya B deki noktalara varirsak orada kaliyoruz. Eger diger noktalardan birindeysek 1/4 er olasilikla komsu noktalardan birine gidiyoruz. Komsu dedigimiz arada cizgi olan yatay ve dikey komsular. Sorumuz su: x bu ag uzerinde bir nokta olsun. p(x) de, rastgele hareketimize bu noktadan baslarsak B'deki noktaya A'dakilerden daha once varma olasiligimiz olsun.

2) Simdi bu agi bir elektrik devresi olarak dusunun, koseleri birlestiren dogrular da iletken tel olsun. B'deki noktaya 1 Volt, A'daki noktalara da 0 volt potansiyel uyguluyoruz. x bu ag uzerindeki bir nokta ise v(x) de x noktasindaki potansiyel olsun.

Soru: v(x) ile p(x) arasindaki iliski nedir?

  ihsan (...)

Ben tam anlayamadım sanırım, şekil çizmen mümkün mü?


How many ways are there to form a 4-digit number divisible by $ 15$ using $ 0,1,2,4,5,6,7,8,9$ as its digits? (Repeated digits are allowed.)

[hide]
I'll assume that you left $ 3$ out on purpose.
There are a total of $ 8*9*9*9\equal{}5832$ 4-digit numbers. If a number is divisible by $ 15$, then it must be divisible by both $ 3$ and $ 5$. For a number to be divisible by $ 5$, its final digit must end in $ 0$ or $ 5$. Therefore, there are $ 8*9*9*2\equal{}1296$ 4-digit numbers divisible by $ 5$. This leaves us with $ 1000, 1005, 1010,..., 9985, 9990, 9995$. If you take any three consecutive numbers in that sequence, exactly one of them is divisible by both $ 3$ and $ 5$. Therefore, you divide $ 1296$ by $ 3$ to obtain $ 432$ as your final answer.
[/hide]

[quote="darkmage2009"]
If you take any three consecutive numbers in that sequence, exactly one of them is divisible by both $ 3$ and $ 5$. Therefore, you divide $ 1296$ by $ 3$ to obtain $ 432$ as your final answer.
[/quote]

The answer is right. But how can we be sure that if taking any three consecutive numbers in that sequence, exactly one of them is divisible by both $ 3$ and $ 5$ ?

Maybe by considering
$ 5 \equiv 2 mod 3$
$ 10 \equiv 1 mod 3$

so, $ x,x\plus{}5,x\plus{}10$ are different in mod 3, therefore one of it must be divisible by 3.
CMIIW

Perhaps, but with the digit $ 3$ missing, I really can't be sure anymore. I suppose the right way to do it would've been to do casework with the sum of the digits.

EDIT:
[hide]
Case 1: 4-digit number ends in "0". Therefore, the remaining 3 digits must have a sum of 0 mod 3. Possible combinations (mod 3):
0,0,0 $ (2*3*3\equal{}18)$
1,1,1 $ (3*3*3\equal{}27)$
2,2,2 $ (3*3*3\equal{}27)$
0,1,2 $ (2*3*3\equal{}18)$
0,2,1 $ (2*3*3\equal{}18)$
1,0,2 $ (3*3*3\equal{}27)$
1,2,0 $ (3*3*3\equal{}27)$
2,0,1 $ (3*3*3\equal{}27)$
2,1,0 $ (3*3*3\equal{}27)$
Total: $ 216$
Case 2: 4-digit number ends in "5". Therefore, the remaining 3 digits must have a sum of 1 mod 3. Possible combinations (mod 3):
0,0,1 $ (2*3*3\equal{}18)$
0,1,0 $ (2*3*3\equal{}18)$
1,0,0 $ (3*3*3\equal{}27)$
0,2,2 $ (2*3*3\equal{}18)$
2,0,2 $ (3*3*3\equal{}27)$
2,2,0 $ (3*3*3\equal{}27)$
1,1,2 $ (3*3*3\equal{}27)$
1,2,1 $ (3*3*3\equal{}27)$
2,1,1 $ (3*3*3\equal{}27)$
Total: $ 216$

$ 216\plus{}216\equal{}432$, which is our desired answer.
[/hide]

Just wondering, is getting 432 using both ways (although the first way has not addressed the fact that 3 is left out in the solution using the period of 3 consecutive numbers) a coincidence?

[hide="A probably safer way:"]
Let $ A \equal{} \{0,6,9\},B \equal{} \{1,4,7\},C \equal{} \{2,5,8\}$ and let $ abcd$ denote the 4-digit number. Then we have $ 2$ possible choices for $ d$, $ 8$ choices for $ a$, and  $ 9$ arbitary choices for $ b$. For any combinaton of our selected $ d,a$ and $ b$, we can choose $ c$ from exactly one of the sets $ A,B,$ or $ C$, giving us $ 3$ choices for $ c$. Thus there are $ 2\cdot 9\cdot 8 \cdot 3 \equal{} 432$ 4-digit numbers satisfying the condition.[/hide]

By the way, I think darkmage2009's method is quite creative and that may pose an even more interesting problem for us.